1. Introduction

Certain problems of modern physics and technology can be effectively described in terms of nonlocal problems with integral conditions for partial differential equations.These nonlocal conditions arise mainly when the data on the boundary cannot be measured directly. Motivated by this, we consider in the rectangular domain the following nonclassical boundary value problem of finding a solution such that where the function and its derivative are bounded on the interval

Here, we assume that the known function satisfies the conditions given in (1.4) and (1.5), that is,

When considering the classical solution of the problem (1.1)–(1.5), along with (1.5), there should be the fulfilled conditions:

Mathematical modelling of different phenomena leads to problems with nonlocal or integral boundary conditions. Such a condition occurs in the case when one measures an averaged value of some parameter inside the domain. This amounts to the specification of the energy or mass contained in a portion of the conductor or porous medium as a function of time. This problems arise in plasma physics, heat conduction, biology and demography, as well as modelling of technological process, see, for example, [1–5]. Boundary-value problems for parabolic equations with integral boundary condition are investigated by Batten [6], Bouziani and Benouar [7], Cannon [8, 9], Cannon, Perez Esteva and van der Hoek [10], Ionkin [11], Kamynin [12], Shi and Shillor [13], Shi [4], Marhoune and Bouzit [14], Marhoune and Hameida [15], Yurchuk [16], and many references therein. The problem with one-variable (resp., two-variable) boundary integral type condition is studied in [5] and by Marhoune and Latrous [17] (resp., in Marhoune [2]).

Mention that in the cited paper [16], the author proved the existence, uniqueness, and continuous dependence of a stronge solution in weighted Sobolev spaces to the problem

under the following conditions:

This last integral condition in the form

arises, for example, in biochemistry in which is a constant, and in this case is known as the conservation of protein [18]. Further, in [5], the author studied a similar problem with the weak integral condition

The same problem with the new integral condition

was investigated in [2]. The present paper is an extension in the same direction. By constructing a suitable multiplicator, we will try to establish existence and uniqueness of solution of problem (1.1)–(1.5). Note that the multivariables integral type condition (1.5) is considerably much weaker and better than that used in [2]. In fact, some physical problems have motivated specialists to consider nonlocal integral condition (1.5), which tells us the integral total effect of the solution over several independent portions , and of interval at certain time that give this effect over the entire or part of this interval.

We associate with (1.1)–(1.5) the operator defined from into where is the Banach space of functions satisfying (1.4) and (1.5), with the finite norm and is the Hilbert space of vector-valued functions obtained by completion of the space with respect to the norm

where

Using the energy inequalities method proposed in [16], we establish two-sided a priori estimates. Then, we prove that the operator is a linear homeomorphism between the spaces and .

2. Two-Sided A Priori Estimates

Theorem 2.1. For any function one has the a priori estimate where the constant is independent of . In fact,

Proof. Using (1.1) and initial condition (1.3), we obtain Combining the inequalities in (2.2), we obtain (2.1) for .

Theorem 2.2. For any function , one has the a priori estimate with the constant and is such that

Before proving this theorem, we need the following lemma.

Lemma 2.3 (see [19]). For one has

Proof of Theorem 2.2. Define We consider for the quadratic formula with the constant satisfying (2.5), obtained by multiplying (1.1) by , by integrating over , where , with , and by taking the real part. Integrating by parts in (2.8) by report to with the use of boundary conditions (1.4) and (1.5), we obtain On the other hand, by using the elementary inequalities we get Again, integrating by parts the second, third, fourth, and fifth terms of the right-hand side of the inequality (2.10) by report to and taking into account the initial condition (1.3) and (2.5) gives Using (2.11) in (2.10), we get By using the -inequalities on the first integral in the left-hand side of (2.12) and Lemma 2.3, we obtain Now, from (1.1), we have Combining inequalities (2.13) and (2.14), we get As the right-hand side of (2.15) is independent of , by replacing the left-hand side by its upper bound with respect to in the interval , we obtain the desired inequality.

3. Solvability of the Problem

From estimates (2.1) and (2.3), it follows that the operator is continuous and its range is closed in Therefore, the inverse operator exists and is continuous from the closed subspace onto , which means that is an homeomorphism from onto . To obtain the uniqueness of solution, it remains to show that . The proof is based on the following lemma.

Lemma 3.1. Let If for and some , one has where then

Proof. From (3.2) we have Now, for given , we introduce the function Integrating by parts with respect to , we obtain which implies that Then, from (3.4), we obtain where If we introduce the smoothing operators with respect to [16], and then these operators provide the solutions of the respective problems: and also have the following properties: for any , the functions and are in such that and Morever, commutes with so and for
Putting in (3.8), where the constant satisfies , and using (3.11), we obtain
Integrating by parts each term in the right-hand side of (3.12) and taking the real parts yield Using -inequalities, we obtain Combining (3.13) and (3.15), we get From (3.16), we deduce that Then, for we obtain We conclude that , hence , which ends the proof of the the lemma.

Theorem 3.2. The range of coincides with .

Proof. Since is a Hilbert space, we have if and only if the relation for arbitrary and , implies that and
Putting in (3.19), we conclude from Lemma 3.1 that , where then .
Taking in (3.19) yields The range of the operator is everywhere dense in Hilbert space with the norm hence,

Acknowledgment

The authors would like to thank the referee for helpful suggestions and comments.