Given this PDE:
$$\frac{d^2Z}{dz^2}+\frac{\sigma'(z)}{\sigma(z)}\frac{dZ}{dz}-\lambda^2Z=0$$ , with $\sigma>0$ and finite throughout the domain, and with $Z$ subjected to the boundary conditions: $$$$ $Z\rightarrow 0$ as $z\rightarrow \infty$; $Z'(0)=0$ $$$$ The paper I am reading says that, when $\lambda$ is large, the solution of this PDE approaches: $$Z(z)=\frac{1}{\sqrt {\sigma(z)}}e^{\pm \lambda z}$$
Could you explain to me why the solution approaches this form when $\lambda$ is large?
Let $Z(z) = f(\lambda z)/\sqrt{\sigma(z)}$. Then \begin{multline} Z'' + \frac{\sigma'}{\sigma}Z' - \lambda^2Z \\= \frac{1}{\sqrt{\sigma}}\left[\lambda^2f'' - \frac{\sigma'}{\sigma}\lambda f' - \left(\frac{\sigma''}{2\sigma} -\frac{3\sigma'^2}{4\sigma^2}\right)f\right] + \frac{\sigma'}{\sigma\sqrt{\sigma}}\left(\lambda f'-\frac{\sigma'}{2\sigma}f\right) - \frac{\lambda^2}{\sqrt{\sigma}}f \\= \frac{1}{\sqrt{\sigma}}\left[\lambda^2 f'' - \left(\lambda^2 + \frac{\sigma''}{2\sigma} - \frac{\sigma'^2}{4\sigma}\right)f\right] \end{multline} So if we define $\phi = \frac{1}{2}\ln\sigma$, $f$ satisfies $$ f'' - f = \frac{\phi''+\phi'^2}{\lambda^2}f $$ If $\lambda^2 \gg \phi'',\phi'^2$ over the whole domain, then to lowest order we will have $f''- f \simeq0$, and so $f(z) \simeq Ce^{\pm z}$, which in turn means $Z(x)\simeq Ce^{\pm\lambda z}/\sqrt{\sigma(z)}$.
Note that this requires $\phi', \phi''$ to stay bounded. This is not a consequence of the boundedness of $\sigma$. For example, $\sigma(z) = \exp[\sin(z^2)]$ is bounded and positive everywhere but the corresponding $\phi'(z) = z\cos(z^2)$ is unbounded.