A conservative numerical scheme for Rosenau-RLW equation based on multiple integral finite volume method

Abstract

A multiple integral finite volume method combined and Lagrange interpolation are applied in this paper to the Rosenau-RLW (RRLW) equation. We construct a two-level implicit fully discrete scheme for the RRLW equation. The numerical scheme has the accuracy of third order in space and second order in time, respectively. The solvability and uniqueness of the numerical solution are shown. We verify that the numerical scheme keeps the original equation characteristic of energy conservation. It is proved that the numerical scheme is convergent in the order of $O(\tau ^{2} + h^{3})$ and unconditionally stable. A numerical experiment is given to demonstrate the validity and accuracy of scheme.

1 Introduction

It is well known that nonlinear partial differential equations exist in many areas of mathematical physics and fluid mechanics. In the nonlinear evolution equations, the Korteweg–de Vries (KdV) and Rosenau-RLW (RLW) equations are two typical cases, given by

$$ u_{t} + u_{xxx} + uu_{x} = 0 $$

(1.1)

and

$$ u_{t} - u_{xxt} + u_{x} + uu_{x} = 0. $$

(1.2)

The KdV equation (1.1) is a nonlinear model used to study the change forms of long waves propagating in a rectangular channel. The RLW equation (1.2) is used to simulate wave motion in media with nonlinear wave steeping and dispersion. The RLW equation was proposed by Peregrine [1, 2] based on the classical KdV equation, and an explanation of different situations of a nonlinear dispersive wave was given in his research.

At the same time, we notice that the motion described by the RLW equation has the same approximate boundary as the KdV equation. It is well known that the KdV equation has corresponding shortcomings. With the aim to overcome these unavoidable shortcomings of the KdV equation, Rosenau [3, 4] introduced an equation

$$ u_{t} + u_{xxxxt} + u_{x} + uu_{x} = 0. $$

(1.3)

The theoretical studies, on the existence, uniqueness and regularity for the solution of (1.3), have been performed by Park [5]. Various numerical techniques have been used to solve the Rosenau equation [6,7,8,9,10,11,12], particularly including the discontinuous Galerkin method, the C1-conforming finite element method [13], the finite difference method and the orthogonal cubic spline collocation method. More detailed solving processes can be obtained in Refs. [14,15,16,17,18,19].

On the other hand, for further understanding more general nonlinear behaviors of the waves, the term $- u_{xxt}$ needs to be considered. So we address

$$ u_{t} - u_{xxt} + u_{xxxxt} + u_{x} + uu_{x} = 0 . $$

(1.4)

This equation is usually called the Rosenau-RLW (RRLW) equation. Zuo et al. [20] proposed a Crank–Nicolson finite difference method for the RRLW equation. Meanwhile, a three-level difference scheme for (1.4) is investigated by Pan et al. [21]. Furthermore, the finite element approximate solution is used to solve (1.4) and related error estimations for both semi-discrete and fully discrete Galerkin methods are established [22,23,24]. The coupling equation of KdV and RRLW is also solved through a three-level average implicit finite difference scheme, showing second-order accuracy in space and time, simultaneously [25]. In Ref. [26], the Galerkin cubic B-spline finite element method is proposed to construct the numerical scheme for the RRLW equation. Pan [27] investigated the C-N scheme of RRLW equation through a more classic finite difference approach, and corresponding solvability and convergence have been proved. In addition, the difference scheme for the general RRLW equation is constructed by Wang [28] with some theoretical proofs.

The main contribution of the current work is to present a two-level implicit numerical scheme for the following RRLW equation with some theoretic analysis:

$$ u_{t} - u_{xxt} + u_{xxxxt} + u_{x} + uu_{x} = 0,\quad ( x,t ) \in ( x_{l},x_{r} ) \times [ 0,T ] $$

(1.5)

with the initial condition

$$ u(x,0)=u_{0}(x),\quad x\in[x_{l},x_{r}] $$

(1.6)

and boundary conditions

$$ u(x_{l},t)=u(x_{r},t)=0,\qquad u_{xx}(x_{l},t)=u_{xx}(x_{r},t)=0,\quad t\in[0,T]. $$

(1.7)

Throughout this paper, we assume that the initial condition $u_{0}(x)$ is sufficiently smooth as required by the error analysis. The system (1.5)–(1.7) is known to satisfy the following conservative law:

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

(1.8)

The contents of this paperis as follows. Firstly, in Sect. 2, we present some notations and lemmas. In Sect. 3, we propose a multiple integral finite volume method which is a tool for discrete partial differential equations. Thus a two-level implicit numerical scheme for the RRLW equation is obtained. Next we discuss the discrete energy conservative laws of the numerical scheme and prove its solvability and uniqueness in Sect. 3. We give prior estimates of the numerical scheme in Sect. 4 and prove that the numerical scheme is convergent and stable in Sect. 5. Finally, the error analysis and energy analysis of numerical examples are given in Sect. 6.

In fact, the numerical discrete scheme with parameters constructed by us shows all the discrete schemes that can be constructed by the finite difference method. By choosing the undetermined parameters, we find the best discrete scheme. The best discrete scheme can preserve the conservation property of the original differential equation well. At the same time, for unknown functions in the original differential equation, this method reduces greatly the requirements for the unknown functions in terms of mathematics. More importantly, we explain the concrete and detailed methods about improving the accuracy of the numerical discrete scheme.

2 Some notations and lemmas

2.1 Some notations

Let h and τ be the uniform step sizes in the spatial direction and temporal direction, respectively, where $f=\frac{x_{r}-x_{l}}{J}$ and $\tau=\frac{T}{N}$.

Denote $x_{j} = x_{l} + jh \ ( 0 \le j \le J )$, $t_{n} = n\tau \ ( 0 \le n \le N )$ and $u_{j}^{n} \approx u ( x_{l} + jh,n\tau )$. Denote the grid $\varOmega _{h} = \{ x_{j}|j = 0,1, \ldots,J \}$ and $Z_{h}^{0} = \{ u_{j}|u_{0} = u_{J} = 0,j = 0,1, \ldots,J\}$. As usual, the following notations will be used:

$$\begin{aligned} &\bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{t}} = \frac{u_{j} ^{n + 1} - u_{j}^{n}}{\tau },\qquad u_{j}^{n + \frac{1}{2}} = \frac{u_{j}^{n + 1} + u_{j}^{n}}{2},\\ &\bigl( u_{j}^{n} \bigr)_{x} = \frac{u_{j + 1}^{n} - u_{j}^{n}}{h}, \qquad \bigl( u_{j}^{n} \bigr)_{\overline{x}} = \frac{u_{j}^{n} - u_{j - 1} ^{n}}{h}, \\ &\bigl( u_{j}^{n} \bigr)_{\widehat{x}} = \frac{u_{j + 1}^{n} - u _{j - 1}^{n}}{2h},\qquad \bigl( u_{j}^{n} \bigr)_{x\overline{x}} = \bigl( u_{j}^{n} \bigr) _{\overline{x}x} = \frac{u_{j + 1}^{n} - 2u_{j}^{n} + u_{j - 1}^{n}}{h ^{2}}, \\ &\bigl(u_{j}^{n + \frac{1}{2}}\bigr)_{xx\overline{x}\overline{x}} = \frac{u_{j - 2}^{n + \frac{1}{2}} - 4u_{j - 1}^{n + \frac{1}{2}} + 6u_{j}^{n + \frac{1}{2}} - 4u_{j + 1}^{n + \frac{1}{2}} + u_{j + 2}^{n + \frac{1}{2}}}{h^{4}},\qquad \bigl(u^{n},v^{n}\bigr) = h\sum _{j} u_{j}^{n} v_{j}^{n}, \\ &\bigl\Vert u^{n} \bigr\Vert = \sqrt{\bigl(u^{n},u^{n} \bigr)},\qquad \bigl\Vert u^{n} \bigr\Vert _{\infty } = \max_{0 \le j \le J} \bigl\vert u_{j}^{n} \bigr\vert . \end{aligned}$$

In this paper we denote by C a positive constant, which may be of different values on different occasions.

2.2 Some lemmas

Lemma 2.1

For any two mesh functions $u,v \in Z_{h}^{0}$, we have

$$ ( u_{x},v ) = - ( u,v_{\overline{x}} ),\qquad ( u_{x\overline{x}},v ) = - ( u_{x},v_{x} ),\qquad ( u_{x\overline{x}},u ) = - ( u_{x},u_{x} ) = - \Vert u_{x} \Vert ^{2}. $$

Furthermore, if $( u_{0}^{n} )_{x\overline{x}} = ( u_{J}^{n} )_{x\overline{x}} = 0$, then $( ( {u}^{n} )_{xx\overline{x}\overline{x}},u^{n} ) = \Vert u_{xx}^{n} \Vert ^{2}$.

Lemma 2.2

$2B - E$ is a positive definite matrix, where matrix E is the identity matrix of order J + 1 and

$$ B = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 7 & 1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 7 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 7 & 1 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 7 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{bmatrix}_{(J + 1) \times (J + 1)}. $$

Proof

We know that

$$ 2B - E = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 13 & 2 & \cdots & 0 & 0 & 0 \\ 0 & 2 & 13 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 13 & 2 & 0 \\ 0 & 0 & 0 & \cdots & 2 & 13 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{bmatrix}_{(J + 1) \times (J + 1)}. $$

Let $B_1=[1],B_2=\left[\begin{array}{cc}1& 0\\ 0& 13\end{array}\right],B_3=\left[\begin{array}{ccc}1& 0& 0\\ 0& 13& 2\\ 0& 2& 13\end{array}\right]$, $B_{J + 1} = 2B - E$. It is obvious that $\vert B_{1} \vert = 1 > 0, \vert B_{2} \vert = 13 > 0, \vert B_{3} \vert = 165 > 0$ and $\{ \vert B_{j} \vert \}\ (j = 4, \ldots,J)$ obeys

$$ \vert B_{j} \vert = 13 \vert B_{j - 1} \vert - \vert B_{j - 2} \vert + \vert B_{j - 3} \vert . $$

We can get

$$\begin{aligned} \begin{aligned} & \vert B_{4} \vert = 13 \vert B_{3} \vert - \vert B_{2} \vert + \vert B_{1} \vert , \\ &\vert B_{5} \vert = 13 \vert B_{4} \vert - \vert B_{3} \vert + \vert B_{2} \vert , \\ &\cdots \\ &\vert B_{J} \vert = 13 \vert B_{J - 1} \vert - \vert B_{J - 2} \vert + \vert B_{J - 3} \vert . \end{aligned} \end{aligned}$$

(2.1)

We assume that $\vert B_{1} \vert , \vert B_{2} \vert , \ldots , \vert B_{J - 1} \vert > 0$. We want to prove $\vert B_{J} \vert > 0$. From (2.1),

$$\begin{aligned} \vert B_{J} \vert &= 12 \bigl( \vert B_{J - 1} \vert + \cdots + \vert B_{3} \vert \bigr) - \vert B_{J - 2} \vert + \vert B_{3} \vert + \vert B_{1} \vert \\ &= 11 \bigl( \vert B_{J - 1} \vert + \cdots + \vert B_{3} \vert \bigr) + \bigl( \vert B_{J - 1} \vert + \vert B _{J - 3} \vert + \cdots + \vert B_{3} \vert \bigr) + 166 > 0 . \end{aligned}$$

(2.2)

By the assumption, we have

$$ \vert B_{J + 1} \vert = \vert B_{J} \vert > 0. $$

So $2B - E$ is a positive definite matrix. □

Lemma 2.3

(Fixed point theorem [29])

H is a finite dimensional inner product space. We assume $g:H \to H$ and g is continuous. If there exists $\alpha > 0$, for any x in H, as long as $\Vert x \Vert = \alpha$, we have $(g(x),x) > 0$. Then there must exist $x^{ *} \in H\ ( \Vert x^{ *} \Vert \le \alpha )$, which obeys $g(x^{ *} ) = 0$.

Lemma 2.4

(Discrete Sobolev’s inequality [29])

For any discrete function $u_{h}$ and for any given $\varepsilon > 0$, there exists a constant $K ( \varepsilon, n )$, depending only on ε and n, such that

$$ \Vert u \Vert _{\infty } \le \varepsilon \bigl\Vert {u}_{x}^{n} \bigr\Vert + K ( \varepsilon, n ) \bigl\Vert {u}^{n} \bigr\Vert . $$

Lemma 2.5

(Discrete Gronwall’s inequality [30])

Suppose that the discrete function $w_{h}$ 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, $C_{n} \ ( n = 1,2, \ldots,N )$ are nonnegative constants. Then

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

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

3 An implicit conservative numerical scheme and its discrete conservative law

3.1 The multiple integral finite volume method

It is necessary to introduce the multiple integral finite volume method (MIFVM) briefly. The method is a new approximation method for solving partial differential equation, which is proposed by Yuesheng Luo. The basic idea is to make the original partial differential equation to be an integral equation by a certain number of integrations in the spatial x direction. The goal is that the integral equation no longer contains the derivative item of the unknown function. In this way, we avoid handling the approximation as regards the derivative term.

Firstly, the number of integrations depends on the order of the highest derivative in the spatial x direction of the unknown function in the partial differential equation. The relationship between m and r satisfies $m = 2^{r} - 1$, where m is the number of integrations and r is the order of the highest derivative of the unknown function in the spatial x direction.

Secondly, is it well known that the original partial differential equation usually is expressed by the derivative of an unknown function and an unknown function, for example $u_{x}, u_{xx}, u_{xxx}, u_{xxxx}$ and u. The MIFVM is to transform the original partial differential equation into an integral equation, which is expressed by $u_{j}^{n}$. $u_{j}^{n} \approx u ( x_{l} + jh,n\tau )$ is the unknown function’s approximate value at the grid node. If there is $u_{j + \varepsilon }^{n}$, which is not at the grid node, in the integral equation, we often use the Lagrange polynomial to deal with it. This step is just an approximation to the original function.

For example, for (1.5), the order of the highest derivative of unknown function is four. By $m = 2^{4} - 1$, we should make 15 times integrations for every item, as shown by

$$ \int _{xxxx} u_{t} - \int _{xxxx} u_{xxt} + \int _{xxxx} u_{xxxxt} + \int _{xxxx} u_{x} + \int _{xxxx} uu_{x} = 0, $$

(3.1)

where

$$\begin{aligned} \int _{xxxx} u &\overset{\mathrm{def}}{=} \int _{x_{j} + \varepsilon _{7}}^{x_{j} + \varepsilon _{8}} \,dx_{f_{2}} \int _{x_{j} + \varepsilon _{6}}^{x_{j} + \varepsilon _{7}} \,dx_{f_{1}} \int _{x_{j} + \varepsilon _{5}}^{x_{j} + \varepsilon _{6}} \,dx_{e_{2}} \int _{x_{j}}^{x_{j} + \varepsilon _{5}} \,dx_{e_{1}} \\ &\quad{} \times \int _{x_{j} - \varepsilon _{4}}^{x_{j}} \,dx_{d_{2}} \int _{x_{j} - \varepsilon _{3}}^{x_{j} - \varepsilon _{4}} \,dx_{d_{1}} \int _{x_{j} - \varepsilon _{2}}^{x_{j} - \varepsilon _{3}} \,dx_{c_{2}} \\ &\quad{} \times\int _{x_{j} - \varepsilon _{1}}^{x_{j} - \varepsilon _{2}} \,dx_{c_{1}} \int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx _{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \\ &\quad{} \times \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} u \,dx, \end{aligned}$$

(3.2)

$\varepsilon _{i} \in R,i = 1,2, \ldots,8$. The center difference is used to deal with the first derivative in the time direction. Then we can get a numerical scheme for the original equation.

3.2 A two-level implicit numerical scheme

In order to get a numerical scheme, which can preserve some properties of the original equation, applying MIFVM to Eq. (1.5) on time $n + \frac{1}{2}$ level, we let $\varepsilon _{1} = - \varepsilon _{4} = - \varepsilon _{5} = \varepsilon _{8} = \sqrt{3} h$, $\varepsilon _{2} = - \varepsilon _{3} = - \varepsilon _{6} = \varepsilon _{7} = \frac{ \sqrt{3}}{3}h$. So we can get

$$\begin{aligned} &\int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx _{f_{2}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{ \sqrt{3}}{3}h} \,dx_{f_{1}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{ \sqrt{3}}{3}h} \,dx_{e_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j}} \,dx_{e _{1}} \int _{x_{j}}^{x_{j} + \sqrt{3} h} \,dx_{d_{2}} \\ &\quad{} \times \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{d_{1}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx _{c_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{\sqrt{3}}{3}h} \,dx _{c_{1}} \\ &\quad{} \times\int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx_{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} u_{t}\,dx \\ &\quad{}- \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{f _{2}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx_{f_{1}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{ \sqrt{3}}{3}h} \,dx_{e_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j}} \,dx_{e _{1}} \int _{x_{j}}^{x_{j} + \sqrt{3} h} \,dx_{d_{2}} \\ &\quad{} \times \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{d_{1}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx _{c_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{\sqrt{3}}{3}h} \,dx _{c_{1}} \\ &\quad{} \times\int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx_{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} u_{xxt}\,dx \\ &\quad{}+ \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{f _{2}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx_{f_{1}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{ \sqrt{3}}{3}h} \,dx_{e_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j}} \,dx_{e _{1}} \int _{x_{j}}^{x_{j} + \sqrt{3} h} \,dx_{d_{2}} \\ &\quad{} \times \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{d_{1}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx _{c_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{\sqrt{3}}{3}h} \,dx _{c_{1}} \\ &\quad{} \times\int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx_{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} u_{xxxxt}\,dx \\ &\quad {}+ \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{f _{2}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx_{f_{1}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{ \sqrt{3}}{3}h} \,dx_{e_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j}} \,dx_{e _{1}} \int _{x_{j}}^{x_{j} + \sqrt{3} h} \,dx_{d_{2}} \\ &\quad{} \times \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{d_{1}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx _{c_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{\sqrt{3}}{3}h} \,dx _{c_{1}} \\ &\quad{} \times\int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx_{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} u_{x}\,dx \\ &\quad{}+ \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{f _{2}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx_{f_{1}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{ \sqrt{3}}{3}h} \,dx_{e_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j}} \,dx_{e _{1}} \int _{x_{j}}^{x_{j} + \sqrt{3} h} \,dx_{d_{2}} \\ &\quad{} \times \int _{x_{j} + \frac{\sqrt{3}}{3}h}^{x_{j} + \sqrt{3} h} \,dx_{d_{1}} \int _{x_{j} - \frac{\sqrt{3}}{3}h}^{x_{j} + \frac{\sqrt{3}}{3}h} \,dx _{c_{2}} \int _{x_{j} - \sqrt{3} h}^{x_{j} - \frac{\sqrt{3}}{3}h} \,dx _{c_{1}} \\ &\quad{} \times\int _{x_{f_{1}}}^{x_{f_{2}}} \,dx_{f} \int _{x_{e_{1}}}^{x_{e_{2}}} \,dx_{e} \int _{x_{d_{1}}}^{x_{d_{2}}} \,dx_{d} \int _{x_{c_{1}}}^{x_{c_{2}}} \,dx_{c} \int _{x_{e}}^{x_{f}} \,dx_{b} \int _{x_{c}}^{x_{d}} \,dx_{a} \int _{x_{a}}^{x_{b}} uu_{x}\,dx = 0. \end{aligned}$$

(3.3)

There are five integration items. For the first item, the approximation of the first-order derivative in the time direction is

$$ u_{t} \bigl( x,t^{{n + \frac{1}{2}}} \bigr) = \frac{u^{n + 1} ( x ) - u^{n} ( x )}{\tau } + O \bigl( \tau ^{2} \bigr). $$

(3.4)

For the third item, we use the five points Lagrange interpolation for the x direction. It is

$$\begin{aligned} u ( x,t ) = {}&\frac{ ( x - x_{j - 1} ) ( x - x_{j} ) ( x - x_{j + 1} ) ( x - x_{j + 2} )}{ ( x_{j - 2} - x_{j - 1} ) ( x_{j - 2} - x_{j} ) ( x_{j - 2} - x_{j + 1} ) ( x_{j - 2} - x_{j + 2} )}u _{j - 2} ( t ) \\ &{}+ \frac{ ( x - x_{j - 2} ) ( x - x_{j} ) ( x - x_{j + 1} ) ( x - x_{j + 2} )}{ ( x_{j - 1} - x_{j - 2} ) ( x_{j - 1} - x_{j} ) ( x_{j - 1} - x_{j + 1} ) ( x_{j - 1} - x_{j + 2} )}u_{j - 1} ( t ) \\ &{}+ \frac{ ( x - x_{j - 2} ) ( x - x_{j - 1} ) ( x - x_{j + 1} ) ( x - x_{j + 2} )}{ ( x_{j} - x_{j - 2} ) ( x_{j} - x_{j - 1} ) ( x _{j} - x_{j + 1} ) ( x_{j} - x_{j + 2} )}u_{j} ( t ) \\ &{}+ \frac{ ( x - x_{j - 2} ) ( x - x_{j - 1} ) ( x - x_{j} ) ( x - x_{j + 2} )}{ ( x _{j + 1} - x_{j - 2} ) ( x_{j + 1} - x_{j - 1} ) ( x_{j + 1} - x_{j} ) ( x_{j + 1} - x_{j + 2} )}u _{j + 1} ( t ) \\ &{}+ \frac{ ( x - x_{j - 2} ) ( x - x_{j - 1} ) ( x - x_{j} ) ( x - x_{j + 1} )}{ ( x _{j + 2} - x_{j - 2} ) ( x_{j + 2} - x_{j - 1} ) ( x_{j + 2} - x_{j} ) ( x_{j + 2} - x_{j + 1} )}u _{j + 2} ( t ) + O\bigl(h^{5} \bigr). \end{aligned}$$

(3.5)

For the other three items, we use three points Lagrange interpolation for the x direction. It is

$$\begin{aligned} u ( x,t ) ={}& \frac{ ( x - x_{j} ) ( x - x _{j + 1} )}{ ( x_{j - 1} - x_{j} ) ( x_{j - 1} - x_{j + 1} )}u_{j - 1} ( t ) + \frac{ ( x - x _{j - 1} ) ( x - x_{j + 1} )}{ ( x_{j} - x _{j - 1} ) ( x_{j} - x_{j + 1} )}u_{j} ( t ) \\ &{}+ \frac{ ( x - x_{j - 1} ) ( x - x_{j} )}{ ( x_{j + 1} - x_{j - 1} ) ( x_{j + 1} - x_{j} )}u _{j + 1} ( t ) + O\bigl(h^{3} \bigr) . \end{aligned}$$

(3.6)

Substituting (3.4), (3.5) and (3.6) into (3.3) and simplifying, we get a two-level implicit scheme for equation (1.5). We obtain

$$\begin{aligned} &\frac{1}{9} \bigl( u_{j - 1}^{n + \frac{1}{2}} + 7u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr)_{\widehat{t}} - \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{x\overline{x}\widehat{t}} + \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{xx\overline{x}\overline{x} \widehat{t}} + \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \\ &\quad{}+ \frac{1}{3} \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \bigl( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr) = 0, \quad 1 \le j \le J - 1, 1 \le n \le N - 1, \end{aligned}$$

(3.7)

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

(3.8)

$$\begin{aligned} &u_{0}^{n} = u_{J}^{n} = 0,\qquad \bigl(u_{0}^{n}\bigr)_{x\overline{x}} = \bigl(u_{J}^{n}\bigr)_{x \overline{x}} = 0,\quad 1 \le n \le N - 1. \end{aligned}$$

(3.9)

3.3 Conservative law of the discrete format

Theorem 3.1

The two-level implicit numerical scheme (3.7) admits the following invariant:

$$ E^{n} = \frac{7}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \frac{2h}{9} \sum _{j = 1}^{J - 1} u_{j}^{n}u_{j + 1}^{n} + \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} = E^{n - 1} = \cdots = E^{0}. $$

Proof

Computing the inner product of (3.7) with $2u^{n + \frac{1}{2}}$ (i.e. $u^{n + 1} + u^{n}$), according to Lemma 2.1, we have

$$\begin{aligned} &\frac{7}{9\tau } \bigl\Vert u^{n + 1} \bigr\Vert ^{2} + \frac{2h}{9 \tau } \sum_{j = 1}^{J - 1} u_{j}^{n + 1}u_{j + 1}^{n + 1} - \frac{7}{9 \tau } \bigl\Vert u^{n} \bigr\Vert ^{2} + \frac{2h}{9\tau } \sum_{j = 1} ^{J - 1} u_{j}^{n}u_{j + 1}^{n} + \frac{1}{\tau } \bigl( \bigl\Vert u _{x}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} \bigr) \\ &\quad{}+ \frac{1}{\tau } \bigl( \bigl\Vert u_{xx}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} \bigr) + \bigl( P, 2u^{n + \frac{1}{2}} \bigr) = 0, \end{aligned}$$

(3.10)

where $P = \frac{1}{3} ( u_{j}^{n + \frac{1}{2}} )_{ \widehat{x}} ( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} )$. Now, computing the last term of the left-hand side in (3.10), we get

$$\begin{aligned} & \bigl( P, 2u^{n + \frac{1}{2}} \bigr) \\ &\quad= \biggl( \frac{2}{3} \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \bigl( u_{j - 1} ^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr),u^{n + \frac{1}{2}} \biggr) \\ &\quad{}= \frac{1}{3}\sum_{j = 1}^{J - 1} \bigl( u_{j + 1}^{n + \frac{1}{2}}u _{j}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}} - u _{j - 1}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}}u_{j - 1}^{n + \frac{1}{2}} - u_{j - 1}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}}u _{j}^{n + \frac{1}{2}} \bigr) \\ &\quad{}= \frac{1}{3} \Biggl[ \sum_{j = 1}^{J - 2} \bigl( u_{j}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}} + u _{j}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}} - u_{j}^{n + \frac{1}{2}}u_{j}^{n + \frac{1}{2}}u_{j + 1} ^{n + \frac{1}{2}} - u_{j}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}}u_{j + 1}^{n + \frac{1}{2}} \bigr) \Biggr] \\ &\qquad{}+ u_{J - 1}^{n + \frac{1}{2}}u_{J - 1}^{n + \frac{1}{2}}u_{J}^{n + \frac{1}{2}} + u_{J - 1}^{n + \frac{1}{2}}u_{J}^{n + \frac{1}{2}}u _{J}^{n + \frac{1}{2}} \\ &\qquad{}- \bigl( u_{0}^{n + \frac{1}{2}}u_{0}^{n + \frac{1}{2}}u_{1}^{n + \frac{1}{2}} + u_{0}^{n + \frac{1}{2}}u_{1} ^{n + \frac{1}{2}}u_{1}^{n + \frac{1}{2}} \bigr) = 0 . \end{aligned}$$

(3.11)

Substituting (3.11) into (3.10), we have

$$\begin{aligned} &\frac{7}{9} \bigl\Vert u^{n + 1} \bigr\Vert ^{2} + \frac{2h}{9}\sum_{j = 1}^{J - 1} u_{j}^{n + 1}u_{j + 1}^{n + 1} - \frac{7}{9} \bigl\Vert u ^{n} \bigr\Vert ^{2} \\ &\quad{}+ \frac{2h}{9}\sum_{j = 1}^{J - 1} u_{j}^{n}u _{j + 1}^{n} + \bigl\Vert u_{x}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert u _{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} =0. \end{aligned}$$

We let

$$ E^{n} = \frac{7}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \frac{2h}{9} \sum _{j = 1}^{J - 1} u_{j}^{n}u_{j + 1}^{n} + \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2}. $$

Then we obtain

$$ E^{n} = \frac{7}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \frac{2h}{9} \sum _{j = 1}^{J - 1} u_{j}^{n}u_{j + 1}^{n} + \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} = E^{n - 1} = \cdots = E^{0}. $$

(3.12)

The proof is completed. □

3.4 Solvability

Next, we shall prove the solvability of the difference scheme (3.7).

Theorem 3.2

The finite difference scheme (3.7) is solvable.

Proof

For the difference scheme (3.7)–(3.9), we assume that $u^{0},u^{1}, \ldots,u^{n}$ ($n \le N - 1$) obey (3.7). Next, we will prove that $u^{n + 1}$ also satisfies (3.7). The operation g is defined as follows:

$$ g(v) = \frac{2}{9}A \bigl( v - u^{n} \bigr) - 2 \bigl(v_{x\overline{x}} - u_{x\overline{x}}^{n}\bigr) + 2v_{xx\overline{x}\overline{x}} - 2u_{{xx \overline{x}\overline{x}}}^{n} + \tau v_{\widehat{x}} + \frac{\tau }{3} ( v_{j - 1} + v_{j} + v_{j + 1} )v_{ \widehat{x}} . $$

(3.13)

It is obvious that g is continuous. Computing the inner product of (3.13) with v, according to Lemma 2.1, we have

$$\begin{aligned} \bigl( g(v),v \bigr) ={}& \frac{2}{9} ( Av,v ) - \frac{2}{9} \bigl( Au^{n},v \bigr) - 2 ( v_{x\overline{x}},v ) + 2 \bigl( u_{x\overline{x}}^{n},v \bigr) + (2v_{xx \overline{x}\overline{x}},v) - \bigl(2u_{xx\overline{x}\overline{x}}^{n},v\bigr) \\ \ge{}& \frac{2}{9}v^{T}W^{T}v + \frac{2}{9}v^{T}L^{T}v - \frac{2}{9} \bigl\Vert Au^{n} \bigr\Vert \Vert v \Vert + 2 \Vert v_{x} \Vert ^{2}\\ &{} - 2 \bigl\Vert u_{x}^{n} \bigr\Vert \Vert v_{x} \Vert + 2 \Vert v_{xx} \Vert ^{2} - 2 \bigl\Vert u_{xx} ^{n} \bigr\Vert \Vert v_{xx} \Vert \\ \ge{}& \frac{2}{9} \bigl( \lambda _{0}v_{0}^{2} + \lambda _{2}v_{1}^{2} + \cdots + \lambda _{J}v_{J}^{2} \bigr) - \frac{1}{9} \bigl\Vert Au^{n} \bigr\Vert ^{2} \\ &{}- \frac{1}{9} \Vert v \Vert ^{2} + \Vert v _{x} \Vert ^{2} - \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \Vert v_{xx} \Vert ^{2} - \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} \\ \ge{}& \frac{1}{9} ( 2\lambda _{\min } - 1 ) \Vert v \Vert ^{2} - \frac{1}{9} \bigl\Vert Au^{n} \bigr\Vert ^{2} - \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} - \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2}, \end{aligned}$$

where

$$\begin{aligned} &A = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 7 & 1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 7 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 7 & 1 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 7 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{bmatrix}_{(J + 1) \times (J + 1)} = W + L, \\ &W = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 7 & 1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 7 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 7 & 1 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 7 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{bmatrix}_{(J + 1) \times (J + 1)}, \\ &L = \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{(J + 1) \times (J + 1)}, \end{aligned}$$

$\lambda _{ \min }$ is the minimum eigenvalue of W. By Lemma 2.2, we know that $2\lambda _{ \min } - 1 > 0$. Let

$$ \Vert v \Vert ^{2} = \frac{1}{2\lambda _{\min } - 1} \bigl( \bigl\Vert Au^{n} \bigr\Vert ^{2} + 9 \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + 9 \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} + 1 \bigr). $$

We have $(g(v),v)>0, \forall v\in Z_{h}^{0}$. According to Lemma 2.3, there is $v^{ *} \in Z_{h}^{0}$ which obeys $g ( v^{ *} ) = 0$. So we let $v^{ *} = \frac{u^{n + 1} + u^{n}}{2}$, then $u^{n + 1}$ obeys (3.7). □

Theorem 3.3

The finite difference scheme (3.7) has a unique solution.

Proof

Let $I^{n + 1} = U^{n + 1} - V^{n + 1}$, where $U^{n + 1}$ and $V^{n + 1}$ are both the solution of scheme (3.7). We want to prove that $I^{n + 1} = 0$. According to (3.7), we get

$$\begin{aligned} &\frac{1}{9} \bigl( \bigl( I_{j - 1}^{n + \frac{1}{2}} \bigr)_{ \widehat{t}} + 7 \bigl( I_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{t}} + \bigl( I_{j + 1}^{n + \frac{1}{2}} \bigr)_{\widehat{t}} \bigr) - \bigl( I_{j}^{n + \frac{1}{2}} \bigr)_{x\overline{x}\widehat{t}} + \bigl( I_{j}^{n + \frac{1}{2}} \bigr)_{xx\overline{x} \overline{x}\widehat{t}} + \bigl( I_{j}^{n + \frac{1}{2}} \bigr) _{\widehat{x}} \\ &\quad{}+ \varphi \bigl(u_{j}^{n + \frac{1}{2}},u_{j}^{n + \frac{1}{2}} \bigr) - \varphi \bigl(v_{j}^{n + \frac{1}{2}},v_{j}^{n + \frac{1}{2}} \bigr) = 0, \end{aligned}$$

(3.14)

where $\varphi (u_{j}^{n + \frac{1}{2}},u_{j}^{n + \frac{1}{2}}) = \frac{1}{3} ( u_{j}^{n + \frac{1}{2}} )_{\widehat{x}} ( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} )$. Computing the inner product of (3.12) with $2I^{n+\frac{1}{2}}(I^{n}+I^{n+1})$, we have

$$\begin{aligned} &\frac{7}{9\tau} \bigl\Vert I^{n+1} \bigr\Vert ^{2} + \frac{2h}{9\tau}\sum_{j=1}^{J-1}I_{j}^{n+1}I_{j+1}^{n+1}-\frac{7}{9\tau} \bigl\Vert I^{n} \bigr\Vert -\frac{2h}{9\tau}\sum _{j=1}^{J-1}I_{j}^{n}I_{j+1}^{n}\\ &\quad{}+ \frac{1}{\tau}\bigl( \bigl\Vert I_{x}^{n+1} \bigr\Vert ^{2} - \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} \bigr)+\frac{1}{\tau}\bigl( \bigl\Vert I_{xx}^{n+1} \bigr\Vert ^{2} - \bigl\Vert I_{xx}^{n} \bigr\Vert ^{2}\bigr)\\ &\quad{}+ \bigl(\varphi\bigl(u_{j}^{n+\frac{1}{2}},u_{j}^{n+\frac{1}{2}} \bigr)- \varphi \bigl(v_{j}^{n+\frac{1}{2}},v_{j}^{n+\frac{1}{2}} \bigr), I^{n+\frac{1}{2}}\bigr)=0. \end{aligned}$$

Due to the result that

$$\begin{aligned} &{ -} \bigl(\varphi \bigl(u_{j}^{n + \frac{1}{2}},u_{j}^{n + \frac{1}{2}} \bigr) - \varphi \bigl(v_{j}^{n + \frac{1}{2}},v_{j}^{n + \frac{1}{2}} \bigr),I^{n + \frac{1}{2}}\bigr) \\ &\quad= - \frac{2}{3}h\sum_{j = 1}^{J - 1} \bigl( I_{j - 1}^{n + \frac{1}{2}} + I_{j}^{n + \frac{1}{2}} + I_{j + 1}^{n + \frac{1}{2}} \bigr) \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}}I_{j} ^{n + \frac{1}{2}}\\ &\qquad{} - \frac{2}{3}h\sum_{j = 1}^{J - 1} \bigl( I_{j - 1} ^{n + \frac{1}{2}} + I_{j}^{n + \frac{1}{2}} + I_{j + 1}^{n + \frac{1}{2}} \bigr) \bigl( v_{j}^{n + \frac{1}{2}} \bigr)_{ \widehat{x}}I_{j}^{n + \frac{1}{2}} \\ &\qquad{}- \frac{2}{3}h\sum_{j = 1}^{J - 1} \bigl( v_{j - 1}^{n + \frac{1}{2}} + v_{j}^{n + \frac{1}{2}} + v_{j + 1}^{n + \frac{1}{2}} \bigr) \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}}I_{j}^{n + \frac{1}{2}} \\ &\qquad{}+ \frac{2}{3}h\sum_{j = 1}^{J - 1} \bigl( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr) \bigl( v_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}}I_{j} ^{n + \frac{1}{2}} \\ &\quad\le \frac{2}{3}Ch\sum_{j = 1}^{J - 1} \bigl( \bigl\vert I_{j - 1}^{n + \frac{1}{2}} \bigr\vert + \bigl\vert I_{j}^{n + \frac{1}{2}} \bigr\vert + \bigl\vert I_{j + 1}^{n + \frac{1}{2}} \bigr\vert \bigr) \bigl\vert I_{j} ^{n + \frac{1}{2}} \bigr\vert + \frac{2}{3}Ch\sum _{j = 1}^{J - 1} \bigl\vert \bigl( I_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \bigr\vert \bigl\vert I_{j}^{n + \frac{1}{2}} \bigr\vert \\ &\quad\le C \bigl( \bigl\Vert I^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I^{n} \bigr\Vert ^{2} + \bigl\Vert I_{x}^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} \bigr), \end{aligned}$$

we have

$$\begin{aligned} &\frac{7}{9\tau } \bigl\Vert I^{n + 1} \bigr\Vert ^{2} + \frac{2h}{9 \tau } \sum_{j = 1}^{J - 1} I_{j}^{n + 1}I_{j + 1}^{n + 1} - \frac{7}{9 \tau } \bigl\Vert I^{n} \bigr\Vert ^{2} - \frac{2h}{9\tau } \sum_{j = 1} ^{J - 1} I_{j}^{n}I_{j + 1}^{n} \\ &\qquad{}+ \frac{1}{\tau } \bigl( \bigl\Vert I_{x}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} \bigr) + \frac{1}{\tau } \bigl( \bigl\Vert I_{xx}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert I_{xx}^{n} \bigr\Vert ^{2} \bigr) \\ &\quad\le C \bigl( \bigl\Vert I^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I^{n} \bigr\Vert ^{2} + \bigl\Vert I_{x}^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} \bigr). \end{aligned}$$

Let

$$ Q^{n} = \frac{7}{9} \bigl\Vert I^{n} \bigr\Vert ^{2} + \frac{2h}{9} \sum _{j = 1}^{J - 1} I_{j}^{n}I_{j + 1}^{n} + \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert I_{xx}^{n} \bigr\Vert ^{2}. $$

So we get

$$ Q^{n + 1} - Q^{n} \le C\tau \bigl(Q^{n + 1} + Q^{n}\bigr). $$

That is,

$$ Q^{n + 1} \le \frac{1 + C\tau }{1 - C\tau } Q^{n} = (1 + \vartheta \tau )Q^{n}, $$

where $\vartheta = \frac{2C}{1 - C\tau }$. Then we have

$$ Q^{n + 1} \le (1 + \vartheta \tau )Q^{n} \le \cdots \le (1 + \vartheta \tau )^{n + 1}Q^{0} \le \cdots \le e^{\vartheta \tau (n + 1)}Q^{0}. $$

Because $Q^{0} = \frac{7}{9} \Vert I^{0} \Vert ^{2} + \frac{2h}{9}\sum_{j = 1}^{J - 1} I_{j}^{0}I_{j + 1}^{0} + \Vert I _{x}^{0} \Vert ^{2} + \Vert I_{xx}^{0} \Vert ^{2} = 0$, $Q^{n + 1} \le e^{\vartheta \tau (n + 1)}Q^{0} = 0$. We have

$$ Q^{n} = \frac{7}{9} \bigl\Vert I^{n} \bigr\Vert ^{2} + \frac{2h}{9} \sum _{j = 1}^{J - 1} I_{j}^{n}I_{j + 1}^{n} + \bigl\Vert I_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert I_{xx}^{n} \bigr\Vert ^{2} = 0. $$

It is well known that

$$ \frac{5}{9} \bigl\Vert I^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I_{x}^{n + 1} \bigr\Vert ^{2} + \bigl\Vert I_{xx}^{n + 1} \bigr\Vert ^{2} \le Q ^{n} = 0. $$

So $\Vert I^{n + 1} \Vert ^{2} = \Vert U^{n + 1} - V^{n + 1} \Vert ^{2} = 0$. We can say that $U^{n + 1} = V^{n + 1}$. □

4 Some prior estimates for the difference solution

In this section, we shall make some prior estimates for the numerical scheme.

Theorem 4.1

Assume that $u_{0} \in H_{0}^{2} [ x_{l},x _{r} ]$ then there is the estimation for the solution of the numerical scheme (3.7)

$$ \bigl\Vert u_{xx}^{n} \bigr\Vert \le C,\qquad \bigl\Vert u_{x}^{n} \bigr\Vert \le C,\qquad \bigl\Vert u^{n} \bigr\Vert _{\infty } \le C. $$

Proof

It follows from (3.12) that

$$ \frac{5}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \Biggl( \frac{2}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \frac{2}{9}\sum _{i = 1} ^{J - 1} u_{i}^{n}u_{i + 1}^{n}h \Biggr) + \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} = C $$

(4.1)

we obtain from (4.1)

$$ \frac{5}{9} \bigl\Vert u^{n} \bigr\Vert ^{2} + \bigl\Vert u_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert u_{xx}^{n} \bigr\Vert ^{2} \le C. $$

So

$$ \bigl\Vert u_{xx}^{n} \bigr\Vert \le C,\qquad \bigl\Vert u_{x}^{n} \bigr\Vert \le C, \qquad \bigl\Vert u^{n} \bigr\Vert \le C. $$

It follows from Lemma 2.4 that $\Vert u^{n} \Vert _{\infty } \le C$. □

Remark 4.1

Theorem 4.1 implies that the scheme (3.7)–(3.9) is unconditionally stable.

5 Convergence and stability of the difference scheme

Now, we consider the truncation error of scheme of (3.7). Firstly, we define the truncation error as follows:

$$\begin{aligned} Er_{j}^{n + \frac{1}{2}} ={}& \frac{1}{9\tau } \bigl( v_{i - 1}^{n + 1} + 7v_{i}^{n + 1} + v_{i + 1}^{n + 1} - v_{i - 1}^{n} - 7v_{i}^{n} - v _{i + 1}^{n} \bigr) + \bigl( v_{i}^{n + \frac{1}{2}} \bigr)_{ \widehat{x}} \\ &{}- \bigl( v_{i}^{n + \frac{1}{2}} \bigr)_{x\overline{x}\widehat{t}} + \bigl( v_{i}^{n + \frac{1}{2}} \bigr)_{xx\overline{x} \overline{x}\widehat{t}} + \frac{1}{3} \bigl( v_{i}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \bigl( v_{i - 1}^{n + \frac{1}{2}} + v_{i}^{n + \frac{1}{2}} + v_{i + 1}^{n + \frac{1}{2}} \bigr). \end{aligned}$$

(5.1)

According to Taylor expansion and Lagrange interpolation, we obtain the following.

Lemma 5.1

Assume that $u_{0} \in H_{0}^{2} [ x_{l},x _{r} ]$ and $u ( x,t ) \in C^{5,2}$, then the truncation errors of the numerical scheme (3.7) satisfy

$$ Er_{j}^{n + \frac{1}{2}} = O \bigl( \tau ^{2} + h^{3} \bigr). $$

(5.2)

Theorem 5.1

Assume that $u_{0} \in H_{0}^{2} [ x_{l},x _{r} ]$ and $u ( x,t ) \in C^{5,2}$, then the solution of the numerical scheme (3.7) converges to the solution of the initial-boundary value problem (1.5)–(1.7) with order $O ( \tau ^{2} + h^{3} )$ by the $\Vert \cdot \Vert _{\infty }$ norm.

Proof

Let $e_{j}^{n + \frac{1}{2}} = u(x_{j},t^{n + \frac{1}{2}}) - u_{j}^{n + \frac{1}{2}}$. Subtracting (3.7) from (5.1), we have

$$\begin{aligned} Er_{j}^{n + \frac{1}{2}} ={}& \frac{1}{9\tau } \bigl( e_{j - 1}^{n + 1} + 7e_{j}^{n + 1} + e_{j + 1}^{n + 1} - e_{j - 1}^{n} - 7e_{j}^{n} - e _{j + 1}^{n} \bigr) + \frac{1}{4} \bigl( \bigl( e_{j}^{n + 1} \bigr) _{\widehat{x}} + \bigl( e_{j}^{n} \bigr)_{\widehat{x}} \bigr) - \bigl( e_{j}^{n + \frac{1}{2}} \bigr)_{x\overline{x}\widehat{t}} \\ &{}+ \bigl( e_{j}^{n + \frac{1}{2}} \bigr)_{xx\overline{x} \overline{x}\widehat{t}} + \frac{1}{3} \bigl( u\bigl(x_{j},t^{n + \frac{1}{2}}\bigr) \bigr)_{\widehat{x}} \bigl( u\bigl(x_{j - 1},t^{n + \frac{1}{2}}\bigr) + u\bigl(x_{j},t^{n + \frac{1}{2}}\bigr) + u\bigl(x_{j + 1},t^{n + \frac{1}{2}} \bigr) \bigr) \\ &{}- \frac{1}{3} \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{\widehat{x}} \bigl( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr). \end{aligned}$$

(5.3)

For simple notation, the last two terms of (5.3) are defined by

$$\begin{aligned} Q ={}& \bigl( u\bigl(x_{j},t^{n + \frac{1}{2}}\bigr) \bigr)_{\widehat{x}} \bigl( u\bigl(x_{j - 1},t^{n + \frac{1}{2}}\bigr) + u\bigl(x_{j},t^{n + \frac{1}{2}}\bigr) + u\bigl(x _{j + 1},t^{n + \frac{1}{2}}\bigr) \bigr) \\ &{}- \frac{1}{3} \bigl( u_{j}^{n + \frac{1}{2}} \bigr)_{ \widehat{x}} \bigl( u_{j - 1}^{n + \frac{1}{2}} + u_{j}^{n + \frac{1}{2}} + u_{j + 1}^{n + \frac{1}{2}} \bigr). \end{aligned}$$

Computing the inner product of (5.3) with $2e^{n + \frac{1}{2}}$ (i.e. $e^{n + 1} + e^{n}$), we get

$$\begin{aligned} &\bigl( Er_{j}^{n + \frac{1}{2}},2e^{n + \frac{1}{2}} \bigr) \\ &\quad= \frac{7}{9\tau } \bigl\Vert e^{n + 1} \bigr\Vert ^{2} + \frac{2h}{9 \tau } \sum_{j = 1}^{J - 1} e_{j}^{n + 1}e_{j + 1}^{n + 1} - \frac{7}{9 \tau } \bigl\Vert e^{n} \bigr\Vert ^{2} - \frac{2h}{9\tau } \sum_{j = 1} ^{J - 1} e_{j}^{n}e_{j + 1}^{n} + \frac{1}{\tau } \bigl( \bigl\Vert e _{x}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert e_{x}^{n} \bigr\Vert ^{2} \bigr) \\ &\qquad{}+ \frac{1}{\tau } \bigl( \bigl\Vert e_{xx}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert e_{xx}^{n} \bigr\Vert ^{2} \bigr) + \bigl( Q,2e^{n + \frac{1}{2}} \bigr). \end{aligned}$$

(5.4)

According to Theorem 3.3, we obtain

$$ - \bigl( Q,2e^{n + \frac{1}{2}} \bigr) \le C \bigl( \bigl\Vert e ^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e^{n} \bigr\Vert ^{2} + \bigl\Vert e_{x}^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e_{x}^{n} \bigr\Vert ^{2} \bigr). $$

(5.5)

In addition, obviously

$$ \bigl( Er_{j}^{n + \frac{1}{2}},2e^{n + \frac{1}{2}} \bigr) = \bigl( Er_{j}^{n + \frac{1}{2}},e^{n + 1} + e^{n} \bigr) \le \bigl\Vert Er^{n + \frac{1}{2}} \bigr\Vert ^{2} + \frac{1}{2} \bigl( \bigl\Vert e^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e^{n} \bigr\Vert ^{2} \bigr). $$

Substituting (5.4) and (5.5) into (5.3), we have

$$\begin{aligned} & \frac{7}{9} \bigl\Vert e^{n + 1} \bigr\Vert ^{2} + \frac{2h}{9} \sum_{j = 1}^{J - 1} e_{j}^{n + 1}e_{j + 1}^{n + 1} - \frac{7}{9} \bigl\Vert e^{n} \bigr\Vert ^{2} - \frac{2h}{9}\sum_{j = 1}^{J - 1} e _{j}^{n}e_{j + 1}^{n} \\ &\qquad{}+ \bigl\Vert e_{x}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert e_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert e_{xx}^{n + 1} \bigr\Vert ^{2} - \bigl\Vert e_{xx}^{n} \bigr\Vert ^{2} \\ &\quad\le \tau \bigl\Vert Er^{n + \frac{1}{2}} \bigr\Vert ^{2} + \frac{ \tau }{2} \bigl( \bigl\Vert e^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e ^{n} \bigr\Vert ^{2} \bigr) \\ &\qquad{} + C\tau \bigl( \bigl\Vert e^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e^{n} \bigr\Vert ^{2} + \bigl\Vert e_{x} ^{n + 1} \bigr\Vert ^{2} + \bigl\Vert e_{x}^{n} \bigr\Vert ^{2} \bigr) . \end{aligned}$$

(5.6)

Let $B^{n} = \frac{7}{9} \Vert e^{n} \Vert ^{2} + \frac{2h}{9}\sum_{j = 1}^{J - 1} e_{j}^{n}e_{j + 1}^{n} + \Vert e _{x}^{n} \Vert ^{2} + \Vert e_{xx}^{n} \Vert ^{2}$, then (5.6) can be rewritten as

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

So

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

If τ is sufficiently small so that it satisfies $1 - C\tau = \sigma > 0$, then

$$ B^{n + 1} - B^{n} \le C\tau \bigl\Vert Er^{n + \frac{1}{2}} \bigr\Vert ^{2} + C\tau B^{n}. $$

(5.7)

Summing up (5.7) from 0 to $n - 1$, we have

$$B^{n}-B^{0}\leq C\tau \sum_{i=0}^{n-1} \bigl\Vert Er^{i+\frac{1}{2}} \bigr\Vert ^{2}+C\tau \sum _{i=0}^{n-1}B^{i}. $$

Noticing, $\tau \sum_{i=0}^{n-1}\Vert Er^{i+\frac{1}{2}}\Vert^{2} \leq n \tau \max_{0\leq i\leq n-1}\Vert Er^{i+\frac{1}{2}}\Vert^{2} \leq T \cdot O (\tau^{2}+h^{2})^{2}$, $e^{0}=0$. We get $B^{0} = 0$.

Hence, from Lemma 2.5, we obtain

$$ B^{n} \le O \bigl( \tau ^{2} + h^{3} \bigr)^{2}. $$

That is,

$$ \frac{7}{9} \bigl\Vert e^{n} \bigr\Vert ^{2} + \bigl\Vert e_{x}^{n} \bigr\Vert ^{2} + \bigl\Vert e_{xx}^{n} \bigr\Vert ^{2} \le O \bigl( \tau ^{2} + h^{3} \bigr)^{2}. $$

So

$$ \bigl\Vert e^{n} \bigr\Vert \le O \bigl( \tau ^{2} + h^{3} \bigr),\qquad \bigl\Vert e_{x}^{n} \bigr\Vert \le O \bigl( \tau ^{2} + h^{3} \bigr),\qquad \bigl\Vert e_{xx}^{n} \bigr\Vert \le O \bigl( \tau ^{2} + h^{3} \bigr). $$

Using Lemma 2.4, we get $\Vert e^{n} \Vert _{\infty } \le O ( \tau ^{2} + h^{3} )$. □

Theorem 5.2

Under the conditions of Theorem 5.1, the solution of the numerical scheme (3.7) is unconditionally stable by the $\Vert \cdot \Vert _{\infty }$ norm.

6 Numerical experiment

In this section, we will calculate some numerical experiments to verify the correctness of our theoretical analysis in the above part.

Consider the following initial boundary value problem of the Rosenau-RLW equation:

$$ u_{t} - u_{xxt} + u_{xxxxt} + u_{x} + uu_{x} = 0,\quad (x,t) \in (0,1) \times (0,T) $$

(6.1)

with the initial condition

$$ u(x,0) = x^{3}(1 - x)^{3}, \quad x \in [0,1] $$

(6.2)

and boundary conditions

$$ u(0,t) = u(1,t) = 0,\qquad u_{xx}(0,t) = u_{xx}(1,t) = 0,\quad t \in [0,T]. $$

(6.3)

Because we do not know the exact solution of (6.1)–(6.3), we consider the numerical solution of fine grid, taking $h = \frac{1}{500}$, as the accurate solution of (6.1)–(6.3). Next we compare the numerical solution of the coarse grid and the numerical solution of fine grid. In order to obtain the error estimation, we consider the solution as a reference solution of the grid. In Table 1, with the time step $\tau = 0.02$, we give the absolute error between numerical solution and accurate solution under a different spatial step h.

Table 1 Absolute error of numerical solution at $\tau = 0.02$

According to the numerical results in Table 1, we can see that numerical format (3.7) is effective.

In Table 2, with $h = \tau = 0.01$, discrete energy values of numerical format (3.7) are given and are compared with the discrete energy values in the literature [20] and [21]. It can be seen that the multiple integral finite volume method preserves the conservation of discrete energy better than the numerical method in the literature [20] and [21]. Figure 1 shows the energy conservation properties of scheme (3.7) with $J = 100, \tau = 0.02, T = 10$.

Figure 1

Energy conservation properties of (3.7) at $J=100$, $\tau=0.02$, $T=10$

Table 2 Discrete energy values at $h=\tau=0.02$

7 Conclusion

In this paper, we have presented a two-level implicit nonlinear numerical scheme for the Rosenau-RLW equation, which has a wide range of applications in physics. The uniqueness, convergence and stability with $O(\tau^{2}+h^{3})$ of the numerical scheme were discussed in detail. The scheme kept the energy conservation characteristic of the original equation. Finally, a numerical experiment shows that our scheme is efficient.