Operator on a Sobolev space [closed]

Question

I'm studying Sobolev spaces using Evans' PDE book. I can't figure out this simple fact. Let $L$ be an operator in this form: $$Lu= \sum{D_i(a_{ij}D_j(u))+\sum{b_iD_i(u)+cu}}.$$

I can't understand why $L \in H^{-1}$, the dual space of $H_0^1$. I'm struggling because to me it would require that $u$ must have 2 derivatives so $u \in H^2$ to be well defined. Am I missing something? Thanks.

EDIT: My problem is about this identity $Lu=f$, $f \in L^2$, in $ \Omega$ plus some boundary conditions for example $u \in H_0^1$.I can define a weak solution as $u \in H_0^1$ such that $B[u,v]=(f,v) $ for all $v \in H_0^1$. Ok that's fine, but when i write $Lu=f $ what does it mean?

Answer 1

I assume $u\in H^1(\Omega)$ for some nice domain $\Omega\subset\mathbb R^n$, and you wonder why $Lu\in H^{-1}(\Omega)$. (Correct me if I'm wrong.) For any $v\in H^1_0$, formal integration by parts gives $$ \langle Lu,v\rangle = \int_\Omega -a_{ij}D_juD_iv+b_iD_iub+cuv. $$ This integral makes perfect sense since $u,v,\nabla u,\nabla v\in L^2(\Omega)$. This is actually how one should define $L$ as an operator $L:H^1\to H^{-1}$, and it should not be hard show that it is continuous if the weights $a,b,c$ are good enough. The reason for taking $v\in H^1_0$ instead of $H^1$ is that it's nice to have a formulation without boundary terms that agrees with the classical one for smooth functions.

Let me add explicitly that you can lose more derivatives than you have. If you have one derivative ($u\in H^1$), you lose two ($u\in H^{-1}$). You can define the operator $L$ between many spaces, not only $H^2\to H^0$. If you end up with $H^s$ with $s<0$, you just need to interpret the derivatives in a distributional sense.

Answer 2

I can't comment yet so let me post as answer. Forget about the coefficients and just take $L=-\Delta$, the Dirichlet Laplacian. Then you are asking why $-\Delta u \in H^{-1}(\Omega)$ when $u \in H^1_0(\Omega)$. It is because the weak Laplacian is defined $$\langle -\Delta u, v \rangle_{H^{-1}, H^1_0} := \int_\Omega \nabla u \nabla v.$$ It is easy to verify that this operator $-\Delta u$ is a bounded linear functional belonging to the dual space of $H^1_0$.

So it seems incorrect in your OP to say " I can't figure out this simple fact." because it is a definition.

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When you say that $-\Delta u =f$ holds as an equality in $H^{-1}$, it means exactly that $$\int_\Omega \nabla u \nabla v = \langle f, v \rangle_{H^{-1}, H^1_0}$$ holds for all $v \in H^1_0$. When you say that $-\Delta u =f$ holds as equality in $L^2$, you are saying that in fact $-\Delta u$ is in the subspace $L^2 \subset H^{-1}$, and therefore in addition to the above, $-\Delta u(x) = f(x)$ holds pointwise a.e. $x \in \Omega$.

I think this probably is not the meaning you intended (you probably meant the weak formulation holds but the right hand side becomes $\int_\Omega fv$ since $f \in L^2$) but you should be aware of it.