I want to prove the Dirichlet principle (Energy is minimum) for this problem: $$-\Delta u=0 \quad in \quad D\\ \frac{\partial u}{\partial n}=h\quad on \quad \partial D $$
Solution: Is it possible to use this : Let u and v be two solutions then w=u-v is also a solution then $$\Delta w=0 \quad in \quad D\\ \frac{\partial w}{\partial n}=0\quad on \quad \partial D $$ We multiply the equation by w and integrate over D and then apply Green first identity: $$\iiint_D w\Delta w dx=0\Rightarrow -\iiint_D \nabla w\nabla w dx+\iint_{\partial D} \frac{\partial w}{\partial n}ds \Rightarrow \\-\iiint_D \nabla w\nabla w dx=0\Rightarrow \iiint_D |\nabla w|^2 dx=0$$ $$|\nabla w|^2= |\nabla u-\nabla v|^2=| \nabla u|^2-2|\nabla u|.|\nabla v|-|\nabla u|^2$$ If we use Young inequality then $$|\nabla u|.|\nabla v|\leq \frac{|\nabla u|^2}{2}+\frac{|\nabla v|^2}{2}\Rightarrow -2|\nabla u|.|\nabla v|\geq |\nabla u|^2+|\nabla v|^2$$ $$|\nabla w|^2\geq |\nabla u|^2+|\nabla u|^2+|\nabla v|^2+|\nabla v|^2=2(|\nabla u|^2+|\nabla v|^2)$$ $$\iiint_D |\nabla w|^2 dx\geq 2 \Big(\iiint_D |\nabla u|^2 dx+\iiint_D |\nabla v|^2 dx\Big)\Rightarrow 2 \Big(\iiint_D |\nabla u|^2 dx+\iiint_D |\nabla v|^2 dx\Big)\leq0 \Rightarrow \frac{1}{2}\Big(\iiint_D |\nabla u|^2 dx+\iiint_D |\nabla v|^2 dx\Big)\leq0 \Rightarrow E(u)+E(v)\leq0 \Rightarrow E(u)\leq-E(v)$$
Is what I did make sense?