On an Initial Boundary Value Problem for a Class of Odd Higher Order Pseudohyperbolic Integrodifferential Equations

This paper is devoted to the study of the well-posedness of an initial boundary value problem for an odd higher order nonlinear pseudohyperbolic integrodifferential partial differential equation. We associate to the equation n nonlocal conditions and n + 1 classical conditions. Upon some a priori estimates and density arguments, we first establish the existence and uniqueness of the strongly generalized solution in a class of a certain type of Sobolev spaces for the associated linear mixed problem. On the basis of the obtained results for the linear problem, we apply an iterative process in order to establish the well-posedness of the nonlinear problem.


Introduction
Classical and nonclassical and local and nonlocal initial boundary value problems for partial differential equations are widely studied and athre being studied nowadays.One of the most important and crucial tools to be applied to partial differential equations is functional analysis.It is the universal language of mathematics.No serious study in partial differential equations, mathematical physics, numerical analysis, mathematical economics, or control theory is conceivable without a broad solicitation to methods and results of the functional analysis and its applications.
The main objective of this research work is to develop one of the powerful methods of functional analysis, namely, the energy inequality method for a certain classes of partial differential equations with nonlocal constraints of convolution type in some functional spaces of Sobolev type.This method, based on the ideas of Petrovski [1], Leray [2], Garding [3], and presented on a method form by Dezin [4], was used to investigate and study different categories of mixed problems related to elliptic, parabolic, and hyperbolic equations [5][6][7][8][9][10][11][12], mixed equations [13][14][15], nonclassical equations [16,17], and operational equations [18,19], with classical conditions of types: Cauchy, Dirichlet, Neumann, and Robinson.
Mixed nonlocal problems are especially inspired from modern physics and technological sciences and they describe many physical and biological phenomena.That is in terms of applications, nonlocal mixed problems are widely applied in medical science, biological processes, chemical reaction diffusion, heat conduction processes, population dynamics, thermoelasticity, control theory, and in so many other domains of research.It is worth to mention that for these types of problems, we cannot measure the data directly on the boundary, but we only know the average value of the solution on the domain.
In Section 2, we pose and set the problem to be solved.In Section 3, we give some notations, introduce the functional frame, and state some important inequalities that will be used in the sequel.Section 4 is devoted to the proof of the uniqueness of the solution of the associated linear problem.In Section 5, we establish and prove the existence of solution of the posed associated linear problem.In the last Section, Section 6, we solve the nonlinear problem.On the basis of the results obtained in Sections 4 and 5, and by using an iterative process, we prove the existence and uniqueness of

Problem Setting
In the rectangle  = (0, ) × (0, ), where 0 <  < ∞ and 0 <  < ∞, we consider the nonlinear higher order pseudohyperbolic differential equation of odd order where In ( 1),  is a given function which will be specified later on and () is a function satisfying the conditions for all  ∈ [0, ] and all constants   ;  = 0, 4 are strictly positive.
To (1), we associate the initial conditions the Dirichlet boundary condition the Neumann boundary conditions and the nonlocal conditions where the data functions  1 and  2 satisfy the compatibility conditions In this paper, we are concerned with the proof of wellposedness of the nonlinear nonlocal initial boundary value problem ( 1)-( 6) in some weighted Sobolev spaces.
The main tools used in our proofs are mainly based on some iterative processes, some priori bounds, and some density arguments.

Functional Framework, Notations, and Some Inequalities
For the investigation of problem ( 1)-( 6), we need the following function spaces.Let  2 () be the usual Hilbert space of square integrable functions and let (0, ) =   2 (0, ) [24] be the Hilbert space of Sobolev type constituted of functions  ∈  2 (0, ) if  = 0 and of functions  such that and with associated norm One denotes by  2 (0, ; (0, )) the set of all abstract strongly measurable functions  on [0, ] into (0, ) such that The space  2 (0, ; (0, )) is a Hilbert space having the inner product One can write problem (1)-( 6) in an operator form  = F = (, 1 ,  2 ), where  = (L, ℓ 1 , ℓ 2 ) is an unbounded operator with domain (), acting from a Banach space  into a Hilbert space  constructed as below.One defines the domain of the operator  as the set The space  is the Hilbert  2 () × (0, ) × (0, ) of multivalued functions F = (,
We establish a priori bound from which we deduce the uniqueness of solution of problem (17).Theorem 3. If the coefficients () satisfy condition (H1), then there exists a positive constant  independent of  such that for all  ∈ ().
Proof.See Appendix B.
We denote by  the closure of the operator  and by () the domain of definition of  and define the strong solution of problem (17) as the solution of the operator equation  = F.
Inequality (18) can be extended to We can deduce from ( 19) that the strong solution of problem ( 17) is unique if it exists and depends continuously on F = (,  1 ,  2 ) ∈  and that the image Im() of the operator  coincides with the set Im().

Solvability of the Associated Linear Problem
Theorem 5. Assume that conditions H1 and H2 are hold.Then problem (17) admits a unique strong solution satisfying  ∈ (0, ; (0, )),   ∈ (0, ; (0, )) and ,   depend continuously on the given data and verify Proof.Since Im() ⊂  is closed and Im() = Im(), then in order to prove the existence of the strong solution, we have to show that Im() = .We first prove it in the following special case: Theorem 6.If conditions of Theorem 3 are satisfied and for Ψ ∈  2 (), we have Proof.We first define the function (, ) by the relation We now consider the equation and define  by Relations ( 23) and ( 24) imply that  is in   () ⊆  0 (), where   () = {/ ∈ (),  = 0 for  ≤ }.
We now have The following lemma shows that Ψ(, ) given by ( 25) is in  2 (  ), where   = (0, ) × (, ).Lemma 7. If conditions of Theorem 6 are satisfied, then the function  defined by the relations (23) and (24) has derivatives up to third order which included are in  2 (  ).

Proof. See Appendix B.
We now continue to prove Theorem 6.We replace Ψ given by ( 25) in ( 21) to get Straight forward successive integration by parts of the two terms in (26) gives Substitution of ( 27) into (26) yields By dropping the second term on the left-hand side (28) and by using conditions H1 and H2, we obtain We now consider the two elementary inequalities Combination of inequalities ( 29)-( 30) leads to where We now introduce a new function  defined by (, ) = ∫     , then (, )/ = (, ) − (, ), and (, )/ = (, ), and we have If we choose  0 > 0 such that 1 − 2( −  0 ) = 1/2, then for all  ∈ [ −  0 , ], inequality (33) implies that Inequality (34) can be written in the form of where It follows from (35) that () exp(4 7 ) ≤ 0 from which it follows that Ψ = 0 almost everywhere in  − 0 .By reiterating the same procedure, we deduce that Ψ = 0 a.e., in .We now continue the proof of Theorem 5.

The Nonlinear Problem
On the basis of the results obtained for the linear case, we are now able to establish the existence and uniqueness results for the nonlinear problem (1)- (6).
By using the above conditions on V and , we can write (45) in the form of On the other hand, we have .
It is obvious that from the partial differential equation in (50) we have and we also have By using conditions on V, evaluation of the right-hand side of (62) gives Combination of ( 62) and (65) leads to Application of Cauchy Shwartz to the two terms of the righthand side of (66) gives It follows from (66)-(67) that On the other hand we have Now taking into account inequalities (68) and (69) and passing to limit inequality (64) as  → ∞, we obtain which is exactly inequality (48).Now since  ∈  1 (0, ;  2 (0, )), then ∫  0 (  (, )/  ) ∈ (), and we conclude that   (, )/  = 0,  =  + 1, 2, almost everywhere.

Theorem 10 .
Assume that condition (44) is fulfilled, then the initial boundary value problem (40) admits a unique solution.