Solution details

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

[math] (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})[/math]n + [math] D_L \frac {\partial^2{}}{\partial{z^2}}[/math] - [math] W \frac {\partial{}}{\partial{z}}[/math] n = 0

Steps to solve Boltzmann equation

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

[math] n(x', y', z') = e^{\lambda_L z'} V(x,y,z) [/math]

so

[math] \nabla'^2 V = \lambda_L^2 V [/math]

where

[math] \nabla'^2 V = \frac {\partial^2{}}{\partial{x'^2}} + \frac {\partial^2{}}{\partial{y'^2}} + \frac {\partial^2{}}{\partial{z^2}}[/math]

and

[math] x' = \frac {D_L}{D} x [/math] [math] y' = \frac {D_L}{D} y [/math]

In spherical coordinates:

[math] \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 [/math] which is symmetric in [math]\phi[/math] direction.

the solution of the zenith angle direction is the Legendre polynomial if it satisfied the following condition:

[math]\frac {1}{sin\theta'} \frac{\partial{}}{\partial{\theta'}} sin\theta'\frac{\partial{V}}{\partial{\theta'}} =\frac {1}{\mu} \frac{d}{d \mu} (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} = -k(k+1) P_k(\mu) [/math]