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International Journal of Analysis

Volume 2013 (2013), Article ID 957163, 8 pages

http://dx.doi.org/10.1155/2013/957163

## Mixed Problem with an Integral Two-Space-Variables Condition for a Class of Hyperbolic Equations

Department of Mathematics and Informatics, The Larbi Ben M'hidi University, Oum El Bouaghi 04000, Algeria

Received 21 August 2012; Revised 25 November 2012; Accepted 27 November 2012

Academic Editor: Baruch Cahlon

Copyright © 2013 Taki-Eddine Oussaeif and Abdelfatah Bouziani. 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.

#### Abstract

This paper is devoted to the proof of the existence and uniqueness of the classical solution of mixed problems which combine Neumann condition and integral two-space-variables condition for a class of hyperbolic equations. The proof is based on a priori estimate “energy inequality” and the density of the range of the operator generated by the problem considered.

#### 1. Introduction

The integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow, and population dynamics.

Cannon was the first who drew attention to these problems with an integral one-space-variable condition [1], and their importance has been pointed out by Samarskii [2]. The existence and uniqueness of the classical solution of mixed problem combining a Dirichlet and integral condition for the equation of heat demonstrated by cannon [1] using the potential method.

Always using the potential method, Kamynin established in [3] the existence and uniqueness of the classical solution of a similar problem with a more general representation.

Subsequently, more works related to these problems with an integral one-space-variable have been published, among them, we cite the work of Benouar and Yurchuk [4], Cannon and Van Der Hoek [5, 6], Cannon-Esteva-Van Der Hoek [7], Ionkin [8], Jumarhon and McKee [9], Kartynnik [10], Lin [11], Shi [12] and Yurchuk [13]. In these works, mixed problems related to one-dimensional parabolic equations of second order combining a local condition and an integral condition was discussed. Also, by referring to the articles of Bouziani [14–16] and Bouziani and Benouar [17–19], the authors have studied mixed problems with integral conditions for some partial differential equations, specially hyperbolic equation with integral condition which has been investigated in Bouziani [20].

The present paper is devoted to the study of problems with a boundary integral two-space-variables condition for second-order hyperbolic equation.

#### 2. Setting of the Problem

In the rectangle , with , we consider the hyperbolic equation: where the coefficient is a real-valued function belonging to such that in the rest of the paper, , , , denote strictly positive constants.

we adjoin to (1) the initial conditions the Neumann condition and the integral condition where and are known functions.

We will assume that the function and satisfy a compatibility conditions with (5), that is,

The presence of integral terms in boundary conditions can, in general, greatly complicate the application of standard functional or numerical techniques specially the integral two-space-variables condition. In order to avoid this difficulty, we introduce a technique to transfer this problem to another classically less complicated one which does not contain integral conditions. For that, we establish the following lemma.

Lemma 1. *Problem (1)–(5) is equivalent to the following problem :
*

*Proof. *Let be a solution of (1)–(5), we prove that

So, by integrating (1) with respect to over and and taking into account (6) and (7), we obtain
and we get
The integral condition (5) has nothing to do with the coefficient . Then, (11) imposes

Let now be a solution of , then we are bound to prove that

So, by integrating (1) with respect to over and taking into account that
we obtain
and by integrating (1) with respect to over and taking in consideration
we obtain
By combining the two preceding (15) and (17) and taking into account (7), we get

#### 3. A Priori Estimate and Its Consequences

In this paper, we prove the existence and the uniqueness of the solution of the problem (1)–(5) and of the operator equation where with domain of difinition consisting of functions such that , , , and satisfies conditions (3) and (4); the operator is considered from to , where is the Banach space consisting of all functions having a finite norm and is the Hilbert space consisting of all elements for which the norm is finite.

Theorem 2. *For any function , we have the inequality
**
where is a positive constant independent of .*

*Proof. *Multiplying the (1) by the following :
and integrating over , where ,(1) by integrating over . Consequently,

Employing integration by parts in (24), and taking into account the boundary conditions in , we obtain

Using the Cauchy’s inequality, we get

Substituting (25) and (26) into (24), we obtain

Then, using the Cauchy’s inequality and according to conditions (2), we get
so, we obtain
where

By virtue of Lemma 7.1 in [21] and by using it twice, we find
where
(2) by integrating over . Consequently,

Employing integration by parts in (33) and taking into account the boundary conditions in , we obtain

By virtue of the Cauchy’s inequality, we obtain

Substituting (34) and (35) into (33) and according to conditions (2), we get
where

By virtue of Lemma 7.1 in [21] and by using it twice, we find
where
(3) by integrating over . Consequently,

Employing integration by parts in (40) and taking into account the boundary conditions in , we obtain

Using the Cauchy’s inequality, we get

Substituting (41) and (42) into (40), we obtain

Then, using the Cauchy’s -inequality and according to conditions (2), we get
so, we obtain
where

By virtue of Lemma 7.1 in [21] and by using it twice, we find
where

Combining inequalities (31), (38), and (47), we obtain
where

The right-hand side of (49) is independent of ; hence replacing the left-hand side by its upper bound with respect to from to , we obtain the desired inequality, where .

Corollary 3. *A solution of the problem (1)–(5) is unique if it exists and depends continuously on .*

#### 4. Solvability of the Problem

To show the existence of solutions, we prove that is dense in for all and for arbitrary .

Theorem 4. *Suppose the conditions of Theorem 2 are satisfied. Then, the problem (1)–(5) admits a unique solution .*

*Proof. *First, we prove that is dense in for the special case where is reduced to , where .

Proposition 5. *Let the conditions of Theorem 4 be satisfied. If for and for all , we have
**
then, vanishes almost everywhere in .*

*Proof. *The scalar product of is defined by
the equality (51) can be written as follows:

If we put
where is a constant such that and , then, satisfies the boundary conditions in . As a result of (53), we obtain the equality

The left-hand side of (55) shows that the mapping
is a continuous linear functional of . From the right-hand side of (55), It is true if the function has the following properties:

In terms of the given function and from the equality (55), we give the function in terms of as follows:
and satisfies the same conditions of the function .

Replacing in (55) by its representation (58), we obtain

Integrating by parts the right-hand side of (59) with respect to and and by taking the conditions of the function yields

According to condition (2), we obtain

Integrating by parts the left-hand side of (61) with respect to , and by taking the conditions of the function , we obtain

Then, by (61) and (62) we obtain
And thus in , hence in . This proves Proposition 5.

We return to the proof of Theorem 4. We have already noted that it is sufficient to prove that the set dense in .

Suppose that for some and for all it holds

Then, we must prove that . Putting in (64), we have

Hence, Proposition 5 implies that . Thus, (64) takes the form

Since the ranges of the trace operators and are independents and the ranges of values and are everywhere dense in the Hilbert space with the norm then the equality (66) implies that , (we recall satisfies a compatibility conditions). Hence implies . Therefore, the proof of Theorem 4 is complete.

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