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! :>
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.