A Robust Preconditioner for Two-dimensional Conservative Space-Fractional Diffusion Equations on Convex Domains

Abstract

Preconditioners are popularly used for speeding up Krylov subspace iterative method for solving linear systems from discretization of fractional differential equations (FDEs) on rectangular domains. Though some recent works have been developed for FDEs in general convex domains, it still has room for improvement in the design of preconditioner. For this sake, in this paper, we theoretically study the preconditioner problem for two-dimensional conservative space-fractional diffusion equations with variable coefficients and propose a robust preconditioner with penalty term which can deal with any convex domains significantly. We further prove that the proposed preconditioner equals to the coefficient matrix plus a low rank matrix and a matrix with small norm under certain conditions. It implies that the new preconditioner can effectively accelerate the convergence rate of Krylov subspace method. Experimental results show the good performance of the robust preconditioner.

Appendix

1.1 A. Proof of Lemma 1

Proof

Define $f_a(x)=-\frac{1}{2}(x+\frac{1}{2})^\alpha +\frac{1}{\alpha +1}[(x+1)^{\alpha +1}-(x+\frac{1}{2})^{\alpha +1}]$, then we have $a_m^{(\alpha )}=f_a(m)-f_a(m-1)$. Observed that

$$\begin{aligned} f'_a(x)= & {} -\frac{\alpha }{2}(x+1)^{\alpha -1}+(x+1)^{\alpha }-\left( x+\frac{1}{2}\right) ^{\alpha }\\= & {} \frac{(2-2t)^{-\alpha }}{t^{1-\alpha }}[t^{1-\alpha }-(1-\alpha )t-\alpha ], \end{aligned}$$

with $t=\frac{x+\frac{1}{2}}{x+1}$, which satisfies $0<t<1$ provided $x\ge 1$.

Denote $q_{a}(t)=t^{1-\alpha }-[(1-\alpha )t+\alpha ]$, then $q'_{a}(t)=(1-\alpha )(t^{-\alpha }-1)>0,$ so that $q_{a}(t)$ is a monotonic increasing function and the root locates at $t=1$, which implies that $q_a(t)<0$ when $0<t<1$. It is clear that $\frac{(2-2t)^{-\alpha }}{t^{1-\alpha }}>0$, hence $f'_a(x)<0$, $t\in (0,1)$, which yields $\{a^{(\alpha )}_m\}$ is negative for $m\ge 1$. Similarly,

$$\begin{aligned} f''_a(x)= & {} -\frac{\alpha (\alpha -1)}{2}(x+1)^{\alpha -2}+\alpha (x+1)^{\alpha -1}-\alpha \left( x+\frac{1}{2}\right) ^{\alpha -1}\\= & {} \frac{\alpha (2-2t)^{1-\alpha }}{t^{2-\alpha }}[t^{2-\alpha }-(2-\alpha )t-(\alpha -1)]. \end{aligned}$$

Let $p_a(t)=t^{2-\alpha }-[(2-\alpha )t+(\alpha -1)]$. Due to $p'_{a}(t)=(2-\alpha )(t^{1-\alpha }-1)<0$ and $p_{a}(1)=0$, we obtain $p_a(t)>0$ when $0<t<1$. Hence $f''_a(x)>0$ for $t\in (0,1)$, which implies $\{a^{(\alpha )}_m\}$ is an increasing sequence for $m\ge 1$.

In the same way, $\{b_m^{(\alpha )}\}$ can be written as $b_m^{(\alpha )} = f_b(m)-f_b(m-1)$, where

$$\begin{aligned} f_b(x)=-\frac{1}{2}\left( x+\frac{1}{2}\right) ^\alpha +\frac{1}{\alpha +1}\left[ \left( x+\frac{1}{2}\right) ^{\alpha +1}-x^{\alpha +1}\right] . \end{aligned}$$

Let $t=\frac{x}{x+\frac{1}{2}}$, then

$$\begin{aligned} f'_b(x)= & {} -(2-2t)^{-\alpha }(t^\alpha -\alpha t-1+\alpha ), \\ f''_b(x)= & {} -\alpha (2-2t)^{1-\alpha }[t^{\alpha -1}-(\alpha -1)t-2+\alpha ]. \end{aligned}$$

Note that $t^\alpha -\alpha t-1+\alpha < 0$ and $t^{\alpha -1}-(\alpha -1)t-2+\alpha >0$ when $0<t<1$. One can similarly prove that $f'_b(x)>0$ and $f''_b(x)<0$. Therefore, $\{b_m^{(\alpha )}\}$ is a positive and decreasing sequence.

As for the sequence $\{c_m^{(\alpha )}\}$, we firstly define $f_c(x)=\frac{1}{2}[(x+\frac{3}{2})^\alpha -(x+\frac{1}{2})^\alpha ]$ and $g_c(x)=\frac{1}{1+\alpha }[(x+\frac{3}{2})^{\alpha +1}+(x+\frac{1}{2})^{\alpha +1}-2(x+1)^{\alpha +1}],$ then

$$\begin{aligned} c_m^{(\alpha )}= & {} f_c(m)-f_c(m-1)+g_c(m)-g_c(m-1). \end{aligned}$$

Define $k_c(x)=x^{\alpha }$, then we know that $k''_c(x)=\alpha (\alpha -1)x^{\alpha -2}<0$ and $k_c(x)$ is concave. Hence, $g_c'(m)=k_c(m+\frac{3}{2})+k_c(m+\frac{1}{2})-2k_c(m+1)<0$. Besides, it is clear that $f'_c(m)<0$, thus $c^{(\alpha )}_{m} < 0$. Let $q_c(x)=x^{\alpha -1}$, then

$$\begin{aligned} q''_c(x)=(\alpha -1)(\alpha -2)x^{\alpha -3}>0, \end{aligned}$$

which implies that $q_c(x)$ is a convex function, i.e.,

$$\begin{aligned}&q_c\left( m+\frac{3}{2}\right) +q_c\left( m-\frac{1}{2}\right) \ge 2q_c\left( m+\frac{1}{2}\right) \quad \mathrm{and}\quad q_c\left( m+\frac{3}{2}\right) \\&\quad +q_c\left( m+\frac{1}{2}\right) \ge 2q_c(m+1). \end{aligned}$$

Therefore,

$$\begin{aligned} f'_c(m)-f'_c(m-1)= & {} \frac{\alpha }{2} \left[ q_c\left( m+\frac{3}{2}\right) -2q_c\left( m+\frac{1}{2}\right) +q_c\left( m-\frac{1}{2}\right) \right] \ge 0, \end{aligned}$$

and

$$\begin{aligned} g''_c(m)= & {} \alpha \left[ q_c\left( m+\frac{3}{2}\right) +q_c\left( m+\frac{1}{2}\right) -2q_c (m+1)\right] \ge 0, \end{aligned}$$

which yields that

$$\begin{aligned} g'_c(m)-g'_c(m-1)\ge 0. \end{aligned}$$

Consequently $\{c_m^{(\alpha )}\}$ is an increasing sequence for $m\ge 1$.

1.2 B. Proof of Lemma 3

Proof

In accordance with the proof of the Lemma 1,

$$\begin{aligned} c_m^{(\alpha )}= & {} f_c(m)-f_c(m-1)+g_c(m)-g_c(m-1)\\= & {} \int ^{m}_{m-1}f'_c(s)ds+\int ^{m}_{m-1}g'_c(s)ds. \end{aligned}$$

Then

$$\begin{aligned} \left| \int ^{m}_{m-1}f'_c(s)ds\right|= & {} \left| \int ^{m}_{m-1}\frac{\alpha }{2}\left[ \left( s+\frac{3}{2}\right) ^{\alpha -1}-\left( s+\frac{1}{2}\right) ^{\alpha -1}\right] ds\right| \\= & {} \frac{|\alpha (\alpha -1)|}{2}\int ^{m}_{m-1}ds \int _{s-1}^{s}\left( k+\frac{3}{2}\right) ^{\alpha -2}dk\\\le & {} \frac{|\alpha (\alpha -1)|}{2}\int ^{m}_{m-1}ds \int _{s-1}^{s}\left( s+\frac{1}{2}\right) ^{\alpha -2}dk\\= & {} \frac{\alpha (1-\alpha )}{2}\int ^{m}_{m-1} \left( s+\frac{1}{2}\right) ^{\alpha -2} ds\\\le & {} \frac{\alpha (1-\alpha )}{2}\int ^{m}_{m-1} \left( m-\frac{1}{2}\right) ^{\alpha -2} ds\\= & {} \frac{\alpha (1-\alpha )}{2}\left( m-\frac{1}{2}\right) ^{\alpha -2} \end{aligned}$$

and

$$\begin{aligned} \left| \int ^{m}_{m-1}g'_c(s)ds\right|= & {} \left| \int ^{m}_{m-1}\left[ \left( s+\frac{3}{2}\right) ^\alpha +\left( s+\frac{1}{2}\right) ^\alpha -2(s+1)^\alpha \right] ds\right| \\= & {} \left| \alpha (\alpha -1)\right| \int ^{m}_{m-1}ds\int _{s-\frac{1}{2}}^{s}dk\int ^{k}_{k-\frac{1}{2}}\left( \eta +\frac{3}{2}\right) ^{\alpha -2}d\eta \\\le & {} \frac{\alpha (1-\alpha )}{4}\left( m-\frac{1}{2}\right) ^{\alpha -2}. \end{aligned}$$

As $m\ge 1$ and $0<\alpha <1$, it holds that

$$\begin{aligned} |c_m^{(\alpha )}|\le & {} \left| \int ^{m}_{m-1}f'(s)ds\right| +\left| \int ^{m}_{m-1}g'(s)ds\right| \\\le & {} \frac{\alpha (1-\alpha )}{2}\left( m-\frac{1}{2}\right) ^{\alpha -2}+\frac{\alpha (1-\alpha )}{4}\left( m-\frac{1}{2}\right) ^{\alpha -2}\\\le & {} \frac{3\alpha (1-\alpha )2^{2-\alpha }}{4}\frac{1}{m^{2-\alpha }}. \end{aligned}$$

With the similar proof, when $m>1$, it is satisfied that

$$\begin{aligned} |b_{m}^{(\alpha )}|\le & {} \left| \frac{\alpha }{2}\int ^{m}_{m-1}{\left( s+\frac{1}{2}\right) ^{\alpha -1}}ds\right| + \left| \int ^{m}_{m-1}{\left( s+\frac{1}{2}\right) ^{\alpha }-s^\alpha }ds\right| \\\le & {} \frac{\alpha }{2}\int ^{m}_{m-1}{\left( m-\frac{1}{2}\right) ^{\alpha -1}}ds + \alpha \int ^{m}_{m-1}ds \int ^{s+\frac{1}{2}}_{s}{s^{\alpha -1}}dk\\\le & {} \frac{\alpha }{2}\left( m-\frac{1}{2}\right) ^{\alpha -1}+ \frac{\alpha }{2}(m-1)^{\alpha -1}\\\le & {} 2^{1-\alpha }\alpha \frac{1}{m^{1-\alpha }}. \end{aligned}$$

It is easy to know that $b_1^{(\alpha )}\le 2^{1-\alpha }\alpha$. In sum, we have

$$\begin{aligned} |b_{m}^{(\alpha )}|\le \frac{2^{1-\alpha }\alpha }{m^{1-\alpha }}, \end{aligned}$$

for all $m\ge 1$ and $0<\alpha <1$.

The proof for sequence $\{a_m^{(\alpha )}\}$ is similar to the one of $\{b_m^{(\alpha )}\}$.

1.3 C. Proof of Lemma 6

Proof

First of all, we consider the matrix $A^{x}_{0}$. For any given $\epsilon >0$, take

$$\begin{aligned} N_{0}=\Bigg \lceil \left( \frac{3\alpha (1-\alpha )2^{2-\alpha }}{4{\epsilon }}\right) ^{\frac{1}{2-\alpha }} \max \{\gamma _x^{\frac{1}{2-\alpha }},(1-\gamma _x)^{\frac{1}{2-\alpha }}\}\Bigg \rceil >0, \end{aligned}$$

from Lemma 1 and Lemma 3, for all $n>N_0$, we have

$$\begin{aligned} \sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{0,m}|= & {} \sum ^{\infty }_{m=n+1}\Big |\gamma _{x}(c_{m-1}^{(\alpha )} - c_{m}^{(\alpha )})\Big | \le \gamma _{x}\sum ^{\infty }_{m=N_0+1}\left( c_{m}^{(\alpha )}-c_{m-1}^{(\alpha )}\right) \\\le & {} -\gamma _{x}c_{N_0}^{(\alpha )} \le \frac{3\gamma _x\alpha (1-\alpha )2^{2-\alpha }}{4 N_{0}^{2-\alpha }}\le \frac{\gamma _x\epsilon }{\max \{\gamma _x,1-\gamma _x\}} \le \epsilon , \end{aligned}$$

and

$$\begin{aligned} \sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{0,-m}|=\sum ^{\infty }_{m=n+1}\Big |(1-\gamma _{x})(c_{m-1}^{(\alpha )} - c_{m}^{(\alpha )})\Big |\le \frac{3(1-\gamma _x)\alpha (1-\alpha )2^{2-\alpha }}{4 N_{0}^{2-\alpha }}\le \epsilon . \end{aligned}$$

For the matrix $A^x_1$, take

$$\begin{aligned} N_1=\Bigg \lceil \left( \frac{2^{1-\alpha }\alpha }{\epsilon }\right) ^{\frac{1}{1-\alpha }} \max \{\gamma _x^{\frac{1}{1-\alpha }},(1-\gamma _x)^{\frac{1}{1-\alpha }}\}\Bigg \rceil >0. \end{aligned}$$

For all $n>N_1$, we get

$$\begin{aligned} \sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{1,m}|= & {} \sum ^{\infty }_{m=n+1}\Big |\gamma _{x}(a_{m-1}^{(\alpha )} - a_{m}^{(\alpha )})\Big | \le \gamma _{x}\sum ^{\infty }_{m=N_1+1}\left( a_{m}^{(\alpha )}-a_{m-1}^{(\alpha )}\right) \\\le & {} -\gamma _{x}a_{N_1}^{(\alpha )} \le \frac{2^{1-\alpha }\gamma _x \alpha }{N_1^{1-\alpha }}\le \frac{\gamma _x\epsilon }{\max \big \{\gamma _x, 1-\gamma _x\}}\le \epsilon , \end{aligned}$$

and

$$\begin{aligned} \sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{1,-m}|= & {} \sum ^{\infty }_{m=n+1}\Big |(1-\gamma _{x})(b_{m+1}^{(\alpha )} - b_{m}^{(\alpha )})\Big | \le (1-\gamma _{x})\sum ^{\infty }_{m=N_1+1}\left( b_{m}^{(\alpha )}-b_{m+1}^{(\alpha )}\right) \\\le & {} (1-\gamma _{x})b_{N_1+1}^{(\alpha )}<(1-\gamma _{x})b_{N_1}^{(\alpha )} \le \frac{2^{1-\alpha }(1-\gamma _x)\alpha }{N_1^{1-\alpha }}\le \epsilon . \end{aligned}$$

For the matrix $A^x_{-1}$, with similar way, take

$$\begin{aligned} N_{-1}=\Bigg \lceil \left( \frac{2^{1-\alpha }(1-\gamma _x)\alpha }{\epsilon }\right) ^{\frac{1}{1-\alpha }} \max \{\gamma _x^{\frac{1}{1-\alpha }},(1-\gamma _x)^{\frac{1}{1-\alpha }}\}\Bigg \rceil >0, \end{aligned}$$

for all $n>N_{-1}$, we obtain $\sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{-1,m}|<\epsilon$ and $\sum ^{\infty }_{m=n+1}|w^{(\alpha )}_{-1,-m}|\le \epsilon$.

In sum, the result of the theorem follows by taking $N_c= \max \{N_{-1},N_0,N_1\}$.

1.4 D. Proof of Lemma 7

Proof

Without loss of generality, we focus on the case when the matrix size $N_x$ is even. With the definition of Strang’s circulant approximation, we have

$$\begin{aligned} A^x_j-s(A^x_j)=U_j^x+E_j^x, \end{aligned}$$

where $j=-1$, 0, 1, and

$$\begin{aligned} U_j^x= & {} T_{N_x}(w^{(\alpha )}_{j,{N_x-1}},\ldots ,w^{(\alpha )}_{j,{N_x-N_c}},w^{(\alpha )}_{j,{N_x-N_c-1}}{-}w^{(\alpha )}_{j,{-N_c-1}},\ldots , w^{(\alpha )}_{j,{\frac{N_x}{2}+1}}{-}w^{(\alpha )}_{j,{-\frac{N_x}{2}+1}},\\&w^{(\alpha )}_{j,{\frac{N_x}{2}}},0,\ldots ,0;0;0,\ldots ,0,w^{(\alpha )}_{j,{-\frac{N_x}{2}}},w^{(\alpha )}_{j,{-\frac{N_x}{2}-1}}{-}w^{(\alpha )}_{j,{\frac{N_x}{2}-1}},\ldots ,\\&w^{(\alpha )}_{j,{-N_x+N_c+1}}{-}w^{(\alpha )}_{j,{N_c+1}},w^{(\alpha )}_{j,{-N_x+N_c}}, \ldots ,w^{(\alpha )}_{j,{1-N_x}}),\\ E_j^x= & {} -T_{N_x}(w^{(\alpha )}_{j,{-1}}, w^{(\alpha )}_{j,{-2}},\ldots ,w^{(\alpha )}_{j,{-N_c}},0,\ldots ,0;0;0,\ldots ,0,w^{(\alpha )}_{j,{N_c}},w^{(\alpha )}_{j,{N_c-1}},\ldots ,w^{(\alpha )}_{j,{1}}). \end{aligned}$$

Obviously, the rank of matrix $E_j^x$ is $2N_c$. On the other hand, since $U_j^x$ is a Toeplitz matrix, with Lemma 6, we have

$$\begin{aligned} \Vert U_j^x\Vert _1\le & {} \max \left\{ \sum _{m=\frac{N_x}{2}}^{N_x-1}|w^{(\alpha )}_{j,m}|+\sum _{m=N_c+1}^{\frac{N_x}{2}-1}|w^{(\alpha )}_{j,-m}|, \sum _{m=\frac{N_x}{2}}^{N_x-1}|w^{(\alpha )}_{j,-m}|+\sum _{m=N_c+1}^{\frac{N_x}{2}-1}|w^{(\alpha )}_{j,m}|\right\} \\\le & {} \max \{2\epsilon ,2\epsilon \}=2\epsilon . \end{aligned}$$

In a similar way, we know that $\Vert U_j^x\Vert _\infty \le 2\epsilon$, therefore,

$$\begin{aligned} \Vert U^x_j\Vert _2\le \sqrt{\Vert U^x_j\Vert _1 \cdot \Vert U^x_j\Vert _\infty }=2\epsilon . \end{aligned}$$

When $N_x$ is odd, the theorem can be proven similarly.

1.5 E. Proof of Lemma 8

Proof

With the assumption that P is invertible, it holds that

$$\begin{aligned} {\hat{P}}-P= & {} P(P^{-1}{\hat{P}}-I_N)\\= & {} P\Big [\big ((I_N-\varTheta )P_r^{-1}+\varTheta \big (P_r+\frac{1}{\eta }I_N\big )^{-1}\big )\big ((I_{N}-\varTheta )P_r+\varTheta \big (P_r+\frac{1}{\eta }I_N\big )\big )-I_N\Big ]\\= & {} P\big [ (I_N-\varTheta )P_r^{-1}P_r -(I_N-\varTheta )P_r^{-1} \varTheta P_r + \varTheta \big (P_r+\frac{1}{\eta }I_N\big )^{-1}\big (P_r+\frac{1}{ \eta }I_N\big )\\&-\varTheta \big (P_r+\frac{1}{\eta }I_N\big )^{-1} (I_N-\varTheta )\big (P_r+\frac{1}{ \eta }I_N\big ) -I_N\big ]\\= & {} P\big [-(I_N-\varTheta )P_r^{-1} \varTheta P_r -\varTheta \big (P_r+\frac{1}{\eta }I_N\big )^{-1} (I_N-\varTheta )\big (P_r+\frac{1}{ \eta }I_N\big )\big ]. \end{aligned}$$

Since

$$\begin{aligned}&\mathrm{rank}(-(I_N-\varTheta )P_r^{-1} \varTheta P_r)\le \min \{\mathrm{rank}(I_N-\varTheta ),\mathrm{rank}(\varTheta )\},\\&\mathrm{rank}\big (-\varTheta \big (P_r+\frac{1}{\eta }I_N\big )^{-1} (I_N-\varTheta )\big (P_r+\frac{1}{ \eta }I_N\big )\big )\le \min \{\mathrm{rank}(\varTheta ),\mathrm{rank}(I_N-\varTheta )\}, \end{aligned}$$

it implies that

$$\begin{aligned} \mathrm{rank}({\hat{P}}-P)\le 2\min \{\mathrm{rank}(\varTheta ), \mathrm{rank}(I_N-\varTheta )\}. \end{aligned}$$