Notes for AKT-140117/0:28:51

To demonstrate how to use the Euler-Lagrange equation in classical mechanics, we solve brachistochrone problem as an example. The problem is described in the blackboard shot, which is to find the path of a particle that minimizes the time [math]\displaystyle{ T }[/math] traveled from point [math]\displaystyle{ P_1 }[/math] to point [math]\displaystyle{ P_2 }[/math] in a uniform gravitational field. In this situation, we assume there is no friction along the path; thus the energy is conserved. Let [math]\displaystyle{ y }[/math] be the vertical coordinate. Then, by the conservation of energy, we have

[math]\displaystyle{ \frac{1}{2}mv^2-\frac{1}{2}mv_i^2=\frac{1}{2}mv^2=mgy. }[/math]

Thus, we have

[math]\displaystyle{ v=\sqrt{2gy}. }[/math]

Then, the time [math]\displaystyle{ T }[/math] may be described as

[math]\displaystyle{ T=\int_{P_1}^{P_2}\frac{ds}{v}, }[/math]

where [math]\displaystyle{ ds }[/math] is the infinitesimal arclength of the path. Then, let [math]\displaystyle{ x }[/math] be the horizontal coordinate, we have [math]\displaystyle{ ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}dx. }[/math] Thus, the above equation would be

[math]\displaystyle{ T=\int_{P_1}^{P_2}\sqrt{\frac{1+y'^2}{2gy}}dx. }[/math]

Now, let [math]\displaystyle{ L=\sqrt{\frac{1+y'^2}{2gy}} }[/math], we apply the Euler-Lagrange equation and obtain

[math]\displaystyle{ \frac{\partial L}{\partial y}=-\frac{1}{2y}\sqrt{\frac{1+y'^2}{2g}}=\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)=\frac{1}{\sqrt{2g}}\frac{2yy''-y'^4-y'^2}{2(y(1+y'^2))^{3/2}}. }[/math]

If we rearrange the equation and integrate, we obtain the equation

[math]\displaystyle{ C=\frac{1}{\sqrt{2gy\left(1+y'^2\right)}} }[/math]

where [math]\displaystyle{ C }[/math] is some constant. Then, we rearrange the equation and obatin

[math]\displaystyle{ \left(1+y'^2\right)y=\frac{1}{2gC^2}=k^2. }[/math]

Then, we can solve this equation with parameterization and obtain the final result

[math]\displaystyle{ x\left(t\right)=\frac{k^2}{2}\left(t-\sin t\right),y\left(t\right)=\frac{k^2}{2}\left(1-\cos t\right) }[/math]