Discretization of SPML

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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

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

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

(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,

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