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I'm struggling to find a geometric, or at least some intuitive understanding of eigenvalues and eigenfunctions in Sturm-Liouville problems (which I've been looking at in a PDE course).

For instance, for the Sturm-Liouville problem:

$$(p(x)\phi')'+q(x)\phi+\lambda\sigma(x)\phi=0$$

with boundary conditions of course, I struggle to see why this rather arbitrarily placed $\lambda$ deserves the high honour of being called an eigenvalue. And then we solve for $\phi$ and call it an eigenfunction. In linear algebra, eigenvalues and eigenvectors of a transformation have a number of nice geometric interpretations, and frankly I feel quite comfortable in seeing their importance in that setting, but I'm unsure why we can prescribe this terminology here.

In the course I'm taking we're making a great deal of fuss about non-negativity of eigenvalues, the orthogonality of eigenfunctions, etc., etc., but while I can follow the various proofs algebraically I must admit I feel quite lost without this basic understanding.

Thanks

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I would like to reference the german Wikipedia page for the Sturm-Liouville problem.

First (to be consistent with that site) let me introduce a slightly different formulation of the problem, which is the PDE \begin{align} -\left( p \cdot \psi' \right)' + q \cdot \psi = \lambda \cdot w \cdot \psi \end{align} so this is identically to your equation, except that one has to negate $q$ and $\psi$ is $\phi$.

For the PDE, one can write the differential operator as \begin{align} L = \frac{1}{w} \left( -\frac{d}{dx} \, p\, \frac{d}{dx} +q \right) \end{align} which is linear, and called the Sturm-Liouville-operator. And for this operator we get \begin{align} L\psi = \lambda \psi \end{align} This is, why we call $\psi$ and eigenfunction and $\lambda$ an eigenvalue.

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