A THREE-POINT BOUNDARY VALUE PROBLEM WITH AN INTEGRAL CONDITION FOR A THIRD-ORDER PARTIAL DIFFERENTIAL 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 operator generated.


Preliminairies
In this paper, we prove the existence and uniqueness of a strong solution of the problem (1.1)- (1.5).For this, we consider the solution of problem (1.1)-(1.5)as a solution of the operator equation where the operator L has domain of definition D(L) consisting of functions u ∈ L 2 (Ω) such that (∂ k+1 u/∂t k ∂x)(x,t) ∈ L 2 (Ω), k = 1,3 and satisfing the conditions (1.4)-(1.5).The operator L is considered from E to F, where E is the Banach space consisting of function u ∈ L 2 (Ω), with the finite norm F is the Hilbert space of functions Ᏺ = ( f ,0,0,0), f ∈ L 2 (Ω), with the finite norm C. Latrous and A. Memou 35 where (2.4)

An energy inequality and its application
Theorem 3.1.For any function u ∈ D(L), the a priori estimate where (3.2) where J x u = x l u(x,t)dx.We consider the quadratic form obtained by multiplying (1.1) by exp(−ct)Mu, with the constant c satisfying (3.2), integrating over Ω = (0,1) × (0,T), and taking the real part: By substituting the expression of Mu in (3.4), integrating with respect to x, and using the Dirichlet and integral conditions, we obtain Re

Integrating by parts −2 Re
T 0 1 l exp(−ct)a(x,t)u(∂ 3 u/∂t 3 )dx dt with respect to t, and using the initial conditions, the final conditions, and the elementary inequalities, we obtain C. Latrous and A. Memou 37 From (1.1), we get Combining this last inequality with (3.6) and using the conditions (3.2) yield which is the desired inequality.
It can be proved in a standard way that the operator L : E → F is closable.Let L be the closure of this operator, with the domain of definition D(L).

Definition 3.2. A solution of the operator equation
The a priori estimate (3.1) can be extended to strong solutions, that is, we have the estimate (3.9) This last inequality implies the following corollaries.

Solvability of problem (1.1)-(1.5)
To prove the solvability of problem (1.1)-(1.5) it is sufficient to show that R(L) is dense in F. The proof is based on the following lemma.
Lemma 4.1.Suppose that the function a(x,t) and its derivatives are bounded.Let where holds, for arbitrary u ∈ D 0 (L), and then w = 0.
Proof.The equality (4.1) can be written as follows: for a given w(x,t), where Following [25], we introduce the smoothing operators with respect to t, (J −1 ) = (I − (∂ 3 /∂t 3 )) −1 , and (J −1 ) * = (I + (∂ 3 /∂t 3 )) −1 which provide the solution of the respective problems: And also, we have the following properties: for any u ∈ L 2 (0,T), the function Substituting the function u in (4.3) by the smoothing function u and using the relation The operator A(t) has a continuous inverse in L 2 (0,1) defined by where Then we have 1 l A −1 (t)u = 0, hence, the function J −1 u = u ε can be represented in the form The adjoint of B (t) has the form where where The left-hand side of (4.13) is a continuous linear functional of u, hence the function h has the derivatives ∂h /∂x, (1 − x)(∂h /∂x) ∈ L 2 (Ω), and the condition h (0,t) = 0 is satisfied.

Corollary 3 . 3 .
A strong solution of (1.1)-(1.5)  is unique and depends continuously on Ᏺ.Corollary 3.4.The range R(L) of L is closed in F and R(L) = R(L).38Three-point boundary value problemCorollary 3.4  shows that to prove that problem (1.1)-(1.5)has a strong solution for arbitrary Ᏺ, it suffices to prove that set R(L) is dense in F.