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Solution to ODE

  1. Jul 18, 2006 #1
    As part of a separable solution to a PDE, I get the following ODE:

    X''-rX=0 (*),

    with -infty<x<infty and the boundary condition X(+/-infty)=0 (X is an odd function here). Thus I have assumed r>0 to avoid the periodic solution, cos. I, therefore, argue that the solution is the symmetric ~exp(-sqrt(r)|x|). This, however, has a discontinuity at x=0 which, seems to me, contrasts with (*) which implies X' and X'' must be continuous across x=0.

    Any ideas? (Many thanks.)
    Last edited: Jul 18, 2006
  2. jcsd
  3. Jul 18, 2006 #2


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    It's a second order ODE, so you should get two linearly independent solutions, and the general solution will be a linear combination of these. However, the requirement that the function goes to zero at both infinity and negative infinity will force a trivial solution (X=0), so you should make sure these are the right BC's. Note that your solution is valid everywhere but x=0, where the derivative is not defined.
  4. Jul 18, 2006 #3
    From the two linearly independent solutions (exp(-ax), exp(ax)) I retain only the one with the negative exponent, because the positive one blows up at infty and patch the other one up by letting y->|y|, so that both the boundary conditions at +/-infty and the evenness of the solution are reproduced. This, nevertheless, as I said, results in a discontinuity at x=0 which appears to be in disagreement with the DE.
  5. Jul 18, 2006 #4


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    The general solution will be of the form A exp(-ax) + B exp(ax), where A and B are constants. You can't have something like A=1 for x>0 and A=0 for x<0, because then A is not a constant. Such a solution would satisfy the DE for x<0 or for x>0, but at x=0, the derivative of A is not zero (it is not defined), and so the function is not a linear combination of the base solutions. As I said, the only solution that is valid everywhere and satisfies the BCs is X=0.
  6. Jul 18, 2006 #5
    If a non-trivial solution that fits the boundary conditions X(+/-infty)=0, X(x)=X(-x) cannot be provided in this way, does this mean that the solution simply cannot be given in terms of elementary functions or that no separable solution exists for the original PDE?

    Very grateful.
  7. Jul 18, 2006 #6


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    No, there is no non-trivial solution. When we say that all solutions are linear combinations of these two base functions, we mean exactly that. It just so happens that every possible solution to this ODE has a nice form. I'm not sure what your PDE is, and it may be as you say that there simply aren't any seperable solutions, but there still could be other solutions that aren't seperable.

    Another option is that you may not care whether some point in the domain satisfies the ODE (for example, if you are looking at a fluid velocity field, the field inside a solid object is not physically meaningful), and then you can uapply different solutions in the different (disconnected) regions (eg, x>0 and x<0, like the function in your first post)
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