Consider the equation , with $u\in H^1$ $$\begin{cases}\Delta u = f & \text{in }\Omega\\ \displaystyle \frac{\partial u}{\partial \nu} = 0 & \text{in } \partial \Omega\end{cases}$$ where $\nu$ is the (unitary) exterior normal to the boundary $\partial \Omega$.
($\Omega$ is a very very nice bounded set)
I want to show that $u$ is a solution of the problem if and only if $\int_{\Omega}fdx=0$.
So the weak form is (using green's):$$\int_{\Omega}Du.Dv \ dx=\int_{\Omega}fv \ dx$$
for all $v\in H^1$.
of course If $u$ is a solution we have that $\int_{\Omega}fdx=0$. (taking $v=1$).
But I can't prove the rest. Some hints will be useful :).
Thanks alot for your time!
Hint: You showed that if the problem is solvable then $\int_{\Omega}fdx=0$, suppose that $\int_{\Omega}fdx=0$ and show that the problem is solvable you need to proof a existence theorem, or use one if you know some.
Edit:
try to use Fredholm alternative
for every $f ∈ L^2(Ω) $the boundary value problem (∗)$ \triangle u = f \,\,in\,\, Ω,\,\,\, \frac{\partial u}{\partial v}= 0 \,\,\,on \,\,\,∂Ω$ has a unique solution $u\in H^1(\Omega)$
OR
) there is a nontrivial solution $u ∈ H^1 (Ω)$ of $\triangle u = 0 $ in Ω, $\frac{\partial u}{\partial v}= 0 $on $\partialΩ$. In case this case we have additionally the system is solvable ⇔ $<h,f>_{L^2}=0$, $v\in L^2 = 0 $ for every weak solution $v ∈ H^1 (Ω)$