# Diff of /trunk/doc/cookbook/example01.tex

revision 2861 by gross, Wed Jan 20 02:47:42 2010 UTC revision 2862 by gross, Thu Jan 21 04:45:39 2010 UTC
# Line 107  we rearrange \refEq{eqn:hddisc}; Line 107  we rearrange \refEq{eqn:hddisc};
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  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.  $t^{(0)}=0$ 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;  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    \label{ESCRIPT SET}
111  u=T^{(n)};  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)}  A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)}
113
# Line 119  T(x,0) = T\hackscore{ref} = 0 Line 119  T(x,0) = T\hackscore{ref} = 0
119  for all $x$ in the domain.  for all $x$ in the domain.
120
121  \subsection{Boundary Conditions}  \subsection{Boundary Conditions}
122  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 Neumann and Dirichlet boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively.  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.
123  A Dirichlet boundary condition is conceptually simpler and is used to prescribe a known value to the unknown - in our example the temperature - on boundary of the region of interest.  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.
124  \editor{LUTZ: This is not correct!!}  For our model problem we want to keep the initial temperature setting on the left side of the
125  For this model Dirichlet boundary conditions exist where we have applied our heat source. As the heat source is a constant, we can simulate its presence on that boundary. This is done by continuously resetting the temperature of the boundary, so that it is the same as the heat source.    iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step.

Neumann boundary conditions describe the radiation or flux that is normal to the boundary surface. This aptly describes our insulation conditions as we do not want to exert a constant temperature as with the heat source. However, we do want to prevent any loss of energy from the system.
126
127  While the flux for this model is zero, it is important to note the requirements for Neumann boundary conditions. For heat diffusion these can be described by specifying a radiation condition which prescribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional  On the other end of the iron rod we want to add an appropriate boundary condition to define insolation to prevent
128  to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$; in general terms this is;  any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing
129    the heat flux $\kappa \frac{\partial T}{\partial x}$  to zero on the right end of the rod
130    In our simplified one dimensional model this is expressed
131    in the form;
132
133   \kappa T\hackscore{,i} \hat{n}\hackscore i = \eta (T\hackscore{ref}-T)  \kappa \frac{\partial T}{\partial x}  = 0
134  \label{eqn:hdbc}
135    or in a more general case as
136    \label{NEUMAN 1}
137    \kappa \nabla T \cdot n  = 0
138
139  and simplified to our one dimensional model we have;  where $n$  is the outer normal field \index{outer normal field} at the surface of the domain.
140    For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the
141    dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of
142    the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal
143    derivative is $T\hackscore{,i} n\hackscore i$.};
144
145  \kappa \frac{\partial T}{\partial dx} \hat{n}\hackscore x = \eta (T\hackscore{ref}-T)  \nabla T \cdot n  = \frac{\partial T}{\partial n} \; .
146
147    A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE.
148
149    The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod
150    and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}.
151    It is a nature of a boundary value problem that it allows to make statements on the solution in the
152    interior of the domain from information known on the boundary only. In most cases
153    we use the term partial differential equation but in fact mean a boundary value problem.
154    It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that
155    at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set.
156
157    Conviniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate.
158    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
159    $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is;
160    \label{NEUMAN 2}
161    n\cdot A \cdot\nabla u = 0
162
163    which is used everywhere on the boundary. Again $n$ denotes the outer normal field.
164    Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}.
165    With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this
166    condition translates into
167    \label{NEUMAN 2}
168    \kappa \frac{\partial T}{\partial x} = 0
169
170  where $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium and $\hat{n}\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field} at the surface of the domain. These two conditions form a boundary value problem that has to be solved for each time step. Due to the perfect insulation in our model we can set $\eta = 0$ which results in zero flux - no energy in or out - we do not need to worry about the Neumann terms of the general form for this example.  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.
171
172
173
174
175  \subsection{A \textit{1D} Clarification}  \subsection{A \textit{1D} Clarification}
176  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;  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;

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