4
$\begingroup$

Let $f=\frac{1}{|x|},x\in\mathbb{R^3}$ and $\Omega=[-b,b]^3$. How to construct a quadrature scheme to solve $$ \int_\Omega f\phi\psi dx\quad ? $$ where $\phi\psi$ is smooth function.

I know there exists a transformation called Duffy transformation which can eliminate the singularity at $x=0$ by using the jacobian. Indeed, let $$\begin{aligned}T&=\{(x_1,x_2,x_3);0\leq x_1\leq 1,0\leq x_2\leq x_1,0\leq x_3\leq x_1\}\\ \hat{T}&=\{(\hat{x}_1,\hat{x}_2,\hat{x}_3);0\leq \hat{x}_1\leq 1,0\leq\hat{x_2}\leq1,0\leq\hat{x_3}\leq 1\}\end{aligned}$$

Define $F:\hat{T}\rightarrow T$ as $$ F(\hat{x}_1,\hat{x}_2,\hat{x}_3)=(\hat{x}_1,\hat{x}_1\hat{x}_2,\hat{x}_1\hat{x}_3) $$ Then $$ \begin{aligned} \int_{T}\frac{1}{|x|}\phi\psi dx&=\int_{\hat{T}}\frac{1}{\sqrt{x_1^2+x_2^2+x_3^2}}\phi\cdot\psi(x)dx =\int_{\hat{T}}\frac{\phi\cdot\psi(\hat{x_1},\hat{x}_1\hat{x}_2,\hat{x}_1\hat{x}_3)}{\sqrt{\hat{x}_1^2+\hat{x}_1^2\hat{x}_2^2+\hat{x}_1^2\hat{x}_3^2}}\hat{x}_1^2d\hat{x}\\ &=\int_\hat{T}\frac{\hat{x}_1\phi\cdot\psi(\hat{x_1},\hat{x}_1\hat{x}_2,\hat{x}_1\hat{x}_3)}{\sqrt{1+\hat{x}_2^2+\hat{x}_3^2}}d\hat{x} \end{aligned} $$ which can be solved by usual quadrature scheme.

On the other hand, someone can also avoid the singularity by omitting a $\varepsilon$ cube which contains the origin.

However, these are not enough for me when I want to study the convergence rate of numerical integration on $\Omega$. I hope the error can be bounded by $O(h)\Vert \phi\Vert_{1,2,\Omega}\Vert \psi\Vert_{1,2,\Omega}$

Is there any other way to integrate a $L^2$ function or a function with singularity at vertex directly with the error bounded by $L^2$ norm?

Thank you very much.

Improve this question
$\endgroup$
10
  • $\begingroup$ If you omit an $\varepsilon$ cube from the domain, call it $E\subset \Omega$, then using Cauchy-Schwarz the error due to this omission is bounded by $\|f\|_E \|g\|_E \le O(\varepsilon^2) \|g\|_\Omega$. Would that be sufficient for your purposes? $\endgroup$ Commented Oct 7, 2017 at 8:43
  • $\begingroup$ Sorry, I wrongly estimated the norm $\|f\|_E$ (used $L^1$ instead of $L^2$). The correct $L^2$ norm is $O(\sqrt{\varepsilon})$. $\endgroup$ Commented Oct 7, 2017 at 8:46
  • $\begingroup$ @IgorKhavkine Well... I think there is a little mistake. $O(\sqrt{\varepsilon})$ might can not be obtained by using holder inequality. $\endgroup$ Commented Oct 7, 2017 at 10:22
  • 1
    $\begingroup$ Why not split g into $g(x)=g(0)+g(x)-g(0)$ and handle the $g(0)$ part separately? $\endgroup$ Commented Oct 8, 2017 at 14:15
  • 1
    $\begingroup$ @MichaelRenardy I don't understand what's you say. Could you explain it a little bit more? $\endgroup$ Commented Oct 8, 2017 at 14:18

1 Answer 1

Reset to default
2
$\begingroup$

Following Michael Renardy's suggestion, write $g=\phi\psi$, and write $$ g(x)=g(0)+x\cdot G(x) , $$ where $G$ is a smooth function. Then the integral becomes $$ \int_\Omega fg = g(0)\int_\Omega f + \sum_i\int_\Omega \frac{x_i}{|x|}G_i(x)d^3x . $$ The integrands in the latter integrals are bounded, and the first integral can be written as $$ \int_\Omega f = \int_{B_r}f + \int_{\Omega\setminus B_r}f , $$ where the first can be computed analytically, and the second has a smooth integrand. Here $B_r$ is some ball centred at the origin.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks for you answer. But the first integral may not be computed analytically sometimes due to the boundary, isn't it? $\endgroup$ Commented Oct 10, 2017 at 2:38
  • $\begingroup$ @whereamI look at the edit $\endgroup$ Commented Oct 10, 2017 at 2:47
  • $\begingroup$ Well.. I think the error of the first part $g(0)\int_\Omega f$ can still not be bounded by $\Vert\phi\Vert_2\Vert\psi\Vert_2$. Anyway, thank you very much, it gives me some ideas. $\endgroup$ Commented Oct 10, 2017 at 2:52
  • $\begingroup$ @whereamI: Even the value of the integral itself cannot be bounded by that, because that would imply that $f$ is bounded. $\endgroup$ Commented Oct 10, 2017 at 2:55
  • $\begingroup$ Why do you think if the integral bounded by that then $f$ is bounded? Could you explain it a little bit more? $\endgroup$ Commented Oct 10, 2017 at 3:01

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.