next up previous contents index

5.1.3 Steady-state solutions with operator adding or splitting

Suppose both operators \bgroup\color{DEFcolor}$A$\egroup and \bgroup\color{DEFcolor}$B$\egroup are zero,

A = B = 0
\end{displaymath} (214)
resulting in individually stationary solutions (eg. hydrostatic and radiative equilibrium),
q_A^{n+1} = q^{n}
\enspace , \enspace \enspace
q_B^{n+1} = q^{n}
\enspace ,
\end{displaymath} (215)
which gives for both operator combination methods
q_{A+B}^{n+1} = q^{n}
\enspace .
\end{displaymath} (216)
However, if the individual operators are non-zero but cancel each other,
B = -A \neq 0
\end{displaymath} (217)
the adding of the operator gives equilibrium Eq. (216), while operator splitting
$\displaystyle q_A^{n,*}$ $\textstyle =$ $\displaystyle q^{n} + \Delta t  A(q^{n})$  
$\displaystyle q_{A+-A}^{n+1}$ $\textstyle =$ $\displaystyle q_A^{n,*} - \Delta t  A(q_A^{n,*})
\enspace .$ (218)
This is not necessarily a stationary solution. Here, operator adding is superior.