r-Modified Crank-Nicholson difference scheme for fractional parabolic PDE

Abstract

The second order of accuracy stable difference scheme for the numerical solution of the mixed problem for the fractional parabolic equation are presented using by r-modified Crank-Nicholson difference scheme. Stability estimate for the solution of this difference scheme is obtained. A procedure of modified Gauss elimination method is used for solving this difference scheme in the case of one-dimensional fractional parabolic partial differential equations. Numerical results for this scheme and the Crank-Nicholson scheme are compared in test examples.

1 Introduction

At present, there is a huge number of theoretical and applied works devoted to the study of fractional differential equations. Solutions of various problems for fractional differential equations can be found, for example, in the monographs of Podlubny [1], Kilbas, Srivastava, and Trujillo [2], Diethelm [3], and in [4–11]. These problems were studied in various directions: qualitative properties of solutions, spectral problems, various statements of boundary value problems, and numerical investigations.

Many problems in fluid flow, dynamical and diffusion processes, control theory, mechanics, and other areas of physics can be reduced fractional partial differential equations.

In [12] the simple connection of fractional derivatives with fractional powers of first order differential operator was presented. This approach is important to obtain the formula for the fractional difference derivative. Presently, many mathematicians apply this approach and operator tools to investigate various problems for fractional partial differential equations which appear in applied problems (see, e.g., [13–20] and the references therein).

In previous paper [17] authors investigated stability estimates for Crank-Nicholson schemes for the Dirichlet problem for the fractional parabolic equation

$$\{\begin{array}{l}\frac{\partial u(t,x)}{\partial t}+D_t^{1/2}u(t,x)-{\sum }_{p=1}^m{(a_p(x)u_{x_p})}_{x_p}+\sigma u(t,x)=f(t,x),\\ x=(x_1,\dots ,x_m)\in \mathrm{\Omega },0<t<T,\\ u(t,x)=0,x\in S,\\ u(0,x)=0,x\in \overline{\mathrm{\Omega }}.\end{array}$$

(1.1)

Here $D_t^{1/2}=D_{0+}^{1/2}$ is the standard Riemann-Liouville’s derivative of order $1/2$ and Ω is the open cube in the m-dimensional Euclidean space

$${\mathbb{R}}^m:\{x\in \mathrm{\Omega }:x=(x_1,\dots ,x_m);0<x_p<1,1\le p\le m\}$$

with boundary S, $\overline{\mathrm{\Omega }}=\mathrm{\Omega }\cup S$, $a_p(x)$ ($x\in \mathrm{\Omega }$) and $f(t,x)$ ($t\in (0,T)$, $x\in \mathrm{\Omega }$) are given smooth functions and $a_p(x)\ge a>0$, $\sigma >0$.

In [18] the authors investigated stability estimates for Crank-Nicholson schemes for the Neumann problem for the fractional parabolic equation

$$\{\begin{array}{l}\frac{\partial u(t,x)}{\partial t}+D_t^{1/2}u(t,x)-{\sum }_{p=1}^m{(a_p(x)u_{x_p})}_{x_p}+\sigma u(t,x)=f(t,x),\\ x=(x_1,\dots ,x_m)\in \mathrm{\Omega },0<t<T,\\ u(t,x)=0,x\in S,\\ \frac{\partial u(t,x)}{\partial \overrightarrow{n}}=0,x\in \overline{\mathrm{\Omega }}.\end{array}$$

(1.2)

The role played by stability inequalities (well posedness) in the study of boundary-value problems for parabolic partial differential equations is well known (see, e.g., [21–26]).

In the present paper, we consider an r-modified Crank-Nicholson difference scheme of the above mentioned two problems (1.1), (1.2). This r-modified scheme is of the second order of accuracy in t and in space variables difference schemes for the approximate solution of problems. The stability estimate for the solution of this difference scheme is established. We use a procedure of a modified Gauss elimination method for solving this difference scheme in the case of one-dimensional fractional parabolic partial differential equations.

2 Stability of difference scheme

Let us define the grid space

$$\{\begin{array}{l}{\overline{\mathrm{\Omega }}}_h=\{x=x_p=(h_1{p}_1,\dots ,h_m{p}_m),p=(p_1,\dots ,p_m),\\ \phantom{{\overline{\mathrm{\Omega }}}_h=}0\le p_j\le M_j,h_j{M}_j=1,j=1,\dots ,m\},\\ {\mathrm{\Omega }}_h={\overline{\mathrm{\Omega }}}_h\cap \mathrm{\Omega },S_h={\overline{\mathrm{\Omega }}}_h\cap S.\end{array}$$

We introduce the Hilbert space $L_{2h}=L_2({\overline{\mathrm{\Omega }}}_h)$ of the grid function ${\phi }^h(x)=\{\phi (h_1{j}_1,\dots ,h_m{j}_m)\}$ defined on $\overline{\mathrm{\Omega }}$, equipped with the norm

$${\parallel {\phi }^h\parallel }_{L_2({\overline{\mathrm{\Omega }}}_h)}={\left(\sum _{x\in {\overline{\mathrm{\Omega }}}_h}\right|{\phi }^h(x){|}^2{h}_1\cdots h_m)}^{1/2}.$$

To the differential operator $A^x$ generated by problem (1.1) or (1.2), respectively, we assign the difference operator $A_h^x$ by the formula

$$A_h^x{u}^h=-\sum _{p=1}^m{(a_p(x)u_{{\overline{x}}_p}^h)}_{x_p,j_p}+\sigma u^h$$

(2.1)

acting in the space of grid functions $u^h(x)$, satisfying the conditions $u^h(x)=0$ or $\frac{\partial u(t,x)}{\partial \overrightarrow{n}}=0$ ($\mathrm{\forall }x\in S_h$). It is known that $A_h^x$ is a self-adjoint positive definite operator in $L_2({\overline{\mathrm{\Omega }}}_h)$. Here,

$$\begin{array}{c}{\phi }_{x_p,j_p}=\frac{1}{h_p}(\phi (h_1{j}_1,\dots ,h_j(j_j+1),\dots ,h_m{j}_m)-\phi (h_1{j}_1,\dots ,h_j{j}_j,\dots ,h_m{j}_m)),\hfill \\ {\phi }_{{\overline{x}}_p,j_p}=\frac{1}{h_p}(\phi (h_1{j}_1,\dots ,h_j{j}_j,\dots ,h_m{j}_m)-\phi (h_1{j}_1,\dots ,h_j(j_j-1),\dots ,h_m{j}_m)).\hfill \end{array}$$

With the help of $A_h^x$, we arrive at the initial boundary value problem

$$\{\begin{array}{l}\frac{d{v}^h(t,x)}{dt}+D_t^{1/2}{v}^h(t,x)+A_h^x{v}^h(t,x)=f^h(t,x),0<t<T,x\in {\mathrm{\Omega }}_h,\\ v^h(0,x)=0,x\in \overline{\mathrm{\Omega }}\end{array}$$

(2.2)

for a finite system of ordinary fractional differential equations.

We denote

$$d=\frac{2}{\sqrt{\pi \tau }},\lambda (q)=\sqrt{q+1/2}-\sqrt{q-1/2},\mu (q)=-\frac{1}{3}({(q+1/2)}^{3/2}-{(q-1/2)}^{3/2}).$$

Applying the second order of the approximation formula

$$D_{t_{k-\frac{\tau }{2}}}^{1/2}{u}_k=d\sum _{i=0}^k{\alpha }_{k,i}{u}_i$$

(2.3)

for

$$D_{t_{k-\frac{\tau }{2}}}^{1/2}u(t_k-\tau /2)=\frac{1}{\mathrm{\Gamma }(1/2)}{\int }_0^{t_k-\tau /2}{(t_k-\tau /2-s)}^{-1/2}{u}^{\prime }(s)ds$$

(see [16]) and the Crank-Nicholson difference scheme for parabolic equations, one can present the second order of accuracy difference scheme with respect to t and to x. Here,

$$\begin{array}{c}{\alpha }_{1,0}=-\sqrt{2}/3,{\alpha }_{1,1}=\sqrt{2}/3,k=1,\hfill \\ {\alpha }_{2,0}=-2\sqrt{6}/5,{\alpha }_{2,1}=\sqrt{6}/5,{\alpha }_{2,2}=\sqrt{6}/5,k=2,\hfill \\ {\alpha }_{k,0}=(k-2)\lambda (k-2)+\mu (k-2),\hfill \\ {\alpha }_{k,1}=(k-3)\lambda (k-3)+\mu (k-3)+(3-2k)\lambda (k-2)-2\mu (k-2),\hfill \\ {\alpha }_{k,i}=(k-i-2)\lambda (k-i-2)+\mu (k-i-2)\hfill \\ \phantom{{\alpha }_{k,i}=}+(2i-2k+1)\lambda (k-i-1)-2\mu (k-i-1)\hfill \\ \phantom{{\alpha }_{k,i}=}+(k-i+1)\lambda (k-i)+\mu (k-i),2\le i\le k-3,\hfill \\ {\alpha }_{k,k-2}=-3\lambda (1)-2\mu (1)+3\lambda (2)+\mu (2)-\frac{1}{6\sqrt{2}},\hfill \\ {\alpha }_{k,k-1}=2\lambda (1)+\mu (1)-\frac{\sqrt{2}}{3},3\le k\le N.\hfill \end{array}$$

Now, we introduce the second order accuracy r-modified Crank-Nicholson difference scheme in the following form:

$$\{\begin{array}{l}\frac{u_k^h(x)-u_{k-1}^h(x)}{\tau }+D_{t_k}^{1/2}{u}_k^h(x)+A_h^x{u}_k^h(x)=f_k^h(x),x\in {\overline{\mathrm{\Omega }}}_h,1\le k\le r,\\ \frac{u_k^h(x)-u_{k-1}^h(x)}{\tau }+D_{t_k}^{1/2}{u}_k^h(x)+\frac{1}{2}{A}_h^x[u_k^h(x)+u_{k-1}^h(x)]\\ =f_k^h(x),x\in {\overline{\mathrm{\Omega }}}_h,r+1\le k\le N,\\ f_k^h(x)=f(t_k-\frac{\tau }{2},x),N\tau =T,t_k=k\tau ,1\le k\le N,\\ u_0^h(x)=0,x\in {\overline{\mathrm{\Omega }}}_h\end{array}$$

(2.4)

for the approximate solution of problem (2.2).

Theorem 2.1 Let τ and $|h|$ be sufficiently small positive numbers. Then the solutions of the difference scheme (2.4) satisfy the following stability estimate:

$$\underset{1\le k\le N}{max}{\parallel u_k^h\parallel }_{L_{2h}}\le C\underset{1\le k\le N}{max}{\parallel f_k^h\parallel }_{L_{2h}},$$

(2.5)

where C does not depend on τ, h, and $f_k^h$, $1\le k\le N$.

Proof Consider the difference scheme (2.4). We have

$$u_k^h(x)=\{\begin{array}{ll}{\sum }_{s=1}^k{R}_1^{k-s+1}\tau [f_s^h(x)-D_{t_s}^{1/2}{u}_s^h(x)],& 1\le k\le r,\\ {\sum }_{s=1}^{k-r}{B}^{k-r}{R}_1^{r-s+1}\tau [f_s^h(x)-D_{t_s}^{1/2}{u}_s^h(x)]\\ +{\sum }_{l=1}^{k-r}{B}^{k-r-l}R\tau [f_{r+l}^h(x)-D_{t_{r+l}}^{1/2}{u}_{r+l}^h(x)],& r+1\le k\le N,\end{array}$$

(2.6)

where

$$\begin{array}{c}{R}_1^{-1}=(I+\tau A_h^x),\hfill \\ R^{-1}=(I+\frac{\tau }{2}{A}_h^x),\hfill \\ B=R(I-\frac{\tau }{2}{A}_h^x).\hfill \end{array}$$

We obtain

$$\underset{1\le k\le N}{max}{\parallel D_{t_k}^{1/2}{u}_k^h\parallel }_{L_{2h}}\le M\underset{1\le k\le N}{max}{\parallel f_k^h\parallel }_{L_{2h}}.$$

(2.7)

Let us write $z_k={\parallel D_{t_k}^{1/2}{u}_k^h\parallel }_{L_{2h}}$. Using (2.6), we have

$$D_{t_k}^{1/2}{u}_k^h(x)=\{\begin{array}{ll}d{\sum }_{i=0}^k{\alpha }_{k,i}[{\sum }_{s=1}^i{R}_1^{i-s+1}\tau (f_s^h(x)-D_{t_s}^{1/2}{u}_s^h(x))],& 1\le k\le r,\\ d{\sum }_{i=0}^k{\alpha }_{k,i}[{\sum }_{s=1}^{i-r}{B}^{i-r}{R}_1^{r-s+1}\tau (f_s^h(x)-D_{t_s}^{1/2}{u}_s^h(x))\\ +{\sum }_{l=1}^{i-r}{B}^{i-r-l}R\tau (f_{r+l}^h(x)-D_{t_{r+l}}^{1/2}{u}_{r+l}^h(x))],& r+1\le k\le N.\end{array}$$

Now, let us estimate $z_k={\parallel D_{t_k}^{1/2}{u}_k^h\parallel }_{L_{2h}}$, $1\le k\le N$. From the triangle inequality, it follows that

$$\begin{array}{rcl}{z}_1& \le & {\parallel {\alpha }_{1,1}{R}_1\parallel }_{L_{2h}\to L_{2h}}({\parallel f_1^h(x)\parallel }_{L_{2h}}+{\parallel D_{\tau }^{1/2}{u}_1\parallel }_{L_{2h}})\sqrt{\tau }\\ \le & M_1\sqrt{\tau }({\parallel f_1^h(x)\parallel }_{L_{2h}}+z_1).\end{array}$$

(2.8)

Applying the triangle inequality and the estimates [24]

$$\begin{array}{c}{\parallel R_1^k\parallel }_{L_{2h}\to L_{2h}}\le \frac{M}{k\tau },\hfill \\ {\parallel B^{i-r}{R}_1^{r-s+1}\parallel }_{L_{2h}\to L_{2h}}\le \frac{M}{k\tau },\hfill \\ {\parallel B^{i-r-l}R\parallel }_{L_{2h}\to L_{2h}}\le M,1\le k\le N,\hfill \end{array}$$

(2.9)

we have

$$\begin{array}{c}{z}_k\le \sum _{i=0}^k\sum _{s=1}^i{\parallel {\alpha }_{k,i}{R}_1^{i-s+1}\parallel }_{L_{2h}\to L_{2h}}({\parallel f_s^h(x)\parallel }_{L_{2h}}+{\parallel D_{t_s}^{1/2}{u}_s^h(x)\parallel }_{L_{2h}})\sqrt{\tau }\hfill \\ \phantom{z_k}\le M_2\sum _{s=1}^{k-1}\left[\frac{1}{\sqrt{(k-s)\tau }}\tau ({\parallel f_s^h(x)\parallel }_{L_{2h}}+z_s)\right]\hfill \\ \phantom{z_k\le }+M_3({\parallel f_k^h(x)\parallel }_{L_{2h}}+z_k)\sqrt{\tau },2\le k\le r,\hfill \end{array}$$

(2.10)

$$\begin{array}{c}{z}_k\le \sum _{i=0}^k\sum _{s=1}^{i-r}{\parallel {\alpha }_{k,i}{B}^{i-r}{R}_1^{r-s+1}\parallel }_{L_{2h}\to L_{2h}}({\parallel f_s^h(x)\parallel }_{L_{2h}}+{\parallel D_{t_s}^{1/2}{u}_s^h(x)\parallel }_{L_{2h}})\sqrt{\tau }\hfill \\ \phantom{z_k\le }+\sum _{i=0}^k\sum _{l=1}^{i-r}{\parallel {\alpha }_{k,i}{B}^{i-r-l}R\parallel }_{L_{2h}\to L_{2h}}({\parallel f_{r+l}^h(x)\parallel }_{L_{2h}}+{\parallel D_{t_{r+l}}^{1/2}{u}_{r+l}^h(x)\parallel }_{L_{2h}})\sqrt{\tau }\hfill \\ \phantom{z_k}\le M_2^{\prime }\sum _{s=1}^{k-1}\left[\frac{1}{\sqrt{(k-s)\tau }}\tau ({\parallel f_s^h(x)\parallel }_{L_{2h}}+z_s)\right]\hfill \\ \phantom{z_k\le }+M_3^{\prime }({\parallel f_k^h(x)\parallel }_{L_{2h}}+z_k)\sqrt{\tau }\hfill \\ \phantom{z_k\le }+M_2^{\prime \prime }\sum _{l=1}^{k-1}\left[\frac{1}{\sqrt{(k-s)\tau }}\tau ({\parallel f_{r+l}^h(x)\parallel }_{L_{2h}}+z_{r+l})\right]\hfill \\ \phantom{z_k\le }+M_3^{\prime \prime }({\parallel f_k^h(x)\parallel }_{L_{2h}}+z_k)\sqrt{\tau },r+1\le k\le N.\hfill \end{array}$$

(2.11)

Hence, applying the difference analog of the integral inequality and inequalities (2.8), (2.10), and (2.11), we get

$$\begin{array}{rcl}{\parallel {\{z_k\}}_1^N\parallel }_{L_{2h}}& =& {\parallel {\left\{D_{\tau }^{\frac{1}{2}}{u}_k\right\}}_1^N\parallel }_{L_{2h}}\\ \le & M{\parallel f^{\tau }\parallel }_{L_{2h}}.\end{array}$$

(2.12)

The proof of estimate (2.5) for the solution of (2.4) follows from (2.6), (2.9), and (2.12). Note that $M_{\ast }$, $M_0$ are independent from τ, h, and $f_k^h$, $1\le k\le N$. Theorem 2.1 is proved. □

3 Numerical analysis

We consider two examples for numerical results.

Example 3.1 We consider the following initial boundary value problem with Dirichlet condition for the one-dimensional fractional parabolic partial differential equation:

$$\{\begin{array}{l}\frac{\partial u(t,x)}{\partial t}+D_t^{1/2}u(t,x)-\frac{\partial }{\partial x}((1+x)\frac{\partial u(t,x)}{\partial x})=f(t,x),\\ f(t,x)=[\frac{3\sqrt{t}}{2}+\frac{3\sqrt{\pi }t}{4}+(1+x){\pi }^2{t}^{3/2}]sin\pi x\\ \phantom{f(t,x)=}-\pi t^{3/2}cos\pi x,0<t<1,0<x<1,\\ u(t,0)=u(t,1)=0,0\le t\le 1,\\ u(0,x)=0,0\le x\le 1.\end{array}$$

(3.1)

It is clear that the exact solution of problem (3.1) is

$$u(t,x)=t^{3/2}sin\pi x.$$

Applying the r-modified Crank-Nicholson difference scheme (2.4), we get

$$\{\begin{array}{l}\frac{u_n^k-u_n^{k-1}}{\tau }+D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k-[(1+x_n)\frac{u_{n+1}^k-2u_n^k+u_{n-1}^k}{h^2}+\frac{u_{n+1}^k-u_{n-1}^k}{2h}]={\phi }_n^k,1\le k\le r,\\ \frac{u_n^k-u_n^{k-1}}{\tau }+D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k-\frac{1}{2}[(1+x_n)\frac{u_{n+1}^k-2u_n^k+u_{n-1}^k}{h^2}+\frac{u_{n+1}^k-u_{n-1}^k}{2h}\\ +(1+x_n)\frac{u_{n+1}^{k-1}-2u_n^{k-1}+u_{n-1}^{k-1}}{h^2}+\frac{u_{n+1}^{k-1}-u_{n-1}^{k-1}}{2h}]={\phi }_n^k,r+1\le k\le N,\\ {\phi }_n^k=f(t_k-\frac{\tau }{2},x_n),t_k=k\tau ,x_n=nh,1\le k\le N,1\le n\le M-1,\\ u_0^k=u_M^k=0,0\le k\le N,\\ u_n^0=0,0\le n\le M,\end{array}$$

where $D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k$ is defined by (2.3) for any n, $1\le n\le M-1$. We can rewrite it in the system of equations with matrix coefficients

$$\{\begin{array}{l}A{U}_{n+1}+B{U}_n+C{U}_{n-1}=D{\phi }_n,1\le n\le M-1,\\ U_0=\tilde{0},\\ U_M=\tilde{0}.\end{array}$$

(3.2)

Here and in the sequel $\tilde{0}$ is the $(N+1)\times 1$ zero matrix,

$$A={\left[\begin{array}{cccccccccccc}0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ 0& a_n& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ 0& 0& a_n& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & a_n& 0& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& a_n& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& z_n& z_n& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& 0& z_n& z_n& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & z_n& z_n& 0\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& z_n& z_n\end{array}\right]}_{(N+1)\times (N+1)},$$

(3.3)

$$B=B_1+B_2,$$

(3.4)

$$\begin{array}{c}{B}_1={\left[\begin{array}{cccccc}{b}_{11}& 0& 0& \cdots & 0& 0\\ b_{21}& b_{22}& 0& \cdots & 0& 0\\ b_{31}& b_{32}& b_{33}& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ b_{N,1}& b_{N,2}& b_{N,3}& \cdots & b_{N,N}& 0\\ b_{N+1,1}& b_{N+1,2}& b_{N+1,3}& \cdots & b_{N+1,N}& b_{N+1,N+1}\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ B_2={\left[\begin{array}{cccccccccccc}1& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ -1/\tau & {\eta }_n& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ 0& -1/\tau & {\eta }_n& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & {\eta }_n& 0& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & -1/\tau & {\eta }_n& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& v_n& w_n& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& 0& v_n& w_n& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & v_n& w_n& 0\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& v_n& w_n\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ C={\left[\begin{array}{cccccccccccc}0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ 0& c_n& 0& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ 0& 0& c_n& \cdots & 0& 0& 0& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & c_n& 0& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& c_n& 0& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& y_n& y_n& 0& \cdots & 0& 0& 0\\ 0& 0& 0& \cdots & 0& 0& y_n& y_n& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & y_n& y_n& 0\\ 0& 0& 0& \cdots & 0& 0& 0& 0& \cdots & 0& y_n& y_n\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ {\phi }_n={\left[\begin{array}{c}{\phi }_n^0\\ {\phi }_n^1\\ {\phi }_n^2\\ ⋮\\ {\phi }_n^{N-1}\\ {\phi }_n^N\end{array}\right]}_{(N+1)\times 1},U_q={\left[\begin{array}{c}{u}_q^0\\ u_q^1\\ u_q^2\\ ⋮\\ u_q^{N-1}\\ u_q^N\end{array}\right]}_{(N+1)\times 1},q=n\pm 1,n,\hfill \\ a_n=-\frac{1+x_n}{h^2}-\frac{1}{2h},c_n=-\frac{1+x_n}{h^2}+\frac{1}{2h},\hfill \\ z_n=-\frac{1}{2}(\frac{1+x_n}{h^2}+\frac{1}{2h}),y_n=-\frac{1}{2}(\frac{1+x_n}{h^2}-\frac{1}{2h}),\hfill \\ {\eta }_n=\frac{1}{\tau }+\frac{2(1+x_n)}{h^2},\hfill \\ v_n=-\frac{1}{\tau }+\frac{1+x_n}{h^2},w_n=\frac{1}{\tau }+\frac{1+x_n}{h^2},\hfill \\ b_{11}=1,b_{21}=-\frac{2\sqrt{2}}{3\sqrt{\pi \tau }},b_{22}=\frac{2\sqrt{2}}{3\sqrt{\pi \tau }},\hfill \\ b_{31}=-\frac{4\sqrt{6}}{5\sqrt{\pi \tau }},b_{32}=\frac{2\sqrt{6}}{5\sqrt{\pi \tau }},b_{33}=\frac{2\sqrt{6}}{5\sqrt{\pi \tau }},\hfill \\ b_{41}=d[1\lambda (1)+\mu (1)],b_{42}=d[-3\lambda (1)-2\mu (1)]-d/6\sqrt{2},\hfill \\ b_{43}=d[2\lambda (1)+\mu (1)]-4d/6\sqrt{2},b_{44}=5d/6\sqrt{2},\hfill \\ b_{51}=d[2\lambda (2)+\mu (2)],b_{52}=d[-5\lambda (2)-2\mu (2)+1\lambda (1)+\mu (1)],\hfill \\ b_{53}=d[3\lambda (2)+\mu (2)-3\lambda (1)-2\mu (1)]-d/6\sqrt{2},\hfill \\ b_{54}=d[2\lambda (1)+\mu (1)]-4d/6\sqrt{2},b_{55}=5d/6\sqrt{2},\hfill \\ b_{ij}=\{\begin{array}{ll}d[(i-3)\lambda (i-3)+\mu (i-3)],& j=1,\\ d[(5-2i)\lambda (i-3)-2\mu (i-3)+(i-4)\lambda (i-4)+\mu (i-4)],& j=2,\\ d[(i-j+1)\lambda (i-j)+\mu (i-j)+(2j-2i+1)\lambda (i-j-1)\\ -2\mu (i-j-1)+(i-j-2)\lambda (i-j-2)+\mu (i-j-2)],& 3\le j\le i-3,\\ d[3\lambda (2)+\mu (2)-3\lambda (1)-2\mu (1)]-d/6\sqrt{2},& j=i-2,\\ d[2\lambda (1)+\mu (1)]-4d/6\sqrt{2},& j=i-1,\\ 5d/6\sqrt{2},& j=i,\\ 0,& i<j\le r+1\end{array}\hfill \end{array}$$

(3.5)

for $i=6,7,\dots ,N+1$, and

$${\phi }_n^k=[\frac{3\sqrt{k\tau }}{2}+\frac{3\sqrt{\pi }k\tau }{4}+{\pi }^2{(k\tau )}^{3/2}(1+nh)]sin(\pi nh)-\pi {(k\tau )}^{3/2}cos(\pi nh).$$

(3.6)

For solving (3.2) we use a modified Gauss elimination method [27]. Hence, we seek a solution of the matrix equation in the following form:

$$U_j={\alpha }_{j+1}{U}_{j+1}+{\beta }_{j+1},j=M-1,\dots ,2,1,$$

(3.7)

where ${\alpha }_j$ ($j=1,2,\dots ,M$) are $(N+1)\times (N+1)$ square matrices and ${\beta }_j$ ($j=1,2,\dots ,M$) are $(N+1)\times 1$ column matrices defined by

$${\alpha }_{j+1}=-{(B+C{\alpha }_j)}^{-1}A,$$

(3.8)

$${\beta }_{j+1}={(B+C{\alpha }_j)}^{-1}(D{\phi }_j-C{\beta }_j),j=1,2,\dots ,M-1,$$

(3.9)

where $j=1,2,\dots ,M-1$, ${\alpha }_1$ is the $(N+1)\times (N+1)$ zero matrix and ${\beta }_1$ is the $(N+1)\times 1$ zero matrix and $U_M=0$.

Example 3.2 We consider the following initial boundary value problem with Neumann condition for the one-dimensional fractional parabolic partial differential equation:

$$\{\begin{array}{l}\frac{\partial u(t,x)}{\partial t}+D_t^{1/2}u(t,x)-\frac{\partial }{\partial x}((1+x)\frac{\partial u(t,x)}{\partial x})+u(t,x)=f(t,x),\\ f(t,x)=(3+t+\frac{16\sqrt{t}}{5\sqrt{\pi }}+{\pi }^{2}t(1+x))t^{2}cos\pi x+\pi t^{3}sin\pi x,0<t<1,0<x<1,\\ u_x(t,0)=u_x(t,1)=0,0\le t\le 1,\\ u(0,x)=0,0\le x\le 1.\end{array}$$

(3.10)

It is clear that the exact solution of problem (3.10) is

$$u(t,x)=t^{3}cos\pi x.$$

Applying the r-modified Crank-Nicholson difference scheme (2.4), we get

$$\{\begin{array}{l}\frac{u_n^k-u_n^{k-1}}{\tau }+D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k-[(1+x_n)\frac{u_{n+1}^k-2u_n^k+u_{n-1}^k}{h^2}+\frac{u_{n+1}^k-u_{n-1}^k}{2h}+u_n^k]={\phi }_n^k,1\le k\le r,\\ \frac{u_n^k-u_n^{k-1}}{\tau }+D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k-\frac{1}{2}[(1+x_n)\frac{u_{n+1}^k-2u_n^k+u_{n-1}^k}{h^2}+\frac{u_{n+1}^k-u_{n-1}^k}{2h}+u_n^k\\ +(1+x_n)\frac{u_{n+1}^{k-1}-2u_n^{k-1}+u_{n-1}^{k-1}}{h^2}+\frac{u_{n+1}^{k-1}-u_{n-1}^{k-1}}{2h}+u_n^{k-1}]={\phi }_n^k,r+1\le k\le N,\\ {\phi }_n^k=f(t_k-\frac{\tau }{2},x_n),t_k=k\tau ,x_n=nh,1\le k\le N,1\le n\le M-1,\\ u_0^0=0,k=0,\\ -\frac{h}{4\tau }{u}_0^{k-1}+[\frac{1}{h}+\frac{h}{2}{D}_t^{1/2}+\frac{h}{2}]u_0^k+\frac{h}{4\tau }{u}_0^{k+1}=\frac{1}{h}{u}_1^k+\frac{h}{2}{\phi }_0^k,1\le k\le N-1,\\ \frac{h}{4\tau }{u}_0^{N-2}+\frac{h}{\tau }{u}_0^{N-1}+[\frac{1}{h}+\frac{3h}{4\tau }+\frac{h}{2}{D}_t^{1/2}+\frac{h}{2}]u_0^N=\frac{1}{h}{u}_1^N+\frac{h}{2}{\phi }_0^N,k=N,\\ 3u_M^k-4u_{M-1}^k+u_{M-2}^k=0,0\le k\le N,\\ u_n^0=0,0\le n\le M,\end{array}$$

where $D_{t_k-\frac{\tau }{2}}^{1/2}{u}_n^k$ is defined by (2.3). We can rewrite it in the form of a system of equations with matrix coefficients

$$\{\begin{array}{l}A{U}_{n+1}+B{U}_n+C{U}_{n-1}=D{\phi }_n,1\le n\le M-1,\\ E{U}_0=F{U}_1+R{\phi }_0,3U_M-4U_{M-1}+U_{M-2}=\tilde{0}.\end{array}$$

Here, A, B, C are defined by (3.3), (3.4), and (3.5):

$$\begin{array}{c}F={\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 0\\ 0& 1/h& 0& \cdots & 0& 0\\ 0& 0& 1/h& \cdots & 0& 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& 0& \cdots & 1/h& 0\\ 0& 0& 0& \cdots & 0& 1/h\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ R={\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 0\\ 0& h/2& 0& \cdots & 0& 0\\ 0& 0& h/2& \cdots & 0& 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& 0& \cdots & h/2& 0\\ 0& 0& 0& \cdots & 0& h/2\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ E={\left[\begin{array}{cccccc}{e}_{11}& 0& 0& \cdots & 0& 0\\ e_{21}& e_{22}& 0& \cdots & 0& 0\\ e_{31}& e_{32}& e_{33}& \cdots & 0& 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ e_{N1}& e_{N2}& e_{N3}& \cdots & e_{NN}& 0\\ e_{N+1,1}& e_{N+1,2}& e_{N+1,3}& \cdots & e_{N+1,N}& e_{N+1,N+1}\end{array}\right]}_{(N+1)\times (N+1)},\hfill \\ e_{11}=1,e_{21}=-\frac{h}{4\tau }-\frac{4h}{3\sqrt{\pi \tau }},e_{22}=\frac{1}{h}+\frac{h}{2}+\frac{4h}{3\sqrt{\pi \tau }},e_{23}=\frac{h}{4\tau },\hfill \\ e_{31}=\frac{2\sqrt{2}h}{15\sqrt{\pi \tau }},e_{32}=\frac{-16\sqrt{2}h}{15\sqrt{\pi \tau }}-\frac{h}{4\tau },e_{33}=\frac{1}{h}+\frac{h}{2}+\frac{14\sqrt{2}h}{15\sqrt{\pi \tau }},e_{34}=\frac{h}{4\tau },\hfill \\ e_{41}=\frac{dh}{2}[(1+1/2)\lambda (1)+\mu (1)],e_{42}=\frac{dh}{2}[-4\lambda (1)-2\mu (1)+1/2\lambda (0)+\mu (0)],\hfill \\ e_{43}=-\frac{h}{4\tau }+\frac{dh}{2}[(2+1/2)\lambda (1)+\mu (1)-2-2(-1/3)],\hfill \\ e_{44}=\frac{1}{h}+\frac{h}{2}+\frac{dh}{2}[(1+1/2)\lambda (0)+\mu (0)],e_{45}=\frac{h}{4\tau },\hfill \\ e_{51}=\frac{dh}{2}[(2+1/2)\lambda (2)+\mu (2)],\hfill \\ e_{52}=\frac{dh}{2}[-2\cdot 3\lambda (2)-2\mu (2)+(1+1/2)\lambda (1)+\mu (1)],\hfill \\ e_{53}=\frac{dh}{2}[(2+1+1/2)\lambda (2)+\mu (2)-2\cdot 2\lambda (1)-2\mu (1)+1/2\lambda (0)+\mu (0)],\hfill \\ e_{54}=-\frac{h}{4\tau }+\frac{dh}{2}[(1+1+1/2)\lambda (1)+\mu (1)-2\lambda (0)-2\mu (0)],\hfill \\ e_{55}=\frac{1}{h}+\frac{h}{2}+\frac{dh}{2}[(1+1/2)\lambda (0)+\mu (0)],e_{56}=\frac{h}{4\tau },\hfill \\ e_{ij}=\{\begin{array}{ll}\frac{dh}{2}[(i-3+1/2)\lambda (i-3)+\mu (i-3)],& j=1,\\ \frac{dh}{2}[-2(i-2)\lambda (i-3)-2\mu (i-3)+(i-4+1/2)\lambda (i-4)+\mu (i-4)],& j=2,\\ \frac{dh}{2}[(i-j+1+1/2)\lambda (i-j)+\mu (i-j)-2(i-j)\lambda (i-j-1)\\ -2\mu (i-j-1)+(i-j-2+1/2)\lambda (i-j-2)+\mu (i-j-2)],& 3\le j\le i-2,\\ -\frac{h}{4\tau }+\frac{dh}{2}[(2+1/2)\lambda (1)+\mu (1)-2\lambda (0)-2\mu (0)],& j=i-1,\\ \frac{1}{h}+\frac{h}{2}+\frac{dh}{2}[(1+1/2)\lambda (0)+\mu (0)],& j=i,\\ \frac{h}{4\tau },& j=i+1,\\ \frac{h}{4\tau }+\frac{dh}{2}[(i-N+2+1/2)\lambda (i-N+1)+\mu (i-N+1)\\ -2(i-N+1)\lambda (i-N)-2\mu (i-N)\\ +(i-N-1+1/2)\lambda (i-N-1)+\mu (i-N-1)],& j=N-1,\\ -\frac{h}{\tau }+\frac{dh}{2}[(2+1/2)\lambda (1)+\mu (1)-2\lambda (0)-2\mu (0)],& j=N,\\ \frac{1}{h}+\frac{h}{2}+\frac{3h}{4\tau }+\frac{dh}{2}[(1+1/2)\lambda (0)+\mu (0)],& j=N+1\\ 0,& j>i+1\end{array}\hfill \end{array}$$

for $i=6,7,\dots ,N+1$, and

$$\begin{array}{c}{\phi }_0^k=(3+k\tau +\frac{16\sqrt{k\tau }}{5\sqrt{\pi }}+{\pi }^{2}k\tau ){(k\tau )}^2,\hfill \\ {\phi }_n^k=[3+k\tau +\frac{16\sqrt{k\tau }}{5\sqrt{\pi }}+{\pi }^2(k\tau )(1+nh)]{(k\tau )}^{2}cos(\pi nh)+\pi {(k\tau )}^{3}sin(\pi nh).\hfill \end{array}$$

(3.11)

For solving this matrix equation we will use the same method as for Example 3.1. Namely, we use (3.7), (3.8), (3.9), and

$$\begin{array}{c}{u}_M={[3I-4{\alpha }_M+{\alpha }_{M-1}{\alpha }_M]}^{-1}\ast [(4I-{\alpha }_{M-1}){\beta }_M-{\beta }_{M-1}],\hfill \\ {\alpha }_1=E^{-1}F,{\beta }_1=E^{-1}R{\phi }_0.\hfill \end{array}$$

Finally, we give the results of the numerical analysis. The numerical solutions are recorded for different values of the modification parameter r, and discretization parameters N and M. Besides $u_n^k$ represents the numerical solutions of these difference schemes at $(t_k,x_n)$. The error is computed by the following formula:

$$E_M^N=\underset{1\le k\le N,1\le n\le M-1}{max}|u(t_k,x_n)-u_n^k|.$$

Table 1 and Table 2 are constructed for $N=M=40\text{ and }80$, respectively. As can be seen from Table 1, the r-modified Crank-Nicholson difference scheme is more accurate than the Crank-Nicholson and Rothe difference schemes. Table 2 shows that the r-modified Crank-Nicholson difference scheme has the same order error as the Crank-Nicholson difference scheme.

Table 1 Error analysis for Dirichlet problem

Table 2 Error analysis for Neumann problem

4 Conclusion

In this study, the second order of accuracy stable difference scheme for the numerical solution of the mixed problem for the fractional parabolic equation is investigated. We have obtained a stability estimate for the solution of this difference scheme. The theoretical statements for the solution of this difference scheme for one-dimensional parabolic equations are supported by numerical examples obtained by computer.