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Tridiagonal system

The tridiagonal matrix to be solved is built using:


  
$\displaystyle \text{odd } k \qquad$ $\textstyle A(k) \eta^{n+{\scriptscriptstyle \frac{1}{2}}}_{i\;j}
+ B(k) u^{n+1}_{i\;j} + C(k) \eta^{n+{\scriptscriptstyle \frac{1}{2}}
}_{i\;j+1} = D(k)$   (7.12)
$\displaystyle \text{even } k \qquad$ $\textstyle A(k) u^{n+1}_{i\;j-1} + B(k) \eta^{n+{
{\scriptscriptstyle \frac{1}{2}}}}_{i\;j} + C(k) u^{n+1}_{i\;j} = D(k)$   (7.13)

An iteration from k=ki to k=kf. 

  
  $\textstyle \text{odd } k$ $\displaystyle \begin{cases}A_k &= -\frac{g \Delta t}{2 \Delta x} \\  B_k
&= 1 +...
...tstyle \frac{1}{2}}
\Delta t \varepsilon \mathcal{D} u^n_{i\;j} \\  \end{cases}$ (7.14)
  $\textstyle \text{even } k$ $\displaystyle \begin{cases}A_k &=-\frac{\Delta t}{2 \Delta x}
H^n_{i\;j-1} \\  ...
...}
v_{i\;j}^n - H^n_{i-{\scriptstyle \frac{1}{2}}\; j} v_{i-1\;j}^n) \end{cases}$ (7.15)



Elias Kaplan M.Sc.
1998-07-22