Apply Green's theorem to prove that, if $V$ and $V'$ be solutions of Laplace's equation such that $V=V'$ at all points of the closed surface $S$, then $V=V'$ throughout the interior of $S$.
Attempt:
Clearly, $\nabla^2V=0=\nabla^2V'$. Let $U=V-V'$, then $\nabla^2 U=0$. We know that $\nabla U=\frac{\partial U}{\partial \bar{n}}\bar{n}$. One can write by Gauss's theory here for $U$ that
$$\iint_S U\frac{\partial U}{\partial \bar{n}}dS=\iiint_V|\nabla U|^2dV$$
I have no idea how to go further as I have to use Gree's th (not div th) and how to understand about the conclusion "at all points of the closed surface S", and "throughout the interior of S".
Although, I know the 2nd identity of Greens as $$\iint_S (V\frac{\partial V}{\partial \bar{n}}-V'\frac{\partial V'}{\partial \bar{n}})dS=\iiint_V (V\nabla^2 V -V' \nabla^2 V')dV=0$$ (as $\nabla^2V=0=\nabla^2V'$).
Using Green's identity
$$\iiint_V u \Delta v dV = - \iiint_V \nabla u \cdot \nabla v \; dV + \iint_S u \frac{\partial v}{ \partial n} dS $$
We define the new function $U = V- V'$ and replace both u, and v in the above
$$\iiint_V U \Delta U dV = - \iiint_V \nabla U \cdot \nabla U \; dV + \iint_S U \frac{\partial U}{ \partial n} dS $$
Which simplifies to
$$\iiint_V U \Delta U dV = - \iiint_V |\nabla U|^2 \; dV + \iint_S U \frac{\partial U}{ \partial n} dS $$
The left hand integral is identically zero since we have that $\Delta U = 0$. Next you have to get rid of the surface integral. You have to make the assumption that since U satisfies the laplace equation, either require that $U=0$, or $\nabla U \cdot n = 0$ on the boundary. This means the surface integral is also zero. After that you have $$\iiint_V |\nabla U|^2 \; dV = 0 $$ Since V is arbitrary the integrand is identically zero $$ \Rightarrow \nabla U = 0$$ Which means that it is a constant, in the dirichlet problem we can choose this constant to be $0$ hence $$U = 0 \rightarrow V - V' = 0$$