Laplace equation and minimum energy A function $\Phi$ in a region $V$ satisfies given Dirichlet BC on the boundary $S$. How to show that $\int_V |\nabla\Phi|^2dV$ is minimum iff $\Phi$ satisfies the Laplace equation $\nabla^2 \Phi=0$ inside $V$? 
 A: The "if" part is wrong. Here is the counterexample:
For $V=[0,\infty)$. Let $\Phi(x)=x$. Then $\Phi(0)=0$, $\Delta\Phi=0$, and 
$\displaystyle\int_V|\Phi|^2dx=\int_0^\infty dx=\infty$ which is not minimizer because 
$\displaystyle\int_V|\Phi|^2dx\geq 0=\int_V|0|^2dx$.
But the "only if" part is true. Here is the proof: 
For any function $\psi$ in the region $V$, we define 
$$f(t)=\int_V|\nabla (\Phi+t\psi)|^2dx.$$
Therefore, $\displaystyle\int_V|\nabla (\Phi+t\psi)|^2dx\geq \int_V|\nabla \Phi|^2dx$ if and only if $f(t)\geq f(0)$ if and only if $t=0$ is a minimum of $f$. This implies that $t=0$ is a critical point of $f$, i.e. 
$$\tag{1}0=f'(0)=\int_V\langle\nabla\Phi,\nabla\psi\rangle dx=-\int_V\psi\Delta\Phi dx$$
where the last equality follows from integration by parts and the condition that $\Phi=0$ on $S=\partial V$. Since $(1)$ is true for arbitrary $\psi$, we must have $\Delta\Phi=0$. 
A: Hint: Calculate the Gateaux Derivative and set it equal to $0$. Then by the fundamental lemma of the calculus of variations, the result follows.
