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Reasons for decoupling the pressure and velocity computations

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Old   April 9, 2017, 01:21
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Lucky
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Hey folks, some food for thought...

To begin with, this pressure-velocity coupling problem starts exactly because we try to decompose the stress tensor into a deviatoric part and non-deviatoric part. Put another way, why do we even solve the Navier-Stokes equations and not the generalized Cauchy momentum transport equation? In even more obvious language, why do we solve for pressure & velocity and not just velocity and then reconstruct the pressure field from the non-deviatoric part of the stress tensor? We already implicitly tried to decompose/decouple pressure out of the stress tensor (for a very good reason, because we need pressure for many things).

We tend to treat the pressure/velocity fields as potentials for each other only in the inviscid case. And we say that unless it is a potential flow, that pressure & velocity are always coupled. However, don't forget that pressure can be interpreted as just a part of the stress tensor. This becomes even more painfully obvious if you drop Navier-Stokes and try to include surface tension.

In a way, we have set up the problem such that it is extremely favorable in many situations to be solved in a segregated manner, because we are looking for this decomposed part of the stress tensor (the pressure).
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Old   April 9, 2017, 04:28
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Filippo Maria Denaro
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To tell the truth, I am not sure that the Newton law can be still extended to the isotropic part in such a way to get a relation only with the velocity.
In any case, for incompressible flow, we denote with "pressure" something that has philosophically nothing to do with the extraction of the isotropic part of the stress tensor. It could be denoted as a,b,c,d functions... The only reason to introduce the gradient of such function is to take into account of the divergence-free constraint and the additional scalar function acts as a lagrangian multiplier.
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Old   April 9, 2017, 05:21
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Originally Posted by FMDenaro View Post
To tell the truth, I am not sure that the Newton law can be still extended to the isotropic part in such a way to get a relation only with the velocity.
Just start with the Cauchy momentum equation. Introduce a constitutive relation for the stress tensor. If you split the stress tensor into a deviatoric and non-deviatoric part, and then apply the hydrostatic condition, you notice the non-deviatoric part is the mechanical pressure and you call the resulting equation the Navier-Stokes equation. It's from here that Stokes realized the pressure in the N-S isn't the thermodynamic pressure and from there imposed the Stokes condition. No magic need be invented.

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In any case, for incompressible flow, we denote with "pressure" something that has philosophically nothing to do with the extraction of the isotropic part of the stress tensor. It could be denoted as a,b,c,d functions... The only reason to introduce the gradient of such function is to take into account of the divergence-free constraint and the additional scalar function acts as a lagrangian multiplier.
This is really a perfect example, and more or less proves my point. In incompressible flows, pressure is whatever it needs to be to satisfy the divergence free constraint. Which really harks back to, it is the non-deviatoric part of the stress tensor in the first place. The difference is that for incompressible flows, the pressure is even less the thermodynamic pressure, it is the magic force that imposes the incompressibility constraint. Once again, we created our own problem.
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Old   April 9, 2017, 05:34
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but still, we have a coupling with this additional function.... The only way I could think is to project the momentum along the velocity vector field, integrate over the domain and only if Int[S] n.vp dS=0 you get the weak form where the "pressure" disappear...
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Old   April 9, 2017, 12:02
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We are in the weeds, kind folks. 8)

My view is that incompressible flow is a non-physical, but truly functional, approximation to a compressible flow with high speed of sound. I think of it as an asymptotic expansion with rho(p) => rhoConstant and c -> inf. In my mind, incompressible flow...for all of its simplicity, it not truly fundamental in the same way that, say, the Boussinesq approximation is also simple and useful, but also not fundamental.

Compressible flow is properly local, as required by phenomena described by differential equation. By introducing an infinite speed of sound, we've broken locality in a significant way. Now, minute deviations in the divergence-free velocity field in one area can instantly effect the flow in the entire rest of the domain. When systems are no longer local, I view them as fundamentally requiring an integro-differential expression to model them. And the incompressible pressure term is just such an expression. It can be viewed as the Green's function for the Laplacian times div(u) integrated over the entire domain. We don't evaluate it that way for efficiency reasons, but one could write the incompressible momentum equation without pressure using that form, and I'd argue that it is irreducible from there without some explicit simplifying assumptions.

But, I am truly just a humble CFD coder. My mathematical analysis day are two full decades behind me.
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Old   April 9, 2017, 12:12
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yes, you are right about the mathematical modelling that simplifies the real case of a very low compressible flow
In my mind, the role of the coupling with the "pressure" is of mathematical nature. If you work with the point-wise differential form you need two PDE equations and two unknowns to close the problems (v,p). Using the weak form you can eliminate the pressure fro the resulting integro-differential equation but, while we know that a weak solution could exist, what about the unicity? From the weak form is a classical issue to determine multiple solutions...
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Old   April 9, 2017, 12:39
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The boundary conditions on our "pressure" variable are baked into the Green's function. That Green's integral result gives us the "pressure," so the momentum equation will use the gradient of it. That eliminates any gauge constant (for the case of no fixed-pressure boundary conditions). The resulting momentum equation is still subject to our normal velocity boundary conditions. That should lead to a properly closed system and a unique solution. Again, this is based more instinct than hard analysis. I am happy to be corrected.
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Old   April 9, 2017, 12:58
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At the best of my knowledge, the weak solution is not unique, despite fixing the BC.s
Actually, that should be a general problem of the NS equations also for the compressible formulation.

In a very poor example, you cannot find a unique function that provide the known value of the Int[a,b] f(x) dx even if you know f(a) and f(b).
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Old   April 9, 2017, 13:04
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Indeed, you are right.

http://www.claymath.org/sites/defaul...vierstokes.pdf

If you solve it, the Clay Institute has a million dollars for you.
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Old   April 9, 2017, 14:30
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Uniqueness is another million dollar problem.

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Originally Posted by FMDenaro View Post
If you work with the point-wise differential form you need two PDE equations and two unknowns to close the problems (v,p).
Yes but we introduced p as a fake variable.

Once again the Cauchy momentum equation gives you one equation with two unknowns, the velocity u and the stress tensor. Introduction of the constitutive Newtonian/Stokes fluid model allows you to re-write the stress tensor in terms of grad(U). So you can say that now you have one equation with one unknown u. The continuity equation is a constraint on u (really the momentum). The incompressible condition is yet another constraint.

I am not arguing that pressure is decoupled. It should be obvious, that if you try to do this decomposition on u into u & p that p must definitely be coupled to the momentum equation because that's where it came from. My point is, you can say that p is coupled to the momentum equation, or you can take a step back and just say we have a momentum equation.

So why did I even bother mentioning any of this? So the original question was why is a predictor-corrector approach a good idea & why is it a bad idea to solve the coupled problem? My analogy, which has already opened a can of worms, is that well why was it a good idea to decompose the stress tensor into a deviatoric and non-deviatoric part in the first place? Why don't we just solve the more general Cauchy momentum equation? We want to solve for a p and that gives rise to certain problems that we must now deal with because of our own bias.
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Old   April 9, 2017, 14:41
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Quote:
Originally Posted by LuckyTran View Post
Uniqueness is another million dollar problem.



Yes but we introduced p as a fake variable.

Once again the Cauchy momentum equation gives you one equation with two unknowns, the velocity u and the stress tensor. Introduction of the constitutive Newtonian/Stokes fluid model allows you to re-write the stress tensor in terms of grad(U). So you can say that now you have one equation with one unknown u. The continuity equation is a constraint on u (really the momentum). The incompressible condition is yet another constraint.
We have to do all this dance of introducing "fake" variable P and other constraints because we are interested in solution to these equations.

The one variable one equation scenario is lovely except that we don't know practical way of solving this one equation.

I guess the discussion is about the practical part of solving these equations other wise we shall look into quantum aspects too and see if this one equation is really one equation or in some case manifests as multiple equations at those scales.

Even after so many things solving navier stokes is not cheap, one shall remember turbulence models are another addition for our convenience and don't really part of navier stokes and flow. They are there to help getting solution that could be used in practical sense.
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Old   April 9, 2017, 14:46
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I wish to add the reasons why one should invest in coupled solver because this was not mentioned before.

The main case for coupled solver is in its robustness and it is in the fact that it is less sensitive to under relaxation factors.

Segregated solvers are nice and cheap but sometimes you end up with cases difficult enough that segregated approach does not cut it. There you have no choice but to go with coupled solvers.

There are many such problems where one can end up. This is why coupled solvers remain important class of solvers. One should know where to use which solver and should not dismiss one for other.
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Old   April 9, 2017, 18:05
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Again, we should not forget that the decoupling between velocity and pressure leads to generate an error that is not present in the coupled counterpart system. And this error is irrespective of the order of space-time discretization but is inherent to the splitting. Often, this error is not sufficiently highlighted but can affect the boundary layer resolution.
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Old   April 10, 2017, 00:39
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And because I cannot stop myself from further muddying the waters, I would like to point out that there is a third way that has not seen a lot of attention here. That is using a segregated/fractional step method to precondition a proper Newton-Krylov iteration of the entire nonlinear system instead of just a linearized coupled form. The preconditioner may suffer from all of the missing cross-terms and the splitting errors that they always do, but the Newton correction is always driven by the full nonlinear residual projected into that Krylov space.

Incompressible flow seems like it might be a better match for NK schemes than fully compressible flows. Newton methods are less forgiving when converging to solutions with real discontinuities like shocks. Incompressible flow solutions are usually safely differentiable, at least up to the second derivative. I don't want this to evolve into a "why don't we use NK instead of SIMPLE/PISO" discussion, but I did want to point out that there is another algorithmic approach that is *potentially* more coupled in a nonlinear sense than linearized [u,v,w,p] coupled matrix solvers. And NK is *potentially* able to offer quadratic convergence rates for per-iteration costs only slightly higher than SIMPLE/PISO.
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Old   April 10, 2017, 01:47
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Quote:
Originally Posted by mprinkey View Post
And because I cannot stop myself from further muddying the waters, I would like to point out that there is a third way that has not seen a lot of attention here. That is using a segregated/fractional step method to precondition a proper Newton-Krylov iteration of the entire nonlinear system instead of just a linearized coupled form. The preconditioner may suffer from all of the missing cross-terms and the splitting errors that they always do, but the Newton correction is always driven by the full nonlinear residual projected into that Krylov space.
In past in year 2011-12 i tried it and had mixed success. Off course that was not full attempt but trying the idea over a weekend.

I believe others have also tried it and there is literature around it.

One can note that if one uses unpreconditioned gmres then to solve this schur all one need is matrix vector product. Which can be done without actually constructing it.

If then one wishes to precondition this gmres then all one need is preconditioner that is also based on matrix vector product.

This is what is going to be my viscoelastic flow solver for wildkatze when i implement. Its on card so in an year you would see it in action too.
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Old   April 10, 2017, 02:01
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In past in year 2011-12 i tried it and had mixed success. Off course that was not full attempt but trying the idea over a weekend.
I exerted about the same level of effort around the same time frame and encountered similar results. I haven't followed the state of the art. I'll need to catch up and I'll look forward to your code release next year. 8)

I wasn't really trying to tout NK as the best of the bunch. I think it still has potential, but I fully recognize that the residual calculation and orthgonalization costs could STILL make it slower than good old SIMPLE. I just think that it belongs in the conversation with linear coupled solvers.
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Old   April 10, 2017, 11:41
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The main case for coupled solver is in its robustness and it is in the fact that it is less sensitive to under relaxation factors.
I don't actually disagree, but one must keep in mind that the pre-conditioner more-or-less plays the role of under relaxation and the choice of a pre-conditioner is like choosing an under-relaxation factor.

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And because I cannot stop myself from further muddying the waters,
I think it was helpful to stir things up quite a bit to really appreciate that the answer is itself muddy. It is too common nowadays to obsess over "definite answers" to every issue.

Btw, Paolo hinted at it but I would like to really highlight it even further. Parallelization and scalability is a very important problem in computing or applied math, and you don't really see this problem when you focus on only the theoretical mathematics. Divide & conquer approaches are naturally well-suited for parallel computations.
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