Changes between Version 6 and Version 7 of ThermalConduction

Timestamp:

08/15/16 12:44:28 (8 years ago)

Author:

Jonathan

Comment:

—

ThermalConduction

@@ -172 +172 @@
 and it is independent of direction, so it is just a scalar.
 
+So in summary we have
+
+|| $A$ || $\frac{\kappa}{\rho c_v  \left ( \lambda + 1 \right ) }$ || 
+|| $B_{ij} $ || $\kappa \delta{ij}$||
+|| $C$ ||  $\left ( 1 - \phi \left ( \lambda+ 1 \right ) \right )$ ||
+|| $D$ ||  $\phi \left ( \lambda + 1 \right )$ ||
+
+and
 
 ||$\alpha_0$ || $- \displaystyle \sum_{i} \frac{B_{\hat{i}} + B_{-\hat{i}}}{\Delta x^2}$ ||
@@ -177 +185 @@
 || $\alpha*_{\pm i \pm j}$ || $ 0$ ||
 
-Also if $\kappa$ is a constant, then we have
+
+And our equation becomes
+
+$T_0 + \Delta t \displaystyle \sum_{\parallel, \perp} AC \left [ \alpha_0 T^{\lambda+1}_0 +  \displaystyle \sum_{\pm j} \alpha_{\pm j} T^{\lambda+1}_{\pm \hat{j}}  \right ] = $
+$T'_0 - \Delta t \displaystyle \sum_{\parallel, \perp} AD \left [  \alpha_0 T^{\lambda}_0T'_0 + \displaystyle \sum_{\pm j} \alpha_{\pm j} T^{\lambda}_{\pm \hat{j}} T'_{\pm \hat{j}}  \right ]$
+
+
+Now in the code, 
+|| $A = A$ ||
+|| $C = \frac{\Delta t A C}{\Delta x^2}$ ||
+|| $D = \frac{\Delta t A D}{\Delta x^2}$ ||
+|| $\mbox{Tlambda} = T_{\pm \hat{j}}^\lambda$ ||
+|| $\mbox{T} = T_{\pm \hat{j}}$ ||
+|| $\mbox{T0} = T $ ||
+|| $\mbox{T0lambda} = T^{\lambda}$ ||
+
+And stencil(0) has the coefficient of $T_0'$, and stencil(1:2*ndim) has the coefficients of $T_{\pm \hat{j}}'$ and source has everything else.
+
+
+
+
+Now if $\kappa$ is a constant, then we have
 
 ||$\alpha_0$ || $- \displaystyle \sum_{i} \frac{2B}{\Delta x^2}$ ||