Solution details

asymptotic solution details for Boltzmann equation for a hole has a uniform electric field

$(\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})$n + $D_L \frac {\partial^2{}}{\partial{z^2}}$ - $W \frac {\partial{}}{\partial{z}}$ n = 0

Steps to solve Boltzmann equation

for the previous equation let consider the asymptotic solution has the form:

$n(x', y', z') = e^{\lambda_L z'} V(x,y,z)$

so

$\nabla'^2 V = \lambda_L^2 V$

where

$\nabla'^2 V = \frac {\partial^2{}}{\partial{x'^2}} + \frac {\partial^2{}}{\partial{y'^2}} + \frac {\partial^2{}}{\partial{z^2}}$

and

$x' = \frac {D_L}{D} x$ $y' = \frac {D_L}{D} y$

In spherical coordinates:

$\frac {1}{r'^2} \frac{\partial{}}{\partial{r'}}r'^2 \frac{\partial{V}}{\partial{r'}} + \frac {1}{r'^2 sin\theta'} \frac{\partial{}}{\partial{\theta}} sin\theta \frac{\partial{V}}{\partial{\theta}} = \lambda_L^2 V$ which is symmetric in $\phi$ direction.

Assuming $V(r',\theta) = R_k(r')P_k(\mu)$the solution of the zenith angle direction is the Legendre polynomial, and can be written as:

$\frac {1}{r'sin\theta} \frac{\partial{}}{\partial{\theta}} sin\theta\frac{\partial{V}}{\partial{\theta}} = R_k(r') \frac{d}{d \mu} \left [ (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} \right]$

and

$\frac{d}{d \mu} \left [ (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} \right]= -k(k+1) P_k(\mu)$

so,

$\frac {1}{r'^2} \frac{d}{dr'}\left (r'^2 \frac{dR_k}{dr'}\right) - \left [ \frac{k(k+1)}{r'^2} +\lambda_L^2 \right]R_k = \frac{d^2 R_k}{dr'^2} +\frac{2}{r'} \frac{dR_k}{dr'}-\left [ \frac{k(k+1)}{r'^2} +\lambda_L^2 \right]R_k = 0$