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Boundary conditions - unique solution

  1. Jan 27, 2015 #1
    I recently solved a differential equation with the solution:

    f(x) = Aexp(ikx) + Bexp(-ikx)

    with the periodic boundary condition f(x+L)=f(x). This condition leads to:

    Aexp(ikx)exp(ikL) + Bexp(-ikx)exp(-ikL) = Aexp(ikx) + Bexp(-ikx) (1)

    Now the way I figured out the constants A and B was that I said, that (1) must be true in particular for x=0 and x=π/2k which gave some equations to solve for A and B.

    But something appears weird to me with this method of using two arbitrary values for x in (1) to determine A and B. Namely how do I know that I should also get this value for A and B if I use two different values of x? I do realize you could maybe show this by a calculation letting x1 and x2 be arbitrary but from a bigger perspective: Is it always true that the solution of a differential equation is uniquely specified by a periodic boundary condition? Maybe I'm just confused about something simple - it's late at night..
     
  2. jcsd
  3. Jan 27, 2015 #2
    Do you require f(x) to be real, or are you also interested in cases where f(x) is complex?

    Chet
     
  4. Jan 27, 2015 #3
    I have no requirement on f being real.
     
  5. Jan 27, 2015 #4
    Actually, your periodicity requirement puts a restriction on k, not A and B: k = 2nπ/L, where n is any integer.

    Chet
     
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