To show unique solution for the Laplace equation The problem is in the top while its weak form at the end, source
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I know that the solution is unique because the boundary condition is Dirichlet. 
But I want to show this.
How can you show the solution here is unique?
 A: $ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $ 
This can be proved assuming that there exists two different solutions of the Laplace equation subjected to Dirichlet boundary conditions, say $u_1$ and $u_2$. Since $u_1$ and $u_2$ are solutions of the problem, then $u_1=0$ in $\mathbf{x} \in \partial \Omega$ and $u_2=0$ in $\mathbf{x} \in \partial \Omega$ so we conclude $u_1 = u_2$ in $\mathbf{x} \in \partial \Omega$.
To see that $u_1 = u_2$ in the enclosed region by $\partial \Omega$, i.e., in $\mathbf{x} \in \Omega$, we shall consider the volume integral of the quantity $(\nabla U)^2$, $U = u_1-u_2$:

$$ \color{blue}{K =  \int_\Omega (\nabla U)^2 \, \mathrm{d}V > 0} \tag{1}$$

If we use the fact that $(\nabla U)^2 = \nabla \cdot( U \nabla U) - U \nabla^2 U \, ,{}^1$ and notice that $\nabla^2 U = \nabla^2(u_1-u_2)=0$ then we can write $(1)$ as follows:
$$K = \int_\Omega \nabla \cdot U \nabla U \, \mathrm{d}V. \tag{2}$$
Using the divergence theorem in $(2)$ we get:
$$ K = \int_{\partial \Omega} U \, \nabla U \, \mathrm{d}S, \tag{3}$$
but $U = 0$ all over $\partial \Omega$ as we have shown before, so $(3)$ readily yields to $K = 0$. This is only possible if, from $(1)$, $U$ is a constant on $\Omega$. But, since the constant is $0$ at least in $\mathbf{x} \in \partial \Omega$, it follows that $U = 0$ in $\mathbf{x} \in \Omega$, proving uniqueness.
Hope this helps.
Cheers!

Edit: just for completeness
${}^1$, this follows from:
$$ \pd{}{x_i} U \pd{U}{x_i} = \pd{U}{x_i} \pd{U}{x_i} + U \pd{U}{x_i \partial x_i}.$$
