Calculus of Variations and the Italian Challenge

One thing I did this summer was a course on the so called ‘Calculus of Variations’. A good example question would go something like this.

Consider the set of all functions y, continuous with continuous derivatives, such that y(0) = 0, and y(1) = 1. Find the function y that minimizes the value of

J is called a functional, taking a function and returning a real number.

Taking, for example, y(x) = x (notice that it satisfies the smoothness requirements, and the boundary conditions), the above integral comes out to 4/3. So we know that the minimum value of the integral is, at most, 4/3.

Suppose you thought that the minimzing function were a quadratic polynomial, . The smoothness conditions are necessarily satisfied. For y(0) = 0, that requires c = 0. y(1) = a + b, so y(1) = 1 requires b = 1 – a. So we’re looking for a function of the form . If you evaluate the integral for that function, you get

If you minimize the above expression with respect to a, you find that the minimum value occurs when a = 5/22, and that J(y) = 347/264, or approximately 1.314, ever so slightly less than 4/3. So,

performs slightly better than y(x) = x … but that isn’t enough to show it’s an absolute minimizer.

You could continue on in this way, checking other functions but – let’s be honest. You’re really shooting blindly, because you don’t know the form of the function. Expanding out polynomial minimizers, it may lead you to the taylor expansion of the actual minimizer, but it would be messy and unsatisfying.

Using techniques in the Calculus of Variations, you can solve the problem with some degree of elegance. Elegance that I won’t put forth in detail here – I really just meant for this to be a brief introduction. But you can turn such a minimization problem into an associated differential equation problem. The nice thing about this is that, in comparison to these beasts, differential equations are relatively well understood. The associated problem is as follows. If y(x) minimizes J(y), then y(x) must satisfy this differential eqation problem, with boundary conditions y(0) = 0, y(1) = 1,

You can solve that using your favorite technique to yield the function,

Evaluating J(y), this yields

And this, you can show, is the absolute minimizer of J. Notice, interestingly, the quadratic solution was actually quite close to being the real minimizer. Neat!

Now, you can do all kinds of nice things with this. For example, describing shapes with functions, you can find the shape a hanging cable of fixed length assumes. You can find the curve that minimizes the amount of time it takes to slide a bead along a wire of that curve (the assumption here is, sliding under gravity). And suppose you were an ancient African queen, promised howevermuch land you could encompass with a bull hide, you might be interested in maximizing the total area you could encompass with a fixed amount of material. Many many things.

Of course, Mathematicians always have to go and take things to the next level, which brings me to the Italian Challenge. Now, as the story was presented to me, in Italy, they have these tests for professorships at universities. However, if the administration has a particular candidate in mind, they’ll tailor the test for their strengths. Some candidate must’ve been an expert at calculus of variations, because we have the following problem.

Consider the set of functions, continuous with a continuous first derivative, over the range of x from 0 to 1. We also have the restriction that y(0) = 0, and y(1) = 1. Given

Prove that J(y) has no minimizer of that set of functions.

I’m fascinated. In essence, given any function y(x), you can always find a function w(x) such that J(y) is greater than J(w).

This is interesting on a number of levels. Notice, for instance, that each term of the integrand is positive. Thus we know that, for any function y(x), J(y) is strictly greater than 0. So the values of J are bound from below, but have no minimum! Very interesting.

The associated differential equation problem goes as follows. If y(x) minimizes J(y), y(x) must satisfy y(0) = 0, y(1) = 1, and

Notice, interestingly, you’re not guaranteed that y”(x) exists or is continuous.

But, if J(y) has no mimimizer, and we know that if y(x) minimizes J(y) it satisfies that differential equation, there must be something wrong with that differential equation. It can’t have a solution – because then J(y) has a minimizer. But why?

I don’t know…