On an Inverse Boundary Value Problem for a Fourth Order Elliptic Equation with Integral Condition

and Applied Analysis 3 By (2) and consistency conditions (13), we have u (0, 0) − h (0) = φ 0 (0) − h (0) = 0, u t (0, T) − h 󸀠 (T) = φ 1 (0) − h 󸀠 (T) = 0, u tt (0, 0) − h 󸀠󸀠 (T) = φ 2 (0) − h 󸀠󸀠 (0) = 0, u t (0, T) − h 󸀠󸀠󸀠 (T) = φ 1 (0) − h 󸀠󸀠󸀠 (T) = 0. (21) 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: u (x, t) = ∞ ∑ k=0 u k (t) cos λ k x (λ k = kπ) , (22) where


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 [1], Lavrent' ev [2,3], Denisov [4], Ivanchov [5], 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.
In the papers [11][12][13][14][15] the inverse boundary value problems were investigated for a second order elliptic equation in a rectangular domain.

Problem Statement and Its Reduction to Equivalent Problem
Consider the following equation: and with the additional condition (0, ) = ℎ ( ) (0 ≤ ≤ ) , (3) all the conditions of (1)- (5) are satisfied in the ordinary sense.
Alongside with inverse boundary value problem, consider the following auxiliary inverse boundary value problem. It is required to determine the pair { ( , ), ( )} of the functions ( , ) and ( ) possessing the properties (1) and (2) of definition of the classic solution of problems (1)-(5) from relations (1)- (3) and The following lemma is valid.
Further, from (11) and (16) we get Abstract and Applied Analysis 3 By (2) and consistency conditions (13), we have From (20) and (21), by Lemma 3 we conclude that condition (5) is fulfilled. The lemma is proved.

Investigation of the Existence and Uniqueness of the Classic Solution of the Inverse Boundary Value Problem
We will look for the first component ( , ) of the solution ( , ), ( ) of problems (1)-(3), (10), and (11) in the following form: where and moreover, Then applying the formal scheme of the Fourier method, from (1), (2) we get (4) ( ) + 4 ( ) = ( ; , ) From (25), (26) we obtain 0 ( ) = 00 + 10 + ( Abstract and Applied Analysis where and moreover, Now, from (11) and (22) we get that (25) 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 ( , ) ≡ 0 on , and determine the norm in this set by the following formula: 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, 5 2, is a linear normalized space. Now prove its completeness. Let Using relations (45) and passing to limit as → ∞ in (46), we get Hence, by arbitrariness of (or the same, passing to limit as → ∞), we get Accept the denotation Since 0 ( , ) = [ 0 ( , )− ( , )]+ ( , ) and by (48) 0 ( , ) − ( , ) ∈ 5 2, and also ( , ) ∈ 5 2, , we get that Then, by (48) for any > 0 there exists a number such that And this means that the sequence ( , ) converges in 5 2, to the element 0 ( , ) ∈ 5 2, . This proves the completeness and consequently the Banach property of the space It is known that 5 2, and 5 are Banach spaces. Now, in the space 5 consider the operator where 0 ( ),̃( ) ( = 1, 2, . . .), and̃( ) are equal to the right hand sides of (28), (29), and (35), respectively.
It is easy to see that Taking into account these relations, by means of simple transformations we find Abstract and Applied Analysis Suppose that the data of problems (1) where 1 ( ) = 0 ( ) 2 (0,1) + 1 ( ) 2 (0,1) + 2 2 ( ) 2 (0,1) 0 ( ) 2 (0,1) + √ 6 4 (5 + 8) From inequalities (57) we get where So we can prove the following theorem. Proof. In the space 5 consider where the components Φ ( , ) ( = 1, 2) of the operator Φ( , ) are defined from the right sides of (32) and (35). Consider the operator Φ( , ) in the ball = from 3 . Similarly to (59), we get that for any , 1 , 2 ∈ the following estimates are valid: Then taking into account (61) in (63) it follows that the operator Φ acts in the ball = and is contractive. Therefore, in the ball = the operator Φ has a unique fixed point { , } that is a unique solution of (62) in the ball = ; that is, it is a unique solution of systems (32), (35) in the ball = .
By means of Lemma 3, a unique solvability of initial problems (1)-(5) follows from the last theorem.