Let $s\in (0,1), N\ge 2$ and $U$ be the unique radially decreasing solution of \begin{equation} \ \ \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. \end{equation} for $p\in \left(1, \frac{N+2s}{N2s} \right).$ It is know from this work of Frank, Lenzmann and Silvestre that there are two constants $0<c<C$ 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.
1 Answer
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) $p<\frac{2N}{N2s}1$, since I think it follows from the Pohozaev identity for the fractional Laplacian that no nonzero solution exists for $p\ge\frac{2N}{N2s}1$.
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 $p\in \left(1,\frac{N+2}{N2}\right)$ 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.

$\begingroup$ As I said in my answer, without the restriction to the range $p\in \left( 1, \frac{2N}{N2s} \right)$ 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. $\endgroup$ Oct 4, 2023 at 12:14

$\begingroup$ Actually, Neither I am not interested in the range of p not the existence of solution, I am only interested in the decay property. $\endgroup$– sadiazOct 4, 2023 at 12:25

1$\begingroup$ Well, for sure you cannot prove any decay property of a solution if no solution exists right? $\endgroup$ Oct 4, 2023 at 12:29

$\begingroup$ As commented in my question. "If U is the unique solution", so it becomes clear that I am not looking at the existence question. The decay is not of the exponential type (fractional Laplace) is well known from the work of Lenzmann, Frank and Sylvester. $\endgroup$– sadiazOct 4, 2023 at 16:22

$\begingroup$ You are right, the decay at infinity is not exponential by this work of LenzmannFrankSilvestre. I suggested an edit to the post with this work so that other people can also understand what is the point of the question. You shouldn't assume everyone knows all the papers in your field, if you implicitly use some research work in your question then cite the reference for everyone. $\endgroup$ Oct 4, 2023 at 17:17