1
$\begingroup$

Let and let $p(|z|)$ be the radial solution of the following equation $$ \Delta p + 4q = 0\quad \text{in } \mathbb{R}^n $$ where

  • $n\geq 2$,
  • $0<\alpha<1$,
  • $q \triangleq q(|z|) = \frac 1{1+ |z|^{2+\alpha}}$.

Question. Is it true that $p(|z|)\rightarrow 0$ as $|z|\rightarrow\infty$ and furthermore $$ p(z) \sim \frac 1{1+ |z|^{\alpha } }\quad \text{in }\, \mathbb{R}^n\; ? $$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The solution decays like $r^{2-n}$ if $n \geq 3$. Setting $u=r^{\frac{1-n}{2}} v$, then $$v''-\frac{(n-1)(n-3)}{4r^2} v +g(r)v=0$$ with $g=4q$. This gives the result if $n=3$ since $rg(r) \in L^1$ and then all solutions of the above equation behave like $1,r$ as $r \to \infty$ so that the decaying solution behaves like $r^{-1}$. If $n >3$ one obtains the result by using the Liouville transformation for the equation satified by $v$, as in Chapter 6 of the book by Olver, Asymptotics and special functions (with some care because the functions $\psi_{f,g}$ does not belong to $L^1$). I can write down the details if somebody is interested but I do not see a short way as above.

EDIT The case $n>3$ goes as follows. Let $f(r)=(n-1)(n-3)r^{-2}$ and $\phi(r)=\int_1^r \sqrt{f(s)}ds=\frac{\sqrt{(n-1)(n-3)}}{2}\log r$. The function $w(s)=f^{\frac 14}(\phi^{-1}(s)) v(\phi^{-1}(s))$ satisies the equation $$w''(s)=\left (1+\frac{\psi_{f,g}(\phi^{-1}(s))}{\sqrt{f(\phi^{-1}(s))}}\right )w(s)$$ with $$\psi_{f,g}(r)=f^{-\frac 14}(-\frac {d^2}{dr^2} +g)f^{-\frac 14}$$ (see the book of Olver for this way). Performimg all computations one obtains $$ w''(s)= \left ( \frac{(n-2)^2}{(n-1)(n-3)} +O(e^{-\gamma s}) \right )w(s)$$ for some $\gamma>0$ (since $g$ decays more than quadratically). Then the decaying solution $w$ behaves like $w(s) \approx e^{-\frac {n-2}{\sqrt{(n-1)(n-3}}s}=r^{1-\frac n2}$. Coming back first to $v$ and then to $u$ one gets $u \approx r^{2-n}$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks! I also write something about it.\begin{gather*} p''(r)+\frac{n-1}{r}p'(r)=-4q, \\ (r^{n-1}p')'=r^{n-1}p''+(n-1)r^{n-2}p'=r^{n-1}(-4q),\\\int_{0}^{\rho}(s^{n-1}p(s)')'\,\mathrm{d} s=\int_{0}^{\rho}s^{n-1}(-4q(s))\,\mathrm{d}s=\rho^{n-1}p'(\rho),\\\rho^{1-n}\int_{0}^{\rho}s^{n-1}(-4q(s))\,\mathrm{d}s=p'(\rho)\Rightarrow \int_r^{\infty}\rho^{1-n}\int_{0}^{\rho}s^{n-1}(-4q(s))\,\mathrm{d}s=\int_r^{\infty}p'(\rho)\,\mathrm{d}\rho \end{gather*}but loss the information about $p(z)$ in the infinity. I also quite interesting about your details! Thanks again! $\endgroup$
    – Davidi Cone
    Commented Jun 16, 2023 at 5:44
  • $\begingroup$ I added (shortly) the case $n>3$ but I did not consider $n=2$. $\endgroup$
    – Giorgio Metafune
    Commented Jun 16, 2023 at 7:05
  • $\begingroup$ Thanks! But it seems $\alpha$ has no effect on the infinity property of the solution $p(r)$? $\endgroup$
    – Davidi Cone
    Commented Jun 18, 2023 at 11:14
  • $\begingroup$ Not at the first oder approximation but it should enter the remainder. $\endgroup$
    – Giorgio Metafune
    Commented Jun 18, 2023 at 13:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .