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.