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2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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4 % Copyright (c) 2003-2009 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
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12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14 We will start by examining a simple one dimensional heat diffusion example. This problem will provide a good launch pad to build our knowledge of \esc and demonstrate how to solve simple partial differential equations (PDEs)\footnote{Wikipedia provides an excellent and comprehensive introduction to \textit{Partial Differential Equations} \url{http://en.wikipedia.org/wiki/Partial_differential_equation}, however their relevance to \esc and implementation should become a clearer as we develop our understanding further into the cookbook.}
15
16 \section{One Dimensional Heat Diffusion in an Iron Rod}
17
18 %\label{Sec:1DHDv0}
19 The first model consists of a simple cold iron bar at a constant temperature of zero \reffig{fig:onedhdmodel}. The bar is perfectly insulated on all sides with a heating element at one end. Intuition tells us that as heat is applied; energy will disperse along the bar via conduction. With time the bar will reach a constant temperature equivalent to that of the heat source.
20 \begin{figure}[h!]
21 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff}}
22 \caption{One dimensional model of an Iron bar.}
23 \label{fig:onedhdmodel}
24 \end{figure}
25 \subsection{1D Heat Diffusion Equation}
26 We can model the heat distribution of this problem over time using the 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}};
27 which is defined as:
28 \begin{equation}
29 \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H
30 \label{eqn:hd}
31 \end{equation}
32 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal
33 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
34 parameters are \textbf{constant}.
35 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 $x$.
36
37 \subsection{PDEs and the General Form}
38 Potentially, it is now possible to solve PDE \refEq{eqn:hd} analytically and this would produce 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 discretised
39 the PDE \refEq{eqn:hd} in time and space so finally we are left with a finite number of equations for a finite number of spatial and time steps in the model. While discretization introduces approximations and a degree of error, we find that a sufficiently sampled model is generally accurate enough for the requirements of the modeler.
40
41 Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will
42 leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time
43 step to progress in time - \esc can help us here.
44
45 For the discretization in time we will use is the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It bases on the
46 approximation
47 \begin{equation}
48 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h}
49 \label{eqn:beuler}
50 \end{equation}
51 for $\frac{\partial T}{\partial t}$ at time $t$
52 where $h$ is the time step size. This can also be written as;
53 \begin{equation}
54 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h}
55 \label{eqn:Tbeuler}
56 \end{equation}
57 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has
58 \begin{equation}
59 \begin{array}{rcl}
60 t^{(n)} & = & t^{(n-1)}+h \\
61 T^{(n)} & = & T(t^{(n-1)}) \\
62 \end{array}
63 \label{eqn:Neuler}
64 \end{equation}
65 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get;
66 \begin{equation}
67 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H
68 \label{eqn:hddisc}
69 \end{equation}
70 Notice that we evaluate the spatial derivative term at current time $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, one can use evaluate the spatial derivative term at the previous time $t^{(n-1)}$. This
71 approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which
72 we are not discussed here but has the major disadvantage that depending on the
73 material parameter as well as the discretiztion 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
74 that the approximation of the temperature will not grow beyond its initial bounds and becomes unphysical.
75 The backward Euler which we use here is unconditionally stable meaning that under the assumption of
76 physically correct problem set-up the temperature approximation remains physical for all times.
77 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler}
78 is sufficiently small so a good approximation of the true temperature is calculated. It is
79 therefore crucial that the user remains critical about his/her results and for instance compares
80 the results for different time and spatial step sizes.
81
82 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear
83 differential equation \refEq{eqn:hddisc} which is only including spatial derivatives. To solve this problem
84 we want to to use \esc.
85
86 \esc interfaces with any given PDE via a general form. For the purpose of this introduction we will illustrate a simpler version of the full linear PDE general form which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{In the form of the \esc users guide which using the Einstein convention is written as
87 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$}
88 is described by;
89 \begin{equation}\label{eqn:commonform nabla}
90 -\nabla\cdot(A\cdot\nabla u) + Du = f
91 \end{equation}
92 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
93 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ;
94 \begin{equation}
95 \nabla = \frac{\partial}{\partial x}
96 \end{equation}
97 and we can write \refEq{eqn:commonform nabla} as;
98 \begin{equation}\label{eqn:commonform}
99 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f
100 \end{equation}
101 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc}
102 we rearrange \refEq{eqn:hddisc};
103 \begin{equation}
104 \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)}
105 \label{eqn:hdgenf}
106 \end{equation}
107 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
108 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant.
109 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} it can be seen that;
110 \begin{equation}\label{ESCRIPT SET}
111 u=T^{(n)};
112 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)}
113 \end{equation}
114
115 % Now that the general form has been established, it can be submitted to \esc. Note that it is necessary to establish the state of our system at time zero or $T^{(n=0)}$. This is due to the time derivative approximation we have used from \refEq{eqn:Tbeuler}. Our model stipulates a starting temperature in the iron bar of 0\textcelsius. Thus the temperature distribution is simply;
116 % \begin{equation}
117 % T(x,0) = \left
118 % \end{equation}
119 % for all $x$ in the domain.
120
121 \subsection{Boundary Conditions}
122 \label{SEC BOUNDARY COND}
123 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.
124 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown - in our example the temperature - on parts of or the entire boundary of the region of interest.
125 For our model problem we want to keep the initial temperature setting on the left side of the
126 iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step.
127
128 On the other end of the iron rod we want to add an appropriate boundary condition to define insulation to prevent
129 any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing
130 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero on the right end of the rod
131 In our simplified one dimensional model this is expressed
132 in the form;
133 \begin{equation}
134 \kappa \frac{\partial T}{\partial x} = 0
135 \end{equation}
136 or in a more general case as
137 \begin{equation}\label{NEUMAN 1}
138 \kappa \nabla T \cdot n = 0
139 \end{equation}
140 where $n$ is the outer normal field \index{outer normal field} at the surface of the domain.
141 For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the
142 dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of
143 the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal
144 derivative is $T\hackscore{,i} n\hackscore i$.};
145 \begin{equation}
146 \nabla T \cdot n = \frac{\partial T}{\partial n} \; .
147 \end{equation}
148 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE.
149
150 The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod
151 and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}.
152 It is a nature of a boundary value problem that it allows to make statements on the solution in the
153 interior of the domain from information known on the boundary only. In most cases
154 we use the term partial differential equation but in fact mean a boundary value problem.
155 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that
156 at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set.
157
158 Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate.
159 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
160 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is;
161 \begin{equation}\label{NEUMAN 2}
162 n\cdot A \cdot\nabla u = 0
163 \end{equation}
164 which is used everywhere on the boundary. Again $n$ denotes the outer normal field.
165 Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}.
166 With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this
167 condition translates into
168 \begin{equation}\label{NEUMAN 2b}
169 \kappa \frac{\partial T}{\partial x} = 0
170 \end{equation}
171 for the right hand side of the rod. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We will discuss the Dirichlet boundary condition later.
172
173 \subsection{Outline of the Implementation}
174 \label{sec:outline}
175 To solve the heat diffusion equation (equation \refEq{eqn:hd}) we will 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 will discuss later in details will have four major steps. Firstly we need to define the domain where we want to
176 calculate the temperature. For our problem this is the iron rod which has a rectangular shape. Secondly we need to define the PDE
177 we need to solve in each time step to get the updated temperature. Thirdly we need to define the 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. As a work flow this takes the form;
178 \begin{enumerate}
179 \item create domain
180 \item create PDE
181 \item while end time not reached:
182 \begin{enumerate}
183 \item set PDE coefficients
184 \item solve PDE
185 \item update time marker
186 \end{enumerate}
187 \item end of calculation
188 \end{enumerate}
189 In the terminology of \pyt the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it defined by it usage and features
190 rather than its actual representation. So we will create a domain object to describe the geometry of our iron rod. The main feature
191 of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature
192 on a domain. In fact the domain object has many more features - most of them you will
193 never use and do not need to understand. Similar a PDE object is defined by the fact that we can define the coefficients of the PDE and solve the PDE. At a
194 later stage you may use more advanced features of the PDE class but you need to worry about them only at the point when you use them.
195
196
197 \begin{figure}[t]
198 \centering
199 \includegraphics[width=6in]{figures/functionspace.pdf}
200 \label{fig:fs}
201 \caption{\esc domain construction overview}
202 \end{figure}
203
204 \subsection{The Domain Constructor in \esc}
205 \label{ss:domcon}
206 It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again
207 from the user's point of view the representation of these spatially distributed values is not relevant.
208
209 There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is
210 a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number
211 elements or cells to be used along the length and height, see Figure~\ref{fig:fs}. Any specially distributed value
212 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}i for details.}. Therefore we will have access to an approximation of the true PDE solution only.
213 The quality of the approximation depends - besides other factors- mainly on the number of elements being used. In fact, the
214 approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of
215 elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage.
216
217 In general, one can thinks about a domain object as a composition of nodes and elements.
218 As shown in Figure~\ref{fig:fs}, an element is defined by the nodes used to describe its vertices.
219 To represent spatial distributed values the user can use
220 the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain.
221 The different approach used to represent values is called \textbf{function space} and is attached to all objects
222 in \esc representing a spatial distributed value such as the solution of a PDE. The three
223 function spaces we will use at the moment are;
224 \begin{enumerate}
225 \item the nodes, called by \verb|ContinuousFunction(domain)| ;
226 \item the elements/cells, called by \verb|Function(domain)| ; and
227 \item the boundary, called by \verb|FunctionOnBoundary(domain)| .
228 \end{enumerate}
229 A function space object such as \verb|ContinuousFunction(domain)| has the method \verb|getX| attached to it. This method returns the
230 location of the so-called \textbf{sample points} used to represent values with the particular function space attached to it. So the
231 call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the nodes used to describe the domain while
232 the \verb|Function(domain).getX()| returns the coordinates of numerical integration points within elements, see
233 Figure~\ref{fig:fs}.
234
235 This distinction between different representations of spatial distributed values
236 is important in order to be able to vary the degrees of smoothness in a PDE problem.
237 The coefficients of a PDE need not be continuous thus this qualifies as a \verb|Function()| type.
238 On the other hand a temperature distribution must be continuous and needs to be represented with a \verb|ContinuousFunction()| function space.
239 An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object.
240 \esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verb|FunctionOnBoundary()|
241 or \verb|Function()|. On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() .
242 These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed.
243
244 Later in this introduction we will discuss how
245 to define specific areas of geometry with different materials which are represented by different material coefficients such the
246 thermal conductivities $kappa$. A very powerful technique to define these types of PDE
247 coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names.
248 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}.
249
250
251 \subsection{A Clarification for the 1D Case}
252 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;
253 \begin{equation}\label{eqn:commonform2D}
254 -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}}
255 -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y}
256 -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x}
257 -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}}
258 + Du = f
259 \end{equation}
260 Notice that for the higher dimensional case $A$ becomes a matrix. It is also
261 important to notice that the usage of the Nabla operator creates
262 a compact formulation which is also independent from the spatial dimension.
263 So to make the general PDE \refEq{eqn:commonform2D} one dimensional as
264 shown in \refEq{eqn:commonform} we need to set
265 \begin{equation}
266 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0
267 \end{equation}
268
269
270 \subsection{Developing a PDE Solution Script}
271 \label{sec:key}
272 \sslist{onedheatdiffbase.py}
273 We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules.
274 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.}
275 that we will require.
276 \begin{python}
277 from esys.escript import *
278 # This defines the LinearPDE module as LinearPDE
279 from esys.escript.linearPDEs import LinearPDE
280 # This imports the rectangle domain function from finley.
281 from esys.finley import Rectangle
282 # A useful unit handling package which will make sure all our units
283 # match up in the equations under SI.
284 from esys.escript.unitsSI import *
285 \end{python}
286 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.
287
288 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 iron rod, we will simply use a rectangular model.
289
290 Using a rectangular model simplifies our rod which would be a \textit{3D} object, into a single dimension. The iron rod will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the rod because a cylinder is symmetrical about its centre. 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.
291 \begin{python}
292 #Domain related.
293 mx = 1*m #meters - model length
294 my = .1*m #meters - model width
295 ndx = 100 # mesh steps in x direction
296 ndy = 1 # mesh steps in y direction - one dimension means one element
297 \end{python}
298 The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as:
299 \begin{python}
300 #PDE related
301 rho = 7874. *kg/m**3 #kg/m^{3} density of iron
302 cp = 449.*J/(kg*K) # J/Kg.K thermal capacity
303 rhocp = rho*cp
304 kappa = 80.*W/m/K # watts/m.Kthermal conductivity
305 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source
306 T0=100 * Celsius # initial temperature at left end of rod
307 Tref=20 * Celsius # base temperature
308 \end{python}
309 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:
310 \begin{python}
311 t=0 * day #our start time, usually zero
312 tend=1. * day # - time to end simulation
313 outputs = 200 # number of time steps required.
314 h=(tend-t)/outputs #size of time step
315 #user warning statement
316 print "Expected Number of time outputs is: ", (tend-t)/h
317 i=0 #loop counter
318 \end{python}
319 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 rod as:
320 \begin{python}
321 #generate domain using rectangle
322 rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy)
323 \end{python}
324 \verb rod now describes a domain in the manner of Section \ref{ss:domcon}. There is an easy way to extract
325 the coordinates of the nodes used to describe the domain \verb|rod| using the
326 domain property function \verb|getX()| . This function sets the vertices of each cell as finite points to solve in the solution. If we let \verb|x| be these finite points, then;
327 \begin{python}
328 #extract data points - the solution points
329 x=rod.getX()
330 \end{python}
331 The data locations of specific function spaces can be returned in a similar manner by extracting the relevant function space from the domain followed by the \verb|.getX()| method.
332
333 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.
334 \begin{python}
335 mypde=LinearSinglePDE(rod)
336 A=zeros((2,2)))
337 A[0,0]=kappa
338 q=whereZero(x[0])
339 mypde.setValue(A=A, D=rhocp/h, q=q, r=T0)
340 \end{python}
341 The argument \verb|q| has not been discussed yet: In fact the arguments \verb|q| and \verb|r| are used to define
342 Dirichlet boundary condition as discussed in Section~\ref{SEC BOUNDARY COND}. In the \esc
343 PDE from the argument \verb|q| indicates by a positive value for which nodes we want to apply a
344 Dirichlet boundary condition, ie. where we want to prescribe the value of the PDE solution
345 rather then using the PDE. The actually value for the solution to be taken is set by the argument \verb|r|.
346 In our case we want to keep the initial temperature $T0$ on the left face of the rode for all times. Notice,
347 that as set to a constant value \verb|r| is assumed to have the same value
348 at all nodes, however only the value at those nodes marked by a positive value by \verb|q| are actually used.
349
350 In order to set \verb|q| we use
351 \verb|whereZero| function. The function returns the value (positive) one for those data points (=nodes) where the argument is equal to zero and otherwise returns (non-positive) value zero.
352 As \verb|x[0]| given the $x$-coordinates of the nodes for the domain,
353 \verb|whereZero(x[0])| gives the value $1$ for the nodes at the left end of the rod $x=x_0=0$ and
354 zero elsewhere which is exactly what we need.
355
356 In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric.
357 Symmetry of a PDE is defined by;
358 \begin{equation}\label{eqn:symm}
359 A\hackscore{jl}=A\hackscore{lj}
360 \end{equation}
361 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;
362 \begin{python}
363 myPDE.setSymmetryOn()
364 \end{python}
365 Next we need to establish the initial temperature distribution \verb|T|. We want to have this initial
366 value to be \verb|Tref| except at the left end of the rod $x=0$ where we have the temperature \verb|T0|. We use;
367 \begin{python}
368 # ... set initial temperature ....
369 T = T0*whereZero(x[0])+Tref*(1-whereZero(x[0]))
370 \end{python}
371 Finally we will initialize 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.
372 \begin{python}
373 while t < tend:
374 i+=1 #increment the counter
375 t+=h #increment the current time
376 mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
377 T=mypde.getSolution() #get the PDE solution
378 totE = integrate(rhocp*T) #get the total heat (energy) in the system
379 \end{python}
380 The last statement in this script calculates the total energy in the system as volume integral
381 of $\rho \c_p T$ over the rod.
382
383 \subsection{Plotting the Total Energy}
384 \sslist{onedheatdiff001.py}
385
386 \esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualization.
387 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.
388 The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots.
389 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/}}
390 which bases the \verb|VTK| toolkit. We will discuss the usage of \verb|VTK| based
391 visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE.
392
393 For our simple problem we have two plotting tasks: Firstly we are interested in showing the
394 behavior of the total energy over time and secondly in how the temperature distribution within the rod is
395 developing over time. Lets start with the first task.
396
397 \begin{figure}
398 \begin{center}
399 \includegraphics[width=4in]{figures/ttrodpyplot150}
400 \caption{Total Energy in Rod over Time (in seconds).}
401 \label{fig:onedheatout1}
402 \end{center}
403 \end{figure}
404
405 The trick is to create a record of the time marks and the corresponding total energies observed.
406 \pyt provides the concept of lists for this. Before
407 the time loop is opened we create empty lists for the time marks \verb|t_list| and the total energies \verb|E_list|.
408 After the new temperature as been calculated by solving the PDE we append the new time marker and total energy
409 to the corresponding list using the \verb|append| method. With these modifications the script looks as follows:
410 \begin{python}
411 t_list=[]
412 E_list=[]
413 # ... start iteration:
414 while t<tend:
415 t+=h
416 mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients
417 T=mypde.getSolution() #get the PDE solution
418 totE=integrate(rhocp*T)
419 t_list.append(t) # add current time mark to record
420 E_list.append(totE) # add current total energy to record
421 \end{python}
422 To plot $t$ over $totE$ we use the \mpl a module contained within \pylab which needs to be loaded before used;
423 \begin{python}
424 import pylab as pl # plotting package.
425 \end{python}
426 Here we are not using the \verb|from pylab import *| in order to name clashes for function names
427 with \esc.
428
429 The following statements are added to the script after the time loop has been completed;
430 \begin{python}
431 pl.plot(t_list,E_list)
432 pl.title("Total Energy")
433 pl.savefig("totE.png")
434 \end{python}
435 The first statement hands over the time marks and corresponding total energies to the plotter.
436 The second statement is setting the title for the plot. The last statement renders the plot and writes the
437 result into the file \verb|totE.png| which can be displayed by (almost) any image viewer. Your result should look
438 similar to Figure~\ref{fig:onedheatout1}.
439
440 \subsection{Plotting the Temperature Distribution}
441 \sslist{onedheatdiff001b.py}
442 For plotting the spatial distribution of the temperature we need to modify the strategy we have used
443 for the total energy. Instead of producing a final plot at the end we will generate a
444 picture at each time step which can be browsed as slide show or composed to a movie.
445 The first problem we encounter is that if we produce an image in each time step we need
446 to make sure that the images previously generated are not overwritten.
447
448 To develop an incrementing file name we can use the following convention. It is convenient to
449 put all image file showing the same variable - in our case the temperature distribution -
450 into a separate directory. As part of the \verb|os| module\footnote{The \texttt{os} module provides
451 a powerful interface to interact with the operating system, see \url{http://docs.python.org/library/os.html}.} \pyt
452 provides the \verb|os.path.join| command to build file and
453 directory names in a platform independent way. Assuming that
454 \verb|save_path| is name of directory we want to put the results the command is;
455 \begin{python}
456 import os
457 os.path.join(save_path, "tempT%03d.png"%i )
458 \end{python}
459 where \verb|i| is the time step counter.
460 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;
461 \begin{verbatim}
462 save_path = "data"
463 \end{verbatim}
464 while a sub-folder of \verb data called \verb onedheatdiff001 would be defined by;
465 \begin{verbatim}
466 save_path = os.path.join("data","onedheatdiff001")
467 \end{verbatim}
468 The second argument of \verb join \xspace contains a string which is the filename or subdirectory name. We can use the operator \verb|%| to increment our file names with the value \verb|i| denoting a incrementing counter. The substring \verb %03d does this by defining the following parameters;
469 \begin{itemize}
470 \item \verb 0 becomes the padding number;
471 \item \verb 3 tells us the amount of padding numbers that are required; and
472 \item \verb d indicates the end of the \verb % operator.
473 \end{itemize}
474 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;
475 \begin{verbatim}
476 rodpyplot.png
477 rodpyplot.png
478 rodpyplot.png
479 ...
480 \end{verbatim}
481 and so on.
482
483 A sub-folder check/constructor is available in \esc. The command;
484 \begin{verbatim}
485 mkDir(save_path)
486 \end{verbatim}
487 will check for the existence of \verb save_path and if missing, make the required directories.
488
489 We start by modifying our solution script from before.
490 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 data points used to represent
491 the temperature as a \pyt list of tuples. As a solution of a PDE
492 the temperature has the \verb|Solution(rod)| function space attribute. We use
493 the \verb|getX()| method to get the coordinates of the data points as an \esc object
494 which is then converted to a \numpy array. The $x$ component is then extracted via an array slice to the variable \verb|plx|;
495 \begin{python}
496 import numpy as np # array package.
497 #convert solution points for plotting
498 plx = Solution(rod).getX().toListOfTuples()
499 plx = np.array(plx) # convert to tuple to numpy array
500 plx = plx[:,0] # extract x locations
501 \end{python}
502
503 \begin{figure}
504 \begin{center}
505 \includegraphics[width=4in]{figures/rodpyplot001}
506 \includegraphics[width=4in]{figures/rodpyplot050}
507 \includegraphics[width=4in]{figures/rodpyplot200}
508 \caption{Temperature ($T$) distribution in rod at time steps $1$, $50$ and $200$.}
509 \label{fig:onedheatout}
510 \end{center}
511 \end{figure}
512
513 For each time step we will generate a plot of the temperature distribution and save each to a file. We use the same
514 techniques provided by \mpl as we have used to plot the total energy over time.
515 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.
516 Finally, the figure is saved to a \verb|*.png| file and cleared for the following iteration.
517 \begin{python}
518 # ... start iteration:
519 while t<tend:
520 ....
521 T=mypde.getSolution() #get the PDE solution
522 tempT = T.toListOfTuples() # convert to a tuple
523 pl.plot(plx,tempT) # plot solution
524 # set scale (Temperature should be between Tref and T0)
525 pl.axis([0,mx,Tref*.9,T0*1.1])
526 # add title
527 pl.title("Temperature across rod at time %e minutes"%(t/minutes))
528 #save figure to file
529 pl.savefig(os.path.join(save_path,"tempT","rodpyplot%03d.png") %i)
530 \end{python}
531 Some results are shown in Figure~\ref{fig:onedheatout}.
532
533 \subsection{Make a video}
534 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.
535 \begin{python}
536 # compile the *.png files to create a *.avi videos that show T change
537 # with time. This operation uses Linux mencoder. For other operating
538 # systems it is possible to use your favorite video compiler to
539 # convert image files to videos.
540
541 os.system("mencoder mf://"+save_path+"/tempT"+"/*.png -mf type=png:\
542 w=800:h=600:fps=25 -ovc lavc -lavcopts vcodec=mpeg4 -oac copy -o \
543 onedheatdiff001tempT.avi")
544 \end{python}
545

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