The Alternating Direction Implicit Method in 2-D

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SUMMARY

The discussion focuses on the Alternating Direction Implicit (ADI) method in 2-D as derived from the textbook "Applied Numerical Analysis, 7th Edition" by Gerald and Wheatley. The equations for row-wise and column-wise traversions are presented, highlighting the iterative nature of the method. Participants clarify that the calculation of the central node value, uO^{(k+1)}, relies on an initial guess and iterative refinement through a recursive loop until convergence is achieved. The necessity of a cut-off procedure for determining convergence is emphasized, along with the importance of understanding stability and convergence intricacies.

PREREQUISITES
  • Familiarity with the Alternating Direction Implicit (ADI) method
  • Understanding of iterative methods in numerical analysis
  • Knowledge of convergence criteria in numerical methods
  • Basic concepts of recursion in computational algorithms
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  • Study the implementation of the ADI method in MATLAB or Python
  • Explore convergence criteria and stability analysis in numerical methods
  • Learn about non-linear systems and their iterative solutions
  • Investigate alternative numerical methods for solving partial differential equations
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Numerical analysts, computational mathematicians, and students studying numerical methods who seek to understand the ADI method and its application in solving 2-D problems.

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

uO^{(k+1)}=uO^{(k)}+\rho(uL-2uO+uR)^{(k+1)}+\rho(uA-2uO+uB)^{(k)}

and for column-wise traversions

uO^{(k+2)}=uO^{(k+1)}+\rho(uA-2uO+uB)^{(k+2)}+\rho(uL-2uO+uR)^{(k+1)}

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 :biggrin:) 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 uO^{(k+1)} if we don't yet have (uL-2uO+uR)^{(k+1)} ?

Any insight will be greatly appreciated!
phyz
 
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Anybody? If I didn't provide enough information, please let me know. If you have access to a worked example, please provide it and I'll figure it out for myself. Even if you have knowledge of a different method that uses a similar iteration scheme, that will also be handy.
 
Okay, I think I understand your predicament, namely that you have an in-part, implicit system, where quantities are, indeed, defined between themselves and other equally unknown quantities.

What you generally do then, for example for non-linear systems, is

1. that you have a method for determining a FIRST CHOICE for the unknown quantity that you put into the problematic right-hand side.

2. On basis of this, you calculate directly what the left-hand side should be (ideally, you are to get out the same quantity!).

3. You now go into a recursive loop:
Your newly calculated quantity is plugged into the right-hand side, calculating a new left-hand side quantity.

4. At the end of each step, you perform a quantity comparison analysis, that utilizes a necessary criterion for Cauchy convergence:
Namely:
That in order for some sequence to converge, the terms themselves must become equal to each other, i.e, their difference must go to zero as you proceed along your sequence.

5. You need some cut-off procedure, like for example:
If my 10 last terms do not vary more than 0.01% from each other, then I am SATISFIED, and say that the loop has converged.

Of course, this is NOT mathematically rigorous, but is how we do it in practice, and just about the only way to do so in practice.

Generally, a deep familiarity with the intricacies of stability and convergence is needed in order to set up really intelligent cut-off procedures.
 
arildno said:
3. You now go into a recursive loop:
Your newly calculated quantity is plugged into the right-hand side, calculating a new left-hand side quantity.

Recursion! Of course! I sometimes forget that most (if not all) of these methods are designed specifically for use with computers!

Thanks arildno! :smile:
 

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