Transformation Methods for the Numerical Integration of Three-Dimensional Singular Functions

Abstract

The implementations of the eXtended Finite Element Method and the Boundary Element Method need to face the challenge of integrating singular functions. Since standard quadrature techniques usually produce inaccurate results, a number of specific algorithms have been developed to address this problem. We present a general framework for the systematic formulation of the three-dimensional case. The classical cubic transformation is also considered, including an analytical optimization of its parameters for improved practical efficiency.

Appendix

A proof of the Theorem in Sect. 3.2.5 is now provided.

Lemma 1

Consider the family of cubic polynomials given by

$$\begin{aligned} P_{r}(\tau )=(1-r)\tau ^{3}-r\tau +\varepsilon , \end{aligned}$$

(31)

with $r\in [0,1)$ and $\varepsilon >0$. Then, $P_{r}$ has exactly one negative root that is a strictly decreasing function of the parameter r.

Furthermore, the polynomials $(1-r)\tau ^{3}+r\tau -\varepsilon$ have exactly one positive root that is a strictly decreasing function of r.

Proof

A direct application of Descartes’ rule of signs shows that $P_{r}(\tau )$ only has a negative root, that will be denoted by $\tau _{1}(r)$, as shown in Fig. 11.

Assuming $0\leqslant r_{1}<r_{2}<1$, it can be readily shown that

$$\begin{aligned} P_{r_{1}}(\tau )-P_{r_{2}}(\tau )=(r_{2}-r_{1})(\tau ^{2}+1)\tau , \end{aligned}$$

(32)

i.e., $P_{r}$ intersect only at $(0,\varepsilon )$ and $P_{r_{1}}(\tau )<P_{r_{2}}(\tau )$ if $\tau <0$. Since $P_{r_{1}}$ vanishes at $\tau _{1}(r_{1})$, it follows from (32) that $P_{r_{2}}(\tau _{1}(r_{1}))>0$. Moreover, $P_{r_{2}}(\tau )\rightarrow -\infty$ as $\tau \rightarrow -\infty$, and it is a consequence of Bolzano’s theorem that $P_{r_{2}}$ has its negative root in $(-\infty ,\tau _{1}(r_{1}))$, i.e., $\tau _{1}(r)$ is strictly decreasing.

The second part of the Lemma is proved in a completely analogous way. $\square$

Fig. 11

Intersection of $P_{r}(\tau )$

Lemma 2

The poles of $\phi _{N}(q(t))$ reach a maximum distance to the real axis for the optimal value $r_{0}$ given in (30).

Proof

The composite kernel

$$\begin{aligned} \phi _{N}(q(t))=\left( (rt+(1-r)t^{3})^{2}+\varepsilon ^{2}\right) ^{-\alpha /2}, \end{aligned}$$

has 6 complex poles given by the roots of the equation

$$\begin{aligned} (1-r)t^{3}+rt\pm i\varepsilon =0. \end{aligned}$$

(33)

Since t is a solution of (33) if $-t$ is a solution, it suffices considering the 3 complex roots of

$$\begin{aligned} (1-r)t^{3}+rt-i\varepsilon =0, \end{aligned}$$

(34)

where $\varepsilon >0$ is assumed without loss of generality. In order to avoid complex coefficients, let $t=i\tau$ in (34) to obtain $P_{r}(\tau )=0$, with $P_{r}$ defined in (31). Hence, the real part of the roots $\tau _{j}$ is taken into account from now on. $\square$

It is clear that $P_{r}$ has two or zero positive roots depending on the sign of $P_{r}(\tau )$ at the local minimum point, $\tau _{m}(r)=\left( \frac{r}{3(1-r)}\right) ^{1/2}$, with

$$\begin{aligned} P_{r}(\tau _{m}(r))=\varepsilon \left( 1-\sqrt{\frac{4}{27\varepsilon ^{2}}\frac{r^{3}}{1-r}}\right) . \end{aligned}$$

(35)

We denote by $r_{0}$ the value of $r\in (0,1)$ for which $P_{r}(\tau _{m}(r_{0}))=0$, and discuss the distance of the closest root to the imaginary axis when $P_{r}(\tau _{m}(r))$ changes its sign:

• (i) $P_{r}(\tau _{m}(r))\geqslant 0$ $(0\leqslant r\leqslant r_{0})$ In this case $P_{r}$ has two complex conjugate roots, denoted by $\tau _{23}(r)$. The two complex roots merge into a double real root in the limit case $P_{r}(\tau _{m}(r))=0$. One of the well-known Vieta’s formulas states that the sum of the three roots of a cubic polynomial equals its quadratic coefficient (with sign changed). Since in our case this coefficient is zero, the real part of the complex roots is $\mathfrak {R}(\tau _{23})=-\frac{\tau _{1}}{2}$, i.e., $\tau _{23}$ is closer than $\tau _{1}$ to the imaginary axis, and, according to Lemma 1 its real part is a positive and strictly increasing function of r. Its maximum $\tau _{0}$ is then reached at $r_{0}$, with

  $$\begin{aligned} \tau _{0}(\varepsilon )=\frac{3\varepsilon }{2r_{0}(\varepsilon )}. \end{aligned}$$

  (36)

• (ii) $P_{r}(\tau _{m}(r))<0$ $(r_{0}<r<1)$ In this case $P_{r}$ has two distinct positive roots, denoted by $\tau _{2}(r)$ and $\tau _{3}(r)$, with $\tau _{2}<\tau _{3}$. Since, according to (32), $P_{r}(\tau _{0})<0$ and $P_{r}(0)=\varepsilon$, it follows from Bolzano’s theorem that $0<\tau _{2}<\tau _{0}$. In other words, the root $\tau _{2}$ is always closer to the imaginary axis than $\tau _{0}$.

We conclude that the distance of the closest pole to the real axis reaches a maximum at $r_{0}$, i.e. when $P_{r}(\tau _{m}(r))=0$ in (35), which is equivalent to the cubic equation

$$\begin{aligned} \frac{4}{27\varepsilon ^{2}}r^{3}+r-1=0. \end{aligned}$$

(37)

Explicit inversion of (37) by means of the well-known classical formulas (see e.g. [35]) leads to the final expression for $r_{0}$ provided in (30). $\square$

Lemma 3

The poles of $\phi _{N}(q(t))$ and $\phi _{N}(\bar{v}(v))$ have imaginary parts bounded below by $\left( \frac{\varepsilon }{2}\right) ^{1/3}$ and $\frac{\varepsilon ^{1/3}}{2}$ respectively.

Proof

From (22) we have that $v=\frac{t(v)-t_{0}}{t_{1}-t_{0}}$ and taking (36) into account it follows that the complex poles of $\phi _{N}(\bar{v}(v))$ have imaginary parts given by

$$\begin{aligned} \mathfrak {I}(v)=\frac{3\varepsilon }{2r_{0}(\varepsilon )(t_{1}(\varepsilon )-t_{0}(\varepsilon ))}. \end{aligned}$$

We next find upper bounds for both factors in the denominator.

The term in brackets in (30) is bounded above by $2^{1/3}$, and thus $r_{0}\leqslant 3\left( \frac{\varepsilon }{2}\right) ^{2/3}$. We notice that (36) implies a lower bound for $\tau _{0}$, namely $\tau _{0}\geqslant \left( \frac{\varepsilon }{2}\right) ^{1/3}.$

On the other hand, according to (21)–(23) we can consider $t_{j}$ as functions of $\bar{v}_{p}$ and try to find the maximum of the function $t_{1}(\bar{v}_{p})-t_{0}(\bar{v}_{p})$, i.e. we impose

$$\begin{aligned} \frac{d}{d\bar{v}_{p}}\left( t_{1}(\bar{v}_{p})-t_{0}(\bar{v}_{p})\right) =\frac{1}{q'(t_{1})}-\frac{1}{q'(t_{0})}=0, \end{aligned}$$

from where the condition $t_{1}^{2}-t_{0}^{2}=0$ can be derived. Since q(t) is strictly increasing, $t_{1}>t_{0}$ and we have that $t_{1}=-t_{0}$. Substituting terms in (23) and summing equations for $j=1,2$, we obtain $\bar{v}_{p}=\frac{1}{2}$. This way, the equation for $t_{1}$ takes the form

$$\begin{aligned} (1-r)t_{1}^{3}+rt_{1}-\frac{1}{2}=0. \end{aligned}$$

(38)

Applying the second part of Lemma 1, the only positive real root of (38) is a decreasing function of r. Its maximum is therefore reached at $r=0$, i.e. $t_{1}\leqslant \left( \frac{1}{2}\right) ^{1/3}$. It follows that $t_{1}-t_{0}=2t_{1}\leqslant 2^{2/3}$, which finishes the proof. $\square$

The Theorem in Sect. 3.2.5 is a consequence of Lemmas 1–3.