Can You Hear the Shape of a Drum? (Part 2)

In the last post, we explained how vibrating strings work.complicatedTo summarize, the string’s position, given as a function f(x,t), is controlled by a differential equation, \dfrac{d^2 f}{dt^2} = \dfrac{d^2 f}{dx^2}.1

5The left hand side of this equation, \dfrac{d^2 f}{dt^2}, is the vertical acceleration of the string. Meanwhile, \dfrac{d^2 f}{dx^2} measures how much the string is curving at one instant of time, as you move from left to right. The more the string is curving, the bigger this is.6The punchline of the last post was that any vibration of the string, including the complicated one above, can be represented by a sum of special “self-similar” solutions to the differential equation. These solutions keep their overall shape, and simply vibrate by scaling up and down.In order to keep their shape, these solutions had to satisfy a special equation, \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x), which says that the second derivative of the function (i.e., how much it is bending) must be equal to a constant -\nu^2 times the function itself. If we want the string to have length \pi, the \nu had to be natural numbers, \nu 1, 2, 3, \cdots. The video above are the solutions when \nu is 1, 2, 3, or 7.

The solutions f(x) of \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x), which represent the initial states of the string, are eigenfunctions of the “transformation” \dfrac{d^2}{dx^2}, with eigenvalues \lambda = -\nu^2.

The key observation is that the eigenvalue of the self-similar solutions controls the speed of the vibration. As you can see above, the \nu=7 solution vibrates seven times as fast as the \nu=1 solution.

The tone you hear from an instrument is related to how quickly the air (and thus the string) is vibrating. The self-similar solutions are the vibrations of the string that produce pure tones — a single note.

Since any vibration is a combination of these basic pure tone vibrations, when the string vibrates, you will hear all the tones of all the pure tone vibrations you combined together, each at different volumes depending on how much of that pure tone vibration you used. For instance, this vibration would have the \nu=1 and \nu=2 tones in equal amounts:sin1and2However, these pure tones are the only tones the string can produce.

In other words, the eigenvalues (of \dfrac{d^2}{dx^2}) control the tones a string can produce, while the tones a string can produce tell you the eigenvalues. The set of all eigenvalues (of \dfrac{d^2}{dx^2} for a particular string) is called the spectrum.

We finally get to the first question: Can you hear the length of a string?hearLengthHow could we hear the length?

If we vibrate a string at random, our vibration will probably2 be a combination of small amounts of all the pure tone vibrations. If we had perfect pitch (and could hear tones both very low, and infinitely high), we could hear all of those notes, and, from them, deduce the spectrum of the string.

37octavesSince the spectrum and the possible tones are equivalent, a more mathematical way to ask the question is: If you know the spectrum of a string, can you somehow calculate its length?3

To answer that, we need to know how the eigenvalues of \dfrac{d^2}{dx^2} change when the length of the string changes.

So far, we’ve assumed the length of the string was \pi, for simplicity. In that case, the eigenfuctions associated to the eigenvalue \lambda = -\nu^2 was f_\nu(x) = \sin(\nu x). While f_\nu(x) is an eigenfunction for any \nu, it could only represent a vibration on our string of length \pi if \nu was 1, 2, 3, etc. That was because we needed both sides of the string to be fixed.sinesIf our string is instead length L, we will still need both sides of the string to be fixed. For that to happen, we will need f_\nu(x) to be zero at the two ends, x=0 and x=L. That will happen when \nu = \dfrac{\pi}{L} n, where n is some natural number. (Notice that if L=\pi, this is exactly what we had before.) In other words, our eigenvalues are -\dfrac{\pi^2}{L^2} n^2.

That means you can hear the length of the string.doubtedMeFor example, the spectrum for a string of length \pi would be \{-1, -4, -9, \cdots\}, and you would hear the associated tones.

The spectrum for a string of length 2\pi would be \left\{-\dfrac{1}{4}, -1, -\dfrac{9}{4}, \cdots \right\}. Those two spectra (plural of spectrum) are quite different. In fact, you can even tell them apart even by just looking at the first eigenvalue.

You can hear the length of a string just by hearing its lowest tone.4

We’re finally prepared to talk about the original question we had: Can you hear the shape of a drum?yippeeAs with strings, the vibration of a drum is controlled by a differential equation. Unsurprisingly, it’s very similar to the one for a vibrating string. The function f(x, y, t) representing the position of the drum head at any particular time is more complicated now, though. It depends on time and both x and y, since the drum head is two dimensional.

The differential equation is \dfrac{d^2 f}{dt^2} = \Delta f. The left hand side, \dfrac{d^2 f}{dt^2} is the vertical acceleration of the drum head, as before.

The right hand side looks different, but mostly just because I’m using notation you may not be familiar with. The symbol \Delta (capital Delta) is called the Laplacian, and is involved in many of the most important differential equations. Since we have two space dimensions for our drum head, \Delta f = \dfrac{d^2 f}{dx^2} + \dfrac{d^2 f}{dy^2}, the sum of the second derivatives in the two space directions.5

That means the Laplacian is really just the same thing we had on the right hand side of the equation for a vibrating string. The quantity \Delta f adds up how much the drum head is bending in both the x and y directions.

So, as before, where the drum head is bending the most, it accelerates the most as well, bringing it back toward the resting position.restEven on a square drum, solutions to this differential equation can be quite complicated.3Dcomplicated.gifBut, as before, every solution is a combination of small amounts of special self similar solutions.

To find self similar solutions, we again need \dfrac{d^2 f}{dt^2} = -\nu^2 f — the acceleration of the drum head must be equal to a (negative) multiple of the current position.

Since this is the same equation as before, it has the same solution. If we had an initial position f(x, y, 0), the position at time t would be f(x, y, t) = f(x, y, 0) \cos(\nu t). The initial position simply vibrates up and down sinusoidally.3DBasic.gifThe frequency of the vibration, and thus the tone produced by the drum, is controlled by the value \lambda = -\nu^2. The larger \nu is, the higher the frequency, and thus the higher the tone produced.

That leaves the hard part of the problem.hardpartIn order to find initial positions of the drum head that will lead to these simple, self-similar solutions, we need to find solutions to the equation \Delta f = -\nu^2 f.

This is again an eigenvalue problem. The Laplacian transforms our function f into a new function. Usually this new function, \Delta f would be unrelated to the original function f, but we are looking for those special functions for which this transformation simply scales the original function by some constant -\nu^2.

This may not seem that much harder than it was for a string, but it turns out to be very hard to solve explicitly.

For a rectangular drum, it’s not too bad. It turns out the eigenfunctions are just products of sine waves–one in the x direction and one in the y direction. For instance, on a square of side length \pi, f(x,y,0) = \sin(2x)\sin(3y) is an eigenfunction with eigenvalue \lambda = -2^2-3^2 = -13.3D23.gifYou can also find explicit solutions for a circular drum, and for some triangles (equilateral, isosceles right, and 30-60-90), but for just about anything more complicated than those examples, we have no idea how to find explicit eigenfunctions and eigenvalues.6

But, fortunately, finding eigenvalues and eigenfunctions was not the question.

The question was “Can you hear the shape of a drum?” In other words, if you already know the eigenvalues (the spectrum), can you figure out the shape of the drum?

It’s been known for a long time that there are some things we can tell about the drum from the spectrum.

The most “obvious” one is the area of the drum head. A bigger drum head makes a lower tone.

Unlike the string, though, the lowest tone (equivalently, the eigenvalue closest to zero) is not sufficient to tell you the area. It’d be a good start, but it’s not enough.inadequateInstead, we have to look at how the eigenvalues are spread out.

For the string, the spectrum for length \pi was \{-1, -4, -9, \cdots\}. For length 2\pi, the spectrum was \left\{-\dfrac{1}{4}, -1, -\dfrac{9}{4}, \cdots \right\}.

The thing to notice is that, for the longer string, the spectrum is packed closer together. Another way to say this is that, for a given number -R, the spectrum for the string of length 2\pi has more eigenvalues closer to zero than -R than the spectrum for the string of length \pi. Though the formula may not be obvious, there is a way to calculate the length of the string based on this way of interpreting the spectrum.lengthMoreEigensFor a drum, a similar idea works. Suppose we knew the spectrum for a drum completely. We then could count how many eigenvalues are closer to zero than -R for any -R. We’ll call that number of eigenvalues N(R). (N for number of eigenvalues.)N10As R grows, the number of eigenvalues closer to zero than -R also grows. In fact, they grow in a very specific way.

Hermann Weyl (said “vile”) showed in 1911 that, for a drum head, N(R) will grow roughly linearly in R. Not only that, the slope of that line predicts the area of the drum head!HearAreaPrecisely, the slope can be measured as \displaystyle\lim_{R\to \infty} \dfrac{N(R)}{R}. (This is saying something like, “See how many times bigger N(R) is than R for large R. If N(R) were precisely a line, this would give the slope. It works just as well for N(R) that are almost a line.)

Once you have that slope, it’s easy to find the area. In fact, A = 4\pi\displaystyle\lim_{R\to \infty} \dfrac{N(R)}{R}!AreaWeyl’s law tells us that knowing the spectrum determines the area.

But the area is not the entire shape.

Besides the area, what other information can we figure out from the spectrum?

Weyl’s law says that N(R) \approx \dfrac{A}{4\pi}R. But we can be more precise than that. If we look at N(R) - \dfrac{A}{4\pi}R (i.e., see how far off our approximation was), that new thing looks about like a multiple of the square root function \sqrt{R}! And, again, the multiple tells us something about our shape.

Weyl conjectured (and it was proven by Victor Ivrii in 1980) that N(R) - \dfrac{A}{4\pi}R \approx \dfrac{P}{4\pi}\sqrt{R}, where P is the perimeter of the drum head. In other words, if we know the spectrum, a more careful analysis would give us the area and the perimeter of the drum.PerimeterThat would be enough to tell you the shape of a drum, if you knew that it was rectangular. You could hear the difference between a square (lots of area, not so much perimeter) and a stretched out rectangle (no so much area, lots of perimeter.)

This is hopeful. This \sqrt{R} level approximation was good, but there’s still more information lurking about in the spectrum. Though no better approximation results have been proven (that I know of), you might hope that you can hear the shape of a drum.

Unfortunately, you can’t.cursesIn 1991, Gordon, Webb and Wolpert found some shapes that are obviously different, but have the same spectrum. There are lots of examples now, but here are two of them:isospectralNotice that they both have the same area and perimeter. They have to, thanks to Weyl’s and Ivrii’s work. But, despite being different, they have the exact same spectrum.

Of course, these drums heads are not exactly normal shapes for drum heads. Most drums that I’ve seen have had drum heads that were at least convex, meaning they don’t cut in.ExoticDrumsIf you assume a few reasonable things about the shape of the drum head, it turns out that you can hear the shape of a drum. Our earlier example of being able to hear the shape of a rectangle is a strong example.

 

Within the last 10 years, Steve Zelditch proved a much better result. We assume the drum head is convex, has no holes, has very smooth boundary7, has at least one mirror symmetry, and a few other smaller technical assumptions.

DifferentDrumsIf we assume the shape follows those rules, Zelditch proved you can hear the shape of a drum. If you know the spectrum, you can reconstruct what shape the drum had.SoundsLikeThat’s it for drums. Next time, we’ll start talking about the limits of proof in mathematics–Gödel’s incompleteness theorem! It may be a meandering path, passing through ideas about how math is done, what axioms are, famous paradoxes in math, and more.

Short Bibliography


  1. This equation is often called the wave equation, since it controls the waves of the string. 
  2. Mathematically, a 100% chance. The probability of choosing a pure tone vibration, or even some finite combination of them, has a 0% chance of happening. 
  3. We are assuming that all strings are the same material, thickness, tautness, etc., so that the only question is what the length of the string is. 
  4. The lowest tone since the lowest eigenvalue \lambda represents the lowest frequency vibration possible, and thus the lowest sounding tone. Though, it should be mentioned one more time that this is assuming all the strings we’re comparing have the same material, thickness, tautness, etc. 
  5. In higher dimensions, we just continue adding second derivatives in each of the directions. 
  6. Though there are very good algorithms for estimating both eigenfunctions and eigenvalues for any shape of drum (and any dimension) on a computer. 
  7. The boundary needs to be analytic, which means you need to be able to take its derivative infinitely many times, and, in addition, the Taylor series for the boundary has to converge. This is a strong condition. For instance, this means there can’t be any corners on the shape. 
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