Second-order PDE with $H^{-2}$ right-hand side

Question

I have a second-order PDE $$ \mathcal{L} u = f \text{ + homogenous Dirichlet boundary conditions} $$ and I can prove that the bilinear form of the weak form of the differential equation fulfills all prerequisites of the Lax-Milgram Theorem, such that there exists a unique weak solution $u \in H^{1}$ for every data $f \in H^{-1}$. My question now is, whether there can be said anything about less regular data $f \in H^{-2}$? Can it be shown that there exists a unique solution $u \in H^0 = L^2$?

Answer 1

How do you define $H^{-2}?$ As the dual of $H^2$? Take any $\varphi \in H^{-2} \setminus H^{-1}$ and then the problem $$ L(u,v)=\varphi(u) $$ is...
1)not defined for $u,v \in L^2$ since the bilinearform requires $u,v \in H^1$. What is the (weak) gradient of an $L^2$ function anyway. As an example: How do you define $$ L(u,v)=\int \nabla u \cdot \nabla v dx $$
for $u,v$ in $L^2$?
2)The RHS $\varphi(u) \in H^{-2} \setminus H^{-1}$ is not a continuous functional for input functions $u \in H^1$ or $u \in L^2$ and hence Lax-Milgram does not apply.

The only remedy you have is to switch to some space of distributions and formulate your problem there, in which case existence gets a lot harder. A good sample problem to test your idea is $$ u''=\delta'(x) $$ on the unit intervall $I=[0,1]$.