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The solution to the finite difference scheme is obtained through
recursive back-substitution.
Equation 31 - is solved by,
| ![\begin{eqnarray}
A_{ij}^{n+\frac{1}{2}}= g_{22}\,(\mathbf{s}^n_{ij})
\quad
C_{...
...athbf{s}^n_{ij})\;\mathbf{S}_{ij}(u^n)\;+\;\mathbf{P}_{ij}\,(u^n).\end{eqnarray}](img60.gif) |
|
| (43) |
The solution for a fixed number j in range
obtained through,
| ![\begin{eqnarray}
u^{n+1/2}_{ij}= \alpha^{n+1/2}_{i+1j}\; u^{n+\frac{1}{2}}_{i+1\...
...
i=1,N-1,&
\;\;\;\;\;\;\;\;
u^{n+\frac{1}{2}}_{Nj} = \phi_1(1,jh),\end{eqnarray}](img62.gif) |
(44) |
where,
Equation 32 - is solved by,
| ![\begin{eqnarray}
A_{ij}^{n+\frac{1}{2}}= g_{22}\,(\mathbf{s}^n_{ij})
\quad
C_{...
...athbf{s}^n_{ij})\;\mathbf{S}_{ij}(v^n)\;+\;\mathbf{P}_{ij}\,(v^n).\end{eqnarray}](img64.gif) |
|
| (45) |
The solution for a fixed j in range
is obtained
through,
| ![\begin{eqnarray}
v^{n+1/2}_{ij}= \alpha^{n+1/2}_{i+1j}\; v^{n+\frac{1}{2}}_{i+1\...
...
i=1,N-1,&
\;\;\;\;\;\;\;\;
v^{n+\frac{1}{2}}_{Nj} = \phi_2(1,jh),\end{eqnarray}](img65.gif) |
(46) |
where,
Equation 33 - is solved by,
| ![\begin{eqnarray}
A_{ij}^{n+\frac{1}{2}}= g_{11}\,(\mathbf{s}^n_{ij})
\quad
C_{...
...^{n+1/2}_{ij}
- g_{11}\,(\mathbf{s}^n_{ij})\;\mathbf{S}_{ij}(u^n).\end{eqnarray}](img67.gif) |
|
| (47) |
The solution for a fixed i in the range
is
obtained
through,
| ![\begin{eqnarray}
u^{n+1}_{ij}= \alpha^{n+1}_{ij+1}\; u^{n+1}_{ij+1}+ \beta^{n+1}_{ij+1}, &
\quad
i=1,N-1,&
\quad
u^{n+1}_{iN} = \phi_1(ih,1),\end{eqnarray}](img69.gif) |
(48) |
where,
Equation 34 - is solved by,
| ![\begin{eqnarray}
A_{ij}^{n+\frac{1}{2}}= g_{11}\,(\mathbf{s}^n_{ij})
\quad
C_{...
...^{n+1/2}_{ij}
- g_{11}\,(\mathbf{s}^n_{ij})\;\mathbf{S}_{ij}(v^n).\end{eqnarray}](img71.gif) |
|
| (49) |
The solution for a fixed i in the range
is
obtained through,
| ![\begin{eqnarray}
v^{n+1}_{ij}= \alpha^{n+1}_{ij+1}\; v^{n+1}_{ij+1}+ \beta^{n+1}_{ij+1}, &
\quad
i=1,N-1,&
\quad
v^{n+1}_{iN} = \phi_2(ih,0),\end{eqnarray}](img72.gif) |
(50) |
where,
An approximate solution of the equations 19 will be
found at some large time TN that satisfies,
| ![\begin{displaymath}
\max_{0\le i,j,\le N} \frac{1}{\tau}
\sqrt{ (u^{n+1}_{ij}-u^n_{ij})^{2} + (v^{n+1}_{ij}-v^n_{ij})^{2}} \le \epsilon.\end{displaymath}](img74.gif) |
(51) |
Next: Step-by-Step Algorithm
Up: REFERENCES
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Stanford Exploration Project
4/5/2006