| Author | Message | |||
     Noah Tracy (noahsark) Junior Member Username: noahsark Post Number: 3 Registered: 11-2003 |
I am working with the attached file. The basics are this- I ask flexpde to solve Div(-kappa*Grad(V))=0. I then plot Div(-kappa*Grad(V) for the solution and it's no where near zero. Is this to be expected? I am surprised. I am using version 3 of flexpde. Thanks so much for your kind assistance. Noah
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 771 Registered: 06-2003 |
The first thing to realize is that the PDE is not the only thing determining the solution. The boundary conditions must also be satisfied. And if the boundary conditions are not compatible with the PDE, the PDE will not be satisfied. In your case, you have imposed a value boundary condition on V on the cylinder axis. This distribution is not in itself a solution of the PDE in z, but requires that V = C1-C2*r^2 in the vicinity of the axis. But the solution is also required to be zero along the top boundary, and the equipotentials must meet the bottom and right boundaries at right angles. I assert that there is no solution that satisfies both Laplace's equation AND the imposed boundary conditions, and that the deviations that you see from div(grad(V))=0 are a manifestation of that fact. Another thing to realize is that the finite element method does not in fact "solve" the PDE system. It merely minimizes the deviation of the approximate solution from the requirements of the PDE. That is, it finds the distribution of nodal values that has the least error in solving the PDE. In this case, "minimum" is not the same as "small" (let alone "zero"). | |||
| Noah Tracy (noahsark) Member Username: noahsark Post Number: 4 Registered: 11-2003 |
Ahhh, I have asked the impossible. Hmm. It's always the boundary conditions, isn't it? I'll have to think about this, then. I was trying to do something simple to clear up problems with a more complicated system. Thanks for your suggestion. |
