# Weak Formulations and Lax Milgram:

I have a question on how to put a PDE into weak form, and more importantly, how to properly choose the space of test functions. I know that for an elliptic problem, we want to start with a problem like $Lu = f$, multiply by a smooth test function $v$, integrate by parts typically and we end up with a bilinear form $a(u,v) = l(v)$, where $a$ is coercive and bounded and $l$ is a bounded functional on some Hilbert space $H$. Then, Lax Milgram will tell us there exists a unique solution $u \in H$ of the above problem for all $v \in H$. My question is: how do we properly choose the Hilbert space $H$?

An example from a book of mine: if we have $-\Delta u = f$ on $\Omega$ with the condition that $u = 0$ on $\partial \Omega$, then we multiply by a smooth function $v$, and integrate by parts to arrive at

$$\int_{\Omega} \nabla u \nabla v dx = \int_{\Omega} fv dx \text{ for all } v \in H^1_0(\Omega).$$ I certainly see that choosing $H^1_0(\Omega)$ sounds reasonable, as that condition ensures that $v$ satisfies the boundary condition and also that the bilinear form $a(u,v)$ makes sense (i.e., the integral of $\nabla u \nabla v$ makes sense). Is this the only possibility for a Hilbert space we could choose of test functions? What if we know ahead of time that our solution $u$ is extremely smooth (let's say our data $f$ is $C^{\infty}$ for example). Would it be permissible, albeit un-necessary to choose our test function space to be $H^2_0(\Omega)$? What are the criteria for choosing this space of test functions? Do we choose the test function space to A: make sure the bilinear form $a(u,v)$ makes sense (i.e., we can differentiate the test functions enough times and they're still in $L^2$) and B: they satisfy the boundary conditions?

I apologize if this question is unclear, and thanks for any help!

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In the case of the Dirichlet problem, for example: $$\tag{(D)} \begin{cases} -\Delta u = f & \Omega \\ u= 0 & \partial \Omega\end{cases},$$ a solution $u$, whatever it is, must be something that realizes $$b(u, v)=0,\quad \forall v \in \text{some test function space},$$ where $$b(u, v)=\int\left(-\Delta u - f\right)v\, dx,$$ whenever this makes sense. Turns out that, if we require $u, v\in H^1_0(\Omega)$, then $b$ takes on a super-nice form, the Lax-Milgram's theorem kicks in, everything goes on smoothly and our lives are beautiful. What if we had taken $H^2_0$ instead? Well, in this case we would have had trouble because, even in the simplest case $f=0$, the quadratic form $$b(u,u)=\int \lvert \nabla u \rvert^2\, dx$$ is not coercive, because it cannot control second derivatives. So the Lax-Milgram's theorem doesn't apply and our lives are miserable.
(A last remark which may possibly contradict everything above. As far as I know, there is an abstract theory of linear operators and quadratic forms on Hilbert spaces which, among other things, proclaims that $H^1_0$ is the "right" domain for the quadratic form $b(u,u)$ when $L$ is the Laplacian. If you really are interested in this you could look for the keywords "form domain of self-adjoint operators" or "Friedrichs extension". I am sure that those things are treated in Reed & Simon's Methods of Modern Mathematical Physics and in Zeidler's Applied Functional Analysis.)