Identifying extrema of a functional 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!
 A: $f(t)=0$ 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 $-\infty$.
A: To find the extrema of
$$
J(f)=\int_{-3}^{-2}\left(f^2(t)+f'(t)\right)\mathrm{d}t
$$
note that
$$
\begin{align}
\delta J(f)
&=\int_{-3}^{-2}\left(2f(t)\,\delta f+\delta f'(t)\right)\mathrm{d}t\\
&=\int_{-3}^{-2}\left(2f(t)+\delta_0(t+2)-\delta_0(t+3)\right)\delta f\,\mathrm{d}t
\end{align}
$$
which indicates a critical function would look like
$$
f(t)=\tfrac12\delta_0(t+3)-\tfrac12\delta_0(t+2)
$$
where $\delta_0$ is the usual Dirac delta function.
To approximate this function, we can test
$$
f_n(x)=\left\{\begin{array}{}
n-n^2(x+3)&\text{if }x\in\left[-3,-3+\frac1n\right]\\
0&\text{if }x\in\left[-3+\frac1n,-2-\frac1n\right]\\
-n-n^2(x+2)&\text{if }x\in\left[-2-\frac1n,-2\right]
\end{array}\right.
$$

Computing, we get
$$
J(f_n)=-\frac43n
$$
We can also test
$$
g_n(x)=\left\{\begin{array}{}
n^2(x+3)&\text{if }x\in\left[-3,-3+\frac1n\right]\\
2n-n^2(x+3)&\text{if }x\in\left[-3+\frac1n,-3+\frac2n\right]\\
0&\text{if }x\in\left[-3+\frac2n,-2-\frac2n\right]\\
-2n-n^2(x+2)&\text{if }x\in\left[-2-\frac2n,-2-\frac1n\right]\\
n^2(x+2)&\text{if }x\in\left[-2-\frac1n,-2\right]
\end{array}\right.
$$

Computing, we get
$$
J(g_n)=\frac43n
$$
Thus, there are no extremal functions since functions can be found that go to both $\pm\infty$.
This is not a good example to show how to determine maxima and minima in Calculus of Variations since extremal functions do not exist.
A: OP's functional is of the form
$$J[f]~=~f(b) -f(a) +\int_a^b \mathrm{d}t~f(t)^2. \tag{A}$$
An infinitesimal variation reads
$$\delta J[f]~=~\delta f(b) -\delta f(a) +2\int_a^b \mathrm{d}t~f(t)~\delta f(t). \tag{B}$$
If we consider a test function of the form
$$ f(t)~=~|t-c|^{-1/3}, \tag{C}$$
then the integral (A) is bounded but we can make $f(a)$ or $f(b)$ go to $+\infty$  by letting $c\to a$ or $c\to b$, respectively. This shows that OP's functional (A) is unbounded from both above and below.
