Convergence criteria

lost.identity · August 2, 2010, 11:55

Hi,

I'm using the SIMPLER method with a certain under-relaxation factor. Initially my convergence criteria was based on L2-norm as follows

$error = \frac{||R||^n_2}{N} \leq 10^{-4}$

where R is the difference between the previous iteration and the current iteration. However I noticed that this criteria only gave the analytical solution when the URF=1, which I later found out that with a URF<1, it gives me false convergence as the difference between the two iteraitons is reduced.

I then decided to use the following criteria instead

$error = \frac{||R||^n_2}{||R||^m_{2_{max}}} \leq 10^{-4}$

where $||R||^m_{2_{max}}$ denotes the maximum error after m iterations. Now I seem to get the correct result, I'm just wondering is this a normal way to determine convergence for such cases?

Thanks.

praveen · August 2, 2010, 12:11

This can be a tricky issue. Suppose you are solving Au=f using an iterative method. Then

$\|u^{n+1}-u^n\|$

is not a good convergence measure. A better one is

$\| A u^{n+1} - f\|$

but even this can be misleading if A is poorly conditioned. Assuming A is not very badly conditioned, a still better one is

$\frac{ \| A u^{n+1} - f\|}{\|f\|}$

lost.identity · August 2, 2010, 12:19

Thank you very much. One of the problem is when I look at papers everyone seems to be using their own convergence criteria. Do you know of any book that explains how this is done? The Kreyszig Engineering maths book gives your first method.

I've seen a paper where it was done as (using your notation)

$\frac{||u^{n+1}-u^n||}{||f||}$

I think this is similar to what you gave at the end but without the coefficient matrix A being multiplied. Is it necessary to multiply it?

praveen · August 2, 2010, 12:23

If your purpose is to solve Au=f, then it is natural to measure convergence through the difference ||Au-f|| since this measures how well the equation is satisfied by the current solution.

Dividing by ||f|| will remove scale effects.

August 2, 2010, 11:55	Convergence criteria	#1
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Hi, I'm using the SIMPLER method with a certain under-relaxation factor. Initially my convergence criteria was based on L2-norm as follows $error = \frac{\|\|R\|\|^n_2}{N} \leq 10^{-4}$ where R is the difference between the previous iteration and the current iteration. However I noticed that this criteria only gave the analytical solution when the URF=1, which I later found out that with a URF<1, it gives me false convergence as the difference between the two iteraitons is reduced. I then decided to use the following criteria instead $error = \frac{\|\|R\|\|^n_2}{\|\|R\|\|^m_{2_{max}}} \leq 10^{-4}$ where $\|\|R\|\|^m_{2_{max}}$ denotes the maximum error after m iterations. Now I seem to get the correct result, I'm just wondering is this a normal way to determine convergence for such cases? Thanks.

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August 2, 2010, 12:11		#2
praveen Super Moderator Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 343 Blog Entries: 6 Rep Power: 18	This can be a tricky issue. Suppose you are solving Au=f using an iterative method. Then $\\|u^{n+1}-u^n\\|$ is not a good convergence measure. A better one is $\\| A u^{n+1} - f\\|$ but even this can be misleading if A is poorly conditioned. Assuming A is not very badly conditioned, a still better one is $\frac{ \\| A u^{n+1} - f\\|}{\\|f\\|}$

August 2, 2010, 12:19		#3
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Thank you very much. One of the problem is when I look at papers everyone seems to be using their own convergence criteria. Do you know of any book that explains how this is done? The Kreyszig Engineering maths book gives your first method. I've seen a paper where it was done as (using your notation) $\frac{\|\|u^{n+1}-u^n\|\|}{\|\|f\|\|}$ I think this is similar to what you gave at the end but without the coefficient matrix A being multiplied. Is it necessary to multiply it?

August 2, 2010, 12:23		#4
praveen Super Moderator Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 343 Blog Entries: 6 Rep Power: 18	If your purpose is to solve Au=f, then it is natural to measure convergence through the difference \|\|Au-f\|\| since this measures how well the equation is satisfied by the current solution. Dividing by \|\|f\|\| will remove scale effects.