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Journal of Mathematics
Volume 2013 (2013), Article ID 457631, 8 pages
A Mixed Problem with an Integral Two-Space-Variables Condition for Parabolic Equation with The Bessel Operator
Department of Mathematics and Informatics, The Larbi Ben M'hidi University, 04000 Oum El Bouaghi, Algeria
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.
The importance of boundary value problems with integral boundary condition has been pointed out by Samarskiĭ  and problems with integral conditions for parabolic equations were treated by Kamynin , Ionkin , Yurchuk , Benouar and Yurchuk , Bouziani , Bouziani and Benouar [7, 8], and Mesloub and Bouziani . Other parabolic problems also arise in plasma physics by Samarskiĭ , heat conduction by Cannon , Ionkin , dynamics of ground waters by Nakhushev , Vodakhova , Kartynnik , and Lin . Regular case of this problem is studied in . The problem where the equation contains an operator of the form , instead of Bessel operator, is treated in . Similar problems for second-order parabolic equations are investegated by the potential method in . Later problems with an integral two-space-variables condition for parabolic equations were treated by Marhoune , Marhoune and Lakhal . Motivated by this, we study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator.
2. Setting of the Problem
In the rectangular domain , with , we consider the following equation: with the initial data Neumann boundary condition and the integral condition where is a known function.
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.
Proof. Let be a solution of (1)–(4), we prove that So, multiplying (1) by and integrating with respect to over and and taking into account (4) and (6), we obtain Then, from (3), we obtain Let now be a solution of , we are bound to prove that So, multiplying (1) by and integrating with respect to over and and taking into account we obtain combining the two preceding equations, and from (6) we get
3. 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 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 the weighted Lebesgue space that consists of all measurable functions equipped with the finite norm where . If , are identified with the standard spaces .
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 , with domain of definition consisting of functions such that , , , , 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 for which the norm is finite.
Theorem 2. For any function , one has the estimate where is a positive constant independent of .
Proof. Multiplying (1) by the following :
and integrating over , where ,
Integration by parts, we obtain
and by using the Cauchy's -inequality, we get
Substituting (21)–(22) into , we obtain
Now, by using the conditions in (23), (24), and (25), respectively, we obtain
Or in , by combining (26), (27), and (28), we get
Using Lemma 1 in , one has where
The right-hand side of (30) is independent of , hence replacing the left-hand side by its upper bound with respect to from to , we obtain the desired inequality, where .
4. Existence of Solution
To show the existance of solutions, we prove that is dense in for all and for arbitrary .
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 , one has then vanishes almost everywhere in .
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 , then, satisfies the boundary conditions in . 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 :
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
So, we obtain
Now, by combining over , we get by taking into accont the boundary condition (40) of the function yields
Then, we obtain And thus in , then 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 (47), we have Hence Proposition 5 implies that . Thus (47) takes the form Since the range of the trace operator is everywhere dense in the Hilbert space with the norm the equality (49) implies that . Hence implies . Therefore, the proof of Theorem 4 is complete.
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