Brachistochrone problem

Two points $(x_1,y_1)$,$(x_2,y_2)$. Find a curve allows a particle to travel from one point to another in the shortest time.

Problem Solve

Using conservation of energy, the time can be written as:

$$\begin{array}{rl}v& =\frac{ds}{dt}=\sqrt{2g(y_1-y)}\\ T& =\int dt\\ & ={\int }_{y_1}^{y_2}\frac{ds}{\sqrt{2g(y_1-y)}}\\ & ={\int }_{x_1}^{x_2}\frac{\sqrt{1+(y^{\prime }{)}^2}}{\sqrt{2g(y_1-y)}}dx\end{array}\text{\hspace{1em}(1)}$$

(1) could be seen in the form of:

$$\begin{array}{rl}T& ={\int }_{x_1}^{x_2}F(y,y^{\prime },x)dx\\ F(y,y^{\prime },x)& =\frac{\sqrt{1+(y^{\prime }{)}^2}}{\sqrt{2g(y_1-y)}}\end{array}\text{\hspace{1em}(2)}$$

(2) is a standard form for Euler-Lagrange equation. The equivalent condition for the time reach its minima is:

$$\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})=\frac{\partial F}{\partial y}\text{\hspace{1em}(3)}$$

Here, we construct a exact differential:

$$\begin{array}{r}\frac{d}{dt}(F-\frac{\partial F}{\partial y^{\prime }}{y}^{\prime })=0\end{array}\text{\hspace{1em}(4)}$$

Thus, we could get:

$$\begin{array}{r}F-\frac{\partial F}{\partial y^{\prime }}{y}^{\prime }=Const\end{array}\text{\hspace{1em}(5)}$$

we can set $y_1=0$ without loss of generality:

$$\begin{array}{rl}dx& =\frac{1}{\sqrt{\frac{k}{y}-1}}dy\\ & =\sqrt{\frac{y}{k-y}}dy\\ x& =\int \sqrt{\frac{y}{k-y}}dy\end{array}$$

with $y=k{\sin }^2\theta$, we can deal with this integral:

$$\begin{array}{rl}x& =2k\int {\sin }^2\theta d\theta \\ & =k\int (1-\cos 2\theta )d\theta \\ & =k\left(\theta -\frac{1}{2}\sin 2\theta \right)+C\\ y& =\frac{k}{2}(1-cos2\theta )\end{array}\text{\hspace{1em}(6)(7)}$$

Conclusion

The result is:

$$\{\begin{array}{rl}x& =k\left(\theta -\frac{1}{2}\sin 2\theta \right)\\ y& =\frac{k}{2}(1-cos2\theta )\end{array}\text{\hspace{1em}(8)}$$

which is a Cycloid.

Related

• Calculus of Variations