Timeline for Reference request : Global boundedness of weak solution for Neumann problem
Current License: CC BY-SA 4.0
8 events
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| Jul 15, 2020 at 11:12 | comment | added | Thomas | I think you have to assume something on $\|u\|_1$ or on $\int u$ otherwise, you have a solution up to a cosntant. This should fix the problem of lower order terms by bootstrap arguments. | |
| Jul 12, 2020 at 7:59 | comment | added | Will Kwon | @JochenGlueck Of course, we need to assume the regularity of $\Omega$, at least Lipschitz domain to guarantee the Sobolev embedding theorem. | |
| Jul 11, 2020 at 23:01 | comment | added | Giorgio Metafune | Take $\|F\|_p=1$ and iterate your inequalities until $\beta>p-1$. Then you can absorbe the first term on the RHS into the second and go on with the usual iteration. I did not check the details but it should work. | |
| Jul 11, 2020 at 22:48 | comment | added | Jochen Glueck | What kind of regularity do you assume for the boundary of $\Omega$? (I have serious doubts whether this is true for arbitary bounded domains.) | |
| Jul 11, 2020 at 22:08 | comment | added | Will Kwon | @GiorgioMetafune Thanks for your answer. I edit my question. In the case of Dirichlet boundary problem, our function is lie on $W^{1,2}_0$, but in the case of Neumann problem, the original function does not lie on that space. This makes me hard to use Sobolev embedding. In the case of De Giorgi approach, it seems okay to obtain the desired result, but I'm not sure whether the desired assertion can be deduced via Moser's argument. | |
| Jul 11, 2020 at 22:05 | history | edited | Will Kwon | CC BY-SA 4.0 |
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| Jul 11, 2020 at 15:08 | comment | added | Giorgio Metafune | You are right, this is usually stated in the case of Dirichlet boundary condition. But what is the problem trying to repeat the usual proof? In the case where $\Omega$ is the whole space $R^n$ both De Giorgi's approach or Moser's work. | |
| Jul 11, 2020 at 13:29 | history | asked | Will Kwon | CC BY-SA 4.0 |