一类具有外部扩散及非线性边界流问题解的爆破

A Class of Blow-Up Solution with External Diffusion and Nonlinear Boundary Flow Problems

摘要: 本文研究了一类具有外部扩散及非线性边界流问题，利用构造辅助函数，结合极值原理，利用经典的微分不等式，分别得到了其在Neumman边界和Dirichlet边界下爆破解存在的充分条件，爆破时刻的上界估计，最后给出了定理在两个非线性问题中的具体应用。

Abstract: In this paper, we study a class of problems with external diffusion and nonlinear boundary flow. By using the construction auxiliary function and the maximum principle, the sufficient conditions for the existence of the blow-up solution under the Neumman boundary and the Dirichlet boundary are obtained respectively by using the classical differential inequality. The upper bound of the “blow-up time” is estimated, and finally the concrete application of the theorem in two nonlinear problems is given.

1. 引言

偏微分方程中，非线性抛物方程式具有重要意义，在非线性科学中，它也是重要的研究方向之一。非线性抛物方程本身的非线性性质使得其解有爆破奇性可以对现实世界中许多扩散现象有一个科学的描述，在化学、物理、工程领域，非线性方程的爆破解问题可以应用到物质扩散与热传导问题，反映系统的不稳定状态 [1] 。在描述自然现象如宇宙黑洞、热传导上、物质的扩散等问题上爆破解具有实际应用 [2] [3] [4] 。

本文我们首先考虑一类非线性热传导模型：

$\{\begin{array}{l}{\left(h\left(u\right)\right)}_t=\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)+g\left(t\right)f\left(u\right)\text{in}D\times \left(0,T\right)\\ \frac{\partial u}{\partial n}=k\left(x,u,t\right)\text{on}\partial D\times \left(0,T\right)\\ u\left(x,0\right)=u_0\left(x\right)>0\text{in}\bar{D}\end{array}$ (1.1)

其中 $D\subset R^N\left(N\ge 2\right)$ 是一个有界区域且具有光滑边界 $\partial D$ ， $\partial u/\partial n$ 是在边界 $\partial D$ 上向外的法向导数， $T$ 是最终时间， $\bar{D}$ 是有界区域 $D$ 的闭包。定义 $R^{+}=\left(0,+\infty \right)$ 。在本文中假定： $u_0\left(x\right)$ 是正的 $C^2\left(\bar{D}\right)$ 函数。 $a$ 是正的 $C^2\left({ℝ}^{+}\times {\overline{ℝ}}^{+}\right)$ 函数， $b$ 是正的 $C^3\left(R^{+}\right)$ 函数， $f$ 是正的 $C^2\left({ℝ}^{+}\right)$ 函数， $g$ 是正的 $C^2\left({\overline{ℝ}}^{+}\right)$ 函数，对于任意 $s\in {ℝ}^{+}$ ， $h^{\prime }\left(s\right)>0$ ， $k$ 是负的 $C^2\left({\overline{ℝ}}^{+}\right)$ 函数。 $\nabla \left(a\left(u,t\right)b\left(x\right)\nabla u\right)$ 为反应项， $g\left(t\right)f\left(u\right)$ 为扩散项， $\frac{\partial u}{\partial n}=k\left(x,u,t\right)$ 通过边界区域向区域内部的热流。该模型可以解释成一个具有外部扩散和非线性边界流的

混合介质中的热传播过程、化学反应过程等 [5] ，本文通过构造辅助函数与极值原理得到了方程爆破解存在的充分条件与爆破时刻的上界。

随后我们考虑这类非线性热传导模型在Dirichlet边界条件下的爆破解：

$\{\begin{array}{l}{\left(h\left(u\right)\right)}_t=\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)+g\left(t\right)f\left(u\right)\text{in}D\times \left(0,T\right)\\ u\left(x,t\right)=k\left(x,u,t\right)\text{on}\partial D\times \left(0,T\right)\\ u\left(x,0\right)=u_0\left(x\right)>0\text{in}\bar{D}\end{array}$ (1.2)

其中 $D\subset R^N\left(N\ge 2\right)$ 是一个有界区域且具有光滑边界 $\partial D$ ， $\partial u/\partial n$ 是在边界 $\partial D$ 上向外的法向导数， $T$ 是最终时间， $\bar{D}$ 是有界区域 $D$ 的闭包。定义 $R^{+}=\left(0,+\infty \right)$ 。在本文中假定： $u_0\left(x\right)$ 是正的 $C^2\left(\bar{D}\right)$ 函数。 $a$ 是正的 $C^2\left({ℝ}^{+}\times {\overline{ℝ}}^{+}\right)$ 函数， $b$ 是正的 $C^3\left(R^{+}\right)$ 函数， $f$ 是正的 $C^2\left({ℝ}^{+}\right)$ 函数， $g$ 是正的 $C^2\left({\overline{ℝ}}^{+}\right)$ 函数，对于任意 $s\in {ℝ}^{+}$ ， $h^{\prime }\left(s\right)>0$ ， $k$ 是正的 $C^2\left({\overline{ℝ}}^{+}\right)$ 函数。该方程为附加非线性边界条件 $u\left(x,t\right)=k\left(x,u,t\right)$ 和适当初值的热方程。该模型揭示了非线性扩散系统中非线性的相互作用 [6] 。本文得到了方程爆破解存在的充分条件与爆破时刻的上界。

2. 爆破解

2.1. 具Neumann边界条件下的爆破解

我们的主要结果是：

定理1.1 设 $u$ 是问题(1.1)的一个解，如果以下条件成立：

(i) 对于任意 $\left(s,t\right)\in {ℝ}^{+}\times {ℝ}^{+}$ ，

${\left(\frac{1}{a\left(s,t\right)}{\left(\frac{a\left(s,t\right)f\left(s\right)}{h^{\prime }\left(s\right)}\right)}_s\right)}_s+\frac{1}{g\left(t\right)}{\left(\frac{a_s\left(s,t\right)}{a\left(s,t\right)}\right)}_t\ge 0$ (2.1.2)

(ii)

${\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}<{\displaystyle {\int }_0^{+\infty }g\left(t\right)\text{d}t},M_0=\underset{\bar{D}}{\max }{u}_0\left(x\right)$ (2.1.3)

(iii)

$\underset{\bar{D}}{\min }\nabla \cdot \left(a\left(u_0,0\right)b\left(x\right)\nabla u_0\left(x\right)\right)\ge 0$ (2.1.4)

(iv) 则 $u\left(x,t\right)$ 是方程的一个爆破解，并且

$u\left(x,t\right)\le P^{-1}\left({\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}\right),\forall \left(x,t\right)\in \bar{D}$ (2.1.5)

其中

$P\left(z\right)={\displaystyle {\int }_z^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s},\forall z\ge M_0$ (2.1.6)

证明：构造辅助函数：

$P\left(x,t\right)=h^{\prime }\left(u\right)u_t-g\left(t\right)f\left(u\right)$ (2.1.7)

则，有

$\nabla P\left(x,t\right)=h^{″}{u}_t\nabla u+h^{\prime }\nabla u_t-g{f}^{\prime }\nabla u$ (2.1.8)

$\Delta P\left(x,t\right)=h^{‴}{\left|\nabla u\right|}^2{u}_t+2h^{″}\nabla u\cdot \nabla u_t+h^{″}{u}_t\Delta u+h^{\prime }\Delta u_t-g{f}^{″}{\left|\nabla u\right|}^2-g{f}^{\prime }\Delta u$ (2.1.9)

由(2.1.6)

$\begin{array}{c}{P}_t={\left[h^{\prime }\left(u\right)u_t-g\left(t\right)f\left(u\right)\right]}_t={\left[{\left(h^{\prime }\left(u\right)\right)}_t-g\left(t\right)f\left(u\right)\right]}_t\\ ={\left[\nabla \left(a\left(u,t\right)b\left(x\right)\nabla u\right)\right]}_t={\left(ab\Delta u+a_{u}b{\left|\nabla u\right|}^2+a\nabla b\cdot \nabla u\right)}_t\\ =a_{u}b{u}_t\Delta u+a_{t}b\Delta u+ab\Delta u_t+a_{uu}b{\left|\nabla u\right|}^2{u}_t+a_{ut}b{\left|\nabla u\right|}^2\\ +2a_{u}b\nabla u\cdot \nabla u_t+a_u{u}_t\nabla b\cdot \nabla u+a_t\nabla b\cdot \nabla u+a\nabla b\cdot \nabla u_t\end{array}$ (2.1.10)

由(2.1.8)和(2.1.9)，我们得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta P-P_t=\left(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b\right){\left|\nabla u\right|}^2{u}_t+\left(\frac{2ab{h}^{″}}{h^{\prime }}-2a_{u}b\right)\nabla u\nabla u_t\\ +\left(\frac{ab{h}^{″}}{h^{\prime }}-a_{u}b\right)u_t\Delta u-\left(\frac{abg{f}^{″}}{h^{\prime }}+a_{ut}b\right){\left|\nabla u\right|}^2-\left(\frac{abg{f}^{\prime }}{h^{\prime }}+a_{t}b\right)\Delta u\\ -a_u{u}_t\nabla b\nabla u-a_t\nabla b\nabla u-a\nabla b\nabla u_t\end{array}$ (2.1.11)

由(2.1.5)我们得：

$\Delta u=\frac{h^{\prime }}{ab}{u}_t-\frac{a_u}{a}{\left|\nabla u\right|}^2-\frac{1}{b}\nabla b\cdot \nabla u-\frac{gf}{ab}$ (2.1.12)

将(2.1.11)代入(2.1.10)我们可以得到：

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta P-P_t=\left(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b+\frac{{\left(a_u\right)}^{2}b}{a}-\frac{a_{u}b{h}^{″}}{h^{\prime }}\right){\left|\nabla u\right|}^2{u}_t+\left(\frac{2ab{h}^{″}}{h^{\prime }}-2a_{u}b\right)\nabla u\cdot \nabla u_t\\ +\left(h^{″}-\frac{a_u{h}^{\prime }}{a}\right){\left(u_t\right)}^2-\frac{a{h}^{″}}{h^{\prime }}{u}_t\nabla b\cdot \nabla u+\left(\frac{a_{u}gf}{a}-\frac{gf{h}^{″}}{h^{\prime }}-g{f}^{\prime }-\frac{a_t{h}^{\prime }}{a}\right)u_t\\ +(\frac{a_{u}bg{f}^{\prime }}{h^{\prime }}+\frac{a_u{a}_{t}b}{a}-\frac{abg{f}^{″}}{h^{\prime }}-a_{ut}b){\left|\nabla u\right|}^2\text{+}\frac{ag{f}^{\prime }}{h^{\prime }}\nabla b\cdot \nabla u+\frac{g^{2}f{f}^{\prime }}{h^{\prime }}\\ +\frac{a_{t}gf}{a}-a\nabla b\cdot \nabla u_t\end{array}$ (2.1.13)

由(2.1.7)，我们有：

$\nabla u_t=\frac{1}{h^{\prime }}\nabla P-\frac{h^{″}}{h^{\prime }}{u}_t\nabla u+\frac{g{f}^{\prime }}{h^{\prime }}\nabla u$ (2.1.14)

将(2.1.13)代入(2.1.12)可得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta P+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla P-P_t\\ =(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b+\frac{{(a_u)}^{2}b}{a}+\frac{a_{u}b{h}^{″}}{h^{\prime }}-\frac{2ab{\left(h^{″}\right)}^2}{{\left(h^{\prime }\right)}^2}){\left|\nabla u\right|}^2{u}_t\\ \text{+}\left(\frac{2ab{h}^{″}g{f}^{\prime }}{{\left(h^{\prime }\right)}^2}-\frac{a_{u}bg{f}^{\prime }}{h^{\prime }}-\frac{abg{f}^{″}}{h^{\prime }}+\frac{a_u{a}_{t}b}{a}-a_{ut}b\right){\left|\nabla u\right|}^2\\ +\left(h^{″}-\frac{a_u{h}^{\prime }}{a}\right){\left(u_t\right)}^2+\left(\frac{a_{u}gf}{a}-\frac{gf{h}^{″}}{h^{\prime }}-g{f}^{\prime }-\frac{a_t{h}^{\prime }}{a}\right)u_t+\frac{g^{2}f{f}^{\prime }}{h^{\prime }}+\frac{a_{t}gf}{a}\end{array}$ (2.1.15)

由(2.1.6)我们得

$u_t=\frac{1}{h^{\prime }}P+\frac{gf}{h^{\prime }}$ (2.1.16)

将(2.1.15)代入(2.1.14)我们可得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta P+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla P+\{ab{\left[\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\right]}_u{\left|\nabla u\right|}^2\\ +\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\left(P+gf\right)+\frac{gf}{h^{\prime }}+\frac{a_t}{a}\}P-P_t=-abg\left\{{\left[\frac{1}{a}{\left(\frac{af}{h^{\prime }}\right)}_u\right]}_u+\frac{1}{g}{\left(\frac{a_u}{a}\right)}_t\right\}{\left|\nabla u\right|}^2\end{array}$ (2.1.17)

由(2.1.1)得等式右边(2.1.16)是非负的，即

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta P+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla P\\ +\{ab{\left[\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\right]}_u{\left|\nabla u\right|}^2+\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\left(P+gf\right)+\frac{gf}{h^{\prime }}+\frac{a_t}{a}\}P-P_t\ge 0\end{array}$ (2.1.18)

由(2.1.5)，我们知道

$P\left(x,0\right)=\nabla \cdot \left(a\left(M_0,0\right)b\left(x\right)\nabla M_0\right)=0$ (2.1.19)

与

$\frac{\partial P}{\partial n}=h^{″}{u}_t\frac{\partial u}{\partial n}+h^{\prime }\frac{\partial u_t}{\partial n}-g{f}^{\prime }\frac{\partial u}{\partial n}=h^{\prime }{\left(\frac{\partial u}{\partial n}\right)}_t=h^{\prime }k\left(x,u,t\right)$ (2.1.20)

结合(2.1.17)~(2.1.19)，通过极值原理，我们可知 $P\ge 0\text{in}\bar{D}\times \left[0,T\right]$

因此

$\frac{h^{\prime }\left(u\right)}{f\left(u\right)}{u}_t\ge g\left(t\right)$ (2.1.21)

设存在 $x_0\in \bar{D}$ ，使得 $u_0\left(0\right)=M_0$ 对(2.1.20)在 $\left[0,t\right]$ 上积分得：

${\displaystyle {\int }_0^t\frac{h^{\prime }\left(u\right)}{f\left(u\right)}{u}_t\text{d}t}={\displaystyle {\int }_{M_0}^{u\left(x_0,t\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_0^{t}g\left(t\right)\text{d}t}$ (2.1.22)

如果 $u$ 在有限时间 $t=T$ 内不爆破，当 $t\to +\infty$ 时(2.1.21)变形为

${\displaystyle {\int }_0^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}{u}_t\text{d}t}\ge {\displaystyle {\int }_0^{+\infty }g\left(t\right)\text{d}t}$ (2.1.23)

与条件矛盾，所以 $u$ 一定在有限时间 $t=T$ 内爆破，即当 $t\to T$ 时， $u\left(x,t\right)\to +\infty$

并且有下式成立

$\underset{t\to T}{\lim }{\displaystyle {\int }_{M_0}^{u\left(x_0,t\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}t}\ge \underset{t\to T}{\lim }{\displaystyle {\int }_0^{t}g\left(t\right)\text{d}t}$

因此

${\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_0^{T}g\left(t\right)\text{d}t}=\eta (T)$

我们可以得到

$T\le {\eta }^{-1}\left({\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\right)$

对每一个固定的 $x\in \bar{D}$ ，对(2.1.20)在 $\left[t,s\right]\left(0<t<s<T\right)$ 上积分得

$P\left(u\left(x,t\right)\right)\ge P\left(u\left(x,t\right)\right)-P\left(u\left(x,s\right)\right)={\displaystyle {\int }_{u\left(x,t\right)}^{u\left(x,s\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_t^{s}g\left(t\right)\text{d}t}$

因此，令 $s\to T$ 有

$P\left(u\left(x,t\right)\right)\ge {\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}$

由于 $P$ 是严格减函数，所以有

$u\left(x,t\right)\le P^{-1}\left({\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}\right)$

这里 $P^{-1}$ 是 $P$ 的反函数，证毕。

2.2. 具Dirichlet边界条件下的爆破解

定理1.2 设 $u$ 是问题(1.2)的一个解，如果以下条件成立：

(i) 对于任意 $\left(s,t\right)\in {ℝ}^{+}\times {ℝ}^{+}$ ，

${\left(\frac{1}{a\left(s,t\right)}{\left(\frac{a\left(s,t\right)f\left(s\right)}{h^{\prime }\left(s\right)}\right)}_s\right)}_s+\frac{1}{g\left(t\right)}{\left(\frac{a_s\left(s,t\right)}{a\left(s,t\right)}\right)}_t\ge 0$ (2.2.2)

(ii)

${\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}<{\displaystyle {\int }_0^{+\infty }g\left(t\right)\text{d}t},M_0=\underset{\bar{D}}{\max }{u}_0\left(x\right)$ (2.2.3)

(iii)

$\underset{\bar{D}}{\min }\nabla \cdot \left(a\left(u_0,0\right)b\left(x\right)\nabla u_0\left(x\right)\right)=0$ (2.2.4)

(iv) 则 $u\left(x,t\right)$ 是方程的一个爆破解，并且

$u\left(x,t\right)\le Q^{-1}\left({\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}\right),\forall \left(x,t\right)\in \bar{D}$ (2.2.5)

其中

$Q\left(z\right)={\displaystyle {\int }_{u\left(x,t\right)}^{u\left(x,s\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s},\forall z\ge M_0$ (2.2.6)

证明：由(2.2.1)可知

$\Delta u+\frac{a}{h^{\prime }}\nabla b\cdot \nabla u-u_t=-\frac{gf}{h^{\prime }}-\frac{a_{u}b}{h^{\prime }}{\left|\nabla u\right|}^2<0$

根据极值原理可知

$u\left(x,t\right)=k\left(x,u,t\right)\ge 0$

构造辅助函数：

$Q\left(x,t\right)=h^{\prime }\left(u\right)u_t-g\left(t\right)f\left(u\right)$ (2.2.7)

则，有

$\nabla Q\left(x,t\right)=h^{″}{u}_t\nabla u+h^{\prime }\nabla u_t-g{f}^{\prime }\nabla u$ (2.2.8)

$\Delta Q\left(x,t\right)=h^{‴}{\left|\nabla u\right|}^2{u}_t+2h^{″}\nabla u\cdot \nabla u_t+h^{″}{u}_t\Delta u+h^{\prime }\Delta u_t-g{f}^{″}{\left|\nabla u\right|}^2-g{f}^{\prime }\Delta u$ (2.2.9)

由(2.6)

$\begin{array}{c}{P}_t={\left[h^{\prime }\left(u\right)u_t-g\left(t\right)f\left(u\right)\right]}_t={\left[{\left(h^{\prime }\left(u\right)\right)}_t-g\left(t\right)f\left(u\right)\right]}_t={\left[\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)\right]}_t\\ ={\left(ab\Delta u+a_{u}b{\left|\nabla u\right|}^2+a\nabla b\cdot \nabla u\right)}_t=a_{u}b{u}_t\Delta u+a_{t}b\Delta u+ab\Delta u_t+a_{uu}b{\left|\nabla u\right|}^2{u}_t\\ +a_{ut}b{\left|\nabla u\right|}^2+2a_{u}b\nabla u\cdot \nabla u_t+a_u{u}_t\nabla b\cdot \nabla u+a_t\nabla b\cdot \nabla u+a\nabla b\cdot \nabla u_t\end{array}$ (2.2.10)

由(2.2.8)和(2.2.9)，我们得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta Q-Q_t=\left(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b\right){\left|\nabla u\right|}^2{u}_t+\left(\frac{2ab{h}^{″}}{h^{\prime }}-2a_{u}b\right)\nabla u\cdot \nabla u_t+\left(\frac{ab{h}^{″}}{h^{\prime }}-a_{u}b\right)u_t\Delta u\\ -\left(\frac{abg{f}^{″}}{h^{\prime }}+a_{ut}b\right){\left|\nabla u\right|}^2-\left(\frac{abg{f}^{\prime }}{h^{\prime }}+a_{t}b\right)\Delta u-a_u{u}_t\nabla b\cdot \nabla u-a_t\nabla b\cdot \nabla u\\ -a\nabla b\cdot \nabla u_t.\end{array}$ (2.2.11)

由(2.2.5)我们得：

$\Delta u=\frac{h^{\prime }}{ab}{u}_t-\frac{a_u}{a}{\left|\nabla u\right|}^2-\frac{1}{b}\nabla b\cdot \nabla u-\frac{gf}{ab}$ (2.2.12)

将(2.2.11)代入(2.2.10)我们可以得到：

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta Q-Q_t=\left(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b+\frac{{\left(a_u\right)}^{2}b}{a}-\frac{a_{u}b{h}^{″}}{h^{\prime }}\right){\left|\nabla u\right|}^2{u}_t+\left(\frac{2ab{h}^{″}}{h^{\prime }}-2a_{u}b\right)\nabla u\cdot \nabla u_t\\ +\left(h^{″}-\frac{a_u{h}^{\prime }}{a}\right){\left(u_t\right)}^2-\frac{a{h}^{″}}{h^{\prime }}{u}_t\nabla b\cdot \nabla u+\left(\frac{a_{u}gf}{a}-\frac{gf{h}^{″}}{h^{\prime }}-g{f}^{\prime }-\frac{a_t{h}^{\prime }}{a}\right)u_t\\ +(\frac{a_{u}bg{f}^{\prime }}{h^{\prime }}+\frac{a_u{a}_{t}b}{a}-\frac{abg{f}^{″}}{h^{\prime }}-a_{ut}b){\left|\nabla u\right|}^2+\frac{ag{f}^{\prime }}{h^{\prime }}\nabla b\cdot \nabla u+\frac{g^{2}f{f}^{\prime }}{h^{\prime }}\\ +\frac{a_{t}gf}{a}-a\nabla b\cdot \nabla u_t\end{array}$ (2.2.13)

由(2.2.7)，我们有：

$\nabla u_t=\frac{1}{h^{\prime }}\nabla Q-\frac{h^{″}}{h^{\prime }}{u}_t\nabla u+\frac{g{f}^{\prime }}{h^{\prime }}\nabla u$ (2.2.14)

将(2.2.13)代入(2.2.12)可得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta Q+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla Q-Q_t\\ =(\frac{ab{h}^{‴}}{h^{\prime }}-a_{uu}b+\frac{{\left(a_u\right)}^{2}b}{a}+\frac{a_{u}b{h}^{″}}{h^{\prime }}-\frac{2ab{\left(h^{″}\right)}^2}{{\left(h^{\prime }\right)}^2}){\left|\nabla u\right|}^2{u}_t\\ +\left(\frac{2ab{h}^{″}g{f}^{\prime }}{{\left(h^{\prime }\right)}^2}-\frac{a_{u}bg{f}^{\prime }}{h^{\prime }}-\frac{abg{f}^{″}}{h^{\prime }}+\frac{a_u{a}_{t}b}{a}-a_{ut}b\right){\left|\nabla u\right|}^2\\ +\left(h^{″}-\frac{a_u{h}^{\prime }}{a}\right){\left(u_t\right)}^2+\left(\frac{a_{u}gf}{a}-\frac{gf{h}^{″}}{h^{\prime }}-g{f}^{\prime }-\frac{a_t{h}^{\prime }}{a}\right)u_t+\frac{g^{2}f{f}^{\prime }}{h^{\prime }}+\frac{a_{t}gf}{a}\end{array}$ (2.2.15)

由(2.2.6)我们得

$u_t=\frac{1}{h^{\prime }}Q+\frac{gf}{h^{\prime }}$ (2.2.16)

将(2.2.15)代入(2.2.14)我们可得

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta Q+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla Q+\{ab{\left[\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\right]}_u{\left|\nabla u\right|}^2\\ +\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\left(Q+gf\right)+\frac{gf}{h^{\prime }}+\frac{a_t}{a}\}Q-Q_t=-abg\left\{{\left[\frac{1}{a}{\left(\frac{af}{h^{\prime }}\right)}_u\right]}_u+\frac{1}{g}{\left(\frac{a_u}{a}\right)}_t\right\}{\left|\nabla u\right|}^2\end{array}$ (2.2.17)

由(2.2.1)得等式右边(2.2.16)是非负的，即

$\begin{array}{l}\frac{ab}{h^{\prime }}\Delta Q+\left[2b{\left(\frac{a}{h^{\prime }}\right)}_u\nabla u+\frac{a}{h^{\prime }}\nabla b\right]\cdot \nabla Q\\ +\{ab{\left[\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\right]}_u{\left|\nabla u\right|}^2+\frac{1}{a}{\left(\frac{a}{h^{\prime }}\right)}_u\left(P+gf\right)+\frac{gf}{h^{\prime }}+\frac{a_t}{a}\}Q-Q_t\ge 0\end{array}$ (2.2.18)

由(2.2.5)，我们知道；

$Q\left(x,0\right)=\nabla \cdot \left(a\left(M_0,0\right)b\left(x\right)\nabla M_0\right)=0$ (2.2.19)

结合(2.2.17)~(2.2.19)，通过极值原理，我们可知 $Q\ge 0,\text{in}\bar{D}\times \left[0,T\right]$

因此

$\frac{h^{\prime }\left(u\right)}{f\left(u\right)}{u}_t\ge g\left(t\right)$ (2.2.21)

设存在 $x_0\in \bar{D}$ ，使得 $u_0\left(0\right)=M_0$ 对(2.2.20)在 $\left[0,t\right]$ 上积分得：

${\displaystyle {\int }_0^t\frac{h^{\prime }\left(u\right)}{f\left(u\right)}{u}_t\text{d}t}={\displaystyle {\int }_{M_0}^{u\left(x_0,t\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_0^{t}g\left(t\right)\text{d}t}$ (2.2.22)

如果 $u$ 在有限时间 $t=T$ 内不爆破，当 $t\to +\infty$ 时(2.2.21)变形为

${\displaystyle {\int }_0^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}{u}_t\text{d}t}\ge {\displaystyle {\int }_0^{+\infty }g\left(t\right)\text{d}t}$ (2.2.23)

与条件矛盾，所以 $u$ 一定在有限时间 $t=T$ 内爆破，即当 $t\to T$ 时， $u\left(x,t\right)\to +\infty$

并且有下式成立

$\underset{t\to T}{\lim }{\displaystyle {\int }_{M_0}^{u\left(x_0,t\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}t}\ge \underset{t\to T}{\lim }{\displaystyle {\int }_0^{t}g\left(t\right)\text{d}t}$

因此

${\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_0^{T}g\left(t\right)\text{d}t}=\eta (T)$

我们可以得到

$T\le {\eta }^{-1}\left({\displaystyle {\int }_{M_0}^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\right)$

对每一个固定的 $x\in \bar{D}$ ，对(2.2.20)在 $\left[t,s\right]\left(0<t<s<T\right)$ 上积分得

$Q\left(u\left(x,t\right)\right)\ge Q\left(u\left(x,t\right)\right)-Q\left(u\left(x,s\right)\right)={\displaystyle {\int }_{u\left(x,t\right)}^{u\left(x,s\right)}\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\ge {\displaystyle {\int }_t^{s}g\left(t\right)\text{d}t}$

因此，令 $s\to T$ 有

$Q\left(u\left(x,t\right)\right)\ge {\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}$

由于 $Q$ 是严格减函数，所以有

$u\left(x,t\right)\le Q^{-1}\left({\displaystyle {\int }_t^{T}g\left(t\right)\text{d}t}\right)$

这里 $Q^{-1}$ 是 $Q$ 的反函数，证毕。

3. 应用举例

这里，我们分别给出定理1.1和定理1.2的应用举例。

例1.1 设 $u$ 是下面问题的一个解：

$\{\begin{array}{l}{\left(u{\text{e}}^u\right)}_t=\nabla \cdot \left(\frac{1}{u+t}{\text{e}}^{{\left|x\right|}^2}\nabla u\right)+\left(1+t\right)\left({\text{e}}^u\cdot u^3\right)\text{in}D\times \left(0,T\right)\\ \frac{\partial u}{\partial n}=-2{\text{e}}^{u-2+t{\left|x\right|}^2}\text{on}\partial D\times \left(0,T\right)\\ u\left(x,0\right)=1+{\left|x\right|}^2\text{in}\bar{D}\end{array}$

其中 $D=\left\{x=\left(x_1,x_2,x_3\right)|{\left|x\right|}^2=x_1^2+x_2^2+x_3^2<1\right\}$ 是 $R^3$ 上的单位球，我们有

$h\left(u\right)=u{\text{e}}^u,a\left(u,t\right)=\frac{1}{u+t},b\left(x\right)={\text{e}}^{{\left|x\right|}^2},g\left(t\right)=1+t,f\left(u\right)={\text{e}}^u{u}^3$

$k\left(x,u,t\right)=-2{\text{e}}^{u-2+t{\left|x\right|}^2}$

$u_0\left(x\right)=1+{\left|x\right|}^2$

假设

$s={\left|x\right|}^2$

可知 $0<s<1$ 且

$\begin{array}{c}\beta =\underset{D}{\min }\nabla \left\{a\left(u_0,0\right)b\left(x\right)\nabla u_0\left(x\right)\right\}=\nabla \cdot \left\{\left(\frac{1}{1+{\left|x\right|}^2}\right){\text{e}}^{{\left|x\right|}^2}\nabla \left(1+{\left|x\right|}^2\right)\right\}\\ =\nabla \cdot \left\{\left(\frac{1}{1+s}\right){\text{e}}^s\nabla \left(1+s\right)\right\}=\frac{s{\text{e}}^s}{{\left(1+s\right)}^2}=0\end{array}$

根据定理2.1可知， $u$ 一定在有限时间 $T$ 爆破，并且有：

$T\le {\eta }^{-1}\left({\displaystyle {\int }_2^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\right)={\eta }^{-1}\left(\frac{3}{4}\right)=0.6514$

$u\left(x,t\right)\le P^{-1}\left(\left(T-t\right)+\frac{{\left(T-t\right)}^2}{2}\right)$

这里：

$P\left(z\right)={\displaystyle {\int }_t^s\frac{h^{\prime }\left(t\right)}{f\left(t\right)}\text{d}t}={\displaystyle {\int }_t^s\frac{1+t}{t^3}\text{d}t},\forall z\ge M_0$

其中 $P^{-1}$ 是 $P$ 的反函数。

例1.2 设 $u$ 是下面问题的一个解：

$\{\begin{array}{l}{\left(u+{\text{e}}^u\right)}_t=\nabla \cdot \left({\text{e}}^{-u-t}{\text{e}}^{{\left|x\right|}^2}\nabla u\right)+t\left(2{\text{e}}^u+2\right)\text{in}D\times \left(0,T\right)\\ \frac{\partial u}{\partial n}={\text{e}}^{\frac{1}{2}\left(u-2\right)}+{\text{e}}^{u-2+t{\left|x\right|}^2}\text{on}\partial D\times \left(0,T\right)\\ u\left(x,0\right)=1+{\left|x\right|}^2\text{in}\text{<mover accent="true"> <mi> D </mi> <mo> ¯ </mo> </mover>}\end{array}$

其中 $D=\left\{x=\left(x_1,x_2,x_3\right)|{\left|x\right|}^2=x_1^2+x_2^2+x_3^2<1\right\}$ 是 $R^3$ 上的单位球，我们有

$h\left(u\right)=u+{\text{e}}^u,a\left(u,t\right)={\text{e}}^{-u-t},b\left(x\right)={\text{e}}^{{\left|x\right|}^2},g\left(t\right)=t,f\left(u\right)=2{\text{e}}^u+2$

$k\left(x,u,t\right)={\text{e}}^{\frac{1}{2}\left(u-2\right)}+{\text{e}}^{u-2+t{\left|x\right|}^2}$

$u_0\left(x\right)=1+{\left|x\right|}^2$

假设

$s={\left|x\right|}^2$

可知 $0<s<1$ 且

$\begin{array}{c}\beta =\underset{D}{\min }\nabla \left\{a\left(u_0,0\right)b\left(x\right)\nabla u_0\left(x\right)\right\}\\ =\nabla \left\{{\text{e}}^{-{\left|x\right|}^2}{\text{e}}^{{\left|x\right|}^2}\nabla \left(1+{\left|x\right|}^2\right)\right\}=\nabla \left\{\nabla \left(1+s\right)\right\}=0\end{array}$

根据定理2.1可知， $u$ 一定在有限时间 $T$ 爆破，并且有：

$T\le {\eta }^{-1}\left({\displaystyle {\int }_2^{+\infty }\frac{h^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s}\right)={\eta }^{-1}\left(2\right)=2$

$u\left(x,t\right)\le P^{-1}\left(\frac{{\left(T-t\right)}^2}{2}\right)$

这里：

$P\left(z\right)={\displaystyle {\int }_t^s\frac{h^{\prime }\left(t\right)}{f\left(t\right)}\text{d}t}={\displaystyle {\int }_t^s\frac{1}{2}\text{d}t},\forall z\ge M_0$

其中 $P^{-1}$ 是 $P$ 的反函数。

基金项目

中央高校理科创新基金项目(No. A0920502051619-113)。