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Sturm Liouville operator

  1. Dec 12, 2013 #1
    Hello, I'm solving the previous exams and I have a problem with an exercise:
    1. The problem statement, all variables and given/known data

    q(x) a real function defined in [0,1] and continuous
    L a sturm Liouville operator :
    Lf(x)=f''(x)+q(x)*f(x)

    f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.

    Is L a symetric operator relative to the scalar product defined as
    (f,g)=∫f(x)*g(x) dx from 0 to 1 ?

    I just want to be sure I have to show that (Lf(x),Lg(x))=(Lg(x),Lf(x)) ( or not equal) ?

    Thanks
     
  2. jcsd
  3. Dec 12, 2013 #2

    Dick

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    No, you want to show (Lf(x),g(x))=(f(x),Lg(x)).
     
  4. Dec 14, 2013 #3
    Hello,
    So if I want to prove that (Lf(x),g(x))=(f(x),Lg(x))
    The operator is applied on functions which properties are f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.
    So when I calculate
    (Lf(x),g(x))
    I can use the fact that g'(1)=g'(0)=0 ?
    Thanks
     
  5. Dec 14, 2013 #4

    Dick

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    I think you have g'(1)=g(0)=0. That's a little different.
     
  6. Dec 15, 2013 #5
    Hello,
    I'm sorry it's f'(0)=f'(1)=0
    So if I want to show that it is symmetric I have to calculate this part
    ∫f''(x)g(x) from 0 to 1
    By integration
    =[ f(x)*f'(x) ] - ∫g'(x)f'(x) dx
    = - ∫g'(x)f'(x) dx
    =-[g'(x)*f(x)] + ∫f(x)*g''(x)dx
    =∫f(x)*g''(x)dx

    So if ∫f''(x)g(x)=∫f(x)*g''(x)dx
    We can rewrite
    (Lf(x),g(x))= ∫f''(x)g(x) + ∫g(x)f(x)g(x) dx
    =∫f(x)*g''(x)dx + ∫g(x)f(x)g(x) dx
    =(f(x),Lg(x)) ?
     
  7. Dec 15, 2013 #6

    Dick

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    Well, yes it's integration by parts. I don't think that's a very clear presentation though.
     
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