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So far I have only the most basic understanding of the Sobolev space, like the existence of a unique weak solution, or what it means in the PDE: $$\Delta u = -f ,\qquad \text{where } f \text{ is a Schwartz function}$$ However, how do I think about a PDE when $f$ only satisfies a weaker condition, like $f \in L^2$. I think I got lost in all the different kinds of Sobolev spaces, and what is allowed in each space. So, just take that $f \in L^2$. Can you kindly show how you examine the existence of a unique solution to the Laplace equation above? Thanks ....

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I assume that $\Omega$ is bounded and has smooth boundary and that you have zero Dirichlet boundary conditions. If $f\in L^2(\Omega)$ (in fact one can take weaker conditions $f\in H^{-1}(\Omega)$ here) then consider the weak formulation of the Poisson equation:

$$\int \nabla u \cdot \nabla v = \int f v, \forall v\in H^1_0(\Omega).$$

Using the Lax-Milgram theorem, one can deduce existence and uniqueness of solutions $u \in H^1_0(\Omega)$ satisfying the weak formulation.

In my opinion, it is also instructive to establish uniqueness by hand since it requires several ubiquitous estimates. Note that

$$\|\nabla u \|^2_{L^2(\Omega)}=\int \nabla u \cdot \nabla u = \int f u \leq \|f\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)}\leq C_\epsilon\|f\|_{L^2(\Omega)}^2+\epsilon \|u\|_{L^2(\Omega)}^2.$$

Using the Poincare inequality for $u$: $\|u\|_{L^2(\Omega)}\leq C\|\nabla u\|_{L^2(\Omega)}$ and taking $\epsilon$ sufficiently small we have

$$\|\nabla u\|_{L^2(\Omega)} \leq C \|f\|_{L^2(\Omega)}.$$

So, assuming there are two weak solutions $u,v\in H^1_0(\Omega)$, consider $w=u-v$ which solves $\Delta w = 0$ with $w=0$ on $\partial \Omega$. Our above estimate establishes that $w\equiv 0$ and so $u=v$.

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  • $\begingroup$ Lax-Milgram theorem also gives uniqueness. It is not needed to prove it separately. $\endgroup$ – Pedro Dec 14 '16 at 3:12
  • $\begingroup$ You are right. In fact, I simply like this proof of uniqueness since it is easy to write and uses several useful estimates. However, I will edit my answer to indicate that Lax-Milgram is sufficient. Thank you! $\endgroup$ – Matt Dec 14 '16 at 3:13
  • $\begingroup$ Yes, it is instructive to prove it separately (especially because in general the uniqueness is not easily given with the existence, which happens if we use the Galerkin method as sketched here). $\endgroup$ – Pedro Dec 14 '16 at 3:31
  • $\begingroup$ So, if $f \in H^{-2}$, then to maintain integrability, I need to go up to $u \in H^2$, right? Now what happens it I need am using a different $p$-norm? $\endgroup$ – Andy Tam Dec 15 '16 at 4:46
  • $\begingroup$ If $f\in H^{-2}$ you can expect the solution to be $u\in L^2$ but your test functions $v$ would come from $v\in H^{2}_0$. I do not know anything interesting to say about different $p$-norms, but I can suggest for example looking up literature on weak solutions of the $p$-Laplacian. There you can expect solutions in $W^{1,p}$. $\endgroup$ – Matt Dec 15 '16 at 14:26

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