…and here it is also (almost) the same in practice. You use the recurrence relation on the fs and the definition of momentum, more than once and at more than one location, to end up with something we know. In other words, find an expression for f8-f7 and don’t be frightened if you have to apply the relations and/or definitions more than one. This is what we did before when you got stuck before. I always think it is best to struggle through a problem, but I can give you step-by-step instructions if you like.
Hi pleb!
If you can provide a solution, please do! I’ve spent so much time with this now and I would really like to see how it is done!
Thanks in advance!
- write down the definition of momentum at j=1 (ie rho*u1=f1-f3+etc);
- use the bounce back condition to replace the unknowns with know fs;
- this gives you f7-f8 at j=0; use the recurrence relation to find these. You should know have something that has f7 and f8 at j=1 only.
- (f8-f7)(j=2)=rho*u2-(f1-f3)-(f5-f6);
- play with this equation: use the recurrence relations to find (f5-f6) at j=2. This will give you f5 and f6 at j=1. These are given by the boundary condition.
- note that we are almost back where we started - not surprising considering we have a recurrence formula. Get to something that has f7 and f8 at j=2. You should thus have something like (f7-f8)(j=2)=momentum terms+force terms +something else involving (f7-f8)(j=2). So you can solve for (f7-f8) at j=2.
- pop this back into the momentum at the wall at solve for u1.
Good luck!
pleb01 Wrote:
- write down the definition of momentum at j=1
(ie rho*u1=f1-f3+etc);- use the bounce back condition to replace the
unknowns with know fs;- this gives you f7-f8 at j=0; use the recurrence
relation to find these. You should know have
something that has f7 and f8 at j=1 only.
from my earlier post above…
rho*u1 = c*(f1(j=1)-f3(j=1)) - 1/(6*tau) * rho*u1 - 1/6*dx/c*rho*G + c*1/tau * (f8(j=1)-f7(j=1)). (A)
- (f8-f7)(j=2)=rho*u2-(f1-f3)-(f5-f6);
- play with this equation: use the recurrence
relations to find (f5-f6) at j=2. This will give
you f5 and f6 at j=1. These are given by the
boundary condition.- note that we are almost back where we started -
not surprising considering we have a recurrence
formula. Get to something that has f7 and f8 at
j=2. You should thus have something like
(f7-f8)(j=2)=momentum terms+force terms +something
else involving (f7-f8)(j=2). So you can solve for
(f7-f8) at j=2.
Please confirm: So the idea is to start a second recursion at j=2 until we can use the BB-rule and then make it back to j=2 and solve for (f7-f8) ??
- pop this back into the momentum at the wall at
solve for u1.
Please confirm: I substitute my expression for (f7-f8) into (A) from above to get an expression for u1 that now includes the slip velocity that we are searching?
Good luck!
Well, it’s the same recurrence relation, of course; but you want to use what’s known, and use any imposed conditions, to find what is unknown (or undetermined). What I wrote above is an attempt to break the procedure down to bite-size pieces, but yes: you essentially manipulate everything as stated above to find u1 in terms of the slip. Once you have u1 you can easily find U_s form equation (15) in the original manuscript. Try it, but be careful with the algebra. If you’re still not making any progress then send me a personal message - I could probably knock up a Maple file very quickly and send it to you.
Hi, pleb!
I made it now, finally (with I little help from Maxima and yours of course)!
However coming to this, Eq. (20), really gives some monster terms and isn’t very elegant compared to the other results in the paper. I wonder if there is a more elegent way in here, too. Do you really have to solve for the f8[2]-f7[2] explicitly?
Anyway I’m pleased now to see that it really works… so thanks a lot!
Good stuff - I’m glad you got there! Just one more quick remark. I don’t think this equation is that much less elegant than the previous result that gives u at the wall (the one that includes u0 and something I think the authors call \bar{u0}). Perhaps my perception/memory is a little unreliable because I haven’t gone through all the algebra in a while, but I seem to recall both expressions having some messy algebra.The techniques, however, are quite similar - would getting the numerical slip for the “modified” bounce back condition appear much simpler had we not found the u0 thing first (ie the thing we can’t use for “normal” bounce back)? Because we still had to find a way of dealing with f7 and f8 at neighbouring nodes. But perhaps this is a matter of personal judgement. The important thing is you got there (this is an important result in the literature), so well done! Is it now time for Irena Ginzburg’s papers…?
Tim
pleb01 Wrote:
Good stuff - I’m glad you got there! Just one more
quick remark. I don’t think this equation is that
much less elegant than the previous result that
gives u at the wall (the one that includes u0 and
something I think the authors call \bar{u0}).
Perhaps my perception/memory is a little
unreliable because I haven’t gone through all the
algebra in a while, but I seem to recall both
expressions having some messy algebra.
Well, I am sure there are different ways to derive Us for the “modified bounce back”, too. But the way I did Eq. (22), I could reuse most of the stuff I had done to get to Eq.s (9) and (10). No need to solve for PDFs (f7-f8) and resubstitute later.
I have a feeling that tells me that there should be a more elegant way for Eq. (20), too.
The techniques, however, are quite similar - would
getting the numerical slip for the “modified”
bounce back condition appear much simpler had we
not found the u0 thing first (ie the thing we
can’t use for “normal” bounce back)? Because we
still had to find a way of dealing with f7 and f8
at neighbouring nodes. But perhaps this is a
matter of personal judgement. The important thing
is you got there (this is an important result in
the literature), so well done! Is it now time for
Irena Ginzburg’s papers…?Tim
That was a good one! What would she say if she could read that? 
But anyway thanks again for all your hints!!
Yet another question: The paper does not state in any way how well these results hold for 3-dimensional models, such as D3Q15, D3Q19 or D3Q27 - only one result from Ginzburg&Adler on 3d FCHC is mentioned. Have there been further papers generalizing the results of that 1997 paper?
Hi,
I’m not aware of any direct generalisations to other lattices. For other 2d models it should be straightforward (but not very interesting). As for 3d models…well, in principle it will be very similar with just the algebra being more annoying (if you do it by hand) because the main ideas are already quite general. However, there is one important difference: in 3d, the flow becomes not one in a channel but one in a circular pipe (Hagen-Poiseuille flow). Since the flow is axisymmetric and still essentially 1D, you’d probably never use a 3d lattice for this and instead use D2Q9 axisymmetric model. Therefore you’ll probably be left with nothing new or exciting. I think Ginzburg and D’humieres have done some work on TRT models (and the whole point of TRT models is to eliminate or reduce the numerical slip error) for general lattices and/or 3d lattices.
Dear, Tim!
After reading another paper I have been coming back to the problem.
In Verhaege et al. “Lattice Boltzmann modeling of microchannel flow in slip flow regime”, they are giving the formula (28)
for the slip velocity of the bounce back scheme with half-way wall.
Obviousl, this is different from the formula given in the He et al 1997 for half-way bounce back, formula (43)
where I have already simplified by substituting the other variables.
I am now wondering why they can possibly end up with a different formula.
Best regards,
Simon
Okay, I think I have got a clue by now. It is that
So the difference is probably how you define momentum / velocity of the LB model in presence of a force term.
Anyone agree?
Yep, you’ve got it!
I am afraid, I have to reopen this thread again. In Ginzburg 94 (http://dx.doi.org/10.1051/jp2:1994123), the perfect choice of a second relaxation time is obtained by expanding the solution of the LBE at a boundary node, for a given lattice link i-,+.
You’ll find this as Eq. (45). However, I am wondering why the second term,
\lambda[sup]-1[/sup] Q[sub]i-,xz[/sub] d[sub]z[/sub]u[sub]x[/sub],
appearing on both sides of the equation does not cancel out. On the right hand side one has Q[sub]i+,xz[/sub], but as it appears to me, it holds
Q[sub]i-,xz[/sub] = Q[sub]i+,xz[/sub],
so it should cancel, right?
But then when you cancel it, it seems impossible to derive the final relation between \lambda and \lamba[sub]2[/sub].
Is there any other resource available, containing a derivation of “magic collision numbers” for the BB scheme?
Not quite. It’s because the left and right hand sides in equation 45 are evaluated at different locations.
The left hand side is evaluated at z, while the right hand side is evaluated at z+1 (or, at z0 and z0+1, if we care about a wall at z=z0). Note that the terms involving Q are multiplied by gradients of velocity. Since u=U*(z^2-(z0+Delta)^2), ie eqn 46, then the bit on the LHS is
Q_{-}Ud/dz (z^2-(z0+Delta)^2)
while the RHS gives
Q_{+}Ud/dz ((z+1)^2-(z0+Delta)^2)
A lot of things cancel (and indeed Q_{-}+Q_{+}=0) but you should have a term left over, the 2z bit (or the derivative of this, rather) on the RHS, allowing you to find eqn 47 etc.
Good luck!
Thank you so much!
But why is it sufficient to look only at one link? I would have expected that one needs to evaluate the whole velocity-moment, i.e., all lattice directions which contribute to u_x.
Best regards!
Simon
No problem, Simon.
It’s because it is considering bounce-back only, with a wall between nodes, so you only need to look at opposite pairs of lattice velocities: incoming and outgoing directions. The equations in that paper hold for any such pair, and there are no particles moving on and along a wall.