Error Analysis and Numerical Investigation of an L1-2 Fourth-Order Difference Scheme for Solving the Time-Fractional Burgers Equation

Abstract

This paper presents a finite difference approach for solving the time-fractional Burgers’ equation, which is a model for nonlinear flow with memory effects. The method leverages the $L1-2$ formula for the fractional derivative and provides a novel linearization strategy to efficiently transform the system into a stable linear problem. Rigorous analysis establishes the existence, uniqueness, and pointwise-in-time convergence of the numerical solution in the $L_2$ norm. The proposed formulation achieves second-order time accuracy and fourth-order spatial accuracy under smooth initial conditions, with numerically verified temporal convergence rates of $O({\tau }^{1+\alpha }+{\tau }^2{t}_n^{\alpha -2})$ for solutions with weak singularities. Critically, numerical findings demonstrate that the method is robust and highly efficient, offering high-resolution solutions at a substantially lower computational cost than equivalent graded-mesh formulations.

1. Introduction

Fractional calculus plays a vital role in various fields, extending the concept of differentiation to non-integer orders. This generalization enables a deeper exploration of complex phenomena in science, engineering, and mathematics. This study presents a numerical analysis of the nonlinear diffusion equation described by the one-dimensional time-fractional Burgers equation

$$\{\begin{array}{}\mathrm{(1a)}& {𝜕}_t^{\alpha }u(x,t)-u_{xx}(x,t)+u(x,t)u_x(x,t)=0,\hspace{1em}\hspace{1em}\hfill & (x,t)\in \mathsf{\Omega }\times (0,T],\hfill \mathrm{(1b)}& u(0,t)=u(L,t)=0,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & 0<t\le T,\hfill \mathrm{(1c)}& u(x,0)=u_0\left(x\right),\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & x\in \mathsf{\Omega }.\hfill \end{array}$$

where $0<\alpha <1$, $\mathsf{\Omega }=(0,L)$, and $u_0\left(x\right)$ are given function. Here, the Caputo fractional derivative of order $\alpha$ is expressed as

$${𝜕}_t^{\alpha }u(x,t)={\displaystyle \frac{1}{\mathsf{\Gamma }(1-\alpha )}}{\int }_0^t{\displaystyle \frac{1}{{(t-s)}^{\alpha }}}{\displaystyle \frac{𝜕u(x,s)}{𝜕s}}ds,$$

(2)

where $\mathsf{\Gamma }\left(x\right)$ is the gamma function. The Burgers equation, originally formulated by Bateman [1], serves as a fundamental model that encapsulates essential concepts in fluid dynamics [2,3]. It has significant applications in various scientific and industrial fields [4,5,6]. The fractional version of the Burgers equation has been utilized to describe the unidirectional propagation of weakly nonlinear acoustic waves in gas-filled pipes [7,8].

Several analytical and semi-analytical attempts for the time-fractional Burgers equation have been reported (e.g., [9,10]). However, as noted in [11], the classical Hopf–Cole transformation cannot be applied in the fractional setting, making closed-form solutions inaccessible. This limitation has motivated the development of numerous numerical approaches. The widely used $L1$ discretization [12], based on piecewise linear interpolation, provides $(2-\alpha )$-order accuracy for the Caputo derivative and has inspired many subsequent methods. Yaseen and Abbas [13] introduced an efficient cubic trigonometric B-spline scheme with a linearization strategy to reduce computational cost while preserving unconditional stability and high accuracy. Zhang and Yang [14] developed a fully discrete finite volume element method for the generalized fractional Burgers equation. Building on the $L1$ framework, several enhanced finite difference schemes have been proposed for their simplicity and efficiency. Li et al. [15] proposed an unconditionally stable linear implicit finite difference scheme with first-order temporal and second-order spatial accuracy. In [16], a nonlinear difference scheme attained second-order spatial and $(2-\alpha )$-order temporal accuracy, and a nonlinear scheme on nonuniform meshes was introduced in [17], achieving $min\{2-\alpha ,r\alpha \}$ temporal accuracy. Furthermore, Zhang et al. [18] proposed a compact high-order spatial scheme, while Peng et al. [19] developed a compact method for mixed-type fractional Burgers equations. For nonlocal fourth-order Burgers models, Tian et al. [20] constructed an $L1$-based implicit scheme on graded meshes with $\alpha$-robust stability and optimal convergence, and Wang et al. [21] later introduced a nonlinear compact scheme using an $L1$ discretization with a double-reduction strategy, establishing its stability and convergence.

Since the introduction of the $L1$ scheme, it has been widely applied to a broad class of nonlinear fractional differential equations. To improve temporal accuracy, several L-type formulas have been developed, including the $L1-2$ method [22], the $L2-1_{\sigma }$ method [23], the $\mathcal{F}L2-1_{\sigma }$ method [24], and related variants [25,26]. For example, Wang et al. [27] employed the $L2-1_{\sigma }$ formula on graded meshes to construct a compact scheme for the time-fractional Burgers equation. Peng et al. [28] proposed a second-order fast finite difference method for a generalized version of the time-fractional Burgers equation, utilizing the nonuniform $L2-1_{\sigma }$ formula with a sum-of-exponentials approach. Guan et al. [29] a numerical scheme combining the $L2-1_{\sigma }$ formula with a radial basis function finite difference method was proposed for the generalized time-fractional Burgers equation, achieving $\mathcal{O}({\tau }^2+h^2)$ with numerical results confirming the method’s efficiency and accuracy. Recently, Dwivedi and Rajeev [30] introduced a fast high-order linearized exponential method that combines the $\mathcal{F}L2-1_{\sigma }$ scheme with a tailored finite point formula based on exponential basis functions for the two-dimensional time-fractional Burgers equation, enhancing computational efficiency while preserving accuracy.

While several time discretizations are available, applying $L1$-type schemes on nonuniform meshes often leads to nonlinear and computationally expensive methods, and linearization typically reduces the temporal accuracy to first order [15]. Phumichot et al. [31] obtained $(2-\alpha )$-order temporal and fourth-order spatial accuracy through linearization. Wang and Sun [32] also introduced linear schemes using exponential basis functions, though with only first-order spatial accuracy. These observations illustrate the difficulty of constructing a linear scheme with both second-order temporal and high-order spatial accuracy. To address this challenge, we develop an efficient linear difference scheme based on the $L1-2$ formula, a higher-order extension of the classical $L1$ method. In contrast to the $L2-1_{\sigma }$ schemes, which require nonlinear treatments, the proposed approach attains second-order temporal and high-order spatial accuracy within a fully linear and stable framework.

Accordingly, this study outlines its key contributions in a comprehensive manner.

• A new linear finite difference scheme for the fractional Burgers equation is developed based on the $L1-2$ method. The scheme incorporates a modified fourth-order linearized formulation for the convection term [33], achieving second-order temporal accuracy together with high-order spatial accuracy.
• Existence, uniqueness, and $L_2$-stability of the scheme are established using the standard Gronwall inequality, overcoming the difficulty posed by the negative convolution weights of the $L1-2$ formula [34].
• Under weak-regularity conditions, the proposed scheme is shown numerically to achieve the temporal accuracy $O({\tau }^{1+\alpha }+{\tau }^2{t}_n^{\alpha -2})$, representing a clear improvement over the classical $L1$ method. Numerical experiments further indicate that the scheme remains accurate in long-time simulations, where uniform meshes offer a practical and efficient alternative to graded meshes.

The remainder of the paper is organized as follows. Section 2 presents the necessary preliminaries, introduces the $L1-2$ discretization, and describes the new linearized fourth-order treatment of the nonlinear convection term. Section 3 establishes existence, uniqueness, $L_2$-stability, and pointwise error estimates for the proposed scheme. Section 4 provides numerical results that confirm the theory and illustrate the method’s performance for both smooth and non-smooth solutions. Concluding remarks are given in Section 5.

2. Formulation and Analysis of the Proposed Scheme

The spatial and temporal step sizes are given by $[0,L]\times [0,T]$ with uniform partitions ${\mathsf{\Omega }}_h=\{x_j=jh|0\le j\le J\}$ and ${\mathsf{\Omega }}_{\tau }=\{t_n=n\tau |0\le n\le N\}$, where N and J are positive integers. The spatial and temporal step sizes are given by $h=L/J$ and $\tau =T/N$, respectively. Based on the boundary conditions (1b), we define the solution space in its discrete form as

$${\mathbb{V}}_h=\left\{\mathbf{U}|\mathbf{U}={[U_0,U_1,\dots ,U_J]}^T,U_0=U_J=0\right\}.$$

The discrete $L_2$ inner product and its corresponding discrete $L_2$-norm are defined as

$$\langle \mathbf{U},\mathbf{V}\rangle =h{\mathbf{U}}^T\mathbf{V},\parallel \mathbf{U}\parallel ={\langle \mathbf{U},\mathbf{U}\rangle }^{{\displaystyle \frac{1}{2}}}.$$

The discrete maximum norm is given by ${\parallel \mathbf{U}\parallel }_{\infty }={max}_{1\le j\le J-1}\left|{\mathbf{U}}_j\right|.$

2.1. Space Discretization

For the spatial discretization, the second derivative is approximated by the fourth-order finite difference formula in [35]. The nonlinear convection term is treated using the fourth-order formulation of [33] together with a linearization step to preserve the overall linearity of the scheme. To support this construction, we introduce the matrices representing the central first- and second-order difference operators used in the discrete spatial derivatives

$$\begin{array}{cccc}\hfill & {\mathbf{D}}_1={\displaystyle \frac{1}{2h}}{\left[\begin{array}{ccccc}0& 1& 0& \dots & 0\\ -1& 0& 1& \dots & 0\\ & \ddots & \ddots & \ddots & \\ 0& \dots & -1& 0& 1\\ 0& \dots & 0& -1& 0\end{array}\right]}_{(J-1)\times (J-1)},\hfill & \hfill & {\mathbf{D}}_2={\displaystyle \frac{1}{h^2}}{\left[\begin{array}{ccccc}-2& 1& 0& \dots & 0\\ 1& -2& 1& \dots & 0\\ & \ddots & \ddots & \ddots & \\ 0& \dots & 1& -2& 1\\ 0& \dots & 0& 1& -2\end{array}\right]}_{(J-1)\times (J-1)}.\hfill \end{array}$$

According to Taylor’s expansion, based on the fourth-order difference formula [35], we note that

$${\mathbf{u}}_{xx}(\mathbf{x},t_n)={\mathbf{H}}^{-1}{\mathbf{D}}_2{\mathbf{U}}^n+O\left(h^4\right),$$

where

$${\mathbf{u}}_{xx}(\mathbf{x},t_n)={\left[u_{xx}(x_1,t^n),u_{xx}(x_2,t^n),\dots ,u_{xx}(x_{J-1},t^n)\right]}^T$$

and

$$\mathbf{H}={\displaystyle \frac{1}{12}}{\left[\begin{array}{ccccc}10& 1& 0& \dots & 0\\ 1& 10& 1& \dots & 0\\ & \ddots & \ddots & \ddots & \\ 0& \dots & 1& 10& 1\\ 0& \dots & 0& 1& 10\end{array}\right]}_{(J-1)\times (J-1)}.$$

Noting that the nonlinear term can be rewritten as $u{u}_x={\displaystyle \frac{1}{3}}(u{u}_x+{\left(u^2\right)}_x)$, we build upon the formulation in [33] and employ a high-order approximation for the nonlinear convection term, derived from the following lemma.

Lemma 1

([33]). Let $w\left(x\right)\in C^5[x_{i-1},x_{i+1}]$ and $W\left(x\right)=w^{″}\left(x\right)$, we have

$$w\left(x_i\right)w^{\prime }\left(x_i\right)=\mathsf{\Psi }{(w,w)}_i-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }{(W,w)}_i+O\left(h^4\right),$$

where  $\mathsf{\Psi }{(w,v)}_i={\displaystyle \frac{1}{3}}\left[w_j\left(D_x{v}_j\right)+D_x\left(w_j{v}_j\right)\right]$ and $D_x{w}_j={\displaystyle \frac{w_{j+1}-w_{j-1}}{2h}}.$

For simplicity, we define the discrete function $\mathsf{\Psi }:{\mathbb{R}}^{J-1}\times {\mathbb{R}}^{J-1}\to {\mathbb{R}}^{J-1}$ as

$$\mathsf{\Psi }(\mathbf{U},\mathbf{V})={\displaystyle \frac{1}{3}}\left[\mathbf{diag}\left(\mathbf{U}\right){\mathbf{D}}_1\mathbf{V}+{\mathbf{D}}_1\left(\mathbf{diag}\left(\mathbf{U}\right)\mathbf{V}\right)\right],$$

where

$$\mathbf{diag}\left(\mathbf{U}\right)={\left[\begin{array}{ccccc}{U}_1& 0& 0& \dots & 0\\ 0& U_2& 0& \dots & 0\\ & \ddots & \ddots & \ddots & \\ 0& \dots & 0& U_{J-2}& 0\\ 0& \dots & 0& 0& U_{J-1}\end{array}\right]}_{(J-1)\times (J-1)}.$$

Although the formulation may lead to nonlinear relation in $\mathbf{U}$ and $\mathbf{V}$, it provides a convenient structure for subsequent linearization during the time discretization step.

2.2. Temporal Approximation

After the spatial discretization, the remaining task is to approximate the temporal fractional derivative and advance the solution in time. To this end, we adopt the $L1-2$ formula [22] to discretize the Caputo fractional derivative in Equation (2). To ensure the overall scheme remains linear, a linearization technique is then introduced for the nonlinear convection term. The discrete fractional derivative is expressed as

$${}_2\mathsf{\Delta }_t^{\alpha }g\left(t_n\right)={\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}\left(c_0^{\left(\alpha \right)}g\left(t_n\right)-\sum _{k=1}^{n-1}\left(c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right)g\left(t_k\right)-c_{n-1}^{\left(\alpha \right)}g\left(t_0\right)\right),$$

where $c_0^{\left(\alpha \right)}=1$, for $n=1$ and for $n\ge 2$, we have

$$c_k^{\left(\alpha \right)}=\left\{\begin{array}{cc}{a}_0^{\left(\alpha \right)}+b_0^{\left(\alpha \right)},\hfill & k=0,\hfill \\ a_k^{\left(\alpha \right)}+b_k^{\left(\alpha \right)}-b_{k-1}^{\left(\alpha \right)},\hfill & k=1,2,\dots ,n-2,\hfill \\ a_k^{\left(\alpha \right)}-b_{k-1}^{\left(\alpha \right)},\hfill & k=n-1,\hfill \end{array}\right.$$

(3)

where $a_k^{\left(\alpha \right)}={(k+1)}^{1-\alpha }-{\left(k\right)}^{1-\alpha }$ and

$$b_k^{\left(\alpha \right)}={\displaystyle \frac{{(k+1)}^{2-\alpha }-{\left(k\right)}^{2-\alpha }}{2-\alpha }}-{\displaystyle \frac{{(k+1)}^{1-\alpha }-{\left(k\right)}^{1-\alpha }}{2}},$$

for $k=1,2\dots ,n.$

Lemma 2

([22]). Assume that $g\left(t\right)\in C^3[0,t_n]$ and let $R\left(g\left(t_n\right)\right)={𝜕}_t^{\alpha }g\left(t\right){|}_{t=t_n}{-}_2{\mathsf{\Delta }}_t^{\alpha }g\left(t_n\right).$ Then we have

$$\begin{array}{cc}\hfill \left|R\right(g\left(t_1\right)\left)\right|& \le {\displaystyle \frac{\alpha }{2\mathsf{\Gamma }(3-\alpha )}}\underset{t_0\le t\le t_1}{max}\left|f^{″}\left(t\right)\right|{\tau }^{2-\alpha },\hfill \\ \hfill \left|R\right(g\left(t_k\right)\left)\right|& \le {\displaystyle \frac{1}{2\mathsf{\Gamma }(1-\alpha )}}\{\underset{t_0\le t\le t_1}{max}\left|f^{″}\left(t\right)\right|{\tau }^3\hfill \\ \hfill & +\left[{\displaystyle \frac{1}{12}}+{\displaystyle \frac{\alpha }{3(1-\alpha )(2-\alpha )}}\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3-\alpha }}\right)\underset{t_0\le t\le t_n}{max}\left|f^{‴}\left(t\right)\right|{\tau }^{3-\alpha }\right]\},k\ge 2.\hfill \end{array}$$

To construct a high-order linear scheme, we employ the linearization $U_i^n=2U_i^{n-1}-U_i^{n-2}+\mathcal{O}\left({\tau }^2\right)$, which provides a second-order temporal prediction. This approximation can be naturally combined with the high-order nonlinear formulation introduced in Lemma 1, allowing the convection term to be expressed in a fully linear form without sacrificing accuracy. As a result, we propose the following $L1-2$ fourth-order difference scheme for the time-fractional Burgers Equation (1a)–(1c):

$$\left\{,\begin{array}{}\mathrm{(4a)}& {\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}({\mathbf{U}}^1-{\mathbf{U}}^0)-{\mathbf{H}}^{-1}{\mathbf{D}}_2\left({\mathbf{U}}^1\right)+\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1)=0,\hfill & n=1,\hfill \mathrm{(4b)}& {}_2\mathsf{\Delta }_t^{\alpha }\left({\mathbf{U}}^n\right)-{\mathbf{H}}^{-1}{\mathbf{D}}_2\left(u_i^n\right)+\mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n)=0,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & n\ge 2,\hfill \mathrm{(4c)}& U_0^n=U_J^n=0,\hfill & 1\le n\le N,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \mathrm{(4d)}& {\mathbf{U}}^0=u_0\left(\mathbf{x}\right).\hfill & \end{array}\right.$$

where ${\widehat{\mathbf{U}}}^n=2{\mathbf{U}}^{n-1}-{\mathbf{U}}^{n-2}$. Notably, the computation of ${\mathbf{U}}^1$ attains a temporal accuracy of $(2-\alpha )-$order, while for $n\ge 2$, it achieves second-order accuracy in time. Furthermore, following the analysis in [36], it can be established that the global convergence order of the Equation (4b)–(4d) remains unchanged by the presence of ${\mathbf{U}}^1$.

Remark 1.

To keep the scheme well posed at the boundary, we adopt the auxiliary condition $u_{xx}=0$ on $𝜕\mathsf{\Omega }$, following [18]. This assumption simplifies boundary closure and ensures that the formulation remains stable and accurate near the boundary. When $u_{xx}\ne 0$, alternative high-order treatments such as Richardson extrapolation or one-sided compact stencils can be incorporated without modifying the interior discretization.

2.3. Preliminary Lemmas

We proceed with an analysis of the existence and uniqueness of the solution, followed by a discussion on the convergence and stability of our numerical scheme. To lay the groundwork, we first review key lemmas required in this section.

Lemma 3

([22]). Let $0<\alpha <1$ and $c_k^{\left(\alpha \right)}$$(0\le k\le n-1$, for $n\ge 3)$ is defined in Equation (3). Then, we have

(i) 

$c_0^{\left(\alpha \right)}>\left|c_1^{\left(\alpha \right)}\right|$;

(ii) 

$c_k^{\left(\alpha \right)}>0$, for $k\ne 1$;

(iii) 

$c_2^{\left(\alpha \right)}\ge c_3^{\left(\alpha \right)}\ge \dots \ge c_{n-1}^{\left(\alpha \right)}$ ;

(iv) 

$c_0^{\left(\alpha \right)}>c_2^{\left(\alpha \right)}$;

(v) 

${\sum }_{k=1}^{n-1}{c}_k^{\left(\alpha \right)}=n^{1-\alpha }$.

Lemma 4.

Let $\mathbf{W},\mathbf{V}\in {\mathbb{V}}_h$, we have

(i) 

$\langle {\mathbf{D}}_1\mathbf{W},\mathbf{V}\rangle =-\langle \mathbf{W},{\mathbf{D}}_1\mathbf{V}\rangle$.

(ii) 

$\langle {\mathbf{D}}_2\mathbf{W},\mathbf{V}\rangle =\langle \mathbf{W},{\mathbf{D}}_2\mathbf{V}\rangle =-\langle {\mathbf{D}}_1^{+}\mathbf{W},{\mathbf{D}}_1^{+}\mathbf{V}\rangle$,

where the forward first-order difference operator matrix ${\mathbf{D}}_1^{+}$ is defined by

$${\mathbf{D}}_1^{+}={\displaystyle \frac{1}{h}}{\left[\begin{array}{ccccc}-1& 1& 0& \dots & 0\\ 0& -1& 1& \dots & 0\\ & \ddots & \ddots & \ddots & \\ 0& \dots & 0& -1& 1\\ 0& \dots & 0& 0& -1\end{array}\right]}_{(J-1)\times (J-1)}.$$

Moreover, we have

$$\langle {\mathbf{D}}_1\mathbf{W},\mathbf{W}\rangle =0,\langle {\mathbf{D}}_2\mathbf{W},\mathbf{W}\rangle =-{\parallel {\mathbf{D}}_1^{+}\mathbf{W}\parallel }^2.$$

The following lemma plays a key role in establishing the boundedness of the numerical solution.

Lemma 5

([37]). Let $\mathbf{U}\in {\mathbb{V}}_h,$ we have

$${\parallel \mathbf{U}\parallel }^2\le \langle {\mathbf{H}}^{-\mathbf{1}}\mathbf{U},\mathbf{U}\rangle ={\parallel \mathbf{R}\mathbf{U}\parallel }^2\le {\displaystyle \frac{3}{2}}{\parallel \mathbf{U}\parallel }^2,$$

where $\mathbf{R}$ is the Cholesky decomposition matrix of $\mathbf{H}$ defined by ${\mathbf{H}}^{-\mathbf{1}}={\mathbf{R}}^{\mathbf{T}}\mathbf{R}.$

Lemma 6

([38]). Let $\mathbf{U},\mathbf{V}\in {\mathbb{V}}_h,$ we have

$$\langle \mathsf{\Psi }(\mathbf{U},\mathbf{V}),\mathbf{V}\rangle =\mathbf{0}.$$

Lemma 7

([12]). Let $\mathbf{U}\in {\mathbb{V}}_h,$ we have

$$\parallel \mathbf{U}\parallel \le {\displaystyle \frac{L}{\sqrt{6}}}\parallel {\mathbf{D}}_1^{+}\mathbf{U}\parallel \mathrm{and}{\parallel \mathbf{U}\parallel }_{\infty }\le {\displaystyle \frac{\sqrt{L}}{2}}\parallel {\mathbf{D}}_1^{+}\mathbf{U}\parallel .$$

Lemma  8

([12]). Let $\mathbf{U}\in {\mathbb{V}}_h,$ we have

$$\parallel {\mathbf{D}}_1^{+}\mathbf{U}\parallel \le {\displaystyle \frac{2}{h}}\parallel \mathbf{U}\parallel \mathrm{and}\parallel {\mathbf{D}}_2\mathbf{U}\parallel \le {\displaystyle \frac{2}{h}}\parallel {\mathbf{D}}_1^{+}\mathbf{U}\parallel .$$

Lemma  9

([39]). For any ${ϵ}_0>0$ and $\mathbf{U}\in {\mathbb{V}}_h,$ it holds that

$${\parallel \mathbf{U}\parallel }_{\infty }^2\le \left({\displaystyle \frac{1}{{ϵ}_0}}+{\displaystyle \frac{1}{L}}\right){\parallel \mathbf{U}\parallel }^2+{ϵ}_0{\parallel {\mathbf{D}}_1^{+}\mathbf{U}\parallel }^2.$$

Lemma  10

(Discrete Gronwall inequality [40]). For any non-negative functions $A_k$ and $B_k$. If

$$\begin{array}{c}\hfill A_k\le D_k+\sum _{m=0}^{k-1}{B}_m{A}_m,k\ge 1,\end{array}$$

where $D_k$ is a non-decreasing and non-negative function, then we have

$$\begin{array}{c}\hfill A_k\le D_{k}exp\left(\sum _{m=0}^{k-1}{B}_m\right),k\ge 1.\end{array}$$

3. Theoretical Analysis of the Proposed Scheme

In this section, we analyze the solvability, $L_2$-stability, and convergence of the fully discrete Equation (4a)–(4d). We show that the scheme attains second-order accuracy in time and fourth-order accuracy in space.

3.1. Existence and Uniqueness

In this section, we establish the solvability of Equation (4a)–(4d). The Browder fixed point theorem is applied to demonstrate existence for $n=1$, and the proof will be start on this theorem before using the linear scheme when $n\ge 2$.

Lemma 11

(Browder fixed point theorem). Let H be a finite dimensional inner product space. Suppose that $g:H\to H$ is continuous and there exist an $\epsilon >0$ such that $\langle g\left(\mathbf{X}\right),\mathbf{X}\rangle >0$ for all $\mathbf{X}\in H$ with $\parallel \mathbf{X}\parallel =\epsilon$. Then there exists ${\mathbf{X}}^{*}\in H$ such that $g\left({\mathbf{X}}^{*}\right)=0$ and $\parallel {\mathbf{X}}^{*}\parallel \le \epsilon$.

Theorem 1. 

(Uniquely Solvability) The $L1-2$ fourth-order difference Equation (4a)–(4d) has a unique solution.

Proof. 

We apply mathematical induction to prove this theorem.

Step 1:

We establish the uniqueness of ${\mathbf{U}}^0$ from the initial condition. Since Equation (4a) is nonlinear, we invoke the Browder fixed point theorem to guarantee the existence of a numerical solution. Here, we define an operator $g:{\mathbb{V}}_h\to {\mathbb{V}}_h$ as the following form

$$g\left(\mathbf{X}\right)={\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}(\mathbf{X}-{\mathbf{U}}^0)-{\mathbf{H}}^{-1}{\mathbf{D}}_2\left(\mathbf{X}\right)+\mathsf{\Psi }(\mathbf{X},\mathbf{X})-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2\mathbf{X},\mathbf{X}).$$

To apply the Browder fixed point theorem, we use Lemma 6 to obtain

$$\begin{array}{cc}\hfill \langle g\left(\mathbf{X}\right),\mathbf{X}\rangle & ={\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}{(\parallel \mathbf{X}\parallel }^2-\langle \mathbf{X},{\mathbf{U}}^0\rangle )+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}\mathbf{X}\parallel }^2+\langle \mathsf{\Psi }(\mathbf{X},\mathbf{X}),\mathbf{X}\rangle -{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2\mathbf{X},\mathbf{X}),\mathbf{X}\rangle \hfill \\ \hfill & \ge {\displaystyle \frac{{\tau }^{-\alpha }}{2\mathsf{\Gamma }(2-\alpha )}}{(\parallel \mathbf{X}\parallel }^2-\parallel {\mathbf{U}}^0{\parallel }^2)+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}\mathbf{X}\parallel }^2.\hfill \end{array}$$

This suggests us that $\langle g\left(\mathbf{X}\right),\mathbf{X}\rangle \ge 0$, for all $\mathbf{X}\in {\mathbb{V}}_h$ with $\parallel \mathbf{X}\parallel =\parallel {\mathbf{U}}^0\parallel +1$. By the Browder fixed point theorem, there exists ${\mathbf{X}}^{*}\in {\mathbb{V}}_h$ which satisfies $g\left({\mathbf{X}}^{*}\right)=0$, establishing the existence of a numerical solution in the case of $n=1$. The uniqueness of the numerical solution will show later (see Remark 2).

Step 2:

We assume that ${\mathbf{U}}^0,{\mathbf{U}}^1,\dots ,{\mathbf{U}}^{n-1}$ satisfy the Equation (4b)–(4d). Here, we assume that ${\mathbf{U}}^n$ and ${\mathbf{W}}^n$ are two solutions for the Equation (4b)–(4d). Consequently, the term ${\mathbf{E}}^n={\mathbf{U}}^n-{\mathbf{W}}^n$ can be obtained by

$${\displaystyle \frac{{\tau }^{-\alpha }{c}_0^{\left(\alpha \right)}}{\mathsf{\Gamma }(2-\alpha )}}{\mathbf{E}}^n-{\mathbf{H}}^{-1}{\mathbf{D}}_2\left({\mathbf{E}}^n\right)+\mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{E}}^n)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{E}}^n)=0,$$

(5)

where Lemma 6 is used. Computing the inner product between Equation (5) and ${\mathbf{E}}^n$, we have

$${\displaystyle \frac{{\tau }^{-\alpha }{c}_0^{\left(\alpha \right)}}{\mathsf{\Gamma }(2-\alpha )}}{\parallel {\mathbf{E}}^n\parallel }^2+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{E}}^n\parallel }^2=0,$$

where Lemma 3 is applied. This gives $\parallel {\mathbf{E}}^n\parallel =0$. That is, Equation (5) has only a trivial solution. Therefore, the Equation (4b)–(4d) determines ${\mathbf{U}}^n$ uniquely. This completes the proof.    □

3.2. Preliminary Bounds

In this section, we establish the boundedness of the numerical solution for the proposed scheme, which is crucial for ensuring its convergence and stability.

Lemma 12.

Let ${\mathbf{U}}^k\in {\mathbb{V}}_h$, for $k=0,1,2,\dots ,n$. For any ${ϵ}_1>0$, we have

$$\begin{array}{cc}\hfill \left|{\langle }_2{\mathsf{\Delta }}_t^{\alpha }{\mathbf{U}}^n,{\mathbf{U}}^n\rangle \right|& \ge {\displaystyle \frac{1}{\mu }}\left(c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}\right){\parallel {\mathbf{U}}^n\parallel }^2\hfill \\ \hfill & -{\displaystyle \frac{1}{{ϵ}_1\mu }}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{U}}^0\parallel }^2\right],\hfill \end{array}$$

where ${\delta }_2^{\left(\alpha \right)}=\left|c_1^{\left(\alpha \right)}-c_2^{\left(\alpha \right)}\right|-(c_1^{\left(\alpha \right)}-c_2^{\left(\alpha \right)})$ and $\mu =2\mathsf{\Gamma }(2-\alpha ){\tau }^{\alpha }$.

Proof. 

By using the Cauchy–Schwarz inequality and Lemma 3, for any ${ϵ}_1>0$, we see that

$$\begin{array}{cc}\hfill \left|{\langle }_2{\mathsf{\Delta }}_t^{\alpha }{\mathbf{U}}^n,{\mathbf{U}}^n\rangle \right|& ={\displaystyle \frac{1}{\mu }}\left|c_0^{\left(\alpha \right)}\langle {\mathbf{U}}^n,{\mathbf{U}}^n\rangle -\sum _{k=1}^{n-1}\left(c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right)\langle {\mathbf{U}}^k,{\mathbf{U}}^n\rangle -c_{n-1}^{\left(\alpha \right)}\langle {\mathbf{U}}^0,u^n\rangle \right|\hfill \\ \hfill & \ge {\displaystyle \frac{1}{\mu }}\left[c_0^{\left(\alpha \right)}\parallel {\mathbf{U}}^n{\parallel }^2-\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{U}}^k\parallel \parallel u^n\parallel -c_{n-1}^{\left(\alpha \right)}\parallel {\mathbf{U}}^0\parallel \parallel {\mathbf{U}}^n\parallel \right]\hfill \\ \hfill & \ge {\displaystyle \frac{1}{\mu }}\left(c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}\right){\parallel {\mathbf{U}}^n\parallel }^2\hfill \\ \hfill & -{\displaystyle \frac{1}{{ϵ}_1\mu }}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{U}}^0\parallel }^2\right],\hfill \end{array}$$

which yields the desired result. The proof is complete.    □

Next, we show that the scheme is unconditionally stable with respect to the initial data, meaning the discrete $L_2$ norm of the numerical solution remains bounded for any temporal and spatial step sizes. This ensures that small perturbations in the initial value do not lead to uncontrolled growth in the solution.

Theorem 2. 

($L_2-$stability) Assuming the initial condition $u_0\in L_2\left(\overline{\mathsf{\Omega }}\right)$, the numerical solution obtained from the $L1-2$ fourth-order difference Equation (4a)–(4d) satisfies

$$\parallel {\mathbf{U}}^n\parallel \le \parallel {\mathbf{U}}^0\parallel .$$

Proof. 

The proof proceeds by mathematical induction. It follows from the $u_0\in L_2\left(\overline{\mathsf{\Omega }}\right)$ that $\parallel {\mathbf{U}}^0\parallel$ is bounded.

Step 1: 

For the first level ${\mathbf{U}}^1$, consider the inner product of Equation (4a) with ${\mathbf{U}}^1$, we get

$${\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}(\parallel {\mathbf{U}}^1{\parallel }^2-\langle {\mathbf{U}}^0,{\mathbf{U}}^1\rangle )+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{U}}^1\parallel }^2+\langle \mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{U}}^1\rangle -{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{U}}^1\rangle =0.$$

By using the Cauchy–Schwartz inequality, we see that

$$\langle {\mathbf{U}}^0,{\mathbf{U}}^1\rangle \le {\displaystyle \frac{1}{2}}\left(\parallel {\mathbf{U}}^0{\parallel }^2+{\parallel {\mathbf{U}}^1\parallel }^2\right),$$

which implies

$${\displaystyle \frac{1}{2}}\parallel {\mathbf{U}}^1{\parallel }^2\le {\displaystyle \frac{1}{2}}\parallel {\mathbf{U}}^1{\parallel }^2+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{U}}^1\parallel }^2\le {\displaystyle \frac{1}{2}}{\parallel {\mathbf{U}}^0\parallel }^2,$$

where Lemma 6 is used. The result is as required.

Step 2: 

As the induction hypothesis, we assume that

$$\parallel {\mathbf{U}}^k\parallel \le \parallel {\mathbf{U}}^0\parallel ,n=0,1,2,\dots ,n-1.$$

(6)

Applying the inner product of Equation (4b) with ${\mathbf{U}}^n$, we derive

$${\langle }_2{\mathsf{\Delta }}_t^{\alpha }{\mathbf{U}}^n,{\mathbf{U}}^n\rangle +{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{U}}^n\parallel }^2+\langle \mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{U}}^n\rangle -{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{U}}^n\rangle =0.$$

By applying Lemmas 6 and 12, we arrive at

$$\left(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}\right){\parallel {\mathbf{U}}^n\parallel }^2+2{ϵ}_1\mathsf{\Gamma }(2-\alpha ){\tau }^{\alpha }{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{U}}^n\parallel }^2\le {\displaystyle \frac{1}{{ϵ}_1}}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{U}}^0\parallel }^2\right],$$

(7)

where ${\delta }_2^{\left(\alpha \right)}=\left|c_1^{\left(\alpha \right)}-c_2^{\left(\alpha \right)}\right|-(c_1^{\left(\alpha \right)}-c_2^{\left(\alpha \right)})$. Here, we consider two cases on the parameter ${\delta }_2^{\left(\alpha \right)}$.

Case 1:

$c_1^{\left(\alpha \right)}\ge c_2^{\left(\alpha \right)}$, we have ${\delta }_2^{\left(\alpha \right)}=0$. By applying Lemmas 3 and 12 with ${ϵ}_1=1$, and the assumption (6), Equation (7) becomes

$$c_0^{\left(\alpha \right)}\parallel {\mathbf{U}}^n{\parallel }^2\le \left[\sum _{k=1}^{n-1}\left(c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right)+c_{n-1}^{\left(\alpha \right)}\right]\parallel {\mathbf{U}}^0{\parallel }^2=c_0^{\left(\alpha \right)}{\parallel {\mathbf{U}}^0\parallel }^2.$$

Case 2:

$c_1^{\left(\alpha \right)}<c_2^{\left(\alpha \right)}$, we have ${\delta }_2^{\left(\alpha \right)}=2(c_2^{\left(\alpha \right)}-c_1^{\left(\alpha \right)})>0$. Therefore, we choose ${ϵ}_1=c_0^{\left(\alpha \right)}/{\delta }_2^{\left(\alpha \right)}$. Introducing ${ϵ}_2>0$, Equation (7) gives

$$\parallel {\mathbf{U}}^n{\parallel }^2\le {ϵ}_2{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{{\delta }_2^{\left(\alpha \right)}}{{\left(c_0^{\left(\alpha \right)}\right)}^2}}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{U}}^0\parallel }^2\right].$$

By applying Lemmas 3 and 10, we have

$$\parallel {\mathbf{U}}^n{\parallel }^2\le {ϵ}_2{\parallel {\mathbf{U}}^0\parallel }^{2}exp\left({\displaystyle \frac{{\delta }_2^{\left(\alpha \right)}(c_0^{\left(\alpha \right)}+2c_2^{\left(\alpha \right)}-2c_1^{\left(\alpha \right)})}{{\left(c_0^{\left(\alpha \right)}\right)}^2}}\right).$$

Finally, we take ${ϵ}_2\le 1/exp\left({\displaystyle \frac{{\delta }_2^{\left(\alpha \right)}(c_0^{\left(\alpha \right)}+2c_2^{\left(\alpha \right)}-2c_1^{\left(\alpha \right)})}{{\left(c_0^{\left(\alpha \right)}\right)}^2}}\right)$. The proof is complete.    □

Before we are going to prove the discrete $H^1-$bounded for the numerical solution. Let us introduce the following constants for simplicity:

$$\begin{array}{cccc}\hfill & {\kappa }_0={\displaystyle \frac{8}{9}}\left({\displaystyle \frac{64}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right){\parallel {\mathbf{U}}^0\parallel }^2,\hfill & \hfill & {\kappa }_1={\displaystyle \frac{32}{9}}\left({\displaystyle \frac{256}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right){\parallel {\mathbf{U}}^0\parallel }^2,\hfill \\ \hfill & {\kappa }_2=8\left(64\parallel {\mathbf{U}}^0{\parallel }^2+{\displaystyle \frac{1}{L}}\right){\parallel {\mathbf{U}}^0\parallel }^2,\hfill & \hfill & {\kappa }_3=32\left(256\parallel {\mathbf{U}}^0{\parallel }^2+{\displaystyle \frac{1}{L}}\right){\parallel {\mathbf{U}}^0\parallel }^2.\hfill \end{array}$$

For any positive constants $\rho$ and ${ϵ}_1$, we define

$${\mu }_0=min\left\{{\displaystyle \frac{1-\rho }{{\kappa }_0+{\kappa }_1}},{\displaystyle \frac{2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-\rho }{2({\kappa }_2+{\kappa }_3)}}\right\},$$

and

$$C_1=max\{{(1-\mu ({\kappa }_0+{\kappa }_1))}^{-1/2},{\left(2{\eta }^{-1}exp\left(2(c_0^{\left(\alpha \right)}-c_{n-1}^{\left(\alpha \right)}+{\delta }_2^{\left(\alpha \right)}){\eta }^{-1}\right)\right)}^{-1/2}\}\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0\parallel ,$$

where $\eta ={ϵ}_1(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-2\mu ({\kappa }_2+{\kappa }_3))$.

Theorem 3. 

($H^1-$bounded) Suppose $u_0\in H_0^1\left(\overline{\mathsf{\Omega }}\right)$. If there exists a positive constants ρ and ${ϵ}_1$ such that ${\mu }_0>0$. Then, when $\mu \le {\mu }_0$, the numerical solution obtained fromthe $L1-2$ fourth-order difference Equation (4a)–(4d) satisfies

$${\parallel {\mathbf{U}}^n\parallel }_1\le C_1.$$

Proof. Step 1: 

By considering the inner product of Equation (4a) with $-{\mathbf{D}}_2{\mathbf{U}}^1$, we arrive at

$${\displaystyle \frac{1}{\mu }}(\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+\langle {\mathbf{U}}^0,{\mathbf{D}}_2{\mathbf{U}}^1\rangle )+{\parallel \mathbf{R}{\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2=\langle \mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle -{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle .$$

(8)

Noting that

$$\langle {\mathbf{U}}^0,{\mathbf{D}}_2{\mathbf{U}}^1\rangle =-\langle {\mathbf{D}}_1^{+}{\mathbf{U}}^0,{\mathbf{D}}_1^{+}{\mathbf{U}}^1\rangle \le {\displaystyle \frac{1}{2}}\left(\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0{\parallel }^2+{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1\parallel }^2\right).$$

We observe that the first nonlinear term can be written as

$$\begin{array}{cc}\hfill \langle \mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle & ={\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1.\hfill \end{array}$$

(9)

First, we observe that

$$\begin{array}{cc}\hfill {\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1& \le {\displaystyle \frac{1}{3}}\parallel {\mathbf{U}}^0\parallel \parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }_{\infty }\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{4}{9}}\parallel {\mathbf{U}}^0{\parallel }^2\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }_{\infty }^2+{\displaystyle \frac{1}{16}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2,\hfill \end{array}$$

(10)

where Theorem 2 is are used. Regarding to the result of Lemma 9 with ${ϵ}_0={\displaystyle \frac{9}{64\parallel {\mathbf{U}}^0{\parallel }^2}}$, we have

$$\parallel {\mathbf{D}}_1{\mathbf{U}}^1{\parallel }_{\infty }^2\le \left({\displaystyle \frac{64}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{9}{64\parallel {\mathbf{U}}^0{\parallel }^2}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

This yields

$${\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1\le {\displaystyle \frac{4}{9}}\parallel {\mathbf{U}}^0{\parallel }^2\left({\displaystyle \frac{64}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{8}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2.$$

Moreover, the second term in the right hand side of Equation (9) can be estimated as

$$\begin{array}{cc}\hfill {\displaystyle \frac{h}{3}}{\left[{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1& \le {\displaystyle \frac{1}{3}}\parallel {\mathbf{D}}_1\left(\mathbf{diag}\left(\left({\mathbf{U}}^1\right)\right){\mathbf{U}}^1\right)\parallel \parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{4}{9}}\parallel {\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{U}}^1\right){\parallel }^2+{\displaystyle \frac{1}{16}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2,\hfill \end{array}$$

where the Cauchy–Schwartz and the Young’s inequality are used. Note that

$$\parallel {\mathbf{D}}_1\left(\mathbf{diag}\left(\left({\mathbf{U}}^1\right)\right){\mathbf{U}}^1\right){\parallel }^2={\displaystyle \frac{1}{2}}\parallel \mathbf{diag}\left(({\mathbf{T}}^{+}+{\mathbf{T}}^{-}){\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1{\parallel }^2\le \parallel {\mathbf{U}}^0{\parallel }^2{\parallel {\mathbf{D}}_1{\mathbf{U}}^1\parallel }_{\infty }^2,$$

(11)

where Theorem 2 is used. Again, by applying Lemma 9 with ${ϵ}_0={\displaystyle \frac{9}{64\parallel {\mathbf{U}}^0{\parallel }^2}}$, we have

$$\parallel {\mathbf{D}}_1{\mathbf{U}}^1{\parallel }_{\infty }^2\le \left({\displaystyle \frac{64}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{9}{64\parallel {\mathbf{U}}^0{\parallel }^2}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

which yields

$${\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1\le {\displaystyle \frac{4}{9}}\left({\displaystyle \frac{64}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{U}}^0{\parallel }^2\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{8}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

Integrating above estimations, the nonlinear term (9) gives

$$\langle \mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle \le {\kappa }_0\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{4}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

(12)

For the second nonlinear term, we see that

$$\begin{array}{cc}\hfill {\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle & ={\displaystyle \frac{h^3}{6}}{\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1.\hfill \end{array}$$

(13)

By utilizing Lemma 8 and Theorem 2, we remark that

$$\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel \le {\displaystyle \frac{2}{h}}\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1\parallel \le {\displaystyle \frac{4}{h^2}}\parallel {\mathbf{U}}^0\parallel .$$

Based on the arguments used in Equation (11), we can conclude that

$${\displaystyle \frac{h^3}{6}}{\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1\le {\displaystyle \frac{16}{9}}\left({\displaystyle \frac{256}{9}}\parallel {\mathbf{U}}^0\parallel +{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{U}}^0{\parallel }^2\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{8}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

(14)

For the second terms of Equation (13), we note that

$$\begin{array}{cc}\hfill {\displaystyle \frac{h^3}{6}}{\left[{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1& \le {\displaystyle \frac{h^2}{6}}\parallel {\mathbf{D}}_1\left(\mathbf{diag}\left(\left({\mathbf{D}}_2{\mathbf{U}}^1\right)\right){\mathbf{U}}^1\right)\parallel \parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{h^4}{9}}\parallel {\mathbf{D}}_1\left(\mathbf{diag}\left(\left({\mathbf{D}}_2{\mathbf{U}}^1\right)\right){\mathbf{U}}^1\right){\parallel }^2+{\displaystyle \frac{1}{16}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.\hfill \end{array}$$

By the same argument with Equation (11), it is easy to derive that

$${\displaystyle \frac{h^3}{6}}{\left[{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_2{\mathbf{U}}^1\le {\displaystyle \frac{16}{9}}\left({\displaystyle \frac{256}{9}}{\parallel {\mathbf{U}}^0\parallel }^2+{\displaystyle \frac{1}{L}}\right)\parallel {\mathbf{U}}^0{\parallel }^2\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{8}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2$$

Integrating above estimation, Equation (13) gives

$${\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{D}}_2{\mathbf{U}}^1\rangle \le {\kappa }_2\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{4}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2.$$

(15)

Therefore, Equation (8) implies

$$\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2+{\displaystyle \frac{1}{2}}\mu {\parallel {\mathbf{D}}_2{\mathbf{U}}^1\parallel }^2\le \parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0{\parallel }^2+\mu ({\kappa }_0+{\kappa }_1){\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1\parallel }^2.$$

We then conclude that

$$\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^1{\parallel }^2\le {\displaystyle \frac{1}{1-\mu ({\kappa }_0+{\kappa }_1)}}{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0\parallel }^2.$$

Step 2: 

Taking the inner product of Equation (4b) with $-{\mathbf{D}}_2{\mathbf{U}}^n$, we have

$$-{\langle }_2{\mathsf{\Delta }}_t^{\alpha }{\mathbf{U}}^n,\mathbf{R}{\mathbf{D}}_2{\mathbf{U}}^n\rangle +{\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2=-\langle \mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle +{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle .$$

By applying Lemmas 4, 5 and 12, we have

$$\begin{array}{cc}\hfill \left(2c_0^{\left(\alpha \right)}-ϵ{\delta }_2^{\left(\alpha \right)}\right){\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^n\parallel }^2+\mu {\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2& \le {\displaystyle \frac{1}{ϵ}}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0\parallel }^2\right]\hfill \\ \hfill & -\mu \langle \mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle -\mu {\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle .\hfill \end{array}$$

(16)

Following the reasoning outlined to derive Equations (12) and (15), applying the fact that $\parallel {\widehat{\mathbf{U}}}^n\parallel \le 3\parallel {\mathbf{U}}^0\parallel$, it can be deduced

$$\begin{array}{cc}\hfill \langle \mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle & \le {\kappa }_2\parallel {\mathbf{D}}_1{\mathbf{U}}^n{\parallel }^2+{\displaystyle \frac{1}{4}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2,\hfill \\ \hfill {\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{D}}_2{\mathbf{U}}^n\rangle & \le {\kappa }_3\parallel {\mathbf{D}}_1{\mathbf{U}}^n{\parallel }^2+{\displaystyle \frac{1}{4}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2.\hfill \end{array}$$

Therefore, Equation (16) can be rewritten as

$$\begin{array}{c}{\displaystyle \frac{(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)})}{2}}{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^n\parallel }^2+{\displaystyle \frac{\mu }{2}}{\parallel {\mathbf{D}}_2{\mathbf{U}}^n\parallel }^2\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le \mu ({\kappa }_2+{\kappa }_3){\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^n\parallel }^2+{\displaystyle \frac{1}{{ϵ}_1}}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0\parallel }^2\right].\end{array}$$

Then, under the condition $\mu \le {\mu }_0$ and the fact $c_{n-1}^{\left(\alpha \right)}<1$, we have

$$\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^n{\parallel }^2\le {\displaystyle \frac{2}{{ϵ}_1(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-2\mu ({\kappa }_2+{\kappa }_3))}}\left[\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0{\parallel }^2+\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^k\parallel }^2\right].$$

Using Lemma 10, together with Lemma 3, we obtain

$$\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^n{\parallel }^2\le 2{\eta }^{-1}{\parallel {\mathbf{D}}_1^{+}{\mathbf{U}}^0\parallel }^{2}exp\left(2(c_0^{\left(\alpha \right)}-c_{n-1}^{\left(\alpha \right)}+{\delta }_2^{\left(\alpha \right)}){\eta }^{-1}\right).$$

This completes the proof.    □

By applying Theorem 3 with Lemma 7, the following corollary can be directly obtained.

Corollary 1.

Under the assumption stated in Theorem 3, the numerical solution obtained fromthe $L1-2$ fourth-order difference Equation (4a)–(4d) satisfies

$${\parallel {\mathbf{U}}^n\parallel }_{\infty }\le {\displaystyle \frac{\sqrt{L}}{2}}{C}_1.$$

3.3. Error Estimate

This subsection focuses on deriving the pointwise error estimate of the proposed scheme. The standard error, measuring the discrepancy between the exact and numerical solutions, is introduced as

$${\mathbf{e}}^n=\mathbf{u}(\mathbf{x},t_n)-{\mathbf{U}}^n:={\mathbf{u}}^n-{\mathbf{U}}^n.$$

Therefore, the system of error equations can be written as

(i)

$n=1$

$${\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}({\mathbf{e}}^1-{\mathbf{e}}^0)-{\mathbf{D}}_2{\mathbf{e}}^1+\mathsf{\Psi }({\mathbf{u}}^1,{\mathbf{u}}^1)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{u}}^1,{\mathbf{u}}^1)-\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1)+{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1)={\mathbf{R}}^1,$$

(17)

(ii)

$n\ge 2$

$${}_2\mathsf{\Delta }_t^{\alpha }\left({\mathbf{e}}^n\right)-{\mathbf{H}}_2^{-1}{\mathbf{D}}_2\left({\mathbf{e}}^n\right)+\mathsf{\Psi }({\widehat{\mathbf{u}}}^n,{\mathbf{u}}^n)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{u}}}^n,{\mathbf{u}}^n)-\mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n)+{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n)={\mathbf{R}}^n.$$

(18)

Through implementation of the Taylor expansion and Lemma 2, there exists positive constants $C_2$ and $C_3$ such that

$$\parallel {\mathbf{R}}^n\parallel \le \left\{\begin{array}{cc}{C}_2({\tau }^{2-\alpha }+h^4),& n=1,\\ C_3({\tau }^2+h^4),& n=2,\dots ,N.\end{array}\right.$$

(19)

For simplicity, we denote that

$$\begin{array}{cc}\hfill {\kappa }_4& ={\displaystyle \frac{2}{9}}{\left(C_4+{\displaystyle \frac{1}{2}}{C}_1\sqrt{L}\right)}^2+{\displaystyle \frac{2}{9}}{\left(C_1+C_4\right)}^2,\hfill \\ \hfill {\kappa }_5& ={\displaystyle \frac{1}{18}}{\left(4C_4+{\displaystyle \frac{1}{2}}{C}_1\sqrt{L}\right)}^2+{\displaystyle \frac{2}{9}}{\left({\displaystyle \frac{1}{8}}{C}_4+\sqrt{L}{C}_1\right)}^2,\hfill \end{array}$$

and

$$C_4=\underset{(x,t)\in \overline{\mathsf{\Omega }}\times (0,T]}{max}\left\{\right|u(x,t)|,|u_x(x,t)\left|\right\}.$$

Lemma 13.

For the numerical solution ${\mathbf{U}}^n$ of the $L1-2$ fourth-order difference Equation (4a)–(4d) and let $u(x,t)$ be the solution of the (1a)–(1c). Under the assumption stated in Theorem 3, the following estimation hold

$$\langle \mathsf{\Psi }({\mathbf{u}}^1,{\mathbf{u}}^1)-\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle \le {\displaystyle \frac{1}{2}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1{\parallel }^2+{\kappa }_4{\parallel {\mathbf{e}}^1\parallel }^2,$$

(20)

and

$$\langle \mathsf{\Psi }({\widehat{\mathbf{u}}}^n,{\mathbf{u}}^n)-\mathsf{\Psi }({\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{e}}^n\rangle \le {\displaystyle \frac{1}{2}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^n{\parallel }^2+9{\kappa }_4{\parallel {\mathbf{e}}^n\parallel }^2.$$

(21)

Proof. 

Consider

$$\begin{array}{cc}\hfill \langle \mathsf{\Psi }({\mathbf{u}}^1,{\mathbf{u}}^1)-\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle & ={\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{u}}^1\right){\mathbf{D}}_1{\mathbf{u}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{u}}^1\right){\mathbf{u}}^1\right)\right]}^T{\mathbf{e}}^1\hfill \\ \hfill & -{\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{e}}^1\hfill \\ \hfill & :={\mathbf{P}}_1+{\mathbf{P}}_2.\hfill \end{array}$$

By applying Corollary 1, for the term ${\mathbf{P}}_1$, we have

$$\begin{array}{cc}\hfill {\mathbf{P}}_1& ={\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{u}}^1\right)\left({\mathbf{D}}_1{\mathbf{e}}^1\right)\right]}^T{\mathbf{e}}^1+{\displaystyle \frac{h}{3}}{\left[\mathbf{diag}\left({\mathbf{e}}^1\right)\left({\mathbf{D}}_1{\mathbf{U}}^1\right)\right]}^T{\mathbf{e}}^1\hfill \\ \hfill & \le {\displaystyle \frac{1}{3}}{C}_4\parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel \parallel {\mathbf{e}}^1\parallel +{\displaystyle \frac{1}{3}}{C}_1\parallel {\mathbf{e}}^1{\parallel }_{\infty }\parallel {\mathbf{e}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}\parallel {\mathbf{D}}_1{\mathbf{e}}^1{\parallel }^2+{\displaystyle \frac{2}{9}}{\left(C_4+{\displaystyle \frac{1}{2}}{C}_1\sqrt{L}\right)}^2{\parallel {\mathbf{e}}^1\parallel }^2.\hfill \end{array}$$

Similarly, we observe that

$$\begin{array}{cc}\hfill {\mathbf{P}}_2& =-{\displaystyle \frac{h}{3}}{\left[\left(\mathbf{diag}\left({\mathbf{u}}^1\right){\mathbf{e}}^1\right)+\left(\mathbf{diag}\left({\mathbf{e}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_1{\mathbf{e}}^1\hfill \\ \hfill & \le {\displaystyle \frac{1}{3}}\left(C_1+C_4\right)\parallel {\mathbf{e}}^1\parallel \parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1{\parallel }^2+{\displaystyle \frac{2}{9}}{\left(C_1+C_4\right)}^2{\parallel {\mathbf{e}}^1\parallel }^2.\hfill \end{array}$$

Based on above estimations, Equation (20) can be obtained. Similarly, for Equation (21), by adopting $\parallel {\widehat{\mathbf{U}}}^n{\parallel }_{\infty }\le 3C_1$ and $\parallel {\widehat{\mathbf{u}}}^n{\parallel }_{\infty }\le 3C_4$, the proof is complete.    □

Lemma 14.

For the numerical solution ${\mathbf{U}}^n$ of the $L1-2$ fourth-order difference Equation (4a)–(4d) and let $u(x,t)$ be the solution of the (1a)–(1c). Under the assumption stated in Theorem 3, the following estimation hold

$${\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{u}}^1,{\mathbf{u}}^1)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle \le {\displaystyle \frac{1}{2}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1{\parallel }^2+{\kappa }_5{\parallel {\mathbf{e}}^1\parallel }^2,$$

(22)

and

$${\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{u}}}^n,{\mathbf{u}}^n)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n,{\mathbf{U}}^n),{\mathbf{e}}^n\rangle \le {\displaystyle \frac{1}{2}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^n{\parallel }^2+9{\kappa }_5{\parallel {\mathbf{e}}^n\parallel }^2.$$

(23)

Proof. 

We observe that

$$\begin{array}{cc}\hfill {\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{u}}^1,{\mathbf{u}}^1)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle & ={\displaystyle \frac{h^3}{6}}{\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{u}}^1\right){\mathbf{D}}_1{\mathbf{u}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{u}}^1\right){\mathbf{u}}^1\right)\right]}^T{\mathbf{e}}^1\hfill \\ \hfill & -{\displaystyle \frac{h^3}{6}}{\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^1\right){\mathbf{D}}_1{\mathbf{U}}^1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{e}}^1\hfill \\ \hfill & :={\mathbf{P}}_3+{\mathbf{P}}_4.\hfill \end{array}$$

For the term ${\mathbf{P}}_3$, we have

$$\begin{array}{cc}\hfill {\mathbf{P}}_3& ={\displaystyle \frac{h^3}{6}}\left({\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{u}}^1\right)\left({\mathbf{D}}_1{\mathbf{e}}^1\right)\right]}^T{\mathbf{e}}^1+{\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{e}}^1\right)\left({\mathbf{D}}_1{\mathbf{U}}^1\right)\right]}^T{\mathbf{e}}^1\right)\hfill \\ \hfill & \le {\displaystyle \frac{h^2}{6}}\left(\parallel {\mathbf{D}}_2{\mathbf{u}}^1{\parallel }_{\infty }\parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel \parallel {\mathbf{e}}^1\parallel +\parallel {\mathbf{D}}_1{\mathbf{U}}^1{\parallel }_{\infty }\parallel {\mathbf{D}}_2{\mathbf{e}}^1\parallel \parallel {\mathbf{e}}^1\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{2}{3}}\parallel {\mathbf{u}}^1{\parallel }_{\infty }\parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel \parallel {\mathbf{e}}^1\parallel +{\displaystyle \frac{1}{6}}\parallel {\mathbf{U}}^1{\parallel }_{\infty }\parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel \parallel {\mathbf{e}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}\parallel {\mathbf{D}}_1{\mathbf{e}}^1{\parallel }^2+{\displaystyle \frac{1}{18}}{\left(4C_4+{\displaystyle \frac{1}{2}}{C}_1\sqrt{L}\right)}^2{\parallel {\mathbf{e}}^1\parallel }^2,\hfill \end{array}$$

where Corollary 1 is applied. In a similar manner, we note that

$$\begin{array}{cc}\hfill {\mathbf{P}}_4& =-{\displaystyle \frac{h^2}{6}}{\left[\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{u}}^1\right){\mathbf{e}}^1\right)+\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{e}}^1\right){\mathbf{U}}^1\right)\right]}^T{\mathbf{D}}_1{\mathbf{e}}^1\hfill \\ \hfill & \le {\displaystyle \frac{1}{24}}\parallel {\mathbf{u}}^1{\parallel }_{\infty }\parallel {\mathbf{e}}^1\parallel \parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel +{\displaystyle \frac{2}{3}}\parallel {\mathbf{U}}^1{\parallel }_{\infty }\parallel {\mathbf{e}}^1\parallel \parallel {\mathbf{D}}_1{\mathbf{e}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{3}}\left({\displaystyle \frac{1}{8}}{C}_4+\sqrt{L}{C}_1\right)\parallel {\mathbf{e}}^1\parallel \parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}\parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1{\parallel }^2+{\displaystyle \frac{2}{9}}{\left({\displaystyle \frac{1}{8}}{C}_4+\sqrt{L}{C}_1\right)}^2{\parallel {\mathbf{e}}^1\parallel }^2.\hfill \end{array}$$

Relying on the above estimations, we directly obtain Equation (22). In a similar manner, for the second estimate, by applying $\parallel {\widehat{\mathbf{U}}}^n{\parallel }_{\infty }\le 3C_1$ and $\parallel {\widehat{\mathbf{u}}}^n{\parallel }_{\infty }\le 3C_4$, Equation (23) estimate follows straightforwardly. Thus, the proof is complete.    □

Remark 2.

To show the uniqueness of the numerical solution in Theorem 1 ($n=1$), we assume that there is another solution ${\mathbf{W}}^1$ satisfying the Equation (4a). Introducing ${\mathbf{E}}^1={\mathbf{U}}^1-{\mathbf{W}}^1$, we note that

$${\displaystyle \frac{1}{\mu }}{\mathbf{E}}^1-{\mathbf{H}}_2^{-1}{\mathbf{D}}_2{\mathbf{E}}^1+\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1)-\mathsf{\Psi }({\mathbf{W}}^1,{\mathbf{W}}^1)-{\displaystyle \frac{h^2}{2}}\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1)-\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{W}}^1,{\mathbf{W}}^1)=0.$$

When Corollary 1 is applied instead of using $C_4$, based on Lemmas 13 and 14, the inner product between above equation and ${\mathbf{E}}^1$ gives

$${\displaystyle \frac{1}{\mu }}\parallel {\mathbf{E}}^1{\parallel }^2+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{E}}^1\parallel }^2\le \parallel {\mathbf{D}}_1^{+}{\mathbf{e}}^1{\parallel }^2+{\kappa }_6{\parallel {\mathbf{E}}^1\parallel }^2,$$

where

$${\kappa }_6={\displaystyle \frac{83}{162}}L{C}_1^2+{\displaystyle \frac{2}{9}}{\left(1+{\displaystyle \frac{1}{2}}\sqrt{L}\right)}^2{C}_1^2.$$

That is, the Equation (4a) determines ${\mathbf{U}}^1$ uniquely provided that ${\kappa }_6\mu <1$.

Accordingly, we derive the error estimate as follows.

Theorem 4.

For the numerical solution ${\mathbf{U}}^n$ of the $L1-2$ fourth-order difference Equation (4a)–(4d) and let $u(x,t)$ be the solution of the (1a)–(1c) with regularity $u(x,t)\in C^{6,3}\left(\overline{\mathsf{\Omega }}\right)$. Under the assumption stated in Theorem 3, if there exists a positive constant ϱ such that

$$max\left\{1-\mu ({\kappa }_4+{\kappa }_5),2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-\mu ({\displaystyle \frac{1}{2}}+9({\kappa }_4+{\kappa }_5))\right\}\ge \varrho >0,$$

denote

$$C_5=max\left\{{\displaystyle \frac{\mathsf{\Gamma }(2-\alpha )}{\sqrt{2\varrho }}}{C}_2,{\left[{\displaystyle \frac{{\mu }_0}{2\varrho }}exp\left({\displaystyle \frac{c_0^{\left(\alpha \right)}+{\delta }_2^{\left(\alpha \right)}}{{ϵ}_1\varrho }}\right)\right]}^{1/2}{C}_3\right\},$$

then we have

$$\parallel {\mathbf{e}}^n\parallel \le C_5({\tau }^2+h^4).$$

Proof. Step 1: 

By computing the inner product of Equation (17) with ${\mathbf{e}}^1$, we establish that

$$\begin{array}{c}{\displaystyle \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}{\parallel {\mathbf{e}}^1\parallel }^2+{\parallel \mathbf{R}{\mathbf{D}}_1^{+}{\mathbf{e}}^1\parallel }^2=\langle {\mathbf{R}}^1,{\mathbf{e}}^1\rangle -\langle \mathsf{\Psi }({\mathbf{u}}^1,{\mathbf{u}}^1)-\mathsf{\Psi }({\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle \hfill \\ \hfill +{\displaystyle \frac{h^2}{2}}\langle \mathsf{\Psi }({\mathbf{D}}_2{\mathbf{u}}^1,{\mathbf{u}}^1)-\mathsf{\Psi }({\mathbf{D}}_2{\mathbf{U}}^1,{\mathbf{U}}^1),{\mathbf{e}}^1\rangle .\end{array}$$

(24)

By using the Young’s inequality, we note that

$$\langle {\mathbf{R}}^1,{\mathbf{e}}^1\rangle \le {\displaystyle \frac{{\tau }^{\alpha }\mathsf{\Gamma }(2-\alpha )}{2}}\parallel {\mathbf{R}}^1{\parallel }^2+{\displaystyle \frac{1}{2{\tau }^{\alpha }\mathsf{\Gamma }(2-\alpha )}}{\parallel {\mathbf{e}}^1\parallel }^2.$$

By applying Lemmas 5, 13 and 14, Equation (24) can be rewritten as

$$\left(1-\mu ({\kappa }_4+{\kappa }_5)\right){\parallel {\mathbf{e}}^1\parallel }^2\le {\displaystyle \frac{{\tau }^{2\alpha }{\mathsf{\Gamma }}^2(2-\alpha )}{2}}{C}_2^2{({\tau }^{2-\alpha }+h^4)}^2.$$

Regarding to Equation (19), we see that this suggests us that, when $\tau$ is sufficient small such that $1-\mu ({\kappa }_4+{\kappa }_5)\ge \varrho >0$, we can conclude that

$$\parallel {\mathbf{e}}^1\parallel \le {\displaystyle \frac{\mathsf{\Gamma }(2-\alpha )}{\sqrt{2\varrho }}}{C}_2({\tau }^2+h^4).$$

Step 2: 

Next, we compute the inner product in Equation (18) with ${\mathbf{e}}^n$, we have

$${\displaystyle \frac{1}{\mu }}\left(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}\right){\parallel {\mathbf{e}}^n\parallel }^2-{\displaystyle \frac{1}{{ϵ}_1\mu }}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{e}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{e}}^0\parallel }^2\right]\le \left|{\langle }_2{\mathsf{\Delta }}_t^{\alpha }{\mathbf{e}}^n,{\mathbf{e}}^n\rangle \right|,$$

(25)

where Lemma 12 is used. Note that

$$\langle {\mathbf{R}}^n,{\mathbf{e}}^n\rangle \le {\displaystyle \frac{1}{2}}\parallel {\mathbf{R}}^n{\parallel }^2+{\displaystyle \frac{1}{2}}\parallel {\mathbf{e}}^n{\parallel }^2\le {\displaystyle \frac{1}{2}}{C}_3^2{({\tau }^2+h^4)}^2+{\displaystyle \frac{1}{2}}{\parallel {\mathbf{e}}^n\parallel }^2.$$

By utilizing Lemmas 5, 13 and 14, Equation (25) gives

$$\begin{array}{c}\left(2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-\mu ({\displaystyle \frac{1}{2}}+9({\kappa }_4+{\kappa }_5))\right){\parallel {\mathbf{e}}^n\parallel }^2\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le {\displaystyle \frac{1}{2}}{C}_3^2\mu {({\tau }^2+h^4)}^2+{\displaystyle \frac{1}{{ϵ}_1}}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{e}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{e}}^0\parallel }^2\right].\end{array}$$

Using Lemma 10 and the assumption $2c_0^{\left(\alpha \right)}-{ϵ}_1{\delta }_2^{\left(\alpha \right)}-\mu ({\displaystyle \frac{1}{2}}+9({\kappa }_4+{\kappa }_5))\ge \varrho >0$, we have

$$\parallel {\mathbf{e}}^n{\parallel }^2\le {\displaystyle \frac{1}{2\varrho }}{C}_3^2\mu {({\tau }^2+h^4)}^2+{\displaystyle \frac{1}{{ϵ}_1\varrho }}\left[\sum _{k=1}^{n-1}\left|c_{n-k-1}^{\left(\alpha \right)}-c_{n-k}^{\left(\alpha \right)}\right|\parallel {\mathbf{e}}^k{\parallel }^2+c_{n-1}^{\left(\alpha \right)}{\parallel {\mathbf{e}}^0\parallel }^2\right].$$

By applying Lemma 10 together with Lemma 3, we arrive at

$$\parallel {\mathbf{e}}^n{\parallel }^2\le {\displaystyle \frac{1}{2\varrho }}{\mu }_{0}exp\left({\displaystyle \frac{c_0^{\left(\alpha \right)}+{\delta }_2^{\left(\alpha \right)}}{{ϵ}_1\varrho }}\right)C_3^2{({\tau }^2+h^4)}^2.$$

The proof is finished.    □

Remark 3.

It is well known that solutions of time–fractional problems often exhibit an initial singularity near $t=0$ [41,42]. Accordingly, we assume that the solution satisfies the following weak regularity conditions:

$$\begin{array}{cc}\hfill {∥{\displaystyle \frac{{𝜕}^{s}u}{𝜕t^s}}∥}_{L^{\infty }}& \le C_6(t^{\alpha -s}+1),s=1,2,3,\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill {∥{\displaystyle \frac{{𝜕}^{k}u}{𝜕x^k}}∥}_{L^{\infty }}& \le C_7,k=0,1,2,\dots ,6.\hfill \end{array}$$

(27)

where $C_6$ and $C_7$ denote positive constants. As reported in many publications [41,42,43,44,45], such weak temporal regularity destroys the validity of classical pointwise error estimates. Combining the error analysis of the $L1$ scheme in [43] with the graded-mesh analysis for $L1-2$ in [45], we conclude that, under weak regularity assumptions, the local truncation error of the $L1-2$ scheme on a uniform mesh reduces to

$$|R\left(g\left(t_n\right)\right)|=\left|{𝜕}_t^{\alpha }{g\left(t\right)|}_{t=t_k}-{}_2\mathsf{\Delta }_t^{\alpha }g\left(t_n\right)\right|\le \left\{\begin{array}{cc}{C}_8{\tau }^{\alpha },\hfill & n=1,\hfill \\ C_9{\tau }^{1+\alpha },\hfill & n=2,\dots ,N.\hfill \end{array}\right.$$

In addition, the limited temporal regularity also affects the linearization of the nonlinear term, leading to

$$\parallel u^n-(2u^{n-1}-u^{n-2})\parallel \le C_{10}{\tau }^2(1+t_n^{\alpha -2}).$$

Consequently, the truncation error of the proposed scheme can be estimated only as

$$\parallel {\mathbf{R}}^n\parallel \le C_{11}({\tau }^{1+\alpha }+{\tau }^2{t}_n^{\alpha -2}+h^4),n=1,2,\dots ,N,$$

where $C_{11}$ is a positive constant. This observation shows that, under weak regularity, the temporal order of the $L1-2$ fourth-order difference scheme on uniform meshes reduces to $O(1+\alpha )$. Such a framework is more consistent with the realistic pointwise behavior of solutions to time-fractional Burgers equations. In particular, away from the initial layer near $t=0$, the contribution of the term ${\tau }^2{t}_n^{\alpha -2}$ becomes negligible, so that the local truncation error is effectively governed by the $O\left({\tau }^{1+\alpha }\right)$ term. However, the pointwise convergence analysis of the scheme is more intricate and requires a deeper investigation of the $L1-2$ formula.

4. Numerical Experiments

We provide numerical results in this section to demonstrate the accuracy and efficiency of the proposed method. The performance is measured using the maximum error norm, defined as follows

$$\begin{array}{cc}\hfill {\mathrm{Error}}^n& ={\left(h\sum _{j=1}^{J-1}|u(x_j,t_n)-u_j^n|\right)}^{1/2}.\hfill \end{array}$$

The rate of convergence is determined using

$${\mathrm{Rate}}_x={log}_2\left({\displaystyle \frac{{\mathrm{Error}}_h}{{\mathrm{Error}}_{h/2}}}\right),{\mathrm{Rate}}_t={log}_2\left({\displaystyle \frac{{\mathrm{Error}}_{\tau }}{{\mathrm{Error}}_{\tau /2}}}\right),$$

where ${\mathrm{Error}}_h$ and ${\mathrm{Error}}_{h/2}$ represent the norm errors associated with grid sizes h and $h/2$, respectively. The corresponding errors for the time grid sizes $\tau$ and $\tau /2$ are denoted as ${\mathrm{Error}}_{\tau }$ and ${\mathrm{Error}}_{\tau /2}$, respectively. To validate the capability of the proposed scheme, we provide numerical examples in two cases: one with a smooth solution and another with a non-smooth solution, demonstrating its reliability and accuracy. To implement the proposed difference scheme, the complete computational procedure is summarized in Algorithm 1, which outlines the main initialization, iteration, and update steps of the method.

Algorithm 1 The algorithm framework of the $L1-2$ fourth-order scheme

Require:  Parameters $L,T,J,N$;    1: Preparation: Divide the mesh in time and space; obtain the initial condition ${\mathbf{U}}^0$ and the source term $\mathbf{F}$ (omit if not applicable);    2: Construct the matrices $\mathbf{H}$, ${\mathbf{D}}_1$, and ${\mathbf{D}}_2$;    3: Calculate $c_0^{\left(\alpha \right)},c_1^{\left(\alpha \right)},\dots ,c_{J-1}^{\left(\alpha \right)}$ and $\mu ={\tau }^{\alpha }\mathsf{\Gamma }(2-\alpha )$; Ensure:  Numerical solution ${\mathbf{U}}^1$    4: Maximum iteration steps: maxstep = 1000;    5: iterative error limit: tor =${10}^{-8}$;    6: initial iteration value ${\mathbf{U}}^{1,0}={\mathbf{U}}^0$;    7: Calculate ${\mathbf{B}}_1=\mathbf{H}\left({\mathbf{U}}^0+\mu \mathbf{F}\right)$;    8: $k=0$;    9: while 1 do  10:    ${\mathbf{A}}_1=-\mu {\mathbf{D}}_2+\mathbf{H}\left[\mathbf{I}+\mu {\displaystyle \frac{1}{3}}\left[\mathbf{diag}\left({\mathbf{U}}^{1,k}\right){\mathbf{D}}_1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{U}}^{1,k}\right)\right)\right]-\mu {\displaystyle \frac{h^2}{6}}\left[\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^{1,k}\right){\mathbf{D}}_1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{D}}_2{\mathbf{U}}^{1,k}\right)\right)\right]\right];$  11:    $k=k+1$;  12:    $s_2={\mathbf{A}}_1\setminus {\mathbf{B}}_1$; $s_1={\mathbf{U}}^{1,k}$; ${\mathbf{U}}^{1,k+1}=s_2$;  13:    if $k\ge \mathrm{maxstep}$ or $\parallel {\mathbf{U}}^{1,k}-{\mathbf{U}}^{1,k}{\parallel }_{\infty }<\mathrm{tor}$ then  14:      return  15:    end if  16: end while Ensure:  Numerical solution ${\mathbf{U}}^n$ ($n\ge 2$)  17: for $n\ge 2$do  18:    Calculate ${\widehat{\mathbf{U}}}^n=2{\mathbf{U}}^{n-1}-{\mathbf{U}}^{n-2}$;  19:    $\mathbf{A}=\mu {\mathbf{D}}_2+\mathbf{H}\left[\mathbf{I}+\mu {\displaystyle \frac{1}{3}}\left[\mathbf{diag}\left({\widehat{\mathbf{U}}}^n\right){\mathbf{D}}_1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\widehat{\mathbf{U}}}^n\right)\right)\right]-\mu {\displaystyle \frac{h^2}{6}}\left[\mathbf{diag}\left({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n\right){\mathbf{D}}_1+{\mathbf{D}}_1\left(\mathbf{diag}\left({\mathbf{D}}_2{\widehat{\mathbf{U}}}^n\right)\right)\right]\right];$  20:    $\mathbf{B}=\mathbf{H}\left({\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}^{\left(\alpha \right)}+c_{n-k}^{\left(\alpha \right)}\right){\mathbf{U}}^k-c_{n-1}^{\left(\alpha \right)}{\mathbf{U}}^0+\mu \mathbf{F}}\right)$;  21:    Calculate ${\mathbf{U}}^{n+1}=\mathbf{A}\setminus \mathbf{B}$;  22: end for

Example 1. (Smooth solution) In this example, we report on computational experiments for the problem (1a)–(1c), considering a smooth solution [18]. The exact solution is given by $u(x,t)=t^{2}sin\left(2\pi x\right)$ with the initial condition $u(x,0)=0$ and the corresponding source term

$$f(x,t)={\displaystyle \frac{2t^{2-\alpha }sin\left(\pi x\right)}{\mathsf{\Gamma }(3-\alpha )}}+4{\pi }^2{t}^{2}sin\left(2\pi x\right)+2{\pi }^2{t}^{4}sin\left(2\pi x\right)cos\left(2\pi x\right).$$

To verify the accuracy of the numerical solutions, the schemes were tested over the computational domain $\mathsf{\Omega }=(0,1)$ with $T=1$. We also presented a comparative analysis of our numerical results against those obtained using the compact difference schemes proposed in [18,31]. For simplicity, we denote both schemes as Scheme I and Scheme II, respectively. Since Scheme I is nonlinear, we use the standard Newton iteration to solve the resulting algebraic system, terminating the process after a maximum of 1000 iterations.

To assess precision in both spatial and temporal dimensions, sufficiently small step sizes were chosen to ensure that errors were dominated by a single factor. For the spatial test, the time discretization was fixed at $N=5000$, and spatial accuracy was examined for $J=8,16,32$. The results in

Table 1. Comparison of estimated errors and temporal convergence order for $J=128$ and different N.

$\mathbf{\alpha }$	N	Present		Scheme I		Scheme II
		${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$	${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$	${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$
0.3	16	$1.0892\times {10}^{-4}$	-	$4.0668\times {10}^{-5}$	-	$2.2250\times {10}^{-4}$	-
	32	$2.7242\times {10}^{-5}$	2.00	$1.3253\times {10}^{-5}$	1.62	$5.6517\times {10}^{-5}$	1.98
	64	$6.8208\times {10}^{-6}$	2.00	$4.2439\times {10}^{-6}$	1.64	$1.4420\times {10}^{-5}$	1.97
0.5	16	$1.0877\times {10}^{-4}$	-	$1.2101\times {10}^{-4}$	-	$2.5272\times {10}^{-4}$	-
	32	$2.7204\times {10}^{-5}$	2.00	$4.3835\times {10}^{-5}$	1.47	$7.1156\times {10}^{-5}$	1.83
	64	$6.8114\times {10}^{-6}$	2.00	$1.5723\times {10}^{-5}$	1.48	$2.1175\times {10}^{-5}$	1.75
0.7	16	$1.0862\times {10}^{-4}$	-	$3.1006\times {10}^{-4}$	-	$3.8538\times {10}^{-4}$	-
	32	$2.7167\times {10}^{-5}$	2.00	$1.2698\times {10}^{-4}$	1.29	$1.4000\times {10}^{-4}$	1.46
	64	$6.8020\times {10}^{-6}$	2.00	$5.1792\times {10}^{-5}$	1.29	$5.4026\times {10}^{-5}$	1.37

, reported for $\alpha =0.3,0.5,$ and $0.7$, show fourth-order spatial convergence for all schemes, with the proposed linear method producing noticeably smaller errors, especially for larger $\alpha$. For the temporal test, we set $J=128$ and used $N=16,32,64$. As detailed in

Table 2. Comparison of estimated errors and spatial convergence order for $N=5000$ and different J.

$\mathbf{\alpha }$	J	Present		Scheme I		Scheme II
		${\mathrm{Error}}_{\mathit{h}}$	${\mathrm{Rate}}_{\mathit{x}}$	${\mathrm{Error}}_{\mathit{h}}$	${\mathrm{Rate}}_{\mathit{x}}$	${\mathrm{Error}}_{\mathit{h}}$	${\mathrm{Rate}}_{\mathit{x}}$
0.3	16	$8.6389\times {10}^{-5}$	-	$8.6390\times {10}^{-5}$	-	$7.6178\times {10}^{-5}$	-
	32	$5.4332\times {10}^{-6}$	3.99	$5.4344\times {10}^{-6}$	3.99	$4.6986\times {10}^{-6}$	4.02
	64	$3.4080\times {10}^{-7}$	3.99	$3.4234\times {10}^{-7}$	3.99	$2.9601\times {10}^{-7}$	3.99
0.5	16	$8.5954\times {10}^{-5}$	-	$8.5972\times {10}^{-4}$	-	$7.5795\times {10}^{-5}$	4.10
	32	$5.4059\times {10}^{-6}$	3.99	$5.4238\times {10}^{-6}$	3.99	$4.6926\times {10}^{-6}$	4.01
	64	$3.3910\times {10}^{-7}$	3.99	$3.5719\times {10}^{-7}$	3.92	$3.1337\times {10}^{-7}$	3.90
0.7	16	$8.5469\times {10}^{-5}$	-	$8.5615\times {10}^{-5}$	3.97	$7.5507\times {10}^{-5}$	4.10
	32	$5.3755\times {10}^{-6}$	3.99	$5.5200\times {10}^{-6}$	3.96	$4.8103\times {10}^{-6}$	3.97
	64	$3.3719\times {10}^{-7}$	3.99	$4.9241\times {10}^{-7}$	3.49	$4.6092\times {10}^{-7}$	3.38

, our scheme maintains superior accuracy as $\alpha$ increases, achieving the expected second-order temporal convergence, whereas Scheme I and Scheme II attain only $(2-\alpha )$-order accuracy.

To evaluate the performance of our scheme on uniform temporal meshes, we repeated the simulations with the same settings but allowed $\alpha$ to vary from 0 to 1. The graphical results are presented in Figure 1 and Figure 2. Our observations indicate that, in the case of a smooth solution, the scheme demonstrates $\alpha -$robust, implying that the numerical solution remains stable and does not blow-up as $\alpha \to 1^{-}$. However, as shown in Figure 2, the errors associated with Scheme I and Scheme II tend to increase with larger values of $\alpha$ increases.

Example 2.(Non-smooth solution) In this example, we set the exact solution

$$u(x,t)=sin\left(\pi x\right)-{\displaystyle \frac{t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}}sin\left(2\pi x\right)$$

with the initial condition $u_0\left(x\right)=sin\left(\pi x\right)$ and the source term

$$\begin{array}{c}f(x,t)={\pi }^{2}sin\left(\pi x\right)-\left(1+4{\pi }^2{\displaystyle \frac{t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}}\right)sin\left(2\pi x\right)\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+\pi \left(sin\left(\pi x\right)-{\displaystyle \frac{t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}}sin\left(2\pi x\right)\right)\left(cos\left(\pi x\right)-{\displaystyle \frac{2t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}}cos\left(2\pi x\right)\right).\end{array}$$

It can be observed that the solution satisfies the weak singularity assumption (26), with $u_t(x,t)$ behaving like $t^{\alpha -1}$ near $t=0$, so that the first derivative becomes unbounded at the initial time. To assess the performance of the scheme under such conditions, we carried out simulations using the same setup as in Example 1, and the numerical results are reported in

Table 3. Temporal error estimates and convergence order for $J=128$ and different N and $\alpha$.

N	$\mathbf{\alpha }=0.3$		$\mathbf{\alpha }=0.5$		$\mathbf{\alpha }=0.7$
	${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$	${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$	${\mathrm{Error}}_{\tau }$	${\mathrm{Rate}}_{\mathit{t}}$
16	$3.5561\times {10}^{-5}$	-	$2.9999\times {10}^{-5}$	-	$2.2445\times {10}^{-5}$	-
32	$1.2169\times {10}^{-5}$	1.55	$7.7830\times {10}^{-6}$	1.95	$5.3801\times {10}^{-6}$	2.06
64	$4.6244\times {10}^{-6}$	1.40	$2.2273\times {10}^{-6}$	1.81	$1.3296\times {10}^{-6}$	2.02

and

Table 4. Spatial error estimates and convergence order for $N=5000$ and different J and $\alpha$.

J	$\mathbf{\alpha }=0.3$		$\mathbf{\alpha }=0.5$		$\mathbf{\alpha }=0.7$
	${\mathbf{Error}}_{\mathit{h}}$	${\mathbf{Rate}}_{\mathit{x}}$	${\mathbf{Error}}_{\mathit{\tau }}$	${\mathbf{Rate}}_{\mathit{x}}$	${\mathbf{Error}}_{\mathit{h}}$	${\mathbf{Rate}}_{\mathit{t}}$
16	$1.1296\times {10}^{-4}$	-	$1.1488\times {10}^{-4}$	-	$1.1091\times {10}^{-4}$	-
32	$7.1132\times {10}^{-6}$	3.99	$7.2425\times {10}^{-6}$	3.99	$6.9926\times {10}^{-6}$	3.99
64	$4.3458\times {10}^{-7}$	4.03	$4.4986\times {10}^{-6}$	4.01	$4.3619\times {10}^{-6}$	4.00

. These results confirm that the proposed scheme preserves fourth-order spatial accuracy. Furthermore, as shown in Table 3, the temporal convergence rate decreases as $\alpha$ becomes smaller, approaching $(1+\alpha )$-order accuracy, in agreement with Remark 3. Nevertheless, in terms of errors, our proposed method demonstrates a superior resolution highlighting its robustness and effectiveness, even with uniform meshes.

Next, we compared our scheme with Scheme I [18], which uses nonuniform meshes with the optimal grading $r=(2-\alpha )/\alpha$. For $\alpha =0.01,0.1,0.2,$ and $0.8$, the results in

Table 5. Comparison of errors and temporal convergence order for $J=1000$ and $\alpha =0.01,0.1,0.2$ and $0.8$.

	N	Present			Scheme I
		${\mathbf{Error}}_{\mathit{\tau }}$	${\mathbf{Rate}}_{\mathit{t}}$	$\mathbf{CPU}$	${\mathbf{Error}}_{\mathit{\tau }}$	${\mathbf{Rate}}_{\mathit{t}}$	$\mathbf{CPU}$
$\alpha =0.01$	10	$1.0387\times {10}^{-4}$	-	1.0771	$1.0722\times {10}^{-2}$	-	7.8648
	20	$3.5031\times {10}^{-6}$	4.89	1.8721	$1.1185\times {10}^{-2}$	−0.06	13.8930
	40	$1.7490\times {10}^{-6}$	1.00	3.1969	$1.2007\times {10}^{-2}$	−0.10	26.8233
$\alpha =0.1$	10	$7.0648\times {10}^{-5}$	-	1.0772	$8.8567\times {10}^{-5}$	-	7.7838
	20	$2.2609\times {10}^{-5}$	1.64	1.7705	$1.7756\times {10}^{-4}$	−1.00	13.9710
	40	$1.0204\times {10}^{-5}$	1.15	3.1969	$3.1495\times {10}^{-4}$	−0.83	26.7718
$\alpha =0.2$	10	$7.5497\times {10}^{-5}$	-	1.2209	$1.9254\times {10}^{-4}$	-	7.6818
	20	$2.6989\times {10}^{-5}$	1.48	1.9863	$5.6746\times {10}^{-5}$	1.76	14.002
	40	$1.0993\times {10}^{-5}$	1.30	3.8010	$1.7285\times {10}^{-5}$	1.71	27.6309
$\alpha =0.8$	10	$4.4491\times {10}^{-4}$	-	1.2152	$1.4676\times {10}^{-4}$	-	7.2779
	20	$1.0338\times {10}^{-5}$	2.11	1.9719	$5.8370\times {10}^{-5}$	1.33	12.8353
	40	$2.4929\times {10}^{-6}$	2.05	3.8709	$2.4340\times {10}^{-5}$	1.26	22.2970

show that our method consistently yields smaller errors, with the advantage most pronounced for small $\alpha$. The proposed method continues to exhibit a steady decrease in error as N increases, with an observed rate close to $O\left({\tau }^{1+\alpha }\right)$, as expected. In contrast, Scheme I achieves its theoretical $(2-\alpha )$ order only for $\alpha =0.2$ and $0.8$; for $\alpha =0.1$ and $0.01$, the convergence deteriorates, and the reported orders fluctuate or even become negative. This loss of accuracy is due to the grading parameter $r=(2-\alpha )/\alpha$, which becomes very large for small $\alpha$, creating a strongly clustered mesh near $t=0$ and extreme step-size ratios at later times, thereby undermining the expected $O\left({\tau }^{2-\alpha }\right)$ accuracy. Finally, the CPU times in Table 5 indicate that the proposed method significantly outperforms the nonlinear Scheme I, offering a notable reduction in computational cost.

The cases $\alpha =0.2$ and $0.8$ are presented in Figure 3 and Figure 4 as representative examples where both schemes exhibit stable temporal accuracy, allowing a meaningful visual comparison of the error structures produced by the two methods. It can be observed that the distribution errors from Scheme I are smoother compared to those of the present scheme in both cases. This behavior corresponds to the nature of Scheme I, which utilizes non-uniform temporal meshes. Although our scheme is implemented on uniform meshes, it effectively captures the numerical solution at the final time, even though the errors are relatively higher around $t=0$. This demonstrates the capability of our method to compensate for the lower temporal convergence order through superior numerical resolution.

To further evaluate the performance of the proposed scheme in long-time simulations, we extend the simulation to a final time of $T=50$ by using $J=50$ and $N=6000$. The fractional order is also set to be $\alpha =0.2$ and $0.8$ to examine the numerical behavior over an extended period, particularly in terms of stability, accuracy, and error propagation. The numerical results are presented in Figure 5. For $\alpha =0.2$, both schemes produced smooth solutions; however, the numerical solution obtained by Scheme I exhibits unstable when $\alpha =0.8$. Additionally, when $\alpha =0.2$, the $L_2$-norm errors at each time step for both schemes were plotted in Figure 6. Although the singularity of the solution still reduces numerical accuracy within the initial layer near $t=0$ (see Remark 3), the proposed scheme achieves significantly improved resolution once $t_n$ is sufficiently far from zero. This finding supports the preference for the $L1-2$ fourth-order difference scheme on uniform meshes over graded meshes in long-time simulations.

Finally, we consider the homogeneous time-fractional Burgers equation with $f(x,t)=0$ and initial data $u_0\left(x\right)=sin\left(\pi x\right)$. Simulations are carried out for $N=20,40,80,160$ on a fixed spatial grid $J=1000$, testing $\alpha =0.01,0.1,0.2,$ and $0.8$. Since no exact solution is available, the result with $N=320$ serves as the reference. Figure 7 compares the $L_2$-errors of our method and Scheme I. The proposed scheme consistently produces much smaller errors for all $\alpha$. When the grading parameter $r=(2-\alpha )/\alpha$ becomes large (small $\alpha$), Scheme I suffers a clear loss of accuracy, while the present uniform-grid method remains stable and convergent. These results demonstrate that the proposed uniform discretization is a reliable alternative to graded meshes.

5. Concluding Remarks

In this paper, we developed an $L1-2$ fourth-order difference scheme for solving the time-fractional Burgers equation. The proposed scheme combines a novel linearized difference formulation for the nonlinear convection term with the $L1-2$ method on uniform temporal meshes. For sufficiently smooth solutions, the method attains second-order accuracy in time, whereas for solutions with weak singularities the temporal accuracy is reduced. In this case, numerical experiments indicate that the scheme achieves a convergence rate of $O({\tau }^{1+\alpha }+{\tau }^2{t}_n^{\alpha -2})$. The scheme achieves fourth-order spatial accuracy, provided that the time step is sufficiently small to prevent temporal errors from dominating. While graded meshes adaptively refine time steps near $t=0$ to mitigate singularities, their efficiency diminishes in long-time simulations due to excessive early-stage refinement, which leads to higher computational costs. In contrast, the proposed uniform-mesh approach ensures balanced time stepping across the entire interval, maintaining accuracy throughout the simulation. Numerical results further indicate that the absolute error for uniform meshes remain comparable to those obtained with graded meshes. Moreover, long-time simulations highlight the robustness of uniform meshes in maintaining stability and accuracy over extended periods. This suggests that incorporating a preconditioning technique [46] or adopting a hybrid graded–uniform mesh strategy could enhance early-time accuracy while maintaining global computational efficiency.

However, the current analysis primarily addresses regular solutions, and extending the theoretical framework to nonregular cases remains a challenging yet important task, as it would provide a more complete understanding of the scheme’s stability and convergence in practical scenarios. We anticipate that the analytical techniques developed in the recent work [47] could be adapted to this setting, but a comprehensive study is still required, particularly for the $L1-2$ formula. Building upon this foundation, future research will aim to extend the proposed method to higher-dimensional fractional models and to accelerate computations through fast iterative solvers. In addition, sum-of-exponentials-based strategies [48,49,50] and parallel implementation [51,52] will be explored to enhance the efficiency of the time-fractional derivative approximation. These developments will also support the analysis and simulation of more general nonlinear time-fractional equations.