Stationary phase approximation

Downward continuation using the DSR equation for a separate constant-offset sections requires the computation of the integral in equation (A-1). This can be done for all offsets by taking the inverse transform of the downward continuation exponential, or for a single offset by finding a stationary phase approximation to the integral  

$$I= \int dk_h \; e^{ik_z(\omega,k_y,k_h)z-ik_h h}.$$ (10)

Integrals of the form

$$I(k)=\int_{-\infty}^{\infty} e^{ik\phi(t)}f(t)\; dt$$

are approximated asymptotically (Zauderer, 1989) when $k \rightarrow \infty$ by

$$I(k) \approx e^{ik\phi (t_0)} f(t_0) e^{{\rm sign} (\phi''(... ...t[{{2\pi} \over {k \mid \phi''(t_0) \mid }} \right]^{1 \over 2}$$

where t₀ is the ``stationary point'' where the derivative of the phase is zero. The approximation described here assumes the second derivative is non-zero, which I will prove is the situation in this case.

The phase of the exponential in this case is

$$\phi(k_h)={-{\rm sign}(\omega) z \left[ \; \sqrt{{\omega^2 \... ...omega^2 \over v^2}-{1 \over 4}(k_y-k_h)^2} \; \right] - k_h h}.$$ (11)

In order to evaluate the stationary point we need to find the roots of the equation

$$\phi ' (k_h)= -h +{\rm sign}(\omega) z \left[ {{k_h+k_y} \ov... ... {\sqrt{{\omega^2 \over v^2}-{1 \over 4}(k_y-k_h)^2}}} \right].$$ (12)

I was not able to find these roots, as it involves an equation of sixth degree. However, I will point out some properties of the phase, which should make the numerical computation simpler than the full integration.

The second derivative of the phase is non-zero, as it consists of the sum of four positive terms multiplied by ${\rm sign}(\omega)$:

$$\begin{array} {lcl} \phi '' (k_h) & = & {{{\rm sign}(\omega... ...\over 4}(k_y+k_h)^2} \right ]^{3 \over 2}}} \right].\end{array}$$ (13)

This ensures that there is no change in curvature, and the phase always has a maximum or a minimum and therefore a stationary point. Indeed, for a fixed pair of values $\omega,k_y$, Figure A-1 shows the phase function for several depth levels. The sign of $\omega$ determines if the phase is positive or negative.

 

App1
Figure 8 Plot of the phase function for a constant pair of values $\omega,k_y$.The different plots correspond to increasing depth levels.
a. Phase corresponding to positive $\omega$.
b. Phase corresponding to negative $\omega$.