When we look at the interval $[0,\pi)$ as an interval in $\mathbb{R}$, we know $\{\sin(nx)\}_n$ forms a basis of $L^2(0,\pi)$ since they are the eigenfunctions of Laplacian (negative second derivative) with Dirichlet boundary condition, with eigenvalues $n^2$.
However if we look at $[0,\pi)$ as a torus, then the eigenfunctions of Laplacian with periodic boundary condition will contain a constant function, $\sin(2nx)$ and $\cos(2nx)$, in some sense this agrees with Fourier theory which is always defined on a torus instead of an interval.
My question is: the $L^2[0,\pi)$ space is different when $[0,\pi)$ is seen as an interval vs a torus, we need more functions to form a basis in the latter case. The possible causes of this could be:
- $L^2[0,\pi)$ w.r.t. torus is a bigger space
- $L^2[0,\pi)$ w.r.t. torus has a stronger $L^2$-norm.
But I think neither of these are true, so what is the cause of this subtle difference.
A more general question, it certainly makes sense to define a "fourier series" with only $\sin(nx)$ terms in the infinite sum when we look at $[0,\pi)$ as an interval. But why do we only care about Fourier analysis on torus?
Fourier looked at many types of problems associated with the heat equation. The most common types all lead to selfadjoint problems with complete orthonormal bases of eigenfunctions \begin{align} & u_{xx}=\lambda u \\ &\mbox{(periodic) } u(0)=u(\pi),\; u'(0)=u'(\pi) \\ &\;\;\;\; 1,\cos(2nx),\sin(2nx)\\ &\mbox{(Dirichlet) } u(0)=0=u(\pi), \\ & \;\;\;\; \sin(nx)\\ &\mbox{(Neumann) } u'(0)=0=u'(\pi) \\ & \;\;\;\; 1,\cos(nx)\\ \end{align} The first one, the periodic case, was used by Fourier to study the heat equation for a circular metal ring (torus.) All of these three generate a complete orthogonal basis for $L^2[0,\pi]$.