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Define $G(x)= \frac{1}{4\pi ||x||}$, suppose that $f(x)$ is known, S is a surface in $\mathbb{R}^3$, and x is fixed, $x \in S$.

I have formulas for computing the following numerically:

$$ p(x) = \int _{S}f(x') \frac{\partial G(x-x')}{\partial n' } dS' $$

Suppose now that the normal derivative in the integral is not taken with respect to $x'$, but with respect to $x$. That is, I want to compute the following:

$$ q(x) = \int _{S} f(x') \frac{\partial G(x-x')}{\partial n} dS' $$

The best would of course be if I could use the values of the first integrals to compute the second ones, but I don't know if this would be possible.

What I tried is the following: We consider the difference

$$ \begin{align} q(x)-p(x) &= \int _{S} f(x') \frac{\partial G(x-x')}{\partial n} dS' - \int _{S} f(x') \frac{\partial G(x-x')}{\partial n'} dS' \\ &= \int _{S} f(x') \left ( \frac{\partial G(x-x')}{\partial n}-\frac{\partial G(x-x')}{\partial n'} \right ) dS' \end{align} $$

Further we have a triangulation of $S$ and assume that $f(x)$ is piecewise constant on each triangle. The surface has the triangles $\Delta_1,...,\Delta_n$ and we assume that $f_i$ is the constant approximation of $f$ on triangle $\Delta_i$ (so $f_i(x) = 0$ if $x$ is not on $\Delta_i$). Then we can write

$$ \begin{align} q(x)-p(x) &= \sum _n \int _{\Delta _n} f_n(x')\left ( \frac{\partial G(x-x')}{\partial n} - \frac{\partial G(x-x')}{\partial n'} \right ) dS' \\ &= \sum _n f_n \int _{\Delta _n} \left ( \frac{\partial G(x-x')}{\partial n} - \frac{\partial G(x-x')}{\partial n'} \right ) dS' \\ \end{align} $$

If now $x$ and $x'$ are on the same triangle, then the normals $n$ and $n'$ are equal and hence $q(x)=p(x)$. But what happens if that's not the case? Are there any other formulas that can be used to connect $p(x)$ and $q(x)$? What confuses me is the fact that we take the normal derivative w.r.t. $n$ and not $n'$ in $q(x)$.

Any input is more than welcome!

Niklas Andersson
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1 Answers1

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I don't believe these are usefully related, i.e., you cannot use $p$ to tell you anything much about $q$. The function $p$ (the double-layer potential) is a weighted integral of $f$, with the weight being the normal derivative of $G$ on the surface. It represents electrostatic potential of a surface that has density charges $\pm f$ on opposite sides. (More precisely, it has dipole density $f$.)

The function $q$, on the other hand, can be written as $$q(x) = \frac{\partial }{\partial n} \int _{S} f(x') G(x-x')\, dS'$$ That is, we integrate $f$ against the weight $G$, thus obtaining its single layer potential, and take the normal derivative of that. So, $q$ represents the normal component of electric force of the surface with charge density $f$.

These are just different animals: dipole potential vs force of a single charge.

  • Thank you for your answer! So I can just move out the normal derivative? I still need to discretize this though, but that is maybe not so problematic? If I triangulate S and assume that f is constant on each triangle, I am basically left with computing what I called q(x) without the f-term. Just compute the integral for fixed x shouldn't be problematic, but how do I take the normal derivative numerically when I only have computed the integral for x on the surface? Should I approximate that one as well by e.g. finite differences, moving my x h units along the normal derivative? – Niklas Andersson Jul 05 '14 at 00:20
  • @NiklasAndersson Yes, $x$ is a parameter in the integral, differentiation with respect to parameter can be done inside or outside, assuming nothing pathological happens with $f$. However, for computational purposes I recommend leaving the derivative inside (where you have it), because then it can be taken analytically. My reason for moving it out was to clarify the meaning of the integral. –  Jul 05 '14 at 01:36
  • I see... In this case I already have a function implemented that computes the values of integral(G(x-x')dS') for fixed x, I should be able to use this to compute approximate values of q(x) using finite differences, right? – Niklas Andersson Jul 06 '14 at 13:14
  • @NiklasAndersson Yes you can. For better results, use a symmetric finite difference formula. –  Jul 06 '14 at 13:40
  • Thank you very much for your help! – Niklas Andersson Jul 06 '14 at 13:56