A high-accuracy compact conservative scheme for generalized regularized long-wave equation

Abstract

In this article, we develop a high-order compact conservative numerical scheme to solve the initial-boundary problem of GRLW equation. The proposed scheme is three-level and linear-implicit based on a finite difference method. A detailed numerical analysis of the scheme is presented including a convergence analysis result. Some numerical examples are provided to show the present scheme is efficient, reliable, and of high accuracy.

1 Introduction

The Cauchy problem of the generalized regularized long-wave (GRLW) equation reads

$$\begin{aligned} &u_{t}-\beta u_{xxt}+u_{x}+\alpha \bigl(u^{p}\bigr)_{x}=0, \end{aligned}$$

(1.1)

$$\begin{aligned} &u(x,0)=u_{0}(x), \end{aligned}$$

(1.2)

where α, β are positive constants and p is a positive integer [1]. The GRLW equation was first put forward by Peregrine [2] and Benjamin et al. [3] as a model for small-amplitude long waves on the surface of water in a channel. Many authors [4–8] have recently studied models for long waves in nonlinear dispersive systems. When $p=2$, (1.1) is usually called the RLW equation. When $p=3$, (1.1) is called a modified regularized long-wave (MRLW) equation. Various numerical techniques have been developed to solve the equation. These partly include the finite difference method, finite element methods, the least squares method, and a collocation method with quadratic B-splines, cubic B-splines and septic splines; we refer to [9–20], and references therein.

In general, the solutions of the system (1.1)-(1.2) decays rapidly to zero for $\vert x\vert \gg0$. Therefore, numerically we can solve the system (1.1)-(1.2) in a compact domain $\Omega =(x_{l},x_{r})$ with $-x_{l}\gg0$ and $x_{r}\gg0$. We can add the boundary conditions to the Cauchy problem (1.1)-(1.2),

$$ u(x_{l},t)=u(x_{r},t)=0. $$

(1.3)

It is well known that the system (1.1)-(1.2) possesses the following conservative law:

$$ E(t)=\Vert u\Vert ^{2}_{L_{2}}+\Vert u_{x} \Vert ^{2}_{L_{2}}=E(0). $$

(1.4)

In [1], Zhang considered a linear conservative scheme for GRLW equation, however, the accuracy of the scheme is only second-order. Recently, there has been growing interest in high-order compact methods to solve the partial differential equations [21–31], where fourth-order compact finite difference approximation solutions for the transient wave equations, a N-carrier system, the Klein-Gordon equation, the Sine-Gordon equation, the one-dimensional heat and advection-diffusion equations, the Schrödinger equation, the Klein-Gordon-Schrödinger equation and the RLW equation were shown, respectively. These numerical methods may give us many enlightenments to design a new numerical scheme for the GRLW equation. For a wide and most complete vision concerning the importance, the breadth, and the interest of the topics covered, we should also recall the study done on the long waves in [32–36].

The main purpose of this paper is to construct a new numerical scheme which has the following advantages:

• 1. Coupling with the Richardson extrapolation, the new scheme is high-accuracy and without refined mesh; it has an accuracy of $O(\tau ^{2}+h^{4})$.

• 2. The new scheme is linearized and preserves the original conservative property.

• 3. The coefficient matrices of the scheme is symmetric and pentadiagonal, and the Thomas algorithm can be employed to solve it effectively.

• 4. Useful numerical examples are given to show the efficiency of the scheme.

The rest of this paper is organized as follows. In Section 2, a high-accuracy linear-compact difference scheme for the GRLW equation is described. In Section 3, we discuss the solvability of the scheme and the estimate of the difference solution. In Section 4, the fourth-order convergence and stability of the scheme are proved by the discrete energy method. Numerical results are reported in Section 5.

2 High-accuracy compact scheme and its discrete conservative law

In this section, we describe a high-order linear-compact difference scheme for (1.1)-(1.2).

Let $h=\frac{x_{r}-x_{l}}{J}$ and $\tau=\frac{T}{N}$ be the uniform step size in the spatial and the temporal direction, respectively. Denote $x_{j}=jh$ ($0\le j\le J$), $t_{n}=n\tau$ ($0\le n\le N$), $u^{n}_{j}\approx u(x_{j},t_{n})$ and $Z^{0}_{h}=\{u=(u_{j})|u_{-1}=u_{0}=u_{J}=u_{J+1}=0, j=0,1,\ldots, J\}$. For simplicity, we introduce the following notations of the difference operators:

$$\begin{aligned}& \delta_{x}u^{n}_{j}=\frac{u^{n}_{j+1}-u^{n}_{j}}{h},\qquad \delta_{\bar{x}}u^{n}_{j}=\frac{u^{n}_{j}-u^{n}_{j-1}}{h},\qquad \delta_{\hat{x}}u^{n}_{j}=\frac{u^{n}_{j+1}-u^{n}_{j-1}}{2h},\qquad \delta_{\ddot{x}}u^{n}_{j}=\frac{u^{n}_{j+2}-u^{n}_{j-2}}{4h}, \\& \delta_{\hat{t}}u^{n}_{j}=\frac{u^{n+1}_{j}-u^{n-1}_{j}}{2\tau}, \qquad \delta _{t}u^{n}_{j}=\frac{u^{n+1}_{j}-u^{n}_{j}}{\tau},\qquad \bar{u}^{n}_{j}=\frac {u^{n+1}_{j}+u^{n-1}_{j}}{2}. \end{aligned}$$

In the paper, C denotes a general positive constant which may have different values in different occurrences.

Based on the notations above, we consider the following high-accuracy linear-compact scheme for the initial-boundary problem (1.1)-(1.3),

$$\begin{aligned}& \begin{aligned}[b] &\delta_{\hat{t}}u^{n}_{j}-\frac{4}{3}\beta \delta_{x}\delta_{\bar{x}}\delta _{\hat{t}}u^{n}_{j}+ \frac{1}{3}\beta\delta_{\hat{x}}\delta_{\hat{x}}\delta _{\hat{t}}u^{n}_{j}+\frac{4}{3} \delta_{\hat{x}}\bigl(u^{n}_{j}\bigr)-\frac{1}{3} \delta _{\ddot{x}}\bigl(u^{n}_{j}\bigr) \\ & \quad {}+\frac{p}{p+1}\alpha\biggl[\frac{4}{3}\bigl\{ \bigl( \delta_{\hat{x}}\bar {u}^{n}_{j}\bigr) \bigl(u^{n}_{j}\bigr)^{p-1}+\delta_{\hat{x}} \bigl[\bigl(u^{n}_{j}\bigr)^{p-1}\bigl( \bar{u}^{n}_{j}\bigr)\bigr]\bigr\} \\ & \quad {}-\frac{1}{3}\bigl\{ \bigl(\delta_{\ddot{x}}\bar{u}^{n}_{j}\bigr) \bigl(u^{n}_{j}\bigr)^{p-1}+\delta_{\ddot {x}} \bigl[\bigl(u^{n}_{j}\bigr)^{p-1}\bigl( \bar{u}^{n}_{j}\bigr)\bigr]\bigr\} \biggr]=0,\\ &\quad 1\le j\le J-1, 1\le n\le N-1, \end{aligned} \end{aligned}$$

(2.1)

$$\begin{aligned}& \begin{aligned}[b] &u^{1}_{j}-\frac{4}{3}\beta\delta_{x} \delta_{\bar{x}}u^{1}_{j}+\frac{1}{3}\beta \delta_{\hat{x}}\delta_{\hat{x}}u^{1}_{j}\\ &\quad =u_{0}(x_{j})- \frac{d^{2} u_{0}}{dx^{2}}(x_{j})-\tau\frac{du_{0}}{dx}(x_{j})-\tau \alpha p u_{0}^{p-1}(x_{j})\frac{du_{0}}{dx}(x_{j}), \end{aligned} \end{aligned}$$

(2.2)

$$\begin{aligned}& u^{0}_{j}=u_{0}(x_{j}),\quad 1\le j\le J, \end{aligned}$$

(2.3)

$$\begin{aligned}& u^{n}_{0}=u^{n}_{J}=0,\quad 0\le n\le N. \end{aligned}$$

(2.4)

For convenience, the last term of (2.1) is defined by

$$ \kappa\bigl(u^{n},\bar{u}^{n}\bigr)=\kappa_{1} \bigl(u^{n},\bar{u}^{n}\bigr)+\kappa_{2} \bigl(u^{n},\bar {u}^{n}\bigr), $$

(2.5)

where

$$\begin{aligned} &\kappa_{1}\bigl(u^{n},\bar{u}^{n}\bigr)= \frac{4p}{3(p+1)}\alpha\bigl\{ \bigl(\delta_{\hat{x}}\bar {u}^{n} \bigr) \bigl(u^{n}\bigr)^{p-1}+\delta_{\hat{x}}\bigl[ \bigl(u^{n}\bigr)^{p-1}\bigl(\bar{u}^{n}\bigr)\bigr] \bigr\} , \\ &\kappa_{2}\bigl(u^{n},\bar{u}^{n}\bigr)=- \frac{p}{3(p+1)}\alpha\bigl\{ \bigl(\delta_{\ddot{x}}\bar {u}^{n} \bigr) \bigl(u^{n}\bigr)^{p-1}+\delta_{\ddot{x}}\bigl[ \bigl(u^{n}\bigr)^{p-1}\bigl(\bar{u}^{n}\bigr)\bigr] \bigr\} . \end{aligned}$$

Theorem 2.1

Suppose $u_{0}\in H^{1}_{0}[x_{l},x_{r}]$, then the scheme (2.1) admits the following invariant:

$$\begin{aligned} E^{n}={}&\frac{1}{2}\bigl(\bigl\Vert u^{n+1}\bigr\Vert ^{2}+\bigl\Vert u^{n}\bigr\Vert ^{2} \bigr)+\frac{2}{3}\beta\bigl(\bigl\Vert \delta _{x}u^{n+1} \bigr\Vert ^{2}+\bigl\Vert \delta_{x}u^{n}\bigr\Vert ^{2}\bigr)-\frac{1}{6}\beta\bigl(\bigl\Vert \delta_{\hat {x}}u^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta_{\hat{x}}u^{n}\bigr\Vert ^{2}\bigr) \\ &{}+\frac{4}{3}h\tau\sum_{j=1}^{J-1} \delta_{x}u^{n}_{j}u^{n+1}_{j}- \frac {1}{3}h\tau\sum_{j=1}^{J-1} \delta_{\ddot {x}}u^{n}_{j}u^{n+1}_{j}=E^{n-1}= \cdots=E^{0}. \end{aligned}$$

(2.6)

Proof

Taking in (2.1) the inner product with $2\bar {u}^{n}$ and using the boundary condition (2.4) yield

$$\begin{aligned} &\frac{1}{2\tau}\bigl(\bigl\Vert u^{n+1}\bigr\Vert ^{2}-\bigl\Vert u^{n-1}\bigr\Vert ^{2}\bigr)+ \frac{2}{3\tau}\beta \bigl(\bigl\Vert \delta_{x}u^{n+1} \bigr\Vert ^{2}-\bigl\Vert \delta_{x}u^{n-1}_{x} \bigr\Vert ^{2}\bigr) \\ &\quad{}-\frac{1}{6\tau}\beta \bigl(\bigl\Vert \delta_{\hat{x}}u^{n+1}\bigr\Vert ^{2}-\bigl\Vert \delta_{\hat{x}}u^{n-1}_{x}\bigr\Vert ^{2} \bigr) \\ &\quad{} +\frac{4}{3}\bigl(\delta_{x}u^{n},2 \bar{u}^{n}\bigr)-\frac{1}{3}\bigl(\delta_{\ddot {x}}u^{n},2 \bar{u}^{n}\bigr)+\bigl(\kappa_{1}\bigl(u^{n}, \bar{u}^{n}\bigr)+\kappa_{2}\bigl(u^{n},\bar {u}^{n}\bigr),2\bar{u}^{n}\bigr)=0. \end{aligned}$$

(2.7)

Notice that

$$ \frac{4}{3}\bigl(\delta_{x}u^{n},2 \bar{u}^{n}\bigr)=\frac{4}{3}h\sum_{j=1}^{J-1} \bigl(\delta _{x}u^{n}_{j}u^{n+1}_{j}-u^{n}_{j} \delta_{x}u^{n-1}_{j}\bigr) $$

(2.8)

and

$$ \frac{1}{3}\bigl(\delta_{\ddot{x}}u^{n},2 \bar{u}^{n}\bigr)=\frac{1}{3}h\sum_{j=1}^{J-1} \bigl(\delta_{\ddot{x}}u^{n}_{j}u^{n+1}_{j}-u^{n}_{j} \delta_{\ddot {x}}u^{n-1}_{j}\bigr). $$

(2.9)

Now, computing the last term of the left-hand side in (2.7), we have

$$\begin{aligned} \bigl(\kappa_{1}\bigl(u^{n},\bar{u}^{n}\bigr),2 \bar{u}^{n}\bigr)={}&\frac{8p}{3(p+1)}\alpha h\sum _{j=1}^{J-1}\bigl\{ \bigl(u^{n}_{j} \bigr)^{p-1}\delta_{\hat{x}}\bigl(\bar{u}^{n}_{j} \bigr)+\delta _{\hat{x}}\bigl[\bigl(u^{n}_{j} \bigr)^{p-1}\bar{u}^{n}_{j}\bigr]\bigr\} \bar{u}^{n}_{j} \\ ={}& \frac{4p}{3(p+1)}\alpha\sum_{j=1}^{J-1} \bigl[\bigl(u^{n}_{j}\bigr)^{p-1}\bar {u}^{n}_{j+1}\bar{u}^{n}_{j}- \bigl(u^{n}_{j+1}\bigr)^{p-1}\bar{u}^{n}_{j+1} \bar {u}^{n}_{j}\bigr] \\ &{}-\frac{4p}{3(p+1)}\alpha\sum_{j=1}^{J-1} \bigl[\bigl(u^{n}_{j-1}\bigr)^{p-1}\bar {u}^{n}_{j}\bar{u}^{n}_{j-1}- \bigl(u^{n}_{j}\bigr)^{p-1}\bar{u}^{n}_{j} \bar {u}^{n}_{j-1}\bigr] \\ ={}&0. \end{aligned}$$

(2.10)

Similarly to the proof of (2.10), we get

$$ \bigl(\kappa_{2}\bigl(u^{n},\bar{u}^{n}\bigr),2 \bar{u}^{n}\bigr)=0. $$

(2.11)

Substituting (2.8)-(2.11) into (2.7). Let

$$\begin{aligned} E^{n}={}&\frac{1}{2}\bigl(\bigl\Vert u^{n+1}\bigr\Vert ^{2}+\bigl\Vert u^{n}\bigr\Vert ^{2} \bigr)+\frac{2}{3}\beta\bigl(\bigl\Vert \delta _{x}u^{n+1} \bigr\Vert ^{2}+\bigl\Vert \delta_{x}u^{n}\bigr\Vert ^{2}\bigr)-\frac{1}{6}\beta\bigl(\bigl\Vert \delta_{\hat {x}}u^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta_{\hat{x}}u^{n}\bigr\Vert ^{2}\bigr) \\ &{}+\frac{4}{3}h\tau\sum_{j=1}^{J-1} \delta_{x}u^{n}_{j}u^{n+1}_{j}- \frac {1}{3}h\tau\sum_{j=1}^{J-1} \delta_{\ddot{x}}u^{n}_{j}u^{n+1}_{j}. \end{aligned}$$

By the definition of $E^{n}$, (2.6) follows. □

3 Solvability and estimate for the difference solution

In this section, we shall discuss the estimate for the difference solution and the solvability of the difference scheme (2.1). For $\forall v^{n}$, $w^{n}\in Z^{0}_{h}$, we define the discrete inner products and norms on $Z^{0}_{h}$ via

$$\begin{aligned}& \bigl(v^{n},w^{n}\bigr)=h\sum _{j=1}^{J-1}v^{n}_{j} {w^{n}_{j}}, \qquad \bigl(\delta_{x}v^{n}, \delta_{x}w^{n}\bigr)_{l}=h\sum _{j=0}^{J-1}\delta _{x}v^{n}_{j} \delta_{x}{w^{n}_{j}}, \qquad \bigl\Vert v^{n}\bigr\Vert ^{2}=\bigl(v^{n},v^{n} \bigr), \\& \bigl\Vert \delta_{x}v^{n}\bigr\Vert =\sqrt{\bigl( \delta_{x}v^{n},\delta_{x}v^{n} \bigr)_{l}},\qquad \bigl\Vert \delta_{\ddot {x}}v^{n}\bigr\Vert =\sqrt{\bigl(\delta_{\ddot{x}}v^{n},\delta_{\ddot{x}}v^{n} \bigr)_{l}},\qquad \bigl\Vert v^{n}\bigr\Vert _{\infty}= \max_{1\le j\le J-1}\bigl\vert v^{n}_{j}\bigr\vert . \end{aligned}$$

To analyze the estimates of difference solution for the scheme (2.1)-(2.4), the following lemmas should be introduced.

Lemma 3.1

([31])

For a mesh function $u\in Z^{0}_{h}$, by the Cauchy-Schwarz inequality, we have

$$\begin{aligned} \Vert \delta_{\ddot{x}}u\Vert ^{2}\leq \Vert \delta_{\hat{x}}u\Vert ^{2}\leq \Vert \delta_{x}u \Vert ^{2}. \end{aligned}$$

Lemma 3.2

(Discrete Sobolev’s inequality [37])

There exist two constants $C_{1}$ and $C_{2}$ such that

$$\begin{aligned} \bigl\Vert u^{n}\bigr\Vert _{\infty}\le C_{1}\bigl\Vert u^{n}\bigr\Vert +C_{2}\bigl\Vert \delta_{x}u^{n}\bigr\Vert ^{2}. \end{aligned}$$

Theorem 3.1

Suppose that $u_{0}\in H^{1}$, then there is the estimation for the solution $u^{n}$ of the scheme (2.1): $\Vert u^{n}\Vert \le C$, $\Vert \delta_{x} u^{n}\Vert \le C$, which yields $\Vert u^{n}\Vert _{\infty}\le C$.

Proof

It follows from (2.6) and the Cauchy-Schwartz inequality that

$$\begin{aligned} &\frac{1}{2}\bigl(\bigl\Vert u^{n+1}\bigr\Vert ^{2}+\bigl\Vert u^{n}\bigr\Vert ^{2}\bigr)+ \frac{2}{3}\beta\bigl(\bigl\Vert \delta _{x}u^{n+1} \bigr\Vert ^{2}+\bigl\Vert \delta_{x}u^{n}\bigr\Vert ^{2}\bigr)-\frac{1}{6}\beta\bigl(\bigl\Vert \delta_{\hat {x}}u^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta_{\hat{x}}u^{n}\bigr\Vert ^{2}\bigr) \\ &\quad \leq C+\frac{4}{3}h\tau\sum_{j=1}^{J-1} \bigl\vert \delta_{x}u^{n}_{j}u^{n+1}_{j} \bigr\vert +\frac {1}{3}h\tau\sum_{j=1}^{J-1} \bigl\vert \delta_{\ddot{x}}u^{n}_{j}u^{n+1}_{j} \bigr\vert \\ &\quad \leq C+\frac{2}{3}\tau\bigl(\bigl\Vert \delta_{x}u^{n} \bigr\Vert ^{2}+\bigl\Vert u^{n+1}\bigr\Vert ^{2}\bigr)+\frac {1}{6}\tau\bigl(\bigl\Vert \delta_{\ddot{x}}u^{n}\bigr\Vert ^{2}+\bigl\Vert u^{n+1}\bigr\Vert ^{2}\bigr). \end{aligned}$$

(3.1)

According to Lemma 3.1, we obtain from (3.1)

$$\begin{aligned} \frac{1}{2}\biggl[\biggl(1-\frac{5}{3}\tau\biggr)\bigl\Vert u^{n+1}\bigr\Vert ^{2}+\bigl\Vert u^{n}\bigr\Vert ^{2}\biggr]+\frac {1}{2}\biggl[\beta\bigl\Vert \delta_{x}u^{n+1}\bigr\Vert ^{2}+\biggl(\beta- \frac{5}{3}\tau\biggr)\bigl\Vert \delta _{x}u^{n} \bigr\Vert ^{2}\biggr]\leq C. \end{aligned}$$

(3.2)

This implies for small τ which satisfies $\beta-\frac{5}{3}\tau> 0$ that we have

$$\begin{aligned} \bigl\Vert u^{n}\bigr\Vert \le C,\qquad \bigl\Vert \delta_{x}u^{n}\bigr\Vert \le C. \end{aligned}$$

(3.3)

Using Lemma 3.2, we obtain

$$\begin{aligned} \bigl\Vert u^{n}\bigr\Vert _{\infty}\le C. \end{aligned}$$

(3.4)

 □

Remark 3.1

Theorem 3.1 implies that the scheme (2.1) is unconditionally stable.

Theorem 3.2

The difference scheme (2.1) is uniquely solvable.

Proof

Let us prove the unique solvability by induction. It is obvious that $u^{0}$ and $u^{1}$ are uniquely determined by (2.3) and (2.2), respectively. Suppose that $u^{0},u^{1},\ldots,u^{n}$ be solved uniquely. Consider $u^{n+1}$ in (2.1) which satisfies

$$\begin{aligned} &\frac{1}{2\tau}u^{n+1}_{j}-\frac{2}{3\tau}\beta \delta_{x}\delta_{\bar {x}}u^{n+1}_{j}+ \frac{1}{6\tau}\delta_{\hat{x}}\delta_{\hat {x}}\bigl(u^{n+1}_{j} \bigr)+\frac{2p}{3(p+1)}\alpha\bigl\{ \bigl(u^{n}_{j} \bigr)^{p-1}\delta_{\hat {x}}u^{n+1}_{j}+ \delta_{\hat{x}}\bigl[\bigl(u^{n}_{j} \bigr)^{p-1}u_{j}^{n+1}\bigr]\bigr\} \\ &\quad {}-\frac{p}{6(p+1)}\bigl\{ \bigl(u^{n}_{j} \bigr)^{p-1}\delta_{\ddot{x}}u^{n+1}_{j}+\delta _{\ddot{x}}\bigl[\bigl(u^{n}_{j}\bigr)^{p-1}u_{j}^{n+1} \bigr]\bigr\} =0. \end{aligned}$$

(3.5)

Taking the inner product of (3.5) with $u^{n+1}$, we obtain

$$\begin{aligned} \frac{1}{2\tau}\bigl\Vert u^{n+1}\bigr\Vert ^{2}+ \frac{2}{3\tau}\beta\bigl\Vert \delta _{x}u^{n+1}\bigr\Vert ^{2}-\frac{1}{6\tau}\beta\bigl\Vert \delta_{\hat {x}}u^{n+1} \bigr\Vert ^{2}+\bigl(I-\mathit{II},u^{n+1}\bigr)=0, \end{aligned}$$

(3.6)

where

$$\begin{aligned}& I=\frac{2p}{3(p+1)}\alpha\bigl\{ \bigl(u^{n}_{j} \bigr)^{p-1}\delta_{\hat {x}}u^{n+1}_{j}+ \delta_{\hat{x}}\bigl[\bigl(u^{n}_{j} \bigr)^{p-1}u_{j}^{n+1}\bigr]\bigr\} , \\& \mathit{II}=\frac {p}{6(p+1)}\bigl\{ \bigl(u^{n}_{j} \bigr)^{p-1}\delta_{\ddot{x}}u^{n+1}_{j}+ \delta_{\ddot {x}}\bigl[\bigl(u^{n}_{j} \bigr)^{p-1}u_{j}^{n+1}\bigr]\bigr\} . \end{aligned}$$

Similarly to the proof of (2.10), we get

$$\begin{aligned} \bigl(I,u^{n+1}\bigr)=0, \qquad \bigl(\mathit{II},u^{n+1}\bigr)=0. \end{aligned}$$

(3.7)

It follows from (3.6)-(3.7) and Lemma 3.1 that

$$\begin{aligned} \frac{1}{2\tau}\bigl\Vert u^{n+1}\bigr\Vert ^{2}+ \frac{1}{2\tau}\beta\bigl\Vert \delta _{x}u^{n+1}\bigr\Vert ^{2}\leq0. \end{aligned}$$

(3.8)

That is, (3.5) has only a trivial solution. Hence, (2.1) determines $u^{n+1}_{j}$ uniquely. This completes the proof of Theorem 3.2. □

4 Convergence and stability of the difference scheme

First, we shall consider the truncation error of the difference scheme (2.8)-(2.10). Let $v^{n}_{j}=u(x_{j},t_{n})$. We define the truncation error as follows:

$$\begin{aligned}& \begin{aligned}[b] &Er^{n}_{j}={}\delta_{\hat{t}}v^{n}_{j}- \frac{4}{3}\beta\delta_{x}\delta_{\bar {x}} \delta_{\hat{t}}v^{n}_{j}+\frac{1}{3}\beta \delta_{\hat{x}}\delta_{\hat {x}}\delta_{\hat{t}}v^{n}_{j}+ \frac{4}{3}\delta_{\hat{x}}\bigl(v^{n}_{j}\bigr)- \frac {1}{3}\delta_{\ddot{x}}\bigl(v^{n}_{j}\bigr) \\ &\phantom{Er^{n}_{j}=}{} +\frac{p}{p+1}\alpha\biggl[\frac{4}{3}\bigl\{ \bigl( \delta_{\hat{x}}\bar {v}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}+\delta_{\hat{x}} \bigl[\bigl(v^{n}_{j}\bigr)^{p-1}\bigl( \bar{v}^{n}_{j}\bigr)\bigr]\bigr\} \\ &\phantom{Er^{n}_{j}=}{} -\frac{1}{3}\bigl\{ \bigl(\delta_{\ddot{x}}\bar{v}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}+\delta_{\ddot {x}} \bigl[\bigl(v^{n}_{j}\bigr)^{p-1}\bigl( \bar{v}^{n}_{j}\bigr)\bigr]\bigr\} \biggr],\\ &\quad 1\le j\le J-1, 1\le n\le N-1, \end{aligned} \end{aligned}$$

(4.1)

$$\begin{aligned}& \begin{aligned}[b] s^{0}_{j}={}&v^{1}_{j}- \frac{4}{3}\beta\delta_{x}\delta_{\bar{x}}v^{1}_{j}+ \frac {1}{3}\beta\delta_{\hat{x}}\delta_{\hat{x}}v^{1}_{j}-u_{0}(x_{j})\\ &{} + \frac{d^{2} u_{0}}{dx^{2}}(x_{j})+\tau\frac{du_{0}}{dx}(x_{j})+\tau \alpha p u_{0}^{p-1}(x_{j})\frac{du_{0}}{dx}(x_{j}),\end{aligned} \end{aligned}$$

(4.2)

$$\begin{aligned}& v^{0}_{j}=u_{0}(x_{j}),\quad 1\le j\le J, \end{aligned}$$

(4.3)

$$\begin{aligned}& v^{n}_{0}=v^{n}_{J}=0,\quad 0\le n\le N. \end{aligned}$$

(4.4)

Using a Taylor expansion, we obtain $\vert Er^{n}\vert +\vert s^{0}\vert =O(\tau^{2}+h^{4})$ holds if $\tau,h\to0$.

Next, we shall discuss the convergence and stability of the scheme (2.1)-(2.4).

Lemma 4.1

(Discrete Gronwall inequality [37])

Suppose that the discrete mesh function $\{w^{n}|n=1,2,\ldots,N; N\tau=T\}$ satisfies the recurrence formula

$$ w^{n}-w^{n-1}\le A\tau w^{n}+B\tau w^{n-1}+C_{n}\tau, $$

where A, B, and $C_{n}$ ($n=1,\ldots,N$) are nonnegative constants. Then

$$ \bigl\Vert w^{n}\bigr\Vert _{\infty}\le\Biggl(w^{0}+ \tau\sum_{k=1}^{N}C_{k} \Biggr)e^{2(A+B)T}, $$

where τ is small, such that $(A+B)\tau\le\frac{N-1}{2N}$ ($N > 1$).

Theorem 4.1

Assume that $u_{0}\in H^{1}$, then the solution $u^{n}$ of the scheme (2.1)-(2.4) converges to the solution of the initial-boundary problem (1.1)-(1.3) and the rate of convergence is $O(\tau^{2}+h^{4})$ by the $\Vert \cdot \Vert _{\infty}$ norm.

Proof

Let $e^{n}_{j}=v^{n}_{j}-u^{n}_{j}$. From (4.1)-(4.4) and (2.1)-(2.4), we have

$$\begin{aligned}& \begin{aligned}[b] Er^{n}_{j}={}&\delta_{\hat{t}}e^{n}_{j}- \frac{4}{3}\beta\delta_{x}\delta_{\bar {x}} \delta_{\hat{t}}e^{n}_{j}+\frac{1}{3}\beta \delta_{\hat{x}}\delta_{\hat {x}}\delta_{\hat{t}}e^{n}_{j}+ \frac{4}{3}\delta_{\hat{x}}\bigl(e^{n}_{j}\bigr)- \frac {1}{3}\delta_{\ddot{x}}\bigl(e^{n}_{j}\bigr) \\ & {}+\frac{p}{p+1}\alpha\biggl[\frac{4}{3}\bigl\{ \bigl( \delta_{\hat{x}}\bar {v}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}+\delta_{\hat{x}} \bigl[\bigl(v^{n}_{j}\bigr)^{p-1}\bigl( \bar{v}^{n}_{j}\bigr)\bigr]\bigr\} \\ & {} -\frac{4}{3}\bigl\{ \bigl(\delta_{\hat{x}}\bar{u}^{n}_{j}\bigr) \bigl(u^{n}_{j}\bigr)^{p-1}+\delta_{\hat {x}} \bigl[\bigl(u^{n}_{j}\bigr)^{p-1}\bigl( \bar{u}^{n}_{j}\bigr)\bigr]\bigr\} \biggr] \\ & {}-\frac{p}{p+1}\alpha\biggl[\frac{1}{3}\bigl\{ \bigl( \delta_{\ddot{x}}\bar {v}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}+\delta_{\ddot{x}} \bigl[\bigl(v^{n}_{j}\bigr)^{p-1}\bigl( \bar{v}^{n}_{j}\bigr)\bigr]\bigr\} \\ & {}-\frac{1}{3}\bigl\{ \bigl(\delta_{\ddot{x}}\bar{u}^{n}_{j}\bigr) \bigl(u^{n}_{j}\bigr)^{p-1}+\delta_{\ddot {x}} \bigl[\bigl(u^{n}_{j}\bigr)^{p-1}\bigl( \bar{u}^{n}_{j}\bigr)\bigr]\bigr\} \biggr], \quad 1\le j\le J-1, 1\le n\le N-1, \end{aligned} \end{aligned}$$

(4.5)

$$\begin{aligned}& s^{0}_{j}=e^{1}_{j}- \frac{4}{3}\beta\delta_{x}\delta_{\bar{x}}e^{1}_{j}+ \frac {1}{3}\beta\delta_{\hat{x}}\delta_{\hat{x}}e^{1}_{j}, \end{aligned}$$

(4.6)

$$\begin{aligned}& e^{0}_{j}=0,\quad 1\le j\le J, \end{aligned}$$

(4.7)

$$\begin{aligned}& e^{n}_{0}=e^{n}_{J}=0,\quad 0\le n\le N. \end{aligned}$$

(4.8)

Taking in (4.5) the inner product with $2\bar{e}^{n}$ (i.e. $e^{n+1}+e^{n-1}$), we obtain

$$\begin{aligned} \bigl(Er^{n},2\bar{e}^{n}\bigr)={}&\delta_{\hat{t}}\bigl\Vert e^{n}\bigr\Vert ^{2}+\frac{4}{3}\beta\delta _{\hat{t}}\bigl\Vert \delta_{x}e^{n}\bigr\Vert ^{2} -\frac{1}{3}\beta\delta_{\hat{t}}\bigl\Vert \delta_{\hat{x}}e^{n}\bigr\Vert ^{2}+ \frac {4}{3}h\sum_{j=1}^{J-1}\bigl( \delta_{x}e^{n}_{j}e^{n+1}_{j}-e^{n}_{j} \delta_{x}e^{n-1}_{j}\bigr) \\ & {}-\frac{1}{3}h\sum_{j=1}^{J-1}\bigl( \delta_{\ddot {x}}e^{n}_{j}e^{n+1}_{j}-e^{n}_{j} \delta_{\ddot {x}}e^{n-1}_{j}\bigr)+\bigl(P_{1}+P_{2}+Q_{1}+Q_{2},2 \bar{e}^{n}\bigr), \end{aligned}$$

(4.9)

where

$$\begin{aligned}& P_{1}=\frac{4p\alpha}{3(p+1)}\bigl[\bigl(\delta_{\hat{x}}\bar {v}^{n}\bigr) \bigl(v^{n}\bigr)^{p-1}-\bigl( \delta_{\hat{x}}\bar{u}^{n}\bigr) \bigl(u^{n} \bigr)^{p-1}\bigr],\\& P_{2}=\frac{4p\alpha}{3(p+1)}\bigl\{ \delta_{\hat{x}}\bigl[\bigl(v^{n}\bigr)^{p-1}\bar {v}^{n}\bigr]-\delta_{\hat{x}}\bigl[\bigl(u^{n} \bigr)^{p-1}\bar{u}^{n}\bigr]\bigr\} , \\& Q_{1}=-\frac{p\alpha}{3(p+1)}\bigl[\bigl(\delta_{\ddot{x}}\bar {v}^{n}\bigr) \bigl(v^{n}\bigr)^{p-1}-\bigl( \delta_{\ddot{x}}\bar{u}^{n}\bigr) \bigl(u^{n} \bigr)^{p-1}\bigr], \\& Q_{2}=-\frac{p\alpha}{3(p+1)}\bigl\{ \delta_{\ddot{x}}\bigl[\bigl(v^{n}\bigr)^{p-1}\bar {v}^{n}\bigr]-\delta_{\ddot{x}}\bigl[\bigl(u^{n} \bigr)^{p-1}\bar{u}^{n}\bigr]\bigr\} . \end{aligned}$$

Computing the sixth term on the right-hand side of (4.9) and using Lemma 3.1 and Theorem 3.1 yield

$$\begin{aligned} \bigl(P_{1},2\bar{e}^{n}\bigr)&=\frac{4p\alpha}{3(p+1)}\bigl( \bigl[\bigl(\delta_{\hat{x}}\bar {v}^{n}\bigr) \bigl(v^{n}\bigr)^{p-1}-\bigl(\delta_{\hat{x}} \bar{u}^{n}\bigr) \bigl(u^{n}\bigr)^{p-1}\bigr],2 \bar{e}^{n}\bigr) \\ &=\frac{8p\alpha}{3(p+1)}\bigl(\bigl[\bigl(\delta_{\hat{x}}\bar {e}^{n}\bigr) \bigl(v^{n}\bigr)^{p-1}+\bigl( \delta_{\hat{x}}\bar {u}^{n}\bigr) \bigl(\bigl(v^{n} \bigr)^{p-1}-\bigl(u^{n}\bigr)^{p-1}\bigr)\bigr], \bar{e}^{n}\bigr) \\ &=\frac{8p\alpha}{3(p+1)}h\Biggl\{ \sum_{j=1}^{J-1} \bigl(\delta_{\hat{x}}\bar {e}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}\bar{e}^{n}_{j} +\sum_{j=1}^{J-1}\bigl(\delta_{\hat{x}} \bar {u}^{n}_{j}\bigr)\bigl[\bigl(v^{n}_{j} \bigr)^{p-1}-\bigl(u^{n}_{j}\bigr)^{p-1} \bigr]\bar{e}^{n}_{j}\Biggr\} \\ &=\frac{8p\alpha}{3(p+1)}h\Biggl(\sum_{j=1}^{J-1} \bigl(\delta_{\hat{x}}\bar {e}^{n}_{j}\bigr) \bigl(v^{n}_{j}\bigr)^{p-1}\bar{e}^{n}_{j} +\sum_{j=1}^{J-1}\bigl(\delta_{\hat{x}} \bar{u}^{n}_{j}\bigr)\Biggl[e_{j}^{n} \sum_{k=0}^{p-2}\bigl(v^{n}_{j} \bigr)^{p-2-k}\bigl(u^{n}_{j}\bigr)^{k} \Biggr]\bar{e}^{n}_{j}\Biggr) \\ &\le C\bigl(\bigl\Vert \delta_{\hat{x}}\bar{e}^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert \bar {e}^{n}\bigr\Vert ^{2}\bigr) \\ &\le C\bigl(\bigl\Vert \delta_{x}e^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n-1} \bigr\Vert ^{2}\bigr), \end{aligned}$$

(4.10)

where the Cauchy-Schwartz inequality and Lemma 3.1 are used.

Similarly, we can also obtain

$$\begin{aligned} &\bigl(P_{2},2\bar{e}^{n}\bigr) \le C\bigl(\bigl\Vert \delta_{x}e^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n-1}\bigr\Vert ^{2} \bigr), \end{aligned}$$

(4.11)

$$\begin{aligned} &\bigl(Q_{1},2\bar{e}^{n}\bigr) \le C\bigl(\bigl\Vert \delta_{x}e^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n-1}\bigr\Vert ^{2} \bigr), \end{aligned}$$

(4.12)

$$\begin{aligned} &\bigl(Q_{2},2\bar{e}^{n}\bigr) \le C\bigl(\bigl\Vert \delta_{x}e^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n-1}\bigr\Vert ^{2} \bigr). \end{aligned}$$

(4.13)

In addition, it is obvious that

$$\begin{aligned}& \bigl(Er^{n},2\bar{e}^{n}\bigr)\le\bigl\Vert Er^{n}\bigr\Vert ^{2} +\frac{1}{2}\bigl(\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n-1}\bigr\Vert ^{2}\bigr), \end{aligned}$$

(4.14)

$$\begin{aligned}& \frac{4}{3}h\sum_{j=1}^{J-1}\bigl( \delta_{x}e^{n}_{j}e^{n+1}_{j}-e^{n}_{j} \delta _{x}e^{n-1}_{j}\bigr)\leq C\bigl(\bigl\Vert \delta_{x}e^{n}\bigr\Vert ^{2}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}\bigr), \end{aligned}$$

(4.15)

$$\begin{aligned}& -\frac{1}{3}h\sum_{j=1}^{J-1}\bigl( \delta_{\ddot {x}}e^{n}_{j}e^{n+1}_{j}-e^{n}_{j} \delta_{\ddot{x}}e^{n-1}_{j}\bigr)\leq C\bigl(\bigl\Vert \delta _{x}e^{n}\bigr\Vert ^{2}+\bigl\Vert \delta_{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}\bigr). \end{aligned}$$

(4.16)

It follows from (4.9)-(4.16) that

$$\begin{aligned} &\delta_{\hat{t}}\bigl\Vert e^{n}\bigr\Vert ^{2}+ \frac{4}{3}\beta\delta_{\hat{t}}\bigl\Vert \delta _{x}e^{n}\bigr\Vert ^{2}-\frac{1}{3} \beta\delta_{\hat{t}}\bigl\Vert \delta_{\hat {x}}e^{n}\bigr\Vert ^{2} \\ &\quad \le\bigl\Vert Er^{n}\bigr\Vert ^{2}+C\bigl(\bigl\Vert \delta_{x}e^{n+1}\bigr\Vert ^{2}+\bigl\Vert \delta_{x}e^{n}\bigr\Vert ^{2} \\ &\qquad {}+\bigl\Vert \delta _{x}e^{n-1}\bigr\Vert ^{2}+\bigl\Vert e^{n+1}\bigr\Vert ^{2}+\bigl\Vert e^{n}\bigr\Vert ^{2}+\bigl\Vert e^{n-1}\bigr\Vert ^{2} \bigr). \end{aligned}$$

(4.17)

Let $B^{n}=\frac{1}{2}(\Vert e^{n+1}\Vert ^{2}+\Vert e^{n}\Vert ^{2})+\frac{\beta}{2}(\Vert \delta _{x}e^{n+1}\Vert ^{2}+\Vert \delta_{x}e^{n}\Vert ^{2})$. Using Lemma 3.1, (4.17) can be written as follows:

$$\begin{aligned} B^{n}-B^{n-1}\le\tau\bigl\Vert Er^{n}\bigr\Vert ^{2}+C\tau\bigl(B^{n}+B^{n-1}\bigr). \end{aligned}$$

(4.18)

According to Lemma 4.1, we can immediately obtain

$$\begin{aligned} B^{n}\le\Bigl(B^{0}+T\sup_{l\leq n\leq N}\bigl\Vert Er^{n}\bigr\Vert ^{2}\Bigr) e^{CT}. \end{aligned}$$

(4.19)

Taking the inner product of (4.6) with $e^{1}$ yields

$$\begin{aligned} \bigl(s^{0},e^{1}\bigr)=\bigl\Vert e^{1}\bigr\Vert ^{2}+\frac{4}{3}\beta\bigl\Vert \delta_{x}e^{1} \bigr\Vert ^{2}-\frac{1}{3}\beta \bigl\Vert \delta_{\ddot{x}}e^{1}\bigr\Vert ^{2}. \end{aligned}$$

(4.20)

This, together with $(s^{0},e^{1})\leq\frac{1}{2}(\Vert s^{0}\Vert ^{2}+\Vert e^{1}\Vert ^{2})$, $\vert s^{0}\vert =O(\tau^{2}+h^{4})$, and Lemma 3.1, gives

$$\begin{aligned} \bigl\Vert e^{1}\bigr\Vert \leq O\bigl(\tau^{2}+h^{4} \bigr), \qquad \bigl\Vert \delta_{x}e^{1}\bigr\Vert \leq O\bigl( \tau^{2}+h^{4}\bigr). \end{aligned}$$

(4.21)

From the discrete initial condition (4.7), we know that $B^{0}=[O(\tau^{2}+h^{4})]^{2}$.

It follows from (4.19) that

$$\begin{aligned} B^{n}\le\bigl[O\bigl(\tau^{2}+h^{4}\bigr) \bigr]^{2}. \end{aligned}$$

(4.22)

Then we have

$$\begin{aligned} \bigl\Vert e^{n}\bigr\Vert \le O\bigl(\tau^{2}+h^{4} \bigr),\qquad \bigl\Vert \delta_{x}e^{n}\bigr\Vert \le O\bigl( \tau^{2}+h^{4}\bigr). \end{aligned}$$

(4.23)

This, together with Lemmas 3.2, gives

$$\begin{aligned} \bigl\Vert e^{n}\bigr\Vert _{\infty}\le O\bigl( \tau^{2}+h^{4}\bigr). \end{aligned}$$

(4.24)

This completes the proof of Theorem 4.1. □

Similarly, we can prove stability of the difference solution.

Theorem 4.2

Under the conditions of Theorem  4.1, the solution of the scheme (2.1)-(2.4) is unconditionally stable by the $\Vert \cdot \Vert _{\infty}$ norm.

5 Numerical experiments

In this section, we give some numerical experiments to demonstrate our theoretical results obtained in the previous sections. We will measure the accuracy of the proposed scheme using the absolute error defined by $e^{n}=\Vert v^{n}-u^{n}\Vert _{\infty}$.

Consider the initial-boundary value problem (1.1)-(1.3). In the numerical experiments, we take $x_{l}=-40$, $x_{r}=60$, $T=10$, and choose three cases $p=2,3,4$, respectively. In order to verify the accuracy $O(\tau^{2}+h^{4})$, we take τ and h small enough to verify the fourth-order accuracy and second-order accuracy in the spatial and temporal directions, respectively. The convergence order figures of $\log(e^{n})\mbox{-}\log(h)$ with h and the ones of $\log (e^{n})\mbox{-}\log(\tau)$ with τ small enough are given in Figures 1-6 for various mesh steps h and τ at $t=10$. From Figures 1-6, it is obvious that the scheme (2.1)-(2.4) is convergent in the maximum norm, and the convergence order is $O(\tau^{2}+h^{4})$.

We show in Theorem 2.1 that the numerical solution $u^{n}$ of the scheme (2.1) satisfies the conservation of discrete energy. In Tables 1-3, the values of $E^{n}$ for the scheme (2.1) are presented for three cases $p=2,3,4$ under various mesh steps h and τ, respectively. It is easy to see from Tables 1-3 that the scheme (2.1) preserves the discrete energy very well, which also shows the accuracy and efficiency of the scheme in this paper.

Case 5.1

Take $p=2$. Consider the following initial-boundary problem of RLW equation:

$$\begin{aligned} &u_{t}-u_{xxt}+u_{x}+uu_{x}=0, \end{aligned}$$

(5.1)

$$\begin{aligned} &u(x,0)=u_{0}(x), \end{aligned}$$

(5.2)

$$\begin{aligned} &u(x_{l},t)=u(x_{r},t)=0. \end{aligned}$$

(5.3)

In computations, we choose the initial condition $u(x,0)=\operatorname {sech}^{2}(\frac{1}{4}x)$ [10]. The convergence order figures and the values of $E^{n}$ are shown in Figures 1-2 and Table 1, respectively.

Figure 1

The spatial convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Figure 2

The temporal convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Table 1 Discrete energy $\pmb{E^{n}}$ of the scheme ( 2.1 ) at different time t when $\pmb{h=0.1,0.05}$ and $\pmb{p=2}$

Case 5.2

Take $p=3$. We consider the following initial-boundary problem of MRLW equation:

$$\begin{aligned} &u_{t}-u_{xxt}+u_{x}+\bigl(u^{3} \bigr)_{x}=0, \end{aligned}$$

(5.4)

$$\begin{aligned} &u(x,0)=u_{0}(x), \end{aligned}$$

(5.5)

$$\begin{aligned} &u(x_{l},t)=u(x_{r},t)=0. \end{aligned}$$

(5.6)

In experiments, we choose the initial condition $u(x,0)=\frac{\sqrt {6}}{3}\operatorname{sech}(\frac{1}{2}x)$ [1]. The convergence order figures and the values of $E^{n}$ are shown in Figures 3-4 and Table 2, respectively.

Figure 3

The spatial convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Figure 4

The temporal convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Table 2 Discrete energy $\pmb{E^{n}}$ of the scheme ( 2.1 ) at different times t when $\pmb{h=0.1,0.05}$ and $\pmb{p=3}$

Case 5.3

Take $p=4$. We consider the initial-boundary problem (1.1)-(1.3) of GRLW equation:

$$\begin{aligned} &u_{t}-u_{xxt}+u_{x}+u^{3}u_{x}=0, \end{aligned}$$

(5.7)

$$\begin{aligned} &u(x,0)=u_{0}(x), \end{aligned}$$

(5.8)

$$\begin{aligned} &u(x_{l},t)=u(x_{r},t)=0. \end{aligned}$$

(5.9)

In the following experiments, we choose the initial condition $u(x,0)=\operatorname{sech}^{\frac{2}{3}}(\frac{3\sqrt{11}}{22}x)$ [1]. The convergence order figures and the values of $E^{n}$ are shown in Figures 5-6 and Table 3, respectively.

Figure 5

The spatial convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Figure 6

The temporal convergence order in maximal norm for $\pmb{u^{n}}$ at $\pmb{t=10}$ with different h and τ computed by the scheme ( 2.1 )-( 2.4 ).

Table 3 Discrete energy $\pmb{E^{n}}$ of the scheme ( 2.1 ) at different time t when $\pmb{(h,\tau)=(0.1,0.1)}$ and (0.05,0.025) for $\pmb{p=4}$