I'm currently working on the problem of finding the function with minimum arc length when the area between itself and the x-axis is fixed.
More formally, we have to find f(t) such that $$ \text{minimizes}~\int_0^T \sqrt{1+f'(t)^2}\, dt \\ \text{subject to}~f(0)= f(T)=0~\text{and} \int_0^T f(t)\,dt = M $$
I tried to solve it using Pontryagin Maximum Principle, but I can't get the closed form solution.
Can anyone help me? Thank you
You can formulate it in a form that can be solved by Pontryagin Maximum Principle using
\begin{align} \min_u &\int_0^T \sqrt{1 + u^2}\,dt, \\ \text{s.t.}& \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} u \\ x_1 \end{bmatrix}, \\ & x_1(0) = x_2(0) = x_1(T) = 0,\ x_2(T) = M. \end{align}
Namely $x_1(t)$ is equivalent to $f(t)$ and thus $f'(t) = u(t)$, and $x_2(T)$ is equivalent to the integral of $f(t)$ from zero to $T$.