Asymptotic Expansion of ODE

  • Thread starter Chewie666
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  • #1
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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
 

Answers and Replies

  • #2
pasmith
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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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