I find the problem of hearing the shape of a drum fascinating. Specifically, given two connected subsets of $\mathbb R^2$ with piecewise-smooth boundaries (or a suitable generalization to a riemannian manifold) it is in general impossible to use the spectrum of the laplacian to uniquely identify the shape of the domain.

As a simple example, reported in the Wikipedia link above, these two shapes

enter image description here

have identical laplacian spectra, as explained more at length by Gordon and Webb in American Scientist 84 no. 1, 46 (1996) (jstor). This can be shown by taking any eigenfunction $\varphi$ on $D_1$, cutting it up into the marked triangle regions, and taking the specified linear combinations on $D_2$ to make a laplacian eigenfunction with the same boundary conditions, correct matching at the internal boundaries, and the same eigenvalue.

What bugs me about this problem is that the spectrum is not really what you hear on a real-world drum. More specifically, the spectrum gives you the frequencies of the fundamentals and overtones that the drum is allowed to produce, but it does not tell you in what proportions it will do so. If you hit a circular drum in the centre, you will excite a large amplitude of the fundamental, resulting in a deeper sound, but if you drum it near the edge you will produce a rapping sound with a stronger proportion of higher overtones.

The set of sounds that a drum can produce is therefore better modelled by the pair of

  • its Laplace-Beltrami spectrum, together with
  • the set $T$ of timbres it can produce.

    By a timbre I mean the sequence of eigenfunction weights when the drum is hit at $x$, i.e. $$\tau(x)=\left( \sqrt{\sum_k|\varphi_{n,k}(x)|^2} \right)_{n=1}^\infty,$$ where the $\varphi_{n,k}$ are the (possibly multiple) orthonormalized eigenfunctions corresponding to the eigenvalue $\lambda_n$, and two timbres are thought of as equivalent if they only differ by a global constant (which trivially corresponds to how hard you drum).

    (This quantity is meant to model the amount of energy in each eigenspace after a point excitation to the wave equation (via e.g. $\Delta \varphi-\partial_t^2\varphi=0$, $\varphi(y,0)=0$, $\partial_t\varphi(y,0)=\delta(x,y)$), and perhaps there are cleaner or equivalent definitions for $\tau(x)$ based on that PDE. However, the definition above is plenty to satisfy the intuition of how the musical timbre depends on the drumpoint from a physicist's perspective.)

    The set of all timbres is then $T=\{\tau(x):x\in D\}$: that is, you're provided all the timbres, but not which place in the geometry they come from. (As pointed out in the comments, initial knowledge of the geometry renders the whole thing moot.)

In this paradigm, two isospectral surfaces $D_1$ and $D_2$ would still be 'acoustically distinguishable' if, say, $D_1$ contains a point $x^*$ which produces a timbre $\tau^{(1)}(x^*)$ that is not matched by the timbre $\tau^{(2)}(y)$ of any point $y\in D_2$. In other words, you can't see the drum, and you can't see where the drummer is hitting it, but if drummer 1 can produce a timbre that drummer 2 cannot, then the drums must be different.

(Alternatively, one might have access to a function $u\mapsto \tau(f(u))$, where $u$ lives in some standard domain $D_0$ like the unit disk and $f:D_0\to D$ is an unknown (homeo/diffeo/etc)-morphism. That is, you get a control that specifies where the drum should be hit, but you don't know exactly what it does. You can then say two drums $D_1$ and $D_2$ are indistinguishable if there's a morphism $g:D_1\to D_2$ such that $\tau^{(2)}(g(x))=\tau^{(1)}(x)$ for all $x\in D_1$, i.e. a mapping from each point $x$ in $D_1$ to an identically-sounding drum point in $D_2$.)

So, to come to my question: has this paradigm been explored in the literature? For the known isospectral-but-not-isomorphic regions (or more complex manifolds), are any pairs known to be acoustically indistinguishable as defined above? Or must acoustically indistinguishable surfaces be isometric? Does this concept have an established name in the literature? If it has been explored, are there good introductory surveys (say, pitched at a level where a physicist might understand them)? If it hasn't, are there fundamental reasons why this is a much harder problem overall?

  • 3
    $\begingroup$ If you know the spectrum of the Laplacian and the restrictions of the eigenfunctions to a tiny region $R\subset D$, then you can determine the shape of that region $R$. www3.nd.edu/~lnicolae/RandMorseSpecGeom.pdf $\endgroup$ – Liviu Nicolaescu Jan 5 '16 at 16:36
  • 2
    $\begingroup$ @LiviuNicolaescu: Should "the shape of that region $R$" be "the shape of $D$"? $\endgroup$ – Jason Starr Jan 5 '16 at 16:38
  • $\begingroup$ By shape I understand the Riemann metric defining the Laplacian. In the case at hand the metric is Euclidean so there is not much to say. $\endgroup$ – Liviu Nicolaescu Jan 5 '16 at 19:12
  • 1
    $\begingroup$ Very nice question and I wonder whether there is any connection to the theory of spectral maximal partitions. $\endgroup$ – Delio Mugnolo Feb 5 '16 at 11:02
  • 1
    $\begingroup$ @EmilioPisanty A good starting point, with a survey of earlier results and more numerical experiments than proofs is journals.cambridge.org/action/… $\endgroup$ – Delio Mugnolo Feb 5 '16 at 19:10

Sometimes you can

Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.

To explore the eigenfunctions, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, shown as thick black lines, though of course the timbres $\tau(x)$ as I defined them contain more information in terms of the relative weights of the eigenfunctions when they're not zero. To begin with, already by the fourth eigenfunction we're able to 'count' the shape of the drum as in Carlo Beenakker's answer (i.e. if you have access to the number of nodal domains for that eigenvalue, then you can immediately tell them apart). The nodes, however, also give quick and clear examples of the $x^*$s we seek.

In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

Specifically, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.

Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.

The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.

  • $\begingroup$ Very, very cool! One comment on the code: The lastest version of Mathematica has a more direct access to the eigensystem via DEigensystem. I haven't used it much, but it's probably faster. $\endgroup$ – Mark McClure Mar 9 '16 at 14:40
  • $\begingroup$ @Mark Thanks =). And thanks for the pointer - as usual, it's hard to keep track of the enormous bag of in-built tricks in Mathematica. $\endgroup$ – Emilio Pisanty Mar 9 '16 at 14:53

As mentioned in the comments, knowing both eigenvalues and eigenfunctions gives you enough information to find the shape of the domain, so to make this problem more challenging one might ask what minimal information on the eigenfunctions one would need to carry out the reconstruction. One answer to this problem was given in Can one count the shape of a drum? (2006).

The "counting" refers to the number $\nu_n$ of nodal domains of the $n$-th eigenvalue $E_n$. (A nodal domain is a connected region in which the eigenfunction has a fixed sign. In one dimension $\nu_n=n$, in higher dimensions $\nu_n\leq n$.) Gnutzmann, Karageorge, and Smilansky conjecture in the cited paper that the sequence $\{\nu_n, E_n\}$, $n=1,2,\ldots$, is sufficient to recover the shape of the domain. The conjecture has been proven for certain classes of isospectral domains (flat tori in three and four dimensions). It does not hold for the discrete Laplacian on a graph (see Isospectral graphs with identical nodal counts).

  • 1
    $\begingroup$ Please see the edited question, which I think represents my intent better. I certainly don't mean that one would have access to the eigenfunctions directly, i.e. one would have access to all the timbres produced by the drum but not geometrical information of where they come from. On the other hand, your nodal-domain-count reference does address the alternative (i.e. you know which timbres are produced near to each other, but not their geometry), since the nodal domains would survive a homeomorphism, but if I read you correctly it's undecided for planar domains. $\endgroup$ – Emilio Pisanty Jan 6 '16 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.