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REFERENCES

APPENDIX A

In equation (1) the values of the constant kz, given by equation (2), have to be real. Imaginary values of kz do not satisfy the downward continuation ordinary differential equation

\begin{displaymath}
{{\partial^2 P} \over {\partial z^2}}=
-k_z^2 P\end{displaymath}

and have to be excluded. Real values of kz from equation (2) require the conditions:

\begin{displaymath}
\begin{array}
{lcl}
\mid \omega \mid & \geq & \mid v_y+v_h \mid
\\ \\ \mid \omega \mid & \geq & \mid v_y-v_h \mid\end{array}\end{displaymath}

which can be reduced to the condition  
 \begin{displaymath}
\begin{array}
{lcl}
\mid \omega \mid & \geq & \mid v_y \mid + \mid v_h \mid\end{array}.\end{displaymath} (16)
After the change of variable from $\omega$ to $\omega_0$ in equation (5)

\begin{displaymath}
\omega \equiv
{\omega_0 { \left[ 1+{ {v^2 k_h^2 } \over 
{ 4 \omega_0^2-v^2 k_y^2}} \right]}^{1 \over 2}},\end{displaymath}

we want to express $\omega_0$ function of $\omega$ and determine the integration boundaries for $\omega_0$.We start with the expression for $\omega$:
\begin{displaymath}
\omega^2 = \omega_0^2 + 
{{\omega_0^2 v_h^2} \over {\omega_0^2-v_y^2}}\end{displaymath} (17)
and after reducing

\begin{displaymath}
\omega^2 \omega_0^2 - \omega^2 v_y^2 = 
\omega_0^4 - \omega_0^2 v_y^2 + \omega_0^2 v_h^2\end{displaymath}

and grouping

\begin{displaymath}
\omega_0^4 - \omega_0^2 (\omega^2 + v_y^2 - v_h^2)
+\omega^2 v_y^2 = 0\end{displaymath}

we have
\begin{displaymath}
\omega_0^2 = {1 \over 2} 
(\omega^2+v_y^2-v_h^2 \pm 
{\sqrt {(\omega^2+v_y^2-v_h^2)^2-4 \omega^2 v_y^2}}).\end{displaymath} (18)
The discriminant $\Delta$ is

\begin{displaymath}
\Delta =
(\omega-v_y-v_h)(\omega-v_y+v_h)
(\omega+v_y-v_h)(\omega+v_y+v_h).\end{displaymath}

From the conditions on kz, $\Delta$ is always positive and therefore $\omega_0^2$ is always real within the $\omega$ limits.

The integration limits for $\omega_0$ are found by starting with the limits for $\omega$ :

\begin{displaymath}
\omega = \mid v_h \mid + \mid v_y \mid\end{displaymath}

and after we square both sides

\begin{displaymath}
\omega^2 = v_h^2+v_y^2+2\mid v_h v_h \mid\end{displaymath}

and replacing $\omega^2$ in the equation for $\omega_0^2$ we have
\begin{displaymath}
\begin{array}
{lcl}
\omega_0^2 & = & \displaystyle{
{1 \over...
 ...\ \\  & = & \displaystyle{
v_y^2+\mid v_h v_y \mid}.\end{array}\end{displaymath} (19)
The integration in $\omega_0$ is done from $(-\infty,-\sqrt{v_y^2+\mid v_h v_y \mid})$ and from $(\sqrt{v_y^2+\mid v_h v_y \mid},\infty)$. After changing the order of integration from $\omega_0$ to kh, the integration boundaries for kh become
\begin{displaymath}
k_h \in (-{2 \over v}{{\omega_0^2-v_y^2} \over {\mid v_y \mid}},
{2 \over v}{{\omega_0^2-v_y^2} \over {\mid v_y \mid}}).\end{displaymath} (20)


previous up next print clean
Next: About this document ... Up: Popovici : PDE for Previous: Acknowledgments
Stanford Exploration Project
11/17/1997