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Series solution to nonlinear differential equation

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data

    By truncating the differential equation below at n=12, derive the form of the solution, obtaining expressions for all the ancoefficients in terms of the parameter [itex]\lambda [/itex].


    2. Relevant equations
    The ODE is:
    [tex] \frac{\mathrm{d^2}\phi }{\mathrm{d} x^2} = \frac{\phi^{3/2}}{x^{1/2}} [/tex]

    I am told that the solutions have the form:

    [tex] \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{3/2} [/tex]


    3. The attempt at a solution
    The way I have solved series equations before was to equate the coefficients of like powers of x. In this case, however, I have those x underneath the root, and I don't know how to go about doing this. I have the terms, via Mathematica. But I don't know how to get relationships between the coefficients.

    Code (Text):
    p[x_] := 1 + \[Lambda]*x + Sum[Subscript[a, n]*x^(n/2), {n, 3, 12}];
    pdd[x_] :=
      Sum[Subscript[a, n]*(n^2 - 2*n)/4 x^((n - 4)/2), {n, 3, 12}];

    LHS = pdd[x] - x^(-1/2)*(p[x])^(3/2)
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Sep 26, 2014 #2

    haruspex

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    Do you mean [tex] \phi = 1+ \lambda x + \sum_{n=2}^{12}a_{n} x^{n} [/tex]?
     
  4. Sep 26, 2014 #3
    No, but I did make an error in the above. It should be:

    [tex] \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{n/2} [/tex]

    But I can't figure out how to edit my original post.
     
  5. Sep 26, 2014 #4

    haruspex

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    That suggests to me the substitution x = u2. Haven't tried it, though.
     
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