Asymptotic classification conjecture for positive singular solutions of poly-harmonic equations

Let N=2mN=2m, let n2mn\geqslant 2m, and let BR=BR0RnB_R^*=B_R\setminus\\{0\\}\subset\mathbb{R}^n. Consider a positive singular solution uC2m(Rn0)u\in C^{2m}(\mathbb{R}^n\setminus\\{0\\}) of

(Δ)mu=up1uin BR,(-\Delta)^m u=|u|^{p-1}u\quad\text{in }B_R^*,

where p[2m,,2m)p\in[2_{m,*},2_m^*), 2m,=nn2m2_{m,*}=\frac{n}{n-2m}, and 2m=n+2mn2m2_m^*=\frac{n+2m}{n-2m}. Write u\overline{u} for the radial asymptotic profile and let K0(n,p)K_0(n,p) and K^N,0(n)\widehat{K}_{N,0}(n) be the constants appearing in the asserted asymptotics. Assume that (Δu)j0(-\Delta u)^j\geqslant0 for every j=1,,m1j=1,\dots,m-1.

Asymptotic classification conjecture. If p(1,2m1)p\in(1,2_m^*-1), then

u(x)=(1+O(x))u(x)as x0.u(x)=(1+\mathcal{O}(|x|))\overline{u}(x)\quad\text{as }x\to0.

Moreover, (a) if p(1,2m)p\in(1,2_m^*), then u(x)xNnu(x)\simeq |x|^{N-n} as x0x\to0; (b) if p=2m,p=2_{m,*}, then

u(x)=(1+o(1))K^0(n)nNNxNn(lnx)NNNas x0,u(x)=(1+\mathrm{o}(1))\widehat{K}_{0}(n)^{\frac{n-N}{N}}|x|^{N-n}(\ln|x|)^{\frac{N-N}{N}}\quad\text{as }x\to0,

where

K^N,0(n)=2m2(m1)!mj=0m1(n2j)(nN)2;\widehat{K}_{N,0}(n)=\frac{2^{m-2}(m-1)!}{m}\prod_{j=0}^{m-1}(n-2j)(n-N)^2;

and (c) if p(2m,,2m1)p\in(2_{m,*},2_m^*-1), then

u(x)=(1+o(1))K0(n,p)1p1xNp1as x0.u(x)=(1+\mathrm{o}(1))K_0(n,p)^{\frac{1}{p-1}}|x|^{-\frac{N}{p-1}}\quad\text{as }x\to0.

This conjecture seeks a complete local asymptotic classification near an isolated singularity for positive singular solutions of subcritical even-order poly-harmonic equations. The parser reports that the super poly-harmonic condition can be removed by an independent proof of X. Huang, Y. Li, and H. Yang, so the stated result is solved.

Sources & referencesView supporting material

Primary source

João Henrique Andrade and Juncheng Wei, “Asymptotics for positive singular solutions to subcritical sixth order equations”, arXiv:2210.15102 (2022).

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