WIDE-ANGLE DEPTH MIGRATION

For increasing the accuracy at higher dips, we can use more terms in the Taylor-series expansion. Resulting in a matrix with thicker bands than the tridiagonal matrix. Using the second-order term in the expansion, we see that the approximated square-root operator takes the form

$$M = -i{\omega \over v}\left({\bf I} + {v^2 \over 2\omega^2 \... ...x^2}{\bf T} - {v^4 \over 8\omega^4 \Delta x^4}{\bf T}^2\right),$$ (21)

where I is the identity matrix, and T represents the tridiagonal matrix that approximates the second partial derivative. Let

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

with

$$a=-{i\over 2}({\omega \over v} -{v \over \omega \Delta x^2}-{6 v^3 \over 8\omega^3 \Delta x^4}),$$

$$b=-i({v \over 2\omega \Delta x^2}+{v^3 \over 2\omega^3 \Delta x^4}),$$

and

$$c=i{v^3 \over 8\omega^3 \Delta x^4}.$$

We split the matrix into five pieces, M=M_e+M_o+M_t1+M_t2+M_t3, where

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

$$M_o = \left( \begin{array} {cccccc} a&0&0&0&\cdots&0\\ 0&a... ...0\\ 0&0&0&a&\cdots&0\\ \cdots& & & & & \\ \end{array} \right),$$

$$M_{t1} = \left( \begin{array} {ccccccccccc} 0&0&c&.&.&.&.&... ...&0&\cdots&0\\ \cdots& & & & & & & & & &\\ \end{array} \right),$$

$$M_{t2} = \left( \begin{array} {ccccccccccc} 0&.&.&.&.&.&.&... ...&0&\cdots&0\\ \cdots& & & & & & & & & &\\ \end{array} \right),$$

and

$$M_{t3} = \left( \begin{array} {ccccccccccc} 0&0&.&.&.&.&.&... ...&.&\cdots&0\\ \cdots& & & & & & & & & &\\ \end{array} \right).$$

In the same manner as the tridiagonal matrix, we can approximate the time-stepping operator as

$$B= \exp(\Delta z M_e)\exp(\Delta z M_o)\exp(\Delta z M_{t1})\exp(\Delta z M_{t2})\exp(\Delta z M_{t3}).$$ (22)

M_t1, M_t2, and M_t3 are block diagonal, and the small block matrix F along the diagonal is defined as

$$F =\Delta z \left( \begin{array} {ccc} 0&0&c\\ 0&0&0\\ c&0&0\end{array}\right).$$

Using the series definition of the exponential function, we see that

$$\exp(F) ={1\over 2} \left( \begin{array} {ccc} \exp(c\Delt... ...a z)&0&\exp(c\Delta z)+\exp(-c\Delta z)\\ \end{array} \right).$$

Exponentiating M_t1, M_t2, and M_t3 amounts to exponentiating F. The terms $\exp(\Delta z M_{t1})$,$\exp(\Delta z M_{t2})$ and $exp(\Delta z M_{t3})$ are also block diagonal. Since the eigenvalues of $\exp(F)$ are 1 and $\exp(\pm c)$with imaginary c, it follows that $\Vert\exp(\Delta z M_{ti})\Vert = 1$.As before, to prove the unconditional stability of the algorithm, we need only show that $\Vert B\Vert\leq 1$. This follows immediately since

$$\Vert B\Vert=\Vert\exp(\Delta z M_e)\exp(\Delta z M_o)\exp(\Delta z M_{t1})\exp(\Delta z M_{t2})\exp(\Delta z M_{t3})\Vert$$

$$\leq\Vert\exp(\Delta z M_e)\Vert\Vert\exp(\Delta z M_o)\Vert... ...t\Vert\exp(\Delta z M_{t2})\Vert\Vert\exp(\Delta z M_{t3})\Vert$$

each of which is equal to 1 according to the preceding argument.

In Figure , we compared the impulse responses of first-order approximation with the second-order approximation in the Taylor-series expansion of the square-root operator. We also superposed semicircles which is the theoretical solution of the extrapolation operator on Fig and the higher order approximation shows better fitting to semicircles than the first-first order approximation.

With the same manner as showed in this section, we can get more accurate operator by taking more terms in a Taylor series expansion of the square-root operator. However, it will produce a matrix with increasing width of the band and thus will cause to an increasing in computation cost.

 

fig3
Figure 3 Impulse response of (a) explicit depth migration with the first-order approximation for the square-root operator and of (b) explicit depth migration with the second-order approximation for the square-root operator.