反应扩散方程和方程组是一类重要的抛物型偏微分方程, 它们有着非常广泛的实际背景, 涉及了如物理, 化学, 以及生物群体动力学等许多方面的科学研究领域.它们是自然界中广泛存在的扩散现象的一种数学抽象.

关于如下形式的拟线性退化抛物方程

$\begin{equation*} \begin{cases} u_t=f(u)(\Delta u+au), \quad x\in{\it \Omega}, t＞0, \\ u(x, t)=0, x\in\partial{\it \Omega}, t＞0, \\ u(x, 0)=u_0(x), x\in{{\it \Omega}} \end{cases} \end{equation*}$

和方程组

$\begin{equation*} \begin{cases}u_t=f(v)(\Delta u+au), x\in{\it \Omega}, t＞0, \\ v_t=g(u)(\Delta v+bv), x\in{\it \Omega}, t＞0, \\u(x, t)=v(x, t)=0, x\in\partial{\it \Omega}, t＞0, \\u(x, 0)=u_0(x), v(x, 0)=v_0(x), \quad x\in{{\it \Omega}} \end{cases} \end{equation*}$

古典解的整体存在和非存在性的研究已经取得了丰富的结果, 其中文献[1-3]讨论的是单方程的情形, 而文献[4-6]讨论了强耦合方程组的情形.他们通过对$a, \ b$以及区域${\it \Omega}$大小关系的讨论, 分别建立了整体解的存在与不存在性条件.

2008年, Chen ^[7]研究了如下形式的拟线性问题

$\begin{equation*} \begin{cases} u_t=u^\alpha v^\beta\Delta u+au^pv^q, \\ v_t=u^\theta v^\eta\Delta v+bu^rv^s. \end{cases} \end{equation*}$

利用一种基于积分估计的方法, 他证明了上述问题是否存在整体解与$(q-\beta)(r-\theta)-(\alpha+1-p)(\eta+1-s)$的符号密切相关.

受上述工作的启发, 本文研究如下形式的扩散方程组解的整体存在与非存在性:

$\begin{equation} \label{1.1} \begin{cases} u_t=v^p[\Delta u+u(a_1-b_1u^r+c_1v^s)], \quad x\in{\it \Omega}, \ t＞0, \\ v_t=u^q[\Delta v+v(a_2+b_2u^l-c_2v^m)], x\in{\it \Omega}, t＞0, \\u(x, t)=v(x, t)=0, x\in\partial{\it \Omega}, \ t＞0, \\ u(x, 0)=u_0(x), \ v(x, 0)=v_0(x), x\in{\it \Omega}, \end{cases} \end{equation}$

(1.1)

其中${\it \Omega}$是$R^N$中具有光滑边界$\partial{\it \Omega}$的有界区域, $a_i, b_i, c_i(i=1, 2), p, q＞0$, $r, s, l, m\geq1$.初值$(u_0, v_0)$满足

$\begin{equation} \label{1.2} \begin{cases}u_0, v_0\in C^1(\overline{{\it \Omega}}), \quad u_0, v_0＞0, \quad x\in{\it \Omega}, \\ u_0=v_0=0, \frac{\partial{u_0}}{\partial n}＜0, \frac{\partial{v_0}}{\partial n}＜0,\quad x\in\partial{\it \Omega}, \end{cases} \end{equation}$

(1.2)

这里$n$代表$\partial{\it \Omega}$上的外法向量.模型(1.1) 可以用来描述生物种群动力学中具有互惠作用的两种群的种群密度的变化规律.其中未知函数$u(x, t), v(x, t)$分别代表两种群在$t$时刻$x$处的种群密度, $a_1, a_2$是各自的净出生率, 系数$b_1, c_2$代表了种群内部的竞争而系数$b_2, c_1$代表种群间的互惠作用.问题的强耦合性使得通常的比较原理一般不再成立.为了克服这一困难, 本文中我们采用Chen在文献[7]中用过的借助积分估计的方法来证明问题(1.1) 整体解的存在与非存在性.我们将证明当种群内部竞争强于种群间的互惠作用时, 问题(1.1) 存在整体古典解而当两种群具有强互惠作用时, 该问题所有的正古典解都不是整体存在的.类似问题的讨论还可以参见文献[8-10].

我们称$(u, v)$是问题(1.1) 的古典解, 如果存在$0＜T\leq\infty, \text{使得}(u, v)\in[C^{2, 1}({\it \Omega}\times(0, T))\cap C(\overline{{\it \Omega}}\times[0, T))]^2$, $(u, v)$在${\it \Omega}\times(0, T)$内满足方程并且连续地满足初边值条件.

本文中我们用$\lambda_1$代表$-\Delta$在${\it \Omega}$内齐次Dirichlet边值问题的第一特征值, $\varphi＞0$为相应的特征函数满足$\max\limits_{x\in\overline{{\it \Omega}}}\varphi(x)=1$.易知$\lambda_1＞0, \ \frac{\partial\varphi}{\partial n}＜0$于$\partial{\it \Omega}$.

文章安排如下:第二部分我们给出正古典解局部存在性的证明.在第三和第四两部分我们分别给出问题整体解存在和不存在的充分条件.

由于在$\partial{\it \Omega}$上$u=v=0$, 所以$(1.1)$式不是严格抛物型的.经典的抛物方程理论^{[11, 12]}不能直接用于证明它的古典解的局部存在性.为了克服这个困难, 我们采用正则逼近的方法.这个方法已被很多学者用来证明退化方程或方程组解的局部存在性(见文献[3, 13, 14]).这里我们只是简要地给出证明框架.考虑如下逼近问题

$\begin{equation}\label{2.1} \begin{cases} u_{\epsilon t}=v_\epsilon^p[\Delta u_\epsilon+u_\epsilon(a_1-b_1u_\epsilon^r+c_1v_\epsilon^s)], x\in{\it \Omega}, \ t＞0, \\ v_{\epsilon t}=u_\epsilon^q[\Delta v_\epsilon+v_\epsilon(a_2+b_2u_\epsilon^l-c_2v_\epsilon^m)], x\in{\it \Omega}, \ t＞0, \\ u_\epsilon(x, t)=v_\epsilon(x, t)=\epsilon, x\in\partial{\it \Omega}, \ t＞0, \\ u_\epsilon(x, 0)=u_0(x)+\epsilon, \ v_\epsilon(x, 0)=v_0(x)+\epsilon, \quad x\in{\it \Omega}. \end{cases} \end{equation}$

(2.1)

由抛物方程的经典理论可知(2.1) 式存在惟一正解$(u_\epsilon, v_\epsilon)\in [C^{2, 1}({\it \Omega}\times(0, T(\epsilon)))\cap C^{1, 0}(\overline{{\it \Omega}}\times[0, T(\epsilon))]^2$ ($0＜T(\epsilon)\leq\infty$), 这里$T(\epsilon)$是解的最大存在时间.此外, 由类似与文献[13]中的方法可得, 对任意$0＜\epsilon\leq\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$, 在$\overline{{\it \Omega}}\times[0, T(\epsilon))$上都有$u_\epsilon, v_\epsilon\geq\epsilon$.

为了讨论$(u_\epsilon, v_\epsilon)$的收敛性, 我们需要对它们做一些上、下界的估计.由假设(1.2) 式可知, 存在两个正常数$k$, $K$使得

$\begin{equation}\label{2.2} k\varphi\leq u_0, v_0\leq K\varphi, \qquad x\in\overline{{\it \Omega}}. \end{equation}$

(2.2)

记$M=\max\{\max\limits_{\overline{{\it \Omega}}}u_0(x), \max\limits_{\overline{{\it \Omega}}}v_0(x)\}$.令$(f(t), g(t))$为下述ODE的惟一解

$\begin{equation}\label{2.3} \begin{cases} f^\prime=fg^p(a_1+c_1g^s), t＞0, \\ g^\prime=f^qg(a_2+b_2f^l), t＞0, \\ f(0)=g(0)=M+\max\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}. \end{cases} \end{equation}$

(2.3)

则$f(t), g(t)\geq M+\max\{(\frac{a_1}{b_1})^{\frac{1}{r}}$, $(\frac{a_2}{c_2})^{\frac{1}{m}}\}$.记$T^*$ ($0＜T^*<\infty$)为它的最大存在时间(注意$T^*<\infty$因为$(f(t), g(t))$在有限时刻爆破).我们可以得到正则化问题解$(u_\epsilon, v_\epsilon)$的上、下界估计, 也就是如下的

命题2.1  设$0＜\epsilon\leq\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$, $(u_\epsilon, v_\epsilon)$是问题(2.1) 的解.则对任意固定的$T$, $0＜T<\min\{T(\epsilon), T^*\}$, 有

$\begin{equation*}u_\epsilon\leq f(t), \ v_\epsilon\leq g(t), \qquad (x, t)\in\overline{{\it \Omega}}\times[0, T]. \end{equation*}$

这表明对任意的$0＜\epsilon\leq\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$, 都有$T(\epsilon)\geq T^*$.

命题2.2  设$0＜\epsilon\leq\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$, $(u_\epsilon, v_\epsilon)$是问题(2.1) 的解, 且正常数$k$满足(2.2) 式.则存在一个不依赖于$\epsilon$的正常数$\rho$使得

$\begin{equation*} u_\epsilon, v_\epsilon\geq ke^{-\rho t}\varphi, \qquad(x, t)\in\overline{{\it \Omega}}\times[0, T_*], \end{equation*}$

这里$T_*=\frac{T^*}{2}$.

这两个命题可以用类似于文献[13]中的方法给予证明, 这里我们略去其细节.

借助于命题2.1和2.2, 并利用标准的局部Schauder估计和对角线方法可知, 存在$\{\epsilon\}$的子列$\{\epsilon^\prime\}$和$u, v\in C^{2+\alpha, 1+\frac{\alpha}{2}}_{loc}({\it \Omega}\times(0, T_*])$使得对任意${\it \Omega}_*\Subset{\it \Omega}$和$0＜\tau＜T_*$都有

$\begin{equation*} (u_{\epsilon^\prime}, v_{\epsilon^\prime})\rightarrow (u, v) \quad \text{在}\quad [C^{2+\beta, 1+\frac{\beta}{2}}(\overline{{\it \Omega}}_*\times[\tau, T_*])]^2\quad(0＜\beta＜\alpha), \quad \quad\epsilon^\prime\rightarrow 0^+. \end{equation*}$

于是$(u, v)$满足方程(1.1).

固定$\epsilon_0:0＜\epsilon_0＜\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$.对任意${\it \Omega}_0\Subset{\it \Omega}$ and $0＜\epsilon^\prime<\epsilon_0$, 再次利用命题2.1, 2.2, 标准的$L^p$估计和嵌入定理可知$u_{\epsilon^\prime}$和$v_{\epsilon^\prime}$的$C^{\alpha, \frac{\alpha}{2}}(\overline{{\it \Omega}}_0\times[0, T_*])$范数对任意$\epsilon^\prime<\epsilon_0$是一致有界的.于是可知当$\epsilon^\prime\rightarrow 0^+$时,

$\begin{equation*} (u_{\epsilon^\prime}, v_{\epsilon^\prime})\rightarrow (u, v) \quad \text{在}\quad [C^{\beta, \frac{\beta}{2}}(\overline{{\it \Omega}}_0\times[0, T_*])]^2\quad(0＜\beta＜\alpha). \end{equation*}$

这表明$u, v \in C({\it \Omega}\times[0, T_*])$.类似于文献[2]中的讨论我们可知$(u, v)$连续到抛物侧边界$\partial{\it \Omega}\times(0, T_*]$.综上可得如下的局部存在性定理.

定理2.1  假设初值$(u_0(x), v_0(x))$满足(1.2) 式.则问题(1.1) 存在一个正古典解

$\begin{equation*} (u, v)\in[C^{2+\alpha, 1+\frac{\alpha}{2}}_{\rm loc}({\it \Omega}\times(0, T_*))\cap C(\overline{{\it \Omega}}\times[0, T_*])]^2, \quad \alpha:0＜\alpha＜1. \end{equation*}$

在这部分我们将给出问题(1.1) 存在整体正古典解的一个充分条件, 我们的结果是如下:

定理3.1 假设$r=s$, $l=m$, $b_1\geq c_1$, $b_2\leq c_2$.则只要$\lambda_1>\max\{a_1, a_2\}$, 问题(1.1) 就有整体正古典解.

证 我们的证明将分三步进行.

第一步 根据特征值对区域${\it \Omega}$的连续依赖性可知, 对任意$\rho\in(0, \lambda_1)$, 可以找到一个区域$D:D\Supset{\it \Omega}$, 使得$\lambda_1-\rho$是$-\Delta$在$D$内齐次Dirichlet边值问题的第一特征值, 即

$\left\{ {\begin{array}{*{20}{l}} { - \Delta \varphi = ({\lambda _1} - \rho )\varphi ,{\rm{ }}\quad \quad x \in D,}\\ {\varphi = 0,\quad \quad \quad \quad \quad \quad \quad x \in \partial D.} \end{array}} \right.$

(3.1)

记$\varphi$为满足$\max_{x\in D}\varphi(x)=1$的第一特征函数.易知在$\overline{{\it \Omega}}$上$\varphi＞0$.对任意正整数$n$, 定义

$\begin{equation}\label{3.2} h_{\epsilon,n}(t)=\int_{\it \Omega}\frac{u_\epsilon^n}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}},\quad\quad w_{\epsilon,n}(t)=\int_{\it \Omega}\frac{v_\epsilon^n}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}, \end{equation}$

(3.2)

这里$k＞0$待定.对(3.2) 式关于$t$求导, 利用方程(2.1) 并分部积分可得

$\begin{align} & h_{\epsilon ,n}^{\prime }(t)=n\int_{\mathit{\Omega }}{\frac{u_{\epsilon }^{n-1}v_{\epsilon }^{p}}{{{\varphi }^{kn}}}}[\Delta {{u}_{\epsilon }}+{{u}_{\epsilon }}({{a}_{1}}-{{b}_{1}}u_{\epsilon }^{s}+{{c}_{1}}v_{\epsilon }^{s})]\underset{\centerdot }{\mathop{\rm{x}}}\, \\ & =-n(n-1)\int_{\mathit{\Omega }}{\frac{u_{\epsilon }^{n-2}v_{\epsilon }^{p}}{{{\varphi }^{kn}}}}|\nabla {{u}_{\epsilon }}{{|}^{2}}\underset{\centerdot }{\mathop{\rm{x}}}\,+k{{n}^{2}}\int_{\mathit{\Omega }}{\frac{u_{\epsilon }^{n-1}v_{\epsilon }^{p}}{{{\varphi }^{kn+1}}}}\nabla {{u}_{\epsilon }}\nabla \varphi \underset{\centerdot }{\mathop{\rm{x}}}\, \\ & +n\int_{\partial \Omega }{\frac{{{\epsilon }^{n-1+p}}}{{{\varphi }^{kn}}}}\frac{\partial {{u}_{\epsilon }}}{\partial n}\underset{\centerdot }{\mathop{\rm{S}}}\,\rm{-}\mathit{np}\int_{\mathit{\Omega }}{\frac{\mathit{u}_{\epsilon }^{\mathit{n}\rm{-1}}\mathit{v}_{\epsilon }^{\mathit{p}\rm{-1}}}{{{\varphi }^{\mathit{kn}}}}}\nabla {{\mathit{v}}_{\epsilon }}\nabla {{\mathit{u}}_{\epsilon }}\underset{\centerdot }{\mathop{\rm{x}}}\, \\ & +n\int_{\mathit{\Omega }}{\frac{u_{\epsilon }^{n}v_{\epsilon }^{p}}{{{\varphi }^{kn}}}}({{a}_{1}}-{{b}_{1}}u_{\epsilon }^{s}+{{c}_{1}}v_{\epsilon }^{s})\underset{\centerdot }{\mathop{\rm{x}}}\,. \\ \end{align}$

(3.3)

为估计(3.3) 式, 我们利用下述恒等式

$\begin{equation}\label{3.4} \varphi^2|\nabla u_\epsilon|^2=|\varphi\nabla u_\epsilon-ku_\epsilon\nabla\varphi|^2+2ku_\epsilon\varphi\nabla u_\epsilon\nabla\varphi-k^2u_\epsilon^2|\nabla\varphi|^2. \end{equation}$

(3.4)

此时(3.3) 式变为

$\begin{eqnarray}\label{3.5} h_{\epsilon, n}^\prime(t) =-n(n-1)\int_{\it \Omega}\frac{u_\epsilon^{n-2}v_\epsilon^p}{\varphi^{kn+2}}|\varphi\nabla u_\epsilon-ku_\epsilon\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -kn(n-2)\int_{\it \Omega}\frac{u_\epsilon^{n-1}v_\epsilon^p}{\varphi^{kn+1}}\nabla u_\epsilon\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +k^2n(n-1)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+2}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+n\int_{\partial{\it \Omega}}\frac{\epsilon^{n-1+p}}{\varphi^{kn}}\frac{\partial u_\epsilon}{\partial n}\rm{\underset{\scriptscriptstyle\centerdot}{S}}\nonumber\\ -np\int_{\it \Omega}\frac{u_\epsilon^{n-1}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla v_\epsilon(\varphi\nabla u_\epsilon-ku_\epsilon\nabla\varphi)\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -nkp\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla v_\epsilon\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}+n\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn}}(a_1-b_1u_\epsilon^s+c_1v_\epsilon^s)\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(3.5)

对(3.5) 式中右端第二项分部积分可得

$\begin{eqnarray}\label{3.6} -n\int_{\it \Omega}\frac{u_\epsilon^{n-1}v_\epsilon^p}{\varphi^{kn+1}}\nabla u_\epsilon\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}} =-\int_{\partial{\it \Omega}}\frac{\epsilon^{n+p}}{\varphi^{kn+1}}\frac{\partial \varphi}{\partial n}\rm{\underset{\scriptscriptstyle\centerdot}{S}}-(kn+1)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+2}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+1}}\Delta\varphi \underset{\centerdot }{\mathop{\rm{x}}}+p\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla\varphi\nabla v_\epsilon \underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(3.6)

借助(3.1) 和(3.6) 式可得

$\begin{eqnarray}\label{3.7} -kn(n-2)\int_{\it \Omega}\frac{u_\epsilon^{n-1}v_\epsilon^p}{\varphi^{kn+1}}\nabla u_\epsilon\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}} +k^2n(n-1)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+2}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ =[k^2n(n-1)-k(n-2)(kn+1)]\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+2}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}-k(n-2)\int_{\partial{\it \Omega}}\frac{\epsilon^{n+p}}{\varphi^{kn+1}}\frac{\partial \varphi}{\partial n}\rm{\underset{\scriptscriptstyle\centerdot}{S}}\nonumber\\ -k(n-2)(\lambda_1-\rho)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}+kp(n-2)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla\varphi\nabla v_\epsilon \underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(3.7)

(3.5) 式中右端第五项可以借助Cauchy不等式估计如下:

$\begin{eqnarray}\label{3.8} np\Big|\int_{\it \Omega}\frac{u_\epsilon^{n-1}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla v_\epsilon(\varphi\nabla u_\epsilon-ku_\epsilon\nabla\varphi)\underset{\centerdot }{\mathop{\rm{x}}}\Big| \le \mathit{n}(\mathit{n}-1)\int_{\it \Omega}\frac{u_\epsilon^{n-2}v_\epsilon^p}{\varphi^{kn+2}}|\varphi\nabla u_\epsilon-ku_\epsilon\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +\frac{np^2}{n-1}\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-2}}{\varphi^{kn}}|\nabla v_\epsilon|^2\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(3.8)

将(3.7) 和(3.8) 式代入(3.5) 式, 可得

$\begin{eqnarray}\label{3.9} h_{\epsilon, n}^\prime(t)\nonumber\\ \le k(nk-n+2)\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn+2}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}} -[k(n-2)(\lambda_1-\rho)-a_1n]\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^p}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -2kp\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-1}}{\varphi^{kn+1}}\nabla\varphi\nabla v_\epsilon \underset{\centerdot }{\mathop{\rm{x}}}-b_1n\int_{\it \Omega}\frac{u_\epsilon^{n+s}v_\epsilon^p}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}-k(n-2)\int_{\partial{\it \Omega}}\frac{\epsilon^{n+p}}{\varphi^{kn+1}}\frac{\partial \varphi}{\partial n}\rm{\underset{\scriptscriptstyle\centerdot}{S}}\nonumber\\ +\frac{np^2}{n-1}\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p-2}}{\varphi^{kn}}|\nabla v_\epsilon|^2\underset{\centerdot }{\mathop{\rm{x}}}+c_1n\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p+s}}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}} +n\int_{\partial{\it \Omega}}\frac{\epsilon^{n-1+p}}{\varphi^{kn}}\frac{\partial u_\epsilon}{\partial n}\rm{\underset{\scriptscriptstyle\centerdot}{S}}. \end{eqnarray}$

(3.9)

利用Young不等式可以对(3.9) 式右端倒数第二项估计如下:

$\begin{eqnarray}\label{3.10} c_1n\int_{\it \Omega}\frac{u_\epsilon^{n}v_\epsilon^{p+s}}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}} \leq\frac{c_1n^2}{n+s}\int_{\it \Omega}\frac{u_\epsilon^{n+s}v_\epsilon^{p}}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}+\frac{c_1ns}{n+s}\int_{\it \Omega}\frac{v_\epsilon^{n+p+s}}{\varphi^{kn}}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(3.10)

由于对任意$0＜T＜T(\epsilon)$, $u_\epsilon, \ v_\epsilon$以及$\nabla v_\epsilon$在$\overline{{\it \Omega}}\times[0, T]$上是有界的, 故我们可以选取充分大的$n$使得在$\overline{{\it \Omega}}\times[0, T]$上有

$\begin{equation}\label{3.11} n\geq2pv_\epsilon^{p-2}(kv_\epsilon\varphi^{-1}|\nabla\varphi\nabla v_\epsilon|+p|\nabla v_\epsilon|^2)\text{且}n\geq v_\epsilon^{p+s}. n\geq u_\epsilon^{q+l} \end{equation}$

(3.11)

注意到$\lambda_1＞\max\{a_1, a_2\}$, 我们可以选取$\rho＞0$充分小使得$\lambda_1-\rho＞\max\{a_1, a_2\}$.然后取$k＜1$但是充分接近$1$使得$k(\lambda_1-\rho)＞\max\{a_1, a_2\}$.这样, 对充分大的$n$可得

$\begin{align} & h_{\epsilon ,n}^{\prime }(t)\le n{{h}_{\epsilon ,n}}(t)-({{b}_{1}}-{{c}_{1}})n\int_{\Omega }{\frac{u_{\epsilon }^{n+s}v_{\epsilon }^{p}}{{{\varphi }^{kn}}}}\underset{\centerdot }{\mathop{\rm{x}}}\,+{{c}_{1}}sn{{w}_{\epsilon ,n}}(t) \\ & +n\int_{\partial \Omega }{\frac{{{\epsilon }^{n-1+p}}}{{{\varphi }^{kn}}}}\frac{\partial {{u}_{\epsilon }}}{\partial n}\underset{\centerdot }{\mathop{\rm{S}}}\,\rm{-}\mathit{k}(\mathit{n}\rm{-2})\int_{\partial \Omega }{\frac{{{\epsilon }^{\rm{n+p}}}}{{{\varphi }^{\rm{kn+1}}}}}\frac{\partial \varphi }{\partial \mathit{n}}\underset{\centerdot }{\mathop{\rm{S}}}\, \\ & \le n{{h}_{\epsilon ,n}}(t)+{{c}_{1}}sn{{w}_{\epsilon ,n}}(t)+{{\delta }_{1}}n\int_{\partial \Omega }{\frac{|\frac{\partial \varphi }{\partial n}|+\varphi |\frac{\partial {{u}_{\epsilon }}}{\partial n}|}{{{\varphi }^{kn+1}}}}\underset{\centerdot }{\mathop{\rm{S}}}\,. \\ \end{align}$

(3.12)

类似的, 可以得到

$ \begin{eqnarray}\label{3.13} w_{\epsilon, n}^\prime(t)\le nw_{\epsilon, n}(t)+b_2\ln h_{\epsilon, n}(t)+\delta_2n\int_{\partial{\it \Omega}}\frac{|\frac{\partial \varphi}{\partial n}|+\varphi|\frac{\partial v_\epsilon}{\partial n}|}{\varphi^{kn+1}}\rm{\underset{\scriptscriptstyle\centerdot}{S}}. \end{eqnarray}$

(3.13)

将(3.12) 和(3.13) 式相加可知对充分大的$n$和任意$t\in[0, T]$有

$\begin{eqnarray}\label{3.14} \frac{d}{dt}(h_{\epsilon, n}(t)+w_{\epsilon, n}(t)) \leq\delta_3n(h_{\epsilon, n}(t)+w_{\epsilon, n}(t))\nonumber\\ +\delta_4n\int_{\partial{\it \Omega}}\frac{1}{\varphi^{kn+1}}(|\frac{\partial \varphi}{\partial n}|+\varphi|\frac{\partial u_\epsilon}{\partial n}|+\varphi|\frac{\partial v_\epsilon}{\partial n}|)\rm{\underset{\scriptscriptstyle\centerdot}{S}}, \end{eqnarray}$

(3.14)

这里$\delta_1, \cdots, \delta_4$是不依赖于$n$和$\epsilon$的正常数.求解(3.14) 式可得

$\begin{eqnarray}\label{3.15} h_{\epsilon, n}(t)+w_{\epsilon, n}(t)\leq(h_{\epsilon, n}(0)+w_{\epsilon, n}(0))e^{\delta_3nt}\nonumber\\ +\int_0^te^{\delta_4n(t-\tau)}\int_{\partial{\it \Omega}}\frac{\delta_4n}{\varphi^{kn+1}}\Big(\Big|\frac{\partial \varphi}{\partial n}\Big|+\varphi\Big|\frac{\partial u_\epsilon}{\partial n}\Big|+\varphi\Big|\frac{\partial v_\epsilon}{\partial n}\Big|\Big)\rm{\underset{\scriptscriptstyle\centerdot}{S}}d\tau. \end{eqnarray}$

(3.15)

对(3.15) 式开$n$次方根并令$n\rightarrow\infty$可得

$\begin{equation}\label{3.16} \max\Big\{\max\limits_{\overline{{\it \Omega}}}\frac{u_\epsilon(x, t)}{\varphi^k(x)}, \max\limits_{\overline{{\it \Omega}}}\frac{v_\epsilon(x, t)}{\varphi^k(x)}\Big\} \leq\max\Big\{\max\limits_{\overline{{\it \Omega}}}\frac{u_0(x)}{\varphi^k(x)}, \max\limits_{\overline{{\it \Omega}}}\frac{v_0(x)}{\varphi^k(x)}\Big\}e^{(\delta_3+\delta_4)t}. \end{equation}$

(3.16)

这表明$T(\epsilon)=\infty$, 且对任意$0＜\epsilon\leq\min\{(\frac{a_1}{b_1})^{\frac{1}{r}}, (\frac{a_2}{c_2})^{\frac{1}{m}}\}$, 都有$u_\epsilon+v_\epsilon\leq c_0e^{Mt}$.

第二步 第一步的结果表明对任意$T<T(\epsilon)$, $(u_\epsilon, v_\epsilon)$在$[0, T]$上都是有界的.利用类似于命题2.2的方法可以证明在${\it \Omega}\times[0, T]$上, $u_\epsilon, v_\epsilon\geq ke^{-\rho t}\varphi$, 这里$\rho$是不依赖于$\epsilon$的正常数.

第三步  由前两步的结论可知, 对任意${\it \Omega}_n\Subset{\it \Omega}$和$0＜\tau<T_n<T$, 存在两个正常数$\sigma(n, t)$和$M(n, t)$使得在$\overline{{\it \Omega}}_n\times[\tau, T_n]$上有$\sigma(n, t)\leq u_\epsilon, v_\epsilon\leq M(n, t)$.再次利用标准的局部Schauder估计和对角线法则可知存在$\{\epsilon\}$的子列$\{\epsilon^\prime\}$和$u, v\in C^{2+\alpha, 1+\frac{\alpha}{2}}_{\rm loc}({\it \Omega}\times(0, T))$使得对任意${\it \Omega}_*\Subset{\it \Omega}$和$0＜\tau＜T_0＜T$

$\begin{equation*} (u_{\epsilon^\prime}, v_{\epsilon^\prime})\rightarrow (u, v) \quad \text{于}\quad [C^{2+\alpha, 1+\frac{\alpha}{2}}(\overline{{\it \Omega}}_*\times[\tau, T_0])]^2, \quad\quad\epsilon^\prime\rightarrow 0^+. \end{equation*}$

于是$(u, v)$在${\it \Omega}\times(0, T)$上满足(1.1) 的方程.

类似于局部存在性时的讨论可知$(u, v)\in [C^{2+\alpha, 1+\frac{\alpha}{2}}_{\rm loc}({\it \Omega}\times(0, T))\cap C(\overline{{\it \Omega}}\times[0, T))]^2$是(1.1) 式的古典解.由$T$的任意性可知, $(u, v)\in [C^{2+\alpha, 1+\frac{\alpha}{2}}_{loc}({\it \Omega}\times(0, \infty))\cap C(\overline{{\it \Omega}}\times[0, \infty))]^2$, 即$(u, v)$是(1.1) 式的整体解.证毕.

注 利用类似于文献[14]中的方法还可以证明, 通过正则化方法得到的解$(u, v)$其实是问题(1.1) 的一个最大解, 即若$(\overline{u}, \overline{v})$是(1.1) 式的任意一个解, 则必有$u\geq\overline{u}, \ v\geq\overline{v}$.这和定理3.1表明此时问题(1.1) 的所有解都是整体解.

本节给出问题(1.1) 不存在整体古典解的一个充分条件, 主要结果是如下的

定理4.1  假设$b_1\leq c_1, \ c_2\leq b_2, r=s, l=m$, $0＜p, \ q<2$.则当$\lambda_1<\min\{a_1, a_2\}$时, 问题(1.1) 不存在非平凡的整体古典解.

证 若不然假设问题(1.1) 存在一个非平凡的整体古典解, 我们将证明它必在有限时刻爆破.根据特征值对区域$\it \Omega$的连续依赖性我们可以找到一个小常数$\theta＞0$和区域$D, \ D\Subset {\it \Omega}$使得

$\begin{equation}\label{4.1} \lambda_1+(4+\lambda_1)\theta+3\theta^2＜\min\{a_1, a_2\}, \ (1+\theta)p\leq2, \ (1+\theta)q\leq2, \end{equation}$

(4.1)

且$\lambda_1+\theta$是$-\Delta$在$D$内齐次Dirichlet边值问题的第一特征值, 即

$ \begin{equation}\label{4.2} \begin{cases} -\Delta\varphi=(\lambda_1+\theta)\varphi,\quad x\in D, \\ \varphi=0,\quad x\in\partial D. \end{cases} \end{equation}$

(4.2)

易知$u$和$v$在$\overline{D}$上是严格正的.

对任意正整数$n$, 定义

$\begin{equation}\label{4.3} h_n(t)=\int_D\frac{\varphi^{kn}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}, \qquad w_n(t)=\int_D\frac{\varphi^{kn}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}, \end{equation}$

(4.3)

这里$k\geq1$待定.对(4.3) 式关于$t$求导, 利用方程(1.1) 和分部积分公式可得

$\begin{eqnarray}\label{4.4} h_n^\prime(t)=-n\int_D\frac{\varphi^{kn}v^p}{u^{n+1}}[\Delta u+u(a_1-b_1u^s+c_1v^s)]\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ =-n(n+1)\int_D\frac{\varphi^{kn}v^p}{u^{n+2}}|\nabla u|^2\underset{\centerdot }{\mathop{\rm{x}}}+np\int_D\frac{\varphi^{kn}v^{p-1}}{u^{n+1}}\nabla v\nabla u \underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +kn^2\int_D\frac{\varphi^{kn-1}v^{p}}{u^{n+1}}\nabla \varphi\nabla u \underset{\centerdot }{\mathop{\rm{x}}} -n\int_D\frac{\varphi^{kn}v^p}{u^{n}}(a_1-b_1u^s+c_1v^s)\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.4)

为了估计(4.4) 式, 我们利用如下恒等式

$ \begin{equation*} \varphi^2|\nabla u|^2=|\varphi\nabla u-ku\nabla\varphi|^2+2ku\varphi\nabla u\nabla\varphi-k^2u^2|\nabla\varphi|^2. \end{equation*}$

此时(4.4) 式变为

$ \begin{eqnarray}\label{4.5} h_n^\prime(t) =-n(n+1)\int_D\frac{\varphi^{kn-2}v^p}{u^{n+2}}|\varphi\nabla u-ku\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}-kn(n+2)\int_D\frac{\varphi^{kn-1}v^{p}}{u^{n+1}}\nabla \varphi\nabla u \underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +k^2n(n+1)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+np\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^{n+1}}\nabla v(\varphi\nabla u-ku\nabla\varphi)\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +knp\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^{n}}\nabla v\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}-n\int_D\frac{\varphi^{kn}v^p}{u^{n}}(a_1-b_1u^s+c_1v^s)\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.5)

对(4.5) 右端第二项利用分部积分可得

$\begin{eqnarray}\label{4.6} -n\int_D\frac{\varphi^{kn-1}v^{p}}{u^{n+1}}\nabla \varphi\nabla u \underset{\centerdot }{\mathop{\rm{x}}} =-\int_D\frac{\varphi^{kn-1}v^p}{u^n}\Delta\varphi \underset{\centerdot }{\mathop{\rm{x}}}-p\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^n}\nabla v\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -(kn-1)\int_D\frac{\varphi^{kn-2}v^{p}}{u^n}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.6)

利用(4.2) 式和上述关系可得

$\begin{align} & -kn(n+2)\int_{D}{\frac{{{\varphi }^{kn-1}}{{v}^{p}}}{{{u}^{n}}\rm{+1}}}\nabla \varphi \nabla u\underset{\centerdot }{\mathop{\rm{X}}}\,\rm{+}{{\mathit{k}}^{\rm{2}}}\mathit{n}(\mathit{n}\rm{+1})\int_{\mathit{D}}{\frac{{{\varphi }^{\mathit{kn}\rm{-2}}}{{\mathit{v}}^{\mathit{p}}}}{{{\mathit{u}}^{\mathit{n}}}}}\rm{ }\!\!|\!\!\rm{ }\nabla \varphi {{\rm{ }\!\!|\!\!\rm{ }}^{\rm{2}}}\underset{\centerdot }{\mathop{\rm{X}}}\, \\ & \rm{=-}\mathit{k}(\mathit{nk}\rm{-}\mathit{n}\rm{-2})\int_{\mathit{D}}{\frac{{{\varphi }^{\mathit{kn}\rm{-2}}}{{\mathit{v}}^{\mathit{p}}}}{{{\mathit{u}}^{\mathit{n}}}}}\rm{ }\!\!|\!\!\rm{ }\nabla \varphi {{\rm{ }\!\!|\!\!\rm{ }}^{\rm{2}}}\underset{\centerdot }{\mathop{\rm{X}}}\,\rm{+}\mathit{k}(\mathit{n}\rm{+2})({{\lambda }_{\rm{1}}}\rm{+}\theta )\int_{\mathit{D}}{\frac{{{\varphi }^{\mathit{kn}}}{{\mathit{v}}^{\mathit{p}}}}{{{\mathit{u}}^{\mathit{n}}}}}\underset{\centerdot }{\mathop{\rm{X}}}\, \\ & \rm{-}\mathit{kp}(\mathit{n}\rm{+2})\int_{\mathit{D}}{\frac{{{\varphi }^{\mathit{kn}\rm{-1}}}{{\mathit{v}}^{\mathit{p}\rm{-1}}}}{{{\mathit{u}}^{\mathit{n}}}}}\nabla \mathit{v}\nabla \varphi \underset{\centerdot }{\mathop{\rm{X}}}\,. \\ \end{align}$

(4.7)

对(4.5) 式中右端第四项可以用Cauchy不等式作如下估计:

$\begin{eqnarray}\label{4.8} \Big|np\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^{n+1}}\nabla v(\varphi\nabla u-ku\nabla\varphi)\underset{\centerdot }{\mathop{\rm{x}}}\Big|\nonumber\\ \le n(n+1)\int_D\frac{\varphi^{kn-2}v^p}{u^{n+2}}|\varphi\nabla u-ku\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+\frac{np^2}{n+1}\int_D\frac{\varphi^{kn}v^{p-2}}{u^n}|\nabla v|^2\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.8)

将(4.7) 和(4.8) 式代入(4.5) 式, 我们得到

$\begin{eqnarray}\label{4.9} h_n^\prime(t) \leq-k(nk-n-2)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+k(n+2)(\lambda_1+\theta)\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +\frac{np^2}{n+1}\int_D\frac{\varphi^{kn}v^{p-2}}{u^n}|\nabla v|^2\underset{\centerdot }{\mathop{\rm{x}}}-2kp\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^n}\nabla v\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -n\int_D\frac{\varphi^{kn}v^p}{u^{n}}(a_1-b_1u^s+c_1v^s)\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.9)

同样对(4.9) 式中右端第四项利用Cauchy不等式可得

$\begin{eqnarray}\label{4.10} \Big|2kp\int_D\frac{\varphi^{kn-1}v^{p-1}}{u^n}\nabla v\nabla\varphi \underset{\centerdot }{\mathop{\rm{x}}}\Big| \leq kp\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+kp\int_D\frac{\varphi^{kn}v^{p-2}}{u^n}|\nabla v|^2\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.10)

将(4.10})式代入(4.9) 式有

$\begin{eqnarray}\label{4.11} h_n^\prime(t)\le k(n+2)(\lambda_1+\theta)\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-k(nk-n-2-p)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ +(kp+\frac{np^2}{n+1})\int_D\frac{\varphi^{kn}v^{p-2}}{u^n}|\nabla v|^2\underset{\centerdot }{\mathop{\rm{x}}} -n\int_D\frac{\varphi^{kn}v^p}{u^{n}}(a_1-b_1u^s+c_1v^s)\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.11)

由于$(u, v)$对所有的$t＞0$都存在, 且在$\overline{D}$上是严格正的, 故对任意给定的$T＞0$, 可以选择$n$充分大使得在$\overline{D}\times[0, T]$上

$\begin{equation*} k(n+2)\theta\geq p(k+p)v^{-2}|\nabla v|^2, \end{equation*}$

于是有

$ \begin{eqnarray}\label{4.12} h_n^\prime(t)\le k(n+2)(\lambda_1+2\theta)\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-k(nk-n-2-p)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -n\int_D\frac{\varphi^{kn}v^p}{u^{n}}(a_1-b_1u^s+c_1v^s)\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ =[k(n+2)(\lambda_1+2\theta)-a_1n]\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}+b_1n\int_D\frac{\varphi^{kn}v^p}{u^{n-s}}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -k(nk-n-2-p)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}-c_1n\int_D\frac{\varphi^{kn}v^{p+s}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.12)

利用Young不等式, 可以对(4.12) 式右端第三项做如下估计:

${{b}_{1}}n\int_{D}{\frac{{{\varphi }^{kn}}{{v}^{p}}}{{{u}^{n-s}}}}\underset{\centerdot }{\mathop{\text{X}}}\,\le {{\mathsf{b}}_{\text{1}}}(\text{n-s})\int_{\text{D}}{\frac{{{\varphi }^{\text{kn}}}{{\text{v}}^{\text{p+s}}}}{{{\text{u}}^{\text{n}}}}}\underset{\centerdot }{\mathop{\text{X}}}\,\text{+}{{\mathsf{b}}_{\text{1}}}\text{s}\int_{\text{D}}{\frac{{{\varphi }^{\text{kn}}}}{{{\text{v}}^{\text{n-p-s}}}}}\underset{\centerdot }{\mathop{\text{X}}}\,.$

(4.13)

将(4.13) 式代入(4.12) 式得

$\begin{eqnarray}\label{4.14} h_n^\prime(t)\leq[k(n+2)(\lambda_1+2\theta)-a_1n]\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}} -[(c_1-b_1)n+b_1s]\int_D\frac{\varphi^{kn}v^{p+s}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -k(nk-n-2-p)\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+b_1s\int_D\frac{\varphi^{kn}}{v^{n-p-s}}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.14)

类似的, 我们可以得到

$\begin{eqnarray}\label{4.15} w_n^\prime(t)\leq[k(n+2)(\lambda_1+2\theta)-a_2n]\int_D\frac{\varphi^{kn}u^q}{v^n}\underset{\centerdot }{\mathop{\rm{x}}} -[(b_2-c_2)n+c_2l]\int_D\frac{\varphi^{kn}u^{q+l}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ -k(nk-n-2-q)\int_D\frac{\varphi^{kn-2}u^q}{v^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}+c_2l\int_D\frac{\varphi^{kn}}{u^{n-q-l}}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.15)

由于$u, v$在$\overline{D}\times[0, T]$上有严格正的上下界, 故我们可以选取$n$充分大使得

$\begin{eqnarray}\label{4.16} b_1s\int_D\frac{\varphi^{kn}}{v^{n-p-s}}\underset{\centerdot }{\mathop{\rm{x}}}\leq\theta n\int_D\frac{\varphi^{kn}u^q}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}, \qquad c_2l\int_D\frac{\varphi^{kn}}{u^{n-q-l}}\underset{\centerdot }{\mathop{\rm{x}}}\leq\theta n\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.16)

取$k=1+\theta$, 将(4.14) 和(4.15) 式相加, 并将(4.16) 式代入其中可得

$\begin{eqnarray}\label{4.17} h_n^\prime(t)+w_n^\prime(t)\leq-\theta n\Big(\int_D\frac{\varphi^{kn}v^p}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}+\int_D\frac{\varphi^{kn-2}v^p}{u^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\Big)\nonumber\\ -\theta n\Big(\int_D\frac{\varphi^{kn}u^q}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}+\int_D\frac{\varphi^{kn-2}u^q}{v^{n}}|\nabla\varphi|^2\underset{\centerdot }{\mathop{\rm{x}}}\Big). \end{eqnarray}$

(4.17)

令$H_n(t)=h_n(t)+w_n(t)$.由(4.17) 式可知, 对任意$0\leq t_1\leq t_2\leq T$有

$\begin{equation*} H_n(t_2)\leq H_n(t_1). \end{equation*}$

对$H_n(t)$开$n$次方根并令$n\rightarrow\infty$可得

$ \begin{equation}\label{4.18} H(t)\stackrel{\triangle}{=}\max\Big\{\max\limits_{\overline{D}}\frac{\varphi^{k}}{u(x, t)}, \max\limits_{\overline{D}}\frac{\varphi^{k}}{v(x, t)}\Big\}\leq \max\Big\{\max\limits_{\overline{D}}\frac{\varphi^{k}}{u_0}, \max\limits_{\overline{D}}\frac{\varphi^{k}}{v_0}\Big\}. \end{equation}$

(4.18)

由于(4.18) 式不依赖于$T$, 故它对所有的$t＞0$都成立.此外由(4.17) 式可知$H(t)$是单调递减的.

记$\lim\limits_{t\rightarrow\infty}H(t)=A$.如果$A＞0$, 则$u$和$v$在$\overline{D}\times[0, \infty)$上是有界的.于是由(4.17) 式可知, 对充分大但是固定的$n$有$H^\prime_n(t)\leq-c_n$.这是一个矛盾, 因为此时$t$不可能趋于无穷大.如果$A=0$, 则对任意$N>1$, 存在充分大的$t_1$使得对所有的$t\geq t_1$都有

$\begin{equation}\label{4.19} \frac{\varphi^{k}}{u(x, t)}, \ \frac{\varphi^{k}}{v(x, t)}\leq H(t)\leq \frac{1}{N}\quad \text{或}\quad u(x, t), v(x, t)\geq N\varphi^k. \end{equation}$

(4.19)

为了得到$(u, v)$在$D$上的正下界, 我们可以重新选择一个区域$D_1$满足$D\Subset D_1\Subset{\it \Omega}$, 使得对于满足$0＜\widetilde{\theta}<\theta$的$\widetilde{\theta}$, $\lambda_1+\widetilde{\theta}$为$-\Delta$在$D$内齐次Dirichlet边值问题的第一特征值.用上述方法可以证明(4.19) 式在$D_1$上也是成立的.于是可以选取$N$充分大可知在$\overline{D}\times[t_1, \infty)$上$u, v\geq1$.不失一般性, 假设$p\leq q$.令$\delta^2=\min_{\overline{D}}\{\varphi^2+|\nabla\varphi|^2\}＞0$, 则由(4.19) 和(4.17) 式可得

$\begin{eqnarray}\label{4.20} H_n^\prime(t)\leq-\theta nN^p\delta^2\int_D\frac{\varphi^{kn-2+kp}}{u^{n}}\underset{\centerdot }{\mathop{\rm{x}}}-\theta nN^q\delta^2\int_D\frac{\varphi^{kn-2+kq}}{v^{n}}\underset{\centerdot }{\mathop{\rm{x}}}\nonumber\\ \leq-\theta nN^p\delta^2\int_D\frac{\varphi^{kn}}{u^{n}}\underset{\centerdot }{\mathop{\rm{x}}}-\theta nN^q\delta^2\int_D\frac{\varphi^{kn}}{v^{n}}\underset{\centerdot }{\mathop{\rm{x}}}. \end{eqnarray}$

(4.20)

第二个不等式成立是因为$kp, kq\leq2$, $\varphi\leq1$.在(4.19) 式中选取$t_1$和$N$使得$\theta N^p\delta^2\geq1$, and $\theta N^\beta\delta^2>1$, 则由(4.20) 式可知对任意$t\geq t_1, $

$\begin{equation*} H^\prime_n(t)\leq-nH_n(t). \end{equation*}$

由此可得对任意$t\geq t_1$, $H_n(t)\leq H_n(t_1)e^{-n(t-t_1)}$, 进而有$H(t)\leq H(t_1)e^{-(t-t_1)}$.类似于(4.19) 式, 我们得到

$\begin{equation}\label{4.21} u(x, t), v(x, t)\geq Ne^{t-t_1}\varphi^k, \quad\forall t\geq t_1. \end{equation}$

(4.21)

定义序列$\{t_i\}$满足$t_{i+1}=t_i+\frac{2^{2-i}}{p}$.易知当$i\rightarrow\infty$时, $t_i\rightarrow t_1+\frac{4}{p}$.将(4.21) 式代入(4.17) 式得

$ H^\prime_n(t)\leq-\theta nN^p\delta^2e^{p(t-t_1)}\int_D\frac{\varphi^{kn-2+kp}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-\theta nN^q\delta^2e^{q(t-t_1)}\int_D\frac{\varphi^{kn-2+kq}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}\\ \leq-ne^{p(t_2-t_1)}\int_D\frac{\varphi^{kn}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-ne^{q(t_2-t_1)}\int_D\frac{\varphi^{kn}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}\\ \leq-n2^{2}H_n(t), \qquad \forall t\geq t_2. $

由此可得

$\begin{equation*} H_n(t)\leq H_n(t_2)e^{-4n(t-t_2)}\leq H_n(t_1)e^{-4n(t-t_2)} \end{equation*}$

或

$\begin{equation}\label{4.22} u(x, t), \ v(x, t)\geq Ne^{4(t-t_2)}\varphi^k\geq N2^{4(t-t_2)}\varphi^k, \qquad t\geq t_2. \end{equation}$

(4.22)

假设对$i\leq2$, 有

$\begin{equation}\label{4.23} u(x, t), v(x, t)\geq N2^{(t-t_i)2^{2i-2}}\varphi^k, \quad \forall t\geq t_i.\quad\text{for}\quad i\leq2. \end{equation}$

(4.23)

重复上面的操作可得

$ H^\prime_n(t)\leq-\theta nN^p\delta^22^{p(t-t_i)2^{2i-2}}\int_D\frac{\varphi^{kn}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-\theta nN^q\delta^22^{q(t-t_i)2^{2i-2}}\int_D\frac{\varphi^{kn}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}\\ \leq-n2^{p(t_{i+1}-t_i)2^{2i-2}}\int_D\frac{\varphi^{kn}}{u^n}\underset{\centerdot }{\mathop{\rm{x}}}-n2^{q(t_{i+1}-t_i)2^{2i-2}} \int_D\frac{\varphi^{kn}}{v^n}\underset{\centerdot }{\mathop{\rm{x}}}\\ \leq-n2^{2^i}H_n(t)\\ \leq-n2^{2i}H_n(t), \qquad \forall t\geq t_{i+1}. $

由此可得

$\begin{equation*} H_n(t)\leq H_n(t_{i+1})e^{-n(t-t_{i+1})2^{2i}}, \end{equation*}$

进而有

$ \begin{equation*} u(x, t), \ v(x, t)\geq N2^{(t-t_{i+1})2^{2i}}\varphi^k, \qquad \forall t\geq t_{i+1}. \end{equation*}$

于是由数学归纳法可知, (4.23) 式对所有的$i\geq2$都成立.取$t=t_0>t_1+\frac{4}{p}$并令$i\rightarrow\infty$可得矛盾.证毕.