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Bibliography

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Appendix A

The 1/6-th trick

Given the filter $D_2 (k)$, defined in formula (10), we can construct an accurate approximation to the second derivative operator $-k^2$ by considering a filter ratio (another Padé-type approximation) of the form
\begin{displaymath}
-k^2 \approx \frac{D_2(k)}{1 + \beta D_2 (k)}\;,
\end{displaymath} (25)

where $\beta$ is an adjustable constant (Claerbout, 1985). The actual Padé coefficient is $\beta=1/12$. As pointed out by Francis Muir, the value of $\beta = 1/4 - 1/\pi^2 \approx 1/6.726$ gives an exact fit at the Nyquist frequency $k = \pi$. Fitting the derivative operator in the $L_1$ norm yields the value of $\beta
\approx 1/8.13$. All these approximations are shown in Figure A-1.

sixth
Figure A-1.
The second-derivative operator in the wavenumber domain and its approximations.
sixth
[pdf] [png] [sage]

Appendix B

Constructing an ``isotropic'' Laplacian operator

The problem of approximating the Laplacian operator in two dimensions not only inherits the inaccuracies of the one-dimensional finite-difference approximations, but also raises the issue of azimuthal asymmetry. For example, the usual five-point filter
\begin{displaymath}
F_5 =
\begin{array}{\vert r\vert r\vert r\vert}
\hline
0 & 1...
...\\ \hline
1 & -4 & 1 \\ \hline
0 & 1 & 0 \\ \hline
\end{array}\end{displaymath} (26)

exhibits a clear difference between the grid directions and the directions at a 45-degree angle to the grid. To overcome this unpleasant anisotropy, we can consider a slightly larger filter of the form
\begin{displaymath}
F_9 = \begin{array}{\vert r\vert r\vert r\vert}
\hline
\alph...
...gamma \\ \hline
\alpha & \gamma & \alpha \\ \hline
\end{array}\end{displaymath} (27)

where the constants $\alpha$ and $\gamma$ are to be defined. The Fourier-domain representation of filter (B-2) is
\begin{displaymath}
F_9 (k_x,k_y) = 4\,\alpha\,[\cos{k_x}\,\cos{k_y} - 1] +
2\,\gamma\,[\cos{k_x}+\cos{k_y}-2]\;,
\end{displaymath} (28)

and the isotropic filter that we can try to approximate is defined analogously to its one-dimensional equivalent, as follows:
\begin{displaymath}
F (k_x,k_y) = 2 (\cos{k} -1) = 2 (\cos{\sqrt{k_x^2+k_y^2}} - 1)\;.
\end{displaymath} (29)

Comparing equations (B-3) and (B-4), we notice that they match exactly, when either of the wavenumbers $k_x$ or $k_y$ is equal to zero, provided that
\begin{displaymath}
\alpha = \frac{1-\gamma}{2}\;.
\end{displaymath} (30)

Therefore, we can reduce the problem to estimating a single coefficient $\gamma$. Another way of expressing this conclusion is to represent filter $F_9$ in equation (B-3) as a linear combination of filter $F_5$ from equation (B-3) and its rotated version (Cole, 1994), as follows:
\begin{displaymath}
F_9 = \gamma\,
\begin{array}{\vert r\vert r\vert r\vert}
\h...
...hline
0 & -2 & 0 \\ \hline
1/2 & 0 & 1/2 \\ \hline
\end{array}\end{displaymath} (31)

With the value of $\gamma=0.5$, filter $F_9$ takes the value
\begin{displaymath}
F_9 = \begin{array}{\vert r\vert r\vert r\vert}
\hline
1/4 &...
...1/2 & -3 & 1/2 \\ \hline
1/4 & 1/2 & 1/4 \\ \hline
\end{array}\end{displaymath} (32)

and corresponds precisely to the nine-point McClellan filter (Hale, 1991a; McClellan, 1973). On the other hand, the value of $\gamma =2/3$ gives the least error in the vicinity of the zero wavenumber $k$. In this case, the filter is
\begin{displaymath}
F_9 = \begin{array}{\vert r\vert r\vert r\vert}
\hline
1/6 &...
... & -10/3 & 2/3 \\ \hline
1/6 & 2/3 & 1/6 \\ \hline
\end{array}\end{displaymath} (33)

Errors of different approximations are plotted in Figure B-1[*]

laplace
laplace
Figure B-1.
The numerical anisotropy error of different Laplacian approximations. Both the five-point Laplacian (plot a) and its rotated version (plot b) are accurate along the axes, but exhibit significant anisotropy in between at large wavenumbers. The nine-point McClellan filter (plot c) has a reduced error, while the filter with $\gamma =2/3$ (plot d) has the flattest error around the origin.
[pdf] [png] [sage]

Under the helix transform, a filter of the general form (B-2) becomes equivalent to a one-dimensional filter with the $Z$ transform

$\displaystyle F_9 (Z)$ $\textstyle =$ $\displaystyle \alpha\,Z^{-N_x-1} + \gamma\,Z^{-N_x} + \alpha\,Z^{-N_x+1} + \gamma\,Z^{-1}
-4\,(\alpha + \gamma)$  
    $\displaystyle +
\gamma\,Z + \alpha\,Z^{N_x-1} + \gamma\,Z^{N_x} + \alpha\,Z^{N_x+1}\;,$ (34)

where $N_x$ is the helix period (the number of grid points in the $x$ dimension). To find the inverse of a convolution with filter (B-9), we factorize the filter into the causal minimum-phase component and its adjoint:
\begin{displaymath}
F_9 (Z) = P (Z) P (1/Z)\;.
\end{displaymath} (35)

To find the coefficients of the filter $P$, any one-dimensional spectral factorization method can be applied. It is important to point out that the result of factorization (neglecting the numerical errors) does not depend on $N_x$. Another approach is to define a residual error vector for the coefficients of Z in equation (B-10) and minimize it for some particular norm. For example, minimizing the $\mathcal{L}_1$ norm when $F_9$ is the McClellan filter (B-7), we discover that the filter $P$, after transforming back to two dimensions, takes the form
\begin{displaymath}
\begin{array}{\vert r\vert r\vert r\vert r\vert r\vert r\ver...
...3 & 0.0808 & 0.2543 & 0.3521 & 0.1553 & & \\ \hline
\end{array}\end{displaymath} (36)

The results of applying a recursive deconvolution with filter (B-11) are shown in Figure B-2. An essentially similar procedure, only with a different set of filters, works for implicit wavefield extrapolation.

inv-laplace
inv-laplace
Figure B-2.
Inverting the Laplacian operator by a helix deconvolution. The top left plot shows the input, which contains a single spike and the causal minimum-phase filter $P$. The top right plot is the result of inverse filtering. As expected, the filter is deconvolved into a spike, and the spike turns into a smooth one-sided impulse. After the second run, in the backward (adjoint) direction, we obtain a numerical solution of Laplace's equation! In the two bottom plots, the solution is shown with grayscale and contours.
[pdf] [png] [scons]


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2014-02-17