Difference between revisions of "Solution details"

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<math> \nabla'^2 V = \lambda_L^2 V </math>
 
<math> \nabla'^2 V = \lambda_L^2 V </math>
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where
  
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<math> \nabla'^2 V = \frac {\partial^2{}}{\partial{x'^2}} + \frac {\partial^2{}}{\partial{y'^2}} + \frac {\partial^2{}}{\partial{z^2}}</math>
  
<math> \nabla'^2 V = \frac {\partial^2{}}{\partial{x'^2}} + \frac {\partial^2{}}{\partial{y'^2}} + \frac {\partial^2{}}{\partial{z^2}}</math>
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and
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<math> x' = \frac {D_L}{D} x </math>
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<math> y' = \frac {D_L}{D} y </math>
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In spherical coordinates:
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<math> \frac {1}{r^'^2} \frac{\partial{}}{\partial{r'}}r'^2 \frac{\partial{V}}{\partial{r'} + \frac {1}{r^'sin\theta'} \frac{\partial{}}{\partial{\theta'} sin\theta'\frac{\partial{V}}{\partial{\theta'}  = \lambda_L^2 </math>

Revision as of 18:15, 25 October 2013

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^'sin\theta'} \frac{\partial{}}{\partial{\theta'} sin\theta'\frac{\partial{V}}{\partial{\theta'} = \lambda_L^2 V [/math]