VTI homogeneous media

The time-processing operators in VTI media, as mentioned earlier, depend primarily on two parameters, $V_{\rm nmo}$ and $\eta$. This dependency becomes exact when the shear wave velocity (V_S0) is set to zero (Alkhalifah, 1997c). Setting V_S0=0, although not practical for TI media, yields remarkably accurate kinematic representations. Errors due to this approximation, for practical VTI models, are kinematically, in a worse-case scenario, less than 0.5 percent, which is far within the limits of seismic accuracy. This acoustic approximation yields simplified equations, including a simplified dispersion relation. Because the vertical P-wave velocity does not have any significance in time-related processing in VTI media (Alkhalifah and Tsvankin, 1995), I will refer to $V_{\rm nmo}$ as the velocity and denote it by the symbol v to simplify comparisons with isotropic media. The zero-offset time-migration dispersion relation for VTI media, when V_S0=0, is given by

$$\cos \theta(t) = 1- \frac{v^2 p_x^2}{1-2 \eta v^2 p_x^2}$$

(Alkhalifah, 1997c). Based on this relation, the normalized DSR equation, for VTI media, has the form  

$${T(p_x,p_h,v,\eta) \over \tau} = 0.5(\sqrt{1-\frac{(p_x+p_h)... ...x-p_h)^2 v^2}{1-2 \eta v^2 (p_x-p_h)^2}}) + p_h \frac{X}{\tau},$$ (33)

Clearly, for $\eta$=0, the above equations reduce to there isotropic counterparts. The stationary point (or points) satisfy the following relation

$$\begin{eqnarray} \frac{\partial T}{\tau \partial p_h} = 0.5(\frac{(p_x-p_h) v^2}... ...(p_x+p_h)^2 v^2}{1-2 \eta v^2 (p_x+p_h)^2}}}) +\frac{X}{\tau} =0.\end{eqnarray}$$ (34)

 

ph1eta2m
Figure 20 Left: Values of p_h as a function p_x calculated numerically (solid curves), and calculated analytically (dashed curves) using equation (B-1). Right: The absolute difference between the two curves on the right. The medium is homogeneous and transversely isotropic with v=2.0 km/s and $\eta$=0.2.. 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).

Following the same approach used in the previous sections for isotropic media, we set p_x=0, and as a result, equation (B-10) reduces to

$$\frac{p_h v^2}{(1-2 \eta v^2 p_h^2)^3 \sqrt{1-\frac{p_h^2 v^2}{1-2 \eta v^2 p_h^2}}} =\frac{X}{\tau}.$$

Squaring both sides and expanding in p_h, we get

$$8 \eta^3 X^2 (1+2 \eta) y^4 - 4 \eta^2 X^2 (3+8 \eta) y^3 + ... ...X^2 (1+4 \eta) y^2 - [X^2 (1+8 \eta)+ \tau^2 v^2] y + X^2 = 0,$$

where y=p_h² v². This is a quartic equation in y, which, although,it has analytical solutions, is best solved numerically. Considering either $\eta$ or y ($\alpha p_h^2$), or both, to be small, we can drop terms of y beyond the quadratic (perturbation theory; Bender and Orszag, 1978) , and as a result, the quartic equation reduces to

a y² +b y +c =0,

where

$$a=6 \eta X^2 (1+4 \eta),$$

$$b=-X^2 (1+8 \eta)- \tau^2 v^2,$$

and

 

c=X². (35)

Because $\eta$ can be small, as small as zero for isotropic media, I prefer to use the following form of solution of the quadratic equation:

$$y = \frac{2 c}{-b \pm \sqrt{b^2-4ac}}$$

(Press et al., 1988). Thus,  

$$p_{h0} = \frac{1}{v} \sqrt{\frac{2 c}{-b \pm \sqrt{b^2-4ac}}},$$ (36)

where the positive-sign root is used. The negative-sign root results in an imaginary p_h solution which is clearly not the solution we are looking for. Another, yet better, approximation is described in Appendix C using perturbation series as well as Shanks transformation. For $\eta$=0 (isotropic media), a=0, and this equation reduces to equation (B-3), which is its isotropic counterpart. This is amazing considering we have dropped some terms in solving the quartic equation. Equation (B-11) is not the exact solution for horizontal reflectors in VTI media. However, as we will see bellow, it is a good approximation; better than similar approximations based on Taylors series expansion of traveltimes around the zero-offset point. Also, to insure we do not get imaginary roots for p_h, b²-4ac must be greater or equal zero. This results in the following condition that must be satisfied:

$$\tau^2 \gt [2 \sqrt{6 \eta (1+4 \eta)} - (1+8 \eta)] \frac{X^2}{v^2},$$

which corresponds to large offset-to-depth ratio. For practical values of $\eta$(<0.3), this condition is far within the typical mute zone. For example, if v=2 km/s, $\eta=0.3$, and X=3 km, $\tau$=1.14 s.

Expanding equation (B-10) using Taylors series, around p_h=0 ($p_x=\frac{1}{v \sqrt{1+2 \eta}}$), and ignoring terms beyond the linear, yields  

$$\frac{X}{\tau} - \frac{ p_h v^2 [1+4 \eta p_x^2 v^2-6 \eta (... ...{1-p_x^2 v^2- 2 \eta p_x^2 v^2}{1-2 \eta p_x^2 v^2})^{3/2}} =0.$$ (37)

As a result, the stationary point is given by  

$$p_h = \frac{\frac{X}{\tau}}{v^2} \frac{(1-2 \eta p_x^2 v^2)^... ...v^2})^{3/2}}{[1+4 \eta p_x^2 v^2-6 \eta (1+2 \eta) p_x^4 v^4]}.$$ (38)

Again, I insert equation (B-11) in place of $\frac{\frac{X}{\tau}}{2 v^2}$, and obtain  

$$p_h = \frac{1}{v} \sqrt{\frac{2 c}{-b \pm \sqrt{b^2-4ac}}} \... ...v^2})^{3/2}}{[1+4 \eta p_x^2 v^2-6 \eta (1+2 \eta) p_x^4 v^4]}.$$ (39)

where a, b, and c are given by equation (B-11). Equation (B-15) is exact for $p_x=\frac{1}{v \sqrt{1+2 \eta}}$ (p_h=0). However, unlike the isotropic medium case, equation (B-15) is an approximation at p_x=0. [Recall that we dropped terms beyond the quadratic in equation (B-13).] Despite the approximation used at p_x=0, I will refer to this equation as the 2-point solution for VTI media.

 

ph2eta2m
Figure 21 Same as in Figure B-1, but with the analytical solution evaluated using the 2-point fitting of equation (B-15).

Figure B-5  shows p_h given by equation (B-15) and compares it with the accurate result calculated numerically from equation (B-10) for three sets of $\frac{X}{\tau}$. Despite the large non-hyperbolic moveout associated with horizontal events in such VTI media (Alkhalifah, 1997b), the approximation at p_x=0, which corresponds to a horizontal event, is rather good. Combine this with the fact that the phase [equation (B-9)] changes slowly as a function of (or rather insensitive to) p_h around the exact solution, the result is a good approximation of the phase for all p_x (slopes).

The accuracy of equation (B-15) can be further enhanced, as in the isotropic case, by fitting it to the exact solution for p_x=p_h; the angular correspondence of this equality depends on the offset-to-depth ratio. Following the same steps used to obtain the p_h for p_x=0, including dropping terms beyond the quadratic in a similar quartic equation to that of equation (B-13), the stationary point is given by  

$$p_{hs} = \frac{1}{v} \sqrt{\frac{2 c}{-b \pm \sqrt{b^2-4ac}}},$$ (40)

where

$$a=96 \eta X^2 (1+4 \eta),$$

$$b=-4 X^2 (1+8 \eta) - \tau^2 v^2,$$

and

c=X².

Equatio (B-16)n has a stricter condition for its validity (that is avoiding imaginary roots) than that of equation (B-11). Specifically,

$$\tau^2 \gt 4 [2 \sqrt{6 \eta (1+4 \eta)} - (1+8 \eta)] \frac{X^2}{v^2},$$

which correspond to double the time for the horizontal reflector solution fitting. For v=2 km/s, $\eta=0.3$, and X=3 km, $\tau$=2.28 s.

Again, inserting equation (B-16) into equation (B-15) requires solving two equations and, similar to the case of isotropic media, the solution is given by  

$$p_h = p_{h0} \frac{(1-2 \eta p_x^2 v^2)^4 (\frac{1-p_x^2 v^... ... \eta p_x^2 v^2-6 \eta (1+2 \eta) p_x^4 v^4]} [a(p_{h0})p_x+b],$$ (41)

where setting p_x=0 gives b=1 and setting p_x=p_hs results in

$$a= \frac{1}{p_{h0}} \frac{1+4 \eta p_x^2 v^2-6 \eta (1+2 \et... ...eta p_x^2 v^2}{ 1-2 \eta p_x^2 v^2})^{3/2}}-\frac{1}{p_{hs}}.$$

 

ph3eta2m
Figure 22 Same as in Figure B-1, but with the analytical solution evaluated using the 3-point fitting of equation (B-17).

Figure  B-6 shows the numerically-driven curves (solid ones) along with the analytical curves calculated using the 3-point fitting of equation B-17. The 3-point equation provides the best estimation to the exact solution for $\frac{X}{\tau}$.