Eigenvalues of the circle over the Laplacian operator

Question

I would like to find the spectrum of a circle. It seems that there are no boundary conditions, but I'm not quite certain. How could I find the spectrum of a circle over the Laplacian operator?

Answer 1

Not a solution, but a sketch. You need to find a manageable expression for $\Delta f$ if $f:S^1 \rightarrow \mathbb{R}$. One option is to work with a (almost onto) chart of $S^1$, e.g. by using an arclength parametrized curve $c:(-\pi,\pi)\rightarrow S^1$ and by looking at the resulting equation for $f\circ c$ instead. The fact that $f$ is assumed to be continuous on the circle imposes a periodicity condition on $g:=f\circ c$. I'm rather sure (but did not check, you should do that) you'll end up with the equation $ g^{\prime\prime} = \lambda g$ for the eigenvalue equation in this setup, which implies together with the periodicity condition that the corresponing solutions are given by $\sin(kx)$ and $\cos(kx)$ with $k\in \mathbb{Z}$, so the spectrum will consist of the squared integers (multiplicities to be calculated, be careful with the sign). (Assuming I did not make a silly mistake on my way to the result).

Edit: maybe I should add that I, without proof, assumed that the spectrum consists of Eigenvalues only.