Bezier Curves Based Numerical Solutions of Delay Systems ...
Transcript of Bezier Curves Based Numerical Solutions of Delay Systems ...
Research ArticleBezier Curves Based Numerical Solutions ofDelay Systems with Inverse Time
F Ghomanjani1 M H Farahi1 A KJlJccedilman2 A V Kamyad1 and N Pariz3
1 Department of Mathematics Ferdowsi University of Mashhad Mashhad Iran2Department of Mathematics Universiti Putra Malaysia 43400 Serdang Selangor Malaysia3 Department of Control Faculty of Engineering Ferdowsi University of Mashhad Mashhad Iran
Correspondence should be addressed to A Kılıcman akilicupmedumy
Received 10 July 2013 Accepted 29 December 2013 Published 27 February 2014
Academic Editor Mufid Abudiab
Copyright copy 2014 F Ghomanjani et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper applied for the first time the Bernsteinrsquos approximation on delay differential equations and delay systems with inversedelay that models these problems The direct algorithm is given for solving this problem The delay function and inverse timefunction are expanded by the Bezier curves The Bezier curves are chosen as piecewise polynomials of degree 119899 and the Beziercurves are determined on any subinterval by 119899 + 1 control points The approximated solution of delay systems containing inversetime is derived To validate accuracy of the present algorithm some examples are solved
1 Introduction
Delay differential equations (DDEs) differ fromODEs in thatthe derivative at any time depends on the solution at priortimes (and in the case of neutral equations on the derivativeat prior times)
DDEs often arise when traditional pointwise model-ing assumptions are replaced by more realistic distributedassumptions for example when the birth rate of predatorsis affected by prior levels of predators or prey rather than byonly the current levels in a predator-prey model
Because the derivative (119905) depends on the solution atprevious time(s) it is necessary to provide an initial historyfunction to specify the value of the solution before time 119905 =
0 In many common models the history is a constant butnonconstant history functions are encountered routinely
For most problems there is a jump derivative discontinu-ity at the initial time In most models the DDE and the initialfunction are incompatible for some derivative order usuallythe first the left and right derivatives at 119905 = 0 are not equalDelay systems containing inverse time are an important classof systems
(119905) = 119909 (119905 minus 1) (0+
) = 1 (0minus
) = 0 (1)
A fascinating property is how such derivative discontinuitiesare propagated in time For the equation and history justdescribed for example the initial first discontinuity is propa-gated as a second degree discontinuity at time 119905 = 1 as a thirddegree discontinuity at time 119905 = 2 and more generally as adiscontinuity in the (119899 + 1)st derivative at time 119905 = 119899
Delay differential equations are type of differential equa-tions where the time derivatives at the current time dependon the solution and possibly its derivatives at previous times(see [1ndash4])
The basic theory concerning the stable factors for exam-ple existence and uniqueness of solutions was presented in[1 3] Since then DDEs have been extensively studied inrecent decades and a great number of monographs have beenpublished including significant works on dynamics of DDEsby Hale and Lunel [5] and on stability by Niculescu [6] Theinterest in study of DDEs is caused by the fact that manyprocesses have time delays and have been models for betterrepresentations by systems of DDEs in science engineeringeconomics and so forth Such systems however are still notfeasible to actively analyze and control precisely thus thestudy of systems of DDEs has actively been conducted overthe recent decades (see [7ndash10])
Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 602641 16 pageshttpdxdoiorg1011552014602641
2 Mathematical Problems in Engineering
Wu et al [11] developed a computational method forsolving an optimal control problem which is governed bya switched dynamical system with time delay Kharatishivili[12] has approached this problem by extending Pontryaginrsquosmaximum principle to time delay systems The actual solu-tion involves a two-point boundary-value problem in whichadvances and delays are presented In addition this solutiondoes not yield a feedback controller Optimal-time controlof delay systems has been considered by Oguztoreli [13]who obtained several results concerning bang-bang controlswhich are parallel to those of LaSalle [14] for nondelay sys-tems For a time-invariant system with an infinite upper limitin the performance measure Krasovskii [15] has developedthe forms of the controller and the performance measureRoss [16] has obtained a set of differential equations for theunknowns in the forms of Krasovskii However Rossrsquos resultsare not applicable to time-varying systems with a finite limitin the performance measure
Basin and Perez [17] presented an optimal regulator fora linear system with multiple state and input delays and aquadratic criterion The optimal regulator equations wereobtained by reducing the original problem to the linear-quadratic regulator design for a system without delays (see[17 18])
This paper aims at solving delay systems containinginverse time of the following form
x (119905) = 119860 (119905) x (119905) + 119862 (119905) (1199091(119905 minus 1205911) sdot sdot sdot 119909
119901(119905 minus 120591119901))
119879
+ 119863 (119905) (1199091(119905119891minus 119905) sdot sdot sdot 119909
119901(119905119891minus 119905))
119879
+ 119866 (119905) u (119905)
x (119905) = 120601 (119905) 119905 isin [minus120591max 1199050]
(2)
where x(119905) = (1199091(119905) sdot sdot sdot 119909
119901(119905))119879
isin R119901 u(119905) = (1199061(119905) sdot sdot sdot
119906119898(119905))119879
isin R119898 are respectively state and control functionswhile 120601(119905) = (120601
1(119905) sdot sdot sdot 120601
119901(119905))119879 is known vector function and
120591119894rsquos (119894 = 1 2 119901) are nonnegative constant time delays
and 120591max = max120591119894 1 le 119894 le 119901 We assume the matrices
119860(119905) = [119886119894119895(119905)]119901times119901
119862(119905) = [119888119894119895(119905)]119901times119901
119863(119905) = [119889119894119895(119905)]119901times119901
and119866(119905) = [119892
119894119895(119905)]119898times119898
are matrix functions We need to imposethe continuity condition on x(119905) and its first derivative wherethese constraints appeared in Section 2
Piecewise polynomial functions are often used to repre-sent the approximate solution in the numerical solution ofdifferential equations (see [19ndash22]) B-splines due to numer-ical stability and arbitrary order of accuracy have becomepopular tools for solving differential equations (where Bezierform is a special case of B-splines)There aremany papers andbooks dealing with the Bezier curves or surface techniques
Harada and Nakamae [23] Nurnberger and Zeilfelder[24] used the Bezier control points in approximated dataand functions Zheng et al [22] proposed the use of controlpoints of the Bernstein-Bezier form for solving differentialequations numerically and also Evrenosoglu and Somali [25]used this approach for solving singular perturbed two-pointboundary-value problems The Bezier curves are used insolving partial differential equations as well Wave and Heat
equations are solved in Bezier form (see [26ndash29]) the Beziercurves are used for solving dynamical systems (see [30]) andalso the Bezier control pointsmethod is used for solving delaydifferential equation (see [31 32])
TheBezier curvesmethodwas presentedwhichwas statedfor solving the optimal control systems with pantographdelays (see [33]) The method was computationally attractiveand also reduced the CPU time and the computer memoryand at the same time keeps the accuracy of the solution Thealgorithm had been successfully applied to the pantographequations Comparing with other methods the results ofnumerical examples demonstrated that thismethodwasmoreaccurate than some existing methods (see [33])
Using Bezier curve Ghomanjani et al [34] had usedleast square method for numerical solutions of time-varyinglinear optimal control problems with time delays in state andcontrol
Some other applications of the Bezier functions andcontrol points are found in [35ndash37] that are used in computeraided geometric design and image compression
The use of the Bezier curves is a novel idea for solvingdelay systems containing inverse time The approach used inthis paper reduces the CPU time and the computer memorycomparing with existingmethods (see the numerical results)Although the method is very easy to use and straightforwardthe obtained results are satisfactory (see the numericalresults) We suggest a technique similar to the one used in[22 25] for solving delay systems containing inverse timeThecurrent paper is organized as follows
In Section 2 Function approximation will be introducedConvergence analysis will be stated in Section 3 In Section 4some numerical examples are solved which show the effi-ciency and reliability of the method Finally Section 5 willgive a conclusion briefly
2 Function Approximation
Divide the interval [1199050 119905119891] into a set of grid points such that
119905119894= 1199050+ 119894ℎ 119894 = 0 1 119896 (3)
where ℎ = (119905119891minus 1199050)119896 and 119896 is a positive integer Let 119878
119895=
[119905119895minus1
119905119895] for 119895 = 1 2 119896 Then for 119905 isin 119878
119895 delay
systems containing inverse time (2) can be decomposed tothe following problem
x119895(119905) = 119860 (119905) x
119895(119905)
+ 119862 (119905) (119909minus1198961
1+119895
1(119905 minus 1205911) sdot sdot sdot 119909
minus119896119901
1+119895
119901(119905 minus 120591119901))
119879
+ 119863 (119905) (1199091198962minus119895
1(119905119891minus 119905) sdot sdot sdot 119909
1198962minus119895
119901(119905119891minus 119905))
119879
+ 119866 (119905) u119895(119905)
x119895(120579) = 120601 (120579) 120579 isin [minus120591max 1199050]
(4)
where x119895(119905) = (119909
119895
1(119905) sdot sdot sdot 119909
119895
119901(119905))119879 and u
119895(119905) = (119906
119895
1(119905) sdot sdot sdot 119906
119895
119898(119905))119879
are respectively vectors of x(119905) and u(119905)which are considered
Mathematical Problems in Engineering 3
in 119905 isin 119878119895 We mention that 119909minus119896
119894
1+119895
119894(119905 minus 120591119894) 1 le 119894 le 119901 is the 119894th
component of (119909minus1198961
1+119895
1(119905minus1205911) sdot sdot sdot 119909
minus119896119901
1+119895
119901(119905minus120591119901))119879where (119905minus120591
119894) isin
[119905minus119896119894
1+119895minus1
119905minus119896119894
1+119895] and 119909
1198962minus119895
119894(119905119891
minus 119905) 1 le 119894 le 119901 is the 119894thcomponent of (1199091198962minus119895
1(119905119891minus 119905) sdot sdot sdot 119909
1198962minus119895
119901(119905119891minus 119905))119879 where (119905
119891minus 119905) isin
[1199051198962minus119895minus1
1199051198962minus119895] Also
119896119894
1=
120591119894
ℎ
120591119894
ℎ
isin N
([
120591119894
ℎ
] + 1)
120591119894
ℎ
notin N
1 le 119894 le 119901
1198962=
119905119891
ℎ
119905119891
ℎ
isin N
([
119905119891
ℎ
] + 1)
119905119891
ℎ
notin N
(5)
where [120591119894ℎ] and [119905
119891ℎ] denote the integer part of 120591
119894ℎ and
119905119891ℎ respectivelyOur strategy is to use Bezier curves to approximate the
solutions x119895(119905) and u
119895(119905) by k
119895(119905) and w
119895(119905) respectively
where k119895(119905) and w
119895(119905) are given below Individual Bezier
curves that are defined over the subintervals are joinedtogether to form the Bezier spline curves For 119895 = 1 2 119896define the Bezier polynomials of degree 119899 that approximaterespectively the actions of x
119895(119905) and u
119895(119905) over the interval
[119905119895minus1
119905119895] as follows
k119895(119905) =
119899
sum
119903=0
a119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
w119895(119905) =
119899
sum
119903=0
b119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
(6)
where
119861119903119899
(
119905 minus 119905119895minus1
ℎ
) = (
119899
119903)
1
ℎ119899(119905119895minus 119905)
119899minus119903
(119905 minus 119905119895minus1
)
119903
(7)
is the Bernstein polynomial of degree 119899 over the interval[119905119895minus1
119905119895] a119895119903and b119895
119903are respectively 119901 and 119898 ordered vectors
from the control points (see [22]) By substituting (6) in (4)1198771119895
(119905) for 119905 isin [119905119895minus1
119905119895] can be defined as follows
1198771119895
(119905) = k119895(119905) minus 119860 (119905) k
119895(119905)
minus 119862 (119905) (Vminus1198961
1+119895
1(119905 minus 1205911) sdot sdot sdot Vminus119896
119901
1+119895
119901(119905 minus 120591119901))
119879
minus 119863 (119905) (V1198962minus1198951
(119905119891minus 119905) sdot sdot sdot V1198962minus119895
119901(119905119891minus 119905))
119879
minus 119866 (119905)w119895(119905)
(8)
Let k(119905) = sum119896
119895=11205941
119895(119905)k119895(119905) and w(119905) = sum
119896
119895=11205942
119895(119905)w119895(119905)
where1205941119895(119905) and120594
2
119895(119905) are respectively characteristic function
of k119895(119905) and w
119895(119905) for 119905 isin [119905
119895minus1 119905119895] Beside the boundary
conditions on k(119905) at each node we need to impose thecontinuity condition on each successive pair of k
119895(119905) to
guarantee the smoothness
Since the differential equation is of first order the conti-nuity of x (or k) and its first derivative gives
k(119904)119895
(119905119895) = k(119904)119895+1
(119905119895) 119904 = 0 1 119895 = 1 2 119896 minus 1 (9)
where k(119904)119895(119905119895) is the 119904th derivative k
119895(119905) with respect to 119905 at
119905 = 119905119895
Thus the vector of control points a119895119903(119903 = 0 1 119899 minus 1 119899)
must satisfy (see the Appendix)
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(10)
According to the definition of the 119905119894= 1199050+ 119894ℎ we get that
119905119895minus 119905119895minus1
= ℎ Therefore
a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
)
(11)
One may recall that a119895119903is a 119901 ordered vector This approach is
called the subdivision scheme (or ℎ-refinement in the finiteelement literature) This method is based on the control-point-based method
Remark 1 By considering the 1198621 continuity of w the follow-ing constraints will be added to constraints in (10)
b119895119899(119905119895minus 119905119895minus1
)
119899
= b119895+10
(119905119895+1
minus 119905119895)
119899
(b119895119899minus b119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (b119895+11
minus b119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(12)
where the so-called b119895119903(119903 = 0 1 119899 minus 1 119899) is an 119898 ordered
vectorNow the residual function can be defined in 119878
119895as follows
119877119895= int
119905119895
119905119895minus1
100381710038171003817100381710038171198771119895
(119905)
10038171003817100381710038171003817
2
119889119905 (13)
where sdot is the Euclidean norm (recall that 1198771119895
(119905) is a 119901
vector where 119905 isin 119878119895)
Our aim is to solve the following problem over 119878 =
⋃119896
119895=1119878119895
min119896
sum
119895=1
119877119895
st a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
) 119895 = 1 2 119896 minus 1
(14)
The mathematical programming problem (14) can be solvedby many subroutine algorithms Here we used Maple 12 tosolve this optimization problem
4 Mathematical Problems in Engineering
Remark 2 Consider the following boundary value problem
y (119905) = 119877 (119905) y (119905) + 119876 (119905) y (119905 minus 120572) + 119878 (119905) z (119905) + a (119905)
z (119905) = 119881 (119905) y (119905) + 119870 (119905) z (119905 + 120572) + 119882 (119905) z (119905) + b (119905)
y (1199050) = y0
y (119905) = 120601 (119905) 119905 isin [minus120572 1199050)
z (119905119891) = z0
z (119905) = 120595 (119905) 119905 isin (119905119891 119905119891+ 120572]
(15)
where y(119905) z(119905) a(119905) b(119905) 120601(119905) and 120595(119905) are the vectorsof appropriate dimensions 119877(119905) 119876(119905) 119878(119905) 119881(119905) 119870(119905) and119882(119905) are the matrices of appropriate dimensions and 120572 isnonnegative constant time delay
Let
x (119905) = [y(119905)119879 z(119905119891minus 119905)
119879
]
119879
(16)
where 119879 is the transpose then
x (119905) = [y119879 (119905) minusz119879 (119905119891minus 119905)]
119879 (17)
satisfies that
x (119905) = 119860 (119905) x (119905) + 119862 (119905) x (119905 minus 120572)
+ 119863 (119905) x (119905119891minus 119905) + u (119905) 119905 isin [119905
0 119905119891]
x (1199050) = x0= [y1198790
z1198790]
119879
(18)
where
119860 (119905) = 119864(2)
11otimes 119877 (119905) minus 119864
(2)
22otimes 119882(119905
119891minus 119905)
119862 (119905) = 119864(2)
11otimes 119876 (119905) minus 119864
(2)
22otimes 119870 (119905
119891minus 119905)
119863 (119905) = 119864(2)
12otimes 119878 (119905) minus 119864
(2)
21otimes 119881 (119905
119891minus 119905)
u (119905) = [a119879 (119905) minusb119879 (119905119891minus 119905)]
119879
(19)
where119864(119891)119894119895
is the119891times119891matrix with 1 at its entry (119894 119895) and zeroselsewhere and otimes is Kronecker product (see eg [4 38 39])
Remark 3 Now the following delay differential equation canbe considered
(119905) = 119891 (119905 119909 (119905) 119909 (119905 minus 120591 (119905 119909 (119905)))) 119905 ge 0 (20)
with initial condition
119909 (119905) = 120601 (119905) 119905 isin [minus120582 0] (21)
where 120582 equiv inf119905 minus 120591(119905 119906) 119905 ge 0 119906 isin R In the case when 120582 isnot finite [minus120582 0] denotes the interval (minusinfin 0]
Furthermore we assume that
120591 (119905 119906) ge 0 forall119905 ge 0 119906 isin R (22)
that is (20) is a delay differential equationThe existence anduniqueness of the solution of initial value problem (20)-(21)was stated in [40]
Equation (20) is converted into a nonlinear programmingproblem (NLP) by applying Bezier control points methodwhereas the MATLAB optimization routine FMINCON isused for solving resulting NLP Numerical example showsthat the proposed method is efficient and very easy to use
Remark 4 Now we limit ourselves to consider the followingnonlinear delay differential equation in the type
119871119909 (119905) = 119865 (119905 119909 (119905) 119909 (120591 (119905))) 1199050le 119905 le 119905
119891 (23)
with the following initial conditions
119909(119896)
(1199050) = 119909119896
0 119896 = 0 1 119899 minus 1
119909 (119905) = 120601 (119905) 119905 le 1199050
(24)
where the differential operator 119871 is defined by 119871(sdot) =
119889119899(sdot)119889119905119899
3 Convergence Analysis
In this section without loss of generality we analyze theconvergence of the control-point-based method applied tothe problem (2) with time delay in state when 119901 = 119898 = 1and the time interval is [0 1] So the following problem isconsidered
119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) =
119889119909 (119905)
119889119905
minus 119860 (119905) 119909 (119905) minus 119862 (119905) 119909 (119905 minus 120591) minus 119866 (119905) 119906 (119905)
minus 119863 (119905) 119909 (1 minus 119905) = 119865 (119905) 119905 isin [0 1]
119909 (119905) = 1199090= 119886 119905 le 0 119909 (1) = 119909
119891= 119887
119906 (119905) = 1199060= 1198861 119905 le 0
(25)
where 119909(119905) isin 119877 119906(119905) isin 119877 and 119886 119887 1198861are given real numbers
and 119860(119905) 119862(119905) 119866(119905) 119863(119905) and 119865(119905) are known polynomialsfor 119905 isin [0 1] The constant time delay 120591 is nonnegative
Without loss of generality we consider the interval [0 1]instead of [119905
0 119905119891] since the variable 119905 can be changed with the
new variable 119911 by 119905 = (119905119891minus 1199050)119911 + 1199050where 119911 isin [0 1]
Lemma 5 For a polynomial in Bezier form
119909 (119905) =
1198991
sum
119894=0
1198861198941198991
1198611198941198991(119905) (26)
we have
sum1198991
119894=01198862
1198941198991
1198991+ 1
ge
sum1198991+1
119894=01198862
1198941198991+1
1198991+ 2
ge sdot sdot sdot
ge
sum1198991+1198981
119894=01198862
1198941198991+1198981
1198991+ 1198981+ 1
997888rarr int
1
0
1199092
(119905) 119889119905 1198981997888rarr +infin
(27)
Mathematical Problems in Engineering 5
where 1198861198941198991+1198981
is the Bezier coefficient of 119909(119905) after degree-elevating to degree 119899
1+ 1198981
Proof See [22 page 245]
The convergence of the approximate solution could bedone in two ways
(1) degree raising the Bezier polynomial approximation(2) subdivision of the time interval
In the following the convergence in each case canbe proven although in numerical examples we used onlysubdivision case (see [32])
31 Degree Raising
Theorem 6 If the problem (25) with inverse time in state hasa unique 119862
1 continuous trajectory solution 119909 1198620 continuouscontrol solution 119906 then the approximate solution obtained bythe control-point-based method converges to the exact solution(119909 119906) as the degree of the approximate solution tends to infinity
Proof Given an arbitrary small positive number 120598 gt
0 by the Weierstrass theorem (see [41]) one can easilyfind polynomials 119876
11198731
(119905) of degree 1198731and 119876
21198732
(119905) ofdegree 119873
2such that 119889119894119876
11198731
(119905)119889119905119894minus 119889119894119909(119905)119889119905
119894infin
le 1205981611988911989411987611198731
(119905 minus 120591)119889119905119894minus 119889119894119909(119905 minus 120591)119889119905
119894infin
le 12059816 119894 = 0 111987621198732
(119905) minus 119906(119905)infin
le 12059816 and 11987611198731
(1 minus 119905) minus 119909(1 minus 119905)infin
le
12059816 where sdot infin
stands for the 119871infin-norm over [0 1]
Especially we have10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
le
120598
16
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
le
120598
16
(28)
In general 11987611198731
(119905) and 11987621198732
(119905) do not satisfy the boundaryconditions After a small perturbation with linear and con-stant polynomials 120572119905 + 120573 120574 respectively for 119876
11198731
(119905) and11987621198732
(119905) we can obtain polynomials 11987511198731
(119905) = 11987611198731
(119905) +
(120572119905 + 120573) and 11987521198732
(119905) = 11987621198732
(119905) + 120574 such that 11987511198731
(119905) satisfiesthe boundary conditions 119875
11198731
(0) = 119886 11987511198731
(1) = 119887 and11987521198732
(0) = 1198861Thus119876
11198731
(0)+120573 = 119886 and11987611198731
(1)+120572+120573 = 119887By using (28) one has
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
=1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
(29)
Since
120572infin
minus10038171003817100381710038171205731003817100381710038171003817infin
le1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
(30)
so
120572infin
le
120598
16
+10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
+
120598
16
=
120598
8
(31)
By the time from 1198861= 11987521198732
(0) = 11987621198732
(0) + 120574
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205741003817100381710038171003817infin
le
120598
16
(32)
Now we have
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987611198731(119905) + 120572119905 + 120573 minus 119909 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987611198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
8
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
+ 120572 minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
le
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 120572infin
le
3120598
16
lt
120598
5
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987621198732(119905) + 120574 minus 119906 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987621198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+10038171003817100381710038171205741003817100381710038171003817infin
le
120598
8
lt
120598
5
(33)
so
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905 minus 120591)
119889119905
minus
119889119909 (119905 minus 120591)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
lt
120598
5
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
lt
120598
5
(34)
Now let 119871119875119873(119905) = 119871(119875
11198731
(119905) 11987521198732
(119905) 11987511198731
(119905minus120591) 11987511198731
(1minus119905)11988911987511198731
(119905)119889119905) = 11988911987511198731
(119905)119889119905minus119860(119905)11987511198731
(119905)minus119862(119905)11987511198731
(119905minus120591)minus
119866(119905)11987521198732
(119905) minus 119863(119905)11987511198731
(1 minus 119905) = 119865(119905) for every 119905 isin [0 1]Thus for119873 ge max119873
1 1198732 an upper bound is found for the
following residual
1003817100381710038171003817119871119875119873(119905) minus 119865 (119905)
1003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
119871(11987511198731(119905) 11987521198732(119905) 11987511198731(119905 minus 120591)
11987511198731(1 minus 119905)
11988911987511198731(119905)
119889119905
) minus 119865 (119905)
100381710038171003817100381710038171003817100381710038171003817infin
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014
Advances in
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Complex AnalysisJournal of
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Advances in
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Discrete MathematicsJournal of
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Stochastic AnalysisInternational Journal of
2 Mathematical Problems in Engineering
Wu et al [11] developed a computational method forsolving an optimal control problem which is governed bya switched dynamical system with time delay Kharatishivili[12] has approached this problem by extending Pontryaginrsquosmaximum principle to time delay systems The actual solu-tion involves a two-point boundary-value problem in whichadvances and delays are presented In addition this solutiondoes not yield a feedback controller Optimal-time controlof delay systems has been considered by Oguztoreli [13]who obtained several results concerning bang-bang controlswhich are parallel to those of LaSalle [14] for nondelay sys-tems For a time-invariant system with an infinite upper limitin the performance measure Krasovskii [15] has developedthe forms of the controller and the performance measureRoss [16] has obtained a set of differential equations for theunknowns in the forms of Krasovskii However Rossrsquos resultsare not applicable to time-varying systems with a finite limitin the performance measure
Basin and Perez [17] presented an optimal regulator fora linear system with multiple state and input delays and aquadratic criterion The optimal regulator equations wereobtained by reducing the original problem to the linear-quadratic regulator design for a system without delays (see[17 18])
This paper aims at solving delay systems containinginverse time of the following form
x (119905) = 119860 (119905) x (119905) + 119862 (119905) (1199091(119905 minus 1205911) sdot sdot sdot 119909
119901(119905 minus 120591119901))
119879
+ 119863 (119905) (1199091(119905119891minus 119905) sdot sdot sdot 119909
119901(119905119891minus 119905))
119879
+ 119866 (119905) u (119905)
x (119905) = 120601 (119905) 119905 isin [minus120591max 1199050]
(2)
where x(119905) = (1199091(119905) sdot sdot sdot 119909
119901(119905))119879
isin R119901 u(119905) = (1199061(119905) sdot sdot sdot
119906119898(119905))119879
isin R119898 are respectively state and control functionswhile 120601(119905) = (120601
1(119905) sdot sdot sdot 120601
119901(119905))119879 is known vector function and
120591119894rsquos (119894 = 1 2 119901) are nonnegative constant time delays
and 120591max = max120591119894 1 le 119894 le 119901 We assume the matrices
119860(119905) = [119886119894119895(119905)]119901times119901
119862(119905) = [119888119894119895(119905)]119901times119901
119863(119905) = [119889119894119895(119905)]119901times119901
and119866(119905) = [119892
119894119895(119905)]119898times119898
are matrix functions We need to imposethe continuity condition on x(119905) and its first derivative wherethese constraints appeared in Section 2
Piecewise polynomial functions are often used to repre-sent the approximate solution in the numerical solution ofdifferential equations (see [19ndash22]) B-splines due to numer-ical stability and arbitrary order of accuracy have becomepopular tools for solving differential equations (where Bezierform is a special case of B-splines)There aremany papers andbooks dealing with the Bezier curves or surface techniques
Harada and Nakamae [23] Nurnberger and Zeilfelder[24] used the Bezier control points in approximated dataand functions Zheng et al [22] proposed the use of controlpoints of the Bernstein-Bezier form for solving differentialequations numerically and also Evrenosoglu and Somali [25]used this approach for solving singular perturbed two-pointboundary-value problems The Bezier curves are used insolving partial differential equations as well Wave and Heat
equations are solved in Bezier form (see [26ndash29]) the Beziercurves are used for solving dynamical systems (see [30]) andalso the Bezier control pointsmethod is used for solving delaydifferential equation (see [31 32])
TheBezier curvesmethodwas presentedwhichwas statedfor solving the optimal control systems with pantographdelays (see [33]) The method was computationally attractiveand also reduced the CPU time and the computer memoryand at the same time keeps the accuracy of the solution Thealgorithm had been successfully applied to the pantographequations Comparing with other methods the results ofnumerical examples demonstrated that thismethodwasmoreaccurate than some existing methods (see [33])
Using Bezier curve Ghomanjani et al [34] had usedleast square method for numerical solutions of time-varyinglinear optimal control problems with time delays in state andcontrol
Some other applications of the Bezier functions andcontrol points are found in [35ndash37] that are used in computeraided geometric design and image compression
The use of the Bezier curves is a novel idea for solvingdelay systems containing inverse time The approach used inthis paper reduces the CPU time and the computer memorycomparing with existingmethods (see the numerical results)Although the method is very easy to use and straightforwardthe obtained results are satisfactory (see the numericalresults) We suggest a technique similar to the one used in[22 25] for solving delay systems containing inverse timeThecurrent paper is organized as follows
In Section 2 Function approximation will be introducedConvergence analysis will be stated in Section 3 In Section 4some numerical examples are solved which show the effi-ciency and reliability of the method Finally Section 5 willgive a conclusion briefly
2 Function Approximation
Divide the interval [1199050 119905119891] into a set of grid points such that
119905119894= 1199050+ 119894ℎ 119894 = 0 1 119896 (3)
where ℎ = (119905119891minus 1199050)119896 and 119896 is a positive integer Let 119878
119895=
[119905119895minus1
119905119895] for 119895 = 1 2 119896 Then for 119905 isin 119878
119895 delay
systems containing inverse time (2) can be decomposed tothe following problem
x119895(119905) = 119860 (119905) x
119895(119905)
+ 119862 (119905) (119909minus1198961
1+119895
1(119905 minus 1205911) sdot sdot sdot 119909
minus119896119901
1+119895
119901(119905 minus 120591119901))
119879
+ 119863 (119905) (1199091198962minus119895
1(119905119891minus 119905) sdot sdot sdot 119909
1198962minus119895
119901(119905119891minus 119905))
119879
+ 119866 (119905) u119895(119905)
x119895(120579) = 120601 (120579) 120579 isin [minus120591max 1199050]
(4)
where x119895(119905) = (119909
119895
1(119905) sdot sdot sdot 119909
119895
119901(119905))119879 and u
119895(119905) = (119906
119895
1(119905) sdot sdot sdot 119906
119895
119898(119905))119879
are respectively vectors of x(119905) and u(119905)which are considered
Mathematical Problems in Engineering 3
in 119905 isin 119878119895 We mention that 119909minus119896
119894
1+119895
119894(119905 minus 120591119894) 1 le 119894 le 119901 is the 119894th
component of (119909minus1198961
1+119895
1(119905minus1205911) sdot sdot sdot 119909
minus119896119901
1+119895
119901(119905minus120591119901))119879where (119905minus120591
119894) isin
[119905minus119896119894
1+119895minus1
119905minus119896119894
1+119895] and 119909
1198962minus119895
119894(119905119891
minus 119905) 1 le 119894 le 119901 is the 119894thcomponent of (1199091198962minus119895
1(119905119891minus 119905) sdot sdot sdot 119909
1198962minus119895
119901(119905119891minus 119905))119879 where (119905
119891minus 119905) isin
[1199051198962minus119895minus1
1199051198962minus119895] Also
119896119894
1=
120591119894
ℎ
120591119894
ℎ
isin N
([
120591119894
ℎ
] + 1)
120591119894
ℎ
notin N
1 le 119894 le 119901
1198962=
119905119891
ℎ
119905119891
ℎ
isin N
([
119905119891
ℎ
] + 1)
119905119891
ℎ
notin N
(5)
where [120591119894ℎ] and [119905
119891ℎ] denote the integer part of 120591
119894ℎ and
119905119891ℎ respectivelyOur strategy is to use Bezier curves to approximate the
solutions x119895(119905) and u
119895(119905) by k
119895(119905) and w
119895(119905) respectively
where k119895(119905) and w
119895(119905) are given below Individual Bezier
curves that are defined over the subintervals are joinedtogether to form the Bezier spline curves For 119895 = 1 2 119896define the Bezier polynomials of degree 119899 that approximaterespectively the actions of x
119895(119905) and u
119895(119905) over the interval
[119905119895minus1
119905119895] as follows
k119895(119905) =
119899
sum
119903=0
a119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
w119895(119905) =
119899
sum
119903=0
b119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
(6)
where
119861119903119899
(
119905 minus 119905119895minus1
ℎ
) = (
119899
119903)
1
ℎ119899(119905119895minus 119905)
119899minus119903
(119905 minus 119905119895minus1
)
119903
(7)
is the Bernstein polynomial of degree 119899 over the interval[119905119895minus1
119905119895] a119895119903and b119895
119903are respectively 119901 and 119898 ordered vectors
from the control points (see [22]) By substituting (6) in (4)1198771119895
(119905) for 119905 isin [119905119895minus1
119905119895] can be defined as follows
1198771119895
(119905) = k119895(119905) minus 119860 (119905) k
119895(119905)
minus 119862 (119905) (Vminus1198961
1+119895
1(119905 minus 1205911) sdot sdot sdot Vminus119896
119901
1+119895
119901(119905 minus 120591119901))
119879
minus 119863 (119905) (V1198962minus1198951
(119905119891minus 119905) sdot sdot sdot V1198962minus119895
119901(119905119891minus 119905))
119879
minus 119866 (119905)w119895(119905)
(8)
Let k(119905) = sum119896
119895=11205941
119895(119905)k119895(119905) and w(119905) = sum
119896
119895=11205942
119895(119905)w119895(119905)
where1205941119895(119905) and120594
2
119895(119905) are respectively characteristic function
of k119895(119905) and w
119895(119905) for 119905 isin [119905
119895minus1 119905119895] Beside the boundary
conditions on k(119905) at each node we need to impose thecontinuity condition on each successive pair of k
119895(119905) to
guarantee the smoothness
Since the differential equation is of first order the conti-nuity of x (or k) and its first derivative gives
k(119904)119895
(119905119895) = k(119904)119895+1
(119905119895) 119904 = 0 1 119895 = 1 2 119896 minus 1 (9)
where k(119904)119895(119905119895) is the 119904th derivative k
119895(119905) with respect to 119905 at
119905 = 119905119895
Thus the vector of control points a119895119903(119903 = 0 1 119899 minus 1 119899)
must satisfy (see the Appendix)
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(10)
According to the definition of the 119905119894= 1199050+ 119894ℎ we get that
119905119895minus 119905119895minus1
= ℎ Therefore
a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
)
(11)
One may recall that a119895119903is a 119901 ordered vector This approach is
called the subdivision scheme (or ℎ-refinement in the finiteelement literature) This method is based on the control-point-based method
Remark 1 By considering the 1198621 continuity of w the follow-ing constraints will be added to constraints in (10)
b119895119899(119905119895minus 119905119895minus1
)
119899
= b119895+10
(119905119895+1
minus 119905119895)
119899
(b119895119899minus b119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (b119895+11
minus b119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(12)
where the so-called b119895119903(119903 = 0 1 119899 minus 1 119899) is an 119898 ordered
vectorNow the residual function can be defined in 119878
119895as follows
119877119895= int
119905119895
119905119895minus1
100381710038171003817100381710038171198771119895
(119905)
10038171003817100381710038171003817
2
119889119905 (13)
where sdot is the Euclidean norm (recall that 1198771119895
(119905) is a 119901
vector where 119905 isin 119878119895)
Our aim is to solve the following problem over 119878 =
⋃119896
119895=1119878119895
min119896
sum
119895=1
119877119895
st a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
) 119895 = 1 2 119896 minus 1
(14)
The mathematical programming problem (14) can be solvedby many subroutine algorithms Here we used Maple 12 tosolve this optimization problem
4 Mathematical Problems in Engineering
Remark 2 Consider the following boundary value problem
y (119905) = 119877 (119905) y (119905) + 119876 (119905) y (119905 minus 120572) + 119878 (119905) z (119905) + a (119905)
z (119905) = 119881 (119905) y (119905) + 119870 (119905) z (119905 + 120572) + 119882 (119905) z (119905) + b (119905)
y (1199050) = y0
y (119905) = 120601 (119905) 119905 isin [minus120572 1199050)
z (119905119891) = z0
z (119905) = 120595 (119905) 119905 isin (119905119891 119905119891+ 120572]
(15)
where y(119905) z(119905) a(119905) b(119905) 120601(119905) and 120595(119905) are the vectorsof appropriate dimensions 119877(119905) 119876(119905) 119878(119905) 119881(119905) 119870(119905) and119882(119905) are the matrices of appropriate dimensions and 120572 isnonnegative constant time delay
Let
x (119905) = [y(119905)119879 z(119905119891minus 119905)
119879
]
119879
(16)
where 119879 is the transpose then
x (119905) = [y119879 (119905) minusz119879 (119905119891minus 119905)]
119879 (17)
satisfies that
x (119905) = 119860 (119905) x (119905) + 119862 (119905) x (119905 minus 120572)
+ 119863 (119905) x (119905119891minus 119905) + u (119905) 119905 isin [119905
0 119905119891]
x (1199050) = x0= [y1198790
z1198790]
119879
(18)
where
119860 (119905) = 119864(2)
11otimes 119877 (119905) minus 119864
(2)
22otimes 119882(119905
119891minus 119905)
119862 (119905) = 119864(2)
11otimes 119876 (119905) minus 119864
(2)
22otimes 119870 (119905
119891minus 119905)
119863 (119905) = 119864(2)
12otimes 119878 (119905) minus 119864
(2)
21otimes 119881 (119905
119891minus 119905)
u (119905) = [a119879 (119905) minusb119879 (119905119891minus 119905)]
119879
(19)
where119864(119891)119894119895
is the119891times119891matrix with 1 at its entry (119894 119895) and zeroselsewhere and otimes is Kronecker product (see eg [4 38 39])
Remark 3 Now the following delay differential equation canbe considered
(119905) = 119891 (119905 119909 (119905) 119909 (119905 minus 120591 (119905 119909 (119905)))) 119905 ge 0 (20)
with initial condition
119909 (119905) = 120601 (119905) 119905 isin [minus120582 0] (21)
where 120582 equiv inf119905 minus 120591(119905 119906) 119905 ge 0 119906 isin R In the case when 120582 isnot finite [minus120582 0] denotes the interval (minusinfin 0]
Furthermore we assume that
120591 (119905 119906) ge 0 forall119905 ge 0 119906 isin R (22)
that is (20) is a delay differential equationThe existence anduniqueness of the solution of initial value problem (20)-(21)was stated in [40]
Equation (20) is converted into a nonlinear programmingproblem (NLP) by applying Bezier control points methodwhereas the MATLAB optimization routine FMINCON isused for solving resulting NLP Numerical example showsthat the proposed method is efficient and very easy to use
Remark 4 Now we limit ourselves to consider the followingnonlinear delay differential equation in the type
119871119909 (119905) = 119865 (119905 119909 (119905) 119909 (120591 (119905))) 1199050le 119905 le 119905
119891 (23)
with the following initial conditions
119909(119896)
(1199050) = 119909119896
0 119896 = 0 1 119899 minus 1
119909 (119905) = 120601 (119905) 119905 le 1199050
(24)
where the differential operator 119871 is defined by 119871(sdot) =
119889119899(sdot)119889119905119899
3 Convergence Analysis
In this section without loss of generality we analyze theconvergence of the control-point-based method applied tothe problem (2) with time delay in state when 119901 = 119898 = 1and the time interval is [0 1] So the following problem isconsidered
119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) =
119889119909 (119905)
119889119905
minus 119860 (119905) 119909 (119905) minus 119862 (119905) 119909 (119905 minus 120591) minus 119866 (119905) 119906 (119905)
minus 119863 (119905) 119909 (1 minus 119905) = 119865 (119905) 119905 isin [0 1]
119909 (119905) = 1199090= 119886 119905 le 0 119909 (1) = 119909
119891= 119887
119906 (119905) = 1199060= 1198861 119905 le 0
(25)
where 119909(119905) isin 119877 119906(119905) isin 119877 and 119886 119887 1198861are given real numbers
and 119860(119905) 119862(119905) 119866(119905) 119863(119905) and 119865(119905) are known polynomialsfor 119905 isin [0 1] The constant time delay 120591 is nonnegative
Without loss of generality we consider the interval [0 1]instead of [119905
0 119905119891] since the variable 119905 can be changed with the
new variable 119911 by 119905 = (119905119891minus 1199050)119911 + 1199050where 119911 isin [0 1]
Lemma 5 For a polynomial in Bezier form
119909 (119905) =
1198991
sum
119894=0
1198861198941198991
1198611198941198991(119905) (26)
we have
sum1198991
119894=01198862
1198941198991
1198991+ 1
ge
sum1198991+1
119894=01198862
1198941198991+1
1198991+ 2
ge sdot sdot sdot
ge
sum1198991+1198981
119894=01198862
1198941198991+1198981
1198991+ 1198981+ 1
997888rarr int
1
0
1199092
(119905) 119889119905 1198981997888rarr +infin
(27)
Mathematical Problems in Engineering 5
where 1198861198941198991+1198981
is the Bezier coefficient of 119909(119905) after degree-elevating to degree 119899
1+ 1198981
Proof See [22 page 245]
The convergence of the approximate solution could bedone in two ways
(1) degree raising the Bezier polynomial approximation(2) subdivision of the time interval
In the following the convergence in each case canbe proven although in numerical examples we used onlysubdivision case (see [32])
31 Degree Raising
Theorem 6 If the problem (25) with inverse time in state hasa unique 119862
1 continuous trajectory solution 119909 1198620 continuouscontrol solution 119906 then the approximate solution obtained bythe control-point-based method converges to the exact solution(119909 119906) as the degree of the approximate solution tends to infinity
Proof Given an arbitrary small positive number 120598 gt
0 by the Weierstrass theorem (see [41]) one can easilyfind polynomials 119876
11198731
(119905) of degree 1198731and 119876
21198732
(119905) ofdegree 119873
2such that 119889119894119876
11198731
(119905)119889119905119894minus 119889119894119909(119905)119889119905
119894infin
le 1205981611988911989411987611198731
(119905 minus 120591)119889119905119894minus 119889119894119909(119905 minus 120591)119889119905
119894infin
le 12059816 119894 = 0 111987621198732
(119905) minus 119906(119905)infin
le 12059816 and 11987611198731
(1 minus 119905) minus 119909(1 minus 119905)infin
le
12059816 where sdot infin
stands for the 119871infin-norm over [0 1]
Especially we have10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
le
120598
16
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
le
120598
16
(28)
In general 11987611198731
(119905) and 11987621198732
(119905) do not satisfy the boundaryconditions After a small perturbation with linear and con-stant polynomials 120572119905 + 120573 120574 respectively for 119876
11198731
(119905) and11987621198732
(119905) we can obtain polynomials 11987511198731
(119905) = 11987611198731
(119905) +
(120572119905 + 120573) and 11987521198732
(119905) = 11987621198732
(119905) + 120574 such that 11987511198731
(119905) satisfiesthe boundary conditions 119875
11198731
(0) = 119886 11987511198731
(1) = 119887 and11987521198732
(0) = 1198861Thus119876
11198731
(0)+120573 = 119886 and11987611198731
(1)+120572+120573 = 119887By using (28) one has
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
=1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
(29)
Since
120572infin
minus10038171003817100381710038171205731003817100381710038171003817infin
le1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
(30)
so
120572infin
le
120598
16
+10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
+
120598
16
=
120598
8
(31)
By the time from 1198861= 11987521198732
(0) = 11987621198732
(0) + 120574
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205741003817100381710038171003817infin
le
120598
16
(32)
Now we have
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987611198731(119905) + 120572119905 + 120573 minus 119909 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987611198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
8
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
+ 120572 minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
le
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 120572infin
le
3120598
16
lt
120598
5
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987621198732(119905) + 120574 minus 119906 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987621198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+10038171003817100381710038171205741003817100381710038171003817infin
le
120598
8
lt
120598
5
(33)
so
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905 minus 120591)
119889119905
minus
119889119909 (119905 minus 120591)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
lt
120598
5
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
lt
120598
5
(34)
Now let 119871119875119873(119905) = 119871(119875
11198731
(119905) 11987521198732
(119905) 11987511198731
(119905minus120591) 11987511198731
(1minus119905)11988911987511198731
(119905)119889119905) = 11988911987511198731
(119905)119889119905minus119860(119905)11987511198731
(119905)minus119862(119905)11987511198731
(119905minus120591)minus
119866(119905)11987521198732
(119905) minus 119863(119905)11987511198731
(1 minus 119905) = 119865(119905) for every 119905 isin [0 1]Thus for119873 ge max119873
1 1198732 an upper bound is found for the
following residual
1003817100381710038171003817119871119875119873(119905) minus 119865 (119905)
1003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
119871(11987511198731(119905) 11987521198732(119905) 11987511198731(119905 minus 120591)
11987511198731(1 minus 119905)
11988911987511198731(119905)
119889119905
) minus 119865 (119905)
100381710038171003817100381710038171003817100381710038171003817infin
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Advances in
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Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 3
in 119905 isin 119878119895 We mention that 119909minus119896
119894
1+119895
119894(119905 minus 120591119894) 1 le 119894 le 119901 is the 119894th
component of (119909minus1198961
1+119895
1(119905minus1205911) sdot sdot sdot 119909
minus119896119901
1+119895
119901(119905minus120591119901))119879where (119905minus120591
119894) isin
[119905minus119896119894
1+119895minus1
119905minus119896119894
1+119895] and 119909
1198962minus119895
119894(119905119891
minus 119905) 1 le 119894 le 119901 is the 119894thcomponent of (1199091198962minus119895
1(119905119891minus 119905) sdot sdot sdot 119909
1198962minus119895
119901(119905119891minus 119905))119879 where (119905
119891minus 119905) isin
[1199051198962minus119895minus1
1199051198962minus119895] Also
119896119894
1=
120591119894
ℎ
120591119894
ℎ
isin N
([
120591119894
ℎ
] + 1)
120591119894
ℎ
notin N
1 le 119894 le 119901
1198962=
119905119891
ℎ
119905119891
ℎ
isin N
([
119905119891
ℎ
] + 1)
119905119891
ℎ
notin N
(5)
where [120591119894ℎ] and [119905
119891ℎ] denote the integer part of 120591
119894ℎ and
119905119891ℎ respectivelyOur strategy is to use Bezier curves to approximate the
solutions x119895(119905) and u
119895(119905) by k
119895(119905) and w
119895(119905) respectively
where k119895(119905) and w
119895(119905) are given below Individual Bezier
curves that are defined over the subintervals are joinedtogether to form the Bezier spline curves For 119895 = 1 2 119896define the Bezier polynomials of degree 119899 that approximaterespectively the actions of x
119895(119905) and u
119895(119905) over the interval
[119905119895minus1
119905119895] as follows
k119895(119905) =
119899
sum
119903=0
a119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
w119895(119905) =
119899
sum
119903=0
b119895119903119861119903119899
(
119905 minus 119905119895minus1
ℎ
)
(6)
where
119861119903119899
(
119905 minus 119905119895minus1
ℎ
) = (
119899
119903)
1
ℎ119899(119905119895minus 119905)
119899minus119903
(119905 minus 119905119895minus1
)
119903
(7)
is the Bernstein polynomial of degree 119899 over the interval[119905119895minus1
119905119895] a119895119903and b119895
119903are respectively 119901 and 119898 ordered vectors
from the control points (see [22]) By substituting (6) in (4)1198771119895
(119905) for 119905 isin [119905119895minus1
119905119895] can be defined as follows
1198771119895
(119905) = k119895(119905) minus 119860 (119905) k
119895(119905)
minus 119862 (119905) (Vminus1198961
1+119895
1(119905 minus 1205911) sdot sdot sdot Vminus119896
119901
1+119895
119901(119905 minus 120591119901))
119879
minus 119863 (119905) (V1198962minus1198951
(119905119891minus 119905) sdot sdot sdot V1198962minus119895
119901(119905119891minus 119905))
119879
minus 119866 (119905)w119895(119905)
(8)
Let k(119905) = sum119896
119895=11205941
119895(119905)k119895(119905) and w(119905) = sum
119896
119895=11205942
119895(119905)w119895(119905)
where1205941119895(119905) and120594
2
119895(119905) are respectively characteristic function
of k119895(119905) and w
119895(119905) for 119905 isin [119905
119895minus1 119905119895] Beside the boundary
conditions on k(119905) at each node we need to impose thecontinuity condition on each successive pair of k
119895(119905) to
guarantee the smoothness
Since the differential equation is of first order the conti-nuity of x (or k) and its first derivative gives
k(119904)119895
(119905119895) = k(119904)119895+1
(119905119895) 119904 = 0 1 119895 = 1 2 119896 minus 1 (9)
where k(119904)119895(119905119895) is the 119904th derivative k
119895(119905) with respect to 119905 at
119905 = 119905119895
Thus the vector of control points a119895119903(119903 = 0 1 119899 minus 1 119899)
must satisfy (see the Appendix)
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(10)
According to the definition of the 119905119894= 1199050+ 119894ℎ we get that
119905119895minus 119905119895minus1
= ℎ Therefore
a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
)
(11)
One may recall that a119895119903is a 119901 ordered vector This approach is
called the subdivision scheme (or ℎ-refinement in the finiteelement literature) This method is based on the control-point-based method
Remark 1 By considering the 1198621 continuity of w the follow-ing constraints will be added to constraints in (10)
b119895119899(119905119895minus 119905119895minus1
)
119899
= b119895+10
(119905119895+1
minus 119905119895)
119899
(b119895119899minus b119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (b119895+11
minus b119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(12)
where the so-called b119895119903(119903 = 0 1 119899 minus 1 119899) is an 119898 ordered
vectorNow the residual function can be defined in 119878
119895as follows
119877119895= int
119905119895
119905119895minus1
100381710038171003817100381710038171198771119895
(119905)
10038171003817100381710038171003817
2
119889119905 (13)
where sdot is the Euclidean norm (recall that 1198771119895
(119905) is a 119901
vector where 119905 isin 119878119895)
Our aim is to solve the following problem over 119878 =
⋃119896
119895=1119878119895
min119896
sum
119895=1
119877119895
st a119895119899= a119895+10
(a119895119899minus a119895119899minus1
) = (a119895+11
minus a119895+10
) 119895 = 1 2 119896 minus 1
(14)
The mathematical programming problem (14) can be solvedby many subroutine algorithms Here we used Maple 12 tosolve this optimization problem
4 Mathematical Problems in Engineering
Remark 2 Consider the following boundary value problem
y (119905) = 119877 (119905) y (119905) + 119876 (119905) y (119905 minus 120572) + 119878 (119905) z (119905) + a (119905)
z (119905) = 119881 (119905) y (119905) + 119870 (119905) z (119905 + 120572) + 119882 (119905) z (119905) + b (119905)
y (1199050) = y0
y (119905) = 120601 (119905) 119905 isin [minus120572 1199050)
z (119905119891) = z0
z (119905) = 120595 (119905) 119905 isin (119905119891 119905119891+ 120572]
(15)
where y(119905) z(119905) a(119905) b(119905) 120601(119905) and 120595(119905) are the vectorsof appropriate dimensions 119877(119905) 119876(119905) 119878(119905) 119881(119905) 119870(119905) and119882(119905) are the matrices of appropriate dimensions and 120572 isnonnegative constant time delay
Let
x (119905) = [y(119905)119879 z(119905119891minus 119905)
119879
]
119879
(16)
where 119879 is the transpose then
x (119905) = [y119879 (119905) minusz119879 (119905119891minus 119905)]
119879 (17)
satisfies that
x (119905) = 119860 (119905) x (119905) + 119862 (119905) x (119905 minus 120572)
+ 119863 (119905) x (119905119891minus 119905) + u (119905) 119905 isin [119905
0 119905119891]
x (1199050) = x0= [y1198790
z1198790]
119879
(18)
where
119860 (119905) = 119864(2)
11otimes 119877 (119905) minus 119864
(2)
22otimes 119882(119905
119891minus 119905)
119862 (119905) = 119864(2)
11otimes 119876 (119905) minus 119864
(2)
22otimes 119870 (119905
119891minus 119905)
119863 (119905) = 119864(2)
12otimes 119878 (119905) minus 119864
(2)
21otimes 119881 (119905
119891minus 119905)
u (119905) = [a119879 (119905) minusb119879 (119905119891minus 119905)]
119879
(19)
where119864(119891)119894119895
is the119891times119891matrix with 1 at its entry (119894 119895) and zeroselsewhere and otimes is Kronecker product (see eg [4 38 39])
Remark 3 Now the following delay differential equation canbe considered
(119905) = 119891 (119905 119909 (119905) 119909 (119905 minus 120591 (119905 119909 (119905)))) 119905 ge 0 (20)
with initial condition
119909 (119905) = 120601 (119905) 119905 isin [minus120582 0] (21)
where 120582 equiv inf119905 minus 120591(119905 119906) 119905 ge 0 119906 isin R In the case when 120582 isnot finite [minus120582 0] denotes the interval (minusinfin 0]
Furthermore we assume that
120591 (119905 119906) ge 0 forall119905 ge 0 119906 isin R (22)
that is (20) is a delay differential equationThe existence anduniqueness of the solution of initial value problem (20)-(21)was stated in [40]
Equation (20) is converted into a nonlinear programmingproblem (NLP) by applying Bezier control points methodwhereas the MATLAB optimization routine FMINCON isused for solving resulting NLP Numerical example showsthat the proposed method is efficient and very easy to use
Remark 4 Now we limit ourselves to consider the followingnonlinear delay differential equation in the type
119871119909 (119905) = 119865 (119905 119909 (119905) 119909 (120591 (119905))) 1199050le 119905 le 119905
119891 (23)
with the following initial conditions
119909(119896)
(1199050) = 119909119896
0 119896 = 0 1 119899 minus 1
119909 (119905) = 120601 (119905) 119905 le 1199050
(24)
where the differential operator 119871 is defined by 119871(sdot) =
119889119899(sdot)119889119905119899
3 Convergence Analysis
In this section without loss of generality we analyze theconvergence of the control-point-based method applied tothe problem (2) with time delay in state when 119901 = 119898 = 1and the time interval is [0 1] So the following problem isconsidered
119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) =
119889119909 (119905)
119889119905
minus 119860 (119905) 119909 (119905) minus 119862 (119905) 119909 (119905 minus 120591) minus 119866 (119905) 119906 (119905)
minus 119863 (119905) 119909 (1 minus 119905) = 119865 (119905) 119905 isin [0 1]
119909 (119905) = 1199090= 119886 119905 le 0 119909 (1) = 119909
119891= 119887
119906 (119905) = 1199060= 1198861 119905 le 0
(25)
where 119909(119905) isin 119877 119906(119905) isin 119877 and 119886 119887 1198861are given real numbers
and 119860(119905) 119862(119905) 119866(119905) 119863(119905) and 119865(119905) are known polynomialsfor 119905 isin [0 1] The constant time delay 120591 is nonnegative
Without loss of generality we consider the interval [0 1]instead of [119905
0 119905119891] since the variable 119905 can be changed with the
new variable 119911 by 119905 = (119905119891minus 1199050)119911 + 1199050where 119911 isin [0 1]
Lemma 5 For a polynomial in Bezier form
119909 (119905) =
1198991
sum
119894=0
1198861198941198991
1198611198941198991(119905) (26)
we have
sum1198991
119894=01198862
1198941198991
1198991+ 1
ge
sum1198991+1
119894=01198862
1198941198991+1
1198991+ 2
ge sdot sdot sdot
ge
sum1198991+1198981
119894=01198862
1198941198991+1198981
1198991+ 1198981+ 1
997888rarr int
1
0
1199092
(119905) 119889119905 1198981997888rarr +infin
(27)
Mathematical Problems in Engineering 5
where 1198861198941198991+1198981
is the Bezier coefficient of 119909(119905) after degree-elevating to degree 119899
1+ 1198981
Proof See [22 page 245]
The convergence of the approximate solution could bedone in two ways
(1) degree raising the Bezier polynomial approximation(2) subdivision of the time interval
In the following the convergence in each case canbe proven although in numerical examples we used onlysubdivision case (see [32])
31 Degree Raising
Theorem 6 If the problem (25) with inverse time in state hasa unique 119862
1 continuous trajectory solution 119909 1198620 continuouscontrol solution 119906 then the approximate solution obtained bythe control-point-based method converges to the exact solution(119909 119906) as the degree of the approximate solution tends to infinity
Proof Given an arbitrary small positive number 120598 gt
0 by the Weierstrass theorem (see [41]) one can easilyfind polynomials 119876
11198731
(119905) of degree 1198731and 119876
21198732
(119905) ofdegree 119873
2such that 119889119894119876
11198731
(119905)119889119905119894minus 119889119894119909(119905)119889119905
119894infin
le 1205981611988911989411987611198731
(119905 minus 120591)119889119905119894minus 119889119894119909(119905 minus 120591)119889119905
119894infin
le 12059816 119894 = 0 111987621198732
(119905) minus 119906(119905)infin
le 12059816 and 11987611198731
(1 minus 119905) minus 119909(1 minus 119905)infin
le
12059816 where sdot infin
stands for the 119871infin-norm over [0 1]
Especially we have10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
le
120598
16
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
le
120598
16
(28)
In general 11987611198731
(119905) and 11987621198732
(119905) do not satisfy the boundaryconditions After a small perturbation with linear and con-stant polynomials 120572119905 + 120573 120574 respectively for 119876
11198731
(119905) and11987621198732
(119905) we can obtain polynomials 11987511198731
(119905) = 11987611198731
(119905) +
(120572119905 + 120573) and 11987521198732
(119905) = 11987621198732
(119905) + 120574 such that 11987511198731
(119905) satisfiesthe boundary conditions 119875
11198731
(0) = 119886 11987511198731
(1) = 119887 and11987521198732
(0) = 1198861Thus119876
11198731
(0)+120573 = 119886 and11987611198731
(1)+120572+120573 = 119887By using (28) one has
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
=1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
(29)
Since
120572infin
minus10038171003817100381710038171205731003817100381710038171003817infin
le1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
(30)
so
120572infin
le
120598
16
+10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
+
120598
16
=
120598
8
(31)
By the time from 1198861= 11987521198732
(0) = 11987621198732
(0) + 120574
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205741003817100381710038171003817infin
le
120598
16
(32)
Now we have
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987611198731(119905) + 120572119905 + 120573 minus 119909 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987611198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
8
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
+ 120572 minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
le
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 120572infin
le
3120598
16
lt
120598
5
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987621198732(119905) + 120574 minus 119906 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987621198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+10038171003817100381710038171205741003817100381710038171003817infin
le
120598
8
lt
120598
5
(33)
so
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905 minus 120591)
119889119905
minus
119889119909 (119905 minus 120591)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
lt
120598
5
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
lt
120598
5
(34)
Now let 119871119875119873(119905) = 119871(119875
11198731
(119905) 11987521198732
(119905) 11987511198731
(119905minus120591) 11987511198731
(1minus119905)11988911987511198731
(119905)119889119905) = 11988911987511198731
(119905)119889119905minus119860(119905)11987511198731
(119905)minus119862(119905)11987511198731
(119905minus120591)minus
119866(119905)11987521198732
(119905) minus 119863(119905)11987511198731
(1 minus 119905) = 119865(119905) for every 119905 isin [0 1]Thus for119873 ge max119873
1 1198732 an upper bound is found for the
following residual
1003817100381710038171003817119871119875119873(119905) minus 119865 (119905)
1003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
119871(11987511198731(119905) 11987521198732(119905) 11987511198731(119905 minus 120591)
11987511198731(1 minus 119905)
11988911987511198731(119905)
119889119905
) minus 119865 (119905)
100381710038171003817100381710038171003817100381710038171003817infin
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
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Complex AnalysisJournal of
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OptimizationJournal of
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Volume 2014
International Journal of
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OperationsResearch
Advances in
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Advances in
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Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
Remark 2 Consider the following boundary value problem
y (119905) = 119877 (119905) y (119905) + 119876 (119905) y (119905 minus 120572) + 119878 (119905) z (119905) + a (119905)
z (119905) = 119881 (119905) y (119905) + 119870 (119905) z (119905 + 120572) + 119882 (119905) z (119905) + b (119905)
y (1199050) = y0
y (119905) = 120601 (119905) 119905 isin [minus120572 1199050)
z (119905119891) = z0
z (119905) = 120595 (119905) 119905 isin (119905119891 119905119891+ 120572]
(15)
where y(119905) z(119905) a(119905) b(119905) 120601(119905) and 120595(119905) are the vectorsof appropriate dimensions 119877(119905) 119876(119905) 119878(119905) 119881(119905) 119870(119905) and119882(119905) are the matrices of appropriate dimensions and 120572 isnonnegative constant time delay
Let
x (119905) = [y(119905)119879 z(119905119891minus 119905)
119879
]
119879
(16)
where 119879 is the transpose then
x (119905) = [y119879 (119905) minusz119879 (119905119891minus 119905)]
119879 (17)
satisfies that
x (119905) = 119860 (119905) x (119905) + 119862 (119905) x (119905 minus 120572)
+ 119863 (119905) x (119905119891minus 119905) + u (119905) 119905 isin [119905
0 119905119891]
x (1199050) = x0= [y1198790
z1198790]
119879
(18)
where
119860 (119905) = 119864(2)
11otimes 119877 (119905) minus 119864
(2)
22otimes 119882(119905
119891minus 119905)
119862 (119905) = 119864(2)
11otimes 119876 (119905) minus 119864
(2)
22otimes 119870 (119905
119891minus 119905)
119863 (119905) = 119864(2)
12otimes 119878 (119905) minus 119864
(2)
21otimes 119881 (119905
119891minus 119905)
u (119905) = [a119879 (119905) minusb119879 (119905119891minus 119905)]
119879
(19)
where119864(119891)119894119895
is the119891times119891matrix with 1 at its entry (119894 119895) and zeroselsewhere and otimes is Kronecker product (see eg [4 38 39])
Remark 3 Now the following delay differential equation canbe considered
(119905) = 119891 (119905 119909 (119905) 119909 (119905 minus 120591 (119905 119909 (119905)))) 119905 ge 0 (20)
with initial condition
119909 (119905) = 120601 (119905) 119905 isin [minus120582 0] (21)
where 120582 equiv inf119905 minus 120591(119905 119906) 119905 ge 0 119906 isin R In the case when 120582 isnot finite [minus120582 0] denotes the interval (minusinfin 0]
Furthermore we assume that
120591 (119905 119906) ge 0 forall119905 ge 0 119906 isin R (22)
that is (20) is a delay differential equationThe existence anduniqueness of the solution of initial value problem (20)-(21)was stated in [40]
Equation (20) is converted into a nonlinear programmingproblem (NLP) by applying Bezier control points methodwhereas the MATLAB optimization routine FMINCON isused for solving resulting NLP Numerical example showsthat the proposed method is efficient and very easy to use
Remark 4 Now we limit ourselves to consider the followingnonlinear delay differential equation in the type
119871119909 (119905) = 119865 (119905 119909 (119905) 119909 (120591 (119905))) 1199050le 119905 le 119905
119891 (23)
with the following initial conditions
119909(119896)
(1199050) = 119909119896
0 119896 = 0 1 119899 minus 1
119909 (119905) = 120601 (119905) 119905 le 1199050
(24)
where the differential operator 119871 is defined by 119871(sdot) =
119889119899(sdot)119889119905119899
3 Convergence Analysis
In this section without loss of generality we analyze theconvergence of the control-point-based method applied tothe problem (2) with time delay in state when 119901 = 119898 = 1and the time interval is [0 1] So the following problem isconsidered
119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) =
119889119909 (119905)
119889119905
minus 119860 (119905) 119909 (119905) minus 119862 (119905) 119909 (119905 minus 120591) minus 119866 (119905) 119906 (119905)
minus 119863 (119905) 119909 (1 minus 119905) = 119865 (119905) 119905 isin [0 1]
119909 (119905) = 1199090= 119886 119905 le 0 119909 (1) = 119909
119891= 119887
119906 (119905) = 1199060= 1198861 119905 le 0
(25)
where 119909(119905) isin 119877 119906(119905) isin 119877 and 119886 119887 1198861are given real numbers
and 119860(119905) 119862(119905) 119866(119905) 119863(119905) and 119865(119905) are known polynomialsfor 119905 isin [0 1] The constant time delay 120591 is nonnegative
Without loss of generality we consider the interval [0 1]instead of [119905
0 119905119891] since the variable 119905 can be changed with the
new variable 119911 by 119905 = (119905119891minus 1199050)119911 + 1199050where 119911 isin [0 1]
Lemma 5 For a polynomial in Bezier form
119909 (119905) =
1198991
sum
119894=0
1198861198941198991
1198611198941198991(119905) (26)
we have
sum1198991
119894=01198862
1198941198991
1198991+ 1
ge
sum1198991+1
119894=01198862
1198941198991+1
1198991+ 2
ge sdot sdot sdot
ge
sum1198991+1198981
119894=01198862
1198941198991+1198981
1198991+ 1198981+ 1
997888rarr int
1
0
1199092
(119905) 119889119905 1198981997888rarr +infin
(27)
Mathematical Problems in Engineering 5
where 1198861198941198991+1198981
is the Bezier coefficient of 119909(119905) after degree-elevating to degree 119899
1+ 1198981
Proof See [22 page 245]
The convergence of the approximate solution could bedone in two ways
(1) degree raising the Bezier polynomial approximation(2) subdivision of the time interval
In the following the convergence in each case canbe proven although in numerical examples we used onlysubdivision case (see [32])
31 Degree Raising
Theorem 6 If the problem (25) with inverse time in state hasa unique 119862
1 continuous trajectory solution 119909 1198620 continuouscontrol solution 119906 then the approximate solution obtained bythe control-point-based method converges to the exact solution(119909 119906) as the degree of the approximate solution tends to infinity
Proof Given an arbitrary small positive number 120598 gt
0 by the Weierstrass theorem (see [41]) one can easilyfind polynomials 119876
11198731
(119905) of degree 1198731and 119876
21198732
(119905) ofdegree 119873
2such that 119889119894119876
11198731
(119905)119889119905119894minus 119889119894119909(119905)119889119905
119894infin
le 1205981611988911989411987611198731
(119905 minus 120591)119889119905119894minus 119889119894119909(119905 minus 120591)119889119905
119894infin
le 12059816 119894 = 0 111987621198732
(119905) minus 119906(119905)infin
le 12059816 and 11987611198731
(1 minus 119905) minus 119909(1 minus 119905)infin
le
12059816 where sdot infin
stands for the 119871infin-norm over [0 1]
Especially we have10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
le
120598
16
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
le
120598
16
(28)
In general 11987611198731
(119905) and 11987621198732
(119905) do not satisfy the boundaryconditions After a small perturbation with linear and con-stant polynomials 120572119905 + 120573 120574 respectively for 119876
11198731
(119905) and11987621198732
(119905) we can obtain polynomials 11987511198731
(119905) = 11987611198731
(119905) +
(120572119905 + 120573) and 11987521198732
(119905) = 11987621198732
(119905) + 120574 such that 11987511198731
(119905) satisfiesthe boundary conditions 119875
11198731
(0) = 119886 11987511198731
(1) = 119887 and11987521198732
(0) = 1198861Thus119876
11198731
(0)+120573 = 119886 and11987611198731
(1)+120572+120573 = 119887By using (28) one has
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
=1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
(29)
Since
120572infin
minus10038171003817100381710038171205731003817100381710038171003817infin
le1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
(30)
so
120572infin
le
120598
16
+10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
+
120598
16
=
120598
8
(31)
By the time from 1198861= 11987521198732
(0) = 11987621198732
(0) + 120574
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205741003817100381710038171003817infin
le
120598
16
(32)
Now we have
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987611198731(119905) + 120572119905 + 120573 minus 119909 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987611198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
8
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
+ 120572 minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
le
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 120572infin
le
3120598
16
lt
120598
5
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987621198732(119905) + 120574 minus 119906 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987621198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+10038171003817100381710038171205741003817100381710038171003817infin
le
120598
8
lt
120598
5
(33)
so
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905 minus 120591)
119889119905
minus
119889119909 (119905 minus 120591)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
lt
120598
5
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
lt
120598
5
(34)
Now let 119871119875119873(119905) = 119871(119875
11198731
(119905) 11987521198732
(119905) 11987511198731
(119905minus120591) 11987511198731
(1minus119905)11988911987511198731
(119905)119889119905) = 11988911987511198731
(119905)119889119905minus119860(119905)11987511198731
(119905)minus119862(119905)11987511198731
(119905minus120591)minus
119866(119905)11987521198732
(119905) minus 119863(119905)11987511198731
(1 minus 119905) = 119865(119905) for every 119905 isin [0 1]Thus for119873 ge max119873
1 1198732 an upper bound is found for the
following residual
1003817100381710038171003817119871119875119873(119905) minus 119865 (119905)
1003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
119871(11987511198731(119905) 11987521198732(119905) 11987511198731(119905 minus 120591)
11987511198731(1 minus 119905)
11988911987511198731(119905)
119889119905
) minus 119865 (119905)
100381710038171003817100381710038171003817100381710038171003817infin
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of
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Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
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Complex AnalysisJournal of
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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Advances in
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
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Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 5
where 1198861198941198991+1198981
is the Bezier coefficient of 119909(119905) after degree-elevating to degree 119899
1+ 1198981
Proof See [22 page 245]
The convergence of the approximate solution could bedone in two ways
(1) degree raising the Bezier polynomial approximation(2) subdivision of the time interval
In the following the convergence in each case canbe proven although in numerical examples we used onlysubdivision case (see [32])
31 Degree Raising
Theorem 6 If the problem (25) with inverse time in state hasa unique 119862
1 continuous trajectory solution 119909 1198620 continuouscontrol solution 119906 then the approximate solution obtained bythe control-point-based method converges to the exact solution(119909 119906) as the degree of the approximate solution tends to infinity
Proof Given an arbitrary small positive number 120598 gt
0 by the Weierstrass theorem (see [41]) one can easilyfind polynomials 119876
11198731
(119905) of degree 1198731and 119876
21198732
(119905) ofdegree 119873
2such that 119889119894119876
11198731
(119905)119889119905119894minus 119889119894119909(119905)119889119905
119894infin
le 1205981611988911989411987611198731
(119905 minus 120591)119889119905119894minus 119889119894119909(119905 minus 120591)119889119905
119894infin
le 12059816 119894 = 0 111987621198732
(119905) minus 119906(119905)infin
le 12059816 and 11987611198731
(1 minus 119905) minus 119909(1 minus 119905)infin
le
12059816 where sdot infin
stands for the 119871infin-norm over [0 1]
Especially we have10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
le
120598
16
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
le
120598
16
(28)
In general 11987611198731
(119905) and 11987621198732
(119905) do not satisfy the boundaryconditions After a small perturbation with linear and con-stant polynomials 120572119905 + 120573 120574 respectively for 119876
11198731
(119905) and11987621198732
(119905) we can obtain polynomials 11987511198731
(119905) = 11987611198731
(119905) +
(120572119905 + 120573) and 11987521198732
(119905) = 11987621198732
(119905) + 120574 such that 11987511198731
(119905) satisfiesthe boundary conditions 119875
11198731
(0) = 119886 11987511198731
(1) = 119887 and11987521198732
(0) = 1198861Thus119876
11198731
(0)+120573 = 119886 and11987611198731
(1)+120572+120573 = 119887By using (28) one has
10038171003817100381710038171003817119887 minus 11987611198731(1)
10038171003817100381710038171003817infin
=1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
10038171003817100381710038171003817119886 minus 119876
11198731(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
(29)
Since
120572infin
minus10038171003817100381710038171205731003817100381710038171003817infin
le1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
16
(30)
so
120572infin
le
120598
16
+10038171003817100381710038171205731003817100381710038171003817infin
le
120598
16
+
120598
16
=
120598
8
(31)
By the time from 1198861= 11987521198732
(0) = 11987621198732
(0) + 120574
100381710038171003817100381710038171198861minus 11987621198732(0)
10038171003817100381710038171003817infin
=10038171003817100381710038171205741003817100381710038171003817infin
le
120598
16
(32)
Now we have
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987611198731(119905) + 120572119905 + 120573 minus 119909 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987611198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+1003817100381710038171003817120572 + 120573
1003817100381710038171003817infin
le
120598
8
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
+ 120572 minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
le
100381710038171003817100381710038171003817100381710038171003817
11988911987611198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 120572infin
le
3120598
16
lt
120598
5
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
=
1003817100381710038171003817100381711987621198732(119905) + 120574 minus 119906 (119905)
10038171003817100381710038171003817infin
le
1003817100381710038171003817100381711987621198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+10038171003817100381710038171205741003817100381710038171003817infin
le
120598
8
lt
120598
5
(33)
so
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
lt
120598
5
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905 minus 120591)
119889119905
minus
119889119909 (119905 minus 120591)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
lt
120598
5
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
lt
120598
5
(34)
Now let 119871119875119873(119905) = 119871(119875
11198731
(119905) 11987521198732
(119905) 11987511198731
(119905minus120591) 11987511198731
(1minus119905)11988911987511198731
(119905)119889119905) = 11988911987511198731
(119905)119889119905minus119860(119905)11987511198731
(119905)minus119862(119905)11987511198731
(119905minus120591)minus
119866(119905)11987521198732
(119905) minus 119863(119905)11987511198731
(1 minus 119905) = 119865(119905) for every 119905 isin [0 1]Thus for119873 ge max119873
1 1198732 an upper bound is found for the
following residual
1003817100381710038171003817119871119875119873(119905) minus 119865 (119905)
1003817100381710038171003817infin
=
100381710038171003817100381710038171003817100381710038171003817
119871(11987511198731(119905) 11987521198732(119905) 11987511198731(119905 minus 120591)
11987511198731(1 minus 119905)
11988911987511198731(119905)
119889119905
) minus 119865 (119905)
100381710038171003817100381710038171003817100381710038171003817infin
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
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Complex AnalysisJournal of
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Volume 2014
International Journal of
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Advances in
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
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Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
le
100381710038171003817100381710038171003817100381710038171003817
11988911987511198731(119905)
119889119905
minus
119889119909 (119905)
119889119905
100381710038171003817100381710038171003817100381710038171003817infin
+ 119860 (119905)infin
1003817100381710038171003817100381711987511198731(119905) minus 119909 (119905)
10038171003817100381710038171003817infin
+ 119862 (119905)infin
1003817100381710038171003817100381711987511198731(119905 minus 120591) minus 119909 (119905 minus 120591)
10038171003817100381710038171003817infin
+ 119866 (119905)infin
1003817100381710038171003817100381711987521198732(119905) minus 119906 (119905)
10038171003817100381710038171003817infin
+ 119863 (119905)infin
1003817100381710038171003817100381711987511198731(1 minus 119905) minus 119909 (1 minus 119905)
10038171003817100381710038171003817infin
le 1198621(
120598
5
+
120598
5
+
120598
5
+
120598
5
+
120598
5
) = 1198621120598
(35)
where 1198621= 1 + 119860(119905)
infin+ 119862(119905)
infin+ 119866(119905)
infin+ 119863(119905)
infinis a
constantSince the residual 119877(119875
119873) = 119871119875
119873(119905)minus119865(119905) is a polynomial
it can be represented by a Bezier form Therefore we have
119877 (119875119873) =
1198981
sum
119894=0
1198891198941198981
1198611198941198981(119905) (36)
Then by Lemma 5 there exists an integer119872(ge 119873) such thatwhen119898
1gt 119872 we have1003816100381610038161003816100381610038161003816100381610038161003816
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
minus int
1
0
(119877 (119875119873))2
119889119905
1003816100381610038161003816100381610038161003816100381610038161003816
lt 120598 (37)
which gives
1
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
lt 120598 + int
1
0
(119877 (119875119873))2
119889119905
le 120598 + 1198622
11205982
(38)
Suppose 119909(119905) and 119906(119905) are approximated solution of (25)obtained by the control-point-based method of degree 119898
2
(1198982ge 1198981ge 119872) Let
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
1198982
sum
119894=0
1198881198941198982
1198611198941198982(119905) 119898
2ge 1198981ge 119872 119905 isin [0 1]
(39)
Define the following norm for difference approximated solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
= int
1
0
1
sum
119895=0
100381610038161003816100381610038161003816100381610038161003816
119889119895119909 (119905)
119889119905119895
minus
119889119895119909 (119905)
119889119905119895
100381610038161003816100381610038161003816100381610038161003816
2
119889119905
+ int
1
0
|119906 (0) minus 119906 (0)| 119889119905
(40)
By (40) Lemma 5 the boundary conditions 119909(0) = 119886 =
11987511198731
(0) = 119909(0) 119909(1) = 119887 = 11987511198731
(1) = 119909(1) and 119906(0) =
1198861= 11987521198732
(0) = 119906(0) one can show that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862( |119909 (0) minus 119909 (0)|
+ |119909 (1) minus 119909 (1)| + |119906 (0) minus 119906 (0)|
+
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817
2
2
)
= 119862int
1
0
1198982
sum
119894=0
(1198881198941198982
1198611198941198982(119905))
2
119889119905
le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
(41)
The last inequality in (41) is obtained by Lemma 5 where119862 isa constant positive number Now
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905)) le
119862
1198982+ 1
1198982
sum
119894=0
1198882
1198941198982
le
119862
1198982+ 1
1198982
sum
119894=0
1198892
1198941198982
le
119862
1198981+ 1
1198981
sum
119894=0
1198892
1198941198981
le 119862 (120598 + 1198622
11205982
)
= 1205981 1198981ge 119872
(42)
where the last inequality in (42) comes from (36) Thiscompletes the proof
32 Subdivision
Theorem 7 Let (119909 119906) be the approximate solution of theproblem (25) with inverse time obtained by the subdivisionscheme of the control-point-based method If (25) has a uniquesolution (119909 119906) and (119909 119906) is smooth enough so that the cubicspline 119879(119909 119906) interpolates to (119909 119906) and converges to (119909 119906) inthe order 119874(ℎ
119902) (119902 gt 2) where ℎ is the maximal width of all
subintervals then (119909 119906) converges to (119909 119906) as ℎ rarr 0
Proof We first impose a uniform partition prod119889= ⋃119894[119905119894 119905119894+1
]
on the interval [0 1] as 119905119894= 119894119889 where 119889 = 1(119899
1+ 1)
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
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OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
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Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 7
Let 119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905) be the cubic
spline over prod119889which is interpolating to (119909 119906) Then for an
arbitrary small positive number 120598 gt 0 there exists a 1205751gt 0
such that
10038171003817100381710038171003817100381710038171003817
119871 (119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus119871(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038171003817100381710038171003817100381710038171003817infin
le 120598
(43)
provided that 119889 lt 1205751 Let 119877(119868
119889(119909(119905) 119906(119905) 119909(119905 minus
120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) = 119871(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus
119905) 119889119909(119905)119889119905)) minus 119865(119905) be the residual For each subinterval[119905119894 119905119894+1
] 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905))
is a polynomial On each interval [119905119894 119905119894+1
] we imposeanother uniform partition prod
119894ℎ= ⋃
119895[119905119894119895 119905119894119895+1
] as119905119894119895
= 119894119889 + 119895ℎ where ℎ = 119889(1198981+ 1) 119895 = 0 119898
1
Express 119877(119868119889(119909(119905) 119906(119905) 119909(119905 minus 120591) 119909(1 minus 119905) 119889119909(119905)119889119905)) in
[119905119894119895minus1
119905119894119895] as
119877(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
=
119897
sum
1199011=0
119903119894119895
1199011
1198611199011119897(119905) 119905 isin [119905
119894119895minus1 119905119894119895]
(44)
By Lemma 3 in [22] there exists a 1205752gt 0 (120575
2le 1205751) such that
when ℎ lt 1205752 we have
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1)
times int
119905119894+1
119905119894
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905 le
120598
119889
(45)
Thus
10038161003816100381610038161003816100381610038161003816100381610038161003816
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
minus (119897 + 1) int
1
0
1198772
(119868119889(119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
10038161003816100381610038161003816100381610038161003816100381610038161003816
119889119905
le 120598
(46)
or
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
lt (119897 + 1) int
1
0
1198772
(119868119889(
119889119909 (119905)
119889119905
119909 (119905) 119906 (119905) 119909 (119905 minus 120591)
119909 (1 minus 119905)
119889119909 (119905)
119889119905
))119889119905 + 120598
lt (119897 + 1) 1205982
+ 120598
(47)
Now combining the partitionsprod119889and allprod
119894ℎgives a denser
partition with the length ℎ for each subinterval Suppose(119909(119905) 119906(119905)) is the approximate solution by the control-point-based method with respect to this partition and denote theresidual over [119905
119894119895minus1 119905119894119895] by
119877(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
= 119871(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
) minus 119865 (119905)
=
119897
sum
1199011=0
119888119894119895
1199011
1198611199011119897(119905)
(48)
Define the following norm for difference approximate solu-tion (119909(119905) 119906(119905)) and exact solution (119909(119905) 119906(119905))
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
=
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119909 (119905) minus 119909 (119905)|2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
10038161003816100381610038161003816100381610038161003816
119889119909 (119905)
119889119905
minus
119889119909 (119905)
119889119905
10038161003816100381610038161003816100381610038161003816
2
119889119905
+
1198991
sum
119894=1
1198981
sum
119895=1
int
119905119894119895
119905119894119895minus1
|119906 (0) minus 119906 (0)| 119889119905
(49)
Then there is a constant 119862 such that
(119909 (119905) 119906 (119905)) minus (119909 (119905) 119906 (119905))
le 119862
10038171003817100381710038171003817100381710038171003817
119877((119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
)
minus(119909 (119905) 119906 (119905) 119909 (119905 minus 120591) 119909 (1 minus 119905)
119889119909 (119905)
119889119905
))
100381710038171003817100381710038171003817100381710038172
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
(50)
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
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Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Mathematical Problems in Engineering
the last inequality in (50) is obtained by Lemma 5 It can beshown that
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119888119894119895
1199011
)
2
le
119862
119897 + 1
1198991
sum
119894=1
1198981
sum
119895=1
(119905119894119895
minus 119905119894119895minus1
)
119897
sum
1199011=0
(119903119894119895
1199011
)
2
le 119862(1205982
+
120598
119897 + 1
) = 1205982
(51)
By Lemma 3 in [22] we conclude that the approximatesolution converges to the exact solution in order 119900(ℎ119902) (119902 gt
2) This completes the proof
4 Numerical Examples
Applying the presented method in Examples 1 2 and 3 theBezier curves are chosen as piecewise polynomials of degree3
Example 8 Consider the delay system containing inversetime described by (see [4])
x (119905) = [1199052+ 1 minus119905
2
0 minus9
] x (119905) + [
1 minus1
9 0] x (119905 minus
1
3
)
+ [
1 0
minus1 1] x (1 minus 119905) + [
4119905 + 3
8119905 + 15] 119906 (119905)
120601 (119905) = [
1199052minus 1
1199052+ 1
] 119905 isin [minus
1
3
0]
(52)
where we have the following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199052+ 1]
119879
(53)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 6 wehave the approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
minus1000000001 + 8333333337 times 10minus9119905 + 09999999669119905
2+ 10minus71199053 0 le 119905 le
1
6
minus09999999988 + 813333333 times 10minus9119905 + 09999999829119905
2
1
6
le 119905 le
1
3
minus09999999997 + 200 times 10minus10
119905 + 1199052
1
3
le 119905 le
1
2
minus09999999927 minus 2202222223 times 10minus8119905 + 1000000017119905
2
1
2
le 119905 le
2
3
minus09999999902 minus 1504444443 times 10minus8119905 + 09999999963119905
2+ 10minus81199053
2
3
le 119905 le
5
6
minus1000000032 + 1120666667 times 10minus7119905 + 09999998702119905
2+ 5 times 10
minus81199053
5
6
le 119905 le 1
1199092(119905) =
1000000001 + 0000011825119905 + 099964476691199052+ 00023693119905
3 0 le 119905 le
1
6
1000000001 + 000001180813339119905 + 099964476631199052+ 00023695119905
3
1
6
le 119905 le
1
3
09999999645 + 000001211131104119905 + 099964396691199052+ 00023702119905
3
1
3
le 119905 le
1
2
1000000063 + 000001151408882119905 + 099964521691199052+ 00023693119905
3
1
2
le 119905 le
2
3
09581187057 + 01594325022119905 + 080408138291199052+ 00783674119905
3
2
3
le 119905 le
5
6
09581181451 + 01594344559119905 + 080407910021199052+ 00783683119905
3
5
6
le 119905 le 1
(54)
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 9
The graphs of approximate trajectories are shown in Figures1 and 2
Example 9 Consider the boundary value problem describedby (see [4])
119910 (119905) = 16119905119910 (119905 minus
1
4
) minus 16119911 (119905) + 81199052
+ 17119905 + 16
(119905) = 64119905119910 (119905) minus 64119911 (119905 +
1
4
) + 511199052
+ 76119905 + 65
119910 (119905) = 1199052
minus 1 minus
1
4
le 119905 le 0
119911 (119905) = 1199053
+ 1 1 le 119905 le
5
4
(55)
From (18) we have (see [4])
x (119905) = [
16119905 0
0 64] x (119905 minus
1
4
) + [
0 minus16
64119905 minus 64 0] x (1 minus 119905)
+ [
81199052+ 17119905 + 16
minus511199052+ 178119905 minus 62
]
120601 (119905) = [
1199052minus 1
minus1199053+ 31199052minus 3119905 + 1
] 119905 isin [minus
1
4
0]
(56)
where x(119905) = [1199091(119905) 1199092(119905)]
119879
= [119910(119905) 119911(1 minus 119905)]
119879 and we havethe following exact solution
x (119905) = [1199091(119905) 1199092(119905)]
119879
= [1199052minus 1 1199053+ 1]
119879
(57)
Let 119906(119905) = 1 Then by (14) and choosing 119899 = 3 119896 = 4 we havethe approximate solution x(119905) = [119909
1(119905) 1199092(119905)]
119879
1199091(119905) =
1199052minus 1 0 le 119905 le
1
4
1199052minus 1
1
4
le 119905 le
1
2
1199052minus 1
1
2
le 119905 le
3
4
minus1000000006 + 20625 times 10minus8119905
+09999999751199052+ 10minus81199053
3
4
le 119905 le 1
1199092(119905) = 119905
3
+ 1
(58)
The graphs of approximate trajectories are shown in Figures3 and 4
0 1
minus02
minus04
minus06
minus08
minus1
minus05 05
t
Approximate x1(t)Exact x1(t)
Figure 1The graph of approximated trajectory 1199091(119905) for Example 1
2
18
16
14
12
minus05 0 05 1
t
Approximate x2(t)Exact x2(t)
Figure 2The graph of approximated trajectory 1199092(119905) for Example 1
Example 10 Consider the time-varying delay systemdescribed by (see [42])
[
1(119905)
2(119905)
] = [
0 1
minus25 minus5119905]
[
[
[
[
1199091(119905 minus
1
4
)
1199092(119905 minus
1
4
)
]
]
]
]
+ [
0
1]
[
1199091(119905)
1199092(119905)
] = [
0
0] 119905 isin [minus
1
4
0]
(59)
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
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Complex AnalysisJournal of
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Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
10 Mathematical Problems in Engineering
The exact solutions are [42]
1199091(119905) =
0 119905 isin [0
1
4
]
1
32
minus
1
4
119905 +
1
2
1199052 119905 isin [
1
4
1
2
]
1
32
minus
19
96
119905 +
3
16
1199052+
5
8
1199053minus
5
12
1199054 119905 isin [
1
2
3
4
]
minus
9641
32768
+
37391
24576
119905 minus
3183
1024
1199052+
7065
2304
1199053minus
135
384
1199054minus
85
96
1199055+
5
18
1199056 119905 isin [
3
4
1]
1199092(119905) =
119905 119905 isin [0
1
4
]
minus
5
384
+ 119905 +
5
8
1199052minus
5
3
1199053 119905 isin [
1
4
1
2
]
775
1536
minus
17
8
119905 +
1295
192
1199052minus
115
24
1199053minus
75
32
1199054+
5
3
1199055 119905 isin [
1
2
3
4
]
87997
132120
minus
1051
1024
119905 minus
95755
49152
1199052+
21515
1536
1199053minus
55325
3072
1199054+
335
96
1199055+
2125
576
1199056minus
25
21
1199057 119905 isin [
3
4
1]
(60)
Here this problem is solved by choosing 119896 = 8 and 119899 = 3 thefollowing approximate solutions 119909
1(119905) and 119909
2(119905) are found In
Tables 1 and 2 exact numerical results of this method andobtained results in [42] are shown respectively
1199091(119905) =
minus0001524977445119905 + 0049811489101199052minus 03456171465119905
3 119905 isin [0
1
8
]
minus0002668294207 + 006251408351119905 minus 046250099861199052+ 1020549487119905
3 119905 isin [
1
8
1
4
]
0006613889339 minus 004887212012119905 minus 0016956181141199052+ 04264897281119905
3 119905 isin [
1
4
3
8
]
001307452454 minus 01005572014119905 + 012087070151199052+ 0303976944119905
3 119905 isin [
3
8
1
2
]
01271590458 minus 07850643303119905 + 14898849611199052minus 0608699230119905
3 119905 isin [
1
2
5
8
]
006579667219 minus 04905249419119905 + 10186219481199052minus 0357358960119905
3 119905 isin [
5
8
3
4
]
03247255416 minus 1526240419119905 + 23995759181199052minus 09711162800119905
3 119905 isin [
3
4
7
8
]
06384881122 minus 2601997790119905 + 36290128981199052minus 1439473220119905
3 119905 isin [
7
8
1]
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 11
1199092(119905) =
1003041110119905 minus 0091233300001199052+ 06082219700119905
3 119905 isin [0
1
8
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
8
1
4
]
0003925049727 + 09088399072119905 + 066237636501199052minus 1401403820119905
3 119905 isin [
1
4
3
8
]
minus002462216250 + 1075221794119905 + 046667461251199052minus 1558091100119905
3 119905 isin [
3
8
1
2
]
03991598156 minus 1467470069119905 + 55520583251199052minus 4948346900119905
3 119905 isin [
1
2
5
8
]
000006281562500 + 04481955219119905 + 24869933881199052minus 3313645600119905
3 119905 isin [
5
8
3
4
]
minus1159405308 + 5086068009119905 minus 36968365821199052minus 05652767300119905
3 119905 isin [
3
4
7
8
]
minus5634050302 + 2042770799119905 minus 21230139421199052+ 6114076730119905
3 119905 isin [
7
8
1]
(61)
Example 11 Consider the following system described by (see[40])
(119905) =
8
119905 + 1
119909 (119905 minus (
119905
2
+
1
2
)) 119905 ge 0
119909 (119905) = (119905 + 1)2
119905 isin [minus
1
2
0]
(62)
Analytic solution of the initial value problem (IVP) is 119909(119905) =
(119905 + 1)2 By choosing 119896 = 1 and 119899 = 16 (degree raising) we
obtain the following solution
119909 (119905) = 1 + 02018032795 times 10minus4
11990512
minus 0015725157561199057
minus 00085727025731199055
+ 0017419590101199056
minus 0000154066590111990511
minus 01834453040 times 10minus5
11990513
+ 1101285958 times 10minus7
11990514
+ 00086693288941199058
+ 1999552507119905
+ 6306939519 times 10minus11
11990516
minus 3928281389
times 10minus9
11990515
minus 00032133472291199059
+ 099935258561199052
+ 0000834273668911990510
+ 00044389856571199053
minus 00026204484421199054
(63)
In Table 3 exact and presented methods are shown respec-tively
Example 12 Consider the following system described by (see[40])
(119905) = 119909 (119905 minus 1 minus
1
119905 + 1
) 119905 ge 0
119909 (119905) =
2
3
(119905 + 2) minus2 le 119905 le minus05
1 minus05 le 119905 le 0
(64)
where the exact solution is 119909(119905) = 1 + (23)119905 + 11990533 minus
(23) log(119905+1) on [0 1] and 119909(119905) = 1minus(23) log 2+119905 on [1 2]By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 + 54244277951199055
minus 16119814461199056
minus 25522508861199052
+ 79639037471199053
+ 03574277875119905 minus 92365174821199054
+ 019289236461199057
(65)
In Table 4 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 13 Consider the following system described by (see[40])
(119905) = minus119909 (119905 minus 120591 (119905)) 119905 isin [0 2]
119909 (0) = 1
120591 (119905) equiv
119905 minus 2 + radic4 minus 2119905 0 le 119905 le 2
0 119905 gt 2
(66)
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of
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Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
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Complex AnalysisJournal of
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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Volume 2014
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OperationsResearch
Advances in
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Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Advances in
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Volume 2014
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Stochastic AnalysisInternational Journal of
12 Mathematical Problems in Engineering
The solution of this problem is
119909 (119905) =
(119905 minus 2)2
4
0 le 119905 le 2
0 119905 gt 2
(67)
By choosing 119896 = 1 and 119899 = 7 (degree raising) we obtain thefollowing solution
119909 (119905) = 1 minus 1000000002119905 + 3207267830 times 10minus9
1199056
+ 02500000112 times 1199052
minus 3416339151 times 10minus10
1199057
minus 1204800000 times 10minus8
1199055
minus 2304000000 times 10minus8
1199053
+ 2296000000 times 10minus8
1199054
(68)
In Table 5 exact numerical results of this method method in[40] error of presented method and error of the method in[40] are shown respectively
Example 14 Consider the following LDDE described by
1198893119909 (119905)
1198891199053
= minus119909 (119905) minus 119909 (119905 minus 03) + 119890minus119905+03
0 le 119905 le 1 (69)
with the initial conditions
119909 (0) = 1
119889119909 (0)
119889119905
= minus1
1198892119909 (0)
1198891199052
= 1 119909 (119905) = 119890minus119905
119905 le 0
(70)
where the exact solution of this example is 119909(119905) = 119890minus119905 Here
this problem is solved by choosing 119896 = 10 and 119899 = 3The graph of error is shown in Figure 5 and the followingapproximate solution 119909(119905) is found
119909 (119905) =
1 minus 119905 + 051199052minus 0172928119905
3 119905 isin [0 01]
09999767558 minus 0999302674119905 + 0493026741199052minus 01496838119905
3 119905 isin [01 02]
09998081522 minus 0996773576119905 + 0480381041199052minus 01286073119905
3 119905 isin [02 03]
09992953871 minus 0991645877119905 + 0463288531199052minus 01096154119905
3 119905 isin [03 04]
09982244623 minus 0983613861119905 + 0443208291199052minus 00928817119905
3 119905 isin [04 05]
09964070488 minus 0972709334119905 + 0421399141199052minus 00783422119905
3 119905 isin [05 06]
09937493164 minus 09594206119905 + 0399251141199052minus 00660377119905
3 119905 isin [06 07]
09903114379 minus 0944686777119905 + 0378202731199052minus 00560146119905
3 119905 isin [07 08]
09863822587 minus 0929952279119905 + 0359784511199052minus 00483403119905
3 119905 isin [08 09]
09825547252 minus 0917193744119905 + 0345608261199052minus 00430898119905
3 119905 isin [09 1]
(71)
Example 15 Consider the second-order linear decay differ-ential equation
(119905) =
3
4
119909 (119905) + 119909 (
119905
2
) minus 1199052
+ 2 0 le 119905 le 1
119909 (0) = 0 (0) = 0
(72)
The exact solution of this problem is 119909(119905) = 1199052 Here this
problem is solved by choosing 119896 = 1 and 119899 = 7 The followingapproximate solution 119909(119905) is found
119909 (119905) = 18828480001199052
minus 50726239991199053
+ 15564000001199054
minus 28142400001199055
+ 30840000001199056
minus 1199059
+ 71199058
minus 006182400000119905
minus 20010000001199057
(73)
In Table 6 exact numerical results of this method error ofpresentedmethod and error of themethod in [43] are shownrespectively
5 Conclusions
Using the Bezier curves the general algorithm is providedfor the delay systems containing inverse time Numericalexamples show that the proposedmethod is efficient and veryeasy to use
Appendix
In this Appendix we specify the derivative of Bezier curveBy (6) we have
k119895(119905) =
119899
sum
119894=0
119886119895
119894119861119894119899
(119905) 119905 isin [0 1] (A1)
where 119861119894119899(119905) = (119899119894(119899 minus 119894))119905
119894(1 minus 119905)
119899minus119894Now we have (see [44])
119889119861119894119899
(119905)
119889119905
= 119899 (119861119894minus1119899minus1
(119905) minus 119861119894119899minus1
(119905)) 0 le 119894 le 119899 (A2)
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 13
minus02
minus04
minus06
minus08
minus1
minus06 minus04 minus02 0 02 04 06 08 1
Approximate x1(t)Exact x1(t)
t
Figure 3The graph of approximated trajectory 1199091(119905) for Example 2
5
4
3
2
minus06 minus04 minus02 0 02 04 06 08 1
t
Approximate x2(t)Exact x2(t)
Figure 4The graph of approximated trajectory 1199092(119905) for Example 2
where 119861minus1119899minus1
(119905) = 119861119899119899minus1
(119905) = 0 and
119861119894minus1119899minus1
(119905) =
(119899 minus 1)
(119894 minus 1) (119899 minus 119894)
119905119894minus1
(1 minus 119905)119899minus119894
119861119894119899minus1
(119905) =
(119899 minus 1)
119894 (119899 minus 119894 minus 1)
119905119894
(1 minus 119905)119899minus119894minus1
(A3)
00014
00012
00010
00008
00006
00004
00002
0
0 02 04 06 08 1
t
Error
Figure 5 The graph of error for Example 7
By using (A2) the first derivative k119895(119905) is shown as
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=1
119899a119895119894119861119894minus1119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119899a119895119894+1
119861119894119899minus1
(119905) minus
119899minus1
sum
119894=0
119899a119895119894119861119894119899minus1
(119905)
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 a119895119894+1
minus a119895119894
(A4)
Now we specify the procedure of derivation of (10) from (9)By (6) we have
k119895(119905) = (
119899
0) a1198950
1
ℎ119899(119905119895minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895119899
1
ℎ119899(119905 minus 119905119895minus1
)
119899
k119895+1
(119905) = (
119899
0) a119895+10
1
ℎ119899(119905119895+1
minus 119905)
119899
+ sdot sdot sdot + (
119899
119899) a119895+1119899
1
ℎ119899(119905 minus 119905119895)
119899
(A5)
by substituting 119905 = 119905119895into (A5) one has
k119895(119905119895) = a119895119899
1
ℎ119899(119905119895minus 119905119895minus1
)
119899
k119895+1
(119905119895) = a119895+10
1
ℎ119899(119905119895+1
minus 119905119895)
119899
(A6)
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Mathematical Problems in Engineering
Table 1 Exact and estimated values of 1199091(119905) for Example 3
119905 Exact 1199091(119905) Present 119909
1(119905) 119909
1(119905) in [42]
000 0000000 00000000000000000 minus0000088
005 0000000 00000050777072400 minus0000046
010 0000000 00000000000000000 0000021015 0000000 minus0000253099630375 0000083020 0000000 minus0000501121553000 minus0000128
025 0000000 minus106875 times 10minus11
minus0000024
030 0001250 00019414196591000 0001400035 0005000 00057172621996375 0004987040 0011250 00116454806360000 0011157045 0020000 00199999999857500 0019968050 0031250 00310107172150000 0031304055 0044971 004479153044625000 0045021060 0061000 006099999990000000 0060991065 0079086 00791835285950000 0079044070 0098917 00989798441000000 0098901075 0120117 01201170006000000 0120143080 0142244 01422502585600000 0142266085 0164728 01647280007500000 0164710090 0186819 01868145712000000 0186803095 0207606 02076060001475000 0207623100 0226030 02260300002000000 0226030
Table 2 Exact and estimated values of 1199092(119905) for Example 3
119905 Exact 1199092(119905) Present 119909
2(119905) 119909
2(119905) in [42]
000 0000000 0000000000000000 0001169005 0050000 0050000000000000 0049923010 0100000 0099999999970000 0100294015 0150000 0150424766127000 0149740020 0200000 0200976855207000 0199902025 0250000 0250636614672500 0250170030 0298229 0298229000000000 0298294035 0345083 0342083000067500 0342098040 0380313 0380416662700000 0380186045 0411667 0411748202343750 0411593050 0434896 0434896000125000 0435025055 0448306 04482677054750000 0448326060 0448532 04485758408000000 0448483065 0432078 0432134688390000 0432080070 0395846 0395846000275000 0395868075 0337199 0337199000906250 0337171080 0254052 0254052000960000 0254038085 0145303 0145637497343750 0145354090 0011316 0011635894970000 0011295095 minus0145872 minus014587200166625 minus0145924
100 minus0322405 minus032240500200000 minus0322386
To preserve the continuity of the Bezier curves at the nodesone needs to impose the condition k
119895(119905119895) = k119895+1
(119905119895) so from
(A6) we have
a119895119899(119905119895minus 119905119895minus1
)
119899
= a119895+10
(119905119895+1
minus 119905119895)
119899
(A7)
Table 3 Exact and estimated values of 119909(119905) for Example 4
119905 Exact Presented method05 225 22499152590316310 4 40000000000000015 625 6249957002587592 9 900000000128046
From (A4) the first derivatives of k119895(119905) and k
119895+1(119905) are
respectively
119889k119895(119905)
119889119905
=
119899minus1
sum
119894=0
119861119894119899minus1
(119905) 119899 (a119895119894+1
minus a119895119894)
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895minus 119905)
119899minus1minus119894
(119905 minus 119905119895minus1
)
119894
times
1
ℎ119899119899 (a119895119894+1
minus a119895119894)
= (
119899 minus 1
0) 119899 (a119895
1minus a1198950)
1
ℎ119899(119905119895minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895
119899minus a119895119899minus1
)
times
1
ℎ119899(119905 minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905)
119889119905
=
119899minus1
sum
119894=0
(
119899 minus 1
119894) (119905119895+1
minus 119905)
119899minus1minus119894
(119905 minus 119905119895)
119894
times
1
ℎ119899119899 (a119895+1119894+1
minus a119895+1119894
)
= (
119899 minus 1
0) 119899 (a119895+1
1minus a119895+10
)
1
ℎ119899(119905119895+1
minus 119905)
119899minus1
+ sdot sdot sdot + (
119899 minus 1
119899 minus 1) 119899 (a119895+1
119899minus a119895+1119899minus1
)
times
1
ℎ119899(119905 minus 119905119895)
119899minus1
(A8)
By substituting 119905 = 119905119895into (A8) we have
119889k119895(119905119895)
119889119905
= 119899 (a119895119899minus a119895119899minus1
)
1
ℎ119899(119905119895minus 119905119895minus1
)
119899minus1
119889k119895+1
(119905119895)
119889119905
= 119899 (a119895+11
minus a119895+10
)
1
ℎ119899(119905119895+1
119905119895)
119899minus1
(A9)
and to preserve the continuity of the first derivative of Beziercurves at nodes by equalizing (A9) we have
(a119895119899minus a119895119899minus1
) (119905119895minus 119905119895minus1
)
119899minus1
= (a119895+11
minus a119895+10
) (119905119895+1
minus 119905119895)
119899minus1
(A10)
where it shows the equality (10)
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 15
Table 4 Exact and estimated values of 119909(119905) for Example 5
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]05 110468992792789 110468992817860 11451 250709000000000 times 10
minus10
12232 times 10minus3
10 153790187962670 153790188062000 15361 993297 times 10minus10
17685 times 10minus3
14 193790187962670 193768171138582 19361 0220168240883 times 10minus3
17685 times 10minus3
15 203790187962670 203790188078453 20362 1157827 times 10minus9
16125 times 10minus3
20 253790187962670 253790188032000 25870 693297 times 10minus10
49096 times 10minus2
Table 5 Exact and estimated values of 119909(119905) for Example 6
119905 Exact Presented method Method in [40] Error of presented method Error of the method in [40]10 025 0250000000017634 0250013 17634 times 10
minus11
128346 times 10minus5
20 00 00 526486 times 10minus7 00 526486 times 10
minus7
Table 6 Exact and estimated values of 119909(119905) for Example 8
119905 Exact Presented method Error Of presented method Error of the method in [43]02 004 00400000000049152 49152 times 10
minus12
173 times 10minus6
04 016 01600000000193540 19354 times 10minus11
110 times 10minus5
06 036 03600000000221180 22118 times 10minus11
126 times 10minus4
08 064 06400000000073730 7373 times 10minus12
707 times 10minus4
Conflict of Interests
The authors declare that they have no conflict of interestsregarding publication of this paper
Acknowledgment
The authors would like to thank the anonymous reviewersfor their careful reading constructive comments and nicesuggestions which have improved the paper very much
References
[1] G Adomian and R Rach ldquoNonlinear stochastic differentialdelay equationsrdquo Journal of Mathematical Analysis and Appli-cations vol 91 no 1 pp 94ndash101 1983
[2] J Baranowski ldquoLegendre polynomial approximations of timedelay systemsrdquo in Proceedings of the 12th International PhDWorkshop p 2326 2010
[3] F Maghami Asl and A G Ulsoy ldquoAnalysis of a system oflinear delay differential equationsrdquo Journal of Dynamic SystemsMeasurement and Control Transactions of the ASME vol 125no 2 pp 215ndash223 2003
[4] X T Wang ldquoNumerical solution of delay systems containinginverse time by hybrid functionsrdquo Applied Mathematics andComputation vol 173 no 1 pp 535ndash546 2006
[5] J K Hale and S M V Lunel Introduction to FunctionalDifferential Equations Springer New York NY USA 1993
[6] S I Niculescu Delay Effects on Stability a Robust ControlApproach Springer New York NY USA 2001
[7] D H Eller and J I Aggarwal ldquoOptimal control of linear time-delay systemsrdquo IEEE Transactions on Automatic Control vol 14no 14 pp 678ndash687 1969
[8] L Gollmann D Kern and H Maurer ldquoOptimal controlproblems with delays in state and control variables subject to
mixed control-state constraintsrdquo Optimal Control Applicationsand Methods vol 30 no 4 pp 341ndash365 2009
[9] N N Krasovskii ldquoOptimal processes in systems with time lagsrdquoin Proceedings of the 2nd International Conference of Interna-tional Federation of Automatic Control Basel Switzerland 1963
[10] R Loxton K L Teo and V Rehbock ldquoAn optimizationapproach to state-delay identificationrdquo IEEE Transactions onAutomatic Control vol 55 no 9 pp 2113ndash2119 2010
[11] C Wu K L Teo R Li and Y Zhao ldquoOptimal control ofswitched systems with time delayrdquoApplied Mathematics Lettersvol 19 no 10 pp 1062ndash1067 2006
[12] G L Kharatishivili ldquoThe maximum principle in the theoryof optimal processes with time lagsrdquo Doklady Akademii NaukSSSR vol 136 no 1 1961
[13] M N Oguztoreli ldquoA time optimal control problem for systemsdescribed by differential difference equationsrdquo SIAM Journal ofControl vol 1 no 3 pp 290ndash310 1963
[14] J P LaSalle ldquoThe time optimal control problemrdquo in Contribu-tions to the Theory of Nonlinear Oscillations vol 5 pp 1ndash24Princeton University Press Princeton NJ USA 1960
[15] N N Krasovskii ldquoOn the analytic construction of an optimalcontrol in a system with time lagsrdquo Prikladnaya Matematika iMekhanika vol 26 no 1 pp 50ndash67 1962
[16] D W Ross Optimal control of systems described by differentialdifference equations [PhD thesis] Department of ElectricalEnergy Stanford University Stanford Calif USA 1968
[17] M Basin and J Perez ldquoAn optimal regulator for linear systemswith multiple state and input delaysrdquo Optimal Control Applica-tions and Methods vol 28 no 1 pp 45ndash57 2007
[18] M Basin and J Rodriguez-Gonzalez ldquoOptimal control forlinear systems with multiple time delays in control inputrdquo IEEETransactions onAutomatic Control vol 51 no 1 pp 91ndash97 2006
[19] M Heinkenschloss ldquoA time-domain decomposition iterativemethod for the solution of distributed linear quadratic optimalcontrol problemsrdquo Journal of Computational and Applied Math-ematics vol 173 no 1 pp 169ndash198 2005
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Mathematical Problems in Engineering
[20] H Juddu ldquoSpectral method for constrained linear-quaraticoptimal controlrdquoMathematics Computers in Simulation vol 58pp 159ndash169 2002
[21] R Winkel ldquoGeneralized bernstein polynomials and Beziercurves an application of umbral calculus to computer aidedgeometric designrdquoAdvances in AppliedMathematics vol 27 no1 pp 51ndash81 2001
[22] J Zheng T W Sederberg and R W Johnson ldquoLeast squaresmethods for solving differential equations using Bezier controlpointsrdquo Applied Numerical Mathematics vol 48 no 2 pp 237ndash252 2004
[23] K Harada and E Nakamae ldquoApplication of the Bezier curve todata interpolationrdquo Computer-Aided Design vol 14 no 1 pp55ndash59 1982
[24] G Nurnberger and F Zeilfelder ldquoDevelopments in bivariatespline interpolationrdquo Journal of Computational and AppliedMathematics vol 121 no 1 pp 125ndash152 2000
[25] M Evrenosoglu and S Somali ldquoLeast squares methods for solv-ing singularly perturbed two-point boundary value problemsusing Bezier control pointsrdquo Applied Mathematics Letters vol21 no 10 pp 1029ndash1032 2008
[26] J V Beltran and J Monterde ldquoBezier solutions of the waveequationrdquo in Computational Science and Its ApplicationsmdashICCSA 2004 vol 3044 of Lecture Notes in Computer Science pp631ndash640 2004
[27] R Cholewa A J Nowak R A Bialecki and L C WrobelldquoCubic Bezier splines for BEM heat transfer analysis of the 2-D continuous casting problemsrdquoComputationalMechanics vol28 no 3-4 pp 282ndash290 2002
[28] B Lang ldquoThe synthesis of wave forms using Bezier curves withcontrol point modulationrdquo in Proceedings of the 2nd CEMSResearch Student Conference Morgan Kaufamann 2002
[29] A T Layton and M Van de Panne ldquoA numerically efficientand stable algorithm for animating water wavesrdquo The VisualComputer vol 18 no 1 pp 41ndash53 2002
[30] M Gachpazan ldquoSolving of time varying quadratic optimal con-trol problems by using Bezier control pointsrdquo Computationaland Applied Mathematics vol 30 no 2 pp 367ndash379 2011
[31] F Ghomanjani and M H Farahi ldquoThe Bezier control pointsmethod for solving delay differential equationrdquo Intelligent Con-trol and Automation vol 3 no 2 pp 188ndash196 2012
[32] F Ghomanjani M H Farahi and M Gachpazan ldquoBeziercontrol points method to solve constrained quadratic optimalcontrol of time varying linear systemsrdquo Computational andApplied Mathematics vol 31 no 3 p 124 2012
[33] F Ghomanjani M H Farahi and A V Kamyad ldquoNumericalsolution of some linear optimal control systems with pan-tograph delaysrdquo IMA Journal of Mathematical Control andInformation 2013
[34] F Ghomanjani M H Farahi and M Gachpazan ldquoOptimalcontrol of time-varying linear delay systems based on the Beziercurvesrdquo Computational and Applied Mathematics 2013
[35] C-H Chu C C L Wang and C-R Tsai ldquoComputer aidedgeometric design of strip using developable Bezier patchesrdquoComputers in Industry vol 59 no 6 pp 601ndash611 2008
[36] G E Farin Curve and Surfaces for Computer Aided GeometricDesign Academic Press New York NY USA 1st edition 1988
[37] Y Q Shi and H Sun Image and Video Compression forMultimedia Engineering CRC 2000
[38] A Kılıcman ldquoOn the matrix convolutional products and theirapplicationsrdquo AIP Conference Proceedings vol 1309 pp 607ndash622 2010
[39] A Kılıcman and Z Al Zhour ldquoKronecker operational matricesfor fractional calculus and some applicationsrdquo Applied Mathe-matics and Computation vol 187 no 1 pp 250ndash265 2007
[40] I Gyori F Hartung and J Turi ldquoOn numerical solutionsfor a class of nonlinear delay equations with time-and state-dependent delaysrdquo in Proceedings of the World Congress ofNonlinear Analysts pp 1391ndash1402 New York NY USA 1996
[41] W Rudin Principles of Mathematical Analysis McGraw-Hill1986
[42] C Hwang and M-Y Chen ldquoAnalysis of time-delay systemsusing the Galerkin methodrdquo International Journal of Controlvol 44 no 3 pp 847ndash866 1986
[43] O A Taiwo andO S Odetunde ldquoOn the numerical approxima-tion of delay differential equations by a decompositionmethodrdquoAsian Journal of Mathematics and Statitics vol 3 no 4 pp 237ndash243 2010
[44] H Prautzsch W Boehm and M Paluszny Bezier and B-SplineTechniques Springer 2001
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Advances in
Mathematical Physics
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
International Journal of
Combinatorics
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of