Numerical Solution of ODEs-IVP
Transcript of Numerical Solution of ODEs-IVP
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Prof. A.A. Adesina, ChSE, UNSW 1
NUMERICAL SOLUTION OFORDINARY DIFFERENTIAL
EQUATIONSINITIAL VALUE PROBLEMS
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GENERAL CONSIDERATIONS
The modelling
ofunsteady-state processes
often lead to ODEs
with
time as the independent variable. In other cases, the independent
variable may be a spatial co-ordinate as in the steady-statetreatment of tubular or columnal systems
such as plug flow and
trickle-bed reactors, packed adsorption and absorption towers oreven tubular heat exchangers.
In general, for an nth
order ODE, n conditions are needed tocompletely solve the problem. If ALL these n specifications are
provided at the same value of the independent variable, say, t orx, for all dependent variables, then the ODE is
said to be an INITIAL
VALUE PROBLEM. While this may imply that the value of t or x is
supposed to be the starting point in the system (e.g. t=0 or x=0),there is no mathematical incongruity if
the value oft or x is the end-
point or some other physically admissible point
in the independent
variable range.
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METHODS FOR 1ST ORDER NONLINEARODE - Euler methods
The general 1st
order nonlinear ODE may be written;
The Euler method, one of the earliest techniques for solving ODE
stipulates
that, if we can represent the LHS of Eqn
(7.1) by its first forward finite
difference, then at any position, i, we have,
0 0 0 0
( 7 . 1 )
w i t h t h e i n i t i a l c o n d i t i o n t h a t
a t ( )O
( ,
R
)
x x y y y
d yf x y
d
x y
x
= = =
=
1
2 2 3 3
(7.2)
but from our previous discussion on finite difference operators, we can write,
..... ...2 6
i i i
i ii i
y y y
h D y h D yy hDy
+ =
= + + + (7.3)
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Euler methods
Euler assumed a truncation of the series in Eqn
(7.3) after the 1st
term, thus,
In fact, replacing the term, Dyi
with ffrom Eqn
(7.1), we get the so-called
explicit (forward) Euler method, written,
It is evident that this method is only marginally accurate since
the error is of
order h
2
.
2
2
1
(7.4)so that upon substitution into Eqn (7.2), we receive,
(7
)
( ) )
(
.5
i i
i i i
y hD y O h
y y hD y O h+
= +
= + +
21 ( , ) ( ) (7.6)i i i iy y hf x y O h+ = + +
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Euler methods
If we use the backward finite difference approximation, we can write,
But from earlier notes, we know that
1 1
1 1
(7.7)so that, in terms of backward differences, we have,
(7.8)
i i i i
i i i
y y y y
y y y
+ +
+ +
= =
= +
2 2 3 3
1 11 1
21 1 1
(7.9)
so that combining with Eqns (7.1) & (7.2) we secure,
.............2 6
( (7.10, ) ( ))
i ii i
i i i i
h D y h D yy hDy
y y hf x y O h
+ +
+ +
+ + +
= +
= + +
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Euler methods
Eqn
(7.10)
is commonly called the implicit Euler method
because it
involves the calculation of the function, f, at the unknown value of
yi+1
. We observe that the error in Eqn (7.10) is also of the order of h2.
In principle, implicit equations cannot be solved individually but must
be set up as sets simultaneous algebraic equations. If these are
linear, then we can use any of the techniques we have encounteredbefore (say, generalised inverse matrix approach) to deliver asolution. However, if they are nonlinear, we may use the Jacobian method for solving the set of nonlinear algebraic equations.Regardless, the computation may be time-consuming depending on
how many integration steps we need to do.
Euler simplified this problem by using Eqn
(7.6) to predict yi+1
and
then employing this predicted value
to get a corrected estimate from
Eqn (7.10).
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Euler methods
Eqns
(7.6) and (7.10) constitute the Euler predictor-corrector pair and is more
accurate because it has an error oforder h3
(which can be readily shown by
adding Eqns (7.3) & (7.9)). Thus, the
Euler Predictor-Corrector method
is:
As a result, the Euler Predictor-Corrector method is preferred over the
explicit or implicit method
for the purpose of computation.
( )
( ) [ ]
2
1 Pr
2
1 1 1
(7.11a( , ) ( )
( , ) ( , ) ( )
)
(7.11b2
)
i i i i
i i i i i iCor
y y hf x y O h
hy y f x y f x y O h
+
+ + +
= + +
= + + +
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Runge-Kutta
Methods
The Runge-Kutta
methods are arguably the most widely used
integration techniques for solving ODEs
with initial conditions
(IVP-ODEs). Although the formal derivation for each method issimilar to the Euler method in the sense that we take advantage of
particular approximations of the Taylor series expansion of the
differential operator (cf. Lecture notes on Finite Difference operators
as recalled in Eqns (7.3) & (7.9)), we will NOT provide formalderivations here but simply summarise the necessary equations to
be used for each RK method as displayed in the Tables below.
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Table 1: Summary of Runge-Kutta integration formulas (2nd-4th order)
Orderof theRKMethod
Formula foryi+1 Formula for RKconstants
Order oftheerrorinvloved
Second
1 1 2
1( )
2i iy y k k
+= + + 1
2 1
( , )
( , )
i i
i i
k hf x y
k hf x h y k
== + +
O(h3)
Third1 1 2 3
1( 4 )
6i i
y y k k k+ = + + + 11
2
3 2 1
( , )
,2 2
( , 2 )
i i
i i
i i
k hf x y
h kk hf x y
k hf x h y k k
= = + +
= + +
O(h4)
Fourth1 1 2 3 4
1( 2 2 )
6i i
y y k k k k+ = + + + + 11
2
23
4 3
( , )
,2 2
,2 2
( , )
i i
i i
i i
i i
k hf x y
h kk hf x y
h kk hf x y
k hf x h y k
= = + + = + +
= + +
O(h5)
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Table 2: Runge-Kutta-Gill integration formulas
Orderof theRKMethod
Formula foryi+1 Formula for RK constants Order oftheerrorinvloved
Runge-Kutta-Gill
1 1 2 3 41 1 12 1 2 16 2 2
i iy y k k k k+ = + + + + +
( ) ( )
( )
1
12
31 2
4 32
( , )
,2 2
,2
2 1 2 22 2
,
2 2
22
i i
i i
i
i
i
i
k hf x y
h kk hf x y
hx
k hf k ky
x h
k hf kky
= = + + +
= + + + = + +
O(h5)
The Runge-Kutta-Gill method is the most widely used 4th
order method and the constants are
selected to reduce the amount of storage required in the solution of a large number of
simultaneous 1st
order ODEs. We will discuss application of RK methods to set of ODEs later.
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Table 3: Summary of Runge-Kutta integration formulas (5th order)
Orderof theRKMethod
Formula foryi+1 Formula for RK constants Order oftheerrorinvloved
Fifth1 1 3 4 5 6
1(7 32 12 32 7 )
90i i
y k k k k k+ = + + + + + 11
2
1 23
34
32 45
3 51 2 46
( , )
,2 2
3
,2 16 16
,2 2
63 3 9,4 16 16 16
6 84 12,
7 7 7 7 7
i i
i i
i i
i i
i i
i i
k hf x y
h kk hf x y
h k k
k hf x y
khk hf x y
kh k k
k hf x y
k kk k kk hf x h y
= = + + = + + + = + + = + + + = + + + + +
O(h6)
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Comments on Runge-Kutta methodselection
In general, the 4th
order RK method is more commonly used
because
it has a good accuracy for most engineering situations. However, forstiff ODEs, it may be necessary to reduce the step-size, h, betweensuccessive integration steps. What is the criterion for implementing a
step-size reduction to avoid error propagation through the integration
procedure? Collatz (1960) has recommended that if,
After each integration step, then the step-size, h, should be
decreased
(to probably half its current value).
3 2
2 1
0 . 1 (7 .12)k k
k k
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Illustrative example
Let us find the solution to the IVP
In this case, clearly,
We divide the interval into 10 steps so that x0
=0.0, x1
=0.1..x10
=1.0
with h=0.1. Thus, y(x0
0=1.0. Using the iterative formulas given on
Table 1, we get,
th
(E1.1)
with over the interval 0 x 1 usingy(0 4 order RK method)=1.0
2
1dy y xdx =
+
( , ) 2 1x y y x+
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Results of the 4th
order Runge-Kutta methods for the Illustrative Example
xi R-K R-K-G Exact value of yi
0.0 1.0 1.0 1.00.1 1.01034164 1.01034164 1.01034260
0.2 1.04280472 1.04280472 1.04280663
0.3 1.09971619 1.09971619 1.09971809
0.4 1.18364811 1.18364811 1.18365002
0.5 1.29744053 1.29744148 1.29744339
0.6 1.44423485 1.44423580 1.44423676
0.7 1.62750244 1.62750340 1.62750530
0.8 1.85107803 1.85107994 1.85108185
0.9 2.11920166 2.11920357 2.11920643
1.0 2.43655777 2.43656063 2.43656349
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Predictor-Corrector Methods
The Runge-Kutta
methods are single-step
methods and are good for
solving IVP because they self-starting. However, they may be
plagued with instabilities for highly stiff ODEs which require smallstep-size to overcome numerical instabilities. Predictor-Correctormethods are preferred
for such situations because they deliver more
stable results. They are, however, nonself-starting
and are regarded
as multi-step techniques.
The most common PC methods are summarised
below.
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Table 4: Summary of commonly used Predictor-Corrector Methods
Type of
PC
method
Predictor Equation Corrector Equation Error
involved
Euler ( )1 Pri i iy y hf+ = + ( ) ( )1 12
i i i iCor
hy y f f+ ++ + O(h
2)
Milnes
4thorder
( ) ( )1 3 1 2Pr
42 2
3i i i i i
hy y f f f+ + + ( ) ( )1 1 1 14
3i i i i i
Cor
hy y f f f+ + + + + O(h
5)
Adams-
Moulton
method
( ) ( )1 1 2 3Pr 55 59 37 924i i i i i ihy y f f f f+ + + ( ) ( )1 1 1 29 19 524i i i i i iCor hy y f f f f+ + + + + O(h5)
Milnes
6th order ( ) 1 21 5Pr3 4
11 14 26310 14 11
i i ii i
i i
f f fhy y
f f +
+ = + + ( ) 1 11 2 37 32 12
245 32 7
i i ii iCor
i i
f f fhy y
f f+ + + + = + + +
O(h7)
Note that: fi = f(xi, yi); fi+1 = f(xi+1, yi+1) etc, in the Table above
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Predictor-Corrector Methods
How many times do we iterate on the Corrector equation? Typically, we
employ a termination criterion to determine when to stop. For example,
we can use:
1, 1, 1
1, 1
i+1m number of iterat
ions
perform
ed so far on
is
(7.12)
where
the toleranc
indicates the
usin
1,2,....
Corrector Equat e faciong the and
y
i m i m
i m
y y
ym
+ + + =
-4
,
a small positive number,
tor
10typi or smcal ally, ler.
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Worked Example
Use the Milnes 4th
order PC method to solve
Solution
Assume h=0.1, then let us calculate the 1st three points (y1 ,y2 ,y3 ) bythe single-step method of 4th order R-K technique since the PC isNOT self-starting. Recall that we need, yi-3
to use the Milnes P-Cmethod.
0.8
y(1.0)=3.
(E1)
with over th ra
0.1( )
nge x0 =1.0 to 2
.0e
dy x ydx = +
0.8( , ) 0.1( )f x y x y+
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Starting with the 4th order RK method and using y(1) =3.0; we have, x0=1.0and y0=3.0
1st integration step
0.8
1 0 0
0.8
2 0 0 1
3 0 0 2
4 0 0 3
1 2 3 4
1 0
( , ) 0.1[0.1(1 3) ] 0.0303143( / 2, / 2 ) 0 .1[0.1(1.05 3.01516) ] 0.0307087
( / 2, / 2 ) 0.0307099
( , ) 0.0311043
12 2 0.0307093
63.0
k h f x yk h f x h y k
k h f x h y k
k h f x h y k
y k k k k
y y y
= = + == + + = + == + + == + + =
= + + + = = + =
11 0.1 1
307093
d
1
an
.x = + =
2
nd
integration step0.8
1 1 1
2 1 1 1
3 1 1 2
4 1 1 3
1 2 3
2
2
4
1
( , ) 0.1[0.1(4.13071) ] 0.0311042
( / 2, / 2) 0.0314985
( / 2, / 2 ) 0.0314997
( , ) 0.0311043
12 2 0.0314991
6
3.03
and
1.1
149
0.
9
1
1
1 .
k h f x y
k h f x h y k
k h f x h y k
k h f x h y k
y k k k
x
k
y y y
= = == + + == + + == + + =
= + + + =
= += =
=+
2
Similar repetitive steps give,
3 31.3, 3.09449x y =
At the 4th integration step, we now turn to the 4th order Milnes
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At the 4 integration step, we now turn to the 4 order Milne s
PC. Hence,
( )4 0 3 2 11
2
3
4 3 4 3 2
0.8
4 4
,
4
4
4 0
4Pr : 2 2
3
0.311042
0.318939
0.326834
0.43.0 (0.622084 0.653668 0.318939) 3.127
: ( 4 )
3( , ) 0.1(1.4 3.12758) 0.334728
0.13.09449 (1.961
583
00) 3.15983
hC or y y f f f
f f x y
y
hy y f f f
f
f
f
y
= + + += = +
= + +=
= = + =
==
= + + =
4 ,1 4 ,0
4
0.8
4 4 4 ,0
4 ,1 3 4 3 2
,0
-4
1st iteration step of the Co rrector Equa tion
f ( , ) 0.1(1.4 3.15986) 0.336636
( 4 )3
Check for converge
0.
nc
13.09449 (1.96291) 3.15992
3
e
L et us assu m e th 1
6
at = 0
f x y
hy
y y
f
y
y f f
= = + == + +
+= + =
5 4st
th
3.15992 3.159861.8988 10 10
3.15986
hence convegrence was met after the 1 iteration so we cannow proceed to the 5 integration step.
=
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Solution of higher order ODEs (Solution toSimultaneous 1
st
order ODEs)
In principle, since it is possible to convert an nth
order ODE
to a set ofn 1st
order ODEs.
The approach to solving an nth
order ODE
is therefore exactly the same as solving n 1st
order ODEs. We would 1st
show how this conversion may be carried out and then
proceed with a discussion of the simlutaneous
solution of n 1st
order ODEs.
1 2 1
1 2 1 01 2 1
1
( , ) ( , ) .... ( , ) ( , ) ( , , ,... ) (7.13)
Let us defi
ne new variables as:
(7.1z 4
n n n n
n nn n n n
d y d y d y dy dy d ya x y a x y a x y a x y g x y
dx dx dx dx dx dx
y
+ + + =
=1
2 1 1 2
2
23 2 1 22
2
21 2 1 22
( , , , ..... )
( , , , .....
)
(7.15)
(7.16)
)
( , , , ..... )
n
n
n
nn n nn
dz dyz f x z z z
dx dx
dz d yz f x z z z
dx dx
dz d yz f x z z zdx dx
= = == = =
= = =M
1
11 1 21
1
(7.17)
(7.18)
Introducing these equ ations into the original equation, w e receive
( , , , ..... )
( , , ,...
n
nn n nn
n n
n
n n
dz d yz f x z z z
dx dx
dz d y dy d yg x ydx dx dx dx
= = =
= = 1 21 )= ( , , , ..... ) (7.19)n nf x z z z
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Solution of higher order ODEs (Solution toSimultaneous 1
st
order ODEs)
Eqns
(7.15) to (7.19) constitute the new n 1st
order ODEs
we now
need to solve. The applicable 4th
order Runge-Kutta
constants are:
( )1, 1 2 3 41 1 2
111 122 1 2
221 223 1 2
12 2 1,2,...
6
( , , ,.... ) 1,2,...
, , ,.... 1,2,...2
(7.20
2 2 2
, , ,.
)
(7.21)
(7.22)
...2 2 2 2
i j ij j j j j
j j i i i in
nj j i i i in
nj j i i i in
z z k k k k j n
k hf x z z z j n
kh k kk hf x z z z j n
kh k kk hf x z z z
+ = + + + + == =
= + + + + = = + + + +( )4 1 31 2 32 3
1,2,...
,
(7.23)
(7, ,..... 1,2 .24. ), ..j j i i i in n
j n
k hf x h z k z k z k j n
== + + + + =
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Comments on Runge-Kutta methodselection
A system ODE is said to be stiff if the so-called stiffness ratio, SR, exceeds
a particular value as indicated below, where SR is defined,
Where i
s
are the eigenvalues
of the ODE system and max |Re(i
)| is the
real part of the maximum eigenvalue of the system. In general, if
a). SR is about an order of magnitude
(10,20, etc), the system is NOT stiff.
b). SR is about three orders of magnitude
(1000, 2000, etc) the system is
STIFF
c). SR is about six orders of magnitude
(106, etc), the system is VERY
STIFF.
The eigenvalues
are obtained from the matrix of the coefficients of the set
of ODEs.
max Re( )
m
in R
(7.e( )
25)i
i
SR=
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Comments on Runge-Kutta methodselection
In principle, a stiff ODE
is one whose general solution contains an
exponential term, e.g. e
x
forsome constant . When is large andnegative, the ODE is especially troublesome because it permits the solution
to decay quickly to zero.
For instance, the solution to:
i
0 0
x
(7.26)
with , the may be obtained as
(7.27)
where
( , )
eigenvalue, ,
can vary in magnitude at each step of the in .tegration
Eqn (7.26)
( )
y x y
f
y
dy
f x ydx
= ==
will be at some point in the range of integration.
Thus, the s
stiff if i
olution wil
s NE
l be
GATIVE
unstable.
f
y
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Real stability regions for selected solution methods
Method Stability boundary
Explicit Euler -2 h 0
Implicit Euler 0 < h < , for < 0;-2 h 0, for > 0
Euler Predictor-Corrector -1.077 h 02
ndorder Runge-Kutta -2 h 0
3rd order Runge-Kutta -2.5 h 04
thorder Runge-Kutta -2.785 h 0
5th
order Runge-Kutta -5.7 h 0
Adams-Moulton -1.285 h 0
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For a set of n simultaneous ODEs, we can obtain theeigenvalues from the Jacobian matrix of the functions, fiwritten;
1 1 1
1 2
2 2 2
1 2
1 2
.......
........
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
........
n
n
n n n
n
f f f
y y y
f f f
y y y
J
f f f
y y y
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Comments on Runge-Kutta methodselection
It is apparent from the foregoing considerations that implicit methods
are recommended for handling the solution to stiff ODEs. This is
why Predictor-Corrector methods are somewhat popular amongchemical engineers
since we often deal with nonlinearity due to the
Arrhenius
or vant
Hoff terms
arising from modeling of non-
isothermal tubular reactors/columns or dynamics of reactive lumped
systems with nonlinear kinetics.
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Example problem
Solve
22 2
2
d yx x y y
d x= + +
for y at x = 2.0 ify = 4.0 at x = 1.0 and
0.5 1.0dy at xdx
= =
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Solution
Let
1
22 2
22
( , , )
( , , )
d yz f x y z
d xd y d z
x x y y f x y zd x d x
= =
= = + + =
Assume h = 0.5. From the initial conditions0 0 01 .0 , 4 .0 , 0 .5x y z= = =
1st
R-K constants
0 0 0
1 1 1
0 0 0
1 1 2
( , , ) 0 .5 0 .5 0 .2 5
( , , ) 0 .5 (1 4 1 6 ) 1 0 .5 0
k h f x y z x
k h f x y z
= = =
= = + + =
2nd
R-K constants
0 0 01 1 1 22 1 1
0
0 1 1
0 1 2
2 1
0 0 01 1 1 22 2 2
2 2
( , , )2 2 2
0 .51 1 .2 5
2 2
0 . 2 54 4 .1 2 5
2 21 0 . 5
0 .5 5 .7 52 2
0 .5 5 .7 5 2 .8 7 5
( , , )2 2 2
0 .5 [ (1 .2 5 ) 4 .1 2 5 1 .2 5 ( 4 .1 2 5 ) ]
1 1 . 8 6 7 2
h k kk h f x y z
hx
ky
kz
k x
h k kk h f x y z
x
= + + +
+ = + =
+ = + =
+ = + =
= =
= + + +
= + +
=
3r R-K constantsh k k
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Prof. A.A. Adesina, ChSE, UNSW 30
0 0 02 1 2 23 1 1
0
0 2 1
0 2 2
3 1
0 0 02 1 2 23 2 2
2 2
( , , )2 2 2
0 . 51 1 . 2 5
2 2
2 . 8 7 54 5 . 4 3 7 5
2 2
1 1 . 8 6 7 20 . 5 6 . 4 3 3 62 2
0 . 5 6 . 4 3 3 6 3 . 2 1 6 8
( , , )2 2 2
0 . 5 [ (1 . 2 5 ) 5 . 4 3 7 5 1 . 2 5 ( 5 . 4 3 7 5 ) ]
1 8 . 9 6 2 9
h k kk h f x y z
hx
ky
kz
k
h k kk h f x y z
x
= + + +
+ = + =
+ = + =
+ = + =
= =
= + + +
= + +
=
4th
R-K constants0 0 0
4 1 1 3 1 3 2
0
0
3 1
0
3 2
4 1
0 0 0
4 2 2 3 1 3 2
2 2
1 1 2 1 3 1 4 1
1 2 2 2
( , , )
1 . 5
7 . 2 1 6 8
1 9 . 4 6 2 9
0 . 5 1 9 . 4 6 2 9 9 . 7 3 1 4 5
( , , )
0 . 5 [ ( 7 . 2 1 6 8 ) 1 . 5 7 . 2 1 6 8 (1 . 5 ) ]
3 2 . 5 7 8 7
1[ 2 2 ] 3 . 6 9 4 1 8
6
1 [ 26
k h f x h y k z k
x h
y k
z k
k x
k h f x h y k z k
x
y k k k k
z k k
= + + +
+ =
+ =
+ =
= =
= + + +
= + +
=
= + + + =
= + 3 2 4 2
( 1 ) 0
( 1 ) 0
( 1 ) 0
2 ] 1 7 . 4 5 6 5
7 . 6 9 4 1 8
1 7 . 9 5 6 5
1 . 5
k k
y y y
z z z
x x h
+ + =
= + =
= + =
= + =
We will now usex(1), y
(1), andz
(1)as the starting point for the computation
of ( 2 ) ( 2 ) ( 2 ), ,x y a n d z .
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2nd
integration step
1st
R-K constants
(1) (1) (1)
11 1
(1) (1) (1)
11 2
2 2
( , , ) 0.5 17.9565 8.97824
( , , )
0.5[(7.69418) (7.69418 1.5) (1.5) ] 36.4958
k hf x y z x
k hf x y z
x
= = =
=
= + + =
2
nd
R-K constants
(1) (1) (1)11 1221 1
(1)
(1) 11
(1) 12
21
(1) (1) (1)11 1222 2
2 2
( , , )2 2 2
1.752
12.18332
36.20442
0.5 36.2044 18.1022
( , , )2 2 2
0.5[(1.75) 1.75 12.1833 (12.1833) ]
92.4997
k khk hf x y z
hx
ky
kz
k x
k khk hf x y z
x
= + + +
+ =
+ =
+ =
= =
= + + +
= + +
=
3rd
R-K constants
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3 R K constants
(1 ) (1 ) ( 1 )2 1 2 23 1 1
(1 )
(1 ) 2 1
(1 ) 2 2
3 1
( 1 ) (1 ) ( 1 )2 1 2 23 2 2
2 2
( , , )2 2 2
1 . 7 52
1 6 . 7 4 5 3
2
6 4 . 2 0 6 32
0 . 5 6 4 .2 0 6 3 3 2 . 1 0 3 1 5
( , , )2 2 2
0 .5 [(1 . 7 5 ) 1 . 7 5 1 6 .7 4 5 3 (1 6 . 7 4 5 3 ) ]
1 5 6 . 3 8 6
k khk h f x y z
hx
ky
kz
k x
k khk h f x y z
x
= + + +
+ =
+ =
+ =
= =
= + + +
= + +
=
4th
R-K constants(1 ) (1 ) ( 1 )
4 1 1 3 1 3 2
(1 )
(1 )
3 1
(1 )
3 2
4 1
( 1 ) ( 1 ) ( 1 )
4 2 2 3 1 3 2
2 2
1 1 2 1 3 1 4 1
( , , )
2 . 0
3 9 . 7 9 7 3
1 7 4 . 3 4 20 . 5 1 7 4 . 3 4 2 8 7 . 1 7 1
( , , )
0 . 5 [ ( 2 ) 2 3 9 . 7 9 7 3 ( 3 9 . 7 9 7 3 ) ]
1 6 6 7 . 4 2
1[ 2 2 ] 3 2 .7 6
6
k h f x h y k z k
x h
y k
z kk x
k h f x h y k z k
x
y k k k k
= + + +
+ =
+ =
+ =
= =
= + + +
= + +
=
= + + + =
1 2 2 2 3 2 4 2
( 2 ) (1 )
( 2 ) (1 )
( 2 ) (1 )
1[ 2 2 ] 3 6 6 . 9 4 8
6
4 0 . 4 5 4 2
3 8 4 . 9 0 4
2 . 0
z k k k k
y y y
z z z
x x h
= + + + =
= + =
= + =
= + =
at x = 2.0, y = 40.4542