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Laplace operator in finite element formulation |
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October 18, 2004, 19:20 |
Laplace operator in finite element formulation
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#1 |
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Hi together,
I try to discretizise the pressure poisson equation (for an incompressible viscous code). Well, the discretization of the LHS (laplace operator for pressure) yields to a symmetric matrix L (multiplying element B matrix and its transpose BT) before the nodal pressure. Now, for getting the pressure, I have tried both: (a) to invert the matrix L or (b) to solve a equation system like Lx=B. PROBLEM: Both failed with the message "L is singular". Do you know, where I did the mistake?? If no mistake, please give the trick how to solve this. Extra information: - I used the same (linear) shape functions for velocity as well for pressure Thanks a lot in advance!! Rafael |
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October 19, 2004, 09:43 |
Re: Laplace operator in finite element formulation
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#2 |
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Take a look at your boundary conditions.
With the pressure equation, the matrix becomes singular when the pressure is not specified at (at least) one node. In an incompressible flow, the pressure is usually known to within a constant anyway, so specifying at a node fixes the pressure levels without modifying the pressure gradients (and thus the velocities). Likely a FEM expert will answer this with FEM-based language, and perhaps a 'standard' way to fix the problem. |
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October 19, 2004, 22:09 |
Re: Laplace operator in finite element formulation
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#3 |
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Yeah Jim,
that was it. I apologize, I did not consider (for a first run) the boundary condition term of pressure poisson equation. Thank you for your help! Regards, Rafael |
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