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more corrections on exmaple 1, up to section 2.1.5
1 ahallam 2401
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3     %
4 jfenwick 2881 % Copyright (c) 2003-2010 by University of Queensland
5 ahallam 2401 % 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 gross 2905 \begin{figure}[h!]
15     \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}}
16 gross 2952 \caption{Example 1: Temperature differential along a single interface between two granite blocks.}
17 gross 2905 \label{fig:onedgbmodel}
18     \end{figure}
19 ahallam 2801
20 gross 2949 \section{Example 1: One Dimensional Heat Diffusion in Granite}
21 gross 2905 \label{Sec:1DHDv00}
22 gross 2878
23 gross 2905 The first model consists of two blocks of isotropic material, for instance granite, sitting next to each other.
24 artak 2957 Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is \verb|T2|.
25 gross 2905 We assume that the system is insulated.
26     What would happen to the temperature distribution in each block over time?
27 artak 2958 Intuition tells us that heat will be transported from the hotter block to the cooler one until both
28 gross 2905 blocks have the same temperature.
30 ahallam 2495 \subsection{1D Heat Diffusion Equation}
31 artak 2957 We can model the heat distribution of this problem over time using one dimensional heat diffusion equation\footnote{A detailed discussion on how the heat diffusion equation is derived can be found at \url{http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}};
32 ahallam 2494 which is defined as:
33 ahallam 2401 \begin{equation}
34     \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H
35     \label{eqn:hd}
36     \end{equation}
37 gross 2861 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal
38     conductivity\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we assume that these material
39     parameters are \textbf{constant}.
40     The heat source is defined by the right hand side of \refEq{eqn:hd} as $q\hackscore{H}$; this can take the form of a constant or a function of time and space. For example $q\hackscore{H} = q\hackscore{0}e^{-\gamma t}$ where we have the output of our heat source decaying with time. There are also two partial derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is the spatial change of temperature. As there is only a single spatial dimension to our problem, our temperature solution $T$ is only dependent on the time $t$ and our position along the iron bar $x$.
41 ahallam 2401
42 gross 2861 \subsection{PDEs and the General Form}
43 artak 2958 Potentially, it is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact solution to our problem. However, it is not always possible or practical to solve a problem this way. Alternatively, computers can be used to solve these kinds of problems. To do this, a numerical approach is required to discretise
44     the PDE \refEq{eqn:hd} in time and space, then we left with a finite number of equations for a finite number of spatial points and time steps in the model. While discretisation introduces approximations and a degree of error, a sufficiently sampled model is generally accurate enough to satisfy the requirements of the modeller.
45 gross 2477
46 artak 2958 Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a steady linear PDE which involves spatial derivatives only and needs to be solved in each time step to progress in time. \esc can help us here.
47 gross 2861
48 artak 2958 For time discretization we use the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is based on the
49 gross 2861 approximation
50     \begin{equation}
51     \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h}
52     \label{eqn:beuler}
53     \end{equation}
54     for $\frac{\partial T}{\partial t}$ at time $t$
55     where $h$ is the time step size. This can also be written as;
56     \begin{equation}
57     \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h}
58     \label{eqn:Tbeuler}
59     \end{equation}
60     where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has
61     \begin{equation}
62     \begin{array}{rcl}
63     t^{(n)} & = & t^{(n-1)}+h \\
64     T^{(n)} & = & T(t^{(n-1)}) \\
65     \end{array}
66     \label{eqn:Neuler}
67     \end{equation}
68     Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get;
69     \begin{equation}
70     \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H
71     \label{eqn:hddisc}
72     \end{equation}
73 artak 2958 Notice that we evaluate the spatial derivative term at current time $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, one can evaluate the spatial derivative term at the previous time $t^{(n-1)}$. This
74     approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages, which
75     are not discussed here. However, this scheme has a major disadvantage, namely depending on the
76     material parameter as well as the discretization of the spatial derivative term, the time step size $h$ needs to be chosen sufficiently small to achieve a stable temperature when progressing in time. The term \textit{stable} means
77     that the approximation of the temperature will not grow beyond its initial bounds and become non-physical.
78     The backward Euler scheme, which we use here, is unconditionally stable meaning that under the assumption of
79     physically correct problem set-up the temperature approximation remains physical for all time steps.
80 gross 2861 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler}
81 artak 2958 is sufficiently small, thus a good approximation of the true temperature is computed. It is
82 gross 2861 therefore crucial that the user remains critical about his/her results and for instance compares
83     the results for different time and spatial step sizes.
85     To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear
86 artak 2958 differential equation \refEq{eqn:hddisc} which only includes spatial derivatives. To solve this problem
87     we want to use \esc.
88 gross 2861
89 artak 2958 In \esc any given PDE described by general form. For the purpose of this introduction we illustrate a simpler version of the general form for full linear PDEs which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{The form in the \esc users guide which uses the Einstein convention is written as
90 gross 2477 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$}
91 ahallam 2606 is described by;
92 gross 2477 \begin{equation}\label{eqn:commonform nabla}
93 jfenwick 2657 -\nabla\cdot(A\cdot\nabla u) + Du = f
94 ahallam 2411 \end{equation}
95 gross 2861 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents
96 ahallam 2495 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ;
97 gross 2477 \begin{equation}
98     \nabla = \frac{\partial}{\partial x}
99     \end{equation}
100 gross 2861 and we can write \refEq{eqn:commonform nabla} as;
101 ahallam 2411 \begin{equation}\label{eqn:commonform}
102 gross 2477 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f
103 ahallam 2411 \end{equation}
104 gross 2861 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc}
105     we rearrange \refEq{eqn:hddisc};
106 ahallam 2411 \begin{equation}
107 ahallam 2645 \frac{\rho c\hackscore p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)}
108 ahallam 2494 \label{eqn:hdgenf}
109     \end{equation}
110 ahallam 2775 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is required for \esc to solve our PDE. This can be done by generating a solution for successive increments in the time nodes $t^{(n)}$ where
111 ahallam 2495 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant.
112 artak 2958 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} one can see that;
113 gross 2862 \begin{equation}\label{ESCRIPT SET}
114 gross 2861 u=T^{(n)};
115 ahallam 2494 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)}
116     \end{equation}
118 ahallam 2495 \subsection{Boundary Conditions}
119 gross 2878 \label{SEC BOUNDARY COND}
120 gross 2862 With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively.
121 artak 2958 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown solution (in our example the temperature) on parts of the boundary or on the entire boundary of the region of interest.
122 gross 2905 We discuss Dirichlet boundary condition in our second example presented in Section~\ref{Sec:1DHDv0}.
123 ahallam 2495
124 gross 2905 We make the model assumption that the system is insulated so we need
125     to add an appropriate boundary condition to prevent
126     any loss or inflow of energy at boundary of our domain. Mathematically this is expressed by prescribing
127     the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified one dimensional model this is expressed
128 gross 2862 in the form;
129 ahallam 2494 \begin{equation}
130 gross 2862 \kappa \frac{\partial T}{\partial x} = 0
131 ahallam 2494 \end{equation}
132 gross 2862 or in a more general case as
133     \begin{equation}\label{NEUMAN 1}
134     \kappa \nabla T \cdot n = 0
135     \end{equation}
136     where $n$ is the outer normal field \index{outer normal field} at the surface of the domain.
137 gross 2905 The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of
138 artak 2958 the temperature $T$. Other notations used here are\footnote{The \esc notation for the normal
139 gross 2862 derivative is $T\hackscore{,i} n\hackscore i$.};
140 ahallam 2645 \begin{equation}
141 gross 2862 \nabla T \cdot n = \frac{\partial T}{\partial n} \; .
142 ahallam 2645 \end{equation}
143 gross 2862 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE.
144 ahallam 2494
145 gross 2905 The PDE \refEq{eqn:hdgenf}
146 artak 2958 and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with the Dirichlet boundary conditions) define a \textbf{boundary value problem}.
147     It is the nature of a boundary value problem to allow making statements about the solution in the
148 gross 2862 interior of the domain from information known on the boundary only. In most cases
149 artak 2958 we use the term partial differential equation but in fact it is a boundary value problem.
150 gross 2862 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that
151     at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set.
153 gross 2878 Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate.
154 gross 2862 For a problem of the form in~\refEq{eqn:commonform nabla} the default condition\footnote{In the form of the \esc users guide which is using the Einstein convention is written as
155     $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is;
156     \begin{equation}\label{NEUMAN 2}
157 gross 2948 -n\cdot A \cdot\nabla u = 0
158 gross 2862 \end{equation}
159     which is used everywhere on the boundary. Again $n$ denotes the outer normal field.
160     Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}.
161     With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this
162     condition translates into
163 gross 2867 \begin{equation}\label{NEUMAN 2b}
164 gross 2862 \kappa \frac{\partial T}{\partial x} = 0
165     \end{equation}
166 artak 2958 for the boundary of the domain. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We discuss the Dirichlet boundary condition later.
167 gross 2862
168 gross 2870 \subsection{Outline of the Implementation}
169     \label{sec:outline}
170 artak 2958 To solve the heat diffusion equation (equation \refEq{eqn:hd}) we write a simple \pyt script. At this point we assume that you have some basic understanding of the \pyt programming language. If not, there are some pointers and links available in Section \ref{sec:escpybas}. The script we discuss later in details have four major steps. Firstly, we need to define the domain where we want to
171     calculate the temperature. For our problem this is the joint blocks of granite which has a rectangular shape. Secondly, we need to define the PDE to solve in each time step to get the updated temperature. Thirdly, we need to define the coefficients of the PDE and finally we need to solve the PDE. The last two steps need to be repeated until the final time marker has been reached. The work flow is as follows:
172 gross 2878 \begin{enumerate}
173     \item create domain
174     \item create PDE
175     \item while end time not reached:
176     \begin{enumerate}
177     \item set PDE coefficients
178     \item solve PDE
179     \item update time marker
180     \end{enumerate}
181     \item end of calculation
182     \end{enumerate}
183 artak 2959 In the terminology of \pyt, the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it is defined by its usage and features
184 gross 2905 rather than its actual representation. So we will create a domain object to describe the geometry of the two
185 artak 2959 granite blocks. Then we define PDEs and spatially distributed values such as the temperature
186     on this domain. Similarly, to define a PDE object we use the fact that one needs only to define the coefficients of the PDE and solve the PDE. The PDE object has advanced features, but these are not required in simple cases.
187 gross 2870
189     \begin{figure}[t]
190     \centering
191     \includegraphics[width=6in]{figures/functionspace.pdf}
192     \label{fig:fs}
193     \caption{\esc domain construction overview}
194     \end{figure}
196     \subsection{The Domain Constructor in \esc}
197     \label{ss:domcon}
198     It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again
199     from the user's point of view the representation of these spatially distributed values is not relevant.
201     There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is
202     a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number
203 gross 2905 elements or cells to be used along the length and height, see \reffig{fig:fs}. Any spatially distributed value
204     and the PDE is represented in discrete form using this element representation\footnote{We will use the finite element method (FEM), see \url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}. Therefore we will have access to an approximation of the true PDE solution only.
205 gross 2870 The quality of the approximation depends - besides other factors- mainly on the number of elements being used. In fact, the
206     approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of
207 gross 2878 elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage.
208 gross 2870
209     In general, one can thinks about a domain object as a composition of nodes and elements.
210 gross 2905 As shown in \reffig{fig:fs}, an element is defined by the nodes used to describe its vertices.
211 gross 2870 To represent spatial distributed values the user can use
212     the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain.
213     The different approach used to represent values is called \textbf{function space} and is attached to all objects
214     in \esc representing a spatial distributed value such as the solution of a PDE. The three
215     function spaces we will use at the moment are;
216     \begin{enumerate}
217     \item the nodes, called by \verb|ContinuousFunction(domain)| ;
218     \item the elements/cells, called by \verb|Function(domain)| ; and
219     \item the boundary, called by \verb|FunctionOnBoundary(domain)| .
220     \end{enumerate}
221     A function space object such as \verb|ContinuousFunction(domain)| has the method \verb|getX| attached to it. This method returns the
222     location of the so-called \textbf{sample points} used to represent values with the particular function space attached to it. So the
223     call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the nodes used to describe the domain while
224     the \verb|Function(domain).getX()| returns the coordinates of numerical integration points within elements, see
225 gross 2905 \reffig{fig:fs}.
226 gross 2870
227 gross 2878 This distinction between different representations of spatial distributed values
228 gross 2870 is important in order to be able to vary the degrees of smoothness in a PDE problem.
229     The coefficients of a PDE need not be continuous thus this qualifies as a \verb|Function()| type.
230     On the other hand a temperature distribution must be continuous and needs to be represented with a \verb|ContinuousFunction()| function space.
231     An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object.
232     \esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verb|FunctionOnBoundary()|
233     or \verb|Function()|. On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() .
234 gross 2878 These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed.
235 gross 2870
236     Later in this introduction we will discuss how
237     to define specific areas of geometry with different materials which are represented by different material coefficients such the
238 gross 2878 thermal conductivities $kappa$. A very powerful technique to define these types of PDE
239 gross 2870 coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names.
240 gross 2878 This is simplifying PDE coefficient and flux definitions. It makes for much easier scripting. We will discuss this technique in Section~\ref{STEADY-STATE HEAT REFRACTION}.
241 gross 2870
243     \subsection{A Clarification for the 1D Case}
244 gross 2931 \label{SEC: 1D CLARIFICATION}
245 gross 2861 It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \esc is inherently designed to solve problems that are greater than one dimension and so \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the \textit{Nabla operator} has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full, \refEq{eqn:commonform nabla} assuming a constant coefficient $A$, takes the form;
246 gross 2477 \begin{equation}\label{eqn:commonform2D}
247     -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}}
248     -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y}
249     -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x}
250     -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}}
251     + Du = f
252     \end{equation}
253 ahallam 2606 Notice that for the higher dimensional case $A$ becomes a matrix. It is also
254 ahallam 2495 important to notice that the usage of the Nabla operator creates
255     a compact formulation which is also independent from the spatial dimension.
256 gross 2861 So to make the general PDE \refEq{eqn:commonform2D} one dimensional as
257     shown in \refEq{eqn:commonform} we need to set
258 ahallam 2606 \begin{equation}
259 ahallam 2494 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0
260 gross 2477 \end{equation}
262 gross 2867
263 ahallam 2495 \subsection{Developing a PDE Solution Script}
264 ahallam 2801 \label{sec:key}
265 gross 2949 \sslist{example01a.py}
266 gross 2870 We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules.
267 gross 2905 By developing a script for \esc, the heat diffusion equation can be solved at successive time steps for a predefined period using our general form \refEq{eqn:hdgenf}. Firstly it is necessary to import all the libraries\footnote{The libraries contain predefined scripts that are required to solve certain problems, these can be simple like sine and cosine functions or more complicated like those from our \esc library.}
268 ahallam 2495 that we will require.
269 ahallam 2775 \begin{python}
270 ahallam 2495 from esys.escript import *
271 ahallam 2606 # This defines the LinearPDE module as LinearPDE
272     from esys.escript.linearPDEs import LinearPDE
273     # This imports the rectangle domain function from finley.
274     from esys.finley import Rectangle
275     # A useful unit handling package which will make sure all our units
276     # match up in the equations under SI.
277     from esys.escript.unitsSI import *
278 ahallam 2775 \end{python}
279 gross 2878 It is generally a good idea to import all of the \modescript library, although if the functions and classes required are known they can be specified individually. The function \verb|LinearPDE| has been imported explicitly for ease of use later in the script. \verb|Rectangle| is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables.
280 gross 2477
281 gross 2905 Once our library dependencies have been established, defining the problem specific variables is the next step. In general the number of variables needed will vary between problems. These variables belong to two categories. They are either directly related to the PDE and can be used as inputs into the \esc solver, or they are script variables used to control internal functions and iterations in our problem. For this PDE there are a number of constants which will need values. Firstly, the model upon which we wish to solve our problem needs to be defined. There are many different types of models in \modescript which we will demonstrate in later tutorials but for our granite blocks, we will simply use a rectangular model.
282 ahallam 2401
283 gross 2905 Using a rectangular model simplifies our granite blocks which would in reality be a \textit{3D} object, into a single dimension. The granite blocks will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the block. There are four arguments we must consider when we decide to create a rectangular model, the model \textit{length}, \textit{width} and \textit{step size} in each direction. When defining the size of our problem it will help us determine appropriate values for our model arguments. If we make our dimensions large but our step sizes very small we will to a point, increase the accuracy of our solution. Unfortunately we also increase the number of calculations that must be solved per time step. This means more computational time is required to produce a solution. In this \textit{1D} problem, the bar is defined as being 1 metre long. An appropriate step size \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| need only be 1, this is because our problem stipulates no partial derivatives in the $y$ direction. Thus the temperature does not vary with $y$. Hence, the model parameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI.
284 ahallam 2775 \begin{python}
285 gross 2905 mx = 500.*m #meters - model length
286     my = 100.*m #meters - model width
287     ndx = 50 # mesh steps in x direction
288     ndy = 1 # mesh steps in y direction
289     boundloc = mx/2 # location of boundary between the two blocks
290 ahallam 2775 \end{python}
291 gross 2905 The material constants and the temperature variables must also be defined. For the granite in the model they are defined as:
292 ahallam 2775 \begin{python}
293 ahallam 2495 #PDE related
294 gross 2905 rho = 2750. *kg/m**3 #kg/m^{3} density of iron
295     cp = 790.*J/(kg*K) # J/Kg.K thermal capacity
296 gross 2878 rhocp = rho*cp
297 gross 2905 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity
298 gross 2878 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source
299 gross 2905 T1=20 * Celsius # initial temperature at Block 1
300     T2=2273. * Celsius # base temperature at Block 2
301 ahallam 2775 \end{python}
302 ahallam 2495 Finally, to control our script we will have to specify our timing controls and where we would like to save the output from the solver. This is simple enough:
303 ahallam 2775 \begin{python}
304 gross 2878 t=0 * day #our start time, usually zero
305     tend=1. * day # - time to end simulation
306 ahallam 2495 outputs = 200 # number of time steps required.
307     h=(tend-t)/outputs #size of time step
308 ahallam 2606 #user warning statement
309     print "Expected Number of time outputs is: ", (tend-t)/h
310     i=0 #loop counter
311 ahallam 2775 \end{python}
312 gross 2905 Now that we know our inputs we will build a domain using the \verb Rectangle() function from \verb finley . The four arguments allow us to define our domain \verb model as:
313 ahallam 2775 \begin{python}
314 ahallam 2606 #generate domain using rectangle
315 gross 2905 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy)
316 ahallam 2775 \end{python}
317 gross 2905 \verb blocks now describes a domain in the manner of Section \ref{ss:domcon}. T
318 ahallam 2658
319 ahallam 2775 With a domain and all our required variables established, it is now possible to set up our PDE so that it can be solved by \esc. The first step is to define the type of PDE that we are trying to solve in each time step. In this example it is a single linear PDE\footnote{in comparison to a system of PDEs which will be discussed later.}. We also need to state the values of our general form variables.
320     \begin{python}
321 gross 2905 mypde=LinearPDE(blocks)
322 gross 2878 A=zeros((2,2)))
323     A[0,0]=kappa
324 gross 2905 mypde.setValue(A=A, D=rhocp/h)
325 ahallam 2775 \end{python}
326 gross 2878 In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric.
327     Symmetry of a PDE is defined by;
328 ahallam 2495 \begin{equation}\label{eqn:symm}
329     A\hackscore{jl}=A\hackscore{lj}
330     \end{equation}
331 gross 2878 Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ as well as the right hand side $Y$ may take any value. From the above definition we can see that our PDE is symmetric. The \verb LinearPDE class provides the method \method{checkSymmetry} to check if the given PDE is symmetric. As our PDE is symmetrical we will enable symmetry via;
332 ahallam 2775 \begin{python}
333 ahallam 2495 myPDE.setSymmetryOn()
334 ahallam 2775 \end{python}
335 gross 2905 Next we need to establish the initial temperature distribution \verb|T|. We need to
336     assign the value \verb|T1| to all sample points left to the contact interface at $x\hackscore{0}=\frac{mx}{2}$
337     and the value \verb|T2| right to the contact interface. \esc
338     provides the \verb|whereNegative| function to construct this. In fact,
339     \verb|whereNegative| returns the value $1$ at those sample points where the argument
340     has a negative value. Otherwise zero is returned. If \verb|x| are the $x\hackscore{0}$
341     coordinates of the sample points used to represent the temperature distribution
342     then \verb|x[0]-boundloc| gives us a negative value for
343     all sample points left to the interface and non-negative value to
344     the right of the interface. So with;
345 gross 2878 \begin{python}
346     # ... set initial temperature ....
347 gross 2905 T= T1*whereNegative(x[0]-boundloc)+T2*(1-whereNegative(x[0]-boundloc))
348 gross 2878 \end{python}
349 gross 2905 we get the desired temperature distribution. To get the actual sample points \verb|x| we use
350     the \verb|getX()| method of the function space \verb|Solution(blocks)|
351     which is used to represent the solution of a PDE;
352 gross 2878 \begin{python}
353 gross 2905 x=Solution(blocks).getX()
354     \end{python}
355     As \verb|x| are the sample points for the function space \verb|Solution(blocks)|
356     the initial temperature \verb|T| is using these sample points for representation.
357     Although \esc is trying to be forgiving with the choice of sample points and to convert
358     where necessary the adjustment of the function space is not always possible. So it is
359     advisable to make a careful choice on the function space used.
361     Finally we will initialise an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system.
362     \begin{python}
363 gross 2878 while t < tend:
364     i+=1 #increment the counter
365     t+=h #increment the current time
366     mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
367     T=mypde.getSolution() #get the PDE solution
368     totE = integrate(rhocp*T) #get the total heat (energy) in the system
369     \end{python}
370     The last statement in this script calculates the total energy in the system as volume integral
371 gross 2905 of $\rho c\hackscore{p} T$ over the block. As the blocks are insulated no energy should be get lost or added.
372     The total energy should stay constant for the example discussed here.
373 ahallam 2401
374 gross 2905 \subsection{Running the Script}
375     The script presented so for is available under
376 gross 2949 \verb|example01a.py|. You can edit this file with your favourite text editor.
377 jfenwick 2911 On most operating systems\footnote{The you can use \texttt{run-escript} launcher is not supported under {\it MS Windows} yet.} you can use the \program{run-escript} command
378 gross 2905 to launch {\it escript} scripts. For the example script use;
379     \begin{verbatim}
380 gross 2949 run-escript example01a.py
381 gross 2905 \end{verbatim}
382     The program will print a progress report. Alternatively, you can use
383     the python interpreter directly;
384     \begin{verbatim}
385 gross 2949 python example01a.py
386 gross 2905 \end{verbatim}
387     if the system is configured correctly (Please talk to your system administrator).
389     \begin{figure}
390     \begin{center}
391     \includegraphics[width=4in]{figures/ttblockspyplot150}
392 gross 2952 \caption{Example 1b: Total Energy in the Blocks over Time (in seconds).}
393 gross 2905 \label{fig:onedheatout1}
394     \end{center}
395     \end{figure}
397 gross 2878 \subsection{Plotting the Total Energy}
398 gross 2949 \sslist{example01b.py}
399 gross 2878
400 gross 2905 \esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualisation.
401 gross 2878 Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation.
402     The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots.
403     For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}}
404 gross 2905 which bases on the \verb|VTK| toolkit. We will discuss the usage of \verb|VTK| based
405 gross 2878 visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE.
407     For our simple problem we have two plotting tasks: Firstly we are interested in showing the
408 gross 2905 behaviour of the total energy over time and secondly in how the temperature distribution within the block is
409 gross 2878 developing over time. Lets start with the first task.
411     The trick is to create a record of the time marks and the corresponding total energies observed.
412     \pyt provides the concept of lists for this. Before
413     the time loop is opened we create empty lists for the time marks \verb|t_list| and the total energies \verb|E_list|.
414     After the new temperature as been calculated by solving the PDE we append the new time marker and total energy
415     to the corresponding list using the \verb|append| method. With these modifications the script looks as follows:
416 ahallam 2775 \begin{python}
417 gross 2878 t_list=[]
418     E_list=[]
419     # ... start iteration:
420     while t<tend:
421     t+=h
422     mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
423     T=mypde.getSolution() #get the PDE solution
424     totE=integrate(rhocp*T)
425     t_list.append(t) # add current time mark to record
426     E_list.append(totE) # add current total energy to record
427 ahallam 2775 \end{python}
428 gross 2878 To plot $t$ over $totE$ we use the \mpl a module contained within \pylab which needs to be loaded before used;
429 ahallam 2775 \begin{python}
430 gross 2878 import pylab as pl # plotting package.
431 ahallam 2775 \end{python}
432 gross 2905 Here we are not using the \verb|from pylab import *| in order to avoid name clashes for function names
433     within \esc.
434 ahallam 2401
435 gross 2878 The following statements are added to the script after the time loop has been completed;
436 ahallam 2775 \begin{python}
437 gross 2878 pl.plot(t_list,E_list)
438     pl.title("Total Energy")
439 gross 2905 pl.axis([0,max(t_list),0,max(E_list)*1.1])
440 gross 2878 pl.savefig("totE.png")
441 ahallam 2775 \end{python}
442 gross 2878 The first statement hands over the time marks and corresponding total energies to the plotter.
443 gross 2905 The second statment is setting the title for the plot. The third statement
444     sets the axis ranges. In most cases these are set appropriately by the plotter.
445     The last statement renders the plot and writes the
446     result into the file \verb|totE.png| which can be displayed by (almost) any image viewer.
447     As expected the total energy is constant over time, see \reffig{fig:onedheatout1}.
448 ahallam 2401
449 gross 2878 \subsection{Plotting the Temperature Distribution}
450 gross 2931 \label{sec: plot T}
451 gross 2949 \sslist{example01c.py}
452 gross 2878 For plotting the spatial distribution of the temperature we need to modify the strategy we have used
453     for the total energy. Instead of producing a final plot at the end we will generate a
454     picture at each time step which can be browsed as slide show or composed to a movie.
455     The first problem we encounter is that if we produce an image in each time step we need
456     to make sure that the images previously generated are not overwritten.
458     To develop an incrementing file name we can use the following convention. It is convenient to
459     put all image file showing the same variable - in our case the temperature distribution -
460     into a separate directory. As part of the \verb|os| module\footnote{The \texttt{os} module provides
461     a powerful interface to interact with the operating system, see \url{http://docs.python.org/library/os.html}.} \pyt
462     provides the \verb|os.path.join| command to build file and
463     directory names in a platform independent way. Assuming that
464     \verb|save_path| is name of directory we want to put the results the command is;
465 ahallam 2775 \begin{python}
466 gross 2878 import os
467     os.path.join(save_path, "tempT%03d.png"%i )
468 ahallam 2775 \end{python}
469 gross 2878 where \verb|i| is the time step counter.
470     There are two arguments to the \verb join command. The \verb save_path variable is a predefined string pointing to the directory we want to save our data in, for example a single sub-folder called \verb data would be defined by;
471     \begin{verbatim}
472     save_path = "data"
473     \end{verbatim}
474 gross 2949 while a sub-folder of \verb data called \verb example01 would be defined by;
475 gross 2878 \begin{verbatim}
476 gross 2949 save_path = os.path.join("data","example01")
477 gross 2878 \end{verbatim}
478 gross 2905 The second argument of \verb join \xspace contains a string which is the file name or subdirectory name. We can use the operator \verb|%| to increment our file names with the value \verb|i| denoting a incrementing counter. The sub-string \verb %03d does this by defining the following parameters;
479 gross 2878 \begin{itemize}
480     \item \verb 0 becomes the padding number;
481     \item \verb 3 tells us the amount of padding numbers that are required; and
482     \item \verb d indicates the end of the \verb % operator.
483     \end{itemize}
484     To increment the file name a \verb %i is required directly after the operation the string is involved in. When correctly implemented the output files from this command would be place in the directory defined by \verb save_path as;
485     \begin{verbatim}
486 gross 2905 blockspyplot.png
487     blockspyplot.png
488     blockspyplot.png
489 gross 2878 ...
490     \end{verbatim}
491     and so on.
493     A sub-folder check/constructor is available in \esc. The command;
494     \begin{verbatim}
495     mkDir(save_path)
496     \end{verbatim}
497     will check for the existence of \verb save_path and if missing, make the required directories.
499     We start by modifying our solution script from before.
500 gross 2905 Prior to the \verb|while| loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First we create the node coordinates of the sample points used to represent
501     the temperature as a \pyt list of tuples or a \numpy array as requested by the plotting function.
502     We need to convert thearray \verb|x| previously set as \verb|Solution(blocks).getX()| into a \pyt list
503     and then to a \numpy array. The $x\hackscore{0}$ component is then extracted via an array slice to the variable \verb|plx|;
504 ahallam 2775 \begin{python}
505 gross 2878 import numpy as np # array package.
506     #convert solution points for plotting
507 gross 2905 plx = x.toListOfTuples()
508 gross 2878 plx = np.array(plx) # convert to tuple to numpy array
509     plx = plx[:,0] # extract x locations
510 ahallam 2775 \end{python}
511 gross 2878
512 ahallam 2645 \begin{figure}
513     \begin{center}
514 gross 2905 \includegraphics[width=4in]{figures/blockspyplot001}
515     \includegraphics[width=4in]{figures/blockspyplot050}
516     \includegraphics[width=4in]{figures/blockspyplot200}
517 gross 2952 \caption{Example 1c: Temperature ($T$) distribution in the blocks at time steps $1$, $50$ and $200$.}
518 ahallam 2645 \label{fig:onedheatout}
519     \end{center}
520     \end{figure}
522 gross 2878 For each time step we will generate a plot of the temperature distribution and save each to a file. We use the same
523     techniques provided by \mpl as we have used to plot the total energy over time.
524     The following is appended to the end of the \verb while loop and creates one figure of the temperature distribution. We start by converting the solution to a tuple and then plotting this against our \textit{x coordinates} \verb plx we have generated before. We add a title to the diagram before it is rendered into a file.
525     Finally, the figure is saved to a \verb|*.png| file and cleared for the following iteration.
526 ahallam 2775 \begin{python}
527 gross 2878 # ... start iteration:
528     while t<tend:
529     ....
530     T=mypde.getSolution() #get the PDE solution
531     tempT = T.toListOfTuples() # convert to a tuple
532     pl.plot(plx,tempT) # plot solution
533     # set scale (Temperature should be between Tref and T0)
534     pl.axis([0,mx,Tref*.9,T0*1.1])
535     # add title
536 gross 2905 pl.title("Temperature across the blocks at time %e minutes"%(t/day))
537 gross 2878 #save figure to file
538 gross 2905 pl.savefig(os.path.join(save_path,"tempT","blockspyplot%03d.png") %i)
539 ahallam 2775 \end{python}
540 gross 2905 Some results are shown in \reffig{fig:onedheatout}.
541 jfenwick 2657
542 ahallam 2606 \subsection{Make a video}
543 gross 2878 Our saved plots from the previous section can be cast into a video using the following command appended to the end of the script. \verb mencoder is Linux only however, and other platform users will need to use an alternative video encoder.
544 ahallam 2775 \begin{python}
545 gross 2878 # compile the *.png files to create a *.avi videos that show T change
546     # with time. This operation uses Linux mencoder. For other operating
547 gross 2905 # systems it is possible to use your favourite video compiler to
548 ahallam 2606 # convert image files to videos.
549 gross 2477
550 ahallam 2606 os.system("mencoder mf://"+save_path+"/tempT"+"/*.png -mf type=png:\
551 gross 2878 w=800:h=600:fps=25 -ovc lavc -lavcopts vcodec=mpeg4 -oac copy -o \
552 gross 2949 example01tempT.avi")
553 ahallam 2775 \end{python}
554 gross 2477

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