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May 14, 2008, 05:11 
Re: Reg. BICGSTAB

#21 
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Yes, it will. Try descretization by finite volume method. lol.
I will give you a quote from Peric's book: "; moreover, its coefficient matrix is symmetric so special solvers for symmetric matrices can be used (e.g ICCG solver from the conjugate gradients family ...... )" Chapter 8. Complex Geometries. I do not need Peric to tell me that it will be symmetric, I have already written code for unstructured grids and I know it is symmetric for incompressible cases. It is unsymmetric for compressible because it has one more term due to density correction. 

May 14, 2008, 06:32 
Re: Reg. BICGSTAB

#22 
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You're wrong, try the suggested 1D example on a nonuniform grid  you'll end up with something of the form
a.P_i1 + b.P_i + c.P_i+1 = f_i with, in general, a /= c => matrix is not symmetric! 

May 14, 2008, 07:09 
Re: Reg. BICGSTAB

#23 
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Ok I am wrong , you won, happy now.


May 14, 2008, 08:00 
Re: Reg. BICGSTAB

#24 
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You need to consider symmetry about the diagonal i.e. does A_i,j = A_j,i ?
An offdiagonal entry in row i column j of a symmetric matrix correspond with another in a different row. Your example is limited to terms in a single row of the matrix, so no conclusion on symmetry can be drawn from your example. Infact, you're just considering if A_(ib),i = A_(i+b),i. That's a very different property and is not generally true. andy 

May 14, 2008, 08:38 
Re: Reg. BICGSTAB

#25 
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You need a(i+1) = c(i) for all the i's which in general is not true on a uniform grid  I just missed the indices off. If h_i is the spacing c(i) is calulated using h_(i1) and h_i while a(i+1) will be calculated with h_i and h_(i+1). This is why it will not be symmetric.


May 14, 2008, 09:02 
Re: Reg. BICGSTAB

#26 
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May 14, 2008, 09:15 
Re: Reg. BICGSTAB

#27 
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Ok... I'm still confused. Please can you point out the error in the following, as I cannot spot it myself!!
In general, for any node index k: Rule 1) a_k references the node/cell to the left of node k and uses h_k and h_(k1) Rule 2) c_k references the node/cell to the right of node k and uses h_k and h_(k+1) Now for symmetry, we require a_(j+1) = c_j, as you say. By applying rule (1) above we find that a_(j+1) uses h_(j+1) and h_j. By appying rule (2) above we find that c_j uses h_j and h_(j+1). So ISTM we are using the same h's. This again indicates symmetry. I hope we can get to the bottom of this!! I certainly thought this matrix should be symmetric  it's a Poisson equation so there's no upwind terms to break the symmetry, but I am open to evidence to the contary. Best regards, Andy 

May 14, 2008, 09:23 
Re: Reg. BICGSTAB

#28 
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You should ask yourself if you're a better mathematician than me before posting such things!
Check my counter example more closely (the matrix is tridiagonal so symmetry requires a(i+1)=c(i) as I stated) or give a completely rigourous proof that the discrete form is always symmetric no matter what numerical technique is used!  hint you will not be able do this because there is a counter example. 

May 14, 2008, 09:38 
Re: Reg. BICGSTAB

#29 
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I pointed you to the wiki because you do not understand the basic definition of symmetric matrix which requires that for the matrix to be symmetric Aij = Aji .
That is the other element that should be equal to Aij does not even fall on same raw. (Aji is not present in the raw that contains Aij, in the matrix). Now what you keep on writing and stubbornly arguing is the comparing the elements on the same raw and telling that since they are not same matrix is not symmetric. I gave you a quote from Peric's book but you still think you are more intellegent than Peric. From the wiki link I pointed you, an simple example of symmetric matrix:  1 2 3   2 4 5   4 5 0  this is symmetric matrix and it is not symmetric according to your logic , since in raw 2 : 2 != 5 Learn basic maths first, will help you a lot in life. 

May 14, 2008, 09:44 
correction

#30 
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the matrix i quoted should read
 1 2 3   2 4 5   3 5 6  Just a small typo. 

May 14, 2008, 09:47 
Re: Reg. BICGSTAB

#31 
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It's the fact that the three h's are not equal for a stretched grid that cause's the asymmetry. You really can't talk about the symmetry of the matrix without considering the finite difference grid.
The point is, for example, c(i) = 2.0/( h(i)*(h(i) + h(i1)) ) while a(i) = 2.0/( h(i1)*(h(i) + h(i1)) ) => a(i+1) = 2.0/( h(i)*( h(i+1) + h(i)) ) /= c(i) in general because the h's are not equal! More complex/accurate discretizations are even less likely to be symmetric. The above is an example of what can happen in the incompressible equations on a staggered grid. 

May 14, 2008, 10:45 
Re: Reg. BICGSTAB

#32 
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Thanks  that makes perfect sense now.
I've programmed this myself in the not so distant past, so I must have known this quite recently... I wonder why I forgot!! Good to revisit it I guess. Best wishes, andy 

May 15, 2008, 04:18 
Re: Reg. BICGSTAB

#33 
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The pressure correction equation is poisson type equation, and it is not discretized the way you have written.
between i to j (the coefficient relating i to j) Aij is given as Aij = Area / distance_ij in 1D case area = 1 , so Aij = 1/distance_ij for j relating to i, Aji = Area / distance_ji Since distance_ij = distance_ji (irrespective of how other elements are speard out, distance between two points is what decide this coefficient. So your non uniformity is grid spacing does not effect this). Aij = Aji and hence matrix is symmetric. If you want to descretisize the way you have written feel free to do so, but softwares like Fluent StarCCM do not do what you have written. (But do as I mentioned above). For incompressible cases, pressure matrix is symmetric. (finite volume methods). 

May 15, 2008, 05:00 
Re: Reg. BICGSTAB

#34 
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No the symmetry of the discretized equation depends on the grid and your choice of discretization  This is why when Ferziger and Peric discuss the pressure projection equation they point out that you can either use a defect correction form (thus lagging any asymmetry) or throw away some terms (namely the nonorthogonal ones).
The discretization I've giving is pretty much standard on staggered grids with grid strecthing when the grid changes are smooth. Convert the Laplacian operator into general nonorthogonal curvilinear coordinates and then convince yourself that it's still a symmetric operator. 

May 15, 2008, 05:26 
Re: Reg. BICGSTAB

#35 
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You want to say that what is written here is wrong:
http://www.cfdonline.com/Wiki/Discr...diffusion_term And the non orthoginal terms go into the source part of equations and not in the matrix itself. Clearly the matrix is symmetric. Of course there is no way you would agree with it, since you have decided to ignore all that is presented to you. Best of luck with your quest in CFD. 

May 15, 2008, 05:54 
Re: Reg. BICGSTAB

#36 
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And you're ignoring the facts (if you move the nonothorgonal terms, and all the pressure terms making up the extended stencilm, from the source term then then they contribute to the matrix and make it asymmetric!)
"Best of luck with your quest in CFD." You really don't know anything about me  I'm not some "student" and have a job writing and developing CFD codes. You may also consider I've got a BSc, MSc and PhD in Mathematics + numerous years of work experience (on top of postdoctoral work). My Last word on this pointless argument: There is no one correct way to do CFD  if there was there would be no need to do research (hence your examples are useless  you need to demonstrate that "all possible discretizations" of the equations leads to a completely symmetric equation (no lagging terms in the iteration) which as I've shown, by counter example, is false). 

May 15, 2008, 06:02 
Re: Reg. BICGSTAB

#37 
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1. Where are non orthogonal terms in 1D case.
2. We are not talking about how you would do things, we are talking about how they are generally done. You could write second derivative using many neighbors and claim that it would be asymetric. But this is not the way it is done. It is done the way I have pointed you to the wiki page. And it comes out to be symmetric. To prove it otherwise you need to prove that what I pointed you is wrong. Go ahead and do it. I am waiting. 3. No matter how many degrees you have, it is the fact that the approach Fluent, StarCCM+ takes for pressure descretisation, the matrix comes out to be symmetric. (for incompressible cases). And this is what we been arguing. So your saying that pressure matrix is not symmetric stands wrong. 

June 24, 2008, 10:25 
Re: Reg. BICGSTAB

#38 
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All, BICGSTAB is good if your matrix is not symmetric pos. def. However, if your matrix is sym. pos. def. you are wasting computational resources by doing the second matrixvector product dealing with the transpose. If you are not seeing oscillations in the relative residual, BICG may be faster if you are set on using such a method.
Furthermore, a good preconditioner is essential when using an of the sparse solvers. The main diagonal preconditioner is Jacobi preconditioning and really only works well when the matrix is really diagonally dominant (i.e., the matrix 1/d_ii is a good approximation of the inverse). Other good preconditioners are Incomplete Cholesky which should be used with sym. pos. def. matrices and Incomplete LU should be used otherwise. There are various implementations of these to consider (i.e., zero fillin, Crout type, conservation of row and column sums). If your matrix is composed of blocks of smaller matrices then using a block Jacobi preconditioner can be a good option as well. Depending on the PDE of interest, there are many specialized preconditioners available that are designed for the sparsity pattern of your system. There are mathematical conferences dedicated to this subject. From my experience, if your matrix is not sym. pos. def. then GMRES(m) and choosing (m<=50) minimizes the storage associated with the search directions in the Krylov subspace is a good method. If you use a good incomplete preconditoner (one that conserves rowsums) you can increase your convergence by an order of magnitude in some cases. However, choosing (m) is somewhat of an art since many matrices that are poorly conditioned can result in stagnation of the relative residual. The idea is to choose (m) to minimize storage and still get rapid convergence. In some instances you can get convergence using (m) restart whereas you do not with GMRES w/out restart. If you have a sym. pos. def. matrix then CG is one of the fastest (since it does a considerable less amount of computation), especially with a good preconditioner like ichol. You can determine how many iterations that CG should take with knowledge of the condition number and a little bit of algebra. In general, the worse the condition number (~ the ratio of the largest to smallest eigenvalue), the more problems you will have finding a good combination of a solver and preconditioner to optimize your cpu effort. There are several other iterative solvers that work well for large sparse systems. I recommend Y. Saad's text and C.T. Kelley's text on the subject. Both give good examples using common sparse systems we see in the discretization of PDEs. Saad's text also includes a detailed description of various types of preconditioners.  Regards, DR 

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