MIXED PROBLEM WITH INTEGRAL BOUNDARY CONDITION FOR A HIGH ORDER MIXED TYPE PARTIAL DIFFERENTIAL EQUATION

In this paper, we study a mixed problem with integral boundary conditions for a high order partial differential equation of mixed type. We prove the existence and uniqueness of the solution. The proof is based on energy inequality, and on the density of the range of the operator generated by the considered problem.


Introduction
In the rectangle , we consider the equation (1) where is bounded for , and has bounded partial derivatives such that and , , for i To equation (1) we add the initial conditions (2) the boundary conditions for , , and integral condition (5) where and are known functions which satisfy the compatibility conditions given in equations ( 3)- (5).

Preliminaries
In this paper, we prove existence and uniqueness of a strong solution of the problem stated in equations ( 1)- (5).For this, we consider the problem in equations ( 1)-( 5) as a solution of the operator equation , where with domain of definition consisting of functions , such that, and, satisfies conditions in problems (3)-( 5); the operator is considered from to , where is the Banach space consisting of all functions satisfying equations ( 3)-( 5), with the finite norm sup (7) and is the Hilbert space of vector-valued functions obtained by completion of the space with respect to the norm (8) where is an arbitrary number such that .We then establish an energy inequality: (9) and we show that the operator has the closure .
A solution of the operator equation is called a strong solution of Definition 1: the problem in equations ( 1)- (5).
From this inequality, we obtain the uniqueness of a strong solution if it exists, and the equality of sets and Thus, to prove the existence of a strong solution of the problem in # # equations ( 1)-( 5 (10).
We note the above lemma holds for weaker conditions on .Remark 1: Lemma 2: For , (11) Proof: Starting from and using elementary inequalities yields (11).
(2) Lemma 3: For satisfying the condition in equation , $ Proof: Integrating the expression exp by parts for using $ elementary inequalities and Lemma 2, we obtain expression (12).where , yields:

An Energy Inequality and Its Consequences
Re exp ( exp From equation ( 1) we have: Using elementary inequalities, we obtain where -By integrating the last term on the right-hand side of expression ( 16) and combining the obtained results with Lemmas 1 and 3, and the inequalities in equations ( 17), (18), and (19), we get: ( exp Using elementary inequalities and condition (14), we obtain: As the left-hand side of equation ( 21) is independent of , by replacing the right-hand side by its upper bound with respect to in the interval , we obtain the desired inequality. ./ Lemma 4: The operator from to admits a closure.
Suppose that is a sequence such that Proof: and We must show that The fact that results directly from 2 the continuity of the trace operators Introducing the operator we note that is dense in the Hilbert space obtained from the completion of with respect to the norm 3 3 Additionally, since where is an arbitrary number such that , then .Proof: The equality (24) can be written as follows: (25) If we introduce the smoothing operators with respect to [14] and , , 8 , 9 then these operators provide the solutions of the problems are in such that and .Moreover, commutes with , so and for .
) by the smoothed function , and using the relation ,