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Asymptotic Expansion of ODE

  1. Jul 8, 2013 #1
    Ahoy!

    I'm trying to approximate [itex]f'(r)[/itex] for the following equation using matched asymptotic expansions

    [itex]-\frac{1}{2}\epsilon ff''=\left[\left(\epsilon+2r\right)f''\right]'[/itex]

    where [itex]\epsilon \ll 1[/itex] and with the boundary conditions [itex] f(0)=f'(0)=0, \quad f'(\infty)=1[/itex]

    The inner expansion which satisfies [itex] f'(0)=0[/itex] is simple enough by choosing an appropriate inner variable.

    My problem is trying to form an outer expansion of the form

    [itex] f'=1+\sigma(\epsilon) f_1+ \dots [/itex]

    where [itex]\sigma[/itex] is found through matching. In my working I find [itex]f_i≈A_i\ln r[/itex] which obviously doesn't satisfy [itex]f'(\infty)=0[/itex] unless the constants equal zero.

    I've tried introducing a stretched variable of the form [itex] \gamma =\epsilon r[/itex] but with no success.

    Any suggestions?

    Cheers
     
  2. jcsd
  3. Jul 26, 2013 #2

    pasmith

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    Homework Helper

    It is obvious from inspection that [itex]f(r) = r[/itex] is a solution of the full ODE for any [itex]\epsilon[/itex]. It satisfies [itex]f(0) = 0[/itex] and [itex]f'(\infty) = 1[/itex]. This suggests that higher order corrections to [itex]f(r) = r[/itex] are unnecessary, except in the boundary layer near the origin since [itex]f'(0) \neq 0[/itex]. This is consistent with your conclusion that the outer corrections must vanish.
     
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