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Appendix D: Analytical expressions for the constant horizontal slowness-squared gradient medium

In this appendix, we study the following medium:
\begin{displaymath}
w (z,x) = w_0 - 2 q_x x = w_0 - 2 \mathbf{q} \cdot \mathbf{x}\;,
\end{displaymath} (55)

where $\mathbf{q} = [0,q_x]^T$.

Similarly to the constant velocity gradient medium, it is convenient to write down the ray-tracing system in the form (Cervený, 2001)

\begin{displaymath}
\left\{ \begin{array}{lcl}
d \mathbf{x} / d \sigma & = & \ma...
... d \sigma = \mathbf{p} \cdot \mathbf{p}\;.
\end{array} \right.
\end{displaymath} (56)

Given equation D-1, $\nabla w = -2 \mathbf{q}$ and thus $d \mathbf{p} / d \sigma = - \mathbf{q}$. After integration over $\sigma$, equation D-2 becomes
\begin{displaymath}
\left\{ \begin{array}{lcl}
\mathbf{x} & = & \mathbf{x_0} + \...
...2 + \vert \mathbf{q} \vert^2 \sigma^3/3\;.
\end{array} \right.
\end{displaymath} (57)

For a particular image ray

\begin{displaymath}
\left\{ \begin{array}{lcl}
\mathbf{x_0} & = & [0, x_0]^T\;, ...
...0 - 2 q_x x_0}, 0]^T\;, \\
t_0 & = & t\;,
\end{array} \right.
\end{displaymath} (58)

the equation for $\mathbf{x}$ in D-3 simplifies to
\begin{displaymath}
\left\{ \begin{array}{lcl}
x = x_0 - q_x \sigma^2/2\;, \\
z = \sigma \sqrt{w_0 - 2 q_x x_0}\;.
\end{array} \right.
\end{displaymath} (59)

Solving equation D-5 for $\sigma$ as a function of $z$ and $x$
\begin{displaymath}
\sigma (z,x) = \left[ \frac{(w_0 - 2 q_x x)
- \sqrt{(w_0 - 2 q_x x)^2 - 4 q_x^2 z^2}}{2 q_x^2}
\right]^{\frac{1}{2}}\;.
\end{displaymath} (60)

Combining equations D-3 through D-6, we find
$\displaystyle x_0 (z,x)$ $\textstyle =$ $\displaystyle x + \frac{1}{2} q_x \sigma^2$  
  $\textstyle =$ $\displaystyle \frac{2 w_0 x + q_x z^2}{w_0 + 2 q_x x + \sqrt{(w_0 - 2 q_x x)^2 - 4 q_x^2 z^2}}\;,$ (61)


$\displaystyle t_0 (z,x)$ $\textstyle =$ $\displaystyle (w_0 - 2 q_x x_0) \sigma + \frac{1}{3} q_x^2 \sigma^3$  
  $\textstyle =$ $\displaystyle \frac{\sqrt{2} z \left[ 2 w_0 - 4 q_x x + \sqrt{(w_0 - 2 q_x x)^2...
..._0 - 2 q_x x + \sqrt{(w_0 - 2 q_x x)^2 - 4 q_x^2 z^2} \right]^{\frac{1}{2}}}\;.$ (62)

According to equation 4, D-7 can give rise to the geometrical spreading:

\begin{displaymath}
Q^2 (z,x) =
\frac{2 [(w_0 - 2 q_x x)^2 - 4 q_x^2 z^2]}
{(w_0...
...- 2 q_x x + \sqrt{(w_0 - 2 q_x x)^2 - 4 q_x^2 z^2} \right]}\;.
\end{displaymath} (63)

It is more convenient to express equations D-1 and D-9 in $\sigma$ and $x_0$ instead of directly in $t_0$ and $x_0$:
\begin{displaymath}
w (\sigma,x_0) = w_0 - 2 q_x x_0 + q_x^2 \sigma^2\;,
\end{displaymath} (64)


\begin{displaymath}
Q^2 (\sigma,x_0) = 1 +
q_x^2 \sigma^2 \left( \frac{1}{w_0 - ...
...x x_0} - \frac{4}{w_0 - 2 q_x x_0 + q_x^2 \sigma^2} \right)\;,
\end{displaymath} (65)

where we must resolve $\sigma = \sigma (t_0,x_0)$. This is done by revisiting equation D-8. For given $t_0$ and $x_0$, $\sigma$ is the root of a depressed cubic function of the following form:
\begin{displaymath}
\sigma^3 + \frac{3 (w_0 - 2 q_x x_0)}{q_x^2} \sigma - \frac{3}{q_x^2} t_0 = 0\;.
\end{displaymath} (66)

After some algebraic manipulations, we find
$\displaystyle \sigma (t_0,x_0)$ $\textstyle =$ $\displaystyle \left[ \frac{3 t_0 + \sqrt{9 t_0^2 + 4 (w_0 - 2 q_x x_0)^3 / q_x^2}}{2 q_x^2} \right]^{\frac{1}{3}}$  
  $\textstyle -$ $\displaystyle \frac{w_0 - 2 q_x x_0}{q_x}
\left[ \frac{2}{3 q_x t_0 + \sqrt{9 q_x^2 t_0^2 + 4 (w_0 - 2 q_x x_0)^3}} \right]^{\frac{1}{3}}\;.$ (67)

Finally, inserting equations D-10 and D-11 into equation 3 results in the Dix velocity:

\begin{displaymath}
v_d (t_0,x_0) = \frac{\sqrt{w_0 - 2 q_x x_0}}{w_0 - 2 q_x x_0 - q_x^2 \sigma^2}\;.
\end{displaymath} (68)


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Next: Bibliography Up: Li & Fomel: Time-to-depth Previous: Appendix C: Analytical expressions

2015-03-25