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trunk/doc/cookbook/twodheatdiff001.tex revision 2658 by ahallam, Thu Sep 10 02:58:44 2009 UTC trunk/doc/cookbook/example03.tex revision 2975 by ahallam, Thu Mar 4 05:44:12 2010 UTC
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4  % Copyright (c) 2003-2009 by University of Queensland  % Copyright (c) 2003-2010 by University of Queensland
5  % Earth Systems Science Computational Center (ESSCC)  % Earth Systems Science Computational Center (ESSCC)
6  % http://www.uq.edu.au/esscc  % http://www.uq.edu.au/esscc
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13    
14  \section{Two Dimensional Heat Diffusion for a basic Magmatic Intrusion}  \begin{figure}[t]
 \sslist{twodheatdiff001.py and cblib.py}  
 %\label{Sec:2DHD}  
  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 cross-section through its centre will effectively model a 3D problem in 2D. New concepts will include non-linear boundaries and the ability to prescribe location specific variables.  
   
 \begin{figure}[h!]  
15  \centerline{\includegraphics[width=4.in]{figures/twodheatdiff}}  \centerline{\includegraphics[width=4.in]{figures/twodheatdiff}}
16  \caption{2D model: granitic intrusion of sandstone country rock.}  \caption{Example 3: 2D model: granitic intrusion of sandstone country rock.}
17  \label{fig:twodhdmodel}  \label{fig:twodhdmodel}
18  \end{figure}  \end{figure}
19    
20  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 x-axis. The domain variables are;  \sslist{example03a.py and cblib.py}
21  \begin{verbatim}  
22    Building upon our success from the 1D models, it is now prudent to expand our domain by another dimension.
23    For this example we use 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 cylindrical dome at the base of some cold sandstone country rock. Assuming that the cylinder is very long
24    we model a cross-section as shown in \reffig{fig:twodhdmodel}. We will implement the same
25    diffusion model as we have use for the granite blocks in \refSec{Sec:1DHDv00}
26    but will add the second spatial dimension and show how to define
27    variables depending on the location of the domain.
28    We use \file{onedheatdiff001b.py} as the starting point for develop this model.
29    
30    \section{Example 3: Two Dimensional Heat Diffusion for a basic Magmatic Intrusion}
31    \label{Sec:2DHD}
32    
33    To expand upon our 1D problem, the domain must first be expanded. In fact, we have solved a two dimensional problem already but essentially ignored the second dimension. In our definition phase we create a square domain in $x$ and $y$\footnote{In \esc the notation
34    $x\hackscore{0}$ and $x\hackscore{1}$ is used for $x$ and $y$, respectively.} that is $600$ meters along each side \reffig{fig:twodhdmodel}. Now we set the number of discrete spatial cells to 150 in both direction and the radius of the intrusion to $200$ meters with the centre located at the $300$ meter mark on the $x$-axis. Thus, the domain variables are;
35    \begin{python}
36  mx = 600*m #meters - model length  mx = 600*m #meters - model length
37  my = 600*m #meters - model width  my = 600*m #meters - model width
38  ndx = 100 #mesh steps in x direction  ndx = 150 #mesh steps in x direction
39  ndy = 100 #mesh steps in y direction  ndy = 150 #mesh steps in y direction
40  r = 200*m #meters - radius of intrusion  r = 200*m #meters - radius of intrusion
41  ic = [300, 0] #centre of intrusion (meters)  ic = [300*m, 0] #coordinates of the centre of intrusion (meters)
42  q=0.*Celsius #our heat source temperature is now zero  qH=0.*J/(sec*m**3) #our heat source temperature is zero
43  \end{verbatim}  \end{python}
44  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:  As before we use
45    \begin{python}
46    model = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy)
47    \end{python}
48    to generate the domain.
49    
50    There are two fundamental changes to the PDE that we have discussed in \refSec{Sec:1DHDv00}. Firstly,
51    because the material coefficients for granite and sandstone are different, we need to deal with
52    PDE coefficients which vary with their location in the domain. Secondly, we need
53    to deal with the second spatial dimension. We can investigate these two modification at the same time.
54    In fact, the temperature model \refEq{eqn:hd} we utilised in \refSec{Sec:1DHDv00} applied for the
55    1D case with a constant material parameter only. For the more general case examined in this chapter, the correct model equation is
56    \begin{equation}
57    \rho c\hackscore p \frac{\partial T}{\partial t} -  \frac{\partial }{\partial x} \kappa \frac{\partial T}{\partial x} -  \frac{\partial }{\partial y} \kappa \frac{\partial T}{\partial y} = q\hackscore H
58    \label{eqn:hd2}
59    \end{equation}
60    Notice, that for the 1D case we have $\frac{\partial T}{\partial y}=0$ and
61    for the case of constant material parameters $\frac{\partial }{\partial x} \kappa = \kappa  \frac{\partial }{\partial x}$ thus this new equation coincides with a simplified model equation for this case. It is more convenient
62    to write this equation using the $\nabla$ notation as we have already seen in \refEq{eqn:commonform nabla};
63    \begin{equation}\label{eqn:Tform nabla}
64    \rho c\hackscore p \frac{\partial T}{\partial t}
65    -\nabla \cdot \kappa \nabla T = q\hackscore H
66    \end{equation}
67    We can easily apply the backward Euler scheme as before to end up with
68    \begin{equation}
69    \frac{\rho c\hackscore p}{h} T^{(n)} -\nabla \cdot \kappa \nabla T^{(n)}  = q\hackscore H +  \frac{\rho c\hackscore p}{h} T^{(n-1)}
70    \label{eqn:hdgenf2}
71    \end{equation}
72    which is very similar to \refEq{eqn:hdgenf} used to define the temperature solution in the simple 1D case.
73    The difference is in the second order derivate term $\nabla \cdot \kappa \nabla T^{(n)}$. Under
74    the light of the more general case we need to revisit the \esc PDE form as
75    shown in \ref{eqn:commonform2D}. For the 2D case with variable PDE coefficients
76    the form needs to be read as
77    \begin{equation}\label{eqn:commonform2D 2}
78    -\frac{\partial }{\partial x} A\hackscore{00}\frac{\partial u}{\partial x}
79    -\frac{\partial }{\partial x} A\hackscore{01}\frac{\partial u}{\partial y}
80    -\frac{\partial }{\partial y} A\hackscore{10}\frac{\partial u}{\partial x}
81    -\frac{\partial }{\partial x} A\hackscore{11}\frac{\partial u}{\partial y}
82    + Du = f
83    \end{equation}
84    So besides the settings $u=T^{(n)}$, $D = \frac{\rho c \hackscore{p}}{h}$ and
85    $f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)}$ as we have used before (see \refEq{ESCRIPT SET}) we need to set
86    \begin{equation}\label{eqn: kappa general}
87    A\hackscore{00}=A\hackscore{11}=\kappa; A\hackscore{01}=A\hackscore{10}=0
88    \end{equation}
89    The fundamental difference to the 1D case is that $A\hackscore{11}$ is not set to zero but $\kappa$,
90    which brings in the second dimension. Important to notice that the coefficients
91    of the PDE may depend on their location in the domain and does not influence the usage of the PDE form. This is very convenient as we can introduce spatial dependence to the PDE coefficients without modification to the way we call the PDE solver.
92    
93    A very convenient way to define the matrix $A$ in \refEq{eqn: kappa general} can be carried out using the
94    Kronecker $\delta$ symbol\footnote{see \url{http://en.wikipedia.org/wiki/Kronecker_delta}}. The
95    \esc function \verb|kronecker| returns this matrix;
96    \begin{equation}
97    \verb|kronecker(model)| = \left[
98    \begin{array}{cc}
99     1 & 0 \\
100     0 & 1 \\
101    \end{array}
102    \right]
103    \end{equation}
104    As the argument \verb|model| represents a two dimensional domain the matrix is returned as $2 \times 2$ matrix
105    (In case of a three-dimensional domain a $3 \times 3$ matrix is returned). The call
106    \begin{python}
107    mypde.setValue(A=kappa*kronecker(model),D=rhocp/h)
108    \end{python}
109    sets the PDE coefficients according to \refEq{eqn: kappa general}.  
110    
111    We need to check the boundary conditions before we turn to the question: how we set $\kappa$. As
112    pointed out in \refEq{NEUMAN 2} makes certain assumptions on the boundary conditions. In our case
113    this assumptions translates to;
114    \begin{equation}
115    -n \cdot \kappa \nabla T^{(n)} =
116    -n\hackscore{0} \cdot \kappa \frac{\partial T^{(n)}}{\partial x} - n\hackscore{1} \cdot  \kappa \frac{\partial T^{(n)}}{\partial y} = 0
117    \label{eq:hom flux}
118    \end{equation}
119    which sets the normal component of the heat flux $- \kappa \cdot (\frac{\partial T^{(n)}}{\partial x}, \frac{\partial T^{(n)}}{\partial y})$ to zero. This means that the regions is insulated which is what we want.
120    On the left and right face of the domain where we have $(n\hackscore{0},n\hackscore{1} ) = (\mp 1,0)$
121    this means $\frac{\partial T^{(n)}}{\partial x}=0$ and on the top and bottom on the domain
122    where we have  $(n\hackscore{0},n\hackscore{1} ) = (\pm 1,0)$ this is $\frac{\partial T^{(n)}}{\partial y}=0$.
123    
124    \section{Setting Variable PDE Coefficients}
125    Now we need to look into the problem of how we define the material coefficients
126    $\kappa$ and $\rho c\hackscore p$ depending on their location in the domain.
127    We can make use of the technique used in the granite block example in \refSec{Sec:1DHDv00}
128    to set up the initial temperature. However,
129    the situation is more complicated here as we have to deal with a
130    curved interface between the two sub-domain.
131    
132    Prior to setting up the PDE, the interface between the two materials must be established.
133    The distance $s\ge 0$ between two points $[x,y]$ and $[x\hackscore{0},y\hackscore{0}]$ in Cartesian coordinates is defined as:
134  \begin{equation}  \begin{equation}
135   (x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2} = s^{2}   (x-x\hackscore{0})^{2}+(y-y\hackscore{0})^{2} = s^{2}
136  \end{equation}  \end{equation}
137  If $[x_{0},y_{0}]$ is the point $c$ the centre of the semi-circle 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 $s-r < 0$ for all intrusion points and $s-r > 0$ for all country rock points. Defining these conditions within the script is quite simple and is done using the following command:  If we define the point $[x\hackscore{0},y\hackscore{0}]$ as $ic$ which denotes the centre of the semi-circle of our intrusion, then for all the points $[x,y]$ in our model we can calculate a distance to $ic$.
138  \begin{verbatim}  All the points that fall within the given radius $r$ of our intrusion will have a corresponding
139    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 $s-r < 0$ for all intrusion points and $s-r > 0$
140    for all country rock points.
141    Defining these conditions within the script is quite simple and is done using the following command:
142    \begin{python}
143   bound = length(x-ic)-r #where the boundary will be located   bound = length(x-ic)-r #where the boundary will be located
144  \end{verbatim}  \end{python}
145  This definition of the boundary can now be used with the \verb wherePositive()  and \verb whereNegative()  commands from before to help define the material constants and temperatures in our domain. By examining the general form we solved in the earlier tutorials, it is obvious that both \verb A  and \verb D  depend on the predefined variables. To set these variables accordingly and complete our PDE we use:  This definition of the boundary can now be used with \verb|whereNegative| command again to help define the material constants and temperatures in our domain.
146  \begin{verbatim}  If \verb|kappai| and \verb|kappac| are the
147  A = (kappai)*whereNegative(bound)+(kappac)*wherePositive(bound)  thermal conductivities for the intrusion material granite and for the surrounding sandstone, then we set;
148  D = (rhocpi/h)*whereNegative(bound)+(rhocpc/h)*wherePositive(bound)  \begin{python}
149    x=Function(model).getX()
150  mypde.setValue(A=A*kronecker(model),D=D,d=eta,y=eta*Tc)  bound = length(x-ic)-r
151  \end{verbatim}  kappa = kappai * whereNegative(bound) + kappac * (1-whereNegative(bound))
152  Our PDE has now been properly established. The initial conditions for temperature are set out in a similar matter:  mypde.setValue(A=kappa*kronecker(model))
153  \begin{verbatim}  \end{python}
154    Notice that we are using the sample points of the \verb|Function| function space as expected for the
155    PDE coefficient \verb|A|\footnote{For the experienced user: use \texttt{x=mypde.getFunctionSpace("A").getX()}.}
156    The corresponding statements are used to set $\rho c\hackscore p$.
157    
158    Our PDE has now been properly established. The initial conditions for temperature are set out in a similar manner:
159    \begin{python}
160  #defining the initial temperatures.  #defining the initial temperatures.
161   T= Ti*whereNegative(bound)+Tc*wherePositive(bound)  x=Solution(model).getX()
162  \end{verbatim}  bound = length(x-ic)-r
163  The iteration process now begins as before, but using our new conditions for \verb D  as defined above.  T= Ti*whereNegative(bound)+Tc*(1-whereNegative(bound))
164    \end{python}
165  \subsection{Contouring escript data}  where \verb|Ti| and \verb|Tc| are the initial temperature
166  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  .  in the regions of the granite and surrounding sandstone, respectively. It is important to
167  \begin{verbatim}  notice that we reset \verb|x| and \verb|bound| to refer to the appropriate
168  # rearrage mymesh to suit solution function space for contouring        sample points of a PDE solution\footnote{For the experienced user: use \texttt{x=mypde.getFunctionSpace("r").getX()}.}.
169  oldspacecoords=model.getX()  
170  coords=Data(oldspacecoords, T.getFunctionSpace())  \begin{figure}[ht]
171  coordX, coordY = toXYTuple(coords)  \centerline{\includegraphics[width=4.in]{figures/heatrefraction001.png}}
172    \centerline{\includegraphics[width=4.in]{figures/heatrefraction030.png}}
173    \centerline{\includegraphics[width=4.in]{figures/heatrefraction200.png}}
174    \caption{Example 3a: 2D model: Total temperature distribution ($T$) at time step $1$, $20$ and $200$. Contour lines show temperature.}
175    \label{fig:twodhdans}
176    \end{figure}
177    
178    \section{Contouring \esc data using \modmpl.}
179    \label{Sec:2DHD plot}
180    To complete our transition from a 1D to a 2D model we also need to modify the
181    plotting procedure. As before we use the  \modmpl to do the plotting
182    in this case a contour plot. For 2D contour plots \modmpl expects that the
183    data are regularly gridded. We have no control over the location and ordering of the sample points
184    used to represent the solution. Therefore it is necessary to interpolate our solution onto a regular grid.
185    
186    In \refSec{sec: plot T} we have already learned how to extract the $x$ coordinates of sample points as
187    \verb|numpy| array to hand the values to \modmpl. This can easily be extended to extract both the
188    $x$ and the $y$ coordinates;
189    \begin{python}
190    import numpy as np
191    def toXYTuple(coords):
192        coords = np.array(coords.toListOfTuples()) #convert to Tuple
193        coordX = coords[:,0] #X components.
194        coordY = coords[:,1] #Y components.
195        return coordX,coordY
196    \end{python}
197    For convenience we have put this function into \file{clib.py} file so we can use this
198    function in other scripts more easily.
199    
200    
201    We now generate a regular $100 \times 100$ grid over the domain ($mx$ and $my$
202    are the dimensions in the $x$ and $y$ directions) which is done using the \modnumpy function \verb|linspace|  .
203    \begin{python}
204    from clib import toXYTuple
205    # get sample points for temperature as  for contouring      
206    coordX, coordY = toXYTuple(T.getFunctionSpace().getX())
207  # create regular grid  # create regular grid
208  xi = np.linspace(0.0,mx,100)  xi = np.linspace(0.0,mx,75)
209  yi = np.linspace(0.0,my,100)  yi = np.linspace(0.0,my,75)
210  \end{verbatim}  \end{python}
211  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.  The values \verb|[xi[k], yi[l]]| are the grid points.
212    
213  \begin{verbatim}  The remainder of our contouring commands reside within a \verb while  loop so that a new contour is generated for each time step. For each time step the solution must be regridded for \modmpl using the \verb griddata  function. This function interprets irregularly located values \verb tempT  at locations defined by \verb coordX  and \verb coordY  as values at the new coordinates of a rectangular grid defined by
214    \verb xi   and \verb yi . The output is \verb zi . It is now possible to use the \verb contourf  function which generates colour filled contours. 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\footnote{see \url{http://matplotlib.sourceforge.net/api/}}. 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.
215    \begin{python}
216  #grid the data.  #grid the data.
217  zi = pl.matplotlib.mlab.griddata(coordX,coordY,tempT,xi,yi)  zi = pl.matplotlib.mlab.griddata(coordX,coordY,tempT,xi,yi)
218  # contour the gridded data, plotting dots at the randomly spaced data points.  # contour the gridded data, plotting dots at the randomly spaced data points.
# Line 79  pl.axis([0,600,0,600]) Line 224  pl.axis([0,600,0,600])
224  pl.title("Heat diffusion from an intrusion.")  pl.title("Heat diffusion from an intrusion.")
225  pl.xlabel("Horizontal Displacement (m)")  pl.xlabel("Horizontal Displacement (m)")
226  pl.ylabel("Depth (m)")  pl.ylabel("Depth (m)")
227  pl.savefig(os.path.join(save_path,"heatrefraction%03d.png") %i)  pl.savefig(os.path.join(save_path,"Tcontour%03d.png") %i)
228  pl.clf()          pl.clf()        
229  \end{verbatim}  \end{python}
230    The function \verb|pl.contour| is used to add labeled contour lines to the plot.
231    The results for selected time steps are shown in \reffig{fig:twodhdans}.
232    
 \begin{figure}[h!]  
 \centerline{\includegraphics[width=4.in]{figures/heatrefraction050}}  
 \caption{2D model: Total temperature distribution ($T$) at time $t=50$.}  
 \label{fig:twodhdmodel}  
 \end{figure}  
233    
234    
235    \section{Advanced Visualization using VTK}
236    
237    \sslist{example03b.py}
238    An alternative approach to \modmpl for visualization is the usage of a package which base on
239    visualization tool kit (VTK) library\footnote{see \url{http://www.vtk.org/}}. There are variety
240    of packages available. Here we use the package \mayavi\footnote{see \url{http://code.enthought.com/projects/mayavi/}} as an example.
241    
242    \mayavi is an interactive, GUI driven tool which is
243    really designed to visualize large three dimensional data sets where \modmpl
244    is not suitable. But it is very useful when it comes to two dimensional problems.
245    The decision of which tool is the best can be subjective and the user should determine which package they require and are most comfortable with. The main difference between using \mayavi (and other VTK based tools)
246    or \modmpl is that the actual visualization is detached from the
247    calculation by writing the results to external files
248    and importing them into \mayavi. In 3D the best camera position for rendering a scene is not obvious
249    before the results are available. Therefore the user may need to try
250    different position before the best is found. Moreover, in many cases a 3D interactive
251    visualization is the only way to really understand the results (e.g. using stereographic projection).
252    
253    To write the temperatures at each time step to data files in the VTK file format one
254    needs to insert a \verb|saveVTK| call into the code;
255    \begin{python}
256    while t<=tend:
257          i+=1 #counter
258          t+=h #current time
259          mypde.setValue(Y=qH+T*rhocp/h)
260          T=mypde.getSolution()
261          saveVTK(os.path.join(save_path,"data.%03d.vtu"%i, T=T)
262    \end{python}
263    The data files, eg. \file{data.001.vtu}, contains all necessary information to
264    visualize the temperature and can directly processed by \mayavi. Notice that there is no
265    regridding required. It is recommended to use the file extension \file{.vtu} for files
266    created by \verb|saveVTK|.
267    
268    \begin{figure}[ht]
269    \centerline{\includegraphics[width=4.in]{figures/ScreeshotMayavi2n1}}
270    \caption{Example 3b: \mayavi start up Window.}
271    \label{fig:mayavi window}
272    \end{figure}
273    
274    \begin{figure}[ht]
275    \centerline{\includegraphics[width=4.in]{figures/ScreeshotMayavi2n2}}
276    \caption{Example 3b: \mayavi data control window.}
277    \label{fig:mayavi window2}
278    \end{figure}
279    After you run the script you will find the
280    result files \file{data.*.vtu} in the result directory \file{data/example03}. Run the
281    command
282    \begin{python}
283    >> mayavi2 -d data.001.vtu -m Surface &
284    \end{python}
285    from the result directory. \mayavi will start up a window similar to \reffig{fig:mayavi window}.
286    The right hand side shows the temperature at the first time step. To show
287    the results at other time steps you can click at the item \texttt{VTK XML file (data.001.vtu) (timeseries)}
288    at the top left hand side. This will bring up a new window similar to the window shown in  \reffig{fig:mayavi window2}. By clicking at the arrows in the top right corner you can move forwards and backwards in time.
289    You will also notice the text \textbf{T} next to the item \texttt{Point scalars name}. The
290    name \textbf{T} corresponds to the keyword argument name \texttt{T} we have used
291    in the \verb|saveVTK| call. In this menu item you can select other results
292    you may have written to the output file for visualization.
293    
294    \textbf{For the advanced user}: Using the \modmpl to visualize spatially distributed data
295    is not MPI compatible. However, the \verb|saveVTK| function can be used with MPI. In fact,
296    the function collects the values of the sample points spread across processor ranks into a single.
297    For more details on writing scripts for parallel computing please consult the \emph{user's guide}.

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