Numerical Solution of Time-Fractional Schrödinger Equation by Using FDM

Abstract

In this paper, we first established a high-accuracy difference scheme for the time-fractional Schrödinger equation (TFSE), where the factional term is described in the Caputo derivative. We used the L1-2-3 formula to approximate the Caputo derivative, and the fourth-order compact finite difference scheme is utilized for discretizing the spatial term. The unconditional stability and convergence of the scheme in the maximum norm are proved. Finally, we verified the theoretical result with a numerical test.

1. Introduction

In 1926, the Schrödinger equation was proposed by Schrödinger, who is a physicist from Austria [1], which combines the concept of matter wave with the wave equation to establish a second-order partial differential equation that describes the motion of microscopic particles, and its general form is as follows:

$$i\hslash \frac{\partial u}{\partial t}=-\frac{{\hslash }^2}{2m}\Delta u+Vu.$$

where u is the wave function, ℏ is Planck constant, V is the potential function, m denotes the mass of the particle, and $\Delta$ represents the Laplace operator. In recent years, there have been many studies on the Schrödinger equation [2,3,4,5,6,7,8,9,10]. Researchers have found that fractional differential operators are non-local compared to integer differential operators and are very suitable for describing real-world processes of change with memory as well as hereditary properties. It has become one of the most important tools for describing all kinds of complex mechanical and physical behaviors. In 2004, Naber substituted the time term of the classical Schrödinger equation with the Caputo time-fractional derivative to propose the time-fractional Schrödinger equation (TFSE) [11], which describes the dependence of particle motion.

The TFSE is an integral-differential equation, and since it’s very difficult to find the analytical solution, it has been a widely discussed hot topic to get a numerical solution of the TFSE with a smaller error and higher order. For example, Wei et al. proposed an LDG finite element method to solve the TFSE, which is implicit and fully discrete [12]. Garrappa R. et al. solved the TFSE based on the Krylov projection methods [13]. Liu et al. obtained the approximation solution of the TFSE based on the reproducing kernel theory and collocation method [14]. Zheng et al. presented a spectral collocation method for solving the TFSE [15].

Some L-type formulas have been exploited to replace the Caputo time-fractional term for discretizing the time derivative term and to reap the approximation solution of the TFSE. For example, Eskar, R. et al. used the L1 and L1-2 formulas to discretize the Caputo derivatives, and the compact difference scheme is exploited for the spatial terms to obtain the finite difference scheme [16]. Fei et al. constructed an implicit scheme by adopting the L2-1${}_{\sigma }$ formula to approximate the Caputo term; the weighted and shifted Grünwald formula is used for the spatial term [17]. Cen et al. also adopted the L2-1${}_{\sigma }$ formula on graded meshes for solving the TFKBE with an initial singularity [18]. Ding et al. solved a nonlinear TFSE by using the quintic non-polynomial spline in the spatial term and the L1 formula in the time term [19]. Mokhtari, R. et al. constructed three finite difference schemes by adopting different L-type formulas to approximate the Caputo derivatives in the time direction and the central difference format in the space direction, respectively. The accuracy of the three schemes are $O({\tau }^{2-\alpha }+h^2),O({\tau }^{3-\alpha }+h^2)$, and $O({\tau }^3+h^2)$ [20], where $0⩽\alpha ⩽1$, and $\tau$ (h) is time (spatial) step size. Hadhoud et al. received the approximation solution of the TFSE by using the L1 formula and proved the conditional stability of the technique [21].

In this paper, we use the L1-2-3 formula to approximate the Caputo derivative, and the fourth-order compact difference scheme is exploited to discretize the spatial derivative term for establishing a high-accuracy difference scheme, where the order in the time direction is 3 and the spatial direction is 4. Furthermore, we will prove the scheme is unconditionally stable and convergent in the maximum norm. At the end of the paper, a numerical test is given to prove the theoretical result.

2. Preliminaries

The following TFSE is considered:

$$\begin{array}{cc}\hfill i\frac{{\partial }^{\alpha }u(x,t)}{\partial t^{\alpha }}=\frac{{\partial }^{2}u(x,t)}{\partial x^2}+f(x,t),& x\in \Omega =(0,L),t\in (0,T],\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill u(x,0)=\phi \left(x\right),& x\in \Omega =[0,L],\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill u(0,t)=u(L,t)=\varphi \left(t\right),& t\in [0,T].\hfill \end{array}$$

(3)

where $i=\sqrt{-1}$, $\alpha \in (0,1)$, T and L are positive real numbers, $u_0\left(x\right)$ and $f(x,t)$ are given functions, $\frac{{\partial }^{\alpha }u(x,t)}{\partial t^{\alpha }}$ is the Caputo derivative of order $\alpha \in (0,1)$, which is defined as follows [20]:

$$\frac{{\partial }^{\alpha }u(·,t)}{\partial t^{\alpha }}=\frac{1}{\Gamma (1-\alpha )}{\int }_0^t\frac{u_s(·,s)}{{(t-s)}^{\alpha }}ds.$$

In order to discretize the continuous problem, we first give a dissected grid of the solution region. Let $h=L/M$ and $\tau =T/N$ be the step sizes in the time and space directions, where M and N are two integers. Then $x_j=jh(j=0,1,2,\cdots ,M)$, $t^n=n\tau (n=0,1,2,\cdots ,N)$. Furthermore, we define a mesh that cover the domain $[0,L]\times [0,T]$. Let $\widehat{U}=u_j^n$ is a grid function on the mesh. For any $u,v\in \widehat{U}$, we introduce the following notations:

$${\delta }_x{u}_{j+1/2}^n=\frac{u_{j+1}^n-u_j^n}{h},{\delta }_x{u}_{j-1/2}^n=\frac{u_j^n-u_{j-1}^n}{h},{\delta }_x^2{u}_j^n=\frac{{\delta }_x{u}_{j+1/2}^n-{\delta }_x{u}_{j-1/2}^n}{h},$$

$$(u,v)=h\sum _{j=1}^{M-1}{u}_j{\overline{v}}_j{,\left|\right|u\left|\right|}^2={(u,u),\left|\right|u\left|\right|}_{\infty }=\underset{1⩽j⩽M-1}{max}\left|u_j\right|,$$

$${(u,v)}_1=h\sum _{j=0}^{M-1}\left({\delta }_x{u}_{j+1/2}\right)\left({\delta }_x{\overline{v}}_{j+1/2}\right){,\left|\right|u\left|\right|}_1^2={(u,u)}_1,$$

where the ${\overline{v}}_j$ and ${\overline{v}}_{j+1/2}$ denote the complex-conjugate of $v_j$ and $v_{j+1/2}$.

From the Taylor expansion, we have:

$$\begin{array}{cc}\hfill {\delta }_x^2{u}_j^n& =\frac{1}{h^2}(u_{j-1}^n-2u_j^n+u_{j+1}^n)\hfill \\ \hfill & =\frac{2}{h^2}(\frac{h^2{u}^{″}(x_j,t_n)}{2!}+\frac{h^4{u}^{\left(4\right)}(x_j,t_n)}{4!})+O\left(h^4\right)\hfill \\ \hfill & =(1+\frac{h^2}{12}{\delta }_x^2)u^{″}(x_j,t_n)+O\left(h^4\right),\hfill \end{array}$$

then, we get:

$$\begin{array}{c}\hfill u^{″}(x_j,t_n)=\frac{{\delta }_x^2}{(1+\frac{h^2}{12}{\delta }_x^2)}{u}_j^n+O\left(h^4\right),\end{array}$$

and we define the compact fourth-order difference formula as follow:

$$\begin{array}{c}\hfill H{u}_j^n=(I+\frac{h^2}{12}{\delta }_x^2)u_j^n.\end{array}$$

Definition 1

([22]). (The L1-2-3 formula). Assuming that $\alpha \in (0,1)$ and $u(x,t)\in C^{6,5}$$(\Omega \times [0,T\left]\right)$. We have

$$\begin{array}{cc}\hfill {}_0^C{\mathcal{D}}_t^{\alpha }u(·,t^n)=& \frac{1}{{\tau }^{\alpha }\Gamma (2-\alpha )}\left[d_0{u}^n-\sum _{l=1}^{n-1}(d_{n-l-1}-d_{n-l})u^l-d_{n-1}{u}^0\right],\hfill \end{array}$$

(4)

where $u^n$ and $u^0$ are approximations of $u(·,t^n)$ and $u(·,t^0)$. And for $n=1$,

$$d_0=1,$$

for $n=2$,

$$d_l=\left\{\begin{array}{cc}{a}_l+b_l,\hfill & l=0\hfill \\ a_l-b_{l-1},\hfill & l=1\hfill \end{array}\right.$$

for $n=3$,

$$d_l=\left\{\begin{array}{cc}{a}_l+b_l+g_l,\hfill & l=0\hfill \\ a_l+b_l-b_{l-1}-2g_{l-1},\hfill & l=1\hfill \\ a_l-b_{l-1}+g_{l-2},\hfill & l=2\hfill \end{array}\right.$$

and for $n⩾4$,

$$d_l=\left\{\begin{array}{cc}{a}_l+b_l+g_l,\hfill & l=0\hfill \\ a_l+b_l-b_{l-1}+g_l-2g_{l-1},\hfill & l=1\hfill \\ a_l+b_l-b_{l-1}+g_l-2g_{l-1}+g_{l-2},\hfill & 2⩽l⩽n-3\hfill \\ a_l+b_l-b_{l-1}-2g_{l-1}+g_{l-2},\hfill & l=n-2\hfill \\ a_l-b_{l-1}+g_{l-2},\hfill & l=n-1\hfill \end{array}\right.$$

with

$$\begin{array}{cc}\hfill a_l=& {(l+1)}^{1-\alpha }-l^{1-\alpha },\hfill \end{array}$$

$$\begin{array}{cc}\hfill b_l=& \frac{{(l+1)}^{2-\alpha }-l^{2-\alpha }}{2-\alpha }-\frac{{(l+1)}^{1-\alpha }-l^{1-\alpha }}{2},\hfill \end{array}$$

$$\begin{array}{cc}\hfill g_l=& \frac{{(l+1)}^{3-\alpha }-l^{3-\alpha }}{(2-\alpha )(3-\alpha )}-\frac{{(l+1)}^{1-\alpha }+2l^{1-\alpha }}{6}-\frac{l^{2-\alpha }}{2-\alpha }.\hfill \end{array}$$

Lemma 1

([20]). If $n⩾4$, then we have:

$$d_0>\left|d_1\right|,$$

$$d_0>d_2⩾d_3⩾\cdots ⩾d_{n-1}>0.$$

Lemma 2

([20]). For $d_j(j=0,1,2)$, we have:

$$d_0>1,$$

$$3d_0+2d_1-2d_2>2,$$

$$d_0+d_1-d_2>1/3.$$

Theorem 1

([22]). Let

$${ϵ}_3\left(u(·,t^n)\right)=\frac{{\partial }^{\alpha }u(·,t^n)}{\partial t^{\alpha }}-{}_0^C{\mathcal{D}}_t^{\alpha }u(·,t^n)$$

if $u(x,t)\in C^{6,5}(\Omega \times [0,T])$, then

$$\begin{array}{cc}\hfill |{ϵ}_3\left(u(·,t^1)\right)|& ⩽\frac{\alpha }{2\Gamma (3-\alpha )}{m}_{tt}{\tau }^{2-\alpha },\hfill \\ \hfill |{ϵ}_3\left(u(·,t^2)\right)|& ⩽\frac{\alpha }{3(1-\alpha )(2-\alpha )\Gamma (1-\alpha )}\left(\frac{1}{2}+\frac{1}{3-\alpha }\right)M_{ttt}{\tau }^{3-\alpha }\hfill \\ \hfill & +\frac{\alpha }{12\Gamma (1-\alpha )}{(t^2-t^1)}^{-\alpha -1}{M}_{tt}{\tau }^3,\hfill \\ \hfill |{ϵ}_3\left(u(·,t^n)\right)|& ⩽\frac{12\alpha }{\Gamma (1-\alpha )}{(t^n-t^1)}^{-\alpha -1}{M}_{tt}{\tau }^3+\frac{\alpha }{8\Gamma (1-\alpha )}{(t^n-t^2)}^{-\alpha -1}{M}_{ttt}{\tau }^4\hfill \\ \hfill & +\frac{\alpha }{\Gamma (1-\alpha )}\left(\frac{1}{2}+\frac{1}{12}\frac{27-10\alpha +{\alpha }^2}{{\prod }_{i=1}^4(\alpha -i)}\right)M_{tttt}{\tau }^{4-\alpha },n⩾3\hfill \end{array}$$

where

$$\begin{array}{c}\hfill m_{tt}=\underset{0⩽t⩽t^1}{max}{u}_{tt}(·,t),M_{tt}=\underset{0⩽t⩽t^1}{max}\left|u_{tt}(·,t)\right|,M_{ttt}=\underset{0⩽t⩽t^2}{max}\left|u_{ttt}(·,t)\right|,M_{tttt}=\underset{0⩽t⩽t^n}{max}\left|u_{tttt}(·,t)\right|.\end{array}$$

Lemma 3

([23]). For any $u,v\in \widehat{U}$, we have $({\delta }_x^{2}u,v)=-{(u,v)}_1$.

Lemma 4

([23]). For any $u\in \widehat{U}$, we have ${\left|\right|u\left|\right|}_{\infty }⩽h^{-1/2}\left|\right|u\left|\right|$.

Lemma 5

([24]). For any $u\in \widehat{U}$, we have ${\left|\right|u\left|\right|}_1^2⩽\frac{4}{h^2}{\left|\right|u\left|\right|}^2$.

Lemma 6.

For any $u\in \widehat{U}$, we have $\frac{2}{3}{\left|\right|u\left|\right|}^2⩽(Hu,u)$.

Proof. 

Using Lemma 3 and Lemma 5, we have:

$$\begin{array}{cc}\hfill (Hu,u)& =((I+\frac{h^2}{12}{\delta }_x^2)u,u)=(u,u)+(\frac{h^2}{12}{\delta }_x^{2}u,u)\hfill \\ \hfill & ={\left|\right|u\left|\right|}^2-\frac{h^2}{12}{(u,u)}_1={\left|\right|u\left|\right|}^2-\frac{h^2}{12}{\left|\right|u\left|\right|}_1^2\hfill \\ \hfill & ⩾{\left|\right|u\left|\right|}^2-\frac{1}{3}{\left|\right|u\left|\right|}^2=\frac{2}{3}{\left|\right|u\left|\right|}^2.\hfill \end{array}$$

□

Lemma 7

([25]). Let $\left\{u^n\right\}$ and $\left\{v^n\right\}$ be nonnegative sequences, and c is a nonnegative constant, for all $n⩾1$, if

$$u^n⩽c+\sum _{l=0}^{n-1}{u}^l{v}^l,$$

then,

$$u^n⩽c\prod _{l=0}^{n-1}(1+v^n)⩽cexp\left(\sum _{l=0}^{n-1}{v}^l\right).$$

Lemma 8

([26]). For any $u\in \widehat{U}$, we have $\left|\right|u\left|\right|⩽\frac{L}{\sqrt{6}}{\left|\right|u\left|\right|}_1$.

Lemma 9

([27]). For any $u\in \widehat{U}$, we have $\left(Hu,v\right)=\left(u,Hv\right)$.

Lemma 10.

For any $u\in \widehat{U}$, we have $\left|\right|Hu\left|\right|⩽\frac{4}{3}\left|\right|u\left|\right|$.

Proof. 

Applying the inverse estimate $\left|\right|{\delta }_x^{2}u\left|\right|⩽\frac{4}{h^2}\left|\right|u\left|\right|$, we have:

$$\begin{array}{c}\hfill \left|\right|Hu\left|\right|=\left|\right|u+\frac{h^2}{12}{\delta }_x^{2}u\left|\right|⩽\left|\right|u\left|\right|+\frac{h^2}{12}\left|\right|{\delta }_x^{2}u\left|\right|⩽\left|\right|u\left|\right|+\frac{1}{3}\left|\right|u\left|\right|=\frac{4}{3}\left|\right|u\left|\right|.\end{array}$$

□

3. Analysis of the Method

3.1. Construction of the Difference Scheme

To solve Equation (1), we discretize the time term by using the L1-2-3 formula, and the compact difference scheme is exploited for the spatial term, then we obtain the finite difference scheme as follows:

$$\begin{array}{cc}\hfill i_0^C{\mathcal{D}}_t^{\alpha }{u}_j^n=H^{-1}{\delta }_x^2{u}_j^n+f_j^n,1& ⩽j⩽M-1,1⩽n⩽N\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill u_j^0={\phi }_j,0& ⩽j⩽M\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill u_0^n=u_M^n={\varphi }^n,0& ⩽n⩽N\hfill \end{array}$$

(7)

where $u_j^n$ is an approximation to $u(x_j,t^n)$, and ${\phi }_j=\phi \left(x_j\right)$, ${\varphi }^n=\varphi \left(t^n\right)$, $f_j^n=f(x_j,t^n)$. Since $f_j^n$ has no effect on the discussion of the study that follows, for convenience, we assume $f_j^n=0$.

3.2. Analysis of Stability

In this section, we will analyze the unconditional stability of the scheme (5) that was established in the previous subsection.

Theorem 2.

Difference scheme (5) is unconditionally stable.

Proof. 

For $n=1$, the inner product of Equation (5) and $H{u}^1$ gives:

$$(i_0^C{D}_t^{\alpha }{u}^1,H{u}^1)=(H^{-1}{\delta }_x^2{u}^1,H{u}^1)=({\delta }_x^2{u}^1,u^1).$$

From the Lemma 3, we have:

$$i{d}_0(H{u}^1,u^1)-i{d}_0(H{u}^1,u^0)=-\mu {(u^1,u^1)}_1=-\mu \left|\right|u^1{\left|\right|}_1^2,$$

where $\mu ={\tau }^{\alpha }\Gamma (2-\alpha )$.

According to the Lemma 6 and Cauchy-Schwarz inequality, we can obtain:

$$\frac{2}{3}\left|\right|u^1{\left|\right|}^2⩽\frac{1}{4}\left|\right|H{u}^1{\left|\right|}^2+\left|\right|u^0{\left|\right|}^2.$$

From Lemma 10, here is:

$$\frac{2}{3}\left|\right|u^1{\left|\right|}^2⩽\frac{1}{3}\left|\right|u^1{\left|\right|}^2+\left|\right|u^0{\left|\right|}^2.$$

Eventually, we can get $\left|\right|u^1\left|\right|⩽\sqrt{3}\left|\right|u^0\left|\right|$.

For $n=2$, we can obtain the following equation by inner product of Equation (5) and $H{u}^2$:

$$(i_0^C{D}_t^{\alpha }{u}^2,H{u}^2)=(H^{-1}{\delta }_x^2{u}^2,H{u}^2)=({\delta }_x^2{u}^2,u^2).$$

From Lemma 3, we have:

$$i{d}_0(H{u}^2,u^2)-i(d_0-d_1)(H{u}^2,u^1)-i{d}_1(H{u}^2,u^0)=-\mu \left|\right|u^2{\left|\right|}_1^2.$$

Further, we have:

$$d_0(H{u}^2,u^2)⩽(d_0-d_1)(H{u}^2,u^1)+d_1(H{u}^2,u^0).$$

Using the Lemma 6 and Cauchy-Schwarz inequality, we can obtain:

$$\begin{array}{c}\hfill \frac{2}{3}{d}_0\left|\right|u^2{\left|\right|}^2⩽(d_0-d_1)(\frac{1}{4}\left|\right|H{u}^2{\left|\right|}^2+\left|\right|u^1{\left|\right|}^2)+d_1(\frac{1}{4}\left|\right|H{u}^2{\left|\right|}^2+\left|\right|u^0{\left|\right|}^2).\end{array}$$

From Lemma 10, we can eventually obtain:

$$\left|\right|u^2{\left|\right|}^2⩽\frac{3(d_0-d_1)}{d_0}\left|\right|u^1{\left|\right|}^2+\frac{3d_1}{d_0}\left|\right|u^0{\left|\right|}^2.$$

Then, for $\eta ⩾0$, we now have:

$$\left|\right|u^2{\left|\right|}^2⩽\eta \left|\right|u^0{\left|\right|}^2+\sum _{l=0}^1{v}^l\left|\right|u^l{\left|\right|}^2,$$

in which $v^0=\frac{3d_1}{d_0}$, and $v^1=\frac{3(d_0-d_1)}{d_0}$.

According to Lemma 1, $v^l>0$, then using Lemma 7, we can obtain:

$$\begin{array}{c}\hfill \left|\right|u^2{\left|\right|}^2⩽\eta exp\left(\sum _{l=0}^1{v}^l\right)\left|\right|u^0{\left|\right|}^2=\eta exp\left(3\right)\left|\right|u^0{\left|\right|}^2,\end{array}$$

choosing $\eta ⩽3/exp\left(3\right)$ gives $\left|\right|u^2\left|\right|⩽\sqrt{3}\left|\right|u^0\left|\right|$.

For $n⩾3$, we can obtain the following equation by inner product of Equation (5) and $H{u}^n$:

$$(i_0^C{D}_t^{\alpha }{u}^n,H{u}^n)=(H^{-1}{\delta }_x^2{u}^n,H{u}^n)=({\delta }_x^2{u}^n,u^n).$$

From Lemma 3, we get:

$$i{d}_0(H{u}^n,u^n)-i\sum _{l=1}^{n-1}(d_{n-l-1}-d_{n-l})(H{u}^n,u^l)-i{d}_{n-1}(H{u}^n,u^0)=-\mu \left|\right|u^n{\left|\right|}_1^2.$$

Furthermore, we can obtain:

$$d_0(H{u}^n,u^n)⩽\sum _{l=1}^{n-1}(d_{n-l-1}-d_{n-l})(H{u}^n,u^l)+d_{n-1}(H{u}^n,u^0).$$

Since only $d_1-d_2$ is unknown positive or negative in $d_{n-l-1}-d_{n-l}$, for $l=1,2,\cdots ,$$n-1$, so we discuss it in two cases.

Case1. If $d_2<d_1$, from Lemma 6 and Cauchy-Schwarz inequality:

$$\begin{array}{c}\hfill \frac{2}{3}{d}_0\left|\right|u^n{\left|\right|}^2⩽\sum _{l=1}^{n-1}(d_{n-l-1}-d_{n-l})(\frac{1}{4}\left|\right|H{u}^n{\left|\right|}^2+\left|\right|u^l{\left|\right|}^2)+d_{n-1}(\frac{1}{4}\left|\right|H{u}^n{\left|\right|}^2+\left|\right|u^0{\left|\right|}^2).\end{array}$$

From Lemma 10, we can obtain:

$$\left|\right|u^n{\left|\right|}^2⩽\frac{3{\sum }_{l=1}^{n-1}(d_{n-l-1}-d_{n-l})}{d_0}\left|\right|u^l{\left|\right|}^2+\frac{3d_{n-1}}{d_0}\left|\right|u^0{\left|\right|}^2.$$

Then, for $\eta ⩾0$, we now have:

$$\left|\right|u^n{\left|\right|}^2⩽\eta \left|\right|u^0{\left|\right|}^2+\sum _{l=0}^{n-1}{v}^l\left|\right|u^l{\left|\right|}^2,$$

in which $v^0=\frac{3d_{n-1}}{d_0}$, and $v^l=\frac{3(d_{n-l-1}-d_{n-l})}{d_0}$ for $l=1,2,\cdots ,n-1$. According to Lemma 1, $v^l>0$, then using Lemma 7:

$$\begin{array}{c}\hfill \left|\right|u^n{\left|\right|}^2⩽\eta exp\left(\sum _{l=0}^{n-1}{v}^l\right)\left|\right|u^0{\left|\right|}^2=\eta exp\left(3\right)\left|\right|u^0{\left|\right|}^2,\end{array}$$

choosing $\eta ⩽3/exp\left(3\right)$ gives $\left|\right|u^n\left|\right|⩽\sqrt{3}\left|\right|u^0\left|\right|$.

Eventually, for $n⩾1$, using Lemma 4, we have:

$$\left|\right|u^n{\left|\right|}_{\infty }⩽\sqrt{h}\left|\right|u^n\left|\right|⩽\sqrt{3h}\left|\right|u^0\left|\right|.$$

Case2. If $d_2>d_1$, then we have:

$$\frac{2}{3}{d}_0\left|\right|u^n{\left|\right|}^2⩽\sum _{l=1,l\ne n-2}^{n-1}(d_{n-l-1}-d_{n-l})(H{u}^n,u^l)+(d_2-d_1)(H{u}^n,u^{n-2})+d_{n-1}(H{u}^n,u^0).$$

From Lemma 6 and Cauchy-Schwarz inequality we can obtain:

$$\begin{array}{cc}\hfill \frac{2}{3}{d}_0\left|\right|u^n{\left|\right|}^2⩽& \sum _{l=1,l\ne n-2}^{n-1}(d_{n-l-1}-d_{n-l})(\frac{1}{8}\left|\right|H{u}^n{\left|\right|}^2+2\left|\right|u^l{\left|\right|}^2)+(d_2-d_1)(\frac{1}{8}\left|\right|H{u}^n{\left|\right|}^2+2\left|\right|u^{n-2}{\left|\right|}^2)\hfill \\ & +d_{n-1}(\frac{1}{8}\left|\right|H{u}^n{\left|\right|}^2+2\left|\right|u^0{\left|\right|}^2).\hfill \end{array}$$

Furthermore, using Lemma 10 and Lemma 2, we have:

$$\begin{array}{cc}\hfill \left|\right|u^n{\left|\right|}^2⩽& \frac{12}{3d_0-2d_2+2d_1}\times (\sum _{l=1,l\ne n-2}^{n-1}(d_{n-l-1}-d_{n-l})\left|\right|u^l{\left|\right|}^2+(d_2-d_1)\left|\right|u^{n-2}{\left|\right|}^2+d_{n-1}\left|\right|u^0{\left|\right|}^2)\hfill \\ \hfill ⩽& 6\sum _{l=1,l\ne n-2}^{n-1}(d_{n-l-1}-d_{n-l})\left|\right|u^l{\left|\right|}^2+6(d_2-d_1)\left|\right|u^{n-2}{\left|\right|}^2+6d_{n-1}\left|\right|u^0{\left|\right|}^2.\hfill \end{array}$$

Then, for $\eta ⩾0$, we now have:

$$\left|\right|u^n{\left|\right|}^2⩽\eta \left|\right|u^0{\left|\right|}^2+\sum _{l=0}^{n-1}{v}^l\left|\right|u^l{\left|\right|}^2,$$

in which $v^0=6d_{n-1}$, $v^{n-2}=6(d_2-d_1)$, and $v^l=6(d_{n-l-1}-d_{n-l})$ for $l=1,2,\cdots ,$$n-3,n-1$.

According to Lemma 1, $v^l>0$, then using Lemma 7, we can obtain:

$$\begin{array}{c}\hfill \left|\right|u^n{\left|\right|}^2⩽\eta exp\left(\sum _{l=0}^{n-1}{v}^l\right)\left|\right|u^0{\left|\right|}^2=\eta exp\left(C\right)\left|\right|u^0{\left|\right|}^2,\end{array}$$

where $C=6(d_0-2d_1+2d_2)$, based on Lemma 1, $C>0$. Choosing $\eta ⩽3/exp\left(C\right)$ gives $\left|\right|u^n\left|\right|⩽\sqrt{3}\left|\right|u^0\left|\right|$.

Eventually, for $n⩾1$, using Lemma 4, we have:

$$\left|\right|u^n{\left|\right|}_{\infty }⩽\sqrt{h}\left|\right|u^n\left|\right|⩽\sqrt{3h}\left|\right|u^0\left|\right|.$$

In conclusion, scheme (5) is unconditionally stable. □

3.3. Analysis of Convergence

In the following, we consider the convergence of the difference scheme (5). The error equation holds:

$$\begin{array}{c}\hfill e_j^n=u(x_j,t^n)-u_j^n,\end{array}$$

(8)

where $u(x_j,t^n)$ denotes the exact solution of Equation (1), while $u_j^n$ denotes the numerical solution.

Theorem 3.

Finite difference scheme (5) is always consistent with 3 order accuracy for $n>2$, where $u\in C^{6,5}(\Omega \times [0,T])$.

Proof. 

The local truncation error of the scheme (5) is:

$$\begin{array}{c}\hfill T(x_j,t^n)=i_0^C{D}_t^{\alpha }u(x_j,t^n)-H^{-1}{\delta }_x^{2}u(x_j,t^n)-f(x_j,t^n),\end{array}$$

(9)

using Taylor expansion and Theorem 1, we have:

$$\begin{array}{cc}\hfill T(x_j,t^n)& =i\frac{{\partial }^{\alpha }u(x_j,t^n)}{\partial t^{\alpha }}-\frac{{\partial }^{2}u(x_j,t^n)}{\partial x^2}-i{ϵ}_3\left(u(x_j,t^n)\right)+O\left(h^4\right)\hfill \\ \hfill & =-i{ϵ}_3\left(u(x_j,t^n)\right)+O\left(h^4\right).\hfill \end{array}$$

Let $T_m={max}_{(x,t)\in \Omega \times I}\left|T(x,t)\right|$, then:

$$T_m⩽\left\{\begin{array}{cc}\frac{M_{tt}}{2}{\tau }^{2-\alpha }+O\left(h^4\right),\hfill & t\in \left[0,t^1\right],\hfill \\ \frac{M_{tt}}{40}{\tau }^{2-\alpha }+\frac{M_{tt}}{3}{\tau }^{3-\alpha }+O\left(h^4\right),\hfill & t\in \left(t^1,t^2\right],\hfill \\ \frac{7M_{tt}}{2}{\tau }^3+\frac{M_{tt}}{25}{\tau }^4+\frac{M_{ttt}}{4}{\tau }^{4-\alpha }+O\left(h^4\right),\hfill & t\in \left(t^2,t^n\right].\hfill \end{array}\right.$$

Obviously, for $n>2,T_m=O({\tau }^3+h^4)$. Eventually, we have the following result:

$$\parallel T_j^n\parallel ⩽C_1({\tau }^3+h^4),$$

where $C_1$ is a positive integer. □

Theorem 4.

Finite difference scheme (5) is convergent if $u\in C^{4,4}(\Omega \times I)$.

Proof. 

Subtracting Equation (5) from Equation (9) leads to:

$$\begin{array}{c}\hfill T_j^n=i_0^C{D}_t^{\alpha }{e}_j^n-H^{-1}{\delta }_x^2{e}_j^n.\end{array}$$

(10)

Multiplying H on both sides of Equation (10), we have:

$$\begin{array}{c}\hfill H{T}_j^n=i_0^C{D}_t^{\alpha }H{e}_j^n-{\delta }_x^2{e}_j^n.\end{array}$$

Taking the inner product with respect to $e_j^n$ and fetching the real part, then the following equation holds:

$$\begin{array}{c}\hfill -({\delta }_x^2{e}^n,e^n)=Re(H{T}^n,e^n).\end{array}$$

By Lemma 3 and Lemma 8, we get:

$$\begin{array}{c}\hfill 6\parallel e^n{\parallel }^2⩽L^2\parallel e^n{\parallel }_1^2=L^{2}Re(H{T}^n,e^n)⩽L^2\left|(H{T}^n,e^n)\right|.\end{array}$$

Using Lemma 9, Cauchy-Schwarz inequality and Lemma 10, we can obtain:

$$\begin{array}{c}\hfill 6\parallel e^n{\parallel }^2⩽L^2|\frac{3}{4L^2}\parallel T^n{\parallel }^2+\frac{L^2}{3}\parallel H{e}^n{\parallel }^2|⩽\frac{L^4}{3}\parallel T^n{\parallel }^2+{\parallel e^n\parallel }^2.\end{array}$$

Further, we can get:

$$\begin{array}{c}\hfill \parallel e^n\parallel ⩽\frac{L^2}{\sqrt{15}}\parallel T^n\parallel ⩽\frac{L^2}{\sqrt{15}}{C}_1({\tau }^3+h^4),\end{array}$$

where Theorem 3 used. Eventually, we have:

$$\begin{array}{c}\hfill \parallel e^n\parallel ⩽C\left({\tau }^3+h^4\right),\end{array}$$

where C is a positive integer. Therefore, for $n⩾1$, finite difference scheme (5) is convergent when $u\in C^{6,5}(\Omega \times [0,T]).$ □

4. Numerical Experiment

Furthermore, two numerical examples are given to demonstrate the theoretical analyses of the scheme (5). The following notations will be used when presenting the result,

$$\begin{array}{c}\hfill L^{\infty }-error=\underset{0⩽j⩽M,0⩽n⩽N}{max}\left|e_j^n\right|.\\ \hfill Order=lo{g}_2\left[\frac{L^{\infty }-error(2h,\tau )}{L^{\infty }-error(h,\tau )}\right].\end{array}$$

Example 1.

The one-dimensional TFSE is considered as follows:

$$\begin{array}{cc}\hfill i\frac{{\partial }^{\alpha }u(x,t)}{\partial t^{\alpha }}=\frac{{\partial }^{2}u(x,t)}{\partial x^2}+f(x,t),& x\in \Omega =(0,2),t\in (0,1],\hfill \\ \hfill u(x,0)=0,& x\in [0,2],\hfill \\ \hfill u(0,t)=u(2,t)=0,& t\in [0,1]\hfill \end{array}$$

where $f(x,t)=\frac{2t^{2-\alpha }}{\Gamma (3-\alpha )}(i-1)sin\pi x+(1+i)t^2{\pi }^{2}sin\pi x$, and the exact solution is given by

$$u(x,t)=(1+i)t^{2}sin\pi x.$$

Table 1. Numerical error and convergence order in spatial direction for Example 1.

$\mathit{\alpha }$	h	${\mathit{L}}^{\mathit{\infty }}$-Error	Order
	1/2	0.0395	-
	1/4	2.2777 × 10${}^{-3}$	4.1107
0.1	1/8	1.4014 × 10${}^{-}$${}^4$	4.0268
	1/16	8.7188 × 10${}^{-}$${}^6$	4.0066
	1/32	5.4430 × 10${}^{-}$${}^7$	4.0017
	1/2	0.0393	-
	1/4	2.2777 × 10${}^{-}$${}^3$	4.1104
0.5	1/8	1.3974 × 10${}^{-}$${}^4$	4.0258
	1/16	8.6938 × 10${}^{-}$${}^6$	4.0066
	1/32	5.4273 × 10${}^{-}$${}^7$	4.0017
	1/2	0.0396	-
	1/4	2.2935 × 10${}^{-}$${}^3$	4.1108
0.9	1/8	1.4073 × 10${}^{-}$${}^4$	4.0265
	1/16	8.7792 × 10${}^{-}$${}^6$	4.0027
	1/32	5.7222 × 10${}^{-}$${}^7$	3.9395

and

Table 2. Numerical error and convergence order in spatial direction for Example 2.

$\mathit{\alpha }$	h	${\mathit{L}}^{\mathit{\infty }}$-Error	Order
	$\pi /$2	0.0240	-
	$\pi /$4	1.4026 × 10${}^{-}$${}^3$	4.0953
0.1	$\pi /$8	8.6104 × 10${}^{-}$${}^5$	4.0260
	$\pi /$16	5.3570 × 10${}^{-}$${}^6$	4.0066
	$\pi /$32	3.3509 × 10${}^{-}$${}^7$	3.9990
	$\pi /$2	0.0181	-
	$\pi /$4	1.0620 × 10${}^{-}$${}^3$	4.0891
0.5	$\pi /$8	6.6241 × 10${}^{-}$${}^5$	4.0028
	$\pi /$16	4.1390 × 10${}^{-}$${}^6$	4.0004
	$\pi /$32	2.5174 × 10${}^{-}$${}^7$	4.0393
	$\pi /$2	0.0117	-
	$\pi /$4	6.9124 × 10${}^{-}$${}^4$	4.0856
0.9	$\pi /$8	4.4740 × 10${}^{-}$${}^5$	3.9500
	$\pi /$16	2.7902 × 10${}^{-}$${}^6$	4.0031
	$\pi /$32	1.7359 × 10${}^{-}$${}^7$	4.0066

indicate the maximum norm errors and the convergence orders in spatial direction. When taking different values of $\alpha (0.1,0.5,0.9)$ for $N=2000$; we can know that the order of convergence in spatial direction is 4.

In Figure 1, we show the errors in the maximum norm for time direction attaining the third order of accuracy for $M=2000$ for $\alpha =0.1$ and $\alpha =0.5$.

Figure 2 (Figure 3) represents the real (imaginary) part of the numerical solution and the exact solution for $\alpha =0.7$, $h=1/100$ and $\tau =1/200$; it can be seen that our resulting numerical solution is very close to the exact solution.

Figure 4 gives the absolute modulus error between the numerical and exact solution when $M=20$ and $N=400$ for different $\alpha (0.2,0.8)$, and we can observe that the error is very small.

Example 2.

The one-dimensional TFSE is considered as follows:

$$\begin{array}{cc}\hfill i\frac{{\partial }^{\alpha }u(x,t)}{\partial t^{\alpha }}=\frac{{\partial }^{2}u(x,t)}{\partial x^2}+f(x,t),& x\in \Omega =(0,2\pi ),t\in (0,1],\hfill \\ \hfill u(x,0)=0,& x\in [0,2\pi ],\hfill \\ \hfill u(0,t)=u(2,t)=t^2,& t\in [0,1]\hfill \end{array}$$

where $f(x,t)=-\frac{2t^{2-\alpha }}{\Gamma (3-\alpha )}sinx+t^{2}cosx+i(\frac{2t^{2-\alpha }}{\Gamma (3-\alpha )}cosx+t^{2}sinx)$, and the exact solution is given by

$$u(x,t)=t^2(cosx+isinx).$$

In Figure 5, we show the errors in the maximum norm for time direction attaining the third order of accuracy for $M=2000$ for $\alpha =0.1$ and $\alpha =0.5$.

In Figure 6 (Figure 7), we plot the real (imaginary) part of the numerical solution and the exact solution for $\alpha =0.3$, $h=\pi /100$ and $\tau =1/200$, it can be seen that our resulting numerical solution gives a great approximation of the exact solution.

Figure 8 gives the absolute modulus error between the numerical and exact solution when $M=20$ and $N=400$ for different $\alpha (0.2,0.8)$, and we can observe that the error is very small.

Ref. [16] has used two L-type formulas to approximate the time fractional derivatives to establish two finite difference schemes, and the convergence orders are fourth order accuracy in the spatial direction and $2-\alpha$ and $3-\alpha$ in the temporal direction, respectively. The convergence order in the time direction for two schemes is shown in the

Table 3. The convergence order in time direction [16].

$\mathit{\alpha }$	$\mathit{\tau }$	Example 1		Example 2
		Order(L1)	Order(L1-2)	Order(L1)	Order(L1-2)
0.1	1/10	-	-	-	-
	1/20	1.768	3.042	1.764	3.010
	1/40	1.787	3.021	1.784	2.981
	1/80	1.802	3.010	1.800	2.965
0.5	1/10	-	-	-	-
	1/20	1.472	2.872	1.448	2.553
	1/40	1.480	2.699	1.468	2.524
	1/80	1.486	2.547	1.478	2.511
0.9	1/10	-	-	-	-
	1/20	1.089	1.397	1.051	2.086
	1/40	1.136	1.995	1.074	2.124
	1/80	1.157	2.162	1.087	2.100

:

By following Figure 1 and Figure 5, we can know that with our method, we can achieve third order accuracy in the time direction, which is higher than [16].

5. Conclusions

In this paper, we first proposed a time-fractional Schrödinger equation with the Caputo time-fractional derivative of order $\alpha \in (0,1)$ for constructing the finite difference scheme to obtain the approximation solution of the equation; we approximated the Caputo derivative using the L1-2-3 formula to discretize the time term, and the spatial term is discretized by the fourth-order compact difference scheme; we then analyzed the unconditional stability of the scheme and also proved that the scheme is convergent in the maximum norm with an accuracy of $O({\tau }^3+h^4)$. At the end of this article, we give a numerical example to verify the theoretical result.