我们考虑具有耦合指数反应项的变系数扩散方程组

(1.1) $ \begin{align} u_{t} = {{\rm{div}}} (a_{1}(x)\nabla u(x, t))+b_{1}(x)e^{\alpha_{1}u+\alpha_{2}v}, \quad (x, t)\in\Omega\times(0, t^{*}), \end{align} $

(1.2) $ \begin{align} v_{t} = {{\rm{div}}} (a_{2}(x)\nabla v(x, t))+b_{2}(x)e^{\beta_{1}v+\beta_{2}u}, \quad (x, t)\in\Omega\times(0, t^{*}), \end{align} $

给出齐次Dirichlet边界条件和初始条件

(1.3) $ \begin{align} u(x, t) = v(x, t) = 0, \quad (x, t)\in\partial\Omega\times(0, t^{*}), \end{align} $

(1.4) $ \begin{align} u(x, 0) = u_{0}(x), \quad v(x, 0) = v_{0}(x), \quad x\in\Omega, \end{align} $

其中$ \Omega\subset{\Bbb R} ^{N}(N\geq1) $为具有光滑边界$ \partial\Omega $的有界区域. $ t^{\ast}\leq+\infty , $当$ t^{\ast}<+\infty $时解发生爆破,反之$ t^{\ast} = +\infty $.对$ i = 1, 2 $, $ \alpha_i, \beta_i>0 $为正常数,变扩散系数$ a_i(x) $为正的适当光滑函数,权函数$ b_i(x)\in C(\overline{\Omega}) $满足

(S$ _{1}) $: $ b_i(x)>0, x\in\Omega $且$ b_i(x) = 0, x\in\partial\Omega, $或者

(S$ _{2}) $: $ b_i(x)\geq d_i>0, \quad \forall x\in\bar{\Omega}. $

同时,非负初始值$ u_{0}(x), v_{0}(x) $为$ C^{1} $类连续函数且满足适当的相容性条件.因此,由抛物方程经典理论可知,问题(1.1)–(1.4)存在唯一的非负古典解.

指数反应模型(1.1)–(1.2)出现于许多应用领域.比如,燃烧理论中,模型(1.1)称为固体燃料点火模型,它描述固体燃料在有界的容器内超临界高活化能热爆炸的无量纲点火模型,见文献[1].我们通过简单的变换,模型(1.1)–(1.2)可转换成一个含有梯度项的反应扩散方程组,其常出现在某些物理模型中,比如弹道沉积过程中生长界面像的演化可以用扩散的Hamilton-Jacobi型方程组来描述,见文献[2].同时,模型(1.1)–(1.2)的稳态方程也有许多应用.此时其可描述热自燃问题^[3],由Lord Kelvin^[4]提出的重力平衡中等温气体球问题,由均匀电流对一物体加热产生的温度分布的问题^[5],和关于Euler湍流的Osanger涡模型^[6].它也是微分几何中常感兴趣的问题^[7],以及一些其它的应用.

近几十年来,许多学者致力于反应-扩散方程解的整体存在性和非存在性、爆破现象及解的渐近性态的研究,我们建议读者参考文献[1, 8-11],综述性论文^[12-13]及其中的相关文献.特别是, Bebernes和Eberly^[1], Fila^{[9, Chapter2]}及Quittner和Souplet^{[10, Chapter2; Chapter3]}等的专著中详细介绍了具有Dirichlet边界条件和常系数(即,加权$ a(x) = a>0 $情形)指数增长反应项的反应-扩散方程及方程组中解的定性性质.粗略地讲,爆破的发生以及类型依赖于非线性项,常数系数$ a_i, b_i, \alpha_i $及$ \beta_i $,初始值以及区域的选取.实际上,与具有乘幂递增反应项的反应-扩散模型相比较,指数增长反应模型的研究文献较少.本文中,我们将对一类具有耦合指数反应项的变系数扩散模型解的爆破现象进行研究,主要导出当爆破发生时爆破时间界的估计.据我们所知,此类问题中研究爆破解爆破时间的上界的方法较多,见文献[14],但是对于爆破时间下界很难确定且方法有限.

最近,爆破时间的下界研究方面也有新的起色,其中研究具有乘幂反应的变系数单个方程中解的爆破时间下界的问题较多,参见文献[15-22] (空变系数情形),文献[23-27] (时变系数情形)及相关文献.除了前述的专著[1, 9-10]外,关于具有指数反应项的常数系数单个方程的研究方面, Tello ^[28]讨论了柯西问题解的长时间行为. Pulkkinen^[29]考虑了Dirichlet初边值问题解的爆破与稳定性. Ioku ^[30]讨论了具有另一类平方指数反应项$ e^{u^2} $的热传导方程柯西问题解的整体存在性.之后, Zhang等^[31]在二维空间中利用位势井方法得到了具有形如$ |u|^{p-2}ue^{k|u|^{\alpha}} $ (其中$ p>2, k\geq0, 1<\alpha<2 $)反应项的半线性扩散方程Dirichlet初边值问题整体解的存在性和非存在性以及衰减估计。紧接着, Dai和Zhang^[32]也采用位势井方法得到了具有广义Lewis函数和平方指数反应项的反应-扩散方程Dirichlet初边值问题解的非存在性与衰减估计.最近, Ma和Fang^[33]研究了具有加权指数反应项的单个扩散方程齐次Dirichlet初边值问题$ u_{t} = \Delta u+a(x)e^{\alpha u}, \ (x, t)\in\Omega\times(0, t^{*}). $他们利用上下解的方法、伯努利方程和修正的微分不等式技巧,在若干个适当测度意义下建立了整体解和爆破解的存在性条件,并在高维空间中导出了爆破时间的上、下界估计值.

但是,关于反应方程组中爆破解的爆破时间界的估计的研究甚少.具有乘幂型反应项的半线性方程组的研究方面, Payne和Song^[34]考虑了如下的局部趋化模型

$ \left\{\begin{array}{ll} u_{t} = \Delta u-a_1(uv, _{i}), _{i}, \quad & (x, t)\in\Omega\times(0, t^*), \\ v_{t} = a_{2}\Delta v-a_{3}v+a_{4}u, \quad &(x, t)\in\Omega\times(0, t^*). \end{array}\right. $

他们在二维和三维空间中得到了具有齐次Neumann边界条件的初边值问题爆破解的爆破时间的下界. Xu和Ye^[35]对大初值及某些参数范围内研究了如下弱耦合局部反应-扩散问题

$ \left\{\begin{array}{ll} u_{t} = \Delta u+u^{p}v^{q}, \quad & (x, t)\in\Omega\times(0, t^*), \\ v_{t} = \Delta v+v^{r}v^{s}, \quad & (x, t)\in\Omega\times(0, t^*). \end{array}\right. $

他们在齐次Dirichlet边界条件下得到了爆破解的确切的爆破时间. Payne和Philippin^[36]研究了如下具有时变系数局部源项的半线性抛物方程组

$ \left\{\begin{array}{ll} u_{t} = \Delta u+k_1(t)f_1(v), \quad & (x, t)\in\Omega\times(0, t^*), \\ v_{t} = \Delta v+k_2(t)f_2(u), \quad & (x, t)\in\Omega\times(0, t^*). \end{array}\right. $

在齐次Dirichlet边界条件下,他们得到了初边值问题的解在有限时刻爆破的条件及爆破时间的上界,并在二维和三维空间中得到了爆破时间的下界估计.但是,他们没有考虑高维情形以及时变系数的行为对解的爆破时间下界的影响且与我们的模型(1.1)–(1.4)不同. Tao和Fang^[37]讨论了具有时变系数源项的弱耦合局部反应-扩散方程组

$ \left\{\begin{array}{ll} u_{t} = \Delta u+k_1(t)u^{p}v^{q}, \quad & (x, t)\in\Omega\times(0, t^*), \\ v_{t} = \Delta v+k_2(t)v^{r}v^{s}, \quad & (x, t)\in\Omega\times(0, t^*). \end{array}\right. $

齐次Dirichlet初边值问题解的爆破规则并在高维空间中$ (N\geq3) $给出了两种不同测度意义下的爆破时间的下界估计值.此外,关于具有非局部反应项的扩散方程组的研究方面,我们参考了文献[38-39].

关于具有指数反应项的半线性扩散方程组研究方面, Kanel和Kirane^[40]研究了如下半线性抛物方程组Neumann初边值问题

$ \left\{\begin{array}{ll} u_{t} = a\Delta u-f(u, v), \quad & (x, t)\in\Omega\times(0, t^*), \\ v_{t} = c\Delta u+d\Delta v+f(u, v), \quad& (x, t)\in\Omega\times(0, t^*). \end{array}\right. $

其中$ f(u, v)\leq k\phi(u)e^{\sigma v}, $$ k, \sigma $为正常数且$ \phi(u) $为非负连续局部Lipschitz函数且满足$ \phi(0) = 0 $.在适当的条件下,他们利用热方程的Neumann函数和局部$ L^{p} $先验估计建立了整体解的存在性以及长时间行为.肖^[41]考虑了具有指数源项的反应-扩散方程组

$ \left\{\begin{array}{ll} u_{t} = \Delta u+e^{av}, \quad &(x, t)\in\Omega\times(0, t^*), \\ v_{t} = \Delta v+e^{bu}, \quad &(x, t)\in\Omega\times(0, t^*). \end{array}\right. $

在齐次Dirichlet边界条件下,他利用上下解方法和Kaplan方法,对大初始值得到了解的生命跨度估计值. Ghoul和Zaag^[42]研究了如下具有指数源项的半线性抛物方程组

$ \left\{\begin{array}{ll} u_{t} = \Delta u+e^{pv}, \quad &(x, t)\in\Omega\times(0, t^*), \\ v_{t} = \mu\Delta v+e^{qu}, \quad &(x, t)\in\Omega\times(0, t^*). \end{array}\right. $

他们在齐次Dirichlet边界条件下得到了爆破解的存在性.

综上所述,具有耦合指数反应项的变系数扩散方程组(1.1)–(1.2)齐次Dirichlet初边值问题中解的爆破现象以及爆破时间界的研究还未得到展开.主要难点在于找到变系数$ a_i(x) $, $ b_i(x) $和指数非线性以及空间的维数等对问题爆破现象的影响.由此启发,结合伯努利方程技巧和构造上下解方法以及修正微分不等式技巧,我们建立问题(1.1)–(1.4)整体解和爆破解存在的充分条件,且在整体空间中导出爆破解的爆破时间界的估计值.

本文的剩余部分结构如下:在第2节中,我们建立问题(1.1)–(1.4)整体解的存在性和非存在性的充分条件.在第3节中,我们得到了爆破时间的上界并证明了解在某些特殊情况下发生无限爆破现象.在第4节中,我们导出了在整体空间$ (N\geq1) $中在若干个不同测度意义下爆破时间的下界.

本节中,我们结合伯努利方程技巧和构造上下解方法,得到了大家熟知的问题(1.1)–(1.4)对小的初值解整体存在,对大初值解在有限时刻发生爆破的结果.

我们先引入两个固定边界薄膜振动问题

(2.1) $ \begin{align} {{\rm{div}}}(a_1(x)\nabla\phi (x))+\lambda\phi (x) = 0, \quad x\in\Omega, \quad\phi(x) = 0, \quad x\in\partial\Omega, \end{align} $

(2.2) $ \begin{align} {{\rm{div}}}(a_2(x)\nabla\psi (x))+\mu\psi (x) = 0, \quad x\in\Omega, \quad\psi(x) = 0, \quad x\in\partial\Omega, \end{align} $

其中$ \lambda, \mu $分别为第一特征值,其对应的特征函数为$ \phi(x), \psi(x) $且满足

$ {\rm(H_{1}):} $$ \max\limits_{x\in\Omega}\phi(x) = \min\{\frac{\lambda}{4e\alpha_{1}l_{1}}, \frac{\lambda}{4e\beta_{1}l_{1}}\}, \quad\max\limits_{x\in\Omega}\psi(x) = \min\{\frac{\mu}{4e\alpha_{2}l_{2}}, \frac{\mu}{4e\beta_{1}l_{2}}\}, $

其中$ l_1 $和$ l_2 $为正常数.

由$ b_1(x), b_2(x), \phi(x) $和$ \psi(x) $的连续性,我们可以选择$ \alpha_1, \beta_1, l_1 $和$ l_2 $使得

$ {\rm(H_{2}):} $$ \phi(x)\geq\frac{b_1(x)}{l_1}, \quad\psi\geq\frac{b_2(x)}{l_2}, \quad x\in\Omega, $

或者

$ {\rm(H_{3}):} $$ \phi(x)\geq(\frac{\alpha_1^{p_1}}{p_1})^{\frac{1}{1-p_1}}(b_1(x))^{\frac{1}{1-p_1}}, \quad \psi(x)\geq(\frac{\beta_1^{p_2}}{p_2})^{\frac{1}{1-p_2}}(b_2(x))^{\frac{1}{1-p_1}}, \quad x\in\Omega, $

其中$ b_i(x) $满足$ {\rm(S_{2})} $, $ p_i>1 $为整数, $ i = 1, 2 $.

则我们可以得到如下主要结论.

定理2.1  假设变扩散系数$ a_i(x)\in C(\bar{\Omega}) $,权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{1})}, $对$ i = 1, 2 $.若对满足$ u_{0}\leq\frac{2l_1e\phi(x)}{\lambda} $和$ v_{0}\leq\frac{2l_2e\psi(x)}{\mu} $的所有非负初始值,特征函数$ \phi(x), \psi(x) $满足$ {\rm(H_{1})} $和$ {\rm(H_{2})}, $则问题(1.1)–(1.4)的非负古典解$ (u, v) $整体存在.相反,若对满足$ u_0(x)\geq q\phi(x) $和$ v_0(x)\geq q\psi(x) $的非负初始值,特征函数$ \phi(x), \psi(x) $满足$ {\rm(H_{3})}, $则问题(1.1)–(1.4)的非负古典解$ (u, v) $在有限时刻$ t^{*} $爆破,其中$ q>\max\{\lambda^{\frac{1}{p_1-1}}, \mu^{\frac{1}{p_2-1}}\} $为正常数且$ p_1, p_2>1 $为整数.

证  我们先证明整体解的存在性.令$ \overline{u}(x, t) = \phi(x)f_1(t), $$ \overline{v}(x, t) = \psi(x)f_2(t), $其中$ f_1(t) $和$ f_2(t) $分别是如下两个常微分方程的解

(2.3) $ \begin{align} \left\{\begin{array}{ll} f'_1(t) = -\lambda f_1(t)+l_1e, \quad t>0, \\ f_1(0) = \frac{2l_1e}{\lambda}, \end{array}\right. \end{align} $

及

(2.4) $ \begin{align} \left\{\begin{array}{ll} f'_2(t) = -\mu f_2(t)+l_2e, \quad t>0, \\ f_2(0) = \frac{2l_2e}{\mu}. \end{array}\right. \end{align} $

则直接计算可得

$ f_1(t) = \frac{l_1e^{1-\lambda t}+l_1e}{\lambda}\leq\frac{2l_1e}{\lambda}, \quad f_2(t) = \frac{l_2e^{1-\mu t}+l_2e}{\mu}\leq\frac{2l_2e}{\mu}. $

利用常微分方程(2.3)及条件$ {\rm(H_{1})} $和$ {\rm(H_{2})}, $我们可导出

$ \begin{eqnarray*} \overline{u}_t-{{\rm{div}}} (a_1(x)\nabla\overline{u})-b_1(x)e^{\alpha_1\overline{u}+\alpha_2\overline{v}} & = & \phi(x)f'_1(t)+ \lambda\phi(x)f_1(t)-b_1(x)e^{\alpha_1\phi(x)f_1(t)+\alpha_2\psi(x)f_2(t)}\\ & \geq&\phi(x)[f'_1(t)+\lambda f_1(t)-l_1e^{\frac{2\alpha_1l_1e}{\lambda}\phi(x)+\frac{2\alpha_2l_2e}{\mu}\psi(x)}]\\ & \geq&\phi[f'_1(t)+\lambda f_1(t)-l_1e] = 0. \end{eqnarray*} $

类似地,我们有

$ \overline{v}_t-{{\rm{div}}} (a_2(x)\nabla\overline{v})-b_2(x)e^{\beta_1\overline{v}+\beta_2\overline{u}}\geq0. $

同时,我们易知$ (\overline{u}, \overline{v}) $满足如下初边值条件

$ \overline{u}(x, t) = 0, \quad \overline{v}(x, t) = 0, \quad x\in\partial\Omega, \quad t>0, $

$ \overline{u}(x, 0) = \frac{2l_1e}{\lambda}\phi(x)\geq u_0(x), \quad \overline{v}(x, 0) = \frac{2l_2e}{\mu}\psi(x)\geq v_0(x), \quad x\in\Omega. $

于是,由比较原理易知$ (\overline{u}, \overline{v}) $是问题(1.1)–(1.4)的一个上解且由$ (\overline{u}, \overline{v}) $整体存在性导出$ (u, v) $整体存在.

紧接着,我们证明解的非整体存在性.令$ \underline{u}(x, t) = \phi (x)\omega_1(t) $, $ \underline{v}(x, t) = \psi(x)\omega_2(t), $其中$ \omega_1(t) $和$ \omega_2(t) $分别为如下两个伯努利方程初值问题的解

(2.5) $ \begin{align} \left\{\begin{array}{ll} \omega'_1(t) = -\lambda\omega_1(t)+(\omega_1(t))^{p_1}, \quad t>0, \\ \omega_1(0) = q, \end{array}\right. \end{align} $

及

(2.6) $ \begin{align} \left\{\begin{array}{ll} \omega'_2(t) = -\mu\omega_2(t)+(\omega_2(t))^{p_2}, \quad t>0, \\ \omega_2(0) = q. \end{array}\right. \end{align} $

则我们直接计算可得问题(2.5)和问题(2.6)的确切解

$ \omega_1(t) = \bigg[q^{1-p_1}e^{(p_1-1)\lambda t}+\frac{1}{\lambda}-\frac{1}{\lambda}e^{(p_1-1)\lambda t}\bigg]^{\frac{1}{1-p_1}}, $

$ \omega_2(t) = \bigg[q^{1-p_2}e^{(p_2-1)\mu t}+\frac{1}{\mu}-\frac{1}{\mu}e^{(p_2-1)\mu t}\bigg]^{\frac{1}{1-p_2}}, $

其中$ p_1, p_2>1 $为整数且$ q>\max\{\lambda^{\frac{1}{p_1-1}}, \mu^{\frac{1}{p_2-1}}\} $为正常数.显然, $ \omega_1(t) $和$ \omega_2(t) $分别在有限时刻$ T_1 $和$ T_2 $爆破,其中

$ T_1 = \frac{1}{(p_1-1)\lambda}\ln(\frac{1}{1-\lambda q^{1-p_1}}), $

$ T_2 = \frac{1}{(p_2-1)\mu}\ln(\frac{1}{1-\mu q^{1-p_2}}). $

现在,我们验证$ (\underline{u}, \underline{v}) $是$ (u, v) $的一个下解.

利用常微分方程(2.5)及条件$ {\rm(H_{3})}, $我们可导出

$ \begin{eqnarray*} \underline{u}_t-{{\rm{div}}} (a_1(x)\nabla\underline{u})-b_1(x)e^{\alpha_1\underline{u}+\alpha_2\underline{v}} & = & \phi(x)(\omega_1(t))^{p_1}-b_1(x)e^{\alpha_1\phi(x)\omega_1(t)+\alpha_2\psi(x)\omega_2(t)}\\ & \leq&\phi(x)(\omega_1(t))^{p_1}-b_1(x)e^{\alpha_1\phi(x)\omega_1(t)}\\ & = &\phi(x)(\omega_1(t))^{p_1}-b_1(x)\sum^{\infty}_{k = 0}\frac{(\alpha_1\phi(x))^k(\omega_1(t))^k}{k!}\\ &\leq& 0. \end{eqnarray*} $

类似地,我们有

$ \underline{v}_t-{{\rm{div}}} (a_2(x)\nabla\underline{v})-b_2(x)e^{\beta_1\underline{v}+\beta_2\underline{u}}\leq0. $

同时,我们易知$ (\underline{u}, \underline{v}) $满足如下初边值条件

$ \underline{u}(x, t) = 0, \quad \underline{v}(x, t) = 0, \quad x\in\partial\Omega, \quad t>0, $

$ \underline{u}(x, 0) = q\phi(x)\leq u_0(x), \quad \underline{v}(x, 0) = q\psi(x)\leq v_0(x), \quad x\in\Omega. $

由比较原理,我们知$ (\underline{u}, \underline{v}) $是问题(1.1)–(1.4)的一个下解且$ (u, v) $在有限时刻$ t^* $发生爆破,其爆破时间$ t^* $的上界为$ t^*\leq\min\{T_1, T_2\}. $

定理2.1证毕.

本节中,结合伯努利方程技巧和泰勒展开式,我们建立了问题(1.1)–(1.4)解在有限或无限时刻发生爆破的充分条件,并在有限时刻发生爆破的情形中估计了$ t^* $的上界.

首先,我们给出Kaplan测度意义下的有限时刻爆破.

定理3.1  假设$ (u, v) $为问题(1.1)–(1.4)的非负古典解且变扩散系数$ a_i(x)\in C^0(\bar{\Omega})\bigcap $$ C^2(\Omega), $权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{2})}, $对$ i = 1, 2. $定义辅助函数

$ \theta(t): = \int_{\Omega}u(x, t)\phi(x)+v(x, t)\psi(x){\rm d}x, $

其中$ \phi(x), \psi(x) $为固定边界薄膜振动问题$ (2.1) $和$ (2.2) $中给出的特征函数且对$ k>1 $的正整数满足

$ \int_{\Omega}(b_1(x))^{\frac{-1}{k-1}}\phi(x) = 1, \quad\int_{\Omega}(b_2(x))^{\frac{-1}{k-1}}\psi(x) = 1. $

若$ (u_0, v_0) $足够大且使得

$ \theta(0): = \int_{\Omega}u_0(x, t)\phi(x)+v_0(x, t)\psi(x){\rm d}x\geq\max\bigg\{c', (m_1+1)(m_1+2)(\frac{m_1!\delta_1}{\delta_2^{m_1}})^{\frac{1}{m_1-1}}\bigg\}, $

其中$ c' $为最小的非负常数使得$ e^{\delta_{2}t>\delta_{1}t}, $$ \forall t>c' $且$ m_1>1 $为整数

$ \delta_1 = \max\{\lambda, \mu\}, \quad\delta_2 = \min\{\alpha_1, \beta_1\}, $

则问题(1.1)–(1.4)的非负古典解$ (u, v) $在测度$ \theta(t) $的意义下有限时刻发生爆破且爆破时间$ t^* $的上界为

$ t^*\leq T_3 = -\frac{1}{(m_1-1)\delta_1}\ln\bigg\{1-\frac{[(m_1+1)(m_1+2)]^{m_1-1}m_1!\delta_1}{\delta_1^{m_1}(\theta(0))^{m_1-1}} \bigg\}. $

证  对$ \theta(t) $直接求导并利用Green公式及泰勒展开式,我们有

(3.1) $ \begin{eqnarray} \theta'(t)& = &\int_{\Omega}({{\rm{div}}}(a_1(x)\nabla u)+b_1(x)e^{\alpha_1u+\alpha_2v})\phi(x){\rm d}x+\int_{\Omega}({{\rm{div}}}(a_2(x)\nabla v)+b_2(x)e^{\beta_1v+\beta_2u})\psi(x){\rm d}x\\ &\geq&-\lambda\int_{\Omega}u\phi(x){\rm d}x-\int_{\Omega}v\psi(x){\rm d}x+\int_{\Omega}e^{\alpha_1u}b_1(x)\phi(x){\rm d}x+\int_{\Omega}e^{\beta_1v}b_2(x)\psi(x){\rm d}x \\ &\geq&-\delta_1\theta(t)+\sum^{\infty}_{k = 1}\frac{1}{k!}\int_{\Omega}(\alpha_1u)^kb_1(x)\phi(x){\rm d}x+\sum^{\infty}_{k = 1}\frac{1}{k!}\int_{\Omega}(\beta_1v)^kb_2(x)\psi(x){\rm d}x+2. \end{eqnarray} $

对(3.1)式中第二项运用Hölder不等式及条件$ \int_{\Omega}(b_1(x))^{\frac{-1}{k-1}}\phi(x) = 1, $我们可得

(3.2) $ \begin{equation} \int_{\Omega}u^kb_1(x)\phi(x){\rm d}x\geq\bigg(\int_{\Omega}u(x, t)\phi(x){\rm d}x\bigg)^k. \end{equation} $

类似地,我们有

(3.3) $ \begin{equation} \int_{\Omega}v^kb_2(x)\psi(x){\rm d}x\geq\bigg(\int_{\Omega}v(x, t)\psi(x){\rm d}x\bigg)^k. \end{equation} $

将(3.2)和(3.3)式代入到(3.1)式,我们导出

(3.4) $ \begin{eqnarray} \theta'(t)&\geq&-\delta_1\theta(t)+\sum^{\infty}_{k = 1}\frac{\alpha_1^k}{k!} \bigg(\int_{\Omega}u(x, t)\phi(x){\rm d}x\bigg)^k +\sum^{\infty}_{k = 1}\frac{\beta_1^k}{k!}\bigg(\int_{\Omega}v(x, t)\psi(x){\rm d}x\bigg)^k+2\\ &\geq&-\delta_1\theta(t)+e^{\delta_2\theta(t)}. \end{eqnarray} $

我们先寻找函数使得

(3.5) $ \begin{equation} \theta(t)\geq\Gamma(t), \quad 0\leq t<t^*. \end{equation} $

为了此目的,设$ \Gamma(t) $为如下常微分方程初值问题的解

(3.6) $ \begin{align} \left\{\begin{array}{ll} \Gamma'(t) = -\delta_1\Gamma(t)+e^{\delta_2\Gamma(t)}, \quad 0\leq t<t_1^*, \\ \Gamma(0) = \theta(0). \end{array}\right. \end{align} $

现在,我们证明解$ \Gamma(t) $满足(3.5)式.定义实值函数$ g $为$ g(t): = -\delta_{1}t+e^{\delta_{2}t}. $利用(3.4)式, $ c' $的条件及初始值的假设,我们有

$ \theta(t)\geq\theta(0)>c'. $

因此,对(3.4)式从$ 0 $到$ t $积分,我们可导出不等式$ \int_{\Omega}\frac{\frac{\rm d}{{\rm d}\tau}\theta(t)}{g(\theta(\tau))}{\rm d}\tau\geq t. $

同时, $ \frac{1}{g} $的不定积分记为$ G, $我们得到

(3.7) $ \begin{align} G(\theta(t))-G(\theta(0))\geq t. \end{align} $

类似地,我们有

(3.8) $ \begin{align} G(\Gamma(t))-G(\Gamma(0)) = t. \end{align} $

结合(3.6)–(3.8)式,我们导出

(3.9) $ \begin{align} G(\theta(t))\geq G(\theta(0))+t = G(\Gamma(0))+t = G(\Gamma(t)). \end{align} $

注意到$ G $在$ (c', \infty) $上单调递增,于是我们可知(3.5)式成立.

其次,对于固定的正整数$ k, $我们考虑实值函数$ h:[0, \infty)\rightarrow{\Bbb R} $且定义为

$ h(x): = -\delta_{1}x+\sum\limits^{k}_{m = 0}\frac{1}{m!}(\delta_{2}x)^{m}. $

令$ P_k(t) $为如下伯努利型方程初值问题的解

(3.10) $ \begin{align} \left\{\begin{array}{ll} P'_k(t) = h(P_k(t)), \quad 0\leq t<t_2^*, \\ P_k(0) = (1-\frac{1}{k+2})\Gamma(0), \end{array}\right. \end{align} $

其中$ k $为任意正整数.

利用(3.5)式类似的证明方法,我们有

$ \Gamma(t)\geq P_k(t), \quad 0\leq t<t_1^*. $

再令$ \rho_m(t) $为如下伯努利方程初值问题的解

(3.11) $ \begin{align} \left\{\begin{array}{ll} \rho'_m(t) = -\delta_1\rho_m(t)+\frac{1}{m!}(\delta_2\rho_m(t))^{m}, \quad 0\leq t\leq t_3^*, \\ \rho_m(0) = (\frac{1}{m+1}-\frac{1}{m+2})\Gamma(0). \end{array}\right. \end{align} $

比较问题(3.10)和(3.11),我们易验证

$ P_k(t)\geq\sum\limits_{m = 0}^{k}\rho_m(t), \quad 0\leq t<t_2^*. $

直接计算问题(3.11)的解,并对$ m = 0 $到$ k $求和,我们有

(3.12) $ \begin{align} \sum\limits_{m = 0}^{k}\rho_m(t) = \sum\limits_{m = 0}^{k} \bigg[\frac{\delta_{1}m!}{\delta_{1}m!-\delta_{2}^{m}(\rho_m(0))^{m-1}(1-e^{-(m-1)\delta_{1}t})} \bigg]^{\frac{1}{m-1}} e^{-\delta_{1}t}\rho_m(0). \end{align} $

利用定理3.1中$ \theta(0) $的条件,我们可导出$ \sum\limits_{m = 0}^{k}\rho_m(t) $的爆破时间为

$ T_3 = -\frac{1}{(m_1-1)\delta_1}\ln \bigg\{1-\frac{[(m_1+1)(m_1+2)]^{m_1-1}m_1!\delta_1}{\delta_1^{m_1}(\theta(0))^{m_1-1}} \bigg\}. $

同时,由$ \theta(t)\geq\Gamma(t)\geq P_k(t)\geq\sum\limits_{m = 0}^{k}\rho_m(t) $易知, $ \theta(t) $的爆破时间的上界为

$ t^*\leq T_3 = -\frac{1}{(m_1-1)\delta_1}\ln \bigg\{1-\frac{[(m_1+1)(m_1+2)]^{m_1-1}m_1!\delta_1}{\delta_1^{m_1}(\theta(0))^{m_1-1}} \bigg\}. $

定理3.1证毕.

特别地,在某些特殊情况下问题(1.1)–(1.4)解对任意非负初始值发生无限爆破现象.

定理3.2  假设$ (u, v) $为问题(1.1)–(1.4)的非负古典解且变扩散系数$ a_i(x)\in C^0(\bar{\Omega})\bigcap $$ C^2(\Omega), $权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{2})}, $对$ i = 1, 2. $若$ \min b_i(x) = \gamma_i, $$ \gamma = \min\{\gamma_1, \gamma_2\}\geq\frac{\delta_1}{\delta_2}, $其中$ \delta_i $如定理$ 3.1 $中定义, $ i = 1, 2, $则问题(1.1)–(1.4)的非负古典解$ (u, v) $在测度$ \theta(t) $的意义下发生无限爆破,即, $ t^* = +\infty. $

证  利用类似于定理3.1中方法,我们得到

(3.13) $ \begin{align} \theta'(t)\geq-\delta_1\theta(t)+\int_{\Omega}e^{\alpha_{1}u}b_1(x)\phi(x){\rm d}x+\int_{\Omega}e^{\beta_{1}v}b_2(x)\psi(x){\rm d}x. \end{align} $

因为$ \alpha_{1}u, \beta_{1}v\geq0, $我们有

(3.14) $ \begin{align} e^{\alpha_{1}u}\geq\alpha_{1}u, \quad e^{\beta_{1}v}\geq\beta_{1}v. \end{align} $

将(3.14)式代入到(3.13)式,并利用定理3.2中的条件,我们可导出

(3.15) $ \begin{align} \theta'(t)\geq(\gamma\delta_2-\delta_1)\theta(t). \end{align} $

对(3.15)式从$ 0 $到$ t $积分,我们有

$ \theta(t)\geq\theta(0)e^{(\gamma\delta_2-\delta_1)t}. $

于是,问题(1.1)–(1.4)的非负古典解$ (u, v) $在测度$ \theta(t) $的意义下发生无限爆破.

定理3.2证毕.

我们利用修正微分不等式技巧,在整体空间上若干个不同测度意义下给出了爆破时间$ t^* $下界.注意到,如果我们在球域或非对称凸区域上考虑抛物问题(1.1)–(1.4),则在某些初值条件下一大类古典解可能发生单点爆破^[1].为了避免此种情形,在下文中我们总假设爆破集的测度为正.

由于$ N\geq3 $的情形中爆破时间$ t^* $下界估计时用到了高维空间中的Sobolev型不等式.于是,本小节中,我们假设$ \Omega\subset{\Bbb R} ^{N}(N\geq3) $为具有光滑边界$ \partial\Omega $的有界凸区域.

定理4.1  假设$ (u, v) $为问题(1.1)–(1.4)的非负古典解且对$ i = 1, 2 $,变扩散系数$ a_i(x)\in C^0(\bar{\Omega})\bigcap C^2(\Omega) $满足

(A$ _{1}): $$ a_i(x)\geq c_i>0, \ \forall x\in\bar\Omega, $和

(A$ _{2}): $$ |\nabla a_i^{\frac{1}{2}}(x)|^{2}\leq M_i, \ \forall x\in\bar{\Omega}, $其中$ M_i\geq0 $为常数,权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{2})}, $

定义辅助函数

$ \varphi(t): = \int_{\Omega}e^{ku}+e^{kv}{\rm d}x, $

其中$ k>\max\{N(\alpha_1+\alpha_2)+\alpha_2, N(\beta_1+\beta_2)+\beta_2\}. $

如果$ (u, v) $在测度$ \varphi(t) $的意义下有限时刻$ t^* $发生爆破,则爆破时间$ t^* $的下界为

$ t^*\geq\int^{\infty}_{\varphi(0)}\frac{{\rm d}\eta}{K_1+K_2\eta+K_3\eta^{\frac{N+1}{N}}+K_4\eta^{\frac{N+2}{N}}}, $

其中$ \varphi(0) = \int_{\Omega}e^{ku_{0}}+e^{kv_0}{\rm d}x, $$ K_1, K_2, K_3 $及$ K_4 $由$ (4.10) $式给出.

证  对直接求导并利用(1.1)–(1.3)及Green公式,我们有

(4.1) $ \begin{eqnarray} \varphi'(t)& = &k\int_{\Omega}u_{t}e^{ku}{\rm d}x+k\int_{\Omega}v_{t}e^{kv}{\rm d}x\\ & = &k\int_{\Omega}({{\rm{div}}}(a_1(x)\nabla u)+b_1(x)e^{\alpha_{1}u+\alpha_{2}v})e^{ku}{\rm d}x\\ &&+k\int_{\Omega}({{\rm{div}}}(a_2(x)\nabla v)+b_2(x)e^{\beta_{1}v+\beta_{2}u})e^{kv}{\rm d}x \\ & = &k\int_{\partial\Omega}a_1(x)\frac{\partial u}{\partial\nu}{\rm d}s-k^2\int_{\Omega}a_1(x)e^{ku}|\nabla u|^2{\rm d}x+k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x \\ && +k\int_{\partial\Omega}a_2(x)\frac{\partial v}{\partial\nu}{\rm d}s-k^2\int_{\Omega}a_2(x)e^{kv}|\nabla v|^2{\rm d}x+k\int_{\Omega}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq&-4\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}u}|^2{\rm d}x+k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x \\ &&-4\int_{\Omega}a_2(x)|\nabla e^{\frac{k}{2}v}|^2{\rm d}x+k\int_{\Omega}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x. \end{eqnarray} $

对(4.1)式中第二项运用Hölder不等式及Young不等式,我们可得

(4.2) $ \begin{eqnarray} \int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_2v}&\leq& \bigg(\int_{\Omega}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x\bigg)^{\frac{k-\alpha_2}{k}} \bigg(\int_{\Omega}e^{kv}{\rm d}x\bigg)^{\frac{\alpha_2}{k}}\\ &\leq&\frac{k-\alpha_2}{k}\int_{\Omega}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x +\frac{\alpha_2}{k}\int_{\Omega}e^{kv}{\rm d}x. \end{eqnarray} $

由$ k>\max\{N(\alpha_1+\alpha_2)+\alpha_2, N(\beta_1+\beta_2)+\beta_2\} $易知$ \frac{(k+\alpha_1)N}{(k-\alpha_2)(N+1)}, \frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{(k-\alpha_2)(N+1)}\in(0, 1). $

因此,对(4.2)式中第一项运用Hölder不等式及Young不等式,我们有

(4.3) $ \begin{eqnarray} &&\int_{\Omega}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x\\ &\leq&\bigg(\int_{\Omega}e^{\frac{k(N+1)}{N}}{\rm d}x\bigg)^{\frac{(k+\alpha_1)N}{(k-\alpha_2)(N+1)}} \bigg(\int_{\Omega}(b_1(x))^{\frac{k(N+1)}{k-\alpha_2-(\alpha_1+\alpha_2)N}}{\rm d}x\bigg)^{\frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{(k-\alpha_2)(N+1)}} \\ & \leq&\frac{(k+\alpha_1)N}{(k-\alpha_2)(N+1)}\int_{\Omega}e^{\frac{k(N+1)}{N}}{\rm d}x +\frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{(k-\alpha_2)(N+1)}\int_{\Omega}(b_1(x))^{\frac{k(N+1)}{k-\alpha_2-(\alpha_1+\alpha_2)N}}{\rm d}x.\qquad \end{eqnarray} $

对(4.3)式中第一项运用Hölder不等式并由条件$ {\rm(A_{1})}, $我们可得

(4.4) $ \begin{eqnarray} \int_{\Omega}e^{\frac{k(N+1)}{N}}{\rm d}x&\leq&\bigg(\int_{\Omega}a_1(x)^{\frac{N}{N-2}}e^{\frac{kN}{N-2}u}{\rm d}x\bigg)^{\frac{N-2}{2N}}\bigg(\int_{\Omega}a_1(x)^{-\frac{N}{N+2}}e^{ku}{\rm d}x\bigg)^{\frac{N+2}{N}} \\ & \leq&\frac{1}{\sqrt{c_1}}\bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{N+2}{N}}\bigg(\int_{\Omega}a_1(x)^{\frac{N}{N-2}}e^{\frac{kN}{N-2}u}{\rm d}x\bigg)^{\frac{N-2}{2N}}. \end{eqnarray} $

现在,对(4.4)式的最后一项,利用高维空间$ \Omega\in{\Bbb R} ^{N}(N\geq3) $中的加权Sobolev嵌入不等式

$ \|a_1(x)^{\frac{1}{2}}e^{\frac{k}{2}u}\|_{L^{\frac{2N}{N-2}}(\Omega)}\leq C_s\|a_1(x)^{\frac{1}{2}}e^{\frac{k}{2}u}\|_{W^{1, 2}(\Omega)}, $

其中$ C_{s} $为Sobolev最优常数,并由条件$ {\rm(A_{2})}, $我们有

(4.5) $ \begin{eqnarray} \bigg(\int_{\Omega}a_1(x)^{\frac{N}{N-2}}e^{\frac{kN}{N-2}u}{\rm d}x\bigg)^{\frac{N-2}{2N}}&\leq& C_{s} \bigg[\bigg(\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}}u|^{2}{\rm d}x\bigg)^{\frac{1}{2}}+ \bigg (\int_{\Omega}|\nabla a_{1}^{\frac{1}{2}}(x)|^{2}e^{ku}{\rm d}x\bigg)^{\frac{1}{2}}\bigg] \\ & \leq& C_{s}\bigg[\bigg(\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}}u|^{2}{\rm d}x\bigg)^{\frac{1}{2}}+ \sqrt{M_1}\bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{1}{2}}\bigg]. \end{eqnarray} $

结合(4.2)–(4.5)式并利用带$ \varepsilon $的Young不等式,我们可导出

(4.6) $ \begin{eqnarray} &&k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &\leq&\frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{N+1}\int_{\Omega}(b_1(x))^{\frac{k(N+1)}{k-\alpha_2-(\alpha_1+\alpha_2)N}}{\rm d}x +\alpha_{2}\int_{\Omega}e^{kv}{\rm d}x \\ &&+\frac{(k+\alpha_1)N\sqrt{M_1}}{\sqrt{c_1}(N+1)}C_{s} \bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{N+1}{N}} +\frac{(k+\alpha_1)N}{2\varepsilon_{1}(N+1)} \bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{N+2}{N}} \\ &&+\frac{(k+\alpha_1)N\varepsilon_{1}C_{s}^{2}}{2c_{1}(N+1)}\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}u}|^{2}{\rm d}x , \end{eqnarray} $

其中$ \varepsilon_{1} $是待定的正常数.类似地,我们有

(4.7) $ \begin{eqnarray} &&k\int_{\Omega}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq&\frac{k-\beta_2-(\beta_1+\beta_2)N}{N+1}\int_{\Omega}(b_2(x))^{\frac{k(N+1)}{k-\beta_2-(\beta_1+\beta_2)N}}{\rm d}x +\beta_{2}\int_{\Omega}e^{ku}{\rm d}x \\ &&+\frac{(k+\beta_1)N\sqrt{M_2}}{\sqrt{c_2}(N+1)}C_{s} \bigg(\int_{\Omega}e^{kv}{\rm d}x\bigg)^{\frac{N+1}{N}} +\frac{(k+\beta_1)N}{2\varepsilon_{2}(N+1)}\bigg(\int_{\Omega}e^{kv}{\rm d}x\bigg)^{\frac{N+2}{N}} \\ &&+\frac{(k+\beta_1)N\varepsilon_{2}C_{s}^{2}}{2c_{2}(N+1)}\int_{\Omega}a_2(x)|\nabla e^{\frac{k}{2}v}|^{2}{\rm d}x , \end{eqnarray} $

其中$ \varepsilon_{2} $是待定的正常数.

将(4.6)和(4.7)式代入(4.1)式整理可得

(4.8) $ \begin{eqnarray} \varphi'(t)&\leq& \bigg(\frac{(k+\alpha_1)N\varepsilon_{1}C_{s}^{2}}{2c_{1}(N+1)}-4 \bigg)\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}u}|^{2}{\rm d}x \\ &&+\bigg(\frac{(k+\beta_1)N\varepsilon_{2}C_{s}^{2}}{2c_{2}(N+1)}-4 \bigg)\int_{\Omega}a_2(x)|\nabla e^{\frac{k}{2}v}|^{2}{\rm d}x \\ &&+\frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{N+1}\int_{\Omega}(b_1(x))^{\frac{k(N+1)}{k-\alpha_2-(\alpha_1+\alpha_2)N}}{\rm d}x \\ && +\frac{(k+\alpha_1)N\sqrt{M_1}}{\sqrt{c_1}(N+1)}C_{s} \bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{N+1}{N}} \\ &&+\frac{k-\beta_2-(\beta_1+\beta_2)N}{N+1}\int_{\Omega}(b_2(x))^{\frac{k(N+1)}{k-\beta_2-(\beta_1+\beta_2)N}}{\rm d}x \\ && +\frac{(k+\beta_1)N\sqrt{M_2}}{\sqrt{c_2}(N+1)}C_{s} \bigg(\int_{\Omega}e^{kv}{\rm d}x\bigg)^{\frac{N+1}{N}} +\frac{(k+\alpha_1)N}{2\varepsilon_{1}(N+1)} \bigg(\int_{\Omega}e^{ku}{\rm d}x\bigg)^{\frac{N+2}{N}} \\ &&+\beta_{2}\int_{\Omega}e^{ku}{\rm d}x+\frac{(k+\beta_1)N}{2\varepsilon_{2}(N+1)}\bigg(\int_{\Omega}e^{kv}{\rm d}x\bigg)^{\frac{N+2}{N}} +\alpha_{2}\int_{\Omega}e^{kv}{\rm d}x. \end{eqnarray} $

取$ \varepsilon_{1} = \frac{8c_1(N+1)}{(K+\alpha_1)NC_{s}^{2}}, \varepsilon_{2} = \frac{8c_2(N+1)}{(K+\beta_1)NC_{s}^{2}} $并由(4.8)式,我们可导出

(4.9) $ \begin{equation} \varphi'(t)\leq K_1+K_2\varphi(t)+K_3(\varphi(t))^{\frac{N+1}{N}}+K_4(\varphi(t))^{\frac{N+2}{N}}, \end{equation} $

其中

(4.10) $ \begin{eqnarray} &&K_1: = \frac{k-\alpha_2-(\alpha_1+\alpha_2)N}{N+1}\int_{\Omega}(b_1(x))^{\frac{k(N+1)}{k-\alpha_2-(\alpha_1+\alpha_2)N}}{\rm d}x \\ &&\qquad\quad +\frac{k-\beta_2-(\beta_1+\beta_2)N}{N+1}\int_{\Omega}(b_2(x))^{\frac{k(N+1)}{k-\beta_2-(\beta_1+\beta_2)N}}{\rm d}x, \\ && K_2: = \max\{\alpha_{2}, \beta_{2}\}, \\ && K_3: = \frac{NC_{s}}{N+1} \bigg((k+\alpha_{1})\sqrt{\frac{M_1}{c_1}}+(k+\beta_{1})\sqrt{\frac{M_2}{c_2}}\bigg), \\ && K_4: = \frac{(k+\alpha_{1})^2N^2C^2_s}{16c_1(N+1)^2}+\frac{(k+\beta_{1})^2N^2C^2_s}{16c_2(N+1)^2}. \end{eqnarray} $

如果$ \lim\limits_{t\to t^{*}}\varphi(t) = \infty, $则对(4.9)式从$ 0 $到$ t $积分,我们有

$ t^*\geq\int^{\infty}_{\varphi(0)}\frac{{\rm d}\eta}{K_1+K_2\eta+K_3\eta^{\frac{N+1}{N}}+K_4\eta^{\frac{N+2}{N}}}. $

定理4.1证毕.

其次,我们对权函数加适当条件,可得到问题(1.1)–(1.4)爆破解在加权测度意义下爆破时间下界的估计值.

定理4.2  假设$ (u, v) $为问题(1.1)–(1.4)的非负古典解且对$ i = 1, 2, $变扩散系数满足$ a_i(x)\in C^{0}(\bar\Omega)\cap C^{1}(\Omega), $权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{2})}, $以及如下条件$ {\rm(S_{3})}, $

$ {\rm(S_{3}):} $$ -b_{i}(x)B_{i}\leq\nabla b_{i}(x)\leq b_{i}(x)B_{i}\Leftrightarrow|\frac{\partial b_{i}(x)}{\partial x_{ij}}|\leq B_{ij}b_{i}(x), x\in\Omega, $

其中$ B_{i} = (B_{i1}, B_{i2}, \cdots , B_{iN}) $为正常数向量, $ j = 1, 2, \cdots, N. $定义辅助函数

$ \Phi(t): = \int_{\Omega}b_1(x)e^{ku}{\rm d}x+\int_{\Omega}b_2(x)e^{kv}{\rm d}x, $

其中$ k>\max\{N(\alpha_1+\alpha_2)+\alpha_2, N(\beta_1+\beta_2)+\beta_2\}. $

如果$ (u, v) $在测度$ \Phi(t) $的意义下有限时刻$ t^* $发生爆破,则爆破时间$ t^* $的下界为

$ t^*\geq\int^{\infty}_{\Phi(0)}\frac{{\rm d}\eta}{K_5+K_6\eta^{r_1}+K_7\eta^{r_2}}, $

其中$ \Phi(0) = \int_{\Omega}b_1(x)e^{ku_{0}}+b_2(x)e^{kv_0}{\rm d}x, $$ K_5, K_6, K_7, r_1 $及$ r_2 $由$ (4.23) $式给出.

证  对$ \Phi(t) $直接求导并利用(1.1)–(1.3), Green公式及条件(A$ _{1}) $和(S$ _{3}), $我们有

(4.11) $ \begin{eqnarray} \Phi'(t)& = &k\int_{\Omega}b_1(x)e^{ku}({{\rm{div}}}(a_1(x)\nabla u)+b_1(x)e^{\alpha_{1}u+\alpha{2}v}){\rm d}x\\ &&+k\int_{\Omega}b_2(x)e^{kv}{\rm d}x({{\rm{div}}} (a_2(x)\nabla v)+b_2(x)e^{\beta_{1}v+\beta_{2}u}) \\ & = &k\int_{\partial\Omega}a_1(x)b_1(x)\frac{\partial u}{\partial\nu}{\rm d}s-k\int_{\Omega}a_1(x)\nabla u\cdot\nabla(b_1(x)e^{ku}){\rm d}x\\ &&+k\int_{\Omega}(b_1(x))^2e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x+k\int_{\partial\Omega}a_2(x)b_2(x)\frac{\partial v}{\partial\nu}{\rm d}s\\ &&-k\int_{\Omega}a_2(x)\nabla v\cdot\nabla(b_2(x)e^{kv}){\rm d}x+k\int_{\Omega}(b_2(x))^2e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq & kc_{1}|B_1|\int_{\Omega}b_1(x)e^{ku}|\nabla u|{\rm d}x-k^2c_1\int_{\Omega}b_1(x)e^{ku}|\nabla u|^2{\rm d}x \\ &&+k\int_{\Omega}(b_1(x))^2e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x+kc_{2}|B_2|\int_{\Omega}b_2(x)e^{kv}|\nabla v|{\rm d}x\\ &&-k^2c_2\int_{\Omega}b_2(x)e^{kv}|\nabla v|^2{\rm d}x +k\int_{\Omega}(b_2(x))^2e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x. \end{eqnarray} $

对(4.11)式的第一项、第三项和第四项,运用Hölder不等式和Young不等式,我们可得

(4.12) $ \begin{eqnarray} && \int_{\Omega}b_1(x)e^{ku}|\nabla u|{\rm d}x\leq\frac{1}{2\sigma_1}\int_{\Omega}b_1(x)e^{ku}{\rm d}x+\frac{\sigma_1}{2}\int_{\Omega}b_1(x)e^{ku}|\nabla u|^2{\rm d}x, \end{eqnarray} $

(4.13) $ \begin{eqnarray} &&\int_{\Omega}(b_1(x))^2e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &\leq&\frac{\alpha_{2}}{k}\int_{\Omega}b_2(x)e^{kv}{\rm d}x+\frac{k-\alpha_2}{k}\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_{2}}}(b_2(x))^{-\frac{\alpha_{2}}{k-\alpha_{2}}}e^{\frac{k(k+\alpha_{1})}{k-\alpha_{2}}u}{\rm d}x, \end{eqnarray} $

及

(4.14) $ \begin{equation} \int_{\Omega}b_2(x)e^{kv}|\nabla v|{\rm d}x\leq\frac{1}{2\sigma_2}\int_{\Omega}b_2(x)e^{kv}{\rm d}x+\frac{\sigma_2}{2}\int_{\Omega}b_2(x)e^{kv}|\nabla v|^2{\rm d}x, \end{equation} $

其中$ \sigma_{1}, \sigma_{2} $是待定的正常数.

利用Hölder不等式估计(4.13)式中第二项,我们有

(4.15) $ \begin{eqnarray} &&\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_{2}}}(b_2(x))^{-\frac{\alpha_{2}}{k-\alpha_{2}}}e^{\frac{k(k+\alpha_{1})}{k-\alpha_{2}}u}{\rm d}x \\ &\leq&\bigg(\int_{\Omega}(b_1(x))^{\frac{N}{N-2}}e^{\frac{kN}{N-2}u}{\rm d}x\bigg)^{\frac{N-2}{2N}}\\ &&\times\bigg(\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x\bigg)^{\frac{N+2}{2N}}. \end{eqnarray} $

现在,对(4.15)式的最后一项,利用高维空间$ \Omega\in{\Bbb R} ^{N}(N\geq3) $中的加权Sobolev嵌入不等式

$ \|b_1(x)^{\frac{1}{2}}e^{\frac{k}{2}u}\|_{L^{\frac{2N}{N-2}}(\Omega)}\leq C_s\|b_1(x)^{\frac{1}{2}}e^{\frac{k}{2}u}\|_{W^{1, 2}(\Omega)}, $

其中$ C_{s} $为Sobolev最优常数,并由Minkowski不等式及条件(S$ _{3}) $我们有

(4.16) $ \begin{eqnarray} &&\bigg(\int_{\Omega}(b_1(x))^{\frac{N}{N-2}}e^{\frac{kN}{N-2}u}{\rm d}x\bigg)^{\frac{N-2}{2N}}\\ &\leq&(1+\frac{B_1}{2})C_{s}\bigg(\int_{\Omega}b_1(x)e^{ku}{\rm d}x\bigg)^{\frac{1}{2}} +\frac{kC_s}{2}\bigg(\int_{\Omega}b_1(x)e^{ku}|\nabla u|^2{\rm d}x\bigg)^{\frac{1}{2}}. \end{eqnarray} $

结合(4.13), (4.15)和(4.16)式并利用带$ \varepsilon $的Young不等式,我们可导出

(4.17) $ \begin{eqnarray} &&k\int_{\Omega}(b_1(x))^2e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &\leq&\alpha_{2}\int_{\Omega}b_2(x)e^{kv}{\rm d}x +\frac{k(k-\alpha_{2})\varepsilon_{3}}{4}C_{s}\int_{\Omega}b_1(x)e^{ku}|\nabla u|^2{\rm d}x\\ &&+(k-\alpha_{2})(1+\frac{B_1}{2})C_{s}\bigg(\int_{\Omega}b_1(x)e^{ku}{\rm d}x\bigg)^{\frac{1}{2}} \\ &&\times \bigg(\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x\bigg)^{\frac{N+2}{2N}}\\ &&+\frac{k(k-\alpha_2)}{4\varepsilon_3}C_{s}\bigg(\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x\bigg)^{\frac{N+2}{N}}, \qquad \end{eqnarray} $

其中$ \varepsilon_{3} $是待定的正常数.类似地,我们有

(4.18) $ \begin{eqnarray} &&k\int_{\Omega}(b_2(x))^2e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq&\beta_{2}\int_{\Omega}b_1(x)e^{ku}{\rm d}x +\frac{k(k-\beta_{2})\varepsilon_{4}}{4}C_{s}\int_{\Omega}b_2(x)e^{kv}|\nabla v|^2{\rm d}x\\ && +(k-\beta_{2})(1+\frac{B_2}{2})C_{s}\bigg(\int_{\Omega}b_2(x)e^{kv}{\rm d}x\bigg)^{\frac{1}{2}} \\ &&\times \bigg(\int_{\Omega}(b_2(x))^{\frac{(3k+\beta_2)N}{(k-\beta_2)(N+2)}}(b_1(x))^{-\frac{2\beta_2N}{(k-\beta_2)(N+2)}} e^{\frac{kN(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}v}{\rm d}x\bigg)^{\frac{N+2}{2N}}\\ && +\frac{k(k-\beta_2)}{4\varepsilon_4}C_{s} \bigg(\int_{\Omega}(b_2(x))^{\frac{(3k+\beta_2)N}{(k-\beta_2)(N+2)}}(b_1(x))^{-\frac{2\beta_2N}{(k-\beta_2)(N+2)}} e^{\frac{kN(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}v}{\rm d}x\bigg)^{\frac{N+2}{N}}, \qquad \end{eqnarray} $

其中$ \varepsilon_{4} $是待定的正常数.将(4.12), (4.14), (4.17), (4.18)式代入(4.1)式,并整理可得

(4.19) $ \begin{eqnarray} \Phi'(t)&\leq& \bigg(\frac{kc_1|B_1|\sigma_{1}}{2}+\frac{k(k-\alpha_{2})\varepsilon_{3}}{4}C_{s}-k^{2}c_{1} \bigg)\int_{\Omega}b_1(x)e^{ku}|\nabla u|^2{\rm d}x \\ && +\bigg(\frac{kc_1|B_1|}{2\sigma_{1}}+\beta_{2}\bigg)\int_{\Omega}b_1(x)e^{ku}{\rm d}x \\ &&+\bigg(\frac{kc_2|B_2|\sigma_{2}}{2}+\frac{k(k-\beta_{2})\varepsilon_{4}}{4}C_{s}-k^{2}c_{2} \bigg)\int_{\Omega}b_2(x)e^{kv}|\nabla v|^2{\rm d}x \\ && +\bigg(\frac{kc_2|B_2|}{2\sigma_{2}}+\alpha_{2}\bigg)\int_{\Omega}b_2(x)e^{kv}{\rm d}x+(k-\alpha_{2})(1+\frac{B_1}{2})C_{s}\bigg(\int_{\Omega}b_1(x)e^{ku}{\rm d}x\bigg)^{\frac{1}{2}} \\ &&\times \bigg(\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x\bigg)^{\frac{N+2}{2N}}\\ &&+(k-\beta_{2})(1+\frac{B_2}{2})C_{s}\bigg(\int_{\Omega}b_2(x)e^{kv}{\rm d}x\bigg)^{\frac{1}{2}} \\ &&\times \bigg(\int_{\Omega}(b_2(x))^{\frac{(3k+\beta_2)N}{(k-\beta_2)(N+2)}}(b_1(x))^{-\frac{2\beta_2N}{(k-\beta_2)(N+2)}} e^{\frac{kN(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}v}{\rm d}x\bigg)^{\frac{N+2}{2N}}\\ &&+\frac{k(k-\alpha_2)}{4\varepsilon_3}C_{s}\bigg(\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x\bigg)^{\frac{N+2}{N}}\\ &&+\frac{k(k-\beta_2)}{4\varepsilon_4}C_{s}\bigg(\int_{\Omega}(b_2(x))^{\frac{(3k+\beta_2)N}{(k-\beta_2)(N+2)}}(b_1(x))^{-\frac{2\beta_2N}{(k-\beta_2)(N+2)}} e^{\frac{kN(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}v}{\rm d}x\bigg)^{\frac{N+2}{N}}.\qquad \end{eqnarray} $

对于充分小的$ \sigma_{1}, \sigma_{2}, $我们取$ \varepsilon_{3}, \varepsilon_{4}, $使得

$ \frac{kc_1|B_1|\sigma_{1}}{2}+\frac{k(k-\alpha_{2})\varepsilon_{3}}{4}C_{s}-k^{2}c_{1} = \frac{kc_2|B_2|\sigma_{2}}{2}+\frac{k(k-\beta_{2})\varepsilon_{4}}{4}C_{s}-k^{2}c_{2} = 0. $

同时,由$ k>\max\{N(\alpha_1+\alpha_2)+\alpha_2, N(\beta_1+\beta_2)+\beta_2\} $易知

$ \frac{N(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}\in(0, 1), \quad \frac{2(k-\alpha_2)-2N(\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}\in(0, 1), $

$ \frac{N(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}\in(0, 1), \quad \frac{2(k-\beta_2)-2N(\beta_1+\beta_2)}{(k-\beta_2)(N+2)}\in(0, 1). $

因此,再利用Hölder不等式得到

(4.20) $ \begin{eqnarray} &&\int_{\Omega}(b_1(x))^{\frac{(3k+\alpha_2)N}{(k-\alpha_2)(N+2)}}(b_2(x))^{-\frac{2\alpha_2N}{(k-\alpha_2)(N+2)}} e^{\frac{kN(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}u}{\rm d}x \\ &\leq&\bigg(\int_{\Omega}b_1(x)e^{ku}{\rm d}x\bigg)^{\frac{N(k+2\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}} \\ && \times\bigg(\int_{\Omega}(b_1(x))^{\frac{N(k-\alpha_1)}{k-\alpha_2-N(\alpha_1+\alpha_2)}}(b_2(x))^{-\frac{\alpha_{2}N}{k-\alpha_2-N(\alpha_1+\alpha_2)}}{\rm d}x\bigg) ^{\frac{2(k-\alpha_2)-2N(\alpha_1+\alpha_2)}{(k-\alpha_2)(N+2)}}, \end{eqnarray} $

(4.21) $ \begin{eqnarray} && \int_{\Omega}(b_2(x))^{\frac{(3k+\beta_2)N}{(k-\beta_2)(N+2)}}(b_1(x))^{-\frac{2\beta_2N}{(k-\beta_2)(N+2)}} e^{\frac{kN(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}v}{\rm d}x \\ &\leq&\bigg(\int_{\Omega}b_2(x)e^{kv}{\rm d}x\bigg)^{\frac{N(k+2\beta_1+\beta_2)}{(k-\beta_2)(N+2)}} \\ && \times\bigg(\int_{\Omega}(b_2(x))^{\frac{N(k-\beta_1)}{k-\beta_2-N(\beta_1+\beta_2)}}(b_2(x))^{-\frac{\beta_{2}N}{k-\beta_2-N(\beta_1+\beta_2)}}{\rm d}x\bigg) ^{\frac{2(k-\beta_2)-2N(\beta_1+\beta_2)}{(k-\beta_2)(N+2)}}. \end{eqnarray} $

将(4.20)和(4.21)式代入到(4.19)式,我们导出

(4.22) $ \begin{equation} \Phi'(t)\leq K_4\Phi(t)+K_6(\Phi(t))^{r_1}+K_7(\Phi(t))^{r_2}, \end{equation} $

其中

(4.23) $ \begin{eqnarray} &&r_1: = \max \bigg\{\frac{k+\alpha_{1}}{k-\alpha_{2}}, \frac{k+\beta_{1}}{k-\beta{2}}\bigg\}>1, \\ &&r_2: = \max\bigg\{\frac{k+2\alpha_{1}+\alpha_2}{k-\alpha_{2}}, \frac{k+2\beta_{1}+\beta_2}{k-\beta{2}}\bigg\}>1, \\ &&K_5: = \max\bigg\{\frac{kc_1|B_1|}{2\sigma_{1}}+\beta_{2}, \frac{kc_2|B_2|}{2\sigma_{2}}+\alpha_{2}\bigg\}, \\ &&K_6: = (k-\alpha_{2})(1+\frac{B_1}{2})C_{s} \bigg(\int_{\Omega}(b_1(x))^{\frac{N(k-\alpha_1)}{k-\alpha_2-N(\alpha_1+\alpha_2)}}(b_2(x))^{-\frac{\alpha_{2}N}{k-\alpha_2-N(\alpha_1+\alpha_2)}}{\rm d}x\bigg) ^{\frac{k-\alpha_2-N(\alpha_1+\alpha_2)}{N(k-\alpha_2)}} \\ &&\qquad\quad+(k-\beta_{2})(1+\frac{B_2}{2})C_{s}\bigg(\int_{\Omega}(b_2(x))^{\frac{N(k-\beta_1)}{k-\beta_2-N(\beta_1+\beta_2)}}(b_2(x))^{-\frac{\beta_{2}N}{k-\beta_2-N(\beta_1+\beta_2)}}{\rm d}x\bigg) ^{\frac{k-\beta_2-N(\beta_1+\beta_2)}{N(k-\beta_2)}}, \\ &&K_7: = \frac{k(k-\alpha_2)}{4\varepsilon_3}C_{s} \bigg(\int_{\Omega}(b_1(x))^{\frac{N(k-\alpha_1)}{k-\alpha_2-N(\alpha_1+\alpha_2)}}(b_2(x))^{-\frac{\alpha_{2}N}{k-\alpha_2-N(\alpha_1+\alpha_2)}}{\rm d}x\bigg) ^{\frac{2(k-\alpha_2)-2N(\alpha_1+\alpha_2)}{N(k-\alpha_2)}} \\ &&\qquad\quad+\frac{k(k-\beta_2)}{4\varepsilon_4}C_{s} \bigg(\int_{\Omega}(b_2(x))^{\frac{N(k-\beta_1)}{k-\beta_2-N(\beta_1+\beta_2)}}(b_2(x))^{-\frac{\beta_{2}N}{k-\beta_2-N(\beta_1+\beta_2)}}{\rm d}x\bigg) ^{\frac{2(k-\beta_2)-2N(\beta_1+\beta_2)}{N(k-\beta_2)}}. \\ \end{eqnarray} $

如果$ \lim\limits_{t\to t^*}\Phi(t) = \infty, $则对(4.22)式从$ 0 $到$ t $积分,我们有

$ t^*\geq\int^{\infty}_{\Phi(0)}\frac{{\rm d}\eta}{K_5+K_6\eta^{r_1}+K_7\eta^{r_2}}, $

定理4.2证毕.

当$ N = 1 $时,我们假设$ \Omega = (0, l), $其中$ l>0 $为常数.问题(1.1)–(1.4)可改写为

(4.24) $ \begin{equation} \left\{\begin{array}{ll} u_{t} = (a_{1}(x)u_{x})_{x}+b_1(x)e^{\alpha_{1}u+\alpha_{2}v}, \quad & (x, t)\in(0, l)\times(0, t^{\ast}), \\ v_{t} = (a_{2}(x)v_{x})_{x}+b_2(x)e^{\beta_{1}v+\beta_{2}u}, \quad & (x, t)\in(0, l)\times(0, t^{\ast}), \\ u(0, t) = u(l, t) = v(0, t) = v(l, t) = 0, \quad &t\in(0, t^{\ast}), \\ u(x, 0) = u_{0}(x), \quad v(x, 0) = v_{0}(x), \quad &x\in(0, l). \end{array}\right. \end{equation} $

我们有如下结论.

定理4.3  假设$ (u, v) $为问题$ (4.24) $的非负古典解且对$ i = 1, 2 $,变扩散系数$ a_i(x)\in C^0(\bar{\Omega})\bigcap C^2(\Omega) $满足(A$ _{1}), $权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{2})}, $定义辅助函数

$ \varphi_{1}(t): = \int_{0}^{l}e^{ku}+e^{kv}{\rm d}x, $

其中

$ k>\max\{\alpha_1+2\alpha_2, \beta_1+2\beta_2\}. $

如果$ (u, v) $在测度$ \varphi_{1}(t) $的意义下有限时刻$ t^* $发生爆破,则爆破时间$ t^* $的下界为

$ t^*\geq\int^{\infty}_{\varphi_{1}(0)}\frac{{\rm d}\eta}{J_1+J_2\eta+J_3\eta^2}, $

其中$ \varphi_{1}(0) = \int_{0}^{l}e^{ku_{0}}+e^{kv_0}{\rm d}x, $$ J_1, J_2 $及$ J_3 $由$ (4.31) $式给出.

证  对$ \varphi_{1}(t) $直接求导并利用(4.24)及Green公式,我们有

(4.25) $ \begin{eqnarray} \varphi'_{1}(t)& = &k\int_{0}^{l}e^{ku}((a_1(x)u_{x})_{x}+b_1(x)e^{\alpha_{1}u+\alpha_{2}v}){\rm d}x+k\int_{0}^{l}e^{kv}((a_2(x)v_{x})_{x}+b_2(x)e^{\beta_{1}v+\beta_{2}u}){\rm d}x\\ &\leq&-k^2\int_{0}^{l}e^{ku}a_1(x)|u_{x}|^2{\rm d}x+k\int_{0}^{l}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &&-k^2\int_{0}^{l}e^{kv}a_2(x)|v_{x}|^2{\rm d}x+k\int_{0}^{l}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq & k\int_{0}^{l}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x+k\int_{0}^{l}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x. \end{eqnarray} $

对(4.25)式中第一项运用Hölder不等式及Young不等式,我们可得

(4.26) $ \begin{eqnarray} \int_{0}^{l}b_1(x)e^{(k+\alpha_1)u+\alpha_2v}{\rm d}x&\leq&\bigg(\int_{0}^{l}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x\bigg)^{\frac{k-\alpha_2}{k}} \bigg(\int_{0}^{l}e^{kv}{\rm d}x\bigg)^{\frac{\alpha_2}{k}}\\ &\leq&\frac{k-\alpha_2}{k} \int_{0}^{1}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x +\frac{\alpha_2}{k}\int_{0}^{l}e^{kv}{\rm d}x. \end{eqnarray} $

由$ k>\max\{\alpha_1+2\alpha_2, \beta_1+2\beta_2\}. $易知$ \frac{k+\alpha_{1}}{2(k-\alpha_2)}, \frac{k-\alpha_{1}-2\alpha_{2}}{2(k-\alpha_2)}\in(0, 1). $

因此,对(4.26)式中第一项运用Hölder不等式和Young不等式,我们有

(4.27) $ \begin{eqnarray} \int_{0}^{l}(b_1(x))^{\frac{k}{k-\alpha_2}}e^{\frac{k(k+\alpha_1)}{k-\alpha_2}u}{\rm d}x&\leq&\bigg(\int_{0}^{l}e^{2ku}{\rm d}x\bigg)^{\frac{k+\alpha_{1}}{2(k-\alpha_2)}} \bigg(\int_{0}^{l}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x\bigg)^{\frac{k-\alpha_{1}-2\alpha_{2}}{2(k-\alpha_2)}} \\ &\leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2(k-\alpha_2)}\int_{0}^{l}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k+\alpha_{1}}{2(k-\alpha_2)}\int_{0}^{l}e^{2ku}{\rm d}x.\\ \end{eqnarray} $

将(4.27)代入(4.26)式,我们有

(4.28) $ \begin{eqnarray} k\int_{0}^{l}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x&\leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{0}^{l}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x\\ && +\frac{k+\alpha_{1}}{2}\int_{0}^{l}e^{2ku}{\rm d}x+\alpha_2\int_{0}^{l}e^{kv}{\rm d}x. \end{eqnarray} $

类似地,我们有

(4.29) $ \begin{eqnarray} k\int_{0}^{l}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x&\leq&\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{0}^{l}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x \\ &&+\frac{k+\beta_{1}}{2}\int_{0}^{l}e^{2kv}{\rm d}x+\beta_2\int_{0}^{l}e^{ku}{\rm d}x. \end{eqnarray} $

将(4.28)和(4.29)代入到(4.25)式,我们导出

(4.30) $ \begin{eqnarray} \varphi'_{1}(t)&\leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{0}^{l}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k+\alpha_{1}}{2}\int_{0}^{l}e^{2ku}{\rm d}x+\alpha_2\int_{0}^{l}e^{kv}{\rm d}x\\ &&+\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{0}^{l}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x +\frac{k+\beta_{1}}{2}\int_{0}^{l}e^{2kv}{\rm d}x+\beta_2\int_{0}^{l}e^{ku}{\rm d}x\\ &\leq & J_1+J_2\varphi_1(t)+J_3\varphi_1(t)^{2}, \end{eqnarray} $

其中

(4.31) $ \begin{eqnarray} &&J_1: = \frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{0}^{l}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{0}^{l}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x, \\ &&J_2: = \max\{\alpha_{2}, \beta_{2}\}, \\ &&J_3: = \max\bigg\{\frac{k+\alpha_{1}}{2}, \frac{k+\beta_{1}}{2}\bigg\}. \end{eqnarray} $

如果$ \lim\limits_{t\to t^{*}}\varphi_{1}(t) = \infty, $则对(4.30)式从$ 0 $到$ t^* $积分,我们有

$ t^*\geq\int^{\infty}_{\varphi_{1}(0)}\frac{{\rm d}\eta}{J_1+J_2\eta+J_3\eta^2}, $

定理4.3证毕.

定理4.4  假设$ (u, v) $为问题(1.1)–(1.4)的非负古典解且对$ i = 1, 2 $,变扩散系数$ a_i(x)\in C^0(\bar{\Omega})\bigcap C^2(\Omega) $满足(A$ _{1}) $和(A$ _{2}), $权函数$ b_i(x)\in C(\bar{\Omega}) $满足$ {\rm(S_{1})} $或$ {\rm(S_{2})}, $定义辅助函数

$ \varphi_{2}(t): = \int_{\Omega}e^{ku}+e^{kv}{\rm d}x, $

其中

$ k>\max\{\alpha_1+2\alpha_2, \beta_1+2\beta_2\}. $

如果$ (u, v) $在测度$ \varphi_{2}(t) $的意义下有限时刻$ t^* $发生爆破,则爆破时间$ t^* $的下界为

$ t^*\geq\int^{\infty}_{\varphi_{2}(0)}\frac{{\rm d}\eta}{J'_1+J'_2\eta+J_3\eta^2}, $

其中$ \varphi_{2}(0) = \int_{\Omega}e^{ku_{0}}+e^{kv_0}{\rm d}x, $$ J'_1, J'_2 $由(4.38)式给出, $ J_3 $由$ (4.31) $式给出.

证  对$ \varphi_{2}(t) $直接求导并利用(1.1)–(1.3)式及Green公式,我们有

(4.32) $ \begin{eqnarray} \varphi'_{2}(t)& = &k\int_{\partial\Omega}a_1(x)\frac{\partial u}{\partial\nu}{\rm d}s-k^2\int_{\Omega}a_1(x)e^{ku}|\nabla u|^2{\rm d}x+k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &&+k\int_{\partial\Omega}a_2(x)\frac{\partial v}{\partial\nu}{\rm d}s-k^2\int_{\Omega}a_2(x)e^{kv}|\nabla v|^2{\rm d}x+k\int_{\Omega}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x\\ &\leq&-4\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}u}|^2{\rm d}x+k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x\\ &&-4\int_{\Omega}a_2(x)|\nabla e^{\frac{k}{2}v}|^2{\rm d}x+k\int_{\Omega}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x. \end{eqnarray} $

由加权Poincaré不等式及条件(A$ _{2}), $我们得到

(4.33) $ \begin{equation} C(\Omega)\int_{\Omega}a_1(x)e^{ku}{\rm d}x\leq\int_{\Omega}a_1(x)|\nabla e^{\frac{k}{2}u}|^2{\rm d}x+M_{1}\int_{\Omega}e^{ku}{\rm d}x, \end{equation} $

及

(4.34) $ \begin{equation} C(\Omega)\int_{\Omega}a_2(x)e^{kv}{\rm d}x\leq\int_{\Omega}a_2(x)|\nabla e^{\frac{k}{2}v}|^2{\rm d}x+M_{2}\int_{\Omega}e^{kv}{\rm d}x, \end{equation} $

其中$ C(\Omega)>0. $

类似于定理4.3的证明过程,我们可得

(4.35) $ \begin{eqnarray} k\int_{\Omega}b_1(x)e^{(k+\alpha_1)u+\alpha_{2}v}{\rm d}x&\leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x \\ &&+\frac{k+\alpha_{1}}{2}\int_{\Omega}e^{2ku}{\rm d}x+\alpha_2\int_{\Omega}e^{kv}{\rm d}x, \end{eqnarray} $

及

(4.36) $ \begin{eqnarray} k\int_{0}^{l}b_2(x)e^{(k+\beta_1)v+\beta_{2}u}{\rm d}x&\leq&\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{\Omega}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x \\ &&+\frac{k+\beta_{1}}{2}\int_{\Omega}e^{2kv}{\rm d}x+\beta_2\int_{\Omega}e^{ku}{\rm d}x. \end{eqnarray} $

将(4.33)–(4.36)式代入到(4.32)式,并由条件(A$ _{1}) $我们可导出

(4.37) $ \begin{eqnarray} \varphi'_{2}(t)&\leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{\Omega}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x\\ &&+\frac{k+\alpha_{1}}{2}\int_{\Omega}e^{2ku}{\rm d}x+\frac{k+\beta_{1}}{2}\int_{\Omega}e^{2kv}{\rm d}x +(\beta_2+4M_1)\int_{\Omega}e^{ku}{\rm d}x \\ &&+(\alpha_2+4M_2)\int_{\Omega}e^{kv}{\rm d}x -4C(\Omega)\int_{\Omega}a_1(x)e^{ku}{\rm d}x-4C(\Omega)\int_{\Omega}a_2(x)e^{kv}{\rm d}x\\ & \leq&\frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{\Omega}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x\\ &&+\frac{k+\alpha_{1}}{2}\int_{\Omega}e^{2ku}{\rm d}x+\frac{k+\beta_{1}}{2}\int_{\Omega}e^{2kv}{\rm d}x +(\beta_2+4M_1)\int_{\Omega}e^{ku}{\rm d}x\\ &&+(\alpha_2+4M_2)\int_{\Omega}e^{kv}{\rm d}x \\ &\leq & J'_{1}+J'_{2}\varphi_{2}(t)+J_{3}(\varphi_{2}(t))^2, \end{eqnarray} $

其中$ J_{3} $如(4.31)式,

(4.38) $ \begin{eqnarray} &&J'_{1}: = \frac{k-\alpha_{1}-2\alpha_{2}}{2}\int_{\Omega}(b_1(x))^{\frac{2k}{k-\alpha_1-2\alpha_2}}{\rm d}x +\frac{k-\beta_{1}-2\beta_{2}}{2}\int_{\Omega}(b_2(x))^{\frac{2k}{k-\beta_1-2\beta_2}}{\rm d}x, \\ &&J'_{2}: = \max\{\alpha_{2}+4M_2, \beta+4M_1\}. \end{eqnarray} $

如果$ \lim\limits_{t\to t^{*}}\varphi_{2}(t) = \infty, $则对(4.37)式从$ 0 $到$ t^* $积分,我们有

$ t^*\geq\int^{\infty}_{\varphi_{2}(0)}\frac{{\rm d}\eta}{J'_1+J'_2\eta+J_3\eta^2}. $

定理4.4证毕.

注4.1  本节定理4.1,定理$ 4.2, $定理$ 4.3, $定理$ 4.4 $中我们的证明方法对具有齐次Neumann边界条件的问题都同样适用.