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Old   March 4, 2010, 22:29
Default Viscous Flux Jacobian
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Who has the experience computing viscous flux Jacobian for incompressible flow?

Suppose we set the primitive variables as [P, U, V, W].

I know hand calculating the convective flux Jacobian analytically is easy. But for viscous flux, the situation looks different. My own hand derivation shows that it should be zero for Cartesian grid. It may not be zero in the case of curvilinear grid.

Am I correct?

Thank you.
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Old   March 9, 2010, 13:58
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The viscous flux Jacobians are not zero for a Cartesian grid.
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Old   March 9, 2010, 19:52
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Quote:
Originally Posted by agd View Post
The viscous flux Jacobians are not zero for a Cartesian grid.
Thank you for your reply.

But are you sure? I mean incompressible flow with constant viscosity here.

suppose in 2D case with kinematic viscosity as miu
primitive variable: Q=[ P, U, V ],
convective flux in X: E=[ U, U*U+P, UV ]
viscous flux in X: Ev=[ 0, 2*miu*ddU/ddX, miu*(ddU/ddY+ddV/ddX) ]

When I do the viscous flux Jacobian analytically: dd(Ev)/ddQ, all derivatives of Ev elements in terms of P,U or V get nothing. And someone also remind me the viscous terms are linear for incompressible N-S eq.

Last edited by bearcat; March 9, 2010 at 21:23.
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Old   March 9, 2010, 20:53
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I am assuming you are trying to develop an implicit algorithm. Since the viscous terms are linear, if you write Q(N+1) = Q(N) + DQ and substitute into the diff. eq., the terms containing the DQ can be carried to the left-hand side. The flux Jacobian (or the contribution to the system matrix) follow from that. In this case it will be mu multiplied by the coefficients of the difference terms.
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Old   March 9, 2010, 21:36
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Quote:
Originally Posted by agd View Post
I am assuming you are trying to develop an implicit algorithm. Since the viscous terms are linear, if you write Q(N+1) = Q(N) + DQ and substitute into the diff. eq., the terms containing the DQ can be carried to the left-hand side. The flux Jacobian (or the contribution to the system matrix) follow from that. In this case it will be mu multiplied by the coefficients of the difference terms.


Yes, if we use a Jacobian-free Newton-Krylov method, your idea will be implemented easily. But I think if we can cancel it out analytically, then we don't have to compute it numerically and save some effort.

The original 2D eq is:

ddQ/ddt+dd(E-Ev)/ddX+dd(F-Fv)/ddY=0 (no source term)

The 3x3 Jacobian matrix often acts as a coefficient in system matrix. My opinion is that all elements in viscous flux Jacobian matrix are zero if no coordinate transform is involved.
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Old   March 9, 2010, 22:35
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You can ignore them, but the advantage of including them is that they contribute to the diagonal dominance of the coefficient matrix.
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Old   March 11, 2010, 01:40
Default This paper maybe is useful for you
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For compressible flow, the below paper maybe is useful.
"Implicit multigrid solution of the compressible Navier-Stokes equations with application to distributed parallel processing",
Tysinger, Thomas Lee, Ph.D. Thesis Cornell Univ., Ithaca, NY,1992.
This paper introduced the viscous jacobian matrix for compressible NS equation. I think this method is also useful for your problem.
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Old   March 11, 2010, 07:03
Default not zero
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[QUOTE=bearcat;249236]
viscous flux in X: Ev=[ 0, 2*miu*ddU/ddX, miu*(ddU/ddY+ddV/ddX) ]
When I do the viscous flux Jacobian analytically: dd(Ev)/ddQ, all derivatives of Ev elements in terms of P,U or V get nothing.

You get something.
For example,

dd(Ev)/ddU = [0, 2*miu*dd/ddX, miu*dd/ddY ]

A discretized version would be

dd(Ev)/ddU = [0, 2*miu*(1/Delta x), miu*(1/Delta y) ]


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Old   March 11, 2010, 13:46
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Quote:
Originally Posted by peter.zhao View Post
For compressible flow, the below paper maybe is useful.
"Implicit multigrid solution of the compressible Navier-Stokes equations with application to distributed parallel processing",
Tysinger, Thomas Lee, Ph.D. Thesis Cornell Univ., Ithaca, NY,1992.
This paper introduced the viscous jacobian matrix for compressible NS equation. I think this method is also useful for your problem.
Thank you.

viscous jacobian for compressible flow appears in many CFD books. But that's an easier and completely different case. For incompressible flow we don't have density as a primitive variable.



[QUOTE=gory;249527]
Quote:
Originally Posted by bearcat View Post
viscous flux in X: Ev=[ 0, 2*miu*ddU/ddX, miu*(ddU/ddY+ddV/ddX) ]
When I do the viscous flux Jacobian analytically: dd(Ev)/ddQ, all derivatives of Ev elements in terms of P,U or V get nothing.

You get something.
For example,

dd(Ev)/ddU = [0, 2*miu*dd/ddX, miu*dd/ddY ]

A discretized version would be

dd(Ev)/ddU = [0, 2*miu*(1/Delta x), miu*(1/Delta y) ]

gory
Thank you for reminding me.You really get the point. At first I think they may look like operators rather than values as Jacobian matrix elements: dd(ddU/ddX)/ddU=dd( )/ddX. Now you tell me dd(ddU/ddX)/ddU=1/ddX=1/deltaX. Can you inform me any reference for this?

Last edited by bearcat; March 11, 2010 at 14:08.
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Old   March 11, 2010, 14:20
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It's what you'll get if you follow the steps I outlined above - it results from the linear nature of the PDE and the linearization of the aforementioned viscous terms.
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Old   March 11, 2010, 18:14
Smile Jacobians
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Quote:
Originally Posted by gory View Post

Thank you for reminding me.You really get the point. At first I think they may look like operators rather than values as Jacobian matrix elements: dd(ddU/ddX)/ddU=dd( )/ddX. Now you tell me dd(ddU/ddX)/ddU=1/ddX=1/deltaX. Can you inform me any reference for this?

I don't know any reference.
I agree with agd that you follow his instruction and you'll get it.

Example:
Consider a simple diffusion equation,
ut = df/dx, where f = du/dx (d = \partial, i.e., one partial).
Then,
ut = A du/dx, where A=df/du=d(du/dx)/du=d/dx.
It is an operator.

If you're interested in the Jacobian (I'm not sure if you are)
of the residual. It's related but a bit different, I think.
I guess it goes like this:

A discretized version of df/dx may be

RHS(j) =(u_{j+1} - 2 u_j + u_{j-1})/(Delta x)^2,

where RHSis a residual vector and RHS(j) is the j-th component.

The Jacobian is a matrix defined as
dRHS/dU where U = [u1, u2, u_{j-1}, u_j, ...],
i.e., the discrete values. So, for example,

(dRHS/U)(j,j) = dRHS(j)/du_j = -2 / (Delta x)^2
(dRHS/U)(j,j-1) = dRHS(j)/du_{j-1}= 1/(Delta x)^2
and so on.

In any case, the Jacobian is not zero for viscous terms, I think.
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