Changes between Version 93 and Version 94 of FluxLimitedDiffusion
 Timestamp:
 03/27/13 23:47:57 (12 years ago)
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FluxLimitedDiffusion
v93 v94 503 503 and plug the result into the first equation to get a matrix equation involving only one variable. 504 504 505 [[latex(\color{purple}{ \left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} + \omega_i v_{x,i} \right ) \right ) E^{n+1}_{i1}  \left ( \psi \phi^n_i \right ) e^{n+1}_i=\left [ 1  \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} + \omega_i v_{x,i} \right ) \right ) E^{n}_{i1} +\bar{\psi}\phi^n_i e^n_i + \theta^n_i})]]506 507 505 [[latex(\color{purple}{\left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} + \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n+1}_{i1} =\left [ 1  \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} + \frac{\omega_i v_{x,i}}{1+\psi \phi^n_i} \right ) \right ) E^{n}_{i1} + \frac{ \phi^n_i}{ 1 +\psi \phi^n_i} e^n_i+ \frac{1}{ 1 +\psi \phi^n_i}\theta^i_n})]] 508 509 510 511 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:512 513 [[latex(\left [ 1 + \psi \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} \right ) E^{n+1}_{i1} =\left [ 1  \bar{\psi} \left( \alpha^n_{i+1/2} + \alpha^n_{i1/2} + \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_{i1/2} \right ) E^{n}_{i1} +\phi^n_i e^n_i + \theta^n_i)]]514 [[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 ] )]]515 516 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.517 506 518 507