Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sturm-Liouville Questions

  1. Jul 8, 2013 #1
    In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.

    The first issue I have is that my book began with the statement that given

    $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$

    the problem [itex]L[y] \ = \ f[/itex] can be re-cast in the form [itex]L[y] \ = \ \lambda y[/itex].

    Now it could be a typo on their part but I see no justification for the way you can just do that!

    More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:

    Given a linear second order ode

    $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$

    it is an exact equation if it is derivable from a differential equation of one order lower, i.e.

    $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$

    The equation is exact iff

    $$a''(x) - b'(x) + c(x) = 0. $$

    If [itex]F[/itex] is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.

    This equation is exact iff

    $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$

    If you expand this out you get the Adjoint operator

    $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$

    If you expand [itex]L^*[/itex] you see that we can satisfy [itex]L \ = \ L^*[/itex] if [itex]a'(x) \ = \ b(x)[/itex] & [itex]a''(x) \ = \ b'(x)[/itex] which then turns [itex]L[y][/itex] into something of the form

    $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$

    Thus we seek an integratiing factor [itex]\alpha(x)[/itex] so that we can satisfy this & the condition this will hold is that [itex]\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}[/itex]

    Then we're dealing with:

    $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$

    But again, by what my book said they magically re-cast this problem as

    $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$

    Then calling

    $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$

    a Sturm-Liouville problem.

    My question is, how can I make sense of everything I wrote above? How can I clean it up & interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues & finding full solutions - what's going on? I want to be able to look at [itex]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/itex] & know how & why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness & integrating factors, thanks for reading!
  2. jcsd
  3. Jul 9, 2013 #2
    First and foremost, its important to keep in mind that the purpose of Strum-Liouville theory is to study the eigenvalues and eigenvectors of an ode. It turn out that you can learn a lot about the solution to an ode, and thus gain a significant amount of physical inside into a variety of problems, by identify and studying the eigenvalues and eigenvectors.

    So when your book writes [itex]Ly = \lambda y[/itex] this is not an identity. Instead it is an assertion that we are only going to study the eigenvalue problem.

    Second, the procedure on applying the integrating factor is a method used to covert a second order ODE into the traditional S-L form.

    [itex]\frac{-1}{\alpha}\int dx \left( p \frac{dy}{dx} + q y\right) = \lambda y[/itex]

    This form is useful because the differential operator is self-adjoint over the inner product space defined by
    [itex] g(x) \cdot f(x) = \int dx g(x) f(x) \alpha(x) [/itex]

    and the above form of the S-L problem allows you to identify the correct weight to use in the inner product.
  4. Jul 10, 2013 #3
    Thank you, I don't see how that explains why [itex]L[y] \ = \ f[/itex] can be re-cast in the form [itex]L[y] \ = \ \lambda y[/itex] though, I know we're studying eigenvalues of the operator but how that justifies [itex]L[y] \ = \ f[/itex] becoming [itex]L[y] \ = \ \lambda y[/itex] is beyond me - what does the author mean when he says that? He does it more than once, even at a crucial step in the derivation of the sturm-liouville problem as a whole, so I think there's more to it than just saying we're studying the operator.

    Also I've never seen a sturm-liouville form written as you have done in terms of integration - do you have a reference?

    Is there nothing more one can say as regards sturm-liouvlle theory when looking at [itex]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/itex] other than to say finding the eigenvalues of the differential operator (homogeneous differential operator?) sheds light on the solutions?
  5. Jul 10, 2013 #4
    That would be because its a typo. It should be of the form:

    [itex]\frac{1}{a}\left(-\frac{d}{dx}\left( p(x) \frac{dy}{dx} \right) + q(x) y \right)=\lambda y[/itex]

    As I understand it, the formal S-L problem is to consider the eigenvalue problem for the operator L. I may be wrong,
    However, under certain conditions the eigenvectors of the L[y] form an orthonormal basis. In this case you can project f onto that basis and the ODE L[y] = f becomes equivalent to a sum of ODEs of the from L[y] = λy.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook