I'm new on Calculus of variations and I don't figure out how to find a minimum (or maximum) for the following functional
$$ J(f) = \int_{-3}^{-2}(f^2(t)+f'(t)) ~dt . \tag{1}$$
I have tried to use the Euler-Lagrange equation
$$\frac{\partial\mathcal{L}}{\partial f}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial f'}\right) = 0, \tag{2}$$ where $\mathcal{L}\left(f(t),f^\prime(t),t\right) = f^2(t)+f'(t).$ However, the only solution that I have found was $f(t) =0$, but I don't know if $f(t)=0$ is a local minimum, local maximum or a saddle path of $J$.
Question: If $f$ satisfies the Euler-Lagrange equation, how to test if $f$ is a minimum, maximum or saddle path ? In ordinary calculus, we can compute the Hessian, but I don't know how to proceed in calculus of variations.
Thanks in advance for any help!
Best Answer
$f(t)=0$ is a saddle, indeed there exist
$f_1(t)=-\frac{3}{38}t$ and $f_2(t)=t$ such that
$J(f_1)=-\frac{3}{76}<0=J(f)<\frac{22}{3}=J(f_2)$.
Since from $g(t)=kt$ it follows that $J(g)=\frac{19}{3}k^2+k$, it means that the functional $J$ is unbounded from above, so it does not have maximum value and supremum of $J$ is $+\infty$.
Moreover there exists a sequence of functions $\{h_n(t)\}_{n\in\mathbb{N}-\{0\}}$ defined as $h_n(t)=\frac{1}{2\left(t+2-\frac{1}{n}\right)}$ such that
$J(h_n)=-\frac{n}{4}\left(1-\frac{1}{n+1}\right)$.
It means that the functional $J$ is also unbounded from below, so it does not have minimum value and infimum of $J$ is $-\infty$.