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Weird differential equation

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the following differential equation, with the initial-value y(2)=2.

    [tex] y' = \sqrt{ \frac{1-y^{2} } { 1-x^{2} } } [/tex]


    2. Relevant equations



    3. The attempt at a solution

    This is a strange ODE. It is continuous when either both |x|<1 and |y|<1 or when both |x|>1 and |y|>1. (Both of the regions guarantee the existence and continuity of the derivative)

    [tex] \frac{y'}{ \sqrt{ 1-y^{2} } } = \frac{1 } { \sqrt{1-x^{2} } } [/tex]

    However, when you separate the variables (to solve it), the radical splits, causing the derivative to be defined only when both |x|<1 and |y|<1. The other region i.e. |x|>1 and |y|>1 is lost as a region in which a solution exists. This is strange.

    By the existence theorem, that the function y'(x,y) is continuous when both |x|>1 and |y|>1 should guarantee the existence of a solution, yet separation of variables forces the loss of a solution in that region.

    What then, is the (explicit) solution for that region, or how might I go about finding it?

    I did end up finding the explicit solution to the IVP to be [itex] y = x [/itex] but to do so I required the computation [itex] arcsin(2)-arcsin(2) = 0 [/itex]. This makes no sense.

    BiP
     
    Last edited: Feb 22, 2013
  2. jcsd
  3. Feb 22, 2013 #2

    mfb

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    sqrt(a*b)=sqrt(a)*sqrt(b) is true only for a,b>0, otherwise your operation is invalid in the real numbers. For |x|, |y|>1, you cannot split them like that - unless you use complex numbers.
     
  4. Feb 22, 2013 #3

    Ray Vickson

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    For |x|,|y|>1 you need
    [tex] \frac{y'}{\sqrt{y^2-1}} = \frac{1}{ \sqrt{x^2-1}}.[/tex]

    For the first case you need integrals of the form ##\int dt/\sqrt{1-t^2},## while for the second case you need ##\int dt/\sqrt{t^2-1}.##
     
  5. Feb 22, 2013 #4
    OK thanks, so then how would the explicit solution be obtained when |x|,|y|>1 ???

    BiP
     
  6. Feb 22, 2013 #5

    mfb

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    Oh, of course, see Ray Vickson's post how to split the sqrt in this case.
     
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