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Homework Help: Diff. eq. and boundary conditions

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I am given the following differential equation:

    X'' - k*X=0.

    I am told that k = -m^2, so the general solution is given by:

    X = a*cos(m*x)+b*sin(m*x),

    where a and b are constants. I am also given boundary conditions:

    1) X(-Pi) = X(Pi)
    2) X'(-Pi) = X'(Pi).

    To satisfy #1, m must be an integer. But in my book the author states that in order to satisfy #1 and #2, m must be a positive integer. But I don't understand why m must be positive, because cosine is even.
  2. jcsd
  3. Sep 25, 2008 #2


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    Science Advisor

    I don't know if you are reading that wrong or the author just said it wrong, but, as you say, m does not have to be a positive integer. But X= a cos(mx)+ b sin(mx)= a cos(-mx)+ (-b)sin(-mx) so changing m for -m just changes the sign on the constant b. We can always assume that m is non-negative, since -m would give nothing new, but cannot assume it is positive: taking m= 0 gives X(x)= a which certainly satisfies both differential equation and boundary conditions.
  4. Sep 25, 2008 #3
    The author says: "For k = -m^2, Nontrivial solutions arise only if m = n for n = 1, 2, 3, ..., and the corresponding solutions are therefore X = a*cos(m*x)+b*sin(m*x).".

    I think the reason why we do not want n = 0 is so there are no trivial solutions, i.e. X = const.
  5. Sep 25, 2008 #4
    Seems to me like an eigenvalue problem with periodic boundary condition.
    in this case -k is the eigenvalue, and because k<0, so the soultion
    X=a \cos(\sqrt{-k} \space x)+b \sin(\sqrt{-k} \space x)

    The periodic boundary condition (heat transfer on a circular ring as a physical example) requires a=\=0 and b=\=0 AND
    sin (\sqrt{-k} \space \pi)=0
    \sqrt{-k} \space \pi =n \pi

    note that \sqrt{-k}>0 so n is positive integer.

    Last edited: Sep 25, 2008
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