[ilia.nikiforov]

Hello all, this is a theoretical question regarding symmetry/slip wall boundary conditions.


There are several topics on these forums, and the consensus seems that, in theory, the slip wall boundary condition and the symmetry boundary conditions are the same. For the velocity, at least, they should both mean that the normal component of the velocity is zero, as is the normal gradient of the tangential component of the velocity.


However, the basic problem of the potential flow around a cylinder seems to violate the gradient requirement, despite the fact that, for example, in OpenFOAM, the surface of the cylinder is specified as a "symmetry" boundary condition:
https://www.openfoam.com/documentati...utorialse3.php


The paradox can be seen from the analytical solution, which the above example agrees with:
https://en.wikipedia.org/wiki/Potent...cular_cylinder


The normal velocity, V_r, on the surface of the cylinder, is indeed zero.
However, the normal component of the gradient of the tangential velocity, d(V_theta)/dr, is nonzero -- for R=1 and U=1, it is 2*r^(-3)*sin(theta), which comes out to 2*sin(theta) at the surface of the cylinder.



I understand that potential flow is solving for the scalar-valued potential, therefore imposing the "zero normal gradient of the tangential velocity" boundary condition is not actually possible, because imposing any additional boundary conditions besides the zero normal velocity would overconstrain the boundary value problem. However, I hoped/expected that the resulting solution would satisfy the condition in question nevertheless.


Is there an explanation for this apparent paradox? Am I misunderstanding something?

[FMDenaro]

Originally Posted by ilia.nikiforov

Hello all, this is a theoretical question regarding symmetry/slip wall boundary conditions.


There are several topics on these forums, and the consensus seems that, in theory, the slip wall boundary condition and the symmetry boundary conditions are the same. For the velocity, at least, they should both mean that the normal component of the velocity is zero, as is the normal gradient of the tangential component of the velocity.


However, the basic problem of the potential flow around a cylinder seems to violate the gradient requirement, despite the fact that, for example, in OpenFOAM, the surface of the cylinder is specified as a "symmetry" boundary condition:
https://www.openfoam.com/documentati...utorialse3.php


The paradox can be seen from the analytical solution, which the above example agrees with:
https://en.wikipedia.org/wiki/Potent...cular_cylinder


The normal velocity, V_r, on the surface of the cylinder, is indeed zero.
However, the normal component of the gradient of the tangential velocity, d(V_theta)/dr, is nonzero -- for R=1 and U=1, it is 2*r^(-3)*sin(theta), which comes out to 2*sin(theta) at the surface of the cylinder.



I understand that potential flow is solving for the scalar-valued potential, therefore imposing the "zero normal gradient of the tangential velocity" boundary condition is not actually possible, because imposing any additional boundary conditions besides the zero normal velocity would overconstrain the boundary value problem. However, I hoped/expected that the resulting solution would satisfy the condition in question nevertheless.


Is there an explanation for this apparent paradox? Am I misunderstanding something?

I am not sure if my answer can help you but

1) I think you should focus not only on the character of the potential flow but only on the fact you are considering a non-simply connected domain.



2) Working on the stream-function problem, you can solve the Laplace problem by setting a constant Dirichlet value on the cylinder (not-permeable wall). However, the circuitation of the velocity on a closed line is not univocally defined until one sets explicitly the value of the integral constraint.

3) The condition "zero normal gradient of the tangential velocity" will be a condition for a second derivative of the stream function.

[praveen]

Slip bc means normal velocity is zero.

Symmetry bc means a lot more. You want to use symmetry conditions if your problem has some symmetry. E.g., if you want to solve flow around a cylinder, perhaps the solution is supposed to be symmetric about the line y = 0. So you want to solve only one half of the domain to reduce computational cost. ALL SCALAR flow quantities satisfy phi(x,-y) = phi(x,y) which means that d(phi)/dy = 0 on y = 0.

The velocity components satisfy

u(x,-y) = u(x,y)
v(x,-y) = -v(x,y)

This implies that (assuming smooth flow)

du/dy = 0 and v=0 on y=0

But on the cylinder surface it is not true that d(phi)/dn = 0 for every quantity phi.

You are referring to potential flow, where the normal velocity component being zero means d(phi)/dn = 0 where now phi is velocity potential. Mathematically this looks like a symmetry bc as explained above. But it does not mean that all other quantities also have zero normal derivatives.

If you are solving for velocity directly, then on a slip surface (like cylinder surface) you have

normal velocity = 0

which is part of the symmetry condition (normal velocity v=0 on symmetry surface as described above) but there is no other restriction on other variables.

[ilia.nikiforov]

Originally Posted by praveen

Slip bc means normal velocity is zero.

Symmetry bc means a lot more. You want to use symmetry conditions if your problem has some symmetry. E.g., if you want to solve flow around a cylinder, perhaps the solution is supposed to be symmetric about the line y = 0. So you want to solve only one half of the domain to reduce computational cost. ALL the flow quantities satisfy phi(x,-y) = phi(x,y) which means that d(phi)/dy = 0 on y = 0.

On the cylinder surface it is not true that d(phi)/dn = 0 for every quantity phi.

You are referring to potential flow, where the normal velocity component being zero means d(phi)/dn = 0 where now phi is velocity potential. Mathematically this looks like a symmetry bc as explained above. But it does not mean that all other quantities also have zero normal derivatives.

I see. In this case, the only quantity being solved for is the velocity potential phi, which is indeed symmetric (has zero normal gradient) on the cylinder. Because v=grad(phi), this is the same as setting normal velocity = 0. However things like gradient of velocity involve second derivatives of phi, which are not specified.


Then, the "symmetry" boundary condition would have a different physical interpretation in non-potential flow where the velocity vector is solved for directly.

[praveen]

I improved my answer since it was not completely correct.

It seems common to use the phrase "symmetry bc" or "mirror bc" to refer to zero Neumann condition on some particular quantity, or even when you only want to enforce the zero normal velocity condition.

I think it is better to use the terms

"slip bc"
"zero Neumann bc"

to be more precise.

[LuckyTran]

Originally Posted by ilia.nikiforov

I see. In this case, the only quantity being solved for is the velocity potential phi, which is indeed symmetric (has zero normal gradient) on the cylinder.

In OpenFOAM you apply the symmetry condition as a user input for U, not phi.

[ilia.nikiforov]

Originally Posted by LuckyTran

In OpenFOAM you apply the symmetry condition as a user input for U, not phi.

I see. But then the paradox I am asking about is still present, no?


I am developing my own method for inviscid flow with vorticity, so I am researching what boundary conditions are commonly used with inviscid flow. There are various terms for it, but "full slip", "free slip", "inviscid wall", "Euler wall", "symmetry", etc. all seem to mean the same thing -- zero normal velocity and zero normal gradient of the tangential velocity (and zero normal pressure gradient, I think).



I was curious to see if the basic potential flow examples followed these B.Cs as well, because potential flow is a special case of general inviscid flow. So I am still surprised and confused why this example doesn't agree with this B.C. Is there any physical meaning to the fact that the tangential velocity component has a gradient in the direction normal to the surface?

[FMDenaro]

Originally Posted by ilia.nikiforov

I see. But then the paradox I am asking about is still present, no?


I am developing my own method for inviscid flow with vorticity, so I am researching what boundary conditions are commonly used with inviscid flow. There are various terms for it, but "full slip", "free slip", "inviscid wall", "Euler wall", "symmetry", etc. all seem to mean the same thing -- zero normal velocity and zero normal gradient of the tangential velocity (and zero normal pressure gradient, I think).



I was curious to see if the basic potential flow examples followed these B.Cs as well, because potential flow is a special case of general inviscid flow. So I am still surprised and confused why this example doesn't agree with this B.C. Is there any physical meaning to the fact that the tangential velocity component has a gradient in the direction normal to the surface?

And why d(t.v)/dn= d(t.Grad phi)/dn should be zero on a boundary??

[ilia.nikiforov]

Originally Posted by FMDenaro

And why d(t.v)/dn= d(t.Grad phi)/dn should be zero on a boundary??

I think the explanation is that this represents zero shear stress at the wall, which makes sense since inviscid fluids cannot undergo shear stress. Which makes it even more puzzling that potential flow around a cylinder appears to violate this condition.



Hopefully someone can explain. I am more interested in understanding the theoretical BVP problem statement and the physical meaning of the boundary condition than specific implementation details.

[FMDenaro]

Originally Posted by ilia.nikiforov

I think the explanation is that this represents zero shear stress at the wall, which makes sense since inviscid fluids cannot undergo shear stress. Which makes it even more puzzling that potential flow around a cylinder appears to violate this condition.



Hopefully someone can explain. I am more interested in understanding the theoretical BVP problem statement and the physical meaning of the boundary condition than specific implementation details.

Not at all! The starting hypothesis is that the fluid is ideal, that is there is no viscosity. That is why the viscous stress is zero, not because the derivative of the velocity is zero.

Remember that the Laplace equation for phi is nothing but the divergence-free constraint and the curl v = 0 is automatically implied.

[LuckyTran]

Originally Posted by ilia.nikiforov

the consensus seems that, in theory, the slip wall boundary condition and the symmetry boundary conditions are the same. For the velocity, at least, they should both mean that the normal component of the velocity is zero, as is the normal gradient of the tangential component of the velocity.

The zero normal gradient of tangential velocity is the slip or no shear condition of a viscous fluid. As you aptly noticed, it seems you cannot apply this condition for a potential flow. And that is true. You don't apply this condition for a inviscid fluid.

Originally Posted by ilia.nikiforov

Is there any physical meaning to the fact that the tangential velocity component has a gradient in the direction normal to the surface?

In the example of flow around a cylinder, this manifests physically as the flow turning around the cylinder.


I think you can go even further and show that applying the constraint of no normal gradient in tangential velocity would actually result in a rotational flow.

[ilia.nikiforov]

Originally Posted by FMDenaro

Not at all! The starting hypothesis is that the fluid is ideal, that is there is no viscosity. That is why the viscous stress is zero, not because the derivative of the velocity is zero.

Remember that the Laplace equation for phi is nothing but the divergence-free constraint and the curl v = 0 is automatically implied.

Essentially the question I have is, "what would be the boundary conditions imposed on the velocity field if the flow around a cylinder problem was solved using the incompressible, steady Euler equation and the zero-divergence equation, retaining all three unknowns -- vx, vy, p?"



I do not think that the no-penetration condition is enough, as it is only one equation. There would need to be a second equation for velocity (and one more for the pressure). I thought the second velocity condition would be the normal gradient of the tangential velocity, but clearly that is not satisfied. What would the required second equation be, then?

[ilia.nikiforov]

Originally Posted by LuckyTran

The zero normal gradient of tangential velocity is the slip or no shear condition of a viscous fluid. As you aptly noticed, it seems you cannot apply this condition for a potential flow. And that is true. You don't apply this condition for a inviscid fluid.



In the example of flow around a cylinder, this manifests physically as the flow turning around the cylinder.

Thank you! This is very helpful. I had a suspicion that the curvature of the surface had something to do with it (I know that the zero normal gradient of tangential velocity condition holds for potential flows along planar walls, for example the flow into a 90* corner analytical case).

[FMDenaro]

Originally Posted by ilia.nikiforov

Essentially the question I have is, "what would be the boundary conditions imposed on the velocity field if the flow around a cylinder problem was solved using the incompressible, steady Euler equation and the zero-divergence equation, retaining all three unknowns -- vx, vy, p?"



I do not think that the no-penetration condition is enough, as it is only one equation. There would need to be a second equation for velocity (and one more for the pressure). I thought the second velocity condition would be the normal gradient of the tangential velocity, but clearly that is not satisfied. What would the required second equation be, then?

Your question has a theoretical answer based on the Hodge deomposition theorem: only one condition is required to be the mathematical problem well posed.


In other words, you have the only unknown that is the velocity vector field under the constraints


Div v = 0
Curl v = 0
+ BC


You can set either the normal or the tangential components but never both.

[ilia.nikiforov]

Originally Posted by FMDenaro

Your question has a theoretical answer based on the Hodge deomposition theorem: only one condition is required to be the mathematical problem well posed.


In other words, you have the only unknown that is the velocity vector field under the constraints


Div v = 0
Curl v = 0
+ BC


You can set either the normal or the tangential components but never both.

Curl v = 0 is only true for potential flows, though, and what allows the problem to be reduced to the Laplace equation of a single scalar unknown. In that case, I understand that only one boundary equation is needed. I am interested in the case when curl v is generally nonzero, and the velocity vector must be retained as a vector unknown. I was only looking at the potential flow around a cylinder as a simple special case.


I understand that the well-posedness of the Euler equations, even in the incompressible steady case is not straightforward, but my intuition is that there must be at least one independent scalar equation for each scalar unknown on each boundary in order to have a chance of a well-posed problem.

[FMDenaro]

Originally Posted by ilia.nikiforov

Curl v = 0 is only true for potential flows, though, and what allows the problem to be reduced to the Laplace equation of a single scalar unknown. In that case, I understand that only one boundary equation is needed. I am interested in the case when curl v is generally nonzero, and the velocity vector must be retained as a vector unknown. I was only looking at the potential flow around a cylinder as a simple special case.

There is some confusion in your reply. The presence of the vorticity field z, that is Curl v = z, does not lead to a potential flow problem (even if you can still introduce the part of the complez potential function by means of the vector function psi such that v = Curl psi).
If you accept that a vorticity exists in the field then you cannot introduce the Laplace equation Div Grad phi = 0.

[FMDenaro]

Have a careful reading here https://www.researchgate.net/publica...ary_conditions

[ilia.nikiforov]

Originally Posted by FMDenaro

There is some confusion in your reply. The presence of the vorticity field z, that is Curl v = z, does not lead to a potential flow problem (even if you can still introduce the part of the complez potential function by means of the vector function psi such that v = Curl psi).
If you accept that a vorticity exists in the field then you cannot introduce the Laplace equation Div Grad phi = 0.

I fully understand this. I do not wish to use the Laplace equation, I wish to use the full equations of motion and continuity for a steady incompressible inviscid fluid (Euler equation + zero-divergence). I was looking at the potential flow around a cylinder as a simple example because it theoretically could be solved using the full equations, even though it is normally solved using the simpler potential flow method.

[FMDenaro]

Originally Posted by ilia.nikiforov

I fully understand this. I do not wish to use the Laplace equation, I wish to use the full equations of motion and continuity for a steady incompressible inviscid fluid (Euler equation + zero-divergence). I was looking at the potential flow around a cylinder as a simple example because it theoretically could be solved using the full equations, even though it is normally solved using the simpler potential flow method.

That means


Div v =0


dv/dt + div (vv) + Grad phi = 0

That is a first order hyperbolic system of equation in the velocity field, the "pressure" phi being only a lagrangian multiplier (just apply the curl to the momentum). Again, only one condition is required on the boundary. Since the fluid is ideal, you cannot set the physical non-slip condition and only the normal velocity is prescribed to fulfill the continuity equation.
To have two BCs you need a second order equation, that is the momentum equation with the diffusive term in the RHS.

[ilia.nikiforov]

Originally Posted by FMDenaro

Have a careful reading here https://www.researchgate.net/publica...ary_conditions

Thank you! I will take a look