# Relation beween weak solution and classical solution .

For $$f \in L^2(\Omega)$$ u is called a weak solution of $$\begin{cases}-\Delta u=f&\text{in \Omega} \\ u=0&\text{in \partial \Omega}\end{cases}$$ if:

1.$$u\in W_0^{1,2}(\Omega)$$

2.$$\int_{\Omega}(\nabla u(x)\nabla \phi(x)-f(x)\phi(x))dx=0$$ for $$\phi \in W_0^{1,2}(\Omega)$$.

Now to prove is :Let $$\Omega \subset\mathbb{R^n}$$ be a bounded domain and $$\partial \Omega \in C^{\infty}$$

$$u \in C^2(\overline \Omega)$$ is a weak solution $$\Rightarrow$$u is a classical solution

Can someone give me a hint , do I have to integrate 2. ?

• Do you know the proof of the opposite implication: $u \in C^2(\overline \Omega)$ is a classical solution $\Longrightarrow$ $u$ is a weak solution? If yes, combine it with the fundamental lemma of calculus of variations. – Michał Miśkiewicz Apr 22 at 21:46
• You may want to do integration by parts – George Dewhirst Apr 23 at 0:31
• For the opposite direction I have $-\Delta u=f \Rightarrow -\int_{\Omega}\Delta u \phi(x)dx=\int_{\Omega}f(x)\phi(x)dx$ $-\int_{\Omega}\Delta u \phi(x)dx=-\int_{\partial \Omega}\nabla u(x) \nu\phi(x)d\sigma_x+\int_{\Omega}\nabla u(x) \nabla \phi(x) dx=\int_{ \Omega}\nabla u(x) \nabla \phi(x)dx$ for $\phi \in W_0^{1,2}(\Omega) .$ – Gol D. Roger Apr 23 at 7:57