# Exact decay for solutions of fractional Laplacian equation

Let $$s\in (0,1), N\ge 2$$ and $$U$$ be the unique radially decreasing solution of \ \ \left\{\begin{aligned} (-\Delta)^s U+ U &=U^p &&\text{ in } \mathbb{R}^N\\ U&>0 &&\text{ in } \mathbb{R}^N\\ U(|x|)&\to 0 &&\text{ as } |x| \to \infty \end{aligned} \right. for $$p\in \left(1, \frac{N+2s}{N-2s} \right).$$ It is know from this work of Frank, Lenzmann and Silvestre that there are two constants $$0 such that $$\frac{c}{1+|x|^{N+2s}} \le U(x) \le \frac{C}{1+|x|^{N+2s}} .$$ Is it possible to show that the limit $$\lim_{|x|\to \infty} (1+ |x|^{N+2s})U(x)$$ actually exists? Thank you.

I don't think so. First, even if you don't specify a precise range for $$p$$ I assume you are considering ($$p>0$$ and) $$p<\frac{2N}{N-2s}-1$$, since I think it follows from the Pohozaev identity for the fractional Laplacian that no nonzero solution exists for $$p\ge\frac{2N}{N-2s}-1$$.
In this range where a positive, radial solution exists I don't think the asymptotic you want holds. Even in the limiting case $$s=1$$, for $$p\in \left(1,\frac{N+2}{N-2}\right)$$ the unique radial, positive solution $$u\in H^1(\mathbb{R}^N)$$ to $$-\Delta u+u=u^p$$ in $$\mathbb{R}^N$$ has exponential decay at infinity. I would be very surprised to see polynomial decay instead for the fractional Laplacian.
• As I said in my answer, without the restriction to the range $p\in \left( 1, \frac{2N}{N-2s} \right)$ no "unique radially decreasing solution of [your problem]" exists. So, if you want to prove some property of a solution, for sure you are interested in the correct range of $p$ where one solution exists. Oct 4, 2023 at 12:14