BOUNDARY VALUE PROBLEM WITH INTEGRAL CONDITIONS FOR A LINEAR THIRD-ORDER EQUATION

We prove the existence and uniqueness of a strong solution for a linear third-order equation with integral boundary conditions. The proof uses energy inequalities and the density of the range of the generated operator.


Preliminaries
In this paper, we prove the existence and uniqueness of a strong solution of problem (1.1).For this, we consider the solution of problem (1.1) as a solution of the operator equation Lu = F, where L is the operator with domain of definition , k = 0, 3 and u satisfies conditions (1.1d) and (1.1e).The operator L is considered from E to F, where E is the Banach space of the functions u, u ∈ L 2 (Ω), with the finite norm M. Denche and A. Memou 555 and F is the Hilbert space of the functions F = (f, 0, 0, 0), f ∈ L 2 (Ω), with the finite norm Then we establish an energy inequality and we show that the operator L has the closure L.
From this inequality, we obtain the uniqueness of a strong solution, if it exists, and the equality of the sets R(L) and R(L).Thus, to prove the existence of a strong solution of problem (1.1) for any F ∈ F, it remains to prove that the set R(L) is dense in F.

An energy inequality and its applications
Theorem 3.1.For any function u ∈ D(L), there exists the a priori estimate where with the constant c satisfying sup 556 Boundary value problem with integral conditions where We consider the quadratic form Integrating the last two terms on the right-hand side by parts with respect to x in (3.7) and using the Dirichlet condition (1.1d), we obtain 2Re 1 0 (3.9)

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Integrating each term by parts in (3.9) with respect to t and using the initial and final conditions (1.1b) and (1.1c), we get Re Again, substituting the expression of Mu in (3.11) and using elementary inequality, we get (3.12) By virtue of (1.1a), we have This last inequality combined with (3.12) yields Thus, this inequality implies M. Denche and A. Memou 559 where Then, (3.17) Thus, we obtain the desired inequality.
Lemma 3.2.The operator L from E to F admits a closure.
Proof.Suppose that (u n ) ∈ D(L) is a sequence such that We need to show that F = 0. We introduce the operator with domain D(£ 0 ) consisting of functions v ∈ W 2,3 2 (Ω) satisfying We note that D(£ 0 ) is dense in the Hilbert space obtained by completing L 2 (Ω) with respect to the norm for any function v ∈ D(£ 0 ), it follows that f = 0.
Theorem 3.1 is valid for a strong solution, then we have the inequality Hence we obtain the following corollary.

Solvability of problem (1.1)
To prove the solvability of problem (1.1), it is sufficient to show that R(L) is dense in F. The proof is based on the following lemma.then w = 0.
Proof.Equality (4.1) can be written as follows: For a given w(x, t), we introduce the function v(x, t) such that From (4.3), we conclude that 1 0 v(x, t)dx = 0, and thus, we have where

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Following [23], we introduce the smoothing operators with respect to t, which provide the solutions of the respective problems We also have the following properties: for any g ∈ L 2 (0, T), the functions Substituting the function u in (4.4) by the smoothing function u ε and using the relation where we obtain Passing to the limit, the equality in the relation (4.10) remains true for all functions u ∈ L 2 (Ω) such that , and satisfying condition (1.1d).
The operator A(t) has a continuous inverse in L 2 (0, 1) defined by where Then, we have Then The adjoint of B ε (t) has the form

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where Consequently, equality (4.10) becomes where The left-hand side of (4.17) is a continuous linear functional of u.Hence the function h ε has the derivatives 2 (Ω) and the following conditions are satisfied: From the equality and since the operator (J −1 ε ) * is bounded in L 2 (Ω), for sufficiently small ε, Similarly, we conclude that (∂/∂x)((1 − x)(∂v * ε /∂x)) exists and belongs to L 2 (Ω), and the following conditions are satisfied: Using the properties of smoothing operators, we have and from we have

Corollary 3 . 3 .Corollary 3 . 4 .
A strong solution of problem (1.1) is unique if it exists, and depends continuously on F. The range R(L) of the operator L is closed in F, and R(L) = R(L).

( 4 .
25) Definition 2.1.A solution of the operator equation Lu = F is called a strong solution of problem (1.1).