0
$\begingroup$

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. ?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Michał Miśkiewicz Apr 22 at 21:46
  • $\begingroup$ You may want to do integration by parts $\endgroup$ – George Dewhirst Apr 23 at 0:31
  • $\begingroup$ 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 $\phi \in W_0^{1,2}(\Omega) .$ $\endgroup$ – Gol D. Roger Apr 23 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.