Showing that $x^*(t)=\sin(t)$ minimizes the functional $J(x(t)) = \int_0^{\pi/2} [\dot x(t)^2 - x(t)^2 ]dt$

Question

We are given the functional $$J(x(t)) = \int_0^{\pi/2} [\dot x(t)^2 - x(t)^2 ]dt$$ with the fixed boundary condition $x(0)=0$ and $x(\frac{\pi}{2})=1$.

Could anyone help me prove that $x^*(t)=\sin(t)$ minimizes $J$?

By looking at the second variation of $J$, I got a negative term there, so I could not conclude whether $x^*(t)$ is a min or max. I also tried to show the convexity of $J$, and it all boiled down to showing that $J\geq0$, and I got stuck there.

Any help is appreciated!

Answer 1

Hint: check that if $x(0)=0$ then $$\int\limits_0^{\pi/2} (\dot x^2(t) - x^2(t) )dt=\int\limits_0^{\pi/2}(\dot x(t)-x(t)\cot(t))^2dt.$$

Answer 2

If you plug $x(t)=\sin(t)$ into the integral you get: $$J(x*)=\int_0^{\pi/2}(\cos^2(t)-\sin^2(t))dt = \int_0^{\pi/2} \cos(2t)dt=\sin(2t)/2|_{0}^{\pi/2}=1/2(\sin(\pi)-\sin(0))=0-0=0$$

Now you are left with showing that $J(x)\ge 0$ for every $x$ that satisfies these boundary conditions.

After integration by parts you get: $\int (\dot{x}^2-x^2)dt=x(t)\dot{x}(t)-\int (x\ddot{x}+x\cdot x)dt$, the boundary term should vanish, and we are left with the equation of $\int x(\ddot{x}+x)dt$, which is zero for the minimizer cause from the Euler-Lagrange equations we get that $\ddot{x}+x=0$ for the minimizer solution.

BTW, this Lagrangian is of course the Lagrangian of a simple harmonic oscillator.