Quadratic spline collocation method and efficient preconditioner for the Helmholtz equation with the Sommerfeld boundary conditions

Abstract

In this paper, using the quadratic spline collocation method (QSC), we numerically solve the Helmholtz equations with the Sommerfeld boundary conditions. By reordering the unknowns, we obtain a $3\times 3$ block linear system. Then, we introduce a two-step preconditioner based on the approximate inverse block polynomial preconditioner. Theoretical analysis show this preconditioner can largely gather the eigenvalues around 1. Numerical examples are presented to test the error of QSC method and check the efficiency of the presented preconditioner.

Appendix

Denote by

$$\begin{aligned} e_1= & {} \left( \begin{array}{cccc} 1&0&\ldots&0 \end{array}\right) _{1\times N}, e_N=\left( \begin{array}{cccc} 0&\ldots&0&1 \end{array}\right) _{1\times N}, \mathbf {0}_{1,N}=\left( \begin{array}{ccc} 0&\ldots&0 \end{array}\right) _{1\times N},\\ f_1= & {} \left( \begin{array}{ccccc} 1/2&1/8&0&\ldots&0 \end{array}\right) _{1\times (N+2)}, f_{N+2}=\left( \begin{array}{ccccc} 0&\ldots&0&1/8&1/2 \end{array}\right) _{1\times (N+2)}. \end{aligned}$$

Let

$$\begin{aligned} \overline{A_0}= & {} \widetilde{A_2},\overline{A_1}=1/h^2T^D,\overline{A_2}=1/8(e_1^T,e_N^T), \overline{A_3}=1/h^2(e_1^T,e_N^T),\\ \overline{A_4}= & {} \widetilde{A_{\sigma }}(1,2:N+1),\\ \overline{A_5}= & {} \widetilde{A_{\delta }}(1,2:N+1),\overline{A_6}=1/8e_1^T,\overline{A_7}=1/8e_N^T, \overline{A_8}=1/h^2e_1^T,\overline{A_9}=1/h^2e_N^T,\\ \overline{B_0}= & {} \widetilde{A_3},\overline{B_1}=1/8T^D+I,\overline{B_2}=\widetilde{A_1}, \overline{B_3}=\widetilde{B_1},\overline{B_4}=(\mathbf {0}_{1,N}^T,1/8T^D+I,\mathbf {0}_{1,N}^T)^T,\\ \overline{B_5}= & {} \widetilde{B_2}, \overline{B_6}=f_1^T, \overline{B_7}=f_{N+2}^T,\overline{B_8}=\widetilde{A_{\sigma }},\overline{B_9}=\widetilde{A}_{\delta }.\\ \overline{S_1}= & {} \overline{A_2}\otimes \overline{B_2}+K\overline{A_2}\otimes \overline{B_3}+\overline{A_3}\otimes \overline{B_3}-h^2/12\overline{A_3}\otimes \overline{B_2},\\ \overline{S_2}= & {} \overline{A_1}\otimes \overline{A_6}+K\overline{B_1}\otimes \overline{A_6}+\overline{B_1}\otimes \overline{A_8}-h^2/12\overline{A_1}\otimes \overline{A_8},\\ \overline{S_3}= & {} \overline{A_1}\otimes \overline{A_7}+K\overline{B_1}\otimes \overline{A_7}+\overline{B_1}\otimes \overline{A_9}-h^2/12\overline{A_1}\otimes \overline{A_9}.\\ \end{aligned}$$

Then, there are

$$\begin{aligned} H_{13}=(\overline{S_1},\overline{S_2},\overline{S_3},\mathbf {0}), H_{23}=(\overline{S_1},\overline{S_2},\overline{S_3},\mathbf {0}), \end{aligned}$$

where $\mathbf {0}$ is a $N^2\times (4N+4)$ zero matrix,

$$\begin{aligned} H_{31}=\left( \begin{array}{c} \overline{A_4}\otimes \overline{B_4}\\ \overline{A_5}\otimes \overline{B_4}\\ \overline{B_1}\otimes \overline{A_4}\\ \overline{B_1}\otimes \overline{A_5}\\ \mathbf {0} \end{array}\right) , \quad H_{32}=\left( \begin{array}{c} \mathbf {0}\\ \overline{A_4}\otimes \overline{B_4}\\ \overline{A_5}\otimes \overline{B_4}\\ \overline{B_1}\otimes \overline{A_4}\\ \overline{B_1}\otimes \overline{A_5} \end{array}\right) , \end{aligned}$$

where $\mathbf {0}$ is a $(4N+4)\times N^2$ zero matrix, and

$$\begin{aligned} H_{31}=\left( \begin{array}{cccccccc} \sigma \overline{B_5}&{}\mathbf {0}&{}\overline{A_4}\otimes \overline{B_6}&{}\overline{A_4}\otimes \overline{B_7}&{}1/2\overline{B_5}&{}\mathbf {0}&{}\mathbf {0}&{}\mathbf {0}\\ \mathbf {0}&{}\delta _{N+2}\overline{B_5}&{}\overline{A_5}\otimes \overline{B_6}&{}\overline{A_5}\otimes \overline{B_7}&{}\mathbf {0}&{}1/2\overline{B_5}&{}\mathbf {0}&{}\mathbf {0}\\ \overline{A_6}\otimes \overline{B_8}&{}\overline{A_7}\otimes \overline{B_8}&{}\sigma _1\overline{B_1}&{}\mathbf {0}&{}\overline{A_6}\otimes \overline{A_0}&{}\overline{A_7}\otimes \overline{A_0}&{}1/2\overline{B_1}&{}\mathbf {0}\\ \overline{A_6}\otimes \overline{B_9}&{}\overline{A_7}\otimes \overline{B_9}&{}\mathbf {0}&{}\delta _{N+2}\overline{B_1}&{}\overline{A_6}\otimes \overline{B_0}&{}\overline{A_7}\otimes \overline{B_0}&{}\mathbf {0}&{}1/2\overline{B_1}\\ -1/2\overline{B_5}&{}\mathbf {0}&{}\mathbf {0}&{}\mathbf {0}&{}\sigma _1\overline{B_5}&{}\mathbf {0}&{} \overline{A_4}\otimes \overline{B_6}&{}\overline{A_4}\otimes \overline{B_7}\\ \mathbf {0}&{}-1/2\overline{B_5}&{}\mathbf {0}&{}\mathbf {0}&{}\mathbf {0}&{}\delta _{N+2}\overline{B_5}&{} \overline{A_5}\otimes \overline{B_6}&{}\overline{A_5}\otimes \overline{B_7}\\ -\overline{A_6}\otimes \overline{A_0}&{}-\overline{A_7}\otimes \overline{A_0}&{}-1/2\overline{B_1}&{}\mathbf {0}&{} \overline{A_6}\otimes \overline{B_8}&{}\overline{A_7}\otimes \overline{B_8}&{}\sigma _1\overline{B_1}&{}\mathbf {0}\\ -\overline{A_6}\otimes \overline{B_0}&{}-\overline{A_7}\otimes \overline{B_0}&{}\mathbf {0}&{} -1/2\overline{B_1}&{} \overline{A_6}\otimes \overline{B_9}&{}\overline{A_7}\otimes \overline{B_9}&{}\mathbf {0}&{}\delta _{N+2}\overline{B_1} \end{array}\right) , \end{aligned}$$

Property 1

For small enough step $h>0$, $H_{33}$ is invertible.

Proof

For convenience, we denote by $D=H_{33}$.

By some computations, we know that all the lines of D are strictly diagonally dominant except the lines $1, N+2,N+3,2N+4,4N+5,5N+6,5N+7,6N+8.$ In these eight lines, we find eight elements are the important factors which seriously damaged the diagonal dominance, namely $D_{1,2N+5}, D_{N+2,3N+5}, D_{N+3,3N+4}, D_{2N+4,4N+4}, D_{4N+5,6N+9},$ $D_{5N+6,7N+9}, D_{5N+7,7N+8}, D_{6N+8,8N+8}.$

Denote by $D_k$ the k column of D, now, we do the following eight elementary column transformations in turn:

$$\begin{aligned}&D_{2N+5}\longrightarrow D_{2N+5}-\sigma _2/\sigma _1C_1,\\&D_{3N+4}\longrightarrow D_{3N+4}-\delta _{N+1}/\delta _{N+2}C_{N+3},\\&D_{3N+5}\longrightarrow D_{3N+5}-\sigma _2/\sigma _1C_{N+2},\\&D_{4N+4}\longrightarrow D_{4N+4}-\delta _{N+1}/\delta _{N+2}C_{2N+4},\\&D_{6N+9}\longrightarrow D_{6N+9}-\sigma _2/\sigma _1C_{4N+5},\\&D_{7N+8}\longrightarrow D_{7N+8}-\delta _{N+1}/\delta _{N+2}C_{5N+7},\\&D_{7N+9}\longrightarrow D_{7N+9}-\sigma _2/\sigma _1C_{5N+6},\\&D_{8N+8}\longrightarrow D_{8N+8}-\delta _{N+1}/\delta _{N+2}C_{6N+8}, \end{aligned}$$

where $a\longrightarrow b$ means replacing a with b.

Then, we can conclude that the new obtained matrix is strictly diagonally dominant, and hence is invertible. $\square$