Exact Controllability Of Wave Equation With Multiplicative Controls

Exact Controllability Of Wave Equation With Multiplicative Controls

Ramdas Baburao Sonawane

, Anil Kumar

, Sarjerao Bhaurao Nimse

Received 17 December 2013

Abstract

We consider the problem of exact controllability of the wave equation with Dirichlet boundary conditions and multiplicative controls. A multiplicative con- trol is a coe¢ cient likeuy. Exact controllability result is stated and proved for some particular targets.

1 Introduction

Let D Rⁿ, n 2 N be a bounded domain with smooth boundary @D. We use the notations QT = D (0; T) and P

T = @D (0; T). Consider the following control problem with multiplicative control:

ytt y =uyt+vy; inQT;

y =g; on P

T; (y(x;0); y_t(x;0)) = (y₀(x); y₁(x)) inD;

(1)

where u; v2L¹(Q_T)are controls, g2 C(P

T), and(y₀; y₁)2H₀¹(D) L²(D).

A problem like this arises, in the context of so-called “smart materials", whose properties can be altered by applying various external factors such as temperature, electrical current or magnetic …eld [1]. Russell [2] developed controllability and stability theory for wave equation. Ball, Marsden and Slemrod [3] consider the problem of global approximate controllability of the rod equation utt+uxxx+k(t)uxx = 0with hinged ends and the wave equation utt uxx+k(t)u = 0 with Dirichlet boundary conditions, wherekis control; which appears to be the …rst work on this subject in the framework of partial di¤erential equations(pdes). In [1, Chapter 6], Khapalov proved, in a constructive way, that the set of equilibrium states like(yd;0)of a vibrating string that can approximately be reached in H₀¹(0;1) L²(0;1) by varying its axial load and the gain of damping is dense in the subspaceH₀¹(0;1) f0g of this space; where as in Chapter 8, Khapalov talks about the extension of this problem to the semilinear case.

It seems that the result for exact controllability of (1) obtained in this work is new.

To prove our result we use the method of connecting the multiplicative controllability

Mathematics Sub ject Classi…cations: 35K05, 35L05, 93B05, 93C20.

yK. A. A. N. M. Sonawane Arts, Com. & Sci. College, Satana, Nashik (M.S.), India - 423 301.

zBITS-Pillani, K.K. Birla Goa Campus, Goa, India - 403 726.

xUniversity of Lucknow, Lucknow (U.P.), India - 226 007

45

with the additive distributed controllability by means of some substitution used for parabolic equation by Fernandez [4].

REMARK 1. The zero state (y0; y1) = (0;0) is the …xed point for the solution mappings of system (1), regardless of the choice of controls u. Hence, it cannot be steered anywhere from this state. Hence, everywhere below we consider only non-zero initial states(y0; y1).

Some preliminary results used to prove our main Theorem are given in section 2.

In section 3, we prove the main theorem.

2 Preliminaries

The exact controllability of system (1) is de…ned as follows:

DEFINITION 1. System (1) is said to be exact controllable in timeT₁>0 if, for every initial data(y₀; y₁)2H₀¹(D) L²(D)and desired pro…ley_d2H₀¹(D), there exist controls u; v2L¹(Q_T)such that (y(x; T); y_t(x; T)) = (y_d;0)in D, for allT T₁.

We will need help of the following well known result from [5], while proving our main theorem.

LEMMA 1. Leta2L¹(Q_T),v2L²(Q_T)and(q₀; q₁)2H₀¹(D) L²(D)be given.

Then the solution qof the linear problem

qtt(x; t) q(x; t) +a(x; t)q(x; t) =v(x; t); inQT;

q = 0; on P

T; (q(x;0); q_t(x;0)) = (q₀(x); q₁(x)) inD;

satis…esq2C([0; T];H₀¹(D)),qt2C([0; T];L²(D))and kq(; t)k²C([0;T];H¹₀(D))+kq_t(; t)k²C([0;T];L²(D))

e^C3kT kq(; T₂)k²H₀¹(D)+kq_t(; T₂)k²L²(D)+kvk²L²(QT) ; where k= 1 +kakL¹(QT).

Also, we will use the following known result about the additive controllability from [6,7].

LEMMA 2. For every (q₀; q₁)2H₀¹(D) L²(D)and a2L¹(Q_T), there exists a control functionv2L²(Q_T)such that the weak solutionqto problem

q_tt(x; t) q(x; t) +a(x; t)q(x; t) =v(x; t); inQ_T;

q = 0; on P

T; (q(x;0); q_t(x;0)) = (q₀(x); q₁(x)) inD;

satis…es(q(x; T); qt(x; T)) = (0;0). Moreover,

kvk²L²(QT) C(kakL¹(QT)) kq0k²H₀¹(D)+kq1k²L²(D) : Now it is time to state and prove the main result.

3 The Main Result

Our main theorem is stated as follows:

THEOREM 1. Let be a function de…ned on Q_T. If _t; 2 W^2;1(D), 0 < , 0 < _t in D, and 0 ; _t = a.e. in D, g 2C(P

T), g = onP

T, then there exists a T( )>0 such that for any non-zero initial state(y₀; y₁)2H₀¹(D) L²(D), there exist multiplicative controlsu; v2L¹(Q_T)such that the corresponding solution to (1) inC([0; T];H₀¹(D))\C¹([0; T];L²(D))satis…es

(y(x; T); y_t(x; T)) = ( (x; T); _t(x; T));

in D, for allT T( ).

PROOF. Let z = y . Then z₀ = z(x;0) = y(x;0) (x;0) = y_{0 0}, z₁ = z_t(x;0) =y_t(x;0) _t(x;0) =y_{1 1}, and hence(z₀; z₁) = (y_{0 0}; y_{1 1}). Thus from (1),zsatis…es

z_tt z =u(z_t+ _t) +v(z+ ) + ; in Q_T;

z = 0; on P

T; (z(x;0); z_t(x;0)) = (z₀; z₁); in D:

(2)

It is well known that given (z₀; z₁)2H₀¹(D) L²(D), u; v 2L¹(Q_T), as given in Theorem 1, the problem (2) admits a unique solutionzin the spaceC¹([0; T];L²(D))\ C([0; T];H₀¹(D)).

In order to prove Theorem 1, it is su¢ cient to prove that system (2) is exactly null controllable. We prove this in the following few steps:

STEP 1. In this step, we prove that givenT₁ >0, there exists M >0, such that the corresponding solution to (4) satis…es

kz(; T₁)k²H₀¹(D)+kz_t(; T₁)k²L²(D) M:

Multiplying the system (4) by z_tand integrating overQ_t=D (0; t), we obtain 0 =

Z

Qt

z_ttz_t zz_t uz²_t uz_{t t} z_t vzz_t vz_t + z_t dxd

= Z

Qt

1 2

d

dt(z²_t+krzk²) uz_t² uz_{t t} z_t vzz_t vz_t + z_t dxd : Supposeu; v are constants andv= 0. Then, fort2(0; T)we have

Z

D

z²_t(x; t) +jrz(x; t)j² dx

= Z

D

z²₁+jrz0j² dx+ 2 Z

Qt

uz_t² dxd + 2 Z

Qt

[uzt t+zt zt]dxd kz₁k²L²(D)+krz₀k²L²(D)+ 2ukz_tk²L²(Qt)+jujkz_tk²L²(Qt)+jujk tk²L²(Qt)

+2kztk²L²(Qt)+k k²L²(Qt)+k k²L²(Qt): (3)

By Poincaré’s inequality we have Z

D

z_t²(x; t) +z²(x; t) dx C₁(u+ 2) Z

Dt

z_t²+z² dxd +C₁ kz₁k²L²(D)+krz₀k²L²(D)

+C1 jujk ^tk²L²(Qt)+k k²L²(Qt)+k k²L²(Qt) ; (4) where constant C1is independent ofz. Hence, by Gronwall’s lemma

kz(; t)k²H₀¹(D)+kz_t(; t)k²L²(D)

e^(u+2)C1t kz₁k²L²(D)+krz₀k²L²(D) +C₁ Z t

0

e^{C1(u+2)(t )}kz₁k²L²(D)d +C1

Z t 0

e^{C1(u+2)(t )}krz0k²L²(D)d +jujC1

Z t 0

e^{C1(u+2)(t )}k ^tk²L²(Qt)d +C1

Z t 0

e^{C1(u+2)(t )}k k²L²(Qt)d +C1

Z t 0

e^{C1(u+2)(t )}k k²L²(Qt)d

= e^(u+2)C1t kz1k²L²(D)+krz0k²L²(D) +1 e^C1(u+2)t

(u+ 2) kz1k²L²(D)

+1 e^C1(u+2)t

(u+ 2) krz0k²L²(D)+juj1 e^C1(u+2)t

(u+ 2) k ^tk²L²(Qt)

+1 e^C1(u+2)t

(u+ 2) k k²L²(Qt)+1 e^C1(u+2)t

(u+ 2) k k²L²(Qt):

Hence, for givenT1 >0, we can selectu=u1 < 2 depending on (z0; z1)and juj is su¢ ciently large such that there exists a constantM >0, such that the corresponding solution of (2) satis…es

kz(; T₁)k²H₀¹(D)+kz_t(; T₁)k²L²(D) M:

STEP 2. In this step, we further prove that for any ₀ >0, we can …nd controls u and v and T₂( ) > 0 su¢ ciently large such that the corresponding solution to (2) satis…es

kz(; T₂)k²H₀¹(D)+kz_t(; T₂)k²L²(D) 0:

As t2W^2;1(D), we have t2C(D)by Sobolev embedding theorem. Also, t>0 in D, there exists a positive constant >0such that t >0 in D, hence0 _t 2 L¹(D). Let zT1 =z(x; T1) and z_T⁰₁ = zt(x; T1). Select u2 =

t + 1 andv = 0 in

(T1; T2). Then, we have

ztt z = (

t 1)zt; in D (T1; T2);

z = 0; on@D (T1; T2);

(z(x; T1); zt(x; T1)) = (zT1; z⁰_T1); in D:

(5)

Let >1 be the eigenvalue of in H₀¹(D). Let t2(T1; T2)andQt=D (T1; t).

Multiplying (5) by ztand integrating inD, we have Z

D

z_t²(x; t) +jrz(x; t)j² dx= Z

D

z²₁+jrz0j² dx+ 2 Z

Qt

( ( 1))z_t²dxd :

As in step 1, using Poincaré’s inequality and Gronwall’s lemma we have

kz(; T₂)k²H¹₀(D)+kz_t(; T₂)k²L²(D) C kz(; T₁)k²H₀¹(D)+kz_t(; T₁)k²L²(D) ; where C=e ^{( 1)(T2 T1)}. Thus by step 1,

kz(; T2)k²H₀¹(D)+kzt(; T2)k²L²(D) M e ^{( 1)(T2 T1)}: Hence, for any ₀>0, there exists aT₂( )>0su¢ ciently large such that

kz(; T2)k²H¹₀(D)+kzt(; T2)k²L²(D) 0: (6) STEP 3. In this step, we achieve the result by means of the controllability re- sult with the traditional additive distributive control. Let z_T2 = z(x; T₂) and z⁰_T2 = zt(x; T1). Consider the following system

z_tt z =u(z_t+ _t) +v(z+ ) + inD (T₂; T₂+ 1);

z = 0; on@D (T₂; T₂+ 1);

(z(x; T₂); z_t(x; T₂)) = (z_T2; z_T⁰₂); inD:

(7)

As 2W^2;1(D), we have 2C(D)by Sobolev embedding theorem. Also, >0inD, there exists a positive constant >0such that >0inD, hence0 2L¹(D).

Letu= 0andv= + 1 +v3. Then (7) becomes

z_tt z+ ( 1)z =v₃(z+ ) in D (T₂; T₂+ 1);

z = 0; on@D (T₂; T₂+ 1);

(z(x; T₂); z_t(x; T₂)) = (z_T2; z⁰_T2); in D:

(8)

From (6) we have

kz(; T₂)k²H¹₀(D)+kz_t(; T₂)k²L²(D) 0; (9) where 0will be …xed later. In place of (8), we consider the following system

ztt z+ ( 1)z =w(x; t) in D (T2; T2+ 1);

z = 0; on@D (T2; T2+ 1);

(z(x; T2); zt(x; T2)) = (zT2; z⁰_T2); in D:

(10)

By Lemma 2, there exists a controlw2L²(D (T2; T2+1))such that the corresponding solution to (10) satis…es

(z(; T2+ 1); zt(; T2+ 1)) = (0;0): (11) Moreover,

kwk²L²(D (T2; T2+1)) C2(d) kz(x; T2)k²H¹₀(D)+kzt(x; T2)k²L²(D) ; (12) where d=k = 1kL¹(QT2). Also, using Lemma 1, we have

kz(; t)k²_{C([T2;T2+1];H}¹_0(D))+kzt(; t)k²(C([T2;T2+1];L²(D))

e^C3(1+d) kz(; T2)k²H₀¹(D)+kzt(; T2)k²L²(D)+kwk²L²(QT2) ; (13)

where Q_T2 =D (T₂; T₂+ 1), and constant C₃ depends only on D. From (12) and (13), we have

kz(; t)k²C([T2;T2+1];H¹₀(D))+kzt(; t)k²(C([T2;T2+1];L²(D))

e^C3(1+d)(1 +C2) kz(; T2)k²H₀¹(D)+kzt(; T2)k²L²(D) ; (14) We now select

0< 1

e^C3(1+d)(1 +C₂) <

e^C3(1+d)(1 +C₂):

Here we may assume that >1, and by (6), selectT2 su¢ ciently large such that (kz(; T2)k²H₀¹(D)+kzt(; T2)k²L²(D)) 0:

Hence, we have

kz(; t)k²C([T2;T2+1];H¹₀(D))+kzt(; t)k²(C([T2;T2+1];L²(D)) < : (15) Thus, we can select the multiplicative control for (8)

v3=

z+ a.e. in D (T2; T2+ 1); (16)

where z is the solution of (8). In view of >0, and (15), we havev3 2L¹(D (T2; T2+ 1)).

Hence, in the time interval(T2; T2+ 1), in view of (16), the solution of (8) with the control v3, i.e., the solution of (7) with the controlsu= 0and v= + 1 +v3 and the solution of (10) with the controlwbecome identical. Hence from (11), we have

(z(; T2+ 1); zt(; T2+ 1)) = (0;0);

where zis the corresponding solution to (7) withu= 0andv= + 1 +v3. By steps 1, 2 and 3, we have for any (z0; z1) 2 H₀¹(D) L²(D), we can select T₂( )>0su¢ ciently large such that the corresponding solutionz to (2) with controls

u= 8>

<

>:

u1 in (0; T1);

t in (T₁; T₂);

0 in (T₂; T₂+ 1);

andv= 8>

<

>:

0 in(0; T1);

0 in(T₁; T₂);

+ 1 +v3 in(T2; T2+ 1);

satis…es

(z(; T₂+ 1); z_t(; T₂+ 1)) = (0;0);

where T( ) =T2+ 1 depends on only. This completes the proof of Theorem 1.

REMARK 2. In step 2, we can take 0 = infx2D . So,kz(; T2)kH₀¹(D) 0. Now in step 3, due to null controllabilityzapproaches0. Thus in (16) we havez+ >0 in D (T2; T2+ 1).

From the proof of Theorem 1 we have the following theorem.

THEOREM 2. Let be a function de…ned onQ_T. If ; _t2W^2;1(D),0> ,0> _t in D, and 0 ; t = a.e. inD, g 2C(P

T), g = on P

T, then there exists a T( )>0such that for any non-zero initial state(y0; y1)2H₀¹(D) L²(D), there exist multiplicative controlsu; v 2L¹(QT)such that the corresponding solution to (1) in C([0; T];H₀¹(D))\C¹([0; T];L²(D))satis…es(y(x; T); yt(x; T)) = ( (x; T); t(x; T)), in D, for allT T( ).

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