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Differential K-operators

The DKO is determined as a solution of the equation

{\bf L} u = \rho {\partial^{2} u \over \partial t^{2}}\end{displaymath}

with boundary conditions that supply the uniqueness of the solution.
We have two problems to solve:
The choice of the operator L.
The choice of the boundary conditions.
Let us take the classical eikonal equation

\vert\nabla \tau \vert^2 = {1\over{v^2({\bf r})}}.\end{displaymath}

This is a characteristic equation for any hyperbolic equation of the type:  
a \Delta u + \sum {b_i} {\partial u \over \partial x_{i}}+c{...
 ...artial t}
+ h u = \rho {\partial ^{2} u \over \partial t^{2}} ,\end{displaymath} (80)
at $v=\sqrt{\frac{a}{\rho}}$.The classical wave-equation

\Delta u = 
\frac{1}{v^2({\bf r})} \frac{\partial^2 {u}}{\partial {t^2}}\end{displaymath}

is the simplest hyperbolic equation of this type. But is it also the best choice?

1. Absence of wave dispersion. Let the coefficient of the equation (80) be constant. After substitution

u = {\overline u}\exp \left(-{1\over 2}
 \sum {\beta_i x_i} + {1\over 2}\alpha t \right)\end{displaymath}

where $\beta_i =b_i/a$ and $\alpha =c/a$. The equation is reduced to

\triangle{\overline u} + {\overline h}{\overline u} -
 {1\over v^2}{\partial^2 {\overline u}\over\partial t^2}
 = 0\end{displaymath}

where ${\overline h} =h/a + v^2\alpha^2/4 +
 \sum {\beta_i^2/4}$.

Looking for the solution in the form

\exp[i (\omega t - \sum_{i=1}^{3}k_i x_i)],\end{displaymath}

we obtain the dispersion equation

\omega_{1,2} =\pm v\sqrt{\sum k_i^2 + {\overline h^2}}\end{displaymath}

Since in nondispersive media $\sum k_i^2 = v^2 \omega^2$,only in the case when ${\overline h} =0$ or

h= -{ (c v^{2} + \sum {b_{i}^{2}\over a}) \over 4}.\end{displaymath}

2. Amplitude equivalence. This term means that while discontinuities propagate, amplitudes A change along rays s following the transport equation (see Chapter 4)

2 {dA\over{ds}} + A\left[ v \Delta \tau + {d\over{ds}} \ln(\rho v^2) \right] = 0\end{displaymath}

which is valid for P and S waves. Using the standard technique of ray theory, it is easy to show that in the case of amplitude equivalence



b_{i}={\partial a \over \partial x_{i}}. \end{displaymath}

For homogeneous media the wave equation is the only one that is nondispersive and amplitude-equivalent simultaneously.

3. Nonscattering vertical propagation. If in the amplitude-equivalent equation  
\sum {\partial \over \partial x_{i}} (\rho v^{2} {\partial u...
 ... \partial x_{i}}) 
=\rho {\partial^{2} u \over \partial t^{2}},\end{displaymath} (81)
the following is valid: v=v(z), $\rho = \rho (z)$ and $v(z) \rho (z) = const.$, then plane waves propagate in z-direction without scattering. This is easy to show if one inserts

u = A f \left( t - \int_{0}^{z} {d \xi \over v(\xi)} \right)\end{displaymath}

into equation (81).

II. Boundary conditions. There are few types of conditions that can be used for the wave-field continuation with the help of equation (80).

1. Mixed problem with boundary and initial conditions

u\vert _{z=0} = u_{0}{\:}{\:}(or {\:}{\:}...
 ... {\partial u \over \partial t}\vert _{t=t^{\pm}} = 0\end{array}\end{displaymath}

where t+ = 0 (for forward continuation), t- = T (for reverse continuation), T > tmax, (where tmax is the duration of the seismic record).

This problem has a unique and stable solution (in a restricted domain). It is restricted at $t \rightarrow + \infty $ (for direct continuation) and $t \rightarrow - \infty$ (for inverse continuation) if h>0 and $ +c \geq 0$ (or $-c \geq 0$).

Representations of the solution:

$\bullet$ Finite-difference technique which can be applied for any inhomogeneous media.

$\bullet$ Kirchhoff's type integral:

u^{(\pm)}({\bf r}, t)=
 ...(\pm)}({\bf r}, {\bf r}_0, t-\tau)\over \partial z}
 d{\bf r}_0\end{displaymath} (82)

where $\sum : z=0$, I+ =(0, t), I(-) =(T-t, T), $g^{(\pm)}({\bf r}, {\bf r}_0, t)$-Green's function which is the solution of the equation  
{\bf L}u - \rho {\partial^2 u\over\partial t^2} =\delta
 ({\bf r}-{\bf r}_0)\delta (t)\end{displaymath} (83)
with the zero initial condition at $t=t^\pm$ and $u\vert_{z=0} =0 $.

For inhomogeneous media Green's function usually is unknown (although one can calculate ray zero-approximation of $g^{(\pm)}({\bf r}, {\bf r}_0, t)$). In the case of wave equation Green's function for homogeneous medium is  
g^{(\pm)}({\bf r}, {\bf r}_0, t) =
 {\delta\left( t\mp {R_+\...
 ... R_+} -
 {\delta\left( t\mp {R_-\over v}\right) \over 4\pi R_-}\end{displaymath} (84)
where $R_\pm =\sqrt{(x-x_0)^2 + (y-y_0)^2 + (z\mp z_0)^2}$.It gives  
u^{(\pm)}({\bf r}, t) =
 & {1\over 2\pi}\... D} u({\bf r}_0, t\mp{R\over v}) \Big ] d{\bf
r}_0\end{array}\end{displaymath} (85)
where $R=\mid {\bf r} - {\bf r_0}\mid, \cos \theta =z/R$.

At big values of R the first term in square brackets can be neglected:  
u^\pm ({\bf r}, t)\simeq {1\over 2\pi}
 ...s\theta\over vR}{\bf D}u({\bf r}_0, t\mp{R\over v})
 d{\bf r}_0\end{displaymath} (86)

These expressions are valid only in 3D case. In 2D case Green's function is  
g^+({\bf r}, {\bf r}_0, t) = {1\over 2\pi}\left[{H(t-{R_+\ov...
 ... {H(t-{R_-\over v})\over\sqrt{t^2 - ({R_-\over v})^2}}
 \right]\end{displaymath} (87)

Inserting this into the equation (82) we obtain for the reverse continuation  
 U^{(-)}({\bf r}, t) =
 & {1\over \pi v}
 ...r v})\over
 \sqrt{\tau-t+{2R\over v}}
 \Big ]dx_0\end{array}\end{displaymath} (88)
where $\dot{u} ={\bf D}u$.

The operator  
I^{1\over 2}f(t) = \int\limits_{-\infty}^t {H(t-\tau)
 \over\sqrt{\pi(\tau - t)}} f(t)\, d\tau\end{displaymath} (89)
is by definition the operator of integration of the order 1/2. It is easy to notice that the operator

\int\limits_t^\infty {H(\tau-t)
 \over\sqrt{\pi(\tau - t)}} f(\tau)\, d\tau \end{displaymath}

is the same operator but acting in reverse time. We can denote this constriction as ${\bf I}_{(-)}^{1/2}$. Now we may also introduce ${\bf D}_{(\pm)}^{1/2} = {\bf D}_{(\pm)}
{\bf I}_{(\pm)}^{1/2}$ where signs + and - show what time (usual or reverse) is used.

Let the field u0 (x0, t) contain a wave at t=t0. The field u(-) will contain the wave at t = t0 - R/v. Then the main contribution in the integral (88) is made by the values of the field u0 at $\tau +R/v \simeq
t_0$, that is, $\tau -t + 2R/v \simeq t_0 - R/v - t_0 + R/v +
2R/v \simeq 2R/v$.

Taking into account all these considerations and neglecting the term with fast attenuating at $R\rightarrow\infty$, we derive from equation (88):  
U^{(-)}({\bf r}, t) \simeq {1\over \sqrt{2\pi}}\int
 ...vR}}{\bf D}_{(-)}^{1/2}
 u_0\left(x_0, t+{R\over v}\right) dx_0\end{displaymath} (90)

$\bullet$ Spectral representation in ($\omega, k$)-domain (the wave equation, homogeneous media):

u^{(-)}({\bf r},t) = {\bf F}^{-1}_{x,y,t} \left\{ {\tilde {u} _{0} e ^{ik_{z}^{\pm} z}} 
\right\}\end{displaymath} (91)

k_{z}^{\pm} = \left\{ \begin{array}
\mp \vert k_{z}\ver...
 ...\geq 0 \\ i \vert k_{z}\vert & k_{z} ^{2} < 0\end{array}\right.\end{displaymath}

k^2_z={\omega^2 \over v^2}-k^2_x-k^2_y\end{displaymath}

2. The Cauchy problem with respect to z in the $(\omega,{\bf k})$ domain for the equation

{d^{2} \tilde{u} \over dz^{2}} + k_{z}^{2} \tilde{u} = 0\end{displaymath} (92)
with the conditions,

\tilde{u} (\omega ,k_{x}, k_{y}; z=0) = ...
 ...\over \partial z})_{z=0}=ik_{z}^{\pm } \tilde{u}_{0}\end{array}\end{displaymath}

yields in homogeneous medium to the spectral form

\tilde{u}^{(\pm )} = \tilde{u}_{0}e^{ik_{z}^{\pm}z} .\end{displaymath}

This gives a stable (regular) solution, but the finite-difference solution of equation (92) is stable only in the restricted domain $z <z^{\ast }$ and unstable at $z \rightarrow
 \infty $.We have the same problem with respect to the Cauchy problem for the wave equation with the conditions:

u\vert _{z=0} = u_{0} \\  \\ {\partial u ...
 ... {\bf F}^{-1}
\left\{ {ik_{z}^{\pm } u_{0}} \right\}\end{array}\end{displaymath}

The natural way to stabilize Cauchy's problem is to cut off all frequencies at k2z < 0.

3. Reconstruction-type wave-field extrapolation. In this case we pose the conditions

u\vert _{z=0} = \tilde u_{0}\end{displaymath}

{\partial u \over \partial z} \vert _{z=0} = \sigma _{0}\end{displaymath} (93)
where condition (93) is connected with the source. The solution of this problem gives mixed-type continuation which is nonregular (at $z \rightarrow
 \infty $) in both the (r,t) and the $(\omega,{\bf k})$ domain. The simplest regularization is the same as above.

There is only one reverse type regular continuation into homogeneous medium which satisfies the condition:

u|z=0 =u0 (94)

Of course we have the whole family of continuation methods if we use instead of equation (94) the condition:

\left[ \alpha u + (1- \alpha ) {\partial u \over \partial z} \right] _{z=0} = u_{0}\end{displaymath}

for any $\alpha , 0< \alpha <1$.

In inhomogeneous media different conditions produce different solutions. Figure [*] shows an example of different solutions: (a) is the reversed continuation of a single event at $t=\tau _{0}$ for mixed conditions, and (b) for the Cauchy problem.

In Figure [*]a we use reverse time -t instead of t (we can do it with accordance with relation (51) in Chapter 5) because it promotes better understanding of reflection-refraction pattern at the interface z=d. In reverse time we have the usual process of wave propagation. Dashed lines show some ghost waves.

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