 # Annotation of /trunk/doc/cookbook/example07.tex

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Pressure wave working, accelleration example, new entries to cookbook for example 7.

 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 or longitudinal style wave motion is not possible. The obvious example is sound 19 waves. 20 21 The acoustic wave equation is; 22 \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 where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. 27 28 29 30 \section{Numerical Solution Stability} 31 Unfortunately, the wave equation is belongs to a class of equations called 32 \textbf{stiff} PDEs. This types of equations can be difficult to solve 33 numerically as they tend to oscilate about the exact solution and can 34 eventually blow up. To counter this problem, explicitly stable schemes like the 35 backwards Euler method are required. In terms of the wave equation, the 36 analytical wave must not propagate faster than the numerical wave is able to, 37 and in general, needs to be much slower than the numerical wave. 38 39 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If 40 a wave enters with a propagation velocity of 100m/s then the travel time for 41 the wave between each node will be 0.01 seconds. The time step, must therefore 42 be significantly less than this. Of the order $10E-4$ would be appropriate. 43 44 This requirement for very small step sizes makes stiff equations difficult to 45 solve numerically due to the large number of time iterations required in each 46 solution. Models with very high velocities and fine meshes will be the worst 47 affected by this problem. 48 49 50 \section{Displacement Solution} 51 \sslist{example07a.py} 52 53 We begin the solution to this PDE with the centred difference formula for the 54 second derivative; 55 \begin{equation} 56 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} 57 \label{eqn:centdiff} 58 \end{equation} 59 substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ 60 in \refEq{eqn:acswave}; 61 \begin{equation} 62 \nabla ^2 p - \frac{1}{c^2h^2} \left[p_{(t+1)} - 2p_{(t)} + p_{(t-1)} \right] 63 = 0 64 \label{eqn:waveu} 65 \end{equation} 66 Rearranging for $p_{(t+1)}$; 67 \begin{equation} 68 p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)} - p_{(t-1)} 69 \end{equation} 70 this can be compared with the general form of the \modLPDE module and it 71 becomes clear that $D=1$, $X=-c^2 h^2 \nabla ^2 p_{(t)}$ and $Y=2p_{(t)} - 72 p_{(t-1)}$. 73 74 \section{Acceleration Solution} 75 \sslist{example07b.py} 76 77 An alternative method is to solve for the acceleration $\frac{\partial ^2 78 p}{\partial t^2}$ directly, and derive the the displacement solution from the 79 PDE solution. \refEq{eqn:waveu} is thus modified; 80 \begin{equation} 81 \nabla ^2 p - \frac{1}{c^2} a = 0 82 \label{eqn:wavea} 83 \end{equation} 84 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p_{(t)}$. 85 After each iteration the displacement is re-evaluated via; 86 \begin{equation} 87 p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a 88 \end{equation} 89 90