How many boundary conditions do we need for the pressure?

kinematic_presser · October 11, 2022, 08:40

Hello guys,
I am still somewhat new to CFD (just starting my masters degree) so please bear with me if my question is dumb.

I am simulating fluid flow in a channel for an assignment. The aim of the assignment is to simulate 2 dimensional incompressible Poiseulle flow, moving from left to right. It is a very simple simulation. There are no obstacles and it is a simple rectangular domain. For the velocity, I enforce a velocity inlet condition at the inlet and a homogeneous Neumann condition at the outlet. I enforce a homogeneous Dirichlet condition (no-slip) at the top and bottom wall. This all makes sense to me as the INSE are second order in u .

But, when I was looking over some example tutorial cases in OpenFOAM, I noticed that their simulations were enforcing a boundary condition for the pressure on both the inlet AND the outlet, and correspondingly, on both the bottom AND the top wall. They typically used a homogeneous Neumann condition on the top and the bottom wall, a homogeneous Neumann condition on the inlet, and a Dirichlet condition on the outlet.

This does not make sense to me from a mathematical point of view, because the INSE are first order in p in both the x and y coordinate, so specifying boundary conditions at only, for example, the bottom and left wall should already be enough to fully specify the solution.

I know that the INSE are invariant under constant pressure shifts p --> p+p0 which gives the system a little extra freedom. I also know that typically the pressure is simulated using the pressure Poisson equation, which is of course second order. But this is a consequence of the INSE, not a governing equation in its own right.

I would appreciate some elucidation on this topic. Thanks.

LuckyTran · October 11, 2022, 13:59

The number and types of boundary condition needed at each boundary depend on the nature of the characteristics at the boundary.

When you learn that a 2nd order PDE needs "only 2 boundary conditions" you are:
1) dealing with an equation that is an independent pde in one dynamic variable (i.e. temperature) that is not coupled with anybody else
2) learning how to solve laplace's equation, where all extrema occur on domain extremes
3) solving on a 1D line or 2D rectangle where there is a clear distinction between x and y coordinates and so you have an inlet and outlet (i.e. left and right boundary) in the x direction and inlet and outlet in the y direction (i.e. top and bottom). And you associate this with "two boundary conditions"

Consider for example that you are solving navier-stokes on a hexagonal domain... clearly "two boundary conditions" is not enough to solve this problem. But hmm, that's strange... suddenly the number of BCs needed depends on the number of boundaries....? Consider also that you can't arbitrarily specify higher 4th order or n-th order derivative on all boundaries to solve this problem. Keep in mind that to a mathematician, an inlet and outlet doesn't mean anything in terms of topology, all boundaries are compact surfaces bounded in

. They don't even count how many of them there are.

kinematic_presser · October 11, 2022, 15:10

Thank you for the response. I am familiar with the traditional polygonal domain example. Allow me to be more precise.

Let

be a connected, open, bounded, convex subset of $\mathbb{R}^n$ , where

=2 or 3. (Think of some sort of ball.) Now draw a line through E. Because of the assumptions we made on

, this line intersects $\partial E$ exactly twice.

Now let's say we are solving some partial differential equation in

. As I understand it, for a second order problem or higher, we need to specify two (EXACTLY two) pieces of information, either both at the same point, or one at either point at which this line intersects $\partial E$ , in order to uniquely specify a solution (assuming a unique solution exists). The most common way is to use a Dirichlet condition at each point.

However, for a first order problem, (again, as I understand it) we require one, and ONLY one piece of information on this line, so the most we can do is to specify some condition at only one of these points of intersection. So, in your hexagonal domain example, boundary data on only one of the sides is enough to fully specify a solution (matching our intuition).

That is essentially what I am asking about. When we specify two boundary conditions along this line for the pressure, are we just hoping that some solution exists that meets the criteria, and praying that our numerical scheme doesn't diverge?

I hope this frames my question better.

LuckyTran · October 11, 2022, 15:39

Numerical scheme diverging or not has nothing to do with the structure of the underlying continuous pde. So no need to discuss what happens numerically when you want a general answer to how many BCs you need for a given PDE.

Again you are using an example of an independent equation in E, Navier stokes is a coupled system of non-linear pde's

And if this line enters and exits the same boundary.... what then? You still can have a posedness problem. Oh but you say it can't intersect itself, well a different point on the same "surface" is also not intersecting itself and can be partitioned to be a different surface via "compact surface in R3." I'm not cheating, that's what differential means. I can always carve out your surface to two smol surfaces (i.e. two points) and reduce this argument to specifying conditions at two points and not two surfaces.

Furthermore, the "two" pieces of information you need to provide also aren't arbitrary for posedness. You can overconstrain a problem by specifying E everywhere depending on what the characteristics at each boundary are (the infamous static vs total properties). Consider what happens at a boundary if you have 0 velocity, + infinite velocity, and - infinite velocity. Really what matters is the Mach number, but I hope that drives the point.

LuckyTran · October 11, 2022, 15:52

I will add that it is straightforward to prove uniqueness in a linear ODE/PDE and there you can easily show that you only need "two boundary conditions" for a 2nd-order ode/pde. But Navier-Stokes is not Laplace's equation. Furthermore, you can use a simple substitution to convert an n-order linear ode/pde into a system of n 1st order ode/pde's, i.e. starting with

define u=y' and then you can rewrite it as

These properties make the analysis easier to generalize to higher orders for linear ODE's and PDE's because if you can show uniqueness for the 1st order case, you can then show it for the 2nd order case and so on by using this trick. But Navier-Stokes is not Laplace's equation.

Let me summarize this by saying all linear 2nd order PDE's look very much alike. Coupled sets of non-linear ones don't necessarily have the same simple rules.

FMDenaro · October 11, 2022, 16:33

Quote:

Originally Posted by kinematic_presser

Hello guys,
I am still somewhat new to CFD (just starting my masters degree) so please bear with me if my question is dumb.

I am simulating fluid flow in a channel for an assignment. The aim of the assignment is to simulate 2 dimensional incompressible Poiseulle flow, moving from left to right. It is a very simple simulation. There are no obstacles and it is a simple rectangular domain. For the velocity, I enforce a velocity inlet condition at the inlet and a homogeneous Neumann condition at the outlet. I enforce a homogeneous Dirichlet condition (no-slip) at the top and bottom wall. This all makes sense to me as the INSE are second order in u .

But, when I was looking over some example tutorial cases in OpenFOAM, I noticed that their simulations were enforcing a boundary condition for the pressure on both the inlet AND the outlet, and correspondingly, on both the bottom AND the top wall. They typically used a homogeneous Neumann condition on the top and the bottom wall, a homogeneous Neumann condition on the inlet, and a Dirichlet condition on the outlet.

This does not make sense to me from a mathematical point of view, because the INSE are first order in p in both the x and y coordinate, so specifying boundary conditions at only, for example, the bottom and left wall should already be enough to fully specify the solution.

I know that the INSE are invariant under constant pressure shifts p --> p+p0 which gives the system a little extra freedom. I also know that typically the pressure is simulated using the pressure Poisson equation, which is of course second order. But this is a consequence of the INSE, not a governing equation in its own right.

I would appreciate some elucidation on this topic. Thanks.

Since you are new in CFD, I suggest to stop a moment using OpenFOAM and just think about a more general statement of the problem.
You have a system of equations, the divergence-free velocity constraint and the momentum equation, they should be solved as a coupled system. The hyperbolic divergence-free constraint is transformed into the elliptic Poisson equation for the "pressure" (that has nothing to do with the thermodynamic pressure) and in order for the problem to be well posed, you have to prescribe the BC along the whole frontier. The BC for the pressure is a direct derivation of the Hodge decomposition, for this reason Neumann BCs for the pressure are prescribed. However, the singularity of the matrix requires that you fulfill with your BC the compatibility condition, that is you fulfill the continuity equation.
Therefore, you are wrong, the pressure equation is a governing equation substituting the continuity.

kinematic_presser · October 11, 2022, 17:43

I don't understand everything you have said, but I gather that the point you are trying to convey is "It's not so simple". I suppose all we can do is assign boundary conditions that match the physics of the problem, and hope for the best. Is that correct?

-----

ADDENDUM: Why this concerns me.

The momentum equation can be rewritten as

$\nabla p=-\boldsymbol \partial_t\boldsymbol u-\boldsymbol u\cdot\nabla\boldsymbol u+\nu\Delta\boldsymbol u$

Where the right hand side side is curl-free. Let's suppose we somehow knew the solution for $\boldsymbol u$ ahead of time, and all we have to do is find the pressure that fulfills our prescribed BCs. So, in general, is it possible to find

? Essentially, we focus on the following:

Given a connected, open, bounded subset $\Omega$ of $\mathbb{R}^n$ (n=2,3) and a curl-free vector field $\boldsymbol v$ , is it always possible to find a scalar field $\phi$ that satisfies
$\nabla\phi=\boldsymbol v~~~~\text{in}~\Omega$
$\phi=h~~~~\text{on}~\partial\Omega$

The answer is NO. For example, consider a two-dimensional case and take $\boldsymbol v(x,y)=(x,y)$ . It is easily shown that all of the $\phi$ that satisfy $\nabla \phi=\boldsymbol v$ are of the form $\phi(x,y)=C+x^2/2+y^2/2$ . Therefore taking for example

there is no solution.

kinematic_presser · October 11, 2022, 17:58

Interesting, and thanks for your remarks. Are you essentially saying that since the INSE are second order in u, they are basically second order in p as well by proxy? I am happy enough with this simple explanation if it is true.

FMDenaro · October 11, 2022, 17:58

Quote:

Originally Posted by kinematic_presser

I don't understand everything you have said, but I gather that the point you are trying to convey is "It's not so simple". I suppose all we can do is assign boundary conditions that match the physics of the problem, and hope for the best. Is that correct?

-----

ADDENDUM: Why this concerns me.

The momentum equation can be rewritten as

$\nabla p=-\boldsymbol \partial_t\boldsymbol u-\boldsymbol u\cdot\nabla\boldsymbol u+\nu\Delta\boldsymbol u$

Where the right hand side side is curl-free. Let's suppose we somehow knew the solution for $\boldsymbol u$ ahead of time, and all we have to do is find the pressure that fulfills our prescribed BCs. So, in general, is it possible to find

? Essentially, we focus on the following:

Given a connected, open, bounded subset $\Omega$ of $\mathbb{R}^n$ (n=2,3) and a curl-free vector field $\boldsymbol v$ , is it always possible to find a scalar field $\phi$ that satisfies
$\nabla\phi=\boldsymbol v~~~~\text{in}~\Omega$
$\phi=h~~~~\text{on}~\partial\Omega$

The answer is NO. For example, consider a two-dimensional case and take $\boldsymbol v(x,y)=(x,y)$ . It is easily shown that all of the $\phi$ that satisfy $\nabla \phi=\boldsymbol v$ are of the form $\phi(x,y)=C+x^2/2+y^2/2$ . Therefore taking for example

there is no solution.

In some sense you are right...the formulation for the incompressible flows is somehow both not simple and not physical ... you are assuming that the pressure wave moves at infinite speed, a fact that is not physically meaningful. But it works and provides a good approximation for low Mach flows.

I think you have to start from the theory of incompressible flows to understand that the "pressure" is nothing but a lagrangian multiplier.

I just try to clarify the point in a different way:

Assume dv/dt=a to be the eulerian acceleration and a* to be convective+diffusive terms.

The system writes as

Div a =0 (Continuity)

a + Grad p = a* (Momentum)

Now you have to consider the Helmholtz-Hodge decomposition theorem to understand that

Div Grad p = Div a*

is nothing but the continuity equation and the BCs are deduced by evaluating the momentum equation on the frontier

dp/dn = (a*-a).n

This way you have that the compatibility equation

Int[V] Div Grad p dV = Int[V] Div a* dV

is satisfied.

That allows to have a solution apart an arbitrary function of time. In case the integral relation is not satisfied, your iterative solver will never converge. Fixing a Dirichlet condition is only a way to produce a wrong convergence, it is not necessary at all.

kinematic_presser · October 11, 2022, 17:59

FYI everyone, I have a pretty extensive background in mathematics, so please feel free to communicate your ideas to me using any and all mathematical language you wish. I actually will appreciate it far better than an argument using words alone.

FMDenaro · October 11, 2022, 18:03

Quote:

Originally Posted by kinematic_presser

FYI everyone, I have a pretty extensive background in mathematics, so please feel free to communicate your ideas to me using any and all mathematical language you wish. I actually will appreciate it far better than an argument using words alone.

You can read my old paper about the topic
https://www.researchgate.net/publica...ary_conditions

sbaffini · October 12, 2022, 06:54

I'm not sure I understand your issue but, from the original NSE one can derive a pressure equation which is a Helmholtz equation. Taking the incompressible limit, this equation collapses into a Poisson equation. I won't go into the details of which boundary conditions these equations need.

Incompressible pressure based methods do just the same, solve for a set of equations which has been artificially forced to give an elliptic nature to the pressure equation.

As an additional note, we also have preconditioned density based methods, which work exactly in the opposite direction, by artificially forcing the equations to always have the compressible nature, even in the full incompressible limit, trough an artificial speed of sound. When using such methods, boundary conditions are indeed assigned following the artificially hyperbolic nature of the equations.

Winston Virtaus · May 29, 2023, 22:02

Quote:

Originally Posted by FMDenaro

Now you have to consider the Helmholtz-Hodge decomposition theorem to understand that

Div Grad p = Div a*

is nothing but the continuity equation and the BCs are deduced by evaluating the momentum equation on the frontier

dp/dn = (a*-a).n

This way you have that the compatibility equation

Int[V] Div Grad p dV = Int[V] Div a* dV

is satisfied.

Would the compatibility equation

Int[V] Div Grad p dV = Int[V] Div a* dV

get satisfied if the boundary condition for p is given as dp/dn=0 and at the boundary a*.n is simply set equal to a.n?

I'm assuming a*.n would need to be otherwise somehow interpolated to the boundary if cell centered FVM was used.

FMDenaro · May 30, 2023, 04:19

Quote:

Originally Posted by Winston Virtaus

Would the compatibility equation

Int[V] Div Grad p dV = Int[V] Div a* dV

get satisfied if the boundary condition for p is given as dp/dn=0 and at the boundary a*.n is simply set equal to a.n?

I'm assuming a*.n would need to be otherwise somehow interpolated to the boundary if cell centered FVM was used.

Yes, the Poisson problem is well posted if, equivalently, you prescribe homogeneous Neumann BC and modify the source term close to the boundaries.

Winston Virtaus · May 30, 2023, 22:33

Very interesting! If this kind of a boundary condition is applied to the conventional

u = u*-deltaT*gradP

decomposition, would this boundary condition allow to have nonzero tangential velocities prescribed at boundary aswell?

Or just a nonzero normal component?

FMDenaro · May 31, 2023, 04:09

Quote:

Originally Posted by Winston Virtaus

Very interesting! If this kind of a boundary condition is applied to the conventional

u = u*-deltaT*gradP

decomposition, would this boundary condition allow to have nonzero tangential velocities prescribed at boundary aswell?

Or just a nonzero normal component?

Only the normal component can be prescribed. The tangential component is a result of the numerical solution.
This is a known issue in the fractional time step method.

Winston Virtaus · May 31, 2023, 20:10

Thanks for the info!

I was also wondering if this kind of a boundary condition would also work with impulsively started flows?

For example if flow in a pipe was initially at rest and and then inlet velocity (say 1 m/s) was given at the the first time step, would solution of the poisson equation still enforce continuity?

FMDenaro · June 1, 2023, 04:36

Quote:

Originally Posted by Winston Virtaus

Thanks for the info!

I was also wondering if this kind of a boundary condition would also work with impulsively started flows?

For example if flow in a pipe was initially at rest and and then inlet velocity (say 1 m/s) was given at the the first time step, would solution of the poisson equation still enforce continuity?

Using the incompressible flow model you cannot simulate a case where the flow enters and propagates in the interior. Pressure wave are assumed to travel at infinite speed.
In other words, you need to prescribe an outlet BC to ensure that the mass entering and the mass leaving are impulsively equal. In such a case the Poisson problem is well posed.

Winston Virtaus · June 1, 2023, 21:20

So in theory a badly posed Poisson problem at the start would continue being badly posed through subsequent time steps and eventually even result in a solver crash?

Would a potential flow solution then be a reasonable starting point for the solver?

FMDenaro · June 2, 2023, 00:31

The solver does not converge since from the first time step. A potential initial solution can be used.

October 11, 2022, 08:40	How many boundary conditions do we need for the pressure?	#1
kinematic_presser New Member K Join Date: Oct 2022 Location: Madison, WI Posts: 19 Rep Power: 4	Hello guys, I am still somewhat new to CFD (just starting my masters degree) so please bear with me if my question is dumb. I am simulating fluid flow in a channel for an assignment. The aim of the assignment is to simulate 2 dimensional incompressible Poiseulle flow, moving from left to right. It is a very simple simulation. There are no obstacles and it is a simple rectangular domain. For the velocity, I enforce a velocity inlet condition at the inlet and a homogeneous Neumann condition at the outlet. I enforce a homogeneous Dirichlet condition (no-slip) at the top and bottom wall. This all makes sense to me as the INSE are second order in u . But, when I was looking over some example tutorial cases in OpenFOAM, I noticed that their simulations were enforcing a boundary condition for the pressure on both the inlet AND the outlet, and correspondingly, on both the bottom AND the top wall. They typically used a homogeneous Neumann condition on the top and the bottom wall, a homogeneous Neumann condition on the inlet, and a Dirichlet condition on the outlet. This does not make sense to me from a mathematical point of view, because the INSE are first order in p in both the x and y coordinate, so specifying boundary conditions at only, for example, the bottom and left wall should already be enough to fully specify the solution. I know that the INSE are invariant under constant pressure shifts p --> p+p0 which gives the system a little extra freedom. I also know that typically the pressure is simulated using the pressure Poisson equation, which is of course second order. But this is a consequence of the INSE, not a governing equation in its own right. I would appreciate some elucidation on this topic. Thanks.

October 11, 2022, 15:10		#3
kinematic_presser New Member K Join Date: Oct 2022 Location: Madison, WI Posts: 19 Rep Power: 4	Thank you for the response. I am familiar with the traditional polygonal domain example. Allow me to be more precise. Let be a connected, open, bounded, convex subset of $\mathbb{R}^n$ , where =2 or 3. (Think of some sort of ball.) Now draw a line through E. Because of the assumptions we made on , this line intersects $\partial E$ exactly twice. Now let's say we are solving some partial differential equation in . As I understand it, for a second order problem or higher, we need to specify two (EXACTLY two) pieces of information, either both at the same point, or one at either point at which this line intersects $\partial E$ , in order to uniquely specify a solution (assuming a unique solution exists). The most common way is to use a Dirichlet condition at each point. However, for a first order problem, (again, as I understand it) we require one, and ONLY one piece of information on this line, so the most we can do is to specify some condition at only one of these points of intersection. So, in your hexagonal domain example, boundary data on only one of the sides is enough to fully specify a solution (matching our intuition). That is essentially what I am asking about. When we specify two boundary conditions along this line for the pressure, are we just hoping that some solution exists that meets the criteria, and praying that our numerical scheme doesn't diverge? I hope this frames my question better. Last edited by kinematic_presser; October 11, 2022 at 15:12. Reason: added in that i dont have proof

October 11, 2022, 15:52		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	I will add that it is straightforward to prove uniqueness in a linear ODE/PDE and there you can easily show that you only need "two boundary conditions" for a 2nd-order ode/pde. But Navier-Stokes is not Laplace's equation. Furthermore, you can use a simple substitution to convert an n-order linear ode/pde into a system of n 1st order ode/pde's, i.e. starting with define u=y' and then you can rewrite it as These properties make the analysis easier to generalize to higher orders for linear ODE's and PDE's because if you can show uniqueness for the 1st order case, you can then show it for the 2nd order case and so on by using this trick. But Navier-Stokes is not Laplace's equation. Let me summarize this by saying all linear 2nd order PDE's look very much alike. Coupled sets of non-linear ones don't necessarily have the same simple rules. Last edited by LuckyTran; October 11, 2022 at 18:58.

October 11, 2022, 17:59	about me	#10
kinematic_presser New Member K Join Date: Oct 2022 Location: Madison, WI Posts: 19 Rep Power: 4	FYI everyone, I have a pretty extensive background in mathematics, so please feel free to communicate your ideas to me using any and all mathematical language you wish. I actually will appreciate it far better than an argument using words alone. sbaffini likes this.

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October 11, 2022, 13:59		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	The number and types of boundary condition needed at each boundary depend on the nature of the characteristics at the boundary. When you learn that a 2nd order PDE needs "only 2 boundary conditions" you are: 1) dealing with an equation that is an independent pde in one dynamic variable (i.e. temperature) that is not coupled with anybody else 2) learning how to solve laplace's equation, where all extrema occur on domain extremes 3) solving on a 1D line or 2D rectangle where there is a clear distinction between x and y coordinates and so you have an inlet and outlet (i.e. left and right boundary) in the x direction and inlet and outlet in the y direction (i.e. top and bottom). And you associate this with "two boundary conditions" Consider for example that you are solving navier-stokes on a hexagonal domain... clearly "two boundary conditions" is not enough to solve this problem. But hmm, that's strange... suddenly the number of BCs needed depends on the number of boundaries....? Consider also that you can't arbitrarily specify higher 4th order or n-th order derivative on all boundaries to solve this problem. Keep in mind that to a mathematician, an inlet and outlet doesn't mean anything in terms of topology, all boundaries are compact surfaces bounded in . They don't even count how many of them there are.

October 11, 2022, 15:39		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	Numerical scheme diverging or not has nothing to do with the structure of the underlying continuous pde. So no need to discuss what happens numerically when you want a general answer to how many BCs you need for a given PDE. Again you are using an example of an independent equation in E, Navier stokes is a coupled system of non-linear pde's And if this line enters and exits the same boundary.... what then? You still can have a posedness problem. Oh but you say it can't intersect itself, well a different point on the same "surface" is also not intersecting itself and can be partitioned to be a different surface via "compact surface in R3." I'm not cheating, that's what differential means. I can always carve out your surface to two smol surfaces (i.e. two points) and reduce this argument to specifying conditions at two points and not two surfaces. Furthermore, the "two" pieces of information you need to provide also aren't arbitrary for posedness. You can overconstrain a problem by specifying E everywhere depending on what the characteristics at each boundary are (the infamous static vs total properties). Consider what happens at a boundary if you have 0 velocity, + infinite velocity, and - infinite velocity. Really what matters is the Mach number, but I hope that drives the point.

October 11, 2022, 17:43		#7
kinematic_presser New Member K Join Date: Oct 2022 Location: Madison, WI Posts: 19 Rep Power: 4	I don't understand everything you have said, but I gather that the point you are trying to convey is "It's not so simple". I suppose all we can do is assign boundary conditions that match the physics of the problem, and hope for the best. Is that correct? ----- ADDENDUM: Why this concerns me. The momentum equation can be rewritten as $\nabla p=-\boldsymbol \partial_t\boldsymbol u-\boldsymbol u\cdot\nabla\boldsymbol u+\nu\Delta\boldsymbol u$ Where the right hand side side is curl-free. Let's suppose we somehow knew the solution for $\boldsymbol u$ ahead of time, and all we have to do is find the pressure that fulfills our prescribed BCs. So, in general, is it possible to find ? Essentially, we focus on the following: Given a connected, open, bounded subset $\Omega$ of $\mathbb{R}^n$ (n=2,3) and a curl-free vector field $\boldsymbol v$ , is it always possible to find a scalar field $\phi$ that satisfies $\nabla\phi=\boldsymbol v~~~~\text{in}~\Omega$ $\phi=h~~~~\text{on}~\partial\Omega$ The answer is NO. For example, consider a two-dimensional case and take $\boldsymbol v(x,y)=(x,y)$ . It is easily shown that all of the $\phi$ that satisfy $\nabla \phi=\boldsymbol v$ are of the form $\phi(x,y)=C+x^2/2+y^2/2$ . Therefore taking for example there is no solution.

October 11, 2022, 17:58		#8
kinematic_presser New Member K Join Date: Oct 2022 Location: Madison, WI Posts: 19 Rep Power: 4	Interesting, and thanks for your remarks. Are you essentially saying that since the INSE are second order in u, they are basically second order in p as well by proxy? I am happy enough with this simple explanation if it is true.

October 12, 2022, 06:54		#12
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	I'm not sure I understand your issue but, from the original NSE one can derive a pressure equation which is a Helmholtz equation. Taking the incompressible limit, this equation collapses into a Poisson equation. I won't go into the details of which boundary conditions these equations need. Incompressible pressure based methods do just the same, solve for a set of equations which has been artificially forced to give an elliptic nature to the pressure equation. As an additional note, we also have preconditioned density based methods, which work exactly in the opposite direction, by artificially forcing the equations to always have the compressible nature, even in the full incompressible limit, trough an artificial speed of sound. When using such methods, boundary conditions are indeed assigned following the artificially hyperbolic nature of the equations.

May 30, 2023, 22:33		#15
Winston Virtaus Member Join Date: Apr 2021 Location: Finland Posts: 41 Rep Power: 5	Very interesting! If this kind of a boundary condition is applied to the conventional u = u-deltaTgradP decomposition, would this boundary condition allow to have nonzero tangential velocities prescribed at boundary aswell? Or just a nonzero normal component?

May 31, 2023, 20:10		#17
Winston Virtaus Member Join Date: Apr 2021 Location: Finland Posts: 41 Rep Power: 5	Thanks for the info! I was also wondering if this kind of a boundary condition would also work with impulsively started flows? For example if flow in a pipe was initially at rest and and then inlet velocity (say 1 m/s) was given at the the first time step, would solution of the poisson equation still enforce continuity?

June 1, 2023, 21:20		#19
Winston Virtaus Member Join Date: Apr 2021 Location: Finland Posts: 41 Rep Power: 5	So in theory a badly posed Poisson problem at the start would continue being badly posed through subsequent time steps and eventually even result in a solver crash? Would a potential flow solution then be a reasonable starting point for the solver?

June 2, 2023, 00:31		#20
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	The solver does not converge since from the first time step. A potential initial solution can be used.