15-DEGREE DEPTH MIGRATION

The first-order approximations of the one-way wave equation by both continued-fraction expansion and Taylor-series expansion result in the same equation: the conventional 15-degree equation. The 15-degree migration equation in the $(\omega, x, z)$domain (Claerbout, 1985) is given by

$${\partial P\over\partial z} = -i({\omega \over v} + {v \over 2\omega} {\partial^2 \over\partial x^2})P.$$ (6)

After a second difference approximation to the second partial derivative and subsequent simplification, the equation (6) becomes

$${\partial P\over\partial z} = M P,$$ (7)

where the matrix M takes the form

$$M = -i{\omega \over v} \left( \begin{array} {ccccc} 1& & &... ... \\ &1&-2&1 & \\ & &1&-2&1 \\ & & &1&-2\end{array} \right).$$

Rearranging,

$$M = \left( \begin{array} {cccccc} 2a&b&0&0&\cdots&0\\ b&2a... ...0\\ 0&b&2a&b&\cdots&0\\ \cdots& & & & & \\ \end{array} \right)$$

with

$$a=-{i\over 2}({\omega \over v} -{v \over \omega \Delta x^2})$$

and

$$b=-i{v \over 2\omega \Delta x^2}.$$

The solution to equation (7) is

$$P(z+\Delta z)=\exp(\Delta z M)P(z).$$ (8)

The calculation of the exponentiated matrix $\exp(\Delta z M)$ requires matrix diagonalization, which is numerically expensive. However, we can approximate the exponential of the matrix by writing (Richardson, et al., 1991)

$$\exp(\Delta z M)= \exp(\Delta z M_e)\exp(\Delta z M_o)+\epsilon(\Delta z^2),$$ (9)

where the matrix M is split into two matrices M=M_e+M_o, with

$$M_e = \left( \begin{array} {cccccc} a&b&.&.&...&.\\ b&a&.&... ...&b&...&.\\ .&.&b&a&...&.\\ ...& & & & & \\ \end{array} \right)$$

and

$$M_o = \left( \begin{array} {cccccc} a&.&.&.&...&.\\ .&a&b&... ....&...&.\\ .&.&.&a&...&.\\ ...& & & & & \\ \end{array} \right).$$

The approximate depth-stepping operator,

$$B= \exp(\Delta z M_e)\exp(\Delta z M_o)$$ (10)

forms the basis for our unconditionally stable, explicit algorithm for migration. To compute the matrix exponentials, notice that both M_e and M_o are block diagonal and we need only consider the exponential of the 2 by 2 matrix

$$E =\Delta z \left( \begin{array} {cc} a&b\\ b&a\end{array} \right).$$

By using the eigenvalue decomposition of the matrix, we see that

$$\exp(E) = Q \exp(\Lambda) Q^{-1}$$

$$={1\over 2} \left( \begin{array} {cc} \exp(\Delta z(a+b))+\... ...))&\exp(\Delta z(a+b))+\exp(\Delta z(a-b))\end{array} \right),$$

where Q represents the matrix whose columns are eigenvectors of the matrix E, and $\Lambda$ is the diagonal eigenvalue matrix.

Since both M_e and M_o are block diagonal, exponentiating them amounts to exponentiating E. Both $\exp(\Delta z M_e)$ and $\exp(\Delta z M_o)$are also block diagonal. Since the eigenvalues of $\exp(E)$ are $\exp(a+b)$and $\exp(a-b)$ with a and b imaginary, it follows that $\Vert\exp(\Delta z M_e)\Vert = 1$ and $\Vert\exp(\Delta z M_o)\Vert = 1$.To prove the unconditional stability of the algorithm we need only show that $\Vert B\Vert\leq 1$. This follows immediately since $\Vert B\Vert=\Vert\exp(\Delta z M_e) \exp(\Delta z M_o)\Vert \leq\Vert\exp(\Delta z M_e)\Vert\Vert\exp(\Delta z M_o)\Vert$, each of which 1 according to the preceding discussion.

In Figure , the impulse responses of both implicit and explicit operators of 15-degree migration are compared for the same extrapolation step $\Delta z/ \Delta x = 1$. In the case of the implicit scheme, we used a Crank-Nicholson implementation which has an accuracy on the order of $\Delta z^2$.For the explicit scheme, we performed a second-order approximation using a split M = M_e + M_o whose accuracy is on the order of $\Delta z$.We see that the impulse response of the explicit method has the shape of an ellipse, which is characteristic of the 15-degree extrapolation operator. The impulse response of the explicit method shows some dispersion than that of the implicit scheme due to a poor approximation of the matrix exponent. The way to get a more accurate approximation is discussed in the following section.

 

fig1
Figure 1 The impulse responses (a) of explicit 15-degree migration with split M = M_e + M_o and (b) of implicit 15-degree migration with Crank-Nicholson approximation. Both extrapolations are performed with the same extrapolation step $\Delta z/ \Delta x = 1$.