# Thread: Find the extremal to the functional and discuss whether they provide a max/min

1. ## Find the extremal to the functional 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 $y(0)=y(1)=0$

2. ## Re: Find the extremal to the functional and discuss whether they provide a max/min

Hi dtwazere.

Calculus of Variations can seem a little daunting, however hopefully a few comments will make things seem a little less scary.

1) You can think of the term "functional" in this case as something that sends functions to real numbers. The definition of a functional is actually a little more technical than that (it is a map from a vector space to its underlying scalar field - see Functional (mathematics) - Wikipedia, the free encyclopedia) - in your case the vector space is the space of continuously differentiable functions with $y(0)=0$ and $y(1)=1$ and the "field" is the real numbers. However, thinking of a functional as something whose domain is a certain set of functions (in your case continuously differentiable functions with $y(0)=0$ and $y(1)=1$) and whose range is the real numbers is good enough for now.

2) You have actually done most of the hard work already by solving the Euler-Lagrange equation to find a possible minimizer. I will denote your possible minimizer by $y_{min};$ i.e. $y_{min}(x)=-\frac{1}{4}x^{2}+\frac{5}{4}x.$ Note that my notation is a little dangerous because we haven't actually proved that $y_{min}$ is actually a minimizer of $J$ yet. The last thing we need to note is that the condition $f(0)=0=f(1)$ was left out in the original post where you mentioned $J(y+f).$ The condition $f(0)=1=f(1)$ is essential; $y_{min}+f$ represents a small pertubation of $y_{min}$, however we don't perturb the endpoints of $y_{min}$, which is where the condition $f(0)=1=f(1)$ comes from.

Now we compute

$J(y_{min}+f)=\int_{0}^{1}[(y_{min}+f)']^{2}-(y_{min}+f)dx$

expanding this we get

$J(y_{min}+f)=\int_{0}^{1}(y_{min}')^{2}-y_{min}dx+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx$

Which becomes

$J(y_{min}+f)=J(y_{min})+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx$

If we integrate by parts on the second term on the RHS we obtain
$2\int_{0}^{1}y_{min}'f'dx=2(y'f |_{0}^{1} -\int_{0}^{1}y_{min}''fdx)$

Now use $f(0)=0=f(1)$ and $y_{min}''=-\frac{1}{2}$ on the previous line to get

$2\int_{0}^{1}y_{min}'f'dx=\int_{0}^{1}fdx$

Using the last line in $J(y_{min}+f)$ we have

$J(y_{min}+f)=J(y_{min})+\int_{0}^{1}fdx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx=J(y_{min})+ \int_{0}^{1}(f')^{2}dx\geq J(y_{min})$

This proves that $y_{min}$ really is a minimizer of $J$ for the class of functions that you're considering.

Does this help? Let me know if anything is unclear.

Good luck!

3. ## Re: Find the extremal to the functional and discuss whether they provide a max/min

This really helps, I think I understand most of what you put, thanks a lot!