Combined Compact Symplectic Schemes for Solving Good Boussinesq Equations

1. Introduction

Boussinesq equations are important mathematical physical models in characterizing ocean mixing, atmospheric convection, and intra-Earth convection, which play a key role in fields such as earth sciences, meteorology, and oceanography fields. The study of Boussinesq equation is of great value because it helps us to better understand the hydrodynamic behavior, especially in terms of thermal convection, ocean currents, and atmospheric phenomena. In addition, the study of the Boussinesq equation is essential for the development of numerical models for weather forecasting, climate research, and oceanography. These studies have also helped uncover the fundamental principles that govern fluid motion and heat transfer, contributing to advances in fields such as engineering, environmental science, and geophysics. The GB equation and its various extensions have been extensively analyzed in the existing literature, such as a closed form solution for the two soliton interaction in [1],a highly complicated mechanism for the solitary waves interaction in [2],and the nonlinear stability and convergence of some simple finite difference schemes in [3]. In recent works concerning the numerical solution of PDE, a significant amount addresses the Schrödinger equation, see [4,5,6,7,8,9]. In [9],using a combined compact difference method to solve Schrödinger equation and this scheme is Structure-Preserving.This method originated from [10],it also can be found in[11,12,13,14]. In addition,many works related to GB equations could be found in [15,16,17,18,19,20,21].Higher order Boussinesq equations have been investigated by Z.L. Zou [22].

In solving PDEs numerically, high-order compact (HOC) schemes are often used to discretize spatial derivatives.For example,HOC schemes have been applied to solve steady convection-diffusion equation [12], nonlinear Schrödinger equation [9], Klein-Gordon-Schrö-dinger equation [23], and good Boussinesq equation [17].Compared with general finite difference schemes, HOC schemes have the advantages of smaller error and higher accuracy under the same calculation amount. However, for PDEs with multiple order spatial derivatives, such as good Boussinesq equation ${\mathrm{u}}_{\mathrm{tt}}=-{\mathrm{u}}_{\mathrm{xxxx}}+{\mathrm{u}}_{\mathrm{xx}}+{\left({\mathrm{u}}^2\right)}_{\mathrm{xx}}$, the advantages of classical HOC schemes are often offset. If multiple HOC schemes are used to discretize multiple spatial derivatives simultaneously, it is necessary to perform multiple matrix inverse operations, which will reduce the computational efficiency and affect the accuracy. In [9], Then combined high-order compact (CHOC) scheme is used to approximate PDEs with multiple order spatial derivatives and achieve some discrete conservation laws.

In this paper, three CHOC schemes of good Boussinesq equation are derived. Applying Taylor analysis to an equality combining the solution u and its first derivative, second derivative, and third derivative yields the first three-point CHOC scheme. This scheme has 6th-order precision and has extensive application. Then similarly, we propose the second three-point 8th-order scheme by using a combination of the first,second,third derivatives of the solution. Since the two schemes have a large amount of matrix operations and complex formulation, the third scheme is designed finally by composing the solution and its second derivative, fourth derivative, which greatly simplifies the matrix operations and ensures certain accuracy. In this scheme, through simpler computations, the relationship between the solution and its fourth-order derivative, as well as the relationship between the solution and its second-order derivative can be directly obtained, which can’t be done by first two schemes in this paper. Finally we will use the three schemes to simulate a motion invariant, and summarize the advantages and disadvantages of these schemes. At the same time, compared with a three-point compact scheme with sixth order accuracy derived by Chu and Fan in 1998 [10], our schemes are more accurate.

In this paper, we consider fully discrete schemes for linear good Boussinesq equation

$${\partial }_t^{2}u={\partial }_x^{2}u-{\partial }_x^{4}u,$$

(1.1)

where $0\le x\le L,t>0,L$ is a constant. The following nonlinear good Boussinesq equation is also numerically solved

$${\partial }_t^{2}u={\partial }_x^{2}u-{\partial }_x^{4}u+{\partial }_x^2\left(u^2\right).$$

(1.2)

We consider initial conditions and periodic boundary conditions as follows

$$u(t,0)=u(t,L),u(0,x)=f_1\left(x\right),u_t(0,x)=f_2\left(x\right),0\le x\le L.$$

(1.3)

2. Establishment of the CHOC Scheme

In this paper, we introduce three schemes for discretization of spatial derivatives. To detail the CHOC scheme, we introduce a uniform grid $x_0<x_1\cdots <x_N$ with $x_j=x_0+jh$ and $h=\frac{x_L-x_0}{N},j=1,2,\cdots ,N.$ First, we introduce the simplest scheme (2.1) and (2.2)

$$\begin{array}{cc}\hfill & {\alpha }_1\left(u_{j+1}^{{ }^{\prime }}+u_{j-1}^{{ }^{\prime }}\right)+u_j^{{ }^{\prime }}+{\beta }_{1}h\left(u_{j+1}^{{ }^{\prime \prime }}-u_{j-1}^{{ }^{\prime \prime }}\right)+{\gamma }_1\frac{u_{j+1}-u_{j-1}}{h}=0,\hfill \\ \hfill & {\alpha }_2\left(\frac{u_{j+1}^{{ }^{\prime }}-u_{j-1}^{{ }^{\prime }}}{h}\right)+u_j^{{ }^{\prime \prime }}+{\beta }_2\left(u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}\right)+{\gamma }_2\frac{u_{j+1}-2u_j+u_{j+1}}{h^2}=0,\hfill \end{array}$$

where ${\alpha }_1,{\beta }_1,{\gamma }_1\mathrm{and}{\alpha }_2,{\beta }_2,{\gamma }_2$ are coefficients to be determined according to the accuracy of the approximation. The three-point CHOC scheme for the combination of first and second derivatives is to relate $u_j,u_j^{{ }^{\prime }},u_j^{{ }^{\prime \prime }}$ to their neighbors $u_{j-1},u_{j-1}^{{ }^{\prime }},u_{j-1}^{{ }^{\prime \prime }}\mathrm{and}{u}_{j+1},u_{j+1}^{{ }^{\prime }},u_{j+1}^{{ }^{\prime \prime }}$. This scheme approximates first-order derivative and second-order derivative of u separately using above combinations by Wang and Kong et al [9].

By inserting Taylor expansion to equation (2.1)and(2.2), we can get the following Table 1 and Table 2.

To make this scheme with sixth order convergence, above coefficients must satisfy the following algebraic equations:

$$\begin{array}{cc}\hfill & \begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{11}=2\left({\alpha }_1+{\gamma }_1\right)+1=0,\hfill \\ y_{12}={\alpha }_1+2\left({\beta }_1+\frac{{\gamma }_1}{3!}\right)=0,\hfill \\ y_{13}=2\left(\frac{{\alpha }_1}{4!}+\frac{{\beta }_1}{3!}+\frac{{\gamma }_1}{5!}\right)=0,\hfill \end{array}\right.\end{array}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{21}=2\left({\alpha }_2+{\beta }_2\right)+{\gamma }_2+1=0,\hfill \\ y_{22}=\frac{{\alpha }_2}{3}+{\beta }_2+\frac{{\gamma }_2}{12}=0,\hfill \\ y_{23}=\frac{{\alpha }_2}{5!}+\frac{{\beta }_2}{4!}+\frac{{\gamma }_2}{6!}=0.\hfill \end{array}\right.\end{array}\hfill \end{array}$$

The solutions of above equations are

$${\alpha }_1=\frac{7}{16},{\beta }_1=-\frac{1}{16},{\gamma }_1=\frac{15}{16},$$

and

$${\alpha }_2=\frac{9}{8},{\beta }_2=-\frac{1}{8},{\gamma }_2=-3.$$

Therefore, schemes(2.1) and (2.2) are in the spacific forms

$$\frac{1}{16}\left(7u_{j+1}^{{ }^{\prime }}+16u_j^{{ }^{\prime }}+7u_{j-1}^{{ }^{\prime }}\right)-\frac{h}{16}\left(u_{j+1}^{{ }^{\prime \prime }}-u_{j-1}^{{ }^{\prime \prime }}\right)=\frac{15}{16h}\left(u_{j+1}-u_{j-1}\right),$$

(2.5)

$$\frac{9}{8h}\left(u_{j+1}^{{ }^{\prime }}-u_{j-1}^{{ }^{\prime }}\right)-\frac{1}{8}\left(u_{j+1}^{{ }^{\prime \prime }}-8u_j^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}\right)=\frac{3}{h^2}\left(u_{j+1}-2u_j+u_{j-1}\right).$$

(2.6)

After conducting a thorough analysis, it is determined that this scheme has a relatively limited applicability. Its usage often necessitates complex matrix operations, and it is insufficient for differential equations involving certain high-order derivatives. For Good Boussinesq equation under study in this paper, a fourth-order spatial derivative is involved. To get the numerical solutions of Good Boussinesq equations, we need the discretization of ${\partial }_x^{4}u$and${\partial }_x^{2}u$. Here, we adopt the combination of function values of u and its first-order derivative, second-order derivative to represent fourth-order spatial derivative

$$u_j^{\left(4\right)}=-\frac{36}{h^4}\left(u_{j+1}-2u_j+u_{j-1}\right)+\frac{21}{h^3}\left(u_{j+1}^{{ }^{\prime }}+u_{j-1}^{{ }^{\prime }}\right)-\frac{3}{h^2}\left(u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}\right).$$

(2.7)

Under periodic boundary conditions, by combining (2.5) and (2.6) we have

$$\frac{1}{16}\left[\begin{array}{ccccc} 16& 7& & & 7\\ 7& 16& 7& & \\ & \ddots & \ddots & \ddots & \\ & & 7& 16& 7\\ 7& & & 7& 16\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime }}\\ u_2^{{ }^{\prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime }}\\ u_N^{{ }^{\prime }}\end{array}\right]-\frac{h}{16}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]=\frac{15}{16h}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]$$

$$\frac{9}{8h}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime }}\\ u_2^{{ }^{\prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime }}\\ u_N^{{ }^{\prime }}\end{array}\right]-\frac{1}{8}\left[\begin{array}{ccccc}-8& 1& & & 1\\ 1& -8& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -8& 1\\ 1& & & 1& -8\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]=\frac{3}{h^2}\left[\begin{array}{ccccc}-2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]$$

where

$$\begin{array}{cc}\hfill {\mathrm{A}}_{11}& =\frac{1}{16}\left[\begin{array}{ccccc} 16& 7& & & 7\\ 7& 16& 7& & \\ & \ddots & \ddots & \ddots & \\ & & 7& 16& 7\\ 7& & & 7& 16\end{array}\right],{\mathrm{A}}_{12}=-\frac{\mathrm{h}}{16}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{A}}_{13}=\frac{15}{16\mathrm{h}}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\hfill \end{array}$$

$$\begin{array}{c}\hfill {\mathrm{A}}_{21}=\frac{9}{8\mathrm{h}}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{A}}_{22}=-\frac{1}{8}\left[\begin{array}{ccccc} -8& 1& & & 1\\ 1& -8& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -8& 1\\ 1& & & 1& -8\end{array}\right],{\mathrm{A}}_{23}=\frac{3}{{\mathrm{h}}^2}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\end{array}$$

Therefore, we can represent it in the following form:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathrm{A}}_{11}{U}_x+{\mathrm{A}}_{12}{U}_{xx}={\mathrm{A}}_{13}U\hfill \\ {\mathrm{A}}_{21}{U}_x+{\mathrm{A}}_{22}{U}_{xx}={\mathrm{A}}_{23}U\hfill \end{array}\right.\end{array}$$

(2.27)

By solving (2.8), we can obtain: $U_x=\mathrm{G}·U,U_{xx}=\mathrm{H}·U,\mathrm{G}={\mathrm{A}}^{-1}\mathrm{B},\mathrm{H}={\mathrm{A}}^{-1}\mathrm{C},\mathrm{where}\mathrm{A}={\mathrm{A}}_{11}{\mathrm{A}}_{22}-{\mathrm{A}}_{12}{\mathrm{A}}_{21},\mathrm{B}={\mathrm{A}}_{22}{\mathrm{A}}_{13}-{\mathrm{A}}_{12}{\mathrm{A}}_{23},\mathrm{C}={\mathrm{A}}_{11}{\mathrm{A}}_{23}-{\mathrm{A}}_{21}{\mathrm{A}}_{13}.$ For (2.7) we have

$$\left[\begin{array}{c} {\mathrm{u}}_1^{\left(4\right)}\\ {\mathrm{u}}_2^{\left(4\right)}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{\left(4\right)}\\ {\mathrm{u}}_{\mathrm{N}}^{\left(4\right)}\end{array}\right]=-\frac{36}{{\mathrm{h}}^4}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1\\ {\mathrm{u}}_2\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}\\ {\mathrm{u}}_{\mathrm{N}}\end{array}\right]+\frac{21}{{\mathrm{h}}^3}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1^{{ }^{\prime }}\\ {\mathrm{u}}_2^{{ }^{\prime }}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{{ }^{\prime }}\\ {\mathrm{u}}_{\mathrm{N}}^{{ }^{\prime }}\end{array}\right]-\frac{3}{{\mathrm{h}}^2}\left[\begin{array}{ccccc} 0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1^{{ }^{\prime \prime }}\\ {\mathrm{u}}_2^{{ }^{\prime \prime }}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{{ }^{\prime \prime }}\\ {\mathrm{u}}_{\mathrm{N}}^{{ }^{\prime \prime }}\end{array}\right],$$

where

$$B_1=-\frac{3}{h^2}\left[\begin{array}{ccccc} 0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right].$$

By substituting $U_x=G·U,U_{xx}=H·U$ into it, above expression can be represented as follows

$${\partial }_x^{4}U=-\frac{12}{{\mathrm{h}}^2}{\mathrm{A}}_{23}·U+\frac{56}{3h^2}{\mathrm{A}}_{21}\mathrm{G}·U+\mathrm{BH}·U.$$

(2.9)

Let$\mathrm{M}=\mathrm{B}·\mathrm{H}+\frac{56}{3{\mathrm{h}}^2}{\mathrm{A}}_{21}·\mathrm{G}-\frac{12}{{\mathrm{h}}^2}{\mathrm{A}}_{23}.$ We will have the following schemes to the spatial derivatives

$$\left\{\begin{array}{c}{\partial }_x^{4}U=\mathrm{M}·U,\hfill \\ {\partial }_x^{2}U=\mathrm{H}·U.\hfill \end{array}\right.$$

(2.10)

Next, we will give second CHOC scheme with eighth order accuracy with the combination of first, second and third derivatives relating $u_j,u_j^{{ }^{\prime }},u_j^{{ }^{\prime \prime }},u_j^{{ }^{\prime \prime \prime }}$ to their neighbors $u_{\mathrm{j}-1},u_{j-1}^{{ }^{\prime }},u_{j-1}^{{ }^{\prime \prime }},u_{j-1}^{{ }^{\prime \prime \prime }}$ and $u_{j+1},u_{j+1}^{{ }^{\prime }},u_{j+1}^{{ }^{\prime \prime }},u_{j+1}^{{ }^{\prime \prime \prime }}$. Generalization of (2.1) and (2.2) to the case of three derivatives jields similarly the next CHOC scheme

$${\alpha }_1\left(u_{j+1}^{{ }^{\prime }}+u_{j-1}^{{ }^{\prime }}\right)+u_j^{{ }^{\prime }}+{\beta }_{1}h\left(u_{j+1}^{{ }^{\prime \prime }}-u_{j-1}^{{ }^{\prime \prime }}\right)+{\omega }_1{h}^2\left(u_{j+1}^{{ }^{\prime \prime \prime }}+u_{j-1}^{{ }^{\prime \prime \prime }}\right)+{\gamma }_1\frac{u_{j+1}-u_{j-1}}{h}=0,$$

(2.11)

$${\alpha }_2\left(\frac{u_{j+1}^{{ }^{\prime }}-u_{j-1}^{{ }^{\prime }}}{h}\right)+u_j^{{ }^{\prime \prime }}+{\beta }_2\left(u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}\right)+{\omega }_{2}h\left(u_{j+1}^{{ }^{\prime \prime \prime }}-u_{j-1}^{{ }^{\prime \prime \prime }}\right)+{\gamma }_2\frac{u_{j+1}-2u_j+u_{j+1}}{h^2}=0,$$

(2.12)

$${\alpha }_3\left(\frac{u_{j+1}^{{ }^{\prime }}+u_{j-1}^{{ }^{\prime }}}{h}\right)+u_j^{{ }^{\prime \prime \prime }}+{\beta }_3\left(\frac{u_{j+1}^{{ }^{\prime \prime }}-u_{j-1}^{{ }^{\prime \prime }}}{h}\right)+{\omega }_3\left(u_{j+1}^{{ }^{\prime \prime \prime }}+u_{j-1}^{{ }^{\prime \prime \prime }}\right)+{\gamma }_3\frac{u_{j+1}-u_{j-1}}{h^3}=0,$$

(2.13)

where ${\alpha }_1,{\beta }_1,{\omega }_1,{\gamma }_1\mathrm{and}{\alpha }_2,{\beta }_2,{\omega }_2,{\gamma }_2\mathrm{and}{\alpha }_3,{\beta }_3,{\omega }_3,{\gamma }_3$ are coefficients to be determined according to the accuracy of the approximation. By Taylor expansion of equation(2.11),(2.12) and (2.13) we can get Table 3, Table 4 and Table 5.

To make these schemes of eighth order, they must satisfy the algebraic equations:

$$\begin{array}{cc}\hfill & \begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{11}=2\left({\alpha }_1+{\gamma }_1\right)+1=0\hfill \\ y_{12}={\alpha }_1+2\left({\beta }_1+{\omega }_1+\frac{{\gamma }_1}{3!}\right)=0\hfill \\ y_{13}=2\left(\frac{{\alpha }_1}{4!}+\frac{{\beta }_1}{3!}+\frac{{\omega }_1}{2!}+\frac{{\gamma }_1}{5!}\right)=0\hfill \\ y_{14}=2\left(\frac{{\alpha }_1}{6!}+\frac{{\beta }_1}{5!}+\frac{{\omega }_1}{4!}+\frac{{\gamma }_1}{7!}\right)=0\hfill \end{array}\right.\end{array}\hfill \end{array}$$

(2.14)

and

$$\begin{array}{cc}\hfill & \begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{21}=2\left({\alpha }_2+{\beta }_2+\frac{{\gamma }_2}{2!}\right)+1=0\hfill \\ y_{22}=2\left(\frac{{\alpha }_2}{3!}+\frac{{\beta }_2}{2!}+{\omega }_2+\frac{{\gamma }_2}{4!}\right)=0\hfill \\ y_{23}=2\left(\frac{{\alpha }_2}{5!}+\frac{{\beta }_2}{4!}+\frac{{\omega }_2}{3!}+\frac{{\gamma }_2}{6!}\right)=0\hfill \\ y_{24}=2\left(\frac{{\alpha }_2}{7!}+\frac{{\beta }_2}{6!}+\frac{{\omega }_2}{5!}+\frac{{\gamma }_2}{8!}\right)=0\hfill \end{array}\right.\end{array}\hfill \end{array}$$

(2.15)

and

$$\begin{array}{cc}\hfill & \begin{array}{c}\hfill \left\{\begin{array}{c}{y}_{31}=2\left({\alpha }_3+{\gamma }_3\right)=0\hfill \\ y_{32}=2\left(\frac{{\alpha }_3}{2!}+{\beta }_3+{\omega }_3+\frac{{\gamma }_3}{3!}\right)+1=0\hfill \\ y_{33}=2\left(\frac{{\alpha }_3}{4!}+\frac{{\beta }_3}{3!}+\frac{{\omega }_3}{2!}+\frac{{\gamma }_3}{5!}\right)=0\hfill \\ y_{34}=2\left(\frac{{\alpha }_3}{6!}+\frac{{\beta }_3}{5!}+\frac{{\omega }_3}{4!}+\frac{{\gamma }_3}{7!}\right)=0\hfill \end{array}\right.\end{array}\hfill \end{array}$$

(2.16)

Their unique solutions are

$${\alpha }_1=\frac{19}{32},{\beta }_1=-\frac{1}{8},{\omega }_1=\frac{1}{96},{\gamma }_1=-\frac{35}{32}$$

and

$${\alpha }_2=\frac{29}{16},{\beta }_2=-\frac{5}{16},{\omega }_2=\frac{1}{48},{\gamma }_2=-4$$

and

$${\alpha }_3=-\frac{105}{16},{\beta }_3=\frac{15}{8},{\omega }_3=-\frac{3}{16},{\gamma }_3=\frac{105}{16}$$

respectively. Therefore, the scheme (2.11),(2.12) and (2.13) has in the following specific form:

$$\frac{1}{32}\left(19u_{j+1}^{{ }^{\prime }}+32u_j^{{ }^{\prime }}+19u_{j-1}^{{ }^{\prime }}\right)-\frac{h}{8}\left(u_{j+1}^{{ }^{\prime \prime }}-u_{j-1}^{{ }^{\prime \prime }}\right)+\frac{h^2}{96}\left(u_{j+1}^{{ }^{\prime \prime \prime }}+u_{j-1}^{{ }^{\prime \prime \prime }}\right)=\frac{35}{32h}\left(u_{j+1}-u_{j-1}\right)$$

(2.17)

$$\frac{\left(29u_{j+1}^{{ }^{\prime }}-29u_{j-1}^{{ }^{\prime }}\right)}{16h}-\frac{\left(5u_{j+1}^{{ }^{\prime \prime }}-16u_j^{{ }^{\prime \prime }}+5u_{j-1}^{{ }^{\prime \prime }}\right)}{16}+\frac{h\left(u_{j+1}^{{ }^{\prime \prime \prime }}-u_{j-1}^{{ }^{\prime \prime \prime }}\right)}{48}=\frac{4\left(u_{j+1}-2u_j+u_{j-1}\right)}{h^2}$$

(2.18)

$$\frac{\left(105u_{j+1}^{{ }^{\prime }}+105u_{j-1}^{{ }^{\prime }}\right)}{-16h^2}+\frac{\left(15u_{j+1}^{{ }^{\prime \prime }}-15u_{j-1}^{{ }^{\prime \prime }}\right)}{8h}-\frac{\left(3u_{j+1}^{{ }^{\prime \prime \prime }}-16u_j^{{ }^{\prime \prime \prime }}+3u_{j-1}^{{ }^{\prime \prime \prime }}\right)}{16}=\frac{105\left(u_{j+1}-u_{j-1}\right)}{-16h^3}$$

(2.19)

This three-point scheme possesses eighth-order accuracy and involve three derivatives, so it is more applicable and allows for greater accuracy in comparison to (2.1) and (2.2).

Next, we adopt the combination of function values of u and its first three derivatives to represent fourth-order spatial derivative

$$u_j^{\left(4\right)}=-\frac{72}{h^4}\left(u_{j+1}-2u_j+u_{j-1}\right)+\frac{183}{4h^3}\left(u_{j+1}^{{ }^{\prime }}-u_{j-1}^{{ }^{\prime }}\right)-\frac{39}{4h^2}\left(u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}\right)+\frac{3}{4h}\left(u_{j+1}^{{ }^{\prime \prime \prime }}-u_{j-1}^{{ }^{\prime \prime \prime }}\right)$$

(2.20)

By Combining (2.17), (2.18), (2.19) and (2.20) we obtain

$$\left[\begin{array}{c} {\mathrm{u}}_1^{\left(4\right)}\\ {\mathrm{u}}_2^{\left(4\right)}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{\left(4\right)}\\ {\mathrm{u}}_{\mathrm{N}}^{\left(4\right)}\end{array}\right]=-\frac{72}{{\mathrm{h}}^4}\left[\begin{array}{ccccc}-2& 1& 0& & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1\\ {\mathrm{u}}_2\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}\\ {\mathrm{u}}_{\mathrm{N}}\end{array}\right]+\frac{183}{4{\mathrm{h}}^3}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1^{{ }^{\prime }}\\ {\mathrm{u}}_2^{{ }^{\prime }}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{{ }^{\prime }}\\ {\mathrm{u}}_{\mathrm{N}}^{{ }^{\prime }}\end{array}\right]-\frac{39}{4{\mathrm{h}}^2}\left[\begin{array}{ccccc}0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right]\left[\begin{array}{c} {\mathrm{u}}_1^{{ }^{\prime \prime }}\\ {\mathrm{u}}_2^{{ }^{\prime \prime }}\\ ⋮\\ {\mathrm{u}}_{\mathrm{N}-1}^{{ }^{\prime \prime }}\\ {\mathrm{u}}_{\mathrm{N}}^{{ }^{\prime \prime }}\end{array}\right]$$

$+\frac{3}{4h}-9mu\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]-9mu\left[\begin{array}{c} u_1^{{ }^{\prime \prime \prime }}\\ u_2^{{ }^{\prime \prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime \prime }}\\ u_N^{{ }^{\prime \prime \prime }}\end{array}\right]$

$$\frac{19}{32}\left[\begin{array}{ccccc}\frac{32}{19}& 1& & & 1\\ 1& \frac{32}{19}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{32}{19}& 1\\ 1& & & 1& \frac{32}{19}\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime }}\\ u_2^{{ }^{\prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime }}\\ u_N^{{ }^{\prime }}\end{array}\right]-\frac{h}{8}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]+\frac{h^2}{96}\left[\begin{array}{ccccc}0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime \prime }}\\ u_2^{{ }^{\prime \prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime \prime }}\\ u_N^{{ }^{\prime \prime \prime }}\end{array}\right]$$

(2.21)

$=\frac{35}{32h}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]-9mu\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]$

$$\frac{29}{16h}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime }}\\ u_2^{{ }^{\prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime }}\\ u_N^{{ }^{\prime }}\end{array}\right]-\frac{5}{16}\left[\begin{array}{ccccc}-\frac{16}{5}& 1& & & 1\\ 1& -\frac{16}{5}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -\frac{16}{5}& 1\\ 1& & & 1& -\frac{16}{5}\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]+\frac{h}{48}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime \prime }}\\ u_2^{{ }^{\prime \prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime \prime }}\\ u_N^{{ }^{\prime \prime \prime }}\end{array}\right]$$

(2.17*)

$=\frac{4}{h^2}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]-9mu\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(2.18*)$

$$\frac{-105}{16h^2}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime }}\\ u_2^{{ }^{\prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime }}\\ u_N^{{ }^{\prime }}\end{array}\right]+\frac{15}{8h}\left[\begin{array}{ccccc}0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]-\frac{3}{16}\left[\begin{array}{ccccc}-\frac{16}{3}& 1& & & 1\\ 1& -\frac{16}{3}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -\frac{16}{3}& 1\\ 1& & & 1& -\frac{16}{3}\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime \prime }}\\ u_2^{{ }^{\prime \prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime \prime }}\\ u_N^{{ }^{\prime \prime \prime }}\end{array}\right]$$

$=-\frac{105}{16h^3}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]-9mu\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(2.19*)$

where

$$\begin{array}{cc}\hfill {\mathrm{B}}_{11}& =\frac{19}{32}\left[\begin{array}{ccccc} \frac{32}{19}& 1& & & 1\\ 1& \frac{32}{19}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{32}{19}& 1\\ 1& & & 1& \frac{32}{19}\end{array}\right],{\mathrm{B}}_{12}=-\frac{h}{8}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{B}}_{13}=\frac{h^2}{96}\left[\begin{array}{ccccc} 0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mathrm{B}}_{14}& =\frac{35}{32h}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{B}}_{21}=\frac{29}{16h}-9mu\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{B}}_{22}=-\frac{5}{16}-9mu\left[\begin{array}{ccccc} -\frac{16}{5}& 1& & & 1\\ 1& -\frac{16}{5}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -\frac{16}{5}& 1\\ 1& & & 1& -\frac{16}{5}\end{array}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mathrm{B}}_{23}& =\frac{h}{48}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{B}}_{24}=\frac{4}{h^2}-9mu\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right],{\mathrm{B}}_{31}=-\frac{105}{16h^2}-9mu\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mathrm{B}}_{32}& =\frac{15}{8h}\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right],{\mathrm{B}}_{33}=-\frac{3}{16}-9mu\left[\begin{array}{ccccc} -\frac{16}{3}& 1& & & 1\\ 1& -\frac{16}{3}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -\frac{16}{3}& 1\\ 1& & & 1& -\frac{16}{3}\end{array}\right],{\mathrm{B}}_{34}=-\frac{105}{16h^3}-9mu\left[\begin{array}{ccccc} 0& 1& & & -1\\ -1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & -1& 0& 1\\ 1& & & -1& 0\end{array}\right]\hfill \end{array}$$

Therefore, we can represent it as follows

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathrm{B}}_{11}{U}_x+{\mathrm{B}}_{12}{U}_{xx}+{\mathrm{B}}_{13}{U}_{xxx}={\mathrm{B}}_{14}U\hfill \\ {\mathrm{B}}_{21}{U}_x+{\mathrm{B}}_{22}{U}_{xx}+{\mathrm{B}}_{23}{U}_{xxx}={\mathrm{B}}_{24}U\hfill \\ {\mathrm{B}}_{31}{U}_x+{\mathrm{B}}_{32}{U}_{xx}+{\mathrm{B}}_{33}{U}_{xxx}={\mathrm{B}}_{34}U\hfill \end{array}\right.\end{array}$$

(2.22)

In light of solving (2.22), we can obtain: $U_x={\mathrm{G}}^{*}·U,U_{xx}={\mathrm{H}}^{*}·U,U_{xxx}=\mathrm{I}·U$,${\mathrm{G}}^{*}={\mathrm{D}}^{-1}{\mathrm{D}}_1,{\mathrm{H}}^{*}={\mathrm{D}}^{-1}{\mathrm{D}}_2,\mathrm{I}={\mathrm{D}}^{-1}{\mathrm{D}}_3.$$\mathrm{D}={\mathrm{B}}_{11}{\mathrm{B}}_{22}{\mathrm{B}}_{33}+{\mathrm{B}}_{12}{\mathrm{B}}_{23}{\mathrm{B}}_{31}+{\mathrm{B}}_{13}{\mathrm{B}}_{21}{\mathrm{B}}_{32}-{\mathrm{B}}_{11}{\mathrm{B}}_{23}{\mathrm{B}}_{32}-{\mathrm{B}}_{12}{\mathrm{B}}_{21}{\mathrm{B}}_{33}-{\mathrm{B}}_{13}{\mathrm{B}}_{22}{\mathrm{B}}_{31},$${\mathrm{D}}_1={\mathrm{B}}_{14}{\mathrm{B}}_{22}{\mathrm{B}}_{33}+{\mathrm{B}}_{14}{\mathrm{B}}_{23}{\mathrm{B}}_{31}+{\mathrm{B}}_{13}{\mathrm{B}}_{21}{\mathrm{B}}_{34}-{\mathrm{B}}_{11}{\mathrm{B}}_{23}{\mathrm{B}}_{34}-{\mathrm{B}}_{14}{\mathrm{B}}_{21}{\mathrm{B}}_{33}-{\mathrm{B}}_{13}{\mathrm{B}}_{24}{\mathrm{B}}_{31},$${\mathrm{D}}_2={\mathrm{B}}_{11}{\mathrm{B}}_{22}{\mathrm{B}}_{34}+{\mathrm{B}}_{12}{\mathrm{B}}_{24}{\mathrm{B}}_{31}+{\mathrm{B}}_{14}{\mathrm{B}}_{21}{\mathrm{B}}_{32}-{\mathrm{B}}_{13}{\mathrm{B}}_{24}{\mathrm{B}}_{31}-{\mathrm{B}}_{14}{\mathrm{B}}_{21}{\mathrm{B}}_{33}-{\mathrm{B}}_{11}{\mathrm{B}}_{23}{\mathrm{B}}_{34},$

${\mathrm{D}}_3={\mathrm{B}}_{11}{\mathrm{B}}_{22}{\mathrm{B}}_{34}+{\mathrm{B}}_{12}{\mathrm{B}}_{24}{\mathrm{B}}_{31}+{\mathrm{B}}_{14}{\mathrm{B}}_{21}{\mathrm{B}}_{32}-{\mathrm{B}}_{11}{\mathrm{B}}_{24}{\mathrm{B}}_{32}-{\mathrm{B}}_{12}{\mathrm{B}}_{21}{\mathrm{B}}_{34}-{\mathrm{B}}_{14}{\mathrm{B}}_{22}{\mathrm{B}}_{31}.$ substituting $U_x={\mathrm{G}}^{*}·U,U_{xx}={\mathrm{H}}^{*}·U,U_{xxx}=\mathrm{I}·U$ into (2.21) gives that

$${\partial }_x^{4}U=-\frac{18}{{\mathrm{h}}^2}{\mathrm{B}}_{24}·U-\frac{366}{h^4}{\mathrm{B}}_{12}{\mathrm{G}}^{*}·U+\frac{936}{h^4}{\mathrm{B}}_{13}{\mathrm{H}}^{*}·U-\frac{6}{h^2}{\mathrm{B}}_{12}\mathrm{I}·U$$

(2.23)

let ${\mathrm{M}}^{*}=\frac{936}{h^4}{\mathrm{B}}_{13}{\mathrm{H}}^{*}-\frac{366}{h^4}{\mathrm{B}}_{12}{\mathrm{G}}^{*}-\frac{18}{h^2}{\mathrm{B}}_{24}-\frac{6}{h^2}{\mathrm{B}}_{12}\mathrm{I}.$we will get the following discrete schemes of the spatial derivative:

$$\left\{\begin{array}{c}{\partial }_x^{4}U={\mathrm{M}}^{*}·U\hfill \\ {\partial }_x^{2}U={\mathrm{H}}^{*}·U\hfill \end{array}\right.$$

(2.24)

For above scheme, we find that matrix operation becomes complicated. To get the discrete form of ${\partial }_x^{2}u$ and ${\partial }_x^{4}u$ according to the (2.11), (2.12) and (2.13), it requires many matrix operations, and subsequent simulation of numerical solution will be more difficult. So we consider constructing a direct combination of ${\partial }_x^{2}u,{\partial }_x^{4}u$ and u to look for third CHOC scheme. This scheme will maintain a 6th-order precision and can easily obtain the discrete forms of ${\partial }_x^{2}u$ and ${\partial }_x^{4}u$, which will be more pertinent and accurate This scheme has the following formulation

$$\begin{array}{cc}\hfill & {\alpha }_1\left(\frac{u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}}{2}\right)+u_j^{{ }^{\prime \prime }}+{\beta }_1{h}^2\left(\frac{u_{j+1}^{\left(4\right)}+u_{j-1}^{\left(4\right)}}{2}\right)={\gamma }_1\frac{u_{j+1}-2u_j+u_{j-1}}{h^2},\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(2.25\right)\hfill \\ \hfill & {\alpha }_2\left(\frac{u_{j+1}^{{ }^{\prime \prime }}-2u_j^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}}{h^2}\right)+u_j^{\left(4\right)}+{\beta }_2\left(\frac{u_{j+1}^{\left(4\right)}+u_{j-1}^{\left(4\right)}}{2}\right)={\gamma }_2\frac{u_{j+2}-4u_{j+1}+6u_j-4u_{j-1}+u_{j-2}}{h^4}.\hspace{1em}\hspace{1em}\hspace{1em}\left(2.26\right)\hfill \end{array}$$

We insert Taylor expansions to (2.25) to obtain

$$\begin{array}{cc}\hfill LHS& =u_j^{{ }^{\prime \prime }}+{\alpha }_1\left(\frac{u_{j+1}^{{ }^{\prime \prime }}+u_{j-1}^{{ }^{\prime \prime }}}{2}\right)+{\beta }_1{h}^2\left(\frac{u_{j+1}^{\left(4\right)}+u_{j-1}^{\left(4\right)}}{2}\right)\hfill \\ \hfill & =u_j^{{ }^{\prime \prime }}+{\alpha }_1\left(u_j^{{ }^{\prime \prime }}+\frac{h^2}{2!}{u}_j^{\left(4\right)}+\frac{h^4}{4!}{u}_j^{\left(6\right)}+o\left(h^6\right)\right)+{\beta }_1{h}^2\left(u_j^{\left(4\right)}+\frac{h^2}{2!}{u}_j^{\left(6\right)}+\frac{h^4}{4!}{u}_j^{\left(8\right)}+o\left(h^6\right)\right)\hfill \\ \hfill & =\left(1+{\alpha }_1\right)u_j^{{ }^{\prime \prime }}+\left(\frac{{\alpha }_1}{2!}+{\beta }_1\right)h^2{u}_j^{\left(4\right)}+\left(\frac{{\alpha }_1}{4!}+\frac{{\beta }_1}{2!}\right)h^4{u}_j^{\left(6\right)}+o\left(h^6\right),\hfill \\ \hfill RHS& ={\gamma }_1\left(\frac{u_{j+1}-2u_j+u_{j-1}}{h^2}\right)\hfill \\ \hfill & ={\gamma }_1\left(u_j^{{ }^{\prime \prime }}+\frac{u_j^{\left(4\right)}}{12}{h}^2+\frac{u_j^{\left(6\right)}}{360}{h}^4\right)+o\left(h^6\right).\hfill \end{array}$$

To make the scheme with sixth order accuracy, the coefficients must satisfy the algebraic equations

$$\left\{\begin{array}{c}1+{\alpha }_1=\frac{{\gamma }_1}{2!}\hfill \\ \frac{{\alpha }_1}{2!}+{\beta }_1=\frac{{\gamma }_1}{4!}\hfill \\ \frac{{\alpha }_1}{4!}+\frac{{\beta }_1}{2!}=\frac{{\gamma }_1}{6!}\hfill \end{array}\right.\Rightarrow \left\{\begin{array}{c}{\alpha }_1=\frac{14}{61}\hfill \\ {\beta }_1=-\frac{3}{244}\hfill \\ {\gamma }_1=\frac{75}{61}\hfill \end{array}\right.$$

Similarly,for(2.26)we have:

$$\begin{array}{cc}\hfill LHS& =u_j^{\left(4\right)}+{\alpha }_2\left(u_j^{\left(4\right)}+\frac{u_j^{\left(6\right)}}{12}{h}^2+\frac{u_j^{\left(8\right)}}{360}{h}^4\right)+{\beta }_2\left(u_j^{\left(4\right)}+\frac{u_j^{\left(6\right)}{h}^2}{2!}+\frac{u_j^{\left(8\right)}{h}^4}{4!}\right)+o\left(h^6\right)\hfill \\ \hfill & =\left(1+{\alpha }_2+{\beta }_2\right)u_j^{\left(4\right)}+\left(\frac{{\alpha }_2}{12}+\frac{{\beta }_2}{2!}\right)h^2{u}_j^{\left(6\right)}+\left(\frac{{\alpha }_2}{360}+\frac{{\beta }_2}{4!}\right)h^4{u}_j^{\left(8\right)}+o\left(h^6\right)\hfill \\ \hfill RHS& ={\gamma }_2\frac{u_{j+2}-4u_{j+1}+6u_j-4u_{j-1}+u_{j-2}}{h^4}+o\left(h^6\right)\hfill \\ \hfill & ={\gamma }_2\left(\frac{2^5-2^3}{4!}{u}_j^{\left(4\right)}+\frac{2^7-2^3}{6!}{h}^2{u}_j^{\left(6\right)}+\frac{2^9-2^3}{8!}{h}^4{u}_j^{\left(8\right)}\right)+o\left(h^6\right)\hfill \end{array}$$

To make the scheme with sixth order accuracy, coefficients must satisfy the algebraic equations

$$\left\{\begin{array}{c}1+{\alpha }_2+{\beta }_2=\frac{2^5-2^3}{4!}{\gamma }_2\hfill \\ \frac{{\alpha }_2}{12}+\frac{{\beta }_2}{2!}=\frac{2^7-2^3}{6!}{\gamma }_2\hfill \\ \frac{{\alpha }_2}{360}+\frac{{\beta }_2}{4!}=\frac{2^9-2^3}{8!}{\gamma }_2\hfill \end{array}\right.⟹\left\{\begin{array}{c}{\alpha }_2=\frac{6}{7}\hfill \\ {\beta }_2=\frac{5}{7}\hfill \\ {\gamma }_2=\frac{18}{7}\hfill \end{array}\right.$$

Under periodic boundary conditions, for (2.25) and (2.26), we obtain

$$\frac{7}{61}\left[\begin{array}{ccccc} \frac{61}{7}& 1& & & 1\\ 1& \frac{61}{7}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{61}{7}& 1\\ 1& & & 1& \frac{61}{7}\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]-\frac{3h^2}{488}\left[\begin{array}{ccccc} 0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right]\left[\begin{array}{c} u_1^{\left(4\right)}\\ u_2^{\left(4\right)}\\ ⋮\\ u_{N-1}^{\left(4\right)}\\ u_N^{\left(4\right)}\end{array}\right]=\frac{75}{61h^2}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]$$

$$\frac{6}{7h^2}\left[\begin{array}{ccccc}-2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\left[\begin{array}{c} u_1^{{ }^{\prime \prime }}\\ u_2^{{ }^{\prime \prime }}\\ ⋮\\ u_{N-1}^{{ }^{\prime \prime }}\\ u_N^{{ }^{\prime \prime }}\end{array}\right]+\frac{5}{14}\left[\begin{array}{ccccc}\frac{14}{5}& 1& & & 1\\ 1& \frac{14}{5}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{14}{5}& 1\\ 1& & & 1& \frac{14}{5}\end{array}\right]\left[\begin{array}{c} u_1^{\left(4\right)}\\ u_2^{\left(4\right)}\\ ⋮\\ u_{N-1}^{\left(4\right)}\\ u_N^{\left(4\right)}\end{array}\right]=\frac{18}{7h^4}\left[\begin{array}{ccccc}6& -4& & & -4\\ -4& 6& -4& & \\ & \ddots & \ddots & \ddots & \\ & & -4& 6& -4\\ -4& & & -4& 6\end{array}\right]\left[\begin{array}{c} u_1\\ u_2\\ ⋮\\ u_{N-1}\\ u_N\end{array}\right]$$

where let

$$\begin{array}{c}\hfill {\mathrm{A}}_{11}^{*}=\frac{7}{61}\left[\begin{array}{ccccc} \frac{61}{7}& 1& & & 1\\ 1& \frac{61}{7}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{61}{7}& 1\\ 1& & & 1& \frac{61}{7}\end{array}\right],{\mathrm{A}}_{12}^{*}=-\frac{3h^2}{488}\left[\begin{array}{ccccc} 0& 1& & & 1\\ 1& 0& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& 0& 1\\ 1& & & 1& 0\end{array}\right],{\mathrm{A}}_{13}^{*}=\frac{75}{61h^2}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right]\end{array}$$

$$\begin{array}{c}\hfill {\mathrm{A}}_{21}^{*}=\frac{6}{7h^2}\left[\begin{array}{ccccc} -2& 1& & & 1\\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\\ 1& & & 1& -2\end{array}\right],{\mathrm{A}}_{22}^{*}=\frac{5}{14}\left[\begin{array}{ccccc} \frac{14}{5}& 1& & & 1\\ 1& \frac{14}{5}& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& \frac{14}{5}& 1\\ 1& & & 1& \frac{14}{5}\end{array}\right],{\mathrm{A}}_{23}^{*}=\frac{18}{7h^4}-9mu\left[\begin{array}{cccccccc}6& -4& 1& & & & 1& -4\\ -4& 6& -4& 1& & & & 1\\ 1& -4& 6& -4& 1& & & \\ & \ddots & \ddots & \ddots & \ddots & \ddots & & \\ & & \ddots & \ddots & \ddots & \ddots & \ddots & \\ & & & \ddots & \ddots & \ddots & \ddots & 1\\ 1& & & & 1& -4& 6& -4\\ -4& 1& & & & 1& -4& 6\end{array}\right]\end{array}$$

Therefore, we can represent it in the following form:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathrm{A}}_{11}^{*}{U}_{xx}+{\mathrm{A}}_{12}^{*}{U}_{xxxx}={\mathrm{A}}_{13}^{*}U\hfill \\ {\mathrm{A}}_{21}^{*}{U}_{xx}+{\mathrm{A}}_{22}^{*}{U}_{xxxx}={\mathrm{A}}_{23}^{*}U\hfill \end{array}\right.\end{array}$$

With (2.27), we can readily derive $U_{xx}$ and $U_{xxxx}$ by expressions of U, respectively. This significantly streamlines the matrix operations. By solving (2.27), we can obtain $U_{xx}={\mathrm{G}}_1·U,U_{xxxx}={\mathrm{H}}_1·U$,where ${\mathrm{G}}_1={\mathrm{A}}_1^{-1}\mathrm{B},{\mathrm{H}}_1={\mathrm{A}}_1^{-1}\mathrm{C},\mathrm{where}{\mathrm{A}}_1={\mathrm{A}}_{11}^{*}{\mathrm{A}}_{22}^{*}-{\mathrm{A}}_{12}^{*}{\mathrm{A}}_{21}^{*},{\mathrm{B}}_1={\mathrm{A}}_{22}^{*}{\mathrm{A}}_{13}^{*}-{\mathrm{A}}_{12}^{*}{\mathrm{A}}_{23}^{*},{\mathrm{C}}_1={\mathrm{A}}_{11}^{*}{\mathrm{A}}_{23}^{*}-{\mathrm{A}}_{21}^{*}{\mathrm{A}}_{13}^{*}.$ For good Boussiensq equations, above scheme has higher spatial accuracy compared to the scheme given in [11].

3. Establishment of the Full Discrete Schemes

Let $\tau$ be temporal stepsize and $t_n=n\tau ,n=0,1,2,\cdots ,M$, where $M=T/\tau$. Denote the approximation of $u(x,t_n)$ by $u^n$. Define the following operators

$$\begin{array}{cc}\hfill {\delta }_t{u}^{n+\frac{1}{2}}& =\frac{u^{n+1}-u^n}{\tau },u^{n+\frac{1}{2}}=\frac{u^n+u^{n+1}}{2}.\hfill \end{array}$$

Let $v=u_t$, the considered good Boussinesq equation can be written as

$${\partial }_{t}v={\partial }_x^{2}u-{\partial }_x^{4}u+{\partial }_x^2\left(u^2\right).$$

Applying CHOC scheme (2.10) to the spatial derivatives of above equation gives that

$$\begin{array}{cc}\hfill v_t& =-M·u+H·u+H·u^2\hfill \\ \hfill & =N·u+H·u^2,\hfill \end{array}$$

where $N=H-M.$ By adopting symplectic midpoint scheme with second-order accuracy to above equation, we have the following full discrete scheme

$$u^{n+1}=u^n+\frac{\tau }{2}\left(v^n+v^{n+1}\right),$$

(3.1)

$$-\frac{\tau }{2}N·u^{n+1}+v^{n+1}=\frac{\tau }{2}N·u^n+v^n+\frac{\tau }{2}H·\left\{{\left(u^2\right)}^n+{\left(u^2\right)}^{n+1}\right\}.$$

(3.2)

Similarly, using CHOC scheme (2.24) for the spatial derivatives yields that

$$\begin{array}{cc}\hfill v_t& =-M^{*}·u+H^{*}·u+H^{*}·u^2\hfill \\ \hfill & =N^{*}·u+H^{*}·u^2,\hfill \end{array}$$

where $N^{*}=H^{*}-M^{*}.$ we obtain the corresponding full discrete scheme

$$-\frac{\tau }{2}{N}^{*}·u^{n+1}+v^{n+1}=\frac{\tau }{2}{N}^{*}·u^n+v^n+\frac{\tau }{2}{H}^{*}·\left\{{\left(u^2\right)}^n+{\left(u^2\right)}^{n+1}\right\}.$$

(3.3)

Applying CHOC scheme (2.27) for the spatial derivatives gives that

$$\begin{array}{cc}\hfill v_t& =-H_1·u+G_1·u+G_1·u^2\hfill \\ \hfill & =N_1·u+G_1·u^2,\hfill \end{array}$$

where $N_1=G_1-H_1.$ we get the following full discrete scheme similarly

$$-\frac{\tau }{2}{N}_1·u^{n+1}+v^{n+1}=\frac{\tau }{2}{N}_1·u^n+v^n+\frac{\tau }{2}{G}_1·\left\{{\left(u^2\right)}^n+{\left(u^2\right)}^{n+1}\right\}.$$

(3.4)

By combining (3.2), (3.3), (3.4) with (3.1), respectively, we always obtain the algebraic equation as follows

$$A·T^{n+1}=B·T^n+F\left(T^{n+1},T^n\right),$$

where A and B are some invertible tridiagonal matrices depending on the corresponding schemes, $T^n={\left[u^n,v^n\right]}^{⊺},$ and F is the corresponding nonlinear term.

For simplicity, we will denote the schemes corresponding to (3.2), (3.3) and (3.4) by CHOC-A, CHOC-B and CHOC-C, respectively.

4. Conservation Laws of CHOC Schemes

Under periodic boundary condition (1.3) with $L=1$, some good Boussinesq equations have certain conservation laws. Below, we consider periodic domain $[0,1]$.

Theorem 4.1. 

Let $\parallel ·\parallel$ denote the standard $L^2$-norm for 1-periodic functions. Then along with ${\partial }_t^{2}u=-{\partial }_x^{4}u$, the quadratic functional

$${\parallel u_t\parallel }^2+{\parallel u_{xx}\parallel }^2$$

(4.1)

is an invariant of motion.

Proof. 

According to (1.3), by multiplying both sides of ${\partial }_t^{2}u=-{\partial }_x^{4}u$ by $u_t$ and integrating it by parts, we have

$$\begin{array}{c}\hfill {\int }_0^1{u}_t·u_{tt}dx=-{\int }_0^1{u}_{xx}·u_{txx}dx.\end{array}$$

(4.2)

For (4.2), we take integration with respect to t to get

$$\begin{array}{cc}\hfill {\int }_0^t{\int }_0^1{u}_t·u_{tt}dxdt& =-{\int }_0^t{\int }_0^1{u}_{xx}{u}_{txx}dxdt,\hfill \\ \hfill \Rightarrow {\int }_0^1{u}_t^2(x,t)dx+{\int }_0^1{u}_{xx}^2(x,t)dx& ={\int }_0^1{u}_t^2(x,0)dx+{\int }_0^1{u}_{xx}^2(x,0)dx.\hfill \end{array}$$

Therefore we get (4.1).

Theorem 4.2. 

For nonlinear Boussinesq equation (1.2), the following conservation law is satisfied

$${\int }_0^1{u}_t\left(x,t\right)dx={\int }_0^1{v}^0\left(x\right)dx.$$

(4.3)

If ${\int }_0^1{v}^0\left(x\right)dx=0,$ then we get the conservation law as follows

$${\int }_0^{1}u\left(x,t\right)dx={\int }_0^1{u}^0\left(x\right)dx.$$

(4.4)

Proof. For $t\in \left[0,T\right]$, integrate with respect to x on both sides of (1.2), we obtain

$$\begin{array}{c}\hfill {\int }_0^1{u}_{tt}dx={\int }_0^1{u}_{xx}-{\partial }_x^{4}u+{\left(u^2\right)}_{xx}dx.\end{array}$$

According to (1.3), the integration on right side of above equation is $0,$ i.e.

$${\int }_0^1{u}_{tt}dx=0.$$

Taking integration with respect to t yields that ${\int }_0^1{u}_t\left(x,t\right)dx={\int }_0^1{v}^0\left(x\right)dx$, that is (4.3). If ${\int }_0^1{v}^0\left(x\right)dx=0,$ then ${\int }_0^1{u}_t\left(x,t\right)dx=0,$ which gives (4.4) by integration with respect to t.

First, we are interested in its discrete version of Theorem 4.1 and Theorem 4.2 under numerical analysis for CHOC schemes. To this goal, we list some important properties of circulant matrices [9]. A matrix written in the form

$$Circ\left(c_0,c_1,c_2\cdots c_{N-1}\right)=\left[\begin{array}{ccccc}{c}_0& c_1& c_2& \cdots & c_{N-1}\\ c_{N-1}& c_0& c_1& \cdots & c_{N-2}\\ \cdots & \ddots & \ddots & \ddots & \cdots \\ c_2& c_3& \ddots & \ddots & c_1\\ c_1& c_2& \cdots & c_{N-1}& c_0\end{array}\right]$$

is said to be circulant matrix.

All of the matrices $A_{ij}$($i,j=1,2.$), $B_{ij}$($i,j=1,2,3.$), $A_{ij}^{*}$($i,j=1,2.$) are circulant. There are a lot of favourable characters of this kind of matrices.

We list some of them in the following which are useful in analyzing our schemes.

Proposition 4.3 ([9,23])If A, B are circulant matrices with the same number of rows and columns, then we have

(i) $A+B,A-B,AB$ are circulant matrices.

(ii) If $A^{-1}$ is well defined, then $A^{-1}$ is also a circulant matrix.

(iii) A, B are commutators, that is AB = BA.

(iv) If A, B are symmetric and positive definite matrices, then AB = BA is a symmetric and positive definite matrix.

(v) The eigenvalues of the circulant matrix above is ${\lambda }_j={\sum }_{k=0}^{N-1}{c}_k{e}^{-ik{\theta }_j}$ with the corresponding eigenvector $w_j=\frac{1}{N}{\left[1,e^{-i{\theta }_j},\cdots ,e^{-i\left(N-1\right){\theta }_j}\right]}^{⊺}.$

By Proposition 4.1, we can get the following proposition.

Proposition 4.4.For the matrices in the CHOC solvers (2.10), (2.24), (2.27), we have:

(i) $A_{11}$, $A_{22}$, $-A_{23}$, $B_{22}$, $B_{33}$, $A_{11}^{*}$, $A_{22}^{*}$, $A_{23}^{*}$ are symmetric and positive definite.

(ii) $A_{12}$, $A_{13}$, $A_{21}$, $B_{12}$, $B_{14}$, $B_{21}$, $B_{23}$, $B_{34}$ are skew-symmetric.

(iii) $B_{11}$, $B_{13}$, $B_{24}$, $B_{31}$, $B_{32}$, $A_{12}^{*}$, $A_{13}^{*}$, $A_{21}^{*}$ are symmetric.

(iv) A, $A^{-1}$, D, $A_1$, $C_1$ are symmetric and positive definite, and $D_1$, $D_3$ are skew-symmetric, $D_2$, $B_1$ are symmetric.

(v) $M^{*},H_1$ are symmetric and positive definite, and G, $G^{*}$, I, are skew-symmetric, H, M, $H^{*}$, $G_1,N,N^{*},N_1$ are symmetric.

(vi) All of them are circulant.

Proof. The conclusions(i)–(iv) and (vi) can be observed or be verified from Proposition 4.1. The result (v) can be derived from the last conclusion in Proposition 4.1 by finding their eigenvalues.

Theorem 4.3.Define

$$w^n=\left(v^n,v^n\right)+\left(u^n,R·u^n\right),$$

where $\left(u,v\right)$ is the standard unitary inner product in discrete level for finite dimensional sequence vectors, $v^n={\left[v_1^n,v_2^n,\cdots ,v_{NX}^n\right]}^{⊺},u^n={\left[u_1^n,u_2^n,\cdots ,u_{NX}^n\right]}^{⊺}.$R is the matrix coefficient in approximating fourth-order spatial derivative expressed by M(2.10), $M^{*}$(2.24), $H_1$(2.27). Then for CHOC solvers (2.10), (2.24), (2.27) to solve ${\partial }_t^{2}u=-{\partial }_x^{4}u$, the numerical solutions satisfy that

$${\omega }^{n+1}=w^n.$$

(4.5)

Moreover, to $M^{*}$(2.24), $H_1$(2.27), there exists C such that $C^{⊺}·C=R$, therefore

$$w^n=\left(v^n,v^n\right)+\left(C·u^n,C·u^n\right),$$

Proof. Since $v=u_t$, we have $v_t=-u_{xxxx}$, then its discrete scheme is

$$\frac{v^{n+1}-v^n}{\tau }=-\frac{R{u}^{n+1}+R{u}^n}{2}.$$

By multiplying $\frac{u^{n+1}-u^n}{\tau }=\frac{v^{n+1}+v^n}{2}$ on both sides of above formula, we have

$$\frac{{\left(v^{n+1}+v^n\right)}^{⊺}\left(v^{n+1}-v^n\right)}{2\tau }=-\frac{{\left(v^{n+1}+v^n\right)}^{⊺}}{2}·\frac{R{u}^{n+1}+R{u}^n}{2}.$$

Therefore we obtain that

$$\begin{array}{cc}\hfill {\left(v^{n+1}\right)}^{⊺}\left(v^{n+1}\right)-{\left(v^n\right)}^{⊺}\left(v^n\right)& =-{\left(u^{n+1}\right)}^{⊺}R·\left(u^{n+1}\right)-{\left(u^n\right)}^{⊺}R·\left(u^n\right),\hfill \\ \hfill {\left(v^{n+1}\right)}^{⊺}\left(v^{n+1}\right)+{\left(u^{n+1}\right)}^{⊺}\left(R·u^{n+1}\right)& ={\left(v^n\right)}^{⊺}\left(v^n\right)+{\left(u^n\right)}^{⊺}\left(R·u^n\right),\hfill \end{array}$$

that is ${\omega }^{n+1}={\omega }^n$. Taking into symmetric positivity of $M^{*}$(2.24), $H_1$(2.27), there exists C such that $C^{⊺}·C=R,$ which yields that

$$\begin{array}{c}\hfill \left(u^n,R·u^n\right)={\left(C·u^n\right)}^{⊺}\left(C·u^n\right).\end{array}$$

Therefore, we obtain that

$$w^n=\left(v^n,v^n\right)+\left(C·u^n,C·u^n\right).$$

Theorem 4.4.To nonlinear good Boussinesq equation (1.2) with periodic boundary condition, the schemes CHOC-A, CHOC-B and CHOC-C satisfy the following discrete conservation law

$$\overline{u^n}=h\sum _{j=1}^{NX}{u}_j^n\equiv \overline{u^0}$$

(4.6)

provided that $\overline{v^0}=0.$

Proof. Firstly, we consider schemes CHOC-A to solve nonlinear system (1.2) and have the following formula

$$u^{n+1}=u^n+\tau \left(\frac{v^n+v^{n+1}}{2}\right),v^{n+1}=v^n+\frac{\tau }{2}N\left(u^{n+1}+u^n\right)+\frac{\tau }{2}H\left[{\left(u^n\right)}^2+{\left(u^{n+1}\right)}^2\right].$$

(4.7)

By calculation, we obtain that

$$v^{n+1}=v^n+\frac{\tau }{2}{\left(E-\frac{{\tau }^2}{4}N\right)}^{-1}\left\{\tau N{v}^n+2N{u}^n+H\left[{\left(u^n\right)}^2+{\left(u^{n+1}\right)}^2\right]\right\}.$$

(4.8)

Construct the following iterative algorithm

$$\begin{array}{cc}\hfill v_{(k+1)}^{n+1}& =v^n+\frac{\tau }{2}{\left(E-\frac{{\tau }^2}{4}N\right)}^{-1}\left\{\tau N{v}^n+2N{u}^n+H\left[{\left(u^n\right)}^2+{\left(u_{\left(k\right)}^{n+1}\right)}^2\right]\right\},\hfill \\ \hfill u_{(k+1)}^{n+1}& =u^n+\tau \left(\frac{v^n+v_{(k+1)}^{n+1}}{2}\right),\hfill \end{array}$$

(4.9)

where $k=0,1,...$ and $u_{\left(0\right)}^{n+1}=u^n.$ Then we get that

$$\underset{k\to \infty }{lim}{u}_{\left(k\right)}^{n+1}=u^{n+1},\underset{k\to \infty }{lim}{v}_{\left(k\right)}^{n+1}=v^{n+1}.$$

(4.10)

Considering $\overline{v^0}=0$ and the symmetry of $N,H$, we can obtain that $\overline{v_{(k+1)}^{n+1}}=\overline{v^n}=0,\overline{u_{(k+1)}^{n+1}}=\overline{u^n}.$ The limit (4.10) yields that $\overline{v^{n+1}}=0,\overline{u^{n+1}}=\overline{u^n}.$ The conservation identity (4.6) for the schemes CHOC-B and CHOC-C can be derived similarity.

For linear good Boussinesq equation (1.1), we have the following equivalent Hamiltonian system

$$\left\{\begin{array}{c}{u}_t=v,\hfill \\ v_t=u_{xx}-u_{xxxx},\hfill \end{array}\right.$$

(4.11)

with the Hamiltonian function $H=-\frac{1}{2}\int \left(v^2+u_x^2+u_{xx}^2+V^2\left(u\right)\right)dx$. Thus we obtain the symlecitic conservation law as follows

$$w\left(t\right)=\int du\wedge dvdx=w\left(0\right).$$

(4.12)

Symplectic schemes for Hamiltonian systems are proven to be more efficient than non-symplectic schemes for long-time numerical computations and are widely applied to practical problems arising in many fields of science and engineering which involves celestial mechanics, quantum physics, statistics and so on (see [13,24,25,26]). Next, we will derive that the considered CHOC schemes are symplectic.

Theorem 4.7.To linear good Boussinesq equation (1.1), the schemes CHOC-A, CHOC-B and CHOC-C are symplectic with the following conservation law

$$w^n=h\sum _{j=1}^{NX}d{u}_j^n\wedge d{v}_j^n=w^0.$$

(4.13)

Proof. To (3.2), scheme $CHOC-A$ has the formula as follows

$$\left[\begin{array}{cc}-\frac{\tau }{2}N& E\\ E& -\frac{\tau }{2}E\end{array}\right]\left[\begin{array}{c}{U}^{n+1}\\ V^{n+1}\end{array}\right]=\left[\begin{array}{cc}\frac{\tau }{2}N& E\\ E& \frac{\tau }{2}E\end{array}\right]\left[\begin{array}{c}{U}^n\\ V^n\end{array}\right]$$

Therefore through tedious calculations, scheme CHOC-A is symplectic. Similarity, the scheme CHOC-B and CHOC-C are also symplectic.