# 计算方法之微扰近似 精选

——黄昆

$\dot{v}=g-\alpha v^2$

$\int dx \frac{1}{1-x^2}= \int dx \frac{1}{2}(\frac{1}{1+x}+\frac{1}{1+x}) =\frac{1}{2} (\ln (1+x)-\ln(1-x))=\frac{1}{2}\ln \frac{1+x}{1-x}$

$v(t)=\sqrt{g/\alpha}\frac{e^{\sqrt{g \alpha}t}- e^{-\sqrt{g \alpha}t}} {e^{\sqrt{g \alpha}t}+ e^{-\sqrt{g \alpha}t}} = \tanh \sqrt{g \alpha}t$

$x=\frac{1}{\sqrt{g \alpha}}\ln(\cosh \sqrt{g \alpha}t)$

$\dot{v}=gt$

$\dot{v}=g-\alpha (gt)^2$

$\dot{v}_1 =-2\sqrt{\alpha g} v_1$

$\ddot {\theta}= - \frac{g}{L} \sin \theta$

$\ddot {\theta}= - \frac{g}{L} \theta$

$T=4 \frac{L/g} K(\sin ^2 \frac{\theta_0}{2})$

$K(x)=\int ^{\frac{\pi}{2}}_0 \frac{1}{\sqrt{1-x \sin ^2 \phi}} d\phi$

$-\frac{g}{L}(1+\alpha \theta ^2 _0)^2 \theta = - \frac{g}{L} \theta( 1- \frac{1}{6}\theta ^2)$

$2\alpha \theta ^2 _0 = \frac{1}{6}\theta ^2=\frac{1}{6} \theta ^2_0 \sin \frac{2\pi t}{(1+\alpha \theta ^2 _0) T_0}\approx \frac{1}{12} \theta ^2_0$

$T=(1+\frac{1}{24})\theta ^2 _0$

$T=(1+\frac{1}{16})\theta ^2 _0$

http://blog.sciencenet.cn/blog-1319915-1143105.html

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GMT+8, 2019-10-14 11:35