西南师范大学学报(自然科学版)   2019, Vol. 44 Issue (10): 23-29.  DOI: 10.13718/j.cnki.xsxb.2019.10.006
0
Article Options
  • PDF
  • Abstract
  • Figures
  • References
  • 扩展功能
    Email Alert
    RSS
    本文作者相关文章
    彭秋颖
    吕颖
    欢迎关注西南大学期刊社
     

  • 带有临界指数的Kirchhoff方程最小能量变号解的存在性    [PDF全文]
    彭秋颖 , 吕颖     
    西南大学 数学与统计学院, 重庆 400715
    摘要:研究了一类带临界指数的Kirchhoff方程 $ -\left(a+b \int_{\mathbb{R}^{3}}|\nabla u|^{2} \mathrm{d} x\right) \Delta u+V(x) u=h(x)|u|^{p-2} u+u^{5} \quad x \in \mathbb{R}^{3} $ 其中ab>0,p∈(4,6).利用Nehari流形和变分法获得了该方程的最小能量变号解.
    关键词Kirchhoff方程    临界指数    Nehari流形    变分法    

    考虑如下带临界指数的Kirchhoff方程:

    $ - \left( {a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} } \right)\Delta u + V(x)u = h(x){\left| u \right|^{p - 2}}u + {u^5}\;\;\;\;x \in {\mathbb{R}^3} $ (1)

    其中ab>0,p∈(4,6).定义

    $ {V^ - }\left( x \right) = \max \left\{ { - V\left( x \right),0} \right\} $

    假设V(x),h(x)满足下列条件:

    $\left(\mathrm{V}_{0}\right) V^{-} \in L^{\frac{3}{2}}\left(\mathbb{R}^{3}\right), \int_{\mathrm{R}^{3}}\left|V^{-}(x)\right|^{\frac{3}{2}} \mathrm{d} x<S^{\frac{3}{2}}$,其中

    $ S = \mathop {\inf }\limits_{u \in {D^{1.2}}\left( {{\mathbb{R}^3}} \right)\backslash \{ 0\} } \frac{{\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} }}{{{{\left( {\int_{{\mathbb{R}^3}} {{{\left| u \right|}^6}{\text{d}}x} } \right)}^{\frac{1}{3}}}}} $

    (V1)存在r>0,Cv>0,使得

    $ \begin{array}{*{20}{c}} {V(x) \leqslant {V_\infty } - {C_v}{{\text{e}}^{ - r\left| x \right|}}}&{{\text{a}}.\;{\text{e}}.\;x \in {\mathbb{R}^3}}&{{V_\infty } = \mathop {\lim }\limits_{|x| \to + \infty } V(x) > 0} \end{array} $

    $\left(\mathrm{h}_{0}\right) h \in L^{\frac{6}{6-p}}\left(\mathbb{R}^{3}\right)$

    (h1)存在θ>0,Ch>0,使得

    $ \begin{array}{*{20}{c}} {h(x) \geqslant {h_\infty } - {C_h}{{{e}}^{ - \theta \left| x \right|}}}&{{\text{a}}.\;{\text{e}}.\;x \in {\mathbb{R}^3}}&{{h_\infty } = \mathop {\lim }\limits_{|x| \to + \infty } h(x) > 0} \end{array} $

    近年来,许多学者研究了带临界指数的Kirchhoff方程(参考文献[1-7]).特别地,文献[8]利用山路引理和反证法得到了带临界指数的Kirchhoff方程的变号解.带临界指数的Kirchhoff方程往往存在紧性的缺失,本文将通过比较方程(1)和其极限方程在Nehari流形子集上的极小值大小,克服该问题.

    由文献[9]的命题2.4,我们可以得到方程(1)的极限问题有一个正解w.令

    $ \alpha = {\left( {a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla w } \right|}^2}{\text{d}}x} } \right)^{\frac{1}{2}}} $

    本文的主要结果为:

    定理1  假设条件(V0),(V1),(h0),(h1)成立,若$r<\theta< \frac{p \sqrt {V_{\infty}} }{\alpha}$,则方程(1)有一个正的基态解.

    定理2  假设条件(V0),(V1),(h0),(h1)成立,若$r<\min \left\{\frac{\sqrt{V_{\infty}}}{\alpha}, \theta\right\}, \theta<\frac{p \sqrt {V_{\infty} }}{\alpha}$,则方程(1)有一个最小能量变号解.

    1 预备知识

    方程(1)对应的能量泛函为

    $ I(u) = \frac{1}{2}\int_{{\mathbb{R}^3}} {\left( {a{{\left| {\nabla u} \right|}^2} + V(x){u^2}} \right){\text{d}}x} + \frac{b}{4}{\left( {\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} } \right)^2} \\ - \frac{1}{p}\int_{{\mathbb{R}^2}} {h(x){{\left| u \right|}^p}{\text{d}}x} - \frac{1}{6}\int_{{\mathbb{R}^3}} {{{\left| u \right|}^6}{\text{d}}x} \;\;\;\;u \in X $

    其中

    $ X = \left\{ {u \in {H^1}\left( {{\mathbb{R}^3}} \right):\int_{{\mathbb{R}^3}} {a{{\left| {\nabla u} \right|}^2}} + V(x){{\left| u \right|}^2}{\text{d}}x < + \infty } \right\} $

    范数为

    $ {\left\| u \right\|^2} = \int_{{\mathbb{R}^3}} {\left( {a{{\left| {\nabla u} \right|}^2} + V(x){{\left| u \right|}^2}} \right){\text{d}}x} $

    因为X连续嵌入到Hilbert空间$H^{1}\left(\mathbb{R}^{3}\right)$中,所以X连续嵌入到空间$L^{q}\left(\mathbb{R}^{3}\right)$中,其中q∈(2,2*).

    因为泛函I(u)∈C1(X$\mathbb{R}$),所以方程(1)的解是能量泛函I(u)的临界点.即u是方程(1)的弱解是指:对∀vX,有

    $ \langle I(u),v\rangle = \int_{{\mathbb{R}^3}} {(a\nabla u\nabla v + V(x)uv){\text{d}}x} + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} \int_{{\mathbb{R}^2}} \\ \nabla u\nabla v{\text{d}}x - \int_{{\mathbb{R}^2}} h (x){\left| u \right|^{p - 1}}v{\text{d}}x - \int_{{\mathbb{R}^3}} {{{\left| u \right|}^5}v{\text{d}}x} $

    考虑方程(1)的极限问题

    $ - \left( {a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} } \right)\Delta u + {V_\infty }u = {h_\infty }{\left| u \right|^{p - 2}}u + {u^5} $ (2)

    对应的能量泛函为

    $ {I_\infty }\left( u \right) = \frac{1}{2}\int_{{\mathbb{R}^3}} {\left( {a{{\left| {\nabla u} \right|}^2} + {V_\infty }{u^2}} \right){\text{d}}x} + \frac{b}{4}{\left( {\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} } \right)^2} \\ - \frac{1}{p}\int_{{\pi ^2}} {{h_\infty }} {\left| u \right|^p}{\text{d}}x - \frac{1}{6}\int_{{\mathbb{R}^3}} {{{\left| u \right|}^6}{\text{d}}x} \;\;\;\;u \in X $

    定义Nehari流形

    $ \mathcal{N} = \left\{ {u \in X\backslash \{ 0\} :\left\langle {I'(u),u} \right\rangle = 0} \right\} $
    $ {\mathcal{N}_\infty } = \left\{ {u \in X\backslash \{ 0\} :\left\langle {{{I'}_\infty }(u),u} \right\rangle = 0} \right\} $
    $ {\mathcal{N}^ \pm } = \left\{ {u \in X\backslash 0} \right\}:\left. {\left\langle {I'(u),{u^ + }} \right\rangle = 0,\left\langle {I'(u),{u^ - }} \right\rangle = 0} \right\} $
    $ \mathcal{N}_R^ \pm = {\mathcal{N}^ \pm } \cap {H^1}\left( {{B_R}\left( 0 \right)} \right)\;\;\;\;\;R > 0 $

    本文的思路是:先讨论方程(1)的正解u和极限方程(2)的正解w的性质;再证明$m_{R}^{\pm}=\inf \limits_{u \in \mathcal{N}_{R}^{\pm}} I(u)$在流形$\mathcal{N}$R±上可达到;由形变引理可得,对∀φH1(BR(0)),有〈I′(uR),φ〉=0;当R→∞时,mR±=m±= $\inf \limits_{u \in \mathcal{N}^{\pm}} I(u)$,且〈I′(u),φ〉=0;最后由m± < m+m证得u±≠0.

    2 主要结果的证明

    引理1  极限方程(2)有一个正的基态解wH1($\mathbb R$3),使得

    $ {I_\infty }\left( w \right) = {m_\infty } = \mathop {\inf }\limits_{u \in {\mathcal{N}_\infty }} {I_\infty }\left( u \right) $

    $ \alpha = {\left( {a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla w} \right|}^2}{\text{d}}x} } \right)^{\frac{1}{2}}} $

    对∀δ∈(0,$\sqrt {V_∞} $),存在C=C(δ)>0,使得对∀x$\mathbb R^3$,有$w(x) \leqslant C \mathrm{e}^{-\frac{\delta}{\alpha}|x|}$.

      正解wH1($\mathbb R^3$)存在性的证明类似于文献[9]的命题2.4.对∀x$\mathbb R^3$,令v(x)=w(αx),则

    $ - \Delta v\left( x \right) = {\alpha ^2} \cdot - \Delta w\left( {\alpha x} \right) = {\alpha ^2}\frac{{{h_\infty }{{\left| {w\left( {\alpha x} \right)} \right|}^{p - 2}}w\left( {\alpha x} \right) + {{\left| {w\left( {\alpha x} \right)} \right|}^5} - {V_\infty }w\left( {\alpha x} \right)}}{{a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla w} \right|}^2}{\text{d}}x} }} $

    $ - \Delta v + {V_\infty }v = {h_\infty }{\left| v \right|^{p - 2}}v + {\left| v \right|^5} $

    由文献[10]可得v(x)≤M·e-δ(|x|-R),所以$w(x) \leqslant C e^{-\frac{\delta}{a}|x|}$.

    定理1的证明   类似于文献[2]和文献[11]的命题6.1,由Ekeland变分原理得到PS序列{un}∈ $\mathcal N$.因为

    $ I\left( {{u_n}} \right) = I\left( {{u_n}} \right) - \frac{1}{p}\left\langle {I'\left( {{u_n}} \right),{u_n}} \right\rangle \geqslant \left( {\frac{1}{2} - \frac{1}{p}} \right){\left\| {{u_n}} \right\|^2} > 0 $

    所以{un}有界.又因un$\rightharpoonup$uX,则I′(u)=0.由于对∀x$\mathbb R^3$,有w(x)≤Ce- $\frac{{δ}}{{α}}$|x|,得m < m,由此得u≠0.最后由Fatou引理证得临界点u满足I(u)=m,根据极大值原理可得u是正解.

    注1   若u是方程(1)的正解,且满足I(u)=m= $\mathop {\inf }\limits_{u \in {\cal N}} I(u)$,则类似于引理1,对任何μ>0,存在C=C(μ)>0,使得对∀x$\mathbb R^3$,有(x)≤Ce-μ|x|.

    引理2  取$u \in {\cal N}_R^ \pm $,令hu(ts)=I(tu++su-),对∀ts≥0,hu在点(1,1)处取得极大值.

      因为$u \in {\cal N}_R^ \pm $,所以〈I′(u),u±〉=0,则

    $ {\left\| {u_n^ \pm } \right\|^2} + b{\int_{{\mathbb{R}^3}} {\left| {\nabla u} \right|} ^2}{\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ \pm }} \right|}^2}} {\text{d}}x = \int_{{\mathbb{R}^3}} h (x){\left| {{u^ \pm }} \right|^p}{\text{d}}x + \int_{{\mathbb{R}^3}} {{{\left| {{u^ \pm }} \right|}^6}} {\text{d}}x > 0 $

    又因

    $ \begin{gathered} I\left( {t{u^ + } + s{u^ - }} \right) = \frac{1}{2}\int_{{\mathbb{R}^3}} {\left[ {a{{\left| {\nabla \left( {t{u^ + }} \right) + \nabla \left( {x{u^ - }} \right)} \right|}^2} + V(x){{\left| {t{u^ + } + s{u^ - }} \right|}^2}} \right]{\text{d}}x} \\ + \frac{b}{4}{\left[ {\int_{{\mathbb{R}^3}} {{{\left| {\nabla \left( {t{u^ + }} \right) + \nabla \left( {s{u^ - }} \right)} \right|}^2}} {\text{d}}x} \right]^2} - \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{p}\int_{{\mathbb{R}^3}} h (x){\left| {t{u^ + } + s{u^ - }} \right|^p}{\text{d}}x - \frac{1}{6}\int_{{\mathbb{R}^3}} {{{\left| {{u^ + } + s{u^ - }} \right|}^6}} {\text{d}}x \hfill \\ \end{gathered} $

    所以$\lim\limits _{|(t, s)| \rightarrow+\infty} h^{u}(t, s)=-\infty$,则hu的极大值点在(t0s0)∈[0,+∞)×[0,+∞)处取得.

    第1步  证明s0t0>0.假设s0=0,因为hu(0,0)=0且hu的极大值点为(t0s0),则t0>0.当s>0足够小时,I(su-)>0,于是

    $ {h^u}\left( {{t_0},0} \right) = I\left( {{t_0}{u^ + }} \right) < I\left( {{t_0}{u^ + }} \right) + I\left( {s{u^ - }} \right) + \frac{b}{2}{s^2}t_0^2\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x = {h^u}\left( {{t_0},s} \right) $

    hu(t0s)在s=0处取得极大值矛盾,因此s0>0,同理t0>0.

    第2步  证明s0t0∈(0,1].由hu(ts)的极大值点是(t0s0)知,I(tu++su-)在(t0s0)处的偏导数为0,即

    $ t_0^2{\left\| {{u^ + }} \right\|^2} + bt_0^4{\left( {\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x} \right)^2} + bs_0^2t_0^2\int_{{\mathbb{R}^3}} \\{{{\left| {\nabla {u^ + }} \right|}^2}{\text{d}}x} \int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x = t_0^p\int_{{\mathbb{R}^3}} h (x){\left| {{u^ + }} \right|^\rho }{\text{d}}x + t_0^6\int_{{\mathbb{R}^3}} {{{\left| {{u^ + }} \right|}^6}} {\text{d}}x $

    假设s0t0,因〈I′(u),u+〉=0,则

    $ t_0^{ - 2}{\left\| {{u^ + }} \right\|^2} + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x \geqslant t_0^{p - 4}\int_{{\mathbb{R}^3}} h (x){\left| {{u^ + }} \right|^\rho }{\text{d}}x + t_0^2\int_{{\mathbb{R}^3}} {{{\left| {{u^ + }} \right|}^6}} {\text{d}}x $ (3)
    $ {\left\| {{u^ + }} \right\|^2} + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x = \int_{{\mathbb{R}^3}} h (x){\left| {{u^ + }} \right|^p}{\text{d}}x + \int_{{\mathbb{R}^3}} {{{\left| {{u^ + }} \right|}^6}} {\text{d}}x > {\left\| {{u^ + }} \right\|^2} $ (4)

    由(3),(4)式得

    $ \left( {t_0^2 - 1} \right){\left\| {{u^ + }} \right\|^2} \geqslant \left( {t_0^{p - 4} - 1} \right)\int_{{\mathbb{R}^3}} h (x){\left| {{u^ + }} \right|^p}{\text{d}}x + \left( {t_0^2 - 1} \right)\int_{{\mathbb{R}^3}} {{{\left| {{u^ + }} \right|}^6}} {\text{d}}x $

    所以t0≤1.假设t0s0,由〈I′(tu++su-),su-〉=0得s0≤1.

    第3步  证明hu在(0,1]2\(1,1)处取不到极大值.

    $ \begin{gathered} {h^u}\left( {{t_0},{s_0}} \right) = I\left( {{t_0}{u^ + } + {s_0}{u^ - }} \right) - \frac{1}{p}\left\langle {I'\left( {{t_0}{u^ + } + {s_0}{u^ - }} \right),{t_0}{u^ + } + {s_0}{u^ - }} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\frac{1}{2} - \frac{1}{p}} \right)\left( {t_0^2{{\left\| {{u^ + }} \right\|}^2} + s_0^2{{\left\| {{u^ - }} \right\|}^2}} \right) + \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\frac{1}{4} - \frac{1}{p}} \right)b{\left[ {\int_{{\mathbb{R}^3}} {{{\left| {\nabla \left( {{t_0}{u^ + } + {s_0}{u^ - }} \right)} \right|}^2}} {\text{d}}x} \right]^2} + \left( {\frac{1}{p} - \frac{1}{6}} \right)\int_{{\mathbb{R}^3}} {{{\left| {{t_0}{u^ + } + {s_0}{u^ - }} \right|}^6}} {\text{d}}x \hfill \\ \end{gathered} $

    t0 < 1或s0 < 1,则

    $ {h^u}\left( {{t_0},{s_0}} \right) < I\left( {{u^ + } + {u^ - }} \right) - \frac{1}{p}\left\langle {I'\left( {{u^ + } + {u^ - }} \right),{u^ + } + {u^ - }} \right\rangle = {h^u}\left( {1,1} \right) $

    引理3  对∀R>0,令mR±= $\inf \limits_{u \in \mathcal{N}_{R}^{\pm}} I(u)$,存在uR${\cal N}_R^ \pm $,使得I(uR)=mR±.

      令{un}⊂ ${\cal N}_R^ \pm $,则I(un)→mR±,〈I′(un),un〉=0.由

    $ m_R^ \pm = I\left( {{u_n}} \right) - \left\langle {I'\left( {{u_n}} \right),{u_n}} \right\rangle \geqslant \left( {\frac{1}{2} - \frac{1}{p}} \right){\left\| {{u_n}} \right\|^2} $

    知{un}有界,则存在uH01(BR(0)),使得un $\rightharpoonup$uH01(BR(0)),unuLp(BR(0)),un(x)→u(x)对a.e. x∈(BR(0))一致成立.

    因为〈I′(un),un±〉=0,所以

    $ \int_{{\mathbb{R}^3}} h (x){\left| {u_n^ \pm } \right|^\rho }{\text{d}}x + \int_{{{\text{R}}^3}} {{{\left| {u_n^ \pm } \right|}^6}} {\text{d}}x = {\left. {\left| {u_n^ \pm } \right|} \right|^2} + b\int_{{\mathbb{R}^3}} \\ {{{\left| {\nabla {u_n}} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {\left| {\nabla u_n^ \pm } \right|} {\text{d}}x \geqslant {\left. {\left| {u_n^ \pm } \right|} \right|^2} $

    又因{un}有界,则存在ρ>0,使得

    $ {\left. {\left| {u_n^ \pm } \right|} \right|^2} + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u_n}} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {\left| {\nabla u_n^ \pm } \right|} {\text{d}}x \geqslant {\left. {\left| {u_n^ \pm } \right|} \right|^2} \geqslant {\rho ^2} $

    由引理2的证明知,hu在(tusu)∈ $\mathbb R^2$处取得极大值,其中tusu>0.因此

    $ \frac{\partial }{{\partial {t_u}}}{h^u}\left( {{t_u},{s_u}} \right) = 0 = \frac{\partial }{{\partial {s_u}}}{h^u}\left( {{t_u},{s_u}} \right) $

    uR=(tuu+suu-)∈ ${\cal N}_R^ \pm $.因un±$\rightharpoonup$u±,由引理2得

    $ \begin{gathered} m_R^ \pm \leqslant I\left( {{t_u}{u^ + } + {s_u}{u^ - }} \right) \leqslant \mathop {\lim \inf }\limits_{n \to \infty } I\left( {{t_u}u_n^ + + {s_u}u_n^ - } \right) = \mathop {\lim \inf }\limits_{n \to \infty } {h^{{u_n}}}\left( {{t_{{u_n}}},{s_{{u_n}}}} \right) \leqslant \hfill \\ \;\;\;\;\;\;\;\;\mathop {\lim \inf }\limits_{n \to \infty } {h^{{u_n}}}\left( {1,1} \right) = \mathop {\lim \inf }\limits_{n \to \infty } I\left( {{u_n}} \right) = m_R^ \pm \hfill \\ \end{gathered} $

    引理4  对任何R>0,存在uR${\cal N}_R^ \pm $使得I(uR)=mR±,且〈I′(uR),φ〉=0,∀φH01(BR(0)).

      由文献[9]的形变引理反证可得.

    引理5   $m^{\pm}=\inf \limits_{u \in \mathcal{N}^{\pm}} I(u), \lim\limits _{R \rightarrow \infty} m_{R}^{\pm}=m^{\pm}$.

      证明较简单,类似于文献[12]的引理4.1.

    引理6  m± < m+m.

      由引理1和注1知I(u)=mI(w)=m.令

    $ {x_n} = (0,0,n) \in {\mathbb{R}^3} $
    $ \begin{array}{*{20}{c}} {{w_n}(x) = w\left( {x + {x_n}} \right)}&{\forall x \in {\mathbb{R}^3}} \end{array} $

    对∀n$\mathbb N$,(ts)∈ ${\left[ {\frac{1}{2}, 2} \right]^2}$,定义ψn(x)=tu(x)-swn(x),其中x$\mathbb R^3$.证存在n0$\mathbb N$,当nn0,(st)∈ ${\left[ {\frac{1}{2}, 2} \right]^2}$时,有I(ψn) < m+m.因为

    $ I\left( {t\bar u - s{w_n}} \right) = I\left( {t\bar u} \right) + {I^\infty }\left( {s{w_n}} \right) + {A_n} + {B_n} + {C_n} + {D_n} + {E_n} + {F_n} $

    其中

    $ {A_n} = \frac{1}{2}{s^2}\int_{{\mathbb{R}^3}} {\left( {V(x) - {V_\infty }} \right)} w_n^2{\text{d}}x \leqslant \frac{1}{2}{s^2}\int_{{\mathbb{R}^3}} {\left( { - {C_v}{{\text{e}}^{ - r\left| x \right|}}} \right)} {{\text{e}}^{ - 2\frac{\delta }{a}\left| {x + {x_n}} \right|}}{\text{d}}x \leqslant - {C_1}{{\text{e}}^{ - rn}} $
    $ {D_n} = \frac{{{S^p}}}{p}\int_{{\mathbb{R}^3}} {\left( {{h_\infty } - h(x)} \right)} w_n^p{\text{d}}x \leqslant \frac{{{S^p}}}{p}\int_{{\mathbb{R}^3}} {\left( {{C_h}{{\text{e}}^{ - \theta |x|}}} \right)} {{\text{e}}^{ - p\frac{\delta }{a}\left| {x + {x_n}} \right|}}{\text{d}}x \leqslant {C_2}{{\text{e}}^{ - {\theta _n}}} $

    由注1,有u(x)≤Ce-μ|x|,取$\gamma<\mu<\frac{\sqrt {V_{\infty}} }{\alpha}$,则

    $ {B_n} = - st\int_{{\mathbb{R}^3}} {\left[ {a\left( {\nabla \bar u \cdot \nabla {w_n}} \right) + V(x)\left( {\bar u \cdot {w_n}} \right)} \right]{\text{d}}x} \leqslant - {C_3}{{\text{e}}^{ - \mu n}} $
    $ {E_n} = - \frac{1}{p}\int_{{\mathbb{R}^3}} h (x)\left( {{{\left| {{\psi _n}} \right|}^p} - |t\bar u{|^p} - {{\left| {s{w_n}} \right|}^p}} \right){\text{d}}x \leqslant - C\int_{{\mathbb{R}^3}} \\ {\left( {|t\bar u{|^{p - 1}}s{w_n} + {{\left| {s{w_n}} \right|}^{p - 1}}t\bar u} \right){\text{d}}x} \leqslant - {C_4}{{\text{e}}^{ - \mu n}} $
    $ {F_n} = - \frac{1}{6}\int_{{\mathbb{R}^3}} {\left( {{{\left| {{\psi _n}} \right|}^6} - |t\bar u{|^6} - {{\left| {s{w_n}} \right|}^6}} \right){\text{d}}x} \leqslant - C\int_{{\mathbb{R}^3}} {\left( {|t\bar u{|^5}s{w_n} + {{\left| {s{w_n}} \right|}^5}t\bar u} \right){\text{d}}x} \leqslant - {C_5}{{\text{e}}^{ - \mu n}} $

    因〈I(wn),wn〉=0,令∫$\mathbb R^3$|▽wn|2dx=Gn>0,有

    $ {G_n} + bG_n^2 = \int_{{\mathbb{R}^3}} {{h_\infty }} {\left| {{w_n}} \right|^\rho }{\text{d}}x + \int_{{\mathbb{R}^3}} {{{\left| {{w_n}} \right|}^6}} {\text{d}}x - \int_{{\mathbb{R}^3}} {{V_\infty }} {\left| {{w_n}} \right|^2}{\text{d}}x < C{{\text{e}}^{ - \partial n}} $

    $ {G_n} \leqslant \frac{{ - a + \sqrt {{a^2} + 4C{{\text{e}}^{ - \delta n}}} }}{{2b}} \to 0\quad n \to \infty $

    所以

    $ \begin{gathered} {C_n} = \frac{b}{4}\left[ {4{t^2}{s^2}\int_{{\mathbb{R}^3}} {|\nabla \bar u{|^2}{\text{d}}x} \int_{{\mathbb{R}^3}} {{{\left| {\nabla {w_n}} \right|}^2}} {\text{d}}x + 2{t^2}{s^2}{{\left( {\int_{{\mathbb{R}^3}} \nabla \bar u \cdot \nabla {w_n}{\text{d}}x} \right)}^2} - } \right. \hfill \\ \;\;\;\;\;\;\;\left. { - 4{t^3}s\int_{{\mathbb{R}^3}} {|\nabla \bar u{|^2}{\text{d}}x} \int_{{\mathbb{R}^3}} \nabla \bar u \cdot \nabla {w_n}{\text{d}}x - 4t{s^3}\int_{{\mathbb{R}^3}} {{{\left| {\nabla {w_n}} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} \nabla \bar u \cdot \nabla {w_n}{\text{d}}x} \right] \leqslant \hfill \\ \;\;\;\;\;\;\;{C_6}{{\text{e}}^{ - \mu n}} + {C_7}{G_n} \hfill \\ \end{gathered} $

    $ I\left( {{\psi _n}} \right) \leqslant m + {m_\infty } - {C_1}{{\text{e}}^{ - rn}} + {C_2}{{\text{e}}^{ - \theta n}} - {C_3}{{\text{e}}^{ - \mu n}} + {C_4}{{\text{e}}^{ - \mu n}} - {C_5}{{\text{e}}^{ - \mu n}} + {C_6}{{\text{e}}^{ - \mu n}} + {C_7}{G_n} $

    n→∞时,m±m+m+o(1).

    下证当(t0s0)∈ ${\left[ {\frac{1}{2}, 2} \right]^2}$时,t0u-s0wn${\cal N}_{}^ \pm $.令

    $ {h^ \pm }(t,s,n) = \left\langle {I'\left( {t\bar u - s{w_n}} \right),{{\left( {t\bar u - s{w_n}} \right)}^ \pm }} \right\rangle $

    由〈I′(u),u 〉=0,得

    $ \begin{array}{*{20}{c}} {{h^ + }\left( {\frac{1}{2},0,n} \right) > 0}&{{h^ + }(2,0,n) < 0} \end{array} $

    由〈I(wn),wn〉=0,得

    $ \begin{array}{*{20}{c}} {{h^ - }\left( {0,\frac{1}{2},n} \right) > 0}&{{h^ - }(0,2,n) < 0} \end{array} $

    由Miranda定理知,当n足够大时,存在(t0s0)∈ ${\left[ {\frac{1}{2}, 2} \right]^2}$,使得h±(t0s0n)=0,即t0u-s0wn${\cal N}_{}^ \pm $.

    定理2的证明   令引理3中R=n,则un${\cal N}_{}^ \pm $.因${\cal N}_{}^ \pm $$\cal N$,所以{un}有界.存在uX,使得un$\rightharpoonup$uX.由引理4知u是临界点,下证u±≠0.令K1K2$\mathbb R$,定义

    $ {K_1} = \mathop {\lim }\limits_{n \to \infty } I\left( {u_n^ + } \right) + \frac{b}{4}\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x $
    $ {K_2} = \mathop {\lim }\limits_{n \to \infty } I\left( {u_n^ - } \right) + \frac{b}{4}\int_{{{\bf{R}}^3}} {{{\left| {\nabla {u^ + }} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u^ - }} \right|}^2}} {\text{d}}x $

    因为I(un)=mn±,所以由引理5、引理6得

    $ {K_1} + {K_2} = \mathop {\lim }\limits_{n \to \infty } I\left( {{u_n}} \right) = {m^ \pm } < m + {m_\infty } $

    现证u+≠0.假设u+=0,则un+$\rightharpoonup$0于X.令

    $ \left\| {u_n^ + } \right\|_*^2 = \int_{{\mathbb{R}^3}} {\left( {a{{\left| {\nabla u_n^ + } \right|}^2} + {V_\infty }{{\left| {u_n^ + } \right|}^2}} \right){\text{d}}x} $

    可得‖un+*2=‖un+2+on(1).存在tn>0,使得tnun+${{\cal N}_\infty }$,即

    $ t_n^2\left\| {u_n^ + } \right\|_*^2 + bt_n^4{\left( {\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ + } \right|}^2}} {\text{d}}x} \right)^2} = t_n^p\int_{{\mathbb{R}^3}} {{h_\infty }} {\left| {u_n^ + } \right|^p}{\text{d}}x + t_n^6\int_{{\mathbb{R}^3}} {{{\left| {u_n^ + } \right|}^6}{\text{d}}x} $ (5)

    因〈I′(un),un+〉=0,则

    $ {\left\| {u_n^ + } \right\|^2} + {b^4}\int_{{\mathbb{R}^3}} {{{\left| {\nabla {u_n}} \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ + } \right|}^2}} {\text{d}}x = \int_{{\mathbb{R}^3}} {h\left( x \right)} {\left| {u_n^ + } \right|^p}{\text{d}}x + \int_{{\mathbb{R}^3}} {{{\left| {u_n^ + } \right|}^6}{\text{d}}x} $ (6)

    由条件(h0)知,对∀ε>0,存在R>0,使得$\int_{|x|>R}|h(x)|^{\frac{6}{6-p}} \mathrm{d} x<\varepsilon$.

    $ \begin{gathered} \int_{{\mathbb{R}^3}} h (x){\left| {u_n^ + } \right|^p}{\text{d}}x = \int_{|x| \leqslant R} h (x){\left| {u_n^ + } \right|^p}{\text{d}}x + \int_{|x| > R} h (x){\left| {u_n^ + } \right|^p}{\text{d}}x \leqslant \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_{|x| \leqslant R} h (x){\left| {u_n^ + } \right|^p}{\text{d}}x + {\left( {{{\int_{|x| > R} {\left| {h(x)} \right|} }^{\frac{6}{{6 - p}}}}{\text{d}}x} \right)^{\frac{{6 - p}}{6}}}{\left( {\int_{|x| > R} {{{\left| {u_n^ + } \right|}^6}} {\text{d}}x} \right)^{\frac{p}{6}}} < \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;o\left( 1 \right) + \varepsilon \cdot {\left( {\int_{|x| > R} {{{\left| {u_n^ + } \right|}^6}} {\text{d}}x} \right)^{\frac{p}{6}}} \hfill \\ \end{gathered} $

    ε的任意性知∫$\mathbb R^3$h(x)|un+|pdx→0,根据条件(h1)可知∫$\mathbb R^3$ h|un+|pdx→0.由(5),(6)式可得

    $ \left( {1 - \frac{1}{{t_n^2}}} \right){\left\| {u_n^ + } \right\|^2} + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ - } \right|}^2}} {\text{d}}x{\int_{{\mathbb{R}^3}} {\left| {\nabla u_n^ + } \right|} ^2}{\text{d}}x = \left( {1 - t_n^2} \right)\int_{{\mathbb{R}^3}} {{{\left| {u_n^ + } \right|}^6}} {\text{d}}x $ (7)

    $\mathop {\lim \sup }\limits_{n \to \infty } t_{n}>1$,则(7)式左边大于0,(7)右边小于0,矛盾,则不成立.

    $\mathop {\lim \sup }\limits_{n \to \infty } t_{n} \le 1$

    $ \begin{gathered} {m_\infty } \leqslant {I_\infty }\left( {{t_n}u_n^ + } \right) = {I_\infty }\left( {{t_n}u_n^ + } \right) - \frac{1}{4}\left\langle {{{I'}_\infty }\left( {{t_n}u_n^ + } \right),{t_n}u_n^ + } \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{{t_n^2}}{4}\left\| {u_n^ + } \right\|_*^2 + \left( {\frac{1}{4} - \frac{1}{p}} \right)t_n^p\int_{{\mathbb{R}^3}} {{h_\infty }} {\left| {u_n^ + } \right|^p}{\text{d}}x + \frac{1}{{12}}t_n^6\int_{{\mathbb{R}^3}} {{{\left| {u_n^ + } \right|}^6}} {\text{d}}x \leqslant \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}\left\| {u_n^ + } \right\|_*^2 + \left( {\frac{1}{4} - \frac{1}{p}} \right)\int_{{\mathbb{R}^3}} {{h_\infty }} {\left| {u_n^ + } \right|^p}{\text{d}}x + \frac{1}{{12}}\int_{{\mathbb{R}^3}} {{{\left| {u_n^ + } \right|}^6}} {\text{d}}x = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{I_\infty }\left( {u_n^ + } \right) - \frac{1}{4}\left\langle {{{I'}_\infty }\left( {u_n^ + } \right),u_n^ + } \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{I_\infty }\left( {u_n^ + } \right) + \frac{b}{4}\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ + } \right|}^2}} {\text{d}}x\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ - } \right|}^2}} {\text{d}}x = {K_1} \hfill \\ \end{gathered} $ (8)

    则可得mK1,又因K1+K2 < m+m,则K2 < m.

    存在sn>0,使得snun-$\cal N$,即〈I′(snun-),snun-〉=0.由

    $ 0 = \left\langle {I'\left( {{u_n}} \right),u_n^ - } \right\rangle = \left\langle {I'\left( {u_n^ - } \right),u_n^ - } \right\rangle + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u_n^ + } \right|}^2}} {\text{d}}x\int_{{{\text{R}}^3}} {{{\left| {\nabla u_n^ - } \right|}^2}} {\text{d}}x $

    得〈I′(un-),un-〉 < 0.因此sn≤1,类似于(8)式可得K2m,得到矛盾,假设不成立,即u+≠0.同理u-≠0.

    我们已经证得u是方程(1)的变号解,下证u是最小能量变号解.

    $ \begin{gathered} {m^ \pm } \leqslant I(u) = I(u) - \frac{1}{4}\left\langle {I'(u),u} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\frac{1}{4}{\left\| u \right\|^2} + \left( {\frac{1}{4} - \frac{1}{p}} \right)\int_{{\mathbb{R}^3}} h (x)|u{|^p}{\text{d}}x + \frac{1}{{12}}{\int_{{\mathbb{R}^3}} {\left| u \right|} ^6}{\text{d}}x \leqslant \hfill \\ \;\;\;\;\;\;\;\;\mathop {\lim \inf }\limits_{n \to \infty } \left[ {I\left( {{u_n}} \right) - \frac{1}{4}\left\langle {I'\left( {{u_n}} \right),{u_n}} \right\rangle } \right] = {m^ \pm } \hfill \\ \end{gathered} $

    I(u)=m±,因此定理2证毕.

    参考文献
    [1]
    FAN H N. Multiple Positive Solutions for a Class of Kirchhoff Type Problems Involving Critical Sobolev Exponents[J]. JMathAnalAppl, 2015, 431(1): 150-168.
    [2]
    LEI C Y, SUO H M, CHU C M, et al. On Ground State Solutions for a Kirchhoff Type Equation with Critical Growth[J]. Comput Math Appl, 2016, 72(3): 729-740.
    [3]
    LI G B, YE H G. Existence of Positive Solutions for Nonlinear Kirchhoff Type Problems in $\mathbb R^3$ with Critical Sobolev Exponent[J]. MathMethods ApplSci, 2014, 37(16): 2570-2584.
    [4]
    LIU J, LIU T, PAN H L. A Result on a Non-Autonomous Kirchhoff Type Equation Involving Critical Term[J]. Appl Math Lett, 2018, 85: 82-87. DOI:10.1016/j.aml.2018.05.026
    [5]
    刘选状, 吴行平, 唐春雷. 一类带有临界指数增长项的Kirchhoff型方程正的基态解的存在性[J]. 西南大学学报(自然科学版), 2015, 37(6): 54-59.
    [6]
    任正娟, 商彦英. 带有临界指数的Kirchhoff方程正解的存在性[J]. 西南大学学报(自然科学版), 2016, 38(4): 78-84.
    [7]
    曾兰, 唐春雷. 带有临界指数的Kirchhoff型方程正解的存在性[J]. 西南师范大学学报(自然科学版), 2016, 41(4): 29-34.
    [8]
    XU L P, CHEN H B. Sign-Changing Solution to Schrödinger-Kirchhoff-Type Equations with Critical Exponent[J]. Adv Differ Equ, 2016, 2016: 121. DOI:10.1186/s13662-016-0828-0
    [9]
    HE X M, ZHOU W M. Existence and Concentration Behavior of Positive Solutions for a Kirchhoff Equation in $\mathbb R^3$[J]. JDifferEqu, 2012, 252(2): 1813-1834.
    [10]
    BERESTYCKI H, LIONS P L. Nonlinear Scalar Field Equations.Ⅰ.Existence of a Ground State[J]. Arch Rational MechAnal, 1983, 82(4): 313-345. DOI:10.1007/BF00250555
    [11]
    WANG L, ZHANG B L, CHENG K. Ground State Sign-Changing Solutions for the Schrödinger-Kirchhoff Equation in $\mathbb R^3$[J]. JMathAnalAppl, 2018, 466(2): 1545-1569.
    [12]
    BATISTA A M, FURTADO M F. Positive and Nodal Solution for a Nonlinear Schrödinger-Poisson System with Sign-Changing Potentials[J]. Nonlinear Anal, 2018, 39: 142-156. DOI:10.1016/j.nonrwa.2017.06.005
    Existence of a Sign-Changing Solution with Minimal Energy for a Kirchhoff Equation with Critical Exponents
    PENG Qiu-ying , LÜ Ying     
    School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
    Abstract: In this paper, the following Kirchhoff equation has been considered with critical exponents $ -\left(a+b \int_{\mathbb{R}^{3}}|\nabla u|^{2} \mathrm{d} x\right) \Delta u+V(x) u=h(x)|u|^{p-2} u+u^{5} \quad x \in \mathbb{R}^{3} $ where a, b>0, p∈(4, 6).By means of Nehari manifold and variational method, the sign-changing solution with minimal energy is obtained.
    Key words: Kirchhoff equation    critical exponent    Nehari manifold    variational method    
    X