Stationary point solutions

To evaluate the integrals in equation (), using the stationary phase method, we need to calculate the maximum of the phase function, which for VTI media is given by  

$$T = \frac{1}{2}(\sqrt{1-\frac{(p_x+p_h)^2 v^2}{1-2 \eta v^2 ... ..._h)^2 v^2}{1-2 \eta v^2 (p_x-p_h)^2}})+2 p_x (x-x_0)+2 p_h h_0.$$ (136)

Since the relation between the source-receiver rayparameters (p_s and p_g) and the offset-midpoint rayparameters (p_h and p_x) is linear, we can evaluate the stationary points by solving for p_s and p_g instead of p_h and p_x,  

$$T = \frac{1}{2}(\sqrt{1-\frac{p_g^2 v^2}{1-2 \eta v^2 p_g^2}... ...v^2}{1-2 \eta v^2 p_s^2}})+2 (p_g+p_s) (x-x_0)+2 (p_g-p_s) h_0.$$ (137)

Setting the derivative of equation  in terms of p_s and p_g to zero provides us with two independent equations that can be solved for p_s and p_g, separately. Physically, this implies that we are solving for the source-to-image-point traveltime and receiver-to-image-point traveltime, separately, which makes complete sense. The p_s and p_g stationary point solutions can be used later to evaluate p_h and p_x.

First, p_s is evaluated by solving  

$$\frac{\partial T}{\partial p_s} ={\frac{{v^2}\,\tau \,{p_s}}... ...{p_s}}^2}}{1 - 2\,{v^2}\,\eta \,{{{p_s}}^2}}}}}} } + {y_s} =0,$$ (138)

where y_s=2(x-x₀-h₀). Similarly, p_g is evaluated by solving  

$$\frac{\partial T}{\partial p_g} ={\frac{{v^2}\,\tau \,{p_g}}... ...{{p_g}}^2}}{1 - 2\,{v^2}\,\eta \,{{{p_g}}^2}}}}}} } + {y_g}=0,$$ (139)

where y_g=2(x-x₀+h₀). Equation  is similar to equation , with y_s replaced by y_g, and p_s replaced by p_g. Therefore, solving for p_s will yield an equation that can be used to solve for p_g as well.

To remove the square root in equation , I move y_s to the other side of the equation and square both sides. This will allow us to right equation  in a polynomial form as follows,

$$\begin{eqnarray} -{{{y_s}}^2}+ {{{p_s}}^2}\,\left( {v^4}\,{{\tau }^2} + 6\,{v^... ...a }^3}\,\left( 1 + 2\,\eta \right) \,{{{p_s}}^8}\, {{{y_s}}^2}=0.\end{eqnarray}$$ (140)

This is a fourth-order polynomial in p_s², which can be solved exactly for the four roots in p_s². However, these four roots are given by highly complicated equations which include square roots as well as powers of the order $\frac{1}{3}$. Such equations are not useful for practical use. Thus, I elect to use Shanks transform to obtain approximations that are almost exact, yet more useful for practical implementations. Also, since Shanks transform is based on perturbation theory, it will provide us with the desired solution of p_s² among the four possible solutions; the solution based on perturbation from an isotropic model.

Using Shanks transform, described in Appendix D, we obtain  

$$p_s^2 =\frac{{y_s^2}\,\left( {y_s^6} + 6\,{v^2}\,{y_s^4}\,... ...\eta \right) \,{{\tau }^4} + 4\,{v^6}\,{{\tau }^6} \right) }.$$ (141)

Again, p_g is given by the same equation but with y_s replaced by y_g. B