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SOLVING THE CAUCHY PROBLEM

To obtain an explicit solution of the Cauchy problem (1)-(3), it is convenient to apply the following simple transform of the function P:

 
 \begin{displaymath}
P(t_n,h,y)=Q(t_n,h,y)\,t_n\,H(t_n)\;.\end{displaymath} (46)

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 evenly extrapolate the function Q to negative times, writing the reverse of (47) 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} (47)
With the change of function (47), equation (1) 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} (48)
Applying the change of variables

 
 \begin{displaymath}
\rho={t_n^2 \over 2}\;,\;\nu={h^2 \over {2\,t_n^2}}\end{displaymath} (49)
and Fourier transform in the midpoint coordinate y

 
 \begin{displaymath}
\widetilde{Q}(\rho,\nu)=\int\,Q(\rho,\nu,y)\,\exp (-iky)\,dy\;,\end{displaymath} (50)
I further transform equation (49) 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} (51)

The initial value conditions (2) and (3) 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 Courant (1962). According to this method, the domain of dependence of each point $\{\rho,\nu\}$ is a part of the hyperbola between the points $\{\rho,{h_1^2 \over {4\,\nu}}\}$ and $\{{h_1^2
\over {4\,\rho}},\nu\}$ (Figure 4). If we let $\Sigma$denote this curve, the solution takes an explicit integral form:

 
offrim
Figure 4
Domain of dependence of a point in the transformed coordinate system.

offrim
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\begin{displaymath}
\widetilde{Q}(\rho,\nu)=
{1 \over 2}\, \widetilde{Q}(\rho,{h...
 ...er 2}\, \widetilde{Q}({h_1^2 \over{4\,\rho}},\nu) + 
\nonumber \end{displaymath}   
\begin{displaymath}
+ {1 \over 2}\, \int_{\Sigma}\left(R(\rho_1,\nu_1;\rho,\nu)\...
 ...o,\nu) \over {\partial \rho_1}}
\right)\,d \rho_1 - 
\nonumber \end{displaymath}   
 
 \begin{displaymath}
- {1 \over 2}\, \int_{\Sigma}\left(R(\rho_1,\nu_1;\rho,\nu)\...
 ...o_1,\nu_1;\rho,\nu) \over {\partial \nu_1}}
\right)\,d \nu_1\;.\end{displaymath} (52)
Here R is the Riemann's function of equation (52), 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} (53)
where J0 is Bessel's function of zero order. Integrating by parts and taking into account the connection of the variables on the curve $\Sigma$, we can simplify formula (55) to the form

 
 \begin{displaymath}
\widetilde{Q}(\rho,\nu)=
\widetilde{Q}_0(\rho,\nu)+
\widetilde{Q}_1(\rho,\nu)\;,\end{displaymath} (54)
where

 
 \begin{displaymath}
\widetilde{Q}_0(\rho,\nu) = 
{\partial \over {\partial \rho}...
 ...o_1,\nu_1;\rho,\nu)\,
\widetilde{Q}(\rho_1,\nu_1)\,
d \rho_1\;,\end{displaymath} (55)
 
 \begin{displaymath}
\widetilde{Q}_1(\rho,\nu) = 
- \int_{\Sigma} R(\rho_1,\nu_1;...
 ...widetilde{Q}(\rho_1,\nu_1) \over {\partial \nu_1}}
\,d \nu_1\;.\end{displaymath} (56)

Applying the explicit expression for the Riemann's function R (56) and performing the inverse transform of both the function and the variables allows us to rewrite equations (57), (58), and (59) 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} (57)
where

 
 \begin{displaymath}
\widetilde{P}_0 = 
{\partial \over {\partial t_n}}\,
\int_{\...
 ...\over t_1^2}\right)\,
\left(t_n^2-t_1^2\right)}\right)\,dt_1\;,\end{displaymath} (58)
 
 \begin{displaymath}
\widetilde{P}_1 = 
\int_{\left(h_1/h\right)\,t_n}^{t_n}
h_1\...
 ...ight)\,
\left(t_n^2-t_1^2\right)}\right)\,{dt_1 \over t_1^2}\;,\end{displaymath} (59)
 
 \begin{displaymath}
\widetilde{P}^{(j)}_1(t_1,k) = 
\int\,P\,^{(j)}_1(t_1,y_1)\exp (-iky_1)\,dy_1\;(j=0,1)\;,\end{displaymath} (60)
 
 \begin{displaymath}
\widetilde{P}(t_n,h,k) = 
\int\,P(t_n,h,y)\exp (-iky)\,dy\;(j=0,1)\;.\end{displaymath} (61)

The inverse Fourier transforms of formulas (61) and (62) are reduced to analytically evaluated integrals Gradshtein and Ryzhik (1994) 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} (62)
where

 
 \begin{displaymath}
P_0(t_n,h,y) = 
{\partial \over {\partial t_n}}\,
\int\!\!\i...
 ...2 \over t_1^2}\right)\,
\left(t_n^2-t_1^2\right)-(y-y_1)^2}}\;,\end{displaymath} (63)
 
 \begin{displaymath}
P_1(t_n,h,y) = 
\int\!\!\int_{\Sigma}
{\left(h_1 / t_1^2\rig...
 ...2 \over t_1^2}\right)\,
\left(t_n^2-t_1^2\right)-(y-y_1)^2}}\;.\end{displaymath} (64)
The range of integration $\Sigma$ in (66) and (67) is defined by inequality

 
 \begin{displaymath}
\Theta(t_1,h_1,y_1;t_n,h,y) \gt 0\;,\end{displaymath} (65)
where $\Theta$ is in turn defined by formula (7). Formulas (65), (66), and (67) coincide with (4), (5), and (6) in the main text.

B


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Next: THE KINEMATICS OF OFFSET Up: Fomel: Offset continuation Previous: REFERENCES
Stanford Exploration Project
11/12/1997