Isotropic homogeneous media

In order to evaluate the stationary point (the point in which the phase is minimum or maximum), we need to solve the equation  

$$\frac{\partial T(p_x,p_h,v,\frac{X}{\tau})}{\tau \partial p_... ...c{(p_x-p_h) v^2}{\sqrt{1-(p_x-p_h)^2 v^2}}) +\frac{X}{\tau} =0.$$ (25)

Equation (B-1) reduces to the condition that the horizontal projection for rays from each of the source and receiver to the specular reflection point (SRP) add up to equal the source-receiver offset, X (Popovici, 1995). Solving equation (B-1) in its full form involves solving a sixth-degree polynomial, which I prefer to avoid doing analytically.

Setting p_x=0 in equation (B-1) yields solutions (stationary points) for reflections from horizontal reflectors, and, therefore, equation (B-1) reduces to  

$$\frac{\partial T(p_h,v,\frac{X}{\tau})}{\tau \partial p_h} \... ...= \frac{p_{h0} v^2}{\sqrt{1-p_{h0}^2 v^2}} +\frac{X}{\tau} =0,$$ (26)

with a stationary point given by  

$$p_{h0} = \frac{\frac{X}{\tau}}{v \sqrt{(\frac{X}{\tau})^2 + v^2}}.$$ (27)

In fact, p_h0 is the exact stationary point solution for horizontal reflections, which can be derived directly from the hyperbolic moveout equation:

$$t^2 = \tau^2 + \frac{X^2}{v^2}.$$

Where $p_h=\frac{\partial t}{\partial X}=\frac{1}{2} \frac{\partial t}{\partial h}$,

$$p_h = \frac{X}{v^2 t} = \frac{X}{v \sqrt{X^2 + \tau^2 v^2}},$$

which is the same as equation (B-3).

Expanding equation (B-1) using Taylor's series around p_h=0, and ignoring terms beyond the linear in p_h, yields

$$\frac{p_h v^2}{(1-p_x^2 v^2)^{3/2}} +\frac{X}{\tau} =0,$$

with a stationary point given by  

$$p_h = \frac{\frac{X}{\tau}}{v} (1-p_x^2 v^2)^{3/2}.$$ (28)

Equation (B-4) corresponds to the exact solution for a vertical reflector, where p_h=0 and $p_x=\frac{1}{v}$. It also provides a good approximation away from $p_x=\frac{1}{v}$. Figure B-1 shows a comparison between the exact p_h solution of equation (B-1) (obtained numerically) and that given by equation (B-4) as a function of p_x for three sets of $\frac{X}{\tau}$. The absolute difference between the two solutions is also displayed. As expected, errors increase with offset, and

 

ph1eta0m
Figure 17 Left: Values of p_h as a function p_x calculated numerically (solid curves), and calculated analytically (dashed curves)using equation (B-4). Right: The absolute difference between the two curves on the left. The medium is homogeneous and isotropic with v=2.0 km/s. The black curve corresponds to $\frac{X}{\tau}$=1.0 km/s, the dark-gray curve corresponds to $\frac{X}{\tau}$=2.0 km/s, and the light gray curve corresponds to $\frac{X}{\tau}$=3.0 km/s.

the accuracy of the approximate solution reduces at p_x=0. In fact, for p_x=0, p_h in equation (B-4) equals $\frac{\frac{X}{\tau}}{2 v}$ which is clearly different from the exact solution given by equation (B-3). Inserting the exact solution, in place of the factor $\frac{\frac{X}{\tau}}{2 v}$, in equation (B-4) yields a new approximate solution for p_h given by  

$$p_h = \frac{\frac{X}{\tau}}{v \sqrt{(\frac{X}{\tau})^2 + v^2}} (1-p_x^2 v^2)^{3/2}.$$ (29)

This solution is exact for p_x=0 and $p_x=\frac{1}{v}$, and therefore, I will refer to it as the 2-point solution (it fits exactly at two points). Figure B-2 shows a comparison between the exact solution of equation (B-1) (obtained numerically) and that given by equation (B-5), again for three sets of $\frac{X}{\tau}$.Clearly, this new curve of p_h better fits the exact solution than that given in Figure B-1. Although the percentage of error in p_h seems to be high, the corresponding error in the phase estimate (which is the key quantity used in the migration) is much lower because $p_{\tau}$, as depicted in Figure 1, is rather insensitive to p_h, especially around the exact solution. In other words, the curvature of the phase function around its minimum is low. Such low curvature, according to equation (A-2), will result in a greater contribution in terms of amplitude. This translates to practically no error when it comes to actual geophysical applications.

 

ph2eta0m
Figure 18 Same as in Figure B-1, but with the analytical solution evaluated using the 2-point fitting of equation (B-5).

Let's find yet another exact solution, that is, the solution when p_s=0 (p_x=p_h). The reflector angular correspondence of this approximation depends on the offset. In this case, equation (B-1) reduces to  

$$\frac{\partial T(p_{h},v,\frac{X}{\tau})}{\tau \partial p_{h... ... \frac{p_{hs} v^2}{\sqrt{1-4 p_{hs}^2 v^2}} +\frac{X}{\tau} =0.$$ (30)

with a solution for p_h given by  

$$p_{hs} = \frac{\frac{X}{\tau}}{v \sqrt{4 (\frac{X}{\tau})^2 + v^2}}.$$ (31)

To insure that p_h given by equation (B-5) has the value of p_hs as p_x=p_h, we must insert, some how, equation (B-7) into equation (B-5). This task is accomplished by forcing p_h, given by equation (B-5), to linearly equal p_hs at p_x=p_h, and, thus, equation (B-5) becomes  

$$p_h = \frac{\frac{X}{\tau}}{v \sqrt{(\frac{X}{\tau})^2 + v^2}} (1-p_x^2 v^2)^{3/2} [a(p_{hs})p_x+b].$$ (32)

The term between the square brackets is to insure this linear fit. To solve for a and b in equation (B-8), p_h must satisfy two conditions: (1) p_h should reduce to the form given by equation (B-3) for p_x=0, and this condition will result in b=1. (2) p_h should reduce to the form given by equation (B-7) for p_x=p_hs. The second condition, after some mathematical manipulation, yields

$$a= \frac{1}{p_{h0} (1-p_x^2 v^2)^{3/2}} - \frac{1}{p_{hs}}.$$

Therefore, p_h given by equation (B-8) is an exact solution of equation (B-1) at three points (p_x=0, p_x=p_hs, and $p_x=\frac{1}{v}$) and a good approximation, as demonstrated by Figure B-3, elsewhere. I will refer to this equation as the 3-point solution.

Figure B-3 shows p_h given by equation (B-8) next to the exact solution for p_h. Clearly, the three point fitting given by equation (B-8) resulted in a good approximation of p_h. The accuracy of this approximation will be better appreciated when we observe the low amount of errors in traveltime calculation induced by this approximation (i.e., Figure 3).

 

ph3eta0m
Figure 19 Same as in Figure B-1, but with the analytical solution evaluated using the 3-point fitting of equation (B-8).