用OHAM方法解一个一阶非线性常微分方程

OHAM方法见以前的博文:OHAM解非线性微分方程基本过程

本文我们用此方法解一个一阶非线性常微分方程,见文献Application of Optimal Homotopy Asymptotic Method for solving nonlinear equations arising in heat transfer



方程为

\begin{equation} (1+\epsilon u)\frac{du}{dx}+u=0,\quad u(0)=1, \quad x\in [0,\infty) \label{ex1eq} \end{equation}

对应于博文OHAM解非线性微分方程基本过程的方程(1)

\begin{equation} \begin{split} L(u(x))=&\frac{du}{dx}+u\\ g(x)=&0\\ N(u(x))=&\epsilon u\frac{du}{dx}\\ u(0)=&1 \end{split} \label{ex1dec} \end{equation}

零阶问题

\begin{equation} \frac{du_0(x)}{dx}+u_0(x)=0,\quad u(0)=1 \label{ex1order0} \end{equation}

解之得

\begin{equation} u_0(x)=e^{-x} \label{ex1order0sol} \end{equation}

一阶问题

\begin{equation} \frac{du_1(x)}{dx}+u_1(x)=C_1\epsilon u_0\frac{du_0}{dx}=-C_1\epsilon e^{-2x},\quad u_1(0)=0 \label{ex1order1} \end{equation}

解之得

\begin{equation} u_1(x)=C_1\epsilon(e^{-2x}-e^{-x}) \label{ex1order1sol} \end{equation}

二阶问题

\begin{equation} \begin{split} \frac{du_2(x)}{dx}+u_2(x)=&\frac{du_1(x)}{dx}+u_1(x)+C_1\left(\frac{du_1(x)}{dx}+u_1(x)\right)+\\ &C_1\epsilon\left(u_0(x)\frac{du_1(x)}{dx}+u_1(x)\frac{du_0(x)}{dx}\right)+\\ &C_2\epsilon u_0(x)\frac{du_0(x)}{dx}\\ =&(1+C_1)C_1\epsilon e^{-2x}+C_1^2\epsilon^2(2e^{-2x}-3e^{-3x})-C_2\epsilon e^{-2x}\\ u_2(0)=&0 \end{split} \label{ex1order2} \end{equation}

解之得

\begin{equation} \begin{split} u_2(x)=&\left [\left (\frac{\epsilon}{2}-1 \right ) C_1^2- C_1-C_2 \right ]\epsilon e^{-x}+\\ &[(1-2\epsilon)C_1^2+C_1+ C_2]\epsilon e^{-2x}+\frac{3}{2}C_1^2\epsilon^2 e^{-3x} \end{split} \label{ex1order2sol} \end{equation}

三阶问题

\begin{equation} \begin{split} \frac{du_3(x)}{dx}+u_3(x)=&\frac{du_2(x)}{dx}+u_2(x)+C_1\left(\frac{du_2(x)}{dx}+u_2(x)\right)+C_2\left(\frac{du_1(x)}{dx}+u_1(x)\right)+\\ &C_1\epsilon\left(u_0(x)\frac{du_2(x)}{dx}+u_2(x)\frac{du_0(x)}{dx}+u_1(x)\frac{du_1(x)}{dx}\right)+\\ &C_2\epsilon\left(u_0(x)\frac{du_1(x)}{dx}+u_1(x)\frac{du_0(x)}{dx}\right)+\\ &C_3\epsilon u_0(x)\frac{du_0(x)}{dx}\\ =&-8 \text{C1}^3 e^{-4 x} \epsilon ^3-e^{-3 x} \epsilon \left(-9 \text{C1}^3 \epsilon ^2+6 \text{C1}^3 \epsilon +6 \text{C1}^2 \epsilon +6 \text{C1} \text{C2} \epsilon \right)\\ &-e^{-2 x} \epsilon \left(2 \text{C1}^3 \epsilon ^2-4 \text{C1}^3 \epsilon +\text{C1}^3-4 \text{C1}^2 \epsilon +2 \text{C1}^2-4 \text{C1} \text{C2} \epsilon +2 \text{C1} \text{C2}+\text{C1}+\text{C2} +\text{C3}\right)\\ u_3(0)=&0 \end{split} \label{ex1order3} \end{equation}

解之得

\begin{equation} \begin{split} u_3(x)=&\frac{8}{3} \text{C1}^3 e^{-4 x}\epsilon ^3+\\ &\frac{1}{6} e^{-3 x} \left(-27 \text{C1}^3 \epsilon ^3+18 \text{C1}^3 \epsilon ^2+18 \text{C1}^2 \epsilon ^2+18 \text{C1} \text{C2} \epsilon ^2\right)+\\ & \frac{1}{6} e^{-2 x} \left(12 \text{C1}^3 \epsilon ^3-24 \text{C1}^3 \epsilon ^2+6 \text{C1}^3 \epsilon -24 \text{C1}^2 \epsilon ^2+12 \text{C1}^2 \epsilon -24 \text{C1} \text{C2} \epsilon ^2+12 \text{C1} \text{C2} \epsilon +6 \text{C1} \epsilon +6 \text{C2} \epsilon +6 \text{C3} \epsilon \right)+\\ &\frac{1}{6} e^{-x} \left(\text{C1}^3 \left(-\epsilon ^3\right)+6 \text{C1}^3 \epsilon ^2-6 \text{C1}^3 \epsilon +6 \text{C1}^2 \epsilon ^2-12 \text{C1}^2 \epsilon +6 \text{C1} \text{C2} \epsilon ^2-12 \text{C1} \text{C2} \epsilon -6 \text{C1} \epsilon -6 \text{C2} \epsilon -6 \text{C3} \epsilon \right) \end{split} \label{ex1order3sol} \end{equation}

(注意:文献中的结果是错的)

方程\eqref{ex1eq}的近似解为

\begin{equation} \bar u(x)=u_0(x)+u_1(x)+u_2(x)+u_3(x) \label{uapprox} \end{equation}

但是,常数$C_1, C_2, C_3$ 还没有确定。根据博文[OHAM解非线性微分方程基本过程](http://www.joyfulphysics.net/index.php/archives/214/)的方程[(12)](http://www.joyfulphysics.net/index.php/archives/214/#mjx-eq-12)、[(13)](http://www.joyfulphysics.net/index.php/archives/214/#mjx-eq-13)、[(14)](http://www.joyfulphysics.net/index.php/archives/214/#mjx-eq-14),得$C_1=-0.6590173149421691, C_2=0.02033521733013163, C_3=0.0033435606129693$

方程\eqref{ex1eq}的精确解为

\begin{equation} u(x)=\mathrm{ProductLog}(e^{1 - x}) \label{exact} \end{equation}

将近似解与精确解作图,如下:

标签: oham, 微分方程

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