Eigenvalues of the “Laplacian” on [0,2$\pi]\subset\mathbb{R}$

for the dirichlet eigenvalue Problem on compact and connected Riemannian manifolds, the eigenvalues of the laplacian consists of a discrete sequence.

On the other hand, if we consider $[0,2\pi]\subset\mathbb{R}$, the Eigenvalues of the laplacian $(-\frac{d^2}{dx^2})$ aren't just a discrete sequence.

What is wrong with the example above? I'm a little bit confused.

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Consider the problem first without the boundary conditions:

$$-u'' = \lambda u$$ If $\lambda \leq 0$, then the general solution is a superposition of real exponential functions which can never satisfy a Dirichlet boundary condition. Otherwise, for $\lambda > 0$, the general solution is given by

$$u(x) = A\sin(\sqrt{\lambda}x) + B\cos(\sqrt{\lambda}x)$$

Now $$u(0) = A\sin(0) + B\cos(0) = B$$ so to satisfy the Dirichlet condition at $x = 0$ we must have $B = 0$. Since eigenfunctions span subspaces of other eigenfunctions we might as well also normalize $A = 1$ so we can consider only $$u(x) = \sin(\sqrt{\lambda}x)$$ But we also require that $u(2\pi) = 0$. This forces $2\pi \sqrt{\lambda}$ to be a integer multiple of $2\pi$ or, in other words, $\lambda \in \{k^2: k = 1,2,3,...\}$. So the eigenvalues do in fact form a discrete sequence.

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It seems to me that $2\pi\sqrt{\lambda}$ must be an integer multiple of $\pi$, not $2\pi$ (since $\sin \pi = 0$ as well). Thus $\lambda \in \{ \frac{k^2}{4} \mid k = 1, 2, 3, \dots \}$. Am I right? (this seems to be confirmed by Paul Garrett's answer) – Giovanni Mascellani Oct 25 '15 at 16:43

Indeed, if we simply ignore what happens at the endpoints, obviously every exponential $x\rightarrow e^{c\cdot x}$ is an eigenfunction for $\Delta=-d^2/dx^2$. In effect, this amounts to looking at the open (=non-compact) interval, and, further, in this situation the operator is not self-adjoint, so no wonder that we cannot argue that its eigenvalues are real, etc., not even mentioning the non-discreteness.

To make a self-adjoint operator, something must be done with the endpoints. There is not a unique choice. One choice is requiring that values and values of derivatives match at the ends of the interval, so, in effect, we are looking at a circle. The eigenfunctions are exponentials whose values match at endpoints are $x\rightarrow e^{inx}$ with integer $n$, as expected, with a discrete set of real eigenvalues.

Or, we can choose the Dirichlet problem, that is, vanishing at the boundary, the endpoints. This gives eigenfunctions $x\rightarrow \sin {n\over 2} x\;$ with integer $n$, with a discrete set of real eigenvalues.

In fact, there is a continuum of distinct boundary conditions which make $\Delta$ self-adjoint: the literal requirement is that in integration by parts the boundary contributions cancel themselves. This gives relations on the values and derivatives at endpoints, met by the two examples mentioned, but met by many more, as well.

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I think your are misunderstanding the word "discrete". Discrete means that all points are isolated. In both cases you have this.

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Well [0,1] is not a manifold because of it's boundary. (at $0$ and $2 \pi$, it's not localy isomorphic to an open subset of $\mathbb{R}$).

Actualy, in order to have a similar theorem for manifold with boundary you may restrict your attention to fonction which vanish on the boundary ! For exemple, if you look for the eigen value of the laplacian when you restrict your attention to fonction that are $0$ at $0$ and $2 \pi$, then you will have a discret spectrum.

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