## A p-Laplace Eigenvalue Problem with Coercive Potentials

He Jingran,, Guo Helin,, Wang Wenqing

Center of Mathematical Sciences and Department of Mathematics, Wuhan University of Technology, Wuhan 430070

 基金资助: 国家自然科学基金.  11931012国家自然科学基金.  11871387

 Fund supported: the NSFC.  11931012the NSFC.  11871387

Abstract

In this paper, we are concerned with the asymptotic behavior of solutions for a p-Laplace eigenvalue problem with coercive potentials. The bottom of the potential (The set of global minimum points of the potential) is an ellipsoid, we prove that the solutions of the problem will blow up at one of the endpoints of the major axis of the ellipsoid as the related parameter tends to a threshold value, and we also give the exact blow-up rate.

Keywords： p-Laplace ; Eigenvalue problem ; Asymptotical behavior ; Blow-up rate

He Jingran, Guo Helin, Wang Wenqing. A p-Laplace Eigenvalue Problem with Coercive Potentials. Acta Mathematica Scientia[J], 2021, 41(5): 1323-1332 doi:

## 1 引言

$$$-\Delta_{p}u+V(x)\left|u\right|^{p-2}u = \mu\left|u\right|^{p-2}u+a\left|u\right|^{s-2}u, \ x \in{{\Bbb R}} ^{n},$$$

$$$V\left(x\right)\in C\left({{\Bbb R}} ^{n}, {{\Bbb R}} ^{+}\right), \ \inf\limits_{x\in{{\Bbb R}} ^{n}}V\left(x\right) = 0\ \mbox{及} \lim\limits_{\left|x\right|\rightarrow \infty}V\left(x\right) = \infty.$$$

$$$e_{a} = \inf\bigg\{E_a\left(u\right):u \in {\cal H}\ \mbox{且}\ \int_{{{\Bbb R}} ^{n}}\left|u\right|^{p} {\rm d}x = 1\bigg\},$$$

$\begin{eqnarray} E_a\left(u\right)& = &\int_{{{\Bbb R}} ^{n}}\left(\left|\nabla u\right|^{p}+V\left(x\right)\left|u\right|^{p}\right) {\rm d}x-\frac{pa}{s}\int_{{{\Bbb R}} ^{n}}\left|u\right|^{s} {\rm d}x\\& = & \int_{{{\Bbb R}} ^{n}}\left(\left|\nabla u\right|^{p}+V\left(x\right)\left|u\right|^{p}\right) {\rm d}x-\frac{na}{n+p}\int_{{{\Bbb R}} ^{n}}\left|u\right|^{s} {\rm d}x, \ u \in {\cal H}. \end{eqnarray}$

$p = 2 $$n = 2 时, 方程(1.1) 实际上就变为与冷原子物理中的Bose-Einstein (BEC) 凝聚现象相关的Gross-Pitaevskii (GP) 方程, 参数 a > 0 表示冷原子之间相互吸引作用的强度, 更多关于BEC相关的物理背景, 见文献[2, 4-8]及其参考文献. 对GP方程的研究已有相当丰富的结果, 特别地, Guo等[4] V\left(x\right) 满足(1.2) 式的条件下, 讨论了方程(1.1) 解的存在性. 更进一步, 为了获得方程(1.1) 解的渐近行为, Guo等[4]研究了 V(x) 具有有限个孤立零点的问题, 即对任意的 x \in {{\Bbb R}} ^{2} , 有 $$V(x) = h(x)\prod\limits_{i = 1}^{m}\left|x-x_{i}\right|^{q_{i}}, \ \mbox{其中}\ x_{i} \neq x_{j}, \ \mbox{若}\ i \neq j, \ \mbox{且}\ 0 < C \le h(x) \le \frac{1}{C}.$$ 在(1.5) 式的条件假设下, 当参数 a 趋于某个门槛值时, 文献[4]证明了方程(1.1) 的解在位势 V(x) 最为平坦的最小值点处集中. 在此基础上, 文献[6]考虑了位势 V(x) 具有无穷多个零点的问题, 即对任意的 x \in {{\Bbb R}} ^{2}, \ A > 0 , 令 $$V(x) = (|x|-A)^2.$$ 在条件(1.6) 下, 当参数 a 趋于某个门槛值时, 文献[6]证明了方程(1.1) 的解在集合 \{x \in {{\Bbb R}} ^{2}: |x| = A\} 中的某个点处集中, 并且发生爆破. 由于圆环有非常好的对称性, 其上的点在几何上难以区分, 故对这类位势难以得到爆破点的进一步信息. 因此, 为了获得解的爆破点的更为具体的信息, Guo等[8]对位势 V(x) 提出了如下假设, 即 $$V(x) = \left(\sqrt{\frac{x_{1}^{2}}{c^{2}}+\frac{x_{2}^{2}}{d^{2}}}-1\right)^{2}, \ \mbox{其中}\ c > d > 0, \ (x_{1}, x_{2})\in{{\Bbb R}} ^2.$$ 在条件(1.7) 下, 当参数 a 趋于某个门槛值时, 文献[8]证明了方程(1.1) 的解在位势 V(x) 底部椭圆的长轴端点之一处集中. 1 < p \le 2, \ n \ge 2 时, 谷龙江等在文献[11]中考虑了 V(x) = \left|x\right|^2 的情况, 文献[11]首先证明了方程对应的约束极小化问题(1.3) 可解当且仅当 $$0 \leq a < a_{\ast} \ \mbox{且}\ a_{\ast} = \left(\int_{{{\Bbb R}} ^{n}}\left|Q_{p}\right|^{p}{\rm d}x\right)^{\frac{p}{n}},$$ 其中 Q_{p}\left( x \right) 是下面极限方程的唯一正解 (1 < p \le 2) , 有 $$-\Delta_{p}u+\frac{p}{n}\left|u\right|^{p-2}u = \left|u\right|^{s-2}u, \ u\in W^{1, p}({{\Bbb R}} ^{n}).$$ 然后还证明了当 a 趋于 a_{\ast} 时, 问题(1.3) 的可达元会在 x = 0 处集中. 后来, 在文献[3]中, Gu等将文献[11]的相关结论推广到了 1 < p < n 的情况, 此时方程(1.9) 是否有唯一解尚不清楚, 文献[3]通过新的思路, 避免了文献[11]中定义 a_{\ast} 时对方程(1.9) 解唯一性的依赖, 并且给出了问题(1.3) 可达元的存在性条件. 此外, 在一般强制性条件(1.2) 下, 当参数 a 趋于 a_{\ast} 时, Gu等[3]还证明了问题(1.3) 的可达元必定在位势 V(x) 的极小值点处爆破, 其主要结果回顾如下. 定理1.1[3, Theorem 1.2] 设 1 < p < n , 且 V(x) 满足条件(1.2), 对任意的 a \in ( 0, a_{*} ) , 令 u_{a} \ge 0 是问题(1.3) 相应的极小可达元, 则 (i) $$\mbox{当}\ a \nearrow a_{*}\ \mbox{时}, \ \varepsilon_{a} \triangleq \left( \int_{{{\Bbb R}} ^{n}}|\nabla u_{a}|^{p} {\rm d}x\right)^{-\frac{1}{p}} \to 0.$$ (ii) 设 \overline{z}_{a}\ \mbox{是}\ u_{a} 的某个全局最大值点, 则 $$\lim\limits_{a \nearrow a_{*}}\mbox{dist}(\overline{z}_{a}, {\cal A}) = 0,$$ 其中 {\cal A} = \{x \in {{\Bbb R}} ^{n}, V(x) = 0\}. (iii) 对于满足当 k \to \infty$$ {a_{k}}\nearrow a_{*}$的任意序列$\{a_{k}\}$, 在子列的意义下, 我们仍记为$\{a_{k}\}$, 有

$$$\lim\limits_{k \to \infty}\varepsilon_{a_{k}}^{\frac{n}{p}}u_{a_{k}}(\varepsilon_{a_{k}}x+\overline{z}_{a_k}) = \frac{Q_{p}(x)}{a_{*}^{n/p^{2}}}\ \mbox{在}\ W^{1, p}\left({{\Bbb R}} ^{n}\right)\ \mbox{中成立},$$$

$$$\lim\limits_{k \to \infty}(a_{*}-a_{k})^{\frac{n}{(2+p)p}}u_{a_{k}}((a_{*}-a_{k})^{\frac{1}{2+p}}x+x_{a_{k}}) = \frac{(\lambda)^{\frac{n}{p}}Q_{p}(\lambda x)}{a_{*}^{n/p^{2}}},$$$

$$$\int_{{{\Bbb R}} ^{n}}|Q_{p}|^{s} {\rm d}x = \left(1+\frac{p}{n}\right)\int_{{{\Bbb R}} ^{n}}|\nabla Q_{p}|^{p} {\rm d}x = \left(1+\frac{p}{n}\right)\int_{{{\Bbb R}} ^{n}}|Q_{p}|^{p} {\rm d}x.$$$

$$$|Q_{p}(x)| \le e^{-\delta \left|x\right|}.$$$

$$$|\nabla f(z)|^{2} \ge \frac{1}{a_{1}^{2}},$$$

取$z = (z_1, z_2, \dots, z_n) \in {{\Bbb R}} ^{n}$满足$V(z) = 0$, 由条件(1.13) 和(1.14) 可知$f(z) = \sqrt{\sum\limits_{i = 1}^{n}\frac{z_{i}^{2}}{a_{i}^{2}}} = 1$, 以及

(i) $Q(x)$是正函数, 且为径向对称函数, 即$x \in {{\Bbb R}} ^{n}, \ Q(x) = Q(|x|)$.

(ii) $Q(x)$在无穷远处是呈指数衰减的, 即存在$\delta > 0, \ \rho > 0$, 使得当$|x| > \rho > 0$时,

$$$\int_{{{\Bbb R}} ^{n}}(y\cdot x)Q^{p}(x) {\rm d}x = 0\ \mbox{及}\ \int_{{{\Bbb R}} ^{n}}|y\cdot x|^{2}Q^{p}(x){\rm d}x = \frac{|y|^2}{n}\int_{{{\Bbb R}} ^{n}}|x|^{2}Q^{p}(x) {\rm d}x.$$$

首先, 当$n = 1 $$2 时, (2.5) 显然成立. 当 n \ge 3 时, 令 $$0 < r \in {{\Bbb R}} , \ \mbox{若}\ 1 \le i\le n-2, \ 0 \le \theta_i \le \pi;\ 0 \le \theta_{n-1} \le 2\pi.$$ 在条件(2.6) 下, 我们对 x = (x_{1}, x_{2}, \dots, x_{n}) \in {{\Bbb R}} ^{n} 使用下面的极坐标变换 $$\left\{\begin{array}{ll}x_1 & = r\cos\theta_1, \\x_2 & = r\sin\theta_1\cos\theta_2, \\\vdots\\x_{n-1} & = r\sin\theta_1\sin\theta_2\dots\sin\theta_{n-2}\cos\theta_{n-1}, \\x_{n} & = r\sin\theta_1\sin\theta_2\dots\sin\theta_{n-2}\sin\theta_{n-1}.{\nonumber\\}\end{array}\right.$$ 注意到, 对于任意的 n \ge 1 , 等式(2.8) 成立, 即 $$\int_0^{\pi}\sin^{n}\theta\cos\theta {\rm d}\theta = 0.$$ 因为 Q(x) 是径向对称的, 从而由(2.8) 可得 此外, 我们对坐标系做旋转变换, 即将坐标系旋转 \theta' 度, 使得将点 y 旋转到点 y' = (|y|, 0, \dots, 0) , 则由 Q(x) 的径向对称性可得 $$\int_{{{\Bbb R}} ^{n}}|y\cdot x|^{2}Q^{p}(x) {\rm d}x = \int_{{{\Bbb R}} ^{n}}|y' \cdot x|^{2}Q^{p}(x) {\rm d}x = \int_{{{\Bbb R}} ^{n}}|y|^{2}x_{1}^{2}Q^{p}(x) {\rm d}x.$$ 类似计算(2.9) 式的方法, 我们进一步可得 于是可以推出 $$n\int_{{{\Bbb R}} ^{n}}|y\cdot x|^{2}Q^{p} {\rm d}x = |y|^2\int_{{{\Bbb R}} ^{n}}|x|^{2}Q^{p} {\rm d}x, {\nonumber\\}$$ 即(2.5) 式得证. 然后, 下面我们将给出极小能量 e_a 的上界估计. 引理2.3 设 p \in (1, 2], \ a > 0, \ Q_{p}(x) 是方程(1.9) 正的唯一的径向对称解, 那么 e_a 有下面的上界估计 $$\limsup\limits_{a \nearrow a_{*}}\frac{e_{a}}{\left(a_{*}-a\right)^{\frac{2}{p+2}}} \le \frac{p+2}{2a_{*}}\lambda^{p},$$ 其中 a_{*} 由(1.8) 式定义, \lambda 由(1.17) 式所给出. 对任意的 \tau > 0$$ y_0 \in Z$, 令

$$$\int_{{{\Bbb R}} ^{n}}|\nabla u_\tau|^{p} {\rm d}x = \tau^{p} \quad \mbox{及} \quad \int_{{{\Bbb R}} ^{n}}|u_\tau|^{s} {\rm d}x = \left(1+\frac{p}{n}\right)\frac{\tau^p}{a_*}.$$$

$\begin{eqnarray} \lim\limits_{\tau \to \infty}\tau^2\int_{{{\Bbb R}} ^{n}}V(x)|u_{\tau}|^{p} {\rm d}x & = &\lim\limits_{\tau \to \infty}\frac{\tau^n}{a_{*}^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}\Big(\frac{f(x)-f(y_0)}{1/\tau} \Big)^{2}Q_{p}^{p}(\tau(x-y_0)) {\rm d}x\\ & = & \lim\limits_{\tau \to \infty}\frac{1}{a_{*}^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}\Big(\frac{f({y/\tau}+y_0)-f(y_0)}{1/\tau} \Big)^{2}Q_{p}^{p}(y) {\rm d}y\\& = &\frac{1}{a_{*}^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|\nabla f(y_{0})\cdot y|^{2}Q_{p}^{p}(y) {\rm d}y\\& = &\frac{|\nabla f(y_0)|^2}{na_{*}^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|y|^{2}Q_{p}^{p}(y) {\rm d}y\\& = &\frac{1}{na_{1}^{2}a_{*}^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|y|^{2}Q_{p}^{p}(y) {\rm d}y. \end{eqnarray}$

$\begin{eqnarray} e_{a} \le E_{a}(u_\tau) & = & \int_{{{\Bbb R}} ^{n}}(|\nabla u_\tau|^{p}+V(x)|u_\tau|^{p}) {\rm d}x - \frac{na}{n+p}\int_{{{\Bbb R}} ^{n}}|u_\tau|^{s} {\rm d}x\\& = &\tau^{p}-\frac{na}{n+p}\left(1+\frac{p}{n}\right)\frac{\tau^p}{a_*} + \int_{{{\Bbb R}} ^{n}}V(x)|u_\tau|^{p} {\rm d}x\\& = &\tau^{p}\left(1-\frac{a}{a_*}\right) + \int_{{{\Bbb R}} ^{n}}V(x)|u_\tau|^{p} {\rm d}x. \end{eqnarray}$

$\begin{eqnarray} \limsup\limits_{a \nearrow a_{*}}\frac{e_{a}(u)}{\left(a_{*}-a\right)^{\frac{2}{p+2}}} \le \frac{\lambda^p}{a_*}+\frac{p\lambda^p}{2a_*} = \frac{p+2}{2a_{*}}\lambda^{p}. \end{eqnarray}$

## 3 定理1.2的证明

(i) 引理2.3说明了$\limsup\limits_{a \nearrow a_{*}}{e_{a}(u)}/{\left(a_{*}-a\right)^{\frac{2}{p+2}}}$存在上界, 为了证明(1.16) 式, 我们还需要证明$\liminf\limits_{a \nearrow a_{*}}{e_{a}(u)}/{\left(a_{*}-a\right)^{\frac{2}{p+2}}}$存在下界, 且上界等于下界.

$$$\mbox{当}\ k \to \infty\ \mbox{时}, \ x_{a_k} \to x_0, \ \mbox{且}\ V(x_0) = 0.$$$

$$$w_{a_k} : = \varepsilon_{a_k}^{\frac{n}{p}}u_{a_k}(\varepsilon_{a_k}x + x_{a_k}),$$$

$$$\mu_k : = \frac{1}{\varepsilon_{a_k}^2}\int_{{{\Bbb R}} ^{n}}V(\varepsilon_{a_k}x + x_{a_k})w_{a_k}^{p}(x) {\rm d}x = \frac{1}{\varepsilon_{a_k}^2}\int_{{{\Bbb R}} ^{n}}V(x)u_{a_{k}}^{p}(x){\rm d}x.$$$

$\begin{eqnarray} e_{a_k} = E_{a_k}(u_{a_k}) &\ge& \varepsilon_{a_k}^{-p}\left(1-\frac{a_{k}}{a_*}\right) + \varepsilon_{a_k}^{2}\mu_k\\ &\ge &\left(\frac{p+2}{2a_*}\right)^{\frac{2}{p+2}}(a_*-a_{k})^{\frac{2}{p+2}}\left(\frac{2+p}{p}\mu_k\right)^{\frac{p}{p+2}}. \end{eqnarray}$

$$$\mbox{当}\ k \to \infty\ \mbox{时}, \ \frac{|f(x_{a_k})-f(x_0)|}{\varepsilon_{a_k}}\to + \infty.$$$

$\begin{eqnarray} \liminf\limits_{k \to \infty}\mu_k & = &\liminf\limits_{k \to \infty}\frac{1}{\varepsilon_{a_k}^2}\int_{{{\Bbb R}} ^{n}}V(\varepsilon_{a_k}x + x_{a_k})w_{a_k}^{p}(x) {\rm d}x\\ & = &\liminf\limits_{k \to \infty}\int_{{{\Bbb R}} ^{n}}\Big(\frac{f(\varepsilon_{a_k}x + x_{a_k}) - f(x_{a_k})}{\varepsilon_{a_k}} + \frac{f(x_{a_k}) - f(x_0)}{\varepsilon_{a_k}}\Big)^{2}w_{a_k}^{p}(x){\rm d}x\\ &\ge &\frac{1}{a_*^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}\Big( x\cdot\nabla f(x_0) + \lim\limits_{k \to \infty}\frac{f(x_{a_k}) - f(x_0)}{\varepsilon_{a_k}}\Big)^{2}Q_{p}^{p}(x) {\rm d}x\\ &\ge& M. \end{eqnarray}$

$$$\liminf\limits_{k \to \infty}\frac{e_{a_k}}{\left(a_{*}-a_{k}\right)^{\frac{2}{p+2}}} \ge \left(\frac{p+2}{2a_*}\right)^{\frac{2}{p+2}}\left(\frac{2+p}{2}M\right)^{\frac{p}{p+2}}.$$$

(3.7) 式显然与(2.11) 式相矛盾. 所以序列$\Big\{\frac{f\left(x_{a_{k}}\right) - f(x_{0})}{\varepsilon_{a_{k}}}\Big\}$是有界序列, 即存在序列$\{a_k\}$的子序列, 我们依然记为$\{a_k\}$和某个$n_0 \in {{\Bbb R}}$, 使得

$$$\mbox{当}\ k \to \infty\ \mbox{时}, \ \frac{f(x_{a_k}) - f(x_0)}{\varepsilon_{a_k}} \to n_0.$$$

$\begin{eqnarray} \liminf\limits_{k \to \infty}\mu_k &\ge &\frac{1}{a_*^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}(x\cdot\nabla f(x_0) + n_0)^{2}Q_{p}^{p} {\rm d}x\\ &\ge &\frac{1}{a_*^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|x\cdot\nabla f(x_0)|^{2}Q_{p}^{p} {\rm d}x\\ & = & \frac{|\nabla f(x_0)|^2}{na_*^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|x|^{2}Q_{p}^{p} {\rm d}x\\ &\ge &\frac{1}{na_1^{2}a_*^{\frac{n}{p}}}\int_{{{\Bbb R}} ^{n}}|x|^{2}Q_{p}^{p} {\rm d}x\\ & = &\frac{p}{2a_{*}}\lambda^{p+2}, \end{eqnarray}$

$$$\liminf\limits_{k \to \infty}\frac{e_{a_k}}{\left(a_{*}-a_{k}\right)^{\frac{2}{p+2}}} \ge \left(\frac{p+2}{2a_*}\right)^{\frac{2}{p+2}}\left(\frac{2+p}{p}\liminf\limits_{k \to \infty}\mu_k\right)^{\frac{p}{p+2}} \ge \frac{p+2}{2a_*}\lambda^{p}.$$$

$$$\liminf\limits_{k \to \infty}\frac{e_{a_k}}{\left(a_{*}-a_{k}\right)^{\frac{2}{p+2}}} = \frac{p+2}{2a_*}\lambda^{p}.$$$

(ii) 对任意的$a \in (0, a_*)$, 设$u_a$为问题(1.3) 的一个非负极小可达元. 我们断言下式成立, 即

$$$\lim\limits_{a \nearrow a_*}\frac{(a_{*}-a)^{\frac{1}{p+2}}}{\varepsilon_a} = \lambda,$$$

$$$\lim\limits_{k \to \infty}\frac{(a_{*}-a_k)^{\frac{1}{p+2}}}{\varepsilon_{a_k}} = r \in {{\Bbb R}} ^{+}\setminus\{\lambda\}.$$$

$\begin{eqnarray} e_{a_k} &\ge &\varepsilon_{a_k}^{-p}\left(1 - \frac{a}{a_*}\right)+\varepsilon_{a_k}^{2}\mu_k\\ & = &\frac{(a_*-a_k)^{\frac{2}{p + 2}}}{a_{\ast}}\left(\frac{(a_{*}-a_k)^{\frac{1}{p + 2}}}{\varepsilon_{a_k}}\right)^{p}+(a_*-a_k)^{\frac{2}{p + 2}}\mu_k \left(\frac{(a_{*}-a_k)^{\frac{1}{p+2}}}{\varepsilon_{a_k}}\right)^{-2}. \end{eqnarray}$

$$$\liminf\limits_{k \to \infty}\frac{e_{a_k}}{\left(a_{*}-a_{k}\right)^{\frac{2}{p+2}}} \ge \frac{r^p}{a_*} + \liminf\limits_{k \to \infty}\frac{\mu_k}{r^2} > \frac{p+2}{2a_*}\lambda^{p},$$$

$$$\lim\limits_{k \to \infty}\frac{(a_{*} - a_{k})^{\frac{1}{p + 2}}}{\varepsilon_{a_k}} = \lambda,$$$

$$$\liminf\limits_{k \to \infty}\mu_k = \frac{p}{2a_{*}}\lambda^{p+2},$$$

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