Emden-Lane边值问题变分迭代法

参考文献:Eur. Phys. J. Plus (2017)132: 251

Emden-Lane边值问题:

\begin{equation} \begin{split} & u''(x)+\frac{\alpha}{x}u'(x)=f(x,u) \\ & u'(0)=0, au(1)+bu'(1)=c \end{split} \label{Emden-Lane} \end{equation}

根据变分迭代法,得校正泛函:

\begin{equation} u_{n+1}(x)=u_n(x)+\int_0^x\lambda(s)\left( u_n''(s)+\frac{\alpha}{s}u_n'(s)-f(s,\widetilde{u}_n) \right) \mathrm ds \label{correctionalfunc} \end{equation}

其中,$\widetilde{u}_n$为限制变分$\delta \widetilde{u}_n=0$。

下面求$\lambda(s)$。

对\eqref{correctionalfunc}变分,

\begin{equation} \begin{split} \delta u_{n+1}(x)=&\delta u_n(x)+\delta \int_0^x\lambda(s)\left( u_n''(s)+\frac{\alpha}{s}u_n'(s)-f(s,\widetilde{u}_n) \right) \mathrm ds \\ =&\delta u_n(x)+\delta \int_0^x\lambda(s)\left( u_n''(s)+\frac{\alpha}{s}u_n'(s) \right) \mathrm ds \\ =&\delta u_n(x)+\delta \int_0^x\frac{\lambda(s)}{s^{\alpha}}\left( s^{\alpha}u_n'(s) \right)' \mathrm ds \\ =&\delta u_n(x)+\delta \int_0^x\left(\frac{\mathrm d}{\mathrm ds}\left[ \lambda(s)u_n'(s)\right]- s^{\alpha}u_n'(s)\frac{\mathrm d}{\mathrm ds}\left[ \frac{\lambda(s)}{s^{\alpha}}\right] \right)\mathrm ds \\ =&\delta u_n(x)+\lambda(s) \delta u_n'(s)\left . \right |_0^x -\delta \int_0^x s^{\alpha}u_n'(s)\frac{\mathrm d}{\mathrm ds}\left[ \frac{\lambda(s)}{s^{\alpha}}\right] \mathrm ds \\ =&\delta u_n(x)+\lambda(x) \delta u_n'(x)-\delta \int_0^x u_n'(s)\left[\lambda'(s)-\frac{\alpha}{s}\lambda(s)\right]\mathrm ds \\ =&\delta u_n(x)+\lambda(x) \delta u_n'(x)-\lambda'(s) \delta u_n(s)\left . \right |_0^x + \delta \int_0^x \lambda''(s)u_n(s)\mathrm ds +\\ &\frac{\alpha}{s}\lambda(s)\delta u_n(s)\left . \right |_0^x - \delta \int_0^x \frac{\alpha}{s^2}[s\lambda'(s)-\lambda(s)]u_n(s) \mathrm ds \\ =& \left [1-\lambda'(x)+\frac{\alpha}{x}\lambda(x)\right]\delta u_n(x)+\lambda(x) \delta u_n'(x)+\\ &\int_0^x \left [\lambda''(s)-\alpha\frac{s\lambda'(s)-\lambda(s)}{s^2}\right]\delta u_n(x)\mathrm ds \end{split} \label{variation} \end{equation}

于是得:

\begin{equation} \begin{split} & 1-\lambda'(x)+\frac{\alpha}{x}\lambda(x)=0 \\ & \lambda(x)=0 \\ & \lambda''(s)-\alpha\frac{s\lambda'(s)-\lambda(s)}{s^2}=0 \end{split} \label{stationary} \end{equation}

解以上方程,得:

\begin{equation} \lambda(s,x)= \begin{cases} & s\ln\left(\frac{s}{x}\right) , \alpha = 1\\ & s\frac{1-\left(\frac{s}{x}\right)^{\alpha-1}}{1-\alpha}, \alpha \neq 1 \end{cases} \label{Lag} \end{equation}

标签: 变分迭代法

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