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My attempt:

I tried to use the Green's representation formula twice.

The Green's reprensentation formula:$u(y)=\int_{\partial \Omega}(u(x)\frac{\partial G(x-y)}{\partial v}-G(x-y)\frac {\partial u(x)}{\partial v})dx+\int_\Omega G(x-y)\triangle u(x)dx$

First apply the formula to $\triangle u(y)$. I get

$\triangle u(y)=\int_{\partial \Omega}(\triangle u(x)\frac{\partial G(x-y)}{\partial v}-G(x-y)\frac {\partial \triangle u(x)}{\partial v})dx+\int_\Omega G(x-y)\triangle\triangle u(x)dx$

By assumption in the problem, $\triangle u(y)=0$.

Apply Green's representation formula again:

$u(y)=\int_{\partial \Omega}(u(x)\frac{\partial G(x-y)}{\partial v}-G(x-y)\frac {\partial u(x)}{\partial v})dx+\int_\Omega G(x-y)\triangle u(x)dx$

I get $u(u)=0$.

My problem:

I feel confused about this step in my proof(the part in bold) and I feel it is insufficient.

Can anyone give me some clues or answers! That will be extremely helpful! Thanks so much! :>

Sherry
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  • I don't see how you conclude $\triangle u(y)=0$ there. For this boundary value problem, uniqueness is usually proved via Lax-Milgram lemma, see this question. –  Oct 12 '14 at 22:51
  • @CareBear Because $u=\frac {\partial u}{\partial v}=0$ on $\partial \Omega$? Then can I conclude that $\triangle u=\frac {\partial \triangle u}{\partial v}=0$ on $\partial \Omega$? Then the first part of the equation is 0 and the second part is also 0, then $\triangle u=0$. – Sherry Oct 12 '14 at 23:22
  • @CareBear I am sorry to bother you but is this correct？ – Sherry Oct 12 '14 at 23:35
  • Absolutely not. It's like in one dimensions: you can have $f(a)=f'(a)=0$ but $f''(a)$ and $f'''(a)$ may well be nonzero. If you knew that the Laplacian of $u$ is zero on the boundary, this would be an easier problem, and then your approach would work. –  Oct 12 '14 at 23:35

1 Answers1

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As I said in comments, I do not find the conclusion $\Delta u(y)=0$ justified. Here is a different approach.

Let $v=\Delta u$. Since both $u$ and its normal derivative vanish on the boundary, Green's second identity implies $$\int_\Omega (v\Delta u - u \Delta v) = 0 $$ which simplifies to $$\int_\Omega (\Delta u)^2 = 0 $$ Thus, $u$ is harmonic. Since it has zero boundary values, it is identically zero.