# How to define fluxes for two dimensional convection-diffusion equation?

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 September 8, 2015, 09:14 How to define fluxes for two dimensional convection-diffusion equation? #1 New Member   Tanmay Agrawal Join Date: May 2015 Location: Taiwan Posts: 29 Rep Power: 10

 September 8, 2015, 11:08 #2 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71 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.

September 9, 2015, 08:21
#3
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Tanmay Agrawal
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 Originally Posted by FMDenaro 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.
May I ask you how do we say whether this PDE is hyperbolic or elliptic or parabolic? Because we have three independent variables here i.e. x, y and t. So if lambda is zero, does not that mean that it should rather be parabolic because A = B = C = 0 and therefore B^2 - 4*A*C equals zero?

 September 9, 2015, 08:31 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71 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.

September 9, 2015, 08:35
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 Originally Posted by FMDenaro 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.
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).

September 9, 2015, 08:42
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Filippo Maria Denaro
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 Originally Posted by tanmayagrawal7 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).
Any good textbook about PDE describes this classification...
The eigenvalue-based classification is described also in some classical CFD texbook such as Chap.3 in:

September 9, 2015, 08:44
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 Originally Posted by FMDenaro Any good textbook about PDE describes this classification... 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
Thanks a lot!! BTW, if my lambda is non-zero positive, what kind of PDE would that be? And also, do you have any experience in mesh refinement(local or adaptive)?

September 9, 2015, 10:22
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 Originally Posted by tanmayagrawal7 Thanks a lot!! BTW, if my lambda is non-zero positive, what kind of PDE would that be? And also, do you have any experience in mesh refinement(local or adaptive)?

second order parabolic equation

September 9, 2015, 10:25
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 Originally Posted by FMDenaro second order parabolic equation
Professor, for this PDE with λ>0. Should not it be elliptic? Because A=C=λ and B is zero? How do we manage the third independent variables i.e. time, in such cases?

September 9, 2015, 10:27
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 Originally Posted by tanmayagrawal7 Professor, for this PDE with λ>0. Should not it be elliptic? Because A=C=λ and B is zero? How do we manage the third independent variables i.e. time, in such cases?

no at all...elliptic is only at the steady state (dT/dt=0)

September 9, 2015, 10:34
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 Originally Posted by FMDenaro no at all...elliptic is only at the steady state (dT/dt=0)
Oh! So for unsteady, I would be treating that y as t rather. Is that right?

September 9, 2015, 10:41
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 Originally Posted by tanmayagrawal7 Oh! So for unsteady, I would be treating that y as t rather. Is that right?
Why?? in the unsteady case you have x,y,t as independent variables, x,y in the steady case ...

September 9, 2015, 10:44
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 Originally Posted by FMDenaro Why?? in the unsteady case you have x,y,t as independent variables, x,y in the steady case ...
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?

September 9, 2015, 10:51
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 Originally Posted by tanmayagrawal7 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?
pk, if you want to study a simple model, check for the 1D counterpart

dT/dt + u dT/dx = d/dx (lambda dT/dx)

September 9, 2015, 11:10
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 Originally Posted by FMDenaro pk, if you want to study a simple model, check for the 1D counterpart dT/dt + u dT/dx = d/dx (lambda dT/dx)
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?

September 9, 2015, 11:15
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 Originally Posted by tanmayagrawal7 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?

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).

September 9, 2015, 11:19
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 Originally Posted by FMDenaro 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).
Thanks a lot for your guidance professor. I now, atleast have some idea about it. I actually want to use mesh refinement schemes on this PDE. I was using the method proposed by Berger and Collela (http://www.sciencedirect.com/science...21999189900351) but they have mentioned it to be for hyperbolic PDEs. Now I am not very sure if I can use the same method for this PDE or not?

September 9, 2015, 11:27
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 Originally Posted by tanmayagrawal7 Thanks a lot for your guidance professor. I now, atleast have some idea about it. I actually want to use mesh refinement schemes on this PDE. I was using the method proposed by Berger and Collela (http://www.sciencedirect.com/science...21999189900351) but they have mentioned it to be for hyperbolic PDEs. Now I am not very sure if I can use the same method for this PDE or not?

No, of course the refinement for hyperbolic PDE is different...
have a look to the book I linked

September 9, 2015, 11:32
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Tanmay Agrawal
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 Originally Posted by FMDenaro No, of course the refinement for hyperbolic PDE is different... have a look to the book I linked
Sure Professor. I will get a hard copy of it tomorrow as some of the pages in the chapter containing multigrid methods are not accessible.
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?

September 9, 2015, 11:35
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Filippo Maria Denaro
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 Originally Posted by tanmayagrawal7 Sure Professor. I will get a hard copy of it tomorrow as some of the pages in the chapter containing multigrid methods are not accessible. 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?
you are working with u and v constant and known? In this case, the solution is a simple linear wave propagating along the path-line, the grid-refinement should move dynamically

 Tags finite difference, finite volume, math, mesh, pde