APPENDIX A

In Cartesian coordinates, the eikonal equation is

$$\tau^2_x(x,z;x_c)+\tau^2_z(x,z;x_c) = s^2(x,z) \eqno(A.1)$$

where x_c=x_s or x_r are shot or receiver position. Taking partial derivative of both sides of equation (A.1) with respect to x_c gives

$$2\tau_x(x,z;x_c)\tau_{xx_c}(x,z;x_c)+2\tau_z(x,z;x_c)\tau_{zx_c}(x,z;x_c) = 0. \eqno(A.2)$$

The right-hand side of the equation is zero because the slowness model is independent of shot-receiver position. Assume the derivatives of the traveltimes are continuous functions of (x,z) and x_c, we have

$$\begin{array} {lllll} p_x(x,z;x_c) & = & \tau_{x_cx}(x,z;x_c... ...x_cz}(x,z;x_c) & = & \tau_{zx_c}(x,z;x_c).\end{array}\eqno(A.3)$$

Combining equations (A.2) and (A.3) gives

$$\tau_x(x,z;x_c)p_x(x,z;x_c)+\tau_z(x,z;x_c)p_z(x,z;x_c) = 0. \eqno(A.4)$$

Using the relations between Cartesian coordinates and polar coordinates:

$$\left\{ \begin{array} {lll} x & = & x_c+r\sin \theta \\ z & = & r\cos \theta,\end{array}\right. \eqno(A.5)$$

we can derive

$$\tau_r(r,\theta ;x_c)p_r(r,\theta;x_c)+{1 \over r^2}\tau_\theta(r,\theta;x_c)p_\theta(r,\theta;x_c) = 0. \eqno(A.6)$$