$H^{1}$-norm error analysis of a robust ADI method on graded mesh for three-dimensional subdiffusion problems

Abstract

This work proposes a robust ADI scheme on graded mesh for solving three-dimensional subdiffusion problems. The Caputo fractional derivative is discretized by L1 scheme, where the graded mesh is used to eliminate the weak singular behavior of the exact solution at the initial time $t=0$. The spatial derivatives are approximated by the finite difference method. Based on the improved discrete fractional $Gr\ddot{o}nwall$ inequality, we prove the stability and $\alpha$-robust $H^{1}$-norm convergence, in which the error bound does not blow up when the order of fractional derivative $\alpha \rightarrow 1^{-}$. The 3D numerical examples are proposed to verify the efficiency and accuracy of the ADI method. The CPU time is also provided, which shows the proposed method is very efficient for 3D subdiffusion problems.

1 Introduction

In this article, we consider the following three-dimensional subdiffusion problems

$$\begin{aligned}{} & {} D^{\alpha }_{t}u(x,y,z,t)\!-\!\Delta u(x,y,z,t)\!=\!f(x,y,z,t),\quad \! (x,y,z)\!\in \Upsilon ,\quad \! t\!\in \!(0,T], \end{aligned}$$

(1)

$$\begin{aligned}{} & {} u(x,y,z,t)=\psi (x,y,z,t),\quad (x,y,z)\in \partial \Upsilon ,\quad t\in (0,T], \end{aligned}$$

(2)

$$\begin{aligned}{} & {} u(x,y,z,0)=\phi (x,y,z),\quad (x,y,z)\in \Upsilon , \end{aligned}$$

(3)

in which $\Upsilon =(0,\mathcal {L}_{x})\times (0,\mathcal {L}_{y})\times (0,\mathcal {L}_{z})$, $\partial \Upsilon$ is the boundary, and f, $\psi$, and $\phi$ are smooth functions. $D^{\alpha }_{t}u$ represents the Caputo fractional derivative, which can be written as

$$\begin{aligned} D^{\alpha }_{t}u(\cdot ,t):=\frac{1}{\Gamma (1-\alpha )}\int ^{t}_{0}(t-\xi )^{-\alpha }\frac{\partial u(\cdot ,\xi )}{\partial \xi }d\xi ,\quad \alpha \in (0,1). \end{aligned}$$

In recent years, the high-dimensional subdiffusion problems have been used in a wide variety of application fields due to the nonlocal natures, such as physics, finance, and chemistry [1,2,3,4], and many numerical schemes for solving subdiffusion equations have been proposed [5,6,7,8,9,10,11,12, 26]. Alikhanov [13] constructed a L2-$1_{\sigma }$ formula for time-fractional diffusion equation with variable coefficients. Kopteva et al. [14] considered an initial-boundary value problem of subdiffusion type. Gao et al. [15] constructed finite difference methods for solving a class of time-fractional subdiffusion equations. Zhai et al. [16] constructed the high-precision unconditionally stable finite difference methods for solving time-space fractional diffusion equation. Wu et al. [17] proposed a new high-order finite difference scheme for solving the 2D time-fractional convection-dominated diffusion equation. Du et al. [19] proposed two difference schemes for the multi-dimensional variable-order time-fractional subdiffusion equations. Zeng et al. [20, 21] firstly proposed two fully discrete schemes for the time-fractional subdiffusion equation with space discretized by the finite element method and time discretized by the fractional linear multistep methods. Wang et al. [23] proposed an effective ADI scheme for solving the two-dimensional time-fractional diffusion equation. Balasim et al. [22] developed a new detailed group iterative scheme for 2D time-fractional advection-diffusion equation. Yang et al. [24] constructed a space-time spectral order sinc-collocation method for the fourth-order nonlocal heat model arising in viscoelasticity. Zeng et al. [25] proposed a new alternating direction implicit Galerkin-Legendre spectral method for the two-dimensional Riesz space fractional nonlinear reaction-diffusion equation. Zeng et al. [26] constructed the second-order accurate schemes for the time-fractional subdiffusion equation with unconditional stability. Zhou et al. [27] constructed a fast efficient ADI scheme for 3D nonlocal evolution equation with different weakly singular kernels. Tian et al. [28] proposed an implicit robust difference method for the modified Burgers model with nonlocal dynamic properties. Zhai et al. [29, 30] presented the high-order compact finite difference method for the three-dimensional time-fractional convection-diffusion equation. Roul et al. [31] proposed compact ADI scheme for the two-dimensional time-fractional convection-diffusion equation.

In this paper, we propose a robust ADI scheme for solving three-dimensional subdiffusion problems. We use L1 scheme on graded mesh to approximate the Caputo fractional derivative, where the graded mesh is accustomed to eliminate the weakly singular behavior of the true solution at $t=0$. By using the standard second-order finite difference for the spatial discretization, we obtain a fully discrete scheme. The stability and $\alpha$-robust $H^{1}$-norm convergence analysis are strictly proved. We obtain a robust ADI scheme based on the ADI algorithm and give two numerical examples. The CPU time is also provided. The main contributions of this paper are shown as follows:

• In this paper, an $\alpha$-robust $H^1$-norm convergence analysis for the proposed method based on non-uniform meshes is developed for the three-dimensional subdiffusion equation. Theoretical results prove that the error bound does not blow up when the order of fractional derivative $\alpha \rightarrow 1^{-}$.

• We construct an ADI algorithm to solve the three-dimensional subdiffusion equation, which greatly reduces the computational cost. We compare the difference scheme proposed in this paper with the scheme in [23] in numerical examples. The numerical results verify that our ADI scheme is effective for both 2D and 3D time-fractional subdiffusion problems.

The remainder of this article is organized as follows: In Sect. 2, we construct the ADI scheme for three-dimensional subdiffusion problems. The stability and $\alpha$-robust $H^{1}$-norm convergence analysis are presented in Sect. 3. In Sect. 4, the 3D numerical example is provided. At last, a brief conclusion is provided in Sect. 5.

2 Establishment of the fully discrete ADI scheme

Set $h_{x}=\frac{\mathcal {L}_{x}}{\chi _{1}}$, $h_{y}=\frac{\mathcal {L}_{y}}{\chi _{2}}$, and $h_{z}=\frac{\mathcal {L}_{z}}{\chi _{3}}$ for the positive integers $\chi _{1}$, $\chi _{2}$, and $\chi _{3}$, and $t_{n}=T(n/N)^{\gamma }$ for a positive integer N, $\gamma \ge 1$ is a grading parameter. Set $\tau _{n}=t_{n}-t_{n-1}$ and $\tau =\max \tau _{n}$ for $1\le n\le N$.

Let $\overline{\Upsilon }_{h}=\{(x_{r},y_{j},z_{k})|0\le r\le \chi _{1},0\le j\le \chi _{2},0\le k\le \chi _{3}\},$ and $\Upsilon _{h}=\overline{\Upsilon }_{h}\cap \Upsilon ,$ $\partial \Upsilon _{h}=\Upsilon _{h}\cap \partial \Upsilon$. Denote $u^{n}_{rjk}$ be the approximation solution of (2)–(3) at the mesh point $(x_{r},y_{j},z_{k},t_{n})$.

For $n=1, 2, \cdots , N$, we use the L1 scheme as follows:

$$\begin{aligned} D^{\alpha }_{N}u^{n}_{rjk}&:=\frac{1}{\Gamma (1-\alpha )}\sum ^{n-1}_{p=0}\frac{u^{p+1}_{rjk}-u^{p}_{rjk}}{\tau _{p+1}} \int ^{t_{p+1}}_{t_{p}}(t_{n}-\xi )^{-\alpha }d\xi \nonumber \\&=d_{n,1}u^{n}_{rjk}-d_{n,n}u^{0}_{rjk}-\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})u^{n-p}_{rjk}, \end{aligned}$$

(4)

where

$$\begin{aligned} d_{n,1}=\frac{\tau ^{-\alpha }_{n}}{\Gamma (2-\alpha )},\quad d_{n,p}=\frac{(t_{n}-t_{n-p})^{1-\alpha }-(t_{n}-t_{n-p+1})^{1-\alpha }}{\tau _{n-p+1}\Gamma (2-\alpha )},\quad 1\le p\le n. \end{aligned}$$

Set $\mu _{n}=d^{-1}_{n,1}=\tau ^{\alpha }_{n}\Gamma (2-\alpha )$, the problem (2)–(3) can be approximated by

$$\begin{aligned}{} & {} D^{\alpha }_{N}u^{n}_{rjk}-\Delta _{h} u^{n}_{rjk}=f^{n}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Upsilon _{h}, 1\le n\le N ,\end{aligned}$$

(5)

$$\begin{aligned}{} & {} u^{n}_{rjk}=\psi (x_{r},y_{j},z_{k},t_{n}),\quad (x_{r},y_{j},z_{k})\in \partial \Upsilon _{h}, 1\le n\le N, \end{aligned}$$

(6)

$$\begin{aligned}{} & {} u^{0}_{rjk}=\phi (x_{r},y_{j},z_{k}),\quad (x_{r},y_{j},z_{k})\in \overline{\Upsilon }_{h}, \end{aligned}$$

(7)

where

$$\begin{aligned}&\Delta _{h} u^{n}_{rjk}=\delta ^{2}_{x}u^{n}_{rjk}+\delta ^{2}_{y}u^{n}_{rjk}+\delta ^{2}_{z}u^{n}_{rjk},\quad \delta ^{2}_{x}u^{n}_{rjk}=\frac{u^{n}_{r+1,j,k}-2u^{n}_{rjk}+u^{n}_{r-1,j,k}}{h^{2}_{x}},\\&\delta ^{2}_{y}u^{n}_{rjk}=\frac{u^{n}_{r,j+1,k}-2u^{n}_{rjk}+u^{n}_{r,j-1,k}}{h^{2}_{y}},\quad \delta ^{2}_{z}u^{n}_{rjk}=\frac{u^{n}_{r,j,k+1}-2u^{n}_{rjk}+u^{n}_{r,j,k-1}}{h^{2}_{z}}. \end{aligned}$$

To get ADI scheme, we add a small term $(\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z} -\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})D^{\alpha }_{N}u^{n}_{rjk}$ into (6), which is order $O(N^{-2\alpha })$. Thus, we can obtain the famous ADI scheme:

$$\begin{aligned} (I+\mu ^{2}_{n}{} & {} \delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})D^{\alpha }_{N}u^{n}_{rjk}-\Delta _{h} u^{n}_{rjk}=f^{n}_{rjk}, \nonumber \\{} & {} (x_{r},y_{j},z_{k})\in \Upsilon _{h},\quad 1\le n\le N ,\end{aligned}$$

(8)

$$\begin{aligned}{} & {} u^{n}_{rjk}=\psi (x_{r},y_{j},z_{k},t_{n}),\quad (x_{r},y_{j},z_{k})\in \partial \Upsilon _{h},\quad 1\le n\le N,\end{aligned}$$

(9)

$$\begin{aligned}{} & {} u^{0}_{rjk}=\phi (x_{r},y_{j},z_{k}),\quad (x_{r},y_{j},z_{k})\in \overline{\Upsilon }_{h}. \end{aligned}$$

(10)

By substituting (4) into (8), we have

$$\begin{aligned}&u^{n}_{rjk}\!-\!\mu _{n}\delta ^{2}_{x}u^{n}_{rjk}\!-\mu _{n}\delta ^{2}_{y}u^{n}_{rjk}\!-\mu _{n}\delta ^{2}_{z}u^{n}_{rjk}\!+\mu _{n}^{2}\delta ^{2}_{x}\delta ^{2}_{y}u^{n}_{rjk}\!+\mu _{n}^{2}\delta ^{2}_{x}\delta ^{2}_{z}u^{n}_{rjk}\!+\mu _{n}^{2}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\\ -&\mu _{n}^{3}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}=\mu _{n}(I+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})(d_{n,n}u^{0}_{rjk}+\\&\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})u^{n-p}_{rjk})+\mu _{n}f^{n}_{rjk}. \end{aligned}$$

Define

$$\begin{aligned} F^{n}_{rjk}=&\mu _{n}(I+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})(d_{n,n}u^{0}_{rjk}+\\&\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})u^{n-p}_{rjk}+\mu _{n}f^{n}_{rjk}, \end{aligned}$$

then (8) can be rewritten as the following ADI form:

$$\begin{aligned} (I-\mu _{n}\delta ^{2}_{x})(I-\mu _{n}\delta ^{2}_{y})(I-\mu _{n}\delta ^{2}_{z})u^{n}_{rjk}=F^{n}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Upsilon _{h},\quad 1\le n\le N . \end{aligned}$$

Next, we introduce two intermediate variables to get the solution $\{u^{n}_{rjk}\}$:

$$\begin{aligned}&u^{n-\frac{1}{3}}_{rjk}=(I-\mu _{n}\delta ^{2}_{z})u^{n}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Upsilon _{h},\quad 1\le n\le N,\\&u^{n-\frac{2}{3}}_{rjk}=(I-\mu _{n}\delta ^{2}_{y})u^{n-\frac{1}{3}}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Upsilon _{h},\quad 1\le n\le N, \end{aligned}$$

from which we can solve the problem (8)–(10) by using the following three algorithm steps.

Algorithm 1

The algorithm of solving $\{u^{n-\frac{2}{3}}_{rjk}\}$.

Algorithm 2

The algorithm of solving $\{u^{n-\frac{1}{3}}_{rjk}\}$.

Algorithm 3

The algorithm of solving $\{u^{n}_{rjk}\}$.

3 Stability and convergence analysis

In this section, we shall prove the stability and convergence of the ADI difference scheme (8)–(10) in $H^{1}$-norm sense. For grid function $u=\{u^{n}_{rjk}|0\le r\le \chi _{1}, 0\le j\le \chi _{2}, 0\le k\le \chi _{3}, 0\le n\le N\}$, we denote

$$\begin{aligned}&\Vert u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}-1}_{r=1}h_{y}\sum ^{\chi _{2}-1}_{j=1}h_{z}\sum ^{\chi _{3}-1}_{k=1}(u^{n}_{rjk})^{2},\\&\Vert \delta _{x}u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}}_{r=1}h_{y}\sum ^{\chi _{2}-1}_{j=1}h_{z}\sum ^{\chi _{3}-1}_{k=1}(\delta _{x}u^{n}_{r-\frac{1}{2},j,k})^{2},\\&\Vert \delta _{x}\delta _{y}u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}}_{r=1}h_{y}\sum ^{\chi _{2}}_{j=1}h_{z}\sum ^{\chi _{3}-1}_{k=1}(\delta _{x}\delta _{y}u^{n}_{r-\frac{1}{2},j-\frac{1}{2},k})^{2},\\&\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}}_{r=1}h_{y}\sum ^{\chi _{2}}_{j=1}h_{z}\sum ^{\chi _{3}}_{k=1}(\delta _{x}\delta _{y}\delta _{z}u^{n}_{r-\frac{1}{2},j-\frac{1}{2},k-\frac{1}{2}})^{2},\\&\Vert \delta _{x}\delta _{y}\delta ^{2}_{z}u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}}_{r=1}h_{y}\sum ^{\chi _{2}}_{j=1}h_{z}\sum ^{\chi _{3}-1}_{k=1}(\delta _{x}\delta _{y}\delta ^{2}_{z}u^{n}_{r-\frac{1}{2},j-\frac{1}{2},k})^{2},\\&\Vert \Delta _{h}u^{n}\Vert ^{2}=h_{x}\sum ^{\chi _{1}-1}_{r=1}h_{y}\sum ^{\chi _{2}-1}_{j=1}h_{z}\sum ^{\chi _{3}-1}_{k=1}(\Delta _{h}u^{n}_{rjk})^{2}, \end{aligned}$$

where $\delta _{x}u_{r-\frac{1}{2},j,k}=\frac{1}{h_{x}}(u_{rjk}-u_{r-1,j,k})$. We can analogously define other norms $\Vert \delta _{y}u^{n}\Vert$, $\Vert \delta _{z}u^{n}\Vert$, $\Vert \delta _{y}\delta _{z}u^{n}\Vert$, $\Vert \delta _{x}\delta _{z}u^{n}\Vert$, $\Vert \delta _{x}\delta _{z}\delta ^{2}_{y}u^{n}\Vert$, and $\Vert \delta _{y}\delta _{z}\delta ^{2}_{x}u^{n}\Vert$.

Set $U_{h}=\{u_{rjk}|u_{rjk}=0 \text { if } (x_{r},y_{j},z_{k})\in \partial \Upsilon _{h} \text { and } (x_{r},y_{j},z_{k})\in \overline{\Upsilon }_{h}\}$, for $\forall u$, $w\in U_{h}$, we define

$$\begin{aligned} \Vert \nabla _{h}u^{n}\Vert ^{2}=&\Vert \delta _{x}u^{n}\Vert ^{2}+\Vert \delta _{y}u^{n}\Vert ^{2}+\Vert \delta _{z}u^{n}\Vert ^{2}, \\ \Vert u^{n}\Vert ^{2}_{H^{1}}=&\Vert u^{n}\Vert ^{2}+\Vert \nabla _{h}u^{n}\Vert ^{2},\\ \Vert u^{n}\Vert ^{2}_{A}=&\Vert \nabla _{h}u^{n}\Vert ^{2}+\mu ^{2}_{n}(\Vert \delta _{x}\delta ^{2}_{y}u^{n}\Vert ^{2}+\Vert \delta _{y}\delta ^{2}_{x}u^{n}\Vert ^{2}+\Vert \delta _{x}\delta ^{2}_{z}u^{n}\Vert ^{2} +\Vert \delta _{z}\delta ^{2}_{x}u^{n}\Vert ^{2}\\&+\Vert \delta _{y}\delta ^{2}_{z}u^{n}\Vert ^{2}+\Vert \delta _{z}\delta ^{2}_{y}u^{n}\Vert ^{2}+3\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert ^{2}) +\mu ^{3}_{n}(\Vert \delta _{x}^{2}\delta _{y}\delta _{z}u^{n}\Vert ^{2}\\&+\Vert \delta _{x}\delta _{y}^{2}\delta _{z}u^{n}\Vert ^{2}+\Vert \delta _{x}\delta _{y}\delta _{z}^{2}u^{n}\Vert ^{2}). \end{aligned}$$

3.1 Stability analysis

In this subsection, we first prove a lemma in order to get the stability analysis.

Lemma 3.1

For any grid functions $u,w\in U_{h}$, we have

$$\begin{aligned}&\!-\!h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}\!+\!\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}\!+\!\delta ^{2}_{x}\delta ^{2}_{z}\!+\!\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}\!-\!\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \Delta _{h}w^{n}_{rjk}\nonumber \\ \le&\Vert u^{n}\Vert _{A}\Vert w^{n}\Vert _{A}, \end{aligned}$$

(11)

where the equality holds when $u=w$.

Proof

By utilizing the Cauchy-Schwartz inequality, we get

$$\begin{aligned}{} & {} -h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \delta ^{2}_{x}w^{n}_{rjk}\\= & {} h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(\delta _{x}u^{n}_{r-\frac{1}{2},j,k})(\delta _{x}w^{n}_{r-\frac{1}{2},j,k})+\mu ^{2}_{n}h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\\{} & {} (\delta _{x}\delta _{y}\delta _{z}u^{n}_{r-\frac{1}{2},j-\frac{1}{2},k-\frac{1}{2}})(\delta _{x}\delta _{y}\delta _{z}w^{n}_{r-\frac{1}{2},j-\frac{1}{2},k-\frac{1}{2}})+\mu ^{2}_{n}h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(\delta _{z}\delta ^{2}_{x}u^{n}_{r,j,k-\frac{1}{2}})\\{} & {} (\delta _{z}\delta ^{2}_{x}w^{n}_{r,j,k-\frac{1}{2}})+\mu ^{2}_{n}h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(\delta _{y}\delta ^{2}_{x}u^{n}_{r,j-\frac{1}{2},k})(\delta _{y}\delta ^{2}_{x}w^{n}_{r,j-\frac{1}{2},k})\\{} & {} +\mu ^{3}_{n}h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(\delta ^{2}_{x}\delta _{y}\delta _{z}u^{n}_{r,j-\frac{1}{2},k-\frac{1}{2}})(\delta ^{2}_{x}\delta _{y}\delta _{z}w^{n}_{r,j-\frac{1}{2},k-\frac{1}{2}})\\\le & {} \Vert \delta _{x}u^{n}\Vert \Vert \delta _{x}w^{n}\Vert +\mu ^{2}_{n}(\Vert \delta _{y}\delta ^{2}_{x}u^{n}\Vert \Vert \delta _{y}\delta ^{2}_{x}w^{n}\Vert +\Vert \delta _{z}\delta ^{2}_{x}u^{n}\Vert \Vert \delta _{z}\delta ^{2}_{x}w^{n}\Vert \\{} & {} +\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )+\mu ^{3}_{n}\Vert \delta ^{2}_{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta ^{2}_{x}\delta _{y}\delta _{z}w^{n}\Vert . \end{aligned}$$

In addition, taking the inner product with $\delta ^{2}_{y}w^{n}_{rjk}$ and $\delta ^{2}_{z}w^{n}_{rjk}$, we will give similar results as follows:

$$\begin{aligned}&-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \delta ^{2}_{y}w^{n}_{rjk}\\ \le&\Vert \delta _{y}u^{n}\Vert \Vert \delta _{y}w^{n}\Vert +\mu ^{2}_{n}(\Vert \delta _{x}\delta ^{2}_{y}u^{n}\Vert \Vert \delta _{x}\delta ^{2}_{y}w^{n}\Vert +\Vert \delta _{z}\delta ^{2}_{y}u^{n}\Vert \Vert \delta _{z}\delta ^{2}_{y}w^{n}\Vert +\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )\\&+\mu ^{3}_{n}\Vert \delta ^{2}_{y}\delta _{x}\delta _{z}u^{n}\Vert \Vert \delta ^{2}_{y}\delta _{x}\delta _{z}w^{n}\Vert ,\\&-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \delta ^{2}_{z}w^{n}_{rjk}\\ \le&\Vert \delta _{z}u^{n}\Vert \Vert \delta _{z}w^{n}\Vert +\mu ^{2}_{n}(\Vert \delta _{x}\delta ^{2}_{z}u^{n}\Vert \Vert \delta _{x}\delta ^{2}_{z}w^{n}\Vert +\Vert \delta _{y}\delta ^{2}_{z}u^{n}\Vert \Vert \delta _{y}\delta ^{2}_{z}w^{n}\Vert +\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )\\&+\mu ^{3}_{n}\Vert \delta ^{2}_{z}\delta _{x}\delta _{y}u^{n}\Vert \Vert \delta ^{2}_{z}\delta _{x}\delta _{y}w^{n}\Vert . \end{aligned}$$

By adding the above three inequalities together, and using the Cauchy-Schwartz inequality once again, we get

$$\begin{aligned}&-\!h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}\!+\!\delta ^{2}_{x}\delta ^{2}_{z}\!+\!\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}\!-\!\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \Delta _{h}w^{n}_{rjk}\\ \le&\Vert \delta _{x}u^{n}\Vert \Vert \delta _{x}w^{n}\Vert +\mu ^{2}_{n}(\Vert \delta _{y}\delta ^{2}_{x}u^{n}\Vert \Vert \delta _{y}\delta ^{2}_{x}w^{n}\Vert +\Vert \delta _{z}\delta ^{2}_{x}u^{n}\Vert \Vert \delta _{z}\delta ^{2}_{x}w^{n}\Vert +\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )\\&+\mu ^{3}_{n}\Vert \delta ^{2}_{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta ^{2}_{x}\delta _{y}\delta _{z}w^{n}\Vert \!+\!\Vert \delta _{y}u^{n}\Vert \Vert \delta _{y}w^{n}\Vert \!+\!\mu ^{2}_{n}(\Vert \delta _{x}\delta ^{2}_{y}u^{n}\Vert \Vert \delta _{x}\delta ^{2}_{y}w^{n}\Vert \!+ \!\Vert \delta _{z}\delta ^{2}_{y}u^{n}\Vert \Vert \delta _{z}\delta ^{2}_{y}w^{n}\Vert \\&+\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )+\mu ^{3}_{n}\Vert \delta ^{2}_{y}\delta _{x}\delta _{z}u^{n}\Vert \Vert \delta ^{2}_{y}\delta _{x}\delta _{z}w^{n}\Vert +\Vert \delta _{z}u^{n}\Vert \Vert \delta _{z}w^{n}\Vert +\mu ^{2}_{n}(\Vert \delta _{x}\delta ^{2}_{z}u^{n}\Vert \cdot \\&\Vert \delta _{x}\delta ^{2}_{z}w^{n}\Vert +\Vert \delta _{y}\delta ^{2}_{z}u^{n}\Vert \Vert \delta _{y}\delta ^{2}_{z}w^{n}\Vert +\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert \Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert )+\mu ^{3}_{n}\Vert \delta ^{2}_{z}\delta _{x}\delta _{y}u^{n}\Vert \Vert \delta ^{2}_{z}\delta _{x}\delta _{y}w^{n}\Vert \\ \le&\left[ \Vert \nabla _{h}u^{n}\Vert ^{2}+\mu ^{2}_{n}(\Vert \delta _{y}\delta ^{2}_{x}u^{n}\Vert ^{2}+\Vert \delta _{x}\delta ^{2}_{y}u^{n}\Vert ^{2}+\Vert \delta _{x}\delta ^{2}_{z}u^{n}\Vert ^{2}+\Vert \delta _{z}\delta ^{2}_{x}u^{n}\Vert ^{2}+\Vert \delta _{z}\delta ^{2}_{y}u^{n}\Vert ^{2}\right. \\&\left. +\Vert \delta _{y}\delta ^{2}_{z}u^{n}\Vert ^{2}+3\Vert \delta _{x}\delta _{y}\delta _{z}u^{n}\Vert ^{2} )+\mu ^{3}_{n}(\Vert \delta ^{2}_{x}\delta _{y}\delta _{z}u^{n}\Vert ^{2}+\Vert \delta ^{2}_{y}\delta _{x}\delta _{z}u^{n}\Vert ^{2}+\Vert \delta ^{2}_{z}\delta _{x}\delta _{y}u^{n}\Vert ^{2} )\right] ^{\frac{1}{2}}\\&\left[ \Vert \nabla _{h}w^{n}\Vert ^{2}+\mu ^{2}_{n}(\Vert \delta _{y}\delta ^{2}_{x}w^{n}\Vert ^{2}+\Vert \delta _{x}\delta ^{2}_{y}w^{n}\Vert ^{2}+\Vert \delta _{x}\delta ^{2}_{z}w^{n}\Vert ^{2}+\Vert \delta _{z}\delta ^{2}_{x}w^{n}\Vert ^{2}+\Vert \delta _{z}\delta ^{2}_{y}w^{n}\Vert ^{2}\right. \\&\left. +\Vert \delta _{y}\delta ^{2}_{z}w^{n}\Vert ^{2}+3\Vert \delta _{x}\delta _{y}\delta _{z}w^{n}\Vert ^{2} )+\mu ^{3}_{n}(\Vert \delta ^{2}_{x}\delta _{y}\delta _{z}w^{n}\Vert ^{2}+\Vert \delta ^{2}_{y}\delta _{x}\delta _{z}w^{n}\Vert ^{2}+\Vert \delta ^{2}_{z}\delta _{x}\delta _{y}w^{n}\Vert ^{2} )\right] ^{\frac{1}{2}}\\ \le&\Vert u^{n}\Vert _{A}\Vert w^{n}\Vert _{A}. \end{aligned}$$

The proof is finished. $\square$

Next, by an improved discrete fractional Gr$\ddot{o}$nwall inequality, we yield the following lemmas.

Lemma 3.2

[34, 35] Suppose that the sequences $\{s_{n}\}^{\infty }_{n=1}$, $\{l_{n}\}^{\infty }_{n=1}$ are nonnegative, if the grid function $\{u^{n}:n=0,1,\cdots ,N\}$ satisfies $u_{0}\ge 0$ and for $n=1,2,\cdots ,N$ satisfying $(D^{\alpha }_{N}u^{n})u^{n}\le s^{n}u^{n}+(l^{n})^{2},$ then it holds that

$$\begin{aligned} u^{n}\le u^{0}+\mu _{n}\sum ^{n}_{i=1}\vartheta _{n,i}(s_{i}+l_{i})+\max _{1\le i\le n}\left\{ l^{i}\right\} , \end{aligned}$$

where $\vartheta _{n,n}=1$, $\vartheta _{n,i}=\sum ^{n-i}_{p=1}\frac{1}{d_{n,p}-d_{n,p+1}}\vartheta _{n-p,i}>0$,   $n=1,2,\cdots ,N$,   $i=1,2,\cdots ,n-1$.

Lemma 3.3

[36] For $n=1,\cdots ,N$, we obtain

$$\begin{aligned} \mu _{n}\sum ^{n}_{i=1}\vartheta _{n,i}\le Ct^{\alpha }_{n}. \end{aligned}$$

Theorem 3.4

Suppose $\{u^{n}_{rjk}|0\le r\le \chi _{1},0\le j\le \chi _{2},0\le k\le \chi _{3}, 1\le n\le N\}$ is the solution of discrete problem (8)–(10), where $u^{n}_{rjk}=0$ if $(x_{r},y_{j},z_{k})\in \partial \Psi _{h}$, then we have

$$\begin{aligned} \Vert u^{n}\Vert _{A}\le \Vert u^{0}\Vert _{A}+C(t^{\alpha }_{n}+1)\max _{1\le i\le n}\{\Vert f^{i}\Vert \}. \end{aligned}$$

(12)

Proof

At first, multiplying both sides of (8) by $-h_{x}h_{y}h_{z}\Delta _{h}u^{n}_{rjk}$, and summing over r, j, k for $(x_{r},y_{j},z_{k})\in \Upsilon _{h}$, we get

$$\begin{aligned}&-d_{n,1}h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}\left( u^{n}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{n}_{rjk}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n}_{rjk}\right) \Delta _{h}u^{n}_{rjk}\\&+\Vert \Delta _{h}u^{n}\Vert ^{2}=d_{n,n}[-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(u^{0}_{rjk}+\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{0}_{rjk}-\\&\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{0}_{rjk})\Delta _{h}u^{n}_{rjk}]+\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})[-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}(u^{n-p}_{rjk}+\\&\mu ^{2}_{n}(\delta ^{2}_{x}\delta ^{2}_{y}+\delta ^{2}_{x}\delta ^{2}_{z}+\delta ^{2}_{y}\delta ^{2}_{z})u^{n-p}_{rjk}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z}u^{n-p}_{rjk})\Delta _{h}u^{n}_{rjk}]-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}f^{n}_{rjk}\Delta _{h}u^{n}_{rjk}. \end{aligned}$$

Then, by Lemma 3.1 and using the Cauchy-Schwartz inequality and Young’s inequality, we obtain

$$\begin{aligned} d_{n,1}\Vert u^{n}\Vert ^{2}_{A}&+\Vert \Delta _{h}u^{n}\Vert ^{2}\le d_{n,n}\Vert u^{0}\Vert _{A}\Vert u^{n}\Vert _{A}+\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})\Vert u^{n-p}\Vert _{A}\Vert u^{n}\Vert _{A} \\&+\frac{1}{4}\Vert f^{n}\Vert ^{2}+\Vert \Delta _{h}u^{n}\Vert ^{2}. \end{aligned}$$

Thus, we yield

$$\begin{aligned} d_{n,1}\Vert u^{n}\Vert ^{2}_{A}\le d_{n,n}\Vert u^{0}\Vert _{A}\Vert u^{n}\Vert _{A}+\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})\Vert u^{n-p}\Vert _{A}\Vert u^{n}\Vert _{A}+\frac{1}{4}\Vert f^{n}\Vert ^{2}, \end{aligned}$$

which obtains

$$\begin{aligned} (D^{\alpha }_{N}\Vert u^{n}\Vert _{A})\Vert u^{n}\Vert _{A}\le \frac{1}{4}\Vert f^{n}\Vert ^{2}. \end{aligned}$$

(13)

Finally by applying Lemma 3.2 to (13) and Lemma 3.3, we have

$$\begin{aligned} \Vert u^{n}\Vert _{A}&\le \Vert u^{0}\Vert _{A}+\mu _{n}\sum ^{n}_{i=1}\vartheta _{n,i}\max _{1\le i\le n}\{\Vert f^{i}\Vert \}+\frac{1}{2}\max _{1\le i\le n}\{\Vert f^{i}\Vert \}\\&\le \Vert u^{0}\Vert _{A}+C(t^{\alpha }_{n}+1)\max _{1\le i\le n}\{\Vert f^{i}\Vert \}. \end{aligned}$$

The proof is finished. $\square$

3.2 Convergence analysis

In this subsection, the $H^{1}$-norm convergence of the ADI scheme (8)-(10) will be considered.

Define the notation

$$\begin{aligned} e^{n}_{rjk}:=u(x_{r},y_{j},z_{k},t_{n})-u^{n}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Psi _{h},\quad 0\le n\le N. \end{aligned}$$

Next, for further analysis, several significant lemmas are presented as follows.

Lemma 3.5

[33] Suppose that $|\partial ^{q}_{t}(\cdot ,t)|\le C(1+t^{\alpha -1})$, where $q=0,1,2$. Then for all $(x_{r},y_{j},z_{k},t_{n})\in \Upsilon$, one has

$$\begin{aligned} |D^{\alpha }_{N}u^{n}_{rjk}-D^{\alpha }_{n}u(x_{r},y_{j},z_{k},t_{n})|\le Ct^{-\alpha }_{n}N^{-\min \{\alpha +1,(2-\alpha )/\gamma \}}. \end{aligned}$$

(14)

Lemma 3.6

[36] Set $\mathcal {E}^{n}=N^{-\gamma }t^{\alpha -1}_{n}$ for $1<\gamma <2-\alpha$, $\mathcal {E}^{n}=N^{\alpha -2}t^{\alpha -1}_{n}[1+\ln (t_{n}/t_{1})]$ for $\gamma =2-\alpha$, $\mathcal {E}^{n}=N^{\alpha -2}t^{\alpha -(2-\alpha )/\gamma }_{n}$ for $\gamma >2-\alpha$. Then for $n=1,2,\cdots ,N$, we have

$$\begin{aligned} \mu _{n}\sum ^{n}_{i=1}i^{-\min \{\alpha +1,(2-\alpha )/\gamma \}}\vartheta _{n,i}\le C\mathcal {E}^{n}, \end{aligned}$$

(15)

in which the constant C is $\alpha$-robust.

Denote

$$\begin{aligned}&R_{t}u^{n}_{rjk}\!:=(I+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\!+\!\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}\!+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}\!-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})D^{\alpha }_{N}u^{n}_{rjk}\!-D^{\alpha }_{t}u(x_{r},y_{j},z_{k}),\\&R_{h}u^{n}_{rjk}:=\Delta u(x_{r},y_{j},z_{k},t_{n})-\Delta _{h}u^{n}_{rjk}, \end{aligned}$$

where the small term $(\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})D^{\alpha }_{N}u^{n}_{rjk}$ have the truncation error $O(N^{-2\alpha })$, $\Vert R_{h}u^{n}\Vert =O(h^{2}_{x}+h^{2}_{y}+h^{2}_{z})$.

Then subtracting (2)–(3) from (8)–(10), the error equation are obtained as follows:

$$\begin{aligned}{} & {} (I+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{y}+\mu ^{2}_{n}\delta ^{2}_{x}\delta ^{2}_{z}+\mu ^{2}_{n}\delta ^{2}_{y}\delta ^{2}_{z}-\mu ^{3}_{n}\delta ^{2}_{x}\delta ^{2}_{y}\delta ^{2}_{z})D^{\alpha }_{N}e^{n}_{rjk}-\Delta _{h}e^{n}_{rjk}\nonumber \\ ={} & {} R_{t}u^{n}_{rjk}+R_{h}u^{n}_{rjk},\quad (x_{r},y_{j},z_{k})\in \Upsilon _{h},\quad 1\le n\le N. \end{aligned}$$

(16)

$$\begin{aligned}{} & {} e^{n}_{rjk}=0\quad (x_{r},y_{j},z_{k})\in \partial \Psi _{h}, \quad 1\le n\le N, \end{aligned}$$

(17)

$$\begin{aligned}{} & {} e^{0}_{rjk}=0,\quad (x_{r},y_{j},z_{k})\in \overline{\Upsilon }_{h}. \end{aligned}$$

(18)

Theorem 3.7

Suppose that $|\partial ^{q}_{t}(\cdot ,t)|\le C(1+t^{\alpha -1})$ for $q=0,1,2$. Then there exists a $\alpha$-robust constant C such that

$$\begin{aligned} \Vert u(\cdot ,t_{n})-u^{n}\Vert _{H^{1}}\le C(h^{2}_{x}+h^{2}_{y}+h^{2}_{z}+N^{-\min \{\gamma \alpha ,2-\alpha ,2\alpha \}}), \end{aligned}$$

(19)

in which the constant C does not blow up when $\alpha \rightarrow 1^{-}$.

Proof

Multiplying $-h_{x}h_{y}h_{z}\Delta _{h}e^{n}_{rjk}$ to both sides of (17) and summing over r, j, k for $(x_{r},y_{j},z_{k})\in \Upsilon _{h}$, we can get

$$\begin{aligned} d_{n,1}&\Vert e^{n}\Vert ^{2}_{A}+\Vert \Delta _{h}e^{n}\Vert ^{2}\le d_{n,n}\Vert e^{0}\Vert _{A}\Vert e^{n}\Vert _{A}+\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})\Vert e^{n-p}\Vert _{A}\Vert e^{n}\Vert _{A}\nonumber \\&-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}R_{t}u^{n}_{rjk}\Delta _{h}e^{n}_{rjk}-h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}R_{h}u^{n}_{rjk}\Delta _{h}e^{n}_{rjk}. \end{aligned}$$

(20)

Since

$$\begin{aligned} -h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}R_{t}u^{n}_{rjk}\Delta _{h}e^{n}_{rjk}\le \Vert \nabla _{h}R_{t}u^{n}\Vert \cdot \Vert \nabla _{h}e^{n}\Vert \le \Vert \nabla _{h}R_{t}u^{n}\Vert \cdot \Vert e^{n}\Vert _{A}, \end{aligned}$$

(21)

and

$$\begin{aligned} -h_{x}h_{y}h_{z}\sum ^{\chi _{1}-1}_{r=1}\sum ^{\chi _{2}-1}_{j=1}\sum ^{\chi _{3}-1}_{k=1}R_{h}u^{n}_{rjk}\Delta _{h}e^{n}_{rjk}\le \Vert \Delta _{h}e^{n}\Vert ^{2}+\frac{1}{4}\Vert R_{h}u^{n}\Vert ^{2}. \end{aligned}$$

(22)

Then substituting (21) and (22) into (20), we have

$$\begin{aligned} d_{n,1}\Vert e^{n}\Vert ^{2}_{A}\le d_{n,n}\Vert e^{0}\Vert _{A}\Vert e^{n}\Vert _{A}&+\sum ^{n-1}_{p=1}(d_{n,p}-d_{n,p+1})\Vert e^{n-p}\Vert _{A}\Vert e^{n}\Vert _{A} \\&+\Vert \nabla _{h}R_{t}u^{n}\Vert \cdot \Vert e^{n}\Vert _{A}+\frac{1}{4}\Vert R_{h}u^{n}\Vert ^{2}, \end{aligned}$$

that is,

$$\begin{aligned} (D^{\alpha }_{N}\Vert e^{n}\Vert _{A})\Vert e^{n}\Vert _{A}\le \Vert \nabla _{h}R_{t}u^{n}\Vert \cdot \Vert e^{n}\Vert _{A}+\frac{1}{4}\Vert R_{h}u^{n}\Vert ^{2}. \end{aligned}$$

(23)

From (23), using Lemma 3.2 and noting that $\Vert e^{0}\Vert _{A}=0$, one has

$$\begin{aligned} \Vert e^{n}\Vert _{A}\le C\mu _{n}\sum ^{n}_{i=1}\vartheta _{n,i}(\Vert \nabla _{h}R_{t}u^{i}\Vert +\frac{1}{4}\Vert R_{h}u^{i}\Vert )+\max _{1\le i\le n}\left\{ \frac{1}{4}\Vert R_{h}u^{i}\Vert \right\} . \end{aligned}$$

(24)

Then utilizing Lemmas 3.3, 3.5, and 3.6, one gets

$$\begin{aligned} \Vert e^{n}\Vert _{A}\le&C\mu _{n}\sum ^{n}_{i=1}\vartheta _{n,i}(h^{2}_{x}+h^{2}_{y}+h^{2}_{z}+t^{-\alpha }_{n}N^{-\min \{\alpha +1,(2-\alpha )/\gamma \}}+t^{-\alpha }_{n}N^{-2\alpha })\nonumber \\&+C(h^{2}_{x}+h^{2}_{y}+h^{2}_{z})\nonumber \\ \le&C(h^{2}_{x}+h^{2}_{y}+h^{2}_{z}+N^{-\min \{\gamma \alpha ,2-\alpha ,2\alpha \}}). \end{aligned}$$

(25)

At last, utilizing the definition of $\Vert \cdot \Vert _{A}$ and from [32, Lemma 2.2], we know $\Vert e^{n}\Vert _{H^{1}}\le C\Vert e^{n}\Vert _{A}$. The proof of the theorem is finished. $\square$

Table 1 $H^{1}$-norm errors and the temporal convergence rates for $\chi =128$ with $\gamma =2-\alpha$ for Example 1

4 Numerical experiment

In this section, we conduct two numerical experiments on the proposed ADI difference scheme, which show that the numerical results are in agreement with the theoretical analysis results. Set parameters $\mathcal {L}_{x}=\mathcal {L}_{y}=\mathcal {L}_{z}=\pi$, $\chi _{1}=\chi _{2}=\chi _{3}$, and $T=1$.

Define the $H^1$-norm error $E(\tau ,h)$ of the numerical solutions by

$$\begin{aligned} E_{H^{1}}(\tau ,h)=\sqrt{\Vert e^{n}\Vert ^{2}+\Vert \nabla _{h}e^{n}\Vert ^{2}}. \end{aligned}$$

The spatial and temporal convergence orders are calculated respectively by

$$\begin{aligned} Rate^{x}=\log _{2}\left[ \frac{E_{H^{1}}(\tau ,2h)}{E_{H^{1}}(\tau ,h)}\right] ,\quad Rate^{t}=\log _{2}\left[ \frac{E_{H^{1}}(2\tau ,h)}{E_{H^{1}}(\tau ,h)}\right] . \end{aligned}$$

Example 1

In the first example, we consider the following problem:

$$\begin{aligned}&D^{\alpha }_{t}u(x,y,z,t)-\Delta u(x,y,z,t)=f(x,y,z,t),\quad (x,y,z)\in \Upsilon ,\quad t\in (0,T],\\&u(x,y,z,t)=0,\quad (x,y,z)\in \partial \Upsilon ,\quad t\in (0,T],\\&u(x,y,z,0)=0,\quad (x,y,z)\in \Upsilon , \end{aligned}$$

where the source term is

$$\begin{aligned} f(x,y,z,t)=\Gamma (1+\alpha )\sin x\sin y\sin z+3t^{\alpha }\sin x\sin y\sin z. \end{aligned}$$

The exact solution is $u(x,y,z,t)=t^{\alpha }\sin x\sin y\sin z$.

Table 2 $H^{1}$-norm errors and the temporal convergence rates with $\gamma =2(2-\alpha )$ for Example 1 when $\chi =128$ and $t=T$

Table 3 $H^{1}$-norm errors and the spatial convergence rates for $N=512$ with $\gamma =\frac{2-\alpha }{\alpha }$ for Example 1

Fig. 1

The temporal convergence orders for Example 1

Fig. 2

The numerical solution surface with $\alpha =0.6$, $\chi =128$, and $N=32$ when $\gamma =2-\alpha$ for Example 1

Fig. 3

The exact solution surface with $\alpha =0.6$, $\chi =128$, and $N=32$ when $\gamma =2-\alpha$ for Example 1

Table 1 lists the numerical results about the $H^{1}$-norm errors, the temporal convergence rates, and the CPU time with different $\alpha$ and N when $\gamma =2-\alpha$ and $\chi _{1}=\chi _{2}=\chi _{3}=\chi =128$. We can clearly see that the ADI scheme (8)–(10) obtain $O(N^{-\min \{\gamma \alpha ,2-\alpha ,2\alpha \}})$ accuracy in time as predicted. In Table 2, we give the $H^1$-norm errors and the temporal convergence rates of different small fractional order $\alpha$ for $\gamma =2(2-\alpha ), \chi =128, and t=T$. It can be seen that when $\alpha$ is a small value, the convergence order of $2\alpha$ is basically satisfied. In Table 3, fixing $N=128$, $\gamma =(2-\alpha )/\alpha$ and choosing different $\alpha$, the second-order accuracy can be clearly observed in space. In addition, the CPU times in seconds of Tables 1 and 3 are given. Figure 1 shows the order of temporal convergence. Figures 2 and 3 show the values of the numerical solutions and the exact solutions obtained at different cross sections of the three-dimensional space when $\alpha =0.6$, $\chi =128$, $N=32$, and $\gamma =2-\alpha$, respectively, where the color represents the calculated amplitude.

In Tables 4, 5, and 6, we compare the $H^{1}$-norm errors and the temporal convergence rates of our 3D ADI scheme (8)–(10) with the 2D ADI scheme in [23] for different $\gamma$. Numerical results show that our ADI scheme is effective for both three-dimensional and two-dimensional, and there is no significant increase in CPU time for 3D.

Table 4 Comparison in time with $\gamma =2-\alpha$ and $\chi =128$ for Example 1

Table 5 Comparison in time with $\gamma =2$ and $\chi =128$ for Example 1

Example 2

In this example, we consider the following problem:

$$\begin{aligned}&D^{\alpha }_{t}u(x,y,z,t)-\Delta u(x,y,z,t)=f(x,y,z,t),\quad (x,y,z)\in \Upsilon ,\quad t\in (0,T],\nonumber \\&u(x,y,z,t)=0,\quad (x,y,z)\in \partial \Upsilon ,\quad t\in (0,T],\nonumber \\&u(x,y,z,0)=\sin x\sin y\sin z,\quad (x,y,z)\in \Upsilon ,, \end{aligned}$$

(26)

where the forcing term is

$$\begin{aligned} f(x,y,z,t)=\left[ 3(t^{\alpha }+t^{3})+\Gamma (1+\alpha )+\frac{6t^{3-\alpha }}{\Gamma (4-\alpha )}\right] \sin x\sin y\sin z. \end{aligned}$$

The exact solution u is unknown; we take different $\alpha$ of the numerical solution to verify the feasibility of the algorithm when spacial node number $\chi =128$, temporal node number $N=32$, and the grid mesh $\gamma =2-\alpha$. From Figs. 4 and 5, we can see that the method still works properly in this case.

Table 6 Comparison in time with $\gamma =\frac{2-\alpha }{\alpha }$ and $\chi =128$ for Example 1

Fig. 4

The numerical solution surface with $\alpha =0.6$, $\chi =128$, and $N=32$ when $\gamma =2-\alpha$ for Example 2

Fig. 5

The numerical solution surface with $\alpha =0.8$, $\chi =128$, and $N=32$ when $\gamma =2-\alpha$ for Example 2

5 Conclusion

This study proposes an efficient ADI scheme for three-dimensional subdiffusion problems on graded mesh. We obtain a fully discrete scheme with space discretized by the standard second-order finite difference and the Caputo fractional derivative discretized by L1 scheme. ADI algorithm is used to transform the three-dimensional problem into three one-dimensional problems, which greatly reduces the computational cost. The theoretical analysis shows that our scheme is unconditionally stable with the spatial convergence order of two and the temporal convergence order of $\min \{\gamma \alpha ,2-\alpha ,2\alpha \}$. We give two numerical examples and also compare our present 3D ADI scheme with the 2D ADI scheme in [23]. Numerical results show that our ADI scheme is very effective in solving the three-dimensional subdiffusion problems. In further work, we will consider applying this method to nonlinear problems, such as phase field equations.