Frequency-adapted Galerkin boundary element methods for convex scattering problems

Abstract

We introduce a class of hybrid boundary element methods for the solution of sound soft scattering problems in the exterior of two-dimensional smooth convex obstacles. To facilitate the applicability of our algorithms throughout the entire frequency spectrum, we have enriched our Galerkin approximation spaces, through incorporation of oscillations in the incident field of radiation, into the algebraic and trigonometric polynomial approximation spaces. The resulting methodologies have three distinctive properties. Indeed, from a theoretical point of view (1) they can be tuned to demand only an $\mathcal {O}(k^{\varepsilon })$ increase (for any $\varepsilon >0)$ in the number of degrees of freedom to maintain a fixed accuracy with increasing wavenumber k, owing to the optimal adaptation of approximation spaces to asymptotic stretching (shrinking) of illuminated and deep shadow regions (shadow boundaries), and (2) they are convergent for each fixed wavenumber k, thanks to the additional approximation spaces in the deep shadow region. Perhaps more importantly, from a practical point of view (3) they give rise to linear systems with significantly enhanced condition numbers and this, in turn, allows for more accurate solutions if desired.

Auxiliary results

Lemma 2

(A partial fraction decomposition) Given $c \ne d$, we have the partial fraction decomposition

$$\begin{aligned} \dfrac{1}{(s-c)^{m} \, (d-s)^{m}} = \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m-j}} \left( \dfrac{1}{(s-c)^{j}} + \dfrac{1}{(d-s)^{j}} \right) \end{aligned}$$

valid for all $s \in \mathbb {R} \backslash \{c,d\}$ and all $m \ge 1$.

Proof

The case $m=1$ corresponds to the decomposition

$$\begin{aligned} \dfrac{1}{(s-c)(d-s)} = \dfrac{1}{d-c} \left( \dfrac{1}{s-c} + \dfrac{1}{d-s} \right) \end{aligned}$$

and its yields, by a straightforward induction, the identities

$$\begin{aligned} \dfrac{1}{(s-c)^{j+1}(d-s)} = \sum _{k=1}^{j+1} \dfrac{1}{(d-c)^{j+2-k}} \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-c)^{j+1}} \dfrac{1}{(d-s)} \end{aligned}$$

and

$$\begin{aligned} \dfrac{1}{(s-c)(d-s)^{j+1}} = \dfrac{1}{(d-c)^{j+1}} \dfrac{1}{(s-c)} + \sum _{k=1}^{j+1} \dfrac{1}{(d-c)^{j+2-k}} \dfrac{1}{(d-s)^{k}} \end{aligned}$$

for all $j \in \mathbb {N} \cup \{ 0 \}$. Therefore, by induction on m, we get

$$\begin{aligned}&\dfrac{1}{(s-c)^{m+1}(d-s)^{m+1}} \\&\quad = \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m-j}} \left[ \dfrac{1}{(s-c)^{j+1}(d-s)} + \dfrac{1}{(s-c)(d-s)^{j+1}} \right] \\&\quad = \sum _{j=1}^{m} \sum _{k=1}^{j+1} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\qquad + \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m+1}} \left[ \dfrac{1}{s-c} + \dfrac{1}{d-s} \right] \end{aligned}$$

which we rewrite as

$$\begin{aligned}&\dfrac{1}{(s-c)^{m+1}(d-s)^{m+1}} \\&\quad = \sum _{j=1}^{m} \sum _{k=1}^{j} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\qquad + \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-(j+1)}} \left[ \dfrac{1}{(s-c)^{j+1}} + \dfrac{1}{(d-s)^{j+1}} \right] \\&\qquad + \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m+1}} \left[ \dfrac{1}{s-c} + \dfrac{1}{d-s} \right] . \end{aligned}$$

Changing the order of summation in the double-sum yields

$$\begin{aligned}&\dfrac{1}{(s-c)^{m+1}(d-s)^{m+1}}\\&\quad = \sum _{k=1}^{m} \sum _{j=k}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\quad \quad + \sum _{k=2}^{m+1} \left( {\begin{array}{c}2m-k\\ m-k+1\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\quad \quad + \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m+1}} \left[ \dfrac{1}{s-c} + \dfrac{1}{d-s} \right] \end{aligned}$$

so that a rearrangement gives

$$\begin{aligned}&\dfrac{1}{(s-c)^{m+1}(d-s)^{m+1}} \\&\quad = 2 \sum _{k=1}^{m} \left( {\begin{array}{c}2m-k-1\\ m-k\end{array}}\right) \dfrac{1}{(d-c)^{2m+1}} \left[ \dfrac{1}{s-c} + \dfrac{1}{d-s} \right] \\&\qquad + \sum _{k=2}^{m} \left[ \left( {\begin{array}{c}2m-k\\ m-k+1\end{array}}\right) + \sum _{j=k}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \right] \\&\qquad \times \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\qquad + \left( {\begin{array}{c}m-1\\ 0\end{array}}\right) \dfrac{1}{(d-c)^{m+1}} \left[ \dfrac{1}{(s-c)^{m+1}} + \dfrac{1}{(d-s)^{m+1}} \right] . \end{aligned}$$

Since [24, page161]

$$\begin{aligned} \sum _{j=k}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) = \left( {\begin{array}{c}2m-k\\ m-k\end{array}}\right) , \end{aligned}$$

we have

$$\begin{aligned} 2 \sum _{k=1}^{m} \left( {\begin{array}{c}2m-k-1\\ m-k\end{array}}\right) = 2 \left( {\begin{array}{c}2m-1\\ m-1\end{array}}\right) = \left( {\begin{array}{c}2m\\ m\end{array}}\right) \end{aligned}$$

and

$$\begin{aligned} \left( {\begin{array}{c}2m-k\\ m-k+1\end{array}}\right) + \sum _{j=k}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right)&= \left( {\begin{array}{c}2m-k\\ m-k+1\end{array}}\right) + \left( {\begin{array}{c}2m-k\\ m-k\end{array}}\right) \\&= \left( {\begin{array}{c}2(m+1) - k -1\\ (m+1)-k\end{array}}\right) . \end{aligned}$$

Therefore

$$\begin{aligned}&\dfrac{1}{(s-c)^{m+1}(d-s)^{m+1}} = \left( {\begin{array}{c}2m\\ m\end{array}}\right) \dfrac{1}{(d-c)^{2m+1}} \left[ \dfrac{1}{s-c} + \dfrac{1}{d-s} \right] \\&\quad + \sum _{k=2}^{m} \left( {\begin{array}{c}2(m+1) - k -1\\ (m+1)-k\end{array}}\right) \dfrac{1}{(d-c)^{2(m+1)-k}} \left[ \dfrac{1}{(s-c)^{k}} + \dfrac{1}{(d-s)^{k}} \right] \\&\quad + \left( {\begin{array}{c}m-1\\ 0\end{array}}\right) \dfrac{1}{(d-c)^{m+1}} \left[ \dfrac{1}{(s-c)^{m+1}} + \dfrac{1}{(d-s)^{m+1}} \right] \end{aligned}$$

and this is easily seen to correspond to the given formula for $m+1$. $\square$

Lemma 3

Suppose that either $[\alpha ,\beta ] \subseteq [t_{1},t_{2}] \subseteq (c,d)$ or $[\alpha ,\beta ] \cap (t_{1},t_{2}) = \varnothing$ and $[c,d] \subseteq (t_1,t_2)$. Then, for any $a,b \in \mathbb {R}$, $n \in \mathbb {N} \cup \{ 0 \}$, $m \in \mathbb {N}$, there holds

$$\begin{aligned}&\int _{\alpha }^{\beta } \dfrac{(s-a)^{n} \, (b-s)^{n}}{(s-c)^{m} \, (d-s)^{m}} \, ds \\&\quad = \sum _{j=1}^{m} \sum _{p=0}^{n} \sum _{q=0}^{n} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) \left( {\begin{array}{c}n\\ q\end{array}}\right) \dfrac{(-1)^{n} \, \mathcal {F}(\alpha ,\beta ;a,b;c,d;n,p,q;j)}{(d-c)^{2m-j}}. \end{aligned}$$

Here we have

$$\begin{aligned} \mathcal {F}(\alpha ,\beta ;a,b;c,d;n,p,q;j)= & {} \left( c-a \right) ^{p} \left( c-b \right) ^{q} \log \left( \dfrac{\beta -c}{\alpha -c} \right) \\&+ \left( a-d \right) ^{p} \left( b-d \right) ^{q} \log \left( \dfrac{d-\alpha }{d-\beta } \right) \end{aligned}$$

when $2n-(p+q+j) = -1$, and

$$\begin{aligned}&\mathcal {F}(\alpha ,\beta ;a,b;c,d;n,p,q;j) \\&\quad = \dfrac{\left( c-a \right) ^{p} \left( c-b \right) ^{q}}{2n-(p+q+j)+1} \left[ \left( \beta -c \right) ^{2n-(p+q+j)+1} - \left( \alpha -c \right) ^{2n-(p+q+j)+1} \right] \\&\qquad + \dfrac{\left( a-d \right) ^{p} \left( b-d \right) ^{q}}{2n-(p+q+j)+1} \left[ \left( d-\alpha \right) ^{2n-(p+q+j)+1} - \left( d-\beta \right) ^{2n-(p+q+j)+1} \right] \end{aligned}$$

when $2n-(p+q+j) \ne -1$.

Proof

Lemma 2 entails

$$\begin{aligned} \int _{\alpha }^{\beta } \dfrac{(s-a)^{n} \, (b-s)^{n}}{(s-c)^{m} \, (d-s)^{m}} \, ds= & {} \sum _{j=1}^{m} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{1}{(d-c)^{2m-j}}\\&\times \int _{\alpha }^{\beta } \left( \dfrac{(s-a)^{n} \, (b-s)^{n}}{(s-c)^{j}} + \dfrac{(s-a)^{n} \, (b-s)^{n}}{(d-s)^{j}} \right) \, ds. \end{aligned}$$

Changing variables, we get

$$\begin{aligned}&\int _{\alpha }^{\beta } \dfrac{(s-a)^{n} \, (b-s)^{n}}{(s-c)^{m} \, (d-s)^{m}} \, ds = \sum _{j=1}^{m} \, \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \dfrac{\left( -1\right) ^{n}}{(d-c)^{2m-j}}\\&\quad \times \left( \int _{\alpha -c}^{\beta -c} t^{-j} \, (t+c-a)^{n} \, (t+c-b)^{n} \, dt + \int _{d-\beta }^{d-\alpha } t^{-j} \, (t+a-d)^{n} \, (t+b-d)^{n} \, dt \right) \end{aligned}$$

so that an appeal to the binomial theorem yields

$$\begin{aligned}&\int _{\alpha }^{\beta } \dfrac{(s-a)^{n} \, (b-s)^{n}}{(s-c)^{m} \, (d-s)^{m}} \, ds = \sum _{j=1}^{m} \, \sum _{p=0}^{n} \, \sum _{q=0}^{n} \left( {\begin{array}{c}2m-j-1\\ m-j\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) \left( {\begin{array}{c}n\\ q\end{array}}\right) \dfrac{\left( -1\right) ^{n}}{(d-c)^{2m-j}}\\&\quad \times \left( (c-a)^{p} (c-b)^{q} \int _{\alpha -c}^{\beta -c} t^{2n-(p+q+j)} \, dt + (a-d)^{p} (b-d)^{q} \int _{d-\beta }^{d-\alpha } t^{2n-(p+q+j)} \, dt \right) . \end{aligned}$$

Thus the result. $\square$