本文主要研究以下非线性Kirchhoff问题的单峰解的局部唯一性

(1.1) $ \begin{equation} -\Big(\epsilon^2 a+ \epsilon b \int_{{{\Bbb R}} ^3} |\nabla u|^2{\rm d}x\Big) \Delta u + u = K(x)|u|^{p-1}u, u>0, x \in {{\Bbb R}} ^3, \end{equation} $

其中$ \epsilon>0 $任意小, $ a $, $ b>0 $, $ 1<p<5 $, $ K:{{\Bbb R}} ^3 \rightarrow {{\Bbb R}} $是连续有界函数.

最早研究峰解的唯一性的是Glangetas,在文献[1]中, Glangetas利用拓扑度理论证明了Oh在文献[2]中得到的集中解的唯一性.之后,相关的结果相继出现,参见文献[3-4].近几年, Deng、Lin和Yan在文献[5]中利用局部Pohozaev恒等式证明了具有给定数量的曲率问题具有无穷多峰的正解的唯一性.在文献[6]中, Cao、Li和Luo将文献[5]中的结果推广到条件更弱的情形.在文献[7]中, Guo、Peng和Yan利用类似的方法证明了多调和临界方程的多峰解的局部唯一性.最近, Peng、Wang和Yan在文献[8]中利用局部Pohozaev恒等式构造了$ {{\Bbb R}} ^n (n \geq 5) $上带位势的扰动临界Schrödinger方程的多峰解.我们研究的Kirchhoff方程,由于非局部项$ (\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)\Delta u $的存在,使得Pohozaev恒等式的建立比一般的Schrödinger方程更复杂,这使得证明单峰解的局部唯一性更加困难.我们主要参考文献[9-19]中关于有界区域和全空间上的Kirchhoff方程已有的结果.

我们研究方程(1.1)主要有以下两方面原因.一方面,相比于被研究颇多的Schrödinger方程, Kirchhoff方程在峰解唯一性方面的结果较少,我们知道的只有文献[20],这引起了我们极大的研究兴趣.另一方面来自于近来关于带奇异位势的Kirchhoff波动方程方面所取得的一系列进步,特别是Li等^[20]运用局部的Pohozeav恒等式得到的单峰解的唯一性结果,这启发我们运用类似的方法来研究方程(1.1).

本文主要结果如下:

设$ K: {{\Bbb R}} ^3 \rightarrow {{\Bbb R}} $满足下列条件:

$ (K_1):K $是有界函数且是属于$ C^1 $的,且有$ \inf_{{{\Bbb R}} ^n} K>0 $;

$ (K_2):\exists x_0\in {{\Bbb R}} ^3, r_0>0 $,使得当$ 0<|x-x_0|<r_0 $有$ K(x)<K(x_0) $,并且对于某些$ 0<\alpha<1 $来说,有$ K \in C^{\alpha} (\bar{B}_{r_0}(x_0)) $.这说明$ K(x) $在$ x = x_0 $处是$ \alpha $阶Hölder连续的,为不失一般性,我们假设$ x_0 = 0, r_0 = 10, K(x_0) = 1. $

$ (K_3):K \in C^1 ({{\Bbb R}} ^n) $,并且存在$ \ m>1, \delta >0 $使得

$ \left\{ \begin{array}{ll} { } K(x) = K(x_0)-\sum\limits_{i = 1}^{3} {c_i} |x_i - x_{0, i}|^m + O (|x-x_0|^{m+1}), \, \, &x \in B_\delta (x_0), \\ { } \frac{\partial K}{\partial x_i} = -mc_i |x_i - x_{0, i}|^{m-2} (x_i - x_{0, i}) + O(|x-x_0|^m), \, \, &x \in B_\delta (x_0), \end{array} \right. $

其中, $ c_i>0 \in {{\Bbb R}} , (i = 1, 2, 3) $.

定理1.1  假如$ K $满足假设$ (K_1) $、$ (K_2) $和$ (K_3) $.如果$ u^{(i)}_\epsilon(i = 1, 2) $是方程(1.1)的两个形如$ u_\epsilon = U (\frac{x-y_\epsilon}{\epsilon}) + \varphi _ \epsilon $ (见命题2.1)的解,那么当$ \epsilon>0 $且任意小时,有$ u^{(1)}_\epsilon \equiv u^{(2)}_\epsilon . $进一步,我们假设$ u_\epsilon = U_{\epsilon, y_\epsilon} + \varphi_\epsilon $是唯一解,则

$ |y_\epsilon - x_0| = o(\epsilon), \, \, \Vert \varphi_\epsilon \Vert_\epsilon = O(\epsilon^{3/2+m(1-\tau)}), $

其中$ 0<\tau<1 $且$ \tau $任意小.

为了后面证明定理1.1,我们需要在这一节中做一些准备工作.首先我们引入一些本文后面用到的记号.

令$ u_{\epsilon, y}(x) = u(\frac{x-y}{\epsilon}) $,其中$ x \in {{\Bbb R}} ^3, \epsilon>0, y = (y_1, y_2, y_3 ) \in {{\Bbb R}} ^3 $.

记$ \langle u, v \rangle_\epsilon = \int_{{{\Bbb R}} ^3} (\epsilon ^2 a \nabla u \cdot \nabla v + uv){\rm d}x, $其中$ u, v \in H^1 ({{\Bbb R}} ^3) $时,我们记

$ H_\epsilon = \{u \in H^1 ({{\Bbb R}} ^3): \|u_\epsilon\| \equiv \langle u, u \rangle_\epsilon^{1/2} < \infty \}. $

为了证明我们的定理1.1,首先我们需要得到下述单峰解的存在性结果.

命题2.1  令$ a, b>0, 1<p<5 $,如果假设$ K $满足条件$ (K_1), (K_2) $,那么存在$ \epsilon_0>0 $使得对任意$ \epsilon \in (0, \epsilon_0) $,方程(1.1)有下述形式的解

$ u_\epsilon = U (\frac{x-y_\epsilon}{\epsilon}) + \varphi_\epsilon , \; \varphi_\epsilon \in H_\epsilon. $

当$ \epsilon \rightarrow 0 $时, $ y_\epsilon \rightarrow x_0 $,且有$ \|\varphi_\epsilon \|_\epsilon = o(\epsilon^{3/2}) $.

命题2.1可由基于变分法的Lyapunor-Schmindt约化方法来证明,证明方法类似文献[20]中定理1.2的存在性定理的证明.这里我们省略证明过程.

令$ u_\epsilon = U_{\epsilon, y_\epsilon}+\varphi_\epsilon $是方程(1.1)的解,结合命题2.1易得

(2.1) $ \begin{equation} \| u_\epsilon \|_\epsilon = O(\epsilon^{3/2}). \end{equation} $

令$ {\bar{u}}_\epsilon(x) = u_\epsilon(\epsilon x+y_\epsilon) $,当$ {\bar{u}}_\epsilon>0 $时,在$ {{\Bbb R}} ^3 $上解方程

(2.2) $ \begin{equation} -\bigg(a+b \int_{{{\Bbb R}} ^3} | \nabla \bar{u}_\epsilon|^2{\rm d}x\bigg) \triangle \bar{u}_\epsilon + \bar{u}_\epsilon = \bar{K}_\epsilon (x) \bar{u}^p_\epsilon, \end{equation} $

其中$ \bar{K}_\epsilon (x) = K_\epsilon (\epsilon x + y_\epsilon) $.结合(2.1)式得到

(2.3) $ \begin{equation} \int_{{{\Bbb R}} ^3}(a|\nabla \bar{u}_\epsilon|^2 + \bar{u}^2_\epsilon){\rm d}x = \epsilon ^{-3} \|u_\epsilon\| ^2 _\epsilon = O(1). \end{equation} $

由假设$ (K_1 ) $, $ \bar{K}_\epsilon $是关于$ \epsilon $一致有界的,且有

$ \gamma \equiv \sup\limits_{x \in {{\Bbb R}} ^3} \bar{K}_\epsilon (x) > 0, $

则$ \bar{u}_\epsilon $在$ {{\Bbb R}} ^3 $上满足

$ \left\{ \begin{array}{ll} -a \Delta\bar{u}_\epsilon + \bar{u}_\epsilon \leq \gamma \bar{u}^p_\epsilon, \\ { } \sup\limits_{\epsilon} \|\bar{u}_\epsilon\|_{H^1 ({{\Bbb R}} ^3)} \leq C < \infty. \end{array} \right. $

由比较原理,我们有

(2.4) $ \begin{equation} u_\epsilon(x) \leq C e \frac{-\eta \|x-y_\epsilon\|}{\epsilon} , x \in {{\Bbb R}} ^3. \end{equation} $

由贝塞尔位势,我们得到

$ \Delta \bar{u}_\epsilon \leq \frac{\gamma \bar{u}^p_\epsilon}{-a \Delta + 1}. $

当$ p<5 $时,由标准的位势能理论和Morse迭代可以证明$ \bar{u}_\epsilon \in L^\infty ({{\Bbb R}} ^3) $且有

(2.5) $ \begin{equation} \| \bar{u}_\epsilon \|_{L^{\infty} ({{\Bbb R}} ^3)} \leq C < \infty . \end{equation} $

由(2.5)式和假设$ (K_1) $,从方程(2.5)可推断出$ \|\triangle \bar{u}_\epsilon\|_{L^{\infty} ({{\Bbb R}} ^3)} \leq C < \infty $关于$ \epsilon>0 $充分小一致成立.

接下来我们给出一个关于方程解的局部Pohozave恒等式.

命题2.2  令$ u $是方程(1.1)的一个正解, $ \Omega $是一个在$ {{\Bbb R}} ^3 $上的有界光滑区域,对$ i = 1, 2, 3 $,我们有

(2.6) $ \begin{eqnarray} \int_{\Omega} \frac{\partial K}{\partial x_i} u^{p+1}{\rm d}x & = &\Big (\epsilon ^2 a + \epsilon b \int_{{{\Bbb R}} ^3} |\nabla u|^2{\rm d}x\Big) \Big (-\frac{p+1}{2} \int_{\partial \Omega} |\nabla u|^2 v_i{\rm d}x + (p+1) \int_{\partial \Omega} \frac{\partial u}{\partial v} \frac{\partial u}{\partial x_i}{\rm d}x\Big) {}\\ && + \int_{\partial \Omega} K u^{p+1}v_i{\rm d}x-\frac{p+1}{2} \int_{\partial \Omega} u^2 v_i{\rm d}x, \end{eqnarray} $

其中$ v = (v_1, v_2, v_3) $是$ \partial \Omega $上的单位外法向量.

证  将方程(1.1)的两边同时乘上$ \partial _{x_i} u $然后在$ \Omega $积分,运用分部积分法即可得到以上等式.我们略去详细过程.

接下来,我们给出关于$ |y_\epsilon| $的估计.

命题2.3  设$ K(x) $满足条件$ (K_1), (K_2) $和$ (K_3) $,令$ u_\epsilon = U_{\epsilon, y_\epsilon}+\varphi_\epsilon $是方程(1.1)的一个解,则有

$ |y_\epsilon| = o(\epsilon), \ \mbox{当$\epsilon \rightarrow 0 $时}. $

证  令

(2.7) $ \begin{equation} \int_{B_d (y_\epsilon)} \frac{\partial K}{\partial x_i} (U_{\epsilon, y_\epsilon} + \varphi_\epsilon)^{p+1}{\rm d}x = : \sum\limits_{i = 1}^{3} I_i , \end{equation} $

其中

$ I_1 = (\epsilon ^2 a + \epsilon b \int_{{\mathbb R}^3} | \nabla u|^2{\rm d}x) (-\frac{p+1}{2} \int_{\partial B_d (y_\epsilon)} |\nabla u|^2 v_i{\rm d}x + (p+1) \int_{\partial B_d (y_\epsilon)} \frac{\partial u}{\partial v} \frac{\partial u}{\partial x_i}{\rm d}x), $

$ I_2 = \int_{\partial B_d (y_\epsilon)} K u^{p+1}v_i{\rm d}x, \; \; I_3 = -\frac{p+1}{2} \int_{\partial B_d (y_\epsilon)} u^2 v_i{\rm d}x. $

由(2.1)式知$ (\epsilon^2 a+\epsilon b \int_{{{\Bbb R}} ^3}|\nabla u|^2 {\rm d}x) = O(\epsilon^2 ) $,由文献[20]中(5.13)式的证明可得

(2.8) $ \begin{equation} \epsilon ^2 \int_{\partial B_d (y_\epsilon)} |\nabla u_\epsilon|^2{\rm d}x = O(\|\varphi_\epsilon\|^2_\epsilon). \end{equation} $

结合(2.8)式,我们可以推断出

$ \begin{array}{ll} I_1 = O(\|\varphi_\epsilon \| ^2_\epsilon), \, \, I_2 = O(\|\varphi_\epsilon \| ^{p+1}_\epsilon), \, \, I_3 = O(\|\varphi_\epsilon \| ^2_\epsilon). \end{array} $

因此$ \sum\limits_{i = 1}^{3} I_i = O(\|\varphi_\epsilon\|^2_\epsilon) $,即(2.7)式的右边$ = O(\|\varphi_\epsilon\|_\epsilon^2) $.

下面我们估计(2.7)式的左边.

由Hölder不等式和嵌入定理可知

(2.9) $ \begin{eqnarray} & &\int_{B_d (y_\epsilon)} \frac{\partial K}{\partial x_i} (U_{\epsilon, y_\epsilon} + \varphi_\epsilon)^{p+1}{\rm d}x {}\\ & = &\int_{B_d (y_\epsilon)} \frac{\partial K}{\partial x_i} U ^{p+1}_{\epsilon, y_\epsilon}{\rm d}x + O(\epsilon^{\frac{3p}{2}}\|\varphi_\epsilon\|_\epsilon) + O(\epsilon^{\frac{3(p-1)}{2}}\|\varphi_\epsilon\|^2_\epsilon)+ \cdots + O(\|\varphi_\epsilon\|^{p+1}_\epsilon). \end{eqnarray} $

由假设$ (K_3) $,我们之前已经假定$ x_0 = 0 $,因此我们可以推出,对每个$ i = 1, 2, 3 $,有

(2.10) $ \begin{eqnarray} & &\int_{B_d (y_\epsilon)} \frac{\partial K}{\partial x_i} U^{p+1}_{\epsilon, y_\epsilon}{\rm d}x{}\\ & = &-mc_i \int_{B_d (y_\epsilon)} |x_i|^{m-2} x_i U^{p+1}_{\epsilon, y_\epsilon}{\rm d}x + O \Big(\int_{B_d (y_\epsilon)} |x|^m U^{p+1}_{\epsilon, y_\epsilon}\Big){} \\ & = &-mc_i\epsilon^3 \int_{B_d(0)} |\epsilon z_i + y_{\epsilon , i}|^{m-2} (\epsilon z_i + y_{\epsilon, i})^{U^{p+1}}{\rm d}x + O(\epsilon^3(\epsilon^m + |y_\epsilon|^m)) {}\\ & = &-mc_i\epsilon^3 \int_{{{\Bbb R}} ^3} |\epsilon z_i + y_{\epsilon , i}|^{m-2} (\epsilon z_i + y_{\epsilon, i})^{U^{p+1}}{\rm d}x + O(\epsilon^3(\epsilon^m + |y_\epsilon|^m)). \end{eqnarray} $

联立(2.9)和(2.10)式得到

(2.11) $ \begin{eqnarray} & &\int_{B_d(y_\epsilon)} \frac{\partial K}{\partial x_i} (U_{\epsilon, y_\epsilon} + \varphi_\epsilon)^{p+1}{\rm d}x {}\\ & = & -mc_i\epsilon^3 \int_{{{\Bbb R}} ^3} |\epsilon z_i + y_{\epsilon , i}|^{m-2}(\epsilon z_i + y_{\epsilon, i})^{U^{p+1}}{\rm d}x + O(\epsilon^3(\epsilon^m + |y_\epsilon|^m)){}\\ &&+ O(\epsilon^{\frac{3p}{2}} \|\varphi_\epsilon\|_\epsilon) + O(\epsilon^{\frac{3(p-1)}{2}} \|\varphi_\epsilon\|^2_\epsilon) + \cdots + O(\|\varphi_\epsilon\|^{p+1}_\epsilon). \end{eqnarray} $

当$ c_i\neq0 $时,由假设$ (K_3) $和(2.7)–(2.11)式得到

(2.12) $ \begin{eqnarray} & & \epsilon ^3 \int_{{{\Bbb R}} ^3} |\epsilon z_i + y_{\epsilon, i}| ^{m-2} (\epsilon z_i + y_{\epsilon, i}) U^{p+1}{\rm d}x {}\\ & = & O(\epsilon^{\frac{3p}{2}} \|\varphi_\epsilon\|_\epsilon) + O(\epsilon^{\frac{3(p-1)}{2}} \|\varphi_\epsilon\|^2_\epsilon)+ O(\|\varphi_\epsilon\|^{p+1}_\epsilon) + \cdots +O(\epsilon^3(\epsilon^m + |y_\epsilon|^m) + O(\|\varphi_\epsilon\|^2_\epsilon)) {}\\ & = & O(\epsilon^3(\epsilon^m + |y_\epsilon|^m) + \|\varphi_\epsilon\|^2_\epsilon). \end{eqnarray} $

由文献[20,命题3.2]的证明以及假设$ (K_3) $可知

(2.13) $ \begin{equation} \|\varphi_\epsilon\| = O(\epsilon^{3/2} (\epsilon^{m-\tau} + |y_\epsilon|^{m(1-\tau)})). \end{equation} $

因此

(2.14) $ \begin{equation} \int_{{{\Bbb R}} ^3}|\epsilon z_i + y_{\epsilon, i}|^{m-2} (\epsilon z_i + y_{\epsilon, i})U^{p+1}{\rm d}x = O(\epsilon^{m-\tau} + |y_\epsilon|^{m(1-\tau)}). \end{equation} $

结合基本不等式和Young不等式,类似文献[20],易证$ |y_\epsilon| = O(\epsilon). $

下面我们证明$ |y_\epsilon| = o(\epsilon) $我们应用反证法进行证明.

假设存在$ \epsilon_k \rightarrow 0, \; y_{\epsilon_k} \rightarrow 0 $,使得$ \frac{y_{\epsilon_k}}{\epsilon_k} \rightarrow A \in {{\Bbb R}} ^3 $ (其中$ A\neq 0) $,那么,由(2.14)式,有

$ \int_{{{\Bbb R}} ^3} |z_i + \frac{y_{\epsilon_k}}{\epsilon_k}|^{m-2} (z_i + \frac{y_{\epsilon_k}}{\epsilon_k}) U^{p+1}{\rm d}x = O(\epsilon^{m-\tau}). $

对上式取极限可得

$ \int_{{{\Bbb R}} ^3} | z + A|^{m-2} (z+A)U^{p+1} (z){\rm d}x = 0. $

另一方面, $ U = U(|z|) $是关于$ |z| $严格单调减的,所以有$ A = 0 $,这与假设相矛盾,假设不成立.证毕.

由命题2.3的结论,结合假设$ (K_3) $,我们还可以得到如下估计

(2.15) $ \begin{equation} \|\varphi_{\epsilon, y}\|_\epsilon = O(\epsilon^{\frac{3}{2} + m(1-\tau)}). \end{equation} $

由于$ m>1 $,可以取$ \tau $充分小使得$ m(1-\tau)>1 $.

这一节我们将证明定理1.1,类似文献[21]的方法我们采用反证法进行证明.

假设$ u^{(i)}_\epsilon = U_\epsilon, y^{(i)}_\epsilon + \varphi^(i)_\epsilon $, $ (i = 1, 2) $为由命题2.1得到的两个不同的解,由(2.4)式知$ u_\epsilon^{(i)} $为$ {{\Bbb R}} ^3 $上的有界解,取

$ \xi\epsilon = \frac{u^{(1)}_\epsilon - u^{(2)}_\epsilon}{\|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^{\infty} ({{\Bbb R}} ^3)}} $

并令$ \bar{\xi}_\epsilon (x) \equiv \xi_\epsilon (\epsilon x + y^{(1)}_\epsilon). $显然$ \|\bar{\xi}_\epsilon \|_{L^{\infty} ({{\Bbb R}} ^3)} = 1 $.

此外,由(2.4)式我们有

(3.1) $ \begin{equation} \bar{\xi}_\epsilon (x) \rightarrow 0, \ \mbox{关于$\epsilon >0 $是一致的,当$ |x| \rightarrow \infty$时.} \end{equation} $

下面我们通过证明当$ \epsilon \rightarrow 0 $时, $ \|\bar{\xi}_\epsilon\|_{L^{\infty} ({{\Bbb R}} ^3)}\rightarrow 0 $得到矛盾.由(3.1)式知,我们只需要证明对任意固定的$ R>0 $,当$ \epsilon\rightarrow 0 $时,有

(3.2) $ \begin{equation} \|\bar{\xi}_\epsilon\|_{L^{\infty} (B_{R}(0))} \rightarrow 0. \end{equation} $

为了证明这个结论,我们需要得到下面一系列的结果.

命题3.1

$ \|\xi_\epsilon\|_\epsilon = O(\epsilon^{3/2}). $

证  由于$ u^{(i)}_\epsilon $, $ (i = 1, 2) $方程(1.1)的解,我们将$ u^{(1)}_\epsilon $和$ u^{(2)}_\epsilon $分别代入方程(1.1),将所得两个等式相加得到

(3.3) $ \begin{eqnarray} & & -(2\epsilon^2b+\epsilon b \int_{{{\Bbb R}} ^3}|\nabla u^{(1)}_\epsilon|^2 + |\nabla u^{(2)}_\epsilon|^2 {\rm d}x)\triangle \xi_\epsilon + 2 \xi_\epsilon {}\\ & = & \epsilon b\Big (\int_{{{\Bbb R}} ^3} \nabla (u^{(1)}_\epsilon + u^{(2)}_\epsilon) \dot \nabla \xi_\epsilon{\rm d}x\Big) \triangle (\nabla u^{(1)}_\epsilon + \nabla u^{(2)}_\epsilon) + 2C_\epsilon (x) \xi_\epsilon , \end{eqnarray} $

其中$ C_\epsilon(x) = pK(x) \int_{0}^{1} [t u^{(1)}_\epsilon (x) + (1-t) u^{(2)}_\epsilon (x) ]^{(p-1)}{\rm d}x $.将(3.3)式两边同时乘以$ \xi_\epsilon $,并在$ {{\Bbb R}} ^3 $上积分,去掉含$ b $的项得到

$ \|\xi_\epsilon\|^2_\epsilon \leq \int_{{{\Bbb R}} ^3} C_\epsilon \xi^2_\epsilon {\rm d}x. $

另一方面,注意到$ C_\epsilon \leq C \sum\limits_{i = 1}^{2} (u^{(1)}_\epsilon)^{p-1} $ (因为$ K(x) $有界).则有

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^3} C_\epsilon \xi^2_\epsilon{\rm d}x &\leq &C \sum\limits_{i = 1}^{2} \int_{{{\Bbb R}} ^3} (u^{(i)}_\epsilon)^{p-1}\xi^2_\epsilon{\rm d}x \\ & \leq & C \sum\limits_{i = 1}^{2} \Big (\int_{{{\Bbb R}} ^3} (u^{(i)}_\epsilon)^6{\rm d}x\Big)^{\frac{p-1}{6}} (\int_{{{\Bbb R}} ^3} (\xi^{2}_\epsilon)^{\frac{6}{7-p}}{\rm d}x)^{\frac{7-p}{6}} \\ & \leq & C \sum\limits_{i = 1}^{2} \|\nabla u^{(i)}_\epsilon\|^{p-1}_{L^2 ({{\Bbb R}} ^3)} \Big(\int_{{{\Bbb R}} ^3} \xi^2_\epsilon{\rm d}x\Big) ^ {\frac{7-p}{6}} = O(\epsilon^{\frac{p-1}{2}})\|\xi_\epsilon\|^{\frac{7-p}{3}}. \end{eqnarray*} $

因此$ \|\xi_\epsilon\|_\epsilon = O(\epsilon^{\frac{p-1}{2}})\|\xi_\epsilon\|_\epsilon^{(7-p)/3}, $即有$ \|\xi_\epsilon\|_\epsilon = O(\epsilon^\frac{3}{2}). $命题3.1证毕.

下面我们考虑$ \xi_\epsilon $的渐近行为.

命题3.2  令$ \bar{\xi}_\epsilon(x) = \xi_\epsilon (\epsilon x + y^{(1)}_\epsilon) $,则存在$ d_i \in {{\Bbb R}} , (i = 1, 2, 3) $在允许子列存在的情况下,当$ \epsilon\rightarrow 0 $时有

$ \bar{\xi}_\epsilon \rightarrow \sum\limits_{i = 1}^{3}d_i \partial_{x_i} U, \ \mbox{在$ C^1_{loc} ({{\Bbb R}} ^3) $上}. $

证明过程类似文献[20,定理6.2],此处省略证明过程.

下面,我们来进一步分析$ d_i $.

引理3.1  若$ d_i $为命题3.2中所找到的实数,则$ d_i = 0, (i = 1, 2, 3). $

证  我们将利用Pohozaev恒等式(2.6)来证明.

对$ u^{(1)}_\epsilon $和$ u^{(2)}_\epsilon $在$ B_d (y^{(1)}_\epsilon) $上分别用(2.6)式,代入$ \xi_\epsilon = \frac{u^{(1)}_\epsilon - u^{(2)}_\epsilon}{\|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^\infty ({{\Bbb R}} ^3)}} $,则有

(3.4) $ \begin{eqnarray} & & \int_{B_d(y^{(1)}_\epsilon)} \frac{\partial K}{\partial x_i} A_\epsilon \xi_\epsilon \| u^{(1)}_\epsilon - u^{(2)}_\epsilon \|_{L^\infty ({{\Bbb R}} ^3)}{\rm d}x {}\\ & = & (\epsilon ^2 a + \epsilon b \int_{{{\Bbb R}} ^3} |\nabla u ^{(1)}_\epsilon|^2{\rm d}x) \int_{\partial B_d (y_\epsilon^{(1)})}(-\frac{1}{2}|\nabla u^{(1)}_\epsilon|^2 v_i + \frac{\partial u^{(1)}}{\partial v})\frac{\partial u^{(1)}}{\partial x_i}{\rm d}x{}\\ & & - (\epsilon^2 a + \epsilon b \int_{{{\Bbb R}} ^3} |\nabla u^{(2)}_\epsilon|^2{\rm d}x)\int_{\partial B_d (y_\epsilon^{(1)})}(-\frac{1}{2}|\nabla u^{(1)}_\epsilon|^2 v_i + \frac{\partial u^{(1)}}{\partial v})\frac{\partial u^{(1)}}{\partial x_i}{\rm d}x{}\\ & & + \int_{\partial B_d(y^{(1)}_\epsilon)} K(x)A_\epsilon \xi_\epsilon \|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^{\infty}({{\Bbb R}} ^3)}v_i{\rm d}x{}\\ & & - \frac{1}{2}\int_{\partial B_d (y_\epsilon^{(1)})} (u^{(1)}_\epsilon - u^{(2)}_\epsilon )\xi_\epsilon \|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^{\infty}({{\Bbb R}} ^3)}v_i{\rm d}x, \end{eqnarray} $

其中$ A_\epsilon(x) = \int_{0}^{1}[tu_\epsilon^{(1)}(x) + (1-t)u^{(2)}_\epsilon (x)]^p {\rm d}t. $下面我们分别估计(3.4)式中的每一项.

注意到

$ \Big(\epsilon^2 a + \epsilon b \int_{{{\Bbb R}} ^3} |\nabla u^{(i)}_\epsilon|^2{\rm d}x\Big) = O(\epsilon^2), i = \{1, 2\}. $

且类似命题2.3我们有

$ \int_{\partial B_d (y^{(1)}_\epsilon)} |\nabla u^{(1)}_\epsilon|^2{\rm d}x = O(\|\nabla \varphi^{(i)}_\epsilon \|^2_{L^2({{\Bbb R}} ^3)}). $

由(2.15)式有

$ \begin{eqnarray*} & &\sum\limits_{i = 1}^{2}(\epsilon^2a + \epsilon b \int_{{{\Bbb R}} ^3}|\nabla u^{(i)}_\epsilon|^2{\rm d}x) \int_{\partial B_d(y^{(1)}_\epsilon)}\Big(-\frac{1}{2}|\nabla u^{(i)}_ \epsilon|^2 v_i + \frac{\partial u^{(i)}_\epsilon}{\partial v} \frac{\partial u^{(i)}_\epsilon}{\partial x_i}\Big){\rm d}x{\nonumber}\\ & = &\sum\limits_{i = 1}^{2} O(\epsilon^2 \| \nabla \varphi^{(i)}_\epsilon\|^2_{L^2({{\Bbb R}} ^3)}) = O(\epsilon^{3+2m(1-\tau)}). \end{eqnarray*} $

类似文献[21]可证明

$ \int_{\partial B_d(y^{(1)}_\epsilon)}K(x)A_\epsilon \xi_\epsilon \|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^\infty({{\Bbb R}} ^3)} v_i{\rm d}x = O(\epsilon^{3+2m(1-\tau)p}) $

和

$ \int_{\partial B_d(y^{(1)}_\epsilon)}(u^{(1)}_\epsilon + u^{(2)}_\epsilon)\xi_\epsilon \|u^{(1)}_\epsilon - u^{(2)}_\epsilon\|_{L^\infty({{\Bbb R}} ^3)} v_i{\rm d}x = O(\epsilon^{3+m(1-\tau)}). $

(3.5) $ \begin{equation} \mbox{故(3.4)式右边} = O(\epsilon^{3+m(1-\tau)}). \end{equation} $

下面我们估计(3.4)式左边.

(3.6) $ \begin{eqnarray} \int_{B_d(y^{(1)}_\epsilon)}\frac{\partial K}{\partial x_i} A_\epsilon \xi_\epsilon {\rm d}x & = & -mc_i \int_{B_d(y^{(1)}_\epsilon)}|x_i|^{m-2}x_i A_\epsilon \xi_\epsilon {\rm d}x + O \Big(\int_{B_d(y_\xi^{(1)})}|x_i|^m A_\epsilon \xi_\epsilon {\rm d}x\Big) {}\\ & = : &J_1 + J_2. \end{eqnarray} $

注意到$ A_\epsilon(x) = O(|u^{(1)}|^p + |u^{(2)}|^p ) $,故

(3.7) $ \begin{eqnarray} J_2 & = &\sum\limits_{i = 1}^{2} O\Big(\epsilon ^3 \int_{B_\frac{d}{\epsilon}(0)}|\epsilon_ {y_i}+ y^{(1)} _{\epsilon, i}|^m |u^{(i)}_\epsilon (\epsilon y + y_\epsilon^ {(1)})|^p |\bar{\xi}_\epsilon| {\rm d}y\Big){}\\ & = &\sum\limits_{i = 1}^{2} O\Big(\epsilon ^{3+m} \int_{B_\frac{d}{\epsilon}(0)}(|y|^m + 1)|u^{(i)}_\epsilon (\epsilon y + y_\epsilon^ {(1)})|^p |\bar{\xi}_\epsilon| {\rm d}y\Big){}\\ & = &O(\epsilon^{3+m}) = o(\epsilon^{2+m}). \end{eqnarray} $

由于$ U $在无穷远处是指数衰减的,且有$ |y_\epsilon| = o(\epsilon) $,由命题3.2,且由$ U $是径向对称的,故

(3.8) $ \begin{eqnarray} J_1 & = & -mc_i \int_{\partial B_d (y^{(1)}_\epsilon)} |x_i|^{m-2}x_i A_\epsilon(x)\xi_\epsilon {\rm d}x {}\\ & = &-mc_i\epsilon^3\int_{B_{\frac{d}{\epsilon}}(0)}|\epsilon y_i + y^{(1)}_{\epsilon, i}|^{m-2} (\epsilon y_i + y^{(1)}_{\epsilon, i})A_\epsilon(\epsilon y + y^{(1)}_\epsilon)\bar{\xi_\epsilon}(y){\rm d}y {}\\ & = &-(p+1)mc_i\epsilon^{m+2}\sum\limits_{j = 1}^{3}d_i \int_{{{\Bbb R}} ^3}|y_i|^{m-2} y_i U^p (y) \partial_{y_i}U{\rm d}y + o(\epsilon^{2+m}){}\\ & = &D_id_i\epsilon^{m+2} + o(\epsilon^{2+m}), \end{eqnarray} $

其中

(3.9) $ \begin{equation} D_i = -(p+1)mc_i\int_{{{\Bbb R}} ^3}|y_i|^{m-2}y_iU^p(y)\partial_{y_i}U{\rm d}y \neq 0. \end{equation} $

由(3.6)、(3.7)和(3.8)式得(3.4)式的左边为

(3.10) $ \begin{equation} D_id_i\epsilon^{m+2} + o(\epsilon^{m+2}). \end{equation} $

联立(3.5)和(3.10)式我们得到$ D_id_i\epsilon^{m+2} + o(\epsilon^{m+2}) = O(\epsilon^{3+m(1-\tau)}). $则$ D_id_i = o(1)\rightarrow 0. $由于$ D_i\neq 0 $,故$ d_i = 0 $.引理3.1证毕.

以下是定理1.1的证明.

证  如果(1.1)式存在两个不同的解$ u^{(i)}_\epsilon, \; i = 1, 2 $.由前面的定义的可知$ \xi_\epsilon, \bar{\xi}_\epsilon $满足

$ \|\bar{\xi}_\epsilon\|_{L^\infty({{\Bbb R}} ^3)} = 1, $

由(3.1)和(3.2)式知,当$ \epsilon \rightarrow 0 $时,有$ \|\bar{\xi}_\epsilon\|_{L^\infty({{\Bbb R}} ^3)} = o(1) $,得到矛盾.定理1.1证毕.

在这一节的最后我们类似文献[20],对定理1.1进行更一般的说明.

引理3.2  令$ a, b>0 $且有$ 1<p<5. $假如$ K $满足条件$ (K_1), \; (K_3) $.如果$ u^{(i)}_\epsilon = U_{\epsilon, y^{(i)}_\epsilon} + \varphi^{(i)}_{\epsilon, y_\epsilon^{(i)}} $, $ (i = 1, 2) $是方程(1.1)的两个解,而且当$ \tau >0 $任意小,满足$ y_\epsilon \rightarrow 0 $且有

$ \|\varphi^{(i)}_{\epsilon, y^{(i)}_\epsilon}\|_\epsilon \leq \epsilon^{3/2} (\epsilon^{\alpha-\tau}+(K(0)-K(y^{(i)}_\epsilon))^{(1-\tau)}), $

则有$ u^{(1)}_\epsilon \equiv u^{(2)}_\epsilon. $进一步,我们假设$ u_\epsilon $是唯一解,然后得到$ |y_\epsilon| = o(\epsilon), \, \, \, \|\varphi_\epsilon\|_\epsilon = O(\epsilon^{\frac{3}{2}+m(1-\tau)}). $