具有加权函数的发散型反应扩散方程的爆破分析

Blow-Up Analysis of Divergent Reaction Diffusion Equations with Weighted Functions

摘要: 反应扩散方程以建立数学模型的方式解决物理、化学、生物学、传染病学和核科学领域中的实际问题，反应扩散方程可以恰当地描述很多自然现象，例如浓度和密度的扩散以及化学药剂的燃烧等众多自然现象。反应扩散方程是一类非线性方程，其非线性项可能来自边界条件、扩散项、反应项或者是三者所组合的不同形式的耦合关系，所有这些非线性项均可影响爆破解，另外还发现无论解的全局存在性还是爆破性都受到加权项的影响。本文考虑了一类具有加权函数和含时系数的发散型反应扩散方程的爆破问题，难点在于找到位置加权和时间加权项与边界源对于爆破解的影响，给出了非线性函数u(x,t)正解全局存在的条件，利用微分不等式技术，并构造恰当的辅助函数，分别得到了爆破时间的上界和下界。

Abstract: The reaction diffusion equation solves practical problems in the fields of physics, chemistry, biology, infectious disease, and nuclear science by establishing mathematical models. The reaction diffusion equation can appropriately describe many natural phenomena, such as diffusion of concentration and density, and combustion of chemical agents. The reaction diffusion equation is a type of nonlin-ear equation, whose nonlinear terms may come from boundary conditions, diffusion terms, reaction terms, or different forms of coupling between the three. All of these nonlinear terms can affect the explosive solution. In addition, it has been found that both the global existence and explosive prop-erty of the solution are affected by the weighted terms. This article considers the blow up problem of a class of divergent reaction diffusion equations with weighted functions and time-dependent co-efficients. The difficulty lies in finding the effects of position weighted and time weighted terms and boundary sources on the blow up solution. The conditions for the global existence of positive solu-tions to nonlinear functions u(x,t) are given. By using differential inequality techniques and con-structing appropriate auxiliary functions, the upper and lower bounds of the blow up time are ob-tained, respectively.

1. 引言

近年来对于含时系数 $k\left(t\right)$ 的研究引起了众多学者的关注，已经取得了很多重要成果。Payne [1] [2] [3] 研究了一类具有时间相关系数的半线性热方程的初边值问题，分别在Dirichlet、Neumann和Robin三种条件下，利用微分不等式技术，研究了数据对解的行为(有限时间爆破、解的整体存在性)的影响，导出了爆破发生时爆破时间的上下界。Marras [4] 考虑了一类在边界上存在空间积分且非线性和边界条件上都存在相关系数的反应扩散方程，引入足够的条件，使解在有限时间内爆破，并推出了爆破时间的上下界。Juntang [5] 在Marras的启发下，研究了具有含时系数和非局域边界条件的非线性反应扩散方程，依靠微分不等式技术，得到了解在有限时间爆破的结论和爆破时间的上下界。Zhang [6] 考虑了在多维空间中的一类具有非齐次Neumann边界条件的非线性发散型抛物方程，建立了非线性函数正解全局存在的新条件，并给出了多维空间中爆破时间的上下界。

对于加权函数 $a\left(x\right)$ 也受到了国内外学者关注与研究，取得了许多研究成果。Song [7] 研究了一类具有加权非线性的抛物型方程的爆破现象，得到了该问题在任意光滑的有界区域 $\Omega \in R_n\left(n\ge 3\right)$ 上解的爆破时间的界和爆破速率的估计，甚至，在某些特殊条件下可以得到爆破时间的精确值。Ma和Fang [8] [9] 研究了一类具有加权非局内吸收项的非线性发散型反应扩散方程的爆破分析，利用辅助函数法和修正的微分不等式技术，建立了满足问题解全局存在和有限时刻爆破的条件，并在高维空间下给出了爆破时间的上下界。

加权函数 $a\left(x\right)\in C^2\left(\Omega \right)\cap C^0\left(\bar{\Omega }\right)$ 满足下列条件之一：

$\begin{array}{l}{A}_1:a\left(x\right)>0,x\in \Omega ;a\left(x\right)=0,x\in \partial \Omega \\ A_2:a\left(x\right)\ge c>0,x\in \bar{\Omega }\\ A_3:a\left(x\right)\equiv 0,x\in \bar{\Omega }\\ A_4:0<c_1<a\left(x\right)<c_2,x\in \bar{\Omega }\end{array}$

其中c为常数。

本文考虑了如下具有加权函数和含时系数的发散型反应扩散方程

$\begin{array}{l}{u}_t={\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{\left(a_{ij}\left(x\right)u_{x_i}\right)}_{x_j}}+k_1\left(t\right)a\left(x\right)f\left(u\right)\text{,}\left(x,t\right)\in \Omega \times \left(0,t^{*}\right)\\ {\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{v}_j}=k_2\left(t\right)g\left(u\right),\left(x,t\right)\in \partial \Omega \times \left(0,t^{*}\right)\\ u\left(x,0\right)=u_0\left(x\right)>0,x\in \Omega \end{array}$ (1.1)

其中 $\Omega$ 是 $R_n\left(n\ge 3\right)$ 中的有界星形区域，边界 $\partial \Omega$ 光滑， ${\left(a_{ij}\left(x\right)\right)}_{n\times n}$ 是一个可微正定矩阵， $k_1\left(t\right)$ 和 $k_2\left(t\right)$ 是一个非负微分函数，如果爆破发生， $t^{*}$ 是爆破时间，否则 $t^{*}=+\infty$ 。

在第一节，我们给出了保证 $u\left(x\right)$ 全局存在的非线性充分条件。在第二节中，在保证解在有限时间爆破的条件下，我们给出了爆破时间 $t^{*}$ 的上界估计。第三节中，考虑 $\sigma =1$ 的特殊情况，我们估计了多维空间中爆破时间 $t^{*}$ 的下界。

2. 解的整体存在性

定理1： $\Omega \in R_n\left(n\ge 3\right)$ 是一个有界星型区域，边界 $\partial \Omega$ 光滑，函数 $f,g$ 和含时系数 $k_1\left(t\right),k_2\left(t\right)$ 满足以下条件：

$f\left(s\right)\{\begin{array}{l}\le -\alpha s^p,s>0\\ =0,s\le 0\end{array},g\left(s\right)\{\begin{array}{l}\le -\beta s^q,s>0\\ >0,s\le 0\end{array}$

$m_1\ge k_1\left(t\right)>0,m_2\ge k_2\left(t\right)>0$

$\underset{x>0}{\sup }a\left(x\right)=L<+\infty$

其中 $\alpha ,\beta >0$ ， $p>q$ ， $2q<p+1$ ， $m_1,m_2>0$ ， $L>0$ 。则问题 的解 $u\left(x,t\right)$ 为正解且不发生爆破，对于所有的 $t>0$ ，都有 $u\left(x,t\right)$ 存在。

引理1： $\Omega \in R_n\left(n\ge 2\right)$ 是一个有界星形区域，假设在 $n-1$ 个正方向上凸，对于任意非负递增的 $C^1$ 函数 $h\left(w\right)$ ，有

${\displaystyle {\int }_{\partial \Omega }h\left(w\right)\text{d}s}\le \frac{n}{{\rho }_0}{\displaystyle {\int }_{\Omega }h\left(w\right)\text{d}x}+\frac{d}{{\rho }_0}{\displaystyle {\int }_{\Omega }{h}^{\prime }\left(w\right)\left|\nabla w\right|\text{d}x}$

其中 ${\rho }_0:=\underset{x\in \partial \Omega }{\min }\left(x\cdot v\right)$ ， $d:=\underset{x\to \partial \Omega }{\max }\left|x\right|$

证明：首先将(1.1)乘u并在 $\Omega$ 上积分，得

$\begin{array}{c}{\displaystyle {\int }_{\Omega }u{u}_t}\text{d}x=\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}\\ ={\displaystyle {\int }_{\Omega }u{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{\left(a_{ij}\left(x\right)u_{x_i}\right)}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }ua\left(x\right)f\left(u\right)\text{d}x}\\ =k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }ug\left(u\right)\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)uf\left(u\right)\text{d}x}\end{array}$ (1.2)

定义辅助函数 $\phi \left(t\right):=\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}$ ，由定理中条件整理(1.2)得

${\phi }^{\prime }\left(t\right)\le \beta k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }{u}^{q+1}}\text{d}s-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}-\alpha k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)u^{p+1}}\text{d}x$ (1.3)

由于 ${\left(a_{ij}\left(x\right)\right)}_{n\times n}$ 是一个可微正定矩阵，所以存在 $\theta >0$ ，有

${\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right){\eta }_{x_i}{\eta }_{x_j}}\ge \theta {\left|\eta \right|}^2,\forall \eta =\left({\eta }_1,{\eta }_2,\cdots ,{\eta }_n\right)\in R^n$

所以得到

${\phi }^{\prime }\left(t\right)\le \beta k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }{u}^{q+1}}\text{d}s-\theta {\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}-\alpha k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)u^{p+1}}\text{d}x$ (1.4)

由引理1知

${\displaystyle {\int }_{\partial \Omega }{u}^{q+1}\text{d}s}\le \frac{n}{{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^{q+1}}\text{d}x+\frac{\left(q+1\right)d}{{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^q\left|\nabla u\right|\text{d}x}$ (1.5)

将(1.5)代入(1.4)整理得到

${\phi }^{\prime }\left(t\right)\le \frac{n\beta k_2\left(t\right)}{{\rho }_0}{\displaystyle {\int }_{\partial \Omega }{u}^{q+1}}\text{d}x-\alpha k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)u^{q+1}\text{d}x}+\frac{\left(q+1\right)d\beta k_2\left(t\right)}{{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^q\left|\nabla u\right|\text{d}x}-\theta {\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}$ (1.6)

利用柯西不等式并取 $\tau =\frac{\left(q+1\right)d\beta k_2}{2{\rho }_0\theta }$

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{u}^q\left|\nabla u\right|\text{d}x}\le \frac{1}{2\tau }{\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}+\frac{\tau }{2}{\displaystyle {\int }_{\Omega }{u}^{2q}\text{d}x}\\ =\frac{{\rho }_0\theta }{\beta \left(q+1\right)d{k}_2}{\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}+\frac{\beta \left(q+1\right)d{k}_2}{4{\rho }_0\theta }{\displaystyle {\int }_{\Omega }{u}^{2q}\text{d}x}\end{array}$ (1.7)

将(1.7)代入(1.6)得

$\begin{array}{c}{\phi }^{\prime }\left(t\right)\le \frac{n\beta k_2}{{\rho }_0}{\displaystyle {\int }_{\partial \Omega }{u}^{q+1}}\text{d}x+\theta {\tau }^2{\displaystyle {\int }_{\Omega }{u}^{2q}\text{d}x}-\alpha k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)u^{q+1}\text{d}x}\\ ={\displaystyle {\int }_{\Omega }\left(\frac{n\beta k_2}{{\rho }_0}\frac{u^q}{u^p}-\frac{\alpha k_{1}a\left(x\right)}{2}\right)u^{P+1}\text{d}x}+{\displaystyle {\int }_{\Omega }\left(\theta {\tau }^2\frac{u^{2q}}{u^{p+1}}-\frac{\alpha k_{1}a\left(x\right)}{2}\right)u^{p+1}\text{d}x}\\ \le {\displaystyle {\int }_{\Omega }\left(\frac{n\beta m_2}{{\rho }_0}\frac{u^q}{u^p}-\frac{\alpha m_{1}L}{2}\right)u^{P+1}\text{d}x}+{\displaystyle {\int }_{\Omega }\left(\theta {\tau }^2\frac{u^{2q}}{u^{p+1}}-\frac{\alpha k_{1}L}{2}\right)u^{p+1}\text{d}x}\end{array}$ (1.8)

如果 $\phi \left(t\right)$ 无界，则上述不等式得到 ${\phi }^{\prime }\left(t\right)\le 0$ ，这是矛盾的，所以我们得到一个结论： $\phi \left(t\right)$ 始终有界。

3. 上界估计

定理2：设 $\Omega \in R_n\left(n\ge 2\right)$ 为有界区域，含时系数 $k_1\left(t\right)$ 和 $k_2\left(t\right)$ 为非负非递减微分函数，非负可积函数f和g满足条件：

$\xi f\left(\xi \right)\ge \gamma F\left(\xi \right),\xi g\left(\xi \right)\ge \gamma G\left(\xi \right),\forall \xi >0,\gamma >2,$

$F\left(\xi \right):={\displaystyle {\int }_0^{\xi }f\left(s\right)\text{d}s},G\left(\xi \right):={\displaystyle {\int }_0^{\xi }g\left(s\right)\text{d}s}$

$k_1\left(t\right)\ge {\sigma }_1>0,k_2\left(t\right)\ge {\sigma }_2>0$

${k^{\prime }}_1\left(t\right)>0,{k^{\prime }}_2\left(t\right)>0$

另外，我们假设 $\theta \left(0\right)>0$

$\theta \left(t\right):=k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }G\left(u\right)\text{d}s}-\frac{1}{2}{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}\text{d}x}}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)F\left(u\right)\text{d}x}$

问题(1.1)的非负经典解 $u\left(x,t\right)$ 在时间 $t^{*}\le T$ 时爆破，其中 $T=2\phi \left(0\right)/\gamma \left(\gamma -2\right)\theta \left(0\right)$ 。

证明：首先对(1.1)乘u并在 $\Omega$ 上积分

$\begin{array}{c}{\displaystyle {\int }_{\Omega }u{u}_t\text{d}x}=\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}\\ ={\displaystyle {\int }_{\partial \Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}\left(a_{ij}\left(x\right)u_{x_i}\right)v_j}\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)uf\left(u\right)\text{d}x}\\ =k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }ug\left(u\right)\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)uf\left(u\right)\text{d}x}\end{array}$ (2.1)

定义 $\phi \left(t\right):=\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}$ ，并由定理1中条件可知(2.1)变为：

$\begin{array}{c}{\phi }^{\prime }\left(t\right)=k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }ug\left(u\right)\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)uf\left(u\right)\text{d}x}\\ \ge \gamma k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }G\left(u\right)\text{d}s}+\gamma k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)F\left(u\right)\text{d}x}-\frac{1}{2}\gamma {\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}\end{array}$ (2.2)

受上式启发，定义 $\theta (\; t\; )$

$\theta \left(t\right):=k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }G\left(u\right)\text{d}s}-\frac{1}{2}{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)F\left(u\right)\text{d}x}$ (2.3)

由(2.2)和(2.3)知

${\phi }^{\prime }\left(t\right)\ge \gamma \theta \left(t\right)$ (2.4)

因为 ${k^{\prime }}_1\left(t\right)>0$ 和 ${k^{\prime }}_2\left(t\right)>0$ ，可得

$\begin{array}{c}{\theta }^{\prime }\left(t\right)=k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }g\left(u\right)u_t}\text{d}s+{k^{\prime }}_2\left(t\right){\displaystyle {\int }_{\partial \Omega }G\left(u\right)\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_{j}t}}\text{d}x}\\ +k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)f\left(u\right)u_t\text{d}x}+{k^{\prime }}_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)F\left(u\right)\text{d}x}\\ \ge k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }g\left(u\right)u_t}\text{d}s-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_{j}t}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)f\left(u\right)u_t\text{d}x}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{u}_t{\left(a_{ij}\left(x\right)u_{x_i}\right)}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)f\left(u\right)u_t\text{d}x}\\ ={\displaystyle {\int }_{\Omega }{u}_t\left[{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{u}_t{\left(a_{ij}\left(x\right)u_{x_i}\right)}_{x_j}}+k_1\left(t\right)a\left(x\right)f\left(u\right)\right]\text{d}x}\\ \ge {\displaystyle {\int }_{\Omega }{u}_t^2\text{d}x}\ge 0\end{array}$ (2.5)

假设 $\theta \left(0\right)>0$ ，则对于 $\forall t\in \left(0,t^{*}\right)$ 都有 $\theta \left(t\right)>0$ 。

由 $\phi \left(t\right):=\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^2}\text{d}x$ ，由(2.5)和Holder不等式可知

${\left({\phi }^{\prime }\left(t\right)\right)}^2\le {\displaystyle {\int }_{\Omega }{u}^2\text{d}x}{\displaystyle {\int }_{\Omega }{u}_t^2\text{d}x}\le 2\phi \left(t\right){\theta }^{\prime }\left(t\right)$ (2.6)

将(2.4)代入(2.6)可知

$\phi \left(t\right){\theta }^{\prime }\left(t\right)\ge \frac{\gamma }{2}{\phi }^{\prime }\left(t\right)\theta \left(t\right)$ (2.7)

两边同乘 ${\phi }^{-\left(\frac{\gamma }{2}+1\right)}$ 可得

${\left(\theta \left(t\right){\phi }^{-\frac{\gamma }{2}}\left(t\right)\right)}^{\prime }={\phi }^{-\left(\frac{\gamma }{2}+1\right)}\left(\phi \left(t\right){\theta }^{\prime }\left(t\right)-\frac{\gamma }{2}{\phi }^{\prime }\left(t\right)\theta \left(t\right)\right)\ge 0$ (2.8)

对(2.8)在 $\left(0,t\right)$ 积分可得

$\theta \left(t\right){\phi }^{-\frac{\gamma }{2}}\left(t\right)\ge \theta \left(0\right){\phi }^{-\frac{\gamma }{2}}\left(0\right):=H$ (2.9)

其中H为常数。

由(2.9)可知

$\theta \left(t\right)\ge H{\phi }^{\frac{\gamma }{2}}\left(t\right)$ (2.10)

将(2.10)代入(2.4)可知

${\phi }^{\prime }\left(t\right)\ge \gamma \theta \left(t\right)\ge \gamma H{\phi }^{\frac{\gamma }{2}}\left(t\right)$ (2.11)

则有

${\phi }^{-\frac{\gamma }{2}}\left(t\right){\phi }^{\prime }\left(t\right)\ge \gamma H$ (2.12)

整理可得

$\phi \left(t\right)\ge \frac{1}{{\left[{\phi }^{-\frac{\gamma }{2}+1}\left(0\right)-\frac{\gamma -2}{2}\gamma Ht\right]}^{\frac{2}{\gamma -2}}}$ (2.13)

$\underset{t\to t^{*}}{\lim }\phi \left(t\right)=+\infty$ (2.14)

则有 $t^{*}\le T=\frac{2\phi \left(0\right)}{\gamma \left(\gamma -2\right)\theta \left(0\right)}$ 。

则定理2证毕。

4. 下届估计

引理2 [6] ：假设 $\Omega \in R^n\left(n\ge 3\right)$ 是一个 $n-1$ 正交方向上的有界星形区域，设 $w\left(x\right)$ 是 $\Omega$ 中定义的一个非负 $C^1$ 函数，则对于任意常数 $\sigma \ge 1$ ，有下列不等式成立：

${\displaystyle \underset{\Omega }{\int }{w}^{\left(1+\frac{1}{2^{n-2}}\right)\sigma }}\text{d}x\le {\left(1+2d\right)}^{n-3}\times {\left[\frac{n}{2{\rho }_0}{\displaystyle \underset{\Omega }{\int }{w}^{\sigma }}\text{d}x+\frac{\sigma }{2}\left(1+\frac{d}{{\rho }_0}\right){\displaystyle \underset{\Omega }{\int }{w}^{\sigma -1}\left|\nabla w\right|\text{d}x}\right]}^{1+\frac{1}{2^{n-2}}}$

其中 ${\rho }_0:=\underset{x\in \partial \Omega }{\min }\left(x\cdot v\right)>0$ ， $d:=\underset{x\in \partial \Omega }{\max }\left|x\right|$

定理3：假设含时系数 $k_1\left(t\right)$ 和 $k_2\left(t\right)$ 是一个非负非递减可微函数，非负可积函数f和g满足下列条件

$\begin{array}{l}\gamma \bar{F}\left(s\right)\le f\left(s\right)\le \lambda s^{1+\frac{1}{2^{n-3}}},\bar{F}\left(s\right)=\frac{1}{s}{\displaystyle {\int }_0^{s}f\left(\xi \right)\text{d}\xi }\\ \gamma \bar{G}\left(s\right)\le k_{2}g\left(s\right)\le \mu s^{1+\frac{1}{2^{n-2}}},\bar{G}\left(s\right)=\frac{1}{s}{\displaystyle {\int }_0^{s}g\left(\xi \right)\text{d}\xi }\end{array}$

$\forall s>0$

$\underset{t>0}{\sup }{k}_1\left(t\right)=M_1<+\infty ,\underset{t>0}{\sup }{k}_2\left(t\right)=M_2<+\infty$

$\underset{x>0}{\sup }a\left(x\right)=L<+\infty$

其中 $\lambda >0$ ， $\mu >0$ ， $\gamma >2$ ， $M_1,M_2,L$ 是常数。加权函数 $a\left(x\right)$ 满足(A₁~A₄)则 $u\left(x,t\right)$ 在有限时刻 $t^{*}$ 发生爆破，且有

$t^{*}\ge {\displaystyle {\int }_{\phi \left(0\right)}^{\infty }\frac{\text{d}\eta }{C_1\eta +C_2{\eta }^{\frac{2^{n-2}+1}{2^{n-2}}}+C_3{\eta }^{\frac{2^{n-2}+1}{2^{n-2}-1}}}}$ 。

由定理2可知 $\phi \left(t\right):=\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}$ 在有限时间 $t^{*}$ 爆破。

对 $\phi \left(t\right)$ 求导并用格林公式，可知

${\phi }^{\prime }\left(t\right)={\displaystyle {\int }_{\Omega }u{u}_t\text{d}x}=k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }ug\left(u\right)\text{d}s}-{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i,j=1}{\overset{n}{\sum }}{a}_{ij}\left(x\right)u_{x_i}{u}_{x_j}}\text{d}x}+k_1\left(t\right){\displaystyle {\int }_{\Omega }a\left(x\right)f\left(u\right)u\text{d}x}$ (3.1)

由定理3中的条件可知

${\phi }^{\prime }\left(t\right)\le \lambda L{k}_1\left(t\right){\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}\text{d}x}-\theta {\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}+\mu k_2\left(t\right){\displaystyle {\int }_{\partial \Omega }{u}^{2+\frac{1}{2^{n-2}}}}\text{d}s$ (3.2)

由引理2知

${\displaystyle {\int }_{\partial \Omega }{u}^{2+\frac{1}{2^{n-2}}}}\text{d}s\le \frac{n}{{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-2}}}\text{d}x}+\frac{\left(2^{n-1}+1\right)d}{2^{n-2}{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^{1+\frac{1}{2^{n-2}}}\left|\nabla u\right|\text{d}x}$ (3.3)

由柯西不等式知

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-2}}}\text{d}x}={\displaystyle {\int }_{\Omega }{u}^{1+\frac{1}{2^{n-2}}}u\text{d}x}\\ \le \frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^2}\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}\text{d}x}+\phi \left(t\right)\end{array}$ (3.4)

${\displaystyle {\int }_{\Omega }{u}^{1+\frac{1}{2^{n-2}}}\left|\nabla u\right|\text{d}x}\le \frac{1}{2\tau }{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}}\text{d}x+\frac{\tau }{2}{\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}$ (3.5)

将(3.3)~(3.5)代入(3.2)得

${\phi }^{\prime }\left(t\right)\le \frac{\mu n{k}_2\left(t\right)}{{\rho }_0}\phi \left(t\right)+J{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}}\text{d}x+\left(\frac{\left(2^{n-1}+1\right)\text{d}\tau \mu k_2\left(t\right)}{2^{n-1}{\rho }_0}-\theta \right){\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}$ (3.6)

其中 $J=\lambda L{k}_1\left(t\right)+\frac{\mu n{k}_2\left(t\right)}{2{\rho }_0}+\frac{\left(2^{n-1}+1\right)d\mu k_2\left(t\right)}{2^{n-1}\tau {\rho }_0}$

由引理知

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{u}^{2+\frac{1}{2^{n-3}}}}\text{d}x={\displaystyle {\int }_{\Omega }{u}^{\left(1+\frac{1}{2^{n-2}}\right)2}}\text{d}x\\ \le {\left(1+2d\right)}^{n-3}{\left(\frac{n}{2{\rho }_0}{\displaystyle {\int }_{\Omega }{u}^2\text{d}x}+\left(1+\frac{d}{{\rho }_0}\right){\displaystyle {\int }_{\Omega }u\left|\nabla u\right|\text{d}x}\right)}^{1+\frac{1}{2^{n-2}}}\\ \le {\left(1+2d\right)}^{n-3}{\left(\frac{n}{{\rho }_0}\phi +\left(1+\frac{d}{{\rho }_0}\right){\left({\displaystyle {\int }_{\Omega }{u}^2}\text{d}x\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{1}{2}}\right)}^{1+\frac{1}{2^{n-2}}}\end{array}$

$\begin{array}{c}={\left(1+2d\right)}^{n-3}{\left(\frac{n}{{\rho }_0}\phi \left(t\right)+\left(1+\frac{d}{{\rho }_0}\right)2^{\frac{1}{2}}{\phi }^{\frac{1}{2}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{1}{2}}\right)}^{1+\frac{1}{2^{n-2}}}\\ \le {\left(1+2d\right)}^{n-3}{2}^{\frac{1}{2n-2}}\left[{\left(\frac{n}{{\rho }_0}\right)}^{1+\frac{1}{2^{n-2}}}{\phi }^{1+\frac{1}{2^{n-2}}}\text{+}{\left(1+\frac{d}{{\rho }_0}\right)}^{1+\frac{1}{2^{n-2}}}{2}^{\frac{2^{n-2}+1}{2^{n-1}}}{\phi }^{\frac{2^{n-2}+1}{2^{n-1}}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{2^{n-2}+1}{2^{n-1}}}\right]\end{array}$ (3.7)

$\begin{array}{c}{\phi }^{\frac{2^{n-2}+1}{2^{n-1}}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{2^{n-2}+1}{2^{n-1}}}={\epsilon }^{\frac{2^{n-2}+1}{2^{n-1}}}{\left({\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{2^{n-2}+1}{2^{n-1}}}{\epsilon }^{-\frac{2^{n-2}+1}{2^{n-1}}}{\phi }^{\frac{2^{n-2}+1}{2^{n-1}}}\\ ={\left(\epsilon {\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}\right)}^{\frac{2^{n-2}+1}{2^{n-1}}}{\left({\epsilon }^{-\frac{2^{n-2}+1}{2^{n-2}-1}}{\phi }^{\frac{2^{n-2}+1}{2^{n-2}-1}}\right)}^{\frac{2^{n-2}-1}{2^{n-1}}}\\ \le \frac{2^{n-2}+1}{2^{n-1}}\epsilon {\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}+\frac{2^{n-2}-1}{2^{n-1}}{\epsilon }^{-\frac{2^{n-2}+1}{2^{n-2}-1}}{\phi }^{\frac{2^{n-2}+1}{2^{n-2}-1}}\end{array}$ (3.8)

将(3.7)和(3.8)代入(3.6)得

${\phi }^{\prime }\left(t\right)\le C_1\phi +C_2{\phi }^{1+\frac{1}{2^{n-2}}}+C_3{\phi }^{\frac{2^{n-2}+1}{2^{n-2}-1}}+C_4{\displaystyle {\int }_{\Omega }{\left|\nabla u\right|}^2\text{d}x}$ (3.9)

其中

$\begin{array}{c}{C}_1=\frac{\mu n{M}_2}{{\rho }_0}\\ C_2=J_1{\left(1+2d\right)}^{n-3}{2}^{\frac{1}{2^{n-2}}}{\left(\frac{n}{{\rho }_0}\right)}^{1+\frac{1}{2^{n-2}}}\\ C_3=J_1{\left(1+2d\right)}^{n-3}{2}^{\frac{1}{2^{n-2}}}{\left(1+\frac{d}{{\rho }_0}\right)}^{1+\frac{1}{2^{n-2}}}{2}^{\frac{2^{n-2}+1}{2^{n-1}}}\frac{2^{n-2}-1}{2^{n-1}}{\epsilon }^{-\frac{2^{n-2}+1}{2^{n-2}-1}}\\ C_4=J_1{\left(1+2d\right)}^{n-3}{\left(1+\frac{d}{{\rho }_0}\right)}^{1+\frac{1}{2^{n-2}}}{2}^{\frac{2^{n-2}+1}{2^{n-1}}}\frac{2^{n-2}+1}{2^{n-1}}\epsilon +\frac{\left(2^{n-1}+1\right)d\tau \mu M_2}{2^{n-1}{\rho }_0}-\theta \end{array}$

其中 $J_1=\lambda L{M}_1+\frac{\mu n{M}_2}{2{\rho }_0}+\frac{\left(2^{n-1}+1\right)d\mu M_2}{2^{n-1}\tau {\rho }_0}$

$C_4$ 等价于 $\tau$ 的二次多项式，如果 $\epsilon$ 足够小，就可以得到正的 $\tau$ ，使得 $C_4=0$ 。

${\phi }^{\prime }\left(t\right)\le C_1\phi +C_2{\phi }^{1+\frac{1}{2^{n-2}}}+C_3{\phi }^{\frac{2^{n-2}+1}{2^{n-2}-1}}:=\Gamma \left(\phi \left(t\right)\right)$ (3.10)

${\left({\displaystyle {\int }_{\phi \left(0\right)}^{\phi \left(t\right)}\frac{\text{d}\eta }{\Gamma \left(\eta \right)}}\right)}^{\prime }=\frac{{\phi }^{\prime }\left(t\right)}{\Gamma \left(\phi \right)}\le 1$ (3.11)

对(3.11)在 $\left[0,t\right]$ 积分得

$t\ge {\displaystyle {\int }_{\phi \left(0\right)}^{\phi \left(t\right)}\frac{\text{d}\eta }{\Gamma \left(\eta \right)}}$ (3.12)

$\underset{t\to t^{\text{*}}}{\lim }\phi \left(t\right)=\infty$

则有

$t^{*}\ge {\displaystyle {\int }_{\phi \left(0\right)}^{\infty }\frac{\text{d}\eta }{\Gamma \left(\eta \right)}}={\displaystyle {\int }_{\phi \left(0\right)}^{\infty }\frac{\text{d}\eta }{C_1\eta +C_2{\eta }^{\frac{2^{n-2}+1}{2^{n-2}}}+C_3{\eta }^{\frac{2^{n-2}+1}{2^{n-2}-1}}}}$ (3.13)

则定理3证毕。

5. 结论

本文主要研究了一类具有加权函数和含时系数的发散型反应扩散方程的爆破问题，通过构造合适的辅助函数和微分不等式技术，结合Cauchy不等式和Holder不等式，得到了保证解全局存在的非线性条件和爆破时间的上下界估计。

基金项目

辽宁省教育厅高校科研项目资助(编号：LJKMZ20220832)。