# Green's Formula as proof for harmony

I showed this to my teacher this morning, and he remarked that the argument was a bit strange...but that it might be correct; he recommended using the ball mean property (which I will do, but I'm curious about this path)...

So, the question is: Does the following argument work? Is it a valid proof?

Partial Differential Equations, Jurgen Jost, pg 31, problem 1.7:

Let $\Omega \subset R^3 \backslash${$0$}$,u:\Omega \rightarrow R$ be harmonic. Show that $$v(x_1,x_2,x_3):=\frac{1}{|x|}u(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})$$ is harmonic in the region $\Omega ' := \{ x \in R^3:(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})\in \Omega\}$.

Proposed solution:

$x=(x_1,x_2,x_3)\\\mu = \frac{x}{|x|}=(\frac{x_1}{|x|},\frac{x_2}{|x|},\frac{x_3}{|x|})\\\bar{\mu}=\frac{\mu}{|x|}=(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})\\\frac{d\bar{\mu}}{d\mu}=\frac{1}{|x|}\\\bar{u}=u(\bar{\mu})\\\frac{\partial \bar{u}}{\partial \mu}=\frac{1}{|x|}*\frac{\partial\bar{u}}{\partial\bar{\mu}}\\v(x)=\frac{\bar{u}}{|x|}$

Since the region $\Omega ' \subset \Omega$, $\Delta \bar{u} = 0$.

Let $\Omega '' \subset \Omega '$ be $\Omega '' := \{ x\in R^3:|x| \gt \epsilon : \epsilon > 0\}$

Green's Formula writes out as:

$$\int_{\Omega ''}\{v\Delta \bar{u}-\bar{u}\Delta v\}dv=\int_{\partial\Omega ''}\{v\frac{\partial \bar{u}}{\partial \mu} - \bar{u}\frac{\partial v}{\partial \mu}\}da$$

Which rewrites as:

$$\int_{\Omega ''}\{\frac{\bar{u}}{|x|}\Delta \bar{u}-\bar{u}\Delta \frac{\bar{u}}{|x|}\}dv=\int_{\partial\Omega ''}\{\frac{\bar{u}}{|x|^2}\frac{\partial \bar{u}}{\partial \bar{\mu}} - \frac{\bar{u}}{|x|^2}\frac{\partial \bar{u}}{\partial \bar{\mu}}\}da$$

Noting that the right side equals zero and the first term on the left side equals zero, it simplifies to:

$$-\int_{\Omega ''} \bar{u}\Delta\frac{\bar{u}}{|x|}dv = 0$$

And the non-trivial solution is $\Delta\frac{\bar{u}}{|x|}=0: x \in \Omega ''$.

And, to qualify the last step (triviality): Either $u$ is identically equal to zero, in which case $\Delta v = 0$, or $u\ne 0$, in which case $\Delta v = 0$ anyways.

Then, since $\Omega ' = lim_{\epsilon \rightarrow 0} \Omega '' : x\in R^3 : \frac{x}{|x|^2} \gt \epsilon \gt 0$, $\Delta v(x)=0\forall x \in \Omega'$.

I realize the above notation is sketchy, I just don't know how to apply a limit and an $\epsilon$/$\delta$ type-argument to sets; it is supposed to prevent the integration over the singularity at $\{0\}$ in order to satisfy the divergence theorem.

$\Omega ''$ and the step above was an attempt to incorporate Tomas's suggestions...but I still don't know if all the steps are valid.

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I have two questions: 1 - How do you know that $\Omega'$ is sufficient smooth, so you can use the green identitie? 2 - What happens if $0\in\partial\Omega'$? –  Tomás Oct 18 '12 at 12:20
(2): if $\{0\}$ is in the domain, $\frac{1}{|x|}$ becomes a singularity. It is likely dumped to guarantee smoothness.  (1): Use of the ball mean property is required for the "correct answer." Any function satisfying the mean value property must be smooth, according to the proof... it is sufficient that it be locally integrable and measurable in $\Omega$. Since $\bar{\mu} \in \Omega$, it is locally integrable...because u is locally integrable in $\Omega$. –  Chris Donlan Oct 18 '12 at 16:25
for (1), i mean the boundary of the domain, not the function. for (2), you have to avoid the singularity to use green identitie, it is possible to do this in your case? –  Tomás Oct 18 '12 at 16:28
as to measurability, looking at this it seems like the set of $x\in \Omega '$ encompasses every vector $\gt 0$...so I guess it is measurable. –  Chris Donlan Oct 18 '12 at 16:31
Sorry, let me explain better. The boundary of $\Omega'$ must be at least an $C^{1}$ manifold. Take a look in the proof of the divergence theorem and you will see it asked this hypoteshis. –  Tomás Oct 18 '12 at 16:34