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I've been suggested to this site by some nice people at mathoverflow.net

Before I get started, let me tell you a little about myself. I’m a fourth year Mechanical Engineering student at the University of Michigan, Dearborn Campus. This said, I am not a mathematician, however I do know a bit about math, and I also enjoy doing math.

Please excuse any redundancies I may make and my “lack of rigor”.

The problem I’m working on now is an interesting one, and it goes as follows:

Suppose we have some unknown function of some real variable x, namely f(x), which is “well-behaved” on some known, finite interval, {a,b}. What is meant by “well-behaved” is that it is at least once differentiable, once integrate-able, smooth, and continuous on {a,b}.

Suppose further that we are given some finite area-under-the-curve A, and some finite arc-length S. Suppose finally that we are given f(x=a) and f(x=b).

The problem is this:

Given the above restraints on f(x), prove that there always exists some unique function f(x) that “fits the bill”.

For example:

Say f(x) = Sin(x), a = pi/2, and b = pi.

We know that:

f(a) = 1, f(b) = 0, A = 1, S = Sqrt[2]*EllipticE[1/2] ≈ 1.9101.

The problem in this case would be to prove that f(x) = Sin(x) is the only function that has the known area A, and known arc-length S.

Side Note: f(x) in this case could also be Cos(x - pi/2), however, this would not constitute two unique functions, f(x).

A corollary question could be posed if a proof to the first problem was given, and is as follows:

Given the same criterion for f(x) as before, devise a method of determining what f(x) would have to be to “fit the bill”. That is, like in our previous example, given: A = 1, S = 1.9101; devise a method of determining that f(x) must equal Sin(x).

Another Side Note:

I have realized (with help from mathoverflow.net) that f(x)=(x-1)^2(x+1) and f(-x) have the same Area and arc-length on [-1,1]. Making a potentially erroneous proposition, let us say that for all polynomials of nth order P(x), P(-x) will have the same Area and arc-length on any [-c,c].

This would prove difficult to prove analytically due to the "ugly" expression for arc-length.

While in general f(x) and f(-x) are unique functions, there must be some way of thinking about them as similar enough to "lump" them into one unique function... Just a thought

Thank you all. I look forward to any insight you can shed on the problem.

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3 Answers 3

As explained by Eric and phv, such a function is not unique in general. However, there are some cases in which it is unique. For example, if the arclength is $\sqrt{(b-a)^2 + (f(b)-f(a))^2}$, then the only continuous function from (a,f(a)) to (b,f(b)) is the straight line. Then there is only one possibility for the are under the curve, (b-a)(f(a)+f(b))/2

For any bigger choice of arclength from (a,f(a)) to (b,f(b)), there is a maximum possible area under the curve and a minimum possible area under the curve. (These are symmetric problems-- if you solve one, you'll solve the other.) For any area between these maximum and minimum values, there will be infinitely many solutions.

It is an interesting question to identify these curves with fixed arclength that give maximum and minimum area under the curve. Can you think of any solutions, or perhaps the general solution? The straight line is the most trivial example.

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One caveat to my claim: if the arclength is large enough, the way to get the most area under the curve will be for the curve to go straight up from (a,f(a)) and (b,f(b)) for some distance. Of course, this curve is then no longer the graph of a function. However, it is the limit of the graphs of functions approaching the maximum bounded area. –  Jonas Kibelbek Feb 10 '11 at 18:19
    
This is a good point, +1. But the new question is how do you find such extreme examples...? –  Eric Naslund Feb 10 '11 at 18:28
    
@Eric: I'm pretty sure they will all be circle arcs. The way to prove this would be calculus of variations. I believe I saw a proof for a similar problem long ago, but I don't know the subject well myself. –  Jonas Kibelbek Feb 10 '11 at 18:56
    
Would an example be a Catenary? A function that minimizes its integral for a fixed interval and arc-length? –  user6871 Feb 10 '11 at 19:00
    
@user6871: A catenary is a good suggestion; it minimizes something (tension in a hanging rope, I think?) for a fixed arclength. But does it minimize area under the curve? Check for yourself: compare it with a circle arc. –  Jonas Kibelbek Feb 10 '11 at 19:07

Consider the infinite family of functions $\sin(nx)$ on $[0,2\pi]$. Each will have an integral of $0$, and will have the same endpoints, namely $0$ at each end. Can we choose a constant so the arclength's are the same? Yes we can.

Take the slightly different family of functions $\frac{1}{n}\sin(nx) $. Again, the endpoints are $0$ and the integrals are the same. But this time the arclength is the same for any $n$. We have $$\operatorname{arclength}(\frac{1}{n}\sin(nx))=\int_0^{2\pi} \sqrt{1+(\cos nx)^2}dx$$ making the substitution $x=\frac{u}{n}$ this becomes $$\frac{1}{n}\int_0^{2n\pi}\sqrt{1+\cos^2 x}dx=\int_0^{2\pi}\sqrt{1+\cos^2 x}dx$$ which is independent of $n$. (and where the last equality follows from the periodicity of $\cos x$)

Thus the answer to your question is no. The example above is an infinite family of analytic functions with the same arc-length, integral and endpoints.

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Perhaps I don't understand the problem, but is this proposition true at all.

Suppose let a=0, b=1, and f(x) equal 1 everywhere except on some short interval (x0, x1) where there is a smooth bump. The bump could be anywhere on the interval.

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I wonder what is the best we can do. What about equivalence classes under translation, or analytic functions. No. What condition will give the desired outcome then. Does Monotonicity? Again no, I have an infinitely differentiable counter example (I can draw it but not write it down). But what about an analytic monotonic function? Are those unique? –  Eric Naslund Feb 10 '11 at 18:03
4  
Let's think mechanically. Suppose we have a tank that is suitably thin (narrow) so as to represent a plane. We fix a non-stretching membrane from one end to the other that is long enough to hang loose. Ignoring the problem of leaking around the edges, we put in enough water so that we can evacuate all the air. The membrane represents the function and the water represents the area. We would be able to flex the membrane into all sorts of smooth shapes. Only when we pump in the max amount of water that will fit beneath the membrane with the shape be strictly determined. –  phv3773 Feb 10 '11 at 19:12

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