Asymptotic Representation of the Solution of an ODE This question is based on the accepted answer of a previous question. Given this ODE:
$$\frac{d^2Z}{dz^2}+\frac{\sigma'(z)}{\sigma(z)}\frac{dZ}{dz}-\lambda^2Z=0$$
, with $\sigma>0$ and finite throughout the domain, and with $Z$ subjected to the boundary conditions:
$$$$
$Z\rightarrow 0$ as $z\rightarrow \infty$; $Z'(0)=0$
$$$$
In the previous question, it was shown by eyeballfrog that when $\lambda$ is large, the solution of this ODE approaches:
$$Z(z)=\frac{1}{\sqrt {\sigma(z)}}e^{\pm \lambda z}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space (1)$$
The paper I am reading says that, because of this convergence, the solution of this ODE can be represented asymptotically (as to $\lambda$) as:
$$Z(z,\lambda)=\frac{e^{\pm \lambda z}}{\sqrt{\sigma(z)}}[1+\sum_{n=1}^\infty \alpha_n(z)\lambda^{-n}]$$
Could you explain to me why $Z$ can be represented this way, given the fact that it converges to the form (1)?
 A: Well first we need to prove the series really does converge. I kind of fudged it in the last answer because it's annoying and as a physicist I know the lowest order is the only one that matters.
So anyways, we go back to the equation
$$
f''(z) - f(z) = \frac{\phi''(z) + \phi'(z)^2}{\lambda^2}f(z)
$$
Now we suppose $f(z) = \sum_{n=0}^\infty \lambda^{-2n} f_n(z)$. This gives
$$
f_0''(z) - f_0(z) + \sum_{n=1}^\infty \lambda^{-2n}\left[f_n''(z) - f_n(z)\right] = \sum_{n=1}^\infty \lambda^{-2n} [\phi''(z) + \phi'(z)^2]f_{n-1}(z)
$$
which can be solved order by order. The lowest order is
$$
f_0''(z) - f_0(z) = 0
$$
which we already know is solved by $f_0(z) = Ce^{\pm z}$. The next orders can be found by iterating
$$
f_n''(z) - f_n(z) = [\phi''(z) + \phi'(z)^2]f_{n-1}(z)
$$
The really annoying part is proving that the boundedness of $\phi'$ and $\phi''$ implies each of the $f_n$ scales as at most $C z^ne^{\pm z}$. I'll leave that as an exercise for the reader. The important thing is that this means $f_n(\lambda z) = C\lambda^n\alpha_n(z)e^{\pm \lambda z}$ for some polynomial-order function $\alpha_n(z)$, and thus
$$
f(\lambda z)= \sum_{n=0}^\infty \lambda^{-2n}f_n(\lambda z) = C e^{\pm \lambda z}\left[1 + \sum_{n=0}^\infty\alpha_n(z)\lambda^{-n}\right]
$$
which immediately leads to
$$
Z(z) = \frac{f(\lambda z)}{\sqrt{\sigma(z)}} = C\frac{e^{\pm\lambda z}}{\sqrt{\sigma(z)}}\left[1 + \sum_{n=0}^\infty\alpha_n(z)\lambda^{-n}\right],
$$
which should at the very least be convergent whenever $\lambda > z$.
