$\frac{d}{dx}(p(x)\frac{dy}{dx})+w(x)y(x)+λr(x)y(x)=0$ is the sturm-liouville equation where λ is the eigenvalue.
According to the sturm-liouville theory, for any two functions $y_n$ and $y_m$, that satisfy the sturm-liouville equation, the integral$\int_a^by_n(x)y_m(x)r(x)dx$ is going to be zero if the condition on the bondary a and b is satisfied. $y_n$ and $y_m$ are two functions corespond to two eigenvalue $λ_n$ and $λ_m$.
Questions
1: According to the statment above about the integral, the set of orthogonal functions is actually not just {$y_n(x)$} but {$y_n(x)$$r^{1/2}(x)$}? $y_n(x)$is not orhognoal to $y_m$(x) unless r(x)=1?
2: Due to the bondary condition on the interval betwenn a and b,only solutions that corespond to certain discrete values of λ in the SLE can be orthogonal to each other in the interval [a,b]. However, if a=-∞ and b=+∞, then the eigenvalues in the SLE can be continuous and the bondary condition becomes unrequired? Any two functions $y_n$ and $y_m$ are orthognal to each other in the interval [-∞,+∞] regardless of how small the difference is between their two eigenvalues? The solutions to the time-independent schrodinger's equation is therefor always orthognal to each other over the infinite domain?
Sturm-Liouville problems with non-trivial weight functions $r$ naturally arise when using coordinate systems other than Cartesian for formulating an operator such as the Laplacian. The weight functions have to do with scale factors in the coordinate transformations. The end result is that it seems to make more sense and help to keep things straight if you work in the weighted Hilbert space. For example, the appropriate space for studying a regular problem $$ Lf = \lambda f, \mbox{ where } \\ Lf = -\frac{1}{r}\left[\frac{d}{dx}\left(p\frac{df}{dx}\right)+wf\right] $$ on a finite interval $[a,b]$ is the weighted space $L_r^2[a,b]$ consisting of all Lebesgue measurable functions on $[a,b]$ such that $$ \|f\|_{r} = \int_{a}^{b}|f(x)|^2r(x)dx < \infty. $$ This is an $L^2$ space with respect to the measure $d\mu=rdx$. If you suppose $p$ is in $C^1$, and $r,w$ are in $C$, and if you assume that $p$, $r$ do not vanish on $[a,b]$, then this is a regular Sturm-Liouville problem. Then, if you restrict $L$ to the domain consisting of all twice absolutely continuous functions $f$ on $[a,b]$ such that $Lf \in L^2_r[a,b]$ and which satisfy separated endpoint conditions $$ \cos\alpha f(a)+\sin\alpha f'(a) = 0 ,\\ \cos\beta f(b) + \sin\beta f'(b) = 0, $$ then the operator $L$ is selfadjoint in the strictest Functional Analysis sense. And $L$ has an orthonormal basis of eigenfunctions $\{ \varphi_n \}_{n=1}^{\infty}$ with corresponding eigenvalues $\lambda_1 < \lambda_2 < \lambda_3 < \cdots$. So, in this weighted space, everything has an expansion in terms of this eigenfunctions. Specifically, $$ f = \sum_{n=1}^{\infty}\langle f,\varphi_n\rangle \varphi_n. $$ The function $f$ must be in the weighted space $L^2_r[a,b]$ and the convergence takes place in the norm of the weighted space.
You can translate everything into $L^2[a,b]$ because $$ U : L^2_r[a,b]\rightarrow L^2[a,b] $$ defined by $Uf=r^{1/2}f$ is a unitary map from the weighted Hilbert space onto the unweighted space.
Once you shift to cases where $r$ or $p$ vanish at one or both endpoints, there may or may not be endpoint conditions required to have a selfadjoint operator. And, there may or may not be $L^2_r$ eigenfunctions. Even if there are eigenfunctions, they may not form a basis. There are eigenfunction expansions associated with the possible selfadjoint problems if you have one regular endpoint, but they may include Fourier integral types of expansions and/or some discrete components. It becomes even more complex if you have two singular endpoints.