# Can two drums almost sound the same?

Let $$D\subset \mathbb R^2$$ be a region and let $$\Lambda=\{\lambda_1,\lambda_2,\dots\}$$ be the set of eigenvalues of the Laplacian $$-\Delta$$ (with boundary condition $$\psi=0$$ on $$\partial D$$).

Mark Kac, in his famous article Can One Hear the Shape of a Drum?, only considers asymptotics of the function $$\sum_{\lambda\in\Lambda}e^{-\lambda t}$$ as $$t\to0^+$$, hence cannot detect if finitely many $$\lambda_i$$'s are modified. This lead me to wonder:

Can there be two planar regions $$D_1$$ and $$D_2$$ with Laplace eigenvalues $$\Lambda_1$$ and $$\Lambda_2$$, respectively, such that there exists a $$\lambda>0$$ such that $$\Lambda_1\cap\mathbb R_{>\lambda}=\Lambda_2\cap\mathbb R_{>\lambda}$$ but $$\Lambda_1\ne\Lambda_2$$? In other words, can two regions be "almost" isospectral without actually being isospectral?

• I don’t know if $\lambda\approx 0$ correspond to low or high wavelength tones, but if it is high wave-length you could rephrase this as drums that young people can distinguish, but old people cannot. Commented Jul 13 at 1:23
• @ChristianRemling I don't understand, that satisfies the condition $\Lambda_1\cap\mathbb R_{>\lambda}=\Lambda_2\cap\mathbb R_{>\lambda}$ but it does not satisfy $\Lambda_1\ne\Lambda_2$. Commented Jul 14 at 16:51
• @KentaSuzuki: It is not a formal argument (hence the scare quotes). What you are asking for feels easier than having the eigenvalues agree completely. Commented Jul 14 at 18:00
• @ChristianRemling On the other hand, it is possible for two distinct number fields to have the same zeta function but it is not possible to have the same zeta function except for finitely many Euler factors and then have those Euler factors different. Commented Jul 16 at 17:25
• Sure: Cor 1 of this paper, see also Thm 1 of that paper for a refinement where infinitely many different eigenvalues are allowed but of low enough density. Commented Jul 17 at 6:20