Error Analysis of a Second-Order Method on Fitted Meshes for a Time-Fractional Diffusion Problem

Abstract

Alikhanov’s high-order scheme for Caputo fractional derivatives of order $\alpha \in (0,1)$ is generalised to nonuniform meshes and analysed for initial-value problems (IVPs) and initial-boundary value problems (IBVPs) whose solutions display a typical weak singularity at the initial time. It is shown that, when the mesh is chosen suitably, the scheme attains order $3-\alpha$ convergence for the 1-dimensional IVP and second-order convergence for the IBVP, for which a spectral method is analysed when the spatial domain is the unit square and the extension of this analysis to other spatial domains and other spatial dimensions and discretisations is outlined. Numerical results demonstrate the sharpness of the theoretical convergence estimates.

Appendices

A Proof of Lemma 1

Proof

Set $r_v^{j+\sigma } = \delta _{t}^\alpha v(t_{j+\sigma })- D_t^\alpha v(t_{j+\sigma })$. From (1) we get

$$\begin{aligned} \varGamma (1-\alpha )r_v^{j+\sigma }&=\sum _{s=1}^j \int _{t_{s-1}}^{t_{s}}(t_{j+\sigma }-\eta )^{-\alpha }\big (I_{2,s}v(\eta )-v(\eta )\big )' \,\mathrm{d}\eta \nonumber \\&\qquad +\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }[\delta _tv(t_j)-v'(\eta )] \,\mathrm{d}\eta . \end{aligned}$$

(38)

When $j=0$, this gives

$$\begin{aligned} |\varGamma (1-\alpha )r_v^{\sigma }|&=\left| \int _{0}^{t_{\sigma }}(t_{\sigma }-\eta )^{-\alpha }[\delta _tv(t_0)-v'(\eta )]\mathrm{d}\eta \right| \\&=\left| \int _{0}^{t_{\sigma }}(t_{\sigma }-\eta )^{-\alpha }\eta ^{\alpha -1}\eta ^{1-\alpha }[\delta _tv(t_0)-v'(\eta )]\mathrm{d}\eta \right| \\&\le \psi _v^{\sigma } \tau _1^{-\alpha } \int _{0}^{t_{\sigma }}(t_{\sigma }-\eta )^{-\alpha }\eta ^{\alpha -1}\mathrm{d}\eta \\&\le \varGamma (1-\alpha )\varGamma (\alpha )\psi _v^{\sigma } t_{\sigma }^{-\alpha } \end{aligned}$$

by a standard formula for Euler’s Beta function [5, Theorem D.6].

For $j\ge 1$, Taylor’s theorem gives

$$\begin{aligned}&\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }[\delta _tv(t_j)-v'(\eta )]\,\mathrm{d}\eta \\&\quad =\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }[\delta _tv(t_j)-v'(t_{j+1/2})]\,\mathrm{d}\eta \\&\qquad -v''(t_{j+1/2})\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }(\eta -t_{j+1/2}) \,\mathrm{d}\eta \\&\qquad -\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }\frac{v'''\big (\xi _j(\eta )\big )}{2}(\eta -t_{j+1/2})^2\mathrm{d}\eta \end{aligned}$$

with $t_{j}< \xi _j(\eta ) < t_{j+\sigma }$. Again invoking Taylor’s theorem, for the first term here we get

$$\begin{aligned}&\left| \int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }[\delta _tv(t_j)-v'(t_{j+1/2})]\mathrm{d}\eta \right| \\&\quad \lesssim \tau _{j+1}^2\sup _{\eta \in (t_j,t_{j+1})}|v'''(\eta )|\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }\mathrm{d}\eta \\&\quad \lesssim \tau _{j+1}^{3-\alpha }\sup _{\eta \in (t_j,t_{j+1})}|v'''(\eta )|=t_{j+\sigma }^{-\alpha }\psi _v^{j+\sigma }. \end{aligned}$$

One also has

$$\begin{aligned} \left| \int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }\frac{v'''\big (\xi _j(\eta )\big )}{2}(\eta -t_{j+1/2})^2\mathrm{d}\eta \right| \lesssim \tau _{j+1}^{3-\alpha }\sup _{\eta \in (t_j,t_{j+1})}|v'''(\eta )|=t_{j+\sigma }^{-\alpha }\psi _v^{j+\sigma }. \end{aligned}$$

On the other hand, integration by parts gives

$$\begin{aligned}&\int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }(\eta -t_{j+1/2})\mathrm{d}\eta \\&\quad =-\frac{1}{1-\alpha }(t_{j+\sigma }-\eta )^{1-\alpha }(\eta -t_{j+1/2})\big |_{t_j}^{t_{j+\sigma }}+\frac{1}{1-\alpha }\int _{t_j}^{t_{j+\sigma }} (t_{j+\sigma }-\eta )^{1-\alpha }\mathrm{d}\eta \\&\quad =\frac{\sigma ^{1-\alpha }\tau _{j+1}^{2-\alpha } (2\sigma +\alpha -2)}{2(1-\alpha )(2-\alpha )} =0. \end{aligned}$$

Combining these results, we obtain

$$\begin{aligned} \left| \int _{t_j}^{t_{j+\sigma }}(t_{j+\sigma }-\eta )^{-\alpha }[\delta _tv(t_j)-v'(\eta )]\mathrm{d}\eta \right| \lesssim t_{j+\sigma }^{-\alpha }\psi _v^{j+\sigma }. \end{aligned}$$

(39)

For $j\ge 1$, one gets

$$\begin{aligned}&\left| \int _{0}^{t_1}(t_{j+\sigma }-\eta )^{-\alpha }\left( I_{2,1}v(\eta )-v(\eta )\right) '\mathrm{d}\eta \right| \\&\quad =\int _{0}^{t_1}(t_{j+\sigma }-\eta )^{-\alpha }\eta ^{\alpha -1}\eta ^{1-\alpha }\left( I_{2,1}v(\eta )-v(\eta )\right) '\mathrm{d}\eta \\&\quad \le \sup _{\eta \in (0,t_1)}\left( \eta ^{1-\alpha }|(I_{2,1}v(\eta )-v(\eta ))'|\right) \int _{0}^{t_1}(t_{j+\sigma }-\eta )^{-\alpha }\eta ^{\alpha -1}\mathrm{d}\eta \\&\quad \le (t_{j+\sigma }-t_1)^{-\alpha }\frac{\tau _1^\alpha }{\alpha }\psi _v^{j,1}\tau _1^{-\alpha } \lesssim t_{j+\sigma }^{-\alpha }\psi _v^{j,1}, \end{aligned}$$

where we used $\tau _1/\tau _2\le \rho$ to get $(t_{j+\sigma }-t_1)^{-\alpha }=t_{j+\sigma }^{-\alpha }(1-t_1/t_{j+\sigma })^{-\alpha }\lesssim t_{j+\sigma }^{-\alpha }$.

It is well known (see, e.g., [2, p. 122]) that for $t\in [t_{s-1},t_{s+1}]$ one has

$$\begin{aligned} v(t)-I_{2,s}v(t)=\frac{v'''(\bar{\xi }_s)}{6}(t-t_{s-1})(t-t_s)(t-t_{s+1}) \text { for some } \bar{\xi }_s(t)\in (t_{s-1},t_{s+1}). \end{aligned}$$

For $j\ge 2$, one uses integration by parts and this interpolation error estimate to get

$$\begin{aligned}&\left| \sum _{s=2}^j\int _{t_{s-1}}^{t_s}(t_{j+\sigma }-\eta )^{-\alpha }(I_{2,s}v(\eta )-v(\eta ))' \,\mathrm{d}\eta \right| \nonumber \\&\quad =\left| \alpha \sum _{s=2}^j\int _{t_{s-1}}^{t_s}(t_{j+\sigma }-\eta )^{-\alpha -1}\big (v(\eta )-I_{2,s}v(\eta )\big ) \,\mathrm{d}\eta \right| \nonumber \\&\quad \lesssim \sum _{s=2}^j\psi _v^{j,s}\tau _{j+1}^\alpha t_{s}^{-\alpha }\int _{t_{s-1}}^{t_s}(t_{j+\sigma }-\eta )^{-\alpha -1}\,\mathrm{d}\eta \nonumber \\&\quad \le \sum _{s=2}^j\psi _v^{j,s}\tau _{j+1}^\alpha \int _{t_{s-1}}^{t_s}\eta ^{-\alpha }(t_{j+\sigma }-\eta )^{-\alpha -1}\,\mathrm{d}\eta \nonumber \\&\quad \lesssim \max _{s=2,\ldots ,j}\{\psi _v^{j,s}\}\tau _{j+1}^\alpha \int _{t_1}^{t_j}\eta ^{-\alpha }(t_{j+\sigma }-\eta )^{-\alpha -1}\,\mathrm{d}\eta \lesssim t_{j+\sigma }^{-\alpha }\max _{s=2,\ldots ,j}\{\psi _v^{j,s}\}; \end{aligned}$$

(40)

here the last inequality follows from

$$\begin{aligned}&\int _{t_1}^{t_j}\eta ^{-\alpha }(t_{j+\sigma }-\eta )^{-\alpha -1}\,\mathrm{d}\eta \\&\quad =\int _{t_1}^{t_j/2}\eta ^{-\alpha }(t_{j+\sigma }-\eta )^{-\alpha -1}\mathrm{d}\eta +\int _{t_j/2}^{t_j}\eta ^{-\alpha }(t_{j+\sigma }-\eta )^{-\alpha -1}\,\mathrm{d}\eta \\&\quad \lesssim (t_{j+\sigma }-t_j/2)^{-\alpha -1}(t_j/2)^{1-\alpha }+(t_j/2)^{-\alpha }(t_{j+\sigma }-t_j)^{-\alpha } \\&\quad \lesssim \tau _{j+1}^{-\alpha }t_{j+\sigma }^{-\alpha }, \end{aligned}$$

where we used $\tau _{j+1}\lesssim t_j$ to get $t_{j+\sigma }\lesssim t_j$.

Combining all the inequalities from (38) to (40) gives the desired result. $\square$

B Proof of Lemma 4

Proof

First, by definition (2) and $\sigma \ge 1-\alpha /2$, one gets

$$\begin{aligned} g_{0,0}=\tau _1^{-1}a_{0,0}=\frac{\sigma ^{1-\alpha }}{\varGamma (2-\alpha )}\tau _{1}^{-\alpha } =\frac{\sigma t_{\sigma }^{-\alpha }}{(1-\alpha )\varGamma (1-\alpha )}>\frac{t_{\sigma }^{-\alpha }}{\varGamma (1-\alpha )}\,. \end{aligned}$$

For $k\ge 1$, (9) follows from Lemma 2 as $g_{k,0}=\tau _1^{-1}(a_{k,0}-b_{k,0})$.

To prove (10), by definition (2), and Lemmas 2 and 3, one has

$$\begin{aligned} (2\sigma -1)g_{1,1}-\sigma g_{1,0}&=(2\sigma -1)\tau _2^{-1}(a_{1,1}+b_{1,0})-\sigma \tau _1^{-1}(a_{1,0}-b_{1,0}) \\&>(2\sigma -1)\tau _2^{-1}a_{1,1}-\sigma \tau _1^{-1}(a_{1,0}-b_{1,0}) \\&>(2\sigma -1)\frac{\sigma ^{1-\alpha }}{\varGamma (2-\alpha )}\tau _{2}^{-\alpha }-\sigma \frac{(\sigma \tau _2)^{-\alpha }}{\varGamma (1-\alpha )} \\&\ge 0, \end{aligned}$$

where we used $\sigma \ge 1-\alpha /2$ and $\varGamma (2-\alpha ) = (1-\alpha )\varGamma (1-\alpha )$ in the final inequality.

Next, consider (11). For $k=1$, inequality (10) and $1\ge \sigma \ge 1-\alpha /2>1/2$ imply that $g_{1,1}>g_{1,0}$. For $k\ge 2$, by Lemmas 2 and 3 one gets

$$\begin{aligned} g_{k,1}&=\tau _{2}^{-1}(a_{k,1}+b_{k,0}-b_{k,1})>\tau _{2}^{-1}(a_{k,1}-b_{k,1}) \\&>\frac{(t_{k+\sigma }-t_{1})^{-\alpha }}{\varGamma (1-\alpha )} >\tau _1^{-1}(a_{k,0}-b_{k,0})=g_{k,0}. \end{aligned}$$

We move on to (13). For $k\ge 3$ and $2\le j\le k-1$, by their definitions

$$\begin{aligned} g_{k,j}=\tau _{j+1}^{-1}(a_{k,j}+b_{k,j-1}-b_{k,j}) \text { and } g_{k,j-1}=\tau _{j}^{-1}(a_{k,j-1}+b_{k,j-2}-b_{k,j-1}). \end{aligned}$$

Lemma 2 yields

$$\begin{aligned} \tau _{j+1}^{-1}\big (a_{k,j}-b_{k,j}\big )>\frac{(t_{k+\sigma }-t_{j})^{-\alpha }}{\varGamma (1-\alpha )}>\tau _{j}^{-1}\big (a_{k,j-1}+(1+\rho _{j})b_{k,j-1}\big ). \end{aligned}$$

Thus, to prove $g_{k,j}>g_{k,j-1}$, it suffices to prove

$$\begin{aligned} \tau _{j+1}^{-1} b_{k,j-1} \ge \tau _{j}^{-1}\big (b_{k,j-2}-(2+\rho _j)b_{k,j-1}\big ). \end{aligned}$$

Invoking Lemma 3, we see that it suffices to prove

$$\begin{aligned} \big (\tau _{j+1}^{-1} + (2+\rho _j)\tau _{j}^{-1}\big )\frac{\tau _j^3}{\tau _{j+1}+\tau _j} \ge \tau _{j}^{-1}\frac{\tau _{j-1}^3}{\tau _{j}+\tau _{j-1}}. \end{aligned}$$

But one can easily rearrange this inequality into (12), which by hypothesis is true. To complete the proof of (13), we need to show that $g_{k,k}>g_{k,k-1}$ for $k\ge 2$. Now

$$\begin{aligned} g_{k,k}=\tau _{k+1}^{-1}(a_{k,k}+b_{k,k-1}) \text { and } g_{k,k-1}=\tau _{k}^{-1}(a_{k,k-1}+b_{k,k-2}-b_{k,k-1}). \end{aligned}$$

But

$$\begin{aligned} \tau _{k+1}^{-1}a_{k,k} = \frac{\sigma (\sigma \tau _{k+1})^{-\alpha }}{\varGamma (2-\alpha )}\ \text { and } \tau _{k}^{-1}\big (a_{k,k-1}+(1+\rho _{k})b_{k,k-1}\big ) < \frac{(\sigma \tau _{k+1})^{-\alpha }}{\varGamma (1-\alpha )} \end{aligned}$$

(41)

by definition (2) and Lemma 2. Hence $\tau _{k+1}^{-1}a_{k,k}>\tau _{k}^{-1}\big (a_{k,k-1}+(1+\rho _{k})b_{k,k-1}\big )$ since $\sigma \ge 1-\alpha /2$ and $\varGamma (2-\alpha ) = (1-\alpha )\varGamma (1-\alpha )$. To finish the proof that $g_{k,k}>g_{k,k-1}$, imitate the earlier case $k\ge 3$ and $2\le j\le k-1$.

Now we turn to inequality (15). By (6), one has

$$\begin{aligned}&(2\sigma -1)g_{k,k}-\sigma g_{k,k-1} \\&\quad =(2\sigma -1)\tau _{k+1}^{-1}(a_{k,k}+b_{k,k-1})-\sigma \tau _k^{-1}(a_{k,k-1}+b_{k,k-2}-b_{k,k-1}) \\&\quad =(2\sigma -1)\tau _{k+1}^{-1}a_{k,k}-\sigma \tau _k^{-1}\big (a_{k,k-1}+(1+\rho _k)b_{k,k-1}\big ) \\&\qquad +\big [(2\sigma -1)\tau _{k+1}^{-1}+\sigma (2+\rho _k)\tau _k^{-1}\big ]b_{k,k-1}-\sigma \tau _k^{-1} b_{k,k-2} \\&\quad \ge \frac{1}{\sigma }\left\{ \left[ \left( 2-\frac{1}{\sigma }\right) \tau _{k+1}^{-1}+(2+\rho _k)\tau _k^{-1}\right] b_{k,k-1}-\tau _k^{-1} b_{k,k-2}\right\} \\&\quad >0, \end{aligned}$$

using (41) and $\sigma \ge 1-\alpha /2$ for the penultimate inequality, and Lemma 3 and (14) for the final inequality. Thus (15) is proved. $\square$