1 Introduction

3 Uniqueness of the solution

To prove the uniqueness of solution, we use the ideas formulated in [?]. Thus, we first deduce an integral representation equivalent to (6). Integrating the equation (1) from the problem over the domain and using (2)-(6), we find

(27)
For simplicity considered in (27), case , .
Under assumptions of Theorem 2, (1)-(6) has a unique solution.
Proof. First, we shall show that uniqueness holds in some . Then it will be shown that .
Let and be solutions of problem (1)-(6) and, in addition, let . For each group of solutions, the representation (27).
Considering the difference, we find












(28)
where are solutions between and , i.e.

By the Theorem 2 we have







For the difference , , problems are obtained
(29)
where , , , .
From (29), by the maximum principle, we find
(30)
(31)

We now estimate the components of relation (28):

where , .
Let . Then and we have



(32)

Dividing (32) by , we get

(33)

For the function , in view of (30) and (31), we now obtain:

or
(34)
where
Using (34) for we have

The integral term can be estimated as follows:




Consider an auxiliary problem

whence, by the maximum principle, we get

We introduce the function




This enables us to write

By the maximum principle, we find

Since , we conclude that

Therefore,

If we pass to the limit as in (33), then the right-hand side of (33) tends to zero, and we arrive at a contradiction. Therefore, and, in addition, , , .
We establish the uniqueness of solution of the problem for any as follows:
Let . If , then the problem is solved. Otherwise, we assume that the parameter is bounded and repeat the reasoning presented above in the interval . As a result, we again arrive at a contradiction.
Theorem 3 is proved.