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Basic sturm liouville boundary problems

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute the eigenvalues/functions of the given regular S-L problem

    f''(x)+λf(x)=0
    0<x<π
    f(0)=0
    f'(π)=0

    2. The attempt at a solution

    First off, why is π not included in the given boundary if it tells you f'(x) at π?

    Now for my attempt:

    assuming λ=0
    f''(x)+0*f(x)=0
    f''(x)=0
    f'(x)=a, and f'(π)=0 so a=0
    f(x)=ax+b, but a=0 so f(x)=b, and f(0)=0 so b also =0
    therefore
    f(x)=ax+b=0+0=0
    so this doesn't really help

    assuming λ>0
    f''(x)+λf(x)=0
    char eqn: r^2+λ=0
    r=±√(λ)i
    f(x)=c1*cos(√(λ)x)+c2*sin(√(λ)x)
    f(0)=0 so plug in 0 ->f(0)=c1*cos(0)+c2*sin(0)c1=c1=0, therefore c1=0
    now we are left with
    f(x)=c2*sin(√(λ)x)
    f'(x)=c2*√(λ)*sin(√(λ)x)
    f'(π)=c2*√(λ)*sin(√(λ)π)=0
    we can conclude than c2=0 or λ=0 but that wouldn't get us anywhere
    instead, we assume
    sin(√(λ)π)=0
    √(λ)π=n*π (where n=1,2,...)
    √(λ)=n
    λ=n^2

    so for λ>0 we have an eigenvalue of λ=n^2 where n=1,2,... and an eigenfunction of f(x)=c2*sin(nx) (plug n^2 into λ)

    the last case is λ<0
    f''(x)+λf(x)=0
    char eqn: r^2+λ=0
    r=±√(λ)
    f(x)=c1*e^(√(λ)x)+c2*x*e^(√(λ)x)
    f(0)=c1*e^0+c2*0*e^0=c1=0, therefore c1=0
    we are left with
    f(x)=c2*x*e^(√(λ)x)
    f'(x)=c2*x*√(λ)*e^(√(λ)x)+c2*e^(√(λ)x)
    f'(π)=c2*π*√(λ)*e^(√(λ)π)+c2*e^(√(λ)π)=0
    we can factor it to make it a little more manageable
    f'(π)=c2*e^(√(λ)π)*(√(λ)π+1)
    once again, we can see that c2 and λ=0, but are trivial
    assume that the last term =0
    √(λ)π+1=0
    √(λ)π=-1
    √(λ)=-1/π
    λ=1/π^2
    (here, can I specify that only the negative root is used and not the positive one?)

    so this gives us an eigenvalue of λ=1/π^2 and an eigenfunction of f(x)=c2*x*e^(±x/π)


    The answer in the back of the book is
    λ=(2n-1)^2 /4
    f(x)=sin((2n-1)x/2) n=1,2.....

    My answer looks nothing like that so what am I misunderstanding?
     
  2. jcsd
  3. Sep 22, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That should be a cosine.

    You don't have a repeated root here. It would be much clearer to state this case as ##\lambda = -\mu^2 < 0## so the equation is ##f''(x) -\mu^2f(x) = 0## so your characteristic equation is ##r^2-\mu^2=0##. This will give you the solution pair ##
    \{e^{\mu x}, e^{-\mu x}\}## but it is easier to work with ## \{ \cosh μx, \sinh μx\}##. Try it and you will find no nontrivial solutions.
     
    Last edited: Sep 22, 2012
  4. Sep 22, 2012 #3
    Wow such careless errors haha, but thanks for finding where I went wrong. I got the answer and everything seems good!
     
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