In a set of practice problems I have a question that states:
Show that the extremal for $E[y]=\int^1_0\cos^2(y'(x))dx$ with $y(0)=0, y(1)=1$ is not the minimizer over $D^1(0, 1)$.
To be honest I don't know how to tackle this one. Solving the E-L equation is out of question, I don't even think there is a closed form solution for this one.
I am also confused as to how the minimizer and the extremal are different, I thought that if something is a minimizer it satisfies the E_L equation and is thus an extremal of the functional.
$L(a,b) = \cos^2 b$, Euler Lagrange gives ${d \over dx}{ \partial L(y(x),y'(x)) \over \partial b} = -2 \cos (2 y'(x))y''(x)=0 $. In particular, $y''(x) = 0$ and so $y(x) = x$ giving $E[y] = \cos^2 1$.
With $y^*(x) = x^2$, we see that $(y^*)'(x) = 2x$ and $E[y^*] = {4+\sin 4 \over 8} < \cos^2 1 = E[y]$.