本文研究如下Kirchhoff型方程

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} { } -\left(a+b \int_{{{\Bbb R}} ^{3}} {|\nabla u{{|}^{2}}}\right) \triangle u+u = \left(1+\varepsilon g(x)\right)u^{p}, x \in {{\Bbb R}} ^{3}, \\ u \in H^{1}\left({{\Bbb R}} ^{3}\right), \end{array} \right. \end{equation} $

其中$ \varepsilon $, $ a $, $ b $,都是正常数, $ 1<p<5 $, $ g(x) \in L^{\infty}\left({{\Bbb R}} ^{3}\right) $.

当$ \varepsilon = 0 $时,方程(1.1)就为如下Kirchhcoff方程

(1.2) $ \begin{equation} \left\{ \begin{array}{ll} { } -\left(a+b \int_{{{\Bbb R}} ^{3}} {|\nabla u{{|}^{2}}}\right) \triangle u+u = u^{p}, x \in {{\Bbb R}} ^{3}, \\ u \in H^{1}\left({{\Bbb R}} ^{3}\right), \end{array} \right. \end{equation} $

这类方程是由Kirchhoff在文献[3]中提出的关于时间$ t $的波动方程

(1.3) $ \begin{equation} \rho\frac{\partial^{2}z}{\partial t^{2}}-\bigg(\frac{P_{0}}{h}+\frac{E}{2L}\int^{L}_{0}\bigg|\frac{\partial z}{\partial x}\bigg|^{2}\bigg)\frac{\partial^{2}z}{\partial x^{2}} = 0 . \end{equation} $

该方程是考虑横向振动引起的弦长变化拓展了弹性弦自由振动的经典D'Alembert's波动方程,其中$ \rho $表示物质密度, $ P_{0} $表示初始张力, $ h $是横截面面积, $ L $是弦长, $ E $表示材料的杨氏模量.Kirchhoff型方程在非牛顿力学、宇宙物理、血浆问题和弹性理论等许多领域都有广泛的应用,因此研究这类问题具有很强的实际意义.文献[6-7]最早研究了这类方程,然后Lions在文章[5]中提出了一个抽象的问题框架, Kirchhoff型方程得到广泛的关注及研究.这类方程最显著的特征是方程中含有$ \left(\int_{{{\Bbb R}} ^{3}} {|\nabla u{{|}^{2}}}\right) \triangle u $项,这样方程不仅依赖$ \triangle u $,而且依赖于$ |\nabla u|^{2} $在整个区间上的积分,因此方程(1.1)不再是点态方程,所以这类问题也具有很强的数学理论研究价值.

通常运用变分方法求问题(1.1)的解,也就是研究相应变分泛函

$ f_{\varepsilon}\left(u\right) = \frac{1}{2}\int_{{{\Bbb R}} ^{3}} \left(a|\nabla u{{|}^{2}}+u^{2} \right){\rm d}x+\frac{b}{4}\left(\int_{{{\Bbb R}} ^{3}}|\nabla u{{|}^{2}{\rm d}x}\right)^{2} -\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(1+\varepsilon g(x))u^{p+1}{\rm d}x $

的临界点.

由于区域的无界性导致Sobolev嵌入紧性的消失以及泛函$ f_{\varepsilon}\left(u\right) $自身的特征,应用山路引理得到上述泛函临界点的存在性比较困难.本文运用Ambrosetti在文章[1]中提出的扰动方法,证明了一定条件下问题(1.1)的多解存在性结果.

定理1.1 假设$ g(x)\in L^{\infty}({{\Bbb R}} ^{3}), \quad g(x)\geq 0 $及$ \lim\limits_{x\rightarrow\infty}g(x) = 0 $,且满足以下条件

(1)存在常数$ T, r>0 $,序列$ \left\{P_{k}\right\}_{k\in {\mathbb R}^3} $, $ \left\{\alpha_{k}\right\}_{k\in {{\Bbb R}} } $,使得

$ \alpha_{k}>0, |P_{k}|\rightarrow\infty, \ \mbox{且} g(x)\geq\alpha_{k}, \forall x\in {B}(P_{k}, r), g(x)\leq T\alpha_{k}, \forall|x|\geq\frac{P_{k}}{2}; $

(2)存在序列$ \left\{r_{k}\right\}_{k} $, $ \left\{R_{k}\right\}_{k} $, $ \left\{\beta_{k}\right\}_{k} $,使得$ r_{k} $, $ R_{k}>0 $, $ 0<r_k<R_k<|P_k| $及$ \beta_{k}\geq0 $, $ r_{k}\rightarrow +\infty $, $ R_{k}-r_{k}\rightarrow +\infty $,且

$ g(x)\leq\beta_{k}, \forall x\in \sum\limits_{k} = \left\{x\in {{\Bbb R}} ^{3}|r_{k}\leq|x|\leq R_{k}\right\}; $

(3)存在$ \mu>1 $, $ \delta\in (0, p+1) $, $ C>0 $,使得对充分大的$ k $满足以下条件

$ \alpha_{k}>\mu C_{3}\beta_{k}+Ce^{-\delta\frac{R_{k}-r_{k}}{2}}, $

这里$ C_{3} $为常数,我们在文后给出.

则存在正数$ \varepsilon_{0} $,当$ 0<\varepsilon<\varepsilon_{0} $,问题(1.1)存在无穷多个解.

在本节中,我们简单介绍扰动方法及预备引理.首先在问题(1.1)中,若$ \varepsilon = 0 $则其转化为了以下问题

(2.1) $ \begin{equation} \left\{ \begin{array}{l}{ } -\left(a+b \int_{{{\Bbb R}} ^{3}} {|\nabla u{{|}^{2}}}\right) \triangle u+u = u^{p}, x \in {{\Bbb R}} ^{3}, \\ u \in H^{1}\left({{\Bbb R}} ^{3}\right), \end{array} \right. \end{equation} $

因此问题(1.1)可看成是问题(2.1)的一个扰动,由变分方法知问题(2.1)的解对应于泛函

$ f_{0}\left(u\right) = \frac{1}{2}\int_{{{\Bbb R}} ^{3}} \left(a|\nabla u{{|}^{2}}+u^{2} \right){\rm d}x+\frac{b}{4}\left(\int_{{{\Bbb R}} ^{3}}|\nabla u{{|}^{2}{\rm d}x}\right)^{2} -\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}u^{p+1}{\rm d}x $

的临界点.

令

$ G\left(u\right) = \int_{{{\Bbb R}} ^{3}}g(x)u^{p+1}{\rm d}x, $

则问题(1.1)的解对应泛函

$ f_{\varepsilon}\left(u\right) = f_{0}\left(u\right) - \frac{\varepsilon}{p+1} G\left(u\right). $

由问题(2.1)的特征,我们知$ f_{0}\left(u\right) $具有以下性质

(1) $ f_{0}\left(u\right)\in {\cal C}^{2}\left(H^{1}\left({{\Bbb R}} ^{3}\right), {{\Bbb R}} \right) $,其极小能量解集是一个非退化的流形$ Z $.

(2)对于任意的$ z\in Z $,线性算子$ f''_{0}(z) $是紧算子,并且$ T_{z}Z = \ker[D^{2}f_{0}(z)] $.

由于问题(1.1)的解就是$ f_{\varepsilon}(u) $的临界点,扰动方法就是把研究$ f_{\varepsilon}(u) $的临界点,转化为求$ f_{\varepsilon}(u) $在流形$ Z $上的约束极值,因此我们构造$ Z $的扰动流形$ \overline{Z} $,并且证明了$ f_{\varepsilon}(u) $限制在流形$ \overline{Z} $的临界点就是$ f_{\varepsilon}(u) $在$ H^{1}\left({{\Bbb R}} ^{3}\right) $上的临界点,这样我们就得到相应问题(1.1)的解.

为了证明本文的主要结果,我们首先引用以下几个引理.

引理2.1 (正对称解的唯一性) 存在唯一的关于坐标原点对称的正对称解$ z\in H^{1}({{\Bbb R}} ^{3}) $满足以下方程

$ -\left(a+b \int_{{{\Bbb R}} ^{3}} {|\nabla U{{|}^{2}}}\right) \triangle U+U = U^{p}, U>0 $

并且这个唯一正径向对称解$ z $具有指数衰减的性质,即

$ z\left(x\right)+|\nabla z\left(x\right)|\leq C e^{-\sigma_{0}|x|}, x\in {{\Bbb R}} ^{3}, $

其中$ \sigma_{0}>0, C>0 $.

我们定义线性算子$ L:L^{2}({{\Bbb R}} ^{3}) \rightarrow L^{2}({{\Bbb R}} ^{3}) $

$ L\upsilon = -\left(a+b \int_{{{\Bbb R}} ^{3}} {|\nabla z{{|}^{2}}}\right) \triangle \upsilon-2b\left(\int_{{{\Bbb R}} ^{3}}\nabla z\cdot\nabla \upsilon\right)\triangle z+\upsilon-pz^{p-1}\upsilon, $

对于所有$ \upsilon\in L^{2}({{\Bbb R}} ^{3}) $.则有以下结论.

引理2.2 对于引理2.1中的正径向对称解$ z $在$ H^{1}({{\Bbb R}} ^{3}) $中是非退化的,并且对于上面我们定义的线性算子$ L $的核空间为

$ {\rm Ker}(L) = {\rm span}\left\{\partial_{x_{1}}z, \partial_{x_{2}}z, \partial_{x_{3}}z\right\}. $

这两个引理的证明参见文献[4].

事实上,若$ z(x) $是问题$ (2.1) $关于坐标原点对称的正解,则

$ Z = \left\{z_{\xi} = z(x-\xi)|\xi \in {{\Bbb R}} ^{3}\right\} $

是问题$ (2.1) $相应泛函$ f_{0} $的一个非退化的极值流形,也就是说$ Z $在$ z $点的切空间就是线性算子$ L $的核空间,即$ T{_{z}}Z = {\rm Ker}(L) $.所以我们需要在$ Z $附近构造与它局部微分同胚的流形$ \overline{Z} $,那么$ \overline{Z} $就是$ f_{\varepsilon}(u) $的一个自然约束条件.

为了寻找$ f_{\varepsilon} $具有$ z+w $形式的临界点,其中$ w\in \left(T{_{z}}Z\right)^{\bot} $,我们给出以下引理.

引理2.3 存在$ \varepsilon_{0}>0 $,当$ 0< \varepsilon < \varepsilon_{0} $,有$ w_{\varepsilon}\in \left(T{_{z}}Z\right)^{\bot} $,且$ \nabla f_{\varepsilon}(z+w_{\varepsilon})\in T{_{z}}Z $.

证 首先对任意给定的方程(2.1)的解$ z $,我们考虑方程

$ \nabla f_{\varepsilon}(z+w) = 0, $

即

(2.2) $ \begin{equation} \nabla f_{0}(z+w)-\varepsilon\nabla G(z+w) = 0, \quad w\in(T_{z}Z)^{\bot}, \end{equation} $

(2.3) $ \begin{equation} \nabla f_{0}(z)+D^{2}f_{0}(z)[w]-\varepsilon\nabla G(z+w)+R(w) = 0. \end{equation} $

因为$ \nabla f_{0}(z) = 0 $,所以方程(2.3)可写为

(2.4) $ \begin{equation} D^{2}f_{0}(z)[w]-\varepsilon\nabla G(z+w)+R(w) = 0, \end{equation} $

这里$ R(w) = o(||w||), \ ||w||\rightarrow0 $.

为了证明引理,我们把方程(2.4)投影到$ (T_{z}Z)^{\bot} $,得到一个辅助方程

(2.5) $ \begin{equation} D^{2}f_{0}(z)[w]-\varepsilon P_{z}\nabla G(z+w)+P_{z}R(w) = 0, \end{equation} $

这里$ P_{z} $是$ (T_{z}Z) $投影算子,由于$ z $是$ f_{0} $的非退化临界点,因此线性算子$ L_{z} = D^{2}f_{0}(z) $在$ (T_{z}(Z))^{\bot} $上是可逆算子.于是我们可以应用隐函数存在性定理证明方程(2.5)对于适当的$ \varepsilon_{0}>0 $,任意的$ z\in Z $存在唯一的$ w_{\varepsilon}(z)\in (T_{z}Z)^{\bot} $,并且$ w_{\varepsilon}(z) $关于$ z $是属于$ C^{1} $.及当$ \varepsilon\rightarrow 0 $时,有

$ ||w_{\varepsilon}(z)||\rightarrow0, \; \; ||D_{z}w_{\varepsilon}(z)||\rightarrow0. $

引理2.3证毕.

定义2.1 $ \overline{Z} = \left\{z_{\xi}+w_{\varepsilon}, _{\xi}|\xi\in {{\Bbb R}} ^{3}\right\} $.

引理2.4 对充分小的$ \varepsilon $, $ \overline{Z} $是一个3 -维流形,且与$ Z $局部微分同胚.如果$ u\in \overline{Z} $且$ \nabla f_{\varepsilon |\overline{Z}}(u) = 0 $,则$ \nabla f_{\varepsilon}(u) = 0 $.

证 令$ u\in\overline{Z} $是$ f_{\varepsilon} $限制在$ \overline{Z} $上的极值点,即$ \nabla f_{\varepsilon}(u)\in (T_{u}Z)^{\bot} $;另一方面,由$ \overline{Z} $的定义和$ T_{z}Z = \ker(D^{2}f_{0}(z)) $,有$ \nabla f_{\varepsilon}(u)\in T_{z}Z $.又因为当$ \varepsilon $充分小时, $ ||w_{\varepsilon}(z)||\rightarrow0, ||D_{z}w_{\varepsilon}(z)||\rightarrow0 $,所以$ (T_{u}Z)^{\bot} $与$ T_{z}Z $非常接近,因此$ \nabla f_{\varepsilon}(u) = 0 $.

记:$ \Phi_{\epsilon}(z)\triangleq f_{\varepsilon}(z+w_{\varepsilon}(z)) $.

对于上述泛函$ \Phi_{\epsilon}(z) $我们可以写成如下形式

$ \Phi_{\varepsilon}(z) = f_{\varepsilon}(z+w_{\varepsilon}(z)) = C_{0}-\frac{\varepsilon}{p+1}G(z)+o(\varepsilon). $

事实上,由于$ z $是方程(2.1)的解,所以有

$ \frac{1}{2}\int_{{{\Bbb R}} ^{3}}[a|\nabla z|^{2}+z^{2}]{\rm d}x+\frac{b}{4}\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z)| ^{2}{\rm d}x\bigg)^{2}-\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}z^{p+1}{\rm d}x = C_{0}, $

从而

$ \int_{{{\Bbb R}} ^{3}}[a\nabla z\nabla w_{\varepsilon}(z)+zw_{\varepsilon}(z)]{\rm d}x = \int_{{{\Bbb R}} ^{3}}z^{p}w_{\varepsilon}(z){\rm d}x-b\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\int_{{{\Bbb R}} ^{3}} \nabla z\nabla w_{\varepsilon}(z){\rm d}x, $

$ \begin{eqnarray*} f_{\varepsilon}(z+w_{\varepsilon}(z))& = &\frac{1}{2}\int_{{{\Bbb R}} ^{3}}(a|\nabla(z+w_{\varepsilon})|^{2} +(z+w_{\varepsilon})^{2}){\rm d}x\\ && +\frac{b}{4}\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z+w_{\varepsilon})|^{2}{\rm d}x\bigg)^{2} -\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(1+\varepsilon g(x))(z+w_{\varepsilon})^{p+1}{\rm d}x \\ & = &\frac{1}{2}\int_{{{\Bbb R}} ^{3}}[a|\nabla z|^{2}+2a\nabla z\nabla w_{\varepsilon}+a|\nabla w_{\varepsilon}|^{2}+z^{2}+2zw_{\varepsilon}+w_{\varepsilon}^{2}]{\rm d}x \\ & &+\frac{b}{4}\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z+w_{\varepsilon})|^{2}{\rm d}x\bigg)^{2} -\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(1+\varepsilon g(x))(z+w_{\varepsilon})^{p+1}{\rm d}x \\ &&-\frac{b}{4}\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\bigg)^{2}+\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}z^{p+1}{\rm d}x \\ & &+\frac{b}{4}\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z+w_{\varepsilon})|^{2}{\rm d}x\bigg)^{2} -\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(1+\varepsilon g(x))(z+w_{\varepsilon})^{p+1}{\rm d}x\\ & = &\frac{1}{2}\int_{{{\Bbb R}} _{3}}[a|\nabla z|^{2}+z^{2}]{\rm d}x+\frac{b}{4} \bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z)|^{2}{\rm d}x\bigg)^{2}-\frac{1}{p+1}\int_{{{\Bbb R}} _{3}}z^{p+1}{\rm d}x \\ &&+\int_{{{\Bbb R}} ^{3}}[a\nabla z\nabla w_{\varepsilon}+zw_{\varepsilon}]{\rm d}x+\frac{1}{2} \int_{{{\Bbb R}} ^{3}}(a|\nabla w_{\varepsilon}|^{2}+w_{\varepsilon^{2}}){\rm d}x \\ &&+\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(z^{p+1}-(z+w_{\varepsilon})^{p+1}){\rm d}x\\ &&+\frac{b}{4} \bigg[\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z+w_{\varepsilon})|^{2}{\rm d}x\bigg)^{2}- \bigg(\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\bigg)^{2}\bigg] \\ &&-\frac{\varepsilon}{p+1}\int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}{\rm d}x+\frac{\varepsilon}{p+1}\int_{{{\Bbb R}} ^{3}}g(x) [z^{p+1}-(z+w_{\varepsilon})^{p+1}]{\rm d}x \\ & = &C_{0}+\int_{{{\Bbb R}} ^{3}}z^{p}w_{\varepsilon}{\rm d}x-b\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x \int_{{{\Bbb R}} ^{3}}\nabla z\nabla w_{\varepsilon}{\rm d}x \\ &&+\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(z^{p+1}-(z+w_{\varepsilon})^{p+1}){\rm d}x \\ &&+\frac{b}{4} \bigg[\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla(z+w_{\varepsilon})|^{2}{\rm d}x\bigg)^{2}- \bigg(\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\bigg)^{2}\bigg] \\ &&-\frac{\varepsilon}{p+1}\int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}{\rm d}x-\varepsilon\int_{{{\Bbb R}} ^{3}}g(x)z^{p}w_{\varepsilon}{\rm d}x\\ & = &C_{0}+\int_{{{\Bbb R}} ^{3}}z^{p}w_{\varepsilon}{\rm d}x-b\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x \int_{{{\Bbb R}} ^{3}}\nabla z\nabla w_{\varepsilon}{\rm d}x \\ &&+\frac{1}{p+1}\int_{{{\Bbb R}} ^{3}}(z^{p+1}-(z+w_{\varepsilon})^{p+1}){\rm d}x\\ && +\frac{b}{4} \bigg[\bigg(\int_{{{\Bbb R}} ^{3}}|\nabla z+\nabla w_{\varepsilon}|^{2}{\rm d}x +\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\bigg)\\ &&\times \bigg(\int_{{{\Bbb R}} ^{3}}|\nabla z +\nabla w_{\varepsilon}|^{2}{\rm d}x -\int_{{{\Bbb R}} ^{3}}|\nabla z|^{2}{\rm d}x\bigg)\bigg]\\ &&-\frac{\varepsilon}{p+1}\int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}{\rm d}x-\varepsilon\int_{{{\Bbb R}} ^{3}}g(x)z^{p}w_{\varepsilon}{\rm d}x \\ & = &C_{0}-\frac{\varepsilon}{p+1}\int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}{\rm d}x+o(\varepsilon) \\ & = &C_{0}-\frac{\varepsilon}{p+1}G(z)+o(\varepsilon). \end{eqnarray*} $

在本节我们运用前面的引理对本文的主要结果进行证明,为了求问题(1.1)的解,就是求相应的变分泛函的临界点,由扰动方法及上述引理可知,就是求$ f_{\varepsilon} $在自然约束条件$ \overline{Z} $上的极值点.由$ \Phi_{\varepsilon}(z_{\xi}) = C_{0}-\frac{\varepsilon}{p+1}G(z_{\xi})+o(\varepsilon) $,可知,为了得到问题(1.1)的解就只需要求$ G(z_{\xi}) $在$ \xi \in {{\Bbb R}} ^{3} $的临界点.

定理3.1 若$ g(x)\in L^{\infty}({{\Bbb R}} ^{3}) $,且$ g(x)\geq0, \lim\limits_{|x| \to \infty}g(x) = 0 $,则$ \lim\limits_{|\xi| \to \infty}G(z_{\xi}) = 0 $.

证 因为

$ \begin{eqnarray*} G(z_{\varepsilon}) = \int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}(x-\xi){\rm d}x = \int_{{{\Bbb R}} ^{3}}g(x+\xi)z^{p+1}(x){\rm d}x = G_{1}(R)+G_{\infty}(R), \end{eqnarray*} $

这里

$ \begin{eqnarray*} G_{1}(R) = \int_{|x|<R}g(x)z^{p+1}(x-\xi){\rm d}x, \; G_{\infty}(R) = \int_{|x|\geq R}g(x)z^{p+1}(x-\xi){\rm d}x. \end{eqnarray*} $

由Hölder不等式,可得

$ \begin{eqnarray*} |G_{\infty}(R)|&\leq&\bigg(\int_{|x|\geq R}|g(x)|{\rm d}x\bigg)\bigg(\int_{|x-\xi|\geq R}z^{p+1}(x){\rm d}x\bigg) \\ & \leq&\bigg(\int_{|x|\geq R}|g(x)|{\rm d}x\bigg)\bigg(\int_{{{\Bbb R}} ^{3}}z^{5}{\rm d}x\bigg) \\ &\leq &C_{1}\int_{|x|\geq R}|g(x)|{\rm d}x. \end{eqnarray*} $

因为$ g(x)\in {L}^{\infty}({{\Bbb R}} ^{3}) $,对于给定的$ \eta>0 $,存在$ R>0 $,我们有

$ |G_{\infty}(R)|\leq C_{1}\int_{|x|\geq R}|g(x)|{\rm d}x\leq\eta. $

对于$ G_{1}(R) $,再次利用H$ \ddot{\rm o} $lder不等式,可得

$ \begin{eqnarray*} |G_{1}(R)|& = &\int_{|x|<R}g(x)z^{p+1}(x-\xi){\rm d}x \\ &\leq& C_{2} \int_{|x|<R}z^{p+1}(x-\xi){\rm d}x \\ &\leq& C_{2}\int_{|x+\xi|<R}z^{5}(x){\rm d}x, \end{eqnarray*} $

这里$ C_{2} $是只依赖于$ g(x) $的常数, $ z $是问题(1.1)的唯一径向对称解,所以对于任意的$ R $,有

$ \lim\limits_{|\xi| \to \infty}G_{1}(R) = 0, $

因此我们有

$ \lim\limits_{|\xi| \to \infty}G(z_{\xi}) = 0. $

定理3.1证毕.

定理3.2 假设$ g(x)\in L^{\infty}({{\Bbb R}} ^{3}), \ g(x)\geq0 $,且满足以下条件

$ \rm(1) $存在常数$ T, r>0 $,序列$ \left\{P_{k}\right\}_{k\in {\mathbb R}^{3}} $, $ \left\{\alpha_{k}\right\}_{k\in {\mathbb R}} $,使得

$ \alpha_{k}>0, |P_{k}|\rightarrow\infty, \ \mbox{且}\ g(x)\geq\alpha_{k}, \forall x\in {B}(P_{k}, r), g(x)\leq T\alpha_{k}, \forall|x|\geq\frac{P_{k}}{2}; $

$ \rm(2) $存在序列$ \left\{r_{k}\right\}_{k} $, $ \left\{R_{k}\right\}_{k} $, $ \left\{\beta_{k}\right\}_{k} $,使得$ r_{k} $, $ R_{k}>0 $, $ 0<r_k<R_k<|P_k| $及$ \beta_{k}\geq0 $, $ r_{k}\rightarrow +\infty $, $ R_{k}-r_{k}\rightarrow +\infty $,且

$ g(x)\leq\beta_{k}, \forall x\in \sum\limits_{k} = \left\{x\in {{\Bbb R}} ^{3}|r_{k}\leq|x|\leq R_{k}\right\}; $

$ \rm(3) $存在$ \mu>1 $, $ \delta\in (0, p+1) $, $ C>0 $,使得对充分大的$ k $满足以下条件

$ \alpha_{k}>\mu C_{3}\beta_{k}+Ce^{-\delta\frac{R_{k}-r_{k}}{2}}, $

这里$ C_{3} = \frac{C_{2}}{C_{1}} $, $ C_{1} = \int_{|x|<r}z^{p+1}(y){\rm d}y $, $ C_{2} = \int_{{{\Bbb R}} ^{3}}z^{p+1}(y){\rm d}y $,则$ G(\xi) $存在无穷多个局部极大值.

证 由条件(1)我们有

$ \begin{eqnarray*} G(P_{k})& = &\int_{{{\Bbb R}} ^{3}}g(x)z^{p+1}(x-P_{k}){\rm d}x \geq\alpha_{k}\int_{B(P_{k}, r)}z^{p+1}(x-P_{k}){\rm d}x \\ & = &\alpha_{k}\int_{B(0, r)}z^{p+1}(x){\rm d}x = C_{1}\alpha_{k}. \end{eqnarray*} $

另一方面,对于任意$ \xi\in {{\Bbb R}} ^{3} $,满足$ |\xi| = \frac{R_{k}+r_{k}}{2} $,则

$ \begin{eqnarray*} G(\xi)& = &\int_{\sum\limits_{k}}z^{p+1}(x-\xi)g(x){\rm d}x+\int_{|x|<r_{k}}z^{p+1}(x-\xi)g(x){\rm d}x+\int_{|x|>R_{k}}z^{p+1}(x-\xi)g(x){\rm d}x \\ &\leq&\beta_{k}\int_{\sum\limits_{k}}z^{p+1}(x-\xi){\rm d}x+C\int_{|x|<r_{k}}z^{p+1}(x-\xi){\rm d}x+C\int_{|x|>R_{k}}z^{p+1}(x-\xi){\rm d}x. \end{eqnarray*} $

由于

$ \int_{\sum\limits_{k}}z^{p+1}(x-\xi){\rm d}x\leq\int_{{{\Bbb R}} ^{3}}z^{p+1}(x-\xi){\rm d}x = C_{2}, $

$ \int_{|x|<r_{k}}z^{p+1}(x-\xi){\rm d}x = \int_{|y+\xi|<r_{k}}z^{p+1}(y){\rm d}y, $

由$ |y+\xi|<r_{k} $和$ |\xi| = \frac{R_{k}+r_{k}}{2} $有

$ |y| = |y+\xi-\xi|\geq|-\xi|-|y+\xi|\geq\frac{R_{k}+r_{k}}{2}-r_{k} = \frac{R_{k}-r_{k}}{2}, $

所以

$ \int_{|y+\xi|<r_{k}}z^{p+1}(y){\rm d}y\leq\int_{|y|\geq\frac{R_{k}-r_{k}}{2}}z^{p+1}(y){\rm d}x = O(e^{-\delta_{1}\frac{R_{k}-r_{k}}{2}}), \quad \delta<\delta_{1}<p+1. $

由$ |y+\xi|>R_{k} $与$ |\xi| = \frac{R_{k}+r_{k}}{2} $有

$ |y| = |y+\xi-\xi|\geq|y+\xi|-|\xi|\geq R_{k}-\frac{R_{k}+r_{k}}{2} = \frac{R_{k}-r_{k}}{2}, $

所以

$ \int_{|x|>R_{k}}z^{p+1}(x-\xi){\rm d}x = \int_{|y+\xi|>R_{k}}z^{p+1}(y){\rm d}y\leq\int_{|y|\geq\frac{R_{k}-r_{k}}{2}}z^{p+1}(y){\rm d}y = O(e^{-\delta_{1}\frac{R_{k}-r_{k}}{2}}). $

由上面可以得出

$ G(\xi)\leq C_{2}\beta_{k}+Ce^{-\delta_{1}\frac{R_{k}-r_{k}}{2}}. $

而$ G(P_{k})\geq C_{1}\alpha_{k} $由条件(3)知$ \alpha_{k}>\mu C_{3}\beta_{k}+Ce^{-\delta\frac{R_{k}-r_{k}}{2}} $, $ C_{3} = \frac{C_{2}}{C_{1}} $,所以

$ G(P_{k})\geq C_{1}\alpha_{k}>\mu C_{3}\beta_{k}+Ce^{-\delta\frac{R_{k}-r_{k}}{2}}>G(\xi). $

因此当$ |\xi| = \frac{R_{k}+r_{k}}{2} $时,有$ G(P_{k})\geq G(\xi) $.所以$ G(\xi) $有无穷多个局部极大值.

定理1.1的证明 由前面的引理可知,问题(1.1)的解就是其能量泛函$ f_{\varepsilon}(z) $的临界点,由扰动方法知泛函$ f_{\varepsilon}(z) $的临界点问题就转化为了研究

$ \Phi_{\varepsilon}(z_{\xi}) = C_{0}-\frac{\varepsilon}{p+1}G(z_{\xi})+o(\varepsilon). $

由前面的所讨论的内容知问题(1.1)的解就只要求$ G(z_{_{\xi}}) $在$ \xi \in {{\Bbb R}} ^{3} $上的临界点.而由定理3.1、3.2,我们可以得出在定理(1.1)的条件下$ G(z_{_{\xi}}) $在$ \xi \in {{\Bbb R}} ^{3} $时存在无穷多个临界点,因此问题(1.1)存在无穷多个解存在.