A fast Fourier-Galerkin method solving boundary integral equations for the Helmholtz equation with exponential convergence

Abstract

A boundary integral equation in general form will be considered, which can be used to solve Dirichlet problems for the Helmholtz equation. The goal of this paper is to develop a fast Fourier-Galerkin method solving these boundary integral equations. To this aim, a scheme for splitting integral operators is presented, which splits the corresponding integral operator into a convolution operator and a compact operator. A truncation strategy is presented, which can compress the dense coefficient matrix to a sparse one having only $\mathcal {O}(n)$ nonzero entries, where n is the order of the Fourier basis functions used in the method. The proposed fast method preserves the stability and optimal convergence order. Moreover, exponential convergence can be obtained under suitable assumptions. Numerical examples are presented to confirm the theoretical results for the approximation accuracy and computational complexity of the proposed method.

Appendix: Technical lemmas used in the proofs

For given p ≥ 0, and q > 0, when $\mathcal {A}+{\mathscr{B}}$ is injective from H^p(I) to H^p+m+ 1(I) with a bounded inversion, and ${\mathscr{B}}\in B(H^{p}(I),H^{p+m+q+1}(I))$, we define

$$ C_{m,p,q}:=\left( 2\|\mathcal{B}\|_{p,m+p+q+1}\|(\mathcal{A}+\mathcal{B})^{-1}\|_{m+p+1,p}\right)^{\frac{1}{q}}. $$

Lemma 8

Let p ≥ 0 and q > 0. If $\mathcal {A}+{\mathscr{B}}$ is injective from H^p(I) to H^p+m+ 1(I) with a bounded inversion, and ${\mathscr{B}}\in B(H^{p}(I),H^{p+m+q+1}(I))$, then there holds that for all n ≥ C_m,p,q, and all w ∈ X_n,

$$ \left\|\left( \mathcal{A}+\widetilde{\mathcal{B}}_{n}\right)w\right\|_{p+m+1}\geq \frac{1}{2}\left\|(\mathcal{A}+\mathcal{B})^{-1}\right\|^{-1}_{p+m+1,p}\|w\|_{p}. $$

Proof

Since ${\mathscr{B}}\in B(H^{p}(I),H^{p+m+q+1}(I))$, for all w ∈ X_n,

$$ \begin{array}{@{}rcl@{}} \left\|\left( \mathcal{B}-\widetilde{\mathcal{B}}_{n}\right)w\right\|_{p+m+1}&=& \|(\mathcal{I}-P_{n})\mathcal{B}w\|_{p+m+1}\\ & \leq & n^{-q}\|\mathcal{B}\|_{p,p+m+q+1}\|w\|_{p}. \end{array} $$

Thus, when n ≥ C_m,p,q, there holds that for all w ∈ X_n,

$$ \begin{array}{@{}rcl@{}} \left\|\left( \mathcal{A}+\widetilde{\mathcal{B}}_{n}\right)w\right\|_{p+m+1}&\geq& \left\|\left( \mathcal{A}+\mathcal{B}\right)w\right\|_{p+m+1}-\left\|\left( \mathcal{B}-\widetilde{\mathcal{B}}_{n}\right)w\right\|_{p+m+1}\\ &\geq& \left\|(\mathcal{A}+\mathcal{B})^{-1}\right\|^{-1}_{p+m+1,p}\|w\|_{p}-n^{-q}\|\mathcal{B}\|_{p,p+m+q+1}\|w\|_{p}\\ &\geq&\frac{1}{2}\left\|(\mathcal{A}+\mathcal{B})^{-1}\right\|^{-1}_{p+m+1,p}\|w\|_{p}. \end{array} $$

□

For all $x\in \mathbb {R}$, $r_{1},r_{2}\in \mathbb {R}$ with r₁ − 2 > r₂ > 0, define

$$ G_{r_{1},r_{2}}(x):=\sum\limits_{k\in\mathbb{Z}}\left( 1+(x-k)^{2}\right)^{-\frac{r_{1}}{2}}(1+k^{2})^{\frac{r_{2}}{2}}. $$

It is easy to see that for all $x\in \mathbb {R}$, $G_{r_{1},r_{2}}(x)<+\infty$. In the next lemma, we discuss the decay rate of function $G_{r_{1},r_{2}}$ as variant x going to infinite.

Lemma 9

If $m_{1},m_{2}\in \mathbb {R}$ with m₁ − 1 > m₂ > 1, then for all $x\in \mathbb {R}$,

$$ G_{m_{1},m_{2}}(x)\leq \left( 1+x^{2}\right)^{-\frac{m_{2}}{2}}\max\left\{\sum\limits_{k\in\mathbb{Z}}\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}},2^{m_{2}}\sum\limits_{k\in \mathbb{Z}}\left( 1+k^{2}\right)^{-\frac{m_{1}-m_{2}}{2}}\right\}. $$

(50)

Proof

Since $G_{m_{1},m_{2}}(x)=G_{m_{1},m_{2}}(-x)$, it is sufficient to prove this lemma by showing that (50) holds when x ≥ 0. Note that for x ≥ 0,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k\geq x}\left( 1+\left( x-k\right)^{2}\right)^{-\frac{m_{1}}{2}}\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}}&\leq&\left( 1+x^{2}\right)^{-\frac{m_{2}}{2}}\sum\limits_{k\geq x}\left( 1+\left( k-\lceil x\rceil\right)^{2}\right)^{-\frac{m_{1}}{2}}\\ &\leq& \left( 1+x^{2}\right)^{-\frac{m_{2}}{2}}\sum\limits_{k\in\mathbb{Z}}\left( 1+k^{2}\right)^{-\frac{m_{1}}{2}} \end{array} $$

(51)

and

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k<0}\left( 1+\left( x-k\right)^{2}\right)^{-\frac{m_{1}}{2}}\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}}&\leq&\left( 1+x^{2}\right)^{-\frac{m_{1}}{2}}\sum\limits_{k<0 }\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}}\\ &\leq&\left( 1+x^{2}\right)^{-\frac{m_{1}}{2}}\sum\limits_{k\in \mathbb{Z} }\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}}. \end{array} $$

(52)

Meanwhile, since for 0 ≤ k < x,

$$ \begin{array}{@{}rcl@{}} \left( 1+(x-k)^{2}\right)\left( 1+k^{2}\right)&&\geq \frac{1}{4}(1+x-k)^{2}(1+k)^{2}=\frac{1}{4}(1+x+(x-k)k)^{2}\\&&\geq\frac{1}{4}\left( 1+x\right)^{2}\geq\frac{1}{4}\left( 1+x^{2}\right), \end{array} $$

there holds

$$ \begin{array}{@{}rcl@{}} \sum\limits_{0\leq k< x}\left( 1+\left( x-k\right)^{2}\right)^{-\frac{m_{1}}{2}}\left( 1+k^{2}\right)^{-\frac{m_{2}}{2}}&\leq&\sum\limits_{0\leq k< x}\left( 1+\left( x-k\right)^{2}\right)^{-\frac{m_{1}-m_{2}}{2}}\left( \left( 1+\left( x-k\right)^{2}\right)\left( 1+k^{2}\right)\right)^{-\frac{m_{2}}{2}}\\ &\leq& 2^{m_{2}}\left( 1+x^{2}\right)^{-\frac{m_{2}}{2}}\sum\limits_{0\leq k< x}\left( 1+\left( \lfloor x\rfloor-k\right)^{2}\right)^{-\frac{m_{1}-m_{2}}{2}}\\ &\leq& 2^{m_{2}}\left( 1+x^{2}\right)^{-\frac{m_{2}}{2}}\sum\limits_{k\in \mathbb{Z}}\left( 1+k^{2}\right)^{-\frac{m_{1}-m_{2}}{2}}. \end{array} $$

(53)

Note that m₁ − 1 > m₂ > 1. Combining (51), (52), and (53), from the definition of $G_{m_{1},m_{2}}$ we obtain the desired estimate (50). □

Lemma 10

Let 0 < r < 1 and α > 0. Then, there is a positive constant c such that for all k ≥ 0,

$$ \sum\limits_{l=0}^{k} r^{l} (k+1-l)^{-\alpha}\leq c (k+1)^{-\alpha}, $$

where c depends on r.

Proof

Define

$$ S_{k}:=\sum\limits_{l=0}^{k} r^{l}\left( \frac{k+1}{k+1-l}\right)^{\alpha}. $$

It is easy to see that ${\sum }_{l=0}^{k} r^{l} (k+1-l)^{-\alpha }=(k+1)^{-\alpha }S_{k}.$ Thus, it is sufficient to prove this lemma by showing that the sequence $\{S_{k}:k\in \mathbb {N}\}$ is bounded. Note that

$$ \begin{array}{@{}rcl@{}} rS_{k} &=& \sum\limits_{l=0}^{k} r^{l+1}\left( \frac{k+1}{k+1-l}\right)^{\alpha} \\ &=& \left( \frac{k+1}{k+2}\right)^{\alpha}\sum\limits_{l=1}^{k+1} r^{l}\left( \frac{k+2}{k+2-l}\right)^{\alpha} = \left( \frac{k+1}{k+2}\right)^{\alpha}(S_{k+1}-1), \end{array} $$

i.e., $S_{k+1}=1+\left (\frac {k+2}{k+1}\right )^{\alpha }rS_{k}$. Since $\lim _{k\rightarrow +\infty } \left (\frac {k+2}{k+1}\right )^{\alpha }=1$ and r < 1, for any r < R < 1 there exists an integer k^∗ such that for all k > k^∗, $\left (\frac {k+2}{k+1}\right )^{\alpha }r<R$. Thus, for all k > k^∗, there is S_k+ 1 = 1 + RS_k. Then, by noting that 0 < R < 1, we know that the sequence $\{S_{k}:k\in \mathbb {N}\}$ is bounded. □

Lemma 11

If $\gamma _{1}, \gamma _{2}\in \tilde {C}_{2\pi }(I)$, then function h, defined by (4), is in $\tilde {C}_{2\pi }(I^{2})$.

Proof

We know that, for any d > c > 0, the function $\log$ is real analytic in [c,d]. Note that function h is in ${C}^{\infty }_{2\pi }(I^{2})$, which implies that $\inf \left \{\frac {|\gamma (t)-\gamma (s)|^{2}}{4e^{-2}\sin \limits ^{2}\frac {s-t}{2}}:t,s\in I\right \}>0$. We rewrite h as,

$$ h(t,s):=\frac{1}{2}\log({h^{2}_{1}}(t,s)+{h^{2}_{2}}(t,s)),\quad (t,s)\in I^{2} $$

where,

$$ h_{1}(t,s):=\left\{ \begin{array}{ll} \cfrac{\gamma_{1}(t)-\gamma_{1}(s)}{2e^{-1}\sin\frac{t-s}{2}}, & t\neq s,\\ e\gamma_{1}^{\prime}(t), & t=s, \end{array} \right. \quad h_{2}(t,s):=\left\{ \begin{array}{ll} \frac{\gamma_{2}(t)-\gamma_{2}(s)}{2e^{-1}\sin\frac{t-s}{2}}, & t\neq s,\\ e\gamma_{2}^{\prime}(t), & t=s. \end{array} \right. $$

Therefore, to prove this lemma, it suffices to show $h_{1}, h_{2} \in \tilde {C}_{2\pi }(I^{2})$.

It is clear that the extension of h₁ is real analytic on $\mathbb {R}^{2}$ on $\{(t,s)\in \mathbb {R}^{2}: t\neq s+2k\pi , k\in \mathbb {Z}\}$. Thus, by the 2π-biperiodicity of h₁, to prove $h_{1}\in \tilde {C}_{2\pi }(I^{2})$, we only need to show that for all $t_{0}\in \mathbb {R}$, the extension of function h₁ can be represented by a convergent power series in some neighborhood of point (t₀,t₀). Define

$$ \tilde{h}(t,s):=\left\{ \begin{array}{ll} \cfrac{\gamma_{1}(t)-\gamma_{1}(s)}{t-s}, & t\neq s,\\ \gamma_{1}^{\prime}(t), & t=s, \end{array} \right. {\text{and}} \bar{h}(t,s):=\left\{ \begin{array}{ll} \cfrac{t-s}{2e^{-1}\sin\frac{t-s}{2}}, & t\neq s,\\ e, & t=s. \end{array} \right. $$

for all $t,s\in \mathbb {R}$. Note that for all $t,s\in \mathbb {R}$ with |t − s| < π/2, $h_{1}(t,s)=\tilde {h}(t,s)\bar {h}(t,s)$, and $\bar {h}$ is real analytic on $\{(t,s)\in \mathbb {R}^{2}:|t-s|<\pi /2\}$. Thus, we only need to show that for all $t_{0}\in \mathbb {R}$, function $\tilde {h}$ can be represented by a convergent power series in some neighborhood of point (t₀,t₀). Since $\gamma _{1}\in \tilde {C}_{2\pi }(I)$, there is an neighborhood B_ε(t₀) := {t ∈ R : |t − t₀| < ε} such that for all t ∈B_ε(t₀),

$$ \gamma_{1}(t)=\sum\limits_{k\in\mathbb{N}_{0}} \frac{1}{k!}\gamma_{1}^{(k)}(t_{0}) (t-t_{0})^{k}. $$

Thus, there holds that for all t,s ∈B_ε(t₀) with t≠s,

$$ \begin{array}{@{}rcl@{}} \tilde{h}(t,s)&=& \frac{1}{t-s}\sum\limits_{k\in\mathbb{N}_{0}}\frac{\gamma_{1}^{(k)}(t_{0}) }{k!} ((t-t_{0})^{k}-(s-t_{0})^{k})\\ &=& \sum\limits_{k\in\mathbb{N}}\cfrac{\gamma_{1}^{(k)}(t_{0}) }{k!}\left( \sum\limits_{l=0}^{k-1}(-1)^{k-l+1}{C_{k}^{l}}(t-t_{0})^{l}(t-s)^{k-l-1}\right). \end{array} $$

(54)

For t,s ∈B_ε(t₀) with t = s, it is easy to see that

$$ \gamma_{1}^{\prime}(t)=\sum\limits_{k\in\mathbb{N}}\frac{\gamma_{1}^{(k)}(t_{0}) }{(k-1)!}(t-t_{0})^{k-1}, $$

and

$$ (t-t_{0})^{k-1}=\frac{1}{k}\sum\limits_{l=0}^{k-1}(-1)^{k-l+1}{C_{k}^{l}}(t-t_{0})^{l}(t-s)^{k-l-1}. $$

Thus, (54) holds for all t,s ∈B_ε(t₀). This ensures $h_{1} \in \tilde {C}_{2\pi }(I^{2})$.

Replacing γ₁ and h₁ in above discussion by γ₂ and h₂ respectively, we obtain that $h_{2} \in \tilde {C}_{2\pi }(I^{2})$. Then, we can claim that $h=\frac {1}{2}\log ({h^{2}_{1}}+{h^{2}_{2}})\in \tilde {C}_{2\pi }(I^{2})$. □

Lemma 12

For 0 < r < 1 and m ≥ 0, there exists a positive constant c such that for all $k,l\in \mathbb {Z}$ with kl ≤ 0,

$$ \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-{m}}\leq c r^{|k+l|}(1+|k+l|)\max\{(1+|k|)^{-m},(1+|l|)^{-m}\}, $$

(55)

and for all $k,l\in \mathbb {Z}$ with kl > 0,

$$ \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-{m}}\leq c r^{|k+l|}(|k+l|). $$

(56)

Proof

By noting that

$$ \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-m} = \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|(-l)-k^{\prime}|+|k^{\prime}+(-k)|}(1+|k^{\prime}|)^{-m}, $$

it is sufficient for us to consider the case with k ≥−l. In this case, there holds that

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-m} \\ &=& r^{k+l}\Big[\sum\limits_{k^{\prime} \leq-l} r^{-2\left( k^{\prime}+l\right)}\left( 1+\left|k^{\prime}\right|\right)^{-m}+\sum\limits_{-l< k^{\prime}< k} (1+|k^{\prime}|)^{-m}+\sum\limits_{k^{\prime}\geq k} r^{2(k^{\prime}-k)}(1+|k^{\prime}|)^{-m}\Big].\\ \end{array} $$

(57)

When kl ≤ 0, we continue the estimate (57) from the following two sub-cases.

• (i) In the first sub-case, let − l ≤ k ≤ 0. There holds that

  $$ \sum\limits_{k^{\prime} \leq-l} r^{-2\left( k^{\prime}+l\right)}\left( 1+\left|k^{\prime}\right|\right)^{-m} \leq(1+|l|)^{-m} \sum\limits_{k^{\prime} \leq-l} r^{-2\left( k^{\prime}+l\right)}=\frac{(1+|l|)^{-m}}{1-r^{2}} \leq \frac{(1+|k|)^{-m}}{1-r^{2}}, $$
  $$ \sum\limits_{-l< k^{\prime}< k} (1+|k^{\prime}|)^{-m}\leq (k+l) (1+|k|)^{-m}, $$

  and by Lemma 10

  $$ \begin{array}{@{}rcl@{}} \sum\limits_{k^{\prime}\geq k} r^{2(k^{\prime}-k)}(1+|k^{\prime}|)^{-m}&\leq& \sum\limits_{k\leq k^{\prime}\leq 0} r^{2(k^{\prime}-k)}(1-k^{\prime})^{-m}+{}\sum\limits_{k^{\prime}> 0} r^{2(k^{\prime}-k)}(1+k^{\prime})^{-m} \\ &\leq& \sum\limits_{u=0}^{-k} r^{2 u}(1-k-u)^{-m}+\sum\limits_{k^{\prime}>0} r^{2\left( k^{\prime}-k\right)}\\ &\leq& c_{1} (1+|k|)^{-m}, \end{array} $$

  where c₁ depends only on r and m. Substituting the above inequalities into (57) shows that there is a positive constant c₂ such that for all $k, l \in \mathbb {Z}$ with − l ≤ k ≤ 0,

  $$ \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-{m}} \leq c_{2} r^{k+l}(k+l+1) (1+|k|)^{-m}. $$

  (58)

• (ii) In the second sub-case, let 0 ≤−l ≤ k. By an argument similar to the proof for (58), we can obtain that there is a positive constant c₃ such that for all $k, l\in \mathbb {Z}$ with 0 ≤−l ≤ k,

  $$ \sum\limits_{k^{\prime}\in\mathbb{Z}} r^{|k-k^{\prime}|+|l+k^{\prime}|}(1+|k^{\prime}|)^{-{m}} \leq c_{3} r^{k+l}(k+l+1) (1+|l|)^{-m}. $$

  (59)

  Combining (58) and (59), we know that when kl ≤ 0 and − l ≤ k, inequality (55) holds.

  When kl > 0, there holds that

  $$ \sum\limits_{k^{\prime}\geq l} r^{2(k^{\prime}-l)}(1+|k^{\prime}|)^{-m}\leq \frac{(1+|l|)^{-m}}{1-r^{2}}, $$
  $$ \sum\limits_{-l< k^{\prime}< k} (1+|k^{\prime}|)^{-m}\leq k+l, $$

  and

  $$ \sum\limits_{k^{\prime}\geq k} r^{2(k^{\prime}-k)}(1+|k^{\prime}|)^{-m}\leq \frac{(1+|k|)^{-m}}{1-r^{2}}. $$

  Thus, combining the above inequalities with (57) implies that (56) holds when kl > 0 and − l ≤ k.

□