Find the extremal to the functional $J(y) = \int_{0}^{1} ((y')^2 -y)dx$ and discuss whether they provide a max/min I am having a hard time getting my head around Functionals and Calculus of Variations,
My question is: Given a functional and using the Euler-Lagrange equation to find an extremal, how do we show that the extremal provides a min/max (if it does)
The question I am working on is
$J(y) = \int_{0}^{1} ((y')^2 -y)dx$ with $y(0)=0, y(1)=1$
I found the extremal to be: $y(x) = \frac{-1}{4}x^2 +\frac{5}{4}x$ which I am told is a minimum to the functional problem.
However I am unsure on what is sufficient to show this, in the notes I have it is shown that:
$J(y+f) = J(y) + \int_{0}^{1}(f')^2dx \geq J(y)$ where f is continuously differentiable on the interval 0,1 with $y(0)=y(1)=0$
Thanks in advance!
 A: Here is one possible strategy: 


*

*Reparametrize the dynamical variable
$$\tag{A} y(x)~=~f(x)+\frac{5x-x^2}{4} , $$
where new variable $f(x)$ measures the deviation from the stationary solution $x\mapsto \frac{5x-x^2}{4}$. It obeys the Dirichlet boundary conditions
$$\tag{B} f(0)~=~0~=~f(1).$$

*Show that 
$$\tag{C} J[y]~=~\int_0^1\!  dx\left(y'(x)^2 -y(x)\right) ~\stackrel{(A)}{=}~...~\stackrel{(B)}{=}~\frac{25}{16}+\underbrace{\int_0^1\!  dx~f'(x)^2}_{\geq 0}.$$

*Conclude from eq. (C) that the stationary solution is a unique global minimum.
A: There are conditions based upon which you decide whether the extremal is a minimizing or a maximizing one.
There are 2 conditions 
1.legendre's condition
2. Weierstrass condition
The first one is a bit simpler to apply provided the necessary conditions have to be looked before hand.
$F_{y'y'}>0$ then it is a minima and more so strong one for all y close p where p is $y'$. And else it is a maxima.
Now here in this question it comes out to $2$.hence a minima
A: Find the necessary condition for a sufficiently smooth curve $y (x)$ to be an extremal of the functional 
$$
J(y)=\int_a^b L((x,y(x),y'(x),y''(x))dx
$$
With boundary conditions $y(a)=y_0$, $y(b)=y_1$, $y'(a)=y'_0$, $y'(b)=y'_1$
