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Motivation: Consider the equation, $$-\Delta u = u^p$$ in $\mathbb{R}^n$ for $n\geq 3$ and $p=2^*-1.$ Then we know that this equation has unique positive solutions given by functions of the form $U_{a,b}(x) = c_n(b/(1+b^2|x-a|^2))^{(n-2)/2}$ upto scaling $b>0$ and translation $a\in \mathbb{R}^n$. It has also been shown that the operator linearized operator is non-degenerate in the sense that any solution of $$-\Delta u = pU_{a,b}^{p-1}u$$ is a linear combination of the derivatives $\partial_a U$ and $\partial_b U$ of $U$ wrt to the parameters $a$ and $b$. The key to proving this result is using separation of variables and then apply Sturm-Liouville Theory.

Question: Now suppose we have equations of the form $$-\operatorname{div}(a(x)\nabla u) = a(x)u^p$$ where $a$ is some weight function (that is non-radial) and say that it also has solutions like the function $U$ up to scaling and translation factor, then what techniques can be used to show that the linearized operator $$-\operatorname{div}(a(x)\nabla v) = pa(x)U^{p-1} v$$ is non-degenerate in the sense that any solution of the above is a linear combination of the derivatives of $U$ wrt to the scaling and translation parameter. Since the weight $a$ is non-radial I don't think a simple separation of variables will work and so probably this problem cannot be resolved by using ODE-based methods. Are there are other well-known tools that could be used here?

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  • $\begingroup$ Is the positive solution unique for the divergence form operator? $\endgroup$
    – GabS
    Commented Oct 20, 2021 at 10:25
  • $\begingroup$ @GabS Yes, it is unique! $\endgroup$
    – Student
    Commented Oct 20, 2021 at 12:05
  • $\begingroup$ Shouldn't the first linearized equation be $-\Delta u = p U_{a, b}^{p-1}u$? The one you wrote does not seem linear to me. $\endgroup$ Commented Oct 22, 2021 at 21:23
  • $\begingroup$ @GiuseppeNegro yeah you are right, I will fix that! $\endgroup$
    – Student
    Commented Oct 22, 2021 at 22:02
  • $\begingroup$ I did make progress on one problem where the weight was not radially symmetric. Turns out that if we write the linearized operator in polar coordinates, one can then perform separation of variables, which allows us to finish the proof. $\endgroup$
    – Student
    Commented Dec 3, 2021 at 20:28

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