Orthogonality of eigenfunctions corresponding to an eigenvalue having multiplicity

Question

A Sturm-Liouville's problem is defined as follows: $$(py')'-qy=-\lambda wy$$ where the functions $p(x)$ , $q(x)$ and $w(x)$ are known for $0<x<L$ and $y(x)$ and $\lambda$ are unknown. It is shown that infinite number of eigenfunctions $y_k$ corresponding to the eigenvalue $\lambda_k$ exist and they have the orthogonality property: $$ n\neq k : \int_0^L w y_k y_n dx = 0 $$ The above equality holds if the two eigenfunctions $y_k$ and $y_n$ correspond to two different eigenvalues $\lambda_k$ and $\lambda_n$. However, some eigenvalues may be multiple; e.g. more than one eigenfunction correspond to a specific eigenvalue. What about two different eigenfunctions that correspond to one eigenvalue that has multiplicity? Are they still orthogonal? Please provide proof.

Answer 1

It's not necessarily true that two independent eigenfunctions corresponding to a common eigenvalue are orthogonal. However, if $\ y_1\ $ and $\ y_2\ $ are any two such independent eigenfunctions, then $\ \alpha y_1+\beta y_2\ $ is an eigenfunction, corresponding to the same eigenvalue, for any pair of scalars $\ \alpha\ $ and $\ \beta\ $. So if $\ y_1\ $ and $\ y_2\ $ are not orthogonal, then you can choose \begin{align} \alpha&=-\frac{\int_0^Ly_1y_2dx}{\int_0^Ly_1^2\ dx}\ \ \text{and}\\ \beta&=1 \end{align} to get $$ \hat{y}_2=\alpha y_1+y_2\ , $$ which is orthogonal to $\ y_1\ $, still independent of it, and still an eigenfunction corresponding to the same eigenvalue.

More generally, if $\ y_1, y_2, \dots\ $ are linearly independent eigenfunctions, all corresponding to the same eigenvalue, but not necessarily mutually orthogonal, then you can apply the Gram-Schmidt procedure to them to get a set $\ y_1,\hat{y}_2,\hat{y}_3,\dots\ $ of linearly independent and mutually orthogonal eigenfunctions corresponding to the same eigenvalue.