西南师范大学学报(自然科学版)   2020, Vol. 45 Issue (2): 20-25.  DOI: 10.13718/j.cnki.xsxb.2020.02.004
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  • 高维空间中带临界指数的非齐次Kirchhoff型方程的正解    [PDF全文]
    吉蕾     
    晋中学院 数理学院, 山西 晋中 030619
    摘要:在高维空间中,研究了一类带临界指数的非齐次Kirchhoff型方程.利用变分方法,当N=4时,获得该方程的两个正解;当N>4时,获得该方程正解的存在性.结论补充并完善了近期相关文献的结果.
    关键词Kirchhoff型方程    临界指数    变分法    正解    

    考虑如下一类具有临界指数的非齐次Kirchhoff型方程:

    $ \left\{ {\begin{array}{*{20}{l}} { - \left( {a + b\int_\mathit{\Omega } {{{\left| {\nabla u} \right|}^2}{\rm{d}}x} } \right)\Delta u = \mu {{\left| u \right|}^{{2^*} - 2}}u + \lambda f(x)}&{x \in \mathit{\Omega }}\\ {u = 0}&{x \in \partial \mathit{\Omega }} \end{array}} \right. $ (1)

    其中Ω$\mathbb{R}$N(N≥4)是非空有界开集,abλ>0为参量,$f \in L^{\frac{2 *}{2^{*}-1}}$(Ω)是非零非负函数.

    近期,如下带临界指数的Kirchhoff型方程被广泛研究:

    $ \left\{ {\begin{array}{*{20}{l}} { - \left( {a + b\int_\mathit{\Omega } {{{\left| {\nabla u} \right|}^2}{\rm{d}}x} } \right)\Delta u = \mu {{\left| u \right|}^{{2^*} - 2}}u + \lambda f(x){u^q}}&{x \in \mathit{\Omega }}\\ {u = 0}&{x \in \partial \mathit{\Omega }} \end{array}} \right. $ (2)

    其中Ω$\mathbb{R}$3(N≥3)是非空有界开集,0≤q < 2*-1.当N=3,q=0时,文献[1-3]先后利用变分法和临界点理论研究了方程(2),其中文献[3]完善了文献[1]的结果,补充了文献[2]的结果.本文在文献[1-3]的基础上研究方程(1)正解的存在情况.当N=4,1≤q < 3时,文献[4]获得了方程(2)正解的存在性结果和正解存在的条件.随后,文献[5]在N≥4,1 < q < 3的情况下获得了方程(2)正解的存在性以及解的多重性,补充完善了文献[4]中相应的结果.文献[6]补充了文献[4]中q=1的结果.当N≥4,0 < q < 1时,文献[7]研究了方程(2)正解的存在性与解的多重性.当N=4,-1 < q < 0时,文献[8]研究了方程(2)正解的存在性和多解性条件.特别地,文献[9]证明了文献[4]提出的部分公开方程.文献[10-11]也研究了Kirchhoff型方程正解的存在性.

    为了寻找方程(1)的正解,定义其对应的能量泛函I

    $ I\left( u \right) = \frac{a}{2}{\left\| u \right\|^2} + \frac{b}{4}{\left\| u \right\|^4} - \frac{\mu }{{{2^*}}}\int_\mathit{\Omega } {{{\left( {{u^ + }} \right)}^2}} {\rm{d}}x - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x $

    其中$u = {\left( {\int_\mathit{\Omega } | \nabla u{|^2}{\rm{d}}x} \right)^{\frac{1}{2}}}$为Sobolev空间H01(Ω)的范数,u±=max{±u,0}.显然,IC1(H01(Ω),$\mathbb{R}$),且∀φH01(Ω)有

    $ \left\langle {I'(u),\varphi } \right\rangle = \left( {a + b{{\left\| u \right\|}^2}} \right)\int_\mathit{\Omega } {\left( {\nabla u,\nabla \varphi } \right){\rm{d}}x} - \mu \int_\mathit{\Omega } {{{\left( {{u^ + }} \right)}^{{2^*} - 1}}} \varphi {\rm{d}}x - \lambda \int_\mathit{\Omega } f (x)\varphi {\rm{d}}x $

    众所周知,方程(1)的解与其能量泛函I的临界点是一一对应的.记$|u|_{p}=\left(\int_{\Omega}|u|^{p} \mathrm{d} x\right)^{\frac{1}{p}}$为空间Lp(Ω)的范数,且S为最佳Soboleve嵌入常数,即

    $ S = \mathop {\inf }\limits_{u \in H_0^1\left( \mathit{\Omega } \right)\backslash \left\{ 0 \right\}} \frac{{{{\left\| u \right\|}^2}}}{{\left| u \right|_{{2^ * }}^2}} $ (3)

    定理1  假设abμ>0,N≥4,$f \in L^{\frac{2 *}{2^{*}-1}}$(Ω)是非零非负函数,则存在λ*>0,使得对∀0 < λ < λ*,方程(1)都存在一个正局部极小解u*,满足I(u*) < 0.

      根据Hölder不等式和(3)式,可得

    $ \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x \le {\left| u \right|_{{2^ * }}}{\left| f \right|_{\frac{{{2^ * }}}{{{2^ * } - 1}}}} \le {\left| f \right|_{\frac{{{2^ * }}}{{{2^ * } - 1}}}}{S^{ - \frac{2}{{{2^*}}}}}\left\| u \right\| $ (4)

    从而结合(3)式和(4)式,有

    $ \begin{array}{l} I\left( u \right) \ge \frac{a}{2}{\left\| u \right\|^2} + \frac{b}{4}{\left\| u \right\|^4} - \frac{\mu }{{{2^*}}}{S^{ - \frac{2}{{{2^*}}}}}{\left\| u \right\|^{{2^*}}} - \lambda {\left| f \right|_{\frac{{{2^ * }}}{{{2^ * } - 1}}}}{S^{ - \frac{2}{{{2^*}}}}}\left\| u \right\| = \\ \;\;\;\;\;\;\;\;\;\;\left\| u \right\|\left( {\frac{a}{2}\left\| u \right\| + \frac{b}{4}{{\left\| u \right\|}^3} - \frac{\mu }{{{2^*}}}{S^{ - \frac{2}{{{2^*}}}}}{{\left\| u \right\|}^{{2^*} - 1}} - \lambda {{\left| f \right|}_{\frac{{{2^ * }}}{{{2^ * } - 1}}}}{S^{ - \frac{2}{{{2^*}}}}}} \right) \end{array} $ (5)

    一方面,当N=4时,2*=4.根据(5)式,可得

    $ I(u) \ge \left\| u \right\|\left( {\frac{a}{2}\left\| u \right\| + \frac{{b{S^2} - \mu }}{4}{{\left\| u \right\|}^3} - \lambda {{\left| f \right|}_{\frac{4}{3}}}{S^{ - \frac{1}{2}}}} \right) $

    当0 < μbS2时,I在空间H01(Ω)上是强制的.从而结合(5)式,容易得证:存在正数Rρλ*,使得对∀0 < λ < λ*都有

    $ {\left. {I(u)} \right|_{{S_R}}} > \rho $ (6)

    其中SR={uH01(Ω):‖u‖=R}.当μ>bS2时,对∀t>0,令

    $ g(t) = \frac{a}{2}t - \frac{{\mu - b{S^2}}}{4}{t^3} - \lambda {\left| f \right|_{\frac{4}{3}}}{S^{ - \frac{1}{2}}} $

    容易得到$g^{\prime}(t)=\frac{a}{2}-\frac{3\left(\mu-b S^{2}\right)}{4} t^{2}$.令g′(t)=0,可得$t_{\max }=\left[\frac{2 a}{3\left(\mu-b S^{2}\right)}\right]^{\frac{1}{2}}$,且

    $ \mathop {\max }\limits_{t \ge 0} g\left( t \right) = g\left( {{t_{\max }}} \right) = \frac{a}{3}{\left[ {\frac{{2a}}{{3\left( {\mu - b{S^2}} \right)}}} \right]^{\frac{1}{2}}} - \lambda {\left| f \right|_{\frac{4}{3}}}{S^{ - \frac{1}{2}}} $

    R=tmax>0,有

    $ {\lambda _*} = \frac{{a{S^{\frac{1}{2}}}}}{{3\left| f \right|\frac{4}{3}}}{\left[ {\frac{{2a}}{{3\left( {\mu - b{S^2}} \right)}}} \right]^{\frac{1}{2}}} > 0 $

    从而,结合(5)式,存在ρ>0,使得对∀0 < λ < λ*都有(6)式成立.

    另一方面,当N≥5时,2* < 4.故对∀μ>0,I在空间H01(Ω)上是强制的.从而,容易得到(6)式成立.

    因此,对∀N≥4,μ>0,存在正数Rρλ*,使得对∀0 < λ < λ*都有(6)式成立.对∀uH01(Ω)且u+≠0,有$\mathop {\lim }\limits_{t \to {0^ + }} \frac{{I(tu)}}{t} = - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x < 0$.因此,对∀uH01(Ω)且u+≠0,当t>0充分小时有I(tu) < 0.从而,$m = \mathop {\inf }\limits_{u \in \overline {{B_R}} } I(u) < 0$有定义.接下来,证明泛函IH01(Ω)中能达到局部极小值m,即存在u*$\overline {{B_R}} $使得I(u*)=m,其中$\overline {{B_R}} $={uH01(Ω)}:‖u‖≤R}为一个闭球.

    根据下确界的定义,存在极小化序列{un}⊂$\overline {{B_R}} $,使得$\lim\limits_{n \rightarrow \infty} I\left(u_{n}\right)=m<0$.从而,对{un}的一个子列(仍记为{un}),存在一个u*H01(Ω),使得当n→∞时有

    $ \left\{ {\begin{array}{*{20}{l}} {{u_n} \to {u_*}}&{x \in H_0^1}\\ {{u_n} \to {u_*}}&{x \in {L^p}\left( \mathit{\Omega } \right)\left( {1 \le p < {2^*}} \right)}\\ {{u_n}(x) \to {u_*}(x)}&{{\rm{a}}{\rm{. e}}{\rm{. }}x \in \mathit{\Omega }} \end{array}} \right. $ (7)

    注意到$\overline {{B_R}}$是闭凸集,从而$\overline {{B_R}}$是弱闭的,因此u*$\overline {{B_R}}$.下证I(u*)=m.根据控制收敛定理,可得

    $ \mathop {\lim }\limits_{n \to \infty } \int_\mathit{\Omega } f (x)u_n^ + {\rm{d}}x = \int_\mathit{\Omega } f (x)u_*^ + {\rm{d}}x $ (8)

    不妨假设wn=un-u*,根据以及Brézis-Lieb引理,可得

    $ \begin{array}{*{20}{c}} {{{\left\| {{u_n}} \right\|}^2} = {{\left\| {{w_n}} \right\|}^2} + {{\left\| {{u_*}} \right\|}^2} + o\left( 1 \right)}\\ {{{\left\| {{u_n}} \right\|}^4} = {{\left\| {{x_n}} \right\|}^4} + {{\left\| {{u_*}} \right\|}^4} + 2{{\left\| {{w_n}} \right\|}^2}{{\left\| {{u_*}} \right\|}^2} + o\left( 1 \right)} \end{array} $ (9)
    $ \int_\mathit{\Omega } {{{\left( {u_n^ + } \right)}^{{2^ * }}}} {\rm{d}}x = \int_\mathit{\Omega } {{{\left( {w_n^ + } \right)}^{{2^ * }}}} {\rm{d}}x + \int_\mathit{\Omega } {{{\left( {u_*^ + } \right)}^{{2^ * }}}} {\rm{d}}x + o\left( 1 \right) $ (10)

    其中o(1)是n→∞时的无穷小量.由(6)式,对∀0 < λ < λ*

    $ \left\{ {\begin{array}{*{20}{l}} {\frac{a}{2}{{\left\| u \right\|}^2} + \frac{b}{4}{{\left\| u \right\|}^4} - \frac{\mu }{{{2^*}}}\int_\mathit{\Omega } {{{\left( {{u^ + }} \right)}^{{2^*}}}} {\rm{d}}x \ge 2\delta }&{\forall u \in {S_R}}\\ {\frac{a}{2}{{\left\| u \right\|}^2} + \frac{b}{4}{{\left\| u \right\|}^4} - \frac{\mu }{{{2^*}}}\int_\mathit{\Omega } {{{\left( {{u^ + }} \right)}^{{2^*}}}} {\rm{d}}x \ge 0}&{\forall u \in \overline {{B_R}} } \end{array}} \right. $ (11)

    由(11)式和m < 0可知,存在ε0>0,使得对∀n$\mathbb{N}$+都有‖un‖≤R-ε0.从而,结合u*$\overline {{B_R}}$以及(9)式可知,当n充分大时有wn$\overline {{B_R}}$.再次用(6)式,可得

    $ \frac{a}{2}{\left\| {{w_n}} \right\|^2} + \frac{b}{4}{\left\| {{w_n}} \right\|^4} - \frac{\mu }{{{2^*}}}\left| {w_n^ + } \right|_{{2^*}}^{{2^*}} \ge 0 $

    再结合(8)-(10)式,有

    $ \begin{array}{l} m = I\left( {{u_n}} \right) + o\left( 1 \right) = I\left( {{u_ * }} \right) + \frac{a}{2}{\left\| {{w_n}} \right\|^2} + \frac{b}{4}{\left\| {{w_n}} \right\|^4} + \frac{b}{2}{\left\| {{w_n}} \right\|^2}{\left\| {{u_ * }} \right\|^2} - \frac{\mu }{{{2^*}}}\left| {w_n^ + } \right|_{{2^*}}^{{2^*}} + o\left( 1 \right) \ge \\ \;\;\;\;\;\;I\left( {{u_*}} \right) + \frac{b}{2}{\left\| {{w_n}} \right\|^2}{\left\| {{u_*}} \right\|^2} + o\left( 1 \right) \ge I\left( {{u_*}} \right) + o\left( 1 \right) \end{array} $

    从而,当n→∞时,有mI(u*).又因为u*∈→$\overline {{B_R}}$,从而有I(u*)≥m.因此I(u*)=m.即u*是方程(1)的非零解.从而,对∀φH01(Ω)有

    $ \left( {a + b{{\left\| {{u_*}} \right\|}^2}} \right)\int_\mathit{\Omega } {\left( {\nabla {u_*},\nabla \varphi } \right){\rm{d}}x} - \mu \int_\mathit{\Omega } {{{\left( {u_*^ + } \right)}^{{2^*} - 1}}} \varphi {\rm{d}}x - \lambda \int_\mathit{\Omega } f (x)\varphi {\rm{d}}x = 0 $

    特别地,取φ=$u_*^ - $,可得

    $ - \left( {a + b{{\left\| {u_*^ - } \right\|}^2}} \right){\left\| {u_*^ - } \right\|^2} - \lambda \int_\mathit{\Omega } f (x)u_*^ - {\rm{d}}x = 0 $

    从而$u_*^ - $=0,即u*≥0在Ω中几乎处处成立.因此,u*是一个非零非负解.再由强极大值原理可得,u*是方程(1)的正解且I(u*)=m < 0.定理1证毕.

    接下来,假设N=4,μ>bS2,研究方程(1)的山路解.首先证明泛函IH01(Ω)上满足局部(PS)c条件.此时,$I(u) = \frac{a}{2}u{^2} + \frac{b}{4}u{^4} - \frac{\mu }{4}\left| {{u^ + }} \right|_4^4 - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x$.

    引理1  假设ab>0,μ>bS2$f \in L^{\frac{4}{3}}$(Ω)是一个非零非负函数,则对任意的$c<\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}-D \lambda^{2}$IH01(Ω)上满足局部(PS)c条件,其中$D=\frac{9|f|_{\frac{4}{3}}^{2}}{16 a S}$.

      假设{un}使得IH01(Ω)上满足局部(PS)c条件,即当n→+∞有

    $ \begin{array}{*{20}{c}} {I\left( {{u_n}} \right) \to c}&{I'\left( {{u_n}} \right) \to c} \end{array} $ (12)

    我们断言:{un}是H01(Ω)上的有界序列.事实上,根据(12)式、(3)式以及Hölder不等式,可得

    $ 1 + c + o\left( 1 \right)\left\| {{u_n}} \right\| \ge I\left( {{u_n}} \right) - \frac{1}{4}\left\langle {I'\left( {{u_n}} \right),{u_n}} \right\rangle = \frac{a}{4}{\left\| {{u_n}} \right\|^2} - \frac{{3\lambda }}{4}\int_\mathit{\Omega } f (x)u_n^ + {\rm{d}}x \ge \frac{a}{4}{\left\| {{u_n}} \right\|^2} - \frac{{3\lambda {{\left| f \right|}_{\frac{4}{3}}}}}{{4{S^{\frac{1}{2}}}}}\left\| {{u_n}} \right\| $

    这就意味着{un}在H01(Ω)上有界.从而存在子列,仍记为{un},以及uH01(Ω)为其弱极限,记wn=un-u.当n→∞时,将u*换成u,则(7)-(10)式都成立.由(8)式和(12)式,可得

    $ a{\left\| {{u_n}} \right\|^2} + b{\left\| {{u_n}} \right\|^4} - \mu \left| {u_n^ + } \right|_4^4 - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x = o\left( 1 \right) $

    进一步,结合(9)-(10)式,可得

    $ \begin{array}{l} a{\left\| u \right\|^2} + a{\left\| {{w_n}} \right\|^2} + b{\left\| u \right\|^4} + b{\left\| {{w_n}} \right\|^4} + 2b{\left\| {{w_n}} \right\|^2}{\left\| u \right\|^2} - \\ \mu \left| {w_n^ + } \right|_4^4 - \mu \left| {{u^ + }} \right|_4^4 - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x = o\left( 1 \right) \end{array} $ (13)

    再次利用(12)式,可得

    $ \mathop {\lim }\limits_{n \to \infty } \left\langle {I'\left( {{u_n}} \right),u} \right\rangle = a{\left\| u \right\|^2} + b{\left\| u \right\|^4} + b{l^2}{\left\| u \right\|^2} - \mu \left| {{u^ + }} \right|_4^4 - \lambda \int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x = 0 $ (14)

    一方面,根据(14)式以及Young不等式,可得

    $ I\left( u \right) = \frac{a}{4}{\left\| u \right\|^2} - \frac{{b{l^2}}}{4}{\left\| u \right\|^2} - \frac{{3\lambda }}{4}\int_\mathit{\Omega } f (x){u^ + }{\rm{d}}x \ge - D{\lambda ^2} - \frac{{b{l^2}}}{4}{\left\| u \right\|^2} $ (15)

    另一方面,根据(13)式和(14)式,可得

    $ a{\left\| {{w_n}} \right\|^2} + b{\left\| {{w_n}} \right\|^4} + b{\left\| {{w_n}} \right\|^2}{\left\| u \right\|^2} - \mu \left| {w_n^ + } \right|_4^4 = o\left( 1 \right) $ (16)
    $ I\left( {{u_n}} \right) = I(u) + \frac{a}{2}{\left\| {{w_n}} \right\|^2} + \frac{b}{4}{\left\| {{w_n}} \right\|^4} + \frac{b}{2}{\left\| {{w_n}} \right\|^2}{\left\| u \right\|^2} - \frac{\mu }{4}\left| {w_n^ + } \right|_4^4 + o\left( 1 \right) $ (17)

    由(3)式,可得$\int_\mathit{\Omega } {{{\left( {w_n^ + } \right)}^4}} {\rm{d}}x\int_\Omega {{{\left| {{w_n}} \right|}^4}} {\rm{d}}x\frac{{{{\left\| {{w_n}} \right\|}^4}}}{{{S^2}}}$.从而根据(16)式,可得al2+bl4+bl2u2$\frac{\mu l^{4}}{S^{2}}$,这就意味着$l^{2} \geqslant \frac{\left(a+b\|u\|^{2}\right) S^{2}}{\mu-b S^{2}}$.从而,结合(16)式和(17)式,可得

    $ \begin{array}{l} I(u) = \mathop {\lim }\limits_{n \to \infty } \left[ {I\left( {{u_n}} \right) - \frac{a}{2}{{\left\| {{w_n}} \right\|}^2} - \frac{b}{4}{{\left\| {{w_n}} \right\|}^4} - \frac{b}{2}{{\left\| {{w_n}} \right\|}^2}{{\left\| u \right\|}^2} + \frac{\mu }{4}\left| {w_n^ + } \right|_4^4} \right] = \\ \;\;\;\;\;\;\;\;\;c - \frac{a}{4}\frac{{\left( {a + b{{\left\| u \right\|}^2}} \right){S^2}}}{{\mu - b{S^2}}} - \frac{{b{l^2}}}{4}{\left\| u \right\|^2} \le c - \frac{{{a^2}{S^2}}}{{4\left( {\mu - b{S^2}} \right)}} - \frac{{b{l^2}}}{4}{\left\| u \right\|^2} < - D{\lambda ^2} - \frac{{b{l^2}}}{4}{\left\| u \right\|^2} \end{array} $

    这与(15)式矛盾.故l≡0.即当n→∞时,unu(xH01(Ω)).即对$\forall c<\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}-D \lambda^{2}$IH01(Ω)上满足局部(PS)c条件.引理1证毕.

    众所周知,$U(x)=\frac{2 \sqrt{2}}{1+|x|^{2}}$(x$\mathbb{R}$4)是临界方程-Δu=u3(x$\mathbb{R}$4)的正解,且‖U2=|U|44=S2.记ηC0(Ω)为截断函数,使得0≤η≤1,|▽η|≤C1.当|x|≤δ时,η(x)=1;当|x|≥2δ时,η(x)=0.令uε(x)=$\varepsilon^{-1} \eta(x) U\left(\frac{x}{\varepsilon}\right)$=$\frac{2 \sqrt{2} \varepsilon \eta(x)}{\varepsilon^{2}+|x|^{2}}$.根据文献[12],可得

    $ \left| {{u_\varepsilon }} \right|_4^2 = \left| U \right|_4^2 + O\left( {{\varepsilon ^4}} \right) = S + O\left( {{\varepsilon ^4}} \right)\;\;\;\;{\left\| {{u_\varepsilon }} \right\|^2} = {\left\| U \right\|^2} + O\left( {{\varepsilon ^2}} \right) = {S^2} + O\left( {{\varepsilon ^2}} \right) $ (18)

    引理2  假设ab>0,μ>bS2$f \in L^{\frac{4}{3}}$(Ω)是非零非负函数且满足条件(F):

    $f \in L^{\frac{4}{3}}$存在δρ>0和1<β<2,使得对∀|x|<ρ$\sup\limits_{t \geqslant 0} I\left(t u_{0}\right)<\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}-D \lambda^{2}$.

    则存在λ*>0以及u0H01(Ω),使得对∀0 < λ < λ*有$\mathop {\sup }\limits_{t \ge 0} $ I(tu0) < $\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}$-2.

      对任意的0 < λ < $\left[\frac{a^{2} S^{2}}{4 D\left(\mu-b S^{2}\right)}\right]^{\frac{1}{2}}$,有$\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}$-2>0. ∀t≥0,定义

    $ I\left( {t{u_\varepsilon }} \right) = \frac{a}{2}{t^2}{\left\| {{u_\varepsilon }} \right\|^2} + \frac{b}{4}{t^4}{\left\| {{u_\varepsilon }} \right\|^4} - \frac{{\mu {t^4}}}{4}\left| {{u_\varepsilon }} \right|_4^4 - \lambda t\int_\mathit{\Omega } f (x){u_\varepsilon }{\rm{d}}x $

    进一步,根据(18)式,可得

    $ I\left( {t{u_\varepsilon }} \right) \le \frac{a}{2}{t^2}{\left\| {{u_\varepsilon }} \right\|^2} + \frac{b}{4}{t^4}{\left\| {{u_\varepsilon }} \right\|^4} - \frac{{\mu {t^4}}}{4}\left| {{u_\varepsilon }} \right|_4^4 \le a{S^2}{t^2} - \frac{{\left( {\mu - b{S^2}} \right){S^2}}}{8}{t^4} $ (19)

    因此,类似于文献[3]中引理2.3的证明可得,存在tε>0使得$\sup _{t \geqslant 0}$ I(tuε)=I(tεuε),且存在与ε无关的正常数t0T0使得t0 < tε < T0.令

    $ {I_\varepsilon }\left( t \right) = \frac{a}{2}{t^2}{\left\| {{u_\varepsilon }} \right\|^2} + \frac{b}{4}{t^4}{\left\| {{u_\varepsilon }} \right\|^4} - \frac{{\mu {t^4}}}{4}\left| {{u_\varepsilon }} \right|_4^4 $

    则有I′ε(t)=atuε2+bt3uε4-μ t3|uε|44.令I′ε(t)=0,则

    $ a{\left\| {{u_\varepsilon }} \right\|^2} + b{t^2}{\left\| {{u_\varepsilon }} \right\|^4} - \mu {t^2}\left| {{u_\varepsilon }} \right|_4^4 = 0 $ (20)

    可得Tε2=$\frac{{a{{\left\| {{u_\varepsilon }} \right\|}^2}}}{{\mu \int_\mathit{\Omega } {u_\varepsilon ^4} {\rm{d}}x - b{{\left\| {{u_\varepsilon }} \right\|}^4}}}$.因此,对∀0 < t < TεI′ε(t)>0.而当t>Tε时有I′ε(t) < 0,且Iε(t)在Tε处达到最大值.从而,根据(20)式,可得

    $ {I_\varepsilon }\left( t \right) \le {I_\varepsilon }\left( {{T_\varepsilon }} \right) = \frac{a}{4}{\left\| {{u_\varepsilon }} \right\|^2}T_\varepsilon ^2 = \frac{{{a^2}{{\left[ {{S^2} + O\left( {{\varepsilon ^2}} \right)} \right]}^2}}}{{4\left[ {\mu {{\left( {S + O\left( {{\varepsilon ^4}} \right)} \right)}^2} - b{{\left( {{S^2} + O\left( {{\varepsilon ^2}} \right)} \right)}^2}} \right]}} = \frac{{{a^2}{S^2}}}{{4\left( {\mu - b{S^2}} \right)}} + O\left( {{\varepsilon ^2}} \right) $ (21)

    根据条件(F)以及uε的定义,对∀0 < ε < $\sqrt{\delta}$,可得

    $ \begin{array}{l} \int_\mathit{\Omega } f (x){u_\varepsilon }{\rm{d}}x \ge C\int_{\left| x \right| < \delta } {\frac{{\varepsilon {{\left| x \right|}^{ - \beta }}}}{{{\varepsilon ^2} + {{\left| x \right|}^2}}}{\rm{d}}x} + \int_{\left| x \right| \ge \Delta } f (x){u_\varepsilon }{\rm{d}}x \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;C\varepsilon \int_0^\delta {\frac{{{r^3}}}{{{r^\beta }\left( {{\varepsilon ^2} + {r^2}} \right)}}{\rm{d}}r} = C{\varepsilon ^{2 - \beta }}\int_0^{\frac{\delta }{\varepsilon }} {\frac{{{t^3}}}{{{t^\beta }\left( {1 + {t^2}} \right)}}{\rm{d}}t} \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;C{\varepsilon ^{2 - \beta }}\left[ {\int_0^1 {\frac{{{t^3}}}{{2{t^\beta }}}{\rm{d}}t} + \int_1^{\frac{\delta }{\varepsilon }} {\frac{{{t^3}}}{{2{t^{\beta + 2}}}}{\rm{d}}t} } \right] \ge C{\varepsilon ^{2 - \beta }} \end{array} $ (22)

    其中C表示不同的正常数.因此,根据(21)式和(22)式,可得

    $ I\left( {t{u_\varepsilon }} \right) \le {I_\varepsilon }\left( {{t_\varepsilon }} \right) - \lambda {t_\varepsilon }\int_\mathit{\Omega } f (x){u_\varepsilon }{\rm{d}}x \le {I_\varepsilon }\left( {{T_\varepsilon }} \right) - \lambda {t_0}\int_\mathit{\Omega } f (x){u_\varepsilon }{\rm{d}}x \le \frac{{{a^2}{S^2}}}{{4\left( {\mu - b{S^2}} \right)}} + {C_1}{\varepsilon ^2} - {C_2}\lambda {\varepsilon ^{2 - \beta }} $

    其中C1C2>0为正常数.取λ=ε且0 < λ < $\left(\frac{C_{2}}{C_{1}+D}\right)^{\frac{1}{\beta-1}}$,由于1 < β < 2,可得C1ε2-C2λε2-β=λ2(C1-C2λ1-β) < -2.取

    $ {\lambda ^*} = \min \left\{ {{{\left[ {\frac{{{a^2}{S^2}}}{{4D\left( {\mu - b{S^2}} \right)}}} \right]}^{\frac{1}{2}}},{{\left( {\frac{{{C_2}}}{{{C_1} + D}}} \right)}^{\frac{1}{{\beta - 1}}}}} \right\} $

    当0 < λ < λ*时,有I(tuε) < $\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}$-2.因此,取u0=uε,当0 < λ < λ*时,有$\mathop {\sup }\limits_{t \ge 0}$≥0 I(tu0) < $\frac{a^{2} S^{2}}{4\left(\mu-b S^{2}\right)}$-2.引理2证毕.

    定理2  假设ab>0,N=4,μ>bS2,$f \in L^{\frac{4}{3}}$(Ω)是非零非负函数,且满足条件(F),则存在λ**>0(λ**λ*),使得对∀0 < λ < λ**,方程(1)还存在一个正山路解u**,且I(u**)>0.

      取λ**=min{λ*λ*},则对∀0 < λ < λ**,定理1以及引理1、引理2均成立.从而,根据(6)式和(19)式,容易得到泛函IH01(Ω)上满足山路几何结构.即当uSR时,有I(u)|SR>ρ>0;存在eH01(Ω)使得‖e‖>RI(e) < 0.定义

    $ \mathit{\Gamma } = \left\{ {\gamma \in C\left( {\left[ {0,1} \right],H_0^1\left( \mathit{\Omega } \right)} \right):\gamma \left( 0 \right) = 0,\gamma \left( 1 \right) = e} \right\} $
    $ c = \mathop {\inf }\limits_{\gamma \in \mathit{\Gamma }} \mathop {\max }\limits_{t \in \left[ {0,1} \right]} I\left( {\gamma \left( t \right)} \right) $

    根据引理1和引理2,存在{un}⊂H01(Ω)使得I(un)→c>ρ>0且I′(un)→0,则序列{un}在H01(Ω)中存在收敛子列(仍记为{un}).不妨假设在H01(Ω)中unu**.根据山路定理,可得$\lim\limits_{n \rightarrow \infty}$ I(un)=I(u**)=c>0且I′(u**)=0.因此u**是方程(1)的非零解.根据u*的证明,同理可证,u**是方程(1)的正解.

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    Positive Solutions for a Class of Nonhomogeneous Kirchhoff Type Equation with Critical Exponent in High Dimension
    JI Lei     
    College of Science, Jingzhong University, Jingzhong Shanxi 030619, China
    Abstract: A class of nonhomogeneous Kirchhoff type equations with critical exponent is considered in high dimension. By the variational method, when N=4, two positive solutions are obtained; while N>4, the existence of positive solutions is obtained. The results complete and improve some results of the correlative references.
    Key words: Kirchhoff type equation    critical exponent    variational method    positive solution    
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