Difference between revisions of "Solution details"

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where <math>  I_{\nu}</math> and  
 
where <math>  I_{\nu}</math> and  
 
<math>  K_{\nu}</math> are the modified Bessel of the first and the second kind.
 
<math>  K_{\nu}</math> are the modified Bessel of the first and the second kind.
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In case of of solving for the density number outside the hole, then the solution contains only the modified Bessel of the second kind <math>  K_{\nu}</math>, Also applying hte boundary conditions , implies that  <math> P_{k} = 0</math>  if k is odd, so the solution can be written as :
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The previous solution bcan be written as
 
  
  

Revision as of 20:25, 26 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 <ref name="Huxley"> Huxley, L. G. H. Leonard George Holden, The diffusion and drift of electrons in gases, John Wiley and sons, 1974 , call number QC793.5.E628 H89 </ref>

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.

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

[math]\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] [/math]

and


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

so,

[math] \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 [/math]


The modified Bessel functions, first and second kind, are the solutions for the previous equation but the boundary conditions determines which one to use, in this case [math] r'\rightarrow 0[/math], [math] n \rightarrow \infty [/math], and [math] n \rightarrow 0 [/math] as [math] r'\rightarrow \infty [/math]. so only the modified Bessel of second kind K_k are the non-zaro terms. so the the general solution for the equation can be written as :

[math] V= R_k (r') Pk(\mu) = \exp{(\lambda_L z)}\sum_{k=0}^{\infty} A_k r'^{-1/2} K_{k+1/2} (\lambda_L r') P_k(\mu) [/math]

Solution Analysis

The general form of the previous equation and its solution are defined as the following:

[math] \frac{d^2 y}{dx} +\frac{1-2\alpha}{x} \frac{dy}{dx}-\left [ \frac{\nu^2 \gamma^2 - \alpha^2}{x^2} + (\beta \gamma x^{\gamma - 1})^2 \right]y = 0 [/math]

and

[math] y = x^{\alpha} I_{\nu} (\beta x^{\gamma}) [/math] and

[math] y = x^{\alpha} K_{\nu} (\beta x^{\gamma}) [/math]

where [math] I_{\nu}[/math] and [math] K_{\nu}[/math] are the modified Bessel of the first and the second kind.

In case of of solving for the density number outside the hole, then the solution contains only the modified Bessel of the second kind [math] K_{\nu}[/math], Also applying hte boundary conditions , implies that [math] P_{k} = 0[/math] if k is odd, so the solution can be written as :







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