Numerical Solutions of the Second-Order One-Dimensional Telegraph Equation Based on Reproducing Kernel Hilbert Space Method

Abstract

We investigate the effectiveness of reproducing kernel method (RKM) in solving partial differential equations. We propose a reproducing kernel method for solving the telegraph equation with initial and boundary conditions based on reproducing kernel theory. Its exact solution is represented in the form of a series in reproducing kernel Hilbert space. Some numerical examples are given in order to demonstrate the accuracy of this method. The results obtained from this method are compared with the exact solutions and other methods. Results of numerical examples show that this method is simple, effective, and easy to use.

1. Introduction

The hyperbolic partial differential equations model the vibrations of structures (e.g., buildings, beams, and machines). These equations are the basis for fundamental equations of atomic physics. In this paper, we consider the telegraph equation of the form 𝜕²𝑢𝜕𝑡²(𝑥,𝑡)+2𝛼𝜕𝑢()𝜕𝑡𝑥,𝑡+𝛽²𝑢()=𝜕𝑥,𝑡₂𝑢𝜕𝑥²(𝑥,𝑡)+𝑓(𝑥,𝑡),0≤𝑥,𝑡≤1,𝛼>𝛽≥0, (1) with initial conditions 𝑢(𝑥,0)=𝜑₁(𝑥),𝑢_𝑡(𝑥,0)=𝜑₂(𝑥), (2) and appropriate boundary conditions 𝑢(0,𝑡)=𝑔₀(𝑡),𝑢(1,𝑡)=𝑔₁(𝑡),𝑡≥0 (3) by using reproducing kernel method (RKM). In recent years, much attention has been given in the literature to the development, analysis, and implementation of stable methods for the numerical solution of (1)–(3) [1–3]. Mohanty carried out a new technique to solve the linear one-space-dimensional hyperbolic equation (1) [4]. High-order accurate method for solving linear hyperbolic equation is presented in [5]. A compact finite difference approximation of fourth order for discretizing spatial derivative of linear hyperbolic equation and a collocation method for the time component are used in [6]. A numerical scheme is developed to solve the one-dimensional hyperbolic telegraph equation using the collocation points and approximating the solution using thin plate splines radial basis function [7]. Several test problems were given, and the results of numerical experiments were compared with analytical solutions to confirm the good accuracy of their scheme. Yao [8] investigated a nonlinear hyperbolic telegraph equation with an integral condition by reproducing kernel space at 𝛼=𝛽=0. Yousefi presented a numerical method for solving the one-dimensional hyperbolic telegraph equation by using Legendre multiwavelet Galerkin method [9]. Dehghan and Lakestani presented a numerical technique for the solution of the second-order one-dimensional linear hyperbolic equation [10]. Lakestani and Saray used interpolating scaling functions for solving (1)–(3) [11]. Dehghan provided a solution of the second-order one-dimensional hyperbolic telegraph equation by using the dual reciprocity boundary integral equation (DRBIE) method [12]. The problem has explicit solution that can be obtained by the method of separation of variables in [13].

In this paper, the problem is solved easily and elegantly by using RKM. The technique has many advantages over the classical techniques. It also avoids discretization and provides an efficient numerical solution with high accuracy, minimal calculation, and avoidance of physically unrealistic assumptions. In the next section, we will describe the procedure of this method.

The theory of reproducing kernels was used for the first time at the beginning of the 20th century by Zaremba in his work on boundary value problems for harmonic and biharmonic functions [14]. Reproducing kernel theory has important application in numerical analysis, differential equations, probability, and statistics. Recently, using the RKM, some authors discussed telegraph equation [15], Troesch's porblem [16], MHD Jeffery-Hamel flow [17], Bratu’s problem [18], KdV equation [19], fractional differential equation [20], nonlinear oscillator with discontinuity [21], nonlinear two-point boundary value problems [22], integral equations [23], and nonlinear partial differential equations [24].

The paper is organized as follows. Section 2 introduces several reproducing kernel spaces. The representation in 𝑊(Ω) and a linear operator are presented in Section 3. Section 4 provides the main results. The exact and approximate solutions of (1)–(3) and an iterative method are developed for the kind of problems in the reproducing kernel space. We have proved that the approximate solution converges to the exact solution uniformly. Numerical experiments are illustrated in Section 5. Some conclusions are given in Section 6.

2. Reproducing Kernel Spaces

In this section, some useful reproducing kernel spaces are defined.

Definition 1 (reproducing kernel function). Let 𝐸≠∅. A function 𝐾∶𝐸×𝐸→ℂ is called a reproducing kernel function of the Hilbert space 𝐻 if and only if (a)𝐾(⋅,𝑡)∈𝐻 for all 𝑡∈𝐸, (b)⟨𝜑,𝐾(⋅,𝑡)⟩=𝜑(𝑡) for all 𝑡∈𝐸 and all 𝜑∈𝐻. The last condition is called “the reproducing property” as the value of the function 𝜑 at the point 𝑡 is reproduced by the inner product of 𝜑 with 𝐾(⋅,𝑡).

Definition 2. Hilbert function space 𝐻 is a reproducing kernel space if and only if for any fixed 𝑥∈𝑋, the linear functional 𝐼(𝑓)=𝑓(𝑥) is bounded [25, page 5].

Definition 3. We define the space 𝐻¹₂[0,1] by 𝐻¹₂[0,1]=𝑢∈𝐴𝐶[0,1]∶𝑢^′∈𝐿²[0,1]. (4) The inner product and the norm in 𝐻¹₂[0,1] are defined by 𝑢,𝑔_𝐻12=𝑢(0)𝑔(0)+¹₀𝑢^′(𝑡)𝑔^′(𝑡)d𝑡,𝑢,𝑔∈𝐻¹₂[],‖𝑢‖0,1_𝐻12=⟨𝑢,𝑢⟩_𝐻¹₂,𝑢∈𝐻¹₂[0,1]. (5)

Lemma 4. The space 𝐻¹₂[0,1] is a reproducing kernel space, and its reproducing kernel function 𝑞_𝑠 is given by [25, page 123] 𝑞_𝑠(𝑡)=1+𝑡,𝑡≤𝑠,1+𝑠,𝑡>𝑠. (6)

Definition 5. We define the space 𝐹³₂[0,𝑇] by 𝐹³₂[0,𝑇]=𝑢∈𝐴𝐶[0,𝑇]∶𝑢^′,𝑢^′′[],𝑢∈𝐴𝐶0,𝑇₍₃₎∈𝐿²[0,𝑇],𝑢(0)=𝑢^′(0)=0. (7) The inner product and the norm in 𝐹³₂[0,𝑇] are defined by 𝑢,𝑔_𝐹32=²_𝑖=0𝑢^(𝑖)(0)𝑔^(𝑖)(0)+^𝑇₀𝑢⁽³⁾(𝑡)𝑔⁽³⁾(𝑡)d𝑡,𝑢,𝑔∈𝐹³₂[],‖𝑢‖0,𝑇_𝐹32=⟨𝑢,𝑢⟩_𝐹³₂,𝑢∈𝐹³₂[0,𝑇]. (8)

Lemma 6. The space 𝐹³₂[0,𝑇] is a reproducing kernel space, and its reproducing kernel function 𝑟_𝑠 is given by [25, page 148] 𝑟_𝑠(𝑡)=14𝑠²𝑡²+1𝑠12²𝑡³−124𝑠𝑡⁴+1𝑡120⁵,1𝑡≤𝑠,4𝑠²𝑡²+1𝑠12³𝑡²−124𝑡𝑠⁴+1𝑠120⁵,𝑡>𝑠. (9)

Definition 7. We define the space 𝑊³₂[0,1] by 𝑊³₂[0,1]=𝑢∈𝐴𝐶[0,1]∶𝑢^′,𝑢^′′[],𝑢∈𝐴𝐶0,1₍₃₎∈𝐿²[0,1],𝑢(0)=𝑢(1)=0. (10) The inner product and the norm in 𝑊³₂[0,1] are defined by 𝑢,𝑔_𝑊32=²_𝑖=0𝑢^(𝑖)(0)𝑔^(𝑖)(0)+¹₀𝑢⁽³⁾(𝑥)𝑔⁽³⁾(𝑥)d𝑥,𝑢,𝑔∈𝑊³₂[],‖𝑢‖0,1_𝑊32=⟨𝑢,𝑢⟩_𝑊³₂,𝑢∈𝑊³₂[0,1]. (11) The space 𝑊³₂[0,1] is a reproducing kernel space, and its reproducing kernel function 𝑅_𝑦 is given by the following theorem.

Theorem 8. The space 𝑊³₂[0,1] is a reproducing kernel space, and its reproducing kernel function 𝑅_𝑦 is given by 𝑅_𝑦(𝑥)=⁵_𝑖∑₌₁𝑐_𝑖𝑦𝑥^𝑖,𝑥≤𝑦,_5𝑖∑₌₀𝑑_𝑖𝑦𝑥^𝑖,𝑥>𝑦, (12) where 𝑐₁𝑦=5𝑦156⁴−1𝑦156⁵−5𝑦26²−5𝑦78³+3𝑐13𝑦,₂𝑦=5𝑦624⁴−1𝑦624⁵+21𝑦104²−5𝑦312³−5𝑐26𝑦,₃𝑦=5𝑦1872⁴−1𝑦1872⁵+7𝑦104²−5𝑦936³−5𝑐78𝑦,₄𝑦5=−𝑦3744⁴+1𝑦3744⁵+5𝑦624²+5𝑦1872³−1𝑐104𝑦,₅𝑦=5𝑦3744⁴−1𝑦18720⁵−1𝑦624²−1𝑦1872³−11156𝑦+,𝑑120₀𝑦=1𝑦120⁵,𝑑₁𝑦1=−𝑦104⁴−1𝑦156⁵−5𝑦26²−5𝑦78³+3𝑑13𝑦,₂𝑦=7𝑦104³+5𝑦624⁴−1𝑦624⁵+21𝑦104²−5𝑑26𝑦,₃𝑦=5𝑦1872⁴−1𝑦1872⁵−5𝑦312²−5𝑦936³−5𝑑78𝑦,₄𝑦5=−𝑦3744⁴+1𝑦3744⁵+5𝑦624²+5𝑦1872³+5𝑑156𝑦,₅𝑦=1𝑦3744⁴−1𝑦18720⁵−1𝑦624²−1𝑦1872³−1156𝑦. (13)

Proof. Let 𝑢∈𝑊³₂[0,1] and let 0≤𝑦≤1. By Definition 7 and integrating by parts two times, we obtain that 𝑢,𝑅_𝑦𝑊32=²_𝑖=0𝑢^(𝑖)(0)𝑅_𝑦^(𝑖)(0)+¹₀𝑢⁽³⁾(𝑥)𝑅_𝑦⁽³⁾(𝑥)(0)𝑅d𝑥=𝑢_𝑦(0)+𝑢^′(0)𝑅^′_𝑦(0)+𝑢^′′(0)𝑅_𝑦^′′(0)+𝑢^′′(1)𝑅_𝑦⁽³⁾(1)−𝑢^′′(0)𝑅_𝑦⁽³⁾(0)−𝑢^′(1)𝑅_𝑦⁽⁴⁾(1)+𝑢^′(0)𝑅_𝑦⁽⁴⁾(0)+¹₀𝑢^′(𝑥)𝑅_𝑦⁽⁵⁾(𝑥)d𝑥. (14) After substituting the values of 𝑅_𝑦(0), 𝑅^′_𝑦(0), 𝑅_𝑦^′′(0), 𝑅_𝑦⁽³⁾(0), 𝑅_𝑦⁽⁴⁾(0), 𝑅_𝑦⁽³⁾(1), and 𝑅_𝑦⁽⁴⁾(1) into the above equation, we get 𝑢,𝑅_𝑦𝑊32=𝑢(0)0+𝑢^′(0)×3513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵+𝑢^′′(0)−5513𝑦−𝑦156³+21𝑦52²+5𝑦312⁴−1𝑦312⁵+𝑢^′′(1)0−𝑢^′′(0)×−5513𝑦−𝑦156³+21𝑦52²+5𝑦312⁴−1𝑦312⁵−𝑢^′(1)0+𝑢^′(0)×−3513𝑦+𝑦78³+5𝑦26²−5𝑦156⁴+1𝑦156⁵+¹₀𝑢^′(𝑥)𝑅_𝑦⁽⁵⁾(𝑥)d𝑥; (15) thus we obtain that 𝑢,𝑅_𝑦𝑊32=¹₀𝑢^′(𝑥)𝑅_𝑦⁽⁵⁾(𝑥)=d𝑥_𝑦0𝑢^′(𝑥)𝑅_𝑦⁽⁵⁾(𝑥)d𝑥+¹_𝑦𝑢^′(𝑥)𝑅_𝑦⁽⁵⁾(𝑥)=d𝑥_𝑦0𝑢^′(𝑥)1−10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵d𝑥+¹_𝑦𝑢^′(𝑥)−10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵=𝑢𝑦(0)×d𝑥−𝑢1−10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵+𝑢(1)𝑦×−−𝑢10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵𝑦(0)×=𝑢−𝑢1−10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵+𝑢(1)−10513𝑦−𝑦78³−5𝑦26²+5𝑦156⁴−1𝑦156⁵. (16) By Definition 7, we have 𝑢(0)=𝑢(1)=0. So 𝑢,𝑅_𝑦𝑊32=𝑢𝑦. (17) This completes the proof.

Definition 9. We define the binary space 𝑊(Ω) by 𝑊(Ω)=𝜕𝑢∶⁴𝑢𝜕𝑥²𝜕𝑡²[]×[],𝜕iscompletelycontinuousinΩ=0,10,1⁶𝑢𝜕𝑥³𝜕𝑡³∈𝐿²(Ω),𝑢()()𝑥,0=0,𝜕𝑢𝑥,0𝑢()().𝜕𝑡=0,0,𝑡=0,𝑢1,𝑡=0 (18) The inner product and the norm in 𝑊(Ω) are defined by 𝑢,𝑔_𝑊=²_𝑖=0¹₀𝜕³𝜕𝑡³𝜕^𝑖𝜕𝑥^𝑖𝑢()×𝜕0,𝑡₃𝜕𝑡³𝜕^𝑖𝜕𝑥^𝑖𝑔(+0,𝑡)d𝑡_2𝑗=0𝜕^𝑗𝜕𝑡^𝑗𝑢(),𝜕𝑥,0^𝑗𝜕𝑡^𝑗𝑔(𝑥,0)_𝑊32+¹₀𝜕³𝜕𝑥³𝜕³𝜕𝑡³𝑢()×𝜕𝑥,𝑡₃𝜕𝑥³𝜕³𝜕𝑡³𝑔(‖𝑢‖𝑥,𝑡)d𝑡d𝑥,_𝑊=⟨𝑢,𝑢⟩_𝑊,𝑢∈𝑊(Ω). (19)(20)

Lemma 10. 𝑊(Ω) is a reproducing kernel space, and its reproducing kernel function 𝐾_(𝑦,𝑠) is given by [25, page 148] 𝐾_(𝑦,𝑠)=𝑅_𝑦𝑟_𝑠. (21)

Definition 11. We define the binary space 𝑊(Ω) by 𝑊(Ω)=[]×[],𝜕𝑢∶𝑢iscompletelycontinuousinΩ=0,10,1²𝑢𝜕𝑥𝜕𝑡∈𝐿²(Ω). (22) The inner product and the norm in 𝑊(Ω) are defined by 𝑊𝑢,𝑔=¹₀𝜕𝑢()𝜕𝜕𝑡0,𝑡𝑔(+𝑢()(𝜕𝑡0,𝑡)d𝑡𝑥,0,𝑔𝑥,0)_𝐻12+¹₀𝜕𝜕𝜕𝑥𝑢()𝜕𝜕𝑡𝑥,𝑡𝜕𝜕𝑥𝑔(‖𝑢‖𝑊=𝜕𝑡𝑥,𝑡)d𝑡d𝑥,⟨⟩𝑊𝑢,𝑢,𝑢∈𝑊(Ω). (23)

Lemma 12. 𝑊(Ω) is a reproducing kernel space, and its reproducing kernel function 𝐺_(𝑦,𝑠) is given as [25, page 148] 𝐺_(𝑦,𝑠)=𝑄_𝑦𝑞_𝑠. (24)

Remark 13. Hilbert spaces can be completely classified: there is a unique Hilbert space up to isomorphism for every cardinality of the base. Since finite-dimensional Hilbert spaces are fully understood in linear algebra and since morphisms of Hilbert spaces can always be divided into morphisms of spaces with Aleph-null (𝜒₀) dimensionality, functional analysis of Hilbert spaces mostly deals with the unique Hilbert space of dimensionality Aleph-null and its morphisms. One of the open problems in functional analysis is to prove that every bounded linear operator on a Hilbert space has a proper invariant subspace. Many special cases of this invariant subspace problem have already been proven [26].

3. Solution Representation in 𝑊(Ω)

In this section, the solution of (1) is given in the reproducing kernel space 𝑊(Ω). On defining the linear operator 𝐿∶𝑊(Ω)→𝑊(Ω) by 𝜕𝐿𝑣=²𝑣𝜕𝑡²−𝜕²𝑣𝜕𝑥²+2𝛼𝜕𝑣𝜕𝑡+𝛽²𝑣(𝑥,𝑡), (25) after homogenizing the initial and boundary conditions, model problem (1)–(3) changes to the problem (),()∈[]×[],𝑣()=𝐿𝑣=𝑀𝑥,𝑡𝑥,𝑡0,10,1𝑥,0𝜕𝑣()()()𝜕𝑡𝑥,0=𝑣0,𝑡=𝑣1,𝑡=0, (26) where 𝑀()=𝜕𝑥,𝑡²𝑍𝜕𝑡²()−𝜕𝑥,𝑡²𝑣𝜕𝑥²(𝑥,𝑡)+2𝛼𝜕𝑍()𝜕𝑡𝑥,𝑡+𝛽²𝑍(𝑥,𝑡)+𝑓(𝑥,𝑡); (27) for convenience, we again write 𝑢 instead of 𝑣 in (26)

Lemma 14. 𝐿 is a bounded linear operator.

Proof. Let 𝑢∈𝑊(Ω) and let (𝑥,𝑡)∈Ω. By Lemma 10, we have 𝑢(𝑥,𝑡)=𝑢,𝐾_{(𝑥,𝑡)𝑊}, (28) and thus 𝐿𝑢(𝑥,𝑡)=𝑢,𝐿𝐾_{(𝑥,𝑡)𝑊},𝜕()=𝜕𝜕𝑥𝐿𝑢𝑥,𝑡𝑢,𝜕𝑥𝐿𝐾_{(𝑥,𝑡)𝑊},𝜕()=𝜕𝜕𝑡𝐿𝑢𝑥,𝑡𝑢,𝜕𝑡𝐿𝐾_{(𝑥,𝑡)𝑊},𝜕𝜕𝜕𝑡()=𝜕𝜕𝑥𝐿𝑢𝑥,𝑡𝑢,𝜕𝜕𝑡𝜕𝑥𝐿𝐾_{(𝑥,𝑡)𝑊}. (29) Hence there exist 𝑎₀,𝑏₀,𝑎₁,𝑏₁>0 such that |𝐿𝑢(𝑥,𝑡)|≤𝑎₀‖𝑢‖_𝑊,𝜕(𝜕𝑡𝐿𝑢𝑥,𝑡)≤𝑏₀‖𝑢‖_𝑊,𝜕(𝜕𝑥𝐿𝑢𝑥,𝑡)≤𝑎₁‖𝑢‖_𝑊,𝜕𝜕𝜕𝑡(𝜕𝑥𝐿𝑢𝑥,𝑡)≤𝑏₁‖𝑢‖_𝑊. (30) Therefore, ‖𝐿𝑢‖²𝑊=¹₀𝜕(𝜕𝑡𝐿𝑢0,𝑡)²+⟨()(d𝑡𝐿𝑢𝑥,0,𝐿𝑢𝑥,0)⟩_𝐻¹₂+¹₀𝜕𝜕𝜕𝑥(𝜕𝑡𝐿𝑢𝑥,𝑡)²=d𝑡d𝑥₁₀𝜕(𝜕𝑡𝐿𝑢0,𝑡)²d𝑡+[𝐿𝑢(0,0)]²+¹₀𝜕(𝜕𝑥𝐿𝑢𝑥,0)²d𝑥+¹₀𝜕𝜕𝜕𝑥(𝜕𝑡𝐿𝑢𝑥,𝑡)²≤𝑎d𝑡d𝑥₂₀+𝑎²₁+𝑏²₀+𝑏²₁‖𝑢‖²_𝑊. (31) This completes the proof.

Now, choose a countable dense subset {(𝑥₁,𝑡₁),(𝑥₂,𝑡₂),…} in Ω and define 𝜑_𝑖=𝐺_{(𝑥𝑖,𝑡𝑖)},Ψ_𝑖=𝐿^∗𝜑_𝑖, (32) where 𝐿^∗ is the adjoint operator of 𝐿. The orthonormal system {Ψ_𝑖}^∞_𝑖=1 of 𝑊(Ω) can be derived from the process of Gram-Schmidt orthogonalization of {Ψ_𝑖}^∞_𝑖=1 as Ψ_𝑖=^𝑖_𝑘=1𝛽_𝑖𝑘Ψ_𝑘. (33)

Theorem 15. Suppose that {(𝑥_𝑖,𝑡_𝑖)}^∞_𝑖=1 is dense in Ω. Then {Ψ_𝑖}^∞_𝑖=1 is a complete system in 𝑊(Ω), and Ψ_𝑖=𝐿𝐾_{(𝑥𝑖,𝑡𝑖)}(𝑥,𝑡). (34)

Proof. We have Ψ_𝑖=𝐿^∗𝜑_𝑖=𝐿^∗𝜑_𝑖,𝐾_{(𝑥,𝑡)𝑊}=𝜑_𝑖,𝐿𝐾_(𝑥,𝑡)𝑊=𝐿𝐾_(𝑥,𝑡),𝐺_{(𝑥𝑖,𝑡𝑖)}𝑊=𝐿𝐾_(𝑥,𝑡)𝑥_𝑖,𝑡_𝑖=𝐿𝐾_{(𝑥𝑖,𝑡𝑖)}(𝑥,𝑡). (35) Clearly, Ψ_𝑖∈𝑊(Ω). For each fixed 𝑢∈𝑊(Ω), if 𝑢,Ψ_𝑖𝑊=0,𝑖=1,2,…, (36) then 0=𝑢,Ψ_𝑖𝑊=𝑢,𝐿^∗𝜑_𝑖𝑊=𝐿𝑢,𝜑_𝑖𝑊=𝐿𝑢𝑥_𝑖,𝑡_𝑖,𝑖=1,2,…. (37) Note that {(𝑥_𝑖,𝑡_𝑖)}^∞_𝑖=1 is dense in Ω. Hence, 𝐿𝑢=0. From the existence of 𝐿⁻¹, it follows that 𝑢=0. The proof is completed.

Theorem 16. If {(𝑥_𝑖,𝑡_𝑖)}^∞_𝑖=1 is dense in Ω, then the solution of (26) is given by 𝑢=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖. (38)

Proof. By Theorem 15, {Ψ_𝑖(𝑥,𝑡)}^∞_𝑖=1 is a complete system in 𝑊(Ω). Thus, 𝑢=^∞_𝑖=1𝑢,Ψ_𝑖𝑊Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑢,Ψ_𝑘𝑊Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑢,𝐿^∗𝜑_𝑘𝑊Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐿𝑢,𝜑_𝑘𝑊Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐿𝑢,𝐺_{(𝑥𝑘,𝑡𝑘)}𝑊Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐿𝑢𝑥_𝑘,𝑡_𝑘Ψ_𝑖=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖. (39) This completes the proof.

Now the approximate solution 𝑢_𝑛 can be obtained from the 𝑛-term intercept of the exact solution 𝑢 and 𝑢_𝑛=^𝑛𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖. (40) Obviously 𝑢_𝑛−𝑢_𝑊⟶0,𝑛⟶∞. (41)

Theorem 17. If 𝑢∈𝑊(Ω), then 𝑢_𝑛−𝑢_𝑊⟶0,𝑛⟶∞. (42) Moreover, a sequence ‖𝑢_𝑛−𝑢‖_𝑊 is monotonically decreasing in 𝑛.

Proof. From (38) and (40), it follows that 𝑢_𝑛−𝑢_𝑊=^∞𝑖_{𝑖=𝑛+1𝑘=1}𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖(𝑥,𝑡)_𝑊. (43) Thus, 𝑢_𝑛(𝑥,𝑡)−𝑢(𝑥,𝑡)_𝑊⟶0,𝑛⟶∞. (44) In addition 𝑢_𝑛−𝑢²_𝑊=^∞𝑖_{𝑖=𝑛+1𝑘=1}𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖(𝑥,𝑡)²_𝑊=^∞_𝑖=𝑛+1^𝑖_𝑘=1𝛽_𝑖𝑘𝑀𝑥_𝑘,𝑡_𝑘Ψ_𝑖(𝑥,𝑡)². (45) Clearly, ‖𝑢_𝑛(𝑥,𝑡)−𝑢(𝑥,𝑡)‖_𝑊 is monotonically decreasing in 𝑛.

4. The Method Implementation

(i)If (26) is linear, then the analytical solution of (26) can be obtained directly by (38). (ii)If (26) is nonlinear, then the solution of (26) can be obtained by the following iterative method.

We construct an iterative sequence 𝑢_𝑛, putting anyfixed𝑢₀∈𝑊³₂[],𝑢0,1_𝑛=^𝑛_𝑖=1𝐴_𝑖Ψ_𝑖, (46) where 𝐴₁=𝛽₁₁𝑀𝑥_𝑘,𝑡_𝑘,𝑢𝑥_𝑘,𝑡_𝑘𝐴₂=²_𝑘=1𝛽_2𝑘𝑀𝑥_𝑘,𝑡_𝑘,𝑢_𝑘−1𝑥_𝑘,𝑡_𝑘,⋮𝐴_𝑛=^𝑛_𝑘=1𝛽_𝑛𝑘𝑀𝑥_𝑘,𝑡_𝑘,𝑢_𝑘−1𝑥_𝑘,𝑡_𝑘. (47) Next we will prove that 𝑢_𝑛 given by the iterative formula (46) converges to the exact solution.

Theorem 18. Suppose that ‖𝑢_𝑛‖ defined by (46) is bounded and (26) has a unique solution. If {(𝑥_𝑖,𝑡_𝑖)}^∞_𝑖=1 is dense in Ω, then 𝑢_𝑛 converges to the analytical solution 𝑢 of (26), and 𝑢=^∞_𝑖=1𝐴_𝑖Ψ_𝑖, (48) where 𝐴_𝑖 is given by (47).

Proof. First, we prove the convergence of 𝑢_𝑛. From (46) and the orthonormality of {Ψ_𝑖}^∞_𝑖=1, we infer that 𝑢_𝑛+1²=^𝑛+1_𝑖=1𝐴²_𝑖=^𝑛_𝑖=1𝐴²_𝑖+𝐴²_𝑛+1=𝑢_𝑛²+𝐴²_𝑛+1≥𝑢_𝑛. (49) By (49), ‖𝑢_𝑛‖ is nondecreasing, and by the assumption, ‖𝑢_𝑛‖ is bounded. Thus ‖𝑢_𝑛‖ is convergent. By (49), there exists a constant 𝑐 such that ^∞_𝑖=1𝐴²_𝑖=𝑐. (50) This implies that 𝐴_𝑖^∞_𝑖=1∈ℓ². (51) If 𝑚>𝑛, then 𝑢_𝑚−𝑢_𝑛²=^𝑚−1_𝑘=𝑛𝑢_𝑘+1−𝑢_𝑘²≤^𝑚−1_𝑘=𝑛𝑢_𝑘+1−𝑢_𝑘²=^𝑚−1_𝑘=𝑛𝐴²_𝑘+1⟶0,𝑚,𝑛⟶∞. (52) The completeness of 𝑊(Ω) shows that there exists 𝑢∈𝑊(Ω) such that 𝑢_𝑛→𝑢 as 𝑛→∞. Now, we prove that 𝑢 solves (26). Taking limits in (40), we get 𝑢=^∞_𝑖=1𝐴_𝑖Ψ_𝑖. (53) Note that 𝐿𝑢=^∞_𝑖=1𝐴_𝑖𝐿Ψ_𝑖,(𝐿𝑢)𝑥_𝑘,𝑡_𝑘=^∞_𝑖=1𝐴_𝑖𝐿Ψ_𝑖𝑥_𝑘,𝑡_𝑘=^∞_𝑖=1𝐴_𝑖𝐿Ψ_𝑖,𝐺_{(𝑥𝑘,𝑡𝑘)}𝑊=^∞_𝑖=1𝐴_𝑖𝐿Ψ_𝑖,𝜑_𝑘𝑊=^∞_𝑖=1𝐴_𝑖Ψ_𝑖,𝐿^∗𝜑_𝑘𝑊=^∞_𝑖=1𝐴_𝑖Ψ_𝑖,Ψ_𝑘𝑊. (54)(55) In view of (47), we have 𝐿𝑢𝑥_𝑙,𝑡_𝑙=𝑀𝑥_𝑙,𝑡_𝑙,𝑢_𝑙−1𝑥_𝑙,𝑡_𝑙. (56) Since {(𝑥_𝑖,𝑡_𝑖)}^∞_𝑖=1 is dense in Ω, for each (𝑦,𝑠)∈Ω, there exists a subsequence {(𝑥_𝑛𝑗,𝑡_𝑛𝑗)}^∞_𝑗=1 such that 𝑥_𝑛𝑗,𝑡_𝑛𝑗⟶𝑦,𝑠,𝑗⟶∞. (57) We know that 𝐿𝑢𝑥_𝑛𝑗,𝑡_𝑛𝑗=𝑀𝑥_𝑛𝑗,𝑡_𝑛𝑗,𝑢_𝑛𝑗−1𝑥_𝑛𝑗,𝑡_𝑛𝑗. (58) Let 𝑗→∞. By the continuity of 𝑓, we have (𝐿𝑢)𝑦,𝑠=𝑀𝑦,𝑠,𝑢𝑦,𝑠, (59) which indicates that 𝑢 satisfies (26).

Remark 19. In the same manner, it can be proved that 𝜕𝑢_𝑛−𝜕𝑥𝜕𝑢𝜕𝑥⟶0,𝑛⟶∞, (60) where 𝜕𝑢=𝜕𝑥^∞_𝑖=1𝐴_𝑖𝜕Ψ_𝑖,𝜕𝑥𝜕𝑢_𝑛=𝜕𝑥^𝑛_𝑖=1𝐴_𝑖𝜕Ψ_𝑖,𝜕𝑥 (61) and 𝐴_𝑖 is given by (47).

5. Numerical Results

To test the accuracy of the present method, some numerical experiments are presented in this section. Using our method, we chose 36 points in Ω and obtained the approximate solution 𝑢₃₆. The comparison between interpolating scaling function method [11] and RKM for different values of 𝛼, 𝛽, and 𝑡 is given in Tables 4 and 5. We solve these examples for a set of points 𝑥₁=0,…,𝑥_𝑖=(𝑖−1)ℎ,…,𝑥_𝑁1=1,ℎ=.𝑁−1 (62) In Tables 7 and 10 we calculate the RMS error by the following formula: RMSerror=∑^𝑁+1_𝑖=1𝑢−𝑢₃₆².𝑁+1 (63) It can be seen from Tables 4 and 5 and 7–10 that the results obtained by the RKM are more accurate than those obtained by the methods in [10, 11]. This indicates that RKM is a reliable method. The CPU time (𝑠) is given in Tables 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Numerical solutions are described in the extended domain [−4,4]×[−3,3]. The comparison of RMS error is given for our method and Chebyshev method.

Table 1 

Numerical results for Example 20 for 𝑡=0.5.

Table 2 

Numerical results for Example 20 for 𝑡=1.0.

Table 3 

A comparison between interpolating scaling function method [11] and RKM for different values of 𝛼, 𝛽, and 𝑡 for Example 20.

Table 4 

A comparison between interpolating scaling function method [11] and RKM for different values of 𝛼, 𝛽, and 𝑡 for Example 20.

Table 5 

Numerical results for Example 21 for 𝑡=0.5.

Table 6 

Numerical results for Example 21 for 𝑡=1.0.

Table 7 

RMS errors for Example 21.

Table 8 

Numerical results for Example 22 for 𝑡=0.5.

Table 9 

Numerical results for Example 22 for 𝑡=1.0.

Table 10 

RMS errors for Example 22.

Example 20. Consider the following telegraph equation with initial and boundary conditions: 𝜕²𝑢𝜕𝑡²(𝑥,𝑡)+2𝛼𝜕𝑢()𝜕𝑡𝑥,𝑡+𝛽²𝑢()=𝜕𝑥,𝑡²𝑢𝜕𝑥²(),𝑢()+𝑓𝑥,𝑡𝑥,0=𝜑₁(𝑥)=sinh(𝑥),𝜕𝑢()𝜕𝑡𝑥,0=𝜑₂(𝑥)(𝑥),𝑢()=−2sinh0,𝑡=𝑔₀(𝑡)𝑢()=0,𝑡≥0,1,𝑡=𝑔₁(𝑡)=exp(−2𝑡)sinh(1),𝑡≥0, (64) where 𝑓(𝑥,𝑡)=3−4𝛼+𝛽²exp(−2𝑡)sinh(𝑥). (65) The exact solution of (64) is given by [11] 𝑢(𝑥,𝑡)=exp(−2𝑡)sinh(𝑥). (66) If we apply 𝑣()()(1)()(𝑥)𝑥,𝑡=𝑢𝑥,𝑡−𝑥sinhexp−2𝑡−1+2𝑡+sinh)(2𝑡−1 (67) to (64), then the following equation (68) is obtained: 𝜕²𝑣𝜕𝑡²−𝜕²𝑣𝜕𝑥²+2𝛼𝜕𝑣𝜕𝑡+𝛽²𝑣()(),𝑣()=𝑥,𝑡=𝑀𝑥,𝑡𝑥,0𝜕𝑣()()()𝜕𝑡𝑥,0=𝑣0,𝑡=𝑣1,𝑡=0, (68) where 𝑀(𝑥,𝑡)=3−4𝛼+𝛽²()(𝑥)(𝑥)(1)()𝑥(𝑥)exp−2𝑡sinh−sinh)(2𝑡−1−2𝛼sinh−2exp−2𝑡+2+4𝛼sinh−𝛽²sinh(1)exp(−2𝑡)−1+2𝑡𝑥+𝛽²sinh(𝑥)(2𝑡−1)−4𝑥exp(−2𝑡)sinh(1). (69) After homogenizing the initial and boundary conditions and using the above method, we obtain Tables 1–4 and Figure 1.

Figure 1 

Graph of numerical results for Example 20 (𝛼=10,𝛽=20,𝑡=0.5).

Example 21. Consider the following telegraph equation with initial and boundary conditions: 𝜕²𝑢𝜕𝑡²+2𝛼𝜕𝑢𝜕𝑡+𝛽²𝑢()=𝜕𝑥,𝑡²𝑢𝜕𝑥²(),𝑢()+𝑓𝑥,𝑡𝑥,0=𝜑₁(𝑥)=sin(𝑥),𝜕𝑢()𝜕𝑡𝑥,0=𝜑₂(𝑥)𝑢()=0,0,𝑡=𝑔₀(𝑡)𝑢()=0,𝑡≥0,1,𝑡=𝑔₁(𝑡)=cos(𝑡)sin(1),𝑡≥0, (70) where 𝑓(𝑥,𝑡)=−2𝛼sin(𝑡)sin(𝑥)+𝛽²cos(𝑡)sin(𝑥). (71) The exact solution of (70) is given by [11] 𝑢(𝑥,𝑡)=cos(𝑡)sin(𝑥). (72) If we apply 𝑣(𝑥,𝑡)=𝑢(𝑥,𝑡)−𝑥sin(1)(cos(𝑡)−1)−sin(𝑥) (73) to (70), then the following (74) is obtained: 𝜕²𝑣𝜕𝑡²()−𝜕𝑥,𝑡²𝑣𝜕𝑥²(𝑥,𝑡)+2𝛼𝜕𝑣𝜕𝑡+𝛽²𝑣()(),𝑣()=𝑥,𝑡=𝑀𝑥,𝑡𝑥,0𝜕𝑣()()()𝜕𝑡𝑥,0=𝑣0,𝑡=𝑣1,𝑡=0, (74) where 𝑀(𝑥,𝑡)=−2𝛼sin(𝑡)sin(𝑥)+𝛽²(𝑡)(𝑥)(𝑥)(𝑡)(1)(𝑡)(1)cossin−sin−2𝛼sinsinh𝑥+𝑥cossin−𝛽²𝑥sin(1)(cos(𝑡)−1)−𝛽²sin(𝑥). (75) After homogenizing the initial and boundary conditions and using the above method, we obtain Tables 5–7 and Figures 2 and 3.

Figure 2 

Graph of numerical results for Example 21 (𝛼=10,𝛽=20,𝑡=0.5).

(a)

(b)

Figure 3 

Plots of numerical results for Example 21 (𝛼=6,𝛽=2).

Example 22. Consider the following telegraph equation with initial and boundary conditions: 𝜕²𝑢𝜕𝑡²+2𝛼𝜕𝑢𝜕𝑡+𝛽²𝑢()=𝜕𝑥,𝑡²𝑢𝜕𝑥²(),𝑢()+𝑓𝑥,𝑡𝑥,0=𝜑₁(𝑥)𝑥=tan2,𝜕𝑢()𝜕𝑡𝑥,0=𝜑₂(𝑥)=1+tan²(𝑥/2)2,𝑢(0,𝑡)=𝑔₀(𝑡)𝑡=tan2𝑢(),𝑡≥0,1,𝑡=𝑔₁(𝑡)=tan1+𝑡2,𝑡≥0, (76) where 𝑓(𝑥,𝑡)=𝛼1+tan²𝑥+𝑡2+𝛽²tan𝑥+𝑡2. (77) The exact solution of (76) is given by [10] 𝑢(𝑥,𝑡)=tan𝑥+𝑡2. (78) If we apply 𝑣()()+()𝑡𝑥,𝑡=𝑢𝑥,𝑡𝑥−1tan2−𝑥tan1+𝑡2𝑥−tan2×()1+𝑡tan𝑥/221+𝑥tan2()1+𝑡tan1/22 (79) to (76), then the following equation (80) is obtained 𝜕²𝑣𝜕𝑡²()−𝜕𝑥,𝑡²𝑣𝜕𝑥²(𝑥,𝑡)+2𝛼𝜕𝑣𝜕𝑡+𝛽²𝑣()(),𝑣()=𝑥,𝑡=𝑀𝑥,𝑡𝑥,0𝜕𝑣()()()𝜕𝑡𝑥,0=𝑣0,𝑡=𝑣1,𝑡=0, (80) where 𝑀(𝑥,𝑡)=𝛼1+tan²𝑥+𝑡2+𝛽²tan𝑥+𝑡2+tan(𝑥/2)2+𝑡4+3𝑡4tan²𝑥2×1+tan²𝑥2+(𝑥−1)𝛼1+tan²𝑡2−𝛼𝑥1+tan²𝑡+12−𝛼tan²𝑥2+𝛼𝑥tan²12+𝛽²()𝑡𝑥−1tan2−𝛽²𝑥tan1+𝑡2−𝛽²𝑥tan2×()1+𝑡tan𝑥/22+𝛽²1𝑥tan2()1+𝑡tan1/22×𝑥−12𝑡tan21+tan²𝑡2−𝑥2tan1+𝑡21+tan²1+𝑡2. (81) After homogenizing the initial and boundary conditions and using the above method, we obtain Tables 8–10 and Figure 4.

(a)

(b)

Figure 4 

Plots of numerical results for Example 22 (𝛼=10,𝛽=5).

Remark 23. In Tables 1–9, we abbreviate the exact solution and the approximate solution by AS and ES, respectively. AE stands for the absolute error, that is, the absolute value of the difference of the exact solution and the approximate solution, while RE indicates the relative error, that is, the absolute error divided by the absolute value of the exact solution.

6. Conclusion

In this study, a second-order one-dimensional telegraph equation with initial and boundary conditions was solved by reproducing kernel Hilbert space method. We described the method and used it in some test examples in order to show its applicability and validity in comparison with exact and other numerical solutions. The obtained results show that this approach can solve the problem effectively and need few computations. The satisfactory results that we obtained were compared with the results that were obtained by [10, 11]. Numerical experiments on test examples show that our proposed schemes are of high accuracy and support the theoretical results. As shown in Tables 7 and 10 our results are better than the results that were obtained by [10]. According to these results, it is possible to apply RKM to linear and nonlinear differential equations with initial and boundary conditions. It has been shown that the obtained results are uniformly convergent and the operator that was used is a bounded linear operator. From the results, RKM can be applied to high dimensional partial differential equations, integral equations, and fractional differential equations without any transformation or discretization, and good results can be obtained.