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Solving the Cauchy problem

 To obtain an explicit solution of the Cauchy problem ([*]-[*]) for equation ([*]), it is convenient to apply the following simple transform of the wavefield P:  
 \begin{displaymath}
P(t_n,h,y)=Q(t_n,h,y)\,t_n\,H(t_n)\;.\end{displaymath} (22)
Here the Heavyside function H is included to take into account the causality of the reflection seismic gathers (note that the time tn=0 corresponds to the direct wave arrival). We can extrapolate Q as an even function to negative times, writing the reverse of (A-22) as follows:  
 \begin{displaymath}
Q(t_n,h,y)=Q(-t_n,h,y)=P(\vert t_n\vert,h,y)/\vert t_n\vert\;.\end{displaymath} (23)
With the change of function (A-22), equation ([*]) transforms to  
 \begin{displaymath}
h \, {\partial^2 Q \over \partial y^2} = h\, {\partial^2 Q \...
 ...artial
h} + t_n \, {\partial Q \over {\partial t_n}}\right)
\;.\end{displaymath} (24)
Applying the change of variables  
 \begin{displaymath}
\rho={t_n^2 \over 2}\;,\;\nu={h^2 \over {2\,t_n^2}}\end{displaymath} (25)
and Fourier transform in the midpoint coordinate y  
 \begin{displaymath}
\widetilde{Q}(\rho,\nu,k)=\int\,Q(\rho,\nu,y)\,\exp (-iky)\,dy\;,\end{displaymath} (26)
I further transform equation (A-24) to the canonical form of a hyperbolic-type partial differential equation with two variables:  
 \begin{displaymath}
{\partial^2 \widetilde{Q} \over {\partial \rho \, \partial \nu}} + 
k^2\,\widetilde{Q} = 0\;.\end{displaymath} (27)

 
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Figure 1
Domain of dependence of a point in the transformed coordinate system.

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view

The initial value conditions ([*]) and ([*]) in the $\{\rho,\nu\}$ space are defined on a hyperbola of the form $\rho\,\nu=\left(h_1 \over 2\right)^2=\mbox{constant}$. Now the solution of the Cauchy problem follows directly from Riemann's method (). According to this method, the domain of dependence of each point $\{\rho,\nu\}$ is a part of the hyperbola between the points $\{\rho,{h_1^2 \over {4\,\rho}}\}$ and $\{{h_1^2
\over {4\,\nu}},\nu\}$ (Figure A-1). If we let $\Sigma$denote this curve, the solution takes an explicit integral form:
   \begin{eqnarray}
\widetilde{Q}(\rho,\nu) & = &
{1 \over 2}\, \widetilde{Q}(\rho,...
 ...\rho_1,\nu_1;\rho,\nu) \over {\partial \nu_1}}
\right)\,d \nu_1\;.\end{eqnarray}
(28)
Here R is the Riemann's function of equation (A-27), which has the known explicit analytical expression  
 \begin{displaymath}
R(\rho_1,\nu_1;\rho,\nu)=
J_0\left(2k\,\sqrt{\left(\rho_1-\rho\right)\,
\left(\nu_1-\nu\right)}\right)\;,\end{displaymath} (29)
where J0 is the Bessel function of zeroth order. Integrating by parts and taking into account the connection of the variables on the curve $\Sigma$, we can simplify equation (A-28) to the form  
 \begin{displaymath}
\widetilde{Q}(\rho,\nu)=
\widetilde{Q}_0(\rho,\nu)+
\widetilde{Q}_1(\rho,\nu)\;,\end{displaymath} (30)
where
      \begin{eqnarray}
\widetilde{Q}_0(\rho,\nu) & = &
{\partial \over {\partial \rho}...
 ...l \widetilde{Q}(\rho_1,\nu_1) \over {\partial \nu_1}}
\,d \nu_1\;.\end{eqnarray} (31)
(32)

Applying the explicit expression for the Riemann function R (A-29) and performing the inverse transform of both the function and the variables allows us to rewrite equations (A-30), (A-31), and (A-32) in the original coordinate system. This yields the integral offset continuation operators in the $\{t_n,h,k\}$ domain  
 \begin{displaymath}
\widetilde{P}(t_n,h,k)=
H(t_n)\,\left(\widetilde{P}_0(t_n,h,k) +
t_n\,\widetilde{P}_1(t_n,h,k)\right)\;,\end{displaymath} (33)
where
      \begin{eqnarray}
\widetilde{P}_0 & = & 
{\partial \over {\partial t_n}}\,
\int_{...
 ...}\right)\,
\left(t_n^2-t_1^2\right)}\right)\,{dt_1 \over t_1^2}\;,\end{eqnarray} (34)
(35)
      \begin{eqnarray}
\widetilde{P}^{(j)}_1(t_1,k) & = & 
\int\,P\,^{(j)}_1(t_1,y_1)\...
 ...lde{P}(t_n,h,k) & = & 
\int\,P(t_n,h,y)\exp (-iky)\,dy\;(j=0,1)\;.\end{eqnarray} (36)
(37)

The inverse Fourier transforms of equations (A-34) and (A-35) are reduced to analytically evaluated integrals () to produce explicit integral operators in the time-and-space domain  
 \begin{displaymath}
P(t_n,h,y)=\mbox{sign}(h-h_1)\,
{H(t_n) \over \pi}\,\left(P_0(t_n,h,y) +
t_n\,P_1(t_n,h,y)\right)\;,\end{displaymath} (38)
where
      \begin{eqnarray}
P_0(t_n,h,y) & = & 
{\partial \over {\partial t_n}}\,
\iint_{\S...
 ..._1^2 \over t_1^2}\right)\,
\left(t_n^2-t_1^2\right)-(y-y_1)^2}}\;.\end{eqnarray} (39)
(40)
The range of integration $\Sigma$ in (A-39) and (A-40) is defined by the inequality  
 \begin{displaymath}
\left({h^2 \over t_n^2}-{h_1^2 \over t_1^2}\right)\,
\left(t_n^2-t_1^2\right)-(y-y_1)^2 \gt 0\;.\end{displaymath} (41)

Equations (A-38), (A-39), and (A-40) coincide with ([*]), ([*]), and ([*]) in the main text.


next up previous print clean
Next: The kinematics of offset Up: Table of Contents Previous: Second-order reflection traveltime derivatives
Stanford Exploration Project
12/30/2000