﻿ 一个非线性反应扩散方程的显式精确解 Explicit Exact Solutions for a Nonlinear Reaction-Diffusion Equation

Vol. 09  No. 01 ( 2020 ), Article ID: 33968 , 5 pages
10.12677/AAM.2020.91012

Explicit Exact Solutions for a Nonlinear Reaction-Diffusion Equation

Liangbo Lv

School of Mathematics and Information Science, Guangzhou University, Guangzhou Guangdong

Received: Dec. 26th, 2019; accepted: Jan. 8th, 2020; published: Jan. 15th, 2020

ABSTRACT

In this paper, a nonlinear reaction-diffusion equation is reduced to the standard heat equation by seeking appropriate nonlinear transformation and solves reduced equations with a variable separation approach, finally, through the inverse transformation of the nonlinear transformation, some new explicit exact solutions can be obtained.

Keywords:Nonlinear Reaction-Diffusion Equation, Variable Separation Approach, Nonlinear Transformation, Explicit Exact Solutions

1. 引言

(1.1)

2. 相应的非线性变换

${u}_{t}=\lambda {u}_{xx}+ku,$ (2.1)

$u=g\left(v\right),$ (2.2)

${u}_{t}={g}^{\prime }\left(v\right){v}_{t},$ (2.3)

${u}_{xx}={g}^{″}\left(v\right){v}_{xx}+{g}^{\prime }\left(v\right){v}_{xx},$ (2.4)

${g}^{\prime }\left(v\right){v}_{t}=\lambda \left({g}^{″}\left(v\right){v}_{x}^{2}+{g}^{\prime }\left(v\right){v}_{xx}\right)+g\left(v\right),$ (2.5)

(2.6)

$\left\{\begin{array}{l}{g}^{″}\left(v\right)/{g}^{\prime }\left(v\right)=\frac{2}{a-v}\\ {g}^{\prime }\left(v\right)/g\left(v\right)=\frac{1}{v}-\frac{1}{v-a}\end{array}$ (2.7)

$u=g\left(v\right)=\frac{v}{v-a},$ (2.8)

${\text{e}}^{-kt}{u}_{t}=\lambda {\text{e}}^{-kt}{u}_{xx}+k{\text{e}}^{-kt}u,$ (2.9)

$w={\text{e}}^{-kt}u,$ (2.10)

${w}_{t}=\lambda {w}_{xx}.$ (2.11)

$v=\frac{a{\text{e}}^{kt}w}{{\text{e}}^{kt}w\text{}-\text{}1},$ (2.12)

3. 显式精确解

$w\left(x,t\right)=X\left(x\right)T\left(t\right),$ (3.1)

${T}^{\prime }\left(t\right)X\left(x\right)=\lambda T\left(t\right){X}^{″}\left(x\right),$ (3.2)

$\frac{{T}^{\prime }\left(t\right)}{\lambda T\left(t\right)}=\frac{{X}^{″}\left(x\right)}{X\left(x\right)},$ (3.3)

${T}^{\prime }\left(t\right)=\lambda {\lambda }_{1}T\left(t\right),$ (3.4)

${X}^{″}\left(x\right)={\lambda }_{1}X\left(x\right).$ (3.5)

${\lambda }_{1}=0$ 时，通过求解方程(3.4)、(3.5)可得：

$T\left(t\right)=b,$ (3.6)

$X\left(x\right)=cx+d,$ (3.7)

${v}_{1}\left(x,t\right)=\frac{a{\text{e}}^{kt}\left({C}_{1}x+{C}_{2}\right)}{{\text{e}}^{kt}\left({C}_{1}x+{C}_{2}\right)-1},$ (3.8)

${\lambda }_{1}<0$ 时，求解常微分方程(3.4)、(3.5)可得通解：

$T\left(t\right)={a}_{1}{\text{e}}^{\lambda {\lambda }_{1}t},$ (3.9)

$X\left(x\right)={b}_{1}\mathrm{sin}\left(\sqrt{-{\lambda }_{1}}x\right)+{c}_{1}\mathrm{cos}\left(\sqrt{-{\lambda }_{1}}x\right),$ (3.10)

${v}_{2}\left(x,t\right)=\frac{a{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t}\left({C}_{1}\mathrm{sin}\left(\sqrt{-{\lambda }_{1}}x\right)+{C}_{2}\mathrm{cos}\left(\sqrt{-{\lambda }_{1}}x\right)\right)}{{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t}\left({C}_{1}\mathrm{sin}\left(\sqrt{-{\lambda }_{1}}x\right)+{C}_{2}\mathrm{cos}\left(\sqrt{-{\lambda }_{1}}x\right)\right)-1},$ (3.11)

${\lambda }_{1}>0$ 时，运用与上面相同的步骤可以求得如下显式精确解：

${v}_{3}\left(x,t\right)=\frac{a\left({C}_{1}{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t+\sqrt{{\lambda }_{1}}x}+{C}_{2}{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t-\sqrt{{\lambda }_{1}}x}\right)}{\left({C}_{1}{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t+\sqrt{{\lambda }_{1}}x}+{C}_{2}{\text{e}}^{\left(k+\lambda {\lambda }_{1}\right)t-\sqrt{{\lambda }_{1}}x}\right)-1},$ (3.12)

$w\left(x,t\right)={C}_{1}{\text{e}}^{{k}_{1}x+\lambda {k}_{1}^{2}t+{\zeta }_{0}}+{C}_{2},$ (3.13)

$w\left(x,t\right)={\text{e}}^{-\lambda {k}_{1}^{2}t}\left({C}_{1}\mathrm{cos}\left({k}_{1}x+{\zeta }_{0}\right)+i{C}_{2}\mathrm{sin}\left({k}_{1}x+{\zeta }_{0}\right)\right)+{C}_{3},$ (3.14)

$w\left(x,t\right)={C}_{1}{x}^{2}+{C}_{2}x+2\lambda {C}_{1}t+{C}_{3},$ (3.15)

${v}_{4}\left(x,t\right)=\frac{a\left({C}_{1}{\text{e}}^{{k}_{1}x+\left(\lambda {k}_{1}^{2}+k\right)t+{\zeta }_{0}}+{C}_{2}{\text{e}}^{kt}\right)}{{C}_{1}{\text{e}}^{{k}_{1}x+\left(\lambda {k}_{1}^{2}+k\right)t+{\zeta }_{0}}+{C}_{2}{\text{e}}^{kt}-1},$ (3.16)

${v}_{5}\left(x,t\right)=\frac{a{\text{e}}^{\left(-\lambda {k}_{1}^{2}+k\right)t}\left({C}_{1}\mathrm{cos}\left({k}_{1}x+{\zeta }_{0}\right)+i{C}_{2}\mathrm{sin}\left({k}_{1}x+{\zeta }_{0}\right)\right)+a{C}_{3}{\text{e}}^{kt}}{{\text{e}}^{\left(-\lambda {k}_{1}^{2}+k\right)t}\left({C}_{1}\mathrm{cos}\left({k}_{1}x+{\zeta }_{0}\right)+i{C}_{2}\mathrm{sin}\left({k}_{1}x+{\zeta }_{0}\right)\right)+{C}_{3}{\text{e}}^{kt}-1},$ (3.17)

${v}_{6}\left(x,t\right)=\frac{a{\text{e}}^{kt}\left({C}_{1}{x}^{2}+{C}_{2}x+2\lambda {C}_{1}t+{C}_{3}\right)}{{\text{e}}^{kt}\left({C}_{1}{x}^{2}+{C}_{2}x+2\lambda {C}_{1}t+{C}_{3}\right)-1}.$ (3.18)

4. 结论与讨论

Explicit Exact Solutions for a Nonlinear Reaction-Diffusion Equation[J]. 应用数学进展, 2020, 09(01): 100-104. https://doi.org/10.12677/AAM.2020.91012

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