Timeline for Existence of solution to weak form of linear equation with boundary integral (parabolic PDE)

7 events

when toggle format	what		by	license	comment
Dec 20, 2014 at 16:09	comment	added	Michael Renardy		Using the fact that u is harmonic, you convert $\int_\Omega \nabla u\nabla v$ to $\int_{\partial\Omega}{\partial u\over\partial n}v.$ You then end up with $z_t=Az$, where $A$ is the Dirichlet-to-Neumann map (well studied in the literature).
Dec 20, 2014 at 13:11	comment	added	DeleMax		@MichaelRenardy Right, but then I'd have to consider $\int_{\partial\Omega}z_t \gamma(v) + \int_\Omega \nabla \gamma^{-1}(z)\nabla v=0 \tag{1}$, for all $v \in L^2(0,T;H^1)$, and substiting $w=\gamma(v)$ gives a different weak formulation which is not equivalent, since the test functions are in $H^1(\Omega)$ for a.e. $t$, and the trace is not invertible on that space (inverse is not surjective). Unless some convenient density result holds, or there is some result about mixed abstract parabolic equations of form (1) I am stuck.
Dec 20, 2014 at 2:09	comment	added	Michael Renardy		Your equation implies that u is harmonic, and with this restriction the trace map is invertible.
S Dec 19, 2014 at 20:10	history	suggested	Andrew		parabolic tag added
Dec 19, 2014 at 20:05	review	Suggested edits
S Dec 19, 2014 at 20:10
Dec 19, 2014 at 17:09	history	edited	DeleMax	CC BY-SA 3.0	edited body
Dec 19, 2014 at 16:31	history	asked	DeleMax	CC BY-SA 3.0