2
$\begingroup$

I am familiar with the notion of an eigenvalue/eigenvector pair for a linear operator. Explicitly: If $V$ is a vector space over a field $K$, and $T$ is a linear operator on $V$, then we call $\lambda \in K$ an eigenvalue if there exists a non-zero vector $v \in V$ such that $Tv = \lambda v.$

I am now studying differential equations and am learning about Sturm- Liouville problems. There we define a linear differential operator $$L[y] = -[p(x) y']' + q(x)y$$ for some functions $p(x)$ and $q(x)$, and $y$ is a function in some appropriate function space (which function space is not entirely clear, but I do not think that is too relevant to my question). We then consider the Sturm-Liouville problem: \begin{equation}\tag{1} \label{SL} L[\phi] = \lambda r(x) \phi \end{equation} for some function $r(x)$, subject to some boundary conditions.

My confusion comes from this $r(x)$. Most sources say if $\lambda$ is some complex number for which there exists a non-trivial solution $\phi$ to \ref{SL}, then $\lambda$ is an eigenvalue and $\phi$ the associated eigenfunction. But I would expect the problem to be \begin{equation}\tag{2} \label{expected} L[\phi] = \lambda \phi . \end{equation} How can the notion of an eigenvalue/eigenvector pair jive with this $r(x)$? Is this just a difference in definitions for Sturm-Liouville theory? Is it a notational convenience? Because we could just define $$L[y] = \frac{1}{r(x)}\left( -[p(x) y']' + q(x)y \right),$$ and then things are how we expect them in \ref{expected}.

$\endgroup$
3
$\begingroup$

It's notational convenience. Both Sturm-Liouville operators/problems,

$$L[y] = -[p(x) y']' + q(x)y, L[\phi]= \lambda r(x) \phi$$

and

$$L[y] = \frac{1}{r(x)}\left( -[p(x) y']' + q(x)y \right), L[\phi]=\lambda \phi$$

lead to the same eigenfunctions $\phi_n$ and orthogonality condition, which is

$$(\lambda_m - \lambda_n)\int_a^b \phi_m(x)\phi_n(x)r(x)dx = 0, \lambda_m \neq \lambda_n$$

$\endgroup$
  • $\begingroup$ The latter formulation just seems more natural, in the context of operator theory. Namely that, you fix $r(x)$ to be part of the operator, so that eigenvalues and eigenvectors are a property of the operator. In the former approach, the eigenvalues and eigenvectors of the operator depend on the choice of $r(x)$, when I would expect them to be intrinsic to the operator. $\endgroup$ – frito_mosquito Aug 24 '17 at 15:09
1
$\begingroup$

The standard way to study Sturm-Liouville operators is in the form that you stated looked natural to you: $$ Lf = -\frac{1}{w}\left[-\frac{d}{dx}\left(p\frac{df}{dx}\right)+q\right]. $$ The operator is studied in the weighted Hilbert space $$ L_w^2(a,b)= \left\{ f : \int_{a}^{b}|f|^2wdx < \infty \right\} $$ This is particularly important when the operator is singular at one or both endpoints of the $(a,b)$. In that case, for $a < c < b$, you are guaranteed at least one $L^2_w(a,c]$ solution and at least one $L^2_w[c,b)$ solution of $Lf=\lambda f$ for every non-real $\lambda$. That's a critical property that won't work out right if you try to study the operator in the non-weighted space.

In physical problems, the weight typically arises as a metric scale factor through some standard change of variables on the PDE from Cartesian. So the weight typically has significance in the physical problem as well, and the proper inner product space will involve the Jacobian of a transformation, and the weights will come from the metric scale factors. So it is important how you define these things if you want to study them in the proper theoretical context. And in the proper context you get eigenfunction expansions that generalize those for the Fourier transform and series. For example, if $a$ is a regular endpoint and $b$ is singular, then you get a Fourier transform $$ \hat{f}(s) = \int_{a}^{b}f(x)\phi_{s}(x)w(x)dx $$ where $\phi_s$ are classical eigenfunction solutions that are not necessarily in $L^2_w[a,b)$, and you get a spectral measure $\mu$ for inversion: $$ f = \int_{-\infty}^{\infty}\hat{f}(s)\phi_s(x)d\mu(s) $$ The convergence of the inversion integral must be in $L^2_w(a,b)$, just as it is in $L^2$ for the Fourier transform. In fact there is a Parseval theorem: $$ \|\hat{f}\|_{L^2_\mu} = \|f\|_{L^2_w}. $$ In this space the expansion "diagonalizes" the operator with $f\in \mathcal{D}(L)$ iff $$ \int_{-\infty}^{\infty}s^2|\hat{f}(s)|^2 d\mu(s) < \infty $$ and, using this spectral density measure, $$ Lf = \int_{-\infty}^{\infty}s\hat{f}(s)\phi_s(x)d\mu(s). $$ None of this works out right if you don't have the right space and the right definition of the operator. This gives you Fourier series expansions, Fourier integral expansions, and even a mix of the two.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.