Difference between revisions of "Solution details"

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In spherical coordinates:
 
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>  
+
<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.
 
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:
 
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>
+
<math>\frac {1}{sin\theta} \frac{\partial{}}{\partial{\theta}}  sin\theta\frac{\partial{V}}{\partial{\theta}} = R_k(r') \frac {1}{\mu} \right [ \frac{d}{d \mu} (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} \left] = -k(k+1) P_k(\mu) </math>

Revision as of 22:17, 25 October 2013

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

(2x2+2x2)n + DL2z2 - Wz n = 0

Steps to solve Boltzmann equation

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

n(x,y,z)=eλLzV(x,y,z)

so

2V=λ2LV

where

2V=2x2+2y2+2z2

and

x=DLDx y=DLDy

In spherical coordinates:

1r2rr2Vr+1r2sinθθsinθVθ=λ2LV which is symmetric in ϕ direction.

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

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