有关非线性椭圆型方程的正解问题,近二十年来已有许多研究并取得了丰硕成果,如文献[1-7].纵观这些成果,绝大多数文献是研究方程存在正整解的充分条件,探讨方程存在正整解的充分必要条件的文章则极少见.本文研究的是$ {\Bbb R} ^N $上一类非线性双调和方程

(1.1) $ \begin{equation} \triangle^2u = f(|x|, u, |\nabla u|), \quad x\in {\Bbb R} ^N, N>2 \end{equation} $

存在正整解的充分必要条件及解的性质.在方程(1.1)中, $ \triangle $是Laplace算子, $ |x| $是Euclidean长度, $ \nabla $是Hamilton算子.方程(1.1)的正的径向整体解是指$ u\in C^4({\Bbb R} ^N)(N > 2) $,满足$ u(x) = u(|x|)(x\in {\Bbb R} ^N) $,且其在$ {\Bbb R} ^N $中逐点满足方程(1.1).如果一个径向对称的函数$ u(x) = y(|x|) $是方程(1.1)的整体解,那么$ y(t)\in C^4[0, \infty) $,且满足下述的常微分方程

(1.2) $ \begin{equation} L^2y(t) = f(t, y(t), |y\prime (t)|), \ t \geq 0 \end{equation} $

及初始条件$ y(0) = \gamma_0, y\prime (0) = 0, (Ly)(0) = \gamma_1, (Ly)\prime(0) = 0 $,其中$ (Ly)(0) $表示$ \lim\limits_{t\to 0} (Ly)(t) $, $ (Ly)\prime(0) $表示$ \lim\limits_{t\to 0} (Ly)\prime(t) $, $ \gamma_0, \gamma_1 $是正的常数, $ L $表示$ N $维Laplace算子$ \triangle $的极坐标形式,即

(1.3) $ \begin{equation} L = \frac{1}{t^{N-1}}\frac{\rm d}{{\rm d}t}(t^{N-1}\frac{\rm d}{{\rm d}t}), \quad t = |x|. \end{equation} $

记$ L^1 = L $, $ L^2 $表示$ 2 $重Laplace算子$ \triangle^2 $的极坐标形式.为了证明本文的结论,引入文献[3-4]曾经用过的积分算子$ \Psi: C[0, \infty)\to C^2[0, \infty) $如下:

(1.4) $ \begin{equation} \Psi h(t)\equiv (\Psi h)(t) = \frac{1}{N-2} \int_0^t \Big[1-(\frac{s}{t})^{N-2}\Big]sh(s){\rm d}s, t\geq 0, N\geq 3, \end{equation} $

其中$ \Psi h(0) $理解为$ \lim\limits_{t\to 0}\Psi h(t) $.可以验证算子$ \Psi $是线性的、单调的(即$ h_1\leq h_2 \Rightarrow \Psi h_1\leq \Psi h_2 $).进一步记$ \Psi^1 = \Psi $,定义$ C[0, \infty) $上的算子$ \Psi^2 $如下:

$ \Psi^2 h(t)\equiv (\Psi^2h)(t) = (\Psi(\Psi h))(t)\equiv \Psi(\Psi h(t)) \geq 0, t\geq 0, h\in C[0, \infty). $

引理1.1  设$ h\in C[0, \infty) $,则$ L(\Psi h(t)) = h(t), t\geq 0 $.即$ u(x) = (\Psi h)(|x|) $是方程$ \triangle u = h(|x|)(x\in {\Bbb R} ^N) $的正的径向对称整体解.

引理1.2  设$ h\in C[0, \infty) $,且$ h\geq 0 $,则$ L^2(\Psi^2h)(t) = h(t), t\geq 0 $.即$ u(x) = (\Psi^2h)(|x|) $是方程$ \triangle^2 u(x) = h(|x|)(x\in {\Bbb R} ^N, N > 2) $的正的径向对称整体解,且有

(1.5) $ \begin{equation} 0\leq (\Psi^2 h)(t)\leq \frac{t^2}{2(N-2)^2}\int_0^t sh(s){\rm d}s, t\geq 0, \end{equation} $

(1.6) $ \begin{equation} 0\leq \frac{\rm d}{{\rm d}t}(\Psi^2 h)(t)\leq \frac{t}{N-2}\int_0^t sh(s){\rm d}s, t\geq 0. \end{equation} $

引理1.1,引理1.2的证明见参考文献[4].

引理1.3 设$ h\in C[0, \infty) $,且$ h\geq 0 $,则

(1.7) $ \begin{equation} \lim\limits_{t\to \infty}\frac{\Psi^2 h(t)}{t^2} = \frac{M_0}{2N}, \end{equation} $

其中

(1.8) $ \begin{equation} M_0 = \frac{1}{N-2}\int_0^\infty th(t){\rm d}t<\infty, \ \ (\mbox{或} \; M_0 = \infty). \end{equation} $

证 先证$ M_0 < \infty $的情形.由(1.4)式有

$ \lim\limits_{t\to \infty}(\Psi h)(t) = \lim\limits_{t\to\infty} \frac{1}{N-2}\int_0^t\Big [1-\Big(\frac{s}{t}\Big)^{N-2}\Big ]sh(s){\rm d}s = \frac{1}{N-2}\int_0^\infty sh(s){\rm d}s = M_0. $

$ \begin{eqnarray*} \lim\limits_{t\to \infty}\frac{(\Psi^2h)(t)}{t^2} & = &\lim\limits_{t\to \infty}\frac{(\Psi(\Psi h))(t)}{t^2} \\ & = &\lim\limits_{t\to \infty}\frac{\frac{1}{N-2}\int_0^t\Big[ 1-\big( \frac{s}{t}\big )^{N-2}\Big] s(\Psi h)(s){\rm d}s }{t^2} \Big (\frac{\infty}{\infty}\Big )\\ & = &\lim\limits_{t\to \infty} \frac{\int_0^t\big( \frac{s}{t}\big) ^{N-1} (\Psi h)(s){\rm d}s}{2t}\\ & = & \lim\limits_{t\to \infty} \frac{\int_0^t s^{N-1}(\Psi h)(s){\rm d}s}{2t^N} \Big(\frac{\infty}{\infty}\Big) \\ & = &\lim\limits_{t\to \infty} \frac{(\Psi h)(t)}{2N} = \frac{M_0}{2N}. \end{eqnarray*} $

上述过程也表明当$ M_0 = \infty $时也成立.

在下面的讨论中引入记号$ G = [0, \infty)\times (0, \infty)\times[0, \infty) $.

定理2.1  假设方程$ (1.1) $中的函数$ f $满足

(ⅰ) $ f:G\to [0, \infty) $是连续的,且不恒等于$ 0 $;

(ⅱ)对固定的$ t\geq 0 $,函数$ f(t, u, v) $关于$ u\in(0, \infty) $非增,关于$ v\in[0, \infty) $非增;则方程$ (1.1) $存在满足下述性质

(2.1) $ \begin{equation} \lim\limits_{|x|\to\infty}\frac{u(x)}{|x|^2} = A , \end{equation} $

(2.2) $ \begin{equation} \lim\limits_{|x|\to\infty}\frac{|\nabla u(x)|}{|x|} = B \end{equation} $

的正整解$ u(x) = u(|x|) $的充分必要条件是下面条件(ⅲ)成立,其中$ A > 0, B > 0 $是仅与$ u $有关的常数;

(ⅲ)存在常数$ c > 0 $,使得

$ \int_0^\infty tf(t, c(1+t^2), 2ct){\rm d}t <\infty . $

证 先证充分性.假设定理的条件(ⅲ)成立,故存在常数$ c > 0 $,使得

(2.3) $ \begin{equation} \int_0^\infty tf(t, c(1+t^2), 2ct){\rm d}t <\infty . \end{equation} $

又由定理的条件(ⅰ)和(ⅱ)知,对固定$ (t, u, v)\in G, \xi^{-1}f(t, \xi u, \xi v) $的关于$ \xi \in (0, \infty) $非增,且有

(2.4) $ \begin{equation} \lim\limits_{\xi \to \infty}\xi^{-1} f(t, \xi u, \xi v) = 0. \end{equation} $

故当$ \xi > c $时有

(2.5) $ \begin{equation} \xi^{-1} tf(t, \xi(1+t^2), 2\xi t)\leq c^{-1} tf(t, c(1+t^2), 2ct). \end{equation} $

由(2.3)、(2.4)和(2.5)式以及Lebesgue控制收敛定理得

$ \lim\limits_{\xi \to \infty}\xi^{-1}\int_0^\infty tf(t, \xi (1+t^2), 2\xi t){\rm d}t = 0 . $

于是可以取充分大的正数$ \xi_0 \in (c, \infty) $,使得$ \xi\in(\xi_0, \infty) $时,恒有

(2.6) $ \begin{equation} \int_0^\infty tf(t, \xi (1+t^2), 2\xi t){\rm d}t<\xi . \end{equation} $

作集合

(2.7) $ \begin{equation} Y = \big\{y\in C^1[0, \infty)|\xi(1+t^2)\leq y(t)\leq 2\xi (1+t^2), 2\xi t\leq y^\prime(t)\leq 4\xi t, \ t\geq 0\big\}. \end{equation} $

考虑$ C^1[0, \infty) $中通常的拓扑,即$ C^1[0, \infty) $中点列$ \{y_i(t)\} $收敛于点$ y(t) $定义为$ \{y_i(t)\} $和$ \{y_i^\prime(t)\} $在$ [0, \infty) $的任意紧子集上分别一致收敛于$ y(t) $和$ y^\prime(t) $.易见$ Y $是$ C^1[0, \infty) $中闭凸子集.定义映照$ \pi:Y\to C^1[0, \infty) $如下:

(2.8) $ \begin{equation} \pi y(t) = \xi(1+t^2)+\Psi^2[f(t, y(t), |y^\prime(t)|)], \ \ t\geq 0 . \end{equation} $

下面证明映照$ \pi $把$ Y $连续地映进$ Y $的一个相对紧子集中去.

(a) $ \pi:Y\to Y $.

事实上,对$ \forall y\in Y $,根据(15)式及定理的条件(ⅰ)、(ⅱ)知

(2.9) $ \begin{equation} 0\leq f(t, y(t), |y^\prime(t)|)\leq f(t, \xi(1+t^2), 2\xi t), \ t\geq 0. \end{equation} $

并令

(2.10) $ \begin{equation} g(t) = f(t, y(t), |y^\prime(t)|), w(t) = f(t, \xi(1+t^2), 2\xi t). \end{equation} $

那么根据(2.9)式有

(2.11) $ \begin{equation} 0\leq g(t)\leq w(t), \ \ t\geq 0. \end{equation} $

由(1.5)、(2.6)、(2.10)和(2.11)式得

(2.12) $ \begin{equation} 0\leq \Psi^2g(t)\leq \frac{t^2}{2(N-2)^2}\int_0^t sg(s){\rm d}s\leq \frac{t^2}{2(N-2)^2}\int_0^\infty sw(s){\rm d}s\leq \frac{\xi t^2}{2(N-2)^2}\leq \xi t^2, \ t\geq 0. \end{equation} $

故由(2.8)和(2.12)式得

(2.13) $ \begin{equation} \xi(1+t^2)\leq \pi y(t) = \xi (1+t^2)+\Psi^2 g(t)\leq \xi (1+t^2)+\xi t^2\leq 2\xi(1+t^2), t\geq 0 . \end{equation} $

利用(1.6)、(2.6)及(2.11)式得

(2.14) $ \begin{equation} 0\leq \frac{\rm d}{{\rm d}t}(\Psi^2 g(t))\leq \frac{t}{N-2}\int_0^t sg(s){\rm d}s \leq \frac{t}{N-2}\int_0^\infty sw(s) {\rm d}s \leq 2\xi t, t\geq 0. \end{equation} $

于是由(2.8)和(2.14)式得

(2.15) $ \begin{equation} 2\xi t \leq \frac{\rm d}{{\rm d}t}[\pi y(t)] = 2\xi t +\frac{\rm d}{{\rm d}t}(\Psi^2 g(t))\leq 2\xi t+2\xi t = 4\xi t, t\geq 0. \end{equation} $

由(2.7)、(2.13)和(2.15)式可推出$ \pi:Y\to Y $.

(b) $ \pi $是连续映照.

设$ y_i(t), y(t)\in Y\subset C^1[0, \infty)(i = 1, 2, 3, \cdots) $且$ \{y_i(t)\} $依$ C^1[0, \infty) $的拓扑收敛于$ y(t) $ (当$ i\to \infty $),要证$ \{\pi y_i(t)\} $依$ C^1[0, \infty) $拓扑收敛于$ \pi y(t) $(当$ i\to \infty $),即要证$ \{\pi y_i(t)\} $和$ \{(\pi y_i(t))^\prime \} $在$ [0, \infty) $的任意紧子集上分别一致收敛于$ \pi y(t) $和$ (\pi y)^\prime (t) $.

设$ [0, t_1] $是$ [0, \infty) $的任意紧子区间.先证$ g_i(t)\equiv f(t, y_i(t), |y_i^\prime(t)|)(i = 1, 2, \cdots) $在$ [0, t_1] $上一致收敛于$ g(t)\equiv f(t, y(t), |y^\prime(t)|) $.事实上,因$ y_i(t), y(t)\in Y $,故当$ 0\leq t\leq t_1 $时有

$ \xi \leq y_i(t)( \mbox{或} \, y(t))\leq 2\xi (1+t_1^2), \ i = 1, 2, \cdots , $

$ 0\leq y_i^\prime(t)( \mbox{或}\, y^\prime(t))\leq 4\xi t_1, \ i = 1, 2, \cdots . $

由定理的条件(ⅰ)知, $ f(t, y(t), y^\prime(t)) $在闭区域$ D = [0, t_1]\times[\xi, 2\xi(1+t_1^2)]\times[0, 4\xi t_1] $上连续,故$ f(t, y(t), y^\prime(t)) $在$ D $上一致连续;再注意到$ \{y_i(t)\} $和$ \{y_i^\prime(t)\} $在$ [0, t_1] $上一致收敛于$ y(t) $和$ y^\prime(t) $,从而可证得$ \{g_i(t)\} $在$ [0, t_1] $上一致收敛于$ g(t) $.由(1.4)式知当$ t\in[0, t_1] $时有

$ |\Psi g_i(t)-\Psi g(t)|\leq \frac{1}{N-2}\int_0^t \Big[1-(\frac{s}{t})^{N-2}\Big]s|g_i(s)-g(s)|{\rm d}s\leq \frac{t_1}{N-2}\int_0^{t_1}|g_i(s)-g(s)|{\rm d}s. $

可见$ \{\Psi g_i(t)\} $在$ [0, t_1] $上一致收敛于$ \Psi g(t) $.进而可证$ \{\Psi^2 g_i(t)\} $和$ \pi y_i(t)\} $在$ [0, t_1] $上分别一致收敛于$ \Psi^2 g(t) $和$ \pi y(t) $.当$ t\in[0, t_1] $时,

$ \begin{eqnarray*} \Big|\frac{\rm d}{{\rm d}t}[\Psi^2 g_i(t)]-\frac{\rm d}{{\rm d}t} [\Psi^2 g(t)]\Big|& = &\Big|\frac{1}{N-2}\frac{\rm d}{{\rm d}t}\int_0^t \Big[1-(\frac{s}{t})^{N-2}\Big]s[\Psi g_i(s)-\Psi g(s)]{\rm d}s\Big|\\& = & \Big|\int_0^t(\frac{s}{t})^{N-1}[\Psi g_i(s)-\Psi g(s)]{\rm d}s \Big|\leq \int_0^{t_1}|\Psi g_i(s)-\Psi g(s)|{\rm d}s, \end{eqnarray*} $

可见$ \Big\{\frac{\rm d}{{\rm d}t}[\Psi^2 g_i(t)]\Big\} $在$ [0, t_1] $上一致收敛于$ \frac{\rm d}{{\rm d}t}[\Psi^2 g(t)] $.进而可推出$ \{(\pi y_i)^\prime (t)\} $在$ [0, t_1] $上一致收敛于$ (\pi y)^\prime (t) $.

综上所述, $ \{\pi y_i(t)\} $和$ \{(\pi y_i)^\prime (t)\} $在$ [0, t_1] $上分别一致收敛于$ \pi y(t) $和$ (\pi y)^\prime (t) $.故$ \{(\pi y)_i(t)\} $在$ C^1[0, \infty) $一致收敛于$ \pi y(t) $,即映照$ \pi $是连续的.

(c) $ \pi Y = \{\pi y(t)|y(t)\in Y\} $是相对紧的.

对任意闭区间$ [0, t_1]\subset [0, \infty) $,从(2.13)和(2.15)式可以看出$ \{\pi y(t)|y\in Y\} $与$ \{(\pi y)^\prime (t)|y\in Y\} $在$ [0, t_1] $上是一致有界,故$ \{\pi y(t)|y \in Y\} $在$ [0, t_1] $上是等度连续的.下面证$ \{(\pi y)^\prime (t)|y(t)\in Y\} $在$ [0, t_1] $上也是等度连续的.注意到

(2.16) $ \begin{eqnarray} (\pi y)^\prime(t)& = & \xi (1+t^2)^\prime +\frac{\rm d}{{\rm d}t}\Big(\frac{1}{N-2}\int_0^t\Big[1-(\frac{s}{t})^{N-2}\Big]s\Psi g(s){\rm d}s\Big)\\ & = &2\xi t +\int_0^t(\frac{s}{t})^{N-1}\Psi g(s){\rm d}s. \end{eqnarray} $

(2.17) $ \begin{eqnarray} (\pi y)^{\prime\prime}(t)& = &\frac{\rm d}{{\rm d}t} \Big[\frac{\rm d}{{\rm d}t}(\pi y(t))\Big] = \frac{\rm d}{{\rm d}t}\Big[2\xi t+\frac{\rm d}{{\rm d}t}(\Psi^2 g(t))\Big]\\ & = &2\xi +\frac{\rm d}{{\rm d}t}\Big( \int_0^t(\frac{s}{t})^{N-1}\Psi g(s){\rm d}s\Big)\\ & = &2\xi+\Psi g(t)+(1-N)\frac{1}{t^N}\int_0^t s^{N-1}\Psi g(s){\rm d}s \\ & = &2\xi+I_1+I_2, \ t\geq 0. \end{eqnarray} $

估计(2.17)式中$ I_1 $、$ I_2 $的值如下:由(1.4)、(2.6)和(2.11)式得

(2.18) $ \begin{eqnarray} 0\leq I_1& = &\frac{1}{N-2}\int_0^t\Big[1-(\frac{s}{t})^{N-2}\Big]s g(s){\rm d}s \leq \frac{1}{N-2}\int_0^\infty sg(s){\rm d}s \\ &\leq&\frac{1}{N-2}\int_0^\infty sw(s){\rm d}s< \xi, \quad t\geq 0. \end{eqnarray} $

(2.19) $ \begin{eqnarray} |I_2|& = &(N-1)\frac{1}{t^N}\int_0^ts^{N-1}\Psi g(s){\rm d}s\leq (N-1)\frac{1}{t^N}\int_0^t s^{N-1}\xi {\rm d}s\\ &\leq &\xi \frac{N-1}{N}\frac{t^N}{t^N}\leq \xi. \end{eqnarray} $

由(2.17)、(2.18)和(2.19)式知$ \{(\pi y)^{\prime\prime}(t)|y\in Y\} $在$ [0, \infty) $的任意紧子区间$ [0, t_1] $上一致有界,故$ \{(\pi y)^\prime(t)|y\in Y\} $在$ [0, t_1] $上也是等度连续的.根据Ascoli-Arzela定理(参见文献[8, p10,定理1.30])可知按照$ C^1[0, t_1] $中的拓扑, $ \pi Y $在$ Y $中是相对紧的.进一步用取对角线的方法,可以证明按$ C^1[0, \infty) $拓扑, $ \pi Y $在$ Y $中是相对紧的.由上面的(a)、(b)、(c)可知Schauder-Tychonoff不动点的条件全部满足,于是映照$ \pi $存在着不动点$ y(t)\in Y $ (参见文献[9, p161]),即有$ y(t) = \pi y(t) $.因此有

(2.20) $ \begin{equation} y(t) = \xi(1+t^2)+\Psi^2g(t), t\geq 0. \end{equation} $

注意到$ L^2 $是四阶求导算子,由引理1.2得

$ L^2 y(t) = L^2(\xi(1+t^2)+\Psi^2 g(t)) = g(t), t\geq 0. $

即$ y = y(t) = y(|x|) = u(x) $是方程(1.1)的径向对称的正整体解.下面证明此整体解$ u(x) = y(|x|) $具有性质(2.1)和(2.2).事实上,由引理1.3中的(1.8)式以及(2.6)、(2.8)和(2.11)式可得

(2.21) $ \begin{equation} \lim\limits_{|x|\to\infty}\frac{u(x)}{|x|^2} = \lim\limits_{t\to\infty}\frac{y(t)}{t^2} = \lim\limits_{t\to\infty}\frac{\xi(1+t^2)+\Psi^2 g(t)}{t^2} = \xi+\lim\limits_{t\to\infty} \frac{\Psi^2g(t)}{t^2} = \xi+\frac{M_0}{2N}\equiv A, \end{equation} $

其中

(2.22) $ \begin{equation} M_0 = \frac{1}{N-1}\int_0^\infty tg(t){\rm d}t\leq \frac{1}{N-2}\int_0^\infty tw(t){\rm d}t \leq \frac{\xi}{N-2}<\infty. \end{equation} $

又由(1.6)、(1.7)、(1.8)以及(2.20)式得

(2.23) $ \begin{eqnarray} \lim\limits_{|x|\to\infty}\frac{|\nabla u(x)|}{|x|}& = &\lim\limits_{t\to\infty}\frac{y^\prime (t)}{t} = \lim\limits_{t\to\infty}\frac{2\xi t+\frac{\rm d}{{\rm d}t}(\Psi^2 g(t))}{t} = 2\xi +\lim\limits_{t\to\infty}\frac{\int_0^t\frac{S^{N-1}}{t^{N-1}}\Psi g(s){\rm d}s}{t}\\ & = &2\xi +\lim\limits_{t\to\infty}\frac{\int_0^t s^{N-1}\Psi g(s){\rm d}s}{t^N}\Big(\frac{\infty}{\infty}\Big) = 2\xi+\lim\limits_{t\to\infty}\frac{\Psi g(t)}{N}\\ & = &2\xi+\frac{M_0}{N}\equiv B, \end{eqnarray} $

其中$ M_0 $的意义同(2.22)式.由(2.21)和(2.23)式看到$ A > 0, B > 0 $是仅与整体解$ u(x) $有关的常数.其实,具有上述性质的正整解有无限多个,这可从集合$ Y $的定义式及$ \xi > \xi_0 $选取的任意性推出.

再证必要性.

设$ u(x) = y(|x|) $是方程(1.1)满足性质(2.1)、(2.2)的径向对称正的整体解.于是$ y(t) $满足方程(1.2),即

(2.24) $ \begin{equation} L^2y(t) = f(t, y(t), |y^\prime (t)|), \ t\geq 0 \end{equation} $

且存在正的常数$ c_1, c_2, c_3, c_4 $和$ t_0 > 0 $,使得当$ t > t_0 $时有

(2.25) $ \begin{equation} c_1(1+t^2)\leq y(t)\leq c_2(1+t^2); \, 2c_3t\leq y^\prime (t)\leq 2c_4 t. \end{equation} $

由(2.24)和(2.25)式及定理的条件(ⅰ)、(ⅱ)得

(2.26) $ \begin{equation} L^2y(t) = f(t, y(t), |y^\prime (t)|)\geq f(t, c_2(1+t^2), 2c_4 t)\geq f(t, c_0(1+t^2), 2c_0 t), \, t>t_0, \end{equation} $

其中$ c_0 = \max\{c_2, c_4\} $.记

(2.27) $ \begin{equation} h(t)\equiv f(t, c_0(1+t^2), 2c_0 t), \end{equation} $

则(2.26)式可以写成

(2.28) $ \begin{equation} L^2 y(t)\geq h(t). \end{equation} $

由(1.3)式有

$ L(Ly(t))\equiv \frac{1}{t^{N-1}}\frac{\rm d}{{\rm d}t}\Big[t^{N-1}\frac{\rm d}{{\rm d}t}(Ly(t))\Big]\geq h(t), t>t_0. $

两边乘$ t^{N-1} $后在$ [t_0, t] $上积分得

$ \int_{t_0}^t \frac{\rm d}{{\rm d}s}\Big[s^{N-1}\frac{\rm d}{{\rm d}s}(Ly(s))\Big ]{\rm d}s\geq \int_{t_0}^t s^{N-1}h(s){\rm d}s, \ t>t_0. $

即

$ t^{N-1}\frac{\rm d}{{\rm d}t}(Ly(t))-\Big[ s^{N-1}\frac{\rm d}{{\rm d}s}(Ly(s))\Big]_{s = t_0}\geq \int_0^t s^{N-1}h(s){\rm d}s -\int_0^{t_0} s^{N-1}h(s){\rm d}s , \ t> t_0. $

将上式记为

(2.29) $ \begin{equation} t^{N-1}\frac{\rm d}{{\rm d}t}(Ly(t))-B_1\geq \int_0^t s^{N-1}h(s){\rm d}s -B_2, \ \ t>t_0, \end{equation} $

其中$ B_1, B_2 $为常数.类似于上面的推导, (2.29)式两边乘$ \frac{1}{t^{N-1}} $再积分得

$ \int_{t_0}^t \frac{\rm d}{{\rm d}s}(Ly(s)){\rm d}s \geq \int_{t_0}^t \frac{1}{s^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r\Big){\rm d}s +\int_{t_0}^t (B_1-B_2)\frac{1}{s^{N-1}}{\rm d}s, \ t>t_0. $

即

(2.30) $ \begin{eqnarray} (Ly)(t)-(Ly)(t_0)&\geq & \int_0^t \frac{1}{S^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r\Big){\rm d}s -\int_0^{t_0}\frac{1}{s^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r\Big){\rm d}s\\ &&+\frac{B_1-B_2}{N-2} \Big(\frac{1}{t_0^{N-2}}-\frac{1}{t^{N-2}}\Big), \ t>t_0. \end{eqnarray} $

注意到$ \max\limits_{t_0\leq t < \infty}\Big |\frac{B_1-B_2}{N-2} \Big(\frac{1}{t_0^{N-2}}-\frac{1}{t^{N-2}}\Big)\Big|\leq \frac{|B_1-B_2|}{N-2}\frac{1}{t_0^{N-2}} $,于是

(3.31) $ \begin{eqnarray} &&(Ly)(t_0)-\int_0^{t_0}\frac{1}{s^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r \Big){\rm d}s +\frac{B_1-B_2}{N-2} \Big(\frac{1}{t_0^{N-2}}-\frac{1}{t^{N-2}}\Big) \\ &\geq&(Ly)(t_0)-\int_0^{t_0}\frac{1}{s^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r\Big){\rm d}s -\frac{|B_1-B_2|}{N-2}\frac{1}{t_0^{N-2}}\equiv A_1, \ t>t_0, \end{eqnarray} $

其中$ A_1 $是常数.利用(2.31)式可把(2.30)式简化成

(2.32) $ \begin{equation} (Ly)(t)\geq \int_0^t\frac{1}{s^{N-1}} \Big(\int_0^s r^{N-1}h(r){\rm d}r\Big){\rm d}s+A_1, t>t_0. \end{equation} $

利用Fubini定理,交换(2.32)式右边积分次序得

(2.33) $ \begin{eqnarray} (Ly)(t)&\geq &\int_0^t r^{N-1}h(r) \Big(\int_r^t \frac{1}{s^{N-1}}{\rm d}s\Big){\rm d}r+A_1\\ & = & \frac{1}{N-2}\int_0^t\Big[1-(\frac{r}{t})^{N-2}\Big]rh(r){\rm d}r+A_1, \ t>t_0. \end{eqnarray} $

也即

(2.34) $ \begin{equation} (Ly)(t)\geq \Psi h(t)+A_1, \ t>t_0, \end{equation} $

其中$ A_1 $是常数.对(1.4)式重复上述步骤,知道存在常数$ A_2 $有

(2.35) $ \begin{eqnarray} y(t)&\geq&\Psi(\Psi h(t)+A_1)+A_2 = (\Psi^2h)(t)+\Psi A_1+A_2\\ & = &(\Psi^2h)(t)+\frac{1}{N-2}\int_0^t \Big[1-(\frac{r}{t})^{N-2}\Big]rA_1{\rm d}r+A_2\\ & = &(\Psi^2h)(t)+A_1\frac{t^2}{2N}+A_2, \ t>t_0. \end{eqnarray} $

于是

(2.36) $ \begin{equation} \frac{y(t)}{t^2}\geq \frac{(\Psi^2h)(t)}{t^2}+\frac{A_1}{2N}+\frac{A_2}{t^2}, \ t>t_0. \end{equation} $

在(2.36)式中令$ t\to\infty $并取极限,有

$ \begin{eqnarray*} A& = &\lim\limits_{t\to\infty }\frac{y(t)}{t^2}\geq \lim\limits_{t\to\infty}\Big[\frac{(\Psi^2h)(t)}{t^2}+\frac{A_1}{2N}+\frac{A_2}{t^2}\Big] = \lim\limits_{t\to\infty}\frac{(\Psi^2h)(t)}{t^2}+\frac{A_1}{2N}. \end{eqnarray*} $

如果定理条件(ⅲ)不成立,则有

(2.37) $ \begin{equation} \int_0^\infty tf(t, c_0(1+t^2), 2c_0t){\rm d}t = \infty. \end{equation} $

由(1.7), (1.8), (2.27)和(2.37)式有

(2.38) $ \begin{equation} \lim\limits_{t\to\infty}\frac{\Psi^2 h(t)}{t^2} = \frac{1}{N-2}\int_0^\infty tf(c_0(1+t^2), 2c_0 t){\rm d}t = \infty. \end{equation} $

由(2.37)和(2.38)式得$ \lim\limits_{t\to\infty}\frac{y(t)}{t^2} = \infty $,这与方程(1.1)的整体解具有性质(2.1)矛盾.必要性得证.

定理2.2 假设方程$ (1.1) $中的函数$ f $满足

(ⅰ) $ f:G\to [0, \infty) $是连续的,且不恒等于$ 0 $;

(ⅱ)对固定的$ t\geq 0 $函数$ f(t, u, v) $关于$ u\in(0, \infty) $非增,关于$ v\in[0, \infty) $非减;

(ⅲ)对固定的$ t\geq 0, \lambda^{-1}f(t, \lambda(1+t^2), 4\lambda t) $关于$ \lambda\in(0, \infty) $非增,且

$ \lim\limits_{\lambda\to\infty}\lambda^{-1}f(t, \lambda(1+t^2), 4\lambda t) = 0. $

则方程$ (1.1) $存在满足性质$ (2.1) $、$ (2.2) $正整解$ u(x) = u(|x|) $的充分必要条件是下面条件(ⅳ)成立;

(ⅳ)存在常数$ c > 0 $,使得

$ \int_0^\infty tf(t, c(1+t^2), 4ct){\rm d}t<\infty. $

证 由(ⅲ)知当$ \lambda > c $时有$ \lambda^{-1}tf(t, \lambda(1+t^2), 4\lambda t)\leq c^{-1}t f(t, c(1+t^2), 4ct), $且对一切$ t\geq 0 $有$ \lim\limits_{\lambda\to\infty}\lambda^{-1}t f(t, \lambda(1+t^2), 4\lambda t) = 0. $

由(ⅳ),应用Lebesgue控制收敛定理得

$ \lim\limits_{\lambda\to\infty}\lambda^{-1}\int_0^\infty t f(t, \lambda(1+t^2), 4\lambda t){\rm d}t = 0. $

于是可以取充分大的正数$ \xi_0\in(c, \infty) $,使得$ \xi\in(\xi_0, \infty) $时,恒有

$ \int_0^\infty tf(t, \xi(1+t^2), 4\xi t){\rm d}t<\xi. $

作集合

$ Y = \big\{y\in C^1[0, \infty)\big|\xi(1+t^2)\leq y(t)\leq 2\xi(1+t^2), 2\xi t\leq y^\prime (t)\leq 4\xi t, \ t\geq 0\big\}. $

其余的证明与定理2.1的证明类似.

定理2.3  假设方程$ (1.1) $中的函数$ f $满足定理2.2中的条件(ⅰ)和(ⅱ),又满足

(ⅲ)$ ^\prime $对固定的$ t\geq 0 $, $ \lambda^{-1}f(t, \lambda(1+t^2), 4\lambda t) $关于$ \lambda\in(0, \infty) $非减,且

$ \lim\limits_{\lambda\to 0}\lambda^{-1}f(t, \lambda(1+t^2), 4\lambda t) = 0; $

则方程$ (1.1) $存在满足性质$ (2.1) $、$ (2.2) $正整解$ u(x) = u(|x|) $的充分必要条件是下面条件(ⅳ)成立;

(ⅳ)存在常数$ c > 0 $,使得

$ \int_0^\infty tf(t, c(1+t^2), 4ct){\rm d}t<\infty. $

证 由于定理2.3与定理2.2的差异仅在于以(ⅲ)$ ^\prime $代替(ⅲ),故只须把定理2.2证明过程的“于是可以取充分大的正数$ \xi_0\in(c, \infty) $,使得$ \xi\in(\xi_0, \infty) $时,恒有$ \int_0^\infty tf(t, \xi(1+t^2), 4\xi t){\rm d}t < \xi $”改为“于是可以取充分小的正数$ \xi_0\in(0, c) $,使得$ \xi\in(0, \xi_0) $时,恒有$ \int_0^\infty tf(t, \xi(1+t^2), 4\xi t){\rm d}t < \xi $”.其余的证明与定理2.1的证明类似.

例3.1  考察双调和方程

(3.1) $ \begin{equation} \triangle^2u = \frac{{\rm e}^{-|x|^2}P_k(|x|)}{(1+u)(1+|\nabla u|)}, x\in {\Bbb R} ^N, N>2, \end{equation} $

其中$ P_k(|x|) $是一个关于$ |x| $的$ k $次多项式($ k\geq 0 $), $ f(t, u, v) = \frac{{\rm e}^{-t^2}P_k(t)}{(1+u)(1+v)} $,易见$ f(t, u, v) $满足定理2.1中的条件(ⅰ)和(ⅱ).对任意给定的整数$ k\geq 0 $,选取$ t^\alpha(\alpha = 2) $,当$ c = 1 $,有

$ \lim\limits_{t\to\infty} t^\alpha tf(t, 1+t^2, 2t) = \lim\limits_{t\to\infty}t^2 t\frac{{\rm e}^{-t^2}P_k(t)}{(1+1+t^2)(1+2t)} = 0. $

因为$ \alpha = 2 > 1 $,由广义积分敛散性判别法则知

$ \int_0^\infty tf(t, c(1+t^2), 4ct){\rm d}t<\infty. $

故定理2.1的条件(ⅲ)成立,于是方程(3.1)存在无穷多个满足性质(2.1)、(2.2)的正整解.

例3.2  考察双调和方程

(3.2) $ \begin{equation} \triangle^2u = \frac{{\rm e}^{-|x|^2}|\nabla u|^2}{1+u^2}, x\in {\Bbb R} ^N, N>2, \end{equation} $

这里$ f(t, u, v) = \frac{{\rm e}^{-t^2}v^2}{1+u^2} $,易见$ f(t, u, v) $满足定理2.2的条件(ⅰ)、(ⅱ),取$ t^\alpha(\alpha = 2), c = 1 $,则

$ \lim\limits_{t\to\infty}t^\alpha t f(t, c(1+t^2), 4ct) = \lim\limits_{t\to\infty}t^2 t\frac{{\rm e}^{-t^2}(4t)^2}{1+(1+t^2)^2} = 0. $

因为$ \alpha = 2 > 1 $,由广义积分敛散性判别法则知

$ \int_0^\infty tf(t, c(1+t^2), 4ct){\rm d}t<\infty. $

故定理2.2的条件(ⅳ)成立,于是方程(3.2)存在无穷多个满足性质(2.1)、(2.2)的正整解.