Timeline for Integrable solutions to an elliptic PDE on divergence form have a definite sign
Current License: CC BY-SA 2.5
6 events
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| Dec 28, 2016 at 16:30 | answer | added | Michael Renardy | timeline score: 1 | |
| Feb 9, 2010 at 21:15 | comment | added | Harald Hanche-Olsen | For a purely radial vector field, there is a purely radial solution, easily found, and with $g=0$ in your notation. For a purely rotational field, I don't know of any solution. You clearly can't get away with $g=0$ in that case. | |
| Feb 9, 2010 at 3:31 | comment | added | Harald Hanche-Olsen |
Regarding my edit: It may be that it is sufficient to assume $f$ grows at most linearly. Some such restriction is necessary, though: I can make a counterexample for $n=1$ with $f$ odd, $f(x)=ax^{a-1}$ for $x>0$ where $a>1$. The odd solution $u$ with $A=1$ in the notation of the question will belong to $L^1$. We find $\int_0^\infty u=\int_0^\infty \int_0^x e^{y^a-x^a}\,dy\,dx$. The part where $y<1$ is no problem, and the other part is handled by noting $y^a-x^a<ay^{a-1}(y-x)$ when $y<x$ and integrating.
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| Feb 9, 2010 at 2:34 | history | edited | Harald Hanche-Olsen | CC BY-SA 2.5 |
Bounded f, fixed proof for n=1
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| Feb 9, 2010 at 0:52 | comment | added | Harald Hanche-Olsen | I might add a bit of intuition for why this is true: The laplacian makes stuff diffuse, while the divergence term transports stuff along the vector field while preserving the total amount of stuff. If a solution has two signs, the diffusion will mix the positive and negative parts together, making them shrink. This intuition is of course formalized in the parabolic proof. | |
| Feb 9, 2010 at 0:48 | history | asked | Harald Hanche-Olsen | CC BY-SA 2.5 |