# Eigenvalues of a Plane Curve Laplace-Beltrami Operator

Given a closed plane curve $C$, which is a one-dimentional manifold, what are the eigenvalues of Laplace-Beltrami operator defined on this curve?

I know that the LB eigenvalue problem for unit circle is equivalent to the regular Laplacian eigenvalue problem for the interval of the length $2\pi$ with periodic Boundary conditions.

In a sense, the LB operator on circle can be viewed as a Laplacian of a function depending on the arc length. Therefore the eigenvalues of LB operator on a circle of arbitrary radius will be equal to the eigenvalues of unit circle LB, rescaled by the square of the radius. Does this also hold for any "good" closed plane curve?

To summarize, let me state again the

QUESTION: Do the eigenvalues of Laplace-Beltrami operator defined on an arbitrary closed one dimensional smooth manifold embedded in $\mathbb{R}^2$ depend only on the (arc) length of this manifold?

I perceive it to be counterintuitive for two distinct closed plane curves of the same total length to always have equal LB eigenvalues, but I could neither prove nor disprove this statement.

• Would you be able to find rigorously the eigenvalues of the circle in using LB operator an Helmholtz equation? – user1050421 May 20 '16 at 11:33

You are right. The reason is that if a closed curve $C$ has length $\ell$, then there is an arc length parametrization $\alpha : S^1_\ell \to C \subset \mathbb R^2$, where $S^1_\ell$ is the standard circle with are length $=\ell$. An arc length parametrization is essentially an isometry from $S^1_\ell$ to $C$, thus they have the same Laplace operator.