﻿ 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

Advances in Applied Mathematics
Vol.06 No.08(2017), Article ID:22808,10 pages
10.12677/AAM.2017.68117

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

Yingtao Li1,2, Zhigang Pan1

1School of Mathematics, Southwest Jiaotong University, Chengdu Sichuan

2College of Mathematics, Sichuan University, Chengdu Sichuan

Received: Nov. 7th, 2017; accepted: Nov. 21st, 2017; published: Nov. 27th, 2017

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.

Keywords:External Diffusion, Nonlinear Boundary, Blow-Up Solution, Blow-Up Time

1西南交通大学数学学院，四川 成都

2四川大学数学学院，四川 成都

Copyright © 2017 by authors and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

$\left\{\begin{array}{l}{\left(h\left(u\right)\right)}_{t}=\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)+g\left(t\right)f\left(u\right)\text{in}\text{\hspace{0.17em}}D×\left(0,T\right)\\ \frac{\partial u}{\partial n}=k\left(x,u,t\right)\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\partial D×\left(0,T\right)\\ u\left(x,0\right)={u}_{0}\left(x\right)>0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\overline{D}\end{array}$ (1.1)

$\left\{\begin{array}{l}{\left(h\left(u\right)\right)}_{t}=\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)+g\left(t\right)f\left(u\right)\text{}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}D×\left(0,T\right)\\ u\left(x,t\right)=k\left(x,u,t\right)\text{on}\text{\hspace{0.17em}}\partial D×\left(0,T\right)\\ u\left(x,0\right)={u}_{0}\left(x\right)>0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\overline{D}\end{array}$ (1.2)

2. 爆破解

2.1. 具Neumann边界条件下的爆破解

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

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

(ii)

${\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s<{\int }_{0}^{+\infty }g\left(t\right)\text{d}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{M}_{0}=\underset{\overline{D}}{\mathrm{max}}{u}_{0}\left(x\right)$ (2.1.3)

(iii)

$\underset{\overline{D}}{\mathrm{min}}\nabla \cdot \left(a\left({u}_{0},0\right)b\left(x\right)\nabla {u}_{0}\left(x\right)\right)\ge 0$ (2.1.4)

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

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

$P\left(z\right)={\int }_{z}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s,\text{}\forall z\ge {M}_{0}$ (2.1.6)

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

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

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

$\begin{array}{c}{P}_{t}={\left[{h}^{\prime }\left(u\right){u}_{t}-g\left(t\right)f\left(u\right)\right]}_{t}={\left[{\left({h}^{\prime }\left(u\right)\right)}_{t}-g\left(t\right)f\left(u\right)\right]}_{t}\text{}\\ ={\left[\nabla \left(a\left(u,t\right)b\left(x\right)\nabla u\right)\right]}_{t}={\left(ab\Delta u+{a}_{u}b{|\nabla u|}^{2}+a\nabla b\cdot \nabla u\right)}_{t}\\ ={a}_{u}b{u}_{t}\Delta u+{a}_{t}b\Delta u+ab\Delta {u}_{t}+{a}_{uu}b{|\nabla u|}^{2}{u}_{t}+{a}_{ut}b{|\nabla u|}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{a}_{u}b\nabla u\cdot \nabla {u}_{t}+{a}_{u}{u}_{t}\nabla b\cdot \nabla u+{a}_{t}\nabla b\cdot \nabla u+a\nabla b\cdot \nabla {u}_{t}\end{array}$ (2.1.10)

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

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

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

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

$\begin{array}{l}\frac{ab}{{h}^{\prime }}\Delta P+\left[2b{\left(\frac{a}{{h}^{\prime }}\right)}_{u}\nabla u+\frac{a}{{h}^{\prime }}\nabla b\right]\cdot \nabla P-{P}_{t}\\ =\left(\frac{ab{h}^{‴}}{{h}^{\prime }}-{a}_{uu}b+\frac{{\left({a}_{u}\right)}^{2}b}{a}+\frac{{a}_{u}b{h}^{″}}{{h}^{\prime }}-\frac{2ab{\left({h}^{″}\right)}^{2}}{{\left({h}^{\prime }\right)}^{2}}\right){|\nabla u|}^{2}{u}_{t}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{+}\left(\frac{2ab{h}^{″}g{f}^{\prime }}{{\left({h}^{\prime }\right)}^{2}}-\frac{{a}_{u}bg{f}^{\prime }}{{h}^{\prime }}-\frac{abg{f}^{″}}{{h}^{\prime }}+\frac{{a}_{u}{a}_{t}b}{a}-{a}_{ut}b\right){|\nabla u|}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({h}^{″}-\frac{{a}_{u}{h}^{\prime }}{a}\right){\left({u}_{t}\right)}^{2}+\left(\frac{{a}_{u}gf}{a}-\frac{gf{h}^{″}}{{h}^{\prime }}-g{f}^{\prime }-\frac{{a}_{t}{h}^{\prime }}{a}\right){u}_{t}+\frac{{g}^{2}f{f}^{\prime }}{{h}^{\prime }}+\frac{{a}_{t}gf}{a}\end{array}$ (2.1.15)

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

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

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

$P\left(x,0\right)=\nabla \cdot \left(a\left({M}_{0},0\right)b\left(x\right)\nabla {M}_{0}\right)=0$ (2.1.19)

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

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

${\int }_{0}^{t}\frac{{h}^{\prime }\left(u\right)}{f\left(u\right)}{u}_{t}\text{d}t={\int }_{{M}_{0}}^{u\left({x}_{0},t\right)}\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\ge {\int }_{0}^{t}g\left(t\right)\text{d}t$ (2.1.22)

${\int }_{0}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}{u}_{t}\text{d}t\ge {\int }_{0}^{+\infty }g\left(t\right)\text{d}t$ (2.1.23)

$\underset{t\to T}{\mathrm{lim}}{\int }_{{M}_{0}}^{u\left({x}_{0},t\right)}\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}t\ge \underset{t\to T}{\mathrm{lim}}{\int }_{0}^{t}g\left(t\right)\text{d}t$

${\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\ge {\int }_{0}^{T}g\left(t\right)\text{d}t=\eta \left(T\right)$

$T\le {\eta }^{-1}\left({\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\right)$

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

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

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

2.2. 具Dirichlet边界条件下的爆破解

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

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

(ii)

${\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s<{\int }_{0}^{+\infty }g\left(t\right)\text{d}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{M}_{0}=\underset{\overline{D}}{\mathrm{max}}{u}_{0}\left(x\right)$ (2.2.3)

(iii)

$\underset{\overline{D}}{\mathrm{min}}\nabla \cdot \left(a\left({u}_{0},0\right)b\left(x\right)\nabla {u}_{0}\left(x\right)\right)=0$ (2.2.4)

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

$u\left(x,t\right)\le {Q}^{-1}\left({\int }_{t}^{T}g\left(t\right)\text{d}t\right),\text{\hspace{0.17em}}\forall \left(x,t\right)\in \overline{D}$ (2.2.5)

$Q\left(z\right)={\int }_{u\left(x,t\right)}^{u\left(x,s\right)}\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s,\text{\hspace{0.17em}}\forall z\ge {M}_{0}$ (2.2.6)

$\Delta u+\frac{a}{{h}^{\prime }}\nabla b\cdot \nabla u-{u}_{t}=-\frac{gf}{{h}^{\prime }}-\frac{{a}_{u}b}{{h}^{\prime }}{|\nabla u|}^{2}<0$

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

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

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

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

$\begin{array}{c}{P}_{t}={\left[{h}^{\prime }\left(u\right){u}_{t}-g\left(t\right)f\left(u\right)\right]}_{t}={\left[{\left({h}^{\prime }\left(u\right)\right)}_{t}-g\left(t\right)f\left(u\right)\right]}_{t}={\left[\nabla \cdot \left(a\left(u,t\right)b\left(x\right)\nabla u\right)\right]}_{t}\\ ={\left(ab\Delta u+{a}_{u}b{|\nabla u|}^{2}+a\nabla b\cdot \nabla u\right)}_{t}={a}_{u}b{u}_{t}\Delta u+{a}_{t}b\Delta u+ab\Delta {u}_{t}+{a}_{uu}b{|\nabla u|}^{2}{u}_{t}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{a}_{ut}b{|\nabla u|}^{2}+2{a}_{u}b\nabla u\cdot \nabla {u}_{t}+{a}_{u}{u}_{t}\nabla b\cdot \nabla u+{a}_{t}\nabla b\cdot \nabla u+a\nabla b\cdot \nabla {u}_{t}\end{array}$ (2.2.10)

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

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

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

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

$\begin{array}{l}\frac{ab}{{h}^{\prime }}\Delta Q+\left[2b{\left(\frac{a}{{h}^{\prime }}\right)}_{u}\nabla u+\frac{a}{{h}^{\prime }}\nabla b\right]\cdot \nabla Q-{Q}_{t}\\ =\left(\frac{ab{h}^{‴}}{{h}^{\prime }}-{a}_{uu}b+\frac{{\left({a}_{u}\right)}^{2}b}{a}+\frac{{a}_{u}b{h}^{″}}{{h}^{\prime }}-\frac{2ab{\left({h}^{″}\right)}^{2}}{{\left({h}^{\prime }\right)}^{2}}\right){|\nabla u|}^{2}{u}_{t}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\frac{2ab{h}^{″}g{f}^{\prime }}{{\left({h}^{\prime }\right)}^{2}}-\frac{{a}_{u}bg{f}^{\prime }}{{h}^{\prime }}-\frac{abg{f}^{″}}{{h}^{\prime }}+\frac{{a}_{u}{a}_{t}b}{a}-{a}_{ut}b\right){|\nabla u|}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({h}^{″}-\frac{{a}_{u}{h}^{\prime }}{a}\right){\left({u}_{t}\right)}^{2}+\left(\frac{{a}_{u}gf}{a}-\frac{gf{h}^{″}}{{h}^{\prime }}-g{f}^{\prime }-\frac{{a}_{t}{h}^{\prime }}{a}\right){u}_{t}+\frac{{g}^{2}f{f}^{\prime }}{{h}^{\prime }}+\frac{{a}_{t}gf}{a}\end{array}$ (2.2.15)

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

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

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

$Q\left(x,0\right)=\nabla \cdot \left(a\left({M}_{0},0\right)b\left(x\right)\nabla {M}_{0}\right)=0$ (2.2.19)

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

${\int }_{0}^{t}\frac{{h}^{\prime }\left(u\right)}{f\left(u\right)}{u}_{t}\text{d}t={\int }_{{M}_{0}}^{u\left({x}_{0},t\right)}\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\ge {\int }_{0}^{t}g\left(t\right)\text{d}t$ (2.2.22)

${\int }_{0}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}{u}_{t}\text{d}t\ge {\int }_{0}^{+\infty }g\left(t\right)\text{d}t$ (2.2.23)

$\underset{t\to T}{\mathrm{lim}}{\int }_{{M}_{0}}^{u\left({x}_{0},t\right)}\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}t\ge \underset{t\to T}{\mathrm{lim}}{\int }_{0}^{t}g\left(t\right)\text{d}t$

${\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\ge {\int }_{0}^{T}g\left(t\right)\text{d}t=\eta \left(T\right)$

$T\le {\eta }^{-1}\left({\int }_{{M}_{0}}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\right)$

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

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

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

3. 应用举例

$\left\{\begin{array}{l}{\left(u{\text{e}}^{u}\right)}_{t}=\nabla \cdot \left(\frac{1}{u+t}{\text{e}}^{{|x|}^{2}}\nabla u\right)+\left(1+t\right)\left({\text{e}}^{u}\cdot {u}^{3}\right)\text{}\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}D×\left(0,T\right)\\ \frac{\partial u}{\partial n}=-2{\text{e}}^{u-2+t{|x|}^{2}}\text{}\text{ }\text{ }\text{on}\text{\hspace{0.17em}}\partial D×\left(0,T\right)\\ u\left(x,0\right)=1+{|x|}^{2}\text{in}\text{\hspace{0.17em}}\overline{D}\end{array}$

$h\left(u\right)=u{\text{e}}^{u},\text{\hspace{0.17em}}a\left(u,t\right)=\frac{1}{u+t},\text{\hspace{0.17em}}b\left(x\right)={\text{e}}^{{|x|}^{2}},\text{\hspace{0.17em}}g\left(t\right)=1+t,\text{\hspace{0.17em}}f\left(u\right)={\text{e}}^{u}{u}^{3}$

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

${u}_{0}\left(x\right)=1+{|x|}^{2}$

$s={|x|}^{2}$

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

$T\le {\eta }^{-1}\left({\int }_{2}^{+\infty }\frac{{h}^{\prime }\left(s\right)}{f\left(s\right)}\text{d}s\right)={\eta }^{-1}\left(\frac{3}{4}\right)=0.6514$

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

$P\left(z\right)={\int }_{t}^{s}\frac{{h}^{\prime }\left(t\right)}{f\left(t\right)}\text{d}t={\int }_{t}^{s}\frac{1+t}{{t}^{3}}\text{d}t,\text{}\forall z\ge {M}_{0}$

$\left\{\begin{array}{l}{\left(u+{\text{e}}^{u}\right)}_{t}=\nabla \cdot \left({\text{e}}^{-u-t}{\text{e}}^{{|x|}^{2}}\nabla u\right)+t\left(2{\text{e}}^{u}+2\right)\text{}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}D×\left(0,T\right)\\ \frac{\partial u}{\partial n}={\text{e}}^{\frac{1}{2}\left(u-2\right)}+{\text{e}}^{u-2+t{|x|}^{2}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\partial D×\left(0,T\right)\\ u\left(x,0\right)=1+{|x|}^{2}\text{in}\text{\hspace{0.17em}}\text{D ¯}\end{array}$

$h\left(u\right)=u+{\text{e}}^{u},\text{\hspace{0.17em}}a\left(u,t\right)={\text{e}}^{-u-t},\text{\hspace{0.17em}}b\left(x\right)={\text{e}}^{{|x|}^{2}},g\left(t\right)=t,\text{\hspace{0.17em}}f\left(u\right)=2{\text{e}}^{u}+2$

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

${u}_{0}\left(x\right)=1+{|x|}^{2}$

$s={|x|}^{2}$

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

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

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

$P\left(z\right)={\int }_{t}^{s}\frac{{h}^{\prime }\left(t\right)}{f\left(t\right)}\text{d}t={\int }_{t}^{s}\frac{1}{2}\text{d}t,\text{\hspace{0.17em}}\forall z\ge {M}_{0}$

A Class of Blow-Up Solution with External Diffusion and Nonlinear Boundary Flow Problems[J]. 应用数学进展, 2017, 06(08): 975-984. http://dx.doi.org/10.12677/AAM.2017.68117

1. 1. 王明新. 非线性抛物方程[M]. 北京: 科学出版社, 1997: 1-3.

2. 2. Juntang, D. and Shengjia, L. (2005) Blow-Up Solutions and Global Solutions for a Class of Quasilinear Parabolic Equations with Robin. Computers and Mathematics with Applications, 60, 670-679.

3. 3. Lingling, Z. and Hui, W. (2014) Global and Blow-Up Solutions for a Class of Nonlinear Parabolic Problems under Robin Boundary Condition. Hindawi Publishing Corporation Abstract and Applied Analysis, 7, 232-256.

4. 4. Juntang, D. and Baozhu, G. (2010) Blow-Up and Global Existence for Nonlinear Parabolic Equations with Neumann Boundary Conditions. Computers and Mathematics with Applications, 60, 234-239.

5. 5. 曹京瑞. 几类非线性抛物方程的整体解和爆破解[D]: [硕士学位论文]. 太原: 太原理工大学数学系, 2016.

6. 6. 叶其孝, 李正元, 王明新, 吴雅萍. 反应扩散方程理论[M]. 北京: 科学出版社, 2011: 1-450.