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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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% |
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% Copyright (c) 2003-2010 by University of Queensland |
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% Earth Systems Science Computational Center (ESSCC) |
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% http://www.uq.edu.au/esscc |
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% |
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% Primary Business: Queensland, Australia |
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% Licensed under the Open Software License version 3.0 |
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% http://www.opensource.org/licenses/osl-3.0.php |
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% |
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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The acoustic wave equation governs the propagation of pressure waves. Wave |
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types that obey this law tend to travel in liquids or gases where shear waves |
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or longitudinal style wave motion is not possible. An obvious example is sound |
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waves. |
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|
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The acoustic wave equation is defined as; |
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\begin{equation} |
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\nabla ^2 p - \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0 |
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\label{eqn:acswave} |
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\end{equation} |
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where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. |
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\section{The Laplacian in \esc} |
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The Laplacian opperator which can be written as $\Delta$ or $\nabla^2$ is |
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calculated via the divergence of the gradient of the object, which is in this |
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example $p$. Thus we can write; |
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\begin{equation} |
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\nabla^2 p = \nabla \cdot \nabla p = \frac{\partial^2 p}{\partial |
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x^2\hackscore{i}} |
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\label{eqn:laplacian} |
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\end{equation} |
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For the two dimensional case in Cartesian coordinates \refEq{eqn:laplacian} |
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becomes; |
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\begin{equation} |
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\nabla^2 p = \frac{\partial^2 p}{\partial x^2} |
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+ \frac{\partial^2 p}{\partial y^2} |
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\end{equation} |
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In \esc the Laplacian is calculated using the divergence representation and the |
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intrinsic functions \textit{grad()} and \textit{trace()}. The fucntion |
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\textit{grad{}} will return the spatial gradients of an object. |
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For a rank 0 solution, this is of the form; |
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\begin{equation} |
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\nabla p = \left[ |
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\frac{\partial p}{\partial x \hackscore{0}}, |
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\frac{\partial p}{\partial x \hackscore{1}} |
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\right] |
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\label{eqn:grad} |
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\end{equation} |
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Larger ranked solution objects will return gradient tensors. For example, a |
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pressure field which acts in the directions $p \hackscore{0}$ and $p |
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\hackscore{1}$ would return; |
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\begin{equation} |
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\nabla p = \begin{bmatrix} |
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\frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} & |
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\frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\ |
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\frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} & |
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\frac{\partial p \hackscore{1}}{\partial x \hackscore{1}} |
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\end{bmatrix} |
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\label{eqn:gradrank1} |
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\end{equation} |
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\refEq{eqn:grad} corresponds to the Linear PDE general form value |
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$X$. Notice however that the gernal form contains the term $X \hackscore{i,j}$, |
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hence for a rank 0 object there is no need to do more than calculate the |
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gradient and submit it to the solver. In the case of the rank 1 or greater |
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object, it is nesscary to calculate the trace also. This is the sum of the |
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diagonal in \refeq{eqn:gradrank1}. |
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Thus when solving for equations containing the Laplacian one of two things must |
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be completed. If the object \verb p is less than rank 1 the gradient is |
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calculated via; |
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\begin{verbatim} |
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gradient=grad(p) |
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\end{verbatim} |
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and if the object is greater thank or equal to a rank 1 tensor, the trace of |
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the gradient is calculated. |
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\begin{verbatim} |
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gradient=trace(grad(p)) |
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\end{verbatim} |
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These valuse can then be submitted to the PDE solver via the general form term |
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$X$. The Laplacian is then computed in the solution process by taking the |
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divergence of $X$. |
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\section{Numerical Solution Stability} |
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Unfortunately, the wave equation belongs to a class of equations called |
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\textbf{stiff} PDEs. These types of equations can be difficult to solve |
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numerically as they tend to oscilate about the exact solution and can |
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eventually fail. To counter this problem, explicitly stable schemes like |
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the backwards Euler method are required. There are two variables which must be |
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considered for stability when numerically trying to solve the wave equation. |
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\begin{equation} \label{eqn:freqvel} |
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f=\frac{v}{\lambda} |
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\end{equation} |
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Velocity is one of these variables. For stability the |
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analytical wave must not propagate faster than the numerical wave is able to, |
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and in general, needs to be much slower than the numerical wave. |
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For example, a line 100m long is discretised into 1m intervals or 101 nodes. If |
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a wave enters with a propagation velocity of 100m/s then the travel time for |
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the wave between each node will be 0.01 seconds. The time step, must therefore |
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be significantly less than this. Of the order $10E-4$ would be appropriate. |
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The wave frequency content also plays a part in numerical stability. The |
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nyquist-sampling theorem states that a signals bandwidth content will be |
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accurately represented when an equispaced sampling rate $f \hackscore{n}$ is |
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equal to or greater than twice the maximum frequency of the signal |
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$f\hackscore{s}$, or; |
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\begin{equation} \label{eqn:samptheorem} |
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f\hackscore{n} \geqslant f\hackscore{s} |
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\end{equation} |
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For example a 50Hz signal will require a sampling rate greater than 100Hz or |
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one sample every 0.01 seconds. The wave equation relies on a spatial frequency, |
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thus the sampling theorem in this case applies to the solution mesh spacing. In |
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this way, the frequency content of the input signal directly affects the time |
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discretisation of the problem. |
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To accurately model the wave equation with high resolutions and velocities |
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means that very fine spatial and time discretisation is necessary for most |
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problems. |
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This requirement makes the wave equation arduous to |
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solve numerically due to the large number of time iterations required in each |
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solution. Models with very high velocities and frequencies will be the worst |
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effected by this problem. |
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\section{Displacement Solution} |
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\sslist{example07a.py} |
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We begin the solution to this PDE with the centred difference formula for the |
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second derivative; |
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\begin{equation} |
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f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} |
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\label{eqn:centdiff} |
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\end{equation} |
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substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ |
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in \refEq{eqn:acswave}; |
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\begin{equation} |
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\nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} + |
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p\hackscore{(t-1)} \right] |
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= 0 |
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\label{eqn:waveu} |
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\end{equation} |
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Rearranging for $p_{(t+1)}$; |
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\begin{equation} |
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p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} - |
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p\hackscore{(t-1)} |
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\end{equation} |
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this can be compared with the general form of the \modLPDE module and it |
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becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and |
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$Y=2p_{(t)} - p_{(t-1)}$. |
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The solution script is similar to other that we have created in previous |
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chapters. The general steps are; |
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\begin{enumerate} |
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\item The necessary libraries must be imported. |
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\item The domain needs to be defined. |
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\item The time iteration and control parameters need to be defined. |
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\item The PDE is initialised with source and boundary conditions. |
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\item The time loop is started and the PDE is solved at consecutive time steps. |
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\item All or select solutions are saved to file for visualisation lated on. |
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\end{enumerate} |
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Parts of the script which warrant more attention are the definition of the |
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source, visualising the source, the solution time loop and the VTK data export. |
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\subsection{Pressure Sources} |
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As the pressure is a scalar, one need only define the pressure for two |
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time steps prior to the start of the solution loop. Two known solutions are |
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required because the wave equation contains a double partial derivative with |
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respect to time. This is often a good opportunity to introduce a source to the |
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solution. This model has the source located at it's centre. The source should |
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be smooth and cover a number of samples to satisfy the frequency stability |
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criterion. Small sources will generate high frequency signals. Here, the source |
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is defined by a cosine function. |
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\begin{verbatim} |
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U0=0.01 # amplitude of point source |
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xc=[500,500] #location of point source |
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# define small radius around point xc |
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src_radius = 30 |
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# for first two time steps |
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u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*whereNegative(length(x-xc)-src_radi |
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us) |
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u_m1=u |
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\end{verbatim} |
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When using a rectangular domain |
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\section{Acceleration Solution} |
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\sslist{example07b.py} |
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An alternative method is to solve for the acceleration $\frac{\partial ^2 |
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p}{\partial t^2}$ directly, and derive the the displacement solution from the |
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PDE solution. \refEq{eqn:waveu} is thus modified; |
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\begin{equation} |
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\nabla ^2 p - \frac{1}{c^2} a = 0 |
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\label{eqn:wavea} |
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\end{equation} |
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and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$. |
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After each iteration the displacement is re-evaluated via; |
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\begin{equation} |
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p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a |
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\end{equation} |
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For \esc, the acceleration solution is prefered as it allows the use of matrix |
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lumping. Lumping or mass lumping as it is sometimes known, is the process of |
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aggressively approximating the density elements of a mass matrix into the main |
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diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix |
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inversion. As a result, Lumping can significantly reduce the computational |
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requirements of a problem. |
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To turn lumping on in \esc one can use the command; |
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\begin{verbatim} |
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mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING) |
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\end{verbatim} |
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It is also possible to check if lumping is set using; |
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\begin{verbatim} |
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print mypde.isUsingLumping() |
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\end{verbatim} |
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\section{Stability Investigation} |
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It is now prudent to investigate the stability limitations of this problem. |
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First, we let the frequency content of the source be very small. If the radius |
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of the source which equals the wavelength is 5 meters, than the frequency is |
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the inverse of the wavelength The velocity is $c=380.0ms^{-1}$ thus the source |
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frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. The sampling |
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frequency must be at least twice this. Assuming a rectangular equispaced grid, |
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the sampling interval is $\Delta x = \frac{1000.0}{400} = 2.5$ and the sampling |
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frequency $f\hackscore{s}=\frac{380.0}{2.5}=152$ this is just equal to the |
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required rate satisfying \refeq{eqn:samptheorem}. |
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