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Given a sequence of functions orthogonal over some interval, which satisfy Dirichlet boundary conditions at that Interval, can we construct a Sturm Liouville problem that gives these as its eigenfunctions?


For example, if we take the sequence of scaled Bessel's functions $J_n (\zeta_i x)$ for all positive integral values of $i$ where $\zeta_i$s are roots of the Bessel's function. Note that here, $n$ is fixed and non - negative.

We already know that these form an orthogonal basis over the weight function $x$ and range $[0,1]$ such that $$\int_0^1 J_n (\zeta x)J_n (\zeta_j x) x dx =\frac12 \delta_{ij}(J_n' (\zeta_i ))^2$$

As the $\zeta_i$s are roots (and the Bessel's functions of the first kind are zero at the origin for positive $n$), we already have $J_n(\zeta_i 0)=J_n(\zeta_i 1)=0$, which is a Dirichlet boundary condition on $[0,1]$.

To me, this seems to be the setup for a Sturm Liouville problem. Can we find $p(x)$ and $q(x)$ such that the functions $y_i(x)=J_n(\zeta_i x)$ satisfy $$\frac{\mathrm d}{\mathrm d x}\left[p(x)\frac{\mathrm dy}{\mathrm d x}\right] + q(x)y(x)=\lambda_i xy(x)$$

Update: Turns out we can, for the example given above. The equation is the radial part of the circular membrane problem, i.e.:

$$x^2y''(x) +xy'(x) +(\lambda^2x^2-n^2)y(x)=0$$

with eigenvalue $\lambda$, where the eigenvalues turn out to be successive roots of $J_n$ and the eigenfunctions are $J_n(\zeta_i x)$ for root $\zeta_i$.

However, can this be done in the general case? If I have a series of functions which:

  • Form an orthogonal basis over an interval with a certain weight function

  • Have Dirichlet BCs on the same interval

can I always construct a Sturm-Liouville problem/ODE for the same wight function and boundary that gives rise to these as eigenfunctions? If not, when is this possible?

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  • $\begingroup$ Well, if the eigenbasis is given, the first route I'd try is assume $p,q$ have an eigenexpansion and reduce the problem to conditions on their expansion coefficients. $\endgroup$ – David H Nov 4 '13 at 12:46
  • $\begingroup$ @DavidH Ah, makes sense. Are we sure that such a Sturm-Liouville problem must exist in the first place? $\endgroup$ – Manishearth Nov 4 '13 at 12:50
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    $\begingroup$ As a physicist, I'm damn certain. Because my eigenfunctions are responses produced by a black-box system which are proportional to the input signals, so clearly the black-box contains either an SL-operator or gremlins. ;) $\endgroup$ – David H Nov 4 '13 at 13:07
  • $\begingroup$ @DavidH Hahaha as a physics student myself, usually I'm certain of these things too, but mathematicians don't seem to like such happy-go-lucky certainty. I was looking for caveats to the general "reverse Sturm Liouville problem". $\endgroup$ – Manishearth Nov 4 '13 at 13:13
  • $\begingroup$ I was just making a question about this very idea and stumbled upon this question. Since it is quite interesting, I've added a somewhat steep bounty in the hopes that someone will have interesting insight on this problem. $\endgroup$ – Cameron Williams Feb 23 '15 at 20:36
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The theory of 'integral equations' as described in chapter 3 of Hildebrand's 'Methods of Applied Mathematics' or Hilbert/Courant's book, shows you that the eigenfunctions derive from a Fredholm integral equation

$$y_n(x) = \lambda_n \int_a^b K(x,z)y_n(z)dz$$

and since the $y_n$ are a basis we have

$$f(x) = \sum_n A_n y_n(x) = \sum_n A_n \lambda_n \int_a^b K(x,z)y_n(z)dz= \int_a^b K(x,z)F(z)dz$$

One can naturally convert a Fredholm integral equation with a symmetric kernel back into a boundary value Sturm-Liouville problem, it's in Hildebrand for boundary conditions more general than just Dirichlet.

To understand this though, we should reverse the process. Given a linear IVP second order ode we can convert it into an equivalent Volterra (Variable endpoint) integral equation (3.2 of Hildebrand). That's a great way to motivate integral equations. If we then try to do the same thing for a boundary value problem, the Fixed endpoint problem gives a Fredholm integral equation, but for a general ode $y'' + ay' + by = f$ we end up with a discontinuous integral kernel. The method motivates the necessity for turning $y'' + ay' + by = f$ into Sturm-Liouville form as the natural way of forming a symmetric integral Kernel that is continuous! This is in Hildebrand 3.3 - 3.4. I think it's more natural to think of Sturm theory as being part of the theory of integral equations rather than differential equations for this reason, as well as the fact all the orthogonality proofs invoke integrals randomly, yet here they arise naturally :)

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  • $\begingroup$ Question: does this work in the event that $a=-\infty$ and $b=\infty$? $\endgroup$ – Cameron Williams Feb 27 '15 at 4:18
  • $\begingroup$ And would you mind pointing to the relevant parts of Chapter 3? That text is rather lengthy. $\endgroup$ – Cameron Williams Feb 27 '15 at 4:31
  • $\begingroup$ If the endpoints are infinite it's called a 'singular integral equation' and it depends on the functions involved, whole books are devoted to this case. So 3.1 to 3.12 would be the basic core of the subject, you can pick it up without the first 2 chapters, and you could remove the example sections to really rush it lol. For comparison you could refer to the chapter on Integral Equations in Arfken's Mathematical Methods. To go more deep there's Hilbert/Courant and a problem book by Krasnov, enjoy. $\endgroup$ – bolbteppa Feb 27 '15 at 9:35
  • $\begingroup$ Hmmm. I don't see any mention of constructing $K$ given a set of orthogonal functions in the text. I've read through it twice now. I would think that $K(x,z) = \sum_{n=0}^{\infty} \varphi_n(x)\varphi_n(z)$ where $\varphi_n$ are the orthogonal functions. However there is no guarantee that this is a well-defined function. In fact, it is probably only understandable as a (Dirac delta) distribution in general. $\endgroup$ – Cameron Williams Mar 2 '15 at 2:15
  • $\begingroup$ Your eigenfunction expansion of the integral Kernel is incorrect, here books.google.ie/… is a correct derivation. It should be obvious that the eigenfunction basis just determines the form of $K$ which determintes the form of the Sturm-Liouville equation you end up with, e.g. Hildebrand shows you how a weight is sometimes required to symmetrize a Kernel just as it's sometimes required o form a Sturm-Liouville operator. $\endgroup$ – bolbteppa Mar 2 '15 at 11:11

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