High-Order Time–Space Compact Difference Methods for Semi-Linear Sobolev Equations

Abstract

In this paper, high-order compact difference methods (HOCDMs) are proposed to solve the semi-linear Sobolev equations (SLSEs), which arise in various physical models, such as porous media flow and heat conduction. First, a two-level numerical method is given by applying the Crank–Nicolson (C-N) method in time and the fourth-order compact difference method in space. This method is shown to achieve second-order accuracy in time and fourth-order accuracy in space. Subsequently, we introduce the Richardson extrapolation technique to improve the temporal accuracy of the two-level method from second order to fourth order. Furthermore, we devise a fully fourth-order method in both time and space by applying the fourth-order difference method to discretize both temporal and spatial derivatives, and we provide a proof of its convergence. Finally, a series of numerical experiments is conducted to verify the effectiveness of the proposed methods.

1. Introduction

Consider the following initial boundary value problems (IBVPs) of SLSEs:

$$\left\{\begin{array}{c}{u}_t-\beta u_{xxt}=\alpha u_{xx}+f(x,t,u),x\in (a,b),t\in (0,T],\hfill \\ u(x,0)=\phi (x),x\in [a,b],u(a,t)={\psi }_1(t),u(b,t)={\psi }_2(t),\hfill \end{array}\right.$$

(1)

where $\alpha$, $\beta$, a, b, and T are given non-negative constants with $a<b$ and functions $\phi (x)$, ${\psi }_1(t)$, ${\psi }_2(t)$, and $f(x,t,u)$ are smooth enough on their respective domains and satisfy the following Lipschitz condition with constant $L>0$:

$$|f(x,t,u_1)-f(x,t,u_2)|\le L|u_1-u_2|,\forall (x,t)\in [a,b]\times (0,T],u_1,u_2\in \mathbb{R}.$$

(2)

Sobolev equations have important applications in various fields, including the non-steady-state flow in fissured rocks [1], moisture movement in soil [2], and heat transfer in heterogeneous media [3]. Therefore, the development of numerical methods and theoretical analysis for SLSEs is of significant practical importance. In particular, when $\beta =0$, problem (1) reduces to IBVPs for semi-linear reaction–diffusion equations. However, real-world problems are often highly complex, making it difficult to obtain exact problem (1) solutions. Therefore, it is of great significance to develop accurate and stable numerical methods.

The numerical methods for Sobolev equations have evolved significantly over the past decades. Early approaches focused on low-order finite difference schemes, while recent studies have emphasized high-order accuracy, stability, and efficiency, incorporating compact schemes to solve complex semi-linear or nonlinear problems. To date, a variety of numerical methods have been proposed to solve the Sobolev equations. In the context of finite element methods, Gu [4] studied characteristic finite element methods. Gao, Cui, and Zhao [5] proposed weak Galerkin finite element methods and proved their numerical stability and convergence. Tran and Duong [6] established an error estimate for semi-discrete finite element methods. Zeng and Luo [7] developed a C-N finite element method and provided proofs of existence, stability, and error estimates. Abbaszadeh and Dehghan [8] investigated the discontinuous Galerkin finite element method to solve generalized Sobolev equations. Wang and Li [9] introduced a three-step backward-differentiation formula-based finite element method and presented a superconvergence analysis. Xu, Zhou, and Zhao [10] studied the H-1-conforming virtual element method. Liu, Chen, and Liang [11] established a unified analysis of both the conforming and nonconforming virtual element methods. For the spectral methods, Tang, Li, and Yin [12] proposed a spacetime spectral method, employing Legendre–Galerkin methods in time and dual Petrov–Galerkin methods in space. Jin and Luo [13] studied a collocation spectral method based on Chebyshev polynomials. Dehghan, Shafieeabyaneh, and Abbaszadeh [14] introduced a spectral element method that combined the Crank-–Nicolson scheme in time with the Legendre spectral element method in space. Kumar and Baskar [15] presented a B-spline quasi-interpolation method. Yu and Wang [16] constructed an efficient spacetime spectral method and proved the convergence. Dehghan, Hooshyarfarzinand, and Abbaszadeh [17] proposed a mesh-free method based on the Pascal polynomial expansion. In addition, regarding finite difference methods, Chen, Duan, and Li [18] developed a Newtoian linearized compact difference method and analyzed its convergence properties. Zhang, Qin, and Sun [19] applied a compact difference method to solve Sobolev equations with Burgers-type nonlinearity. Li and Run [20] proposed a block-centered difference method and proved its stability and convergence. Xia and Luo [21] devised a Crank–Nicolson iterative scheme and established its unconditional stability and absolute convergence. Wang and Fu [22] presented two linearized C-N-block-centered methods and provided corresponding error analyses and stability proofs. Haq and Ali [23] developed a numerical scheme based on polynomials and the finite difference method. Mishra and Pany [24] constructed two first-order completely discrete schemes based on backward Euler methods. Singh et al. [25] designed a hybrid high-order method for a semi-linear Sobolev model on polygonal meshes. Chen et al. [26] proposed a two-grid finite difference method. Zhang, Qin, and Zhang [27] gave two linearized compact difference schemes and presented the maximum error estimates.

It is worth noting that the above-mentioned studies primarily focused on finite element methods, spectral methods, and finite difference methods with relatively low temporal accuracy. Moreover, to the best of our knowledge, the numerical methods in references [18,19,20,21,22,23,24,25,26,27] mainly use compact difference methods in the spatial direction and low-accuracy methods in the time direction, and nonlinear terms are calculated using linearization methods. The application of compact difference operators for first-order derivatives in the temporal direction to enhance temporal accuracy has not yet been explored. Therefore, we aim to develop a class of HOCDMs to solve problem (1).

The paper is organized as follows: In Section 2, a two-level HOCDM is constructed, and its convergence is established. In Section 3, we introduce the Richardson extrapolation technique to improve the temporal accuracy of the two-level HOCDM developed in Section 2. In Section 4, a three-level HOCDM with fourth-order accuracy in both time and space for problem (1) is proposed, and its convergence is also proven. Finally, Section 5 presents numerical experiments that validate the theoretical accuracy and efficiency of the proposed methods.

2. Two-Level High-Order Compact Difference Method

We define the spatial grid points as $x_i=a+ih(0\le i\le N)$ and the temporal grid points as $t_n=n\tau (0\le n\le M)$. Additionally, we introduce the temporal off-step points ($t_{n+\frac{1}{2}}=(n+{\displaystyle \frac{1}{2}})\tau (0\le n\le N-1)$), where $h={\displaystyle \frac{b-a}{N}}(N\in \mathbb{N})$ and $\tau ={\displaystyle \frac{T}{M}}(M\in \mathbb{N})$ denote the spatial and temporal step sizes, respectively. Moreover, we define ${\Omega }_h=\{x_i|1\le i\le N-1\}$, $\partial {\Omega }_h=\{x_i|i=0,N\}$, ${\overline{\Omega }}_h={\Omega }_h\cup \partial {\Omega }_h$, ${\overline{\Omega }}_{\tau }=\{t_n|0\le n\le M\}$, ${\overline{\Omega }}_{h\tau }={\overline{\Omega }}_h\times {\overline{\Omega }}_{\tau }$, and $U_i^n=u(x_i,t_n)$. For the grid function $W:=\{w_i^n|0\le i\le N,0\le n\le M\}$, we introduce the following operators:

$${\delta }_t{w}_i^{n+\frac{1}{2}}={\displaystyle \frac{1}{\tau }}(w_i^{n+1}-w_i^n),{\mu }_t{w}_i^{n+\frac{1}{2}}={\displaystyle \frac{1}{2}}(w_i^{n+1}-w_i^n),D_t{w}_i^n={\displaystyle \frac{w_i^{n+1}-w_i^{n-1}}{2\tau }},$$

$${\delta }_x{w}_i^n={\displaystyle \frac{1}{h}}(w_i^n-w_{i-1}^n),{\delta }_x^2{w}_i^n={\displaystyle \frac{1}{h}}({\delta }_x{w}_{i+1}^n-{\delta }_x{w}_i^n),$$

$${\mathcal{A}}_t{w}_i^n=\left\{\begin{array}{l}{\displaystyle \frac{1}{6}}(w_i^{n+1}+5w_i^n+w_i^{n-1}),1\le n\le M-1,\hfill \\ w_i^n,n=0orM,\hfill \end{array}\right.$$

$${\mathcal{A}}_x{w}_i^n=\left\{\begin{array}{l}{\displaystyle \frac{1}{12}}(w_{i+1}^n+10w_i^n+w_{i-1}^n),1\le i\le N-1,\hfill \\ w_i^n,i=0orN.\hfill \end{array}\right.$$

In addition, the following lemma will also play a crucial role in the method construction:

Lemma 1

([28]). Assume that function $G(x)\in C^6([x_{i-1},x_{i+1}])$. Then,

$${\displaystyle \frac{1}{12}}[G^{\prime \prime }(x_{i-1})+10G^{\prime \prime }(x_i)+G^{\prime \prime }(x_{i+1})]-{\displaystyle \frac{1}{h^2}}[G(x_{i-1})-2G(x_i)+G(x_{i+1})]={\displaystyle \frac{h^4}{240}}{G}^{(6)}(x_i)+\mathcal{O}(h^6).$$

Lemma 2

([28]). Suppose $W(t)\in C^5[t_{n-1},t_{n+1}]$. Then, we have

$${\displaystyle \frac{1}{6}}[W^{\prime }(t_{n-1})+4W^{\prime }(t_n)+W^{\prime }(t_{n+1})]={\displaystyle \frac{1}{2\tau }}[W(t_{n+1})-W(t_{n-1})]+{\displaystyle \frac{{\tau }^4}{180}}{W}^{(5)}(t_n)+\mathcal{O}({\tau }^5).$$

2.1. The Two-Level High-Order Compact Difference Method

2.1.1. The Numerical Method Derivation

Setting the problem (1) at point $(x,t)=(x_i,t_{n+\frac{1}{2}})$, for $0\le i\le N-1$ and $0\le n\le M-1$, we have

$${\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})-\beta {\displaystyle \frac{{\partial }^{3}u}{\partial x^2\partial t}}(x_i,t_{n+\frac{1}{2}})=\alpha {\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+\frac{1}{2}})+f(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+{\displaystyle \frac{1}{2}}})).$$

(3)

By applying the Taylor expansion, the following equalities are obtained:

$${\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})={\delta }_t{U}_i^{n+\frac{1}{2}}-{\displaystyle \frac{{\tau }^2}{24}}{\displaystyle \frac{{\partial }^{3}u}{\partial t^3}}(x_i,t_{n+\frac{1}{2}})+\mathcal{O}({\tau }^4),$$

(4)

$${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+\frac{1}{2}})=\frac{1}{2}\left[{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)+{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+1})\right]-{\displaystyle \frac{{\tau }^2}{8}}{\displaystyle \frac{{\partial }^{4}u}{\partial x^2\partial t^2}}(x_i,t_{n+\frac{1}{2}})+\mathcal{O}({\tau }^4),$$

(5)

$$\begin{array}{cc}\hfill & f(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}}))=\frac{1}{2}\left[f(x_i,t_n,U_i^n)+f(x_i,t_{n+1},U_i^{n+1})\right]\hfill \\ \hfill & -{\displaystyle \frac{{\tau }^2}{8}}\left\{{\displaystyle \frac{{\partial }^{2}f}{\partial t^2}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}}))+{\displaystyle \frac{{\partial }^{2}f}{\partial u^2}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_n)){\left[{\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})\right]}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{\partial f}{\partial u}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}})){\displaystyle \frac{{\partial }^{2}u}{\partial t^2}}(x_i,t_{n+\frac{1}{2}})+2{\displaystyle \frac{{\partial }^{2}f}{\partial u\partial t}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}})){\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})\right\}\hfill \\ \hfill & +\mathcal{O}({\tau }^4).\hfill \end{array}$$

(6)

For $1\le i\le N-1$ and $0\le n\le M-1$, substituting (4)–(6) into (3) yields

$$\begin{array}{cc}\hfill & {\delta }_t{U}_i^{n+\frac{1}{2}}-\beta {\delta }_t{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+\frac{1}{2}})={\displaystyle \frac{\alpha }{2}}\left[{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)+{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+1})\right]+\frac{1}{2}[f(x_i,t_n,U_i^n)\hfill \\ \hfill & +f(x_i,t_{n+1},U_i^{n+1})]+r_i^{n+\frac{1}{2}},\hfill \end{array}$$

(7)

where

$$\begin{array}{c}{r}_i^{n+\frac{1}{2}}={\displaystyle \frac{{\tau }^2}{24}}{\displaystyle \frac{{\partial }^{3}u}{\partial t^3}}(x_i,t_{n+\frac{1}{2}})+{\displaystyle \frac{\beta {\tau }^2}{24}}{\displaystyle \frac{{\partial }^{5}u}{\partial t^3\partial x^2}}(x_i,t_{n+\frac{1}{2}})-{\displaystyle \frac{\alpha {\tau }^2}{8}}{\displaystyle \frac{{\partial }^{4}u}{\partial x^2\partial t^2}}(x_i,t_{n+\frac{1}{2}})\hfill \\ -{\displaystyle \frac{{\tau }^2}{8}}\{{\displaystyle \frac{{\partial }^{2}f}{\partial t^2}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}}))+{\displaystyle \frac{{\partial }^{2}f}{\partial u^2}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}})){[{\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})]}^2\hfill \\ +{\displaystyle \frac{\partial f}{\partial u}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}})){\displaystyle \frac{{\partial }^{2}u}{\partial t^2}}(x_i,t_{n+\frac{1}{2}})+2{\displaystyle \frac{{\partial }^{2}f}{\partial u\partial t}}(x_i,t_{n+\frac{1}{2}},u(x_i,t_{n+\frac{1}{2}})){\displaystyle \frac{\partial u}{\partial t}}(x_i,t_{n+\frac{1}{2}})\hfill \\ +\mathcal{O}({\tau }^4).\hfill \end{array}$$

(8)

Applying the operator ${\mathcal{A}}_x$ to both sides of (7), we obtain

$$\begin{array}{cc}\hfill & {\mathcal{A}}_x{\delta }_t{U}_i^{n+\frac{1}{2}}-\beta {\delta }_t{\mathcal{A}}_x{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+\frac{1}{2}})={\displaystyle \frac{\alpha }{2}}{\mathcal{A}}_x\left[{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)+{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_{n+1})\right]\hfill \\ \hfill & +{\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x_i,t_n,U_i^n)+f(x_i,t_{n+1},U_i^{n+1})]+{\mathcal{A}}_x{r}_i^{n+\frac{1}{2}},1\le i\le N-1,0\le n\le M-1.\hfill \end{array}$$

(9)

It follows from Lemma 1 that

$${\mathcal{A}}_x{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_k)={\delta }_x^2{U}_i^k+{\displaystyle \frac{h^4}{240}}{\displaystyle \frac{{\partial }^{6}u}{\partial x^6}}(x_i,t_k)+\mathcal{O}(h^6).$$

(10)

For $1\le i\le N-1$ and $0\le n\le M-1$, substituting (10) into (9) yields

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\mathcal{A}}_x{\delta }_t{U}_i^{n+\frac{1}{2}}-\beta {\delta }_t{\delta }_x^2{U}_i^{n+\frac{1}{2}}=\alpha {\mu }_t{\delta }_x^2{U}_i^{n+\frac{1}{2}}+\frac{1}{2}{\mathcal{A}}_x[f(x_i,t_n,U_i^n)+f(x_i,t_{n+1},U_i^{n+1})]+R_i^{n+\frac{1}{2}},\end{array}\end{array}$$

(11)

where

$$\begin{array}{cc}\hfill & R_i^{n+\frac{1}{2}}={\mathcal{A}}_x{r}_i^{n+\frac{1}{2}}+{\displaystyle \frac{\alpha h^4}{480}}\left[{\displaystyle \frac{{\partial }^{6}u}{\partial x^6}}(x_i,t_n)+{\displaystyle \frac{{\partial }^{6}u}{\partial x^6}}(x_i,t_{n+1})\right]+{\displaystyle \frac{\beta h^4}{240}}{\displaystyle \frac{{\partial }^{7}u}{\partial t\partial x^6}}(x_i,t_{n+\frac{1}{2}})\hfill \\ \hfill & +\mathcal{O}({\tau }^4+{\tau }^2{h}^4+h^6).\hfill \end{array}$$

(12)

Since $u(x,t)\in C^{6,3}({\overline{\Omega }}_{h\tau })$, there exists a constant ($c_0>0$) such that

$$|R_i^{n+\frac{1}{2}}|\le c_0({\tau }^2+h^4),1\le i\le N-1,0\le n\le M-1.$$

(13)

In (11), by omitting the remainder term ($R_i^{n+\frac{1}{2}}$) and replacing $U_i^k$ with its numerical approximation ($u_i^k$) for $k=n$, $n+\frac{1}{2}$, and $n+1$, the following HOCDM for problem (1) is derived:  

$$\begin{array}{cc}\hfill & {\mathcal{A}}_x{\delta }_t{u}_i^{n+\frac{1}{2}}-\beta {\delta }_t{\delta }_x^2{u}_i^{n+\frac{1}{2}}=\alpha {\mu }_t{\delta }_x^2{u}_i^{n+\frac{1}{2}}+\frac{1}{2}{\mathcal{A}}_x[f(x_i,t_n,u_i^n)\hfill \\ \hfill & +f(x_i,t_{n+1},u_i^{n+1})],1\le i\le N-1,0\le n\le M-1,\hfill \end{array}$$

(14)

where, from (1), we have

$$u_i^0=\phi (x_i),0\le i\le N;u_0^n={\psi }_1(t_n),u_N^n={\psi }_2(t_n),1\le n\le M.$$

(15)

By estimating (13), we indicate that methods (14) and (15) have the local accuracy ($\mathcal{O}({\tau }^2+h^4)$). Moreover, the method (14) can be efficiently implemented using the Picard iteration algorithm. The Picard iteration is convergent, and a detailed analysis of the convergence of this iterative method can be found in [29]. Then, $f(x,t,u)$ is approximated using the Newtonian linearization technique, and the numerical method is presented in [18]. The detailed implementation process of methods (14) and (15) is described in Algorithm 1.

Algorithm 1 The computational implementation of HOCDM (14) and (15)

Step 1: $obtaininitialvalues{u}_i^0(0\le i\le N)andboundaryvalues$: $u_0^n,u_N^n,(1\le n\le M)$ $using$(15); Step 2: $forn=0,1,\dots ,M-1$,                 $compute{u}_i^{n}byapplyingtheThomasalgorithmandPicarditerationalgorithmto(14)$,               $end$

2.1.2. The Convergence of the Two-Level HOCDM

In this section, we conduct an error analysis for methods (14) and (15). First, let $V=\left\{v_i\right.\left|0\le i\le N,\right.\left.v_0=v_N=0\right\}$ be the grid function defined on ${\overline{\Omega }}_h$ and define the following inner product and norms for $v,\tilde{v}\in V$:

$$\parallel v\parallel =\sqrt{h\sum _{i=1}^{N-1}{(v_i)}^2}{,|v|}_1=\sqrt{h\sum _{i=1}^N{\left({\displaystyle \frac{v_i-v_{i-1}}{h}}\right)}^2},{\parallel v\parallel }_{\infty }=\underset{1\le i\le N-1}{max}|v_i|,(v,\tilde{v})=h\sum _{i=1}^{N-1}{v}_i{\tilde{v}}_i.$$

In the following convergence analysis, we will utilize the subsequent lemmas.

Lemma 3

(cf. [30]). For all $v,w\in V$, the following inequalities hold:

$$(i){\parallel v\parallel }_{\infty }\le {\displaystyle \frac{\sqrt{b-a}}{2}}{|v|}_1,(ii)\parallel v\parallel \le {\displaystyle \frac{b-a}{\sqrt{6}}}{|v|}_1,(iii)\sum _{i=1}^{N-1}({\delta }_x^2{v}_i)w_i=-\sum _{i=1}^N({\delta }_x{v}_i)({\delta }_x{w}_i).$$

Lemma 4

(cf. [31]). For all $v\in V$, the following inequalities hold:

$$(i)\sum _{i=1}^{N-1}({\mathcal{A}}_x{v}_i)v_i\le \sum _{i=1}^{N-1}{(v_i)}^2,(ii)\sum _{i=1}^{N-1}({\mathcal{A}}_x{v}_i)v_i\ge {\displaystyle \frac{2}{3}}\sum _{i=1}^{N-1}{(v_i)}^2.$$

Lemma 5

(cf. [31]). Assume that there are constants $A,B\ge 0$ such that the non-negative sequence $\{F_k:k\ge 0\}$ satisfies

$$F_{k+1}\le A+B\tau \sum _{i=1}^k{F}_i,\forall k\ge 0.$$

Then, $F_{k+1}\le Aexp(Bk\tau )$ for all $k\ge 0$.

Write $e^n={(e_1^n,e_2^n,\dots ,e_{N-1}^n)}^T$ with $e_i^n=U_i^n-u_i^n(0\le i\le N,0\le n\le M)$. By the initial and boundary conditions in (15), $e_i^0=e_0^n=e_N^n=0(0\le i\le N,1\le n\le M)$. With the above arguments, the error analysis result of the method can be presented as follows:

Theorem 1.

Assume that $u(x,t)\in C^{6,3}({\overline{\Omega }}_{h\tau })$, and Lipschitz condition (2) holds. Then, we have

$${\parallel e^n\parallel }_{\infty }\le c_1({\tau }^2+h^4),1\le n\le M,$$

(16)

where $c_1>0$ is a constant independent of τ and h.

Proof. 

Subtracting (11) from (14) yields

$$\begin{array}{cc}\hfill & {\mathcal{A}}_x{\delta }_t{e}_i^{n+\frac{1}{2}}-\beta {\delta }_x^2{\delta }_t{e}_i^{n+\frac{1}{2}}=\alpha {\mu }_t{\delta }_x^2{e}_i^{n+\frac{1}{2}}+{\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x_i,t_n,U_i^n)-f(x_i,t_n,u_i^n)\hfill \\ \hfill & +f(x_i,t_{n+1},U_i^{n+1})-f(x_i,t_{n+1},u_i^{n+1})]+R_i^{n+\frac{1}{2}},1\le i\le N-1,0\le n\le M-1,\hfill \end{array}$$

(17)

where $u_i^n(1\le i\le N-1,0\le n\le M-1)$ represent the numerical solutions computed using methods (14) and (15) for problem (1), and the corresponding implementation is provided in Algorithm 1.

Taking the inner product of both sides in (17), with ${\delta }_t{e}^{n+\frac{1}{2}}$, it becomes

$$\begin{array}{cc}\hfill & ({\mathcal{A}}_x{\delta }_t{e}^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}})-(\beta {\delta }_x^2{\delta }_t{e}^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}})=(\alpha {\mu }_t{\delta }_x^2{e}^{n+\frac{1}{2}},{\delta }_t{e}^{km+l+\frac{1}{2}})\hfill \\ \hfill & +({\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x,t_n,U^n)-f(x,t_n,u^n)+f(x,t_{n+1},U^{n+1})-f(x,t_{n+1},u^{n+1})],{\delta }_t{e}^{n+\frac{1}{2}})\hfill \\ \hfill & +(R^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}}),0\le n\le M-1.\hfill \end{array}$$

(18)

Applying inequality $(ii)$ in Lemma 4, we have

$$({\mathcal{A}}_x{\delta }_t{e}^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}})\ge {\displaystyle \frac{2}{3}}{\parallel {\delta }_t{e}^{n+\frac{1}{2}}\parallel }^2,0\le n\le M-1.$$

(19)

Using $(iii)$ in Lemma 3, it holds that

$$-(\beta {\delta }_x^2{\delta }_t{e}^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}})=\beta |{\delta }_t{e}^{n+\frac{1}{2}}{|}_1^2,$$

(20)

$$(\alpha {\mu }_t{\delta }_x^2{e}^{n+\frac{1}{2}},{\delta }_t{e}^{km+l+\frac{1}{2}})=-{\displaystyle \frac{\alpha }{2}}(|e^{n+1}{|}_1^2-|e^n{|}_1^2).$$

(21)

By the Lipschitz condition (2), inequality $(ii)$ in Lemma 3, and inequality $(i)$ in Lemma 4, we have

$$\begin{array}{cc}\hfill & ({\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x,t_n,U^n)-f(x,t_n,u^n)+f(x,t_{n+1},U^{n+1})-f(x,t_{n+1},u^{n+1})],{\delta }_t{e}^{n+\frac{1}{2}})\hfill \\ \hfill \le & h\sum _{i=1}^{N-1}\left[{\displaystyle \frac{L}{2}}(|e_i^n| + |e_i^{n+1}|)\right]{\delta }_t{e}_i^{n+\frac{1}{2}}\le {\displaystyle \frac{L^{2}h}{2}}\sum _{i=1}^{N-1}{\left({\displaystyle \frac{|e_i^n| + |e_i^{n+1}|}{2}}\right)}^2 + \frac{1}{2}\parallel {\delta }_t{e}^{n+\frac{1}{2}}{\parallel }^2\hfill \\ \hfill \le & {\displaystyle \frac{L^2}{4}}(\parallel e^n{\parallel }^2+\parallel e^{n+1}{\parallel }^2)+\frac{1}{2}\parallel {\delta }_t{e}^{n+\frac{1}{2}}{\parallel }^2\hfill \\ \hfill \le & {\displaystyle \frac{L^2{(b-a)}^2}{24}}(|e^n{|}_1^2+|e^{n+1}{|}_1^2)+\frac{1}{2}{\parallel {\delta }_t{e}^{n+\frac{1}{2}}\parallel }^2,0\le n\le M-1,\hfill \end{array}$$

(22)

where the following common inequality is also used:

$$\mu \nu \le {\displaystyle \frac{1}{2\zeta }}{\mu }^2+{\displaystyle \frac{\zeta }{2}}{\nu }^2,\forall \mu ,\nu \in \mathbb{R},\zeta >0.$$

(23)

By (13) and (23), for $0\le n\le M-1$, it holds that

$$(R^{n+\frac{1}{2}},{\delta }_t{e}^{n+\frac{1}{2}})=h\sum _{i=1}^{N-1}{R}_i^{n+\frac{1}{2}}{\delta }_t{e}_i^{n+\frac{1}{2}}\le {\displaystyle \frac{1}{6}}\parallel {\delta }_t{e}^{n+\frac{1}{2}}{\parallel }^2+{\displaystyle \frac{3c_0^2(b-a)}{2}}{({\tau }^2+h^4)}^2.$$

(24)

The combination of (18)–(22) and (24) implies

$${\displaystyle \frac{\alpha }{2\tau }}(|e^{n+1}{|}_1^2-|e^n{|}_1^2)\le {\displaystyle \frac{L^2{(b-a)}^2}{24}}(|e^n{|}_1^2+|e^{n+1}{|}_1^2)+{\displaystyle \frac{3c_0^2(b-a)}{2}}{({\tau }^2+h^4)}^2,0\le n\le M-1.$$

(25)

Summing for n from 0 to $\tilde{n}\le M-1$ on both sides of (25) implies that

$$\left[{\displaystyle \frac{\alpha }{2\tau }}-{\displaystyle \frac{L^2{(b-a)}^2}{24}}\right]|e^{\tilde{n}+1}{|}_1^2\le {\displaystyle \frac{L^2{(b-a)}^2}{12}}\sum _{n=1}^{\tilde{n}}{|e^n|}_1^2+{\displaystyle \frac{3(\tilde{n}+1)c_0^2(b-a)}{2}}{({\tau }^2+h^4)}^2.$$

(26)

Multiplying (26) by ${\displaystyle \frac{2\tau }{\alpha }}$ gives

$${\displaystyle \frac{12\alpha -L^2{(b-a)}^2\tau }{12\alpha }}|e^{\tilde{n}+1}{|}_1^2\le {\displaystyle \frac{L^2{(b-a)}^2\tau }{6\alpha }}\sum _{n=1}^{\tilde{n}}{|e^n|}_1^2+{\displaystyle \frac{3c_0^2(b-a)}{\alpha }}{({\tau }^2+h^4)}^2.$$

(27)

Since there exist constants $\tilde{\tau }$ and $\gamma >0$, we have

$$12\alpha -L^2{(b-a)}^2\tau >\gamma ,\tau \in (0,\tilde{\tau }].$$

(28)

Based on (28) and inequality (27), it can be estimated that

$$|e^{\tilde{n}+1}{|}_1^2\le {\displaystyle \frac{2L^2{(b-a)}^2\tau }{12\alpha -L^2{(b-a)}^2\tau }}\sum _{n=1}^{\tilde{n}}{|e^n|}_1^2+{\displaystyle \frac{36c_0^2(b-a)}{12\alpha -L^2{(b-a)}^2\tau }}{({\tau }^2+h^4)}^2.$$

(29)

For the above inequality (29),

$$|e^{\tilde{n}+1}{|}_1^2\le {\gamma }_1\tau \sum _{n=0}^{\tilde{n}}|e^n{|}_1^2+{\gamma }_2{({\tau }^2+h^4)}^2,0\le \tilde{n}\le M-1,$$

(30)

where ${\displaystyle {\gamma }_1=\frac{2L^2{(b-a)}^2}{\gamma }}$ and ${\displaystyle {\gamma }_2=\frac{36c_0^2(b-a)}{\gamma }}$.

Applying Lemma 5 to (30), we have

$$|e^{\tilde{n}+1}{|}_1^2\le {\gamma }_{2}exp({\gamma }_1){({\tau }^2+h^4)}^2,0\le \tilde{n}\le M-1.$$

(31)

It follows, then, from (31) and inequality $(i)$ in Lemma 3 that

$$\parallel e^{\tilde{n}+1}{\parallel }_{\infty }\le c_1({\tau }^2+h^4),0\le \tilde{n}\le M-1,$$

(32)

where ${\displaystyle c_1=\frac{\sqrt{(b-a){\gamma }_{2}exp({\gamma }_1)}}{2}}$. Also, it holds that $1\le \tilde{n}+1\le M$. Therefore error estimate (16) is proved by (32).    □

3. Richardson Extrapolation Technique

In the previous section, we proposed a two-level numerical method for solving problem (1) and demonstrated that this method achieves a computational accuracy of $\mathcal{O}({\tau }^2+h^4)$. To improve the temporal accuracy of the two-level numerical method, we now apply the Richardson extrapolation technique.

Let ${\widehat{e}}^j=U^j-{(u_E)}^j$, $U^j={(u_1^j,u_2^j,\dots ,u_{N-1}^j)}^T$, ${(u_E)}^j={({(u_E)}_1^j,{(u_E)}_2^j,\dots ,{(u_E)}_{N-1}^j)}^T$ and

$${(u_E)}_i^{n+1}=\left\{\begin{array}{c}{\displaystyle \frac{4}{3}}{u}_i^{n+2}({\displaystyle \frac{\tau }{2}},h)-{\displaystyle \frac{1}{3}}{u}_i^{n+1}(\tau ,h),1\le i\le N-1,0\le n\le M-1,\hfill \\ {\psi }_1(t_{n+1}),i=0,0\le n\le M-1,\hfill \\ {\psi }_2(t_{n+1}),i=N,0\le n\le M-1,\hfill \end{array}\right.$$

(33)

where $u_i^j(\tau ,h)$ denote the numerical solution at $(x_i,t_j)$, obtained by applying the two-level numerical methods (14) and (15), with step sizes of $\tau$ and h, to problem (1). By combining (14) and (15) with (33), a two-level numerical method with Richardson extrapolation is derived. For convenience, we refer to this method as the Richardson extrapolation method. The detailed implementation process is described in Algorithm 2.

Algorithm 2 The computational implementation of the Richardson extrapolation method ((14) and (15), (33))

Step 1: $obtaininitialvalues{(u_E)}_i^0(0\le i\le N)andboundaryvalues$: ${(u_E)}_0^n,{(u_E)}_N^n,(1\le n\le M)using$ (33); Step 2: $forn=0,1,...,M-1$, $compute{u}_i^{n+\frac{1}{2}}({\displaystyle \frac{\tau }{2}},h)and{u}_i^{n+1}(\tau ,h)using$(14) and (15),  $compute{(u_E)}_i^{n+1}using$ (33);             $end$

Theorem 2.

Assume that $u(x,t)\in C^{6,6}({\overline{\Omega }}_{h\tau })$ and that Lipschitz condition (2) is satisfied. Then, we have

$$\parallel {\widehat{e}}^n{\parallel }_{\infty }\le c_3({\tau }^4+h^4),1\le n\le M,$$

(34)

where $c_3>0$ is a constant independent of τ and h.

Proof. 

Write $f_1(x,t,u)={\displaystyle \frac{\partial f(x,t,u)}{\partial u}}$ and

$$\begin{array}{cc}\hfill \widehat{r}(x,t)=& {\displaystyle \frac{1}{24}}{\displaystyle \frac{{\partial }^{3}u}{\partial t^3}}(x,t)-{\displaystyle \frac{\alpha }{8}}{\displaystyle \frac{{\partial }^{4}u}{\partial x^2\partial t^2}}(x,t)-{\displaystyle \frac{1}{8}}\left[{\displaystyle \frac{{\partial }^{2}f}{\partial t^2}}(x,t,u(x,t))+{\displaystyle \frac{{\partial }^{2}f}{\partial u^2}}(x,t,u(x,t)){\left({\displaystyle \frac{\partial u}{\partial t}}(x,t)\right)}^2\right.\hfill \\ & \left.+f_1(u,x,t){\displaystyle \frac{{\partial }^{2}u}{\partial t^2}}(x,t)+2{\displaystyle \frac{{\partial }^{2}f}{\partial u\partial t}}(x,t,u(x,t)){\displaystyle \frac{\partial u}{\partial t}}(x,t)\right].\hfill \end{array}$$

Using the above notations, along with (12) and a Taylor expansion, we have

$$R_i^{n+\frac{1}{2}}={\tau }^2{\mathcal{A}}_x\widehat{r}(x_i,t_{n+\frac{1}{2}})+\mathcal{O}({\tau }^4+h^4),0\le n\le M-1.$$

(35)

Let $\omega (x,t)$ be the solution of the following problem:

$$\left\{\begin{array}{c}{\omega }_t(x,t)-\beta {\omega }_{xxt}(x,t)=\alpha {\omega }_{xx}(x,t)+\nu (x,t,\omega (x,t))+\widehat{r}(x,t),x\in (a,b),t\in (0,T],\hfill \\ \omega (x,0)=0,x\in [a,b];\omega (a,t)=\omega (b,t)=0,t\in (0,T],\hfill \end{array}\right.$$

(36)

where $\nu (\omega (x,t),x,t)=\omega (x,t)f_1(x,t,u(x,t))$. Then, in a manner similar to that for the derivation of (11), we can obtain an equivalent form of problem (36) at point $(x_i,t_{n+\frac{1}{2}})$ as follows:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\mathcal{A}}_x{\delta }_t{\omega }_i^{n+\frac{1}{2}}-\beta {\delta }_t{\delta }_x^2{\omega }_i^{n+\frac{1}{2}}=& \alpha {\mu }_t{\delta }_x^2{\omega }_i^{n+\frac{1}{2}}+{\displaystyle \frac{{\mathcal{A}}_x}{2}}[\nu (x_i,t_n,{\omega }_i^n)+\nu (x_i,t_{n+1},{\omega }_i^{n+1})]\hfill \\ & +{\mathcal{A}}_x\widehat{r}(x_i,t_{n+\frac{1}{2}})+{\widehat{R}}_i^{n+\frac{1}{2}},1\le i\le N-1,0\le n\le M-1,\hfill \end{array}\end{array}$$

(37)

where ${\omega }_i^j=\omega (x_i,t_j)$, and ${\widehat{R}}_i^{n+\frac{1}{2}}=\mathcal{O}({\tau }^2+h^4)$. Also, when setting ${\varrho }_i^j=U_i^j-{\tau }^2{\omega }_i^j(j=n,n+1)$ by a Taylor expansion, it holds that

$$f(x_i,t_j,{\varrho }_i^j)=f(x_i,t_j,U_i^j)-{\tau }^2{\omega }_i^j{f}_1(x_i,t_j,U_i^j)+\mathcal{O}({\tau }^4).$$

(38)

Multiplying (37) by $-{\tau }^2$ and then using (17) and (38), for $1\le i\le N-1$ and $0\le n\le M-1$, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\mathcal{A}}_x{\delta }_t{\widehat{\varrho }}_i^{n+\frac{1}{2}}+\beta {\delta }_t{\delta }_x^2{\widehat{\varrho }}_i^{n+\frac{1}{2}}=& \alpha {\mu }_t{\delta }_x^2{\widehat{\varrho }}_i^{n+\frac{1}{2}}+{\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x_i,t_n,{\varrho }_i^n)-f(x_i,t_n,u_i^n)\hfill \\ & +f(x_i,t_{n+1},{\varrho }_i^{n+1})-f(x_i,t_{n+1},u_i^{n+1})]+{\tilde{R}}_i^{n+\frac{1}{2}},\hfill \end{array}\end{array}$$

(39)

where ${\widehat{\varrho }}_i^j={\varrho }_i^j-u_i^j=e_i^j-{\tau }^2{\omega }_i^j(j=n+\frac{1}{2})$, and

$${\tilde{R}}_i^{n+\frac{1}{2}}=R_i^{n+\frac{1}{2}}-{\tau }^2{\mathcal{A}}_x\widehat{r}(x_i,t_{n+\frac{1}{2}})-{\tau }^2{\widehat{R}}_i^{n+\frac{1}{2}}+\mathcal{O}({\tau }^4)=\mathcal{O}({\tau }^4+h^4).$$

(40)

With (39) and (40), condition $u(x,t)\in C^{6,6}({\overline{\Omega }}_{h\tau })$, and the similar proof technique for (16), we can derive the estimate of vector (${\widehat{\varrho }}^{n+1}:={({\widehat{\varrho }}_1^{n+1},{\widehat{\varrho }}_1^{n+1},\dots ,{\widehat{\varrho }}_{N-1}^{n+1})}^T$) as follows:

$$\parallel {\widehat{\varrho }}^{n+1}{\parallel }_{\infty }\le c_4({\tau }^4+h^4),0\le n\le M-1,$$

(41)

where $c_4>0$ is a constant independent of $\tau$ and h. It follows from (33) that

$$\begin{array}{cc}\hfill & U_i^{n+1}-{(u_E)}_i^{n+1}={\displaystyle \frac{4}{3}}\left[U_i^{n+1}-u_i^{n+2}({\displaystyle \frac{\tau }{2}},h)-{\displaystyle \frac{{\tau }^2}{4}}{\omega }_i^{n+1}\right]-{\displaystyle \frac{1}{3}}[U_i^{n+1}\hfill \\ \hfill & -u_i^{n+1}(\tau ,h)-{\tau }^2{\omega }_i^{n+1}],1\le i\le N-1,0\le n\le M-1.\hfill \end{array}$$

(42)

The combination of (41) and (42) implies that

$$\parallel U^{n+1}-{(u_E)}^{n+1}{\parallel }_{\infty }\le c_3({\tau }^4+h^4),0\le n\le M-1.$$

(43)

Since $1\le n+1\le M$, the error estimate in (34) is now concluded by (43). This completes the proof. □

Theorem 2 indicates that the computational accuracy of the two-level HOCDM in time can be improved to the fourth order, while the spatial accuracy of the improved method remains fourth order when Richardson extrapolation is applied, provided that certain conditions are satisfied.

4. The Three-Level HOCDM

4.1. Derivation of the Three-Level HOCDM

Considering problem (1) at points $(x_i,t_n)$, for $1\le i\le N-1$ and $1\le n\le M-1$. It holds that

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial u}{\partial t}}(x_i,t_n)-\beta {\displaystyle \frac{{\partial }^{3}u}{\partial x^2\partial t}}(x_i,t_n)=\alpha {\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)+f(x_i,t_n,u(x_i,t_n)).\end{array}$$

(44)

Applying the operator ${\mathcal{A}}_t$ to both sides of (44), for $1\le i\le N-1$ and $1\le n\le M-1$, we have

$$D_{t}u(x_i,t_n)-\beta D_t{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)=\alpha {\mathcal{A}}_t{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}(x_i,t_n)+{\mathcal{A}}_{t}f(x_i,t_n,u(x_i,t_n))+{\overline{r}}_i^n.$$

(45)

Also, it follows from Lemma 2 that

$${\mathcal{A}}_t{\displaystyle \frac{\partial u}{\partial t}}(x_i,t_n)=D_{t}u(x_i,t_n)+{\displaystyle \frac{{\tau }^4}{180}}{\displaystyle \frac{{\partial }^{5}u}{\partial t^5}}(x_i,t_n)+\mathcal{O}({\tau }^5),$$

(46)

where

$${\overline{r}}_i^n=-{\displaystyle \frac{{\tau }^4}{180}}{\displaystyle \frac{{\partial }^{5}u}{\partial t^5}}(x_i,t_n)+{\displaystyle \frac{\beta {\tau }^4}{180}}{\displaystyle \frac{{\partial }^{7}u}{\partial t^5\partial x^2}}(x_i,t_n)+\mathcal{O}({\tau }^5).$$

Applying the operator ${\mathcal{A}}_x$ to both sides of (45) and using Lemma 1, for $1\le i\le N-1$ and $1\le n\le M-1$, we acquire

$$D_t{\mathcal{A}}_x{U}_i^n-\beta D_t{\delta }_x^2{U}_i^n=\alpha {\mathcal{A}}_t{\delta }_x^2{U}_i^n+{\mathcal{A}}_t{\mathcal{A}}_{x}f(x_i,t_n,u(x_i,t_n))+{\overline{R}}_i^n,$$

(47)

where

$${\overline{R}}_i^n={\mathcal{A}}_x{\overline{r}}_i^n+{\displaystyle \frac{\alpha h^4}{240}}{\mathcal{A}}_t{\displaystyle \frac{{\partial }^{6}u}{\partial x^6}}(x_i,t_n)+{\displaystyle \frac{\beta h^4}{240}}{\displaystyle \frac{{\partial }^{7}u}{\partial t\partial x^6}}(x_i,t_n)+\mathcal{O}({\tau }^5+{\tau }^4{h}^4+h^6).$$

(48)

For the above ${\overline{R}}_i^n$, assuming that $u(x,t)\in C^{6,5}({\overline{\Omega }}_{h\tau })$, there exists a constant ($c_5>0$) such that

$$|{\overline{R}}_i^n|\le c_5({\tau }^4+h^4).$$

(49)

Omitting the truncation errors (${\overline{R}}_i^n$) and replacing $U_i^n$ with ${\overline{u}}_i^n$ in (48), for $1\le i\le N-1$ and $1\le n\le M-1$, it holds that

$$D_t{\mathcal{A}}_x{\overline{u}}_i^n-\beta D_t{\delta }_x^2{\overline{u}}_i^n=\alpha {\mathcal{A}}_t{\delta }_x^2{\overline{u}}_i^n+{\mathcal{A}}_t{\mathcal{A}}_{x}f(x_i,t_n,{\overline{u}}_i^n),$$

(50)

with the initial and boundary conditions as follows:

$${\overline{u}}_i^0=\phi (x_i),0\le i\le N;{\overline{u}}_0^n={\psi }_1(t_n),{\overline{u}}_N^n={\psi }_2(t_n),1\le n\le M.$$

(51)

Since the numerical method ((50) and (51)) is a three-level high-order compact difference method, it is necessary to provide a two-level starting algorithm with fourth-order accuracy in both time and space. Therefore, we will use methods (14) and (15) and (33) given in Section 3 as the starting algorithm of the method ((50) and (51)). The details are as follows: Considering the problem (1) at point $(x_i,t_{\frac{1}{2}})$, using the same method ((14) and (15)), for $1\le i\le N-1$ and $1\le n\le M-1$, we have

$$\begin{array}{cc}\hfill & {\mathcal{A}}_x{\delta }_t{\widehat{u}}_i^{\frac{1}{2}}-\beta {\delta }_t{\delta }_x^2{\widehat{u}}_i^{\frac{1}{2}}=\alpha {\mu }_t{\delta }_x^2{\widehat{u}}_i^{\frac{1}{2}}+{\displaystyle \frac{{\mathcal{A}}_x}{2}}[f(x_i,t_{\frac{1}{2}},{\widehat{u}}_i^0)+f(x_i,t_{\frac{1}{2}},{\widehat{u}}_i^1)],\hfill \end{array}$$

(52)

with initial and boundary conditions as follows:

$${\widehat{u}}_i^0={\overline{u}}_i^0,0\le i\le M_x;{\widehat{u}}_0^n={\psi }_1(t_n),{\widehat{u}}_N^n={\psi }_2(t_n),n=0,1,2.$$

(53)

Next, each ${\overline{u}}_i^1$ is obtained using the following Richardson extrapolation formula:

$${\overline{u}}_i^1=\left\{\begin{array}{c}{\displaystyle \frac{4}{3}}{\widehat{u}}_i^2({\displaystyle \frac{\tau }{2}},h)-{\displaystyle \frac{1}{3}}{\widehat{u}}_i^1(\tau ,h),1\le i\le N-1,\hfill \\ {\psi }_1(t_n),i=0,n=0,1,2,\hfill \\ {\psi }_2(t_n),i=N,n=0,1,2.\hfill \end{array}\right.$$

(54)

For the three-level HOCDM ((50)–(54)), we first obtain ${\overline{u}}_i^1$ using the starting algorithm ((52)–(54)). Then, ${\overline{u}}_i^{n+1}$ is computed using (50) and (51). Moreover, according to (49) and Theorem 2, the three-level HOCDM ((50)–(54)) has the local accuracy ($\mathcal{O}({\tau }^4+h^4)$). Because the Richardson extrapolation method ((52)–(54)) can achieve fourth-order accuracies in both time and space, it needs to compute the numerical solutions $u_i^{n+1}(\tau ,h)$ and $u_i^{n+2}({\displaystyle \frac{\tau }{2}},h)$. However, the three-level HOCDM ((50)–(54)) only uses the Richardson extrapolation technique at point ${\overline{u}}_i^1$, which also makes the computational cost of the three-level method superior to the Richardson extrapolation method. The detailed implementation process is described in Algorithm 3.

Algorithm 3 The computational implementation of the three-level HOCDM ((50)–(54))

Step 1: $obtaininitialvalues{\overline{u}}_i^0(0\le i\le N)andboundaryvalues$: ${\overline{u}}_0^n,{\overline{u}}_N^n,(1\le n\le M)$ $using$ (51); Step 2: $forn=0,1,...,M-1$  $compute{\overline{u}}_i^{1}using$ (52)–(54),  $compute{\overline{u}}_i^{n+1}using$ (50);              $end$

4.2. The Convergence of the Three-Level HOCDM

In this section, we discuss the convergence of the three-level high-order compact difference method ((50)–(54)). First, define ${\overline{e}}^n=({\overline{e}}_1^n,{\overline{e}}_2^n,...,{\overline{e}}_{N-1}^n)$. Based on the above arguments, the convergence theorem of the three-level high-order compact difference method ((50)–(54)) can be established as follows:

Theorem 3.

Assume that $u(x,t)\in C^{6,6}({\overline{\Omega }}_{h\tau })$ and that Lipschitz condition (2) holds. Then, the three-level high-order compact difference method ((50)–(54)) satisfies the following error estimate:

$$\parallel {\overline{e}}^n{\parallel }_{\infty }\le c_6({\tau }^4+h^4),1\le n\le M,$$

(55)

where $c_6>0$ is a constant independent of τ and h.

Proof. 

By subtracting (50) from (47), the error equations for $1\le i\le N-1$ and $0\le n\le M-1$ are as follows:

$$\begin{array}{cc}\hfill & D_t{\mathcal{A}}_x{\overline{e}}_i^{kn}-\beta D_t{\delta }_x^2{\overline{e}}_i^n=\alpha {\mathcal{A}}_t{\delta }_x^2{\overline{e}}_i^n+{\mathcal{A}}_t{\mathcal{A}}_x[f(x_i,t_n,U_i^n)-f(x_i,t_n,{\overline{u}}_i^n)]+{\overline{R}}_i^n.\hfill \end{array}$$

(56)

Applying the inner product with (56) using $D_t{\overline{e}}^n$, we get

$$\begin{array}{cc}\hfill & (D_t{\mathcal{A}}_x{\overline{e}}^n,D_t{\overline{e}}^n)=(\beta D_t{\delta }_x^2{\overline{e}}^n,D_t{\overline{e}}^n)+(\alpha {\mathcal{A}}_t{\delta }_x^2{\overline{e}}^n,D_t{\overline{e}}^n)\hfill \\ \hfill & +({\mathcal{A}}_t{\mathcal{A}}_x[f(x,t_n,U^n)-f(x,t_n,{\overline{u}}^n)],D_t{\overline{e}}^n)+({\overline{R}}^n,D_t{\overline{e}}^n)=\sum _{i=1}^4{\overline{H}}_i.\hfill \end{array}$$

(57)

Next, we give the estimations for each term in (57). According to $(ii)$ in Lemma 4, the left-hand side of (57) is estimated as follows:

$$(D_t{\mathcal{A}}_x{\overline{e}}^n,D_t{\overline{e}}^n)\ge {\displaystyle \frac{2}{3}}{\parallel D_t{\overline{e}}^n\parallel }^2,0\le n\le M-1.$$

(58)

It follows from $(iii)$ in Lemma 3 and $(ii)$ in Lemma 4 that

$${\overline{H}}_1\le -{\displaystyle \frac{2\beta }{3}}{|D_t{\overline{e}}^n|}_1^2,0\le n\le M-1.$$

(59)

Based on $(iii)$ in Lemma 3 and (23), for $0\le n\le M-1$, it holds that

$$\begin{array}{cc}\hfill {\overline{H}}_2=& \alpha \sum _{i=1}^{N-1}{\mathcal{A}}_t{\delta }_x^2{\overline{e}}_i^n{D}_t{\overline{e}}_i^n\hfill \\ \hfill =& \alpha h\sum _{i=1}^{N-1}{\delta }_x^2({\displaystyle \frac{1}{6}}{\overline{e}}_i^{n+1}+{\displaystyle \frac{2}{3}}{\overline{e}}_i^n+{\displaystyle \frac{1}{6}}{\overline{e}}_i^{n-1})D_t{\overline{e}}_i^n\hfill \\ \hfill =& {\displaystyle \frac{\alpha h}{6}}\sum _{i=1}^{N-1}{\delta }_x^2({\overline{e}}_i^{n+1}+{\overline{e}}_i^{n-1})D_t{\overline{e}}_i^n+{\displaystyle \frac{2\alpha h}{3}}\sum _{i=1}^{N-1}{\delta }_x^2{\overline{e}}_i^n{D}_t{\overline{e}}_i^n\hfill \\ \hfill =& {\displaystyle \frac{-\alpha h}{6}}\sum _{i=0}^{N-1}{\delta }_x({\overline{e}}_i^{n+1}+{\overline{e}}_i^{n-1}){\delta }_x{D}_t{\overline{e}}_i^n-{\displaystyle \frac{2\alpha h}{3}}\sum _{i=0}^{N-1}{\delta }_x{\overline{e}}_i^n{\delta }_x{D}_t{\overline{e}}_i^n\hfill \\ \hfill \le & {\displaystyle \frac{-\alpha h}{6}}\sum _{i=0}^{N-1}{\delta }_x({\overline{e}}_i^{n+1}+{\overline{e}}_i^{n-1}){\delta }_x{\displaystyle \frac{{\overline{e}}_i^{n+1}-{\overline{e}}_i^{n-1}}{2\tau }}+{\displaystyle \frac{{\alpha }^2}{6\beta }}|{\overline{e}}^n{|}_1^2+{\displaystyle \frac{2\beta }{3}}{|D_t{\overline{e}}^n|}_1^2\hfill \\ \hfill =& {\displaystyle \frac{\alpha }{12\tau }}(|{\overline{e}}^{n-1}{|}_1^2-|{\overline{e}}^{n+1}{|}_1^2)+{\displaystyle \frac{{\alpha }^2}{6\beta }}|{\overline{e}}^n{|}_1^2+{\displaystyle \frac{2\beta }{3}}{|D_t{\overline{e}}^n|}_1^2.\hfill \end{array}$$

(60)

Combining Lipschitz condition (2) and (23), we get

$$\begin{array}{cc}\hfill & {\overline{H}}_3=h\sum _{i=1}^{N-1}{\mathcal{A}}_t{\mathcal{A}}_x[f(x_i,t_n,U_i^n)-f(x_i,t_n,{\overline{u}}_i^n)]D_t{\overline{e}}_i^n\hfill \\ \hfill \le & h\sum _{i=1}^{N-1}\left\{{\displaystyle \frac{1}{6}}[f(x_i,t_{n+1},U_i^{n+1})-f(x_i,t_{n+1},{\overline{u}}_i^{n+1})]\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{2}{3}}[f(x_i,t_n,U_i^n)-f(x_i,t_n,{\overline{u}}_i^n)]+{\displaystyle \frac{1}{6}}[f(x_i,t_{n-1},U_i^{n-1})-f(x_i,t_{n-1},{\overline{u}}_i^{n-1})]\right\}{D}_t{\overline{e}}_i^n\hfill \\ \hfill \le & h\sum _{i=1}^{N-1}\left[{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n+1}|+{\displaystyle \frac{2}{3}}L|{\overline{e}}_i^n|+{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n-1}|\right]D_t{\overline{e}}_i^n\hfill \\ \hfill \le & {\displaystyle \frac{h}{4\kappa }}\sum _{i=1}^{N-1}{\left[{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n+1}|+{\displaystyle \frac{2}{3}}L|{\overline{e}}_i^n|+{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n-1}|\right]}^2+\kappa {\parallel D_t{\overline{e}}^n\parallel }^2.\hfill \end{array}$$

(61)

Letting $\kappa ={\displaystyle \frac{1}{3}}$ and using $(ii)$ in Lemma 3, for $0\le n\le M-1$, (61) can be estimated as follows:

$$\begin{array}{cc}\hfill {\overline{H}}_3\le & {\displaystyle \frac{3h}{4}}\sum _{i=1}^{N-1}{\left[{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n+1}|+{\displaystyle \frac{2}{3}}L|{\overline{e}}_i^n|+{\displaystyle \frac{1}{6}}L|{\overline{e}}_i^{n-1}|\right]}^2+{\displaystyle \frac{1}{3}}{\parallel D_t{\overline{e}}^n\parallel }^2\hfill \\ \hfill \le & 3\left[{\displaystyle \frac{L^2}{36}}\parallel {\overline{e}}^{n+1}{\parallel }^2+{\displaystyle \frac{L^2}{36}}\parallel {\overline{e}}^{n-1}{\parallel }^2+{\displaystyle \frac{4L^2}{9}}{\parallel {\overline{e}}^n\parallel }^2\right]+{\displaystyle \frac{1}{3}}{\parallel D_t{\overline{e}}^n\parallel }^2\hfill \\ \hfill \le & {\displaystyle \frac{L^2{(b-a)}^2}{72}}|{\overline{e}}^{n+1}{|}_1^2+{\displaystyle \frac{L^2{(b-a)}^2}{72}}|{\overline{e}}^{n-1}{|}_1^2+{\displaystyle \frac{2L^2{(b-a)}^2}{9}}|{\overline{e}}^n{|}_1^2+{\displaystyle \frac{1}{3}}{\parallel D_t{\overline{e}}^n\parallel }^2.\hfill \end{array}$$

(62)

In addition, using (23) and (49), for $0\le n\le M-1$, this implies that

$$\begin{array}{c}\hfill {\overline{H}}_4\le {\displaystyle \frac{3}{4}}\parallel {\overline{R}}^n{\parallel }^2+{\displaystyle \frac{1}{3}}\parallel D_t{\overline{e}}^n{\parallel }^2\le {\displaystyle \frac{3(b-a)c_5^2}{4}}{({\tau }^4+h^4)}^2+{\displaystyle \frac{1}{3}}{\parallel D_t{\overline{e}}^n\parallel }^2.\end{array}$$

(63)

Inserting (58)–(60), (62) and (63) into (57), for $0\le n\le M-1$, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\alpha }{12\tau }}(|{\overline{e}}^{n+1}{|}_1^2-|{\overline{e}}^{n-1}{|}_1^2)\le {\displaystyle \frac{{\alpha }^2}{6\beta }}|{\overline{e}}^n{|}_1^2+{\displaystyle \frac{L^2{(b-a)}^2}{72}}(|{\overline{e}}^{n+1}{|}_1^2+|{\overline{e}}^{n-1}{|}_1^2)\hfill \\ \hfill & +{\displaystyle \frac{2L^2{(b-a)}^2}{9}}{|{\overline{e}}^n|}_1^2+{\displaystyle \frac{3(b-a)c_5^2}{4}}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(64)

Summing up the above (64), for n from 1 to $\overline{n}(\le M-1)$, (64) yields

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\alpha }{12\tau }}(|{\overline{e}}^{\overline{n}+1}{|}_1^2+|{\overline{e}}^{\overline{n}}{|}_1^2-|{\overline{e}}^1{|}_1^2)\le \left[{\displaystyle \frac{{\alpha }^2}{6\beta }}+{\displaystyle \frac{2L^2{(b-a)}^2}{9}}\right]\sum _{l_1=1}^{\overline{n}}{|{\overline{e}}^{l_1}|}_1^2\hfill \\ \hfill & +{\displaystyle \frac{L^2{(b-a)}^2}{72}}\sum _{l_1=1}^{\overline{n}}(|{\overline{e}}^{l_1+1}{|}_1^2+|{\overline{e}}^{l_1-1}{|}_1^2)+{\displaystyle \frac{3(b-a)c_5^2\overline{n}}{4}}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(65)

Merging similar terms in (65), for $0\le \overline{n}\le M-1$, we find that

$$\begin{array}{cc}\hfill & \left[{\displaystyle \frac{\alpha }{12\tau }}-{\displaystyle \frac{L^2{(b-a)}^2}{72}}\right]|{\overline{e}}^{\overline{n}+1}{|}_1^2\le \left[{\displaystyle \frac{{\alpha }^2}{6\beta }}+{\displaystyle \frac{L^2{(b-a)}^2}{4}}\right]\sum _{l_1=1}^{\overline{n}}|{\overline{e}}^{l_1}{|}_1^2+{\displaystyle \frac{\alpha }{12\tau }}{|{\overline{e}}^1|}_1^2\hfill \\ \hfill & +{\displaystyle \frac{3(b-a)c_5^2\overline{n}}{4}}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(66)

Multiplying both sides of (66) by $\tau$, we obtain

$$\begin{array}{cc}\hfill & \left[{\displaystyle \frac{\alpha }{12}}-{\displaystyle \frac{L^2{(b-a)}^2\tau }{72}}\right]|{\overline{e}}^{\overline{n}+1}{|}_1^2\le \left[{\displaystyle \frac{{\alpha }^2}{6\beta }}+{\displaystyle \frac{L^2{(b-a)}^2}{4}}\right]\tau \sum _{l_1=1}^{\overline{n}}|{\overline{e}}^{l_1}{|}_1^2+{\displaystyle \frac{\alpha }{12}}{|{\overline{e}}^1|}_1^2\hfill \\ \hfill & +{\displaystyle \frac{3(b-a)c_5^2}{4}}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(67)

Using theorem 2, we deduce that there exists a constant ($c_7>0$) such that

$$|{\overline{e}}^1{|}_1^2\le c_7{({\tau }^4+h^4)}^2.$$

(68)

Substituting (68) into (67) yields

$$\begin{array}{c}[{\displaystyle \frac{\alpha }{12}}-{\displaystyle \frac{L^2{(b-a)}^2\tau }{72}}]|{\overline{e}}^{\overline{n}+1}{|}_1^2\le [{\displaystyle \frac{{\alpha }^2}{6\beta }}+{\displaystyle \frac{L^2{(b-a)}^2}{4}}]\tau \sum _{l_1=1}^{\overline{n}}|{\overline{e}}^{l_1}{|}_1^2+[{\displaystyle \frac{\alpha c_7}{12}}\hfill \\ +{\displaystyle \frac{3(b-a)c_5^2}{4}}]{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(69)

Since there exists constants $\overline{\tau }$ and $\lambda >0$, it becomes

$$6\alpha -L^2{(b-a)}^2\tau >\lambda ,\overline{\tau }>\tau ,$$

(70)

Based on (70), the inequality (69) is estimated as follows:

$$\begin{array}{cc}\hfill & |{\overline{e}}^{\overline{n}+1}{|}_1^2\le {\displaystyle \frac{[12{\alpha }^2+18\beta L^2{(b-a)}^2]\tau }{\beta [6\alpha -L^2{(b-a)}^2\tau ]}}\sum _{l_1=1}^{\overline{n}}{|{\overline{e}}^{l_1}|}_1^2+{\displaystyle \frac{54(b-a)c_5^2+6\alpha c_7}{6\alpha -L^2{(b-a)}^2\tau }}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(71)

For the inequality (71), we deduce that

$$\begin{array}{cc}\hfill & |{\overline{e}}^{\overline{n}+1}{|}_1^2\le {\displaystyle \frac{[12{\alpha }^2+18\beta L^2{(b-a)}^2]\tau }{\beta \lambda }}\sum _{l_1=1}^{\overline{n}}{|{\overline{e}}^{l_1}|}_1^2+{\displaystyle \frac{54(b-a)c_5^2+6\alpha c_7}{\lambda }}{({\tau }^4+h^4)}^2.\hfill \end{array}$$

(72)

Putting ${\lambda }_1={\displaystyle \frac{[12{\alpha }^2+18\beta L^2{(b-a)}^2]}{\beta \lambda }}$ and ${\lambda }_2={\displaystyle \frac{54(b-a)c_5^2+6\alpha c_7}{\lambda }}$, (72) is converted to

$$|{\overline{e}}^{\overline{n}+1}{|}_1^2\le {\lambda }_1\tau \sum _{l_1=1}^{\overline{n}}{|{\overline{e}}^{l_1}|}_1^2+{\lambda }_2{({\tau }^4+h^4)}^2.$$

(73)

The application of Lemma 5 to (73) gives

$$|{\overline{e}}^{\overline{n}+1}{|}_1^2\le {\lambda }_{2}exp({\lambda }_1){({\tau }^4+h^4)}^2,$$

(74)

and by (74) and inequality $(i)$ in Lemma 3, we get

$$\parallel {\overline{e}}^{\overline{n}+1}{\parallel }_{\infty }\le c_6{({\tau }^4+h^4)}^2,$$

(75)

where $c_6={\displaystyle \frac{\sqrt{(b-a){\lambda }_{2}exp({\lambda }_1)}}{2}}$. Thus, the error estimate (55) holds for $1\le \overline{n}+1\le M$. This completes the proof. □

5. Numerical Experiments

In the previous sections, we proposed three high-order compact difference schemes for solving the semi-linear Sobolev equation (1):

Algorithm 1—the two-level HOCDM ((14) and (15));

Algorithm 2—the Richardson extrapolation method ((14) and (15) and (33));

Algorithm 3—the three-level HOCDM ((50)–(54));

For comparison, we also consider the method from reference [18], referred to as Algorithm 4. The global errors and corresponding convergence orders for each method are computed using the following formulae:

$$E_{\infty }(h,\tau )=\underset{1\le n\le M}{max}{\parallel \ast \parallel }_{\infty },p_{\tau }={\displaystyle \frac{log[E_{\infty }^{\ast }(h,{\tau }_1)/E_{\infty }(h,{\tau }_2)]}{log[{\tau }_1/{\tau }_2]}},$$

$$p_h={\displaystyle \frac{log[E_{\infty }^{\ast }(h_1,\tau )/E_{\infty }^{\ast }(h_2,\tau )]}{log[h_1/h_2]}},p={\displaystyle \frac{log[E_{\infty }^{\ast }(h_3,{\tau }_3)/E_{\infty }^{\ast }(h_4,{\tau }_4)]}{log[h_3/h_4]}},$$

where $E_{\infty }^{\ast }(h,\tau )$ denotes the maximum norm of the error for the given methods, and $(h_1,h_2)$ and $({\tau }_1,{\tau }_2)$ are pairs of successive spatial and temporal step sizes, respectively. And all the calculations in the article are programmed in Matlab2023a. We conduct our computations using double-precision arithmetic and perform them on a private AMD Ryzen 5 5600G computer with Radeon Graphics/3.9 GHz and 8 GB of memory. In addition, we selected the Picard iterative convergence criterion as ${10}^{-12}$.

Example 1.

Consider the following IBVPs of SLSEs:

$$\left\{\begin{array}{c}{u}_t-u_{xxt}=u_{xx}+\sqrt{u}+f(x,t),x\in (0,1),t\in (0,3],\hfill \\ u(x,0)={\displaystyle \frac{1}{1+e^{\sqrt{2}x}}},x\in [0,1],\hfill \\ u(0,t)={\displaystyle \frac{1}{1+e^{2t}}},u(1,t)={\displaystyle \frac{1}{1+e^{\sqrt{2}(1-t\sqrt{2})}}},t\in (0,3].\hfill \end{array}\right.$$

(76)

Example 1 admits an exact solution ($u(x,t)={\displaystyle \frac{1}{1+e^{\sqrt{2}(x-t\sqrt{2})}}}$). When $\alpha =1$, $\beta =1$, $a=0$, and $b=1$, the convergence conditions required for Algorithms 1–3 are fulfilled. Consequently, according to Theorems 1–3, Algorithm 1 has second-order accuracy in time and fourth-order accuracy in space, while Algorithms 2 and 3 attain fourth-order accuracies in both time and space.

To verify the temporal convergence orders of Algorithms 1–3, we solve Example 1 using these algorithms with a fixed spatial step size ($h=1/100$) and varying temporal step sizes (${\displaystyle \tau =\frac{1}{4\times i}}$) for $(i=2,3,4,5,6)$. The corresponding global errors and temporal convergence orders are presented in

Table 1. When $h={\displaystyle \frac{1}{100}}$, the global errors and temporal convergence orders of Example 1.

$\mathsf{\tau }$	Algorithm 1		Algorithm 2		Algorithm 3
	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathsf{\tau }}$	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathsf{\tau }}$	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathsf{\tau }}$
$1/8$	2.4000 × ${10}^{-3}$	–	2.2309 × ${10}^{-5}$	–	4.1279 × ${10}^{-4}$	–
$1/12$	1.0000 × ${10}^{-3}$	2.05	4.2737 × ${10}^{-6}$	4.08	5.5682 × ${10}^{-5}$	4.94
$1/16$	5.7672 × ${10}^{-4}$	2.02	1.3382 × ${10}^{-6}$	4.04	1.5581 × ${10}^{-5}$	4.43
$1/20$	3.6789 × ${10}^{-4}$	2.01	5.4221 × ${10}^{-7}$	4.02	6.5961 × ${10}^{-6}$	3.85
$1/24$	2.5503 × ${10}^{-4}$	2.01	2.6240 × ${10}^{-7}$	4.01	2.8509 × ${10}^{-6}$	4.60

. As observed from Table 1, Algorithm 1 has second-order accuracy in time, while Algorithms 2 and 3 have fourth-order accuracies in time. Subsequently, to examine the spatial convergence orders, we fix the temporal step size at ${\displaystyle \tau =\frac{1}{20,000}}$ and the spatial step size is ${\displaystyle h=\frac{1}{4\times i}}$ for $(i=2,3,4,5,6)$. The computed results are summarized in

Table 2. When $\tau ={\displaystyle \frac{1}{20000}}$, the global errors and spatial convergence orders of Example 1.

h	Algorithm 1		Algorithm 2		Algorithm 3
	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathit{h}}$	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathit{h}}$	${\mathit{E}}_{\infty }$	${\mathit{p}}_{\mathit{h}}$
$1/8$	4.7675 × ${10}^{-7}$	–	4.7711 × ${10}^{-7}$	–	4.7711 × ${10}^{-7}$	–
$1/12$	9.3562 × ${10}^{-8}$	4.02	9.3928 × ${10}^{-8}$	4.01	9.3929 × ${10}^{-8}$	4.01
$1/16$	2.9319 × ${10}^{-8}$	4.03	2.9654 × ${10}^{-8}$	4.01	2.9681 × ${10}^{-8}$	4.00
$1/20$	1.1783 × ${10}^{-8}$	4.09	1.2155 × ${10}^{-8}$	4.00	1.2152 × ${10}^{-8}$	4.00
$1/24$	5.4931 × ${10}^{-9}$	4.19	5.8431 × ${10}^{-9}$	4.02	5.8620 × ${10}^{-9}$	4.00

, which shows that all three algorithms exhibit fourth-order spatial accuracies. Since Algorithms 1–3 employ the same discretization method as that for the second-order spatial derivatives, their spatial errors are comparable. Furthermore, Figure 1b–d illustrates the numerical solutions obtained using Algorithms 1–3, respectively, and, visually, they are almost indistinguishable from the exact solution plotted in Figure 1a. The corresponding global error surfaces are depicted in Figure 2a–c. It was found that the global errors of Algorithms 1–3 were ${10}^{-5}$, ${10}^{-9}$, and ${10}^{-8}$, respectively.

Example 2.

Consider the following IBVPs of SLSEs:

$$\left\{\begin{array}{c}{u}_t-u_{xxt}=u_{xx}+u(1-u)+g(x,t),x\in (0,1),t\in (0,1],\hfill \\ u(x,0)=0,x\in [0,1],\hfill \\ u(0,t)=t^3+t,u(1,t)=0,t\in (0,1].\hfill \end{array}\right.$$

(77)

where

$$\begin{array}{cc}\hfill g(x,t)=& ({\displaystyle \frac{{\pi }^2}{4}}+1)(3t^2+1)cos({\displaystyle \frac{\pi }{2}}x)-(1-{\displaystyle \frac{{\pi }^2}{4}})(t^3+t)cos({\displaystyle \frac{\pi }{2}}x)\hfill \\ & +{(t^3+t)}^2{cos}^2({\displaystyle \frac{\pi }{2}}x)\hfill \end{array}$$

Example 2 admits a sufficiently smooth exact solution ($u(x,t)=(t^3+t)cos({\displaystyle \frac{\pi }{2}}x)$).

Table 3. Global errors and convergence orders of Example 2.

$\mathsf{\tau }$	Algorithm 4 ($\mathsf{\tau }={\mathit{h}}^2)$		Algorithm 2 ($\mathsf{\tau }=\mathit{h}$)		Algorithm 3 ($\mathsf{\tau }=\mathit{h}$)
	${\mathit{E}}_{\infty }$	$\mathit{p}$	${\mathit{E}}_{\infty }$	$\mathit{p}$	${\mathit{E}}_{\infty }$	$\mathit{p}$
$1/5$	3.4186 × ${10}^{-5}$	–	1.5330 × ${10}^{-5}$	–	1.5263 × ${10}^{-5}$	–
$1/10$	2.2410 × ${10}^{-6}$	3.93	9.5549 × ${10}^{-7}$	4.00	9.7483 × ${10}^{-7}$	3.97
$1/20$	1.4007 × ${10}^{-7}$	4.00	6.0070 × ${10}^{-8}$	3.99	6.1049 × ${10}^{-8}$	4.00
$1/40$	8.7545 × ${10}^{-9}$	4.00	3.7538 × ${10}^{-9}$	4.00	3.8101 × ${10}^{-9}$	4.00

presents the global errors and convergence orders for the numerical methods in reference [18] (denoted as Algorithm 4), as well as for Algorithms 2 and 3. Since Algorithm 4 is second-order accurate in time and fourth-order accurate in space, it is implemented with $\tau =h^2$ to balance the temporal and spatial discretization errors. The results in Table 3 indicate that Algorithms 2 and 3 achieve fourth-order accuracies in both time and space, while Algorithm 4 maintains second-order temporal accuracy and fourth-order spatial accuracy. Furthermore, Algorithms 2 and 3 produce significantly smaller errors compared to those produced by Algorithm 4, indicating their superior performance. In addition, we solve Example 2 using Algorithms 1–3 with ${\displaystyle \tau =h=\frac{1}{32}}$. The global errors at different time levels are plotted in Figure 3a–c. These figures further confirm that all three algorithms maintain high accuracy for Example 2.

6. Conclusions

In this paper, we have proposed a class of HOCDMs for solving SLSEs. A two-level scheme was first constructed using the C-N method in time and a fourth-order compact difference scheme in space, which has second-order temporal and fourth-order spatial accuracies. To further improve the time accuracy, the Richardson extrapolation technique was employed, successfully improving the temporal accuracy from second to fourth order. Moreover, a fully fourth-order accurate scheme in both time and space was developed by applying fourth-order discretizations to both temporal and spatial derivatives. And the convergence was also analyzed. Finally, numerical experiments were provided to confirm the accuracy and efficiency of the proposed methods.