Approximation of integral operators by Green quadrature and nested cross approximation

Abstract

We present a fast algorithm that constructs a data-sparse approximation of matrices arising in the context of integral equation methods for elliptic partial differential equations. The new algorithm uses Green’s representation formula in combination with quadrature to obtain a first approximation of the kernel function, and then applies nested cross approximation to obtain a more efficient representation. The resulting ${\mathcal H}^2$-matrix representation requires ${\mathcal O}(n k)$ units of storage for an $n\times n$ matrix, where k depends on the prescribed accuracy.

Appendix A: Proofs of technical lemmas

Proof of Lemma 3

By definition of the maximum norm, we can find $\iota \in \{1,\ldots ,d\}$ such that $2\delta _t = \mathop {\mathrm{diam}}\nolimits _\infty ({\mathcal B}_t) = b_{t,\iota }-a_{t,\iota }$ and $b_{t,\kappa }-a_{t,\kappa }\le 2\delta _t$ holds for all $\kappa \in \{1,\ldots ,d\}$. Most of our claims are direct consequences of this estimate, only the last claim of (16) requires a closer look. Let $x\in \partial \omega _t$ and $y\in {\mathcal F}_t$ be given with

$$\begin{aligned} \Vert x-y\Vert _\infty = \mathop {\mathrm{dist}}\nolimits _\infty (\partial \omega _t, {\mathcal F}_t). \end{aligned}$$

Let $\widehat{x}\in {\mathcal B}_t$ be a point in ${\mathcal B}_t$ that has minimal distance to x. By construction, we have

$$\begin{aligned} \Vert x-\widehat{x}\Vert _\infty \le \delta _t, \end{aligned}$$

and the triangle inequality in combination with (5) yields

$$\begin{aligned} \mathop {\mathrm{dist}}\nolimits _\infty (\partial \omega _t, {\mathcal F}_t)&= \Vert x-y\Vert _\infty \ge \Vert \widehat{x}-y\Vert _\infty - \Vert \widehat{x}-x\Vert _\infty \ge \mathop {\mathrm{dist}}\nolimits _\infty ({\mathcal B}_t, B_s) - \delta _t\\&\ge \mathop {\mathrm{diam}}\nolimits _\infty ({\mathcal B}_t) - \mathop {\mathrm{diam}}\nolimits _\infty ({\mathcal B}_t)/2 = \mathop {\mathrm{diam}}\nolimits _\infty ({\mathcal B}_t)/2 = \delta _t, \end{aligned}$$

and this is the required estimate. $\square$

Proof of Lemma 4

Let $\kappa \in {\mathbb {N}}_0^d$ be a multiindex. We consider the function

$$\begin{aligned} g_x : [-1,1]^d&\rightarrow {\mathbb {R}},&{\hat{z}}&\mapsto (\partial ^\kappa g)(x,\Phi _t({\hat{z}})) \end{aligned}$$

and aim to prove

$$\begin{aligned} \partial ^\nu g_x({\hat{z}})&= s^\nu (\partial ^{\nu +\kappa } g)(x,\Phi _t({\hat{z}}))&\text { for all } {\hat{z}}\in [-1,1]^d,\ \nu \in {\mathbb {N}}_0^d \end{aligned}$$

(31)

with the vector

$$\begin{aligned} s := \begin{pmatrix} (b_{t,1}-a_{t,1}+2\delta _t)/2\\ \vdots \\ (b_{t,d}-a_{t,d}+2\delta _t)/2 \end{pmatrix} \in {\mathbb {R}}^d. \end{aligned}$$

We proceed by induction: for the multiindex $\nu =0$, the identity (31) is trivial.

Let $m\in {\mathbb {N}}_0$ and assume that (31) has been proven for all multiindices $\nu \in {\mathbb {N}}_0^d$ with $|\nu |\le m$. Let $\nu \in {\mathbb {N}}_0^d$ be a multiindex with $|\nu |=m+1$. Then we can find $\mu \in {\mathbb {N}}_0^d$ with $|\mu |=m$ and $i\in \{1,\ldots ,d\}$ such that $\nu =(\mu _1,\ldots ,\mu _{i-1},\mu _i+1,\mu _{i+1},\ldots ,\mu _d)$. This implies

$$\begin{aligned} \partial ^\nu g_x({\hat{z}})&= \frac{\partial }{\partial {\hat{z}}_i} (\partial ^{\mu +\kappa } g_x)({\hat{z}})&\text { for all } {\hat{z}}\in [-1,1]^d. \end{aligned}$$

Applying the induction assumption and the chain rule yields

$$\begin{aligned} \partial ^\nu g_x({\hat{z}})&= \frac{\partial }{\partial {\hat{z}}_i} \partial ^\mu g_x({\hat{z}}) = \frac{\partial }{\partial {\hat{z}}_i} s^\mu (\partial ^{\mu +\kappa } g)(x,\Phi _t({\hat{z}}))\\&= s^\mu \frac{b_{t,i}-a_{t,i}+2\delta _t}{2} \left( \frac{\partial }{\partial {\hat{z}}_i} \partial ^{\mu +\kappa } g\right) (x,\Phi _t({\hat{z}}))\\&= s^\nu (\partial ^{\nu +\kappa } g)(x,\Phi _t({\hat{z}})) \qquad \text { for all } {\hat{z}}\in [-1,1]^d \end{aligned}$$

since $D\Phi _t$ is a diagonal matrix due to (6). The induction is complete.

By definition (7), we have

$$\begin{aligned} \gamma _\iota ({\hat{z}})&= \Phi _t({\hat{z}}_1,\ldots ,{\hat{z}}_{\lceil \iota /2 \rceil -1}, \pm 1, {\hat{z}}_{\lceil \iota /2 \rceil }, \ldots , {\hat{z}}_{d-1}), \end{aligned}$$

therefore (31) implies

$$\begin{aligned} \partial ^{\hat{\nu }} {\hat{f}}_1({\hat{z}})&= s^\nu (\partial _y^\nu g)(x,\gamma _\iota ({\hat{z}}))&\text { for all } {\hat{z}}\in Q,\ \hat{\nu }\in {\mathbb {N}}_0^{d-1} \end{aligned}$$

(32a)

with

$$\begin{aligned} \nu := (\hat{\nu }_1, \ldots , \hat{\nu }_{\lceil \iota /2\rceil -1}, 0, \hat{\nu }_{\lceil \iota /2\rceil }, \ldots , \hat{\nu }_{d-1}). \end{aligned}$$

The exterior normal vector on the surface $\gamma _\iota (Q)$ is the $\iota /2$th canonical unit vector if $\iota$ is even and the negative $(\iota +1)/2$th canonical unit vector if it is uneven, so we obtain also

$$\begin{aligned} \partial ^{\hat{\nu }} {\hat{g}}_2({\hat{z}})&= \pm s^\nu (\partial _y^{\nu +\kappa } g)(x,\gamma _\iota ({\hat{z}}))&\text { for all } {\hat{z}}\in Q,\ \hat{\nu }\in {\mathbb {N}}_0^{d-1}, \end{aligned}$$

(32b)

where $\kappa$ is the $\lceil \iota /2 \rceil$th canonical unit vector in ${\mathbb {N}}_0^d$.

Exchanging the roles of x and y in these arguments yields

$$\begin{aligned} \partial ^{\hat{\nu }} {\hat{f}}_2({\hat{z}})&= s^\nu (\partial _x^\nu g)(\gamma _\iota ({\hat{z}}),y),\end{aligned}$$

(32c)

$$\begin{aligned} \partial ^{\hat{\nu }} {\hat{g}}_1({\hat{z}})&= \pm s^\nu (\partial _x^{\nu +\kappa } g)(\gamma _\iota ({\hat{z}}),y)&\text { for all } {\hat{z}}\in Q,\ \hat{\nu }\in {\mathbb {N}}_0^{d-1}. \end{aligned}$$

(32d)

Now we only have to combine the equations (32) with

$$\begin{aligned} |s^\nu |&\le (2\delta _t)^{|\nu |} \qquad \text { for all } \nu \in {\mathbb {N}}_0^d \end{aligned}$$

and the asymptotic smoothness (15) to obtain

$$\begin{aligned} |\partial ^{\hat{\nu }} {\hat{f}}_1({\hat{z}})|&\le (2\delta _t)^{|\hat{\nu }|} C_\mathrm{as} \hat{\nu }! \frac{c_0^{|\hat{\nu }|}}{\Vert x-\gamma _\iota ({\hat{z}})\Vert ^{\sigma +|\hat{\nu }|}}\\&\le \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^\sigma } \left( \frac{2 \delta _t c_0}{\delta _t} \right) ^{|\hat{\nu }|} = \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^\sigma } (2 c_0)^{|\hat{\nu }|},\\ |\partial ^{\hat{\nu }} {\hat{f}}_2({\hat{z}})|&\le (2\delta _t)^{|\hat{\nu }|} C_\mathrm{as} \hat{\nu }! \frac{c_0^{|\hat{\nu }|}}{\Vert \gamma _\iota ({\hat{z}})-y\Vert ^{\sigma +|\hat{\nu }|}}\\&\le \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^\sigma } \left( \frac{2 \delta _t c_0}{\delta _t} \right) ^{|\hat{\nu }|} = \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^\sigma } (2 c_0)^{|\hat{\nu }|},\\ |\partial ^{\hat{\nu }} {\hat{g}}_1({\hat{z}})|&\le (2\delta _t)^{|\hat{\nu }|} C_\mathrm{as} \hat{\nu }! \frac{c_0^{|\hat{\nu }|+1}}{\Vert x-\gamma _\iota ({\hat{z}})\Vert ^{\sigma +1+|\hat{\nu }|}}\\&\le \frac{C_\mathrm{as} c_0 \hat{\nu }!}{\delta _t^{\sigma +1}} \left( \frac{2 \delta _t c_0}{\delta _t} \right) ^{|\hat{\nu }|} = \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^{\sigma +1}} (2 c_0)^{|\hat{\nu }|},\\ |\partial ^{\hat{\nu }} {\hat{g}}_2({\hat{z}})|&\le (2\delta _t)^{|\hat{\nu }|} C_\mathrm{as} \hat{\nu }! \frac{c_0^{|\hat{\nu }|+1}}{\Vert \gamma _\iota ({\hat{z}})-y\Vert ^{\sigma +1+|\hat{\nu }|}}\\&\le \frac{C_\mathrm{as} c_0 \hat{\nu }!}{\delta _t^{\sigma +1}} \left( \frac{2 \delta _t c_0}{\delta _t} \right) ^{|\hat{\nu }|} = \frac{C_\mathrm{as} \hat{\nu }!}{\delta _t^{\sigma +1}} (2 c_0)^{|\hat{\nu }|}, \end{aligned}$$

where Lemma 3 provides the lower bounds $\delta _t\le \Vert x-\gamma _\iota ({\hat{z}})\Vert$ and $\delta _t\le \Vert \gamma _\iota ({\hat{z}})-y\Vert$ and we have taken advantage of $(\nu +\kappa )!=\nu !=\hat{\nu }!$ due to $\nu _{\lceil \iota /2\rceil }=0$. $\square$