Changes between Version 93 and Version 94 of FluxLimitedDiffusion

Timestamp:

03/27/13 23:47:57 (12 years ago)

Author:

Jonathan

Comment:

—

FluxLimitedDiffusion

@@ -503 +503 @@
 and plug the result into the first equation to get a matrix equation involving only one variable.
 
-   [[latex(\color{purple}{ \left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2} + \epsilon^n_i \right ) \right ] E^{n+1}_i - \left ( \psi \left ( \alpha^n_{i+1/2} - \omega_i v_{x,i} \right ) \right ) E^{n+1}_{i+1} - \left ( \psi \left ( \alpha^n_{i-1/2} + \omega_i v_{x,i} \right ) \right ) E^{n+1}_{i-1} - \left ( \psi \phi^n_i \right ) e^{n+1}_i=\left [ 1 - \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2} + \epsilon^n_i \right ) \right ] E^n_i + \left ( \bar{\psi} \left ( \alpha^n_{i+1/2} - \omega_i v_{x,i} \right ) \right ) E^{n}_{i+1} + \left ( \bar{\psi} \left ( \alpha^n_{i-1/2} + \omega_i v_{x,i} \right ) \right ) E^{n}_{i-1} +\bar{\psi}\phi^n_i e^n_i + \theta^n_i})]]   
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    [[latex(\color{purple}{\left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2} + \frac{\epsilon^n_i}{ 1 +\psi \phi^n_i}\right ) \right ] E^{n+1}_i - \left ( \psi \left ( \alpha^n_{i+1/2}- \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n+1}_{i+1} - \left ( \psi \left ( \alpha^n_{i-1/2} + \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n+1}_{i-1} =\left [ 1 - \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2}  + \frac{\epsilon^n_i }{ 1 +\psi \phi^n_i} \right ) \right ] E^n_i + \left ( \bar{\psi} \left ( \alpha^n_{i+1/2} - \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n}_{i+1} + \left ( \bar{\psi} \left ( \alpha^n_{i-1/2} + \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n}_{i-1} + \frac{ \phi^n_i}{ 1 +\psi \phi^n_i}  e^n_i+ \frac{1}{ 1 +\psi \phi^n_i}\theta^i_n})]]   
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-If we ignore the change in the Planck function due to heating during the implicit solve, it is equivalent to setting [[latex(\psi \phi = 0)]]  This gives the following equations:
-
-   [[latex(\left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2} + \epsilon^n_i \right ) \right ] E^{n+1}_i - \left ( \psi \alpha^n_{i+1/2} \right ) E^{n+1}_{i+1} - \left ( \psi \alpha^n_{i-1/2} \right ) E^{n+1}_{i-1} =\left [ 1 - \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i-1/2} + \epsilon^n_i \right ) \right ] E^n_i+ \left ( \bar{\psi} \alpha^n_{i+1/2} \right ) E^{n}_{i+1} + \left ( \bar{\psi} \alpha^n_{i-1/2} \right ) E^{n}_{i-1} +\phi^n_i e^n_i + \theta^n_i)]]   
-   [[latex(e^{n+1}_i = e^n_i + \epsilon^n_i  \left [ \left ( \psi E^{n+1}_i + \bar{\psi} E^{n}_i \right ) - \frac{4 \pi}{c} B \left ( T^n_i \right )  \right ] )]]   
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-In this case the first equation decouples and can be solved on it's own, and then the solution plugged back into the second equation to solve for the new energy.