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

Abstract

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 [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.

The inverse problems with an integral predetermination condition for parabolic equations were investigated in [6–10].

In the papers [11–15] 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: in the domain an inverse boundary problem with the boundary conditions the integral condition and with the additional condition where are the given functions and are the required functions.

Definition 1. The classic solution of problems (1)–(5) is the pair of the functions and possessing the following properties:(1)the function is continuous in together with all its derivatives contained in (1);(2)the function is continuous on ;(3)all the conditions of (1)–(5) are satisfied in the ordinary sense.
For investigating problems (1)–(5), at first consider the following problem: where is a given function, is a desired function, and under the solution of problems (6) and (7) we will understand a function satisfied in equation (6) and conditions (7).
The following lemma is valid.

Lemma 2 (see [16, 17]). Let the function be such that Furthermore, 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 of the functions and possessing the properties and of definition of the classic solution of problems (1)–(5) from relations (1)–(3) and

The following lemma is valid.

Lemma 3. Let , let , and let the following consistency conditions be fulfilled: Then the following statements are true.(1)Each classic solution of problems (1)–(5) is the solution of problems (1)–(3), (10), and (11) as well.(2)Each solution of problems (1)–(3), (10), and (11) such that is the classic solution of problems (1)–(5).

Proof. Let be a solution of problems (1)–(5). Integrating equation (1) with respect to from 0 to 1, we have Hence, by means of and (3) we obtain (10).
Substituting in (1), we find
Further assuming and differentiating (5) four times, we have
Taking into account last relation and condition (5) in (16) we obtain (11).
Now suppose that 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
By (2) and (12), it is obvious that
Since by Lemma 2, problems (18), (19) have only a trivial solution, then ; that is, condition (4) is fulfilled.
Further, from (11) and (16) we get
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.

3. 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 where From (25), (26) we obtain where and moreover,

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 where

Now, from (11) and (22) we have

For obtaining an equation for the second component of the solution of problems (1)–(3), (10), and (11), substitute expression (29) into (34): where

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

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

Lemma 4. If is any solution of problems (1)–(3), (10), and (11), then the function satisfies systems (28), (29) in .

Proof. Let be any solution of problems (1)–(3), (10), and (11). Then, having multiplied the both sides of (1) by the function , integrating the obtained equality with respect to from 0 to 1, and using the relations we get that (25) is satisfied.
Similarly, from (2) we get that condition (26) is fulfilled.
Thus, is a solution of problems (25), (26). Hence, it directly follows that the function    satisfies    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 the set of all the functions of the form considered in , where each of the functions is continuous on and

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 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, is a linear normalized space. Now prove its completeness. Let be any sequence fundamental in . Then for any there exists a number such that

Consequently, for any fixed ,

This means that for the sequence and for any fixed , the sequences are fundamental in and consequently by the completeness of they converge in the space :

Further, by (43), for any fixed number ,

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 and by (48) and also , we get that

Then, by (48) for any there exists a number such that

And this means that the sequence converges in to the element . This proves the completeness and consequently the Banach property of the space .

Denote by the space of the vector functions with the norm

It is known that and are Banach spaces.

Now, in the space consider the operator where , and 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 Suppose that the data of problems (1)–(3), (10), and (11) satisfy the following conditions:(1), and .(2), , and .(3).

Then from (56), we get where

From inequalities (57) we get where

So we can prove the following theorem.

Theorem 6. Let conditions (1)–(3) be fulfilled and

Then problems (1)–(3), (10), and (11) have a unique solution in the ball of the space .

Proof. In the space consider where the components of the operator are defined from the right sides of (32) and (35).
Consider the operator in the ball from . Similarly to (59), we get that for any the following estimates are valid:
Then taking into account (61) in (63) it follows that the operator acts in the ball and is contractive. Therefore, in the ball the operator 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 .
The function as an element of the space is continuous and has continuous derivatives in .
From (25) it in easy to see that Hence it follows that is continuous in .
It is easy to verify that (1) and conditions (2), (3), (10), and (11) are satisfied in the ordinary sense.
Consequently, is a solution of problems (1)–(3), (10), and (11), and by Lemma 4 it is unique in the ball . 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:

Then problems (1)–(5) have a unique classic solution in the ball of the space .

Conflict of Interests

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