A Weighted ADI Scheme for Subdiffusion Equations

Abstract

A weighted ADI scheme is proposed for solving two-dimensional anomalous diffusion equations with the fractional Caputo derivative. The Alikhanov formula (J Comput Phys 280:424–438, 2015) with a weaker assumption is applied to approximate the fractional derivative and a high-order perturbed term of temporal order $1+2\alpha$ is added to the pure implicit approach. By using the discrete energy method, it is proven that the ADI scheme is stable and convergent with the temporal order of $\min \{1+2\alpha ,2\}$ such that it achieves second-order time accuracy when $\frac{1}{2}\le \alpha <1$. Numerical experiments are included to support the theoretical analysis. Application of suggested method to the solution which lacks the smoothness near the initial time is examined by employing a class of nonuniform meshes refined near the singular point.

Appendix: Proof of Lemma 2.3

We evaluate the error $\mathfrak {R}^{n+\sigma }$ by splitting it into three parts $\mathfrak {R}^{n+\sigma }=\mathfrak {R}_1^{n}+\mathfrak {R}_2^{n}+\mathfrak {R}_3^{n+\sigma }$, where

$$\begin{aligned}&\mathfrak {R}_1^{n} =\frac{1}{\Gamma (1-\alpha )}\int _{t_{0}}^{t_{1}}\frac{g'(s)-\left[ \Pi _{2,k}g(s)\right] '}{(t_{n+\sigma }-s)^{\alpha }}\,\mathrm {d}{s},\\&\mathfrak {R}_2^{n} =\frac{1}{\Gamma (1-\alpha )}\sum _{k=2}^{n}\int _{t_{k-1}}^{t_{k}}\frac{g'(s)-\left[ \Pi _{2,k}g(s)\right] '}{(t_{n+\sigma }-s)^{\alpha }}\,\mathrm {d}{s},\\&\mathfrak {R}_3^{n+\sigma }=\frac{1}{\Gamma (1-\alpha )}\int _{t_{n}}^{t_{n+\sigma }}\frac{g'(s)-\left[ \Pi _{1,n}g(s)\right] '}{(t_{n+\sigma }-s)^{\alpha }}\,\mathrm {d}{s},\quad n\ge 0. \end{aligned}$$

(i) Estimates of $\mathfrak {R}_3^{n+\sigma }$ and $\mathfrak {R}^{\sigma }$. Applying the formula of Taylor expansion, one gets

$$\begin{aligned} g'(s)-\left[ \Pi _{1,n}g(s)\right] '&=g'(s)-\delta _tg^{n+\frac{1}{2}}=g'(s)-g'(t_{n+\frac{1}{2}})+g'(t_{n+\frac{1}{2}})-\delta _tg^{n+\frac{1}{2}}\\&=g''(t_{n+\frac{1}{2}})(s-t_{n+\frac{1}{2}})+\int _{t_{n+\frac{1}{2}}}^sg'''(\lambda )(s-\lambda )\,\mathrm {d}\lambda \\&\quad +\left[ g'(t_{n+\frac{1}{2}})-\delta _tg^{n+\frac{1}{2}}\right] \end{aligned}$$

such that $\mathfrak {R}_3^{n+\sigma }$ can be split into three parts $\mathfrak {R}_{3\ell }^{n+\sigma }$ $(\ell =1,2,3)$ corresponding to three terms of the right hand side. If $\sigma =(2-\alpha )/2$, integration by parts yields (also see [1]),

$$\begin{aligned} \mathfrak {R}_{31}^{n+\sigma }=&\frac{g''(t_{n+\frac{1}{2}})}{\Gamma (1-\alpha )}\int _{t_{n}}^{t_{n+\sigma }}\frac{(s-t_{n+\frac{1}{2}})}{(t_{n+\sigma }-s)^{\alpha }}\,\mathrm {d}{s} =\frac{g''(t_{n+\frac{1}{2}})}{\Gamma (3-\alpha )}\left( \sigma -\frac{2-\alpha }{2}\right) \tau \, t_{\sigma }^{1-\alpha }=0. \end{aligned}$$

Noticing that $t_{n+\frac{1}{2}}<t_{n+\sigma }<t_{n+1}$, we exchange the order of integrations,

$$\begin{aligned} \mathfrak {R}_{32}^{n+\sigma }&=\frac{1}{\Gamma (1-\alpha )}\int _{t_{n}}^{t_{n+\sigma }} \left[ \int _{t_{n+\frac{1}{2}}}^sg'''(\lambda )(s-\lambda )\,\mathrm {d}\lambda \right] \frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }}\nonumber \\&=\frac{t_{\sigma }^{1-\alpha }}{\Gamma (2-\alpha )}\int _{t_n}^{t_{n+\frac{1}{2}}}g'''(s)(s-t_{n})\,\mathrm {d}{s}\nonumber \\&\quad +\frac{1}{\Gamma (2-\alpha )}\int _{t_{n}}^{t_{n+\frac{1}{2}}}(t_{n+\sigma }-s)^{1-\alpha }\int _{t_{n+\frac{1}{2}}}^sg'''(\lambda )\,\mathrm {d}\lambda \,\mathrm {d}{s}\nonumber \\&\quad +\frac{1}{\Gamma (2-\alpha )}\int _{t_{n+\frac{1}{2}}}^{t_{n+\sigma }}(t_{n+\sigma }-s)^{1-\alpha }\int _{t_{n+\frac{1}{2}}}^sg'''(\lambda )\,\mathrm {d}\lambda \,\mathrm {d}{s} \triangleq \sum _{\ell =1}^3\mathfrak {R}_{32\ell }^{n+\sigma }. \end{aligned}$$

(6.1)

Applying the condition (2.3), it is not difficult to obtain

$$\begin{aligned} \left| \mathfrak {R}_{321}^{n+\sigma }\right| \le&\frac{1}{\Gamma (2-\alpha )}\int _{t_n}^{t_{n+\frac{1}{2}}}t_{\sigma }^{1-\alpha }\left| g'''(s)\right| (s-t_{n})\,\mathrm {d}{s} \le \frac{C_g{\sigma }^{1-\alpha }\tau ^{2-\alpha }}{2\alpha \Gamma (2-\alpha )}(t_{n+\frac{1}{2}}^{\alpha }-t_{n}^{\alpha }),\\ \left| \mathfrak {R}_{322}^{n+\sigma }\right| \le&\frac{1}{\Gamma (2-\alpha )}\int _{t_{n}}^{t_{n+\frac{1}{2}}}(t_{n+\sigma }-s)^{1-\alpha } \int ^{t_{n+\frac{1}{2}}}_s\left| g'''(\lambda )\right| \,\mathrm {d}\lambda \,\mathrm {d}{s} \le \frac{C_g{\sigma }^{1-\alpha }\tau ^{2-\alpha }}{2\alpha \Gamma (2-\alpha )}(t_{n+\frac{1}{2}}^{\alpha }-t_{n}^{\alpha }),\\ \left| \mathfrak {R}_{323}^{n+\sigma }\right| \le&\frac{1}{\Gamma (2-\alpha )}\int _{t_{n+\frac{1}{2}}}^{t_{n+\sigma }}(t_{n+\sigma }-s)^{1-\alpha } \int _{t_{n+\frac{1}{2}}}^s\left| g'''(\lambda )\right| \,\mathrm {d}\lambda \,\mathrm {d}{s} \le \frac{C_g{\sigma }^{1-\alpha }\tau ^{2-\alpha }}{\alpha \Gamma (2-\alpha )}(t_{n+\sigma }^{\alpha }-t_{n+\frac{1}{2}}^{\alpha }), \end{aligned}$$

and then

$$\begin{aligned} \left| \mathfrak {R}_{32}^{n+\sigma }\right| \le&\sum _{\ell =1}^3\left| \mathfrak {R}_{32\ell }^{n+\sigma }\right| \le \frac{C_g{\sigma }^{1-\alpha }\tau ^{2}}{\alpha \Gamma (2-\alpha )}\big [(n+\sigma )^{\alpha }-n^{\alpha }\big ],\quad n\ge 0. \end{aligned}$$

(6.2)

Now consider the third part $R_{33}^{n+\sigma }$. Applying the formula of Taylor expansion with integral remainder and the condition (2.3), one has

$$\begin{aligned} \big |\delta _tg^{n+\frac{1}{2}}-g'(t_{n+\frac{1}{2}})\big |\le&\frac{1}{2\tau }\Big |\int _{t_{n+\frac{1}{2}}}^{t_{n+1}}g'''(\lambda )(t_{n+1}-\lambda )^2\,\mathrm {d}\lambda +\int _{t_{n}}^{t_{n+\frac{1}{2}}}g'''(\lambda )(\lambda -t_{n})^2\,\mathrm {d}\lambda \Big |\nonumber \\ \le&\frac{C_g\tau }{4}\,\int _{t_{n}}^{t_{n+1}}\left| g'''(\lambda )\right| \,\mathrm {d}\lambda = \frac{C_g\tau ^{1+\alpha }}{4\alpha }\big [(n+1)^{\alpha }-n^{\alpha }\big ],\quad n\ge 0, \end{aligned}$$

(6.3)

and then

$$\begin{aligned}&\left| \mathfrak {R}_{33}^{n+\sigma }\right| \le \frac{\left| g'(t_{n+\frac{1}{2}})-\delta _tg^{n+\frac{1}{2}}\right| }{\Gamma (1-\alpha )}\int _{t_{n}}^{t_{n+\sigma }}\frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }} \le \frac{C_g\sigma ^{1-\alpha }\tau ^2}{4\alpha \Gamma (2-\alpha )}\big [(n+1)^{\alpha }-n^{\alpha }\big ]. \end{aligned}$$

(6.4)

We apply the above estimates (6.2) and (6.4) to get

$$\begin{aligned} \left| \mathfrak {R}^{\sigma }\right| =\left| \mathfrak {R}_3^{\sigma }\right| \le&\sum _{\ell =1}^3\left| \mathfrak {R}_{3\ell }^{\sigma }\right| \le \frac{5\sigma ^{1-\alpha }C_g\tau ^2}{4\alpha \Gamma (2-\alpha )}. \end{aligned}$$

Thus the estimate (2.4) is valid for $n=0$. For $n\ge 1$, the estimates (6.2) and (6.4) yield

$$\begin{aligned} \left| \mathfrak {R}_3^{n+\sigma }\right| \le&\sum _{\ell =1}^3\left| \mathfrak {R}_{3\ell }^{n+\sigma }\right| \le \frac{5\sigma ^{1-\alpha }C_g\tau ^{2}}{4\Gamma (2-\alpha )},\quad n\ge 1. \end{aligned}$$

(6.5)

(ii) Estimates of $\mathfrak {R}_1^{n}$ and $\mathfrak {R}^{1+\sigma }$. Applying the formula of Taylor expansion, one has

$$\begin{aligned} g'(s)-\left( \Pi _{2,1}g(s)\right) ' =&\frac{s-t_{\frac{1}{2}}}{\tau }\int _{s}^{t_{\frac{3}{2}}}(t_{\frac{3}{2}}-\lambda )g'''(\lambda )\,\mathrm {d}\lambda +\frac{t_{\frac{3}{2}}-s}{\tau }\int _{t_{\frac{1}{2}}}^{s}(\lambda -t_{\frac{1}{2}})g'''(\lambda )\,\mathrm {d}\lambda \\&-\frac{s-t_{\frac{1}{2}}}{\tau }\left[ \delta _tg^{\frac{3}{2}}-g'(t_{\frac{3}{2}})\right] -\frac{t_{\frac{3}{2}}-s}{\tau }\left[ \delta _tg^{\frac{1}{2}}-g'(t_{\frac{1}{2}})\right] ,\quad t_0<s<t_1. \end{aligned}$$

Thus we can split $\mathfrak {R}_{1}^{n}$ into four parts, $\mathfrak {R}_1^{n}=\sum _{\ell =1}^4\mathfrak {R}_{1\ell }^{n}$, where $\mathfrak {R}_{1\ell }^{n}$ $(\ell =1,2,3,4)$ corresponds to the four terms at the right hand side. Using the condition (2.3), one gets

$$\begin{aligned} \left| \mathfrak {R}_{11}^{n}\right|&\le \frac{C_g}{2\Gamma (1-\alpha )}\int _{0}^{t_{1}}\int _{0}^{t_{\frac{3}{2}}}(t_{\frac{3}{2}}-\lambda )\lambda ^{\alpha -1}\,\mathrm {d}\lambda \frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }}\\&\le \frac{(3/2)^{1+\alpha }C_g\tau ^{1+\alpha }}{2\alpha \,\Gamma (1-\alpha )}\int _{0}^{t_{1}}\frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }}\\&\le \frac{9C_g\tau ^{2}}{8\alpha \,\Gamma (2-\alpha )}\left[ (n+\sigma )^{1-\alpha }-(n-1+\sigma )^{1-\alpha }\right] \le \frac{9\sigma ^{-\alpha }C_g\tau ^{2}}{8\alpha \,\Gamma (1-\alpha )},\quad n\ge 1,\\ \left| \mathfrak {R}_{12}^{n}\right|&\le \frac{3}{2\Gamma (1-\alpha )}\int _{0}^{t_{1}}\Big |\int _{t_{\frac{1}{2}}}^{s}(\lambda -t_{\frac{1}{2}})g'''(\lambda )\,\mathrm {d}\lambda \Big | \frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }}\\&\le \frac{3C_g}{2\alpha \,\Gamma (1-\alpha )}\int _{0}^{t_{1}}\int _{0}^{t_1}t_{\frac{1}{2}}\lambda ^{\alpha -1}\,\mathrm {d}\lambda \frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }} \le \frac{3\sigma ^{-\alpha }C_g\tau ^{2}}{4\alpha \,\Gamma (1-\alpha )},\quad n\ge 1. \end{aligned}$$

Applying the estimate (6.3), it is easy to obtain

$$\begin{aligned} \left| \mathfrak {R}_{13}^{n}\right| \le&\frac{\left| g'(t_{\frac{3}{2}})-\delta _tg^{\frac{3}{2}}\right| }{2\Gamma (1-\alpha )}\int _{t_{0}}^{t_{1}}\frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }} \le \frac{\sigma ^{-\alpha }C_g\tau ^{2}}{8\alpha \,\Gamma (1-\alpha )},\quad n\ge 1,\\ \left| \mathfrak {R}_{14}^{n}\right| \le&\frac{3\left| g'(t_{\frac{1}{2}})-\delta _tg^{\frac{1}{2}}\right| }{2\Gamma (1-\alpha )}\int _{t_{0}}^{t_{1}}\frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{\alpha }} \le \frac{3\sigma ^{-\alpha }C_g\tau ^{2}}{8\alpha \,\Gamma (1-\alpha )},\quad n\ge 1, \end{aligned}$$

and then

$$\begin{aligned} \left| \mathfrak {R}_1^{n}\right| \le&\sum _{\ell =1}^4\left| \mathfrak {R}_{1\ell }^{n}\right| \le \frac{3\sigma ^{-\alpha }C_g\tau ^{2}}{\alpha \,\Gamma (1-\alpha )},\quad n\ge 1. \end{aligned}$$

(6.6)

Then it follows from (6.5) and (6.6) that

$$\begin{aligned} \left| \mathfrak {R}^{1+\sigma }\right| \le \left| \mathfrak {R}_1^{1}\right| +\left| \mathfrak {R}_3^{1+\sigma }\right| \le \left( \frac{5}{4}+\frac{3}{\alpha }\right) \frac{\sigma ^{1-\alpha }C_g\tau ^{2}}{\Gamma (2-\alpha )}. \end{aligned}$$

It implies that the claimed estimate (2.4) is valid for $n=1$.

(iii) Estimates of $\mathfrak {R}_2^{n}$ and $\mathfrak {R}^{n+\sigma }$ for $n\ge 2$. One can apply the formula of integration by parts and the local error of quadratic interpolating polynomial $\Pi _{2,k}g(s)$ to get

$$\begin{aligned} \mathfrak {R}_2^{n} \triangleq&\frac{1}{\Gamma (1-\alpha )}\sum _{k=2}^{n}\int _{t_{k-1}}^{t_{k}}\frac{\left[ g(s)-\Pi _{2,k}g(s)\right] '}{(t_{n+\sigma }-s)^{\alpha }}\,\mathrm {d}{s} =-\frac{\alpha }{\Gamma (1-\alpha )}\sum _{k=2}^{n}\int _{t_{k-1}}^{t_{k}}\frac{g(s)-\Pi _{2,k}g(s)}{(t_{n+\sigma }-s)^{1+\alpha }}\,\mathrm {d}{s}\\ =&-\frac{\alpha }{6\Gamma (1-\alpha )}\sum _{k=2}^{n} \int _{t_{k-1}}^{t_{k}}\frac{g'''(\vartheta _k)(s-t_{k-1})(s-t_{k})(s-t_{k+1})}{(t_{n+\sigma }-s)^{1+\alpha }}\,\mathrm {d}{s}. \end{aligned}$$

Reminding that $\int _{t_{k-1}}^{t_{k}}(s-t_{k-1})(t_{k}-s)(t_{k+1}-s)\,\mathrm {d}{s}=\tau ^4/4$, we apply the condition (2.3) and the mean-value law of integrals to find

$$\begin{aligned} \left| \mathfrak {R}_2^{n}\right| \le&\frac{\alpha }{6\Gamma (1-\alpha )}\sum _{k=2}^{n} \frac{C_gt_{k-1}^{\alpha -1}}{(t_{n+\sigma }-\nu _k)^{1+\alpha }} \int _{t_{k-1}}^{t_{k}}(s-t_{k-1})(t_{k}-s)(t_{k+1}-s)\,\mathrm {d}{s}\nonumber \\ \le&\frac{\alpha C_g\tau ^{3}t_1^{\alpha -1}}{24\Gamma (1-\alpha )}\sum _{k=2}^{n} \frac{\tau }{(t_{n+\sigma }-\nu _k)^{1+\alpha }}\le \frac{\alpha C_g\tau ^{2+\alpha }}{24\Gamma (1-\alpha )}\int _{t_{1}}^{t_{n}} \frac{\,\mathrm {d}{s}}{(t_{n+\sigma }-s)^{1+\alpha }}\nonumber \\ =&\frac{C_g\tau ^{2}}{24\Gamma (1-\alpha )}\left[ \sigma ^{-\alpha }-(n-1+\sigma )^{-\alpha }\right] \le \frac{C_g\sigma ^{-\alpha }\tau ^{2}}{24\Gamma (1-\alpha )} \le \frac{C_g\sigma ^{1-\alpha }\tau ^{2}}{12\Gamma (2-\alpha )},\quad n\ge 2, \end{aligned}$$

(6.7)

where $\nu _k\in (t_{k-1},t_k)$, $1\le k\le n$. Then we apply (6.5), (6.6) and (6.7) to get

$$\begin{aligned} \left| \mathfrak {R}^{n+\sigma }\right| \le \left| \mathfrak {R}_1^{n}\right| +\left| \mathfrak {R}_2^{n}\right| +\left| \mathfrak {R}_3^{n+\sigma }\right| \le \frac{4\sigma ^{1-\alpha }C_g\tau ^{2}}{\alpha \,\Gamma (2-\alpha )},\quad n\ge 2. \end{aligned}$$