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Sampling theorem, stability requirements, Laplacian discussion, Pressure wave PDE

 1 ahallam 3003 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2010 by University of Queensland 5 % Earth Systems Science Computational Center (ESSCC) 6 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 11 % 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 13 14 15 16 The acoustic wave equation governs the propagation of pressure waves. Wave 17 types that obey this law tend to travel in liquids or gases where shear waves 18 ahallam 3004 or longitudinal style wave motion is not possible. An obvious example is sound 19 ahallam 3003 waves. 20 21 ahallam 3004 The acoustic wave equation is defined as; 22 ahallam 3003 \begin{equation} 23 \nabla ^2 p - \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0 24 \label{eqn:acswave} 25 \end{equation} 26 ahallam 3004 where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. 27 ahallam 3003 28 ahallam 3004 \section{The Laplacian in \esc} 29 The Laplacian opperator which can be written as $\Delta$ or $\nabla^2$ is 30 calculated via the divergence of the gradient of the object, which is in this 31 example $p$. Thus we can write; 32 \begin{equation} 33 \nabla^2 p = \nabla \cdot \nabla p = \frac{\partial^2 p}{\partial 34 x^2\hackscore{i}} 35 \label{eqn:laplacian} 36 \end{equation} 37 For the two dimensional case in Cartesian coordinates \refEq{eqn:laplacian} 38 becomes; 39 \begin{equation} 40 \nabla^2 p = \frac{\partial^2 p}{\partial x^2} 41 + \frac{\partial^2 p}{\partial y^2} 42 \end{equation} 43 ahallam 3003 44 ahallam 3004 In \esc the Laplacian is calculated using the divergence representation and the 45 intrinsic functions \textit{grad()} and \textit{trace()}. The fucntion 46 \textit{grad{}} will return the spatial gradients of an object. 47 For a rank 0 solution, this is of the form; 48 \begin{equation} 49 \nabla p = \left[ 50 \frac{\partial p}{\partial x \hackscore{0}}, 51 \frac{\partial p}{\partial x \hackscore{1}} 52 \right] 53 \label{eqn:grad} 54 \end{equation} 55 Larger ranked solution objects will return gradient tensors. For example, a 56 pressure field which acts in the directions $p \hackscore{0}$ and $p 57 \hackscore{1}$ would return; 58 \begin{equation} 59 \nabla p = \begin{bmatrix} 60 \frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} & 61 \frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\ 62 \frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} & 63 \frac{\partial p \hackscore{1}}{\partial x \hackscore{1}} 64 \end{bmatrix} 65 \label{eqn:gradrank1} 66 \end{equation} 67 ahallam 3003 68 ahallam 3004 \refEq{eqn:grad} corresponds to the Linear PDE general form value 69 $X$. Notice however that the gernal form contains the term $X \hackscore{i,j}$, 70 hence for a rank 0 object there is no need to do more than calculate the 71 gradient and submit it to the solver. In the case of the rank 1 or greater 72 object, it is nesscary to calculate the trace also. This is the sum of the 73 diagonal in \refeq{eqn:gradrank1}. 74 75 Thus when solving for equations containing the Laplacian one of two things must 76 be completed. If the object \verb p is less than rank 1 the gradient is 77 calculated via; 78 \begin{verbatim} 79 gradient=grad(p) 80 \end{verbatim} 81 and if the object is greater thank or equal to a rank 1 tensor, the trace of 82 the gradient is calculated. 83 \begin{verbatim} 84 gradient=trace(grad(p)) 85 \end{verbatim} 86 87 These valuse can then be submitted to the PDE solver via the general form term 88 $X$. The Laplacian is then computed in the solution process by taking the 89 divergence of $X$. 90 91 ahallam 3003 \section{Numerical Solution Stability} 92 ahallam 3004 Unfortunately, the wave equation belongs to a class of equations called 93 \textbf{stiff} PDEs. These types of equations can be difficult to solve 94 ahallam 3003 numerically as they tend to oscilate about the exact solution and can 95 ahallam 3004 eventually fail. To counter this problem, explicitly stable schemes like 96 the backwards Euler method are required. There are two variables which must be 97 considered for stability when numerically trying to solve the wave equation. 98 99 \begin{equation} \label{eqn:freqvel} 100 f=\frac{v}{\lambda} 101 \end{equation} 102 103 104 Velocity is one of these variables. For stability the 105 ahallam 3003 analytical wave must not propagate faster than the numerical wave is able to, 106 and in general, needs to be much slower than the numerical wave. 107 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If 108 a wave enters with a propagation velocity of 100m/s then the travel time for 109 the wave between each node will be 0.01 seconds. The time step, must therefore 110 be significantly less than this. Of the order $10E-4$ would be appropriate. 111 112 ahallam 3004 The wave frequency content also plays a part in numerical stability. The 113 nyquist-sampling theorem states that a signals bandwidth content will be 114 accurately represented when an equispaced sampling rate $f \hackscore{n}$ is 115 equal to or greater than twice the maximum frequency of the signal 116 $f\hackscore{s}$, or; 117 \begin{equation} \label{eqn:samptheorem} 118 f\hackscore{n} \geqslant f\hackscore{s} 119 \end{equation} 120 For example a 50Hz signal will require a sampling rate greater than 100Hz or 121 one sample every 0.01 seconds. The wave equation relies on a spatial frequency, 122 thus the sampling theorem in this case applies to the solution mesh spacing. In 123 this way, the frequency content of the input signal directly affects the time 124 discretisation of the problem. 125 126 To accurately model the wave equation with high resolutions and velocities 127 means that very fine spatial and time discretisation is necessary for most 128 problems. 129 This requirement makes the wave equation arduous to 130 ahallam 3003 solve numerically due to the large number of time iterations required in each 131 ahallam 3004 solution. Models with very high velocities and frequencies will be the worst 132 effected by this problem. 133 ahallam 3003 134 \section{Displacement Solution} 135 \sslist{example07a.py} 136 137 We begin the solution to this PDE with the centred difference formula for the 138 second derivative; 139 \begin{equation} 140 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} 141 \label{eqn:centdiff} 142 \end{equation} 143 substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ 144 in \refEq{eqn:acswave}; 145 \begin{equation} 146 ahallam 3004 \nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} + 147 p\hackscore{(t-1)} \right] 148 ahallam 3003 = 0 149 \label{eqn:waveu} 150 \end{equation} 151 Rearranging for $p_{(t+1)}$; 152 \begin{equation} 153 ahallam 3004 p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} - 154 p\hackscore{(t-1)} 155 ahallam 3003 \end{equation} 156 this can be compared with the general form of the \modLPDE module and it 157 ahallam 3004 becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and 158 $Y=2p_{(t)} - p_{(t-1)}$. 159 ahallam 3003 160 ahallam 3004 The solution script is similar to other that we have created in previous 161 chapters. The general steps are; 162 \begin{enumerate} 163 \item The necessary libraries must be imported. 164 \item The domain needs to be defined. 165 \item The time iteration and control parameters need to be defined. 166 \item The PDE is initialised with source and boundary conditions. 167 \item The time loop is started and the PDE is solved at consecutive time steps. 168 \item All or select solutions are saved to file for visualisation lated on. 169 \end{enumerate} 170 171 Parts of the script which warrant more attention are the definition of the 172 source, visualising the source, the solution time loop and the VTK data export. 173 174 \subsection{Pressure Sources} 175 As the pressure is a scalar, one need only define the pressure for two 176 time steps prior to the start of the solution loop. Two known solutions are 177 required because the wave equation contains a double partial derivative with 178 respect to time. This is often a good opportunity to introduce a source to the 179 solution. This model has the source located at it's centre. The source should 180 be smooth and cover a number of samples to satisfy the frequency stability 181 criterion. Small sources will generate high frequency signals. Here, the source 182 is defined by a cosine function. 183 \begin{verbatim} 184 U0=0.01 # amplitude of point source 185 xc=[500,500] #location of point source 186 # define small radius around point xc 187 src_radius = 30 188 # for first two time steps 189 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*whereNegative(length(x-xc)-src_radi 190 us) 191 u_m1=u 192 \end{verbatim} 193 When using a rectangular domain 194 195 ahallam 3003 \section{Acceleration Solution} 196 \sslist{example07b.py} 197 198 An alternative method is to solve for the acceleration $\frac{\partial ^2 199 p}{\partial t^2}$ directly, and derive the the displacement solution from the 200 PDE solution. \refEq{eqn:waveu} is thus modified; 201 \begin{equation} 202 \nabla ^2 p - \frac{1}{c^2} a = 0 203 \label{eqn:wavea} 204 \end{equation} 205 ahallam 3004 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$. 206 ahallam 3003 After each iteration the displacement is re-evaluated via; 207 \begin{equation} 208 ahallam 3004 p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a 209 ahallam 3003 \end{equation} 210 211 ahallam 3004 For \esc, the acceleration solution is prefered as it allows the use of matrix 212 lumping. Lumping or mass lumping as it is sometimes known, is the process of 213 aggressively approximating the density elements of a mass matrix into the main 214 diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix 215 inversion. As a result, Lumping can significantly reduce the computational 216 requirements of a problem. 217 ahallam 3003 218 ahallam 3004 To turn lumping on in \esc one can use the command; 219 \begin{verbatim} 220 mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING) 221 \end{verbatim} 222 It is also possible to check if lumping is set using; 223 \begin{verbatim} 224 print mypde.isUsingLumping() 225 \end{verbatim} 226 227 \section{Stability Investigation} 228 It is now prudent to investigate the stability limitations of this problem. 229 First, we let the frequency content of the source be very small. If the radius 230 of the source which equals the wavelength is 5 meters, than the frequency is 231 the inverse of the wavelength The velocity is $c=380.0ms^{-1}$ thus the source 232 frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. The sampling 233 frequency must be at least twice this. Assuming a rectangular equispaced grid, 234 the sampling interval is $\Delta x = \frac{1000.0}{400} = 2.5$ and the sampling 235 frequency $f\hackscore{s}=\frac{380.0}{2.5}=152$ this is just equal to the 236 required rate satisfying \refeq{eqn:samptheorem}. 237 238 239 240