On the numerical solution of a singular second-order thermoelastic system

In this article, we study an initial boundary value problem for a coupled system of thermoelasticity. Some existence and uniqueness results are given. The homotopy analysis method is employed to obtain numerical schemes for solving the posed problem. The efficiency of the derived methods is illustrated through several examples with graphical representations.


Introduction
Linear thermoelastic systems have been studied by many researchers. For example, results dealing with existence and asymptotic behaviors were studied in some literature; 1-6 controllability in Bardos et al. 7 and Zuazua; 8 propagation of singularities in Hrusa and Messaoudi 9 and Racke and Wang; 10 and regularity, decay, and blow up of solutions in Messaoudi 11 and Rivera and Racke. 12 In this article, we are interested in studying the numerical solutions of an initial boundary value problem with Dirichlet conditions for a system of linear thermoelasticity with a coupling parameter. We first give some existence and uniqueness results. Then, we proceed to numerical solutions using the homotopy analysis method (HAM). The precise statement of the given initial boundary value problem is as follows: Let T .0, O = (0, 1), and I = (0, T), and consider the following coupling system System (1) describes a model for the symmetric deformation and temperature distribution in the unit disk. When b = 0, the system decouples, and we have the wave and the heat equations. The appearance of Bessel operator in the first and second equations in system (1) comes from the fact that Laplace operator can be written in that form when we focus for radial solutions and in this case, the functions f , g, u 0 , u 1 , and v 0 must be radial. In this mathematical model, the functions u, v, f , and g represent the displacement, difference absolute temperature, external force, and the heat supply, respectively, and a, b, m are positive constants.
System (1) is associated with the initial conditions and the boundary conditions Mathematics Department, College of Sciences, King Saud University, Riyadh, Saudi Arabia We assume that the data satisfy the compatibility conditions: u 0 (1) = 0, u 1 (1) = 0, and v 0 (1) = 0. Our main concern as mentioned previously is to study numerical solutions of problems (1)- (3). For this purpose, we apply the HAM to derive numerical solutions of system (1) along with the boundary and initial conditions given in equations (2) and (3).
In fact, we obtain a family of series solutions of the given coupled thermoelasticity system by using the initial conditions (partial t-solution). By this method, we could control the convergence region and the rate of the series solutions.
The HAM was initially proposed by Liao 13 in 1992 (see also  ). Since then, it has been employed to solve many types of ordinary and partial differential equations characterizing many problems in applied sciences and engineering (see Bataineh et al. 17 and Hayat et al. 18 and the references therein). A modification of the HAM to solve systems of partial differential equations is presented in Bataineh et al. 17 The rest of the article is organized as follows: in section ''Some existence and unique results,'' we give some existence and uniqueness results. In section ''Application of the HAM,'' we apply the method and give some applications by treating several examples to test the efficiency of the method.

Some existence and uniqueness results
We first introduce some function spaces needed in the sequel. Let L 2 r (O) be the weighted L 2 (O) Hilbert space equipped with the inner product (u, v) L 2 r (Q) = Ð Q xuvdxdt: The space H 1 2, r (O) denotes the Hilbert space obtained by completing C 1 (O) with respect to the norm and with associated inner product The set C(I, Y ) with I = ½0, T , is the space of continuous mappings from I onto a Banach space Y : Our problems (1)-(3) can be considered as the problem of solving the equation GY = F , where Y = (u, v) and F = (F 1 , F 2 ) = (ff , u 0 , u 1 g, fg, v 0 g), and G is the operator given by The operator G : B ! H is considered as an unbounded operator acting on a Banach space B into a Hilbert space H, and with domain denoted by D(G) which is the set of functions Y = (u, v) 2 L 2 (I, L 2 r (O)) 2 for which u t , v t , u tt , u x , v x , u xx , v xx , u tx , v tx 2 L 2 (I, L 2 r (O)) and satisfies the boundary condition (3): The Banach space B is obtained by the closure of D(G) in the norm The set H is the Hilbert space fL 2 is finite. The associated inner product in H is defined by We observe that the mappings Before proceeding to numerical solutions of the given problems (1) À (3), we may state and prove in a short way some existence and uniqueness results. For more details, see literature. [19][20][21] We first establish a priori estimate for the solution from which the uniqueness of the solution follows. Then, based on the density of the range of the operator generated by the problem in consideration, we prove the existence of the solution of problems (1) À (3).
holds for all functions Y = (u, v) 2 D(G).
Proof. We multiply the first equation in system (1) by the operator N 1 = xu t and the second equation by N 2 = xv, then taking into account the boundary and initial conditions, we evaluate the integrals ð Q t xu t udxdt ; Àa The following elementary inequality is essential in our proof By adding side to side the inequalities (5) and (6), using Cauchy-e inequality, discarding the positive fourth term on the left-hand side of (5), applying Gronwall's lemma in Garding,22 and then passing to the supremum with respect to t over ½0, T in the resulted inequality, we obtain in terms of norms where : Then, inequality (4) follows with C = d 1=2 e dT =2 : Let ImG denote the range or the image of the operator G, since we have no information about ImG except that ImG & H, hence, we have to make an extension to G denoted by G, so that we have and ImG = H: It can be shown in a very classical way that the operator G : B ! H has a closure G: In view of inequality (8), we deduce that a strong solution of problems (1) À (3) is unique and depends continuously on the elements F 1 = ff , u 0 , u 1 g and F 2 = fg, u 0 g, and that the image of G is closed in H and coincides with the closure of the image of G. That is, ImG = ImG: Theorem 2. Problems (1) À (3) admit a unique strong solution satisfying and the norms u Proof. To prove the existence of the solution, it is sufficient to show that ImG = H (G is surjective). This means that we have to show that the orthogonal complement of the image of G reduces to zero, that is, fImGg ? = f0g. In fact, this is equivalent to (F , F Ã ) H = 0. Moreover, since H is a Hilbert space, the equality ImG = H holds if and only if from the identity , and there follows that F Ã = 0: For the moment, we assume that the following theorem is valid then, v vanishes almost everywhere in Q.
First, define the functions E j by the relation Let (xu tt , xv t ) be the solution of the system and let We have Replacing the functions w 1 and w 2 , given by (15), in the relation (10), we obtain In the presence of relations (13) and (14), and the boundary conditions (3), equation (16) becomes where Q s = O 3 s, T : By using Cauchy-e inequality and discarding the third and fourth terms on the left-hand side of the obtained inequality, we obtain Integrating (18), we have From the last inequality (19), we deduce that v = (w 1 , w 2 ) = (0, 0) almost everywhere in Q s : Proceeding in this way step by step along cylinders of height s, we prove that v = 0 almost everywhere in Q:

Application of the HAM
First, let us introduce the basic idea of HAM.
Consider a system of differential equations where N i are known operators and u i (x, t) are unknown functions. Then, the zeroth-order deformation equations for system (20) are where q 2 ½0, 1 is an embedding parameter, L i are auxiliary linear operators, u i, 0 (x, t) are initial approximations of the unknown functions u i (x, t), f i (x, t; q) are unknown functions, and h i are nonzero auxiliary parameters used to control and adjust the convergence region, and their permissible values can be determined through the h-curves.
Liu and colleagues [23][24][25] showed that the generalized Taylor series is exactly the usual Taylor series expansion but at another point. This means that we can use the Taylor series expansion at another point to obtain the same result as in the HAM. These results clarified the meaning of the auxiliary parameter h.
In view of equation (21), it is clear that when q = 0 and q = 1, one has Thus, as q deforms from 0 to 1, the functions f i (x, t; q) vary from the initial approximations u i, 0 (x, t) to the exact solutions u i (x, t).
Using Taylor series, one may have where As proved by Liao, 14 the convergence of the power series (22) at q = 1 depends on the choice of the auxiliary parameters, the auxiliary operators, the initial guesses, and the auxiliary functions. If these objects are chosen properly, then for q = 1, equation (22) implies Define the vectors Then, the mth-order deformation equations are defined as follows (see Bataineh et al. 17 ) where Now, the components u i, m (x, t) for m ! 1 can be computed recursively by inverting the linear operators L i in equation (24) along with the conditions from the original problem. For more details, see literature. [13][14][15][16][17] To apply the HAM to the systems (1) À (3), we choose appropriate initial approximations u 0 (x, t) and v 0 (x, t) satisfying the conditions (2)and (3), and the linear operators which satisfy where c i , i = 1, 2, are constants of integration. We also define the two operators N 1 and N 2 by Now, the zeroth-order deformation equations can be constructed as and in view of equation (24), the mth-order deformation equations are and h 1 and h 2 are auxiliary parameters. Now, applying the inverse of the operator L, namely, where c i , i = 1, 2, . . . , 4 are constants of integration which can be determined using the conditions (2)and (3), which lead to c i = 0 for i = 1, 2, . . . , 4: Hence, the series solutions of the systems (1)-(3) are given by

Numerical examples
To test the efficiency of the method, we apply the iterative schemes (27) and (28) to two examples. In the first one, we consider a homogeneous system, and in the second one, we deal with a nonhomogeneous system. In both cases, we consider the same parameter values, namely, a = b = m = 1, and the auxiliary parameters as h 1 = h 2 = h.

Conclusion
Some well-posed results of a linear singular thermoelastic system are obtained. Moreover, the system is solved numerically by HAM. The derived schemes are tested numerically using two examples, one of them is homogeneous and the second one is nonhomogeneous. These examples show that the truncated series solutions converge rapidly just after few terms, showing the efficiency and effectiveness of the HAM method.

Declaration of conflicting interests
The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.