- #1
phyzmatix
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Good day everybody!
In my Numerical Methods textbook (Applied Numerical Analysis, 7ed, Gerald and Wheatley) the authors derive two equations for the ADI method to be used in an iteration scheme. For row-wise traversions, they get
[tex]uO^{(k+1)}=uO^{(k)}+\rho(uL-2uO+uR)^{(k+1)}+\rho(uA-2uO+uB)^{(k)}[/tex]
and for column-wise traversions
[tex]uO^{(k+2)}=uO^{(k+1)}+\rho(uA-2uO+uB)^{(k+2)}+\rho(uL-2uO+uR)^{(k+1)}[/tex]
Where u represents the nodes, A, B, L, R are above, below, left and right respectively, O is the node in the centre (current), k represents the iteration and rho is an acceleration factor.
I understand that we alternate between these equations for successive iterations (hence the name ) but what I don't get is that it seems to me that the value we're trying to calculate is dependent on itself, e.g how do we determine [tex]uO^{(k+1)}[/tex] if we don't yet have [tex](uL-2uO+uR)^{(k+1)}[/tex] ?
Any insight will be greatly appreciated!
phyz
In my Numerical Methods textbook (Applied Numerical Analysis, 7ed, Gerald and Wheatley) the authors derive two equations for the ADI method to be used in an iteration scheme. For row-wise traversions, they get
[tex]uO^{(k+1)}=uO^{(k)}+\rho(uL-2uO+uR)^{(k+1)}+\rho(uA-2uO+uB)^{(k)}[/tex]
and for column-wise traversions
[tex]uO^{(k+2)}=uO^{(k+1)}+\rho(uA-2uO+uB)^{(k+2)}+\rho(uL-2uO+uR)^{(k+1)}[/tex]
Where u represents the nodes, A, B, L, R are above, below, left and right respectively, O is the node in the centre (current), k represents the iteration and rho is an acceleration factor.
I understand that we alternate between these equations for successive iterations (hence the name ) but what I don't get is that it seems to me that the value we're trying to calculate is dependent on itself, e.g how do we determine [tex]uO^{(k+1)}[/tex] if we don't yet have [tex](uL-2uO+uR)^{(k+1)}[/tex] ?
Any insight will be greatly appreciated!
phyz