Optimized Overlapping Schwarz Waveform Relaxation for a Class of Time-Fractional Diffusion Problems

Abstract

For parabolic PDEs with integer-order temporal derivative, if we use the Schwarz waveform relaxation (SWR) algorithm with Robin transmission conditions as the solver, the so-called equioscillation principle is an important concept to get a good Robin parameter, which has a significant effect on the convergence rate of the algorithm. Surprisingly, as we show in this paper such a principle may result in rather disappointing Robin parameter for the SWR algorithm when we use it to solve time-fractional PDEs. For a class of time-fractional diffusion equations, by analyzing a new min–max problem we get much better Robin parameter, which is found very close to the best one that we can make through numerical optimizations and numerical experiments. To use the SWR algorithm in practice, we apply the kernel reduction technique proposed recently by Baffet and Hesthaven to treat the convolutions with kernel function of the form $\mathscr {K}_{\gamma }(t)=t^{-\gamma }/\varGamma (1-\gamma )$, where $\gamma \in (0, 1)$. For time-fractional PDEs with this kind of kernel function, the kernel reduction technique results in efficient one-step numerical schemes. Numerical results obtained by using this technique confirm our theoretical conclusions very well.

The Proof of Lemma 2.4

Proof

For the function $\mathscr {R}$ defined by (2.9b), we have

$$\begin{aligned} \partial _q\mathscr {R}(y,q)=4y\frac{q^2-(1+\phi )y^2}{\left[ (y+q)^2+\phi y^2\right] ^2}e^{-y}= 4y\frac{(q+\sqrt{1+\phi }y)(q-\sqrt{1+\phi }y)}{\left[ (y+q)^2+\phi y^2\right] ^2}e^{-y}. \end{aligned}$$

(5.1)

Hence, for $y\in [y_0, y_1]$ the solution $q_\mathrm{opt}$ of (2.9a) satisfies $\sqrt{1+\phi }y_0\le q_\mathrm{opt}\le \sqrt{1+\phi }y_1$, since otherwise, if $q_\mathrm{opt}<\sqrt{1+\phi }y_0$ (resp. $q_\mathrm{opt}>\sqrt{1+\phi }y_1$), we can further uniformly reduce $\mathscr {R}$ by increasing (resp. decreasing) the value of $q_\mathrm{opt}$. Hence, it suffices to study

$$\begin{aligned} \underset{q\in [q_0, q_1]}{\min }\underset{y\in [y_0, y_1]}{\max }\mathscr {R}(y,q), \end{aligned}$$

(5.2)

where $q_0=y_0\sqrt{1+\phi }$ and $q_1=y_1\sqrt{1+\phi }$.

A routine calculation shows that, for a given $q\in [q_0, q_1]$ the function $\mathscr {R}$ has at most two local extremums, located at

$$\begin{aligned} \bar{y}_{\pm }(q)=\frac{1}{1+\phi }\sqrt{(1-\phi )q^2+2(1+\phi )q\pm 2q\sqrt{d(q)}}, \end{aligned}$$

where $d(q)=-\phi q^2-2\phi (1+\phi )q+(1+\phi )^2$ is the discriminant. Since $\mathscr {R}(y,q)\rightarrow 0$ as $y\rightarrow +\infty$, the larger one $\bar{y}_+(q)$ must be a local maximizer. This local maximizer does not always exist and by letting $d(q)>0$ we know that it exists for $q\in \left( 0, q_d\right)$, where $q_d=(1+\phi )\left( \sqrt{1+{1}/{\phi }}-1\right)$. By using $\mathscr {R}(+\infty ,q)=0$ and $\mathscr {R}(0,q)=1$, we know that for $q=q_d$, i.e., $d(q)=0$, the function $\mathscr {R}$ is strictly decreasing for $y>0$.

Case (A) If $y_0\ge \sqrt{1+\phi }\left( \sqrt{1+{1}/{\phi }}-1\right)$. In this case, we have $d(q)\le 0$ for $q\in [q_0, q_1]$. Hence, $\mathscr {R}_{\max }(q):={\max }_{y\in [y_0, y_1]}\mathscr {R}(y,q)=\mathscr {R}_{\dag }(q)$, with $\mathscr {R}_{\dag }(q):=\max \left\{ \mathscr {R}(y_0,q), \mathscr {R}(y_1,q)\right\}$. From (5.1), it is easy to know that the function $\mathscr {R}(y_0,q)$ (resp. $\mathscr {R}(y_1,q)$) is an increasing (resp. decreasing) function of q when $q\in [q_0, q_1]$. Then, it can be shown that for $q\in [q_0, q_1]$ the function $\mathscr {R}_{\dag }$ attains its global minimum at $q=q_{\dag }$, where $q_{\dag }$ is given by (2.10b); this implies $q_\mathrm{opt}=q_{\dag }$. We have

$$\begin{aligned} \mathscr {R}_{\dag }(q):=\max \left\{ \mathscr {R}(y_0,q), ~~\mathscr {R}(y_1,q)\right\} = {\left\{ \begin{array}{ll} \mathscr {R}(y_1,q), &{}\text {if}~q\in [q_0, q_{\dag }),\\ \mathscr {R}(y_0,q), &{}\text {if}~q\in (q_{\dag }, q_1], \end{array}\right. } \end{aligned}$$

(5.3)

where, for any real numbers $I_1$ and $I_2$, the set $[I_1, I_2)$ (or $(I_1, I_2]$) is empty if $I_1\ge I_2$.

Case (B) $y_0<\sqrt{1+\phi }\left( \sqrt{1+{1}/{\phi }}-1\right) <y_1$. In this case, it is clear that $q_1>q_d>q_0$. We claim $\bar{y}_+(q)>y_0$ for $q\in \left[ q_0, q_d\right]$. Note that, for any $q\in [q_0, q_1]$ we have

$$\begin{aligned} \bar{y}_+(q)>y_0\Leftrightarrow (1-\phi )q^2+2(1+\phi )q+2q\sqrt{d(q)}>(1+\phi )^2y_0^2. \end{aligned}$$

Hence, it is sufficient to prove $(1-\phi )q_0^2+2(1+\phi )q_0>(1+\phi )^2y_0^2$. Since

$$\begin{aligned} (1-\phi )q_0^2+2(1+\phi )q_0>(1+\phi )^2y_0^2\Leftrightarrow y_0< {\sqrt{1+\phi }}/{\phi }, \end{aligned}$$

it is sufficient to prove $\sqrt{1+\phi }\left( \sqrt{1+{1}/{\phi }}-1\right) <{\sqrt{1+\phi }}/{\phi }$. By noticing

$$\begin{aligned} \forall \phi \in (0, 1),~ \sqrt{{1}/{\phi }+1}>1\Rightarrow \sqrt{1+\phi }\left( \sqrt{1+ {1}/{\phi }}-1\right) <{\sqrt{1+\phi }}/{\phi }, \end{aligned}$$

we finally get $\bar{y}_+(q)>y_0$ for $q\in \left[ q_0, q_d\right]$. Now, by using $q_d<q_1$ and the continuity of $\max _{y\in [y_0, y_1]}\mathscr {R}(y,q)$ with respect to q, we have

$$\begin{aligned} \mathscr {\bar{R}}(q_d)\le \mathscr {R}^{\dag }(q_d). \end{aligned}$$

(5.4)

We next claim ${d\mathscr {\bar{R}}(q)}/{dq}<0$ for $q\in [q_0, q_d]$. From (5.1), we have

$$\begin{aligned} \begin{aligned} \frac{d\mathscr {\bar{R}}(q)}{dq}&=\frac{\partial \mathscr {R}(\bar{y}_+,q)}{\partial q}+\frac{\partial \mathscr {R}(\bar{y}_+,q)}{\partial y}\times \frac{d\bar{y}_+(q)}{dq}\\&=4\bar{y}_+\frac{(q+\sqrt{1+\phi } \bar{y}_+)(q-\sqrt{1+\phi }\bar{y}_+)}{\left[ (\bar{y}_++q)^2+\phi \bar{y}_+^2\right] ^2}e^{-\bar{y}_+}, \end{aligned} \end{aligned}$$

where we used the fact that $\bar{y}_+$ is the local maximizer of $\mathscr {R}$, that is $\frac{\partial \mathscr {R}(\bar{y}_+,q)}{\partial y}=0$. Then, it holds that

$$\begin{aligned} q-\sqrt{1+\phi }\bar{y}_+(q)<0\Leftrightarrow \phi q-1-\phi<\sqrt{d(q)}\Leftarrow \phi q-1-\phi<0\Leftarrow \phi q_d-1-\phi <0, \end{aligned}$$

where the last inequality holds for any $\phi \in (0, 1)$, because

$$\begin{aligned} \phi q_d-1-\phi<0\Leftrightarrow \phi \left( \sqrt{1+{1}/{\phi }}-1\right) -1<0\Leftrightarrow \sqrt{\phi }<\sqrt{1+\phi }. \end{aligned}$$

Therefore, ${d\mathscr {\bar{R}}(q)}/{dq}<0$ and this implies that the function $\mathscr {\bar{R}}(q)$ is decreasing for $q\in [q_0, q_d]$. Based on the above analysis, we next consider the following two cases.

${{\textcircled {{1}}}}$ $q_{\dag }<q_d$, ${\bar{y}}(q_{\dag })<y_1$ and $\mathscr {\bar{R}}(q_{\dag })>\mathscr {R}^{\dag }(q_{\dag })$. In this case, it must hold $q_{\dag }<q_1$, since otherwise we have $q_d>q_1$ and this leads to contradiction. Hence, it holds that $\mathscr {R}^{\dag }(q)=\mathscr {R}(y_0, q)$ for $q\in [q_{\dag }, q_1]$. Then, by using ${\bar{y}}(q_{\dag })<y_1$, $\mathscr {\bar{R}}(q_{\dag })>\mathscr {R}^{\dag }(q_{\dag })$ and (5.4), we know that the equation $\mathscr {R}(y_0, q)=\mathscr {\bar{R}}(q)$ has a unique root $q_0^*\in (q_{\dag }, q_d)$ and that $\mathscr {R}_{\max }(q)\left( :=\max _{y\in [y_0, y_1]}\mathscr {R}(y,q)\right)$ attains its global minimum at $q_0^*$, because $\mathscr {\bar{R}}(q)$ and $\mathscr {R}(y_0,q)$) are respectively decreasing and increasing functions of q.

${{\textcircled {{2}}}}$ If any one of the conditions in ${{\textcircled {{1}}}}$ is not satisfied, we claim

$$\begin{aligned} \max _{y\in [y_0, y_1]}\mathscr {R}(y,q_{\dag })=\mathscr {R}^{\dag }(q_{\dag }). \end{aligned}$$

(5.5)

Indeed, this can be proved by using (5.4) and the fact that “both $\mathscr {\bar{R}}(q)$ and $\mathscr {R}(y_1,q)$ are decreasing functions, but $\mathscr {R}(y_0, q)$ is an increasing function”. For example, if $q_{\dag }=q_d$, by using (5.4) we know that (5.5) holds. For the case $q_{\dag }>q_d$ or ${\bar{y}}(q_{\dag })\ge y_1$, (5.5) can be proved similarly. Now, let $\mathbb {Q}=\left\{ q: q\in [q_0, q_d]\text { and }\bar{y}_+(q)<y_1\right\}$ and $\mathbb {Q}_0=[q_0,q_1]/\mathbb {Q}$. Then, we have

$$\begin{aligned} \begin{aligned}&\min _{q\in [q_0, q_1]}\max _{y\in [y_0, y_1]}\mathscr {R}(y,q)=\min \left\{ \min _{q\in \mathbb {Q}}\max \left\{ \mathscr {\bar{R}}(q), \mathscr {R}^{\dag }(q)\right\} , \min _{q\in \mathbb {Q}_0}\mathscr {R}^{\dag }(q)\right\} \\&\quad \ge \min \left\{ \min _{q\in \mathbb {Q}} \mathscr {R}^{\dag }(q), \min _{q\in \mathbb {Q}_0}\mathscr {R}^{\dag }(q)\right\} =\min _{q\in [q_0, q_1]}\mathscr {R}^{\dag }(q)=\mathscr {R}^{\dag }(q_{\dag }). \end{aligned} \end{aligned}$$

(5.6)

This, together with (5.5), implies that the function $\mathscr {R}_{\max }(q)\left( =\max _{y\in [y_0, y_1]}\mathscr {R}(y,q)\right)$ can reach the lower bound $\mathscr {R}^{\dag }(q_{\dag })$; hence, the conclusion $q_\mathrm{opt}=q_{\dag }$ follows.

Case (C) $y_1\le \sqrt{1+\phi }\left( \sqrt{1+{1}/{\phi }}-1\right)$. In this case, we have $q_d\ge q_1$ and therefore the local maximizer $\bar{y}_{+}(q)$ exists for all $q\in [q_0, q_1]$. The first case in (2.12) has been analyzed in Case (B). For $\bar{y}_+\left( q_{\dag }\right) < y_1$, $\mathscr {\bar{R}}(q_{\dag })>\mathscr {{R}}^{\dag }(q_{\dag })$ and $\bar{y}_+(q_1)\ge y_1$, the analysis is also similar to Case (B). If $\bar{y}_+\left( q_{\dag }\right) < y_1$, $\mathscr {\bar{R}}(q_{\dag })>\mathscr {{R}}^{\dag }(q_{\dag })$, $\bar{y}_+(q_1)<y_1$ and $\mathscr {\bar{R}}(q_1)<\mathscr {{R}}^{\dag }(q_1)$, we have

$$\begin{aligned} \mathscr {R}_{\max }(q_{\dag })=\max _{y\in [y_0, y_1]}\mathscr {R}(y, q_{\dag })=\mathscr {\bar{R}}(q_{\dag }),~~\mathscr {R}_{\max }(q_1)=\max _{y\in [y_0, y_1]}\mathscr {R}(y, q_1)=\mathscr {R}^{\dag }(q_1). \end{aligned}$$

This implies that, the equation $\mathscr {\bar{R}}(q)=\mathscr {R}^{\dag }(q)$ has a unique solution $q^*_0\in (q_{\dag }, q_1)$, which is the global minimizer, because $\mathscr {\bar{R}}(q)$ and $\mathscr {R}^{\dag }(q)$ are respectively decreasing and increasing functions for $q\in (q_{\dag }, q_1)$. For the last case, if $q_{\dag }=q_1$, by using the fact that “both $\mathscr {\bar{R}}(q)$ and $\mathscr {R}(y_1,q)$ are decreasing functions”, it is easy to get $q_\mathrm{opt}=q_1$; otherwise, it shall holds that $\mathscr {\bar{R}}(q)\ge \mathscr {R}^{\dag }(q)$ for all $q\in [q_{\dag }, q_1]$ and because of the decreasing property of $\mathscr {\bar{R}}(q)$, we also have $q_\mathrm{opt}=q_1$. $\square$