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Dec 13, 2011 at 14:16 comment added greg I have looked at ODE methods in the sense that I have played around with radial functions and taken derivatives and....but no real explicit radial methods... What I suspect is that there is no radial solution $v$ of $L(v)=0$. The reason I think this is that in this case a change of variables can remove the term $ r^\alpha$ from all the equations, and then there are known results. Let me explain where this is coming from, which i do below, since maybe I don't even need what I am asking for.
Dec 13, 2011 at 13:57 comment added timur @greg, have you looked at ODE methods?
Dec 13, 2011 at 13:42 history edited timur CC BY-SA 3.0
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Dec 13, 2011 at 3:40 comment added greg I meant in the particular case where $L = -\Delta - r^\alpha u(r)^{p-1}$ and where $u$ is a positive solution of the given PDE. If the first eigenvalue was non-negative then one would obtain $ \int | \nabla \psi|^2 \ge \int p r^\alpha u^{p-1} \psi^2$ for all $ \psi $ which are zero on the boundary. Putting $ \psi=u$ into this inequality would show that $ u=0$.
Dec 13, 2011 at 3:18 comment added timur The first eigenvalue of $L=-\Delta-V$ is not negative, if $V$ is small enough. There is a smallness condition involving the width of the domain etc. Intuitively, the first eigenvalue of $-\Delta$ is strictly positive, so one still has a room for subtracting something small.
Dec 13, 2011 at 1:46 comment added greg Thanks for response. I believe here you are refering to the result that says if the $L^\frac{N}{2}$ norm of the quantity in question is small with regards to Sobolev imbedding constant, then $L$ has the maximum principle? In any case I don't think this will work since the first eigenvalue of $L$ is negative. I will look into your other commments. Thanks again.
Dec 12, 2011 at 19:50 history edited timur CC BY-SA 3.0
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Dec 12, 2011 at 19:44 history edited timur CC BY-SA 3.0
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Dec 12, 2011 at 19:36 history edited timur CC BY-SA 3.0
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Dec 12, 2011 at 19:30 history answered timur CC BY-SA 3.0