Does solenoidal field orthogonal to dilatational field?

TurbJet · April 3, 2019, 02:04

Hi,

Currently I am reading a paper about compressible homogeneous isotropic turbulence. In this paper, the authors claim that

$\langle u^s_iu^d_i\rangle = 0$

where <> stands for volume averaging, and superscript s & d represent solenoidal/dilatational velocity fields, which can be obtained from Helmholtz decomposition.

I really don't see how this identity is valid. Solenoidal and dilatational components are orthogonal in spectral space, that's for sure. But I don't think it is equivalent to be orthogonal in physical space. Or I am misunderstanding this identity; it's equal to zero because some other reason?

Can anyone give me some hints?

Appreciate it.

FMDenaro · April 3, 2019, 04:17

Quote:

Originally Posted by TurbJet

Hi,

Currently I am reading a paper about compressible homogeneous isotropic turbulence. In this paper, the authors claim that

$\langle u^s_iu^d_i\rangle = 0$

where <> stands for volume averaging, and superscript s & d represent solenoidal/dilatational velocity fields, which can be obtained from Helmholtz decomposition.

I really don't see how this identity is valid. Solenoidal and dilatational components are orthogonal in spectral space, that's for sure. But I don't think it is equivalent to be orthogonal in physical space. Or I am misunderstanding this identity; it's equal to zero because some other reason?

Can anyone give me some hints?

Appreciate it.

Yes, they are not orthogonal each other in pointwise sense.

The orthogonality is in the sense of the generalized inner product not in the standard inner product implied by the Pythagorean theorem.
I explained this issue after Eq.(3) here https://www.researchgate.net/publica...ary_conditions

TurbJet · April 3, 2019, 14:12

Quote:

Originally Posted by FMDenaro

Yes, they are not orthogonal each other in pointwise sense.

The orthogonality is in the sense of the generalized inner product not in the standard inner product implied by the Pythagorean theorem.
I explained this issue after Eq.(3) here https://www.researchgate.net/publica...ary_conditions

Thanks for the reference. It is really helpful. But I still have few questions:

1. In Case 1, Eq.(3), for the last 2 lines, how you reach final conclusion from 2nd line?

2. Under Eq.(3), you mentioned

Quote:

Observe that the orthogonality would be verified with suitable periodical boundary conditions

Could you explain this a little bit? I can totally see the orthogonality is true with your given boundary conditions in Eq.(2), but I don't see why periodic BC would also lead to orthogonality.

FMDenaro · April 3, 2019, 14:24

Quote:

Originally Posted by TurbJet

Thanks for the reference. It is really helpful. But I still have few questions:

1. In Case 1, Eq.(3), for the last 2 lines, how you reach final conclusion from 2nd line?

2. Under Eq.(3), you mentioned

Could you explain this a little bit? I can totally see the orthogonality is true with your given boundary conditions in Eq.(2), but I don't see why periodic BC would also lead to orthogonality.

- The HHD can be seen in two ways, the second bein the case of prescribed tangential boundary condition. In practice, the well know vorticity-stream function formulation is nothing but the second case of the HHD.
- If you see the surface integral it vanishes if either the normal component is zero or it is prescribed with the periodicity. In such case the integrals along two faces with opposite outward unit normal will compensate

TurbJet · April 4, 2019, 18:31

Quote:

Originally Posted by FMDenaro

- The HHD can be seen in two ways, the second bein the case of prescribed tangential boundary condition. In practice, the well know vorticity-stream function formulation is nothing but the second case of the HHD.
- If you see the surface integral it vanishes if either the normal component is zero or it is prescribed with the periodicity. In such case the integrals along two faces with opposite outward unit normal will compensate

1. For the first question, actually I was trying to ask Case 1 in your Eq.3, the one with 2 divergence, the one from (2)_4. How do you reduce two divergence to 1 surface integral? I guess you didn't assume periodicity here, so I guess the 1st divergence
$\int\nabla\cdot(\phi w_2)dV - ...$
may or may not be zero.

2. See if I am correct. Let's start with your Case 2 in Eq.3. When we reach the 2nd line, i.e.,
$\int\nabla\cdot(b\times\nabla\phi)dV$
with periodicity, the entire divergence will be zero. Correct?
And so the solenoidal part will be orthogonal to the dilatational part.

FMDenaro · April 4, 2019, 18:50

- Case 1 has w2 that is divergence-free by definition then you convert the volume integral of the div to a surface integral of the normal flux. That is zero for periodic conditions

- The same is in Case 2, you use Gauss to convert the volume integral to a surface integral wherein you can use the prescribed BCs to assess that the integral is zero.

Is that the answer you asked or I do not understand your question?

TurbJet · April 4, 2019, 19:01

Quote:

Originally Posted by FMDenaro

- Case 1 has w2 that is divergence-free by definition then you convert the volume integral of the div to a surface integral of the normal flux. That is zero for periodic conditions

- The same is in Case 2, you use Gauss to convert the volume integral to a surface integral wherein you can use the prescribed BCs to assess that the integral is zero.

Is that the answer you asked or I do not understand your question?

- Oh right, divergence-free, I forgot about that.
- I just want to be sure that I understand your Case 2-Eq.3 correctly.

So, in the paper I mentioned at the very beginning, the authors claim that

$\langle u_i^su_i^d\rangle = 0$

where superscript s & d stand for solenoidal/dilatational components. Since you mentioned above that solenoidal/dilatational components are orthogonal in the global sense. I just want to be sure does this averaging operation count as "global"?

FMDenaro · April 5, 2019, 03:52

Quote:

Originally Posted by TurbJet

- Oh right, divergence-free, I forgot about that.
- I just want to be sure that I understand your Case 2-Eq.3 correctly.

So, in the paper I mentioned at the very beginning, the authors claim that

$\langle u_i^su_i^d\rangle = 0$

where superscript s & d stand for solenoidal/dilatational components. Since you mentioned above that solenoidal/dilatational components are orthogonal in the global sense. I just want to be sure does this averaging operation count as "global"?

well, generally the brackets are used as denomination of an ensemble averaging but if the paper addresses them as volume integral then the orthogonality is implied, provided that the correct BCs are associated.

TurbJet · April 6, 2019, 22:32

Quote:

Originally Posted by FMDenaro

well, generally the brackets are used as denomination of an ensemble averaging but if the paper addresses them as volume integral then the orthogonality is implied, provided that the correct BCs are associated.

Yes, in the paper, it's volume averaging, and periodic BCs are applied in all directions.

TurbJet · April 7, 2019, 19:33

Quote:

Originally Posted by FMDenaro

well, generally the brackets are used as denomination of an ensemble averaging but if the paper addresses them as volume integral then the orthogonality is implied, provided that the correct BCs are associated.

BTW, is there any benchmark cases that I can test my HHD code? I only need cases with simple periodic BCs.

FMDenaro · April 8, 2019, 04:15

Quote:

Originally Posted by TurbJet

BTW, is there any benchmark cases that I can test my HHD code? I only need cases with simple periodic BCs.

I used the Taylor solution

TurbJet · April 9, 2019, 01:42

Quote:

Originally Posted by FMDenaro

I used the Taylor solution

The solution to the Taylor-Green vortex?

Also, just a side question: I decompose the field in spectral space via the projection tensor

$P_{ij} = \delta_{ij} - \frac{\kappa_i\kappa_j}{\kappa^2}$

I apply this tensor on the velocity field and it gives back the solenoidal part; and then I use ifft to invert back to physical space.

However, the resulting solenoidal field I got, if I take the real part of the solution, the divergence is significant, not close to zero; but if I include the complex part as well, the divergence is perfectly zero (in the sense of machine precision). But definitely the solution should be real; I am afraid if I directly throw away the complex part, the solution won't be correct.

I check my code many times and I don't see any problem with it. So could you possibly give me some hint?

FMDenaro · April 9, 2019, 03:53

Quote:

Originally Posted by TurbJet

The solution to the Taylor-Green vortex?

Also, just a side question: I decompose the field in spectral space via the projection tensor

$P_{ij} = \delta_{ij} - \frac{\kappa_i\kappa_j}{\kappa^2}$

I apply this tensor on the velocity field and it gives back the solenoidal part; and then I use ifft to invert back to physical space.

However, the resulting solenoidal field I got, if I take the real part of the solution, the divergence is significant, not close to zero; but if I include the complex part as well, the divergence is perfectly zero (in the sense of machine precision). But definitely the solution should be real; I am afraid if I directly throw away the complex part, the solution won't be correct.

I check my code many times and I don't see any problem with it. So could you possibly give me some hint?

yes, you can use the 2D or the 3D formulation of the Taylor solution.

I am not sure to understand what you are doing. First, the HHD allows you to determine the gradient of the pressure, the absolute pressure function being not unique. Then, the divergence-free field is the sum of the original vector field you want to decompose and the computed pressure gradient.
Could you better detail what are you doing? In spectral methods the divergence-free constraint is imposed.

TurbJet · April 9, 2019, 08:15

Quote:

Originally Posted by FMDenaro

yes, you can use the 2D or the 3D formulation of the Taylor solution.

I am not sure to understand what you are doing. First, the HHD allows you to determine the gradient of the pressure, the absolute pressure function being not unique. Then, the divergence-free field is the sum of the original vector field you want to decompose and the computed pressure gradient.
Could you better detail what are you doing? In spectral methods the divergence-free constraint is imposed.

I guess you are talking about doing the decomposition in the physical space by solving the Poisson equation to get the solenoidal/dilatational components. BTW, the velocity field I have is from Compressible Homogeneous Isotropic Turbulence, so I think there should not have pressure gradient at all.

What I am doing is following Pope (Turbulent Flows, Chp.6, Sec.4). The basic idea is doing this decomposition in spectral space. The velocity in spectral space can be decomposed into the components which parallel to the wave vector, and the other one will be orthogonal to it. The one orthogonal to the wave vector is the solenoidal velocity.

To perform this decomposition in spectral space, one can do the projection of the velocity to get this orthogonal component in spectral space by (suppose $\hat{\cdot}$ stands for the Fourier coefficients)

$\hat{u}^{\bot}_i = P_{ij}\hat{u}_j$

in which the projection tensor $P_{ij}$ is defined as

$P_{ij} = \delta_{ij} - \frac{\kappa_i\kappa_j}{\kappa^2}$

where $\kappa_i$ is the wave vector components. Then can subtract the solenoidal parts from original field to get the dilatational parts.

My problem is homogeneous isotropic turbulence, so have periodic BCs in all 3 directions. The basic algorithm I am using is
1. Perform FFT on velocity to get its counterpart in Fourier space;
2. Doing the projection mentioned above to get solenoidal velocity field $\hat{u}_i^{\bot}$ in spectral space
3. Subtract $\hat{u}_i^{\bot}$ from $\hat{u}_i$ to get the dilatational parts $\hat{u}_i^{\parallel}$
4. Apply iFFT on $\hat{u}_i^{\bot}$ and $\hat{u}_i^{\parallel}$ to get the physical solution of solenoidal and dilatational velocity fields.

But my issue is, the solenoidal field I get is a complex field; if I throw out the imaginary parts, the real parts are not exactly divergence-free; if I keep the imaginary parts, then the divergence will be machine-zero.

Apparently, the solution should be real, so at some point I need to toss out the imaginary part anyway. But I am afraid if I do, the solution would not be exactly divergence-free.

FMDenaro · April 9, 2019, 12:28

Some things are still not clear to me.

1) For compressible flow there is no divergence-free constraint, however at a steady state the mass flux rho*v is solenoidal, not the single velocity field. Are you searching for that field?
2) the pressure gradient exists always as a term in the momentum equation
3) When you consider a Fourier representation of a real function f(x), you have to sum the products of the complex numbers, that is F(k)*exp(i*k*x), over all the wavenumbers and use the property of the coefficient to be conjugate. It will result in a real function.

TurbJet · April 9, 2019, 18:25

Quote:

Originally Posted by FMDenaro

Some things are still not clear to me.

1) For compressible flow there is no divergence-free constraint, however at a steady state the mass flux rho*v is solenoidal, not the single velocity field. Are you searching for that field?
2) the pressure gradient exists always as a term in the momentum equation
3) When you consider a Fourier representation of a real function f(x), you have to sum the products of the complex numbers, that is F(k)*exp(i*k*x), over all the wavenumbers and use the property of the coefficient to be conjugate. It will result in a real function.

1. I would say no. It is just the velocity field u I am looking at. For compressible Homogeneous Isotropic Turbulence, many people decompose the velocity field into solenoidal velocity and dilatational velocity (so that they can look into the energy transfer between these two fields, for example), i.e.,

$\bold{u} = \bold{u}_{solenoidal} + \bold{u}_{dilatational}$

I am not looking for some fields that is inherently divergence-free in the compressible flow. I only want to separate the velocity field.

2. Maybe I am not correct on this pressure problem. But it does not concern me right now.

3. I am not sure if I understand this response of yours. What I meant complex, is about when I evaluating the divergence of the solenoidal velocity field (after I do the decomposition) via spectral method, namely, something like

$\frac{\partial u}{\partial x} = \mathcal{F}^{-1}(ik\hat{F}_\kappa)$

where $F^{-1}$ is the inverse Fourier transform. And then the divergence

$div(x, y, z) = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}$

Then I look at the field div, it's complex, and its real part is not exactly divergence-free.

FMDenaro · April 10, 2019, 03:39

Quote:

Originally Posted by TurbJet

1. I would say no. It is just the velocity field u I am looking at. For compressible Homogeneous Isotropic Turbulence, many people decompose the velocity field into solenoidal velocity and dilatational velocity (so that they can look into the energy transfer between these two fields, for example), i.e.,

$\bold{u} = \bold{u}_{solenoidal} + \bold{u}_{dilatational}$

I am not looking for some fields that is inherently divergence-free in the compressible flow. I only want to separate the velocity field.

2. Maybe I am not correct on this pressure problem. But it does not concern me right now.

3. I am not sure if I understand this response of yours. What I meant complex, is about when I evaluating the divergence of the solenoidal velocity field (after I do the decomposition) via spectral method, namely, something like

$\frac{\partial u}{\partial x} = \mathcal{F}^{-1}(ik\hat{F}_\kappa)$

where $F^{-1}$ is the inverse Fourier transform. And then the divergence

$div(x, y, z) = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}$

Then I look at the field div, it's complex, and its real part is not exactly divergence-free.

du/dx+dv/dy+dw/dz = i*(kx*u+ky*v+kz*w)

for real kx,ky,kz that has only the imaginary part, the real part being zero, when you set the divergence-free constraint you get a condition only on it
kx*u+ky*v+kz*w=0

TurbJet · April 10, 2019, 21:39

Quote:

Originally Posted by FMDenaro

du/dx+dv/dy+dw/dz = i*(kx*u+ky*v+kz*w)

for real kx,ky,kz that has only the imaginary part, the real part being zero, when you set the divergence-free constraint you get a condition only on it
kx*u+ky*v+kz*w=0

Seems like you directly evaluate the divergence in the spectral space. But what I do is evaluating each derivative using spectral method, e.g.,

$\widehat{\frac{du}{dx}} = ik_x\hat{u}$

then inverse it to get its result in physical space, i.e.,

$\frac{du}{dx} = \mathcal{F}^{-1}(\widehat{\frac{du}{dx}}) = \mathcal{F}^{-1}(ik_x\hat{u})$

Then I sum each derivative up in physical space to get the divergence. Is this not the way to do it? Should I just compute the divergence in spectral space?

FMDenaro · April 11, 2019, 03:46

Quote:

Originally Posted by TurbJet

Seems like you directly evaluate the divergence in the spectral space. But what I do is evaluating each derivative using spectral method, e.g.,

$\widehat{\frac{du}{dx}} = ik_x\hat{u}$

then inverse it to get its result in physical space, i.e.,

$\frac{du}{dx} = \mathcal{F}^{-1}(\widehat{\frac{du}{dx}}) = \mathcal{F}^{-1}(ik_x\hat{u})$

Then I sum each derivative up in physical space to get the divergence. Is this not the way to do it? Should I just compute the divergence in spectral space?

I do not understand what you are doing... the derivative is

du/dx = i*kx*u

where u is the real function not the Fourier coefficients! It is due to the eigenfunction property

TurbJet · April 11, 2019, 04:03

Quote:

Originally Posted by FMDenaro

I do not understand what you are doing... the derivative is

du/dx = i*kx*u

where u is the real function not the Fourier coefficients! It is due to the eigenfunction property

I don't understand. If u and kx is real, wouldn't this result in imaginary du/dx?

April 3, 2019, 02:04	Does solenoidal field orthogonal to dilatational field?	#1
TurbJet Senior Member Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 10	Hi, Currently I am reading a paper about compressible homogeneous isotropic turbulence. In this paper, the authors claim that $\langle u^s_iu^d_i\rangle = 0$ where <> stands for volume averaging, and superscript s & d represent solenoidal/dilatational velocity fields, which can be obtained from Helmholtz decomposition. I really don't see how this identity is valid. Solenoidal and dilatational components are orthogonal in spectral space, that's for sure. But I don't think it is equivalent to be orthogonal in physical space. Or I am misunderstanding this identity; it's equal to zero because some other reason? Can anyone give me some hints? Appreciate it.

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April 4, 2019, 18:50		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	- Case 1 has w2 that is divergence-free by definition then you convert the volume integral of the div to a surface integral of the normal flux. That is zero for periodic conditions - The same is in Case 2, you use Gauss to convert the volume integral to a surface integral wherein you can use the prescribed BCs to assess that the integral is zero. Is that the answer you asked or I do not understand your question?

April 9, 2019, 12:28		#15
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	Some things are still not clear to me. 1) For compressible flow there is no divergence-free constraint, however at a steady state the mass flux rhov is solenoidal, not the single velocity field. Are you searching for that field? 2) the pressure gradient exists always as a term in the momentum equation 3) When you consider a Fourier representation of a real function f(x), you have to sum the products of the complex numbers, that is F(k)exp(ikx), over all the wavenumbers and use the property of the coefficient to be conjugate. It will result in a real function.