How can I find an outer expansion for f'(r) in this ODE?

[Chewie666]

Ahoy!

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

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

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

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

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

f'=1+\sigma(\epsilon) f_1+ \dots

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

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

Any suggestions?

Cheers

 

[pasmith]

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