Changes between Version 10 and Version 11 of FluxLimitedDiffusion
- Timestamp:
- 03/18/13 19:42:29 (12 years ago)
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FluxLimitedDiffusion
v10 v11 79 79 For now we will assume that [[latex(\kappa_{0P})]] and [[latex(\kappa_{0R})]] are constant over the implicit update. In this case we can solve the radiation energy equation: 80 80 81 [[latex(\frac{\partial E}{\partial t} = \nabla \cdot \frac{c\lambda}{\kappa_{0R}} \nabla E + \kappa_{0P} (4 \pi B-cE))]] 81 [[latex(\frac{\partial E}{\partial t} = \nabla \cdot \mathbf{F} + \kappa_{0P} (4 \pi B-cE) = \nabla \cdot \frac{c\lambda}{\kappa_{0R}} \nabla E + \kappa_{0P} (4 \pi B-cE))]] 82 83 where [[latex(\mathbf{F} = \frac{c\lambda}{\kappa_{0R}} \nabla E)]] 82 84 83 85 Which we can discretize for (1D) as … … 86 88 87 89 where 90 91 [[latex(\frac{\Delta t}{\Delta x}\mathbf{F}^n_{i+1/2} = \alpha^n_{i+1/2} \left ( E^{n+1}_{i+1} - E^{n+1}_i \right ) )]] 92 93 and 88 94 89 95 [[latex(\epsilon^n_i=\frac{c\Delta t}{ \kappa^n_{0P,i}})]] … … 144 150 145 151 == Constant radiative flux == 146 Having a contant radiative flux 147 means that 148 [[latex(\alpha_g \frac{\Delta x}{\Delta t} \left ( E^{n+1}_i-E^{n+1}_g \right ) = F_0)]] 152 To have a constant radiative flux we must have 153 [[latex(\alpha_g \left ( E^{n+1}_i-E^{n+1}_g \right ) = F_0 \frac{\Delta t}{\Delta x})]] 154 155 Which we can solve for 156 [[latex(E^{n+1}_g = E^{n+1}_i - \frac{F_0 \Delta t}{\alpha_g \Delta x})]] 157 158 but when we plug this into the coefficient matrix the terms with [[latex(\alpha_g)]] cancel and we just get [[latex(F_0 \frac{\Delta t}{\Delta x})]] in the source vector
