An Aggregation-Based Two-Grid Method for Multilevel Block Toeplitz Linear Systems

Abstract

This paper presents an aggregation-based two-grid method for solving a multilevel block Toeplitz system. Different from the existing multigrid methods for multilevel block Toeplitz systems, we aggregate a given multilevel block Toeplitz matrix to a new multilevel Toeplitz matrix in such a way that a very sparse coarse grid matrix is constructed in practice. Then, we give an asymptotically tight bound of the convergence rate and provide an algorithm for selecting the optimal prolongation vector and the relaxation factor for our method. Numerical experiments on artificial examples are provided for visualizing the correctness of our analysis, while experiments associated with practical examples show the efficiency of our method in terms of computing time.

Appendix

In the appendix, we discuss the properties of the function $\textbf{f}_p(\varvec{\theta })$ defined as (44) and $\alpha ^{*}(\varvec{\theta })$ defined as (21) in Lemma 1.

Lemma 2

Let $\textbf{f}(\varvec{\theta }):[-\pi , \pi ]^{d} \rightarrow \mathbb {C}^{s \times s}$ be the matrix-valued function which satisfies the AS.1-AS.3 and $p_s\in \mathbb {R}^s$. The matrix-valued function

$$\begin{aligned} \textbf{f}_p(\varvec{\theta }) = {\left\{ \begin{array}{ll} \textbf{f}(\varvec{\theta }) - \frac{\textbf{f}(\varvec{\theta })p_sp_s^T\textbf{f}(\varvec{\theta })}{p_s^T\textbf{f}(\varvec{\theta })p_s}, &{} \quad \textbf{f}(\varvec{\theta })p_s\ne 0\;\\ \textbf{f}(\varvec{\theta }) , &{}\quad \textbf{f}(\varvec{\theta })p_s = 0\; \end{array}\right. } \end{aligned}$$

is semi-positive definite and continuous for $\varvec{\theta }\in [-\pi , \pi ]^{d}$. And $p_s$ is the zero eigenvector of $\textbf{f}_p(\varvec{\theta })$ for all $\varvec{\theta } \in [-\pi , \pi ]^{d}$.

Proof

Based on the Cauchy–Schwarz inequality, we can deduce that the function $\textbf{f}_p(\varvec{\theta })$ is semi-positive definite. Further, we have to discuss the continuity of $\textbf{f}_p$.

If $\textbf{f}(\varvec{\theta })p_s\ne 0$ for all $\varvec{\theta }\in [-\pi , \pi ]^{d}$, we have that

$$\begin{aligned} \textbf{f}_p(\varvec{\theta }) = \textbf{f}(\varvec{\theta }) - \frac{\textbf{f}(\varvec{\theta })p_sp_s^T\textbf{f}(\varvec{\theta })}{p_s^T\textbf{f}(\varvec{\theta })p_s} \end{aligned}$$

is continuous. And If there exists $\varvec{\theta }_0$ such that $\textbf{f}(\varvec{\theta }_0)p_s=0$, we have to prove that

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{\textbf{f}(\varvec{\theta })p_sp_s^T\textbf{f}(\varvec{\theta })}{p_s^T\textbf{f}(\varvec{\theta })p_s} = 0. \end{aligned}$$

(43)

The equation (43) is equal to for all $v\in \mathbb {C}^s$,

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{|v^H\textbf{f}(\varvec{\theta })p_s|^2}{p_s^T\textbf{f}(\varvec{\theta })p_s} = 0. \end{aligned}$$

(44)

Using the Cauchy–Schwarz inequality, we obtain that

$$\begin{aligned} \frac{|r(\varvec{\theta })^H\textbf{f}(\varvec{\theta })p_s|^2}{p_s^H\textbf{f}(\varvec{\theta })p_s} \le \Vert r(\varvec{\theta })\Vert ^2\frac{\Vert \textbf{f}(\varvec{\theta })p_s\Vert ^2}{p_s^H\textbf{f}(\varvec{\theta })p_s} = \Vert r(\varvec{\theta })\Vert ^2 \cdot \frac{p_s^H\textbf{f}(\varvec{\theta })^2p_s}{p_s^H\textbf{f}(\varvec{\theta })p_s}. \end{aligned}$$

Thus, if we prove that

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{p_s^H\textbf{f}(\varvec{\theta })^2p_s}{p_s^H\textbf{f}(\varvec{\theta })p_s} = 0, \end{aligned}$$

the equation (44) holds. Since $\textbf{f}(\varvec{\theta })$ is Hermitian, its spectral decomposition can be formed as:

$$\begin{aligned} \textbf{f}(\varvec{\theta }) = Q(\varvec{\theta })\Lambda (\varvec{\theta })Q(\varvec{\theta })^H = Q(\varvec{\theta })\begin{bmatrix} \lambda _0(\varvec{\theta }) &{} \\ &{} \Lambda _1(\varvec{\theta }) \end{bmatrix}Q(\varvec{\theta })^H \end{aligned}$$

where $Q(\varvec{\theta })$ and $\Lambda (\varvec{\theta })$ represents the eigenvector and eigenvalue matrix, respectively and

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \lambda _0(\varvec{\theta }) = 0. \end{aligned}$$

Since $\textbf{f}(\varvec{\theta }_0)p_s = 0$,

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} Q(\varvec{\theta })^Hp_s = [1,0,\ldots ,0]^T. \end{aligned}$$

Thus,

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{p_s^H\textbf{f}(\varvec{\theta })^2p_s}{p_s^H\textbf{f}(\varvec{\theta })p_s} = \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{p_s^HQ(\varvec{\theta })\Lambda ^2(\varvec{\theta })Q(\varvec{\theta })^Hp_s}{p_s^HQ(\varvec{\theta })\Lambda (\varvec{\theta })Q(\varvec{\theta })^Hp_s} = \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0} \frac{ \lambda _0^2(\varvec{\theta })}{ \lambda _0(\varvec{\theta })} = 0. \end{aligned}$$

(45)

Moreover, we have that for all $\varvec{\theta } \in [-\pi , \pi ]^{d}$

$$\begin{aligned} \textbf{f}_p(\varvec{\theta })p_s = {\left\{ \begin{array}{ll} \textbf{f}(\varvec{\theta })p_s - \frac{\textbf{f}(\varvec{\theta })p_sp_s^T\textbf{f}(\varvec{\theta })}{p_s^T\textbf{f}(\varvec{\theta })p_s}p_s,&{} \quad \textbf{f}(\varvec{\theta })p_s\ne 0\\ \textbf{f}(\varvec{\theta })p_s ,&{}\quad \textbf{f}(\varvec{\theta })p_s = 0 \end{array}\right. }\quad = 0 \end{aligned}$$

which means $p_s$ is the zero eigenvector of $\textbf{f}_p(\varvec{\theta })$. So far, the proof is complete. $\square$

Lemma 3

The optimizer $\alpha ^*(\varvec{\theta })$ in the proof of Lemma 1 has the form:

$$\begin{aligned} \alpha ^*(\varvec{\theta }) = {\left\{ \begin{array}{ll} -\frac{p_s^H\textbf{f}(\varvec{\theta })r(\varvec{\theta })}{p_s^H\textbf{f}(\varvec{\theta })p_s} ,&{} \quad \textbf{f}(\varvec{\theta })p_s\ne 0\;\\ -p_s^Hr(\varvec{\theta }) ,&{}\quad \textbf{f}(\varvec{\theta })p_s = 0\; \end{array}\right. },\quad r(\varvec{\theta })\in \mathcal {S}^{(\textbf{m})} \end{aligned}$$

where $\textbf{f}(\varvec{\theta }):[-\pi , \pi ]^{d} \rightarrow \mathbb {C}^{s \times s}$ be the matrix-valued function which satisfies the AS.1-AS.3, $p_s\in \mathbb {R}^s$ and

$$\begin{aligned} \mathcal {S}^{(\textbf{m})} = \left\{ r^{(\textbf{m})}(\varvec{\theta })= \sum _{\textbf{k} = -\mathbf {[\frac{m}{2}]}}^{\mathbf {m-[\frac{m}{2}]-1}}r_{\textbf{k}}e^{\imath \textbf{k}\cdot \varvec{\theta }} \in \mathbb {C}^s\ |\ r^{(\textbf{m})}(\varvec{\theta }) \text { is not identically zero}\,\ r_{\textbf{k}}^HD_sp_s = 0 \right\} , \end{aligned}$$

Then, we have that $\alpha ^*(\varvec{\theta })$ is continuous and bounded in $[-\pi , \pi ]^{d}$.

Proof

First, we prove $\alpha ^*(\varvec{\theta })$ is continuous in $[-\pi , \pi ]^{d}$. Since $\textbf{f}(\varvec{\theta })$ and $r(\varvec{\theta })$ are continuous, we only need to consider the continuity of $\alpha ^*(\varvec{\theta })$ at $\varvec{\theta }_0$ which satisfies $f(\varvec{\theta }_0)p_s = 0$. Then, we need to prove

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0}-\frac{p_s^H\textbf{f}(\varvec{\theta })r(\varvec{\theta })}{p_s^H\textbf{f}(\varvec{\theta })p_s} = -p_s^Hr(\varvec{\theta }_0). \end{aligned}$$

As the analysis in (45), since $\textbf{f}(\varvec{\theta })$ has the eigenvalue decomposition as

$$\begin{aligned} \textbf{f}(\varvec{\theta }) = Q(\varvec{\theta })\Lambda (\varvec{\theta })Q(\varvec{\theta })^H = Q(\varvec{\theta })\begin{bmatrix} \lambda _0(\varvec{\theta }) &{} \\ &{} \Lambda _1(\varvec{\theta }) \end{bmatrix}Q(\varvec{\theta })^H, \end{aligned}$$

we have

$$\begin{aligned} \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0}-\frac{p_s^H\textbf{f}(\varvec{\theta })r(\varvec{\theta })}{p_s^H\textbf{f}(\varvec{\theta })p_s}&= \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0}-\frac{p_s^HQ(\varvec{\theta })\Lambda (\varvec{\theta })Q(\varvec{\theta })^Hr(\varvec{\theta })}{p_s^HQ(\varvec{\theta })\Lambda (\varvec{\theta })Q(\varvec{\theta })^Hp_s}\\&= \lim _{\varvec{\theta }\rightarrow \varvec{\theta }_0}-\frac{\lambda _0(\varvec{\theta })e_1^T Q(\varvec{\theta })r(\varvec{\theta }) }{\lambda _0(\varvec{\theta })} \\&= -e_1^T Q(\varvec{\theta }_0)r(\varvec{\theta }_0)\\&= -p_s^Hr(\varvec{\theta }_0). \end{aligned}$$

Thus, we have $\alpha ^*(\varvec{\theta })$ is continuous in $[-\pi , \pi ]^{d}$. Since $[-\pi , \pi ]^{d}$ is a closed interval, $\alpha ^*(\varvec{\theta })$ is bounded. $\square$