带时空积分源项的耦合抛物方程的爆破

Blow-Up of a Coupled Parabolic Equation with the Time-Space Integral Source Term

摘要: 本文研究一个带时空积分源项的耦合抛物方程。通过线性变换，利用固定模问题的特征函数方法给出四个带系统解的特殊函数的导数之间的关系，并且构造合适辅助函数得到系统解发生爆破的充分条件；根据时空积分源项在时间初始时为零的特性构造出恰当的爆破的下解，给出系统解爆破的充分条件以及爆破时刻的上界；选取合适的整体存在的上解给出系统解整体存在的充分条件。

Abstract: In this paper, a Coupled Parabolic Equation with time-space integral source term is studied. Through linear transformation, the relationship between the derivatives of four special functions with system solutions is given by using the characteristic function method of fixed mode problem, and the sufficient conditions for the blow-up of system solutions are obtained by constructing appropriate auxiliary functions; According to the characteristic that the source term of space-time integration is zero at the beginning of time, an appropriate blow-up lower solution is constructed, and the sufficient conditions for the blow-up of the system solution and the upper bound of the blow-up time are given; The sufficient conditions for the global existence of the solution of the system are given by selecting the appropriate upper solution of the global existence.

1. 引言

在二十世纪六十年代，Kaplan在[1]研究了如下拟线性抛物方程的爆破问题：

$\frac{\partial u}{\partial t}-{\displaystyle \sum _{i,j=1}^n{a}_{i,j}}\left(x,t,u,\nabla u\right)\frac{{\partial }^{2}u}{\partial x_i\partial x_j}=F\left(x,t,u,\nabla u\right)$ 。自此非线性反应扩散方程解的爆破理论逐渐步入人们的视野，

随后Fujita在[2]中研究Cauchy问题$u_t=\Delta u+u^p$ 解的爆破行为。自此涌现出了众多学者们对非线性反应扩散方程解的爆破理论的研究工作，例如方程解的整体存在或者发生爆破，爆破速率等等。解的爆破理论对于研究化学反应模型，生物种群模型，热传导模型等领域发挥着重要的作用，除了上述爆破性质，许多学者还关注了当方程爆破解存在时，爆破时刻或者爆破点的刻画，Levine在[3]中首次给出方程解爆破时刻的上界，Weissler在[4]中首次给出方程解爆破时刻下界。过去几十年来，随着研究的不断深入，学者们发现边值条件，初值条件，空间的维数等对非线性抛物方程爆破有着紧密的联系，随着研究的抛物方程模型越来越复杂，边值条件具体分为Dirichlet [5]，Neumann [6]，Robin [7]三种。且对于非线性项的系数也变得多种多样，从常系数到时变系数或者权重函数相较于常系数模型，变系数模型包含更多的未知参数，可以用来描述更复杂的物理现象。对于非线性项中含有变系数的抛物方程，学者们通过变系数的性质来构造合适的辅助函数进行处理。另外复杂源项出现了带空间积分源项[8]，带时间积分源项[9]，带时空积分源项[10]三种，使得非线性抛物方程的建模与分析变得更加具有挑战性。

文献[8]研究了如下非局部反应扩散方程；

$\{\begin{array}{ll}{u}_t=\Delta u+k_1\left(t\right)u^p{\displaystyle {\int }_{\Omega }{v}^q}\text{d}x,\hfill & \left(x,t\right)\in \Omega \times \left(0,t^{*}\right),\hfill \\ v_t=\Delta v+k_2\left(t\right)v^p{\displaystyle {\int }_{\Omega }{u}^q}\text{d}x,\hfill & \left(x,t\right)\in \Omega \times \left(0,t^{*}\right),\hfill \\ \beta \frac{\partial u}{\partial \upsilon }+\alpha u=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left(0,t^{*}\right),\hfill \\ \beta \frac{\partial v}{\partial \upsilon }+\alpha v=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left(0,t^{*}\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \Omega ,\hfill \\ v\left(x,0\right)=v_0\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$

其中$\Omega \subset R^N$ 是一个具有光滑边界$\partial \Omega$ 的有界区域，$k_1$ ，$k_2$ 是正函数并且$\alpha$ ，$\beta \ge 0$ ，初始数据$u_0\left(x\right)$ ，$v_0\left(x\right)$ 是非负函数满足相容性条件。作者分别在Dirichlet，Neumann，Robin边界条件下，通过找到爆破的下解给出爆破时间的上界，利用辅助函数法给出爆破时间的下界，再构造辅助函数给出整体解存在的充分条件。

文献[9]研究了如下半线性抛物方程：

$\{\begin{array}{ll}{u}_t-\Delta u=u^q{\displaystyle {\int }_0^t{u}^p}\text{d}x,\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ u\left(x,t\right)=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$

其中$\Omega \subset R^n$ 是一个有界区域且$\partial \Omega \subset C^2$ ，$p$ ，$q\ge 0$ ，初始数据$u_0\left(x\right)$ 是非负连续函数且消失在边界。作者给出存在性定理以及比较原理，在不同假设条件下构造合适的下解和上解分别证明爆破解和整体解的存在性。

文献[10]研究了如下带加权函数的时空积分源项的抛物方程：

$\{\begin{array}{ll}{u}_t=\Delta u+a\left(x\right)f\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }\beta }}\left(x\right)g\left(u\left(x,s\right)\right)\text{d}x\text{d}s\right),\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ u\left(x,t\right)=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$

其中$\Omega \subset R^n$ 是一个有界区域且具有光滑边界，$a\left(x\right)$ ，$u_0\left(x\right)\in C^2\left(\Omega \right)\cap C^0\left(\overline{\Omega }\right)$ ，$a\left(x\right)$ ，$u_0\left(x\right)>0$ ，$x\in \Omega$ 。$a\left(x\right)$ ，$u_0\left(x\right)=0$ ，$x\in \partial \Omega$ ，$\beta \left(x\right)\in C\left(\overline{\Omega }\right)$ ，$\beta \left(x\right)\ge 0$ ，$f\left(s\right)=s^p$ 。作者利用加权函数的一致连续性给出$a\left(x\right)$ 的估计，找到一个u的下解估计$v_j$ ，利用固定模问题的第一特征函数和Green公式给出${\displaystyle {\int }_{B}u}\phi \text{d}x$ 的等价函数，最后利用反证法找到一个序列求出u的等价函数。构造辅助函数，利用等价函数得到导数的估计，最后通过积分计算出爆破速率。

本文研究了如下带时空积分源项的耦合抛物方程：

$\{\begin{array}{ll}{u}_t=\Delta u+u^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{v}^q}}\text{d}x\text{d}s,\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ v_t=\Delta v+v^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{u}^s}}\text{d}x\text{d}s,\hfill & \left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ u\left(x,t\right)=v\left(x,t\right)=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),v\left(x,0\right)=v_0\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$ (1)

其中$\Omega \subset R^N\left(N\ge 1\right)$ 是一个具有$C^2$ 边界的有界区域且，$p$ ，$q$ ，$r$ ，$s>0$ ，初值$u_0\left(x\right)$ ，$v_0\left(x\right)$ 是连续函数且满足相容性条件。

时空积分源项描述了系统在时间和空间上的累积效应。这种非局部性可以捕捉到系统历史状态对当前状态的影响。例如，在核反应堆动力学中，温度的变化会影响中子通量，而中子通量的变化又反过来影响温度。这种反馈机制可以通过时空积分源项来描述。而在核反应堆中，温度和中子通量之间的非线性反馈可能导致温度的急剧上升，从而引发反应堆失控，因此带时空积分源项的抛物方程的爆破现象在理论和应用中都具有重要意义。

2. 爆破解

令$\lambda$ 是下面齐次Dirichlet边界条件下的第一特征值并且$\phi \left(x\right)$ 是对应的特征函数：

$\{\begin{array}{ll}-\Delta \phi =\lambda \phi ,\hfill & x\in \Omega ,\hfill \\ \phi =0,\hfill & x\in \partial \Omega .\hfill \end{array}$

则有$\lambda >0$ ，$\underset{x\in \Omega }{\min }\phi \left(x\right)\le \phi \left(x\right)\le \underset{x\in \Omega }{\max }\phi \left(x\right)$ ，$x\in \Omega$ 且${\displaystyle {\int }_{\Omega }\phi }\left(x\right)\text{d}x=1$ 。

定理1：设$\left(u,v\right)$ 是问题(1)的非负经典解. 若$p<1$ ，$r<1$ ，$qs>\left(1-p\right)\left(1-r\right)$ ，则$\left(u,v\right)$ 在有限时刻爆破。

证明：令$w\left(x,t\right)=u^{1-p},z\left(x,t\right)=v^{1-r}$ ，

通过计算可以得到

$\frac{1}{1-p}{w}^{\frac{p}{1-p}}{w}_t=\frac{1}{1-p}{w}^{\frac{p}{1-p}}\Delta w+\frac{p}{{\left(1-p\right)}^2}{w}^{\frac{2p-1}{1-p}}{\left|\nabla w\right|}^2+w^{\frac{p}{1-p}}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{z}^{\frac{q}{1-r}}}}\text{d}x\text{d}\tau$ ，

$\frac{1}{1-r}{z}^{\frac{r}{1-r}}{z}_t=\frac{1}{1-r}{z}^{\frac{r}{1-r}}\Delta z+\frac{r}{{\left(1-r\right)}^2}{z}^{\frac{2r-1}{1-r}}{\left|\nabla z\right|}^2+z^{\frac{r}{1-r}}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{w}^{\frac{s}{1-p}}}}\text{d}x\text{d}\tau$ 。

故有

$\begin{array}{l}{w}_t\ge \Delta w+\left(1-p\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{z}^{\alpha }}}\text{d}x\text{d}\tau ,\\ z_t\ge \Delta z+\left(1-r\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{w}^{\beta }}}\text{d}x\text{d}\tau .\end{array}$ (2)

其中$\alpha =\frac{q}{1-r}>1,\beta =\frac{s}{1-p}>1$ 。

记

$f_1\left(t\right)={\displaystyle {\int }_{\Omega }w}\left(x,t\right)\phi \left(x\right)\text{d}x,g_1\left(t\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{z}^{\alpha }}}\left(x,t\right)\phi \left(x\right)\text{d}x\text{d}\tau ,$

$f_2\left(t\right)={\displaystyle {\int }_{\Omega }z}\left(x,t\right)\phi \left(x\right)\text{d}x,g_2\left(t\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{w}^{\beta }}}\left(x,t\right)\phi \left(x\right)\text{d}x\text{d}\tau .$

则$f_i\left(t\right)$ 和$g_i\left(t\right)$ 是单调递增函数，则存在一个正常数$t_1$ 使得$f_i\left(t\right)>1$ ，并且$g_i\left(t\right)>1,i=1,2,\forall t\in \left(t_1,t^{*}\right)$ 。

通过使用Green公式有

$\begin{array}{c}{f^{\prime }}_1\left(t\right)={\displaystyle {\int }_{\Omega }{w}_t}\left(x,t\right)\phi \left(x\right)\text{d}x\ge {\displaystyle {\int }_{\Omega }\left(\Delta w+\left(1-p\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{z}^{\alpha }}}\left(x,t\right)\text{d}x\text{d}\tau \right)\phi \left(x\right)\text{d}x}\\ ={\displaystyle {\int }_{\Omega }w}\left(x,t\right)\Delta \phi \left(x\right)\text{d}x+\left(1-p\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{z}^{\alpha }}}\left(x,t\right)\text{d}x\text{d}\tau \\ \ge -\lambda f_1\left(x,t\right)+{\mu }_1{g}_1\left(t\right),\\ {f^{\prime }}_2\left(t\right)\ge -\lambda f_2\left(x,t\right)+{\mu }_2{g}_2\left(t\right),\end{array}$ (3)

其中${\mu }_1=\frac{1-p}{\underset{x\in \Omega }{\max }\phi \left(x\right)},{\mu }_2=\frac{1-r}{\underset{x\in \Omega }{\max }\phi \left(x\right)}$ 。由Jensen不等式得

$\begin{array}{l}{g^{\prime }}_1\left(t\right)={\displaystyle {\int }_{\Omega }{z}^{\alpha }}\left(x,t\right)\phi \left(x\right)\text{d}x\ge {\left({\displaystyle {\int }_{\Omega }z}\left(x,t\right)\phi \left(x\right)\text{d}x\right)}^{\alpha }={\left(f_2\left(t\right)\right)}^{\alpha },\hfill \\ {g^{\prime }}_2\left(t\right)\ge {\left(f_1\left(t\right)\right)}^{\beta }.\hfill \end{array}$ (4)

令$\gamma \in \left(\max \left\{\frac{1}{\alpha },\frac{1}{\beta }\right\},1\right)$ ，则对所有得$ϵ>0$ 都存在一个常数$C>0$ 使得

$C{\left({g^{\prime }}_1\left(t\right)\right)}^{\gamma }\ge C{\left(f_2\left(t\right)\right)}^{\alpha \gamma }\ge \left(3\lambda +1\right)f_2\left(t\right)-ϵ,$

$C{\left({g^{\prime }}_2\left(t\right)\right)}^{\gamma }\ge C{\left(f_1\left(t\right)\right)}^{\beta \gamma }\ge \left(3\lambda +1\right)f_1\left(t\right)-ϵ.$

故

(5)

接下来，令$l\in \left(0,\gamma \right)$ ，并使用Young不等式去处理${\left({g^{\prime }}_i\left(t\right)\right)}^{\gamma }$ ，

$\begin{array}{c}\left(C+1\right){\left({g^{\prime }}_i\left(t\right)\right)}^{\gamma }=\frac{\left(C+1\right){\left({g^{\prime }}_i\left(t\right)\right)}^{\gamma }}{g_i^l\left(t\right)}{g}_i^l\left(t\right)\le \gamma \frac{{\left(C+1\right)}_{}^{\frac{1}{\gamma }}{g^{\prime }}_i\left(t\right)}{g_i^{\frac{l}{\gamma }}\left(t\right)}+(1-\gamma )g_i^{\frac{l}{1-\gamma }}\left(t\right)\\ =\frac{{\gamma }^2{\left(C+1\right)}_{}^{\frac{1}{\gamma }}}{\gamma -l}\left({g^{\prime }}_i^{\theta }\left(t\right)\right)+\left(1-\gamma \right)g_i^{\frac{l}{1-\gamma }}\left(t\right).\end{array}$

其中$i=1,2,\theta =1-\frac{l}{\gamma }\in \left(0,1\right)$ 。选择$ϵ=1-\gamma <\min \left\{{\mu }_1,{\mu }_2\right\}$ ，和足够小得$l$ 使得$\frac{l}{1-\gamma }<1$ 。

记$A\left(t\right)=C_1\left(f_1\left(t\right)+g_1^{\theta }\left(t\right)+f_2\left(t\right)+g_2^{\theta }\left(t\right)\right),t\in \left(t_1,t^{*}\right)$ ，其中$C_1=\max \left\{3,\frac{{\gamma }^2{\left(C+1\right)}_{}^{\frac{1}{\gamma }}}{\gamma -l}\right\}$ 。令

$m=\min \left\{\alpha \gamma ,\beta \gamma ,\frac{1}{\theta }\right\}$ 。使用不等式$a^n+b^n\ge 2^{1-n}{\left(a+b\right)}^n,\forall a,b>0,n>1$ ，有

$\begin{array}{c}{A}^{\prime }\left(t\right)=C_1\left({f^{\prime }}_1\left(t\right)+\left(g_1^{\theta }\left(t\right)\right){}^{\prime }+{f^{\prime }}_2\left(t\right)+\left(g_2^{\theta }\left(t\right)\right){}^{\prime }\right)\\ \ge C_1{f^{\prime }}_1\left(t\right)+\left(C+1\right){\left({g^{\prime }}_1\left(t\right)\right)}^{\gamma }-ϵg_1^{\frac{l}{1-\gamma }}\left(t\right)+C_1{f^{\prime }}_2\left(t\right)+\left(C+1\right){\left({g^{\prime }}_2\left(t\right)\right)}^{\gamma }-ϵg_2^{\frac{l}{1-\gamma }}\left(t\right)\\ \ge {\left(f_2\left(t\right)\right)}^{\alpha \gamma }+{\left(f_1\left(t\right)\right)}^{\beta \gamma }+3{\mu }_2{g}_2\left(t\right)+3{\mu }_1{g}_1\left(t\right)\\ +f_2\left(t\right)+f_1\left(t\right)-2ϵ-ϵg_1^{\frac{l}{1-\gamma }}\left(t\right)-ϵg_2^{\frac{l}{1-\gamma }}\left(t\right)\\ \ge {\left(f_1\left(t\right)\right)}^{\beta \gamma }+f_1\left(t\right)+{\left(f_2\left(t\right)\right)}^{\alpha \gamma }+f_2\left(t\right)+{\mu }_1{g}_1^{1-m\theta }\left(t_1\right)g_1^{m\theta }\left(t\right)\\ +{\mu }_2{g}_2^{1-m\theta }\left(t_1\right)g_2^{m\theta }\left(t\right)+{\mu }_1{g}_1\left(t_1\right)-ϵ+{\mu }_2{g}_2\left(t_1\right)-ϵ\\ \ge f_1^m\left(t\right)+f_2^m\left(t\right)+{\mu }_1{g}_1^{1-m\theta }\left(0\right)g_1^{m\theta }\left(t\right)+{\mu }_2{g}_2^{1-m\theta }\left(0\right)g_2^{m\theta }\left(t\right)\\ \ge C_2\left(f_1^m\left(t\right)+f_2^m\left(t\right)+g_1^{m\theta }\left(t\right)+g_2^{m\theta }\left(t\right)\right)\\ \ge C_2{2}^{1-m}\left[{\left(f_1\left(t\right)+f_1\left(t\right)\right)}^m+{\left(g_1^{\theta }\left(t\right)+g_2^{\theta }\left(t\right)\right)}^m\right]\\ \ge C_2{2}^{2-2m}{\left(f_1\left(t\right)+f_1\left(t\right)+g_1^{\theta }\left(t\right)+g_2^{\theta }\left(t\right)\right)}^m\\ =C_2{2}^{2-2m}{A}^m\left(t\right).\end{array}$ (6)

其中$C_2=\min \left\{1,{\mu }_1{g}_1^{1-m\theta }\left(t_1\right),{\mu }_2{g}_2^{1-m\theta }\left(t_1\right)\right\}$ 。从0到t积分(6)式，有

$A\left(t\right)\ge {\left[J^{1-m}\left(0\right)-\left(m-1\right)C_2{2}^{2-2m}t\right]}^{-\frac{1}{m-1}}.$

所以$A\left(t\right)$ 在有限时刻$t^{*}\le \frac{1}{\left(m-1\right)C_2{2}^{2-2m}{J}^{1-m}\left(0\right)}$ 爆破。

定理2：令$\left(u,v\right)$ 是问题(1)的经典解。若$p,r>1,q,s\ge 1$ 且初值$u_0,v_0$ 足够大，则$\left(u,v\right)$ 在有限时刻爆破。

证明：构造一个下解如下：

$\left(\underset{\_}{u},\underset{\_}{v}\right)=\left(c\phi \left(x\right)\left[\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1}{1-\sigma t}}+{\text{e}}^{ϵ-2\lambda t}\right],c\phi \left(x\right)\left[\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1}{1-\sigma t}}+{\text{e}}^{ϵ-2\lambda t}\right]\right)$ ，有$\left(\underset{\_}{u},\underset{\_}{v}\right)$ 在时刻$t^{*}=\frac{1}{\sigma }$ 处爆破。

通过计算可以得到

${\underset{\_}{u}}_t-\Delta \underset{\_}{u}={\underset{\_}{v}}_t-\Delta \underset{\_}{v}=c\phi \left(x\right)\left[\left(\frac{2\sigma }{{\left(1-\sigma t\right)}^3}+\frac{\sigma }{{\left(1-\sigma t\right)}^4}+\frac{\lambda }{{\left(1-\sigma t\right)}^2}\right){\text{e}}^{\frac{1}{1-\sigma t}}-\lambda {\text{e}}^{ϵ-2\lambda t}\right],$

所以可以选择足够大的$ϵ$ 使得$\left(3\delta +\lambda \right)\text{e}-\lambda {\text{e}}^{ϵ}\le 0$ 。因为${\underset{\_}{u}}_t-\Delta \underset{\_}{u}$ 关于变量$t$ 是增的且在时刻$\frac{1}{\sigma }$ 处趋于

无穷大，则存在一个随着$ϵ$ 的增加而增加的时刻$t_0$ 并且满足${\underset{\_}{u}}_t-\Delta \underset{\_}{u}=0$ 。

当$t\in \left(0,t_0\right)$ 时，

${\underset{\_}{u}}_t-\Delta \underset{\_}{u}\le 0\le {\underset{\_}{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{v}}^q}}\text{d}x\text{d}\tau ,$

${\underset{\_}{v}}_t-\Delta \underset{\_}{v}\le 0\le {\underset{\_}{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{u}}^s}}\text{d}x\text{d}\tau .$

当$t\in \left(t_0,t^{*}\right)$ 时，选择足够大的$ϵ$ 使得$\min \left\{{\text{e}}^{\frac{q}{1-\sigma t_0}},{\text{e}}^{\frac{s}{1-\sigma t_0}}\right\}\ge \frac{2\left(3\sigma +\lambda \right)}{{\left(1-\sigma t_0\right)}^2}$ 和$\frac{1}{1-\sigma t}>\text{ln}2+1$ 成立，故可以得到

$\begin{array}{l}{\underset{\_}{u}}_t-\Delta \underset{\_}{u}\le c\phi \left(x\right)\left(\frac{2\sigma }{{\left(1-\sigma t\right)}^3}+\frac{\sigma }{{\left(1-\sigma t\right)}^4}+\frac{\lambda }{{\left(1-\sigma t\right)}^2}\right){\text{e}}^{\frac{1}{1-\sigma t}}\\ \le c\phi \left(x\right)\frac{3\sigma +\lambda }{{\left(1-\sigma t\right)}^4}{\text{e}}^{\frac{1}{1-\sigma t}}\\ \le \frac{1}{2}c\phi \left(x\right)\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1+q}{1-\sigma t}}\\ {\underset{\_}{v}}_t-\Delta \underset{\_}{v}\le \frac{1}{2}c\phi \left(x\right)\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1+s}{1-\sigma t}}\end{array}$

因为$p,r>1,q,s\ge 1$ ，可以计算

$\begin{array}{c}{\underset{\_}{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{v}}^q}}\text{d}x\text{d}\tau \ge c^{p+q}{\phi }^p\left(x\right){\left(\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1}{1-\sigma t}}\right)}^p{\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x{\displaystyle {\int }_0^t{\left(\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{1}{1-\sigma t}}\right)}^q}\text{d}\tau \\ \ge c^{p+q}{\phi }^p\left(x\right)\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{p}{1-\sigma t}}{\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x\frac{1}{\sigma }\left({\text{e}}^{\frac{1}{1-\sigma t}}-\text{e}\right)\\ \ge c^{p+q}{\phi }^p\left(x\right)\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{p+1}{1-\sigma t}}{\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x\frac{1}{2\sigma },\end{array}$

${\underset{\_}{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{u}}^s}}\text{d}x\text{d}\tau \ge c^{r+s}{\phi }^r\left(x\right)\frac{1}{{\left(1-\sigma t\right)}^2}{\text{e}}^{\frac{r+1}{1-\sigma t}}{\displaystyle {\int }_{\Omega }{\phi }^s}\left(x\right)\text{d}x\frac{1}{2\sigma }$

令$\sigma =\min \left\{c^{p+q-1}{\underset{\_}{\phi }}^{p-1}\left(x\right){\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x,c^{r+s-1}{\underset{\_}{\phi }}^{r-1}\left(x\right){\displaystyle {\int }_{\Omega }{\phi }^s}\left(x\right)\text{d}x\right\}$ ，可以得到

${\underset{\_}{u}}_t-\Delta \underset{\_}{u}\le {\underset{\_}{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{v}}^q}}\text{d}x\text{d}\tau ,$

${\underset{\_}{v}}_t-\Delta \underset{\_}{v}\le {\underset{\_}{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\underset{\_}{u}}^s}}\text{d}x\text{d}\tau .$

另一方面，假设初值$\left(u_0\left(x\right),v_0\left(x\right)\right)$ 足够大且满足

$\begin{array}{l}\min u\left(x,0\right)\ge c\overline{\phi }\left(x\right)\left[\text{e}+{\text{e}}^{ϵ}\right],\\ \min v\left(x,0\right)\ge c\overline{\phi }\left(x\right)\left[\text{e}+{\text{e}}^{ϵ}\right],\\ {\underset{\_}{u}|}_{\partial \Omega }=0\le {u|}_{\partial \Omega },\\ {\underset{\_}{v}|}_{\partial \Omega }=0\le {v|}_{\partial \Omega }.\end{array}$

则$\left(\underset{\_}{u},\underset{\_}{v}\right)$ 是$\left(u,v\right)$ 的下解并且意味着$\left(u,v\right)$ 在有限时刻爆破。

定理3：令$\left(u,v\right)$ 是问题(1)的经典解。若$p,r\ge 1$ ，且$q,s>0$ ，且初值$u_0,v_0$ 足够小，则$\left(u,v\right)$ 整体存在。

证明：如果$p=r=1$ 。令$\left(\overline{u},\overline{v}\right)=\left(C{\text{e}}^{-\zeta t}\phi \left(x\right),C{\text{e}}^{-\zeta t}\phi \left(x\right)\right)$，$\beta =\min \left\{{\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x,{\displaystyle {\int }_{\Omega }{\phi }^s}\left(x\right)\text{d}x\right\}$ 。

$\begin{array}{l}{\overline{u}}_t-\Delta \overline{u}-{\overline{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{v}}^q}}\text{d}x\text{d}\tau =-\zeta C{\text{e}}^{-\zeta t}\phi \left(x\right)+\lambda C{\text{e}}^{-\zeta t}\phi \left(x\right)-C^q\phi \left(x\right){\text{e}}^{-\zeta t}{\displaystyle {\int }_0^t{\text{e}}^{-\zeta q\tau }}\text{d}\tau {\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x\\ =C{\text{e}}^{-\zeta t}\phi \left(x\right)\left(-\zeta +\lambda -C^q\beta \right)\\ {\overline{v}}_t-\Delta \overline{v}-{\overline{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{u}}^s}}\text{d}x\text{d}\tau =C{\text{e}}^{-\zeta t}\phi \left(x\right)\left(-\zeta +\lambda -C^s\beta \right).\end{array}$

选择足够小的$\zeta <\lambda$ 和C满足$\lambda \ge \max \left\{\zeta +C^q\beta ,\zeta +C^s\beta \right\}$ ，接下来选择足够小的$\left(u_0\left(x\right),v_0\left(x\right)\right)$ 满足$\max \left\{\max u_0\left(x\right),\max v_0\left(x\right)\right\}\le C\text{e}\underset{\_}{\phi }\left(x\right)$ ，

则$\left(\overline{u},\overline{v}\right)$ 是$\left(u,v\right)$ 的上解。故$\left(u,v\right)$ 整体存在。

若$p,r\ge 1$ 。令$\left(\overline{u},\overline{v}\right)=\left(\frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }},\frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }}\right)$ ，选择合适的$\alpha$ 满足

$\max \left\{\frac{1}{p+q-1},\frac{1}{r+s-1}\right\}\le \alpha \le \min \left\{\frac{1}{q},\frac{1}{s}\right\}$ ，

令A足够大使得

$\lambda \ge \max \left\{\frac{\alpha }{A+t}+\frac{\beta {\overline{\phi }}^{p-1}\left(x\right)}{1-q\alpha }{\left(A+t\right)}^{-\left(p+q-1\right)\alpha +1},\frac{\alpha }{A+t}+\frac{\beta {\overline{\phi }}^{r-1}\left(x\right)}{1-s\alpha }{\left(A+t\right)}^{-\left(r+s-1\right)\alpha +1}\right\}$ 。

通过计算得：

$\begin{array}{l}{\overline{u}}_t-\Delta \overline{u}-{\overline{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{v}}^q}}\text{d}x\text{d}\tau \\ =-\alpha \frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha +1}}+\lambda \frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }}-{\phi }^p\left(x\right){\left(A+t\right)}^{-p\alpha }{\displaystyle {\int }_0^t{\left(A+\tau \right)}^{-q\alpha }}\text{d}\tau {\displaystyle {\int }_{\Omega }{\phi }^q}\left(x\right)\text{d}x\\ =\frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }}\left(-\frac{\alpha }{A+t}+\lambda -\beta {\phi }^{p-1}\left(x\right){\left(A+t\right)}^{-\left(p-1\right)\alpha }{\displaystyle {\int }_0^t{\left(A+\tau \right)}^{-q\alpha }}\text{d}\tau \right)\\ \ge \frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }}\left(-\frac{\alpha }{A+t}+\lambda -\frac{\beta {\overline{\phi }}^{p-1}\left(x\right)}{1-q\alpha }{\left(A+t\right)}^{-\left(p+q-1\right)\alpha +1}\right)\ge 0,\end{array}$

${\overline{v}}_t-\Delta \overline{v}-{\overline{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{u}}^s}}\text{d}x\text{d}\tau \ge \frac{\phi \left(x\right)}{{\left(A+t\right)}^{\alpha }}\left(-\frac{\alpha }{A+t}+\lambda -\frac{\beta {\overline{\phi }}^{r-1}\left(x\right)}{1-s\alpha }{\left(A+t\right)}^{-\left(r+s-1\right)\alpha +1}\right)\ge 0.$

另一方面假设初值$\left(u_0\left(x\right),v_0\left(x\right)\right)$ 足够小且满足

$\begin{array}{l}\max u\left(x,0\right)\le \frac{c\underset{\_}{\phi }\left(x\right)}{A},\\ \max v\left(x,0\right)\le \frac{c\underset{\_}{\phi }\left(x\right)}{A},\\ {\overline{u}|}_{\partial \Omega }=0\ge {u|}_{\partial \Omega },\\ {\overline{v}|}_{\partial \Omega }=0\ge {v|}_{\partial \Omega }.\end{array}$

所以$\left(\underset{\_}{u},\underset{\_}{v}\right)$ 是$\left(u,v\right)$ 的上解并且意味着$\left(u,v\right)$ 整体存在。

定理4：令$\left(u,v\right)$ 是问题(1)的经典解。假设$p+q\le 1,$ 且$r+s\le 0$ ，则$\left(u,v\right)$ 对任意初值$u_0,v_0$ 都是整体存在的。

证明：令$\left(\overline{u},\overline{v}\right)=\left(B{\text{e}}^{Ht},B{\text{e}}^{Ht}\right)$

$\begin{array}{l}{\overline{u}}_t-\Delta \overline{u}-{\overline{u}}^p{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{v}}^q}}\text{d}x\text{d}\tau =HB{\text{e}}^{Ht}-B^{p+q}{\text{e}}^{Hpt}\left|\Omega \right|{\displaystyle {\int }_0^t{\text{e}}^{Hq\tau }}\text{d}\tau \\ =HB{\text{e}}^{Ht}-\frac{B^{p+q}\left|\Omega \right|}{Hq}\left({\text{e}}^{H\left(p+q\right)t}-1\right)\\ \ge HB{\text{e}}^{Ht}-\frac{B^{p+q}\left|\Omega \right|}{Hq}{\text{e}}^{H\left(p+q\right)t},\\ {\overline{v}}_t-\Delta \overline{v}-{\overline{v}}^r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\overline{u}}^s}}\text{d}x\text{d}\tau \ge HB{\text{e}}^{Ht}-\frac{B^{r+s}\left|\Omega \right|}{Hq}{\text{e}}^{H\left(r+s\right)t}.\end{array}$

选择$B\ge \max \left\{\max u_0\left(x\right),\max v_0\left(x\right)\right\}$ 和$H=\max \left\{{\left(\frac{\left|\Omega \right|}{q{B}^{1-p-q}}\right)}^{\frac{1}{2}},{\left(\frac{\left|\Omega \right|}{s{B}^{1-r-s}}\right)}^{\frac{1}{2}}\right\}$ ，则$\left(\overline{u},\overline{v}\right)$ 是$\left(u,v\right)$ 的上解，进一步说明$\left(u,v\right)$ 是整体存在的。

基金项目

山西省研究生创新项目(2023KY262)。

NOTES

^*通讯作者。