Symmetry results of solutions for elliptic equations

71 Views Asked by At

In the celebrated paper of Gidas, Ni, and Nirenberg, see

http://web.math.unifi.it/users/magnanin/Dott/BibliografiaCorso/GidasNiNirenberg79.pdf

,certain symmetry results of positive solutions of second order elliptic equations are presented. Theorem 1. for example, in their article asserts that the solution must be radial and is decreasing in $r$. Does this result automatically imply that the solution $u=u(r)$ cannot take infinite value at $r=0$? That is, could we conclude that the solution $u=u(r)$ must be bounded for $0\leq r\leq R$?

1

There are 1 best solutions below

0
On

The statement of Theorem 1 says: "let $u>0$ be a positive solution in $C^2 (\overline{\Omega})$". This means: they assume there is such a thing, and prove it's radially symmetric. The assumption $u\in C^2 (\overline{\Omega})$ already implies $u$ is bounded.

So: in order to apply the theorem, you have to already know that $u$ is bounded (among other things).