Friedrichs' inequality in anisotropic Sobolev spaces
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From: Narek Saribekyan on 2 May 2010 15:05 Is there any result on Friedrichs' inequality in anisotropic Sobolev spaces H0^(p, q) (vanishes on boundary with derivatives up to orders p-1 and q-1 respectively)? I want to prove that the bilinear form a(u,v) induced from weak formulation of homogeneous Dirichlet problem (on a rectangle) for Lu = (-1)^p du^2p/dx^2p + (-1)^q du^2q/dy^2q is coercive (so I can apply Lax-Milgram theorem). a(u,v)=(D^p u, D^p v)+(D^q u, D^q v) a(u,u) >= c||u||^2 ? Is L an elliptic operator? Thanks very much, Narek
From: Chip Eastham on 2 May 2010 15:58
On May 2, 3:05 pm, Narek Saribekyan <narek.saribek...(a)gmail.com> wrote: > Is there any result on Friedrichs' inequality in anisotropic Sobolev > spaces H0^(p, q) (vanishes on boundary with derivatives up to orders > p-1 and q-1 respectively)? > I want to prove that the bilinear form a(u,v) induced from weak > formulation of homogeneous Dirichlet problem (on a rectangle) for Lu = > (-1)^p du^2p/dx^2p + (-1)^q du^2q/dy^2q is coercive (so I can apply > Lax-Milgram theorem). > > a(u,v)=(D^p u, D^p v)+(D^q u, D^q v) > > a(u,u) >= c||u||^2 ? > > Is L an elliptic operator? > > Thanks very much, > Narek I think it suffices to know a density result, that C^inf_0 functions are dense in H^{p,q}_0, because you can explicitly show coercivity by integration by parts with a C^inf_0 function. regards, chip |