Changes between Version 16 and Version 17 of u/johannjc/scratchpad5

Timestamp:

10/02/15 17:18:36 (9 years ago)

Author:

Jonathan

Comment:

—

u/johannjc/scratchpad5

@@ -166 +166 @@
 
 $B_{ij}=\kappa \delta{ij}$
+
+is diagonal so the cross terms vanish
+
+and $\alpha_{\pm i, \pm j} = 0$
+
+and it is independent of direction, so it is just a scalar.
+
+|| $E_j$ || $A \left ( \partial_j \kappa \right )$ ||
+|| $F$ || $A \kappa$ ||
+||$\alpha_0 $ ||$ -\frac{2F\delta_{ii} \Delta t}{\Delta x^2}$ ||
+||$\alpha_{\pm j}$ || $ \pm \frac{E_j \Delta t}{2 \Delta x} +  \frac{F\Delta t}{\Delta x^2}$ ||
+||$\alpha_{\pm i, \pm j}$ || $ 0$ ||
+
+Also if $\kappa$ is a constant, the $E_j$ terms drop out.
+
+
+=== Explicit Case ===
+
+For the Explicit case, we just set $\phi = 0$ which sets $D = 0$ and $C=1$ and we have
+
+$T_0 -  \displaystyle \sum_{\parallel, \perp} \left [ \alpha_0 T^{\lambda+1}_0 +  \displaystyle \sum_{\pm j} \alpha_{\pm j} T^{\lambda+1}_{\pm \hat{j}} +  \sum_{\pm i, \pm j,i \ne j} \alpha_{\pm i, \pm j} T^{\lambda+1}_{\pm \hat{i} \pm \hat{j}} \right ] = T'_0$