Dear all,
I came across the figure attached from a text in CFD using FVM. As per the text, the final convergence rate decreases with mesh size. However, I am unable to understand the information from the plot.
From what I understand, the rate of convergence will decrease in the residual/number of iterations. This means that the plot with maximum slope on the residual-iteration plot will have a maximum convergence which, from the figure, is for the finest mesh. However, it doesn't make sense to me. Where I am going wrong?
Thanks
Perhaps I misunderstand you question, but from the plot I would say that the 10x10 has the highest convergence rate, because the residual velocity drops faster than for the others. That matches the statement that the rate of convergence decreases with mesh-size.
Hi,
Oh yes. This was so simple. I was thinking 10 x 10 as cell size. utter misinterpretation.
apologies.
deewakar
dear colleague
in my humble opinion you need to compare answers matrix too.
while you are using fvm, they will converge to a mathematically true answer, but thy haven't same physical meaning.
does this three runs have same answer in quality?
regards
Dear Mahdi Izadi,
You're right about the physical meaning. The convergence in this plot is just a numerical matter. Whether it can pysically resolve the morphology of the fluid is another issue.
Dear Deewakar,
I'm not sure what software you are using or what is your solution method. But from the figures you provided it seems you are not using a very good solver. From the number of grid points you are using and heavy dependance of convergence vs number of iterations I assume you should be using some sort of basic iterative solvers maybe incomplete LU or similar methods.
For example compare the first two figures provided here (simulations with 32^3 nodes from my thesis a few years ago) as you can see using incomplete decomposition methods take significantly more number of iterations than preconditioned-conjugate gradient methods.
However still preconditioned conjugate gradient methods are not exactly scalable (see the third figure with 64^3 and 128^3 points). In this figure you can see that even preconditioned CG method is not scalable (although dependance is much weaker) and only Multigrid methods can provide you with the required scalability (see that both 64 and 128^3 grid point require 4 iteration to converge). So if you are planning to perform some real simulation think about at least using preconditioned CG method which for your size of problem should converge with probably less than 10 iterations.
Good luck
Sina
Dear Sina
Thanks for your kind response. It was not from my own work. As mentioned in ques it was figure from FVM book by Malasekkra though I realized after posting question and Allard reply that I had misinterpreted the figure.
But your reply was till enlightening. Thank you
I would to ask something reg the plot. I guess on y axis you have the residual. Is this normalized residual ? How have you defined residual here?
Thanks
Dear Deewakar,
||r|| is the norm of the residual vector which is defined as b-Ax^k. This is not deficult to calculate but the more tricky part is how to use it for stopping the iterations.
A simple stopping criteria such as ||r^k|| < tol ||r^0|| is not always appropriate (for example this criteria depends heavily on the initial guess think about starting from zero). A better choice would be ||r^k|| < tol (||A|| . ||x^k|| + ||b||) but of course is more expensive to calculate. For time dependant calculation however, since the iteration would start from the previous time step solution then ||r^k||
Hi
I got your point to calculate ||r||. But how do in practical I am still perplexed about.
Say I start with some initial guess (steady state problem) with X=X0.
Using the values A matrix is formed and then we solve for X using AX=b.Say this is the first iteration Now comes the confusion, since I am to calculate residual r=b-AX, where X was calculated using AX=b, which means X satisfies b-AX, so how to residual.
Intuitively I am bale to understand but implementation wise there is some confusion.
Looking forward for kind response.
Thanks
your r^0 would then be b-Ax^0 (x^0 is initial guess input to the linear solver) . You should then update the residual as the process proceeds as r^k=b-Ax^(k-1). Then you can use if(||r^k||= maxit) to stop.
For a theoretical discussion as why these stopping criteria are sufficient look at intermidiate linear algebra book such as Matrix Computations by Golub and look for forward and backward error estimation.
Hi
Thanks for kind response. The expression r^k=b-Ax^(k-1) was something I wanted to confirm.
Not only the size of the mesh but the mesh quality is equally important to achieve good convergence. All these thing go into understanding the flow physics in engineering problems and to optimize them or to make them efficient. Grid resolution becomes ans import aspect in the key regions of the domain. The plot only talks about the convergence base on grid size which is incomplete.
I think that the plot that you added is related with convergence velocity vs mesh size. Sometimes you can obtain a converged solution more fastest with coarser meshes than with fine meshes, but other problem is the quality of the obtained results... For more details you can check a serie of papers from Coleman about this subject, this serie is a bit older, and for structured meshes, but is very interesting to understand the problem. Also, as Mr Varpe say, the mesh quality is a critical issue too, both to obtain a converged solution as good results.
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