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2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2010 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % Primary Business: Queensland, Australia
9 % Licensed under the Open Software License version 3.0
10 % http://www.opensource.org/licenses/osl-3.0.php
11 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
14 \begin{figure}[t]
15 \centerline{\includegraphics[width=4.in]{figures/twodheatdiff}}
16 \caption{Example 3: 2D model: granitic intrusion of sandstone country rock.}
17 \label{fig:twodhdmodel}
18 \end{figure}
20 \sslist{example03a.py and cblib.py}
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.
30 \section{Example 3: Two Dimensional Heat Diffusion for a basic Magmatic Intrusion}
31 \label{Sec:2DHD}
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
37 my = 600*m #meters - model width
38 ndx = 150 #mesh steps in x direction
39 ndy = 150 #mesh steps in y direction
40 r = 200*m #meters - radius of intrusion
41 ic = [300*m, 0] #coordinates of the centre of intrusion (meters)
42 qH=0.*J/(sec*m**3) #our heat source temperature is zero
43 \end{python}
44 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.
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.
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}.
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 $\frac{\partial T^{(n)}}{\partial y}=0$.
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.
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}
135 (x-x\hackscore{0})^{2}+(y-y\hackscore{0})^{2} = s^{2}
136 \end{equation}
137 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 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
144 \end{python}
145 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 If \verb|kappai| and \verb|kappac| are the
147 thermal conductivities for the intrusion material granite and for the surrounding sandstone, then we set;
148 \begin{python}
149 x=Function(model).getX()
150 bound = length(x-ic)-r
151 kappa = kappai * whereNegative(bound) + kappac * (1-whereNegative(bound))
152 mypde.setValue(A=kappa*kronecker(model))
153 \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 $\rho c\hackscore p$.
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.
161 x=Solution(model).getX()
162 bound = length(x-ic)-r
163 T= Ti*whereNegative(bound)+Tc*(1-whereNegative(bound))
164 \end{python}
165 where \verb|Ti| and \verb|Tc| are the initial temperature
166 in the regions of the granite and surrounding sandstone, respectively. It is important to
167 notice that we reset \verb|x| and \verb|bound| to refer to the appropriate
168 sample points of a PDE solution\footnote{For the experienced user: use \texttt{x=mypde.getFunctionSpace("r").getX()}.}.
170 \begin{figure}[ht]
171 \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}
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.
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.
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
208 xi = np.linspace(0.0,mx,75)
209 yi = np.linspace(0.0,my,75)
210 \end{python}
211 The values \verb|[xi[k], yi[l]]| are the grid points.
213 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.
217 zi = pl.matplotlib.mlab.griddata(coordX,coordY,tempT,xi,yi)
218 # contour the gridded data, plotting dots at the randomly spaced data points.
219 pl.matplotlib.pyplot.autumn()
220 pl.contourf(xi,yi,zi,10)
221 CS = pl.contour(xi,yi,zi,5,linewidths=0.5,colors='k')
222 pl.clabel(CS, inline=1, fontsize=8)
223 pl.axis([0,600,0,600])
224 pl.title("Heat diffusion from an intrusion.")
225 pl.xlabel("Horizontal Displacement (m)")
226 pl.ylabel("Depth (m)")
227 pl.savefig(os.path.join(save_path,"Tcontour%03d.png") %i)
228 pl.clf()
229 \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}.
235 \section{Advanced Visualization using VTK}
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.
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).
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|.
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}
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.
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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