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The acoustic wave equation in two dimensional space is
|  |
(9) |
where x and z are the surface and depth axis, respectively.
By approximating the derivatives by finite differences, equation (8) becomes

|  |
(10) |
where i, j and k represent the time, depth and surface indices,
respectively.
By rearranging for time-extrapolation, we get the following equation
|  |
(11) |

where I assumed
.
The matrix representation of this operator
is the same as equation(4) except that the size of each matrix is now larger
than the corresponding matrix in equation(4)
because I need to express the two-dimensional
wavefield with an abstract vector form.
For a space of (z,x)=(5,6),
the matrix
becomes a block tridiagonal matrix as follows :
| ![\begin{displaymath}
{\bf T}=
\left[
\begin{array}
{cccccc}
{\bf R_1}&\bf A&\bf 0...
...\bf 0&\bf 0&\bf 0&\bf 0&{\bf A}&{\bf R_6}\\ \end{array}\right],\end{displaymath}](img26.gif) |
(12) |
where
is a diagonal matrix having the following elements

The matrix
represents a tri-diagonal matrix :
| ![\begin{displaymath}
{\bf R}_k=
\left[
\begin{array}
{ccccc}
2(1-2\alpha_{1,k})&\...
...k}\\ 0&0&0&\alpha_{5,k}&2(1-2\alpha_{5,k})\\ \end{array}\right]\end{displaymath}](img30.gif) |
(13) |
The conjugate operator can easily be obtained
by substitution of the transposed form of the new matrix
into the equation(7).
The code for 2-D algorithm, which has passed dot-product
test, can be found in the CD-rom version of this report.
Next: Conclusions
Up: Ji: Conjugate RTM
Previous: Example
Stanford Exploration Project
11/17/1997