Talk:Sturm-Liouville theory

Numbered equations

Hi Daniel,

I had to transform the templates provided in the WP article for numbered equations into plain old wikimarkup. I attempted to bring these templates over from WP, but they called many other templates and when I got them all over, the combined result didn't work. I decided to use a simple bit of html that defined a span with right justification and also defined an anchor. To reference the equation you then only need to insert a mediawiki markup referencing the anchor. It wouldn't be hard to turn this all into two templates if you think that would be useful. Dan Nessett 19:28, 2 September 2009 (UTC)

Redacted comment. I see it now.

Error

It seems to me that the article contains an error. Consider the definition:

$Lu={1 \over w(x)}\left(-{d \over dx}\left[p(x){du \over dx}\right]+q(x)u\right)$

Contrary to what is implied in the article, the operator L thus defined is not self-adjoint, unless 1/w(x) commutes with the operator to its right. This is in general not the case. The proper way to transform is (L in the next equation is w(x) times L in the previous equation):

$Lu=\lambda \,w\;u\;\Longrightarrow \;w^{-1/2}Lw^{-1/2}w^{1/2}u=\lambda \,w^{1/2}u\;\Longrightarrow \;{\tilde {L}}\,{\tilde {u}}=\lambda \,{\tilde {u}}$

with

${\tilde {L}}\equiv w^{-1/2}Lw^{-1/2}\quad {\hbox{and}}\quad {\tilde {u}}\equiv w^{1/2}u$

Since w(x) is positive-definite w(x)^−½ is well-defined and real. The operator ${\tilde {L}}$ is self-adjoint.

--Paul Wormer 08:09, 14 October 2009 (UTC)

I am getting out of my field of expertise here, but let me ask some questions. First, the section defines L and then posits, " L gives rise to a self-adjoint operator." I am not sure what this means, but one way of interpreting it is L itself is not self-adjoint, but some transformation of it is. That is, L may not itself have real eigenvalues. This statement is followed by, "This can be seen formally by using integration by parts twice, where the boundary terms vanish by virtue of the boundary conditions. It then follows that the eigenvalues of a Sturm–Liouville operator are real and that eigenfunctions of L corresponding to different eigenvalues are orthogonal." This is, admittedly, vague. However, later in the section is the statement, "As a consequence of the Arzelà–Ascoli theorem this integral operator is compact and existence of a sequence of eigenvalues α_n which converge to 0 and eigenfunctions which form an orthonormal basis follows from the spectral theorem for compact operators." So, on the surface, it seems the integral operator plays some role in the argument.

That said, let me note that from the perspective of someone who has very little understanding of S-L theory, this section is obscure and would benefit from a complete rewrite. Dan Nessett 17:09, 14 October 2009 (UTC)

Paul, the factor 1 / w is cancelled by w in the definition of the inner product (I think). Peter Schmitt 19:07, 14 October 2009 (UTC)

[unindent]

Let me answer first Dan and then Peter. I forget about the real function q and define the operator Λ for real p(x)

$\Lambda \;{\stackrel {\mathrm {def} }{=}}\;-{\frac {d}{dx}}p(x){\frac {d}{dx}}$

The operator acts on a space of functions u, v, for which holds (integration with weight 1)

$\int _{a}^{b}\;u(x){\frac {dv(x)}{dx}}\,dx=-\int _{a}^{b}\;{\frac {du(x)}{dx}}v(x)\,dx+\left[u(x)v(x)\right]_{a}^{b}=-\int _{a}^{b}\;{\frac {du(x)}{dx}}v(x)\,dx.$

Here it was used that u and v satisfy periodic boundary conditions (are equal on the boundaries a and b). Then on the space of functions with this boundary condition the derivative is anti-Hermitian (anti-self-adjoint):

$\langle u\;|\;{\frac {dv}{dx}}\rangle =-\langle {\frac {du}{dx}}\;|\;v\rangle \;\Longrightarrow \;{\frac {d}{dx}}^{\dagger }=-{\frac {d}{dx}}.$

Apply the rule valid for arbitrary operators (capitals) and functions (lower case):

$\left(AuB\right)^{\dagger }=B^{\dagger }{\bar {u}}A^{\dagger }$

where the bar indicates complex conjugate. Then clearly, because p(x) is real, we have

$\Lambda ^{\dagger }=-\left({\frac {d}{dx}}\right)^{\dagger }p(x)\left({\frac {d}{dx}}\right)^{\dagger }=\Lambda$

so that Λ is Hermitian (self-adjoint). If we divide by real w on the left we get

$\left({\frac {1}{w(x)}}\Lambda \right)^{\dagger }=\Lambda {\frac {1}{w(x)}}.$

Unless Λ and w commute (and they don't) it is clear that the new operator is non-Hermitian.

Why do I integrate with weight 1 and not with weight w? This is because the original S-L problem has the form

$\Lambda v(x)=\lambda w(x)v(x),\qquad \lambda \in \mathbb {R} .$

Multiply by u(x) and integrate with weight 1

$\int _{a}^{b}\;u(x)\Lambda v(x)dx=\lambda \int _{a}^{b}u(x)w(x)v(x)dx,$

which is what we want. If we would introduce weight w it would appear squared on the right, which is what we don't want.

Two more comments.

I noticed the error because I'm very familiar with the generalized matrix eigenvalue problem (I sometimes felt during my working life that all I did was setting up and solving such problems.)

$\mathbf {H} \mathbf {C} =\mathbf {S} \mathbf {C} {\boldsymbol {\epsilon }}\quad {\hbox{with}}\quad {\boldsymbol {\epsilon }}=\mathrm {diag} (\epsilon _{1},\ldots ,\epsilon _{n})$

The matrix H is Hermitian, S is Hermitian and positive-definite (has non-zero positive eigenvalues) and ε contains the eigenvalues on its diagonal. Beginning students sometimes divide this problem by S, but that is not the way to solve this. One way is transforming by S^−½. Exactly as I did before.

The second comment: I own the classic two-volume book by Courant and Hilbert (2nd German edition). In volume I, chapter V, §3.3 they discuss the S-L problem and introduce the transformation (in their notation) z(x) = v(x) √ρ [where ρ(x) ≡ w(x)]. They give the transformed S-L operator Λ without proof. It took me about two hours to show that their form is indeed equal to the form that I started this discussion with:

${\tilde {\Lambda }}\,z(x)=\left[\rho ^{-1/2}\Lambda \rho ^{-1/2}\right]z(x)=-\left[{\frac {d}{dx}}{\frac {p(x)}{\rho (x)}}{\frac {d}{dx}}\right]z(x)\;-\;{\frac {1}{\sqrt {\rho (x)}}}{\frac {d{\big (}p(x)f(x){\big )}}{dx}}$

with

$f(x)\equiv {\frac {d\left({\frac {z(x)}{\sqrt {\rho (x)}}}\right)}{dx}}$

It took me so long because Courant-Hilbert use a notation in which it is very unclear how far a differential must work. I hope that I avoided this unclarity by introducing f(x). The second term of the transformed Λ is a function, no differentiations are left "free" to act to the right.

--Paul Wormer 09:20, 15 October 2009 (UTC)

Paul, I freely admit that you know much more about partial differential equations and differential operators than I do. This is one of the topics I tried to avoid most of the time. And I believe that your transformation works (though I did not check it in detail).

But I think you misunderstood me. The operator is said to be selfadjoint with respect to the inner product

$\langle f,g\rangle =\int _{a}^{b}{\overline {f(x)}}g(x)w(x)\,dx.$

not the inner product

$\langle f,g\rangle =\int _{a}^{b}{\overline {f(x)}}g(x)\,dx.$

and in the first definition the 1 / w cancels with the w on both sides (your L):

$\langle u\mid {\frac {1}{w}}Lv\rangle =\langle {\frac {1}{w}}Lu\mid v\rangle$

You may want to check this in the book (p.121) cited in the bibliography (accidently, by one of my colleagues here in Vienna).

Peter Schmitt 12:21, 15 October 2009 (UTC)