For two-dimensional or three-dimensional elliptic equations, when will the stiffness matrix be asymmetric and positive definite? This affected the solution efficiency so much that I had to choose an appropriate preconditioner based on the stiffness matrix.

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    $\begingroup$ Did you mean to use the word asymmetric in the body of your Question? It seems contrary to the title. Symmetry is an immediate consequence of using the same ordered basis for test and trial functions. $\endgroup$
    – hardmath
    Oct 10, 2023 at 14:34
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    $\begingroup$ Both symmetry and positive definiteness follow from the PDE you're trying to solve. If the differential operator is symmetric and positive definite, then so will the matrix. $\endgroup$ Oct 10, 2023 at 14:48
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    $\begingroup$ It is worth noting that symmetry and positive-definiteness of the matrix generally only follows from those properties of the operator if the trial and test bases are the same. Mixed element and Petrov-Galerkin discretizations can violate this $\endgroup$
    – whpowell96
    Oct 10, 2023 at 20:00
  • $\begingroup$ Thank you so much! Everyone's answers were so professional that I may not fully understand them. For the Stokes equation, discretizing the Laplacian produces the stiffness matrix. I have not done any in-depth understanding. I am curious whether the spatial changes in viscosity and boundary conditions will cause asymmetry in the stiffness matrix. Because when I used petsc to solve the Uzawa equation, some Krylov methods such as symmlq and nash did not seem to get the correct results (both require the matrix to be symmetric and positive definite), so I posted the above question! $\endgroup$
    – Darcy
    Oct 11, 2023 at 4:55
  • $\begingroup$ Boundary conditions can change the system tho $\endgroup$ Oct 11, 2023 at 14:27


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