Changes between Version 1 and Version 2 of u/johannjc/scratchpad2

Timestamp:

09/29/15 11:47:49 (9 years ago)

Author:

Jonathan

Comment:

—

u/johannjc/scratchpad2

@@ -19 +19 @@
 which can be expanded as 
 
-$n \chi_\perp \partial_i \partial_i T + n \left ( \partial_B \chi_\perp \right ) \left (\partial_i B \right ) \left ( \partial_i T \right ) + 2 \chi_\perp \left ( \partial_i n \right ) \left (\partial_i T \right )$
+$n \chi_\perp \partial_i \partial_i T + n \left ( \partial_B \chi_\perp \right ) \left (\partial_i B \right ) \left ( \partial_i T \right ) + 2 \chi_\perp \left ( \partial_i n \right ) \left (\partial_i T \right ) + n \left (\partial_T \chi_\perp \right )\left (\partial_i T \right ) \left ( \partial_i T \right )$ 
 
 
@@ -45 +45 @@
 Now if we group terms in the expansion 
 
-$ A = B_i \partial_i T + C_{ij} \partial_i \partial_j T + D_{ij} \left ( \partial_i T \right ) \left ( \partial_j T \right ) + E \partial_i \partial_i T$
+$ A = B_i \partial_i T + C_{ij} \partial_i \partial_j T + D_{ij} \left ( \partial_i T \right ) \left ( \partial_j T \right ) + E \partial_i \partial_i T + F \left ( \partial_i T \right ) \left ( \partial_i T \right )$
 
 we can work out coefficients for the various stencil terms (note $D_{ij}+D_{ji} = 2D_{ij}$ since $D_{ij}$ is symmetric)
@@ -52 +52 @@
 $\alpha_0 = -\frac{1}{\Delta t} -\frac{2C_{ii}}{\Delta x^2} - \frac{2\delta_{ii}E}{\Delta x^2}$
 
-$\alpha_{\pm \hat{i}} = \pm\frac{B_i}{2\Delta x} + \frac{C_{ii}}{\Delta x^2} \pm \frac{2D_{ij} \left ( T'_{\hat{j}} - T'_{-\hat{j}} \right )}{4 \Delta x^2} + \frac{E}{\Delta x^2} $
+$\alpha_{\pm \hat{i}} = \pm\frac{B_i}{2\Delta x} + \frac{C_{ii}}{\Delta x^2} \pm \frac{2D_{ij} \left ( T'_{\hat{j}} - T'_{-\hat{j}} \right )}{4 \Delta x^2} + \frac{E}{\Delta x^2} \pm \frac{2F\left ( T'_{\hat{i}} - T'_{-\hat{i}} \right )}{4 \Delta x^2}$
 
 $\alpha_{\pm \hat{i} \pm \hat{j}} = \pm \pm \frac{C_{ij}}{4\Delta x^2}\left ( 1 - \delta_{ij} \right )$
 
-and the right hand side
+and the right hand side is
 
-$A = -\frac{T'}{\delta t} + \frac{D_{ij}}{4 \Delta x^2}\left ( \partial_i T' \right ) \left ( \partial_j T' \right )$
+$A = -\frac{T'}{\delta t} + \frac{D_{ij}}{4 \Delta x^2}\left ( \partial_i T' \right ) \left ( \partial_j T' \right ) + \frac{F}{4\Delta x^2} \left (\partial_i T' \right ) \left (\partial_i T' \right ) $
+
+Now for the fun part...
+{{{
+#!latex
+$
+\begin{array}{rl}
+B_i  = &   \left ( \partial_j n b_j \right ) \left (\chi_\parallel - \chi_\perp \right ) b_i - n b_j \left ( \partial_n \chi_\perp \right ) \left ( \partial_j n \right ) b_i - n b_j \left (\partial_B \chi_\perp \right ) \left ( \partial_j B \right ) b_i  \\
+ & + \left (\partial_i n \right ) \chi_\perp + n \left ( \partial_n \chi_\perp \right ) \left (\partial_i n \right ) + n \left ( \partial_B \chi_\perp \right) \left ( \partial_i B \right ) \\
+C_{ij} = & n b_j \left ( \chi_\parallel - \chi_\perp \right ) b_i \\
+D_{ij} = & n b_j \partial_T \left ( \chi_\parallel - \chi_\perp \right ) b_i \\
+E = & n \chi_\perp \\
+F = & n \partial_T \chi_\perp  \\
+\end{array}
+$
+}}}
+
+Note that if we take the coefficients in $B_i \cdot b_i$ all of the perpendicular terms drop out - except for one involving changes in the direction of b $-n \left (\partial_i b_i \right ) \chi_\perp $