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Self-adjoint boundary value (Sturm-Liouville)

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Under what condition on the constant ##c## and ##c'## are the boundary conditions ##f(b) = cf(a)## and ##f'(b)=c'f'(a)## self-adjoint for the operator ##L(f) = (rf')'+pf## on ##[a,b]##? (Assume that ##r,p## are real.)

    2. Relevant equations
    The boundary conditions are self-adjoint (relative L) if
    ##\left[r(f'\bar g-f\bar g ')\right]_a^b=0##
    for all ##f,g## satisfying the boundary conditions.
    (Assuming that ##L## is formally self-adjoint as it is in this case.)

    3. The attempt at a solution
    ##\left[r(f'\bar g-f\bar g ')\right]_a^b = r(b)\left( f'(b)\bar g (b) - f(b)\bar g '(b)\right) -r(a)\left( f'(a)\bar g(a)-f(a)\bar g '(a)\right) = f'(a)\bar g (a)\left(c'\bar c r(b)-r(a)\right) + f(a)\bar g(a)\left(-r(b)c\bar c'+r(a)\right)##

    So a solution would be ##c'\bar c = c\bar c' = \frac{r(a)}{r(b)}##.

    The answer to the question says only ##c\bar c' = \frac{r(a)}{r(b)}## but I got that ##c'\bar c## also has to have the same value. Am I correct here or did i make a mistake?
     
  2. jcsd
  3. Sep 30, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    [itex]c'\bar c[/itex] is the complex conjugate of [itex]\bar c' c[/itex]. The two are equal as [itex]\frac{r(a)}{r(b)}[/itex] is real and therefore equal to its complex conjugate.
     
  4. Sep 30, 2015 #3
    Thanks!
     
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