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Research Article On the Control of Coefficient Function in a Hyperbolic Problem with Dirichlet Conditions Seda ELret Araz 1 and Murat SubaGi 2 1 Siirt University, Department of Elementary Mathematics Education, Siirt 56100, Turkey 2 Atat¨ urk University, Department of Mathematic, Erzurum 25240, Turkey Correspondence should be addressed to Seda ˙ gret Araz; [email protected] Received 6 February 2018; Revised 2 April 2018; Accepted 4 April 2018; Published 23 September 2018 Academic Editor: Omar Abu Arqub Copyright © 2018 Seda ˙ gret Araz and Murat Subas ¸i. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. is paper presents theoretical results about control of the coefficient function in a hyperbolic problem with Dirichlet conditions. e existence and uniqueness of the optimal solution for optimal control problem are proved and adjoint problem is used to obtain gradient of the functional. However, a second adjoint problem is given to calculate the gradient on the space 1 2 (0, ). Aſter calculating gradient of the cost functional and proving the Lipschitz continuity of the gradient, necessary condition for optimal solution is constructed. 1. Introduction Hyperbolic boundary value problems have appeared as mathematical modelling of physical phenomena like small vibration of a string, in the fields of science and engineering. ere has been much attention to studies related to optimal control problems involving hyperbolic problems [1]. ere have been many studies about optimal control for hyperbolic systems which are considered [2–4]. Some of these important studies can be summarized as follows. Hasanov [5] has considered problem of controlling the function {(, ); ()} for the following problem: = ( () ) + (, ) , (, ) ∈ (0, ) × (0, ) (, 0) = 0 () , (, 0) = 1 () , ∈ (0, ) − (0) (0, ) = () , () (, ) = 0 ∈ (0, ) (1) with the conditions (, ) = () , (, ) = V () , (2) using the functionals 1 () = ∫ 0 [ (, ; ) − ()] 2 , 2 () = ∫ 0 [ (, ; ) − V ()] 2 , 3 () = 1 () + 2 () . (3) Hindawi International Journal of Differential Equations Volume 2018, Article ID 7417590, 6 pages https://doi.org/10.1155/2018/7417590

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Research ArticleOn the Control of Coefficient Function in a Hyperbolic Problemwith Dirichlet Conditions

Seda ELret Araz 1 and Murat SubaGi 2

1Siirt University Department of Elementary Mathematics Education Siirt 56100 Turkey2Ataturk University Department of Mathematic Erzurum 25240 Turkey

Correspondence should be addressed to Seda Igret Araz sedaarazsiirtedutr

Received 6 February 2018 Revised 2 April 2018 Accepted 4 April 2018 Published 23 September 2018

Academic Editor Omar Abu Arqub

Copyright copy 2018 Seda Igret Araz and Murat Subasi This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

This paper presents theoretical results about control of the coefficient function in a hyperbolic problem with Dirichlet conditionsThe existence and uniqueness of the optimal solution for optimal control problem are proved and adjoint problem is used toobtain gradient of the functional However a second adjoint problem is given to calculate the gradient on the space11988212 (0 119897) Aftercalculating gradient of the cost functional and proving the Lipschitz continuity of the gradient necessary condition for optimalsolution is constructed

1 Introduction

Hyperbolic boundary value problems have appeared asmathematical modelling of physical phenomena like smallvibration of a string in the fields of science and engineeringThere has been much attention to studies related to optimalcontrol problems involving hyperbolic problems [1] Therehave been many studies about optimal control for hyperbolicsystems which are considered [2ndash4]

Some of these important studies can be summarized asfollows

Hasanov [5] has considered problem of controlling thefunction 119908 fl 119865(119909 119905) 119891(119905) for the following problem

119906119905119905 = (119896 (119909) 119906119909)119909 + 119865 (119909 119905) (119909 119905) isin (0 119897) times (0 119879)

119906 (119909 0) = 1199060 (119909) 119906119905 (119909 0) = 1199061 (119909)

119909 isin (0 119897)

minus119896 (0) 119906119909 (0 119905) = 119891 (119905) 119896 (119897) 119906119909 (119897 119905) = 0

119905 isin (0 119879)(1)

with the conditions

119906 (119909 119879) = 120583 (119909) 119906119905 (119909 119879) = V (119909) (2)

using the functionals

1198691 (119908) = int119897

0[119906 (119909 119879 119908) minus 120583 (119909)]2 119889119909

1198692 (119908) = int119897

0[119906119905 (119909 119879119908) minus V (119909)]2 119889119909

1198693 (119908) = 1198691 (119908) + 1198692 (119908)

(3)

HindawiInternational Journal of Differential EquationsVolume 2018 Article ID 7417590 6 pageshttpsdoiorg10115520187417590

2 International Journal of Differential Equations

Majewski [6] has controlled the function119906(119909 119910) isin 1198712(119875R119872)for hyperbolic equation

1205972119911120597119909120597119910 (119909 119910)

= 119891(119909 119910 119911 (119909 119910) 120597119911120597119909 (119909 119910) 120597119911120597119910 (119909 119910) 119906 (119909 119910))

119911 (119909 0) = 119911 (0 119910) = 0 forall119909 119910 isin [0 1]

(4)

using the functional

119869119896 (119911 (sdot) 119906 (sdot)) = int119875119865119896 (119909 119910 119911 (119909 119910) 119906 (119909 119910)) 119889119909 119889119910

119896 = 0 1 2 (5)

Yeloglu and Subası [7] have dealt with determination pair119908 fl 119891(119909 119905) ℎ(119909) in the following problem

119901 (119909) 119906119905119905 = (119896 (119909) 119906119909)119909 + 119891 (119909 119905) (119909 119905) isin Ω119879119906 (119909 0) = 119892 (119909) 119906119905 (119909 0) = ℎ (119909)

119909 isin (0 119897)119906 (0 119905) = 0119906 (119897 119905) = 0

119905 isin (0 119879]

(6)

for the functional

119869120572 (119908) = int119897

0[119906 (119909 119879119908) minus 119910 (119909)]2 119889119909 + 120572 1199082119882 (7)

Kroner [8] has specified the function 119906(119905) isin 1198712(0 119879) fornonlinear hyperbolic equation

119910119905119905 minus 119860 (119906 119910) = 119891119910 (0) = 1199100 (119906) 119910119905 (0) = 1199101 (119906)

(8)

using the functional

119869 (119906 1199101) = int11987901198691 (1199101 (119905)) 119889119905 + 1198692 (1199101 (119879)) + 120572

2 1199062119880 (9)

Tagiyev [9] has studied the problem of controlling thecoefficients V = (119896(119909) 119902(119909 t)) isin 119871infin(Ω) times 119871infin(119876119879) for linearhyperbolic equation

12059721199061205971199052 minus

119899sum119894=1

120597120597119909119894 (119896119894 (119909)

120597119906120597119909119894) + 119902 (119909 119905) 119906 = 119891 (119909 119905)

(119909 119905) isin 119876119879119906|119905=0 = 1205930 (119909)

12059711990612059711990510038161003816100381610038161003816100381610038161003816119905=0 = 1205931 (119909) 119906|119878119879 = 0

(10)

using the functional

119869 (V) = 1205720 int119876119879

1003816100381610038161003816119906 (119909 119905 V) minus 1199110 (119909 119905)10038161003816100381610038162 119889119909119889119905

+ 1205721 intΩ

1003816100381610038161003816119906 (119909 119879 V) minus 1199111 (119909)10038161003816100381610038162 119889119909(11)

2 Statement of the Problem

In this study we deal with the process of vibration in finitehomogeneous string occupying the interval (0 119897) As thecontrol function we take the transverse elastic force which isin the coefficient of the vibration problem Also we proposethe usage of a more regular space than the space of squareintegrable functions in the cost functional In general thisprocess exposes some difficulties in the stage of acquiringthe gradient This study offers a second adjoint problem toovercome this case

In the domainΩ fl (119909 119905) isin (0 119897) times (0 119879) we consider thefunctional

119869120572 (119902) = int119897

0[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909 + 120572 1003817100381710038171003817119902 minus 119903100381710038171003817100381721198821

2(0119897)

(12)

on the set

119876 = 119902 (119909) 119902 (119909) isin 11988212 (0 119897) 0 lt 1199021 le 119902 (119909)le 1199022 1003817100381710038171003817119902 (119909)10038171003817100381710038171198821

2(0119897) le 1199023

(13)

subject to the hyperbolic problem

119906119905119905 minus 119906119909119909 + 119902 (119909) 119906 = 0 (119909 119905) isin Ω (14)

119906 (119909 0) = 1205931 (119909) 119906119905 (119909 0) = 1205932 (119909)

119909 isin (0 119897)(15)

119906 (0 119905) = 0119906 (119897 119905) = 0

119905 isin (0 119879) (16)

Here 119910(119909) isin 1198712(0 119897) is the desired target function to which119906(119909 119879)must be close enough The function 119903(119909) isin 11988212 (0 119897) isan initial guess for optimal solution 120572 gt 0 is regularizationparameter 1199021 1199022 1199023 gt 0 are given positive numbers

The initial status functions are in the following spaces

1205931 (119909) isin 11988212 (0 119897) 1205932 (119909) isin 1198712 (0 119897) (17)

The aim of this study is to deal with the problem of

119869120572lowast = inf119902isin119876

119869120572 (119902) = 119869120572 (119902lowast) (18)

under conditions (12)-(17)

International Journal of Differential Equations 3

Namely we want to control the transverse elastic force onthe space11988212 (0 119897) and the solution 119906(119909 119879) corresponding tothis control function must be close enough to 119910(119909) in 1198712(0 119897)In order to get a stable solution we choose the space11988212 (0 119897)which is more regular than 1198712(0 119897)

The inner product and norm in 11988212 (0 119897) are definedrespectively as

(119891 119892)11988212(0119897) = int

119897

0[119891 (119909) 119892 (119909) + 119889119891 (119909)

119889119909 119889119892 (119909)119889119909 ] 1198891199091003817100381710038171003817119891100381710038171003817100381721198821

2(0119897) = int

119897

0([119891 (119909)]2 + [119889119891 (119909)119889119909 ]2)119889119909

(19)

The paper is organized as follows in Section 3 we obtain thegeneralized solution for hyperbolic problem In Section 4 weprove the existence and uniqueness of the optimal solutionIn Section 5 we obtain the adjoint problem for the optimalcontrol problem and find the gradient of the functional Themain contribution of this paper is executed in this sectionBecause the controls are chosen in the space11988212 (0 119897) gettingthe gradient of the functional necessitates finding a secondadjoint problem In the last section we demonstrate theLipschitz continuity of the gradient and state the necessarycondition for optimal solution

3 Solvability of the Problem

In this section we first give the definition of the generalizedsolution for hyperbolic problem

The generalized solution of problem (14)-(15) is the

function 119906 isin o119882112 (Ω) satisfying the following integralequality

intΩ[minus119906119905120578119905 + 119906119909120578119909 + 119902 (119909) 119906120578] 119889119909 119889119905

= int11989701205932 (119909) 120578 (119909 0) 119889119909

(20)

for forall120578 isin o119882112 (Ω) 120578(119909 119879) = 0It can be seen in [10] that solution in the sense of (20)

exists is unique and satisfies the following inequality

max0le119905le119879

(119906 (sdot 119905)21198712(0119897) + 1003817100381710038171003817119906119905 (sdot 119905)100381710038171003817100381721198712(0119897)+ 1003817100381710038171003817119906119909 (sdot 119905)100381710038171003817100381721198712(0119897)) le 1198881 (100381710038171003817100381712059311003817100381710038171003817211988212 (0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(21)

where 1198881 = max31198880 311988801199021 and 1198880 = max1 1199022 or119906211988211

2(Ω)

le 1198882 (100381710038171003817100381712059311003817100381710038171003817211988212(0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(1198882 = 1198881119879)(22)

Since 1205931 and 1205932 are given functions it can be written asfollows

1199062119882112(Ω)

le 1198883 (23)

Now we give an increment 120575119902(119909) isin 11988212 (0 119897) to the controlfunction 119902(119909) such as 119902+120575119902 isin 119876Then the difference function120575119906 = 120575119906(119909 119905) = 119906(119909 119905 119902+120575119902)minus119906(119909 119905 119902) is the solution of thefollowing difference initial-boundary problem

120575119906119905119905 minus 120575119906119909119909 + [119902 (119909) + 120575119902 (119909)] 120575119906 + 120575119902 (119909) 119906 = 0 (24)

120575119906 (119909 0) = 0120575119906119905 (119909 0) = 0 (25)

120575119906 (0 119905) = 0120575119906 (119897 119905) = 0 (26)

By considering (23) we obtain that the solution of abovedifference initial-boundary problem holds the followinginequality

max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) le 1198884 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (27)

Here 1198884 = (119905321198973)1198883 is independent from 1205751199024 Existence and Uniqueness of theOptimal Solution

To demonstrate the existence and the uniqueness of optimalsolution for problem (12)-(17) it is enough to show thatconditions of the following theorem given by Goebel [11]hold

Theorem 1 Let119867 be a uniformly convexBanach space and theset 119876 be a closed bounded and convex subset of 119867 If 120572 gt 0and 120573 ge 1 are given numbers and the functional 119869(119902) is lowersemicontinuous and bounded from below on the set 119876 thenthere is a dense set 119866 of119867 that the functional

119869120572 (119902) = 119869 (119902) + 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817120573119867 (28)

takes its minimum on the set 119876 for forall119903 isin 119866 If 120573 gt 1 thenminimum is unique

Before showing that these conditions have been satisfiedwe prove that the functional

119869 (119902) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909 (29)

is continuous For this we write the following increment ofthe functional

120575119869 (119902) = 119869 (119902 + 120575119902) minus 119869 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] [120575119906 (119909 119879)] 119889119909

+ int1198970[120575119906 (119909 119879)]2 119889119909

(30)

4 International Journal of Differential Equations

Since 119910(119909) isin 1198712(0 119897) if we consider inequalities (22) and(27) we conclude that this increment satisfies the followingcontinuity inequality on the set 119876

1003816100381610038161003816120575119869 (119902)1003816100381610038161003816 le 1198885 (1003817100381710038171003817120575119902100381710038171003817100381711988212(0119897) + 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)) (31)

Here 1198885 is independent of 120575119902Thanks to this inequality we can say that this functional

is also lower semicontinuous and bounded frombelow on theset 119876

On the other hand the set11988212 (0 119897) is a uniformly convexBanach space [12] the set119876 is a closed bounded and convexsubset of11988212 (0 119897) and 120573 = 2

Therefore the conditions of above theorem hold andoptimal solution to the problem (18) is unique

5 Adjoint Problem and Gradient ofthe Functional

In this section we write the Lagrange functional usedfor finding adjoint problem before we show the Frechetdifferentiability of the functional 119869120572(119902) on the set119876 Lagrangefunctional to the problem is

119871 (119906 119902 120578) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909

+ 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817211988212(0119897)

+ int1198790int1198970[119906119905119905 minus 119906119909119909 + 119902 (119909) 119906] 120578 119889119909 119889119905

(32)

Thefirst variation of this functional according to the function119906 is obtained such as

120575119871 = int11989702 [119906 (119909 119879) minus 119910 (119909) minus 120578119905 (119909 119879)] 120575119906 (119909 119879) 119889119909

+ int1198970int1198790(120578119905119905 minus 120578119909119909 + 119902 (119909) 120578) 120575119906 119889119905 119889119909 = 0

(33)

By means of stationary condition 120575119871 = 0 the followingadjoint boundary value problem is found

120578119905119905 minus 120578119909119909 + 119902 (119909) 120578 = 0 (34)

120578 (119909 119879) = 0120578119905 (119909 119879) = 2 [119906 (119909 119879) minus 119910 (119909)] (35)

120578 (0 119905) = 0120578 (119897 119905) = 0 (36)

For forall120574 isin o119882112 (Ω) the function 120578 isin 1198621([0 119879] 1198712(0 119897)) cap1198620([0 119879]11988212 (0 119897)) which satisfies the following equality

int1198790int1198970[minus120578119905120574119905 + 120578119909120574119909 + 119902 (119909) 120578120574] 119889119909 119889119905

= int1198970120578119905 (119909 0) 120574 (119909 0) 119889119909

minus int11989702 [119906 (119909 119879) minus 119910 (119909)] 120574 (119909 119879) 119889119909

(37)

is the solution of adjoint boundary value problem (34)-(36)This solution satisfies the following inequality

100381710038171003817100381712057810038171003817100381710038171198712(0119897) le 1198886 1003817100381710038171003817119906 (119909 119879) minus 119910 (119909)10038171003817100381710038171198712(0119897) forall119905 isin [0 119879] (38)

Now we can pass the calculation of the gradient In orderto do this we must evaluate the increment of the functional119869120572(119902) The increment can be written such as

120575119869120572 (119902) = 119869120572 (119902 + 120575119902) minus 119869120572 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

+ int1198970(120575119906)2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(39)

The difference problem (24)-(26) and the adjoint problem(34)-(36) give together the equality of

2int1198970[119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

= int1198790int1198970[120575119902120575119906120578 + 120575119902119906120578] 119889119909 119889119905

(40)

Inserting (40) in (39) we have

120575119869120572 (119902) = int119879

0int1198970119906120578120575119902 119889119909119889119905 + int119879

0int1198970120578120575119906120575119902 119889119909119889119905

+ int1198970(120575119906 (119909 119879))2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(41)

By (27) and (38) the second and third integrals of the aboveequality give the following inequality

int1198790int1198970120578120575119906120575119902 119889119909 119889119905 + int119897

0(120575119906 (119909 119879))2 119889119909

le 1198887 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(42)

The statement (41) can be rewritten as

120575119869120572 (119902) = ⟨119906120578 120575119902⟩1198712(Ω) + 2120572 ⟨119902 minus 119903 120575119902⟩11988212 (0119897)+ 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(43)

International Journal of Differential Equations 5

or

120575119869120572 (119902) = ⟨int119879

0119906120578 119889119905 120575119902⟩

1198712(0119897)

+ 2120572 ⟨119902 minus 119903 120575119902⟩11988212(0119897) + 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(44)

In order to pass the inner product in 11988212 (0 119897) we rearrange(44) such as

120575119869120572 (119902) = ⟨120585 + 2120572 (119902 minus 119903) 120575119902⟩11988212(0119897)

+ 119900 (10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897))

(45)

Here function 120585(119909) is the solution of the second adjointproblem

minus12058510158401015840 + 120585 = int1198790119906120578119889119905

1205851015840 (0) = 01205851015840 (119897) = 0

(46)

Therefore we have the following gradient

1198691015840120572 (119902) = 120585 + 2120572 (119902 minus 119903) (47)

6 Lipschitz Continuity of the Gradient

In this section we introduce a theorem about Lipschitzcontinuity of the gradient By this means we can express thenecessary condition for optimal solution

Theorem 2 Gradient 1198691015840120572(119902) satisfies the following Lipschitzinequality

100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 1198888 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (48)

Here 1198888 is independent from 120575119902Hence it has been proven that the gradient 1198691015840120572(119902) is

continuous on the set 119876 and it can be seen that it holds theLipschitz condition with constant 1198888 gt 0Proof Increment of the functional 1198691015840120572(119902) by giving the incre-ment of 120575119902 to the control 119902 isin 119876 is obtained

1198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902) = 120585120575 + 2120572 (119902 + 120575119902 minus 119903) minus 120585+ 2120572 (119902 minus 119903) = 120575120585 + 2120572120575119902 (49)

where the function 120575120585(119909) is the solution of the incrementproblem

12057512058510158401015840 (119909) minus 120575120585 (119909) = int1198790(119906120575120575120578 + 120575119906120578) 119889119905 (50)

Taking the norm of (49) in the space11988212 (0 119897) we acquire thefollowing inequality belonging to the functional 1205751198691015840120572(119902)

100381710038171003817100381710038171205751198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 2 10038171003817100381710038171205751205851003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (51)

There is a solution of problem (50) in 11988212 (0 119897) and thissolution satisfies the following inequality

1003817100381710038171003817120575120585 (119909)100381710038171003817100381711988212(0119897) le

100381710038171003817100381710038171003817100381710038171003817int119879

0(119906120575120575120578 + 120575119906120578) 119889119905

1003817100381710038171003817100381710038171003817100381710038171198712(0119897) (52)

The function120575120578 (119909 119905) = 120578120575 (119909 119905) minus 120578 (119909 119905)

= 120578 (119909 119905 119902 + 120575119902) minus 120578 (119909 119905 119902) (53)

in the right hand side of inequality (52) is the solution of thefollowing problem

12059721205751205781205971199052 minus 1205972120575120578

1205971199092 + (119902 (119909) + 120575119902 (119909)) 120575120578 + 120575119902 (119909) 120578 = 0(119909 119905) isin Ω

120575120578 (119909 119879) = 0120575120578119905 (119909 119879) = 2120575119906 (119909 119879) 120575120578 (0 119905) = 120575120578 (119897 119905) = 0

(54)

and this function holds the following inequality

max0le119905le119879

(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897)) le 1198889 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (55)

Here 1198889 is independent of 120575119902So the function 119906120575 that takes place in the right hand side

of (52) holds the same inequality given as follows1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) le 1198883 (56)

Hence inequality (52) has the following property1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 2 1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) max

0le119905le119879(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897))

+ 2 1003817100381710038171003817120578100381710038171003817100381721198712(Ω) max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) (57)

If inequalities (27) (38) (55) and (56) about functions 119906120575120575120578( 119905) 120578 and 120575119906( 119905) are written in (57) then the followingassessment is obtained1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 11988810 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (58)

Here 11988810 is independent of 120575119902Considering inequality (58) the following is written100381710038171003817100381710038171205751198691015840120572 (119902)1003817100381710038171003817100381721198821

2(0119897)

le 2 1003817100381710038171003817120575120585 (119909)1003817100381710038171003817211988212(0119897)

+ 81205722 1003817100381710038171003817120575119902 (119909)1003817100381710038171003817211988212(0119897)

le 211988810 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(59)

So the following inequality for the gradient 1198691015840120572(119902) is obtained100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (60)

Once we take as 1198888 = 11988811 then the proof is obtained

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

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Page 2: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

2 International Journal of Differential Equations

Majewski [6] has controlled the function119906(119909 119910) isin 1198712(119875R119872)for hyperbolic equation

1205972119911120597119909120597119910 (119909 119910)

= 119891(119909 119910 119911 (119909 119910) 120597119911120597119909 (119909 119910) 120597119911120597119910 (119909 119910) 119906 (119909 119910))

119911 (119909 0) = 119911 (0 119910) = 0 forall119909 119910 isin [0 1]

(4)

using the functional

119869119896 (119911 (sdot) 119906 (sdot)) = int119875119865119896 (119909 119910 119911 (119909 119910) 119906 (119909 119910)) 119889119909 119889119910

119896 = 0 1 2 (5)

Yeloglu and Subası [7] have dealt with determination pair119908 fl 119891(119909 119905) ℎ(119909) in the following problem

119901 (119909) 119906119905119905 = (119896 (119909) 119906119909)119909 + 119891 (119909 119905) (119909 119905) isin Ω119879119906 (119909 0) = 119892 (119909) 119906119905 (119909 0) = ℎ (119909)

119909 isin (0 119897)119906 (0 119905) = 0119906 (119897 119905) = 0

119905 isin (0 119879]

(6)

for the functional

119869120572 (119908) = int119897

0[119906 (119909 119879119908) minus 119910 (119909)]2 119889119909 + 120572 1199082119882 (7)

Kroner [8] has specified the function 119906(119905) isin 1198712(0 119879) fornonlinear hyperbolic equation

119910119905119905 minus 119860 (119906 119910) = 119891119910 (0) = 1199100 (119906) 119910119905 (0) = 1199101 (119906)

(8)

using the functional

119869 (119906 1199101) = int11987901198691 (1199101 (119905)) 119889119905 + 1198692 (1199101 (119879)) + 120572

2 1199062119880 (9)

Tagiyev [9] has studied the problem of controlling thecoefficients V = (119896(119909) 119902(119909 t)) isin 119871infin(Ω) times 119871infin(119876119879) for linearhyperbolic equation

12059721199061205971199052 minus

119899sum119894=1

120597120597119909119894 (119896119894 (119909)

120597119906120597119909119894) + 119902 (119909 119905) 119906 = 119891 (119909 119905)

(119909 119905) isin 119876119879119906|119905=0 = 1205930 (119909)

12059711990612059711990510038161003816100381610038161003816100381610038161003816119905=0 = 1205931 (119909) 119906|119878119879 = 0

(10)

using the functional

119869 (V) = 1205720 int119876119879

1003816100381610038161003816119906 (119909 119905 V) minus 1199110 (119909 119905)10038161003816100381610038162 119889119909119889119905

+ 1205721 intΩ

1003816100381610038161003816119906 (119909 119879 V) minus 1199111 (119909)10038161003816100381610038162 119889119909(11)

2 Statement of the Problem

In this study we deal with the process of vibration in finitehomogeneous string occupying the interval (0 119897) As thecontrol function we take the transverse elastic force which isin the coefficient of the vibration problem Also we proposethe usage of a more regular space than the space of squareintegrable functions in the cost functional In general thisprocess exposes some difficulties in the stage of acquiringthe gradient This study offers a second adjoint problem toovercome this case

In the domainΩ fl (119909 119905) isin (0 119897) times (0 119879) we consider thefunctional

119869120572 (119902) = int119897

0[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909 + 120572 1003817100381710038171003817119902 minus 119903100381710038171003817100381721198821

2(0119897)

(12)

on the set

119876 = 119902 (119909) 119902 (119909) isin 11988212 (0 119897) 0 lt 1199021 le 119902 (119909)le 1199022 1003817100381710038171003817119902 (119909)10038171003817100381710038171198821

2(0119897) le 1199023

(13)

subject to the hyperbolic problem

119906119905119905 minus 119906119909119909 + 119902 (119909) 119906 = 0 (119909 119905) isin Ω (14)

119906 (119909 0) = 1205931 (119909) 119906119905 (119909 0) = 1205932 (119909)

119909 isin (0 119897)(15)

119906 (0 119905) = 0119906 (119897 119905) = 0

119905 isin (0 119879) (16)

Here 119910(119909) isin 1198712(0 119897) is the desired target function to which119906(119909 119879)must be close enough The function 119903(119909) isin 11988212 (0 119897) isan initial guess for optimal solution 120572 gt 0 is regularizationparameter 1199021 1199022 1199023 gt 0 are given positive numbers

The initial status functions are in the following spaces

1205931 (119909) isin 11988212 (0 119897) 1205932 (119909) isin 1198712 (0 119897) (17)

The aim of this study is to deal with the problem of

119869120572lowast = inf119902isin119876

119869120572 (119902) = 119869120572 (119902lowast) (18)

under conditions (12)-(17)

International Journal of Differential Equations 3

Namely we want to control the transverse elastic force onthe space11988212 (0 119897) and the solution 119906(119909 119879) corresponding tothis control function must be close enough to 119910(119909) in 1198712(0 119897)In order to get a stable solution we choose the space11988212 (0 119897)which is more regular than 1198712(0 119897)

The inner product and norm in 11988212 (0 119897) are definedrespectively as

(119891 119892)11988212(0119897) = int

119897

0[119891 (119909) 119892 (119909) + 119889119891 (119909)

119889119909 119889119892 (119909)119889119909 ] 1198891199091003817100381710038171003817119891100381710038171003817100381721198821

2(0119897) = int

119897

0([119891 (119909)]2 + [119889119891 (119909)119889119909 ]2)119889119909

(19)

The paper is organized as follows in Section 3 we obtain thegeneralized solution for hyperbolic problem In Section 4 weprove the existence and uniqueness of the optimal solutionIn Section 5 we obtain the adjoint problem for the optimalcontrol problem and find the gradient of the functional Themain contribution of this paper is executed in this sectionBecause the controls are chosen in the space11988212 (0 119897) gettingthe gradient of the functional necessitates finding a secondadjoint problem In the last section we demonstrate theLipschitz continuity of the gradient and state the necessarycondition for optimal solution

3 Solvability of the Problem

In this section we first give the definition of the generalizedsolution for hyperbolic problem

The generalized solution of problem (14)-(15) is the

function 119906 isin o119882112 (Ω) satisfying the following integralequality

intΩ[minus119906119905120578119905 + 119906119909120578119909 + 119902 (119909) 119906120578] 119889119909 119889119905

= int11989701205932 (119909) 120578 (119909 0) 119889119909

(20)

for forall120578 isin o119882112 (Ω) 120578(119909 119879) = 0It can be seen in [10] that solution in the sense of (20)

exists is unique and satisfies the following inequality

max0le119905le119879

(119906 (sdot 119905)21198712(0119897) + 1003817100381710038171003817119906119905 (sdot 119905)100381710038171003817100381721198712(0119897)+ 1003817100381710038171003817119906119909 (sdot 119905)100381710038171003817100381721198712(0119897)) le 1198881 (100381710038171003817100381712059311003817100381710038171003817211988212 (0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(21)

where 1198881 = max31198880 311988801199021 and 1198880 = max1 1199022 or119906211988211

2(Ω)

le 1198882 (100381710038171003817100381712059311003817100381710038171003817211988212(0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(1198882 = 1198881119879)(22)

Since 1205931 and 1205932 are given functions it can be written asfollows

1199062119882112(Ω)

le 1198883 (23)

Now we give an increment 120575119902(119909) isin 11988212 (0 119897) to the controlfunction 119902(119909) such as 119902+120575119902 isin 119876Then the difference function120575119906 = 120575119906(119909 119905) = 119906(119909 119905 119902+120575119902)minus119906(119909 119905 119902) is the solution of thefollowing difference initial-boundary problem

120575119906119905119905 minus 120575119906119909119909 + [119902 (119909) + 120575119902 (119909)] 120575119906 + 120575119902 (119909) 119906 = 0 (24)

120575119906 (119909 0) = 0120575119906119905 (119909 0) = 0 (25)

120575119906 (0 119905) = 0120575119906 (119897 119905) = 0 (26)

By considering (23) we obtain that the solution of abovedifference initial-boundary problem holds the followinginequality

max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) le 1198884 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (27)

Here 1198884 = (119905321198973)1198883 is independent from 1205751199024 Existence and Uniqueness of theOptimal Solution

To demonstrate the existence and the uniqueness of optimalsolution for problem (12)-(17) it is enough to show thatconditions of the following theorem given by Goebel [11]hold

Theorem 1 Let119867 be a uniformly convexBanach space and theset 119876 be a closed bounded and convex subset of 119867 If 120572 gt 0and 120573 ge 1 are given numbers and the functional 119869(119902) is lowersemicontinuous and bounded from below on the set 119876 thenthere is a dense set 119866 of119867 that the functional

119869120572 (119902) = 119869 (119902) + 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817120573119867 (28)

takes its minimum on the set 119876 for forall119903 isin 119866 If 120573 gt 1 thenminimum is unique

Before showing that these conditions have been satisfiedwe prove that the functional

119869 (119902) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909 (29)

is continuous For this we write the following increment ofthe functional

120575119869 (119902) = 119869 (119902 + 120575119902) minus 119869 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] [120575119906 (119909 119879)] 119889119909

+ int1198970[120575119906 (119909 119879)]2 119889119909

(30)

4 International Journal of Differential Equations

Since 119910(119909) isin 1198712(0 119897) if we consider inequalities (22) and(27) we conclude that this increment satisfies the followingcontinuity inequality on the set 119876

1003816100381610038161003816120575119869 (119902)1003816100381610038161003816 le 1198885 (1003817100381710038171003817120575119902100381710038171003817100381711988212(0119897) + 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)) (31)

Here 1198885 is independent of 120575119902Thanks to this inequality we can say that this functional

is also lower semicontinuous and bounded frombelow on theset 119876

On the other hand the set11988212 (0 119897) is a uniformly convexBanach space [12] the set119876 is a closed bounded and convexsubset of11988212 (0 119897) and 120573 = 2

Therefore the conditions of above theorem hold andoptimal solution to the problem (18) is unique

5 Adjoint Problem and Gradient ofthe Functional

In this section we write the Lagrange functional usedfor finding adjoint problem before we show the Frechetdifferentiability of the functional 119869120572(119902) on the set119876 Lagrangefunctional to the problem is

119871 (119906 119902 120578) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909

+ 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817211988212(0119897)

+ int1198790int1198970[119906119905119905 minus 119906119909119909 + 119902 (119909) 119906] 120578 119889119909 119889119905

(32)

Thefirst variation of this functional according to the function119906 is obtained such as

120575119871 = int11989702 [119906 (119909 119879) minus 119910 (119909) minus 120578119905 (119909 119879)] 120575119906 (119909 119879) 119889119909

+ int1198970int1198790(120578119905119905 minus 120578119909119909 + 119902 (119909) 120578) 120575119906 119889119905 119889119909 = 0

(33)

By means of stationary condition 120575119871 = 0 the followingadjoint boundary value problem is found

120578119905119905 minus 120578119909119909 + 119902 (119909) 120578 = 0 (34)

120578 (119909 119879) = 0120578119905 (119909 119879) = 2 [119906 (119909 119879) minus 119910 (119909)] (35)

120578 (0 119905) = 0120578 (119897 119905) = 0 (36)

For forall120574 isin o119882112 (Ω) the function 120578 isin 1198621([0 119879] 1198712(0 119897)) cap1198620([0 119879]11988212 (0 119897)) which satisfies the following equality

int1198790int1198970[minus120578119905120574119905 + 120578119909120574119909 + 119902 (119909) 120578120574] 119889119909 119889119905

= int1198970120578119905 (119909 0) 120574 (119909 0) 119889119909

minus int11989702 [119906 (119909 119879) minus 119910 (119909)] 120574 (119909 119879) 119889119909

(37)

is the solution of adjoint boundary value problem (34)-(36)This solution satisfies the following inequality

100381710038171003817100381712057810038171003817100381710038171198712(0119897) le 1198886 1003817100381710038171003817119906 (119909 119879) minus 119910 (119909)10038171003817100381710038171198712(0119897) forall119905 isin [0 119879] (38)

Now we can pass the calculation of the gradient In orderto do this we must evaluate the increment of the functional119869120572(119902) The increment can be written such as

120575119869120572 (119902) = 119869120572 (119902 + 120575119902) minus 119869120572 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

+ int1198970(120575119906)2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(39)

The difference problem (24)-(26) and the adjoint problem(34)-(36) give together the equality of

2int1198970[119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

= int1198790int1198970[120575119902120575119906120578 + 120575119902119906120578] 119889119909 119889119905

(40)

Inserting (40) in (39) we have

120575119869120572 (119902) = int119879

0int1198970119906120578120575119902 119889119909119889119905 + int119879

0int1198970120578120575119906120575119902 119889119909119889119905

+ int1198970(120575119906 (119909 119879))2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(41)

By (27) and (38) the second and third integrals of the aboveequality give the following inequality

int1198790int1198970120578120575119906120575119902 119889119909 119889119905 + int119897

0(120575119906 (119909 119879))2 119889119909

le 1198887 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(42)

The statement (41) can be rewritten as

120575119869120572 (119902) = ⟨119906120578 120575119902⟩1198712(Ω) + 2120572 ⟨119902 minus 119903 120575119902⟩11988212 (0119897)+ 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(43)

International Journal of Differential Equations 5

or

120575119869120572 (119902) = ⟨int119879

0119906120578 119889119905 120575119902⟩

1198712(0119897)

+ 2120572 ⟨119902 minus 119903 120575119902⟩11988212(0119897) + 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(44)

In order to pass the inner product in 11988212 (0 119897) we rearrange(44) such as

120575119869120572 (119902) = ⟨120585 + 2120572 (119902 minus 119903) 120575119902⟩11988212(0119897)

+ 119900 (10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897))

(45)

Here function 120585(119909) is the solution of the second adjointproblem

minus12058510158401015840 + 120585 = int1198790119906120578119889119905

1205851015840 (0) = 01205851015840 (119897) = 0

(46)

Therefore we have the following gradient

1198691015840120572 (119902) = 120585 + 2120572 (119902 minus 119903) (47)

6 Lipschitz Continuity of the Gradient

In this section we introduce a theorem about Lipschitzcontinuity of the gradient By this means we can express thenecessary condition for optimal solution

Theorem 2 Gradient 1198691015840120572(119902) satisfies the following Lipschitzinequality

100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 1198888 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (48)

Here 1198888 is independent from 120575119902Hence it has been proven that the gradient 1198691015840120572(119902) is

continuous on the set 119876 and it can be seen that it holds theLipschitz condition with constant 1198888 gt 0Proof Increment of the functional 1198691015840120572(119902) by giving the incre-ment of 120575119902 to the control 119902 isin 119876 is obtained

1198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902) = 120585120575 + 2120572 (119902 + 120575119902 minus 119903) minus 120585+ 2120572 (119902 minus 119903) = 120575120585 + 2120572120575119902 (49)

where the function 120575120585(119909) is the solution of the incrementproblem

12057512058510158401015840 (119909) minus 120575120585 (119909) = int1198790(119906120575120575120578 + 120575119906120578) 119889119905 (50)

Taking the norm of (49) in the space11988212 (0 119897) we acquire thefollowing inequality belonging to the functional 1205751198691015840120572(119902)

100381710038171003817100381710038171205751198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 2 10038171003817100381710038171205751205851003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (51)

There is a solution of problem (50) in 11988212 (0 119897) and thissolution satisfies the following inequality

1003817100381710038171003817120575120585 (119909)100381710038171003817100381711988212(0119897) le

100381710038171003817100381710038171003817100381710038171003817int119879

0(119906120575120575120578 + 120575119906120578) 119889119905

1003817100381710038171003817100381710038171003817100381710038171198712(0119897) (52)

The function120575120578 (119909 119905) = 120578120575 (119909 119905) minus 120578 (119909 119905)

= 120578 (119909 119905 119902 + 120575119902) minus 120578 (119909 119905 119902) (53)

in the right hand side of inequality (52) is the solution of thefollowing problem

12059721205751205781205971199052 minus 1205972120575120578

1205971199092 + (119902 (119909) + 120575119902 (119909)) 120575120578 + 120575119902 (119909) 120578 = 0(119909 119905) isin Ω

120575120578 (119909 119879) = 0120575120578119905 (119909 119879) = 2120575119906 (119909 119879) 120575120578 (0 119905) = 120575120578 (119897 119905) = 0

(54)

and this function holds the following inequality

max0le119905le119879

(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897)) le 1198889 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (55)

Here 1198889 is independent of 120575119902So the function 119906120575 that takes place in the right hand side

of (52) holds the same inequality given as follows1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) le 1198883 (56)

Hence inequality (52) has the following property1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 2 1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) max

0le119905le119879(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897))

+ 2 1003817100381710038171003817120578100381710038171003817100381721198712(Ω) max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) (57)

If inequalities (27) (38) (55) and (56) about functions 119906120575120575120578( 119905) 120578 and 120575119906( 119905) are written in (57) then the followingassessment is obtained1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 11988810 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (58)

Here 11988810 is independent of 120575119902Considering inequality (58) the following is written100381710038171003817100381710038171205751198691015840120572 (119902)1003817100381710038171003817100381721198821

2(0119897)

le 2 1003817100381710038171003817120575120585 (119909)1003817100381710038171003817211988212(0119897)

+ 81205722 1003817100381710038171003817120575119902 (119909)1003817100381710038171003817211988212(0119897)

le 211988810 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(59)

So the following inequality for the gradient 1198691015840120572(119902) is obtained100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (60)

Once we take as 1198888 = 11988811 then the proof is obtained

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

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Page 3: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

International Journal of Differential Equations 3

Namely we want to control the transverse elastic force onthe space11988212 (0 119897) and the solution 119906(119909 119879) corresponding tothis control function must be close enough to 119910(119909) in 1198712(0 119897)In order to get a stable solution we choose the space11988212 (0 119897)which is more regular than 1198712(0 119897)

The inner product and norm in 11988212 (0 119897) are definedrespectively as

(119891 119892)11988212(0119897) = int

119897

0[119891 (119909) 119892 (119909) + 119889119891 (119909)

119889119909 119889119892 (119909)119889119909 ] 1198891199091003817100381710038171003817119891100381710038171003817100381721198821

2(0119897) = int

119897

0([119891 (119909)]2 + [119889119891 (119909)119889119909 ]2)119889119909

(19)

The paper is organized as follows in Section 3 we obtain thegeneralized solution for hyperbolic problem In Section 4 weprove the existence and uniqueness of the optimal solutionIn Section 5 we obtain the adjoint problem for the optimalcontrol problem and find the gradient of the functional Themain contribution of this paper is executed in this sectionBecause the controls are chosen in the space11988212 (0 119897) gettingthe gradient of the functional necessitates finding a secondadjoint problem In the last section we demonstrate theLipschitz continuity of the gradient and state the necessarycondition for optimal solution

3 Solvability of the Problem

In this section we first give the definition of the generalizedsolution for hyperbolic problem

The generalized solution of problem (14)-(15) is the

function 119906 isin o119882112 (Ω) satisfying the following integralequality

intΩ[minus119906119905120578119905 + 119906119909120578119909 + 119902 (119909) 119906120578] 119889119909 119889119905

= int11989701205932 (119909) 120578 (119909 0) 119889119909

(20)

for forall120578 isin o119882112 (Ω) 120578(119909 119879) = 0It can be seen in [10] that solution in the sense of (20)

exists is unique and satisfies the following inequality

max0le119905le119879

(119906 (sdot 119905)21198712(0119897) + 1003817100381710038171003817119906119905 (sdot 119905)100381710038171003817100381721198712(0119897)+ 1003817100381710038171003817119906119909 (sdot 119905)100381710038171003817100381721198712(0119897)) le 1198881 (100381710038171003817100381712059311003817100381710038171003817211988212 (0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(21)

where 1198881 = max31198880 311988801199021 and 1198880 = max1 1199022 or119906211988211

2(Ω)

le 1198882 (100381710038171003817100381712059311003817100381710038171003817211988212(0119897) + 10038171003817100381710038171205932100381710038171003817100381721198712(0119897))

(1198882 = 1198881119879)(22)

Since 1205931 and 1205932 are given functions it can be written asfollows

1199062119882112(Ω)

le 1198883 (23)

Now we give an increment 120575119902(119909) isin 11988212 (0 119897) to the controlfunction 119902(119909) such as 119902+120575119902 isin 119876Then the difference function120575119906 = 120575119906(119909 119905) = 119906(119909 119905 119902+120575119902)minus119906(119909 119905 119902) is the solution of thefollowing difference initial-boundary problem

120575119906119905119905 minus 120575119906119909119909 + [119902 (119909) + 120575119902 (119909)] 120575119906 + 120575119902 (119909) 119906 = 0 (24)

120575119906 (119909 0) = 0120575119906119905 (119909 0) = 0 (25)

120575119906 (0 119905) = 0120575119906 (119897 119905) = 0 (26)

By considering (23) we obtain that the solution of abovedifference initial-boundary problem holds the followinginequality

max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) le 1198884 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (27)

Here 1198884 = (119905321198973)1198883 is independent from 1205751199024 Existence and Uniqueness of theOptimal Solution

To demonstrate the existence and the uniqueness of optimalsolution for problem (12)-(17) it is enough to show thatconditions of the following theorem given by Goebel [11]hold

Theorem 1 Let119867 be a uniformly convexBanach space and theset 119876 be a closed bounded and convex subset of 119867 If 120572 gt 0and 120573 ge 1 are given numbers and the functional 119869(119902) is lowersemicontinuous and bounded from below on the set 119876 thenthere is a dense set 119866 of119867 that the functional

119869120572 (119902) = 119869 (119902) + 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817120573119867 (28)

takes its minimum on the set 119876 for forall119903 isin 119866 If 120573 gt 1 thenminimum is unique

Before showing that these conditions have been satisfiedwe prove that the functional

119869 (119902) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909 (29)

is continuous For this we write the following increment ofthe functional

120575119869 (119902) = 119869 (119902 + 120575119902) minus 119869 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] [120575119906 (119909 119879)] 119889119909

+ int1198970[120575119906 (119909 119879)]2 119889119909

(30)

4 International Journal of Differential Equations

Since 119910(119909) isin 1198712(0 119897) if we consider inequalities (22) and(27) we conclude that this increment satisfies the followingcontinuity inequality on the set 119876

1003816100381610038161003816120575119869 (119902)1003816100381610038161003816 le 1198885 (1003817100381710038171003817120575119902100381710038171003817100381711988212(0119897) + 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)) (31)

Here 1198885 is independent of 120575119902Thanks to this inequality we can say that this functional

is also lower semicontinuous and bounded frombelow on theset 119876

On the other hand the set11988212 (0 119897) is a uniformly convexBanach space [12] the set119876 is a closed bounded and convexsubset of11988212 (0 119897) and 120573 = 2

Therefore the conditions of above theorem hold andoptimal solution to the problem (18) is unique

5 Adjoint Problem and Gradient ofthe Functional

In this section we write the Lagrange functional usedfor finding adjoint problem before we show the Frechetdifferentiability of the functional 119869120572(119902) on the set119876 Lagrangefunctional to the problem is

119871 (119906 119902 120578) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909

+ 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817211988212(0119897)

+ int1198790int1198970[119906119905119905 minus 119906119909119909 + 119902 (119909) 119906] 120578 119889119909 119889119905

(32)

Thefirst variation of this functional according to the function119906 is obtained such as

120575119871 = int11989702 [119906 (119909 119879) minus 119910 (119909) minus 120578119905 (119909 119879)] 120575119906 (119909 119879) 119889119909

+ int1198970int1198790(120578119905119905 minus 120578119909119909 + 119902 (119909) 120578) 120575119906 119889119905 119889119909 = 0

(33)

By means of stationary condition 120575119871 = 0 the followingadjoint boundary value problem is found

120578119905119905 minus 120578119909119909 + 119902 (119909) 120578 = 0 (34)

120578 (119909 119879) = 0120578119905 (119909 119879) = 2 [119906 (119909 119879) minus 119910 (119909)] (35)

120578 (0 119905) = 0120578 (119897 119905) = 0 (36)

For forall120574 isin o119882112 (Ω) the function 120578 isin 1198621([0 119879] 1198712(0 119897)) cap1198620([0 119879]11988212 (0 119897)) which satisfies the following equality

int1198790int1198970[minus120578119905120574119905 + 120578119909120574119909 + 119902 (119909) 120578120574] 119889119909 119889119905

= int1198970120578119905 (119909 0) 120574 (119909 0) 119889119909

minus int11989702 [119906 (119909 119879) minus 119910 (119909)] 120574 (119909 119879) 119889119909

(37)

is the solution of adjoint boundary value problem (34)-(36)This solution satisfies the following inequality

100381710038171003817100381712057810038171003817100381710038171198712(0119897) le 1198886 1003817100381710038171003817119906 (119909 119879) minus 119910 (119909)10038171003817100381710038171198712(0119897) forall119905 isin [0 119879] (38)

Now we can pass the calculation of the gradient In orderto do this we must evaluate the increment of the functional119869120572(119902) The increment can be written such as

120575119869120572 (119902) = 119869120572 (119902 + 120575119902) minus 119869120572 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

+ int1198970(120575119906)2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(39)

The difference problem (24)-(26) and the adjoint problem(34)-(36) give together the equality of

2int1198970[119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

= int1198790int1198970[120575119902120575119906120578 + 120575119902119906120578] 119889119909 119889119905

(40)

Inserting (40) in (39) we have

120575119869120572 (119902) = int119879

0int1198970119906120578120575119902 119889119909119889119905 + int119879

0int1198970120578120575119906120575119902 119889119909119889119905

+ int1198970(120575119906 (119909 119879))2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(41)

By (27) and (38) the second and third integrals of the aboveequality give the following inequality

int1198790int1198970120578120575119906120575119902 119889119909 119889119905 + int119897

0(120575119906 (119909 119879))2 119889119909

le 1198887 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(42)

The statement (41) can be rewritten as

120575119869120572 (119902) = ⟨119906120578 120575119902⟩1198712(Ω) + 2120572 ⟨119902 minus 119903 120575119902⟩11988212 (0119897)+ 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(43)

International Journal of Differential Equations 5

or

120575119869120572 (119902) = ⟨int119879

0119906120578 119889119905 120575119902⟩

1198712(0119897)

+ 2120572 ⟨119902 minus 119903 120575119902⟩11988212(0119897) + 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(44)

In order to pass the inner product in 11988212 (0 119897) we rearrange(44) such as

120575119869120572 (119902) = ⟨120585 + 2120572 (119902 minus 119903) 120575119902⟩11988212(0119897)

+ 119900 (10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897))

(45)

Here function 120585(119909) is the solution of the second adjointproblem

minus12058510158401015840 + 120585 = int1198790119906120578119889119905

1205851015840 (0) = 01205851015840 (119897) = 0

(46)

Therefore we have the following gradient

1198691015840120572 (119902) = 120585 + 2120572 (119902 minus 119903) (47)

6 Lipschitz Continuity of the Gradient

In this section we introduce a theorem about Lipschitzcontinuity of the gradient By this means we can express thenecessary condition for optimal solution

Theorem 2 Gradient 1198691015840120572(119902) satisfies the following Lipschitzinequality

100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 1198888 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (48)

Here 1198888 is independent from 120575119902Hence it has been proven that the gradient 1198691015840120572(119902) is

continuous on the set 119876 and it can be seen that it holds theLipschitz condition with constant 1198888 gt 0Proof Increment of the functional 1198691015840120572(119902) by giving the incre-ment of 120575119902 to the control 119902 isin 119876 is obtained

1198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902) = 120585120575 + 2120572 (119902 + 120575119902 minus 119903) minus 120585+ 2120572 (119902 minus 119903) = 120575120585 + 2120572120575119902 (49)

where the function 120575120585(119909) is the solution of the incrementproblem

12057512058510158401015840 (119909) minus 120575120585 (119909) = int1198790(119906120575120575120578 + 120575119906120578) 119889119905 (50)

Taking the norm of (49) in the space11988212 (0 119897) we acquire thefollowing inequality belonging to the functional 1205751198691015840120572(119902)

100381710038171003817100381710038171205751198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 2 10038171003817100381710038171205751205851003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (51)

There is a solution of problem (50) in 11988212 (0 119897) and thissolution satisfies the following inequality

1003817100381710038171003817120575120585 (119909)100381710038171003817100381711988212(0119897) le

100381710038171003817100381710038171003817100381710038171003817int119879

0(119906120575120575120578 + 120575119906120578) 119889119905

1003817100381710038171003817100381710038171003817100381710038171198712(0119897) (52)

The function120575120578 (119909 119905) = 120578120575 (119909 119905) minus 120578 (119909 119905)

= 120578 (119909 119905 119902 + 120575119902) minus 120578 (119909 119905 119902) (53)

in the right hand side of inequality (52) is the solution of thefollowing problem

12059721205751205781205971199052 minus 1205972120575120578

1205971199092 + (119902 (119909) + 120575119902 (119909)) 120575120578 + 120575119902 (119909) 120578 = 0(119909 119905) isin Ω

120575120578 (119909 119879) = 0120575120578119905 (119909 119879) = 2120575119906 (119909 119879) 120575120578 (0 119905) = 120575120578 (119897 119905) = 0

(54)

and this function holds the following inequality

max0le119905le119879

(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897)) le 1198889 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (55)

Here 1198889 is independent of 120575119902So the function 119906120575 that takes place in the right hand side

of (52) holds the same inequality given as follows1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) le 1198883 (56)

Hence inequality (52) has the following property1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 2 1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) max

0le119905le119879(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897))

+ 2 1003817100381710038171003817120578100381710038171003817100381721198712(Ω) max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) (57)

If inequalities (27) (38) (55) and (56) about functions 119906120575120575120578( 119905) 120578 and 120575119906( 119905) are written in (57) then the followingassessment is obtained1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 11988810 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (58)

Here 11988810 is independent of 120575119902Considering inequality (58) the following is written100381710038171003817100381710038171205751198691015840120572 (119902)1003817100381710038171003817100381721198821

2(0119897)

le 2 1003817100381710038171003817120575120585 (119909)1003817100381710038171003817211988212(0119897)

+ 81205722 1003817100381710038171003817120575119902 (119909)1003817100381710038171003817211988212(0119897)

le 211988810 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(59)

So the following inequality for the gradient 1198691015840120572(119902) is obtained100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (60)

Once we take as 1198888 = 11988811 then the proof is obtained

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 4: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

4 International Journal of Differential Equations

Since 119910(119909) isin 1198712(0 119897) if we consider inequalities (22) and(27) we conclude that this increment satisfies the followingcontinuity inequality on the set 119876

1003816100381610038161003816120575119869 (119902)1003816100381610038161003816 le 1198885 (1003817100381710038171003817120575119902100381710038171003817100381711988212(0119897) + 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)) (31)

Here 1198885 is independent of 120575119902Thanks to this inequality we can say that this functional

is also lower semicontinuous and bounded frombelow on theset 119876

On the other hand the set11988212 (0 119897) is a uniformly convexBanach space [12] the set119876 is a closed bounded and convexsubset of11988212 (0 119897) and 120573 = 2

Therefore the conditions of above theorem hold andoptimal solution to the problem (18) is unique

5 Adjoint Problem and Gradient ofthe Functional

In this section we write the Lagrange functional usedfor finding adjoint problem before we show the Frechetdifferentiability of the functional 119869120572(119902) on the set119876 Lagrangefunctional to the problem is

119871 (119906 119902 120578) = int1198970[119906 (119909 119879 119902) minus 119910 (119909)]2 119889119909

+ 120572 1003817100381710038171003817119902 minus 1199031003817100381710038171003817211988212(0119897)

+ int1198790int1198970[119906119905119905 minus 119906119909119909 + 119902 (119909) 119906] 120578 119889119909 119889119905

(32)

Thefirst variation of this functional according to the function119906 is obtained such as

120575119871 = int11989702 [119906 (119909 119879) minus 119910 (119909) minus 120578119905 (119909 119879)] 120575119906 (119909 119879) 119889119909

+ int1198970int1198790(120578119905119905 minus 120578119909119909 + 119902 (119909) 120578) 120575119906 119889119905 119889119909 = 0

(33)

By means of stationary condition 120575119871 = 0 the followingadjoint boundary value problem is found

120578119905119905 minus 120578119909119909 + 119902 (119909) 120578 = 0 (34)

120578 (119909 119879) = 0120578119905 (119909 119879) = 2 [119906 (119909 119879) minus 119910 (119909)] (35)

120578 (0 119905) = 0120578 (119897 119905) = 0 (36)

For forall120574 isin o119882112 (Ω) the function 120578 isin 1198621([0 119879] 1198712(0 119897)) cap1198620([0 119879]11988212 (0 119897)) which satisfies the following equality

int1198790int1198970[minus120578119905120574119905 + 120578119909120574119909 + 119902 (119909) 120578120574] 119889119909 119889119905

= int1198970120578119905 (119909 0) 120574 (119909 0) 119889119909

minus int11989702 [119906 (119909 119879) minus 119910 (119909)] 120574 (119909 119879) 119889119909

(37)

is the solution of adjoint boundary value problem (34)-(36)This solution satisfies the following inequality

100381710038171003817100381712057810038171003817100381710038171198712(0119897) le 1198886 1003817100381710038171003817119906 (119909 119879) minus 119910 (119909)10038171003817100381710038171198712(0119897) forall119905 isin [0 119879] (38)

Now we can pass the calculation of the gradient In orderto do this we must evaluate the increment of the functional119869120572(119902) The increment can be written such as

120575119869120572 (119902) = 119869120572 (119902 + 120575119902) minus 119869120572 (119902)= int11989702 [119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

+ int1198970(120575119906)2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(39)

The difference problem (24)-(26) and the adjoint problem(34)-(36) give together the equality of

2int1198970[119906 (119909 119879) minus 119910 (119909)] (120575119906) 119889119909

= int1198790int1198970[120575119902120575119906120578 + 120575119902119906120578] 119889119909 119889119905

(40)

Inserting (40) in (39) we have

120575119869120572 (119902) = int119879

0int1198970119906120578120575119902 119889119909119889119905 + int119879

0int1198970120578120575119906120575119902 119889119909119889119905

+ int1198970(120575119906 (119909 119879))2 119889119909 + 2120572 ⟨119902 minus 119903 120575119902⟩1198821

2(0119897)

(41)

By (27) and (38) the second and third integrals of the aboveequality give the following inequality

int1198790int1198970120578120575119906120575119902 119889119909 119889119905 + int119897

0(120575119906 (119909 119879))2 119889119909

le 1198887 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(42)

The statement (41) can be rewritten as

120575119869120572 (119902) = ⟨119906120578 120575119902⟩1198712(Ω) + 2120572 ⟨119902 minus 119903 120575119902⟩11988212 (0119897)+ 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(43)

International Journal of Differential Equations 5

or

120575119869120572 (119902) = ⟨int119879

0119906120578 119889119905 120575119902⟩

1198712(0119897)

+ 2120572 ⟨119902 minus 119903 120575119902⟩11988212(0119897) + 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(44)

In order to pass the inner product in 11988212 (0 119897) we rearrange(44) such as

120575119869120572 (119902) = ⟨120585 + 2120572 (119902 minus 119903) 120575119902⟩11988212(0119897)

+ 119900 (10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897))

(45)

Here function 120585(119909) is the solution of the second adjointproblem

minus12058510158401015840 + 120585 = int1198790119906120578119889119905

1205851015840 (0) = 01205851015840 (119897) = 0

(46)

Therefore we have the following gradient

1198691015840120572 (119902) = 120585 + 2120572 (119902 minus 119903) (47)

6 Lipschitz Continuity of the Gradient

In this section we introduce a theorem about Lipschitzcontinuity of the gradient By this means we can express thenecessary condition for optimal solution

Theorem 2 Gradient 1198691015840120572(119902) satisfies the following Lipschitzinequality

100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 1198888 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (48)

Here 1198888 is independent from 120575119902Hence it has been proven that the gradient 1198691015840120572(119902) is

continuous on the set 119876 and it can be seen that it holds theLipschitz condition with constant 1198888 gt 0Proof Increment of the functional 1198691015840120572(119902) by giving the incre-ment of 120575119902 to the control 119902 isin 119876 is obtained

1198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902) = 120585120575 + 2120572 (119902 + 120575119902 minus 119903) minus 120585+ 2120572 (119902 minus 119903) = 120575120585 + 2120572120575119902 (49)

where the function 120575120585(119909) is the solution of the incrementproblem

12057512058510158401015840 (119909) minus 120575120585 (119909) = int1198790(119906120575120575120578 + 120575119906120578) 119889119905 (50)

Taking the norm of (49) in the space11988212 (0 119897) we acquire thefollowing inequality belonging to the functional 1205751198691015840120572(119902)

100381710038171003817100381710038171205751198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 2 10038171003817100381710038171205751205851003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (51)

There is a solution of problem (50) in 11988212 (0 119897) and thissolution satisfies the following inequality

1003817100381710038171003817120575120585 (119909)100381710038171003817100381711988212(0119897) le

100381710038171003817100381710038171003817100381710038171003817int119879

0(119906120575120575120578 + 120575119906120578) 119889119905

1003817100381710038171003817100381710038171003817100381710038171198712(0119897) (52)

The function120575120578 (119909 119905) = 120578120575 (119909 119905) minus 120578 (119909 119905)

= 120578 (119909 119905 119902 + 120575119902) minus 120578 (119909 119905 119902) (53)

in the right hand side of inequality (52) is the solution of thefollowing problem

12059721205751205781205971199052 minus 1205972120575120578

1205971199092 + (119902 (119909) + 120575119902 (119909)) 120575120578 + 120575119902 (119909) 120578 = 0(119909 119905) isin Ω

120575120578 (119909 119879) = 0120575120578119905 (119909 119879) = 2120575119906 (119909 119879) 120575120578 (0 119905) = 120575120578 (119897 119905) = 0

(54)

and this function holds the following inequality

max0le119905le119879

(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897)) le 1198889 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (55)

Here 1198889 is independent of 120575119902So the function 119906120575 that takes place in the right hand side

of (52) holds the same inequality given as follows1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) le 1198883 (56)

Hence inequality (52) has the following property1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 2 1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) max

0le119905le119879(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897))

+ 2 1003817100381710038171003817120578100381710038171003817100381721198712(Ω) max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) (57)

If inequalities (27) (38) (55) and (56) about functions 119906120575120575120578( 119905) 120578 and 120575119906( 119905) are written in (57) then the followingassessment is obtained1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 11988810 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (58)

Here 11988810 is independent of 120575119902Considering inequality (58) the following is written100381710038171003817100381710038171205751198691015840120572 (119902)1003817100381710038171003817100381721198821

2(0119897)

le 2 1003817100381710038171003817120575120585 (119909)1003817100381710038171003817211988212(0119897)

+ 81205722 1003817100381710038171003817120575119902 (119909)1003817100381710038171003817211988212(0119897)

le 211988810 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(59)

So the following inequality for the gradient 1198691015840120572(119902) is obtained100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (60)

Once we take as 1198888 = 11988811 then the proof is obtained

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 5: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

International Journal of Differential Equations 5

or

120575119869120572 (119902) = ⟨int119879

0119906120578 119889119905 120575119902⟩

1198712(0119897)

+ 2120572 ⟨119902 minus 119903 120575119902⟩11988212(0119897) + 119900 (1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897))

(44)

In order to pass the inner product in 11988212 (0 119897) we rearrange(44) such as

120575119869120572 (119902) = ⟨120585 + 2120572 (119902 minus 119903) 120575119902⟩11988212(0119897)

+ 119900 (10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897))

(45)

Here function 120585(119909) is the solution of the second adjointproblem

minus12058510158401015840 + 120585 = int1198790119906120578119889119905

1205851015840 (0) = 01205851015840 (119897) = 0

(46)

Therefore we have the following gradient

1198691015840120572 (119902) = 120585 + 2120572 (119902 minus 119903) (47)

6 Lipschitz Continuity of the Gradient

In this section we introduce a theorem about Lipschitzcontinuity of the gradient By this means we can express thenecessary condition for optimal solution

Theorem 2 Gradient 1198691015840120572(119902) satisfies the following Lipschitzinequality

100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 1198888 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (48)

Here 1198888 is independent from 120575119902Hence it has been proven that the gradient 1198691015840120572(119902) is

continuous on the set 119876 and it can be seen that it holds theLipschitz condition with constant 1198888 gt 0Proof Increment of the functional 1198691015840120572(119902) by giving the incre-ment of 120575119902 to the control 119902 isin 119876 is obtained

1198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902) = 120585120575 + 2120572 (119902 + 120575119902 minus 119903) minus 120585+ 2120572 (119902 minus 119903) = 120575120585 + 2120572120575119902 (49)

where the function 120575120585(119909) is the solution of the incrementproblem

12057512058510158401015840 (119909) minus 120575120585 (119909) = int1198790(119906120575120575120578 + 120575119906120578) 119889119905 (50)

Taking the norm of (49) in the space11988212 (0 119897) we acquire thefollowing inequality belonging to the functional 1205751198691015840120572(119902)

100381710038171003817100381710038171205751198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 2 10038171003817100381710038171205751205851003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (51)

There is a solution of problem (50) in 11988212 (0 119897) and thissolution satisfies the following inequality

1003817100381710038171003817120575120585 (119909)100381710038171003817100381711988212(0119897) le

100381710038171003817100381710038171003817100381710038171003817int119879

0(119906120575120575120578 + 120575119906120578) 119889119905

1003817100381710038171003817100381710038171003817100381710038171198712(0119897) (52)

The function120575120578 (119909 119905) = 120578120575 (119909 119905) minus 120578 (119909 119905)

= 120578 (119909 119905 119902 + 120575119902) minus 120578 (119909 119905 119902) (53)

in the right hand side of inequality (52) is the solution of thefollowing problem

12059721205751205781205971199052 minus 1205972120575120578

1205971199092 + (119902 (119909) + 120575119902 (119909)) 120575120578 + 120575119902 (119909) 120578 = 0(119909 119905) isin Ω

120575120578 (119909 119879) = 0120575120578119905 (119909 119879) = 2120575119906 (119909 119879) 120575120578 (0 119905) = 120575120578 (119897 119905) = 0

(54)

and this function holds the following inequality

max0le119905le119879

(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897)) le 1198889 10038171003817100381710038171205751199021003817100381710038171003817211988212 (0119897) (55)

Here 1198889 is independent of 120575119902So the function 119906120575 that takes place in the right hand side

of (52) holds the same inequality given as follows1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) le 1198883 (56)

Hence inequality (52) has the following property1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 2 1003817100381710038171003817119906120575100381710038171003817100381721198712(Ω) max

0le119905le119879(1003817100381710038171003817120575120578 ( 119905)100381710038171003817100381721198712(0119897))

+ 2 1003817100381710038171003817120578100381710038171003817100381721198712(Ω) max0le119905le119879

(120575119906 ( 119905)21198712(0119897)) (57)

If inequalities (27) (38) (55) and (56) about functions 119906120575120575120578( 119905) 120578 and 120575119906( 119905) are written in (57) then the followingassessment is obtained1003817100381710038171003817120575120585 (119909)100381710038171003817100381721198821

2(0119897) le 11988810 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897) (58)

Here 11988810 is independent of 120575119902Considering inequality (58) the following is written100381710038171003817100381710038171205751198691015840120572 (119902)1003817100381710038171003817100381721198821

2(0119897)

le 2 1003817100381710038171003817120575120585 (119909)1003817100381710038171003817211988212(0119897)

+ 81205722 1003817100381710038171003817120575119902 (119909)1003817100381710038171003817211988212(0119897)

le 211988810 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) + 81205722 1003817100381710038171003817120575119902100381710038171003817100381721198821

2(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897)

(59)

So the following inequality for the gradient 1198691015840120572(119902) is obtained100381710038171003817100381710038171198691015840120572 (119902 + 120575119902) minus 1198691015840120572 (119902)10038171003817100381710038171003817211988212(0119897)

le 11988811 10038171003817100381710038171205751199021003817100381710038171003817211988212(0119897) (60)

Once we take as 1198888 = 11988811 then the proof is obtained

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 6: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

6 International Journal of Differential Equations

7 The Necessary Condition forOptimal Solution

After showing Lipschitz continuity of the gradient it can besaid that the gradient 1198691015840120572(119902) is continuous on the set 119876 andit holds the Lipschitz constant 1198888 gt 0 The fact that thefunctional 119869120572(119902) is continuously differentiable on the set 119876and the set 119876 is convex in that case the following inequalityis valid according to theorem in [13]

⟨1198691015840120572 (119902lowast) 119902 minus 119902lowast⟩11988212(0119897)

ge 0 forall119902 isin 119876 (61)

Therefore the following inequality is written for optimalcontrol problem

⟨120585 + 2120572 (119902lowast minus 119903) 119902 minus 119902lowast⟩11988212(0119897) ge 0 forall119902 isin 119876 (62)

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] H F Guliyev and K S Jabbarova ldquoAn optimal control problemfor weakly nonlinear hyperbolic equationsrdquo Acta MathematicaHungarica vol 131 no 3 pp 197ndash207 2011

[2] G M Bahaa ldquoBoundary control problem of infinite orderdistributed hyperbolic systems involving time lagsrdquo IntelligentControl and Automation vol 3 pp 211ndash221 2012

[3] A Kowalewski I Lasiecka and J Sokołowski ldquoSensitivity anal-ysis of hyperbolic optimal control problemsrdquo ComputationalOptimization and Applications vol 52 no 1 pp 147ndash179 2012

[4] S Serovajsky ldquoOptimal control for systems described by hyper-bolic equation with strong nonlinearityrdquo Journal of AppliedAnalysis and Computation vol 3 no 2 pp 183ndash195 2013

[5] A Hasanov ldquoSimultaneous determination of the source termsin a linear hyperbolic problem from the final overdetermina-tion weak solution approachrdquo IMA Journal of Applied Mathe-matics vol 74 no 1 pp 1ndash19 2008

[6] M Majewski ldquoStability analysis of an optimal control problemfor a hyperbolic equationrdquo Journal of Optimization Theory andApplications vol 141 no 1 pp 127ndash146 2009

[7] T Yeloglu and M Subası ldquoSimultaneous control of the sourceterms in a vibrational string problemrdquo Iranian Journal of Scienceamp Technology Transaction A vol 34 no A1 2010

[8] A Kroner ldquoAdaptive finite elementmethods for optimal controlof second order hyperbolic equationsrdquo Computational Methodsin Applied Mathematics vol 11 no 2 pp 214ndash240 2011

[9] R K Tagiyev ldquoOn optimal control of the hyperbolic equationcoefficientsrdquo Automation and Remote Control vol 73 no 7 pp1145ndash1155 2012

[10] O A Ladyzhenskaya Boundary Value Problems in Mathemati-cal Physics Springer-Verlag New York USA 1985

[11] M Goebel ldquoOn existence of optimal controlrdquo MathematischeNachrichten vol 93 pp 67ndash73 1979

[12] K Yosida Functional Analysis 624 Springer-Verlag New YorkUSA 1980

[13] F P Vasilyev Ekstremal Problemlerin Cozum Metotları 400Nauka 1981

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 7: On the Control of Coefficient Function in a Hyperbolic ...downloads.hindawi.com/journals/ijde/2018/7417590.pdf · InternationalJournalofDierentialEquations Namely,wewanttocontrolthetransverseelasticforceon

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom