/[escript]/trunk/doc/cookbook/TEXT/onedheatdiff001.tex
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revision 2401 by ahallam, Wed Apr 29 04:23:07 2009 UTC revision 2411 by ahallam, Fri May 8 03:13:15 2009 UTC
# Line 82  Because we have a symmetrical problem we Line 82  Because we have a symmetrical problem we
82  \begin{verbatim}  \begin{verbatim}
83   myPDE.setSymmetryOn()   myPDE.setSymmetryOn()
84  \end{verbatim}  \end{verbatim}
85  To input the PDE into \verb escript it must be compared with the general form equation of a linear PDE which is described by:  To input the PDE into \esc it must be compared with the general form\footnote{Available in section ?? of the users guide to \esc }. For most simple PDEs however, the general form is over complicated and confusing. Thus for this example, we will use a simplified version that suits our heat diffusion problem. This simpler form is described by;
86  \begin{equation}\label{eqn:genfrm}  \begin{equation}\label{eqn:simpfrm}
87  -(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}-(B\hackscore{j} u)\hackscore{,j}+C\hackscore{l} u\hackscore{,l}+D u =-X\hackscore{j,j}+Y \; .  -(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =+Y
88    \end{equation}
89    This can be written in the form;
90    \begin{equation}    
91    -\nabla.(A.\nabla u) + Du = f
92    \end{equation}
93    or;
94    \begin{equation}\label{eqn:commonform}
95    -A\frac{\partial^{2}u}{\partial x^{2}\hackscore{i}} + Du = f
96    \end{equation}
97    When comparing equations \eqref{eqn:hd} and \eqref{eqn:commonform} we see that;
98    \begin{equation}
99    A = \kappa . \delta; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H}
100    \end{equation}
101    Additionally we must also consider the boundary conditions of our PDE. They take the form:
102    \begin{equation}
103    \eta \hackscore{j} A\hackscore{jl} u\hackscore{,l} + du = y
104  \end{equation}  \end{equation}
105    
106    

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