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cookbook review final final 3.1 - artak and tony corrections
1 ahallam 2401
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4 jfenwick 2881 % Copyright (c) 2003-2010 by University of Queensland
5 ahallam 2401 % Earth Systems Science Computational Center (ESSCC)
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8     % Primary Business: Queensland, Australia
9     % Licensed under the Open Software License version 3.0
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14 gross 2905 \begin{figure}[h!]
15     \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}}
16 ahallam 2979 \caption{Example 1: Temperature differential along a single interface between
17     two granite blocks.}
18 gross 2905 \label{fig:onedgbmodel}
19     \end{figure}
20 ahallam 2801
21 gross 2949 \section{Example 1: One Dimensional Heat Diffusion in Granite}
22 gross 2905 \label{Sec:1DHDv00}
23 gross 2878
24 ahallam 2979 The first model consists of two blocks of isotropic material, for instance
25     granite, sitting next to each other.
26     Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is
27     \verb|T2|.
28 gross 2905 We assume that the system is insulated.
29     What would happen to the temperature distribution in each block over time?
30 ahallam 2979 Intuition tells us that heat will be transported from the hotter block to the
31     cooler one until both
32 gross 2905 blocks have the same temperature.
34 ahallam 2495 \subsection{1D Heat Diffusion Equation}
35 ahallam 2979 We can model the heat distribution of this problem over time using one
36     dimensional heat diffusion equation\footnote{A detailed discussion on how the
37     heat diffusion equation is derived can be found at
38     \url{
39     http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}};
40 ahallam 2494 which is defined as:
41 ahallam 2401 \begin{equation}
42 ahallam 2979 \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2}
43     T}{\partial x^{2}} = q\hackscore H
44 ahallam 2401 \label{eqn:hd}
45     \end{equation}
46 ahallam 2979 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and
47     $\kappa$ is the thermal
48     conductivity\footnote{A list of some common thermal conductivities is available
49     from Wikipedia
50     \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we
51     assume that these material
52 gross 2861 parameters are \textbf{constant}.
53 ahallam 2979 The heat source is defined by the right hand side of \refEq{eqn:hd} as
54     $q\hackscore{H}$; this can take the form of a constant or a function of time and
55     space. For example $q\hackscore{H} = q\hackscore{0}e^{-\gamma t}$ where we have
56     the output of our heat source decaying with time. There are also two partial
57     derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the
58     change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is
59     the spatial change of temperature. As there is only a single spatial dimension
60     to our problem, our temperature solution $T$ is only dependent on the time $t$
61     and our signed distance from the the block-block interface $x$.
62 ahallam 2401
63 gross 2861 \subsection{PDEs and the General Form}
64 ahallam 2979 It is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact
65     solution to our problem. However, it is not always practical to solve the
66     problem this way. Alternatively, computers can be used to find the solution. To
67     do this, a numerical approach is required to discretise
68     the PDE \refEq{eqn:hd} across time and space, this reduces the problem to a
69     finite number of equations for a finite number of spatial points and time steps.
70     These parameters together define the model. While discretisation introduces
71     approximations and a degree of error, a sufficiently sampled model is generally
72     accurate enough to satisfy the accuracy requirements for the final solution.
73 gross 2477
74 ahallam 2979 Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a
75     steady linear PDE which involves spatial derivatives only and needs to be solved
76     in each time step to progress in time. \esc can help us here.
77 gross 2861
78 ahallam 2979 For time discretization we use the Backwards Euler approximation
79     scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is
80     based on the
81 gross 2861 approximation
82     \begin{equation}
83     \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h}
84     \label{eqn:beuler}
85     \end{equation}
86     for $\frac{\partial T}{\partial t}$ at time $t$
87     where $h$ is the time step size. This can also be written as;
88     \begin{equation}
89     \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h}
90     \label{eqn:Tbeuler}
91     \end{equation}
92 ahallam 2979 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one
93     has
94 gross 2861 \begin{equation}
95     \begin{array}{rcl}
96     t^{(n)} & = & t^{(n-1)}+h \\
97     T^{(n)} & = & T(t^{(n-1)}) \\
98     \end{array}
99     \label{eqn:Neuler}
100     \end{equation}
101     Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get;
102     \begin{equation}
103 ahallam 2979 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2}
104     T^{(n)}}{\partial x^{2}} = q\hackscore H
105 gross 2861 \label{eqn:hddisc}
106     \end{equation}
107 ahallam 2979 Notice that we evaluate the spatial derivative term at the current time
108     $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively,
109     one can evaluate the spatial derivative term at the previous time $t^{(n-1)}$.
110     This
111     approach is called the \textbf{forward Euler} scheme. This scheme can provide
112     some computational advantages, which
113     are not discussed here. However, the \textbf{forward Euler} scheme has a major
114     disadvantage. Namely, depending on the
115     material parameters as well as the domain discretization of the spatial
116     derivative term, the time step size $h$ needs to be chosen sufficiently small to
117     achieve a stable temperature when progressing in time. Stabiliy is achieved if
118     the temperature does not grow beyond its initial bounds and become
119     non-physical.
120     The backward Euler scheme, which we use here, is unconditionally stable meaning
121     that under the assumption of a
122     physically correct problem set-up the temperature approximation remains physical
123     for all time steps.
124     The user needs to keep in mind that the discretization error introduced by
125     \refEq{eqn:beuler}
126     is sufficiently small, thus a good approximation of the true temperature is
127     computed. It is
128     therefore very important that any results are viewed with caution. For example,
129     one may compare the results for different time and spatial step sizes.
130 gross 2861
131     To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear
132 ahallam 2979 differential equation \refEq{eqn:hddisc} which only includes spatial
133     derivatives. To solve this problem
134 artak 2958 we want to use \esc.
135 gross 2861
136 ahallam 2979 In \esc any given PDE can be described by the general form. For the purpose of
137     this introduction we illustrate a simpler version of the general form for full
138     linear PDEs which is available in the \esc user's guide. A simplified form that
139     suits our heat diffusion problem\footnote{The form in the \esc users guide which
140     uses the Einstein convention is written as
141 gross 2477 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$}
142 ahallam 2606 is described by;
143 gross 2477 \begin{equation}\label{eqn:commonform nabla}
144 jfenwick 2657 -\nabla\cdot(A\cdot\nabla u) + Du = f
145 ahallam 2411 \end{equation}
146 ahallam 2979 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The
147     symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del
148     operator} represents
149     the spatial derivative of its subject - in this case $u$. Lets assume for a
150     moment that we deal with a one-dimensional problem then ;
151 gross 2477 \begin{equation}
152     \nabla = \frac{\partial}{\partial x}
153     \end{equation}
154 gross 2861 and we can write \refEq{eqn:commonform nabla} as;
155 ahallam 2411 \begin{equation}\label{eqn:commonform}
156 gross 2477 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f
157 ahallam 2411 \end{equation}
158 ahallam 2979 if $A$ is constant. To match this simplified general form to our problem
159     \refEq{eqn:hddisc}
160 gross 2861 we rearrange \refEq{eqn:hddisc};
161 ahallam 2411 \begin{equation}
162 ahallam 2979 \frac{\rho c\hackscore p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial
163     x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)}
164 ahallam 2494 \label{eqn:hdgenf}
165     \end{equation}
166 ahallam 2979 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is
167     required for \esc to solve our PDE. This can be done by generating a solution
168     for successive increments in the time nodes $t^{(n)}$ where
169     $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed
170     to be constant.
171     In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$.
172     Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} one can see
173     that;
174 gross 2862 \begin{equation}\label{ESCRIPT SET}
175 gross 2861 u=T^{(n)};
176 ahallam 2979 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho
177     c\hackscore p}{h} T^{(n-1)}
178 ahallam 2494 \end{equation}
180 ahallam 2495 \subsection{Boundary Conditions}
181 gross 2878 \label{SEC BOUNDARY COND}
182 ahallam 2979 With the PDE sufficiently modified, consideration must now be given to the
183     boundary conditions of our model. Typically there are two main types of boundary
184     conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary
185     conditions\footnote{More information on Boundary Conditions is available at
186     Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}},
187     respectively.
188     A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to
189     prescribe a known value to the unknown solution (in our example the temperature)
190     on parts of the boundary or on the entire boundary of the region of interest.
191     We discuss Dirichlet boundary condition in our second example presented in
192     Section~\ref{Sec:1DHDv0}.
193 ahallam 2495
194 ahallam 2979 However, for this example we have made the model assumption that the system is
195     insulated, so we need
196 gross 2905 to add an appropriate boundary condition to prevent
197 ahallam 2979 any loss or inflow of energy at the boundary of our domain. Mathematically this
198     is expressed by prescribing
199     the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified
200     one dimensional model this is expressed
201 gross 2862 in the form;
202 ahallam 2494 \begin{equation}
203 gross 2862 \kappa \frac{\partial T}{\partial x} = 0
204 ahallam 2494 \end{equation}
205 gross 2862 or in a more general case as
206     \begin{equation}\label{NEUMAN 1}
207     \kappa \nabla T \cdot n = 0
208     \end{equation}
209 ahallam 2979 where $n$ is the outer normal field \index{outer normal field} at the surface
210     of the domain.
211     The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$.
212     In fact, the term $\nabla T \cdot n$ is the normal derivative of
213     the temperature $T$. Other notations used here are\footnote{The \esc notation
214     for the normal
215 gross 2862 derivative is $T\hackscore{,i} n\hackscore i$.};
216 ahallam 2645 \begin{equation}
217 gross 2862 \nabla T \cdot n = \frac{\partial T}{\partial n} \; .
218 ahallam 2645 \end{equation}
219 ahallam 2979 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary
220     condition} for the PDE.
221 ahallam 2494
222 gross 2905 The PDE \refEq{eqn:hdgenf}
223 ahallam 2979 and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with
224     the Dirichlet boundary conditions) define a \textbf{boundary value problem}.
225     It is the nature of a boundary value problem to allow making statements about
226     the solution in the
227     interior of the domain from information known on the boundary only. In most
228     cases
229     we use the term partial differential equation but in fact it is a boundary value
230     problem.
231     It is important to keep in mind that boundary conditions need to be complete and
232     consistent in the sense that
233     at any point on the boundary either a Dirichlet or a Neuman boundary condition
234     must be set.
235 gross 2862
236 ahallam 2979 Conveniently, \esc makes default assumption on the boundary conditions which the
237     user may modify where appropriate.
238     For a problem of the form in~\refEq{eqn:commonform nabla} the default
239     condition\footnote{In the form of the \esc users guide which is using the
240     Einstein convention is written as
241 gross 2862 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is;
242     \begin{equation}\label{NEUMAN 2}
243 gross 2948 -n\cdot A \cdot\nabla u = 0
244 gross 2862 \end{equation}
245 ahallam 2979 which is used everywhere on the boundary. Again $n$ denotes the outer normal
246     field.
247     Notice that the coefficient $A$ is the same as in the \esc
248     PDE~\ref{eqn:commonform nabla}.
249     With the settings for the coefficients we have already identified in
250     \refEq{ESCRIPT SET} this
251 gross 2862 condition translates into
252 gross 2867 \begin{equation}\label{NEUMAN 2b}
253 gross 2862 \kappa \frac{\partial T}{\partial x} = 0
254     \end{equation}
255 ahallam 2979 for the boundary of the domain. This is identical to the Neuman boundary
256     condition we want to set. \esc will take care of this condition for us. We
257     discuss the Dirichlet boundary condition later.
258 gross 2862
259 gross 2870 \subsection{Outline of the Implementation}
260     \label{sec:outline}
261 ahallam 2979 To solve the heat diffusion equation (\refEq{eqn:hd}) we write a simple \pyt
262     script. At this point we assume that you have some basic understanding of the
263     \pyt programming language. If not, there are some pointers and links available
264     in Section \ref{sec:escpybas}. The script (discussed in \refSec{sec:key}) has
265     four major steps. Firstly, we need to define the domain where we want to
266     calculate the temperature. For our problem this is the joint blocks of granite
267     which has a rectangular shape. Secondly, we need to define the PDE to solve in
268     each time step to get the updated temperature. Thirdly, we need to define the
269     coefficients of the PDE and finally we need to solve the PDE. The last two steps
270     need to be repeated until the final time marker has been reached. The work flow
271     described in \reffig{fig:wf}.
272 ahallam 2975 % \begin{enumerate}
273     % \item create domain
274     % \item create PDE
275     % \item while end time not reached:
276     % \begin{enumerate}
277     % \item set PDE coefficients
278     % \item solve PDE
279     % \item update time marker
280     % \end{enumerate}
281     % \item end of calculation
282     % \end{enumerate}
284     \begin{figure}[h!]
285     \centering
286     \includegraphics[width=1in]{figures/workflow.png}
287     \caption{Workflow for developing an \esc model and solution.}
288     \label{fig:wf}
289     \end{figure}
291 ahallam 2979 In the terminology of \pyt, the domain and PDE are represented by
292     \textbf{objects}. The nice feature of an object is that it is defined by its
293     usage and features
294     rather than its actual representation. So we will create a domain object to
295     describe the geometry of the two
296     granite blocks. Then we define PDEs and spatially distributed values such as the
297     temperature
298     on this domain. Similarly, to define a PDE object we use the fact that one needs
299     only to define the coefficients of the PDE and solve the PDE. The PDE object has
300     advanced features, but these are not required in simple cases.
301 gross 2870
303     \begin{figure}[t]
304     \centering
305     \includegraphics[width=6in]{figures/functionspace.pdf}
306 ahallam 2975 \caption{\esc domain construction overview}
307 gross 2870 \label{fig:fs}
308     \end{figure}
310     \subsection{The Domain Constructor in \esc}
311     \label{ss:domcon}
312 ahallam 2979 Whilst it is not strictly relevant or necessary, a better understanding of
313     how values are spatially distributed (\textit{e.g.} Temperature) and how PDE
314     coefficients are interpreted in \esc can be helpful.
315 gross 2870
316 ahallam 2979 There are various ways to construct domain objects. The simplest form is a
317     rectangular shaped region with a length and height. There is
318     a ready to use function for this named \verb rectangle(). Besides the spatial
319     dimensions this function requires to specify the number of
320     elements or cells to be used along the length and height, see \reffig{fig:fs}.
321     Any spatially distributed value
322     and the PDE is represented in discrete form using this element
323     representation\footnote{We use the finite element method (FEM), see
324     \url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}.
325     Therefore we will have access to an approximation of the true PDE solution
326     only.
327     The quality of the approximation depends - besides other factors- mainly on the
328     number of elements being used. In fact, the
329     approximation becomes better when more elements are used. However, computational
330     cost grows with the number of
331     elements being used. It is therefore important that you find the right balance
332     between the demand in accuracy and acceptable resource usage.
333 gross 2870
334 ahallam 2979 In general, one can think about a domain object as a composition of nodes and
335     elements.
336     As shown in \reffig{fig:fs}, an element is defined by the nodes that are used to
337     describe its vertices.
338 gross 2870 To represent spatial distributed values the user can use
339 ahallam 2979 the values at the nodes, at the elements in the interior of the domain or at the
340     elements located at the surface of the domain.
341     The different approach used to represent values is called \textbf{function
342     space} and is attached to all objects
343     in \esc representing a spatial distributed value such as the solution of a PDE.
344     The three
345 artak 2963 function spaces we use at the moment are;
346 gross 2870 \begin{enumerate}
347     \item the nodes, called by \verb|ContinuousFunction(domain)| ;
348     \item the elements/cells, called by \verb|Function(domain)| ; and
349     \item the boundary, called by \verb|FunctionOnBoundary(domain)| .
350     \end{enumerate}
351 ahallam 2979 A function space object such as \verb|ContinuousFunction(domain)| has the method
352     \verb|getX| attached to it. This method returns the
353     location of the so-called \textbf{sample points} used to represent values of the
354     particular function space. So the
355     call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the
356     nodes used to describe the domain while
357     the \verb|Function(domain).getX()| returns the coordinates of numerical
358     integration points within elements, see
359 gross 2905 \reffig{fig:fs}.
360 gross 2870
361 ahallam 2979 This distinction between different representations of spatially distributed
362     values
363     is important in order to be able to vary the degrees of smoothness in a PDE
364     problem.
365     The coefficients of a PDE do not need to be continuous, thus this qualifies as a
366     \verb|Function()| type.
367     On the other hand a temperature distribution must be continuous and needs to be
368     represented with a \verb|ContinuousFunction()| function space.
369     An influx may only be defined at the boundary and is therefore a \verb
370     FunctionOnBoundary() object.
371     \esc allows certain transformations of the function spaces. A \verb
372     ContinuousFunction() can be transformed into a \verb|FunctionOnBoundary()|
373     or \verb|Function()|. On the other hand there is not enough information in a
374     \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() .
375     These transformations, which are called \textbf{interpolation} are invoked
376     automatically by \esc if needed.
377 gross 2870
378 artak 2963 Later in this introduction we discuss how
379 ahallam 2979 to define specific areas of geometry with different materials which are
380     represented by different material coefficients such the
381     thermal conductivities $\kappa$. A very powerful technique to define these types
382     of PDE
383     coefficients is tagging. Blocks of materials and boundaries can be named and
384     values can be defined on subregions based on their names.
385     This is a method for simplifying PDE coefficient and flux definitions. It makes
386     scripting much easier and we will discuss this technique in
388 gross 2870
390     \subsection{A Clarification for the 1D Case}
391 gross 2931 \label{SEC: 1D CLARIFICATION}
392 ahallam 2979 It is necessary for clarification that we revisit our general PDE from
393     \refeq{eqn:commonform nabla} for two dimensional domain. \esc is inherently
394     designed to solve problems that are greater than one dimension and so
395     \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem.
396     In the case of two spatial dimensions the \textit{Nabla operator} has in fact
397     two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial
398     y})$. Assuming the coefficient $A$ is constant, the \refEq{eqn:commonform nabla}
399     takes the following form;
400 gross 2477 \begin{equation}\label{eqn:commonform2D}
401     -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}}
402     -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y}
403     -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x}
404     -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}}
405     + Du = f
406     \end{equation}
407 ahallam 2606 Notice that for the higher dimensional case $A$ becomes a matrix. It is also
408 ahallam 2495 important to notice that the usage of the Nabla operator creates
409     a compact formulation which is also independent from the spatial dimension.
410 artak 2963 To make the general PDE \refEq{eqn:commonform2D} one dimensional as
411 gross 2861 shown in \refEq{eqn:commonform} we need to set
412 ahallam 2606 \begin{equation}
413 ahallam 2494 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0
414 gross 2477 \end{equation}
416 gross 2867
417 ahallam 2495 \subsection{Developing a PDE Solution Script}
418 ahallam 2801 \label{sec:key}
419 gross 2949 \sslist{example01a.py}
420 ahallam 2979 We write a simple \pyt script which uses the \modescript, \modfinley and \modmpl
421     modules.
422     By developing a script for \esc, the heat diffusion equation can be solved at
423     successive time steps for a predefined period using our general form
424     \refEq{eqn:hdgenf}. Firstly it is necessary to import all the
425     libraries\footnote{The libraries contain predefined scripts that are required to
426     solve certain problems, these can be simple like sine and cosine functions or
427     more complicated like those from our \esc library.}
428 ahallam 2495 that we will require.
429 ahallam 2775 \begin{python}
430 ahallam 2495 from esys.escript import *
431 ahallam 2606 # This defines the LinearPDE module as LinearPDE
432     from esys.escript.linearPDEs import LinearPDE
433     # This imports the rectangle domain function from finley.
434     from esys.finley import Rectangle
435     # A useful unit handling package which will make sure all our units
436     # match up in the equations under SI.
437     from esys.escript.unitsSI import *
438 ahallam 2775 \end{python}
439 ahallam 2979 It is generally a good idea to import all of the \modescript library, although
440     if the functions and classes required are known they can be specified
441     individually. The function \verb|LinearPDE| has been imported explicitly for
442     ease of use later in the script. \verb|Rectangle| is going to be our type of
443     domain. The module \verb unitsSI provides support for SI unit definitions with
444     our variables.
445 gross 2477
446 ahallam 2979 Once our library dependencies have been established, defining the problem
447     specific variables is the next step. In general the number of variables needed
448     will vary between problems. These variables belong to two categories. They are
449     either directly related to the PDE and can be used as inputs into the \esc
450     solver, or they are script variables used to control internal functions and
451     iterations in our problem. For this PDE there are a number of constants which
452     need values. Firstly, the domain upon which we wish to solve our problem needs
453     to be defined. There are different types of domains in \modescript which we
454     demonstrate in later tutorials but for our granite blocks, we simply use a
455     rectangular domain.
456 ahallam 2401
457 ahallam 2979 Using a rectangular domain simplifies our granite blocks (which would in reality
458     be a \textit{3D} object) into a single dimension. The granite blocks will have a
459     lengthways cross section that looks like a rectangle. As a result we do not
460     need to model the volume of the block due to symmetry. There are four arguments
461     we must consider when we decide to create a rectangular domain, the domain
462     \textit{length}, \textit{width} and \textit{step size} in each direction. When
463     defining the size of our problem it will help us determine appropriate values
464     for our model arguments. If we make our dimensions large but our step sizes very
465     small we increase the accuracy of our solution. Unfortunately we also increase
466     the number of calculations that must be solved per time step. This means more
467     computational time is required to produce a solution. In this \textit{1D}
468     problem, the bar is defined as being 1 metre long. An appropriate step size
469     \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| need only be 1, this
470     is because our problem stipulates no partial derivatives in the $y$ direction.
471     Thus the temperature does not vary with $y$. Hence, the model parameters can be
472     defined as follows; note we have used the \verb unitsSI convention to make sure
473     all our input units are converted to SI.
474 ahallam 2775 \begin{python}
475 gross 2905 mx = 500.*m #meters - model length
476     my = 100.*m #meters - model width
477     ndx = 50 # mesh steps in x direction
478     ndy = 1 # mesh steps in y direction
479     boundloc = mx/2 # location of boundary between the two blocks
480 ahallam 2775 \end{python}
481 ahallam 2979 The material constants and the temperature variables must also be defined. For
482     the granite in the model they are defined as:
483 ahallam 2775 \begin{python}
484 ahallam 2495 #PDE related
485 gross 2905 rho = 2750. *kg/m**3 #kg/m^{3} density of iron
486     cp = 790.*J/(kg*K) # J/Kg.K thermal capacity
487 gross 2878 rhocp = rho*cp
488 gross 2905 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity
489 gross 2878 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source
490 gross 2905 T1=20 * Celsius # initial temperature at Block 1
491     T2=2273. * Celsius # base temperature at Block 2
492 ahallam 2775 \end{python}
493 ahallam 2979 Finally, to control our script we will have to specify our timing controls and
494     where we would like to save the output from the solver. This is simple enough:
495 ahallam 2775 \begin{python}
496 gross 2878 t=0 * day #our start time, usually zero
497     tend=1. * day # - time to end simulation
498 ahallam 2495 outputs = 200 # number of time steps required.
499     h=(tend-t)/outputs #size of time step
500 ahallam 2606 #user warning statement
501     print "Expected Number of time outputs is: ", (tend-t)/h
502     i=0 #loop counter
503 ahallam 2775 \end{python}
504 ahallam 2979 Now that we know our inputs we will build a domain using the
505     \verb Rectangle() function from \verb finley . The four arguments allow us to define our domain
506     \verb model as:
507 ahallam 2775 \begin{python}
508 ahallam 2606 #generate domain using rectangle
509 gross 2905 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy)
510 ahallam 2775 \end{python}
511 artak 2963 \verb blocks now describes a domain in the manner of Section \ref{ss:domcon}.
512 ahallam 2658
513 ahallam 2979 With a domain and all our required variables established, it is now possible to
514     set up our PDE so that it can be solved by \esc. The first step is to define the
515     type of PDE that we are trying to solve in each time step. In this example it is
516     a single linear PDE\footnote{in comparison to a system of PDEs which we discuss
517     later.}. We also need to state the values of our general form variables.
518 ahallam 2775 \begin{python}
519 gross 2905 mypde=LinearPDE(blocks)
520 gross 2878 A=zeros((2,2)))
521     A[0,0]=kappa
522 gross 2905 mypde.setValue(A=A, D=rhocp/h)
523 ahallam 2775 \end{python}
524 ahallam 2979 In a many cases it may be possible to decrease the computational time of the
525     solver if the PDE is symmetric.
526 gross 2878 Symmetry of a PDE is defined by;
527 ahallam 2495 \begin{equation}\label{eqn:symm}
528     A\hackscore{jl}=A\hackscore{lj}
529     \end{equation}
530 ahallam 2979 Symmetry is only dependent on the $A$ coefficient in the general form and the
531     other coefficients $D$ as well as the right hand side $Y$. From the above
532     definition we can see that our PDE is symmetric. The \verb LinearPDE class
533     provides the method \method{checkSymmetry} to check if the given PDE is
534     symmetric. As our PDE is symmetrical we enable symmetry via;
535 ahallam 2775 \begin{python}
536 ahallam 2495 myPDE.setSymmetryOn()
537 ahallam 2775 \end{python}
538 ahallam 2979 Next we need to establish the initial temperature distribution \verb|T|. We need
539     to
540     assign the value \verb|T1| to all sample points left to the contact interface at
541     $x\hackscore{0}=\frac{mx}{2}$
542 gross 2905 and the value \verb|T2| right to the contact interface. \esc
543     provides the \verb|whereNegative| function to construct this. In fact,
544 ahallam 2979 \verb|whereNegative| returns the value $1$ at those sample points where the
545     argument
546     has a negative value. Otherwise zero is returned. If \verb|x| are the
547     $x\hackscore{0}$
548 gross 2905 coordinates of the sample points used to represent the temperature distribution
549     then \verb|x[0]-boundloc| gives us a negative value for
550     all sample points left to the interface and non-negative value to
551     the right of the interface. So with;
552 gross 2878 \begin{python}
553     # ... set initial temperature ....
554 gross 2905 T= T1*whereNegative(x[0]-boundloc)+T2*(1-whereNegative(x[0]-boundloc))
555 gross 2878 \end{python}
556 ahallam 2979 we get the desired temperature distribution. To get the actual sample points
557     \verb|x| we use
558 gross 2905 the \verb|getX()| method of the function space \verb|Solution(blocks)|
559     which is used to represent the solution of a PDE;
560 gross 2878 \begin{python}
561 gross 2905 x=Solution(blocks).getX()
562     \end{python}
563 ahallam 2979 As \verb|x| are the sample points for the function space
564     \verb|Solution(blocks)|
565     the initial temperature \verb|T| is using these sample points for
566     representation.
567     Although \esc is trying to be forgiving with the choice of sample points and to
568     convert
569     where necessary the adjustment of the function space is not always possible. So
570     it is
571 gross 2905 advisable to make a careful choice on the function space used.
573 ahallam 2979 Finally we initialise an iteration loop to solve our PDE for all the time steps
574     we specified in the variable section. As the right hand side of the general form
575     is dependent on the previous values for temperature \verb T across the bar this
576     must be updated in the loop. Our output at each time step is \verb T the heat
577     distribution and \verb totT the total heat in the system.
578 gross 2905 \begin{python}
579 gross 2878 while t < tend:
580     i+=1 #increment the counter
581     t+=h #increment the current time
582     mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
583     T=mypde.getSolution() #get the PDE solution
584     totE = integrate(rhocp*T) #get the total heat (energy) in the system
585     \end{python}
586 ahallam 2979 The last statement in this script calculates the total energy in the system as
587     volume integral
588     of $\rho c\hackscore{p} T$ over the block. As the blocks are insulated no energy
589     should be get lost or added.
590 gross 2905 The total energy should stay constant for the example discussed here.
591 ahallam 2401
592 gross 2905 \subsection{Running the Script}
593 ahallam 2975 The script presented so far is available under
594 gross 2949 \verb|example01a.py|. You can edit this file with your favourite text editor.
595 ahallam 2979 On most operating systems\footnote{The you can use \texttt{run-escript} launcher
596     is not supported under {\it MS Windows} yet.} you can use the
597     \program{run-escript} command
598 gross 2905 to launch {\it escript} scripts. For the example script use;
599     \begin{verbatim}
600 gross 2949 run-escript example01a.py
601 gross 2905 \end{verbatim}
602     The program will print a progress report. Alternatively, you can use
603     the python interpreter directly;
604     \begin{verbatim}
605 gross 2949 python example01a.py
606 gross 2905 \end{verbatim}
607 ahallam 2979 if the system is configured correctly (Please talk to your system
608     administrator).
609 gross 2905
610     \begin{figure}
611     \begin{center}
612     \includegraphics[width=4in]{figures/ttblockspyplot150}
613 gross 2952 \caption{Example 1b: Total Energy in the Blocks over Time (in seconds).}
614 gross 2905 \label{fig:onedheatout1}
615     \end{center}
616     \end{figure}
618 gross 2878 \subsection{Plotting the Total Energy}
619 gross 2949 \sslist{example01b.py}
620 gross 2878
621 ahallam 2979 \esc does not include its own plotting capabilities. However, it is possible to
622     use a variety of free \pyt packages for visualisation.
623     Two types will be demonstrated in this cookbook;
624     \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and
625     \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation.
626     The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}}
627     and is good for basic graphs and plots.
628     For more complex visualisation tasks in particular, two and three dimensional
629     problems we recommend the use of more advanced tools. For instance, \mayavi
630     \footnote{\url{http://code.enthought.com/projects/mayavi/}}
631 ahallam 2975 which is based upon the \verb|VTK| toolkit. The usage of \verb|VTK| based
632 ahallam 2979 visualization is discussed in Chapter~\ref{Sec:2DHD} which focusses on a two
633     dimensional PDE.
634 gross 2878
635 ahallam 2979 For our simple granite block problem, we have two plotting tasks. Firstly, we
636     are interested in showing the
637     behaviour of the total energy over time and secondly, how the temperature
638     distribution within the block is
639 gross 2878 developing over time. Lets start with the first task.
641 ahallam 2979 The trick is to create a record of the time marks and the corresponding total
642     energies observed.
643 gross 2878 \pyt provides the concept of lists for this. Before
644 ahallam 2979 the time loop is opened we create empty lists for the time marks \verb|t_list|
645     and the total energies \verb|E_list|.
646     After the new temperature has been calculated by solving the PDE we append the
647     new time marker and the total energy value for that time
648     to the corresponding list using the \verb|append| method. With these
649     modifications our script looks as follows:
650 ahallam 2775 \begin{python}
651 gross 2878 t_list=[]
652     E_list=[]
653     # ... start iteration:
654     while t<tend:
655     t+=h
656     mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
657     T=mypde.getSolution() #get the PDE solution
658     totE=integrate(rhocp*T)
659     t_list.append(t) # add current time mark to record
660     E_list.append(totE) # add current total energy to record
661 ahallam 2775 \end{python}
662 ahallam 2979 To plot $t$ over $totE$ we use \mpl a module contained within \pylab which needs
663     to be loaded before used;
664 ahallam 2775 \begin{python}
665 gross 2878 import pylab as pl # plotting package.
666 ahallam 2775 \end{python}
667 ahallam 2979 Here we are not using the \verb|from pylab import *| in order to avoid name
668     clashes for function names
669 gross 2905 within \esc.
670 ahallam 2401
671 ahallam 2979 The following statements are added to the script after the time loop has been
672     completed;
673 ahallam 2775 \begin{python}
674 gross 2878 pl.plot(t_list,E_list)
675     pl.title("Total Energy")
676 gross 2905 pl.axis([0,max(t_list),0,max(E_list)*1.1])
677 gross 2878 pl.savefig("totE.png")
678 ahallam 2775 \end{python}
679 ahallam 2979 The first statement hands over the time marks and corresponding total energies
680     to the plotter.
681 gross 2905 The second statment is setting the title for the plot. The third statement
682 ahallam 2979 sets the axis ranges. In most cases these are set appropriately by the plotter.
684 gross 2905 The last statement renders the plot and writes the
685 ahallam 2979 result into the file \verb|totE.png| which can be displayed by (almost) any
686     image viewer.
687     As expected the total energy is constant over time, see
688     \reffig{fig:onedheatout1}.
689 ahallam 2401
690 gross 2878 \subsection{Plotting the Temperature Distribution}
691 gross 2931 \label{sec: plot T}
692 gross 2949 \sslist{example01c.py}
693 ahallam 2979 For plotting the spatial distribution of the temperature we need to modify the
694     strategy we have used
695     for the total energy. Instead of producing a final plot at the end we will
696     generate a
697     picture at each time step which can be browsed as a slide show or composed into
698     a movie.
699     The first problem we encounter is that if we produce an image at each time step
700     we need
701 gross 2878 to make sure that the images previously generated are not overwritten.
703 ahallam 2979 To develop an incrementing file name we can use the following convention. It is
704     convenient to
705     put all image file showing the same variable - in our case the temperature
706     distribution -
707     into a separate directory. As part of the \verb|os| module\footnote{The
708     \texttt{os} module provides
709     a powerful interface to interact with the operating system, see
710     \url{http://docs.python.org/library/os.html}.} \pyt
711 gross 2878 provides the \verb|os.path.join| command to build file and
712     directory names in a platform independent way. Assuming that
713 ahallam 2979 \verb|save_path| is name of directory we want to put the results the command
714     is;
715 ahallam 2775 \begin{python}
716 gross 2878 import os
717     os.path.join(save_path, "tempT%03d.png"%i )
718 ahallam 2775 \end{python}
719 gross 2878 where \verb|i| is the time step counter.
720 ahallam 2979 There are two arguments to the \verb join command. The
721     \verb save_path variable is a predefined string pointing to the directory we want to save our
722     data, for example a single sub-folder called \verb data would be defined by;
723 gross 2878 \begin{verbatim}
724     save_path = "data"
725     \end{verbatim}
726 gross 2949 while a sub-folder of \verb data called \verb example01 would be defined by;
727 gross 2878 \begin{verbatim}
728 gross 2949 save_path = os.path.join("data","example01")
729 gross 2878 \end{verbatim}
730 ahallam 2979 The second argument of \verb join \xspace contains a string which is the file
731     name or subdirectory name. We can use the operator \verb|%| to use the value of
732     \verb|i| as part of our filename. The sub-string \verb %03d indicates that we
733     want to substitude a value into the name;
734 gross 2878 \begin{itemize}
735 jfenwick 2961 \item \verb 0 means that small numbers should have leading zeroes;
736 ahallam 2979 \item \verb 3 means that numbers should be written using at least 3 digits;
737     and
738 jfenwick 2961 \item \verb d means that the value to substitute will be an integer.
739 gross 2878 \end{itemize}
740 ahallam 2979 To actually substitute the value of \verb|i| into the name write \verb %i after
741     the string.
742     When done correctly, the output files from this command would be place in the
743     directory defined by \verb save_path as;
744 gross 2878 \begin{verbatim}
745 jfenwick 2961 blockspyplot001.png
746     blockspyplot002.png
747     blockspyplot003.png
748 gross 2878 ...
749     \end{verbatim}
750     and so on.
752     A sub-folder check/constructor is available in \esc. The command;
753     \begin{verbatim}
754     mkDir(save_path)
755     \end{verbatim}
756 ahallam 2979 will check for the existence of \verb save_path and if missing, make the
757     required directories.
758 gross 2878
759 artak 2964 We start by modifying our solution script.
760 ahallam 2979 Prior to the \verb|while| loop we will need to extract our finite solution
761     points to a data object that is compatible with \mpl. First we create the node
762     coordinates of the sample points used to represent
763     the temperature as a \pyt list of tuples or a \numpy array as requested by the
764     plotting function.
765     We need to convert the array \verb|x| previously set as
766     \verb|Solution(blocks).getX()| into a \pyt list
767     and then to a \numpy array. The $x\hackscore{0}$ component is then extracted via
768     an array slice to the variable \verb|plx|;
769 ahallam 2775 \begin{python}
770 gross 2878 import numpy as np # array package.
771     #convert solution points for plotting
772 gross 2905 plx = x.toListOfTuples()
773 gross 2878 plx = np.array(plx) # convert to tuple to numpy array
774     plx = plx[:,0] # extract x locations
775 ahallam 2775 \end{python}
776 gross 2878
777 ahallam 2645 \begin{figure}
778     \begin{center}
779 gross 2905 \includegraphics[width=4in]{figures/blockspyplot001}
780     \includegraphics[width=4in]{figures/blockspyplot050}
781     \includegraphics[width=4in]{figures/blockspyplot200}
782 ahallam 2979 \caption{Example 1c: Temperature ($T$) distribution in the blocks at time steps
783     $1$, $50$ and $200$.}
784 ahallam 2645 \label{fig:onedheatout}
785     \end{center}
786     \end{figure}
788 ahallam 2979 We use the same techniques provided by \mpl as we have used to plot the total
789     energy over time.
790     For each time step we generate a plot of the temperature distribution and save
791     each to a file.
792     The following is appended to the end of the \verb while loop and creates one
793     figure of the temperature distribution. We start by converting the solution to a
794     tuple and then plotting this against our \textit{x coordinates} \verb plx we
795     have generated before. We add a title to the diagram before it is rendered into
796     a file.
797     Finally, the figure is saved to a \verb|*.png| file and cleared for the
798     following iteration.
799 ahallam 2775 \begin{python}
800 gross 2878 # ... start iteration:
801     while t<tend:
802     ....
803     T=mypde.getSolution() #get the PDE solution
804     tempT = T.toListOfTuples() # convert to a tuple
805     pl.plot(plx,tempT) # plot solution
806     # set scale (Temperature should be between Tref and T0)
807     pl.axis([0,mx,Tref*.9,T0*1.1])
808     # add title
809 gross 2905 pl.title("Temperature across the blocks at time %e minutes"%(t/day))
810 gross 2878 #save figure to file
811 gross 2905 pl.savefig(os.path.join(save_path,"tempT","blockspyplot%03d.png") %i)
812 ahallam 2775 \end{python}
813 gross 2905 Some results are shown in \reffig{fig:onedheatout}.
814 jfenwick 2657
815 ahallam 2606 \subsection{Make a video}
816 ahallam 2979 Our saved plots from the previous section can be cast into a video using the
817     following command appended to the end of the script. The \verb mencoder command
818     is Linux only, so other platform users need to use an alternative video encoder.
819 ahallam 2775 \begin{python}
820 gross 2878 # compile the *.png files to create a *.avi videos that show T change
821     # with time. This operation uses Linux mencoder. For other operating
822 gross 2905 # systems it is possible to use your favourite video compiler to
823 ahallam 2606 # convert image files to videos.
824 gross 2477
825 ahallam 2606 os.system("mencoder mf://"+save_path+"/tempT"+"/*.png -mf type=png:\
826 gross 2878 w=800:h=600:fps=25 -ovc lavc -lavcopts vcodec=mpeg4 -oac copy -o \
827 gross 2949 example01tempT.avi")
828 ahallam 2775 \end{python}
829 gross 2477

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