The Brachystochrone Problem

We can use the Euler equation to solve a number of problems that involve finding the shape of a curve that is an extremum. In the classic problem posed by Jacob Bernoulli in the 17th century, we are asked to consider a particle moving from position $\left(x_{1},y_{1}\right)$ to position $\left(x_{2},y_{2}\right)$ in a constant force field. Starting the particle at rest, find the path that minimizes its transit time.

Solution: Choosing a coordinate system so that the initial point is at the origin, we let the force field be directed along the positive $x$ axis. As the force on the particle is constant, and friction is ignored, the field is conservative and the total energy of the particle is defined by Hamilton's principle of conservation to be $T+U=$const. That is, of all possible paths the particle could take, the one that minimizes the time interval is also the one that minimizes the time integral of the difference between the kinetic and potential energies.

Starting from rest, $T+U=0$. The kinetic energy is $T=\frac{1}{2} m v^{2}$, and the potential energy is $U = -F x = -m g x$, where $g$ is the acceleration imparted by the force. To find the total time interval, we must calculate the distance divided by the velocity at each point. The velocity in a gravitational field is calculated from Newton's laws to be

$$v = \sqrt{2 g x}$$

and the instantaneous change is distance is

$$(dx^{2} + dy^{2})^{1/2}$$

Thus

$$t = \int_{(x_{1},y_{1})}^{(x_{2},y_{2})} \frac{ds}{v} = \int \frac{(dx^{2} + dy^{2})^{1/2}}{(2gx)^{1/2}}$$

e wish to minimize the time of transit. Realizing that $2g$ will not affect the final solution, we can remove it and simplify our functional by dividing the numerator by $dx$, giving us

$$f = \left( \frac{1+y^{'2}}{x}\right)^{1/2}$$

And since $\partial f / \partial y = 0$, the Euler equation becomes

$$\frac{d}{dx} \frac{\partial f}{\partial y^{'}} = 0$$

$\frac{\partial f}{\partial y^{'}}$ must then equal some constant. Thinking about this intuitively

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