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Example

The simplest case to study analytically is that of vertically-incident waves in laterally homogeneous media. In this case, all operators become functions of the scalar variable s (unknown layer slowness). If, for a particular temporal frequency $\omega$ and the layer thickness $\Delta z$, we measure the slowness in units of $1/(\omega\,\Delta z)$, the wave continuation operator is simply the phase shift  
 \begin{displaymath}
P(s) = e^{i\,s}\;,\end{displaymath} (25)
and the fundamental nonlinear equation takes the form  
 \begin{displaymath}
 F_\xi(s) = \mathcal W\left({z+\Delta z} \right)\,e^{i\,(1-\xi)\,s} - \mathcal W\left({z } \right)\,e^{-i\,\xi\,s} = 0\;.\end{displaymath} (26)
Noting that  
 \begin{displaymath}
\mathcal W\left({z } \right)= \mathcal W\left({z+\Delta z} \right)\,e^{i\,s^{\star}}\;,\end{displaymath} (27)
where $s^{\star}$ is the true slowness, and that the convergence of Newton's method does not depend on scaling the equation by a constant, we can modify equation (26) to the simpler form  
 \begin{displaymath}
\hat{F}_\xi(s) = F_\xi(s)\,\frac{e^{(\xi-1)\,s^{\star}}}{\ma...
 ...ht)} = 
e^{i\,(1-\xi)\,\Delta s} - e^{-i\,\xi\,\Delta s} = 0\;,\end{displaymath} (28)
where $\Delta s = s - s^{\star}$. The obvious solution of Equation (28) is $\Delta s = 0$. Our task is to find the convergence limits and their dependence on $\xi$.

After a number of algebraic and trigonometric simplifications, the operator G from equation (23) takes the form of the function  
 \begin{displaymath}
\hat{G}_\xi(s) = s - \frac{\sin \left(\Delta s\right)}
 {1 -...
 ...(1 - \xi)\,\xi + 2\,(1 - \xi)\,\xi\,\cos \left(\Delta s\right)}\end{displaymath} (29)
Its derivative is  
 \begin{displaymath}
\hat{G}_\xi'(s) = 2\,\sin^2 \left(\frac{\Delta s}{2}\right)\...
 ...i + 
2\,(1 - \xi)\,\xi\,\cos \left(\Delta s\right)\right]^2}\;.\end{displaymath} (30)
The method will converge in the region around $\Delta s = 0$, where the absolute value of $\hat{G}_\xi'(s)$ is strictly smaller than one. This region (as a function of $\Delta s$ and $\xi$) is plotted in Figure 3. We can see that the convergence region has a finite extent. Its width is the same for $\xi=0$, $\xi=1$, and $\xi=1/2$. Indeed,  
 \begin{displaymath}
\hat{G}_0'(s) = \hat{G}_1'(s) = 
2\,\sin^2{\left(\frac{\Delta s}{2}\right)}\end{displaymath} (31)
and  
 \begin{displaymath}
\hat{G}_{1/2}'(s) = - \tan^2{\left(\frac{\Delta s}{2}\right)}\;.\end{displaymath} (32)
In both cases, the absolute value of the derivative is smaller than one if $\Delta s < \frac{\pi}{2}$. If we take $\omega=2\,\pi \times
100\,\mbox{Hz}$ and $\Delta z = 10\,\mbox{m}$, then the convergence radius is $\Delta s = 0.25\,\mbox{s}/\mbox{km}$. At small $\Delta z$, 
 \begin{displaymath}
\hat{G}_0'(s) = \hat{G}_1'(s) \approx 
\frac{\left(\Delta s\right)^2}{2}\end{displaymath} (33)
and  
 \begin{displaymath}
\hat{G}_{1/2}'(s) \approx \frac{\left(\Delta s\right)^2}{4}\;.\end{displaymath} (34)
The convergence rate is of the same order (cubic) but faster in the case of the bilinear method ($\xi=1/2$), because of the twice smaller constant. Here is an example of iterations starting with s0=2 and converging to $s^{\star} = \pi$. The Born iteration:
\begin{eqnarraystar}
s_1 & = & 2.9093 \\ s_2 & = & 3.13951 \\ s_3 & = & 3.14159\end{eqnarraystar}
The bilinear iteration:
\begin{eqnarraystar}
s_1 & = & 3.28419 \\ s_2 & = & 3.14135 \\ s_3 & = & 3.14159\end{eqnarraystar}
A faster convergence can be achieved at some other values of $\xi$.Examining the Taylor series of $\hat{G}_\xi'(s)$ around $\Delta s = 0$: 
 \begin{displaymath}
\hat{G}_\xi'(s) \approx \left[1 - 6\,(1-\xi)\,\xi\right]\,
\frac{\left(\Delta s\right)^2}{2}\;,\end{displaymath} (35)
we find that the order of convergence is optimized for $\xi = \frac{1}{2} \pm \frac{\sqrt{3}}{6}$. In this case,  
 \begin{displaymath}
\hat{G}_{1/2 \pm \sqrt{3}/6}'(s)
= \frac{4\,\sin^4{\left(\fr...
 ...right)}\right]^2}
\approx \frac{\left(\Delta s\right)^4}{36}\;,\end{displaymath} (36)
and the convergence is fifth order! The example iterations with the optimal value of $\xi$ are:
\begin{eqnarraystar}
s_1 & = & 3.12903 \\ s_2 & = & 3.14159 \\ s_3 & = & 3.14159\end{eqnarraystar}
The radius of convergence with the optimal value of $\xi$ is $\Delta s < \frac{2}{3}\,\pi$.

 
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Figure 3
Convergence region for the Newton-Gauss method in the vertical plane-wave example. Left: 3-D projection. Right: contours. The non-white region on the right plot corresponds to the convergence area. Horizontal axis: $\Delta s$. Vertical axis: $\xi$.
view

Of course, this analysis does not apply directly to the case of non-vertical wave propagation and laterally inhomogenous slowness fields. For reflection wavefields at multiple offsets, the symmetry between downward and upward continuation is broken, as is clear from the experimental results of this paper. However, the simple analysis points to the potential benefits of modifying the Born approximation in the wave-equation velocity estimation.


next up previous print clean
Next: Physical interpretation Up: Newton's method and WEMVA Previous: Newton's method and WEMVA
Stanford Exploration Project
6/7/2002