A Mixed Problem with an Integral Two-Space-Variables Condition for Parabolic Equation with The Bessel Operator

Received 27 October 2012; Revised 12 March 2013; Accepted 29 March 2013 Academic Editor: Mohsen Tadi Copyright © 2013 Bouziani Abdelfatah et al. 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. We study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator. The existence and uniqueness of the solution in functional weighted Sobolev space are proved. The proof is based on a priori estimate “energy inequality” and the density of the range of the operator generated by the problem considered.

We shall assume that the functions  and  satisfy the compatibility condition with (4), that is, The presence of integral terms in boundary condition can, in general, greatly complicate the application of standard functional or numerical techniques, specially the integral two-space-variables condition.Then to avoid this difficulty, we introduce a technique for transfering this problem to another classically less complicated one that does not contain integral conditions.For that, we establish the following lemma.
Lemma 1. Problem (1)-( 4) is equivalent to the following problem (PR): Proof.Let (, ) be a solution of ( 1)-( 4), we prove that So, multiplying (1) by  and integrating with respect to  over (0, ) and (, 1) and taking into account ( 4) and ( 6), we obtain Then, from (3), we obtain Let now (, ) be a solution of (PR), we are bound to prove that So, multiplying (1) by  and integrating with respect to  over (0, ) and (, 1) and taking into account we obtain combining the two preceding equations, and from (6) we get

A Priori Estimate
The method used here is one of the most efficient functional analysis methods in solving partial differential equations with integral conditions, the so-called a priori estimate method or the energy-integral method.This method is essentially based on the construction of multiplicators for each specific given problem, which provides the a priori estimate from which it is possible to establish the solvability of the posed problem.More precisely, the proof is based on an energy inequality and the density of the range of the operator generated by the abstract formulation of the stated problem.But here we use the energy inequality method for the equivalent problem (PR) given in Lemma 1; so to investigate the posed problem, we introduce the needed function spaces.We introduce function spaces needed in our investigation.We denote by  2  (Ω) the weighted Lebesgue space that consists of all measurable functions  equipped with the finite norm where () = ( 2 + ).If () = 1,  2  (Ω) are identified with the standard spaces  2 (Ω).
In this paper, we prove the existence and the uniqueness for solution of the problem (1)-( 4) as a solution of the operator equation where  = (L, ℓ), with domain of definition  consisting of functions  ∈  2  (Ω) such that /, /,  2 / 2 ,  2 / ∈  2  (Ω), and  satisfy condition (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 F = (, ) for which the norm is finite.

Theorem 2. For any function 𝑢 ∈ 𝐵, one has the estimate
where  is a positive constant independent of .

Existence of Solution
To show the existance of solutions, we prove that () is dense in  for all  ∈  and for arbitrary F = (, ) ∈ .Proof.First we prove that () is dense in  for the special case where () ≡  is reduced to  0 (), where  0 () = {,  ∈ () : ℓ = 0}.Proposition 5. Let the conditions of Theorem 4 be satisfied.if for  ∈  2 (Ω) and for all  ∈  0 (), one has then  vanishes almost everywhere in Ω.
Proof.The scalar product of  is defined by the equality (32) can be written as follows: if we put where  is a strictly positive constant and , /, (/)(I  (  )/) ∈  2 (Ω), then,  satisfies the boundary conditions in (PR).As a result of (34), we obtain the equality The left-hand side of (36) shows that the mapping is a continuous linear functional of .From the right-hand side of (36) there follows that is true if the function  has the following properties: In terms of the given function  and from the equality (36), we give the function  in terms of  as follows: and  satisfies the same conditions of the function  in (PR): Replacing  in (36) by its representation (39) and integrating by parts each term of (36) and by taking the conditions (40) and (41), we obtain    (45) > 0,  +  = 1,  ∈ (0, ) ,