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

Regular Sturm-Liouville, one-dimensional eigenfunctspace

  1. Jun 12, 2015 #1
    Sturm-Liouville problem:
    [tex]\frac{d}{dx} (r(x) \frac{dX}{dx}) + q(x)X(x) + \lambda p(x) X(x) = 0 \quad x \in \left[a, b \right][/tex]
    [tex] a_1 X(a) + a_2X'(a) = 0[/tex]
    [tex]b_1 X(b) + b_2X'(b) = 0[/tex]
    [tex]r, r', q, p \in \mathbb{C} \forall x \in in \left[a, b \right][/tex]

    Under the additional condition that r(a)>0 (or r(b) > 0), the S.L BVP cannot have two linearly independent eigenfunctions corresponding to the same eigenvalue.

    X, Y two eigenfunctions corresponding to the same eigenvalue. Then Z = Y(a)X(x) - X(a)Y(x) is also a solution and it can be shown that [skipping how to get z(a), z'(a) as there's no real problem there for me]

    [tex]z(a) = z'(a) = 0 \implies z(x) = 0 \forall x \in \left[a, b \right][/tex]
    QED, catfoots and other scribbles.

    I don't get the final implication at all. Why does knowing that z(a) and z'(a) = 0 at x = a show that z(x) is identically zero over the interval in question?
    Last edited: Jun 12, 2015
  2. jcsd
  3. Jun 12, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    z is a solution to the SL problem with boundary conditions z(a) and z'(a)=0. The trivial solution f(x)=0 also is. Uniqueness leads to z=0.
  4. Jun 12, 2015 #3
    Explains why it wasn't elaborated further, feel silly now. Thank you, need to remind myself of BVP's it seems.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Regular Sturm-Liouville, one-dimensional eigenfunctspace