A Galerkin method for retarded boundary integral equations with smooth and compactly supported temporal basis functions

Abstract

We consider retarded boundary integral formulations of the three-dimensional wave equation in unbounded domains. Our goal is to apply a Galerkin method in space and time in order to solve these problems numerically. In this approach the computation of the system matrix entries is the major bottleneck. We will propose new types of finite-dimensional spaces for the time discretization. They allow variable time-stepping, variable order of approximation and simplify the quadrature problem arising in the generation of the system matrix substantially. The reason is that the basis functions of these spaces are globally smooth and compactly supported. In order to perform numerical tests concerning our new basis functions we consider the special case that the boundary of the scattering problem is the unit sphere. In this case explicit solutions of the problem are available which will serve as reference solutions for the numerical experiments.

Appendix A: Technical estimates

In this section we want to estimate the $n$th derivative of the function $f$ as defined in (3.5). Therefore let

$$\begin{aligned} h( z) :=\text{ erf}( z) \quad \text{ and} \quad g( x) :=\text{ arctanh}x=\frac{1}{2}\ln \frac{1+x}{1-x} \end{aligned}$$

such that $f:=h\circ 2g$. Note that [1, (7.1.19)] implies

$$\begin{aligned} h^{( n+1) }( z) =( -1) ^{n}\frac{2}{\sqrt{\pi }}H_{n}( z) \text{ e}^{{-z^{2}}}\quad n=0,1,2, \ldots \end{aligned}$$

where $H_{n}$ are the Hermite polynomials. Hence,

$$\begin{aligned} f^{( n+1) }( x)&= \left( \frac{d}{dx}\right) ^{n}\left( \frac{4}{\sqrt{\pi }( 1-x^{2}) }\text{ e}^{{-4g^{2}}( x) }\right) \nonumber \\&= \frac{4}{\sqrt{\pi }}\sum _{\ell =0}^{n}\genfrac(){0.0pt}1{n}{\ell }\left( \frac{1}{1-x^{2}}\right) ^{( \ell ) }(\text{ e}^{{-4g^{2}}( x) }) ^{( n-\ell ) }. \end{aligned}$$

(8.1)

 

Lemma 8.1

(Derivatives of $g$) It holds

$$\begin{aligned} \left( \frac{1}{1-x^{2}}\right) ^{( \ell ) }=\frac{\ell !p_{\ell }( x) }{( 1-x^{2}) ^{\ell +1}}\quad \forall x\in ( -1,1), \end{aligned}$$

where

$$\begin{aligned} p_{\ell }( x) :=\frac{( x+1) ^{\ell +1}-( x-1) ^{\ell +1}}{2}. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \left|\, g^{( \ell ) }( x) \right|\le \left\{ \begin{array}{cc} \frac{1}{2}\ln \dfrac{4}{1-x^{2}}&\ell =0\\&\\ \dfrac{( \ell -1) !2^{\ell -1}}{( 1-x^{2}) ^{\ell }}&\ell \in \mathbb N _{\ge 1} \end{array} \right. \quad \forall x\in ( -1,1), \end{aligned}$$

(8.2)

as well as the more generous estimate

$$\begin{aligned} \vert \, g^{( \ell ) }( x) \vert \le q( x) \frac{\ell !2^{\ell -1}}{( 1-x^{2}) ^{\ell } }\quad \forall \ell \in \mathbb N _{0} \end{aligned}$$

(8.3)

with $q( x) = \ln \frac{4}{1-x^{2}}$.

 

Lemma 8.2

(Derivative of composite functions) For $n\ge 1$ and $x\in ( -1,1)$ we have

$$\begin{aligned} ( \text{ e}^{{-4g^{2}}( x) }) ^{( n) }=\text{ e}^{{-4g^{2}}( x) }\sum _{k=1}^{n}A_{n,k}( x) ( -1) ^{k}H_{k}( 2g( x) ), \end{aligned}$$

(8.4a)

where

$$\begin{aligned} A_{n,k}(x) =\frac{2^k}{k!}\sum _{\nu =1}^{k}(-1) ^{k-\nu }\genfrac(){0.0pt}1{k}{\nu }g^{k-\nu }(x) (g^{\nu }) ^{(n) }(x) \end{aligned}$$

(8.4b)

and

$$\begin{aligned} ( g^{\nu }) ^{( n) }&= \sum _{\ell _{\nu -1}=0}^{n} \sum _{\ell _{\nu -2}=0}^{\ell _{\nu -1}}\cdots \sum _{\ell _{1}=0}^{\ell _{2}}\genfrac(){0.0pt}1{n}{\ell _{\nu -1}}\genfrac(){0.0pt}1{\ell _{\nu -1}}{\ell _{n-2}}\cdots \genfrac(){0.0pt}1{\ell _{2}}{\ell _{1}}g^{( n-\ell _{\nu -1}) }g^{( \ell _{\nu -1}-\ell _{\nu -2}) }\nonumber \\&\cdots g^{( \ell _{2}-\ell _{1}) }g^{( \ell _{1}) }. \end{aligned}$$

(8.5)

 

Proof

The representation (8.4a) and (8.4b) follows from [27, formulae (2), (7)], while (8.5) is proved by induction using Leibniz’ product rule for differentiation. $\square$

 

Lemma 8.3

(Estimate of derivatives of composite functions) For $n\ge 1$ and $x\in ( -1,1)$ we have

$$\begin{aligned} \left|\left( \text{ e}^{{-g^{2}}( x) }\right) ^{\left( n\right) }\right|\le \frac{5}{2}\kappa n!\text{ e}^{{-2g^{2}}\left( x\right)}\left( \frac{C_1 q(x) }{1-x^{2}}\right) ^{n} \end{aligned}$$

(8.6)

with $\kappa \approx 1.086435$ and $C_{1}=6\sqrt{2}\text{ e}$.

 

Proof

From (8.3) and (8.5) we conclude for all $n\ge 1$, $\nu \ge 1$, and $x\in (-1,1)$

$$\begin{aligned} \left|( g^{\nu }) ^{( n) }( x) \right|&\le n!2^{n-\nu }\frac{q^{\nu }( x) }{( 1-x^{2}) ^{n}}\sum _{\ell _{\nu -1}=0}^{n}\sum _{\ell _{\nu -2}=0}^{\ell _{\nu -1}}\cdots \sum _{\ell _{1}=0}^{\ell _{2}}1\nonumber \\&= n!2^{n-\nu }\frac{q^{\nu }(x) }{(1-x^{2}) ^{n} }\genfrac(){0.0pt}{}{n+\nu -1}{\nu -1}. \end{aligned}$$

(8.7)

Thus, from (8.4b) we get that

$$\begin{aligned} \vert A_{n,k}(x) \vert&\le \frac{2^k n!}{k!} \frac{q^{k}(x) }{(1-x^{2}) ^{n}}\sum _{\nu =1} ^{k}\genfrac(){0.0pt}1{k}{\nu }2^{n-\nu }\genfrac(){0.0pt}{}{n+\nu -1}{\nu -1}\nonumber \\&\le \frac{2^{n}n!}{k!}\frac{q^{k}(x) }{(1-x^{2}) ^{n}}\left( \frac{n+k}{k}\right) ^{k}\sum _{\nu =1}^{k} \genfrac(){0.0pt}1{k}{\nu }2^{k-\nu }\nonumber \\&\le \frac{2^{n}n!}{k!}\frac{1}{(1-x^{2}) ^{n}}\left( \frac{3(n+k) q(x) }{k}\right) ^{k}. \end{aligned}$$

(8.8)

From [1, (22.14.17)] we obtain

$$\begin{aligned} H_{k}(2 g(x) ) \le \text{ e}^{{2g^{2}} ( x)}\kappa 2^{k/2}\sqrt{k!}. \end{aligned}$$

The combination of (8.4a) and (8.4b), (8.5), (8.7) and (8.8) results in the estimate for the $n$th derivative of $\text{ e}^{-4g^{2}(x) }$:

$$\begin{aligned} \left|\left( \text{ e}^{{-4g^{2}}(x) }\right) ^{(n) }\right|&\le \kappa 2^{n}n!\frac{\text{ e}^{{-2g^{2}}(x)}}{( 1-x^{2}) ^{n}}\sum _{k=1}^{n}\frac{1}{\sqrt{k!}}\left( \frac{3\sqrt{2} (n+k) q(x) }{k}\right) ^{k}\\&\le \kappa n!\text{ e}^{{-2g^{2}}(x)}\left( \frac{6\sqrt{2}q(x) }{1-x^{2}}\right) ^{n}\sum _{k=1}^{n} \frac{1}{\sqrt{k!}}\left( \frac{n+k}{k}\right) ^{k} \\ \nonumber&\le \kappa n!\text{ e}^{{-2g^{2}}(x)}\left( \frac{6\sqrt{2}\text{ e} q(x) }{1-x^{2}}\right) ^{n}\sum _{k=1}^{n} \frac{1}{\sqrt{k!}} \\&\le \frac{5}{2}\kappa n!\text{ e}^{{-2g^{2}}(x)}\left( \frac{6\sqrt{2}\text{ e} q(x) }{1-x^{2}}\right) ^{n}. \end{aligned}$$

$\square$

 

Theorem 8.4

(Estimate of $n$th derivative of $f$) We have

$$\begin{aligned} |f^{(n+1) }(x)| \le C_{2} C_1^n n! \frac{q(x)^n}{(1-x^2)^{n+1}}\text{ e}^{-2g^2(x)} \end{aligned}$$

with $C_{2}=\frac{10\kappa }{\sqrt{\pi }} \frac{C_1\ln (4)}{C_1\ln (4)-2}$.

 

Proof

From (8.1), (8.2) and (8.6) we get

$$\begin{aligned} |f^{(n+1) }(x)|&\le \frac{10\kappa }{\sqrt{\pi }} \sum _{l=0}^{n}\genfrac(){0.0pt}{}{n}{l}\frac{l! 2^l}{(1-x^2)^{l+1}} (n-l)! \left( \frac{C_1 q(x)}{1-x^2} \right)^{n-l}\text{ e}^{-2g^2(x)}\\&\le \frac{10\kappa }{\sqrt{\pi }} C_1^n n! \frac{q(x)^n}{(1-x^2)^{n+1}}\text{ e}^{-2g^2(x)} \sum _{l=0}^{n} \left(\frac{ 2}{C_1 q(x)} \right)^{l}\\&\le \frac{10\kappa }{\sqrt{\pi }} \frac{C_1\ln (4)}{C_1\ln (4)-2} C_1^n n! \frac{q(x)^n}{(1-x^2)^{n+1}}\text{ e}^{-2g^2(x)}, \end{aligned}$$

which leads to the desired result. $\square$

 

Lemma 8.5

For $x\in (-1,1)$ and $\alpha \ge 2$, we have

$$\begin{aligned} \left\Vert \frac{\text{ e}^{{-2g^{2}}(x)}}{(1-x^{2}) ^{\alpha }} \right\Vert_\infty \le \text{ e}^{\sigma _\alpha } \end{aligned}$$

with

$$\begin{aligned} \sigma _\alpha :=\frac{1}{4}\alpha ^2+\frac{1}{2} -\ln \left(\frac{1}{2}\alpha +\frac{1}{2}\sqrt{\alpha ^2-4}\right) . \end{aligned}$$

 

Proof

We set

$$\begin{aligned} \frac{\text{ e}^{{-2g^{2}}(x)}}{(1-x^{2}) ^{\alpha }} = \text{ e}^{s_n(x)}, \end{aligned}$$

where

$$\begin{aligned} s_n(x) := -2\text{ arctanh}(x)^2 - \alpha \text{ ln}(1-x^2). \end{aligned}$$

With the definition of $\text{ arctanh}(x)$ we get

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

Since $s_n(x)$ is symmetric we assume $0\le x < 1$ and get

$$\begin{aligned} s_n(x)&\le -\frac{1}{2}[\ln (1-x)]^2 - \alpha \text{ ln}(1-x) + \text{ ln}(1+x)\text{ ln}(1-x) =:\tilde{s}_n(x) . \end{aligned}$$

$\tilde{s}_n(x)$ is strictly increasing in the interval $[0,0.5]$ for arbitrary $\alpha \in \mathbb R _{\ge 2}$. Therefore we may restrict to find an upper bound for $\tilde{s}_n(x)$ in the interval $[0.5,1[$. With the inequality $\ln (1+x)\ln (1-x)\le -\ln (-\ln (1-x))$ we get

$$\begin{aligned} \tilde{s}_n(x)&\le -\frac{1}{2}[\ln (1-x)]^2 - \alpha \text{ ln}(1-x) -\ln (-\ln (1-x)) =:\hat{s}_n(x) \end{aligned}$$

in $[0.5,1[$. The derivative of $\hat{s}_n(x)$ is given by

$$\begin{aligned} \hat{s}_n^\prime (x)=\frac{[\ln (1-x)]^2+\alpha \ln (1-x)+1}{(1-x)\ln (1-x)} \end{aligned}$$

which has the root

$$\begin{aligned} x_0 = 1-\text{ e}^{-\theta _\alpha }, \end{aligned}$$

where $\theta _\alpha := \frac{1}{2}\alpha +\frac{1}{2}\sqrt{\alpha ^2-4}$. Inserting this above shows that

$$\begin{aligned} s_n(x)\le \alpha \theta _\alpha - \frac{1}{2} \theta ^2_\alpha -\ln {\theta _\alpha } \end{aligned}$$

which leads to the desired result after some straightforward manipulations. $\square$

 

Lemma 8.6

It holds

$$\begin{aligned} \int _{-1}^1 \left(\ln \frac{4}{1-t^2}\right)^n dt \le 16 n! \end{aligned}$$

for $n\in \mathbb N$.

 

Proof

We first note that

$$\begin{aligned}&\int _{-1}^1|\ln (1-t)|^i |\ln (1+t)|^{k-i} dt \\&\quad = \int _{-1}^0 |\ln (1-x)|^i |\ln (1+t)|^{k-i} dt + \int _{0}^1 |\ln (1-x)|^i |\ln (1+t)|^{k-i} dt\\&\quad \le (\ln 2)^i\int _{-1}^0 |\ln (1+t)|^{k-i} dt + (\ln 2)^{k-i}\int _{0}^1 |\ln (1-t)|^i dt\\&\quad = (\ln 2)^i\int _{0}^1 |\ln (t)|^{k-i} dt + (\ln 2)^{k-i}\int _{0}^1 |\ln (t)|^i dt\\&\quad = (\ln 2)^i (k-i)!+ (\ln 2)^{k-i}i!, \end{aligned}$$

where we used [17, (2.711)] in the last step. With these computations we get

$$\begin{aligned} \int _{-1}^1 \left|\left( \ln \frac{4}{1-t^2}\right)^n \right|dt&\le \sum _{k=0}^n \genfrac(){0.0pt}{}{n}{k}\int _{-1}^1 |\ln (1-t^2)|^k(\ln 4)^{n-k} dt\\&\le \sum _{k=0}^n \sum _{i=0}^k \genfrac(){0.0pt}{}{n}{k} \genfrac(){0.0pt}{}{k}{i}(\ln 4)^{n-k} \int _{-1}^1 |\ln (1-t)|^i |\ln (1+t)|^{k-i} dt\\&\le \sum _{k=0}^n \sum _{i=0}^k \genfrac(){0.0pt}{}{n}{k} \genfrac(){0.0pt}{}{k}{i}(\ln 4)^{n-k}( (\ln 2)^i (k-i)!+ (\ln 2)^{k-i}i!)\\&\le \sum _{k=0}^n \genfrac(){0.0pt}{}{n}{k}(\ln 4)^{n-k}\left(k!\sum _{i=0}^k \frac{(\ln 2)^i}{i!} +k!\sum _{i=0}^k \frac{(\ln 2)^{k-i}}{(k-i)!}\right)\\&\le 4\sum _{k=0}^n \genfrac(){0.0pt}{}{n}{k}(\ln 4)^{n-k}k!\\&\le 4 n! \sum _{k=0}^n \frac{(\ln 4)^{n-k}}{(n-k)!} \le 16 n!. \end{aligned}$$

$\square$