For context, I'm studying the paper Coulomb blockade in superconducting quantum point contacts by Averin from 1998. Specifically, I am trying to find how he obtains equation 11 from equation 10, which gives the Landau Zener probability of ending up in a specific branch of the Josephson potential of a superconducting QPC.
Equation 10, describing the Schrodinger equation of the problem in the specific limit under consideration, is given by a system of two coupled first order ODE's: \begin{equation} 2\sqrt{\frac{E_C}{\Delta}} \frac{\partial\psi_s}{\partial x} = -s x \psi_s/2 + \sqrt{R} \psi_{-s} \end{equation} where $s\pm1$.
If I (for convenience) now take $A = 2\sqrt{\frac{E_C}{\Delta}}$ and $B = \sqrt{R}$, then substituting the differential equations one finds a 2nd order ODE that is solved by parabolic cylinder functions. Specifically, \begin{equation} \psi_{-1} = c_1 D_{\frac{-B^2-A}{A}}\left(\frac{x}{\sqrt{A}}\right)+c_2 D_{\frac{B^2}{A}}\left(\frac{i x}{\sqrt{A}}\right) \end{equation}
and
\begin{equation} \psi_1 = -\frac{1}{B}\left(x c_2 D_{\frac{B^2}{A}}\left(\frac{i x}{\sqrt{A}}\right)+\sqrt{A} \left[c_1 D_{-\frac{B^2}{A}}\left(\frac{x}{\sqrt{A}}\right)+i c_2 D_{\frac{B^2+A}{A}}\left(\frac{i x}{\sqrt{A}}\right)\right]\right) \\ \end{equation}
Now, in his paper he then says that by evaluating the asymptotes of these functions, one can find the probability $w$ for the state $s=1$ starting at $x \rightarrow -\infty$ to end up in the state $s=-1$ at $x\rightarrow \infty$. This leads to equation 11, which he writes as \begin{equation} w = \frac{1}{\Gamma(\lambda)}\sqrt{\frac{2\pi}{\lambda}}\left(\frac{\lambda}{e}\right)^\lambda \end{equation}
where $\lambda = \frac{R^2}{2\sqrt{E_C/\Delta}}$, and using our substitutions, we can identify $\lambda = B^2/A$, which we already saw occur in the solution for $\psi_{-1}$.
In terms of mathematics, this statement means that we need to set $|\psi_1(-\infty)|^2 = 1$ and $|\psi_{-1}(-\infty)|^2 = 0$, and then evaluate $w = |\psi_{-1}(\infty)|^2 = 1 - |\psi_{1}(\infty)|^2$. But how does one do this? I need to pick $c_1$ and $c_2$ to satisfy these boundary conditions at $-\infty$, but so far I haven't managed to figure out how to do that.
Some code for evaluating the ODE's:
ff = f[x] /. Solve[A*g'[x] == x*g[x]/2 + B*f[x], f[x]];
dff = D[ff, x];
gg = g[x] /.
DSolve[{A*f'[x] == -x*f[x]/2 + B*g[x] /. {f[x] -> ff,
f'[x] -> dff}}, g[x], x]
dgg = D[gg, x];
ff = FullSimplify[ff /. {g[x] -> gg, g'[x] -> dgg}]