考虑下列一类非线性Schrödinger方程

(1.1) $ \begin{eqnarray} {\rm i}u_{t} = -\Delta u +\overrightarrow{\alpha}\cdot \nabla u+\beta u+\frac{\gamma u}{\sqrt{1+|u|^{2}}}+ \frac{\kappa u} {2\sqrt{1+|u|^{2}}}\Delta\sqrt{1+|u|^{2}}, \end{eqnarray} $

其中$ u = u(t, x) $是未知波函数, $ \overrightarrow{\alpha} $是复向量, $ \beta $是复常数, $ \gamma $和$ \kappa $是实常数.该方程来源于高功率超短激光脉冲在等离子体中的自沟道效应模型,详细物理背景参见文献[1-3].对该方程Cauchy问题解的局部适定性的讨论,参见文献[4-5].

在方程(1.1)中令$ {\overrightarrow{\alpha}} = \overrightarrow{0} $, $ \beta = 0 $.考虑方程(1.1)的驻波解,即形如$ u(t, x) = {\rm e}^{{\rm i}\lambda t}v(x) $的解,其中$ \lambda $为实参数, $ v(x) $是实函数.把它代入方程(1.1)便得到一类含有变分结构的非线性椭圆型方程

(1.2) $ \begin{eqnarray} -\Delta v+ \frac{\kappa v} {2\sqrt{1+v^{2}}}\Delta\sqrt{1+v^{2}}+\frac{\gamma v}{\sqrt{1+v^{2}}}+\lambda v = 0, \ \ \ \ x\in {\Bbb R} ^N. \end{eqnarray} $

当$ N = 2 $, $ \kappa = 2 $, $ \gamma = 1 $, $ -1<\lambda<0 $时, Colin^[6]证明了方程(1.2)存在一个基态解$ v(x)\in C^2( {\Bbb R} ^2) $, $ v(x) $关于$ r = |x| $是轴对称的且单调递减.该基态解在平移的意义下是唯一的.此外,他在文中猜测当$ N\geq3 $时,方程(1.2)没有非平凡解.

本文讨论方程(1.2)当$ N = 1 $时基态解的存在性及渐近行为.这里我们采用ODE的方法,故我们可以对参数$ \kappa $, $ \gamma $和$ \lambda $的相互关系以及取值范围有较详细的刻画.此外,我们还进一步讨论了振荡解和孤波解的存在性.

以下均考虑$ N = 1 $的情况.此时,方程(1.2)可写成二阶非线性ODE

(1.3) $ \begin{equation} \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v''-\frac{\kappa v }{2(1+v^{2})^{2}}v'^{2}-\left(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}\right)v = 0. \end{equation} $

我们有下列结论:

定理1.1  假设$ -\gamma<\lambda <0, $或者$ 0<\lambda <-\gamma, $$ \kappa<\frac{\lambda^{2}}{2\gamma^{2}+2\gamma\lambda}+2 $,则方程$ (1.3) $有一个基态解$ v $,且具有性质:

(ⅰ) $ v(x)>0 $, $ v(-x) = v(x) $;

(ⅱ)当$ x>0 $时,有$ v'(x)<0 $;

(ⅲ) $ \lim\limits_{|x|\rightarrow \infty}v(x) = 0 $.

此外,在平移意义下,该基态解是唯一的且$ \max\limits_{x\in {\Bbb R} }v(x) = \frac{2}{|\lambda|}\sqrt{\gamma^{2}+\gamma\lambda}. $

注1.1  在方程$ (1.3) $两边同时乘以$ v' $,除以$ \frac{(2-\kappa) v^{2}+2}{2(1+v^{2})} $,再关于$ x $积分,容易证明当$ 0\leq \kappa<2 $, $ -\gamma\leq \lambda $且$ \lambda>0 $,或者$ \lambda<0 $且$ \lambda\leq- \gamma $时,方程$ (1.3) $没有满足性质(ⅰ)–(ⅲ)的基态解.

定理1.2  假设$ \kappa\in{\Bbb R} $且下列条件之一成立:

(ⅰ) $ \lambda\gamma<0 $且$ \lambda+\gamma<0 $;

(ⅱ) $ \lambda = 0 $, $ \gamma<0 $;

(ⅲ) $ \lambda<0 $, $ \gamma\leq0 $.

则方程$ (1.3) $有一个振荡解$ v $,满足$ v(x) = v(-x) $且有无穷多个零点.

上述两个定理的证明类似于ODE方法中的打靶方法,关键点是对方程(1.3)初值条件中初始值的选取.而这依赖于$ \lambda $和$ \gamma $的取值范围.

本文最后讨论下列方程

(1.4) $ \begin{eqnarray} {\rm i}u_{t} = - u_{xx} +\alpha u_x+\beta u+\frac{\gamma u}{\sqrt{1+|u|^{2}}}+ \frac{\kappa u} {2\sqrt{1+|u|^{2}}}(\sqrt{1+|u|^{2}})_{xx}, \end{eqnarray} $

其中$ \alpha = \alpha_{1}+{\rm i}\alpha_{2}, $$ \beta = \beta_{1}+{\rm i}\beta_{2} $, $ \alpha_i $, $ \beta_i $$ (i = 1, 2) $为实常数.

我们考虑方程(1.4)的孤波解,即在方程(1.4)中令

$ u(t, x) = {\rm e}^{{\rm i}\lambda (x+ct)}v(x+{\rm d}t), $

$ \lambda $, $ c $, $ d $为实参数.进一步,假设$ d+2\lambda = \alpha_{2}, $$ \alpha_{1}\lambda+\beta_{2} = 0 $,代入方程(1.4)得

(1.5) $ \begin{equation} \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v''-\frac{\kappa v}{2(1+v^{2})^{2}}v'^{2}+\tau v'-\left(\frac{\gamma}{\sqrt{1+v^{2}}}+\mu\right)v = 0, \end{equation} $

其中$ \tau = -\alpha_{1}\neq 0 $, $ \mu = \lambda(c+\lambda-\alpha_{2})+\beta_{1} $.

定理1.3  假设$ \kappa<2 $, $ \gamma, $$ \mu\in {\Bbb R} $, $ \tau\not = 0 $.则方程$ (1.5) $有一个非零解$ v $.进一步,若$ \gamma\mu<0 $,且$ |\mu|<|\gamma| $,则初值问题$ (1.5) $与$ v(0) = \frac{2}{|\mu|}\sqrt{\gamma^{2}+\gamma \mu} $, $ v'(0) = 0 $有一个孤波解$ v $满足

$ 0<v(x)<\frac{2}{|\mu|}\sqrt{\gamma^{2}+\gamma \mu}, $

其中$ x $满足条件$ x\cdot\mbox{sgn}\tau>0 $.

为了证明本文定理,需要下面的Cauchy-Kowalevski局部存在性定理^[7]:

定理2.1 (Cauchy-Kowalevski定理)  设$ D $表示实或复有限维向量空间, $ F:D^{2}\longrightarrow D $是实解析函数,则初值问题

$ v'' = F(v, v'), \ \ v(x_{0}) = a, \ \ v'(x_{0}) = b $

在$ x_{0} $的某个邻域$ I $内存在唯一解.

定理1.1的证明  令$ \theta = \frac{\gamma}{\lambda}, $当$ -\gamma<\lambda <0, $有$ \theta<-1 $.现考虑方程(1.3)的初值条件

(2.1) $ \begin{equation} v(0) = 2\sqrt{\theta^{2}+\theta}, \ \ v'(0) = 0. \end{equation} $

首先证明上述柯西问题(1.3)和(2.1)在$ {\Bbb R} $上存在唯一解.由于$ \kappa<\frac{\lambda^{2}}{2\gamma^{2}+2\gamma\lambda}+2 $,故$ 1-\frac{\kappa v(0)^{2}}{2(1+v(0)^{2})}>0. $因此,存在$ \varepsilon>0 $,使得当$ x\in[0, \varepsilon) $时,有$ 1-\frac{\kappa v(x)^{2}}{2(1+v(x)^{2})}>0 $.由定理2.1可知,方程(1.3)和满足初值条件(2.1)存在局部唯一解$ v(x) $, $ x\in I\subset {\Bbb R} $.下面证: $ 0<v(x)<2\sqrt{\theta^{2}+\theta}, $$ \forall x\in I, $$ x\neq0 $.这暗含了$ I = {\Bbb R} $.

先证$ v(x)>0, $$ \forall x\in I, $$ x\neq0. $由于$ v(0)>0 $,假如存在$ x_{1}\in I $使得$ v(x_{1}) = 0 $且$ v(x)>0, \forall x\in[0, x_{1}). $在方程(1.3)两边乘以$ 2v' $再关于$ x $积分,代入初始条件(2.1)可得

(2.2) $ \begin{equation} \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2} = 2\gamma(\sqrt{1+v^{2}}-1)+\lambda v^{2}. \end{equation} $

把$ v(x_{1}) = 0 $代入(2.2)式,得$ v'(x_{1}) = 0. $由唯一性知$ v(x)\equiv 0 $是问题(1.3)的一个解,这与$ v(0) = 2\sqrt{\theta^{2}+\theta}>0 $矛盾.

下面证$ v(x)<2\sqrt{\theta^{2}+\theta} $, $ \forall x\in I, $$ x\neq0. $把初始条件(2.1)代入方程(1.3),得$ v''(0)<0 $,即$ x = 0 $是$ v(x) $的一个极大值点.若存在某个点,不妨设为$ x_{2}\in I, $$ x_{2}>0 $使得$ v(x_{2})\geq2\sqrt{\theta^{2}+\theta} $,则一定存在某个$ x_3 $满足$ 0<x_{3}<x_{2}, $$ 0<v(x_{3})<2\sqrt{\theta^{2}+\theta} $且$ v'(x_{3}) = 0 $.代入(2.2)式得, $ 2\gamma(\sqrt{1+v(x_3)^{2}}-1)+\lambda v(x_3)^{2} = 0, $即$ v(x_{3}) = 2\sqrt{\theta^{2}+\theta} $或者$ v(x_{3}) = 0 $,矛盾.

(ⅰ)由于$ v(-x) $也是问题的解,由唯一性可知, $ v(-x) = v(x) $.

(ⅱ)由于$ v'(0) = 0 $, $ v''(0)<0 $,故存在$ r_{1}>0 $使得$ v'(x)<0, $$ x\in(0, r_{1}) $.若存在$ x_{4}>r_{1} $使得$ v'(x_{4}) = 0 $,则由(2.2)式知$ v(x_{4}) = 2\sqrt{\theta^{2}+\theta} $或者$ v(x_{4}) = 0 $,这与$ 0<v(x)<2\sqrt{\theta^{2}+\theta} $, $ x\in {\Bbb R} $矛盾.因此$ v'(x)<0, $$ x>0 $.这也意味着$ \max\limits_{x\in {\Bbb R} }v(x) = -\frac{2}{\lambda}\sqrt{\gamma^{2}+\gamma\lambda} $.

(ⅲ)由上述讨论知$ \lim\limits_{x\rightarrow \infty}v(x) $存在.不妨设$ \lim\limits_{x\rightarrow \infty}v(x) = m $,则有$ 0\leq m<2\sqrt{\theta^{2}+\theta} $.由洛必达法则知, $ 1 = \lim\limits_{x\rightarrow \infty}{\frac{v(x)+x}{x}} = \lim\limits_{x\rightarrow \infty}(v'(x)+1) $,即有$ \lim\limits_{x\rightarrow \infty}v'(x) = 0 $.再由(2.2)式可得$ m = 0. $

最后证明在平移意义下,该基态解(即满足性质(ⅰ)–(ⅲ))是唯一的.假如$ w(x) $是方程(1.3)的另一个解,对$ w(x) $作平移变换,使其在$ x = 0 $处取极大值,即$ w'(0) = 0 $.设$ w(0) = w_{0}>0 $,则$ w(-x) = w(x) $.类似等式(2.2)的证明,有

(2.3) $ \begin{equation} \left[1- \frac{\kappa w^{2}}{2(1+w^{2})}\right]w'^{2} = 2\gamma(\sqrt{1+w^{2}}-1)+\lambda w^{2}-2\gamma(\sqrt{1+w_{0}^{2}}-1)-\lambda w_{0}^{2}. \end{equation} $

由于$ \lim\limits_{x\rightarrow \infty}w(x) = 0 $,再次由洛必达法则, $ 1 = \lim\limits_{x\rightarrow \infty}{\frac{w(x)+x}{x}} = \lim\limits_{x\rightarrow \infty}(w'(x)+1) $可得$ \lim\limits_{x\rightarrow \infty}w'(x) = 0 $.对等式(2.3)两边同时取极限有

(2.4) $ \begin{equation} 2\gamma(\sqrt{1+w_{0}^{2}}-1)+\lambda w_{0}^{2} = 0. \end{equation} $

解(2.4)式得$ w_0 = 2\sqrt{\theta^{2}+\theta} $.由唯一性, $ w(x)\equiv v(x) $.

当$ 0<\lambda <-\gamma, $证明过程和上述讨论类似,故省略.

在证明定理1.2之前,先介绍一个引理.

引理2.1^[8]  设$ q\in C([a, \infty)) $,若$ \lim\limits_{|x|\rightarrow \infty}x^{2}q(x)>\frac{1}{4} $.那么方程$ y''+q(x)y = 0 $是振荡的,即任一非零解在$ [a, \infty) $内有无穷多个零点.

定理1.2的证明  类似于定理1.1的证明,方程(1.3)满足初值条件$ v(0) = v_{0}>0, $$ v'(0) = 0 $存在唯一局部解$ v(x) $, $ x\in I $,满足$ v(-x) = v(x) $.其中$ v_0 $的选取遵循$ \lambda+\frac{\gamma}{\sqrt{1+v_{0}^{2}}}<0 $且$ 1-\frac{\kappa v_{0}^{2} }{2(1+v_{0}^{2})}>0 $的原则.在条件(ⅰ)–(ⅲ)中任意一个成立的前提下,这样的$ v_0 $是可以找到的.

下面证明该局部解可以延拓到$ {\Bbb R} $上.事实上,只要我们证明$ |v(x)|\leq v_{0}, $$ \forall x\in I $成立即可.这暗含了局部解可以延拓成整体解.由于

$ v''(0) = \frac{\lambda+\frac{\gamma}{\sqrt{1+v_{0}^{2}}}}{1-\frac{\kappa v_{0}^{2} }{2(1+v_{0}^{2})}}v_{0}<0. $

故$ v(x) $在$ x = 0 $处取局部极大值.若存在$ y_1\in I, $不妨设$ y_1>0 $且满足$ v(y_1)>v_{0} $,则必存在$ 0<y_{2}<y_{1} $,使得$ v(y_{2}) = v_{0} $, $ v(x)\leq v_0 $, $ \forall x\in (0, y_2) $且对于$ \delta>0 $充分小,有$ v(x)>v_{0}, $$ x\in (y_{2}, y_{2}+\delta) $.类似等式(2.3)的证明,用$ v(y_2) $和$ v_0 $分别替换$ w $和$ w_0 $,得$ v'(y_{2}) = 0 $.若$ v''(y_{2})\neq 0, $则$ x = y_{2} $为极大值或极小值,但这与$ y_{2} $的选取矛盾.从而$ v''(y_{2}) = 0. $代入方程(1.3),即有$ (\lambda+\frac{\gamma}{\sqrt{1+v_{0}}})v_{0} = 0 $,也即有$ v_{0} = 0 $,矛盾.因此, $ v(x)\leq v_{0}, $$ \forall x\in I $.类似地,可以证明$ v(x)\geq -v_{0}, $$ \forall x\in I $.

(ⅰ)当$ \kappa\leq 0 $时,方程(1.3)可转化为

$ v''-\frac{1}{(1-\frac{\kappa v^{2}}{2(1+v^{2})})}\left[\frac{\kappa}{2(1+v^{2})^{2}}v'^{2}+\left(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}\right)\right] v = 0. $

令

$ q(x) = -\frac{1}{(1-\frac{\kappa v^{2}}{2(1+v^{2})})}\left[\frac{\kappa}{2(1+v^{2})^{2}}v'^{2}+(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}})\right], $

则

$ q(x)\geq \frac{2}{\kappa-2}\left(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}\right). $

在条件(ⅰ)–(ⅲ)其中之一条件下,一定存在一个常数$ c>0 $使得$ q(x)\geq c>0 $.从而

$ \lim\limits_{|x|\rightarrow \infty}x^{2}q(x) = +\infty. $

(ⅱ)当$ \kappa>0 $时,将方程(1.3)变形为

(2.5) $ \begin{eqnarray} \left[\left(1-\frac{\kappa v^{2}}{2(1+v^{2})}\right)v'\right]' +\left[\frac{\kappa}{2(1+v^{2})^{2}}v'^{2} -\left(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}\right)\right]v = 0. \end{eqnarray} $

令$ p(x) = 1-\frac{\kappa v^{2}}{2(1+v^{2})}, $$ g(x) = \frac{\kappa}{2(1+v^{2})^{2}}v'^{2}-(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}) $,作变量替换$ t = t(x) = \int^{x}_{0}\frac{1}{p(y)}{\rm d}y $则$ \frac{{\rm d}v}{{\rm d}t} = p(x)\frac{{\rm d}v}{{\rm d}x}, $代入方程(2.5)可得

$ \begin{eqnarray*} w''(t)+p(x(t))g(x(t))w(t) = 0, \end{eqnarray*} $

这里$ w(t) = v(x(t)) $.令$ q(t) = p(x(t))g(x(t)) $,则

$ q(t) = \left(1-\frac{\kappa v^{2}}{2(1+v^{2})}\right)\left[\frac{\kappa}{2(1+v^{2})^{2}}v'^{2} -\left(\lambda+\frac{\gamma}{\sqrt{1+v^{2}}}\right)\right]\geq \left(1-\frac{\kappa v_0^{2}}{2(1+v_0^{2})}\right)c_1, $

其中$ c_1>0 $.从而$ \lim\limits_{|t|\rightarrow \infty}t^{2}q(t) = +\infty $.

结合(ⅰ)和(ⅱ),由引理2.1可知, $ v(x) $是振荡的且在$ {\Bbb R} $上有无穷多个零点.

定理1.3的证明  考虑方程$ (1.5) $满足初值条件$ v(0) = v_0, v'(0) = 0 $.利用定理2.1可知局部解存在唯一,现证该局部解可以延拓到$ {\Bbb R} $上.为此,我们对解做一个先验估计.我们只考虑$ x>0 $的情况. $ x<0 $的情况可利用$ w(x) = v(-x) $类似证明.

(1)当$ \tau<0 $时,在方程$ (1.5) $两边乘以$ v $并在$ [0, x] $上积分,整理后可得

$ \left[1- \frac{\kappa v^{2}}{2(1+v^{2})}\right]vv' = \frac{|\tau|}{2}(v^{2}-v_{0}^{2})+ \int^{x}_{0}\left(\mu+\frac{\gamma}{\sqrt{1+v^{2}}}\right)v^{2}{\rm d}y+\int^{x}_{0}\left[1-\frac{\kappa v^{2}(2+v^{2})}{2(1+v^{2})^{2}}\right]v'^{2}{\rm d}y. $

由于对一切$ \kappa<2 $恒有

$ -\frac{\kappa v^{2}}{2(1+v^{2})^{2}}\leq 1-\frac{\kappa v^{2}}{2(1+v^{2})}, $

于是

$ \bigg[1-\frac{\kappa v^{2}(2+v^{2})}{2(1+v^{2})^{2}}\bigg]v'^{2}\leq \Big(2-\frac{\kappa v^{2}}{1+v^{2}}\Big)v'^{2}. $

下面对$ \gamma, $进行讨论:

(ⅰ) $ \gamma\leq0 $,则$ \mu+\gamma \leq \mu+\frac{\gamma}{\sqrt{1+v^{2}}}\leq\mu; $

(ⅱ) $ \gamma>0 $,则$ \mu\leq \mu+\frac{\gamma}{\sqrt{1+v^{2}}}\leq \mu+\gamma. $

令$ m = \max{\{\mu, \gamma+\mu, 1\}} $, $ n = \min{\{\mu, \gamma+\mu, 1\}} $,则由(ⅰ)–(ⅳ)有

(2.6) $ \begin{equation} 2vv'\leq\frac{1}{A}\left(|\tau|v^{2}+\int^{x}_{0}2m v^{2}{\rm d}y+ \int^{x}_{0}\left(4-\frac{2\kappa v^{2}}{1+v^{2}}\right)v'^{2}{\rm d}y\right), \end{equation} $

这里当$ \kappa\leq 0 $时, $ A = 1 $;当$ 0<\kappa<2 $时, $ A = \frac{2-\kappa}{2} $.

对$ (2.6) $式两边在$ [0, x] $上积分,得

(2.7) $ \begin{equation} v^{2}-v_{0}^{2}\leq\frac{\left(|\tau|+2mx\right)}{A}\int^{x}_{0}v^{2}{\rm d}y+ \frac{4x}{A}\int^{x}_{0}\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}{\rm d}y. \end{equation} $

另一方面,在方程(1.5)两边乘以$ 2v' $并在$ [0, x] $上积分然后代入初值可得

(2.8) $ \begin{equation} \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2} = -2\tau\int^{x}_{0}v'^{2}{\rm d}y+2\gamma(\sqrt{1+v^{2}}-1)+\mu v^{2} -2\gamma(\sqrt{1+ v_{0}^{2}}-1)-\mu v_{0}^{2}. \end{equation} $

由上述对$ \gamma, $$ \mu $的讨论可知

$ 2\gamma(\sqrt{1+v^{2}}-1)+\mu v^{2} = (\frac{2\gamma}{\sqrt{1+v^{2}}+1}+\mu)v^{2}\leq mv^{2}, $

$ -2\gamma(\sqrt{1+v_{0}^{2}}-1)-\mu v_{0}^{2}\leq -nv_{0}^{2}. $

根据(2.8)式,有

(2.9) $ \begin{equation} \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}\leq\frac{2|\tau|}{A}\int^{x}_{0}\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}{\rm d}y+m v^{2}-nv_{0}^{2}. \end{equation} $

在不等式$ (2.7) $两边乘以$ 2m $再与$ (2.9) $式相加,可得

$ \begin{eqnarray*} && mv^{2}+\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}\\ &\leq&\frac{2|\tau|+4mx}{A}\int^{x}_{0}mv^{2}{\rm d}y+\frac{2|\tau|+8mx}{A}\int^{x}_{0}\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}{\rm d}y+(2m-n)v_{0}^{2}\\ &\leq & \frac{2|\tau|+8mx}{A}\int^{x}_{0}\left\{mv^{2}+\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}\right\}{\rm d}y+(2m-n)v_{0}^{2}, \end{eqnarray*} $

在任意有限区间$ [0, b] $利用Gronwall不等式可得

$ mv^{2}+\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}\leq\left(2m-n\right)v_{0}^{2} {\rm e}^{\frac{2|\tau|+8mx}{A}x}. $

(2)当$ \tau\geq0 $时,类似(1)的讨论,有

$ 2vv'\leq\frac{1}{A}\left(\tau v_{0}^{2}+\int^{x}_{0}2m v^{2}{\rm d}y+ \int^{x}_{0}\left[4-\frac{2\kappa v^{2}}{(1+v^{2})}\right]v'^{2}{\rm d}y\right). $

积分可得

$ v^{2}-v_{0}^{2}\leq\frac{1}{A}\left(\tau v_{0}^{2}x+2mx\int^{x}_{0}v^{2}{\rm d}y+ 4x\int^{x}_{0}\left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}{\rm d}y\right). $

由(2.8)式有

$ \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}\leq m v^{2}-nv_{0}^{2}. $

从而

$ v^{2}\leq \frac{6mx}{A}\int^{x}_{0}v^{2}{\rm d}y+\left(\frac{\tau x+A-4nx}{A}\right)v_{0}^{2}. $

在任意有限区间$ [0, b] $上,利用Gronwall不等式可得

$ v^{2}\leq\left(\frac{\tau x+A-4nx}{A}\right)v_{0}^{2}{\rm e}^{\frac{6mx}{A}x}. $

综合上述两种情况可知,对$ \forall x\in [0, T), $$ v(x) $均有界,从而该局部解可以延拓为全局解,即方程$ (1.5) $的解$ v $在$ {\Bbb R} $上存在.

进一步,不妨考虑$ \mu>0 $,我们证$ 0<v(x)<\frac{2}{\mu}\sqrt{\gamma^{2}+\gamma \mu} $. $ \mu<0 $的情况可类似证明.由条件可知$ \gamma<0 $,且$ 0<\mu<-\gamma $.考虑方程(1.5)满足初值条件$ v(0) = \frac{2}{\mu}\sqrt{\gamma^{2}+\gamma \mu} $, $ v'(0) = 0 $的解$ v $.

若$ \tau>0 $,如果存在$ x^{*}>0 $,使得$ v(x^{*}) = 0 $或$ v(x^{*}) = x_{0} $代入方程(2.8)有

$ v'^{2}(x^{*})+2\tau\int^{x^{*}}_{0}v'(y)^{2}{\rm d}y = 0, $

或者

$ \left[1-\frac{\kappa v^{2}}{2(1+v^{2})}\right]v'^{2}(x^{*})+2\tau\int^{x^{*}}_{0}v'^{2}{\rm d}y = 0. $

因此在$ [0, x^{*}] $上均有$ v'(y)\equiv 0 $,但这与$ v''(0)<0 $矛盾.若$ \tau<0 $,可类似证明.