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Non-linear ode

  1. Mar 13, 2007 #1
    hey does anybody have any idea how to solve this equation?

    E/m= xd^2x/dt^2

    with initial conditions x=a and dx/dt(a)=0

    its non-linear and so I don't have any idea what to do with it, and maple won't give me an answer.
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2
    Yeah, it's a tricky Emden-Fowler equation. The only thing I can think of is to multiply by x' and integrate to get

    [tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{x} + C [/tex]

    Now, swap for x from the original equation, i.e.,

    [tex]x = \frac{E}{m x^{\prime \prime}}[/tex]

    such that

    [tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{\frac{E}{m x^{\prime \prime}}} + C [/tex]

    rearrange to get

    [tex]x^{\prime \prime} = A e^{-\frac{m x^{\prime 2}}{2 E}} [/tex]

    where A is a constant. Now make the change of variable

    [tex]x^{\prime} = i \sqrt{\frac{2 E}{m}} z [/tex]

    to get

    [tex]i \sqrt{\frac{2E}{m}} z^{\prime} e^{-z^2} = A [/tex]

    but of course

    [tex]\frac{ d }{dt} erf{(z)} = \sqrt{\frac{2}{\pi}} z^{\prime} e^{-z^2}[/tex]

    so integrating gives you

    [tex]erf{(z)} = \alpha t + \beta [/tex]

    where alpha and beta are constants that may or may not be complex

    So there's a solution of the form

    [tex]x(t) = i\sqrt{\frac{2 E}{m}} \int{erf^{-1}{(\alpha t + \beta)} dt}[/tex]

    Note: the inverse erf function is integrable -- see this Wolfram page.
    Last edited: Mar 13, 2007
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