Research Article The Concepts of Well-Posedness and ...Research Article The Concepts of...

Research ArticleThe Concepts of Well-Posedness and Stability inDifferent Function Spaces for the 1D LinearizedEuler Equations

Stefan Balint1 and Agneta M Balint2

1 Department of Computer Science West University of Timisoara Boulevard Vasile Parvan 4 300223 Timisoara Romania2Department of Physics West University of Timisoara Boulevard Vasile Parvan 4 300223 Timisoara Romania

Correspondence should be addressed to Agneta M Balint balintphysicsuvtro

Received 30 August 2013 Revised 8 December 2013 Accepted 15 December 2013 Published 12 January 2014

Academic Editor Abdullah Alotaibi

Copyright copy 2014 S Balint and A M Balint This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers the stability of constant solutions to the 1D Euler equation The idea is to investigate the effect of differentfunction spaces on the well-posedness and stability of the null solution of the 1D linearized Euler equations It is shown thatthe mathematical tools and results depend on the meaning of the concepts ldquoperturbationrdquo ldquosmall perturbationrdquo ldquosolution of thepropagation problemrdquo and ldquosmall solution that is solution close to zerordquo which are specific for each function space

1 Motivation ofthe Mathematical Considerations

Due to the practical importance of the sound attenuation incase of the turbofan aircraft engines in the last years morethan six hundred papers reporting experimental and theoret-ical results on the subject were publishedThepapers refereedhere and those refereed herein concern acoustic perturbationpropagation in a gas flowing through a lined duct and rep-resent just a very small part of the literature concerning thesubject

For describing an instantaneous acoustic perturbationpropagation the authors consider the solution of the nonlin-ear Euler equations (without source) governing the gas flowAfter that the nonlinear Euler equations are linearized atthe specified solution and the homogeneous linearized Eulerequations are derived It is commonly accepted that thoseequations govern the propagation of an instantaneous acous-tic perturbation (called frequently also initial value perturba-tions) More precisely it is assumed that if at the moment letus say 119905 = 0 an instantaneous acoustic perturbation occursthen its propagation is described by that solution of thehomogeneous linearized Euler equations which satisfies on

the duct wall the boundary condition and at 119905 = 0 it is equalto the perturbation in discussion

For describing a source-produced permanent time har-monic perturbation propagation the mathematical objectsdescribing the perturbation are added as right-hand mem-bers to the homogeneous linearized Euler equations It iscommonly accepted that the so-obtained nonhomogeneouslinearized Euler equations govern the propagation of thesource-produced permanent time harmonic perturbation indiscussion More precisely it is assumed that if starting at themoment let us say 119905 = 0 a source begins to produce per-manent time harmonic perturbation then the propagationof this perturbation is described by that solution of the non-homogeneous linearized Euler equations which is equal tozero for 119905 le 0 and verifies on the duct wall the boundaryconditions

It turns that in [1] for a large class of impedancelining models (ie boundary conditions) the above pre-sented initial-boundary value problem was declared ldquoill-posedrdquo because the set of the exponential growth rates ofthe solutions of the homogeneous linearized Euler equationssatisfying the considered boundary conditions is unboundedfrom above Later in [2ndash4] considerable efforts were made

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 872548 10 pageshttpdxdoiorg1011552014872548

2 Abstract and Applied Analysis

for modifying the boundary conditions in order to make theproblem ldquowell posedrdquo in the sense defined in [1] Here it has tobe mentioned that in [1] a concept of ldquowell posed differentialequationrdquo is defined This is different from the concept ofldquowell-posed problemrdquo usually in mathematics [5 6] andintroduced by Hadamard long time ago

Definition 1 FollowingHadamard for a given class of instan-taneous initial value perturbations (permanent source-pro-duced time harmonic perturbations resp) one calls theperturbation propagation problem well posed if there isa unique solution to the problem and the solution variescontinuously with the initial data (source amplitude resp)

In [7] it was shown that the concept introduced in [1] isconfusing because the equation considered in [1] dependingon the function space can be ill-posed or can be well posedIn fact the concept introduced in [1] is the transcription ofthe behavior of an example of a semigroup of contractions ofclass 119862

0which act in a precise function space (see [1 8] pages

240-241 referred in [1]) In this situation according to [8] theresolvent operator of the infinitesimal generator exists in ahalf plane of the form Re 119911 gt 120583 and the Laplace transformcan be considered for every solution of the homogeneousequation

The objective of the present work is to underline thatthe precise description of the function space is crucial evenin the case of the 1D gas flow model where the wall andthe lining effect are absent For the achievement of thisobjective the linear stability analysis of the constant 1D gasflow with respect to the initial value and source-producedpermanent time harmonic perturbations is presented in fourdifferent function spaces revealing significant differencesFor instance in the topological function space119883

4the propa-

gation problem of the initial value perturbation is well posedbut that of the source produced perturbation is ill posed inthe topological function space 119883

1in which the origin has

not absorbing neighborhoods stability and strictly positivegrowth rate coexists in contrast to the case of topologicalvector spaces119883

21198833

2 The 1D Gas Flow Model

In the 1D gas flow model the nonlinear Euler equations gov-erning the flow of an inviscid compressible nonheat con-ducting isentropic perfect gas according to [9] are

120597119906

120597119905+ 119906 sdot

120597119906

120597119909+1

120588sdot120597119901

120597119909= 0

120597120588

120597119905+ 119906 sdot

120597120588

120597119909+ 120588 sdot

120597119906

120597119909= 0

(1)

Here 119905 is time 119906 is velocity along the 119874119909 axis 119901 is pressureand 120588 is density Equation (1) are considered for 119909 isin 1198771015840 and119905 ge 0 It is assumed that 119901 120588 and the absolute temperature 1198791015840satisfy the equation of state of the perfect gas

119901 = 120588 sdot 119877 sdot 1198791015840 (2)

with 119877 = 119888119901minus 119888V 119888119901 119888V being the specific heat capacities at

constant pressure and constant volume respectively Let

119906 equiv 1198800= const gt 0 120588 equiv 120588

0= const gt 0 119901 equiv 119901

0= const gt 0

be a constant solution of the system of partial differentialequations (SPDE) (1) According to (2) 119901

0= 1205880sdot119877sdot1198791015840

0and the

associated isentropic sound speed 1198880verifies 1198882

0= 120574sdot(119901

01205880) =

120574 sdot 119877 sdot 1198791015840

0 where 120574 = 119888

119901119888V

Linearizing (1) at 119906 = 1198800 120588 = 120588

0 119901 = 119901

0and using the

perturbations 1199011015840 1205881015840 of 1199010 1205880satisfying

(120597

120597119905+ 1198800sdot120597

120597119909) (1199011015840minus 1198882

01205881015840) = 0 (3)

the following system of homogeneous linear partial differ-ential equations for the perturbations 1199061015840 1199011015840 of 119880

0 1199010 is

obtained

1205971199061015840

120597119905+ 1198800sdot1205971199061015840

120597119909+1

1205880

sdot1205971199011015840

120597119909= 0

1205971199011015840

120597119905+ 1198800sdot1205971199011015840

120597119909+ 120574 sdot 119901

0sdot1205971199061015840

120597119909= 0

(4)

It is assumed that if at 119905 = 0 an acoustic perturbation occursthen its propagation is given by that solution of (4) which at119905 = 0 is equal to the perturbation in discussion It is assumedalso that if at 119905 = 0 a source begins to produce permanenttime harmonic acoustic perturbation then the propagationof this perturbation is given by that solution of the system ofnon-homogeneous linearized Euler equations

1205971199061015840

120597119905+ 1198800sdot1205971199061015840

120597119909+1

1205880

sdot1205971199011015840

120597119909= 1198761

1205971199011015840

120597119905+ 1198800sdot1205971199011015840

120597119909+ 120574 sdot 119901

0sdot1205971199061015840

120597119909= 1198762

(5)

which is equal to zero for 119905 le 0Here 119876

1 1198762(functions or distributions) describe the

source-produced permanent time harmonic acoustic pertur-bations being equal to zero for 119905 le 0 and periodic in 119905 for119905 gt 0

Definition 2 The constant solution 119906 = 1198800 119901 = 119901

0 120588 = 120588

0

of (1) is linearly stable with respect to the initial value per-turbation (to the source-produced permanent time harmonicperturbation resp) if the ldquosolution 1199061015840(119909 119905) 1199011015840(119909 119905) of (4)rdquo (of(5) resp) ldquois small = close to zerordquo all time 119905 ge 0 providedldquo1199061015840(119909 0) 1199011015840(119909 0)rdquo ((119876

1 1198762) resp) ldquois small = close to zerordquo

In other words the constant solution of (1) is linearlystable if and only if the null solution of (4) is stable

These stability concepts are not necessarily equivalent tothe hydrodynamic stability defined in [10]

The precise meaning of the concepts ldquoperturbationrdquoldquosmall perturbationrdquo ldquosolution of the propagation problemrdquoand ldquosmall solutionrdquo has to be designed by other definitionsand there is some freedom here

Using this freedom in the following we present four dif-ferent function spaces in order to reveal that in each of themthemeaning of ldquoperturbationrdquo ldquosmall perturbationrdquo ldquosolutionto the propagation problemrdquo and ldquosmall solutionrdquo is specificand the results and mathematical tools are specific as well

Abstract and Applied Analysis 3

3 The First Function Space

The set 1198831of the perturbations of the initial value (instan-

taneous perturbations) is the topological function space [11]of the couples 119867 = (119865 119866) of continuously differentiablefunctions 119865 119866 1198771015840 rarr 119877

1015840 with respect to the usual algebraicoperations and topology generated by the uniform conver-gence on 1198771015840 [12]

A neighborhood of the origin119874 is a set1198810of couples from

1198831having the property that there exists 120576 gt 0 such that if for

119867 = (119865 119866) isin 1198831we have |119865(119909)| lt 120576 and |119866(119909)| lt 120576 for any

119909 isin 1198771015840 then119867 = (119865 119866) isin 119881

0

The set 1198811205760defined by

119881120576

0= 119867 = (119865 119866) isin 119883

1 |119865 (119909)| lt 120576 |119866 (119909)| lt 120576 forall119909 isin 119877

1015840

(6)

is a neighborhood of the originThemeaning of the concept ldquothe perturbation119867 = (119865 119866)

isin 1198831is smallrdquo is that there exists 120576 small such that119867 isin 119881

120576

0

For a perturbation 119867 = (119865 119866) isin 1198831(ie 1199061015840(119909 119900) =

119865(119909) 1199011015840(119909 119900) = 119866(119909)) the couple of functions 1198671015840(119909 119905) =

(1199061015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) =

119865 [119909 minus (1198800minus 1198880) 119905] + 119865 [119909 minus (119880

0+ 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] minus 119866 [119909 minus (119880

0minus 1198880) 119905]

211988801205880

1199011015840(119909 119905) = 119888

01205880sdot119865 [119909 minus (119880

0+ 1198880) 119905] minus 119865 [119909 minus (119880

0minus 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] + 119866 [119909 minus (119880

0minus 1198880) 119905]

2

(7)

is continuously differentiable and verifies (4) and the initialcondition

(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(8)

Proposition 3 In the class of the continuously differentiablefunctions the couple of functions (1199061015840(119909 119905) 1199011015840(119909 119905)) given by(7) is the unique point wise (classical) solution of the initialvalue problem (4) and (8)

Proof If (1199061015840(119909 119905) 1199011015840(119909 119905)) is a point wise (classical) solutionof the initial value problem (4) (8) then V1015840 = 1199061015840minus(1119888

01205880) sdot1199011015840

and 1199021015840 = 1199061015840 + (111988801205880) sdot 1199011015840 satisfy the equalities

(120597V1015840120597119905) + (1198800minus 1198880) sdot (120597V1015840120597119909) = 0

V1015840 (119909 0) = 119865 (119909) minus (111988801205880) sdot 119866 (119909)

(1205971199021015840120597119905) + (119880

0+ 1198880) sdot (120597119902

1015840120597119909) = 0

1199021015840(119909 0) = 119865 (119909) + (1119888

01205880) sdot 119866 (119909)

(9)

Hence by using the method of characteristics we obtain

V1015840 (119909 119905) = 119865 (119909 minus (1198800 minus 1198880) 119905) minus1

11988801205880

sdot 119866 (119909 minus (1198800minus 1198880) 119905)

1199021015840(119909 119905) = 119865 (119909 minus (119880

0+ 1198880) 119905) +

1

11988801205880

sdot 119866 (119909 minus (1198800+ 1198880) 119905)

(10)

Taking into account the equalities 1199061015840 = (12)(V1015840 + 1199021015840) 1199011015840 =(119888012058802)(1199021015840minus V1015840) we deduce that 1199061015840(119909 119905) 1199011015840(119909 119905) are given

by the formula (7) So it was shown that if (1199061015840(119909 119905) 1199011015840(119909 119905))is a point wise solution of (4) and (8) then necessarily1199061015840(119909 119905) 119901

1015840(119909 119905) are given by (7) The fact that the couple of

functions given by (7) is a solution of (4) (8) is obtained byverification

In other words when the set of the perturbations of theinitial value is 119883

1 then the initial value problem (4) and (8)

has a unique classical (point wise) solution [9] given by (7)If a sequence of perturbations 119867

119899= (119865119899 119866119899) isin 119883

1

converges in 1198831to 119867 = (119865 119866) isin 119883

1 then for any 119905 ge 0

(fixed) the sequence of the corresponding solutions 1198671015840119899(sdot 119905)

belongs to 1198831and converges in 119883

1to the solution 1198671015840(sdot 119905)

corresponding to119867Thismeans that when the set of the initial value perturba-

tions is the function space 1198831 then the initial value problem

(4) and (8) is well posed in sense of Hadamard [5 6] on [0 119879]for any 119879 gt 0

In the following the couple of functions defined by (7) isconsidered to be the solution to the problem of propagationof the initial value perturbation for data in119883

1

The linear stability of the constant flow 1198800 1199010 1205880would

mean that for any 120576 gt 0 there exists 120575 = 120575(120576) such that forany 119867 = (119865 119866) isin 119881

120575(120576)

0the corresponding solution 1198671015840(119909 119905)

(given by (7)) satisfies1198671015840(sdot 119905) isin 1198811205760for any 119905 ge 0

Concerning the linear stability the following statementholds

Proposition 4 For any 120576 gt 0 and 120575(120576) = 1205762max (1+11988801205880 1+

(111988801205880)) if119867 isin 119881

120575(120576)

0 then1198671015840(sdot 119905) isin 119881120576

0for 119905 ge 0

Proof If119867 = (119865 119866) isin 119881120575(120576)

0 then |119865(119909)| lt 120575(120576) and |119866(119909)| lt

120575(120576) for any 119909 isin 1198771015840 Since 120575(120576) le 1205762(1 + 11988801205880) and 120575(120576) le

1205762(1+(111988801205880)) it follows that the following inequalities hold

|119865(119909)| lt 1205762(1 + 11988801205880) |119865(119909)| lt 1205762(1 + (1119888

01205880)) |119866(119909)| lt

1205762(1 + 11988801205880) and |119866(119909)| le 1205762(1 + (1119888

01205880)) for any 119909 isin 1198771015840

By using (7) we obtain that for any 119909 isin 1198771015840 and 119905 ge 0 thefollowing inequalities hold

100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 2 sdot

120576

2 (1 + 11988801205880)+

1

211988801205880

sdot 2 sdot120576

2 (1 + (111988801205880))= 120576


100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 211988801205880sdot

120576

2 (1 + (111988801205880))

+1

2sdot

2120576

2 (1 + 11988801205880)= 120576

(11)

Hence1198671015840(sdot 119905) isin 1198811205760for any 119905 ge 0

This means that the null solution of (4) is stable that isthe constant solution 119906 = 119880

0 119901 = 119901

0 120588 = 120588

0of (1) is linearly

stable with respect to the initial value perturbation with datafrom119883

1

On the other hand for initial data from 1198831the set of the

exponential growth rates of the solutions of the initial valueproblem (4) and (8) is the whole real axis and there exist alsosolutions whose exponential growth rate is equal to +infin (forinstance that of the solution corresponding to the initial data119865(119909) = 119866(119909) = exp(1199092))

Here the exponential growth rate of1198671015840(119909 119905) = (1199061015840(119909 119905)1199011015840(119909 119905)) is defined as

maxsup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816

119905 sup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816

119905 (12)

The stability of the null solution of (4) (ie linear stabilityof the constant solution) in the presence of solutions havingstrictly positive or +infin growth rate can be surprising Thatis because usually the stability of the null solution of theinfinite dimensional linear evolutionary equations is analyzedin a function space in which the Hille-Yosida theory canbe applied [5 6 8 13] that is Banach space or locallyconvex and sequentially complete topological vector spacesIn the topological function space 119883

1the origin possesses

neighborhoods which are not absorbing This is the mainreasonwhile stability and strictly positive growth rate coexist

For initial value perturbations from1198831 the very popular

Briggs-Bers stability analysis [14 15] in case of the problem(4) (8) cannot be applied That is because in this contextdispersion relations cannot be derived (for an arbitrary solu-tion the Fourier transform with respect to 119909 and the Laplacetransform with respect to 119905 in this context do not exist) Soit is impossible to analyze the imaginary parts of the zeros ofthe dispersion relations as requires the Briggs-Bers stabilitycriterion

We consider now source-produced permanent time har-monic perturbations whose amplitudes belong to 119883

1 More

precisely perturbations for which the right-hand members1198761 1198762 appearing in (5) are of the form

1198761 (119909 119905) = 119865 (119909) sdot ℎ (119905) sdot sin120596119891119905

1198762(119909 119905) = 119866 (119909) sdot ℎ (119905) sdot sin120596

119891119905

(13)

Here (119865 119866) isin 1198831and represent the amplitude of the per-

turbation ℎ(119905) is Heaviside function and 120596119891gt 0 is the

angular frequency Such a perturbation will be denoted by119875(119865 119866 120596

119891)

For the perturbation 119875(119865 119866 120596119891) the couple of functions

1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) = ℎ (119905)

sdot int

119905

0

[ (119865 [119909 minus (1198800minus 1198880) (119905 minus 120591)]

+119865 [119909 minus (1198800+ 1198880) (119905 minus 120591)]) times (2)

minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]

minus119866 [119909 minus (1198800minus 1198880) (119905 minus 120591)]) times (2119888

01205880)minus1]

sdot sin120596119891120591119889120591

1199011015840(119909 119905) = ℎ (119905)

sdot int

119905

0

[11988801205880sdot (119865 [119909 minus (119880

0+ 1198880) (119905 minus 120591)]

minus119865 [119909 minus (1198800minus 1198880) (119905 minus 120591)]) times (2)

minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]

+119866 [119909 minus (1198800minus 1198880) (119905 minus 120591)]) times (2)

minus1]

sdot sin120596119891120591119889120591

(14)

is continuously differentiable and verifies (5) and the condi-tion

1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (15)

Proposition 5 In the class of the continuously differentiablefunctions the couple of functions (1199061015840(119909 119905) 1199011015840(119909 119905)) given by(14) is the unique point wise solution of the problem (5) (13)and (15)

Proof The proof of this proposition is similar to the proof ofProposition 3

In the following the couple of functions defined by (14) isconsidered to be the solution of the problemof propagation ofthe source-produced permanent time harmonic perturbationin case of the function space119883

1

Proposition 6 For 119879 gt 0 120576 gt 0 and 120575 = 120575(120576 119879) = 1205762119879max (1 + 119888

01205880 1 + (1119888

01205880)) gt 0 if 119867 = (119865 119866) isin 119881

120575

0 then

1198671015840(∘ 119905) isin 119881

120576

0 for any 119905 isin [0 119879]


This means that in this function space the propagationproblem (5) and (13) of the source-produced permanent timeharmonic perturbation is well posed on [0 119879] for any 119879 gt 0

Concerning the linear stability of the constant flow thefollowing statement holds


Proposition 7 In the case of source produced permanent timeharmonic perturbations whose amplitude belongs to the func-tion space 119883

1the constant flow is linearly unstable

Proof The linear instability of the constant flow 1198800 1199010 1205880

means that there exists 120576 gt 0 such that for any120575 gt 0 there exist119867 = (119865 119866) isin 119881

120575

0and 119905 ge 0 such that the corresponding

solution1198671015840(119909 119905) (given by (14)) satisfies1198671015840(sdot 119905) notin 1198811205760

The instability can be seen considering for instance 120576 = 1the perturbation119875

120575(119865 119866 120596

119891)= ℎ(119905)sdot(120575sdotsin 119909sdotsin120596

119891119905 0)where

120575 gt 0 is an arbitrary real number and 120596119891= 1198880minus 1198800 Note that

by choosing 120575 small the above perturbation can be made assmall as we wish According to (14) the propagation of thisperturbation is given by

1199061015840(119909 119905) = ℎ (119905) sdot

120575

4

sdot minus 119905 sdot cos (119909 + 120596119891119905) +

sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) +

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

1199011015840(119909 119905) = ℎ (119905) sdot

11988801205880120575

4

sdot 119905 sdot cos (119909 + 120596119891119905) minus

sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) minus

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

(16)

Hence for a given 120575 if 119905 is sufficiently large then1198671015840(sdot 119905) notin 11988110

(that is because (1199061015840(sdot 119905) 1199011015840(sdot 119905)) is as large as we wish)

The Briggs-Bers stability analysis [14 15] with respectto source-produced permanent time harmonic perturbationwhose amplitude belongs to 119883

1 cannot be applied That is

because in this context dispersion relations cannot be derived(there exist perturbations whose amplitude has no Fouriertransform for instance 119865(119909) = 119866(119909) = exp(119909)) So it isimpossible to analyze the imaginary parts of the zeros ofthe dispersion relations as requires the Briggs-Bers stabilitycriterion

It follows that the above instability cannot be obtained byBriggs-Bers stability analysis

4 The Second Function Space

The set 1198832of the perturbations of the initial value is the

normed space [16] of the couples119867 = (119865 119866) of continuouslydifferentiable and bounded functions 119865 119866 119877

1015840rarr 1198771015840 with

respect to the usual algebraic operations and norm definedby

119867 = maxsup119909isin1198771015840

|119865 (119909)| sup119909isin1198771015840

|119866 (119909)| (17)

The set 1198811205760 defined by

119881120576

0= 119867 = (119865 119866) isin 119883

2 119867 lt 120576 (18)

is a neighborhood of the origin and the meaning of theconcept ldquothe perturbation 119867 = (119865 119866) isin 119883

1is smallrdquo is that

there exists 120576 gt 0 small such that119867 isin 119881120576

0

For an initial data 119867 = (119865 119866) isin 1198832the couple of

functions 1198671015840(119909 119905) = (1199061015840(119909 119905) 119901

1015840(119909 119905)) given by (7) is the

unique bounded classical solution of (4) and (8)If a sequence of initial data119867

119899= (119865119899 119866119899) isin 1198832converges

in 1198832to 119867 = (119865 119866) isin 119883

2 then for any 119905 ge 0 (fixed) the

sequence of the corresponding solutions1198671015840119899(sdot 119905) converges in

1198832to the solution1198671015840(sdot 119905) corresponding to119867This means that for the set of initial data 119883

2 the initial

value problem (4) and (19)

(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(19)

is well posed in sense of Hadamard [5 6] on [0 119879] for any119879 gt 0


2


mean that for any 120576 gt 0 there exists 120575 = 120575(120576) such that for any119867 isin 119883

2if 119867 lt 120575(120576) the corresponding solution 1198671015840(119909 119905)

(given by (7)) satisfies 1198671015840(sdot 119905) lt 120576 for any 119905 ge 0Concerning the linear stability the following statement

holds

Proposition 8 For any 120576 gt 0 and119867 isin 1198832if 119867 lt 1205762max

(1 + 11988801205880 1 + 1119888

01205880) then 1198671015840(sdot 119905) lt 120576 for 119905 ge 0


Proof The proof is similar to that of Proposition 4


0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2

For initial data from 1198832the exponential growth rate

given by (12) of the solution of the initial value problem (4)and (19) is equal to zero However for initial values from 119883

2

the Briggs-Bers stability analysis in case of the problem (4)and (19) cannot be applied That is because for instance theFourier transformwith respect to 119909 of the initial value 119865(119909) =119866(119909) equiv 1 does not exist So dispersion relations cannot bederived and the stability criterion cannot be used Thereforethe above stability results cannot be obtained by Briggs-Bersstability analysis

For source-produced permanent time harmonic pertur-bations whose amplitudes belong to 119883

2 the right-hand

members11987611198762 appearing in (5) are those given by (13) with

(119865 119866) isin 1198832

For the perturbation 119875(119865 119866 120596119891)((119865 119866) isin 119883

2) the couple

of functions1198671015840(119909 119905) = (1199061015840(119909 119905) 1199011015840(119909 119905)) given by (14) is theunique solution of the problem (5) (13) and (15) satisfying

1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)

Concerning the continuous dependence on the amplitude ofthe perturbation the following statement holds

Proposition 9 For any 119879 gt 0 120576 gt 0 and 120575 = 120575(120576 119879) = 1205762119879max (1+ 119888

01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]

Proof The proof is similar to the proof of Proposition 6

This means that the problem (5) and (20) is well posed on[0 119879] for any 119879 gt 0

When the amplitude of the source-produced perturba-tion belongs to the function space119883

2the couple of functions

defined by (14) is considered to be the solution of the problemof propagation of the source-produced permanent time har-monic perturbation


Proposition 10 The constant flow of (1) is linearly unstablewith respect to the source-produced permanent time harmonicperturbations whose amplitude belongs to the function space1198832

Proof It turns that the perturbations 119875120575(119865 119866 120596

119891) = ℎ(119905) sdot

(120575 sdot sin119909 sdot sin120596119891119905 0) considered in the function space119883

1are

appropriate to show the above statement


2 can not be applied That

is because in this context dispersion relations can not bederived (there exist perturbations whose amplitude has noFourier transform for instance the perturbation ℎ(119905) sdot (sin 119909 sdot

sin120596119891119905 0) has no Fourier transform with respect to 119909) So it

is impossible to analyze the imaginary parts of the zeros ofthe dispersion relations as requires the Briggs-Bers stabilitycriterion Therefore the above instability result cannot beobtained by Briggs-Bers stability analysis

5 The Third Function Space

In this case the set 1198833of the initial data is the normed space

of the couples 119867 = (119865 119866) of functions 119865 119866 isin 1198712(1198771015840) [17]

with respect to the usual algebraic operations and the normdefined by

1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)

It has to be emphasized that the functions 119865 119866 isin 1198712(1198771015840) aredefined up to addition of a measure zero function

Themeaning of the concept ldquothe perturbation119867 = (119865 119866)

isin 1198833is smallrdquo is that there exists 120576 gt 0 small such that

1198671198833

lt 120576For an initial data 119867 = (119865 119866) isin 119883

3the couple of func-

tions1198671015840(119909 119905) = (1199061015840(119909 119905) 1199011015840(119909 119905)) given by (7) is called gen-eralized solution of (4) A generalized solution is unique upto addition of a measure zero solution In general1198671015840(119909 119905) isnot differentiable but 1198671015840(sdot 119905) isin 119883

3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)

When 119865 119866 are continuously differentiable then1198671015840(119909 119905) sat-isfies (4) in classical sense

It has to be noted that there exist continuous functions119865 119866 which are not differentiable at every point [17] Thegeneralized solution which corresponds to an initial data ofthis type is continuous but it is not differentiable at everypoint

Concerning the continuous dependence on the initialdata we remark that for 119905 fixed1198671015840(sdot 119905) satisfies

100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)

Inequality (23) implies that if the sequence of initial data119867119899=

(119865119899 119866119899) isin 119883

3converges in 119883

3to 119867 = (119865 119866) isin 119883

3 then for

any 119905 ge 0 the sequence of the corresponding solutions1198671015840119899(sdot 119905)

converges in 1198833to 1198671015840(sdot 119905) corresponding to 119867 = (119865 119866) isin

1198833This means continuous dependence on the initial dataTherefore in the case of the set of initial data 119883

3 the

initial value problem (4) and (22) is well posed in sense ofHadamard on [0 119879] for every 119879 gt 0

In the following the couple of functions defined by (7) upto addition of a measure zero solution is considered to bethe solution to the problem of propagation of the initial valueperturbation for data in119883

3

As concerns stability the following statement holds



(1 + 11988801205880 1 + 1119888



This means that the null solution of (4) is stable that isthe constant solution = 119880

0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3


given by (12) of the solution of the initial value problem (4)and (22) is equal to zero

For every generalized solution the Laplace transform isdefined for Re 119911 gt 0 and the Fourier transform with respectto 119909 exists So dispersion relation can be derived and thestability can be analyzed by analyzing the imaginary partsof the zeros of the dispersion relation This means that bothsteps of the Briggs-Bers stability analysis can be undertakenHowever the above stability result was obtained directly notby Briggs-Bers stability analysis

What can be strange for engineers in this function spaceis the presence of solutions which are continuous but are notdifferentiable at every pointWhich kind of propagation doesrepresent such a solution

For source-produced permanent time harmonic per-turbation whose amplitudes belong to 119883

3 the right-hand


(119865 119866) isin 1198833


3) the couple

of functions1198671015840(119909 119905) = (1199061015840(119909 119905) 1199011015840(119909 119905)) given by (14) up toaddition of a measure zero solution will be called generalizedsolution of the propagation problem (5) and (24)

1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)


Proposition 12 For any 119879 gt 0 120576 gt 0 and 120575 = 120575(120576 119879) = 1205762119879max(1+119888

01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]



The linear instability of the constant flow of (1) withrespect to this kind of perturbations can be shown similarlyas in the function space119883

1 by considering the perturbations

119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800

Since the Fourier transform with respect to 119909 and theLaplace transform with respect to 119905 of any solution existdispersion relation can be derived and the stability analysiscould be undertaken by analyzing the imaginary parts ofzeros of dispersion relations (both steps of the Briggs-Bersstability analysis can be undertaken in this function space)However in this paper the instability was derived directly

6 The Fourth Function Space

The set1198834of the perturbations of the initial data is the locally

convex vector space of the couples 119867 = (119865 119866) of tempereddistributions 119865 119866 isin 1198781015840(1198771015840) [8] with respect to the usual alge-braic operations and seminorms 119902

119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)

where 119861 are bounded sets in the space of the rapidly decreas-ing functions 119878(1198771015840) and 119902

119861is the seminorm on 1198781015840(1198771015840) defined

by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)

is a neighborhood of the origin and the meaning of the con-cept ldquothe perturbation119867 = (119865 119866) isin 119883

4is smallrdquo is that there

exists 120576 gt 0 small and 119861 bounded such that 119867 = (119865 119866) isin

119881120576119902119861

0If the initial value perturbation is a couple of tempered

distributions 119867 = (119865 119866) isin 1198834 then a solution of (4) is a

family1198671015840(119905) = (1199061015840(119905) 1199011015840(119905)) of couples of tempered distribu-tions which satisfies (4) for 119905 ge 0 and the initial condition

1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)

Proposition 13 If the initial value problem (4) and (27) has asolution1198671015840(119905) = (1199061015840(119905) 1199011015840(119905)) then its Fourier transform withrespect to 119909 is a family of couples of tempered distributions1015840(119905) = (

1015840(119905) 1199011015840(119905)) 119905 ge 0 which satisfies the following equa-

tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)

and the initial condition

1015840(0) = 119865 119901

1015840(0) = 119866 (29)

Proof By computation

Remark that the problem (4) and (27) is well posed if andonly if the problem (28) and (29) is well posed

Concerning the problem (28) and (29) the following state-ment holds

Proposition 14 The problem (28) and (29) is well posed onany finite interval and its solutions is given by

1015840(119905) =

119890minus119894(1198800minus1198880)119896119905+ 119890minus119894(1198800+1198880)119896119905

2

sdot 119865 +119890minus119894(1198800+1198880)119896119905minus 119890minus119894(1198800minus1198880)119896119905

211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880

119890minus119894(1198800+1198880)119896119905minus 119890minus119894(1198800minus1198880)119896119905

2

sdot 119865 +119890minus119894(1198800+1198880)119896119905+ 119890minus119894(1198800minus1198880)119896119905

2sdot 119866

(30)


Proof The existence and uniqueness are easy to see We willshow only the continuous dependence on the initial datain case of the problem (28) and (29) For that consider119879 gt 0 120576 gt 0 and a semi norm 119902

119861in 1198834 After that con-

sider a bounded set 1198611015840 sub 119878(1198771015840) which contains the followingbounded sets


2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)

for any 119905 isin [0 119879] the semi norm 1199021198611015840 in 119883

4and 120575 = 120576(1 +

1199021198611015840(119865 119866)) gt 0 Remark that for any (119865 119866) isin 119883

4with 119902

1198611015840

(119865 119866) lt 120575 we have 119902119861(1015840(119905) 1199011015840(119905)) lt 120576 for 119905 isin [0 119879]

Proposition 15 The problem (4) and (27) is well posed

Proof The inverse Fourier transform of (1015840(119905) 1199011015840(119905)) = 1015840(119905)given by (30) is the unique solution of (4) and (27) Thecontinuous dependence of the solutions of (4) and (27) is acorollary of the continuous dependence on the initial dataof the solution of (28) and (29) and of the continuity of theFourier and inverse Fourier transforms

So the problem (4) and (27) is well posed

Concerning the stability the following statement holds

Proposition 16 The null solution of (28) is unstable

Proof The instability of the null solution of (28) means thatthere exist 120576

0gt 0 and a semi norm 119902

1198610

such that for any realnumber 120575 gt 0 and any semi norm 119902

119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861

0and 119905 ge 0 such that the corresponding solution

1198671015840(119905) (given by (30)) satisfies1198671015840(119905) notin 1198811205760 1199021198610

0 Let us consider

now 1205760= 1 119861

0mdasha bounded set of rapidly decreasing func-

tions which contains the function 1205930(119896) = exp(minus1198962) and

the tempered distribution defined by 1198650(120593) = 120593

1015840(0) Remark

that for an arbitrary real number 120575 gt 0 and an arbitrarybounded set 119861 of rapidly decreasing functions there exists areal number 120576 gt 0 such that 119902

119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)

It follows that (120576 sdot 1198650 0) isin 119881

120575119902119861

0 On the other hand for the

solution (1015840(119905) 1199011015840(119905)) of the initial value problem (28) (29)with 119865 = 120576119865

0and 119866 = 0 the following relations hold

1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)

Therefore for 119905 sufficiently large we have 1199021198610

(1015840(119905) 1199011015840(119905)) gt

1 = 1205760

Hence the null solution of (28) is unstable Since theinverse Fourier transform is continuous the null solution of(4) is also unstable in119883

4


4 the right-hand

members 1198761 1198762 appearing in (5) are given by

1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834

The unknown tempered distributions 1199061015840(119905) 1199011015840(119905) have tosatisfy (5) and the condition

1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)

If the problem (5) and (34) has a solution 1198671015840(119905) = (1199061015840(119905)

1199011015840(119905)) isin 119883

4 then the Fourier transforms 1015840(119905) V1015840(119905) are tem-

pered distributions and satisfy

1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2

times [119890119894120596119891119905minus 119890minus119894(1198800minus1198880)119896119905

119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

minus119890119894120596119891119905minus 119890minus119894(1198800minus1198880)119896119905

119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2

times [119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800minus1198880)119896119905

119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)

It can be seen that for arbitrary 119865 119866 isin 1198781015840(1198771015840) 1015840(119905) V1015840(119905)

given by (36) are not necessarily tempered distributionsThis contradiction shows that the problem (5) and (34) is illposed

7 Conclusions

(i) The obtained results show that the well-posedness andlinear stability concepts of the constant gas flow in 1D flowmodel are highly dependent on the function space Thereexists function space in which the constant gas flow is stableand another function space inwhich the same flow is unstablewith respect to the initial value perturbations There existsfunction space in which the propagation problem of thesource produced perturbation is well posed and anotherfunction space in which the propagation problem is ill posed

(ii) For the stability of the constant gas flow with respectto the initial value perturbations it is not necessary that theset of the exponential growth rate of the solutions of thehomogeneous linearized Euler equations be bounded fromabove There exists function space in which the constant gasflow is stable with respect to the initial value perturbationsalthough the set of the exponential growth rate of thesolutions of the homogeneous Euler equations is the wholereal axis So the stability of the null solution cannot be deniedjust because the set of the exponential growth rate of thesolutions is not bounded from above

(iii) The stability of the constant gas flow with respect tothe initial value perturbations is not a sufficient condition forthe stability of the same solution with respect to source pro-duced permanent time harmonic perturbations There existsfunction space in which the constant gas flow is stable withrespect to the initial value perturbations and it is unstablewith respect to the source produced perturbations

(iv) In two of the considered function spaces the Briggs-Bers stability analysis cannot be applied because dispersionrelations cannot be derived So the stability or instabilityresults obtained directly in this function spaces cannot beobtained by Briggs-Bers stability analysis

(v) Continuous dependence on the initial data (on thesource amplitude resp) is an expression of stability on a finite

interval of time and stability means continuous dependenceon the infinite interval of time [0 +infin) In practice theduration of the propagation is finite For the eight situationsconsidered in this paper the continuous dependence on anyfinite interval of time in seven situations is validThe stabilityon [0 +infin) is valid only in three situations So the continuousdependence on any finite interval of time is less dependenton the function space Taking into account the above factsfrom practical point of view the question is the continuousdependence or the stability is important

(vi) When theoretical results are tested against experi-mental results the computed mathematical variable has tocorrespond to the experimentally measured quantity Forinstance in this paper there are four different mathematicalvariables for expressing that the perturbation and the corre-sponding solution of the propagation problem are small thatis close to zero Which one of them does correspond to anexperimentally measured quantity when the duct is infinitelylong

(vii) The results obtained could be useful in a better un-derstanding of some apparently strange results published inthe literature concerning the sound propagation in a gas flow-ing through a lined duct since there are neither set rules norunderstanding of the ldquorightrdquo way to model the phenomenon

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Thisworkwas supported by a grant of the RomanianNationalAuthority for Scientific Research CNCS-UEFISCDI projectno PN-II-ID-PCE-2011-3-0171 The results have been pre-sented orally at Sz-Nagy Centennial Conference June 24ndash28 2013 Szeged Hungary (httpwwwmathu-szegedhuSzNagy100programphp)

References

[1] E J Brambley ldquoFundamental problems with the model ofuniform flow over acoustic liningsrdquo Journal of Sound andVibration vol 322 no 4-5 pp 1026ndash1037 2009

[2] E J Brambley ldquoA well-posed modified myers boundary con-ditionrdquo in Proceedings of the 16th AIAACEAS AeroacousticsConference (31st AIAA Aeroacoustics Conference) June 2010

[3] Y Renou and Y Auregan ldquoOn a modified Myers boundarycondition to match lined wall impedance deduced from severalexperimentalmethods in presence of a grazing flowrdquo inProceed-ings of the 16th AIAACEASAeroacoustics Conference p 14 June2010

[4] E J Brambley ldquoWell-posed boundary condition for acousticliners in straight ducts with flowrdquo AIAA Journal vol 49 no 6pp 1272ndash1282 2011

[5] G E Ladas and V Lakshmikantham Differential Equations inAbstract Spaces Academic Press New York NY USA 1972

[6] S G Krein Differential Equations in Banach Spaces IzdldquoNaukardquo Moskow 1967 (Russian)


[7] A M Balint St Balint and R Szabo ldquoLinear stability of a nonslipping gas flow in a rectangular lined duct with respect toperturbations of the initial value by indefinitely differentiabledisturbances having compact supportrdquoAIPConference Proceed-ings vol 1493 p 1023 2012

[8] K Yosida Functional Analysis vol 123 Springer Berlin Ger-many 6th edition 1980

[9] B L Rozdjestvenskii and N N Ianenko Quasilinear Systems ofEquations andTheir Applications in Gasdynamics Izd ldquoNaukardquoMoskow Russia 1978 (Russian)

[10] P G Drazin and W N Reid Hydrodynamic Stability Cam-bridge University Press Cambridge UK 1995

[11] S H Saperstone Semidynamical Systems in Infinite-Dimensional Spaces vol 37 Springer New York NY USA1981

[12] N Bourbaki Elements de Mathematique Premier Partie LivreIII Topologie Generale chapitre 1 structure topologique Her-mann Paris France 1966

[13] E Hille and R S Phillips Functional Analysis and Semi-GroupsAmerican Mathematical Society Providence RI USA 1957

[14] A Bers ldquoSpace-time evolution of plasma-instabilitiesmdashabsolute and convectiverdquo in Handbook of Plasma Physics I AA Galeev and R M Sudan Eds pp 452ndash484 North HollandAmsterdam The Netherlands 1983

[15] R J Briggs Electron Stream Interaction with Plasmas ResearchMonograph no 29 MIT Press Cambridge Mass USA 1964

[16] N Dunford and J T Schwartz Linear Operators Part 1 GeneralTheory Wiley-Interscience New York NY USA 1957

[17] R Riesz and B Sz-Nagy Lecons drsquoAnalyse FonctionnelleAkademiai Kiado Budapest Hungary 1953

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MathematicsJournal of


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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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for modifying the boundary conditions in order to make theproblem ldquowell posedrdquo in the sense defined in [1] Here it has tobe mentioned that in [1] a concept of ldquowell posed differentialequationrdquo is defined This is different from the concept ofldquowell-posed problemrdquo usually in mathematics [5 6] andintroduced by Hadamard long time ago

Definition 1 FollowingHadamard for a given class of instan-taneous initial value perturbations (permanent source-pro-duced time harmonic perturbations resp) one calls theperturbation propagation problem well posed if there isa unique solution to the problem and the solution variescontinuously with the initial data (source amplitude resp)

In [7] it was shown that the concept introduced in [1] isconfusing because the equation considered in [1] dependingon the function space can be ill-posed or can be well posedIn fact the concept introduced in [1] is the transcription ofthe behavior of an example of a semigroup of contractions ofclass 119862

0which act in a precise function space (see [1 8] pages

240-241 referred in [1]) In this situation according to [8] theresolvent operator of the infinitesimal generator exists in ahalf plane of the form Re 119911 gt 120583 and the Laplace transformcan be considered for every solution of the homogeneousequation

The objective of the present work is to underline thatthe precise description of the function space is crucial evenin the case of the 1D gas flow model where the wall andthe lining effect are absent For the achievement of thisobjective the linear stability analysis of the constant 1D gasflow with respect to the initial value and source-producedpermanent time harmonic perturbations is presented in fourdifferent function spaces revealing significant differencesFor instance in the topological function space119883

4the propa-

gation problem of the initial value perturbation is well posedbut that of the source produced perturbation is ill posed inthe topological function space 119883

1in which the origin has

not absorbing neighborhoods stability and strictly positivegrowth rate coexists in contrast to the case of topologicalvector spaces119883

21198833

2 The 1D Gas Flow Model

In the 1D gas flow model the nonlinear Euler equations gov-erning the flow of an inviscid compressible nonheat con-ducting isentropic perfect gas according to [9] are

120597119906

120597119905+ 119906 sdot

120597119906

120597119909+1

120588sdot120597119901

120597119909= 0

120597120588

120597119905+ 119906 sdot

120597120588

120597119909+ 120588 sdot

120597119906

120597119909= 0

(1)

Here 119905 is time 119906 is velocity along the 119874119909 axis 119901 is pressureand 120588 is density Equation (1) are considered for 119909 isin 1198771015840 and119905 ge 0 It is assumed that 119901 120588 and the absolute temperature 1198791015840satisfy the equation of state of the perfect gas

119901 = 120588 sdot 119877 sdot 1198791015840 (2)

with 119877 = 119888119901minus 119888V 119888119901 119888V being the specific heat capacities at

constant pressure and constant volume respectively Let

119906 equiv 1198800= const gt 0 120588 equiv 120588

0= const gt 0 119901 equiv 119901

0= const gt 0

be a constant solution of the system of partial differentialequations (SPDE) (1) According to (2) 119901

0= 1205880sdot119877sdot1198791015840

0and the

associated isentropic sound speed 1198880verifies 1198882

0= 120574sdot(119901

01205880) =

120574 sdot 119877 sdot 1198791015840

0 where 120574 = 119888

119901119888V

Linearizing (1) at 119906 = 1198800 120588 = 120588

0 119901 = 119901

0and using the

perturbations 1199011015840 1205881015840 of 1199010 1205880satisfying

(120597

120597119905+ 1198800sdot120597

120597119909) (1199011015840minus 1198882

01205881015840) = 0 (3)

the following system of homogeneous linear partial differ-ential equations for the perturbations 1199061015840 1199011015840 of 119880

0 1199010 is

obtained

1205971199061015840

120597119905+ 1198800sdot1205971199061015840

120597119909+1

1205880

sdot1205971199011015840

120597119909= 0

1205971199011015840

120597119905+ 1198800sdot1205971199011015840

120597119909+ 120574 sdot 119901

0sdot1205971199061015840

120597119909= 0

(4)

It is assumed that if at 119905 = 0 an acoustic perturbation occursthen its propagation is given by that solution of (4) which at119905 = 0 is equal to the perturbation in discussion It is assumedalso that if at 119905 = 0 a source begins to produce permanenttime harmonic acoustic perturbation then the propagationof this perturbation is given by that solution of the system ofnon-homogeneous linearized Euler equations

1205971199061015840

120597119905+ 1198800sdot1205971199061015840

120597119909+1

1205880

sdot1205971199011015840

120597119909= 1198761

1205971199011015840

120597119905+ 1198800sdot1205971199011015840

120597119909+ 120574 sdot 119901

0sdot1205971199061015840

120597119909= 1198762

(5)

which is equal to zero for 119905 le 0Here 119876

1 1198762(functions or distributions) describe the

source-produced permanent time harmonic acoustic pertur-bations being equal to zero for 119905 le 0 and periodic in 119905 for119905 gt 0

Definition 2 The constant solution 119906 = 1198800 119901 = 119901

0 120588 = 120588

0

of (1) is linearly stable with respect to the initial value per-turbation (to the source-produced permanent time harmonicperturbation resp) if the ldquosolution 1199061015840(119909 119905) 1199011015840(119909 119905) of (4)rdquo (of(5) resp) ldquois small = close to zerordquo all time 119905 ge 0 providedldquo1199061015840(119909 0) 1199011015840(119909 0)rdquo ((119876

1 1198762) resp) ldquois small = close to zerordquo

In other words the constant solution of (1) is linearlystable if and only if the null solution of (4) is stable

These stability concepts are not necessarily equivalent tothe hydrodynamic stability defined in [10]

The precise meaning of the concepts ldquoperturbationrdquoldquosmall perturbationrdquo ldquosolution of the propagation problemrdquoand ldquosmall solutionrdquo has to be designed by other definitionsand there is some freedom here

Using this freedom in the following we present four dif-ferent function spaces in order to reveal that in each of themthemeaning of ldquoperturbationrdquo ldquosmall perturbationrdquo ldquosolutionto the propagation problemrdquo and ldquosmall solutionrdquo is specificand the results and mathematical tools are specific as well









119909 isin 1198771015840 then119867 = (119865 119866) isin 119881

0


119881120576

0= 119867 = (119865 119866) isin 119883

1 |119865 (119909)| lt 120576 |119866 (119909)| lt 120576 forall119909 isin 119877

1015840

(6)



120576

0



(1199061015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) =


0+ 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] minus 119866 [119909 minus (119880

0minus 1198880) 119905]

211988801205880

1199011015840(119909 119905) = 119888

01205880sdot119865 [119909 minus (119880

0+ 1198880) 119905] minus 119865 [119909 minus (119880

0minus 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] + 119866 [119909 minus (119880

0minus 1198880) 119905]

2

(7)


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(8)



01205880) sdot1199011015840


(120597V1015840120597119905) + (1198800minus 1198880) sdot (120597V1015840120597119909) = 0

V1015840 (119909 0) = 119865 (119909) minus (111988801205880) sdot 119866 (119909)

(1205971199021015840120597119905) + (119880

0+ 1198880) sdot (120597119902

1015840120597119909) = 0

1199021015840(119909 0) = 119865 (119909) + (1119888

01205880) sdot 119866 (119909)

(9)



11988801205880

sdot 119866 (119909 minus (1198800minus 1198880) 119905)

1199021015840(119909 119905) = 119865 (119909 minus (119880

0+ 1198880) 119905) +

1

11988801205880

sdot 119866 (119909 minus (1198800+ 1198880) 119905)

(10)








119899= (119865119899 119866119899) isin 119883

1










1



120575(120576)





(111988801205880)) if119867 isin 119881

120575(120576)

0 then1198671015840(sdot 119905) isin 119881120576

0for 119905 ge 0

Proof If119867 = (119865 119866) isin 119881120575(120576)

0 then |119865(119909)| lt 120575(120576) and |119866(119909)| lt



|119865(119909)| lt 1205762(1 + 11988801205880) |119865(119909)| lt 1205762(1 + (1119888

01205880)) |119866(119909)| lt

1205762(1 + 11988801205880) and |119866(119909)| le 1205762(1 + (1119888

01205880)) for any 119909 isin 1198771015840


100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 2 sdot

120576

2 (1 + 11988801205880)+

1

211988801205880

sdot 2 sdot120576

2 (1 + (111988801205880))= 120576


100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 211988801205880sdot

120576

2 (1 + (111988801205880))

+1

2sdot

2120576

2 (1 + 11988801205880)= 120576

(11)



0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


1




maxsup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816

119905 sup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816

119905 (12)







1 More


1198761 (119909 119905) = 119865 (119909) sdot ℎ (119905) sdot sin120596119891119905

1198762(119909 119905) = 119866 (119909) sdot ℎ (119905) sdot sin120596

119891119905

(13)




119891)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) = ℎ (119905)

sdot int

119905

0



minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


01205880)minus1]

sdot sin120596119891120591119889120591

1199011015840(119909 119905) = ℎ (119905)

sdot int

119905

0

[11988801205880sdot (119865 [119909 minus (119880

0+ 1198880) (119905 minus 120591)]


minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


minus1]

sdot sin120596119891120591119889120591

(14)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (15)




1


01205880 1 + (1119888

01205880)) gt 0 if 119867 = (119865 119866) isin 119881

120575

0 then

1198671015840(∘ 119905) isin 119881

120576

0 for any 119905 isin [0 119879]









120575




120575(119865 119866 120596


119891119905 0)where



1199061015840(119909 119905) = ℎ (119905) sdot

120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) +

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

1199011015840(119909 119905) = ℎ (119905) sdot

11988801205880120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) minus

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

(16)










1015840rarr 1198771015840 with


119867 = maxsup119909isin1198771015840

|119865 (119909)| sup119909isin1198771015840

|119866 (119909)| (17)


119881120576

0= 119867 = (119865 119866) isin 119883

2 119867 lt 120576 (18)




0


functions 1198671015840(119909 119905) = (1199061015840(119909 119905) 119901

1015840(119909 119905)) given by (7) is the


119899= (119865119899 119866119899) isin 1198832converges

in 1198832to 119867 = (119865 119866) isin 119883




2 the initial


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(19)



2





holds


(1 + 11988801205880 1 + 1119888





0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2



2



2 the right-hand


(119865 119866) isin 1198832


2) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)



01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]









119891) = ℎ(119905) sdot


1are











1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)




1198671198833




3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)




100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)


(119865119899 119866119899) isin 119883


3to 119867 = (119865 119866) isin 119883

3 then for




3 the



3




(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















119909 isin 1198771015840 then119867 = (119865 119866) isin 119881

0


119881120576

0= 119867 = (119865 119866) isin 119883

1 |119865 (119909)| lt 120576 |119866 (119909)| lt 120576 forall119909 isin 119877

1015840

(6)



120576

0



(1199061015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) =


0+ 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] minus 119866 [119909 minus (119880

0minus 1198880) 119905]

211988801205880

1199011015840(119909 119905) = 119888

01205880sdot119865 [119909 minus (119880

0+ 1198880) 119905] minus 119865 [119909 minus (119880

0minus 1198880) 119905]

2

+119866 [119909 minus (119880

0+ 1198880) 119905] + 119866 [119909 minus (119880

0minus 1198880) 119905]

2

(7)


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(8)



01205880) sdot1199011015840


(120597V1015840120597119905) + (1198800minus 1198880) sdot (120597V1015840120597119909) = 0

V1015840 (119909 0) = 119865 (119909) minus (111988801205880) sdot 119866 (119909)

(1205971199021015840120597119905) + (119880

0+ 1198880) sdot (120597119902

1015840120597119909) = 0

1199021015840(119909 0) = 119865 (119909) + (1119888

01205880) sdot 119866 (119909)

(9)



11988801205880

sdot 119866 (119909 minus (1198800minus 1198880) 119905)

1199021015840(119909 119905) = 119865 (119909 minus (119880

0+ 1198880) 119905) +

1

11988801205880

sdot 119866 (119909 minus (1198800+ 1198880) 119905)

(10)








119899= (119865119899 119866119899) isin 119883

1










1



120575(120576)





(111988801205880)) if119867 isin 119881

120575(120576)

0 then1198671015840(sdot 119905) isin 119881120576

0for 119905 ge 0

Proof If119867 = (119865 119866) isin 119881120575(120576)

0 then |119865(119909)| lt 120575(120576) and |119866(119909)| lt



|119865(119909)| lt 1205762(1 + 11988801205880) |119865(119909)| lt 1205762(1 + (1119888

01205880)) |119866(119909)| lt

1205762(1 + 11988801205880) and |119866(119909)| le 1205762(1 + (1119888

01205880)) for any 119909 isin 1198771015840


100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 2 sdot

120576

2 (1 + 11988801205880)+

1

211988801205880

sdot 2 sdot120576

2 (1 + (111988801205880))= 120576


100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 211988801205880sdot

120576

2 (1 + (111988801205880))

+1

2sdot

2120576

2 (1 + 11988801205880)= 120576

(11)



0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


1




maxsup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816

119905 sup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816

119905 (12)







1 More


1198761 (119909 119905) = 119865 (119909) sdot ℎ (119905) sdot sin120596119891119905

1198762(119909 119905) = 119866 (119909) sdot ℎ (119905) sdot sin120596

119891119905

(13)




119891)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) = ℎ (119905)

sdot int

119905

0



minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


01205880)minus1]

sdot sin120596119891120591119889120591

1199011015840(119909 119905) = ℎ (119905)

sdot int

119905

0

[11988801205880sdot (119865 [119909 minus (119880

0+ 1198880) (119905 minus 120591)]


minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


minus1]

sdot sin120596119891120591119889120591

(14)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (15)




1


01205880 1 + (1119888

01205880)) gt 0 if 119867 = (119865 119866) isin 119881

120575

0 then

1198671015840(∘ 119905) isin 119881

120576

0 for any 119905 isin [0 119879]









120575




120575(119865 119866 120596


119891119905 0)where



1199061015840(119909 119905) = ℎ (119905) sdot

120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) +

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

1199011015840(119909 119905) = ℎ (119905) sdot

11988801205880120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) minus

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

(16)










1015840rarr 1198771015840 with


119867 = maxsup119909isin1198771015840

|119865 (119909)| sup119909isin1198771015840

|119866 (119909)| (17)


119881120576

0= 119867 = (119865 119866) isin 119883

2 119867 lt 120576 (18)




0


functions 1198671015840(119909 119905) = (1199061015840(119909 119905) 119901

1015840(119909 119905)) given by (7) is the


119899= (119865119899 119866119899) isin 1198832converges

in 1198832to 119867 = (119865 119866) isin 119883




2 the initial


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(19)



2





holds


(1 + 11988801205880 1 + 1119888





0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2



2



2 the right-hand


(119865 119866) isin 1198832


2) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)



01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]









119891) = ℎ(119905) sdot


1are











1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)




1198671198833




3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)




100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)


(119865119899 119866119899) isin 119883


3to 119867 = (119865 119866) isin 119883

3 then for




3 the



3




(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816lt1

2sdot 211988801205880sdot

120576

2 (1 + (111988801205880))

+1

2sdot

2120576

2 (1 + 11988801205880)= 120576

(11)



0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


1




maxsup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199061015840(119909 119905)

10038161003816100381610038161003816

119905 sup119909

lim119905rarrinfin

ln 100381610038161003816100381610038161199011015840(119909 119905)

10038161003816100381610038161003816

119905 (12)







1 More


1198761 (119909 119905) = 119865 (119909) sdot ℎ (119905) sdot sin120596119891119905

1198762(119909 119905) = 119866 (119909) sdot ℎ (119905) sdot sin120596

119891119905

(13)




119891)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) given by

1199061015840(119909 119905) = ℎ (119905)

sdot int

119905

0



minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


01205880)minus1]

sdot sin120596119891120591119889120591

1199011015840(119909 119905) = ℎ (119905)

sdot int

119905

0

[11988801205880sdot (119865 [119909 minus (119880

0+ 1198880) (119905 minus 120591)]


minus1

+ (119866 [119909 minus (1198800+ 1198880) (119905 minus 120591)]


minus1]

sdot sin120596119891120591119889120591

(14)


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (15)




1


01205880 1 + (1119888

01205880)) gt 0 if 119867 = (119865 119866) isin 119881

120575

0 then

1198671015840(∘ 119905) isin 119881

120576

0 for any 119905 isin [0 119879]









120575




120575(119865 119866 120596


119891119905 0)where



1199061015840(119909 119905) = ℎ (119905) sdot

120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) +

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

1199011015840(119909 119905) = ℎ (119905) sdot

11988801205880120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) minus

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

(16)










1015840rarr 1198771015840 with


119867 = maxsup119909isin1198771015840

|119865 (119909)| sup119909isin1198771015840

|119866 (119909)| (17)


119881120576

0= 119867 = (119865 119866) isin 119883

2 119867 lt 120576 (18)




0


functions 1198671015840(119909 119905) = (1199061015840(119909 119905) 119901

1015840(119909 119905)) given by (7) is the


119899= (119865119899 119866119899) isin 1198832converges

in 1198832to 119867 = (119865 119866) isin 119883




2 the initial


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(19)



2





holds


(1 + 11988801205880 1 + 1119888





0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2



2



2 the right-hand


(119865 119866) isin 1198832


2) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)



01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]









119891) = ℎ(119905) sdot


1are











1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)




1198671198833




3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)




100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)


(119865119899 119866119899) isin 119883


3to 119867 = (119865 119866) isin 119883

3 then for




3 the



3




(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














120575




120575(119865 119866 120596


119891119905 0)where



1199061015840(119909 119905) = ℎ (119905) sdot

120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) +

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

1199011015840(119909 119905) = ℎ (119905) sdot

11988801205880120575

4


sin 2120596119891119905

2120596119891

sdot cos (119909 + 120596119891119905) minus

sin2120596119891119905

120596119891

sdot sin (119909 + 120596119891119905)

+ [

sin 2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

minussin 21198880119905

21198880

]

sdot cos (119909 + (120596119891minus 21198880) 119905)

+ [

sin2 (120596119891minus 1198880) 119905

2 (120596119891minus 1198880)

+sin21198880119905

1198880

]

sdot sin (119909 + (120596119891minus 21198880) 119905)

(16)










1015840rarr 1198771015840 with


119867 = maxsup119909isin1198771015840

|119865 (119909)| sup119909isin1198771015840

|119866 (119909)| (17)


119881120576

0= 119867 = (119865 119866) isin 119883

2 119867 lt 120576 (18)




0


functions 1198671015840(119909 119905) = (1199061015840(119909 119905) 119901

1015840(119909 119905)) given by (7) is the


119899= (119865119899 119866119899) isin 1198832converges

in 1198832to 119867 = (119865 119866) isin 119883




2 the initial


(1199061015840(119909 0) 119901

1015840(119909 0)) = 119867

1015840(119909 0) = 119867 (119909) = (119865 (119909) 119866 (119909))

(19)



2





holds


(1 + 11988801205880 1 + 1119888





0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2



2



2 the right-hand


(119865 119866) isin 1198832


2) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)



01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]









119891) = ℎ(119905) sdot


1are











1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)




1198671198833




3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)




100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)


(119865119899 119866119899) isin 119883


3to 119867 = (119865 119866) isin 119883

3 then for




3 the



3




(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


2



2



2 the right-hand


(119865 119866) isin 1198832


2) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (20)



01205880 1 + (1119888

01205880)) gt 0 if 119867

1198832

lt 120575 then 1198671015840(sdot 119905)1198832

lt 120576 for any 119905 isin [0 119879]









119891) = ℎ(119905) sdot


1are











1198671198833

= max(int1198771015840

|119865 (119909)|2119889119909)

12

(int

1198771015840

|119866 (119909)|2119889119909)

12

(21)




1198671198833




3and 1198671015840(119909 0) satisfy the

condition

1198671015840(119909 0) = (119906

1015840(119909 0) 119901

1015840(119909 0)) = (119865 119866) (22)




100381710038171003817100381710038171198671015840(sdot 119905)

100381710038171003817100381710038171198833

le max1 + 11988801205880 1 +

1

11988801205880

sdot100381710038171003817100381710038171198671015840(sdot 0)

100381710038171003817100381710038171198833

(23)


(119865119899 119866119899) isin 119883


3to 119867 = (119865 119866) isin 119883

3 then for




3 the



3




(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(1 + 11988801205880 1 + 1119888




0 119901 = 119901

0 120588 = 120588

0of (1) is linearly


3






3 the right-hand


(119865 119866) isin 1198833


3) the couple


1198671015840(119909 119905) = (119906

1015840(119909 119905) 119901

1015840(119909 119905)) = (0 0) for 119905 le 0 (24)



01205880 1+(1119888

01205880)) gt 0 if 119867

1198833

lt 120575 then 1198671015840(sdot 119905)1198833

lt

120576 for any 119905 isin [0 119879]





119865(119909) = 120575 sin1199091 + 1199092 119866(119909) = 0 and 120596119891= 1198880minus 1198800





119861defined by

119902119861(119867) = max 119902

119861(119865) 119902

119861(119866) (25)



by 119902119861(119865) = sup

120593isin119861|119865(120593)|

The set 1198811205761199021198610

defined by

119881120576119902119861

0= 119867 = (119865 119866) isin 119883

4 119902119861(119867) lt 120576 (26)




119881120576119902119861




1198671015840(0) = (119906

1015840(0) 119901

1015840(0)) = 119867 = (119865 119866) isin 119883

4 (27)



tions1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= 0

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= 0

(28)


1015840(0) = 119865 119901

1015840(0) = 119866 (29)





1015840(119905) =


2


211988801205880

sdot 119866

1199011015840(119905) = 119888

01205880


2


2sdot 119866

(30)






2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














2sdot 119861


211988801205880

sdot 119861

11988801205880


2sdot 119861

(31)


4and 120575 = 120576(1 +


4with 119902

1198611015840









1198610


119861in 1198834there exist 119867 =

(119865 119866) isin 119881120575119902119861



0 Let us consider

now 1205760= 1 119861




1015840(0) Remark


119861(120576 1198650) lt 120575 (119881120575119902119861

0is absorbing)


120575119902119861




1015840(119905) = minus119894120576119880

0119905120593 (0) + 120593

1015840(0)

1199011015840(119905) = minus119894120576119888

2

01205880119905120593 (0)

1199021198610

(1015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161015840(119905) (1205930)10038161003816100381610038161003816= 1205761198800119905

1199021198610

(1199011015840(119905)) = 120576 sdot sup

120593isin1198610

100381610038161003816100381610038161199011015840(119905) (120593)

10038161003816100381610038161003816ge 120576

sdot100381610038161003816100381610038161199011015840(119905) (1205930)10038161003816100381610038161003816= 1205761198882

01205880119905

1199021198610

(1015840(119905) 1199011015840(119905)) ge 120576119905max 119880

0 1198882

01205880

(32)


(1015840(119905) 1199011015840(119905)) gt

1 = 1205760


4


4 the right-hand


1198761= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865 119876

2= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866 (33)

where119867 = (119865 119866) isin 1198834


1199061015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0 (34)


1199011015840(119905)) isin 119883



1205971015840

120597119905+ 119894119896119880

0sdot 1015840+ 119894119896

1

1205880

sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119865

1205971199011015840

120597119905+ 1205741199010119894119896 sdot 1015840+ 119894119896119880

0sdot 1199011015840= ℎ (119905) sdot 119890

119894120596119891119905sdot 119866

1015840(119905) = 0 119901

1015840(119905) = 0 for 119905 le 0

(35)

Hence we got

1015840(119905) =

ℎ (119905)

2


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

+119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

211988801205880

[119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866


1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1199011015840(119905) =

11988801205880sdot ℎ (119905)

2


119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119865

+ℎ (119905)

2[

119890119894120596119891119905minus 119890minus119894(1198800+1198880)119896119905

119894 [(1198800+ 1198880) sdot 119896 + 120596

119891]


119894 [(1198800minus 1198880) sdot 119896 + 120596

119891]

] sdot 119866

(36)



7 Conclusions











Acknowledgment


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article The Concepts of Well-Posedness and ...Research Article The Concepts of...

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