It’s that time of year again! The Putnam math competition is this Saturday, 12/1. Here’s a hard calculus problem from last year’s Putnam. As always, I’ll try and justify my thinking processes that led me to the solution (even though they are indirect).

**Putnam 2011/A5:**

Let and be twice continuously differentiable functions with the following properties:

- for every ;
- for every , and ;
- for every , the vector is either 0 or parallel to the vector .
Prove that there exists a constant such that for every and any , we have

**Gut reaction: Yikes!**

This problem looks very intimidating. I never got beyond this point on the Putnam. Decided to work on the other problems instead.

But let’s solve it.

**Our first step is to justify to ourselves why such a convoluted statement might be true.**

We’ll do this by looking at a *specific example* of a that satisfies the conditions. *Why is each condition essential?* Why would satisfy the conclusion? Could we generalize from our example?

**Let’s find a F.**

The simplest thing to do would be to take . (Technically we have to take something else around but let’s not worry about this; we’re just trying to get some intuition.) Then is parallel to . Integrating against and , we find that one possibility for is . The problem claims that . Is this true? Let’s suppose all the are at least 1 (we didn’t define very well around ). Suppose . Then add up to by *telescoping*. So one of the differences must be at most . We can just take .

**What was essential here?**

The fact that the integral of is , which converges as . In general, the condition , or , forces the antiderivative to converge as . One possibility for is . Then our argument would go through as above: again, we have by telescoping

where . Thus letting we find that .

**But we’re just getting started…**

We now have some idea why the statement would be true. However, we have not solved the problem at all! We took a such that . We need to prove it for all such that . This means the gradient could be scaled by an arbitrary number depending on :

.

This would mess up any way to write as a difference of a function in and as above. What do we do?

(Thus we see that to solve our problem, we have to give at least a partial answer to the question: given only that the gradient field of a function is *parallel to* a given vector field, rather than the gradient field itself, how do we recover the function?)

**A condition on gradient fields gives a condition on h**

It’s clear that we need to understand what could be. Let’s find some condition on . We know that in order for some vector field to be a gradient field, we must have

(This fact comes from the fact that for a twice continuously differentiable function , .)

Applying this to gives

**We can’t “solve” this PDE… or can we?**

I got stuck here for a very long time. The basic problem is that it seems like “too many” functions solve this equation. After all, we have a PDE in two variables, but we only have 1 equation relating them. How can we hope to say *anything *about ?

The idea is that would have a *very general form*. We take as inspiration the wave equation

which has as a general solution the traveling wave where is *any* “nice” function. The equation is not enough for us to give a very explicit solution, but we’ve still managed to describe all solutions: they must be functions in terms of the quantity !

Can we do the same here? Is a function in some expression? What would be a natural choice? The equation for looks actually a lot like the gradient equation for . Of course, is a solution. Playing around a bit, we find is a solution—look familiar? Let’s guess that a function in , i.e., we’ll try to prove for some function .

This is the key observation for the problem.

**h is a function in G(u)-G(v) because G(u)-G(v)=k are the level lines**

Saying that for some function is the same as saying that the value of depends *only* on the value of . Why would this be the case? (If the above ideas were kind of unmotivated, hopefully our proof now will be enlightening!) Another way of saying this is that *doesn’t change *if we move along the curve for constant.

In fact, (1) is exactly telling us that doesn’t change along a certain direction: it says that

i.e., the directional derivative of in the direction is 0. Now *are exactly the integral curves of *!*** Indeed, implicitly differentiating gives .

Suppose we move along from to with unit speed in the direction. Let this path be . How does change? Using (2),

.

Since is connected (see below), this shows is a function of . We can write

for some function .

***(We should really check some things here. Note is an at most single-valued function of . Indeed, fixing a , is a strictly increasing function of because is the integral of . There is at most one value of that works. Moreover, the curve is connected. Indeed, if and are points on the curve, then for any we have , so by the Intermediate Value Theorem, there is a solution to . Moreover, continuity considerations show that depends on continuously. Or you can probably just argue by taking the derivative.)

**Now finish.**

We’ve gotten through the hard part: we know what form has to take. We know

Since we have an expression for the gradient field, we can now find up to a constant, in terms of this function . Let be the antiderivative of . Then

for some . But the condition actually gives (we have by choice of antiderivative).

Now suppose . Again, we note is increasing and

This means that for some . Then

Because has continuous derivative, on this derivative has a maximum, say . Then . Hence

We let and we are done.

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