考虑如下问题解的存在性

(1.1) $ \begin{equation} -\Big(a-b\int_\Omega|\nabla u|^2{\rm d}x\Big)\Delta u = u^q, x\in\Omega, \end{equation} $

这里常数$ a, b $不同时为零, $ q\neq-1 $, $ \Omega\subset{{\Bbb R}} ^N $.它的研究与多个数学物理问题紧密相关:

(1)奇异的Yamabe问题:是否能够在紧黎曼流形$ ({\cal M}, g) $上找到一个完备的度量$ \tilde{g} $,使得$ \tilde{g} $在$ {\cal M} $上逐点保角地等价于$ g $,并且它的标准曲率等于非零常数?事实上,当$ ({\cal M}, g) $是$ {{\Bbb R}} ^N $中的球并且$ g $是标准度量时,假设$ \lambda $是一个与$ N $和曲率有关的正数,则奇异的Yamabe问题等价于求如下问题之一的解^[1]

(1.2) $ \begin{equation} \Delta u\pm \lambda u^q = 0. \end{equation} $

(2)带吸收项的流体问题(如波动方程和扩散方程^[2]):文献[3]将$ u_{tt}-\Delta u = \lambda u^q $满足临界指数时的爆破解推广到边界;文献[4]利用截断技巧考虑$ u_t+\Delta u+\lambda u^q = 0 $,文献[5]基于变分和筹数理论分析$ u_t = \Delta u+\lambda |u|^{q-1}u $,文献[6]通过行波解变换研究$ u_t\pm\Delta u = \lambda u^q $,诸类文献均在不同指数和区域条件下获得解的存在性和多重性.而文献[7]将问题$ i\partial_tu+\Delta_M u \pm \lambda u^q = 0 $推广到星形图$ (M, {\cal G}) $上.更多关于问题$ u_t(\mbox{或}u_{tt})\pm\Delta u+\lambda u^q = 0 $的研究参见文献[2, 8-10]及相关引用.事实上,文献[3-10]的稳态方程均归结为问题(1.2),而问题(1.2)符号为正^{[1, 11-16]}和符号为负^[17-24]时已有大量关于解的存在性和多重性结果.

(3) Kirchhoff型振动问题:德国物理学家Kirchhoff^[25]首次提出如下弹性弦振动方程

$ \varrho h \frac{\partial^2u}{\partial t^2} -\Big( p_0+\frac{Eh}{2L}\int_0^L\Big|\frac{\partial u}{\partial x}\Big|^2{\rm d}x \Big)\frac{\partial^2 u}{\partial x^2} = f_1\big(\frac{\partial u}{\partial x}\big)+f_2(x, u). $

其初始问题中外力$ f_1 $, $ f_2 $恒为零, $ t\geq0 $代表时间,弦长$ 0<x<L $, $ \varrho $是材质密度, $ E $是杨氏模量, $ h $是截面面积, $ p_0 $是轴向初始拉力,而解$ u = u(x, t) $代表弦的横向位移.这类问题后来被推广到含有对流项($ f_1\not\equiv0 $, $ f_1\equiv0 $表示只有吸收项)、同号源(如$ f_2(x, u) = |u|^{q-1}u $)、异号源(如$ f_2(x, u) = -|u|^{q-1}u $)和混合源(如$ f_2(x, u) = \lambda|u|^{q-1}u-|u|^{q-2}u $)等其他问题中,更多的研究参见文献[25]的被引状况.特别地,其稳态方程的解被称为驻波解.考虑方程(1.1)的边值问题, $ a>0>b $时为Kirchhoff问题的稳态方程,文献[26]利用山路引理和反证法获得唯一弱解的存在性,文献[27]利用达到函数获得临界指数时存在无穷多解,文献[28]利用Ljusternik-Schnirelman极小极大方法获得无穷多弱解的存在性,文献[29]利用Nehari流型分别获得无穷多解和极小能量解的存在性.而$ a, b>0 $的情形首次出现在文献[30]中,作者利用变分方法获得至少一个非平凡非正解和一个非平凡非负解的存在性,文献[31]利用变分方法和泛函扰动获得至少两个正解,文献[32]利用达到函数获得无穷多解,并且文献[32]列举纳米材料研究中涉及负有效模量的问题用以指出这是含负模量的Kirchhoff问题,文献[33]将问题(1.1)满足$ a, b>0 $的情形称作新Kirchhoff型问题并研究了$ q $为次临界函数的情形.关于问题(1.1)当$ a>0 $且$ b\in{{\Bbb R}} $时的研究结果,除文献[25-33]外还可以参见文献[34-36]及相关的引用.

注意到前述文献中$ q $的情形各异,而所获得的结果中,主要假设存在解的前提下获得解的估计,或是通过与临界点相关的各种方法获得问题(1.1)或(1.2)弱解的存在性及多重性,本文将直接构造出这类问题解的表达式,从而获得多重古典解.主要结果如下.

定理1.1 记$ \Omega = [c_1, d_1]\times[c_2, d_2]\times\cdots\times[c_N, d_N] $是$ {{\Bbb R}} ^N(N\geq1) $中的有界矩体,那么当$ q\neq-1 $时问题(1.1)存在如下类型的古典解

Ⅰ 设$ q = 1 $,当$ a\leq0<b $或者$ ab>0 $时:指数函数型.

Ⅱ $ q = 1 $且不满足$ a\leq0\leq b $时:三角函数型.

特别地, $ ab>0 $;或$ b<0<a<\big[\pi^2\sum\limits_{i = 1}^N(d_i-c_i)^{-2}\big]^{-1} $;或$ b = 0 $且$ a\sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(c_i-d_i)^2} = 1 $时,都可以满足零边值条件.其中$ n_i\; (i = 1, \cdots, N) $是任意$ N $个非零整数.

Ⅲ 设$ q = 0 $,对任意不同时为零的$ a, b\in{{\Bbb R}} $,定义$ u^0\equiv1 $时:幂函数型.

Ⅳ $ q<-1 $不满足$ a\leq0\leq b $;及$ q>-1, q\neq1 $且不满足$ a\geq0\geq b $时:幂函数型.

注1.1 定理Ⅱ和Ⅳ包含问题(1.2)的解,定理的证明中将给出各种解的表达式.

证明主要结果前,有必要叙述如下事实.

命题2.1 对泛函$ f: L^2(\Omega)\to(-\infty, +\infty) $, $ u\mapsto f(u) $,如果取定$ u\in L^2(\Omega) $,则$ f(u) $是一个常数.因此,如果$ u $是问题$ -\lambda \Delta u = u^q $的古典解且$ \nabla u\in L^2(\Omega)\bigcap C(\Omega) $,则当$ (a-b\int_\Omega|\nabla u|^2{\rm d}x) = \lambda $时$ u $是问题(1.1)的古典解.

证 Ⅰ 设$ N\geq1 $,令$ v: v(x) = \prod\limits_{i = 1}^N \beta_i^{x_i} $,其中$ \beta_i>0, \beta_i\neq1\; (i = 1, \cdots, N) $,那么

(2.1) $ \begin{equation} \left. \begin{array}{l} { } \frac{\partial v(x)}{\partial x_i} = (\ln\beta_i)\prod\limits_{i = 1}^N \beta_i^{x_i} = \ln\beta_i\cdot v(x), \\ \frac{\partial^2 v(x)}{\partial x_i^2} = (\ln\beta_i) \frac{\partial v(x)}{\partial x_i} = (\ln\beta_i)^2\cdot v(x), \\ \Delta v(x) = \sum\limits_{i = 1}^N \frac{\partial^2 v(x)}{\partial x_i^2} = \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big] \cdot v(x). \end{array}\right\} \end{equation} $

因此对任意的$ t\in {{\Bbb R}} $有$ \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big]^{-1}\Delta (tv) = tv $.根据(2.1)式有

$ \begin{eqnarray*} |\nabla v(x)|^2 = \sum\limits_{i = 1}^N\Big(\frac{\partial v(x)}{\partial x_i}\Big)^2 = \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big]\Big[\prod\limits_{i = 1}^N \beta_i^{2x_i}\Big], \end{eqnarray*} $

显然$ \nabla v\in L^2(\Omega)\bigcap C(\Omega) $,从而$ \int_\Omega |\nabla v(x)|^2{\rm d}x $是一个正常数(设为$ \Lambda $).事实上

$ \begin{eqnarray*} \int_\Omega |\nabla v(x)|^2{\rm d}x & = &\Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big] \Big[ \prod\limits_{i = 1}^N\int_{c_i}^{d_i} \beta_i^{2x_i} {\rm d}x_i\Big] = \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big] \prod\limits_{i = 1}^N\Big[ \frac{\beta_i^{2x_i}}{\ln\beta_i^2}\Big|_{c_i}^{d_i}\Big] \\ & = & \frac{1}{2^N} \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big] \prod\limits_{i = 1}^N\Big[ \frac{\beta_i^{2x_i}}{\ln\beta_i}\Big|_{c_i}^{d_i}\Big] \\ & = & \frac{1}{2^N} \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big]\Big[\prod\limits_{i = 1}^N \ln\beta_i \Big]^{-1}\prod\limits_{i = 1}^N \Big(\beta_i^{2d_i}-\beta_i^{2c_i} \Big) = \Lambda . \end{eqnarray*} $

如命题1所述,如果存在实数$ t $满足

(2.2) $ \begin{eqnarray} -\Big(a-b\int_\Omega|\nabla (tv)|^2{\rm d}x\Big) = \Big[\sum\limits_{i = 1}^N(\ln\beta_i)^2\Big]^{-1} = b\Lambda t^2-a, \end{eqnarray} $

则$ tv $是问题

(2.3) $ \begin{equation} -\Big(a-b\int_\Omega|\nabla u|^2{\rm d}x\Big)\Delta u = u, x\in\Omega \end{equation} $

的解.实际上当$ a\leq0<b $或$ ab>0 $时通过调节$ \beta_i $可使(2.2)始终有实值解可以表示为

$ t_\pm = \pm\Bigg[ \frac{a\sum\limits_{i = 1}^N(\ln\beta_i)^2 +1}{b\Lambda\sum\limits_{i = 1}^N(\ln\beta_i)^2} \Bigg]^{\frac{1}{2}}, $

从而

(2.4) $ \begin{eqnarray} u(x) = \pm\Bigg[ \frac{a\sum\limits_{i = 1}^N(\ln\beta_i)^2 +1}{b\Lambda\sum\limits_{i = 1}^N(\ln\beta_i)^2} \Bigg]^{\frac{1}{2}} \cdot \prod\limits_{i = 1}^N \beta_i^{x_i} : = C\prod\limits_{i = 1}^N \beta_i^{x_i} \end{eqnarray} $

是方程(1.1)的解.因此根据$ \beta_i>0, \beta_i\neq1\ (i = 1, \cdots, N) $的任意性可知问题(2.3)存在无穷多解.同理可证$ a\in{{\Bbb R}} $且$ b<0 $时问题(2.3)存在无穷多解形如(2.4)式.

例2.1 取$ \beta_i\equiv e $,则问题(2.3)有解可以表示为如下形式

$ u(x) = \pm 2^\frac{N}{2} \Big[ \frac{bN^2}{a N+1} \prod\limits_{i = 1}^N \big|e^{2d_i}-e^{2c_i} \big| \Big)^{-\frac{1}{2}} \cdot \exp \Big[\sum\limits_{i = 1}^N x_i\Big]. $

例2.2 设$ \Omega = [0, 1]^{N+1} $,取$ \{\beta_i\equiv e\}_{i = 1}^N $, $ \beta_{N+1} = e^k(k\neq0) $,则$ v(x) = e^{kx_{N+1}}\prod\limits_{i = 1}^N e^{x_i} $,

$ \sum\limits_{i = 1}^{N+1}(\ln\beta_i)^2 = N+k^2 , $

$ \Lambda = \int_\Omega |\nabla v|^2{\rm d}x = \frac{\big( N+k^2 \big)\big( e^2-1 \big)^{N} \big( e^{2k}-1 \big)}{2^{N+1}k}, $

$ \begin{eqnarray*} t_\pm& = &\pm\Bigg[ \frac{a \big( N+k^2 \big)+1}{N+k^2} \cdot \frac{2^{N+1}k}{b\big(N +k^2 \big)\big( e^2-1 \big)^{N} \big( e^{2k}-1 \big)} \Bigg]^\frac{1}{2} \\ & = &\pm\Bigg[ \frac{ 2^{N+1}k \big[1+ a(N+k^2)\big] } {b\big( N+k^2 \big)^2\big( e^2-1 \big)^{N} \big( e^{2k}-1 \big)} \Bigg]^\frac{1}{2}. \end{eqnarray*} $

问题(2.3)有解可以表示为

$ \begin{eqnarray*} u(x) = \pm\Bigg[ \frac{ 2^{N+1}k \big[1+ a(N+k^2)\big] }{b\big( N+k^2 \big)^2\big( e^2-1 \big)^{N} \big( e^{2k}-1 \big)} \Bigg]^\frac{1}{2} \cdot \exp \Big[kx_{N+1}\sum\limits_{i = 1}^N x_i\Big]. \end{eqnarray*} $

因此根据$ k\neq0 $的任意性可知问题(2.3)存在无穷多解.

证 Ⅱ 令$ v(x) = \prod\limits_{i = 1}^N\sin (\alpha_ix_i-\beta_i) $, $ \alpha_i\neq0, \beta_i\ (i = 1, \cdots, N) $为任意常数,那么

(2.5) $ \begin{equation} \left. \begin{array}{ll} \frac{\partial v(x)}{\partial x_i} = \alpha_i \cos (\alpha_ix_i-\beta_i) \cdot\prod\limits_{j = 1, j\neq i}^N \sin (\alpha_jx_j-\beta_j), \\ \frac{\partial^2 v(x)}{\partial x_i^2} = -\alpha_i^2 \prod\limits_{i = 1}^N\sin (\alpha_ix_i-\beta_i) = -\alpha_i^2\cdot v(x), \\ \Delta v(x) = \sum\limits_{i = 1}^N \frac{\partial^2 v(x)}{\partial x_i^2} = -\Big[\sum\limits_{i = 1}^N\alpha_i^2\Big]\cdot v(x). \end{array}\right\} \end{equation} $

$ a>0 = b $时总可选无穷多$ \alpha_i $满足$ a = (\sum\limits_{i = 1}^N\alpha_i^2)^{-1} $.因此问题(2.3)有无穷多解.

$ b\neq0 $时,根据(2.5)式可得

$ \begin{eqnarray*} |\nabla v(x)|^2 & = &\sum\limits_{i = 1}^N\Big(\frac{\partial v(x)}{\partial x_i}\Big)^2 = \sum\limits_{i = 1}^N\Big[\alpha_i \cos (\alpha_ix_i-\beta_i) \prod\limits_{j = 1, j\neq i}^N \sin (\alpha_jx_j-\beta_j)\Big]^2 \\ & = &\sum\limits_{i = 1}^N\Big[\alpha_i^2 \cos^2 (\alpha_ix_i-\beta_i) \prod\limits_{j = 1, j\neq i}^N \sin^2 (\alpha_jx_j-\beta_j)\Big] \\ & = &\sum\limits_{i = 1}^N\Big[ \alpha_i^2\cdot\frac{1+\cos2(\alpha_ix_i-\beta_i)}{2} \prod\limits_{j = 1, j\neq i}^N \frac{1-\cos2(\alpha_jx_j-\beta_j)}{2} \Big], \end{eqnarray*} $

显然$ \nabla v\in L^2(\Omega)\bigcap C(\Omega) $,因此$ \int_\Omega |\nabla v(x)|^2{\rm d}x $是一个正常数(仍设为$ \Lambda $).与(2.2)式类似, $ a, b>0 $时对任意的$ \alpha_i>(Na)^{-\frac{1}{2}} $, $ a>0>b $时对所有$ |\alpha_i|<(Na)^{-\frac{1}{2}} $以及$ a, b<0 $时对任意的$ \alpha_i\in{{\Bbb R}} $,都可以得出

$ \begin{eqnarray*} u(x): = \pm\Bigg[\frac{a\sum\limits_{i = 1}^N\alpha_i^2-1}{b\Lambda\sum\limits_{i = 1}^N\alpha_i^2} \Bigg]^\frac{1}{2} \cdot \prod\limits_{i = 1}^N\sin (\alpha_ix_i-\beta_i) : = C \prod\limits_{i = 1}^N\sin (\alpha_ix_i-\beta_i) \end{eqnarray*} $

是问题(2.3)的解.再根据$ \alpha_i $的可变性得出$ a>0 $且$ b\in {{\Bbb R}} $或者$ b<0 $且$ a\in {{\Bbb R}} $时问题(2.3)存在无穷多古典解.即“不满足$ a\leq0\leq b $ "时,问题(2.3)存在无穷多古典解.

特别地,对任意的非零整数$ n_i $,如果取$ \alpha_i\equiv \frac{n_i\pi}{d_i-c_i} $,取$ \beta_i\equiv \frac{n_ic_i\pi}{d_i-c_i} $,那么

$ v(x) = \prod\limits_{i = 1}^N\sin \frac{n_i\pi(x_i-c_i)}{d_i-c_i}, \alpha_i^2 = \frac{n_i^2\pi^2}{(d_i-c_i)^2}, \Lambda = |\Omega|\sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(d_i-c_i)^2}, $

当$ ab>0 $时总存在$ n_{i_0}>0\ (i = 1, \cdots, N) $使得$ |n_{i}|>n_{i_0}\ (i = 1, \cdots, N) $时$ t_\pm $是实数; $ b<0<a<\big[\pi^2\sum\limits_{i = 1}^N(d_i-c_i)^{-2}\big]^{-1} $时显然$ t_\pm $是实数.因此可得出

$ \begin{eqnarray*} u(x): = \pm\Bigg[ \frac{a\sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(d_i-c_i)^2}-1}{b \big( \sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(d_i-c_i)^2} \big )^2} \Bigg]^{\frac{1}{2}} \cdot \prod\limits_{i = 1}^N\sin \frac{n_i\pi(x_i-c_i)}{d_i-c_i} \end{eqnarray*} $

是问题

(2.6) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a-b\int_\Omega|\nabla u|^2{\rm d}x\Big)\Delta u = u, &x\in\Omega, \\ u = 0, &x\in\partial\Omega \end{array}\right. \end{equation} $

的解,并且由$ |n_i|\geq |n_{i_0}| $的任意性知$ ab>0 $或$ a = 0>b $时有无穷多解. $ a>0 = b $时问题(2.6)退化为特征值问题$ -\Delta u = a^{-1}u $,此时的特征值恰好是$ \sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(c_i-d_i)^2} $, $ n_i $$ (i = 1, \cdots, N) $是任意$ N $个非零整数.因此$ b = 0 $且$ a\sum\limits_{i = 1}^N\frac{n_i^2\pi^2}{(c_i-d_i)^2} = 1 $时,对任意常数$ t $,直接可以验证

$ \begin{eqnarray*} u(x): = t\prod\limits_{i = 1}^N\sin \frac{n_i\pi(x_i-c_i)}{d_i-c_i} \end{eqnarray*} $

都是问题(2.6)的解,因此问题(2.6)有无穷多解.

例2.3 设$ \Omega = [c_1, c_1+ \pi]\subset{{\Bbb R}} $,则对任意的整数$ n, m $和实数$ t $,当$ b = 0 $且满足$ a = \frac{1}{n^2\pi^2} $时,取$ \alpha_1\equiv n $,取$ \beta_1\equiv n c_1+ 2m\pi $,那么可得

$ u_{t}(x) = \pm t\sin \left[ n\pi (x - c_1 ) + 2m\pi \right] $

是问题(2.6)的解.即对任意整数$ n\neq0 $来说零边值问题$ -\frac{1}{n^2\pi^2}\Delta u = u $有无穷多解.

证 Ⅲ 令$ v(x) = \sum\limits_{i = 1}^N(\alpha_ix_i^2+\beta_ix_i+\gamma_i) $, $ \alpha_i, \beta_i , \gamma_i\; (i = 1, \cdots, N) $为任意常数, $ \alpha_i, \sum\limits_{i = 1}^N\alpha_i\neq0 $,那么

$ \frac{\partial v(x)}{\partial x_i} = 2\alpha_i x_i +\beta_i, \frac{\partial^2 v(x)}{\partial x_i^2} = 2\alpha_i, \Delta v(x) = \sum\limits_{i = 1}^N \frac{\partial^2 v(x)}{\partial x_i^2} = \sum\limits_{i = 1}^N2\alpha_i, $

$ |\nabla v(x)|^2 = \sum\limits_{i = 1}^N\Big(\frac{\partial v(x)}{\partial x_i}\Big)^2 = \sum\limits_{i = 1}^N (4\alpha_i^2x_i^2+4\alpha_i\beta_i x_i+\beta_i^2), \int_\Omega |\nabla v(x)|^2{\rm d}x : = \Lambda. $

$ q = 0 $时,如果$ b = 0 $,则可选无穷多$ \alpha_i $使得$ a = \big(\sum\limits_{i = 1}^N2\alpha_i\big)^{-1} $;如果$ b\neq0 $,总可选无穷多$ \alpha_i, \beta_i , \gamma_i $使得$ t_\pm = \pm \Big[\frac{1+ 2a\sum\limits_{i = 1}^N\alpha_i}{2b\Lambda\sum\limits_{i = 1}^N \alpha_i}\Big]^\frac{1}{2} $是方程$ bt^2 \Lambda-a = \big(\sum\limits_{i = 1}^N2\alpha_i\big)^{-1} $的实值解,从而

$ u(x) = \pm \Bigg[\frac{1+ 2a\sum\limits_{i = 1}^N\alpha_i}{2b\Lambda\sum\limits_{i = 1}^N \alpha_i}\Bigg]^\frac{1}{2} \cdot\sum\limits_{i = 1}^N(\alpha_ix_i^2+\beta_ix_i+\gamma_i) $

即为$ -\Big( a-b\int_\Omega|\nabla u|^2{\rm d}x\Big)\Delta u = 1 $的一族解,并且有无穷个.

证 IV 设$ i = 1, \cdots, N $,任取常数$ \alpha_i\neq0 $, $ \epsilon>0 $, $ p(p\neq-1) $, $ \mu $是依赖于常数$ \alpha_i $和$ p $的待定正常数,令$ u_\epsilon = \mu\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i \Big)^{-p} $.为保证$ u_\epsilon $始终有意义,所取的$ \epsilon $和$ \alpha_i $必须使$ \big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i \big)>0 $对任意$ x = (x_1, x_2, \cdots, x_N)\in\Omega $都成立,而实际上这个条件不难达到,由于$ \Omega $是有界矩体,对任意$ x\in\Omega $,只需要取$ N $对常数”$ \epsilon_i, \alpha_i $"使得$ \epsilon = \sum\limits_{i = 1}^N\epsilon_i $且$ \epsilon_i>\alpha_ix_i $对每个$ x_i\in[c_i, d_i] $$ (i = 1, 2, \cdots, N) $都成立即可,当然也可取$ \epsilon_i\gg\alpha_ix_i $.那么

(2.7) $ \begin{equation} \left. \begin{array}{lll} \frac{\partial u_\epsilon}{\partial x_i} = -p\alpha_i\mu\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{-p-1}, \\ \nabla u_\epsilon = \Bigg( \frac{-p\alpha_1\mu}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{p+1}}, \cdots, \frac{-p\alpha_N\mu}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{p+1}} \Bigg), \\ \frac{\partial^2 u_\epsilon}{\partial x_i^2} = p(p+1)\alpha_i^2\mu\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{-p-2}, \\ \Delta u_\epsilon = \sum\limits_{i = 1}^N\frac{\partial^2 u_\epsilon}{\partial x_i^2} = p(p+1)\mu\sum\limits_{i = 1}^N\alpha_i^2\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{-(p+2)}. \end{array}\right\} \end{equation} $

下面记$ [s] $为$ q+1 $的符号, $ q+1>0 $时$ [s] $表示$ "+" $, $ q+1<0 $时$ [s] $表示$ "-" $,因此$ [s](q+1)>0 $.用$ C_{i1} $, $ C_{i2} $表示只与$ \Omega, \alpha_i\;(i = 1, \cdots, N), N $有关而与变量无关的常数,上一行到下一行表示的可能不是同一个数. $ \widehat{\Omega} $表示$ \Omega $划去第$ N $维, $ \widehat{x}\in\widehat{\Omega} $. $ 2^* $表示临界指数, $ N = 1, 2 $时$ 2^* = +\infty $, $ N\geq3 $时$ 2^* = \frac{2N}{N-2} $.

情形A $ p\notin[-1, 0] $时总有$ p(p+1)>0 $.令

$ \frac{p(p+1)\mu\sum\limits_{i = 1}^N\alpha_i^2}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{p+2}} = \Bigg[ \frac{\mu}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^p} \Bigg]^q, $

那么$ p+2 = pq $且$ p(p+1)\mu\sum\limits_{i = 1}^N\alpha_i^2 = \mu^q $,由此可知

(2.8) $ \begin{equation} \left. \begin{array}{l} p = \frac{2}{q-1}, p(p+1) = \frac{2(q+1)}{(q-1)^2}>0, \mu = \Bigg[\frac{2[s](q+1)\sum\limits_{i = 1}^N\alpha_i^2}{(q-1)^2}\Bigg]^\frac{1}{q-1} . \end{array}\right. \end{equation} $

由于$ p\notin[-1, 0] $, (2.8)式意味着$ q>-1 $且$ q\neq1 $.因此$ \Delta u = u^q $具有如下的一族古典解

(2.9) $ \begin{eqnarray} u_\epsilon = \mu\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{-p} = \Bigg[\frac{2[s](q+1)\sum\limits_{i = 1}^N\alpha_i^2}{(q-1)^2}\Bigg]^\frac{1}{q-1} \Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i \Big)^\frac{2}{1- q}. \end{eqnarray} $

特别地,当$ p = -2 $时,根据(2.8)–(2.9)式有$ q = 0, \Delta u_\epsilon = u_\epsilon^q = 1 $是定理Ⅲ的一种类型,此时

(2.10) $ \begin{eqnarray} \mu = \Big(2\sum\limits_{i = 1}^N\alpha_i^2\Big)^{-1}, u_\epsilon = \Big(2\sum\limits_{i = 1}^N\alpha_i^2\Big)^{-1}\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^2. \end{eqnarray} $

当$ p<-2 $时$ -1<q<0 $,当$ -2<p<-1 $时$ 0<q<1 $,分别对应了奇异指数与次线性指数;当$ p>0 $时$ q>1 $,与超线性次临界指数,临界指数及超临界指数所对应.

情形B $ p\in(-1, 0) $时总有$ p(p+1)<0 $,令

$ -\frac{p(p+1)\mu\sum\limits_{i = 1}^N\alpha_i^2}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{p+2}} = \Bigg[ \frac{\mu}{(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^p} \Bigg]^q, $

则$ p+2 = pq $且$ -p(p+1)\mu\sum\limits_{i = 1}^N\alpha_i^2 = \mu^q $.此时有

(2.11) $ \begin{equation} \left. \begin{array}{l} -1<p = \frac{2}{q-1}<0, p(p+1) = \frac{2(q+1)}{(q-1)^2}<0, \mu = \Bigg[\frac{2[s](q+1)\sum\limits_{i = 1}^N\alpha_i^2}{(q-1)^2}\Bigg]^\frac{1}{q-1}, \end{array}\right. \end{equation} $

从而得出$ q<-1 $,对应了强奇异指数情形.因此$ q<-1 $时(2.9)式满足$ -\Delta u_\epsilon = u_\epsilon^q $.

由于$ q\neq1 $,根据(2.7)–(2.11)式可得

$ \begin{eqnarray*} |\nabla u_\epsilon|^2 = \frac{p^2\mu^2\sum\limits_{i = 1}^N\alpha_i^2} {(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{2(p+1)}} = \sum\limits_{i = 1}^N\alpha_i^2 \frac{\Bigg[\frac{2}{q-1} \big[\frac{[s]2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{(q-1)^2}\big]^\frac{1}{q-1}\Bigg]^2} {(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^{\frac{2(q+1)}{q-1}}}. \end{eqnarray*} $

注意到$ q\in(-1, 1)\bigcup(1, 2^*-1)\bigcup(2^*-1, +\infty) $时$ j(q-1)-2(q+1)\neq0 $对任意的$ j = 1, \cdots, N $总成立; $ q = 2^*-1 $且$ N\geq3 $时$ j(q-1)-2(q+1)\neq0 $对任意的$ j = 1, \cdots, N-1 $也成立,但对$ j = N $有$ j(q-1)-2(q+1) = 0 $.从而

$ \begin{eqnarray*} &&\int_\Omega|\nabla u_\epsilon|^2{\rm d}x = \Big[\sum\limits_{i = 1}^N\alpha_i^2\Big] \Big[\frac{2}{q-1}\Big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2} {[s](q-1)^2}\Big]^\frac{1}{q-1}\Big]^2 \int_{\widehat{\Omega}} \int_{c_N}^{d_N} \Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{-\frac{2(q+1)}{q-1}}{\rm d}x_N {\rm d}\widehat{x} \nonumber\\ & = &\frac{(q-1)\alpha_N^{-1}}{q-1-2(q+1)}\Big[\sum\limits_{i = 1}^N\alpha_i^2\Big] \Big[ \frac{2}{q-1}\Big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{[s](q-1)^2}\Big]^\frac{1}{q-1} \Big]^2 \int_{\widehat{\Omega}} \Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^{\frac{q-1-2(q+1)}{q-1}}\Big|_{c_N}^{d_N} {\rm d}\widehat{x} \nonumber\\ & = &\frac{(q-1) \alpha_N^{-1}}{q-1-2(q+1)}\Big[\sum\limits_{i = 1}^N\alpha_i^2\Big] \Big[ \frac{2}{q-1}\Big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{[s](q-1)^2}\Big]^\frac{1}{q-1} \Big]^2 \nonumber\\ & & \times \int_{\widehat{\Omega}} \left[ \Big(\epsilon+\alpha_N{d_N}+\sum\limits_{i = 1}^{N-1}\alpha_ix_i\Big)^{\frac{q-1-2(q+1)}{q-1}} - \Big(\epsilon+\alpha_N{c_N}+\sum\limits_{i = 1}^{N-1}\alpha_ix_i\Big)^{\frac{q-1-2(q+1)}{q-1}} \right]{\rm d}\widehat{x} \nonumber\\ & = &\Big[\prod\limits_{j = 1}^{N-1}\frac{(q-1) \alpha_{N-j+1}^{-1}}{j(q-1)-2(q+1)} \Big] \Big[\sum\limits_{j = 1}^N\alpha_i^2\Big] \Big[ \frac{2}{q-1}\Big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{[s](q-1)^2}\Big]^\frac{1}{q-1} \Big]^2 \nonumber \\ & & \times \int_{c_1}^{d_1} \Big[\sum\limits_{i = 1}^{2^{N-1}} C_{i1}\Big(C_{i2}+\alpha_1x_1\Big)^{\frac{(N-1)(q-1)-2(q+1)}{q-1}}\Big] {\rm d}x_1 \nonumber\\ & = &\left\{\begin{array}{ll} \Big[\prod\limits_{j = 1}^{N}\frac{(q-1) \alpha_{N-j+1}^{-1}}{j(q-1)-2(q+1)} \Big] \Big[\sum\limits_{i = 1}^N\alpha_i^2\Big] \Big[ \frac{2}{q-1}\Big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{[s](q-1)^2}\Big]^\frac{1}{q-1} \Big]^2 \\ \times\Big[\sum\limits_{i = 1}^{2^{N-1}} C_{i1}\big(C_{i2}+\alpha_1x_1\big)^{\frac{N(q-1)-2(q+1)}{q-1}}\Big|_{c_1}^{d_1}\Big] = :{ \tilde{\Lambda}_1 \Big[\sum\limits_{i = 1}^{2^N} C_{i1} C_{i2}^{\frac{N(q-1)-2(q+1)}{q-1}}\Big] : = \Lambda}, \\ q\in(-\infty, -1)\bigcup(-1, 1)\bigcup(1, 2^*-1)\bigcup(2^*-1, +\infty) ;\\ \mbox{或}\ \Big[\frac{1}{\alpha_1}\prod\limits_{j = 1}^{N-1}\frac{(q-1) \alpha_{N-j+1}^{-1}}{j(q-1)-2(q+1)} \Big] \Big[\sum\limits_{i = 1}^N\alpha_i^2\Big] \Big[ \frac{2}{q-1}\Big[\frac{2(q+1) \sum\limits_{i = 1}^N\alpha_i^2}{[s](q-1)^2}\Big]^\frac{1}{q-1} \Big]^2 \\ \times\Big[\sum\limits_{i = 1}^{2^{N-1}} [C_{i1}\ln(C_{i2}+\alpha_1x_1)] \Big|_{c_1}^{d_1} \Big] = :{ \tilde{\Lambda}_2\sum\limits_{i = 1}^{2^N} C_{i1}\ln C_{i2} }: = \Lambda, q = 2^*-1 (N\geq3). \end{array}\right. \end{eqnarray*} $

由于$ \Omega $各向有界,总可根据$ \alpha_i $和$ \epsilon $的选取,使得上式中的$ \Lambda $取到$ (0, +\infty) $的任意值.

情形C $ p = 0 $时$ u_\epsilon $为常数且$ \Delta u_\epsilon = 0 $, $ p = -1 $时$ \Delta u_\epsilon = 0 $,两者都无讨论意义.

现考虑如下关于$ t $的方程

(2.12) $ \begin{eqnarray} t^2b\Lambda -a = [s]t^{q-1} = \left\{\begin{array}{ll} t^{q-1} ,  & q>-1; \\ -t^{q-1}, & q<-1, \end{array}\right. \end{eqnarray} $

此时方程(2.12)的解等价于函数$ g(t) = t^2b\Lambda -a-[s]t^{q-1} $的零点.直接利用连续函数的零点存在定理可得:当满足$ a\leq0\leq b $且$ q<-1 $;或者$ a\geq0\geq b $且$ q>-1 $这两种情形时, $ g(t) $在$ (0, +\infty) $都不存在零点,而其他情形都存在($ b\neq0 $时依赖于$ \Lambda $的)零点,亦即方程(2.12)始终有实值解$ t_+>0 $,因此问题(1.1)存在无穷多幂函数型古典解,其解析式形如

$ \begin{eqnarray*} u(x) = \frac{\big[\frac{2(q+1)\sum\limits_{i = 1}^N\alpha_i^2}{(q-1)^2}\big]^\frac{1}{q-1}t_+} {(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)^\frac{2}{q-1}} = C_{\epsilon, \alpha_i, q, \Omega}\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^\frac{2}{1- q}, \end{eqnarray*} $

这里$ C_{\epsilon, \alpha_i, q, \Omega} $表示常数,它的取值只与$ \epsilon, \alpha_i, q, \Omega $有关.

另外,对任意的$ \varepsilon >0 $, $ N\geq3 $时,直接可以验证问题$ - \Delta u = u^\frac{N+2}{N-2} $有解

(2.13) $ \begin{equation} u_{\varepsilon}: = \frac{[N(N-2)]^{\frac{N-2}{4}}\varepsilon^{\frac{N-2}{2}}} {(\varepsilon^2+|x|^2)^{\frac{N-2}{2}}}. \end{equation} $

根据文献[37-38]等所述,存在常数$ S>0 $使得

$ \int_{{{\Bbb R}} ^N}|\nabla u_{\varepsilon}|^2{\rm d}x = \int_{{{\Bbb R}} ^N}|u_{\varepsilon}|^\frac{2N}{N-2}{\rm d}x = S^\frac{N}{2}. $

因此必然存在常数$ S_{\varepsilon, \Omega}\in(0, S) $使得

$ \int_\Omega|\nabla u_{\varepsilon}|^2{\rm d}x = S_{\varepsilon, \Omega}^\frac{N}{2}: = \Lambda. $

由于$ t>0 $时$ - t^\frac{4}{N-2}\Delta (tu_\varepsilon) = (tu_\varepsilon)^\frac{N+2}{N-2} $,考虑关于$ t $的方程$ a- t^2b\Lambda = t^\frac{4}{N-2} $,显然$ a>0 $, $ b\geq0 $时它有正实数解(记为$ t_+ $).因此$ q = \frac{N+2}{N-2} $时

$ \begin{eqnarray*} u(x) = t_+u_\varepsilon = C_{\varepsilon, N, \Omega}\big(\varepsilon^2+|x|^2\big)^{\frac{2-N}{2}} \end{eqnarray*} $

是问题(1.1)的古典解,再由$ \varepsilon>0 $的任意性得出无穷多解.

本文主要在各向有界的矩体$ \Omega $上研究并获得了如下问题古典解的存在性与解析式

(3.1) $ \begin{equation} -\Big(a-b\int_\Omega|\nabla u|^2{\rm d}x\Big)\Delta u = u^q, x\in\Omega\subset {{\Bbb R}} ^N (N\geq1), \end{equation} $

这里$ q\neq-1 $,常数$ a, b $不同时为零,而右端为$ \lambda u^q $时两端除以$ \lambda $即与该问题等价.目标是考虑古典解,因此在涉及函数梯度时需要考虑其连续性,并且在非局部积分中需要考虑积分值是否有意义.特别当$ a $与$ b $同号时对所有正整数$ N $来说,存在如下形式的古典解

$ \begin{eqnarray*} u(x) = \left\{\begin{array}{lll}\left. \begin{array}{lll} C_{[\alpha_i, \beta_i, \gamma_i, N, \Omega]}\sum\limits_{i = 1}^N(\alpha_ix_i^2+\beta_ix_i+\gamma_i),   q = 0;\\ C_{[\beta_i, N, \Omega]}\prod\limits_{i = 1}^N \beta_i^{x_i} \ \mbox{或} \ C_{[n_i, \alpha_i, \beta_i, N, \Omega]}\prod\limits_{i = 1}^N\sin (\alpha_ix_i-\beta_i),  q = 1;\\ C_{[\epsilon, \alpha_i, q, N, \Omega]}\Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i\Big)^\frac{2}{1- q},   q\neq\pm 1; \end{array}\right\} &N\geq1; \\ { C_{[\varepsilon, N, \Omega]}\Big(\varepsilon^2+|x|^2\Big)^{\frac{2-N}{2}}, } { }  q = \frac{N+2}{N-2}, &N\geq3. \end{array}\right. \end{eqnarray*} $

这里$ C_{[\cdot]} $表示$ C $是一个只与$ {[\cdot]} $中的信息有关而与变量无关的常数,解的数量主要取决于$ {[\cdot]} $中各个量的取值.当$ a $与$ b $不同号时对应的解参见文中定理及其证明.

思考 指数扰动已成为研究解存在性问题中的一种基本工具,并且已有大量成果^[39-40],但基于本文出现的问题,我们提出如下关于指数扰动的问题,希望能与读者讨论.

问题 根据证明过程可知,对任意正实数$ \lambda, \mu $, $ -\Delta u = \lambda u^{-(1+\frac{1} {\mu})} $与$ \Delta u = \lambda u^{-(1-\frac{1}{\mu})} $都有古典解,注意到$ { \lim_{\mu\to+\infty}}(1+\frac{1}{\mu}) = { \lim_{\mu\to+\infty}}(1-\frac{1}{\mu}) = 1 $,但$ q = -(1\pm\frac{1}{\mu}) $与$ q = -1 $不同.另外,在证明定理IV的过程中$ N\geq3 $时如果$ q = \frac{N-2}{N+2} $,则对应出现$ \int_\Omega|\nabla u|^2{\rm d}x $的值形如$ S_1 = C_1\ln(C_2+\alpha x_1)\Big|_{c}^{d} $,而$ q\neq\frac{N-2}{N+2} $时的值形如$ S_2 = C_1(C_2+\alpha x_1)^{\frac{N(q-1)-2(q+1)}{q-1}}\Big|_{c}^{d} $,通常情况下$ S_1\neq S_2 $.那么,对$ {\cal L}u = \frac{1}{u}+f(x, u) $或者$ {\cal L}u = u^\frac{N-2}{N+2}+f(x, u) $ (这里$ {\cal L} $是算子, $ f(x, u) $是连续函数),通过指数扰动获得的极限函数是否是真实的弱解(或解)?

例3.1 在$ {{\Bbb R}} ^N $的有界矩体$ \Omega $中,对任意的$ \lambda, \varepsilon>0 $, $ q(q-1)\neq0 $,令$ u = \lambda^\frac{1}{q-1}v $,那么

$ \pm\Delta v = \lambda v^q \Leftrightarrow \pm\Delta \big(\lambda^\frac{1}{q-1}u\big) = \big(\lambda^\frac{1}{q-1} u\big)^q \Leftrightarrow \pm\Delta u = u^q. $

即求解$ \pm\Delta u = \pm\lambda u^q $等价于求解$ \pm\Delta u = u^q $.对$ \pm\Delta u = u^{-(1\mp\theta)} $,不同的$ \theta>0 $对应的解也不同.取$ \theta = \frac{1}{n} $,则存在序列$ \{ u_n\}_{n = 1}^{\infty} $使得$ \Delta u_n = u_n^{-(1-\frac{1}{n})} $,尽管$ \lim_{n\to\infty} \Delta u_n = \lim u_n^{-1} $,这种方式还是找不到$ u $使得$ \Delta u = u^{-1} $ (对$ -\Delta u_n = u_n^{-(1+\frac{1}{n})} $也有类似的结果).事实上,可取适当大的正数$ \epsilon $使得$ \epsilon+\sum\limits_{i = 1}^N\alpha_ix_i>0 $总成立,此时$ \{ u_n\}_{n = 1}^{\infty} $可取

$ u_{\alpha, n} = \Big[\frac{2n\sum\limits_{i = 1}^N\alpha_i^2}{(2n-1)^2}\Big]^\frac{n}{1-2n} \Big(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i \Big)^\frac{2n}{2n-1}, n = 1, 2, \cdots, \infty, \alpha = (\alpha_1, \cdots, \alpha_N). $

由于$ \Omega $有界,对$ x\in\Omega $的任意$ x_i $,总存在$ M>0 $使得$ 0<(\epsilon+\sum\limits_{i = 1}^N\alpha_ix_i)<M<+\infty $.注意$ \epsilon $和$ \alpha_i $的可变性导致$ u_{\alpha, n}\; (n = 1, \cdots, +\infty) $都有无穷多个.取$ \alpha_i\equiv1 $,此时

$ \begin{eqnarray*} \lim\limits_{n\to\infty}u_n & = &\lim\limits_{n\to\infty} \left[\Big[\frac{2Nn}{(2n-1)^2}\Big]^\frac{n}{1-2n} \Bigl(\epsilon+\sum\limits_{i = 1}^Nx_i \Bigr)^\frac{2n}{2n-1} \right]\\ & \geq& \frac{M}{M^2+1} \lim\limits_{n\to\infty}\Big[\frac{2Nn}{(2n-1)^2}\Big]^\frac{n}{1-2n} = +\infty, \\ \lim\limits_{n\to\infty}\Delta u_n & = &\lim\limits_{n\to\infty} \left[\Big[\frac{2Nn}{(2n-1)^2}\Big]^\frac{n}{1-2n} \Bigl(\epsilon+\sum\limits_{i = 1}^Nx_i \Bigr)^\frac{2n}{2n-1} \right]^\frac{1-n}{n}\\ & \leq& \frac{M^2+1}{M}\lim\limits_{n\to\infty}\Big[\frac{2Nn}{(2n-1)^2}\Big]^\frac{1-n}{1-2n} = 0, \end{eqnarray*} $

注意这是因为$ \frac{M}{M^2+1}<\min\big\{M, \frac{1}{M}\big\} \leq\max\big\{M, \frac{1}{M}\big\}<\frac{M^2+1}{M} $.从上式可以看出极限函数不存在,而且显然满足$ \lim_{n\to\infty} \Delta u_n = \lim_{n\to\infty} u_n^{-1} = 0 $,但这种方式还是找不到$ u $使得$ \Delta u = u^{-1} $.同理,根据恒等式$ -\Delta u_n = u_n^{-(1+\frac{1}{n})} $知,指数扰动的方式也不能获得$ -\Delta u = u^{-1} $的解.

展望 右端项替换为常数亦或是$ q = 0 $的情形,我们认为当$ \Omega\not\ni\{0\} $是以$ 0 $为心的有界开球时,问题(1.1)亦即问题(3.1)还存在满足球面零值条件的解,结果正进一步验证中.