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Three Eigenvalues do not Determine a Triangle

Speaker: Siddharth Mulherkar
Date of Talk: May 5, 2025

1. The Problem

Suppose \(D\) is a bounded domain in \(\mathbb{R}^2\), and suppose one knows the set of all eigenvalues of the Laplacian on \(D\): \[\begin{cases} - \Delta u_k = \lambda _k u_k & \textrm{on }D, \\ u_k = 0 & \textrm{on }\partial D. \end{cases}\] Knowing \(0 < \lambda _1 \leq \lambda _2 \leq \lambda _3 \leq \cdots\), is it possible to determine \(D\), up to isometric symmetries?

Imagining \(D\) is an artisanal drum, one can phrase the question as: “Can you hear the shape of a drum?” The answer is no: in 1992, two nonisometric polygonal domains with the same spectrum of the Laplacian were exhibited.

One can repeat this question on simpler domains: if \(T\) is a triangle, does the spectrum of \(\Delta \) determine \(T\)? This time, the answer is yes, though early proofs needed to leverage the entire spectrum to determine \(T\).

But triangles are parameterised by \(3\) dimensions of information, so surely we don’t need the whole spectrum to find \(T\). It was shown that, given a triangle \(T\), only the first \(N(T)\) eigenvalues suffice to determine \(T\); naturally, one expects that any \(3\) eigenvalues will do. This is not the case:

This was done with a computer-assisted proof.

2. The Thematic Approach

This is a seemingly intractible problem for computers — one must assess equality of two quantities with an existence statement. The following approach is thematic, as just like the other talks this quarter, one applies some heavy theoretical machinery to convert this into a computer-provable problem.

In this case, the authors first parameterised the space of all triangles (up to isometry) by the square \(\left[ 0, 1 \right]^2\). To each triangle, one defines \[\begin{align*} \xi _{41}(T) = \frac{\lambda _4(T)}{ \lambda _1(T)} && \textrm{and}&& \xi _{21}(T) = \frac{\lambda _2(T)}{\lambda _1(T)}.\end{align*}\] Thus the problem reduces to find two triangles \(T\) and \(T’\) such that \(\xi _{41}(T) = \xi _{41}\left( T’ \right)\) and \(\xi _{21} \left( T \right) = \xi _{21} \left( T’ \right)\).

Now apply Miranda’s theorem, a generalisation of the intermediate value theorem and a variant of Brouwer’s fixed point theorem: all one has to do is check that both \(\xi _{41}\) and \(\xi _{21}\) must “surround” a common value at two different points. By using some neat geometric properties, this can be sped up farther. The point is that this big theorem reduces the problem to checking the values of \(\xi _{41}\) and \(\xi _{21}\) on a couple of line segments in \(\left[ 0, 1 \right]^2\).

3. Delicacies and Inefficiencies

There’s another computational problem: how do you actually compute \(\xi _{41}\) and \(\xi _{21}\)? One approach is to use the finite element method after discretising a triangular domain into many little triangles. This gives lower bounds on \(\lambda _k(T)\), but this ends up being inefficient both as an inequality and as a computational process; in fact, the authors use this to determine a lower bound on \(\lambda _5\).

Instead, the authors proceeded by the “method of particular solutions”. One restricts their view to the subspace spanned by Fourier-Bessel functions, and one solves \(- \Delta \widetilde{u} = \widetilde{\lambda } \widetilde{u}\) with only an approximate boundary condition. It can be shown that \(\widetilde{u} \to \lambda _k\) for some unknown \(k\).

When combined with the lower bound on \(\lambda _5\) from the FEM, this will determine one of the first four eigenvalues. Moreover, one can determine how accurate \(\widetilde{\lambda }\) is in terms of these \(\widetilde{u}\)’s. Once you find four different \(\widetilde{\lambda }\)’s, each sufficiently accurate, you can determine \(\lambda _1,\ldots, \lambda _4\).

(There is a problem with repeated eigenvalues, and apparently many numerical methods — including the above approach — begin to fail.)