Timeline for Stokes problem; naive question on the regularity of pressure term
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2013 at 17:54 | answer | added | Michael Renardy | timeline score: 1 | |
| Nov 1, 2013 at 16:34 | answer | added | username | timeline score: 1 | |
| Mar 5, 2013 at 2:11 | comment | added | Craig | I am still somewhat confused. Evans uses some procedure to show: given $f$ there exists $u$ and $P$ which satisfy the PDE system. Using the method outlined above we see that $ \Delta P = div(f)$ in some weak sense. So there are no issues with existence of $P$ (or possibly I am missing your question). Thanks for replies. | |
| Mar 3, 2013 at 19:56 | comment | added | Daniel Spector | My question seems dumb because it asks a basic question about solvability of the PDE for $P$ you mention. So there is the weak formulation. Are the hypothesis sufficient to apply Lax-Milgram for existence, or is something missing? (i.e. can one really apply the existence theory for Poisson's equation in this setting?) | |
| Mar 1, 2013 at 15:38 | comment | added | Craig | Daniel. Solving which weak PDE? The Poisson equation for $P$ or the the system involving $u,P,f$? In either case the existence is known I am just pointing out that, that it appears one can get some added interior regularity for $P$ for basiscally no extra work and I was suprised (and hence I assume my reasoning must be flawed) that this wasn't pointed out in Evans. Furthermore using this added regularity for $P$ one can get $H^2_{loc} regularity for $u$. | |
| Mar 1, 2013 at 13:02 | comment | added | Daniel Spector | Ok. Now about solving this weak PDE with no boundary condition for $P$? What would you do? | |
| Mar 1, 2013 at 9:03 | comment | added | Craig | Thanks for your responce Delio. If we write $ u=(u^i)$ then we $ - \Delta u^i + P_{x_i}= f^i$ and then take $ \partial_{x_i}$ of the $i^{th}$ equation and sum over $i$ and then reverse the Laplacian and the first order partials. We can make this rigourous by fixing $ \phi $ to be smooth and compactly supported and then we can multiply the equation for $ u^i$ by $ \phi_{x_i}$ and integrate by parts (never using any more than the fact that $ u \in H_0^1$ and $u$ is divergence free) to arrive at $ \int_\Omega \nabla P \cdot \nabla \phi = \int_\Omega f \cdot \nabla \phi$. | |
| Mar 1, 2013 at 7:41 | comment | added | Delio Mugnolo | Craig, can you explain why you expect $\Delta u$ to be divergence free, to? which identity of vector calculus is in action here? | |
| Mar 1, 2013 at 4:36 | comment | added | Craig | I forgot to add that $u$ is divergence free. Sorry. | |
| Mar 1, 2013 at 4:34 | history | asked | Craig | CC BY-SA 3.0 |