Need help in solving an equation involving volume, single and double layer potentials

Question

Let be $V \subset \mathbb{R}^n $, $ 3\leq n$ an open set, where you can apply Gauß's Theorem.

To show is, that for all $ U \in C^{(1)} ( \bar{V} ) \cap C^{(2)} (V) $ with bounded 2nd derivatives the following equation for $y \in V $:

$$ (n-2) \omega_{n-1} U(y) = \int_{ \partial V} \left[\frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x)-U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}}\right] d\sigma(x)- \int_V \frac{ \Delta U(x)}{ |x-y|^{n-2}} dx $$

where

• $ \nu_x $ is the outer normal unit vector on $x\in \partial V$ and

• $w_{n-1} := \frac{n \pi^{n/2}}{ \Gamma( \frac{n}{2} +1) }$

Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$. Therefore, instead of integrating over $V$ , first integrate over $V_{\epsilon}:= V$ \ $ K_{\epsilon}(y) $ and use the limes $\epsilon \rightarrow 0+ $

Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.

Any help is therefore very appreciated !!

Answer 1

Here is a rough sketch on how I think this equation can be shown.

As you stated, we should work on the domain $V_\epsilon$ instead of $V$ first. The other important idea is to use Green's seocond identity.

So we apply Green's second identity to the domain $V_\epsilon$. If we also use that $\Delta W(x)=0$ for $x\neq y$ then the equation becomes $$ (n-2) \omega_{n-1} U(y) = \int_{ \partial K_\epsilon} \left[\frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x)-U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}}\right] d\sigma(x). $$ Note that the terms involving integrals over $V$ or $\partial V$ disappeared.

Now we will work with the right-hand side and consider the convergence $\epsilon\to0$. First we consider the first term of the integral. Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$ are bounded by a constant $C$ if $x$ is close to $y$. We have $$ \left|\int_{ \partial K_\epsilon} frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x) d\sigma(x)\right| \leq C \int_{ \partial K_\epsilon} \frac{1}{|x-y|^{n-2}} d\sigma(x) = C \epsilon^{2-n} \omega_{n-1} \epsilon^{n-1}\to 0, $$ where we used that the surface area of a ball with radius $\epsilon>0$ is given by $\omega_{n-1}\epsilon^{n-1}$.

So it remains to consider the other term. First, we can calculate that $$ \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} = -(n-2) \frac{1}{|x-y|^{n-1}}. $$ Then we have $$ \int_{ \partial K_\epsilon} U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} d\sigma(x) = -(n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(x) d\sigma(x). $$ It remains to consider the difference to the left-hand side of the original equation. $$ \left| (n-2)\omega_{n-1} U(y) -(n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(x) d\sigma(x) \right| =\left| (n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(y)-U(x) d\sigma(x) \right| \leq \sup_{x\in \partial K_\epsilon} |U(y)-U(x)| (n-2)\epsilon^{-(n-1)}\omega_{n-1} \epsilon^{n-1} \to0 $$ where the $\sup$-term converges to $0$ because $U$ is continuous.

Answer 2

I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity $$ \int\limits_V \big[U(x)\Delta W(x) - W(x)\Delta U(x)\big]\mathrm{d}x = \int\limits_{\partial V}\bigg[U(x)\frac{ \partial W(x)}{ \partial \nu_x}-W(x)\frac{ \partial U(x)}{ \partial \nu_x}\bigg] \mathrm{d}\sigma_x\label{1}\tag{1} $$ However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,\epsilon)\triangleq V\setminus B(y,\epsilon)$, where $B(y,\epsilon)$ is the $n$-dimensional ball of radius $\epsilon>0$ centered at $y\in\Bbb R^n$. It is sufficient to recall that, for $n\geq 3$, $$ \Delta \frac{1}{|y-x|^{n-2}}=-(n-2)\omega_{n-1} \delta(x-y)=-(n-2)\omega_{n-1} \delta(y-x)\label{2}\tag{2} $$ where $\delta$ is the usual Dirac distribution supported at $y\in\Bbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting $$ W(x)=\frac{1}{|x-y|^{n-2}}\quad\text{for an arbitrary }y\in\Bbb R^n $$ in formula \eqref{1}, at the left side we get $$ \begin{split} \int\limits_V \big[U(x)\Delta W(x) - W(x)\Delta U(x)\big]\mathrm{d}x &= \int\limits_V \bigg[U(x)\Delta \frac{1}{|x-y|^{n-2}} - \frac{\Delta U(x)}{|x-y|^{n-2}}\bigg]\mathrm{d}x \\ &= \int\limits_V \bigg[U(x)\Delta \frac{1}{|x-y|^{n-2}} - \frac{\Delta U(x)}{|x-y|^{n-2}}\bigg]\mathrm{d}x\\ &= -(n-2)\omega_{n-1}\int\limits_V U(x)\delta(y-x)\mathrm{d}x - \int\limits_V\frac{\Delta U(x)}{|x-y|^{n-2}}\mathrm{d}x\\ &= -(n-2)\omega_{n-1}U(y)- \int\limits_V\frac{\Delta U(x)}{|x-y|^{n-2}}\mathrm{d}x \end{split} $$ while at the second side we simply have $$ \int\limits_{\partial V}\bigg[U(x)\frac{ \partial W(x)}{ \partial \nu_x}-W(x)\frac{ \partial U(x)}{ \partial \nu_x}\bigg] \mathrm{d}\sigma_x =\int\limits_{ \partial V} \left[U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} - \frac{1}{|x-y|^{n-2}} \frac{ \partial U(x)}{ \partial \nu_x}\right] \mathrm{d}\sigma(x) $$ From these two relations, you immediately get the sought for formula.

How to prove directly, without using distribution theory, the following equality, key relation of the whole proof? $$ \int\limits_V U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x=-(n-2)\omega_{n-1}U(y)\quad\forall y\in V\label{3}\tag{3} $$ I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of \eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression $$ \lim_{\epsilon\to 0} \int\limits_{V_\epsilon} U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x, $$ since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e. $$ \Delta \frac{1}{|x-y|^{n-2}}=0\text{ on }V(y,\epsilon)\implies \int\limits_{V_\epsilon} U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x=0\quad\forall\epsilon>0.\label{a}\tag{*} $$ However we can use \eqref{a} and \eqref{1} by defining, for each $B(y,\epsilon)\Subset V$, $$ \begin{split} \int\limits_{V} U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x & \triangleq \lim_{\epsilon\to 0} \int\limits_{B(y,\epsilon)} U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x \\ \triangleq \lim_{\epsilon\to 0}\,& \left[ \,\int\limits_{B(y,\epsilon)}\frac{\Delta U(x)}{|x-y|^{n-2}}\mathrm{d}x + \int\limits_{ \partial B(y,\epsilon)} \left[U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} - \frac{1}{|x-y|^{n-2}} \frac{ \partial U(x)}{ \partial \nu_x}\right] \mathrm{d}\sigma(x)\right] \end{split} $$ and, since $$ \int\limits_{B(y,\epsilon)}\frac{\Delta U(x)}{|x-y|^{n-2}}\mathrm{d}x\underset{\epsilon \to 0}{\longrightarrow} 0 $$ because $U\in C^2(V)\iff \Delta U\in C^0(V)$ and $|x-y|^{2-n} \in L^1_\mathrm{loc}(\Bbb R^n)$ for $n \ge 2$, we finally get $$ \int\limits_{V} U(x)\Delta \frac{1}{|x-y|^{n-2}}\mathrm{d}x \triangleq \lim_{\epsilon\to 0} \int\limits_{ \partial B(y,\epsilon)} \left[U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} - \frac{1}{|x-y|^{n-2}} \frac{ \partial U(x)}{ \partial \nu_x}\right] \mathrm{d}\sigma(x) $$ Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $x\mapsto (r,\theta_1,\ldots,\theta_{n-1})$: for the first one we have $$ \begin{split} \lim_{\epsilon\to 0} \int\limits_{ \partial B(y,\epsilon)} U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} \mathrm{d}\sigma(x)&=\lim_{\epsilon\to 0} \int\limits_{ \partial B(y,\epsilon)} U \frac{ \partial }{\partial r} \frac{1}{r^{n-2}}\mathrm{d}\sigma\\ &=-\lim_{\epsilon\to 0} \frac{n-2}{\epsilon^{n-1}} \int\limits_{ \partial B(y,\epsilon)} U \mathrm{d}\sigma\\ &=-(n-2) \omega_{n-1} \lim_{\epsilon\to 0} \left[\frac{1}{\omega_{n-1}\epsilon^{n-1}}\int\limits_{ \partial B(y,\epsilon)} U \mathrm{d}\sigma \right]\\ &= -(n-2) \omega_{n-1} U(y) \end{split} $$ since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $\partial B(y,\epsilon)$. For the second one we have $$ \begin{split} \lim_{\epsilon\to 0} \int\limits_{ \partial B(y,\epsilon)} \frac{1}{|x-y|^{n-2}} \frac{ \partial U(x)}{ \partial \nu_x}\mathrm{d}\sigma(x) &= \lim_{\epsilon\to 0} \int\limits_{ \partial B(y,\epsilon)} \frac{1}{r^{n-2}} \frac{ \partial U }{\partial r} \mathrm{d}\sigma \\ &= \lim_{\epsilon\to 0} \frac{1}{\epsilon^{n-2}}\int\limits_{ \partial B(y,\epsilon)} \frac{ \partial U}{\partial r} \mathrm{d}\sigma \\ &= \omega_{n-1} \lim_{\epsilon\to 0} \epsilon \cdot \left[\frac{1}{\omega_{n-1} \epsilon^{n-1}}\int\limits_{ \partial B(y,\epsilon)} \frac{ \partial U}{\partial r} \mathrm{d}\sigma\right] \\ &=0 \end{split} $$ since the integral within the square brackets is the spherical mean of the $C^1$ function $\nabla U(x)$ over the sphere $\partial B(y,\epsilon)$. This last step proves equation \eqref{3} and thus the formula in the OP question.

Notes

• In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).

• The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of \eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula \eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of \eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.

[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.

[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.