Problem with BICGSTAB

Amir_Ghasemi · July 28, 2011, 02:39

Hello all,

I have written a BICGSTAB solver for Poisson equation in SPH (Smoothed particle hydrodynamics) method. As all the boundaries are Neumann kind, the resulted matrix is singular, to remove singularity, pressure of one point in the domain is fixed. The solution converge satisfactory but results show inhomogeneity(sharp peak) near the fixed point which is nonphysical.

Any idea that why the issue arise is appreciated.

I have one idea by myself but I am not sure how correct it is. There is some compatibility condition which should be satisfied between Neumann BCs and source term of the Poisson equation. Such constraint dictate that sum of the source terms should equal zero (in case of Neumann pressure BCs and impermeable boundaries). Now when we fix one point in the domain, what will happen to this compatibility condition?, should it be satisfied any more and if yes, how we can do this?

Thank you so much.

Amir_Ghasemi · July 31, 2011, 11:42

Hello all,

I have just found what the problem was. May matrix was singular with constant vector in null space and its determinate was zero and it had emerged from discretizing PPE with Neumann boundary. One should be aware that these are not enough to decide to remove one equation (or fixing the pressure of one point). you can do this, if the system of equation has infinite solutions. Ax = b has infinite solution if det(A) = 0 and b is in the range of A, if these condition exist you can remove one equation and be sure that the result of new system also satisfy the original one and in may system this was not the case.

Ghasemi

TheBoyce · October 13, 2011, 10:48

Hello Ghasemi,

I have the same problem as you initially described in the first post. How did you solve this and obtain a solution without the non-physical pressure gradients?

Amir_Ghasemi · October 14, 2011, 14:16

Hello TheBoyce,

As I have said, I think that before fixing the pressure of one node you should become sure that your matrices possess infinite solution, if this is not the case you obtain non physical distribution around the fixed point. I was dealing with a Poisson eq. in SPH method and I changed my pressure boundary condition to get a symmetric matrix with constant vector in it's null space. for this kind of matrix, sum of your source terms should be zero. I have done this with subtracting each source term from the average.

Again I emphasize that zero determination does not always lead to infinite solution, no solution is also possible. in first case you can fix a point, in second case, fixing a point is not allowed.

July 28, 2011, 02:39	Problem with BICGSTAB	#1
Amir_Ghasemi New Member Amir Join Date: Oct 2010 Posts: 16 Rep Power: 16	Hello all, I have written a BICGSTAB solver for Poisson equation in SPH (Smoothed particle hydrodynamics) method. As all the boundaries are Neumann kind, the resulted matrix is singular, to remove singularity, pressure of one point in the domain is fixed. The solution converge satisfactory but results show inhomogeneity(sharp peak) near the fixed point which is nonphysical. Any idea that why the issue arise is appreciated. I have one idea by myself but I am not sure how correct it is. There is some compatibility condition which should be satisfied between Neumann BCs and source term of the Poisson equation. Such constraint dictate that sum of the source terms should equal zero (in case of Neumann pressure BCs and impermeable boundaries). Now when we fix one point in the domain, what will happen to this compatibility condition?, should it be satisfied any more and if yes, how we can do this? Thank you so much.

July 31, 2011, 11:42		#2
Amir_Ghasemi New Member Amir Join Date: Oct 2010 Posts: 16 Rep Power: 16	Hello all, I have just found what the problem was. May matrix was singular with constant vector in null space and its determinate was zero and it had emerged from discretizing PPE with Neumann boundary. One should be aware that these are not enough to decide to remove one equation (or fixing the pressure of one point). you can do this, if the system of equation has infinite solutions. Ax = b has infinite solution if det(A) = 0 and b is in the range of A, if these condition exist you can remove one equation and be sure that the result of new system also satisfy the original one and in may system this was not the case. Ghasemi Last edited by Amir_Ghasemi; August 4, 2011 at 02:28.

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October 13, 2011, 10:48		#3
TheBoyce New Member Tom Join Date: Mar 2011 Posts: 12 Rep Power: 15	Hello Ghasemi, I have the same problem as you initially described in the first post. How did you solve this and obtain a solution without the non-physical pressure gradients?

October 14, 2011, 14:16		#4
Amir_Ghasemi New Member Amir Join Date: Oct 2010 Posts: 16 Rep Power: 16	Hello TheBoyce, As I have said, I think that before fixing the pressure of one node you should become sure that your matrices possess infinite solution, if this is not the case you obtain non physical distribution around the fixed point. I was dealing with a Poisson eq. in SPH method and I changed my pressure boundary condition to get a symmetric matrix with constant vector in it's null space. for this kind of matrix, sum of your source terms should be zero. I have done this with subtracting each source term from the average. Again I emphasize that zero determination does not always lead to infinite solution, no solution is also possible. in first case you can fix a point, in second case, fixing a point is not allowed.