Zero-offset case

Unfortunately equation (59) is valid only for common source and common receiver gathers. Let us take, for instance, the common offset pattern in the 2D case :

$$s=X-{l\over 2}$$

$$r=X+{l\over 2}$$

at fixed l. In this case we have datum 

$$\tau_{0}(X)=\tau(X-{l\over 2},X+{l\over 2})$$

where $\tau(s,r)$ is the arrival time of a reflected wave propagating from source s to receiver r. Then

$${{\partial\tau_{0}}\over \partial X}={{\partial\tau}\over \partial s}+{{\partial\tau}\over \partial r}.$$

We can't separately determine values ${\partial\tau\over \partial s}$ and ${\partial \tau \over \partial r}$ and, consequently, can't determine initial conditions for eiconal $\tau$.We see that the reverse eikonal's continuation with datum $\tau_{0}$ can't reconstruct the true location of the seismic rays. The problem simplifies greatly when l=0 (zero offset pattern). In this case incidence and reflected rays coincide, and due to the principle of reciprocity,

$${{\partial\tau}\over \partial s}={{\partial\tau}\over \partial r}$$

so

$${{\partial\tau_{0}}\over \partial X}= 2{{\partial\tau}\over \partial r}$$

and  

$${{\partial ({1\over 2}{\tau_{0}})}\over \partial X}={{\partial\tau}\over \partial r}.$$ (61)

This means that the solution $\tau^{(-)}$ of equation (61) with initial condition  

$$\tau{\vert}_{\sum}={1\over 2}{\tau_{0}}$$ (62)

will give the true ray pattern although the travel-times will be different. This is not bad. Since the time of propagation from the reflector R to surface $\Sigma$ along the normal ray equals ${\tau_{0}\over 2}$, the condition  

$$\tau^{(-)}{({\bf r})}=0$$ (63)

determines the location of the reflector. We could obtain the same result by solving the half-velocity eikonal equation  

$${\vert\nabla \tau\vert}^{2}={1\over {({v\over 2})^{2}}}$$ (64)

with initial datum  

$$\tau\vert _{\Sigma}=\tau_{0}{(X)}.$$ (65)

Usually this scheme is obtained from some thought experiment (i.e., the exploding reflector hypothesis). Let us find the forward eikonal's continuation in the medium with velocity ${v({\bf r})\over 2}$ and initial condition  

$$\tau\vert _{R}=0.$$ (66)

This corresponds to a thought experiment with an impulsive source that is distributed along the surface R at t=0). It is easy to understand that the solution $\tau^{+}({\bf r})$ of this problem will coincide on the surface $\Sigma$ with $\tau_{0}{(X)}$. It is clear that the reverse eikonal's continuation $\tau^{(-)}({\bf r})$ of $\tau_{0}{(X)}$in a medium with half-velocity has the isochron $\tau^{(-)}=0$ on the surface R.