Differences between $-\Delta: H_0^1(\Omega)\to H^{-1}(\Omega)$ and $-\Delta: H^2(\Omega)\cap H_0^1(\Omega)\to L^2(\Omega)$

analysis functional-analysis partial-differential-equations sobolev-spaces

1,069

$H^{-1}=(H^1_0)^*$, so $−\Delta:H^1_0(Ω) \rightarrow H^{−1}(Ω)$ means $-\Delta u \in H^{−1}(Ω)$ for any $u \in H^1_0(Ω)$, and that is obviously right. Because $-\Delta u \in H^{−1}(Ω)$ means $(-\Delta u,v)$ is a scalar, for any $v \in H^1_0(Ω)$. this also satisfies the definition of weak solution.

Therefore, $\Delta u \in L^2(\Omega)$, you mean this function is in $L^2$, while, if you say $\Delta u \in H^{-1}(Ω)$, you mean $\Delta u$ is operator on $H^1_0(Ω)$. These are both correct, I believe if you know something about distribution, this would be easy to understand.

I don't know if I catch your point.

1,069

ningbo polymer

Updated on August 01, 2022

Comments

ningbo polymer 10 months

I'll try to explain what I want to know: Let $\Omega\subset\mathbb{R}^n$ be a bounded domain.

When we look to $-\Delta: H^2(\Omega)\cap H_0^1(\Omega)\to L^2(\Omega)$, the meaning of $-\Delta$ is precisely "the sum of second partial derivatives with Dirichlet boundary conditions" since the partial (weak) derivatives of a $ H^2(\Omega)$ exist and lie on $L^2(\Omega)$. So when we say "let's solve $-\Delta u=f$ !" this equation means what it means. To solve this we can define the weak problem "find an $u\in H_0^1(\Omega)$ such that $\langle \nabla u, \nabla v\rangle_0=\langle f,v\rangle_0$ for all $v\in H_0^1(\Omega)$", solve it applying Riesz Representation Theorem and then use regularity theory to show that $u\in H^2(\Omega)$.

It's not the case with $-\Delta: H_0^1(\Omega)\to H^{-1}(\Omega)$. What's the meaning of $-\Delta u$ for $u\in H_0^1(\Omega)$?
- user127096 about 9 years
  
  $-\Delta u$ is a bounded linear functional on $H_0^1$, defined by $v\mapsto \int_\Omega \nabla u\cdot \nabla v$.
ningbo polymer about 9 years

Thanks for you answer, chuck. But what do you mean with $(-\Delta u,v)$? Is this the pairing $H^{-1}$,$H_0^1$? If yes, you are saying $-\Delta$ is simply the Riesz duality map, right? (I think I'll edit the question to include more details)
chuck about 9 years

@ningbo polymer $(-\Delta u, v)$ is the pair $(H^{-1},H^1_0)$ because you can view $-\Delta u$ (actually, you should write $(-\Delta u,.)$, but just like we do in Representation theory, we can regard it as $-\Delta u$) as all the continuous linear functionals on $H^1_0$, which means $-\Delta u$ should be in $H^1_0$. Then, $-\Delta$ maps u which is in $H^1_0$ into $H^{-1}$