曲线积分的计算 - Eufisky - The lost book
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曲线积分的计算

Eufisky posted @ 2017年9月19日 16:05 in 数学分析 with tags 积分计算 , 825 阅读

写出圆周的单层位势$$U(a,b)=\int_{x^2+y^2=R^2}\ln \frac 1{\sqrt{(x-a)^2+(y-b)^2}}ds,\quad \text{其中}\,  a^2+b^2\neq R^2$$


解.不妨设$R>0$,否则考察$-R$.令$x=R\cos\theta,y=R\sin\theta$,则$$ds=\sqrt{\left[x'(\theta)\right]^2+\left[y'(\theta)\right]^2}d\theta=Rd\theta.$$因此

\begin{align*}U(a,b)&=R\int_0^{2\pi}\ln \frac 1{\sqrt{(R\cos\theta-a)^2+(R\sin\theta-b)^2}}d\theta\\&=R\int_0^{2\pi}\ln \frac 1{\sqrt{R^2+a^2+b^2-2aR\cos\theta-2bR\sin\theta}}d\theta\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\sin(\theta+\varphi)\right)d\theta,\quad \text{其中}\, \tan\varphi=\frac ab\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_\pi^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_0^\pi\ln \left(R^2+a^2+b^2+2R\sqrt{a^2+b^2}\cos\theta\right)d\theta.\end{align*}

中间几步注意到了对积分变量进行诸如$u=\theta+c$的变换改变积分上下限,而$\sin x$和$\cos x$的最小周期为$2\pi$.因此仍等于$0$到$2\pi$上的积分.

再由比较常见的Poisson积分公式有$$\int_0^\pi\ln (a\pm b\cos x)dx=\pi\ln\frac{a+\sqrt{a^2-b^2}}{2},\quad a\geq b\geq 0$$此公式证明只需把积分看成是关于$b$的函数,对$b$求导即可.

由此得$$\int_0^\pi\ln \left(a^2\pm 2ab\cos x+b^2\right)dx=\begin{cases}2\pi\ln a,&a\geq b\geq 0\\2\pi\ln b,&b\geq a\geq 0\end{cases}$$因此所求积分为

$$U(a,b)=\begin{cases}-2\pi |R|\ln |R|,& a^2+b^2<R^2\\-\pi |R|\ln \left(a^2+b^2\right),& a^2+b^2>R^2\end{cases}$$


在惯性系内一不受外力作用的刚性飞行器绕固定点转动的动态可用Euler方程描述\begin{align*}J_1\dot\omega_1&=(J_2-J_3)\omega_2\omega_3,\\J_2\dot\omega_2&=(J_3-J_1)\omega_3\omega_1,\\J_3\dot\omega_3&=(J_1-J_2)\omega_1\omega_2.\end{align*}其中$\omega_1,\omega_2,\omega_3$为刚体转动角速度的投影, $J_1,J_2,J_3$为惯性主轴的转动惯量且$J_1,J_2,J_3$均大于$0$.

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