﻿ 含Hardy-Sobolev临界指数的分数阶Kirchhoff型方程多重解的存在性

# 含Hardy-Sobolev临界指数的分数阶Kirchhoff型方程多重解的存在性Existence of Multiple Solutions for Fractional Kirchhoff Equations with Hardy-Sobolev Critical Exponents

Abstract:

Abstract: In this paper, we study a class of critical fractional elliptic problems of Kirchhoff type：

Abstract:
where and q∈(1,2) are constants, and is the Hardy-Sobolev exponent in 3. For a suitable function f(x), we use Nehari manifold and fibering maps to prove the existence of multiple solutions.

1. 引言

${\left(-\Delta \right)}^{s}u-\mu \frac{u}{{|x|}^{2s}}={u}^{{2}_{s}^{*}-1},\text{}u\in {\stackrel{˙}{H}}^{s}\left({ℝ}^{N}\right),$

$\left[a+b{\left({\int }_{{ℝ}^{3}}{|\nabla u|}^{2}-\mu \frac{{u}^{2}}{{|x|}^{2}}\text{d}x\right)}^{\frac{2-\alpha }{2}}\right]\left(-\Delta u-\mu \frac{u}{{|x|}^{2}}\right)=\frac{{|u|}^{{2}^{*}\left(\alpha \right)-2}u}{{|x|}^{\alpha }}+\lambda \frac{f\left(x\right){|u|}^{q-2}u}{{|x|}^{\beta }},$

$\begin{array}{l}\left[a+b{\left({\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(u\left(x\right)-u\left(y\right)\right)}^{2}}{{|x-y|}^{3+2s}}\text{d}x\text{d}y-{\int }_{{ℝ}^{3}}\mu \frac{{u}^{2}}{{|x|}^{2s}}\text{d}x\right)}^{\frac{s}{3-2s}}\right]\left[{\left(-\Delta \right)}^{s}u-\mu \frac{u}{{|x|}^{2s}}\right]\\ ={|u|}^{{2}_{s}^{*}-2}u+\lambda f\left(x\right){|u|}^{q-2}u,\end{array}$ (1.1)

(F) $0\lneqq f\left(x\right)\in {L}^{\infty }\left({ℝ}^{3}\right)$ 且存在 ${R}_{0}>0$ 使得 $suppf\in {B}_{{R}_{0}}\left(0\right)$

1) 当 $\lambda \in \left(0,{\Lambda }_{*}\right)$ 时，问题(1.1)存在至少一个正解；

2) 当 $\lambda \in \left(0,{\Lambda }_{**}\right)$ 时，问题(1.1)存在至少两个正解；

2. Nehari流形和纤维映射

2.1. 预备知识和一些记号

$\begin{array}{l}{‖u‖}_{{\stackrel{˙}{H}}^{s}\left({ℝ}^{3}\right)}={\left({\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(u\left(x\right)-u\left(y\right)\right)}^{2}}{{|x-y|}^{3+2s}}\text{d}x\text{d}y\right)}^{\frac{1}{2}},\\ {\left(u,v\right)}_{{\stackrel{˙}{H}}^{s}\left({ℝ}^{3}\right)}={\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{\left(u\left(x\right)-u\left(y\right)\right)\left(v\left(x\right)-v\left(y\right)\right)}{{|x-y|}^{3+2s}}\text{d}x\text{d}y.\end{array}$

${\Lambda }_{N,S}{\int }_{{ℝ}^{3}}\frac{{u}^{2}\left(x\right)}{{|x|}^{2s}}\text{d}x\le \frac{{C}_{N,s}}{2}{\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(u\left(x\right)-u\left(y\right)\right)}^{2}}{{|x-y|}^{N+2s}}\text{d}x\text{d}y$，其中 ${\Lambda }_{N,S}={2}^{2s}{\Gamma }^{2}\left(\frac{N+2s}{4}\right)/{\Gamma }^{2}\left(\frac{N-2s}{4}\right)$,

$\left(u,v\right)={\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{\left(u\left(x\right)-u\left(y\right)\right)\left(v\left(x\right)-v\left(y\right)\right)}{{|x-y|}^{3+2s}}\text{d}x\text{d}y-{\int }_{{ℝ}^{3}}\mu \frac{uv}{{|x|}^{2s}}\text{d}x,$

$‖u‖={\left({\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(u\left(x\right)-u\left(y\right)\right)}^{2}}{{|x-y|}^{3+2s}}\text{d}x\text{d}y-{\int }_{{ℝ}^{3}}\mu \frac{{u}^{2}}{{|x|}^{2s}}\text{d}x\right)}^{\frac{1}{2}}.$

${S}_{\mu ,s}=\mathrm{inf}\left\{{\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(u\left(x\right)-u\left(y\right)\right)}^{2}}{{|x-y|}^{3+2s}}\text{d}x\text{d}y-{\int }_{{ℝ}^{3}}\mu \frac{{u}^{2}}{{|x|}^{2s}}\text{d}x:u\in {\stackrel{˙}{H}}^{s}\left({ℝ}^{3}\right)和{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x=1\right\},$ (2.1)

${\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{\left(U\left(x\right)-U\left(y\right)\right)}^{2}}{{|x-y|}^{3+2s}}\text{d}x\text{d}y-{\int }_{{ℝ}^{3}}\mu \frac{{U}^{2}}{{|x|}^{2s}}\text{d}x={\int }_{{ℝ}^{3}}{U}^{{2}_{S}^{*}}\text{d}x={S}_{\mu ,S}^{\frac{3}{2s}}.$ (2.2)

$J\left(u\right)=\frac{a}{2}{‖u‖}^{2}+\frac{3-2s}{6-2s}b{‖u‖}^{\frac{6-2s}{3-2s}}-\frac{1}{{2}_{s}^{*}}{\int }_{{ℝ}^{3}}{u}^{{2}_{s}^{*}}\text{d}x-\frac{\lambda }{q}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x.$

$\begin{array}{c}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x\le {|f|}_{\infty }{\left({\int }_{{B}_{{R}_{0}}\left(0\right)}1\text{d}x\right)}^{\frac{{2}_{s}^{*}-q}{{2}_{s}^{*}}}{\left({\int }_{{B}_{{R}_{0}}\left(0\right)}{|u|}^{{2}_{s}^{*}}\text{d}x\right)}^{\frac{q}{{2}_{s}^{*}}}\\ \triangleq {|f|}_{\infty }{|{B}_{{R}_{0}}\left(0\right)|}^{\frac{{2}_{s}^{*}-q}{{2}_{s}^{*}}}{\left({\int }_{{B}_{{R}_{0}}\left(0\right)}{|u|}^{{2}_{s}^{*}}\text{d}x\right)}^{\frac{q}{{2}_{s}^{*}}}\\ \triangleq {|f|}_{\infty }{C}_{{R}_{0},s,q}{\left({\int }_{{B}_{{R}_{0}}\left(0\right)}{|u|}^{{2}_{s}^{*}}\text{d}x\right)}^{\frac{q}{{2}_{s}^{*}}}\triangleq {|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖u‖}^{q}.\end{array}$ (2.3)

${\lambda }_{1}\triangleq \frac{4s\sqrt{2ab}{S}_{\mu ,s}^{\frac{q}{2}}}{\left(3-2s\right)\left({2}_{s}^{*}-q\right){|{f}^{+}|}_{\infty }{C}_{{R}_{0},s,q}}{\left[\frac{2\sqrt{ab\left(2-q\right)\left(\frac{6-2s}{3-2s}-q\right)}{S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3s}{3-2s}\left(\frac{6-3s}{3-2s}-q\right)}>0,$

${\lambda }_{2}\triangleq \frac{a\left({2}_{s}^{*}-2\right){S}_{\mu ,s}^{\frac{q}{2}}}{\left({2}_{s}^{*}-q\right){|{f}^{+}|}_{\infty }{C}_{{R}_{0},s,q}}{\left[\frac{a\left(2-q\right){S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3-2s}{4s}\left(2-q\right)}>0,$

${\lambda }_{3}\triangleq \frac{1}{2}{\left(a{S}_{\mu ,s}\right)}^{\frac{3\left(2-q\right)}{4s}}{\left(\frac{s}{3{C}_{0}}\right)}^{\frac{2-q}{2}}>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\lambda }_{4}\triangleq {\left(\frac{{t}_{1}{\int }_{{ℝ}_{3}}f\left(x\right){|U|}^{q}\text{d}x}{{C}_{0}q}\right)}^{\frac{2-q}{q}}>0,$

${\Lambda }_{1}\triangleq \mathrm{max}\left\{{\lambda }_{1},{\lambda }_{2}\right\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Lambda }_{2}\triangleq \mathrm{max}\left\{q{\lambda }_{1}/\sqrt{2\frac{6-2s}{3-2s}},\frac{q{\lambda }_{2}}{2}\right\},$

${\Lambda }_{*}\triangleq \mathrm{max}\left\{{\Lambda }_{1},{\lambda }_{3},{\lambda }_{4}\right\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Lambda }_{**}\triangleq \mathrm{max}\left\{{\Lambda }_{2},{\lambda }_{3},{\lambda }_{4}\right\}.$

2.2. 预备引理

$\mathcal{N}=\left\{u\in {\stackrel{˙}{H}}^{s}\left({ℝ}^{3}\right)\\left\{0\right\}:〈{J}^{\prime }\left(u\right),u〉=0\right\},$

$a{‖u‖}^{2}+b{‖u‖}^{\frac{6-2s}{3-2s}}-{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x=0\text{ }\text{ }且\text{ }\text{ }u\ne 0.$

$J\left(u\right)=J\left(u\right)-\frac{1}{{2}_{s}^{*}}〈{J}^{\prime }\left(u\right),u〉\ge \frac{as}{3}{‖u‖}^{2}-\left(\frac{1}{q}-\frac{1}{{2}_{s}^{*}}\right){|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖u‖}^{q},$

${\phi }_{u}\left(t\right)=\frac{a}{2}{t}^{2}{‖u‖}^{2}+\frac{3-2s}{6-2s}b{t}^{\frac{6-2s}{3-2s}}{‖u‖}^{\frac{6-2s}{3-2s}}-\frac{{t}^{{2}_{s}^{*}}}{{2}_{s}^{*}}{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\frac{\lambda }{q}{t}^{q}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x,$

${{\phi }^{\prime }}_{u}\left(t\right)=at{‖u‖}^{2}+b{t}^{\frac{3}{3-2s}}{‖u‖}^{\frac{6-2s}{3-2s}}-{t}^{{2}_{s}^{*}-1}{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\lambda {t}^{q-1}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x,$

${{\phi }^{″}}_{u}\left(t\right)=a{‖u‖}^{2}+\frac{3b}{3-2s}{t}^{\frac{2s}{3-2s}}{‖u‖}^{\frac{6-2s}{3-2s}}-\left({2}_{s}^{*}-1\right){t}^{{2}_{s}^{*}-2}{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\lambda \left(q-1\right){t}^{q-2}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x.$

$t{{\phi }^{\prime }}_{u}\left(t\right)=a{t}^{2}{‖u‖}^{2}+b{t}^{\frac{6-2s}{3-2s}}{‖u‖}^{\frac{6-2s}{3-2s}}-{t}^{{2}_{s}^{*}}{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\lambda {t}^{q}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x,$

$\begin{array}{l}{\mathcal{N}}^{+}=\left\{u\in \mathcal{N}:{{\phi }^{″}}_{u}\left(t\right)>0\right\},\\ {\mathcal{N}}^{0}=\left\{u\in \mathcal{N}:{{\phi }^{″}}_{u}\left(t\right)=0\right\},\\ {\mathcal{N}}^{-}=\left\{u\in \mathcal{N}:{{\phi }^{″}}_{u}\left(t\right)<0\right\}.\end{array}$

${{\phi }^{″}}_{u}\left(1\right)=a{‖u‖}^{2}+\frac{3b}{3-2s}{‖u‖}^{\frac{6-2s}{3-2s}}-\left({2}_{s}^{*}-1\right){\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x-\lambda \left(q-1\right){\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x$

$=a\left(2-q\right){‖u‖}^{2}+b\left(\frac{6-2s}{3-2s}-q\right){‖u‖}^{\frac{6-2s}{3-2s}}-\left({2}_{s}^{*}-q\right){\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x$ (2.4)

$=\frac{4as}{2s-3}{‖u‖}^{2}+\frac{2bs}{2s-3}{‖u‖}^{\frac{6-2s}{3-2s}}+\left({2}_{s}^{*}-q\right)\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x.$ (2.5)

$\begin{array}{c}2\sqrt{ab\left(2-q\right)\left(\frac{6-2s}{3-2s}-q\right)}{‖u‖}^{\frac{6-3s}{3-2s}}\le a\left(2-q\right){‖u‖}^{2}+b\left(\frac{6-2s}{3-2s}-q\right){‖u‖}^{\frac{6-2s}{3-2s}}\\ =\left({2}_{s}^{*}-q\right){\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x\\ \le \left({2}_{s}^{*}-q\right){S}_{\mu ,s}^{-\frac{{2}_{s}^{*}}{2}}{‖u‖}^{{2}_{s}^{*}},\end{array}$

$\begin{array}{c}\frac{4s}{3-2s}\sqrt{2ab}{‖u‖}^{\frac{6-3s}{3-2s}}\le \frac{4as}{3-2s}{‖u‖}^{2}+b\frac{2bs}{3-2s}{‖u‖}^{\frac{6-2s}{3-2s}}\\ =\left({2}_{s}^{*}-q\right)\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x\\ \le \left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖u‖}^{q},\end{array}$

${\left[\frac{2\sqrt{ab\left(2-q\right)\left(\frac{6-2s}{3-2s}-q\right)}{S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3s}{3-2s}}\le ‖u‖\le {\left[\frac{\left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}}{\frac{4s}{3-2s}\sqrt{2ab}{S}_{\mu ,s}^{\frac{q}{2}}}\right]}^{\frac{3-2s}{6-3s-q\left(3-2s\right)}}.$ (2.6)

$\begin{array}{c}a\left(2-q\right){‖u‖}^{2}\le a\left(2-q\right){‖u‖}^{2}+b\left(\frac{6-2s}{3-2s}-q\right){‖u‖}^{\frac{6-2s}{3-2s}}\\ =\left({2}_{s}^{*}-q\right){\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x\\ \le \left({2}_{s}^{*}-q\right){S}_{\mu ,s}^{-\frac{{2}_{s}^{*}}{2}}{‖u‖}^{{2}_{s}^{*}},\end{array}$

$\begin{array}{c}\frac{4as}{3-2s}{‖u‖}^{2}\le \frac{4as}{3-2s}{‖u‖}^{2}+\frac{2bs}{3-2s}{‖u‖}^{\frac{6-2s}{3-2s}}\\ =\left({2}_{s}^{*}-q\right)\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x\\ \le \left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖u‖}^{q},\end{array}$

${\left[\frac{a\left(2-q\right){S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3-2s}{4s}}\le ‖u‖\le {\left[\frac{\left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}}{a\frac{4s}{3-2s}{S}_{\mu ,s}^{\frac{q}{2}}}\right]}^{\frac{1}{2-q}}.$ (2.7)

1) 若 $0<\lambda <{\Lambda }_{1}=\mathrm{max}\left\{{\lambda }_{1},{\lambda }_{2}\right\}$，则 ${m}^{+}<0$

2) 若 $0<\lambda <{\Lambda }_{2}\triangleq \mathrm{max}\left\{q{\lambda }_{1}/\sqrt{2\frac{6-2s}{3-2s}},\frac{q{\lambda }_{2}}{2}\right\}$，则存在与b无关的 ${d}_{0}$ 使得 ${m}^{-}>{d}_{0}$。特别地， ${m}^{+}=m<0<{m}^{-}$

$a\left(2-q\right){‖u‖}^{2}+b\left(\frac{6-2s}{3-2s}-q\right){‖u‖}^{\frac{6-2s}{3-2s}}>\left({2}_{s}^{*}-q\right){\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x$,

$\begin{array}{c}J\left(u\right)=-\frac{a\left(2-q\right)}{2q}{‖u‖}^{2}-\frac{b\left[6-2s-q\left(3-2s\right)\right]}{q\left(6-2s\right)}{‖u‖}^{\frac{6-2s}{3-2s}}+\frac{{2}_{s}^{*}-q}{{2}_{s}^{*}q}{\int }_{{ℝ}^{3}}{|u|}^{{2}_{s}^{*}}\text{d}x\\ <\frac{a\left(2-q\right)\left(2-{2}_{s}^{*}\right)}{2q{2}_{s}^{*}}{‖u‖}^{2}-\frac{2bs\left[6-2s-q\left(3-2s\right)\right]}{q{2}_{s}^{*}\left(6-2s\right)\left(3-2s\right)}{‖u‖}^{\frac{6-2s}{3-2s}}<0,\end{array}$

2) 我们分两种情形进行讨论。

$‖u‖>{\left[\frac{2\sqrt{ab\left(2-q\right)\left(\frac{6-2s}{3-2s}-q\right)}{S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3s}{3-2s}}.$ (2.8)

$\begin{array}{c}J\left(u\right)=\frac{a\left({2}_{s}^{*}-2\right)}{2\cdot {2}_{s}^{*}}{‖u‖}^{2}+\frac{s}{\left(3-s\right){2}_{s}^{*}}b{‖u‖}^{\frac{6-2s}{3-2s}}-\frac{\lambda \left({2}_{s}^{*}-q\right)}{q{2}_{s}^{*}}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x\\ \ge {‖u‖}^{q}\left[\frac{2\sqrt{ab\left({2}_{s}^{*}-2\right)s}}{{2}_{s}^{*}\sqrt{6-2s}}{‖u‖}^{\frac{6-3s}{3-2s}-q}-\frac{\lambda \left({2}_{s}^{*}-q\right)}{q{2}_{s}^{*}}{|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}\right].\end{array}$ (2.9)

$‖u‖>{\left[\frac{a\left(2-q\right){S}_{\mu ,s}^{\frac{{2}_{s}^{*}}{2}}}{{2}_{s}^{*}-q}\right]}^{\frac{3-2s}{4s}}.$ (2.10)

$\begin{array}{c}J\left(u\right)=\frac{a\left({2}_{s}^{*}-2\right)}{2\cdot {2}_{s}^{*}}{‖u‖}^{2}+\frac{s}{\left(3-s\right){2}_{s}^{*}}b{‖u‖}^{\frac{6-2s}{3-2s}}-\frac{\lambda \left({2}_{s}^{*}-q\right)}{q{2}_{s}^{*}}{\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x\\ \ge {‖u‖}^{q}\left[\frac{a\left({2}_{s}^{*}-2\right)}{2\cdot {2}_{s}^{*}}{‖u‖}^{2-q}-\frac{\lambda \left({2}_{s}^{*}-q\right)}{q{2}_{s}^{*}}{|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}\right].\end{array}$ (2.11)

3. 定理1.1的证明

1) 若 $0<\lambda <{\Lambda }_{1}$，则 $J\left(u\right)$${\left(PS\right)}_{m}$ 序列 $\left\{{u}_{n}\right\}\subset \mathcal{N}$

2) 若 $0<\lambda <{\Lambda }_{2}$，则 $J\left(u\right)$${\left(PS\right)}_{{m}^{-}}$ 序列 $\left\{{u}_{n}\right\}\subset {\mathcal{N}}^{-}$

$\begin{array}{l}{c}_{\mu ,s}^{*}=\frac{as}{3}{S}_{\mu ,s}{\left(\frac{b{S}_{\mu ,s}^{\frac{3-s}{3-2s}}+\sqrt{{b}^{2}{S}_{\mu ,s}^{\frac{6-2s}{3-2s}}+4a{S}_{\mu ,s}}}{2}\right)}^{\frac{3-2s}{s}}\\ \text{}+\frac{bs\left(3-2s\right)}{3\left(6-2s\right)}{S}_{\mu ,s}^{\frac{3-s}{3-2s}}{\left(\frac{b{S}_{\mu ,s}^{\frac{3-s}{3-2s}}+\sqrt{{b}^{2}{S}_{\mu ,s}^{\frac{6-2s}{3-2s}}+4a{S}_{\mu ,s}}}{2}\right)}^{\frac{3-s}{s}},\end{array}$ (3.1)

${C}_{0}$ 由引理3.3给出的正常数。

$\begin{array}{c}c+1+o\left(1\right)‖{u}_{n}‖\ge J\left({u}_{n}\right)-\frac{3-2s}{6-2s}〈{J}^{\prime }\left({u}_{n}\right),{u}_{n}〉\\ \ge \frac{as}{6-2s}{‖{u}_{n}‖}^{2}-\frac{6-2s-q\left(3-2s\right)}{\left(6-2s\right)q}\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖{u}_{n}‖}^{q},\end{array}$

${|{D}_{s}{u}_{n}\left(x\right)|}^{2}={\int }_{{ℝ}^{3}}\frac{{|{u}_{n}\left(x\right)-{u}_{n}\left(y\right)|}^{2}}{{|x-y|}^{3+2s}}\text{d}y.$

$\begin{array}{l}{|{D}_{s}{u}_{n}|}^{2}⇀\text{d}\stackrel{˜}{\mu }\ge {|{D}_{s}u|}^{2}+\underset{j\in \Gamma }{\sum }{\mu }_{{x}_{j}}{\delta }_{{x}_{j}}+{\mu }_{0}{\delta }_{0},\\ {u}_{n}^{2}{|x|}^{-2}⇀\text{d}\gamma ={u}^{2}{|x|}^{-2}+{\gamma }_{0}{\delta }_{0},\\ {|{u}_{n}|}^{{2}_{s}^{*}}⇀\text{d}\nu ={|u|}^{{2}_{s}^{*}}+\underset{j\in \Gamma }{\sum }{\nu }_{{x}_{j}}{\delta }_{{x}_{j}}+{\nu }_{0}{\delta }_{0},\end{array}$

$\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}{\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{|{u}_{n}\left(x\right)-{u}_{n}\left(y\right)|}^{2}}{{|x-y|}^{3+2s}}{\phi }^{\epsilon }\text{d}y\text{d}x=\underset{\epsilon \to 0}{\mathrm{lim}}{\int }_{{ℝ}^{3}}{\phi }^{\epsilon }\text{d}\stackrel{˜}{\mu }\ge {\mu }_{0},$

$\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}{\int }_{{ℝ}^{3}}{u}_{n}^{2}{|x|}^{-2}{\phi }^{\epsilon }\text{d}x=\underset{\epsilon \to 0}{\mathrm{lim}}{\int }_{{ℝ}^{3}}{\phi }^{\epsilon }\text{d}\gamma ={\gamma }_{0},$

$\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}{\int }_{{ℝ}^{3}}{u}_{n}^{{2}_{s}^{*}}{\phi }^{\epsilon }\text{d}x=\underset{\epsilon \to 0}{\mathrm{lim}}{\int }_{{ℝ}^{3}}{\phi }^{\epsilon }\text{d}\nu ={\nu }_{0},$ (3.2)

$\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}{\int }_{{ℝ}^{3}}f\left(x\right){|{u}_{n}|}^{q}{\phi }^{\epsilon }\text{d}x=0,$

$\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}{\int }_{{ℝ}^{3}}{u}_{n}\left(x\right){\int }_{{ℝ}^{3}}\frac{\left({u}_{n}\left(x\right)-{u}_{n}\left(y\right)\right)\left({\phi }^{\epsilon }\left(x\right)-{\phi }^{\epsilon }\left(y\right)\right)}{{|x-y|}^{3+2s}}\text{d}y\text{d}x=0.$

$\begin{array}{l}0=\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}〈{J}^{\prime }\left({u}_{n}\right),{u}_{n}{\phi }^{\epsilon }〉\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{\epsilon \to 0}{\mathrm{lim}}\underset{n\to \infty }{\mathrm{lim}}\left\{\left(a+b{‖{u}_{n}‖}^{\frac{2s}{3-2s}}\right)\left({\int }_{{ℝ}^{3}}{\int }_{{ℝ}^{3}}\frac{{|{u}_{n}\left(x\right)-{u}_{n}\left(y\right)|}^{2}}{{|x-y|}^{3+2s}}{\phi }^{\epsilon }\text{d}y\text{d}x+{\int }_{{ℝ}^{3}}{u}_{n}\left(x\right){\int }_{{ℝ}^{3}}\frac{\left({u}_{n}\left(x\right)-{u}_{n}\left(y\right)\right)\left({\phi }^{\epsilon }\left(x\right)-{\phi }^{\epsilon }\left(y\right)\right)}{{|x-y|}^{3+2s}}\text{d}y\text{d}x\\ \text{ }\text{ }\begin{array}{c}\text{ }\\ \stackrel{}{\text{ }}\end{array}-{\int }_{{ℝ}^{3}}{u}_{n}^{2}{|x|}^{-2}{\phi }^{\epsilon }\text{d}x\right)-{\int }_{{ℝ}^{3}}{|{u}_{n}|}^{{2}_{s}^{*}}{\phi }^{\epsilon }\text{d}x-\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|{u}_{n}|}^{q}{\phi }^{\epsilon }\text{d}x\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge a\left({\mu }_{0}-\mu {\gamma }_{0}\right)+b{\left({\mu }_{0}-\mu {\gamma }_{0}\right)}^{\frac{3-s}{3-2s}}-{\nu }_{0}.\end{array}$

${S}_{\mu ,s}^{-3/\left(3-2s\right)}{\left({\mu }_{0}-\mu {\gamma }_{0}\right)}^{2s/\left(3-2s\right)}-b{\left({\mu }_{0}-\mu {\gamma }_{0}\right)}^{s/\left(3-2s\right)}-a\ge 0,$

$\left({\mu }_{0}-\mu {\gamma }_{0}\right)\ge {S}_{\mu ,s}{\left(\frac{b{S}_{\mu ,s}^{\frac{3-s}{3-2s}}+\sqrt{{b}^{2}{S}_{\mu ,s}^{\frac{6-2s}{3-2s}}+4a{S}_{\mu ,s}}}{2}\right)}^{\frac{3-2s}{s}},$

$\begin{array}{c}c+{o}_{n}\left(1\right)=J\left({u}_{n}\right)-\frac{3-2s}{6-2s}〈{J}^{\prime }\left({u}_{n}\right),{u}_{n}〉\\ =\frac{as}{6-2s}{‖{u}_{n}‖}^{2}+\frac{s\left(3-2s\right)}{3\left(6-2s\right)}{\int }_{{ℝ}^{3}}{u}_{n}^{{2}_{s}^{*}}\text{d}x-\lambda \frac{6-2s-q\left(3-2s\right)}{q\left(6-2s\right)}{\int }_{{ℝ}^{3}}f\left(x\right){|{u}_{n}|}^{q}\text{d}x\\ \ge \frac{as}{6-2s}\left({\mu }_{0}-\mu {\gamma }_{0}+{‖u‖}^{2}\right)+\frac{s\left(3-2s\right)}{3\left(6-2s\right)}{\nu }_{0}-\lambda \frac{6-2s-q\left(3-2s\right)}{q\left(6-2s\right)}{|f|}_{\infty }{C}_{{R}_{0},s,q}{S}_{\mu ,s}^{-\frac{q}{2}}{‖u‖}^{q}\\ \ge \frac{as}{6-2s}\left({\mu }_{0}-\mu {\gamma }_{0}\right)+\frac{s\left(3-2s\right)}{3\left(6-2s\right)}\left[a\left({\mu }_{0}-\mu {\gamma }_{0}\right)+b{\left({\mu }_{0}-\mu {\gamma }_{0}\right)}^{\frac{3-s}{3-2s}}\right]-{C}_{0}{\lambda }^{\frac{2}{2-q}}\\ =\frac{as}{3}\left({\mu }_{0}-\mu {\gamma }_{0}\right)+\frac{bs\left(3-2s\right)}{3\left(6-2s\right)}{\left({\mu }_{0}-\mu {\gamma }_{0}\right)}^{\frac{3-s}{3-2s}}-{C}_{0}{\lambda }^{\frac{2}{2-q}}\ge {C}_{\mu ,s}^{*}-{C}_{0}{\lambda }^{\frac{2}{2-q}},\end{array}$

${\int }_{{ℝ}^{3}}f\left(x\right){|{u}_{n}|}^{q}\text{d}x\to {\int }_{{ℝ}^{3}}f\left(x\right){|u|}^{q}\text{d}x.$

$\begin{array}{c}o\left(1\right)=〈{J}^{\prime }\left({u}_{n}\right)-{J}^{\prime }\left(u\right),{u}_{n}-u〉\\ =\left(a+b{‖{u}_{n}‖}^{\frac{2s}{3-2s}}\right)\left({u}_{n},{u}_{n}-u\right)-\left(a+b{‖u‖}^{\frac{2s}{3-2s}}\right)\left(u,{u}_{n}-u\right)+o\left(1\right)\\ =\left(a+b{‖{u}_{n}‖}^{\frac{2s}{3-2s}}\right)\left({u}_{n}-u,{u}_{n}-u\right)+b\left({‖{u}_{n}‖}^{\frac{2s}{3-2s}}-{‖u‖}^{\frac{2s}{3-2s}}\right)\left(u,{u}_{n}-u\right)+o\left(1\right)\\ \ge a{‖{u}_{n}-u‖}^{2}+o\left(1\right),\end{array}$

$\underset{t\ge 0}{\mathrm{sup}}J\left(tu\right)<{c}_{\mu ,s}^{*}-{C}_{0}{\lambda }^{\frac{2}{2-q}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{0}=\frac{as\left(2-q\right)}{q\left(3-s\right)}{\left[\frac{\left[6-2s-q\left(3-2s\right)\right]{|f|}_{\infty }{C}_{{R}_{0},s,q}}{2as{S}_{\mu ,s}^{\frac{q}{2}}}\right]}^{\frac{2}{2-q}}.$

$\begin{array}{c}g\left(t\right)=\frac{a}{2}{t}^{2}{‖U‖}^{2}+\frac{3-2s}{6-2s}b{t}^{\frac{6-2s}{3-2s}}{‖U‖}^{\frac{6-2s}{3-2s}}-\frac{{t}^{{2}_{s}^{*}}}{{2}_{s}^{*}}{\int }_{{ℝ}^{3}}{|U|}^{{2}_{s}^{*}}\text{d}x\\ \triangleq {C}_{1}{t}^{2}+{C}_{2}{t}^{\frac{6-2s}{3-2s}}-{C}_{3}{t}^{{2}_{s}^{*}},\text{}t\ge 0.\end{array}$

${g}^{\prime }\left(t\right)=2{C}_{1}t+\frac{6-2s}{3-2s}{C}_{2}{t}^{\frac{3}{3-2s}}-{2}_{s}^{*}{C}_{3}{t}^{\frac{3+2s}{3-2s}},\text{}t\ge 0.$

${g}^{\prime }\left(t\right)=0$ 时，它有唯一解，该解为

$\stackrel{˜}{t}={S}_{\mu ,s}^{-\frac{3-2s}{4s}}{\left(\frac{b{S}_{\mu ,s}^{\frac{3-s}{3-2s}}+\sqrt{{b}^{2}{S}_{\mu ,s}^{\frac{6-2s}{3-2s}}+4a{S}_{\mu ,s}}}{2}\right)}^{\frac{3-2s}{2s}}.$

$\begin{array}{c}\underset{t\ge 0}{\mathrm{max}}g\left(t\right)=g\left(\stackrel{˜}{t}\right)={C}_{1}{\stackrel{˜}{t}}^{2}+{C}_{2}{\stackrel{˜}{t}}^{\frac{6-2s}{3-2s}}-\frac{2{C}_{1}{\stackrel{˜}{t}}^{2}+\frac{6-2s}{3-2s}{C}_{2}{\stackrel{˜}{t}}^{\frac{6-2s}{3-2s}}}{{2}_{s}^{*}}\\ =\frac{2s}{3}{C}_{1}{\stackrel{˜}{t}}^{2}+\frac{s}{3}{C}_{2}{\stackrel{˜}{t}}^{\frac{6-2s}{3-2s}}={c}_{\mu ,s}^{*},\end{array}$

$J\left(tU\right)=g\left(t\right)-\frac{{t}^{q}}{q}\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|U|}^{q}\text{d}x\le {c}_{\mu ,s}^{*}-\frac{{t}^{q}}{q}\lambda {\int }_{{ℝ}^{3}}f\left(x\right){|U|}^{q}\text{d}x.$ (3.3)

$\underset{0\le t\le {t}_{1}}{\mathrm{max}}J\left(tU\right)<{c}_{\mu ,s}^{*}-{C}_{0}{\lambda }^{\frac{2}{2-q}}.$

1) ${u}_{\lambda }$ 是(1.1)的正解且 $J\left({u}_{\lambda }\right)=m={m}^{+}$

2) 当 $\lambda \to {0}^{+}$ 时， $‖{u}_{\lambda }‖\to 0$

$J\left({u}_{n}\right)\to m+o\left(1\right)$${J}^{\prime }\left({u}_{n}\right)\to o\left(1\right)$

${m}^{+}\le J\left({t}_{\lambda }^{+}{u}_{\lambda }\right)

2) 因为 ${u}_{\lambda }\in {\mathcal{N}}^{+}$，由(2.6)和(2.7)可得，

$‖{u}_{\lambda }‖<\mathrm{min}\left\{{\left[\frac{\left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}}{\frac{4s}{3-2s}\sqrt{2ab}{S}_{\mu ,s}^{\frac{q}{2}}}\right]}^{\frac{3-2s}{6-3s-q\left(3-2s\right)}},{\left[\frac{\left({2}_{s}^{*}-q\right)\lambda {|f|}_{\infty }{C}_{{R}_{0},s,q}}{a\frac{4s}{3-2s}{S}_{\mu ,s}^{\frac{q}{2}}}\right]}^{\frac{1}{2-q}}\right\}$,

1) $J\left({U}_{\lambda }\right)={m}^{-}$

2) ${U}_{\lambda }$ 是(1.1)的一个正解。

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