最速降线 与 第一次接触泛函

数学

现在看得很多内容都提到了变分和泛函，本科只学了高数，现在也看一下变分的内容

斯涅尔定律

当光穿过不同介质时，由于在不同介质之间的传播速率不同，会发生折射现象，其入射光和折射光在同一平面，而且与法线夹角满足：

$$\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$$

证明折射定律可以使用费马原理

费马原理：
“最短时间原理”：光线传播的路径是需时最少的路径
“平稳时间原理”：光沿着所需时间为平稳的路径传播

光从$Q$点传播到$P$点所需要的时间$T$为

$$T = \frac{\sqrt{x^2+a^2}}{v_1} + \frac{\sqrt{b^2+(l-x)^2}}{v_2}$$

根据费马原理，光线传播路径是所需时间为极值的路径，则

$$\begin{align*} \frac{\mathrm{d}T}{\mathrm{d}x} & = \frac{x}{v_1\sqrt{x^2+a^2}} + \frac{-(l-x)}{v_2\sqrt{b^2+(l-x)^2}} \\ & = \frac{\sin\theta_1}{v_1} - \frac{\sin\theta_2}{v_2} \\ & = 0 \end{align*}$$

最速降线

由上面的折射定律可知，在速度v已知的情况下，要满足所用时间最少，需要保证

$$\frac{\sin \theta}{v} = Constant$$

根据能量守恒定律

$$\frac{1}{2}mv^2 = mgy \\ v = \sqrt{2gy}$$

在每一点速度方向与法线夹角$\theta$，它与该点的微元$(\mathrm{d}x,\mathrm{d}y)$满足

$$\begin{align*} \sin\theta & = \frac{\mathrm{d}x}{\sqrt{(\mathrm{d}y)^2+(\mathrm{d}x)^2}} \\ & = \frac{1}{\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}} \\ & = \frac{1}{\sqrt{1+{y'}^2}} \end{align*}$$

组合公式得

$$\frac{\sin\theta}{v} = \frac{1}{\sqrt{2gy(1+{y'}^2)}} = Constant$$

整理得，

$$y(1+{y'}^2) = C$$

泛函与变分法

函数是数域到数域的映射，$y=f(x)$，而$\Delta$表示两个数之间的差值$\Delta x =x_1-x_0, \quad\Delta y = y_1-y_0$，而泛函是定义域是函数值域是实数的映射，是从函数组成的一个向量空间到实数的一个映射，$\delta$表示两个函数之间的差值$\delta y = f_1-f_0$，$\delta y$也是一个函数。

设$\mathbf{C}$是一个由函数组成的集合，对于$\mathbf{C}$中的任意一个元素$y(x)$，数集$\mathbf{B}$都有一个元素$J$与之对应，称$J$是$y(x)$的泛函数

$$J = J[y(x)]$$

假设我寻找连接$(x_0,y_0),(x_1,y_1)$两点的一条曲线$y(x)$并且这条曲线能够使泛函$J[y(x)]$取得极值

泛函经常用积分来表示，而我们则要求该积分的极值

$$J[y(x)] = \int_{x_0}^{x_1}F(x,y,y')\mathrm{d}x$$

而求解该极值的方法就是变分法。$\delta y$为$y$的变分，当$y$变为$y+\delta y$时，对应的泛函变为$F(x,y+\delta y,y'+\delta y')$，当$\delta y$趋于无穷小时，可以用一个一阶线性来逼近$x$处的曲线

$$F(x,y+\delta y,y'+\delta y') = F(x,y,y') + \frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y'}\delta y' \\ \delta F = \frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y'}\delta y'$$

那么泛函$J[y(x)]$的变分为

$$\delta J(y) = \int_{x_0}^{x_1}\delta F\mathrm{d}x = \int_{x_0}^{x_1} \frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y'}\delta y' \mathrm{d}x$$

接下来证明泛函$J(y)$取极值的必要条件是

$$\delta J(y) = 0$$

我们有一函数$\varphi(x)$满足

$$\delta y(x) = \varepsilon \varphi(x)$$

那么在两端点值函数的改变量为0，所以

$$\delta y(x_0) = \varphi(x_0) = \delta y(x_1) = \varphi(x_1) = 0$$

这样泛函

$$I(\epsilon) = J[y(x)+\varepsilon\varphi(x)] = \int_{x_0}^{x_1}F(x,y(x)+\varepsilon\varphi(x),y'(x)+\varepsilon\varphi'(x))\mathrm{d}x$$

我们考虑，这时如果$y(x)$能够使泛函满足极值，也就是说，这时$\varepsilon = 0$，对$\varepsilon$求偏导，并且当$\varepsilon=0$时偏导为$0$

$$\left. \frac{\partial I(\varepsilon)}{\partial \epsilon} \right |_{\varepsilon = 0} = 0$$

$$\begin{align*} \frac{\partial I(\varepsilon)}{\partial \epsilon} & = \int_{x_0}^{x_1}\frac{\partial F}{\partial \varepsilon}\mathrm{d}x \\ & = \int_{x_0}^{x_1}\frac{\partial F}{\partial y}\varphi(x) + \frac{\partial F}{\partial y'}\varphi'(x)\mathrm{d}x \\ & = 0 \end{align*}$$

然后用$\varepsilon$乘以上式，并依据$\delta y = \varepsilon \varphi(x)$

$$\begin{align*} & \int_{x_0}^{x_1}\frac{\partial F}{\partial y}\varphi(x) + \frac{\partial F}{\partial y'}\varphi'(x)\mathrm{d}x \\ = & \int_{x_0}^{x_1}\frac{\partial F}{\partial y}\varepsilon\varphi(x) + \frac{\partial F}{\partial y'}\varepsilon\varphi'(x)\mathrm{d}x \\ = & \int_{x_0}^{x_1}\frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y'}\delta y'\mathrm{d}x \\ = & \delta J(y) \\ = & 0 \end{align*}$$

接下来需要证明

$$\delta y' = (\delta y)'$$

• A: $y$
• B: $y+\delta y$
• C: $y+\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{dx} = y+y'\mathrm{d}x$
• D: C->D: $y+y'\mathrm{d}x + \delta(y+y'\mathrm{d}x) = y+y'\mathrm{d}x+ \delta y +\delta y'\mathrm{d}x$
• D: B->D: $y+\delta y +\frac{\mathrm{d}}{\mathrm{d}x}(y+\delta y)\mathrm{d}x = y+\delta y + y'\mathrm{d}x + \frac{\mathrm{d}\delta y}{\mathrm{d}x}\mathrm{d}x$

两式相等得，

$$\delta y' = \frac{\mathrm{d}\delta y}{\mathrm{d}x} = (\delta y)'$$

再根据分布积分

$$\begin{align*} &\int_{x_0}^{x_1}\frac{\partial F}{\partial y'}\delta y'\mathrm{d}x \\ = & \int_{x_0}^{x_1}\frac{\partial F}{\partial y'}(\delta y)'\mathrm{d}x \\ = & \left. \frac{\partial F}{\partial y'} \delta y \right |_{x_0}^{x_1} - \int_{x_0}^{x_1}\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'} \right) \delta y\mathrm{d}x \\ = & - \int_{x_0}^{x_1}\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'} \right) \delta y\mathrm{d}x \qquad // \delta y(x_0) = \delta y(x_1) = 0 \end{align*}$$

带入原式得到泛函的变分

$$\begin{align*} \delta J(y) & = \int_{x_0}^{x_1} \frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y'}\delta y' \mathrm{d}x \\ & = \int_{x_0}^{x_1} \frac{\partial F}{\partial y}\delta y - \frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'} \right) \delta y \mathrm{d}x \\ & = \int_{x_0}^{x_1} \left[ \frac{\partial F}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'} \right)\right] \delta y \mathrm{d}x \\ & = 0 \end{align*}$$

由于$\delta y$可取任意函数形式（也可以用反证法来证），就可以得到欧拉-拉格朗日方程(E-L方程)

$$\frac{\partial F}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'} \right) = 0$$

接下来证一个引理

若$F(x,y,y') = F(y,y')$，则欧拉方程可改写成

$$\frac{\partial F}{\partial y'}y'-F = C$$

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y'}y' \right) & = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y'} \right)y' + \frac{\partial F}{\partial y'}\cdot \frac{\mathrm{d}}{\mathrm{d}x}y' \\ &= \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y'} \right)y' + \frac{\partial F}{\partial y'}y'' \end{align*}$$

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}F & = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial y'}\frac{\partial y'}{\partial x} \\ & = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}y' + \frac{\partial F}{\partial y'}y'' \end{align*}$$

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial F}{\partial y'}y' -F \right) & = y'\left[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\partial F}{\partial y'}\right) - \frac{\partial F}{\partial y} \right] - \frac{\partial F}{\partial x} \\ & = 0 \qquad //\text{中括号为欧拉方程，F与x无关} \end{align*}$$

故，

$$\frac{\partial F}{\partial y'}y' -F = C$$

变分法表达最速降线

现在回到最速降线问题，并用变分法解决

已知

$$v = \sqrt{2gy}$$

$$\begin{align*} \mathrm{d}s & = \sqrt{(\mathrm{d}y)^2+(\mathrm{d}x)^2} \\ & = \sqrt{1+(\frac{\mathrm{d}y}{\mathrm{d}x})^2}\mathrm{d}x \\ & = \sqrt{1+{y'}^2}\mathrm{d}x \\ v & = \frac{\mathrm{d}s}{\mathrm{d}t} \\ \mathrm{d}t & = \frac{\mathrm{d}s}{v} \\ & = \sqrt{\frac{1+{y'}^2}{2gy}}\mathrm{d}x \end{align*}$$

因此可以得到时间$T$的泛函$J(y)$

$$F(x,y,y') = F(y,y') = \sqrt{\frac{1+{y'}^2}{2gy}} \\ J(y) = \int_a^b\sqrt{\frac{1+{y'}^2}{2gy}}\mathrm{d}x$$

由于最速下降曲线问题满足上面的引理，直接用该引理（也可以直接求，过程类似）

$$\begin{align*} & \frac{\partial F}{\partial y'}y' -F \\ & = \frac{1}{2} \left( \frac{1+{y'}^2}{2gy} \right)^{-\frac{1}{2}}\frac{2y'}{2gy}y' - \left( \frac{1+{y'}^2}{2gy} \right)^{\frac{1}{2}} \\ & = \frac{{y'}^2}{(1+{y'}^2)^{\frac{1}{2}}(2gy)^{\frac{1}{2}}} - \frac{(1+{y'}^2)^{\frac{1}{2}}}{(2gy)^{\frac{1}{2}}} \\ & = \frac{{y'}^2-1-{y'}^2}{(1+{y'}^2)^{\frac{1}{2}}(2gy)^{\frac{1}{2}}} \\ & = \frac{1}{(1+{y'}^2)^{\frac{1}{2}}(2gy)^{\frac{1}{2}}} \\ & = C \end{align*}$$

整理可得

$$y(1+{y'}^2)=C'$$

求解最速降线

上面只是用两种方法得到最速降线的表达形式，现在来解出该方程，也就是求解该微分方程

上式的形式是一个三角函数，设$x$是关于$\theta$的函数$x=x(\theta)$，$y'=\cot\theta = \frac{\cos\theta}{\sin\theta}$，由于$y'=\frac{\mathrm{d}y}{\mathrm{d}x}$是关于$x$的微分，故不能由$y'$直接求出$y$关于$\theta$的方程，把$y'$带入上式得

$$y(1+{y'}^2) = y\frac{1}{\sin^2\theta} = C\\ y=C\sin^2\theta=\frac{C}{2}(1-\cos{2\theta})$$

再有

$$y' = \frac{\mathrm{d}y}{\mathrm{d}x} =\left. \frac{\mathrm{d}y}{\mathrm{d}\theta} \middle/ \frac{\mathrm{d}x}{\mathrm{d}\theta}\right.\\ \frac{\mathrm{d}x}{\mathrm{d}\theta} = \left. \frac{\mathrm{d}y}{\mathrm{d}\theta} \middle/ y'\right. = \frac{C\sin{2\theta}}{\cot\theta} = 2C\sin^2\theta= C(1-\cos{2\theta}) \\ \int \mathrm{d}x = \int C(1-\cos{2\theta})\mathrm{d}\theta \\ x+C_1 = C(\theta - \frac{1}{2}\sin{2\theta})+C_2 \\ x = \frac{C}{2}(2\theta - \sin{2\theta})+C_1$$

由此得到方程

$$\left\{\begin{align*} x & = \frac{C}{2}(2\theta - \sin{2\theta})+C_1 \\ y & = \frac{C}{2}(1-\cos{2\theta}) \end{align*}\right.$$

由于$\frac{\mathrm{d}y}{\mathrm{d}x} = \cot\theta$，所以这个$\theta$就是切线与法线夹角，所以端点值$x(0)=y(0)$，得到$C_1 = 0$，并把$t = 2\theta$，就可以得到最速降线的参数方程

$$\left\{\begin{align*} x & = \frac{C}{2}(t - \sin{t}) \\ y & = \frac{C}{2}(1-\cos{t}) \end{align*}\right.$$

由此可知，最速降线是一条摆线，是由一个圆沿一条直线运动时，圆边界上一定点所形成的轨迹

$A$点为该定点，$AC$为速度方向也是该摆线在$A$点的切线，由此$\angle BAC = 90^{\circ}$，$\theta$为上面的速度与法线的夹角，可以得到$\angle C = \angle CBD = \theta$，设该圆半径为$r$，则

$$\begin{align*} AB & = 2r\sin \theta \\ AD & = AB\sin\theta = 2r\sin^2\theta = r(1-\cos{2\theta}) \end{align*}$$

AD为A点的纵坐标，比较上面两式得到生成摆线圆的半径

$$r = \frac{C}{2}$$

最后证明两条最速降线的性质

性质一：生成最速降线轨迹的圆是匀速的

要证明这条性质，我们只需要找到$B$点的横坐标$b$与$\mathrm{d}t$之间的关系即可

$$b = x+y\cot\theta\\ \mathrm{d}b = \mathrm{d}x+\mathrm{d}y\cot \theta = \frac{(\mathrm{d}x)^2+(\mathrm{d}y)^2}{\mathrm{d}x}$$

对摆线式子进行$\theta$求导得

$$\left\{\begin{align*} \mathrm{d}x & = \frac{C}{2}(2 - 2\cos{2\theta})\mathrm{d}\theta \\ \mathrm{d}y & = \frac{C}{2}2\sin{2\theta}\mathrm{d}\theta \end{align*}\right.$$

$$\begin{align*} \mathrm{d}t & = \frac{\mathrm{d}s}{v} \\ & = \left( \frac{(\mathrm{d}x)^2+(\mathrm{d}y)^2}{2gy} \right)^{\frac{1}{2}} \\ & = \left( \frac{ C^2(1-2\cos2\theta+ \cos^22\theta+\sin^22\theta) (\mathrm{d}\theta)^2 }{2g\frac{C}{2}(1-\cos2\theta)} \right)^{\frac{1}{2}} \\ & = \left( \frac{2-2\cos2\theta}{\frac{Cg}{2}(2-2\cos2\theta)} \right)^{\frac{1}{2}}C\mathrm{d}\theta \\ & = \left( \frac{2C}{g} \right)^{\frac{1}{2}}\mathrm{d}\theta \\ \mathrm{d}\theta & = \left( \frac{g}{2C} \right)^{\frac{1}{2}}\mathrm{d}t \\ \mathrm{d}t & = \frac{\mathrm{d}s}{v} \qquad //用\mathrm{d}b表示\\ & = \left( \frac{(\mathrm{d}x)^2+(\mathrm{d}y)^2}{2gy} \right)^{\frac{1}{2}} \\ & = \left( \frac{\mathrm{d}x\mathrm{d}b}{2gy} \right)^{\frac{1}{2}} \\ & = \left( \frac{ \frac{C}{2}(2 - 2\cos{2\theta})\mathrm{d}\theta \mathrm{d}b}{2g\frac{C}{2}(1-\cos2\theta)} \right)^{\frac{1}{2}} \\ & = \left( \frac{\mathrm{d}\theta\mathrm{d}b}{g} \right)^{\frac{1}{2}} \\ (\mathrm{d}t)^2 & = \frac{\mathrm{d}\theta\mathrm{d}b}{g} \\ & = \frac{\left( \frac{g}{2C} \right)^{\frac{1}{2}}\mathrm{d}t\mathrm{d}b}{g} \\ & = (2gC)^{-\frac{1}{2}}\mathrm{d}t\mathrm{d}b \\ \mathrm{d}b & = (2gC)^{\frac{1}{2}}\mathrm{d}t \end{align*}$$

故圆按速度$v_o = (2gC)^{\frac{1}{2}}$匀速前进

性质二：等时性

最速降线的等时性是说，无论从最速降线的哪一点无速度的开始，到达最底端所用的时间相同

假设小球（啊哈）从高度$y_0$下降到$2r$，根据能量守恒得，

$$mg(y-y_0) = \frac{1}{2}mv^2\\ v = \sqrt{2g(y-y_0)}$$

$$\begin{align*} \mathrm{d}t & = \frac{\mathrm{d}s}{v}\\ & = \left( \frac{(\mathrm{d}x)^2+(\mathrm{d}y)^2}{2g(y-y_0)} \right)^{\frac{1}{2}} \\ & = \left( \frac{2C(2-2\cos2\theta)}{g(2\cos2\theta_0-2\cos2\theta)} \right)^{\frac{1}{2}}\mathrm{d}\theta \\ & = \left( \frac{2C(1-\cos2\theta)}{g(\cos2\theta_0-\cos2\theta)} \right)^{\frac{1}{2}}\mathrm{d}\theta \\ & = \left( \frac{2C2\sin^2\theta}{2g(\cos^2\theta_0-\cos^2\theta)} \right)^{\frac{1}{2}}\mathrm{d}\theta \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}} \frac{\sin\theta}{\cos^2\theta_0\sqrt{1-\frac{\cos^2\theta}{\cos^2\theta_0}}} \mathrm{d}\theta \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}} \frac{1}{\sqrt{1-u^2}}\mathrm{d}u \qquad //u = \frac{\cos\theta}{\cos\theta_0} \\ t & = \left( \frac{2C}{g}\right)^{\frac{1}{2}} \int_{\theta_0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-u^2}}\mathrm{d}u \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}} \int_{1}^{0}\frac{1}{\sqrt{1-u^2}}\mathrm{d}u \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}}\int_{1}^{0}\frac{1}{\cos t}\mathrm{d}\sin t \qquad // u=\sin t \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}}\int_{\frac{\pi}{2}}^{\pi}\frac{\cos t}{\cos t}\mathrm{d} t \\ & = \left( \frac{2C}{g}\right)^{\frac{1}{2}} \left.t\right|_{\frac{\pi}{2}}^{\pi} \\ & = \pi\sqrt{\frac{C}{2g}} \\ & = \pi \sqrt{\frac{r}{g}} \end{align*}$$

故所用时间与$y_0$或$\theta_0$无关

再说一点

用上面的方法求解出来的最速下降曲线是一种摆线，而只给出了摆线的参数方程，而该参数方程可以表示很多很多条摆线，我们要做的就是把端点值带入求解摆线的参数，但这个参数好像求不出来

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Reference

• 变分法及其应用
• 最速降线问题
• 古典变分法
• 泛函与变分原理导引
• 變分法上的最速降線之研究