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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