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How to define fluxes for two dimensional convection-diffusion equation? |
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September 8, 2015, 10:14 |
How to define fluxes for two dimensional convection-diffusion equation?
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#1 |
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September 8, 2015, 12:08 |
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#2 |
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Filippo Maria Denaro
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The definition of F and G are correct and they define the fluxes in the exact continous form. Note that in the equation you wrote you assumed u=v=1. The general convective fluxes are u*T and v*T
However, the discretization you wrote is not the only one you can write in a FVM. You wrote a second order central discretization, which combined with the first order explicit time discretization has relevant stability constraint to work with. If lambda vanishes (the equation will be hyperbolic in such case) your discretization is unconditionally unstable. |
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September 9, 2015, 09:21 |
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#3 | |
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September 9, 2015, 09:31 |
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#4 |
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Filippo Maria Denaro
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Classification of PDE is a mathematical tool used for second order PDE (or system of PDE), you have to analyse the nature of the eigenvalues.
In your case if lambda =0 you have a first order equation that has no other classification than hyperbolic. |
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September 9, 2015, 09:35 |
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#5 |
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May you please let me know about any easy resource from where I can judge about the nature of eigenvalues? (Sorry my mathematics background is not very strong). Since I am using a mesh refinement procedure which was designed for hyperbolic PDEs, I am getting some undesired results (probably because my PDE is not hyperbolic as lambda is non-zero positive quantity).
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September 9, 2015, 09:42 |
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#6 | |
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Filippo Maria Denaro
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The eigenvalue-based classification is described also in some classical CFD texbook such as Chap.3 in: https://books.google.it/books?hl=it&...namics&f=false |
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September 9, 2015, 09:44 |
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September 9, 2015, 11:22 |
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#8 |
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Filippo Maria Denaro
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September 9, 2015, 11:25 |
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#9 |
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September 9, 2015, 11:27 |
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#10 |
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Filippo Maria Denaro
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September 9, 2015, 11:34 |
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#11 |
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September 9, 2015, 11:41 |
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#12 |
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Filippo Maria Denaro
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September 9, 2015, 11:44 |
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#13 |
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I mean, just for checking the nature of PDE. To check the nature of unsteady PDE here as parabolic, we are not considering the T_yy term, because space is already considered in T_xx term. Is that right Professor?
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September 9, 2015, 11:51 |
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#14 | |
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Filippo Maria Denaro
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Quote:
dT/dt + u dT/dx = d/dx (lambda dT/dx) |
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September 9, 2015, 12:10 |
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#15 |
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Professor, so if I have any second order PDE in three variables x, y and t and if I have to find the nature of PDE. I should always check it's 1D counterpart?
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September 9, 2015, 12:15 |
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#16 | |
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Filippo Maria Denaro
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Not exactly the same because the number of eigenvelues depends on the number of equations but you can see similar infos... for example the 1D Euler system is hyperbolic (3 real distinct eigenvalues) as same as it would be for the 3D case (5 eigenvalues). |
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September 9, 2015, 12:19 |
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#17 | |
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September 9, 2015, 12:27 |
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#18 | |
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Filippo Maria Denaro
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No, of course the refinement for hyperbolic PDE is different... have a look to the book I linked |
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September 9, 2015, 12:32 |
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#19 | |
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Quote:
And may be simultaneously to check my mesh refinement code, I can put lambda = 0 and then PDE would be first order hyperbolic and I can use that refinement scheme. Is that right? |
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September 9, 2015, 12:35 |
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#20 | |
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Filippo Maria Denaro
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Tags |
finite difference, finite volume, math, mesh, pde |
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