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Question
Does lacking positive definiteness of the matrix of coefficients in a system of equations, make using iterative solvers impractical?

Description
Using the finite volume method, I have obtained a system of equations for the dependent variable (the unknown field). The problem is that I want to use iterative solvers to solve this system of equations; however, the way that I have discretised the governing equation results in a matrix of coefficients which is not positive definite.

May this cause a problem during the solution procedure? Actually I want to know how definiteness of a matrix relates to the choice that should be made regarding the solver. I am also looking for a book that directs attention to this topic.

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    $\begingroup$ What are "iterative solvers" for you? Do you know about GMRES and MINRES, which are iterative solvers for nonsymmetric and symmetric indefinite matrices? $\endgroup$
    – Federico Poloni
    Jun 8, 2020 at 12:06
  • $\begingroup$ I use bi-conjugate gradient stabilized. $\endgroup$
    – Naghi
    Jun 8, 2020 at 12:31
  • $\begingroup$ @FedericoPoloni Please post your comment as an answer in order that I accept it. $\endgroup$
    – Naghi
    Jun 9, 2020 at 11:45
  • $\begingroup$ Alish: Wolfgang Bangerth already gave an answer on the same lines; you can just accept his. $\endgroup$
    – Federico Poloni
    Jun 9, 2020 at 13:52
  • $\begingroup$ @FedericoPoloni Ok, I will, but I guessed that the reputations may deserve you, because you mentioned the answer earlier. $\endgroup$
    – Naghi
    Jun 9, 2020 at 15:28

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No, positive definiteness (and symmetry) are only precondition to using the Conjugate Gradient method. But there are plenty of other iterative methods such as MinRes and GMRES that can be used for indefinite and non-symmetric matrices.

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    $\begingroup$ Even CG can be used for certain semi-definite cases, when the projection of the RHS onto the null space of the matrix is zero - see core.ac.uk/download/pdf/82779418.pdf $\endgroup$
    – Ian Bush
    Jun 10, 2020 at 8:25
  • $\begingroup$ @IanBush -- yes, that is true. But it only pertains to the very special case of a linear system whose eigenvalues are positive or zero. If you have negative eigenvalues, you are screwed. $\endgroup$
    – Wolfgang Bangerth
    Jun 10, 2020 at 14:00
  • $\begingroup$ Yes - semi-definite systems. I agree if the case has negative eigenvalues you have to look elsewhere $\endgroup$
    – Ian Bush
    Jun 10, 2020 at 15:43

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