# Variational formulation-exercice

I have the following exercice: let $\Omega$ an open to $\mathbb{R}^n$, and let $v \in H^1(\Omega)$ and $u \in H^1_0(\Omega)$.

1. Prove that $$-\langle \Delta u, v \rangle_{H^{-1},H^1_0}= \displaystyle\int_{\Omega} \nabla u(x) \cdot \nabla v(x) dx$$ I don't understand how we can answer this question. I'm lost. Help me please.
• What is the LHS ? – anonymus Mar 22 '17 at 10:27
• sorry but i don't understand. Where is LH in my question? Please. – user415040 Mar 22 '17 at 10:31
• I mean : what is your definition of the term $-\langle \Delta u, v \rangle_{H^{-1},H^1_0}$ ? – anonymus Mar 22 '17 at 11:27

Since $u$ does not have second order derivatives, you can define $\Delta u$ only in the sense of distributions, that is, you first consider $$T_u(\phi):=\int_\Omega u\phi\,dx$$ for $\phi\in C^\infty_c(\Omega)$, and then take its distributional derivatives $$\frac{\partial T_u}{\partial x_i}(\phi)=-\int_\Omega u\frac{\partial \phi}{\partial x_i}\,dx$$ and $$\frac{\partial^2 T_u}{\partial x_i^2}(\phi)=\int_\Omega u\frac{\partial^2 \phi}{\partial x_i^2}\,dx.$$ Then $$\Delta T_u(\phi)=\int_\Omega u\Delta \phi\,dx.$$ If you now integrate by parts you get $$\Delta T_u(\phi)=-\int_\Omega \nabla u\cdot\nabla \phi\,dx$$ and by Holder's inequality $$|\Delta T_u(\phi)|\le ||\nabla u||_{L^2}||\nabla \phi||_{L^2}\le ||\nabla u||_{L^2}||\phi||_{H^1}$$ This is telling you that $\Delta T_u$ is a linear continuous operator from $H^1_0(\Omega)$ (you can extend it by density of $C^\infty_c(\Omega)$ to $H^1_0(\Omega)$), and so it belongs to $H^{-1}$.
• What do you mean? I don't understand your question. I just explained how you can interpret the left hand side (LHS) of your equation $\langle \Delta u,v\rangle_{H^{-1},H^1_0}$. The expression $\langle \Delta u,v\rangle_{H^{-1},H^1_0}$ just means the linear operator $\Delta T_u$ applied to $v$. – Gio67 Mar 22 '17 at 11:36
• But why you consider $T_u=\int u \phi$? you can write this only if $u \in L^1_{loc}$. But hear, $u$ isn't in $L^1_{loc}$, it's in $H^1_0$ – user415040 Mar 22 '17 at 11:43
• A function in $H^1_0$ belongs to $L^2$ and so it also belongs to $L^1_{loc}$ – Gio67 Mar 22 '17 at 14:28