13 


14 
\section{Two Dimensional Heat Diffusion for a basic Magmatic Intrusion} 
\section{Two Dimensional Heat Diffusion for a basic Magmatic Intrusion} 
15 
%\label{Sec:2DHD} 
%\label{Sec:2DHD} 
16 
Building upon our success from the 1D models it is now prudent to expand our domain by another dimension. For this example we will be using a very simple magmatic intrusion as the basis for our model. The simulation will be a single event where some molten granite has formed a hemisphericle dome at the base of some cold sandstone country rock. A hemisphere is semetric so taking a crosssection through its centre will effectively model a 3D problem in 2D. New concepts will include nonlinear boundaries and the abilitity to prescribe location specific variables. 
Building upon our success from the 1D models it is now prudent to expand our domain by another dimension. For this example we will be using a very simple magmatic intrusion as the basis for our model. The simulation will be a single event where some molten granite has formed a hemisphericle dome at the base of some cold sandstone country rock. A hemisphere is symmetric so taking a crosssection through its centre will effectively model a 3D problem in 2D. New concepts will include nonlinear boundaries and the ability to prescribe location specific variables. 
17 


18 
\begin{figure}[h!] 
\begin{figure}[h!] 
19 
\centerline{\includegraphics[width=4.in]{figures/twodheatdiff}} 
\centerline{\includegraphics[width=4.in]{figures/twodheatdiff}} 
21 
\label{fig:twodhdmodel} 
\label{fig:twodhdmodel} 
22 
\end{figure} 
\end{figure} 
23 


24 
To expand upon our 1D problem, the domain must first be expanded. This will be done in our definition phase by creating a square domain in $x$ and $y$ that is 600 meters along eacy side \reffig{fig:twodhdmodel}. The number of discrete spatial cells will be 100. The radius of the intrusion will be 200 meters And the location of the centre of the intrusion will be at the 300 meter mark on the xaxis. The domain variables are; 
To expand upon our 1D problem, the domain must first be expanded. This will be done in our definition phase by creating a square domain in $x$ and $y$ that is 600 meters along each side \reffig{fig:twodhdmodel}. The number of discrete spatial cells will be 100. The radius of the intrusion will be 200 meters And the location of the centre of the intrusion will be at the 300 meter mark on the xaxis. The domain variables are; 
25 
\begin{verbatim} 
\begin{verbatim} 
26 
mx = 600 # model lenght 
mx = 600*m #meters  model length 
27 
my = 600 # model width 
my = 600*m #meters  model width 
28 
ndx = 100 # steps in x direction 
ndx = 100 #mesh steps in x direction 
29 
ndy = 100 # steps in y direction 
ndy = 100 #mesh steps in y direction 
30 
r = 200 # radius of intrusion 
r = 200*m #meters  radius of intrusion 
31 
ic = [300, 0] #centre of intrusion 
ic = [300, 0] #centre of intrusion (meters) 
32 

q=0.*Celsius #our heat source temperature is now zero 
33 
\end{verbatim} 
\end{verbatim} 
34 
The next step is to define our variables for each material in the model in a manner similar to the previous tutorial. Note that each material has its own unique set of values. The time steps and set up for the domain remain as before. Prior to setting up the PDE the boundary between the two materials must be established. The distance $s$ between two points in car cartesian coordinates is defined as: 
The next step is to define our variables for each material in the model in a manner similar to the previous tutorial. Note that each material has its own unique set of values. The time steps and set up for the domain remain as before. Prior to setting up the PDE the boundary between the two materials must be established. The distance $s$ between two points in car Cartesian coordinates is defined as: 
35 
\begin{equation} 
\begin{equation} 
36 
(x_{1}x_{0})^{2}+(y_{1}y_{0})^{2} = s^{2} 
(x_{1}x_{0})^{2}+(y_{1}y_{0})^{2} = s^{2} 
37 
\end{equation} 
\end{equation} 
38 
If $[x_{0},y_{0}]$ is the point $c$ the centre of the semicircle that defines our intrusion then for all the points $[x,y]$ in our solution space we can define a distance to $c$. Hence, and points that fall within the radius $r$ of our intrusion will have a coresponding value $s < r$ and all those belonging to the country rock will have a value $s > r$. By subtracting $r$ from both of these conditions we find $sr < 0$ for all intrusion points and $sr > 0$ for all country rock points. Defining these conditions into the code is quite simple and is done using the following command: 
If $[x_{0},y_{0}]$ is the point $c$ the centre of the semicircle that defines our intrusion then for all the points $[x,y]$ in our solution space we can define a distance to $c$. Hence, and points that fall within the radius $r$ of our intrusion will have a corresponding value $s < r$ and all those belonging to the country rock will have a value $s > r$. By subtracting $r$ from both of these conditions we find $sr < 0$ for all intrusion points and $sr > 0$ for all country rock points. Defining these conditions within the script is quite simple and is done using the following command: 
39 
\begin{verbatim} 
\begin{verbatim} 
40 
bound = length(xic)r #where the boundary will be located 
bound = length(xic)r #where the boundary will be located 
41 
\end{verbatim} 
\end{verbatim} 
50 
\begin{verbatim} 
\begin{verbatim} 
51 
T= Ti*whereNegative(bound)+Tc*wherePositive(bound) #defining the initial temperatures. 
T= Ti*whereNegative(bound)+Tc*wherePositive(bound) #defining the initial temperatures. 
52 
\end{verbatim} 
\end{verbatim} 
53 
The itteration process now begins as before, but using our new conditions for \verb D as defined above. 
The iteration process now begins as before, but using our new conditions for \verb D as defined above. 
54 


55 
\subsection{Contouring escript data} 
\subsection{Contouring escript data} 
56 

It is possible to contour our solution using \modmpl . Unfortunately the \modmpl contouring function only accepts regularly gridded data. As our solution is not regularly gridded, it is necessary to interpolate our solution onto a regular grid. First we extract the model coordinates using \verb getX these are then transformed to a numpy array and into individual $x$ and $y$ arrays. We also need to generate our regular grid which is done using the \modnumpy function \verb linspace . 
57 
\begin{verbatim} 
\begin{verbatim} 
58 
# rearrage mymesh to suit solution function space 
# rearrage mymesh to suit solution function space for contouring 
59 
oldspacecoords=model.getX() 
oldspacecoords=model.getX() 
60 
coords=Data(oldspacecoords, T.getFunctionSpace()) 
coords=Data(oldspacecoords, T.getFunctionSpace()) 
61 
coords = np.array(coords.toListOfTuples()) 
coordX, coordY = toXYTuple(coords) 

coordX = coords[:,0] 


coordY = coords[:,1] 

62 
# create regular grid 
# create regular grid 
63 
xi = np.linspace(0.0,600.0,100) 
xi = np.linspace(0.0,mx,100) 
64 
yi = np.linspace(0.0,600.0,100) 
yi = np.linspace(0.0,my,100) 
65 
\end{verbatim} 
\end{verbatim} 
66 

The remainder of our contouring commands reside within the \verb while loop so that a new contour is generated for each time step. For each time step the solution much be regridded for \modmpl using the \verb griddata function. This function interprets an irregular grid and solution from \verb tempT , \verb xi and \verb yi this is transformed to the new coordinates defined by \verb coordX and \verb coordY with an output \verb zi . It is now possible to use the \verb contourf function which generates a colour filled contour. The colour gradient of our plots is set via the command \verb pl.matplotlib.pyplot.autumn() , other colours are listed on the \modmpl web page. Our results are then contoured, visually adjusted using the \modmpl functions and then saved to file. \verb pl.clf() clears the figure in readiness for the next time iteration. 
67 


68 

\begin{verbatim} 
69 

#grid the data. 
70 

zi = pl.matplotlib.mlab.griddata(coordX,coordY,tempT,xi,yi) 
71 

# contour the gridded data, plotting dots at the randomly spaced data points. 
72 

pl.matplotlib.pyplot.autumn() 
73 

pl.contourf(xi,yi,zi,10) 
74 

CS = pl.contour(xi,yi,zi,5,linewidths=0.5,colors='k') 
75 

pl.clabel(CS, inline=1, fontsize=8) 
76 

pl.axis([0,600,0,600]) 
77 

pl.title("Heat diffusion from an intrusion.") 
78 

pl.xlabel("Horizontal Displacement (m)") 
79 

pl.ylabel("Depth (m)") 
80 

pl.savefig(os.path.join(save_path,"heatrefraction%03d.png") %i) 
81 

pl.clf() 
82 

\end{verbatim} 
83 


84 

\begin{figure}[h!] 
85 

\centerline{\includegraphics[width=4.in]{figures/heatrefraction050}} 
86 

\caption{2D model: Total temperature distribution ($T$) at time $t=50$.} 
87 

\label{fig:twodhdmodel} 
88 

\end{figure} 