AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 858516 10.1155/2014/858516 858516 Research Article On an Inverse Boundary Value Problem for a Fourth Order Elliptic Equation with Integral Condition Mehraliyev Yashar T. Janno Jaan Baku State University 1148 Baku Azerbaijan bsu.edu.az 2014 2432014 2014 22 12 2013 16 02 2014 24 3 2014 2014 Copyright © 2014 Yashar T. Mehraliyev. This 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.

An inverse boundary value problem for a fourth order elliptic equation is investigated. At first the initial problem is reduced to the equivalent problem for which the existence and uniqueness theorem of the solution is proved. Further, using these facts, the existence and uniqueness of the classic solution of the initial problem are proved.

1. Introduction

The inverse problems are favorably developing section of up-to-date mathematics. Recently, the inverse problems are widely applied in various fields of science.

Different inverse problems for various types of partial differential equations have been studied in many papers. First of all we note the papers of Tikhonov , Lavrent’ev [2, 3], Denisov , Ivanchov , and their followers.

The goal of our paper is to prove the uniqueness and existence of the solution of a boundary value problem for a fourth order elliptic equation with integral condition.

The inverse problems with an integral predetermination condition for parabolic equations were investigated in .

In the papers  the inverse boundary value problems were investigated for a second order elliptic equation in a rectangular domain.

2. Problem Statement and Its Reduction to Equivalent Problem

Consider the following equation: (1) u t t t t ( x , t ) + u x x x x ( x , t ) = a ( t ) u ( x , t ) + f ( x , t ) in the domain D T = { ( x , t ) : 0 x 1,0 t T } an inverse boundary problem with the boundary conditions (2) u ( x , 0 ) = φ 0 ( x ) , u t ( x , T ) = φ 1 ( x ) , u t t ( x , 0 ) = φ 2 ( x ) ,    u t t t ( x , T ) = φ 3 ( x ) ( 0 x 1 ) , (3) u x ( 0 , t ) = u x ( 1 , t ) = u x x x ( 0 , t ) = 0 , ( 0 t T ) , the integral condition (4) 0 1 u ( x , t ) d x = 0    ( 0 t T ) , and with the additional condition (5) u ( 0 , t ) = h ( t ) ( 0 t T ) , where f ( x , t ) , φ i ( x ) ( i = 0,3 ¯ ) , h ( t ) are the given functions and u ( x , t ) ,    a ( t ) are the required functions.

Definition 1.

The classic solution of problems (1)–(5) is the pair { u ( x , t ) , a ( t ) } of the functions u ( x , t ) and a ( t ) possessing the following properties:

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

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

all the conditions of (1)–(5) are satisfied in the ordinary sense.

For investigating problems (1)–(5), at first consider the following problem: (6) y ( 4 ) ( t ) = a ( t ) y ( t ) ( 0 t T ) , (7) y ( 0 ) = 0 , y ( T ) = 0 , y ′′ ( 0 ) = 0 , y ′′′ ( T ) = 0 , where a ( t ) C [ 0 , T ] is a given function, y = y ( t ) is a desired function, and under the solution of problems (6) and (7) we will understand a function y ( t ) C 4 [ 0 , T ] satisfied in [ 0 , T ] equation (6) and conditions (7).

The following lemma is valid.

Lemma 2 (see [<xref ref-type="bibr" rid="B16a">16</xref>, <xref ref-type="bibr" rid="B16b">17</xref>]).

Let the function a ( t ) C [ 0 , T ] be such that (8) a ( t ) C [ 0 , T ] R = c o n s t . Furthermore, (9) 5 12 T 4 R < 1 . Then problems (6), (7) have only a trivial solution.

Alongside with inverse boundary value problem, consider the following auxiliary inverse boundary value problem. It is required to determine the pair { u ( x , t ) , a ( t ) } of the functions u ( x , t ) and a ( t ) possessing the properties ( 1 ) and ( 2 ) of definition of the classic solution of problems (1)–(5) from relations (1)–(3) and (10) u x x x ( 1 , t ) = 0 ( 0 t T ) , (11) h ( 4 ) ( t ) + u x x x x ( 0 , t ) = a ( t ) h ( t ) + f ( 0 , t ) ( 0 t T ) .

The following lemma is valid.

Lemma 3.

Let φ i ( x ) C [ 0,1 ] ( i = 0,3 ¯ ) h ( t ) C 4 [ 0 , T ] , h ( t ) 0    ( 0 t T ) , f ( x , t ) C ( D T ) , let 0 1 f ( x , t ) d x = 0    ( 0 t T ) , and let the following consistency conditions be fulfilled: (12) 0 1 φ i ( x ) d x = 0 ( i = 0,3 ¯ ) , (13) φ 0 ( 0 ) = h ( 0 ) , φ 1 ( 0 ) = h ( T ) , φ 2 ( 0 ) = h ′′ ( 0 ) , φ 3 ( 0 ) = h ′′′ ( T ) . Then the following statements are true.

Each classic solution u ( x , t ) ,    a ( t ) of problems (1)–(5) is the solution of problems (1)–(3), (10), and (11) as well.

Each solution u ( x , t ) , a ( t ) of problems (1)–(3), (10), and (11) such that (14) 5 12 T 4 a ( t ) C [ 0 , T ] < 1 ,

is the classic solution of problems (1)–(5).

Proof.

Let u ( x , t ) , a ( t ) be a solution of problems (1)–(5). Integrating equation (1) with respect to x from 0 to 1, we have (15) d 4 d t 4 0 1 u ( x , t ) d x + u x x x ( 1 , t ) - u x x x ( 0 , t ) = a ( t ) 0 1 u ( x , t ) d x + 0 1 f ( x , t ) d x ( 0 t T ) . Hence, by means of 0 1 f ( x , t ) d x = 0 ( 0 t T ) and (3) we obtain (10).

Substituting x = 0 in (1), we find (16) u t t t t ( 0 , t ) + u x x x x ( 0 , t ) = a ( t ) u ( 0 , t ) + f ( 0 , t ) ( 0 t T ) .

Further assuming h ( t ) C 4 [ 0 , T ] and differentiating (5) four times, we have (17) u t t t t ( 0 , t ) = h ( 4 ) ( t ) ( 0 t T ) .

Taking into account last relation and condition (5) in (16) we obtain (11).

Now suppose that u ( x , t ) , a ( t ) is a solution of problems (1)–(3), (10), and (11); moreover, (14) is fulfilled. Then, taking into account (3) and (10) in (15) we find (18) d 4 d t 4 0 1 u ( x , t ) d x - a ( t ) 0 1 u ( x , t ) d x = 0 ( 0 t T ) .

By (2) and (12), it is obvious that (19) 0 1 u ( x , 0 ) d x = 0 1 φ 0 ( x ) d x = 0 , 0 1 u t ( x , T ) d x = 0 1 φ 1 ( x ) d x = 0 , 0 1 u t t ( x , 0 ) d x = 0 1 φ 2 ( x ) d x = 0 , 0 1 u t t t ( x , T ) d x = 0 1 φ 3 ( x ) d x = 0 .

Since by Lemma 2, problems (18), (19) have only a trivial solution, then 0 1 u ( x , t ) d x = 0 ( 0 t T ) ; that is, condition (4) is fulfilled.

Further, from (11) and (16) we get (20) d 4 d t 4 ( u ( 0 , t ) - h ( t ) ) = a ( t ) ( u ( 0 , t ) - h ( t ) ) ( 0 t T ) .

By (2) and consistency conditions (13), we have (21) u ( 0,0 ) - h ( 0 ) = φ 0 ( 0 ) - h ( 0 ) = 0 , u t ( 0 , T ) - h ( T ) = φ 1 ( 0 ) - h ( T ) = 0 , u t t ( 0,0 ) - h ′′ ( T ) = φ 2 ( 0 ) - h ′′ ( 0 ) = 0 , u t ( 0 , T ) - h ′′′ ( T ) = φ 1 ( 0 ) - h ′′′ ( T ) = 0 .

From (20) and (21), by Lemma 3 we conclude that condition (5) is fulfilled. The lemma is proved.

3. Investigation of the Existence and Uniqueness of the Classic Solution of the Inverse Boundary Value Problem

We will look for the first component u ( x , t ) of the solution u ( x , t ) , a ( t ) of problems (1)–(3), (10), and (11) in the following form: (22) u ( x , t ) = k = 0 u k ( t ) cos λ k x ( λ k = k π ) , where (23) u k ( t ) = m k 0 1 u ( x , t ) cos λ k x d x ( k = 0,1 , 2 , ) , and moreover, (24) m k = { 1 , k = 0 2 , k = 1,2 , .

Then applying the formal scheme of the Fourier method, from (1), (2) we get (25) u k ( 4 ) ( t ) + λ k 4 u k ( t ) = F k ( t ; u , a ) ( k = 0,1 , 2 , ; 0 t T ) , (26) u k ( 0 ) = φ 0 k , u k ( T ) = φ 1 k , u k ′′ ( 0 ) = φ 2 k , u k ′′′ ( T ) = φ 3 k ( k = 0,1 , 2 , ) , where (27) F k ( t ; u , a ) = f k ( t ) + a ( t ) u k ( t ) , f k ( t ) = m k 0 1 f ( x , t ) cos λ k x d x , φ i k = m k 0 1 φ i ( x ) cos λ k x d x ( i = 0,3 ¯ ; k = 0,1 , 2 , ) . From (25), (26) we obtain (28) u 0 ( t ) = φ 00 + t φ 10 + ( 1 2 t 2 - t T ) φ 20 + ( 1 6 t 3 - 1 2 T 2 t ) φ 30 + 0 T G 0 ( t , τ ) F 0 ( τ ; u ) d τ , (29) u k ( t ) = 1 α k ( T ) × { 0 T 2 ρ k [ - ( 4 ρ k 4 + 1 ) ( c h 2 ρ k T + cos 2 ρ k T ) + c h ρ k t cos ρ k ( 2 T - t ) + c h ρ k ( 2 T - t ) cos ρ k t ( 4 ρ k 4 + 1 ) ] φ 0 k + [ - 2 t ρ k ( 4 ρ K 4 + 1 ) ( c h 2 ρ k T + cos 2 ρ k T ) + 2 ρ k 2 ( s h ρ k ( T + t ) cos ρ k ( T - t ) c h ρ k ( T - t ) sin ρ k ( T + t ) + c h ρ k ( T + t ) sin ρ k ( T - t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) + s h ρ k ( T + t ) cos ρ k ( T - t ) ) 2 t ρ k ] φ 1 k + [ - 1 ρ k ( t 2 - 2 t T ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) - 2 ρ k 3 × ( s h ρ k ( 2 T - t ) sin ρ k t + s h ρ k t ( 2 T - t ) × sin ρ k ( 2 T - t ) ) 1 ρ k ] φ 2 k + [ - 1 ρ k ( t 3 3 - t T 2 ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) + 1 2 ρ k 4 × ( - s h ρ k ( T + t ) cos ρ k ( T - t ) s h ρ k ( T - t ) sin ρ k ( T + t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) - s h ρ k ( T + t ) cos ρ k ( T - t ) ) ( t 3 3 - t T 2 ) ] φ 3 k + 0 T G k ( t , τ ) F k ( τ ; u , a ) d τ } , where (30) G 0 ( t , τ ) = { - t 3 6 + ( T τ - τ 2 2 ) t , t [ 0 , τ ] , - τ 3 6 + ( T t - t 2 2 ) τ , t [ τ , T ] , G k ( t , τ ) = { 1 α k ( T ) [ c h ρ k ( 2 T + t - τ ) sin ρ k ( t - τ ) h + c h ρ k ( 2 T - t - τ ) sin ρ k ( t + τ ) - s h ρ k ( 2 T + t - τ ) cos ρ k ( t - τ ) + s h ρ k ( 2 T - t - τ ) cos ρ k ( t + τ ) - c h ρ k ( t + τ ) sin ρ k ( 2 T - t - τ ) + c h ρ k ( t - τ ) sin ρ k ( 2 T + t - τ ) - s h ρ k ( t + τ ) cos ρ k ( 2 T - t - τ ) - s h ρ k ( t - τ ) cos ρ k ( 2 T + t - τ ) ] , t [ 0 , τ ] , 1 α k ( T ) [ - c h ρ k ( 2 T - t + τ ) sin ρ k ( t - τ ) + c h ρ k ( 2 T - t - τ ) sin ρ k ( t + τ ) - s h ρ k ( 2 T - t + τ ) cos ρ k ( t - τ ) + s h ρ k ( 2 T - t - τ ) cos ρ k ( t + τ ) - c h ρ k ( t + τ ) sin ρ k ( 2 T - t - τ ) + c h ρ k ( t - τ ) sin ρ k ( 2 T - t + τ ) - s h ρ k ( t + τ ) cos ρ k ( 2 T - t - τ ) + s h ρ k ( t - τ ) cos ρ k ( 2 T - t + τ ) ] , t [ τ , T ] , and moreover, (31) α k ( T ) = - 8 ρ k 3 ( c h 2 ρ k T + cos 2 ρ k T ) , ρ k = 2 2 λ k .

After substituting the expressions from (28), (29) into (22), for determining the component of the solution of problems (1)–(3), (10), and (11) we get (32) u ( x , t ) = φ 00 + t φ 10 + ( 1 2 t 2 - t T ) φ 20 + ( 1 6 t 3 - 1 2 T 2 t ) φ 30 + 0 T G 0 ( t , τ ) F 0 ( τ ; u ) d τ + k = 1 { 1 α k ( T ) { 0 T 2 ρ k [ - ( 4 ρ K 4 + 1 ) ( c h 2 ρ k T + cos 2 ρ k T ) + c h ρ k t cos ρ k ( 2 T - t ) + c h ρ k ( 2 T - t ) cos ρ k t ( 4 ρ K 4 + 1 ) ] φ 0 k + [ - 2 t ρ k ( 4 ρ K 4 + 1 ) ( c h 2 ρ k T + cos 2 ρ k T ) + 2 ρ k 2 ( s h ρ k ( T + t ) cos ρ k ( T - t ) c h ρ k ( T - t ) sin ρ k ( T + t )       + c h ρ k ( T + t ) sin ρ k ( T - t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) + s h ρ k ( T + t ) cos ρ k ( T - t ) ) 2 t ρ k ] φ 1 k + [ - 1 ρ k ( t 2 - 2 t T ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) - 2 ρ k 3 × ( sin ρ k ( 2 T - t ) s h ρ k ( 2 T - t ) sin ρ k t + s h ρ k t ( 2 T - t ) × sin ρ k ( 2 T - t ) ) 1 ρ k ] φ 2 k + [ - 1 ρ k ( t 3 3 - t T 2 ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) + 1 2 ρ k 4    × ( s h ρ k ( T + t ) cos ρ k ( T - t ) s h ρ k ( T - t ) sin ρ k ( T + t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) - s h ρ k ( T + t ) cos ρ k ( T - t ) ) 1 ρ k ] φ 3 k + 0 T G k ( t , τ ) F k ( τ ; u , a ) d τ } } cos λ k x , where (33) F k ( t ; u , a ) = f k ( t ) + a ( t ) u k ( t ) = m k 0 1 ( f ( x , t ) + a ( t ) u ( x , t ) ) cos λ k x d x    ( k = 0,1 , 2 , ) .

Now, from (11) and (22) we have (34) a ( t ) = h - 1 ( t ) { h ( 4 ) ( t ) - f ( 0 , t ) + k = 1    λ k 4 u k ( t ) } .

For obtaining an equation for the second component a ( t ) of the solution { u ( x , t ) , a ( t ) } of problems (1)–(3), (10), and (11), substitute expression (29) into (34): (35) a ( t ) = h - 1 ( t ) { h ( 4 ) ( t ) - f ( 0 , t ) } + k = 1 λ k 4 { 0 T 1 α k ( T ) × { 0 T 2 ρ k [ - ( 4 ρ K 4 + 1 ) ( c h 2 ρ k T + cos 2 ρ k T ) + c h ρ k t cos ρ k ( 2 T - t ) + c h ρ k ( 2 T - t ) cos ρ k t ( 4 ρ K 4 + 1 ) ] φ 0 k + [ - 2 t ρ k ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) + 2 ρ k 2 ( s h ρ k ( T + t ) cos ρ k ( T - t ) c h ρ k ( T - t ) sin ρ k ( T + t ) + c h ρ k ( T + t ) sin ρ k ( T - t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) + s h ρ k ( T + t ) × cos ρ k ( T - t ) ) 2 t ρ k ] 2 t ρ k φ 1 k + [ - 1 ρ k ( t 2 - 2 t T ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) - 2 ρ k 3 × ( sin ρ k ( 2 T - t ) s h ρ k ( 2 T - t ) sin ρ k t + s h ρ k t ( 2 T - t ) × sin ρ k ( 2 T - t ) ) 1 ρ k ] φ 2 k + [ - 1 ρ k ( t 3 3 - t T 2 ) ( 4 ρ K 4 + 1 ) × ( c h 2 ρ k T + cos 2 ρ k T ) + 1 2 ρ k 4 × ( s h ρ k ( T + t ) cos ρ k ( T - t ) s h ρ k ( T - t ) sin ρ k ( T + t ) - s h ρ k ( T - t ) cos ρ k ( T + t ) - s h ρ k ( T + t ) × cos ρ k ( T - t ) ) ( t 3 3 - t T 2 ) ] φ 3 k + 0 T G k ( t , τ ) F k ( τ ; u , a ) d τ } } , where (36) F k ( t ; u , a ) = f k ( t ) + a ( t ) u k ( t ) = 2 0 1 ( f ( x , t ) + a ( t ) u ( x , t ) ) cos λ k x d x ( k = 1,2 , ) .

Thus, problems (1)–(3), (10), and (11) were reduced to systems (32), (35) with respect to the unknown functions u ( x , t ) and a ( t ) .

The following lemma is important for studying the uniqueness of the solution of problems (1)–(3), (10), and (11).

Lemma 4.

If { u ( x , t ) , a ( t ) } is any solution of problems (1)–(3), (10), and (11), then the function (37) u k ( t ) = m k 0 1 u ( x , t ) cos λ k x d x ( k = 0,1 , 2 , ) satisfies systems (28), (29) in [ 0 , T ] .

Proof.

Let { u ( x , t ) , a ( t ) } be any solution of problems (1)–(3), (10), and (11). Then, having multiplied the both sides of (1) by the function m k cos λ k x ( k = 0,1 , 2 , ) , integrating the obtained equality with respect to x from 0 to 1, and using the relations (38) m k 0 1 u t t ( x , t ) cos λ k x d x = d 4 d t 4 ( m k 0 1 u ( x , t ) cos λ k x d x ) = u k ( 4 ) ( t ) ( k = 0,1 , 2 , ) , m k 0 1 u x x x x ( x , t ) cos λ k x d x = λ k 4 ( m k 0 1 u ( x , t ) cos λ k x d x ) = λ k 4 u k ( t ) ( k = 0,1 , 2 , ) , we get that (25) is satisfied.

Similarly, from (2) we get that condition (26) is fulfilled.

Thus, u k ( t )    ( k = 0,1 , 2 , ) is a solution of problems (25), (26). Hence, it directly follows that the function   u k ( t )    ( k = 0,1 , 2 , )   satisfies   [ 0 , T ]   in  systems (28), (29). The lemma is proved.

Remark 5.

From Lemma 4 it follows that for proving the uniqueness of the solution of problems (1)–(3), (10), and (11), it suffices to prove the uniqueness of the solution of systems (32), (35).

In order to investigate problems (1)–(3), (10), and (11), consider the following spaces.

Denote B 2 , T 5 the set of all the functions of the form (39) u ( x , t ) = k = 0 u k ( t ) cos λ k x ( λ k = π k ) considered in D T , where each of the functions u k ( t ) ( k = 0,1 , 2 , ) is continuous on [ 0 , T ] and (40) J T ( u ) u 0 ( t ) C [ 0 , T ] + ( k = 1 ( λ k 5 u k ( t ) C [ 0 , T ] ) 2 ) 1 / 2 < .

In this set, we determine the operation of addition and multiplication by the number (real) in the usual way: under the zero element of this set we will understand the function u ( x , t ) 0 on D T , and determine the norm in this set by the following formula: (41) u ( x , t ) B 2 , T 5 = J T ( u ) .

Prove that all these spaces are Banach spaces. Indeed, the validity of the first two axioms of the norms is obvious, and validity of the third axiom of the norm is easily established by means of the summation inequality of Minkovsky; consequently, B 2 , T 5 is a linear normalized space. Now prove its completeness. Let (42) u n ( x , t ) = k = 0 u k , n ( t ) cos λ k x ( n = 1,2 , ) be any sequence fundamental in B 2 , T 5 . Then for any ε > 0 there exists a number n ε such that (43) u n ( x , t ) - u m ( x , t ) B 2 , T 5 = u 0 , n ( t ) - u 0 , m ( t ) C [ 0 , T ] + ( k = 1 ( λ k 5 u k , n ( t ) - u k , m ( t ) C [ 0 , T ] ) 2 ) 1 / 2 < ε n , m n ε .

Consequently, for any fixed k    ( k = 1,2 , ) , (44) u 0 , n ( t ) - u 0 , m ( t ) C [ 0 , T ] < ε , u k , n ( t ) - u k , m ( t ) C [ 0 , T ] < ε n , m n ε .

This means that for the sequence { u 0 , n ( t ) } n - 1 and for any fixed k    ( k = 1,2 , ) , the sequences { u 0 , n ( t ) } n - 1 are fundamental in C [ 0 , T ] and consequently by the completeness of C [ 0 , T ] they converge in the space C [ 0 , T ] : (45) u 0 , n ( t ) C [ 0 , T ] u 0,0 ( t ) C [ 0 , T ]    as    n , u k , n ( t ) C [ 0 , T ] u k , 0 ( t ) C [ 0 , T ]    as n .

Further, by (43), for any fixed number N , (46) u 0 , n ( t ) - u 0 , m ( t ) C [ 0 , T ] + ( k = 1 N ( λ k 5 u k , n ( t ) - u k , m ( t ) C [ 0 , T ] ) 2 ) 1 / 2 < ε n , m n ε .

Using relations (45) and passing to limit as m in (46), we get (47) u 0 , n ( t ) - u 0,0 ( t ) C [ 0 , T ] + ( k = 1 N ( λ k 5 u k , n ( t ) - u k , m ( t ) C [ 0 , T ] ) 2 ) 1 / 2 < ε n n ε .

Hence, by arbitrariness of N (or the same, passing to limit as N ), we get (48) u 0 , n ( t ) - u 0,0 ( t ) C [ 0 , T ] + ( k = 1 ( λ k 5 u k , n ( t ) - u k , 0 ( t ) C [ 0 , T ] ) 2 ) 1 / 2 < ε n n ε .

Accept the denotation (49) u 0 ( x , t ) = k = 0 u k , 0 ( t ) cos λ k x .

Since u 0 ( x , t ) = [ u 0 ( x , t ) - u n ε ( x , t ) ] + u n ε ( x , t ) and by (48) u 0 ( x , t ) - u n ε ( x , t ) B 2 , T 5 and also u n ε ( x , t ) B 2 , T 5 , we get that (50) u 0 ( x , t ) B 2 , T 5 .

Then, by (48) for any ε > 0 there exists a number n ε such that (51) u n ( x , t ) - u 0 ( x , t ) B 2 , T 5 ε n n ε .

And this means that the sequence u n ( x , t ) converges in B 2 , T 5 to the element u 0 ( x , t ) B 2 , T 5 . This proves the completeness and consequently the Banach property of the space B 2 , T 5 .

Denote by E T 5 the space B 2 , T 5 × C [ 0 , T ] of the vector functions z ( x , t ) = { u ( x , t ) , a ( t ) } with the norm (52) z ( x , t ) E T 5 = u ( x , t ) B 2 , T 5 + a ( t ) C [ 0 , T ] .

It is known that B 2 , T 5 and E T 5 are Banach spaces.

Now, in the space E T 5 consider the operator (53) Φ ( u , a ) = { Φ 1 ( u , a ) , Φ 2 ( u , a ) } , where (54) Φ 1 ( u , a ) = u ~ ( x , t ) k = 0 u ~ k ( t ) cos λ k x , Φ 2 ( u , a ) = a ~ ( t ) , u ~ 0 ( t ) , u ~ k ( t ) ( k = 1,2 , ) , and a ~ ( t ) are equal to the right hand sides of (28), (29), and (35), respectively.

It is easy to see that (55) c h 2 ρ k T + cos 2 ρ k T 1 2 c h 2 ρ k T , c h ρ k t c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( 2 T - t ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t T ) , c h ρ k ( T - t ) c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( T + t ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t T ) , s h ρ k t c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( 2 T - t ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t T ) , s h ρ k ( T - t ) c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( T + t ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t T ) , c h ρ k ( 2 T + t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( 2 T - t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t τ T ) , s h ρ k ( 2 T + t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( 2 T - t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t τ T ) , c h ρ k ( t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t τ T ) , s h ρ k ( t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 t τ T ) , c h ρ k ( 2 T - t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( 2 T - t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 τ t T ) , s h ρ k ( 2 T - t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( 2 T - t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 τ t T ) , c h ρ k ( t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , c h ρ k ( t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 τ t T ) , s h ρ k ( t + τ ) c h 2 ρ k T + cos 2 ρ k T 2 , s h ρ k ( t - τ ) c h 2 ρ k T + cos 2 ρ k T 2 ( 0 τ t T ) .

Taking into account these relations, by means of simple transformations we find (56) u ~ 0 ( t ) C [ 0 , T ] | φ 00 | + T | φ 01 | + T 2 | φ 02 | + T 3 | φ 03 | + 7 3 T 3 T ( 0 t | f 0 ( τ ) | 2 d τ ) 1 / 2 + 7 3 T 4 a ( t ) C [ 0 , T ] u 0 ( t ) C [ 0 , T ] , ( k = 1 ( λ k 5 u ~ k ( t ) C [ 0 , T ] ) 2 ) 1 / 2 9 6 4 ( k = 1 ( λ k 5 | φ 0 k | ) 2 ) 1 / 2 + 6 4 ( 5 T + 8 ) ( k = 1 ( λ k 5 | φ 1 k | ) 2 ) 1 / 2 + 6 4 ( 5 T 2 + 2 2 ) ( k = 1 ( λ k 5 | φ 2 k | ) 2 ) 1 / 2 + 6 4 ( 5 T 3 + 8 ) ( k = 1 ( λ k 5 | φ 3 k | ) 2 ) 1 / 2 + 16 3 T ( 0 T k = 1 ( λ k 2 | f k ( τ ) | ) 2 d τ ) 1 / 2 + 16 3 T a ( t ) C [ 0 , T ] ( k = 1 ( λ k 5 u k ( t ) C [ 0 , T ] ) 2 ) 1 / 2 , a ~ ( t ) C [ 0 , T ] h - 1 ( t ) C [ 0 , T ] × { h ( 4 ) ( t ) - f ( 0 , t ) C [ 0 , T ] + ( k = 1 λ k - 2 ) 1 / 2 × [ 9 4 ( k = 1 ( λ k 5 | φ 0 k | ) 2 ) 1 / 2 + 1 4 ( 5 T + 8 ) × ( k = 1 ( λ k 5 | φ 1 k | ) 2 ) 1 / 2 + 1 4 ( 5 T 2 + 2 2 ) × ( k = 1 ( λ k 5 | φ 2 k | ) 2 ) 1 / 2 + 1 4 ( 5 T 3 + 8 ) × ( k = 1 ( λ k 5 | φ 3 k | ) 2 ) 1 / 2 + 8 2 T × ( 0 T k = 1 ( λ k 2 | f k ( τ ) | ) 2 d τ ) 1 / 2 + 8 2 T a ( t ) C [ 0 , T ] × ( k = 1 ( λ k 5 u k ( t ) C [ 0 , T ] ) 2 ) 1 / 2 ] } . Suppose that the data of problems (1)–(3), (10), and (11) satisfy the following conditions:

φ i ( x ) C 4 [ 0,1 ] , φ i ( 5 ) ( x ) L 2 ( 0,1 ) , and φ i ( 0 ) = φ i ( 1 ) = φ i ( 0 ) = φ i ( 1 ) = φ i ( 0 ) = φ i ( 1 ) = φ i ( 4 ) ( 0 ) = φ i ( 4 ) ( 1 ) = 0 ( i = 0,3 ¯ ) .

f ( x , t ) , f x ( x , t ) C ( D T ) , f x x ( x , t ) L 2 ( D T ) , and f x ( 0 , t ) = f x ( 1 , t ) = 0 ( 0 t T ) .

h ( t ) C 4 [ 0 , T ] , h ( t ) 0 ( 0 t T ) .

Then from (56), we get (57) u ~ ( x , t ) B 2 , T 5 A 1 ( T ) + B 1 ( T ) a ( t ) C [ 0 , T ] u ( x , t ) B 2 , T 5 , a ~ ( t ) C [ 0 , T ] A 2 ( T ) + B 2 ( T ) a ( t ) C [ 0 , T ] u ( x , t ) B 2 , T 5 , where (58) A 1 ( T ) = φ 0 ( x ) L 2 ( 0,1 ) + T φ 1 ( x ) L 2 ( 0,1 ) + T 2 φ 2 ( x ) L 2 ( 0,1 ) + T 3 φ 3 ( x ) L 2 ( 0,1 ) + 7 3 T 3 T f ( x , t ) L 2 ( D T ) + 9 6 4 φ 0 ( 6 ) ( x ) L 2 ( 0,1 ) + 6 4 ( 5 T + 8 ) × φ 1 ( 5 ) ( x ) L 2 ( 0,1 ) + 6 4 ( 5 T 2 + 2 2 ) × φ 2 ( 5 ) ( x ) L 2 ( 0,1 ) + 6 4 ( 5 T 3 + 8 ) φ 3 ( 5 ) ( x ) L 2 ( 0,1 ) + 16 3 T f x x ( x , t ) L 2 ( D T ) , B 1 ( T ) = 16 3 T + 7 3 T 4 , A 2 ( T ) = h - 1 ( t ) C [ 0 , T ] × { h ( 4 ) ( t ) - f ( 0 , t ) C [ 0 , T ] + ( k = 1    λ k - 2 )    1 / 2    × [ 1 4 φ 0 ( 5 ) ( x ) L 2 ( 0,1 ) + 1 4 ( 5 T + 8 ) × φ 1 ( 5 ) ( x ) L 2 ( 0,1 ) + 1 4 ( 5 T 2 + 2 2 ) × φ 2 ( 5 ) ( x ) L 2 ( 0,1 ) + 1 4 ( 5 T 3 + 8 ) × φ 3 ( 5 ) ( x ) L 2 ( 0,1 ) + 8 2 T × f x x ( x , t ) L 2 ( D T ) 1 4 ] ( k = 1    λ k - 2 )    1 / 2    } , B 2 ( T ) = h - 1 ( t ) C [ 0 , T ] ( k = 1 λ k - 2 ) 1 / 2 8 2 T .

From inequalities (57) we get (59) u ~ ( x , t ) B 2 , T 5 + a ~ ( t ) C [ 0 , T ] A ( T ) + B ( T ) a ( t ) C [ 0 , T ] u ( x , t ) B 2 , T 5 , where (60) A ( T ) = A 1 ( T ) + A 2 ( T ) , B ( T ) = B 1 ( T ) + B 2 ( T ) .

So we can prove the following theorem.

Theorem 6.

Let conditions (1)–(3) be fulfilled and (61) ( A ( T ) + 2 ) 2 B ( T ) < 1 .

Then problems (1)–(3), (10), and (11) have a unique solution in the ball K = K R ( z E T 5 R = A ( T ) + 2 ) of the space E T 5 .

Proof.

In the space E T 5 consider (62) z = Φ z , where the components Φ i ( u , a ) ( i = 1,2 ) of the operator Φ ( u , a ) are defined from the right sides of (32) and (35).

Consider the operator Φ ( u , a ) in the ball K = K R from E T 3 . Similarly to (59), we get that for any z , z 1 , z 2 K R the following estimates are valid: (63) Φ z E T 5 A ( T ) + B ( T ) a ( t ) C [ 0 , T ] u ( x , t ) B 2 , T 5 A ( T ) + B ( T ) R 2 A ( T ) + B ( T ) ( A ( T ) + 2 ) 2 , Φ z 1 - Φ z 2 E T 5 B ( T ) R ( u 1 ( x , t ) - u 2 ( x , t ) B 2 , T 5 a 1 ( t ) - a 2 ( t ) C [ 0 , T ] + u 1 ( x , t ) - u 2 ( x , t ) B 2 , T 5 ) B ( T ) ( A ( T ) + 2 ) z 1 - z 2 E T 3 .

Then taking into account (61) in (63) it follows that the operator Φ acts in the ball K = K R and is contractive. Therefore, in the ball K = K R the operator Φ has a unique fixed point { u , a } that is a unique solution of (62) in the ball K = K R ; that is, it is a unique solution of systems (32), (35) in the ball K = K R .

The function u ( x , t ) as an element of the space B 2 , T 5 is continuous and has continuous derivatives u x ( x , t ) , u x x ( x , t ) , u x x x ( x , t ) u x x ( x , t ) , u x x x x ( x , t ) in D T .

From (25) it in easy to see that (64) ( k = 1 ( λ k u k ( 4 ) ( t ) C [ 0 , T ] ) 2 ) 1 / 2 2 ( k = 1 ( λ k 5 u k ( t ) C [ 0 , T ] ) 2 ) 1 / 2 + 2 a ( t ) u x ( x , t ) + f x ( x , t ) C [ 0 , T ] L 2 ( 0,1 ) . Hence it follows that u t t t t ( x , t ) is continuous in D T .

It is easy to verify that (1) and conditions (2), (3), (10), and (11) are satisfied in the ordinary sense.

Consequently, u ( x , t ) , a ( t ) is a solution of problems (1)–(3), (10), and (11), and by Lemma 4 it is unique in the ball K = K R . The theorem is proved.

By means of Lemma 3, a unique solvability of initial problems (1)–(5) follows from the last theorem.

Theorem 7.

Let all the conditions of Theorem 6 be fulfilled: (65) 0 1 f ( x , t ) d x = 0 ( 0 t T ) , 0 1 φ i ( x ) d x = 0 ( i = 0,3 ¯ ) , φ 0 ( 0 ) = h ( 0 ) , φ 1 ( 0 ) = h ( T ) , φ 2 ( 0 ) = h ′′ ( 0 ) , φ 3 ( 0 ) = h ′′′ ( T ) , 5 12 ( A ( T ) + 2 ) T 4 < 1 .

Then problems (1)–(5) have a unique classic solution in the ball K = K R ( z E T 5 R = A ( T ) + 2 ) of the space E T 5 .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Tikhonov A. I. On the stability of inverse problems Doklady Akademii Nauk SSSR 1943 39 5 195 198 MR0009685 ZBL0061.23308 Lavrent'ev M. M. On an inverse problem for the wave equation Doklady Akademii Nauk SSSR 1964 157 3 520 521 MR0200619 Lavrent'ev M. M. Romanov V. G. Shishatsky S. P. Ill-posed Problems of Mathematical Physics and Analysis 1980 Moscow, Russia Nauka MR591674 Denisov A. M. Introduction to Theory of Inverse Problems 1994 East Lansing, Mich, USA MSU Ivanchov M. I. Inverse Problems for Equations of Parabolic Type 2003 Lviv, Ukraine VNTL Publishers MR2406459 Īvanchov M. I. Pabirivs'ka N. V. Simultaneous determination of two coefficients in a parabolic equation in the case of nonlocal and integral conditions Ukrainian Mathematical Journal 2001 53 5 589 596 10.1023/A:1012570031242 MR1854549 ZBL0991.35102 Īvanchov M. I. An inverse problem with a free boundary for the heat equation Ukrainian Mathematical Journal 2003 55 7 901 910 10.1023/B:UKMA.0000010607.28568.a7 MR2073860 Prilepko A. I. Kostin A. B. Some inverse problems for parabolic equations with final and integral observation Matematichesky Sbornik 1992 183 4 49 68 10.1070/SM1993v075n02ABEH003394 MR1183397 Prilepko A. I. Tkachenko D. S. Properties of solutions of a parabolic equation and the uniqueness of the solution of the inverse source problem with integral overdetermination Zhurnal Vychislitelnoi Matematiki and Matematicheskoy Fiziki 2003 43 4 562 570 MR1993685 ZBL1078.35138 Kamynin V. L. On an inverse problem of determining the right-hand side of a parabolic equation with the integral overdetermination condition Matematicheskie Zametki 2005 77 4 522 534 10.1007/s11006-005-0047-6 MR2178019 Solov'ev V. V. Inverse problems of source determination for the Poisson equation on the plane Zhurnal Vychislitelnoi Matematiki and Matematicheskoy fiziki 2004 44 5 862 871 MR2096929 Solov'ev V. V. Inverse problems for elliptic equations on the plane. I Differentsial'nye Uravneniya 2006 42 8 1106 1114 10.1134/S0012266106080118 MR2294175 Hajiyev M. M. An inverse problem for a degenerate elliptic equation Application of Functional Analysis Methods in Mathematical Physics Equations 1987 Novosibirsk, Russia Nauk SSSR 66 71 MR1078456 Mehraliyev Y. T. Inverse boundary value problem for a second order elliptic equation with additional integral condition Vestnik Udmurtskogo Universiteta Matematika 2012 1 32 40 MR3015158 ZBL1276.35071 Mehraliyev Y. T. On solvability of an inverse boundary value problem for a second order elliptic equation Vestnic Tverskogo Gosudarstvennogo 2011 23 25 38 MR3015158 ZBL1276.35071 Mehraliyev Y. T. An inverse boundary value problem for a fourth order elliptic equation Proceedings of Institute of Mathematics and Mechanics. National Academy of Sciences of Azerbaijan 2012 36 75 90 MR3015158 ZBL1276.35071 Khudaverdiyev K. I. Veliyev A. A. Investigation of One-Dimensional Mixed Problem for a Class of Pseudohyperbolic equations of Third Order with Nonlinear Operator Right Side 2010 Baku, Azerbaijan Chashyoglu