PDE in H^-1

1. Jan 8, 2005

Feynman

Hello,
How can i proof the existance of a solution of a PDE on H^(-1)( Omega)?

2. Aug 22, 2009

Reb

You mean (...must have meant) that, given the pde (P) $$\texttt{L}u=f$$ on an appropriate space, the right hand side belongs to $$H^{-1}$$. The most appropriate space for the solutions is, in this case, $$L^{2}$$.

The reason for requiring $$f\in H^{-1}$$ is that now the problem (P) can be put into variational formulation, and then the methods of functional analysis can be applied: Say, for linear problems, the Lax-Milgram Theorem. Or, for nonlinear parabolic problems, monotonicity methods.