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Discretization of SPML

Take

$\displaystyle \frac{\partial p_x}{\partial t}+d(x)p_x = v^2 \frac{\partial v_x}{\partial x}
$

for an example. Using the 2nd-order approximation in Eq. (31), we expand it at the time $ (k+\frac{1}{2})\Delta t$ and the point $ [ix\Delta x, iz\Delta z]$

\begin{displaymath}\begin{split}\frac{p_x^{k+1}[ix, iz]-p_x^{k}[ix, iz]}{\Delta ...
...-v_x^{k+\frac{1}{2}}[ix-\frac{1}{2}, iz]}{\Delta x} \end{split}\end{displaymath} (41)

That is to say,

$\displaystyle p_x^{k+1}[ix,iz]=\frac{1-0.5\Delta td[ix]}{1+0.5\Delta t d[ix]}p_...
...v_x^{k+1\frac{1}{2}}[ix+\frac{1}{2},iz]-v_x^{k+\frac{1}{2}}[ix-\frac{1}{2},iz])$ (42)

At time $ k\Delta t$ and $ [ix+\frac{1}{2},iz]$ , we expand

$\displaystyle \frac{\partial v_x}{\partial t}+d(x)v_x = \frac{\partial p}{\partial x}
$

as

$\displaystyle \frac{v_x^{k+\frac{1}{2}}[ix+\frac{1}{2}, iz]-v_x^{k-\frac{1}{2}}...
...c{1}{2}}[ix+\frac{1}{2}, iz]}{2}=\frac{p^{k}[ix+1, iz]-p^{k}[ix, iz]}{\Delta x}$ (43)

Thus, we have

$\displaystyle v_x^{k+\frac{1}{2}}[ix+\frac{1}{2},iz]=\frac{1-0.5\Delta td[ix]}{...
...\frac{1}{1+0.5\Delta t d[ix]}\frac{\Delta t}{\Delta x}(p^k[ix+1,iz]-p^k[ix,iz])$ (44)

In summary,

\begin{equation*}\left\{ \begin{split}&v_x^{k+\frac{1}{2}}[ix+\frac{1}{2},iz]=\f...
...1}[ix,iz]=p_x^{k+1}[ix,iz]+p_z^{k+1}[ix,iz]\\ \end{split} \right.\end{equation*} (45)

If we define:

$\displaystyle b'=\frac{1-0.5\Delta t d}{1+0.5\Delta t d}, b=\mathrm{exp}(-\Delta t d)$ (46)

we can easily find that $ b'$ is a good approximation of $ b$ up to 2nd order, allowing for the 2nd order Pade expansion:

$\displaystyle \mathrm{exp}(z)\approx \frac{1+0.5z}{1-0.5z}$ (47)

Then, we have

$\displaystyle 1-b\approx 1-b'=\frac{\Delta t d}{1+0.5\Delta t d}$ (48)


next up previous [pdf]

Next: Discretization of NPML Up: Discretization Previous: Higher-order approximation of staggered-grid

2021-08-31