ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC EQUATIONS WITH NONLOCAL BOUNDARY CONDITIONS IN NONCLASSICAL FUNCTION SPACES

The aim of this paper is to prove the existence, uniqueness, and continuous dependence upon the data of a generalized solution for certain singular parabolic equations with initial and nonlocal boundary conditions. The proof is based on an a priori estimate established in nonclassical function spaces, and on the density of the range of the operator corresponding to the abstract formulation of the considered problem.


1.
Introduction.This paper is devoted to the solvability of a certain singular parabolic problem with a nonlocal boundary condition.It can be a part in the contribution of the development of the a priori estimates method for solving such problems.The questions related to these problems are so miscellaneous that the elaboration of a general theory is still premature.Therefore, the investigation of these problems requires at every time a separate study.
This work can be considered as a continuation of the results of Yurchuk [12], Benuar and Yurchuk [1], Bouziani [2,3,5,4,6], Bouziani and Benouar [7,8], and Mesloub and Bouziani [9], in so far as, on the one hand, the studied equation is parabolic and, on the other hand, the boundary condition is of integral type.
The remainder of the paper is divided into four sections.In Section 2, we give the statement of the problem.Then in Section 3, we first introduce the appropriate function spaces needed in our investigation, the abstract formulation of the problem and the sense of the generalized solution are presented in Section 3.2, and some properties of special smoothing operators are considered in Section 3.3.The uniqueness and the continuous dependence upon the data of a solution are established in Section 4. In Section 5, the existence of the generalized solution is proved.

Statement of the problem.
In the rectangle Q = (0,b) × (0,T ), we consider the singular parabolic equation where b and T are fixed but arbitrary positive numbers, and a(t) is a known function satisfying the following assumption.
We pose the following problem for (2.1): given the data f , Φ, µ, and M, find a function z = z(x, t) subject to the initial condition and the weighted integral condition b 0 We transform problem (2.1), (2.3), (2.4), and (2.5) with inhomogeneous boundary conditions into a problem with homogeneous boundary conditions.For this, we put Then, problem (2.1), (2.3), (2.4), and (2.5) can be transformed as follows: find a function u = u(x, t) satisfying (2.9)

Function spaces.
We first introduce appropriate function spaces.We denote by C 0 (0,b) the vector space of continuous functions with compact support in (0,b).Since such functions are Lebesgue integrable with respect to dx, we can define on C 0 (0,b) the bilinear form ((•, •)) x given by where * x g = b x g(ξ, t)dξ.We recall that ((•, •)) x is a scalar product on C 0 (0,b) for which C 0 (0,b) is not complete.Thus we are led to introduce its completion.p,ρ (0,T )) to be the completion of the space C 0 (0,b) (resp., C 0 (0,T )) for the norm x (0,b) and L 2 ρ (0,T ), respectively; that is, by the norms of functions u from L 2 x (0,b) and L 2 ρ (0,T ) we understand the nonnegative numbers: In this paper, we also use other weighted spaces such as L 2 σ (0,b), L 2 s (0,b), and L 2 r (0,T ), where σ (x) = x 2 , s(x) = √ x, and r (t) = ρ(t) = e ct/2 , which are Hilbert spaces of (classes of) weighted square integrable functions with finite norms: Let H be a Hilbert space with a norm • H .We denote by L 2 (0,T ; H) (resp., L 2 r (0,T ; H)) the set of all measurable abstract functions u(•,t) from (0,T ) into H such that respectively, We write B 1 2,ρ (0,T ; H) for the space of functions from (0,T ) into H which are weighted Bouziani space for the measure dt.It is a Hilbert space for the norm The following inequalities are well known and are frequently used in this paper.We list them here for convenience.Lemma 3.6.For x ∈ (0,b), the following inequalities hold: x (0,b) . (3.14) We are now in a position to give the abstract formulation corresponding to the problem (2.7), (2.8), and (2.9).

Abstract formulation.
We consider problem (2.7), (2.8), and (2.9) as the solution of the abstract equation where L is the operator which maps u(x, t) to the pair of elements ᏸu and u, so that Lu = (ᏸu, u). (3.16) We consider L as an unbounded operator with domain D(L) consisting of all functions u belonging to L 2 (0,T ; B and satisfying conditions (2.9).We complete D(L) in the norm 1/2 : We write F for the Hilbert space L 2 (0,T ; is finite.We consider the operator L with the above domain as a mapping from B into F .Now, we can introduce the concept of a generalized solution of problem (2.7), (2.8), and (2.9).Let L be the closure of the operator L. is called a generalized solution of problem (2.7), (2.8), and (2.9).
To prove the solvability of problem (2.7), (2.8), and (2.9) in the sense of Definition 3.7, we establish the a priori estimate (3.21) It follows from (3.21) that there is a bounded inverse L −1 on the range R(L) of L. However, since we have no information concerning R(L) except that R(L) ⊂ F , we must extend L, so that an a priori estimate like (3.21) holds for the extension.For this, we prove that L admits a closure.Thus we extend (3.21) to u ∈ D( L) by passing to the limit.It follows that the closure procedure for L reduces to the closure of the range R(L) in F , so that R(L) = R( L), and a bounded inverse L−1 exists on R( L), so the uniqueness of a generalized solution.For existence, it remains to prove that R(L) does not have an orthogonal complement in F .

Smoothing operators.
We consider the operators defined by the relations where * t g = T t g(x, τ)dτ.These operators, first proposed by Yurchuk in abstract form in [11], are used as smoothing operators with respect to t [12].They furnish the solutions of the problems respectively.These operators have, for all v ∈ L 2 (0,T ,L 2 (0,b)), the following properties: (P1) the functions where A (t)v = a (t)(∂/∂x)(x(∂u/∂x)).For the proof of these properties, see, for instance, [4].

Uniqueness and continuous dependence.
In this section, we first establish an a priori estimate.The uniqueness and the continuous dependence of the solution upon the data then are direct corollary of it.Theorem 4.1.Under Assumption 2.1, the solution of problem (2.7), (2.8), and (2.9) satisfies the following a priori estimate: where c is a positive constant independent of u.
Proof.We consider the scalar product in L 2 (0,τ; B The standard integration by parts of the second and last terms on the left-hand side of (4.2) leads to dt  In light of the Cauchy inequality and inequality (3.13), the first two terms and the last term in the right-hand side of (4.4) are then majorized as follows: dt. dt. (4.6) Observing that dt, (4.7) it follows by using (3.13) and (3.14) where We eliminate the last term on the right-hand side of (4.8).To do that we use [3, Lemma 3.1] to obtain We show that the operator L admits a closure, that is, the closure of the graph Proof.For the proof, the reader should refer to [10].
We extend the a priori estimate (4.1) to u ∈ D( L) by passing to the limit, that is, From (4.11) we conclude the following corollaries.
5. The existence of the solution.Now we want to prove the solvability of our problem.Our existence theorem reads as follows.Proof.Corollary 4.4 shows that, to prove that (2.7), (2.8), and (2.9) has a generalized solution for each (f , ϕ) ∈ F , it is sufficient to show that R(L) is dense in F .For this we need the following proposition.
Applying property (P6) to the right-hand side of (5.3), we get According to property (P2), it follows that The standard integration by parts with respect to t of the left-hand side of (5.5) leads to The operator A(t) with boundary conditions (2.9) has, on L 2 (0,b), a continuous inverse.Hence, it is easy to see that (5.7) The calculations of A −1 (t), Λ ε (t), and Λ * ε (t) are straightforward but somewhat tedious.We only give their definitions ( The left-hand side of (5.7) shows that the mapping if the function K ε has the following properties: and satisfies Therefore, we conclude from (5.9) and (5.11) that (5.12) For each fixed x ∈ [0,b] and sufficiently small ε, the operator has a continuous inverse operator on L 2 (0,T ).Thus from (5.12) we obtain If we substitute (5.15) into (5.2),we obtain (5.17 and integrating by parts by taking into account (5.16), we get  Returning to the proof of Theorem 5.1.Since F is a Hilbert space, the density of R(L) in F is equivalent to the property that orthogonality of a vector W = (ω, ω 0 ) ∈ F to the range R(L), that is, the identity

(4. 10 )
Since the right-hand side here does not depend on τ; we take the upper bound of the left-hand side on τ from 0 to T ; hence (4.1) holds with c = c

Proposition 4 . 2 .
The operator L : B → F with domain D(L) has a closure.

ex 3
−ct x 4 ca(t) − a (t) ∂ t e cτ v ∂x a(t)v ∂ t e cτ v ∂x dx dt.

ee
the Cauchy inequality to the last term of the above equality gives −cT a(T )x4 ∂ T e ct v ∂x −ct x 4 ca(t) − a (t) − 36x 2 a 2 thus v ≡ 0, hence ω ≡ 0 almost everywhere in Q.This proves Proposition 5.2.
Hence on B, the following mapping is defined and continuous: