## Sufficient and Necessary Condition for the Existence of Positive Entire Solutions of a Nonlinear Biharmonic Equations on ${{\mathbb{R}}^{N}}$

Ou Xiaohang,

Abstract

The aim of this paper is to study the nonlinear biharmonic equations of the following form $\triangle^2u=f(|x|, u, |\nabla u|)(x\in \mathbb{R} ^N, N>2)$. The Sufficient and necessary condition for the existence of positive entire solutions is proved, and some properties of the solutions are obtained.

Keywords： Nonlinear biharmonic equation ; Positive entire solution ; Close convex subset ; Equicontinuity ; Fixed point theorem

Ou Xiaohang. Sufficient and Necessary Condition for the Existence of Positive Entire Solutions of a Nonlinear Biharmonic Equations on ${{\mathbb{R}}^{N}}$. Acta Mathematica Scientia[J], 2019, 39(3): 535-544 doi:

## 1 引言与引理

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

$$$L^2y(t) = f(t, y(t), |y\prime (t)|), \ t \geq 0$$$

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

$L^1 = L$, $L^2$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:

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

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

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

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

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

先证$M_0 < \infty$的情形.由(1.4)式有

## 2 主要定理

(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

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

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

(ⅲ)存在常数$c > 0$,使得

先证充分性.假设定理的条件(ⅲ)成立,故存在常数$c > 0$,使得

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

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

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

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

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

$$$\pi y(t) = \xi(1+t^2)+\Psi^2[f(t, y(t), |y^\prime(t)|)], \ \ t\geq 0 .$$$

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

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

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

$$$c_1(1+t^2)\leq y(t)\leq c_2(1+t^2); \, 2c_3t\leq y^\prime (t)\leq 2c_4 t.$$$

$$$L^2y(t) = f(t, y(t), |y^\prime (t)|)\geq f(t, c_2(1+t^2), 2c_4 t)\geq f(t, c_0(1+t^2), 2c_0 t), \, t>t_0,$$$

$$$h(t)\equiv f(t, c_0(1+t^2), 2c_0 t),$$$

$$$L^2 y(t)\geq h(t).$$$

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

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

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

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

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

$$$(Ly)(t)\geq \Psi h(t)+A_1, \ t>t_0,$$$

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

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

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

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

(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

(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

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

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