# 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 $y(0)=y(1)=0$

Here is one possible strategy:

1. 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 $$\tag{B} f(0)~=~0~=~f(1).$$

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

3. Conclude from eq. (C) that the stationary solution is a unique global minimum.

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 $y(a)=y_0$, $y(b)=y_1$, $y'(a)=y'_0$, $y'(b)=y'_1$

• how is this an answer? I don't get it.. – Franco Mar 14 '15 at 22:28

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