On solvability of an inverse boundary value problem for the Boussinesq–love equation Текст научной статьи по специальности «Математика»

УДК 517.95

On Solvability of an Inverse Boundary Value Problem for the Boussinesq-Love Equation

Yashar T. Mehraliyev*

Department of differential and inteqral equations,

Baku State University, Khalilov, 23, Baku, AZ1148

Azerbaijan

Received 28.05.2013, received in revised form 10.07.2013, accepted 20.09.2013 In the paper an inverse boundary value problem for the Boussinesq-Love equation with an integral condition of the first kind is investigated. First, the given problem is reduced to an equivalent problem in a certain sense. Then, using the Fourier method the equivalent problem is reduced to solving the system of integral equations. The existence and uniqueness of a solution to the system of integral equation is proved by the contraction mapping principle. This solution is also the unique solution to the equivalent problem. Finally, by equivalence, the theorem of existence and uniqueness of a classical solution to the given problem is proved.

Keywords: inverse boundary value problem, the Boussinesq-Love equation, Fourier method, classical solution.

Introduction

There are many cases where the needs of the practice bring about the problems of determining coefficients or the right hand side of differential equations from some knowledge of its solutions. Such problems are called inverse boundary value problems of mathematical physics. Inverse boundary value problems arise in various areas of human activity such as seismology, mineral exploration, biology, medicine, quality control in industry etc., which makes them an active field of contemporary mathematics.

Inverse problems for various types of PDEs have been studied in many papers. Among them we should mention the papers of A.N. Tikhonov [1], M.M. Lavrentyev [2, 3], V.K. Ivanov [4] and their followers. For a comprehensive overview, the reader should see the monograph by A.M. Denisov [5].

In this paper, following [6-9], we prove existence and uniqueness of the solution to an inverse boundary value problem for the Boussinesq-Love equation modeling the longitudinal waves in an elastic bar with the transverse inertia.

1. Problem statement and its reduction to an equivalent problem

Consider for the Boussinesq-Love equation [10]

Utt(x,t) - Uttxx(x,t) - autxx(x,t) - puxx(x,t) = a(t)u(x,t) + f (x,t) (1)

in the domain DT = {(x,t) : 0 ^ x ^ 1, 0 ^ t ^ T} an inverse boundary problem with the initial conditions

*[email protected] © Siberian Federal University. All rights reserved

u(x, 0) = y(x), ut(x, 0) = ^(x) (0 < x < 1), (2)

the periodic condition

u(0,t) = u(1,t) (0 < t < T), (3)

the non-local integral condition

I u(x,t)dx = 0 (0 < t < T) (4)

Jo

and with the additional condition

u(x0,t) = h(t) (0 < t < T), (5)

where x0 G (0,1), a > 0, ft > 0 are the given numbers, f(x,t), y(x), ^(x), h(t) are the given functions, and u(x,t), a(t) are the required functions.

The condition (4) is a non-local integral condition of first kind, i.e. the one not involving values of unknown functions at the domain's boundary points.

Definition. A classical solution to the problem (1)-(5) is a pair {u(x, t), a(t)} of the functions u(x,t) and a(t) with the following properties

1) the function u(x,t) is continuous in DT together with all its derivatives contained in equation (1);

2) the function a(t) is continuous on [0, T];

3) all the conditions (1)-(5) are satisfied in the ordinary sense.

The following lemma holds.

Lemma 1. Let f(x,t) G C(DT), i f(x,t)dx = 0 (0 < t < T), y>(x),^(x) G Cx[0,1], h(t) G

0

C2 [0,T], h(t) = 0 (0 < t < T) and

1 y/(0) = ^'(1), V'(0) = V'(1),

/ <^>(x)dx = 0, / ^(x)dx = 0, y(xo) = h(0), ^(xo) = h'(0). Jo Jo

Then the problem of finding a classical solution to the problem (1)-(5) is equivalent to the problem of finding functions u(x,t) and a(t) with the properties 1) and 2) of the definition of the classical solution from the relations (1)-(3) and satisfying

ux(0,t)= Ux(1,t) (0 < t < T), (6)

h''(t) - uttxx(x0,t) - autxx(x0,t) - ftuxx(xo,t) = a(t)h(t) + f(xo,t) (0 < t < T). (7) Proof. Let {u (x,t), a (t)} be a classical solution to the problem (1)-(5). Integrating equation (1) with respect to x from 0 to 1, we have

d2 /•1 , , , d2........ d

d2 r1 d2 d

—2 u(x,t)dx - —^ (ux(1,t) - ux(0,t)) - a — (ux(1,t) - ux(0,t)) -

dt y 0 dt dt / q \

/•1 /»1 (8)

-ft (ux(1,t) - ux(0,t)) = a(t) / u(x, t)dx + / f (x, t)dx (0 < t < T).

00

Taking into account that / f (x,t)dx = 0 (0 < t < T) and (4), we find that

0

d2 d -dt2 (ux(1,t) - ux(0,t)) - a- (ux(1,t) - ux(0,t)) -

-ft (ux(1,t) - ux(0,t)) = 0 (0 < t < T).

By (2) and p'(0) = p'(1), ^'(0) = ^'(1) we obtain

Ux(1,0) - Ux(0, 0) = p'(1) - p'(0) = 0,

utx(1, 0) - utx(0, 0) = V'(1) - ^'(0) = 0. (10)

Since the problem (9), (10) has only a trivial solution, we have ux(1,t) - ux(0,t) = 0, i.e. the condition (6) is fulfilled.

Assume now that h(t) G C2[0,T]. Differentiating (5) twice, we get

ut(xo,t) = h'(t), utt(xo,t) = h''(t) (0 < t < T). (11)

It follows from (1) that

utt(xo,t) - uttxx(xo, t) - autxx(xo,t) - ^uKx(xo,t) = a(t)u(xo,t) + f (xo,t) (0 < t < T). (12)

Hence, taking into account (5) and (11), we conclude that (7) is fulfilled.

Now suppose that {u (x,t), a (t)} is a solution to the problem (1)-(3), (6), (7), then from (8) and (6) we find that

r-1 !■ 1

d2 f1 f1

u(x,t)dx - a(tW u(x,t)dx = 0 (0 < t < T). (13)

dt2 Jo Jo

By (2) and / p(x)dx = 0, / ^(x)dx = 0, it is obvious that

oo

/ u(x, 0)dx = p(x)dx = 0, / ut(x, 0)dx = ^(x)dx = 0. (14)

o o o o

Since the problem (13), (14) has only a trivial solution, / u(x,t)dx = 0 (0 ^ t ^ T), i.e. the

o

condition (4) is fulfilled.

From (7) and (12) we obtain

d2

—(u(xo, t) - h(t)) = a(t)(u(xo, t) - h(t)) (0 < t < T). (15)

dt2

By (2) and p(xo) = h(0), ^(xo) = h'(0) we have

f u(xo, 0) - h(0) = p(xo) - h(0) = 0,

S (16)

[ ut(xo, 0) - h'(0) = ^(xo) - h'(0) = 0. From (15) and (16) we conclude that the condition (5) is fulfilled. The lemma is proved. □

2. Existence and uniqueness of the classical solution to the inverse boundary value problem

It is known [11] that the system

1, cos A1x, sin A1x,..., cos Akx, sin Akx,... (17)

is a basis in L2(0,1), where Ak = 2kn (k = 1,2,...). Therefore, it is obvious that for each solution {u(x, t), a(t)} to the problem (1)-(3), (6), (7) its first component u(x,t) has the form:

u(x, t) = ^^ u1k(t) cos Akx + ^^ u2k(t) sin Akx (Ak = 2nk), (18)

k=o k=1

where

r-1

uio(t) = / u(x, t)dx,

l Jo 1

uifc (t) = 2 / u(x,t)cos Afc xdx, «22 (t) = 2 / u(x, t) sin Ak xdx (k = 1, 2,...).

oo

(20)

/0 Jo

Then, applying the formal scheme of the Fourier method, from (1) h (2) we have

ui0(t) = F10(t; u, a) (0 < t < T), (19)

(1 + Ak) uik (t) + aA|uifc (t) + uifc (t) = Fifc (t; u, a) (0 < t < T; i = 1, 2; k = 1, 2,...),

uio(0) = <£>10, u1o(0) = -010, (21)

uifc (0) = <ifc, uifc(0) = -ifc (i = 1,2; k = 1, 2,...), (22)

F1fc (t; u, a) = a(t)u1fc(t) + /u(t) (k = 0,1,...),

/10(t) = /f,1 /(x,t)dx, /1fc(t) = 2 /01 /(x, t) cos Akxdx (k = 1, 2,...), r-1 1

' (x )dx,

where

<10 = <(x)dx, —10 = —(x)dx 1 ./0 1

= 2 / <(x)cosAkxdx, —1k = 2 / —(x)cosAkxdx (k = 1, 2,...), ./0 ./0 1

F2k (t; u, a) = a(t)u2fc(t) + /2k(t), /2ft(t) = 2 /(x,t)sin Afcxdx (k = 1, 2,...),

1 1 0

<2k = 2 / <(x) sin Ak xdx, —2k = 2 / — (x) sin Ak xdx (k = 1, 2,...).

00

It is obvious that A2 < 1 + A2 < 2A|. Therefore

a2 a2A? a2

■8 --T -

a2

Now suppose that —— ft > 0. Solving the problem (19)-(22), we find 8

u10(t) = <10 + t—10 + / (t - t)F10(t; u,a)dr (0 < t < T), (23)

0

«¿2 (t) = — [(M2fce^1ki - MifceM2fci) ^2 + 4 - ) fe + 72

+ J Fifc (t; u, a) (>2k(i-T> - e^1fc(i-T^ dr

(i = 1, 2; 0 < t < T),

(24)

where

= -2(TTÄf) + (-1)i ^

Y2 = M22 - Mifc = 2Afc<

' a2A2fc ß

4(1 + A2 )2 1 + A2

1 a2A2 ß

/4(1 + A2 )2 1 + A2

After substituting the expressions ui2(t) (k = 0,1,...) and u22(t) (k =1,2,...) into (18), for the component u(x,t) of the solution {u(x, t), a(t)} to the problem (1)—(3), (6), (7) we get

u (x,t) = ^10 + t^io + (t - t)Fio(t; u, a)dr+

Jo

o i 1

+ E — [(№eMlfc4 - MifceM2fc4 W + (eM2fc4 - e^4)^ + fc=i ^ 72

+ ^ Fik(t; u, a)(eM2fc(4-r) - eWfc(i-T))dr | cos Akx+ o f 1

+ E — IWeMlfc4 - MifceM2fc+ (eM2fc4 - e^4)^ +

fc=i 7*

+ ^0 F2k(t; u, a)(eM2fc(i-r) - e^1fc(i-r))dT | sin Akx.

Differentiating (24) twice, we get

4M = — ["ik"2k(eMlfci - eM2fci)^jfc + ("2keM2fct - "ik ,t Yk t 'ik C ; « a)("2k e ' - "1' eMlfc(i-T))

/0

t k

+ / Fifc(t; u,a)("2fce^2k(t-T) - "ke^1k(i-T>)dr

0

(i = 1, 2),

'k(t) = — ["ik"2k("ikeMlfct - "2fceM2fci)^jfc + ("2keM2fci - "ke^lfct)V>ik+ t Yk

+ / Fik(r; u,a)("2ke^(i-T) - "1k(t-T})d

0

By (20) and (29) we have

+ Fik (t; u,a) (i = 1, 2).

(26)

0

Now, from (7) and (18) we have

a (t) = [h (t)]-i {h''(t) - f (xo, t) + ^ [A2uik (t) + «A2u^ (t) + ßA2«k (t)] cos AkXo+

k=i

+ Z [AkM2'k (t) + aAk«2k (t) + ßAk«2k (t)] sin Akxo I .

k=i

(27)

(28)

(29)

A'<fc(t) + aA2uifc(t) + ^A2uifc(t) = Fifc(t; u, a) - <fc (t) = = - — [MifcM2fc(MifceMlfc4 - M2fceM2fc4)<pifc + (M2fceM2fci - ^e^4)^ + (30)

+ / Fifc(t; u, a)(M2fce^2k(i-r) - M2fce^1k(i-r))dT (i = 1, 2) .

0

To obtain the equation for the second component a(t) of the solution {u (x,t), a (t)} to the problem (1)-(3), (6), (7), we substitute expression (30) into (27) and have

a (t) = [h (t)]- {h''(t) - f (xo,t) -

oo 1

- Z) _ [^ifc^2fc(MifceM1fci - M2fceM2fci)^ifc + (M2fceM2fci - M2fce^1fci)^ifc+

k=i Yk

cos Ak xo-

+ / Fik(t; u,a)("2ke^2k(i-T) - "2ke^1k(t-T))dT Jo

oo i

- Z ~ ["ik"2k("ikeMlfct - "2keM2fct)^2k + ("2keM2fct - "2keMlfct)^2k+ k=i 7fc t

+ J F2k(t; u,a)("2keM2fc(t-T) - "?keMlfc(t-T))dr sin Akxoj .

u

Thus, the problem (1)—(3), (6), (7) is reduced to solving the system (26), (31) with respect to the unknown functions u(x,t) and a(t).

Similarly to [9], it is possible to prove the following lemma.

Lemma 2. If {u (x,t), a (t)} is any solution to the problem (1)-(3), (6), (7), then the functions

uio(t) = / u(x, t)dx, Jo

uik (t) = 2 / u(x,t)cos Ak xdx, U2k (t) = 2 / u(x, t) sin Ak xdx (k = 1, 2,...).

oo

satisfy the system (23), (24) in [0, T].

Remark. It follows from lemma 2 that to prove the uniqueness of the solution to the problem (1)-(3), (6), (7), it suffices to prove the uniqueness of the solution to the system (26), (31). In order to investigate the problem (1) - (3), (6), (7), consider the following spaces

1. Denote by T [8] the set of all functions u(x,t) of the form

o O

u(x,t) ^^ uik(t) cos Akx + U2k(t) sin Akx (Ak = 2nk),

k=o k=i

defined on DT such that the functions u1k(t) (k = 0,1,...), u2k(t) (k = 1,2,...) are continuous on [0, T] and

(OO \ 2 \ 2

kC (Ak ||uik(t)yc[o,t])M + (Ak ||u2k(t)yc[o,T])2J <

The norm on this set is given by

||u(x,t)||B|T = jT (u).

2. Denote by ET the space T x C[0, T] of the vector-functions z(x, t) = {u(x, t), a(t)} with the norm

||z||ET = ||u(x,t)||B|,T + ||a(t)||C[o,T] .

It is known that T and ET are Banach spaces. Now, in the space ET consider the operator

$(u, a) = {$i(u, a), $2(u, a)} ,

where

OO

$i(u, a) = u(x, t) = uik(t) cos Akx + U2k(t) sin Akx,

k=o k=i

$2(u, a) = a(t),

uio(t), uik(t), k =1,2,...) and a(t) equal to the right hand sides of (23), (24), and (31), respectively.

It is easy to see that

Mifc

< 0, eMifci < 1, e^fc(i-T} < 1 (i = 1, 2; k = 1, 2,...; 0 < t < T, 0 < t < t),

MlfcM22 — î + a2 ^ ß'

72 -A Vl(02-ß)

-ßl V2V8

< . = = Y0 (k — 1' 2'...).

Taking into account these relations, by means of simple transformations we find

i

' rT /0

l|Uio(i)llc[o.T] ^ l^iol + T l^ioi |/io(r)|2 dr +T2 ya(i)yo[oiT]yuio(i)yo[o,T]' (32)

]T(A2 |Ui2(i)||C[o,T])2 < 4a7„ £ (Ak 1)2 + 47o E (A2 l^ik|)2 + \2=i J \2=i / \2 = i /

/ T œ \ 2 / œ \ 2

+47ov/T / E(A2 |/i2 (T)|)2d^ +47oT ||a(i)lc[o,TM E (A2 H^MHc^t] )2

(33)

2=i / \2=i |Ri)HC[o,T] < ||[h(t)] i|c[o,T] {||h''(t) - /(xo't)||C[o,T] +

1

2 2

+ E A-2 ]T

k

V2=i

(œ \ 2 / œ \ 2

E (A2 |^i2 |)M +2a27J £ (A2 |Vi2 |)M +

T œ \ 2 (34)

+2a2Yov^T I / ]T(A32 |/i2(r)|)2dr +

œ

+2a27o T ||a(t)||C[o,T] E Hwi2(i)|C[o,T])

k=1

Suppose that the data of the problem (1)-(3), (6), (7) satisfy the following conditions

a2

1. a > 0, ft > 0,--ft > 0.

8

2. ^(x) G C2[0, 1], ^"'(x) G ¿2(0,1),^(0) = ^(1), y>'(0) = ^(1), ^"(0)= ^"(1).

3. V(x) G C2[0,1], V""(x) G ¿2(0, 1), V(0) = V(1), V'(0) = ^'(1), ^''(0) = ^''(1).

4. f(x,t), fx(x,t), fxx(x,t) G C(Dt), fxxx(x,t) G ¿2(Dt),

f (0, t) = f (1,t),fx (0, t) = fx (1,t),fxx (0, t) = fxx(1,t) (0 < t < T).

5. h(t) G C2[0,T], h(t) = 0 (0 < t < T). Then, from (32)-(34), we get

||u(x,t)||B3 T < Ai(T) + Bi(T) |Kt)||C[oiT]||u(x,t)||£|, t , (35)

||S(t)|| C[o,T ] < A(T) + B2(T) ||a(t)||

C [o,T ] ||u(x,t)||B|T , (36)

where

Ai(T) = ||y(x)||L2(o,i) + T ||^(x)|L2(o,i) + TVT ||f (x,t)|i2(DT) + +8a7o (x)|L2(o,i) +8Yo H^'MH^^i) + 87^V/T ||fxxx(x, t)^ (Dt ),

Bi(T ) = T2 + 87oT,

1

1

œ

2

A2(T )= [h(t)]

+ £ A-

Vfc = 1

4aß7o 2

C[0,T ] ^(3)(x)

||h"(t) - f (xo,i))Hc[o,T] +

+4a Yo

¿2(0,1)

+ 4a27oVT yfxxx(x,i)yL2(DT)]}

^(3)(x)

¿2 (o, 1)

+

B2(T ) = 4a2 Yo [h(t)]

C[o,T ]

5>.

Vfc = 1

-2

T.

It follows from the inequalities (35), (36) that

C[0,T] ||u(x,t)|B3 ,

where A(T) = Ai(T) + A2(T), B(T) = Bi(T) + B2(T). Now we can prove the following theorem.

Theorem 1. Lei the conditions (1)-(5) be fulfilled and

(A(T) +2)2B(T) < 1.

(37)

(38)

Then the problem (1)-(3), (6), (7) has a unique solution in the ball K = KR(||z||E3 ^ R = A(T) + 2) of the space .

Proof. In the space consider the equation

z = $z, (39)

where z = {u, a} and the components $j(w, a) (i = 1, 2) of the operator $(u, a) are given by the right hand sides of the equations (26), (31). Consider the operator $(u, a) in the ball K = KR from E3. Similarly to (37), we see that for any z, z1, z2 G KR the following estimates hold:

||$z||E3 < A(T) + B(T) ||a(t)H C[o,T] ||u(x,t)||B3 T j

(40)

||$zi - < B(T)R(||ai(t) - a2(t)yc[0 T] + ||ui(x,t) - U2(x,t)||B 3 T). (41)

Then, it follows from (38) together with the estimates (40) and (41) that the operator $ acts in the ball K = KR and is contractive. Therefore, in the ball K = KR the operator $ has a unique fixed point {u, a}, that is a unique solution to the equation (39), i.e. a unique solution to the system (26), (31).

u

derivatives ux(x,t) and uxx(x,t) in .

Now from (28) it is obvious that u^fc(t)(i = 1, 2; k = 1, 2,...) is continuous in [0, T] and from the same relation we get

The function u(x, t), as an element of the space B|T, is continuous and has continuous

£(A* Hui* (t)!c[o,T])M < 2v^Yo (A* |)2 + 2^2a £ (A3 |)2 +

Vfc = 1

Vfc=1

vfc=1

/ t

J 3= (A* |fik(t)|)2dr J +^v/2aT ||a(i)||c[o,T] ( 3= (A* ||uifc(i)||C[o,T])2

Vfe=1

1

2

1

2

OO

1

*

OO

OO

OO

1

2

1

2

or

'to \ 2

E (Ak iKkMllciG/T])2 < 2V2ß7Q ll^///(x)yL2(0ji) + 2v^a ||^///(x)yL2(Qjl) +

vk=i )

1

/ TO \ 2

+2av/2T ||/xxx (x,t)||L2(DT) + ||a(t)||C[Q,T] E(Ak (t)|| C[Q,T ])

Vfc=1

Hence, ut(x, t), utx(x, t), wixx(x,t) is continuous in

Next, from (29) it follows that w"fc(t)(i = 1, 2; k = 1, 2,...) is continuous in [0, T] and consequently we have:

1 1 /to \ 2 /TO \ 2

11K'fc(t)IU,Ti)2 < 2a V (Af

(E (A| ||<fc(i)|C[Q,T])^ < 2a (E (Ak |uifc(i)|C[Q,T])^ +

1

/TO \ 2

+2ß ]T (Af |uifc(t)|C[oiT])M +2 ||f (x,t) + a (t) u (x, t)||C[Q ^

¿2(0,1)

From the last relation it is obvious that wtt(x,t), uttx(x, t), wiixx(x,t) is continuous in . It is easy to verify that the equation (1) and conditions (2), (3), (6), (7) are satisfied in the ordinary sense. Consequently, {u (x, t) ,a (t)} is a solution to the problem (1)-(3), (6), (7), and by Lemma 2 it is unique in the ball K = KR. □

By Lemma 1 the unique solvability of the initial problem (1)-(5) follows from the theorem. Theorem 2. Let all the conditions of Theorem 1 be fulfilled and

/ <^>(x)dx = 0, / -0(x)dx = 0, y(xo) = h(0), ^(Xq ) = h' (0).

QQ

Then the problem (1)-(5) has a unique classical solution in the ball K = KR(||z||E3 ^ R = A(T)+2) of the space E3 . T

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[10] Dzh.Uizem, Linear and Nonlinear Waves, M., Mir, 1977 (in Russian).

[11] B.M.Budak, A.A.Samarskii, A.N.Tikhonov, A Collection of Problems in Mathematical Physics, M., Nauka, 1972 (in Russian).

О разрешимости одной обратной краевой задачи для уравнения Бусинеска—Лява

Яшар Т. Мегралиев

В 'работе исследована одна обратная краевая задача для уравнения Буссинеска-Лява. Сначала исходная задача сводится к эквивалентной в определенном смысле задаче. С помощью метода Фурье эквивалентная задача сводится к решению системы интегральных уравнений. Далее, с помощью метода сжатых отображений доказываются существование и единственность решения системы интегральных уравнений, которая также является единственным решением эквивалентной задачи. Пользуясь эквивалентностью, доказываются существование и единственность классического решения исходной задачи.

Ключевые слова: обратная краевая задача, уравнения Буссинеска-Лява, метод Фурье, классическое решение.