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最速降线求法的简介

2022-07-13 13:04 作者:现代微积分  | 我要投稿

此篇文章主要科普下最速降线的求解方法。之所以说是科普是因为内容门槛有些高)有一半以上都是借助于《泛函分析》这本书的内容以及我查阅的诸多文献的,所以就当大部分是“复述”好了[doge]。(我花费了一上午才大致看明白“泛函”和“变分”的基本概念,看来探索真理任重道远呀)

如图,一小球(视为质点)由A点静止释放,沿一光滑路径运动至点C,不计一切阻力,则取何种路径时运动时间最短?

当路径y=y(x)改变时,对应的时间t也会改变,因此路径可以映射到时间上,换言之时间是关于路径的函数t=t(y)

以A为坐标原点,建立左手系(图如所示)

设路径所在曲线为y=y(x)

根据功能关系,有:mgy%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2

解得:v%3D%5Csqrt%7B2gy%7D

v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20得:

dt%3D%5Cfrac%7Bds%7D%7Bv%7D%20%3D%5Cfrac%7B%5Csqrt%7B(dx)%5E2%2B(dy)%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D%20%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20dx%7D%7B%5Csqrt%7B2gy%7D%20%7D%20

(ps:ds%3D%5Csqrt%7B(dx)%5E2%2B(dy)%5E2%7D%20为弧微分,即微小的一段弧可近似用直线替代,用勾股定理求之)

取其在区间[x₁,x₂]上的积分得:

T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%7D%7B%5Csqrt%7B2gy%7D%20%7D%20dx%20

选取不同的y(曲线),时间t也会不同

输入一个函数y,则输出一个时间t,这是一种由函数空间到数域的映射,称之为一种泛函

若我们需要找到一个y(曲线),使得输出的时间t最小,也就是要找到这个泛函极值

求解该问题则需要运用变分法

由上式可知,T是关于x,y,y'的函数,即:

T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20F(x%2Cy%2Cy')dx

设y(曲线)取得关于T的泛函极小值(默认其一阶导存在)

我们对这一曲线附加一微小的扰动(且两个端点处微扰值为0,即定端变分),则曲线形式发生轻微的改变使得T略有增大。当微扰→0时,总时间的改变→0,记作:

%5Cdelta%20T%3DY(y%2B%5Cdelta%20y)-T(y)

ps:符号δ和微分dx的d思想类似,但后者的微扰对应为一个数x,而前者的微扰对应为一族函数,因此为了区别此便将其记作δ

上式展开得:

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%5BF(x%2Cy%2B%5Cdelta%20y%2Cy'%2B%5Cdelta%20y')%20-F(x%2Cy%2Cy')%5Ddx

F(x%2Cy%2B%5Cdelta%20y%2Cy'%2B%5Cdelta%20y')%3DF(x%2Cy%2Cy')%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y'

%3DF(x%2Cy%2Cy')%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y'%7D%20(%5Cdelta%20y)'

取后两项积分得微扰的累计时间δT

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20(%5Cdelta%20y)')dx

考虑到端点处变分值为0,对上式第2项采用分部积分得:

%20%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D(%5Cdelta%20y)'dx%3D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%5Cdelta%20y%7C%5E%7Bx_2%7D_%7Bx_1%7D%20-%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y)dx%20%3D-%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y)dx%20

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D)%20%5Cdelta%20y%20%20dx

令δT→0,则可让T收敛于所设泛函极值

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D)%20%5Cdelta%20y%20%20dx%3D0

由于δy为任取的一阶微小量,则上述积分恒为0需满足括号内恒为0,即:

%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%3D0

上式即著名的E-L方程(欧拉-拉格朗日方程)

43下面用E-L方程求解原题的最佳解泛函(即最速曲线)

据表达式F(x%2Cy%2Cy')%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D%20可知,F是不含变量x的函数

%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%3D0,由E-L方程可得,泛函T出现极值时,有:

F-y'%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%3DC(C为常数)

代入F表达式整理得:

%5Cbegin%7Barray%7D%0A%5C%5C%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D-y'%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%5B1%2B(y')%5E2%5D%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%20%5Ccdot%202y'%7D%7B%5Csqrt%7B2gy%7D%20%7D%20%0A%5C%5C%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D-%5Cfrac%7B(y')%5E2%7D%7B%5Csqrt%7B2gy%7D%5Ccdot%20%5Csqrt%7B1%2B(y')%5E2%7D%20%20%7D%20%0A%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2gy%7D%5Ccdot%20%5Csqrt%7B1%2B(y')%5E2%7D%20%20%7D%20%0A%5C%5C%3DC%0A%5Cend%7Barray%7D

y(1%2B(y')%5E2)%3D%5Cfrac%7B1%7D%7B2gC%5E2%7D%3Dk%20

y'%3D%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Csqrt%7B%5Cfrac%7Bk-y%7D%7By%7D%20%7D%20%20

接下来就是求解微分方程了

分离变量得:%5Csqrt%7B%5Cfrac%7Bk-y%7D%7By%7D%20%7D%20%20dy%3Ddx

换元令y%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20),则

dy%3D%5Cfrac%7Bk%7D%7B2%7D%20sin%5Ctheta%20d%5Ctheta%20

代入得:%5Csqrt%7B%5Cfrac%7B%5Cfrac%7Bk%7D%7B2%7D(1-cos%5Ctheta%20)%20%7D%7Bk-%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)%7D%20%7D%20%20dy%3Ddx

%5Csqrt%7B%5Cfrac%7B1-cos%5Ctheta%20%7D%7B1%2Bcos%5Ctheta%20%7D%20%7D%20%5Ccdot%20%5Csqrt%7B%5Cfrac%7B1-cos%5Ctheta%20%7D%7B1-cos%5Ctheta%20%7D%20%7D%5Ccdot%20%5Cfrac%7Bk%7D%7B2%7Dsin%5Ctheta%20d%5Ctheta%20%3Ddx

dx%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)d%5Ctheta%20

两边积分可得:

x%3D%5Cfrac%7Bk%7D%7B2%7D%20(%5Ctheta%20-sin%5Ctheta%20)%2BC

质点在初始位置时,y%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)%3D0,θ=0

此时x=0,解得C=0

得轨迹参数方程:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D%5Clambda%20(%5Ctheta%20-sin%5Ctheta%20)%5C%5Cy%3D%5Clambda%20(1-cos%5Ctheta%20)%0A%5Cend%7Bmatrix%7D%5Cright.%20

(其中θ为参数)

此参数方程与摆线的参数方程相同,因此最速降线也是一条摆线

求出了该最佳泛函解,我们可以利用光学中的“光线传播时间最短”(费马原理)以及斯涅尔定律(光的折射定律),通过取空介微元结合摆线的性质加以验证。由于篇幅原因,验证的解析这里就不多赘述了。

数学家们的游戏希望有朝一日眼前的你也能参与其中[doge]!

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