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I have a question that asks:

Find the extremal of the functional $$J(x)=\int^{\pi}_02x\sin(t)-\dot x^2 dt$$ with $x(0)=x(\pi)=0$. I found $x(t)=\sin(t)$

It then asks to

Show that this extremal provides the global maximum of $J$

I am not sure how to show this. Do I want to look at $J''(x)$ or at $x''(t)$

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  • $\begingroup$ I believe Euler-Lagrange Equation is the one you need to look into. For the global optimal, you may need further justifications en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation $\endgroup$ – Yan Zhu Sep 29 '13 at 18:38
  • $\begingroup$ So I solved the E-L equation to solve for x(t). But how do I show it is a maximum? $\endgroup$ – yankeefan11 Sep 30 '13 at 16:02
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Looking at $J''$ is the right impulse ($x''$ would not help), but the second variation in infinite-dimensional spaces is hard to handle. It's better to take advantage of the fact that $J$ is algebraically simple: it's quadratic.

Let $x_0$ be the solution of Euler-Lagrange equation. You want to show that $J(x_0+h)- J(x_0)\le 0$ for every function $h$ in your space. Just plug it in and simplify, suing the fact that $x_0 $ solves the Euler-Lagrange equation. You should get $J(x_0+h)- J(x_0 ) = -\int_0^\pi \dot h^2\,dt$.

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