Difference between revisions of "Solution details"

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so  
 
so  
  
<math> \nabla^2 V = \labmda_L^2 V </math>
+
<math> \nabla^2 V = \lambda_L^2 V </math>

Revision as of 18:02, 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]