Estimation of quadrature errors for layer potentials evaluated near surfaces with spherical topology

Abstract

Numerical simulations with rigid particles, drops, or vesicles constitute some examples that involve 3D objects with spherical topology. When the numerical method is based on boundary integral equations, the error in using a regular quadrature rule to approximate the layer potentials that appear in the formulation will increase rapidly as the evaluation point approaches the surface and the integrand becomes sharply peaked. To determine when the accuracy becomes insufficient, and a more costly special quadrature method should be used, error estimates are needed. In this paper, we present quadrature error estimates for layer potentials evaluated near surfaces of genus 0, parametrized using a polar and an azimuthal angle, discretized by a combination of the Gauss-Legendre and the trapezoidal quadrature rules. The error estimates involve no unknown coefficients, but complex-valued roots of a specified distance function. The evaluation of the error estimates in general requires a one-dimensional local root-finding procedure, but for specific geometries, we obtain analytical results. Based on these explicit solutions, we derive simplified error estimates for layer potentials evaluated near spheres; these simple formulas depend only on the distance from the surface, the radius of the sphere, and the number of discretization points. The usefulness of these error estimates is illustrated with numerical examples.

Appendices

Appendix A Derivations for the Gauss-Legendre error

We consider a sphere of radius a and the associated Gauss-Legendre error as defined in Eq. 57 with $E_{fac}^{GL}(\varvec{x},\varphi )$ defined in Eq. 58. Under the map $t=-\cos \theta$, the squared distance function evaluates as

$$R^2(t,\varphi ,\varvec{x})=a^2-2a(\sqrt{1-t^2}(x\cos \varphi + y \sin \varphi )+tz)+x^2+y^2+z^2. $$

and $G_{\varvec{\gamma }, 1}\left( t,\varphi ,\varvec{x} \right) =(\partial R^2/\partial t)^{-1}$. The root $\theta _0(\varphi ,\varvec{x})$ is given in Eq. 44, and $t_0(\varphi ,\varvec{x})=-\cos (\theta _0(\varphi ,\varvec{x}))$.

We start by considering an evaluation point at the symmetry axis, i.e., $\varvec{x}=(0,0,z)$, $z \ne a$. For this case we get $\theta _{0}=\pm i \ln (|z|/a)$ as given in Corollary 1. We get $t_0=(\delta +1/\delta )/2$, where we let $\delta =a/|z|$ if $|z|>a$, and $\delta =|z|/a$ if $|z|<a$, such that $\delta >1$. Hence, $\left|\sqrt{t_0^2-1}\right|=(\delta -1/\delta )/2$, and $\left|{t_0+\sqrt{t_0^2-1}}\right|=\delta$. Finally, we have $G_{\varvec{\gamma }, 1}\left( t,\varphi ,\varvec{x} \right) =1/(2a|z|)$, and combined this yields the expression for $E_{fac}^{GL}(\varvec{x},\varphi )$ given in Eq. 59.

Now, we instead consider an evaluation point at the equator, $\varvec{x} =(0,y,0)$. We could, however, equally well pick $\varvec{x}= (x,0,0)$, or $\varvec{x}= (x,y,0)$ and would obtain the same final result with $\Vert \varvec{x}\Vert =\rho$. With $\varvec{x} =(0,y,0)$ we get

$$\begin{aligned} E_{fac}^{GL}(\varvec{x},\varphi ) =\frac{1}{(2a\big \vert {y}\big |\big |{\sin \varphi }\big |)^p} \frac{\Big |{\sqrt{t_0(\varphi )^2-1}}\Big |}{\big |{t_0(\varphi )\big |}^p} |{\frac{1}{t_0(\varphi ) + \sqrt{t_0(\varphi )^2-1}}|}^{2n+1}. \end{aligned}$$

(A1)

The root $t_0(\varphi )=-\cos (\theta _0(\varphi ))$ where $\theta _0(\varphi )$ is defined in Lemma 3, in Eq. 43. With $\varvec{x} =(0,y,0)$, $\lambda$ in that expression simplifies to $\lambda =(|{y}|/a+a/|{y}|)/(2|{\sin \varphi }|=(\delta +1/\delta )/ (2 |{\sin \varphi }|)$ where we let $\delta =|{y}|/a$ if $|{y}|>a$ and $\delta =a/|{y}|$ if $|{y}|<a$, such that $\delta >1$. The expression for $\lambda ^2-1$ then becomes the same as in Eq. 64, but with $\varphi$ instead of $\theta$. The peak of the error is at the closest point to $\varvec{x}=(0,y,0)$, i.e., at $\varphi =\pi /2$, and also here, we ignore the last term in the expression for $\lambda ^2-1$. With this we get that $\lambda +\sqrt{\lambda ^2-1} \approx \delta /|{\sin \varphi }|$. Introducing $\tilde{\delta }=\delta /|{\sin \varphi }|$, and noting that the square roots are evaluated at points away from the branch cut, we have

$$\begin{aligned} t_{0}=i(\tilde{\delta }-1/\tilde{\delta })/2 \qquad \sqrt{t_{0}^{2}-1}=i(\tilde{\delta }+1/\tilde{\delta })/2 \qquad t_{0}+\sqrt{t_{0}^{2}-1}=i\tilde{\delta }. \end{aligned}$$

Similarly to the derivation based on the trapezoidal error, we evaluate all terms in Eq. A1 except the last term at $\varphi =\pi /2$. We then have

$$\begin{aligned} \begin{aligned} E_{fac}^{GL}(\varvec{x},\varphi ) \approx \frac{1}{2} \frac{1}{(a\big |{y}\big |)^{p}} \frac{\big |{\delta +1/\delta }\big |}{\big |{\delta -1/\delta }\big |^p} \frac{\big |{\sin \varphi }\big |^{2n+1}}{\delta ^{2n+1}}. \end{aligned} \end{aligned}$$

(A2)

Inserting into Eq. 57 and using

$$\begin{aligned} \int _{0}^{2\pi }\big |{\sin \varphi }\big |^{2n_t+1} \text {d} \varphi =\int _{0}^{2\pi }\big |{\cos \varphi }\big |^{2n_t+1} \text {d} \varphi =2 \int _{-\pi /2}^{\pi /2} (\cos \varphi )^{2n_t+1} \text {d} \varphi \end{aligned}$$

we can identify the integral in Eq. 67. With $f=a^2$ the total result becomes $E^{GL}(\varvec{\gamma },a^{2},p,2n_t,\varvec{x})$ as given in Eq. 69.

Appendix B Further reflections on the quadrature error behavior

In Error estimate 4, we derived a simplified estimate for the quadrature error for the integral,

$$\begin{aligned} u(\varvec{x}) = \int _{S} \frac{\sigma (\varvec{y})}{\big \Vert {\varvec{y} - \varvec{x}}\big \Vert ^{2p}} \text {d} S(\varvec{y}), \end{aligned}$$

(B3)

with $\sigma \equiv 1$, where S is a sphere of radius a, discretized with the $n_t =n/2$ Gauss-Legendre rule in the t-direction (under the cosine map), and the $n_\varphi =n$-point trapezoidal rule in the $\varphi$ direction. The evaluation point $\varvec{x} =(x,y,z)\in \mathbb {R}^3$ can be close to, but not on S.

We have in Section 7.1 verified that the simplified error estimate is very accurate for $p=1/2$ (see Fig. 4b), and similar results are found for $p=3/2$. Hence, we are in a position to explore this estimate a bit more, trying to extrapolate some simple rule to understand at which (rescaled) distance special quadrature techniques are needed.

Now let $h=2\pi a/n$, this is the grid size in the $\varphi$ direction at the equator, and let the evaluation point $\varvec{x}$ be such that $\Vert \varvec{x}\Vert =a+qh$, $q >0$. Then, the simplified error estimate in Eq. 63 yields

$$\begin{aligned} E_{sphere}(\Vert \varvec{x}\Vert ,a,p,n)= a^{2(1-p)} E(p,q,n) \end{aligned}$$

(B4)

where

$$\begin{aligned} E(p,q,n)= \frac{8\pi }{ (4 \pi )^p \Gamma (p)} n^{2p-1} \frac{n!!}{(n+1)!!} \frac{1}{q^p} \frac{1}{(1+\pi q/n)^p} \frac{1}{(1+2\pi q/n)^n}. \end{aligned}$$

(B5)

Assume now that for each n, we want to know the value of q needed to achieve $E(p,q,n)=TOL$, for different tolerances TOL and choices of p. That means, how far from the surface, expressed in the multiples of the grid size, must the evaluation point be to reach this tolerance. Since we have an explicit formula, we can find the solution numerically, using, e.g., Newton’s method, and the results are shown in Fig. 9 for $p=1/2$ and 3/2.

Fig. 9

Plot of q versus n, with q such that the error in B5 is equal to TOL at evaluation points with $\Vert \varvec{x}\Vert =a+qh=a+2\pi a/n$

From this figure, we can deduce that we can not formulate any simple rule for what q should be to achieve a certain accuracy, as the curves we get are not flat lines. We could say that to achieve an error smaller than $a^{2(1-p)} \cdot 10^{-4}$, $q=2$ would suffice for $p=1/2$ and $q=2.5$ for $p=3/2$, as long as n is at least 20. For more strict error levels, these statements become less useful, as the needed q changes more with n. The estimate we have considered is for the simplest case of a sphere with density $\sigma \equiv 1$, and it is even more difficult to construct a simple rule for a different surface shape and a variable density.