Hi Liviu, I kinda suspected that 1 was the reason but wasn't quite sure. Thanks for the quick responce.

I am quite familar with the sub/supersolution approach, and was unable to get it to work. What are your ideas on the second approach? thanks for all your help.

and I guess I was supposed to edit my original question and not answer my own question... still trying to figure out how to use this site.

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.

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$.

I must be missing something but I don't see where the $ u(r) \le 1$ is helping anything. As mentioned below I know the first eigenvalue of $L$ is negative, so i think this will cause a problem. I will dig up reference. Thanks.

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.