Abstract

This paper deals with the existence and uniqueness of solutions of a new class of Moore-Gibson-Thompson equation with respect to the nonlocal mixed boundary value problem, source term, and nonnegative memory kernel. Galerkin’s method was the main used tool for proving our result. This work is a generalization of recent homogenous work.

1. Introduction

In this contribution, we are interested to study the existence and uniqueness of solutions of the following problem

Here, and are physical parameters, and is the speed of sound. The convolution term reflects the memory effect of materials due to vicoelasticity, is a given function, and is the relaxation function satisfying

(H1) is a nonincreasing function satisfying where , , and .

(H2) satisfying

(H3)

The phenomena resulting from sound waves (diffraction, interference, reflection) in terms of modeling are very important. As the existence of the third derivative is very important, especially in the field of thermodynamics (EIT), the study of these models is considered the beginning of an in-depth understanding of both convergent and good behavior. From the results extracted, the equation of MGT resulted in nonlinear acoustics, for much depth, see ([1–7]) and especially [8] where equation of MGT appeared for the first time. Also, nonlinear problems of great importance can be considered [9], where Galerkin’s method was applied in solving them, for more depth ([2, 3, 10–13]). Recently, in [14], the authors studied the equation of MGT with memory. Likewise, in [1], the authors used Galerkin’s method to demonstrate the ability to solve a mixed problem of MGT equation in the absence of viscous elasticity and memory. Based on work [9] and the works we mentioned earlier, we want to prove the existence and uniqueness of a weak solution to the problem (1).

We divide this paper into the following: in the second part, we put some definitions and appropriate spaces. Then, we apply Galerkin’s method to prove the existence, and in the fourth part, we demonstrate the uniqueness.

2. Preliminaries

We will define the spaces: and by where

Consider the equation where and stands for the inner product in , is supposed to be a solution of (1) and . Evaluation of the inner product in [9] gives

We give two useful inequalities: (i)Gronwall inequality. Let the nonnegative integrable functions on the interval with the nondecreasing function . If , we have where , hence, (ii)Trace inequality (see [15]). If where Ω is a bounded domain in with smooth boundary , then for any , where .

Definition 1. We call a generalized solution to the problem (1) for each function that fulfills the equation (9) for each .

3. Solvability of the Problem

In this section, we use Galerkin’s method to prove the existence of a generalized solution of our problem.

Theorem 2. If , , , and , then there is at least one generalized solution in to problem (1).

Proof. Let be a fundamental system in , such that .
First, we will give an approximate solution of the problem (1) in the form where are constants given by the conditions, for and can be determined from the relations substitution of (13) into (15), and we find for . From (15) it follows that Let Then, (17) can be written as By differentiating (two times) with respect to , it gives We find a system of differential equations of fifth order with respect to , constant coefficients, and the initial conditions (21). Hence, we obtain a Cauchy problem of linear differential equations with smooth coefficients that is uniquely solvable. Thus, ∀n, ∃u^N (x) satisfying (15).
Now, we prove that u^N is sequence bounded. To do this, we multiply each equation of (15) by the appropriate summing over from 1 to . Hence, by integration the result equality with respect to from 0 to , and , it gives Evaluation of the terms on the LHS of (22) gives So, Thus, Taking into account the equalities (23)–(30) in (22), we end up with Now, multiplying the equations of (15) by , collect them from 1 to and integrating the result with respect to from 0 to , and , we find With the same reasoning in (22), we find A substitution of equalities (33)–(40) in (22) gives Multiplying (32) by and using (41), we get where
With the help of Cauchy and the trace inequalities, we can estimate all the terms in the RHS of (42) that gives Combining inequalities (45)-(60) and equality (44) and make use of the following inequality where and we have where Choosing and sufficiently large By using (2)-(4), the relation (64) reduces to where Using the inequality of Gronwall to (67) and integrating the result from 0 to τ that gives where We deduce from (69) that Hence, is sequence bounded in , and we can extract from it a subsequence for which we use the same notation which converges weakly in to a limit function , and we have to show that is a generalized solution of (1). Since in and in , then .
Now to prove that (15) holds, we multiply each of the relations (15) by a function , . Hence, collect them the obtained equalities ranging from to and integrating the result over on . If we let , then we have for all of the form and
Since Thus, the limit function satisfies (15) for every .
We define the totality of all functions of the form by , with , .
But is dense in , hence the relation (15) holds . Then, we have shown that the limit function is a generalized solution of problem (1) in .