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2nd ODE, unique solutions

  1. Aug 26, 2010 #1
    I've been given a 2nd ODE in the form

    y'' + p(x)y' + q(x)y = 0

    The equation does not satisfy the test for a unique solution at x_0 = 0, because p and q are not continuous at x_0 (both p and q have x in the denominator, so a value of 0 makes the function discontinuous).

    I've two questions.

    1) If the uniqueness-existence theorem is not satisfied, can I conclude that the theorem does not guarantee a unique solution, but failure of the test does not exclude there being a unique solution? I.e. a unique solution is possible, but its existence cannot be predicted using the theorem.

    2) I've found a general solution to the 2nd ODE of

    y_h = c_1*f(x) + c_2*g(x)

    I'm given initial conditions of y(0) =0 and y'(0) = 1.

    using these values, I get c_2 = 4c_1 - 4.

    The other equation I obtain cancels to 0 = 0

    From this, can I conclude that no unique solution exists (for those initial values) to the 2nd ODE, since I'm not able to obtain unique values for c_1 and c_2 ?

  2. jcsd
  3. Aug 26, 2010 #2
    I suppose you refer to the existence and uniqueness theorem for the Cauchy problem. If the DE does not satisfy the conditions of that theorem, then you can't apply it, but yes, this doesn't mean that you can by luck find a solution, or you can be even luckyer, and it could happen that that solution is unique. However the theorem says that you can find an unique solution in a neighbourhood of the initial point, so it is essentially a local theorem, not a global one. In your case, since the DE is singular at x = 0, it seems to me that it's quite impossible to find a solution whose domain includes that point. Unless you found a solution that contains an x factor that cancels with the x in the denominator of p and q? Ignoring this, yes, if you found two solutions then... you proved that the solution is not unique! There are many examples of this, and the DE doesn't even need to be discontinuous (the Cauchy theorem imposes also other conditions).
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