I am looking for a full--proof to refer to. I have an extracted copy of Boyce & Di Prima's book on differential equations "Elementary Differential Equations", specifically on the section on Sturm-Liouville problems.

In the book, is says "we assume without proof that this problem actually has eigenvalues and eigenfunctions" and adds that "The proof may be found in the references by Sagan(chapter 5) or Birkhoff and Rota(Chapter 10)"

I cannot find which book by either of those authors it is referring to. So I have searched online but while there are copious amounts of proofs for eigenvalues in the context of matrices, none were found for Sturm-Liouville problems.

No one seems to have done it online; is it unbelievably long or complicated? If anyone knows either which book the reference is talking about or, websites which provide the proof, it would be great if I can be directed there.

*I don't know what the best tag for this question is, so please edit/modify if you have better tags to put on

  • 1
    $\begingroup$ Birkhoff, G., and Rota, G.-C., Ordinary Differential Equations (4th ed.) (New York: Wiley, 1989). Sagan, H., Boundary and Eigenvalue Problems in Mathematical Physics (New York: Wiley, 1961; New York: Dover, 1989). $\endgroup$
    – Moo
    Feb 21, 2016 at 13:34
  • $\begingroup$ Have you studied compact operators? $\endgroup$ Feb 22, 2016 at 4:52
  • $\begingroup$ no, i haven't studied that $\endgroup$
    – Melba1993
    Feb 25, 2016 at 18:29
  • $\begingroup$ How about equicontinuous families of functions? Or Complex Analysis? If you respond, be sure to use the @TrialAndError to ping me. $\endgroup$ Feb 28, 2016 at 20:36

1 Answer 1


The Sturm-Liouville problem is precisely the linear transformation problem- the only difference is that you need to show that the Sturm-Liouville operator is a self-adjoint linear operator. What is the definition of "adjoint" in the case of second order differential operators?

  • $\begingroup$ so, given a linear operator $L$(lagrange's identity, in the Sturm Liouville case), an adjoint operator would be $L^*$ satisying $(Lu,v)=(u,L^*v)$ for some functions $u,v$, yes? And self adjoint would mean $L=L^*$. I'm just wondering if you mean that $L$ is taken as some matrix..? It would be great if you can elaborate some more on what you've hinted $\endgroup$
    – Melba1993
    Feb 21, 2016 at 13:35

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