An introduction to moment-based boundary conditions for the lattice Boltzmann equation.

1. Understanding lattice Boltzmann boundary
conditions through moments
Tim Reis
School of Computing and Mathematical Sciences
University of Greenwich
London, UK
OpenLB Spring School
Mannheim 2019
Collaborators: Sam Bennett (formerly Cambridge), Paul Dellar
(Oxford), Andreas Hantsch (ILK Dresden), Rebecca Allen
(SINTEF), Seemaa Mohammed (Plymouth), Dave Graham
(Plymouth), Zainab Bu Sinnah (Plymouth), Ivars Krastins
(Greenwich)

2. What is this lecture about?
• It is not a lecture about bounce-back
• It is not a lecture about any particular implementation of
boundary conditions
• It is not about deciding which method is universally the best
• We will look at ways to think about boundary conditions
and how to implement them in the LB framework
• Hopefully, this will give us a clear perspective of LB
boundary conditions, allow you to decide their merits and
shortcomings, and hence when they are most appropriate.
• Ideally, provide some inspiration for you to make some
improvements

3. This is not a lecture on bounce-back. . .
. . . but let’s get this straight:
bounce-back is very useful!!
It is usually the go-to implementation - it is “simple”, second
roder accuarte, and can be adapted to complex geometries
But it is not immune from criticism.

4. Boundary value problems
Boundary conditions are an essential part of the solution to
partial differential equations
They are of paramount importance in determining the
well-posedness of a given problem

5. Dirichlet problems
For example, consider Laplace’s equation on a bounded
domain Ω ⊂ R2
2
u =
∂2u
∂x2
+
∂2u
∂y2
= 0
Dirichlet conditions: Given a function F : ∂Ω → R, we require
u(x) = F(x), x ∈ ∂Ω

6. Neumann problems
For example, consider Laplace’s equation on a bounded
domain Ω ⊂ R2
2
u =
∂2u
∂x2
+
∂2u
∂y2
= 0
Neumann conditions: Given a function F : ∂Ω → R, we
require
∂u(x)
∂n
= F(x), x ∈ ∂Ω,
where n is the outward normal.

7. Example: Navier–Stokes equations
∂u
∂t
+ u · u = −
1
ρ
P + ν 2
u,
· u = 0
At a rigid wall:
• The normal velocity of the fluid = normal velocity of the wall
(no penetration);
• Also, the tangential velocity of the fluid = tangential velocity
of the wall (no–slip)
That is, we usually say: u = 0 on ∂Ω.
Of course, we might want to apply slip conditions, conditions to
the pressure or stress, or inlet/outlet conditions.

8. What does this means for LBE?
The lattice Boltzmann equation (LBE) is often used to
numerically solve these equations and must be accompanied
by relevant boundary conditions
But how can we implement boundary conditions accurately at
lattice grid points?
Lattice Boltzmann computations are performed in the
particle basis, fi, but the problems we are trying to solve
usually impose boundary conditions in terms of
macroscopic scalars, vectors, and tensors (i.e the moment
basis (e.g ρ, ρu, etc.)) and their derivatives.

9. Contents
This lecture will mainly focus on the connection between LB
boundary conditions and LB moments
1. Bounce-back and some perspective
2. Brief overview of some other "particle-based" LB boundary
conditions
3. Moment-based boundary conditions
• The basic idea and plan
• Interpretation of other methods in terms of moments
• Analysis and results, including slip and no-slip flow (if time
permits!)

10. Perspective
For ease of explanation I will start the discussion with the
algorithm written in its most common form (that is for fi, rather
than fi - see my introductory lecture)

11. The stream–collide algorithm
The LBE is a “kinetic" algorithm
fi(x + ci, t + 1) = fi(x, t) −
1
τ
fi − f
(eq)
i
that is used to solved macroscopic PDEs.
Usually more interested in the moments of fi than fi
Moments give us the macroscopic (hydrodynamic) quantities of
interest ( i fi, i fici, etc.)

12. We usually use something like
fi(x + ci, t + 1) = fi(x, t) −
1
τ
fi − f
(eq)
i
to numerically solve
∂fi
∂t
+ ci · fi = −
1
τ
fi − f
(eq)
i
because it approximates
∂u
∂t
+ u · u = −
1
ρ
P + ν 2
u,
· u = 0.

13. Reminder
We MUST supply boundary conditions for the algorithm:
fi(x + ci, t + 1) = fi(x, t) −
1
τ
fi − f
(eq)
i
These should be consistent with the underlying PDE that we
are solving numerically:
∂fi
∂t
+ ci · fi = −
1
τ
fi − f
(eq)
i
and accurately capture the phenomena described by the
problems we are interested in:
∂u
∂t
+ u · u = −
1
ρ
P + ν 2
u,
· u = 0.

14. Reminder
Lattice Boltzmann computations are performed in the
particle basis, fi, but the problems we are trying to solve
usually impose boundary conditions in terms of
macroscopic scalars, vectors, and tensors (i.e the moment
basis (e.g ρ, ρu, etc.)) and their derivatives.

15. Incoming distributions are unknown
fi(x + ci, t + 1) = fi(x, t) −
1
τ
fi − f
(eq)
i
After streaming, the fi at boundary nodes pointing into the flow
domain are unknown
(Before streaming, there is nothing to stream-in to the flow)

16. On-node bounce-back
Bounce-back reverses the direction of distributions leaving the
domain
f2(xb, t) = f4(xb, t), f5(xb, t) = f7(xb, t), f6(xb, t) = f8(xb, t)
For on-node bounce back, the boundary is located on lattice
sites

17. Outline algorithm
Using “outgoing" distribution f4 and “incoming" f2 at a south wall
at j = 1 as an example
collide
f4(j = 2) = f4(j = 2) − 1
τ (f4 − f
(eq)
4 )|(j=2)
f2(j = 2) = f2(j = 2) − 1
τ (f2 − f
(eq)
2 )|(j=2)
stream
f4(j = 1) = f4(j = 2)
f2(j = 3) = f2(j = 2)
bounce-back
f2(j = 2) = f4(j = 1)
Note: No collision on wall (at j = 1) but “modified bounce-back"
allows collisions on wall
It takes an entire time step to complete the process
(f2(j = 2, t + 1) = f4(j = 1))

18. “Half-way" bounce back
Assumes the wall to be located half-way between grid points.
This means the boundary process can be completed in one
time step
f2(x, t + 1) = f4 (x, t), f5(x, t + 1) = f7 (x, t), f6(x, t + 1) = f8 (x, t)
where fi denote post-collisional quantities

19. Outline algorithm
Using “outgoing" distribution f4 and “incoming" f2 at a south wall
at j = 1 as an example
Collide
f4(j = 2) = f4(j = 2) − 1
τ (f4 − f
(eq)
4 )|(j=2)
f2(j = 2) = f2(j = 2) − 1
τ (f2 − f
(eq)
2 )|(j=2)
bounce-back
f2(j = 1) = f4(j = 2)
stream
f4(j = 1) = f4(j = 2)
f2(j = 2) = f2(j = 1)

20. Some notes
On-node and half-way bounce back are really different
versions of the same method!
On-node first order accurate for the velocity field while halfway
is second order accurate (pressure is first order) [Junk and
Yang (2003), He et al (1997)]
Lots of extensions to deal with flow in irregular, heterogeneous,
geometries
Halfway bounce back offers accuracy, stability, and geometric
flexibility. It is often considered (certainly by me) to be most
reliable implementation of lattice Boltzmann boundary
conditions for many engineering applications
But . . .

21. Notes
What if you want to implement something other than no-slip|
And even if you are hoping for no-slip,
These are plots of the solution to Poiseuille flow, analytical and
LB prediction, at Re = 100 (left) and Re = 1.
Viscosity-dependent numerical (unphysical) slip error!

22. Motivation for alternative ideas
The Bounce–Back method is commonly used and is often very
successful.
However,
• it introduces a “numerical slip error" that in proportional to
the grid spacing1 and the viscosity of the fluid and thus
large for low Reynolds number flows (e.g flows in porous
media) [Ginzbourg and Adler (1994), He et al. (1997)]
• there is not much freedom to impose conditions other than
the no-slip.

23. Motivation for alternative ideas
Non-equilibrium bounce-back [Zou and He (1997)] can be
unstable and uses an arbitrary closure condition.
Kinetic/Maxwell’s diffusive slip boundary conditions [Ansumali
and Karlin (2002)] are only first order accurate in general and
still vulnerable to the “numerical slip error" [Verhaeghe et al
(2009)].
Note that Maxwell boundary conditions were first applied to
discrete velocity Boltzmann models by [Boadwell (1964)] and
analysed further by [Gatignol (1977)].

24. Understanding lattice Boltzmann boundary
conditions through moments

25. A simple example
Imagine we want to solve the 1D advection–diffusion equation
∂φ
∂t
+ u
∂φ
∂x
= D
∂2φ
∂x2
which can be written
∂φ
∂t
+
∂ψ
∂x
= 0
where ψ = uφ − D ∂φ
∂x .
The (macroscopic) quantities of interest are
φ and ψ.
So how many distribution functions do we need? At least two!

26. The D1Q2 model for diffusion
Consider a 1D LB model with two directions: f1 and f2
The conserved quantity is φ = f1 + f2
and it’s flux (non–conserved) is ψ = f2 − f1
We have two basis: the ”particle" basis
f =
f1
f2
and the "moment" basis
m =
φ
ψ

27. Moments of the D1Q2 model
f =
f1
f2
, m =
φ
ψ
Since φ = f1 + f2 and ψ = f2 − f1 we can write
m =
φ
ψ
=
1 1
−1 1
f1
f2
= M
f1
f2
.
As we have a linear transformation between f and m, the
transformation matrix M is invertible and thus
f = M−1
m

28. Macroscopic equation
It can be shown that
fi(x + ci, t + 1) = fi(x, t) −
1
τ
fi − f
(eq)
i
approximates
∂φ
∂t
+ u
∂φ
∂x
= D
∂2φ
∂x2
+ O(τ2
), D ∝ τ
What would the boundary conditions need to be?
Eg: ∂x φ = 0
Conditions on fi or its moments?

29. Lattice for 2D Navier-Stokes
D2Q9 has 9 independent moments. Six are hydrodynamic
ρ =
i
fi, ρuα =
i
ficiα, Παβ =
i
ficiαβ,
and the other three are pseudo–kinetic
Qxxy =
i
fic2
ix ciy , Qxyy =
i
ficix c2
iy , Rxxyy =
i
fic2
ix c2
iy .

30. D2Q9 moments
The relationship between f and m is
m = Mf
where f = (f0, f1, . . . , f8)T
and
M =














1 1 1 1 1 1 1 1 1
0 1 0 −1 0 1 −1 −1 1
0 0 1 0 −1 1 1 −1 −1
0 1 0 1 0 1 1 1 1
0 0 1 0 1 1 1 1 1
0 0 0 0 0 1 −1 1 −1
0 0 0 0 0 1 1 −1 −1
0 0 0 0 0 1 −1 −1 1
0 0 0 0 0 1 1 1 1















31. D2Q9 LBE
We know that the D2Q9 LBE approximates the (weakly)
compressible Navier–Stokes equations,
but what boundary conditions should it satisfy?
As an example, consider a solid wall with a no–slip condition

32. Moments at a wall
We need the fluid to have zero velocity relative to the wall:
ux = uy = 0
Navier-Stokes has zero tangential stress at the wall:
Sxx ∝
∂ux
∂x
= 0, =⇒ Πxx = Π
(eq)
xx =
ρ
3
+ ρu2
x =
ρ
3
Do (should) lattice boundary conditions satisfy these
constraints?

33. Counting moments
For the D1Q2 lattice we had 2 independent moments
For the D2Q9 lattice we had 9 independent moments
In general, a n–velocity lattice has n moments
Note that we needed enough moments to define the
conservation quantities and their fluxes
Point of view:
If we accept that we use the LBE to solve macroscopic
equations then the moments should also satisfy the boundary
conditions.

34. Moments at a south wall
m = Mf = M














f0
f1
f2
f3
f4
f5
f6
f7
f8














=














ρ
ρux
ρuy
Πxx
Πyy
Πxy
Qxxy
Qyyx
Rxxyy














Three unknowns on the left =⇒ three constraints needed on
the right

35. Reminder: NSE, DBE, LBE
From the previous lecture we know that the discrete Boltzmann
equation
∂fi
∂t
+ ci · fi = Ωi(f)
has embedded within it the Navier-Stokes equations.
Thus we develop an (indirect) algorithm for the NSE by
discretising the DBE is space and time

36. From discrete Boltzmann to lattice Boltzmann
Integrating the discrete Boltzmann equation
∂fi
∂t
+ ci · fi = Ωi(f)
along a characteristic for time ∆t gives
fi(x + ci∆t, t + ∆t) − fi(x, t) =
∆t
0
Ωi(x + cis, t + s) ds,
Approximating the integral by the trapezium rule yields
fi(x +ci∆t, t+∆t)−fi(x, t) =
∆t
2
Ωi(x +ci∆t, t+∆t)
+ Ωi(x, t) +O ∆t3
.
This is an implicit system.

37. Change of Variables
To obtain a second order explicit LBE at time t + ∆t define [He
et al. (1998)]
fi(x, t) = fi(x, t) +
∆t
2τ
fi(x, t) − f
(0)
i (x, t) .
The new algorithm is
fi(x + ci∆t, t + ∆t) − fi(x, t) = −
∆t
τ + ∆t/2
fi(x, t) − f
(0)
i (x, t)
Note: The “standard" LBE algorithm is obtained from the DBE
using an Euler discretisation (first order in time)

38. Barred moments
fi(x, t) = fi(x, t) +
∆t
2τ
fi(x, t) − f
(0)
i (x, t) .
Note that conserved quantities are obtained easily from fi:
ρ =
i
fi =
i
fi; ρu =
i
fici =
i
fici;
but care must be taken for non-conserved moments, e.g
ΠΠΠ =
i
ficici,
=
i
ficici +
∆t
2τ
i
fi − f
(0)
i cici,
= ΠΠΠ +
∆t
2τ
ΠΠΠ − ΠΠΠ(0)
.

39. A quick note on forcing
A body force Ri in the discrete Boltzmann equation
∂fi
∂t
+ ci · fi = −
1
τ
fi − f
(0)
i + Ri
should have the following moments:
i
Ri = 0,
i
Rici = FFF,
i
Ricici = FFFu + uFFF,
otherwise we do not obtain the correct viscous stress tensor
from the Chapman-Enskog analysis. An example is
Ri = Wi (3(ci − u) + 9(ci · u)ci) · F.

40. The PDE
∂fi
∂t
+ ci · fi = −
1
τ
fi − f
(0)
i + Ri
is discretised using the Trapezium rule and the change of
variable
fi(x, t) = fi(x, t) +
∆t
2τ
fi(x, t) − f
(0)
i (x, t) −
∆t
2
Ri.
This yields the LBE
fi(x+ci∆t, t+∆t)−fi(x, t) = −
∆t
τ + ∆t/2
fi − f
(0)
i +
τ∆t
τ + ∆t/2
Ri
Note:
ρu =
i
fici = ρu −
∆t
2
F.

41. Kinetic–style lattice Boltzmann boundary
conditions

42. The (on-node) bounce–back method
In general, there are three unknown populations at a boundary.
At a south wall, we don’t know f2, f5, f6.
Bounce–back says
f2 = f4, f5 = f7, f6 = f8

43. Bounce–back moments
Let’s look at ρux , ρuy and Πxx (ignore force for simplicity)
ρux = ρux =
i
ficix = f1 − f3 + f5 − f6 − f7 + f8
= f1 − f3 + f7 − f8 − f7 + f8
= f1 − f3.
Therefore bounce–back with the BGK model does NOT satisfy
the no–slip condition.
The non–zero momentum ρux = f1 − f3 is NOT physical slip!
The error occurs for diffuse reflection boundary conditions, too.

44. Bounce–back moments
Let’s look at ρux , ρuy and Πxx .
ρuy = ρuy =
i
ficiy = f2 − f4 + f5 + f6 − f7 + f8
= 0
Therefore the condition uy = 0 IS satisfied.
Πxx =
i
fic2
ix = f1 + f3 + f5 + f6 + f7 + f8,
= f1 + f3 + 2(f7 + f8),
= something strange
(in fact, it does not give the correct stress condition but this
requires further analysis)

45. Notes on the bounce–back method
It gives a grid–dependent velocity error at the wall (numerical
slip)
This is first order (in ∆x) for an on-node implementation and
second order for “half-way" bounce-back [He et al. (1997)]
The precise location where the velocity is zero is a function of
the relaxation time [Ginzbourg and Adler (1994)]
The error is large for large τ (or small Reynolds numbers)
Can be removed/reduced MRT and TRT [Ginzbourg and Adler
(1994), Verhaeghe et al (2009)]
Bounce-back is simple to implement and stable. Often very
accurate

46. Zou and He boundary conditions
Sometimes called non–equilibrium bounce–back [Zou and He
(1997)]
The Chapman-Enskog expansion says fi = f
(eq)
i + f
(ne)
i where
f
(ne)
i = −τWi (3cici − I) : u + ( u)T
Zou and He say
f
(ne)
2 = f
(ne)
4 ,
f
(ne)
5 = f
(ne)
7 +
1
2
(f1 − f3),
f
(ne)
6 = f
(ne)
8 −
1
2
(f1 − f3)

47. Zou and He wall moments
If we look at the hydrodynamic moments at the wall after
applying Zou and He’s method we find that
ρux = ρuy = 0
so the no–slip condition IS satisfied.
However,
Πxx =
i
fic2
ix = f1 + f3 + f5 + f6 + f7 + f8
= 2(f7 + f8),
so the stress condition is NOT satisfied.

48. Remarks
The Zou and He method is not very intuitive
It does satisfy the no–slip condition
It does not guarantee mass conservation
It is less stable than bounce–back
It’s difficult to extend to 3D
Difficult to deal with corners
What do we do for the “fi" scheme (i.e second order time
implementation)?

49. Moment–Method for Boundary Conditions

50. Moment–Method: The basic idea
To try and impose boundary conditions on
momentum and viscous stress only.

51. Moments at a wall
Let’s once again consider a south boundary
After streaming, we don’t know f2, f5, f6.
(equivalently, f2, f5, f6).
Let’s look at ALL the moments at this wall.

52. Moments at a south wall
Hydrodynamic moments
ρ =
i
fi = f0 + f1 + f2 + f3 + f4 + f5 + f6 + f7 + f8,
ρux =
i
ficix = f1 − f3 + f5 − f6 − f7 + f8,
ρuy =
i
ficiy = f2 − f4 + f5 + f6 − f7 − f8,
Πxx =
i
fic2
ix = f1 + f3 + f5 + f6 + f7 + f8,
Πyy =
i
fic2
iy = f2 + f4 + f5 + f6 + f7 + f8,
Πxy =
i
ficix ciy = f5 − f6 + f7 − f8
Terms in red are unknown at the boundary after streaming

53. Moments at a south wall
Non-hydrodynamic (“ghost") moments
Qxxy =
i
fic2
ix ciy = f5 + f6 − f7 − f8,
Qxyy =
i
ficix c2
iy = f5 − f6 − f7 + f8,
Rxxyy =
i
fic2
ix c2
iy = f5 + f6 + f7 + f8.
Terms in red are unknown at the boundary after streaming
Notice the three different combinations of the unknowns

54. Moment–grouping at a south boundary
THE PLAN:
Formulate the boundary conditions in the moment basis, and
then transform them into into boundary conditions for the
distribution functions [Bennett (2010, thesis)]
Moments Unknowns
ρ, ρuy , Πyy f2 + f5 + f6
ρux , Πxy , Qxyy f5 − f6
Πxx , Qxxy , Rxxyy f5 + f6
We can pick one constraint from each group. A natural choice is
ρuy = 0,
ρux = 0,
Πxx = P =
ρ
3
=⇒
∂u
∂x
= 0.

55. Second order implementation
ρuy = 0,
ρux = 0,
Πxx = P =
ρ
3
=⇒
∂u
∂x
= 0.
Our conditions are on the (physical) moments of fi
However, our algorithm is for fi where
fi = fi +
∆t
2τ
fi − f
(0)
i −
∆t
2
Ri.
We must convert our conditions on physical moments to
conditions on numerical moments.

56. Second order implementation
ρuy = 0,
ρux = 0,
Πxx =
ρ
3
and
fi = fi +
∆t
2τ
fi − f
(0)
i −
∆t
2
Ri.
so
ρuy = −
∆t
2
Fx ,
ρux = −
∆t
2
Fx ,
Πxx =
ρ
3
Caution: Care must be taken with the fi → fi transform

57. Solving for the unknowns
For simplicity, consider the case where Fy = 0.
Noting that
ρ = ρuy + f0 + f1 + f3 + 2(f4 + f7 + f8),
with uy = 0, we solve
0 = f2 − f4 + f5 + f6 − f7 − f8,
−
∆t
2
Fx = f1 − f3 + f5 − f6 + f7 + f8,
ρ
3
= f1 + f3 + f5 + f6 + f7 + f8.

58. 0 = f2 − f4 + f5 + f6 − f7 − f8,
−
∆t
2
Fx = f1 − f3 + f5 − f6 + f7 + f8,
ρ
3
= f1 + f3 + f5 + f6 + f7 + f8.
to obtain
f2 = f1 + f3 + f4 + 2 f7 + f8 −
ρ
3
,
f5 = −f1 − f8 +
ρ
6
−
1
2
Fx ,
f6 = −f3 − f7 +
ρ
6
+
1
2
Fx .

59. An example: open boundaries
At a west vertical boundary the moment grouping is
Moments Unknowns
ρ, ρux , Πxx f1 + f5 + f8
ρuy , Πxy , Qxxy f5 − f8
Πyy , Qxyy , Rxxyy f5 + f8
If we want this to be an inflow boundary then we can impose
ρ = ρintlet ,
ρuy = 0,
Πyy = ρinlet /3.

60. Treatment of the corner nodes
How many unknowns do we have at a corner? 5
At a southwest corner, we don’t know f1, f2, f5, f6, f8.
We can impose the inlet and wall boundaries simultaneously
Impose ρ, ρux , ρuy , Πxx , Πxy

61. A quick comparison
Z+H gives errors at the corners
Difference between exact solution and the LB prediction with
bounce-back on walls and Z-H conditions at inlet/outlet.
Fluctuations down to machine precision with moment-based
conditions [Bennett (2010)].
Moral: Beware of corners!!

62. Lid-driven cavity
Flow in a square box, no–slip on all walls.
Top boundary moves with velocity Ulid .
Interesting to look at the streamfunction: ux = −∂y ψ, uy = ∂x ψ
Flow characteristics depend on Re = Ulid L/ν.

63. Moment–based boundary conditions: moving wall
Moments Unknowns
ρ, ρuy , Πyy f4 + f7 + f8
ρux , Πxy , Qxyy f8 − f7
Πxx , Qxxy , Rxxyy f7 + f8
We can pick one constraint from each group. A natural choice is
ρuy = 0,
ρux = Ulid ,
Πxx = Π
(0)
xx =
ρ
3
+ ρU2
lid =⇒
∂u
∂x
= 0.

64. Lid-driven cavity flow: Re = 7500

65. Lid-driven cavity flow: the numbers
Primary
Re = 400
Present Λ = 1/4 0.1139 0.5547 0.6055
Ghia (1982) et al. 0.1139 0.5547 0.6055
Sahin and Owens (2003) 0.1139 0.5536 0.6075
Re = 1000
Present Λ = 1/4 0.1189 0.5313 0.5664
Ghia(1982) 0.1179 0.5313 0.5625
Sahin and Owens (2003) 0.1188 0.5335 0.5639
Botella (1998) et al. 0.1189 0.4692 0.5652
Re = 7500
Present Λ = 1/4 0.1226 0.5117 0.5352
Ghia (1982) et al. 0.1200 0.5117 0.5322
Sahin and Owens (2003) 0.1223 0.5134 0.5376
Note: TRT model. Second order convergence of L2 error norm
for global velocity and pressure fields

66. Convergence Example: Re = 1000
N ||u||2 ||P||2
N=33 0.084382 0.00023177
N=65 0.01754295 0.0000492165
N=129 0.00416137 0.00001268
N=257 0.00083492 0.0000026609

67. Natural Convection
Flow is driven by density variation
∂u
∂t
+ u · u = − P + Pr 2
u + RaPrg,
∂θ
∂t
+ u · θ = 2
θ,

68. LBE Implementation
Two DBEs: One for NSE one for scalar transport
Force acting in vertical direction (Fy )
Moment-Method boundary conditions. Example south wall
f2 = f1 + f3 + f4 + 2(f7 + f8) −
ρ
3
−
∆t
2
Fy ,
f5 = −f1 = f8 +
ρ
6
,
f6 = −f3 − f7 +
ρ
6
.
where ρ = ρuy + f0 + f1 + f3 + 2(f4 + f7 + f8)
and by the f0 → fi transform, ρuy = −Fy /2 at the wall (since
uy = 0)

69. D2Q5 lattice model for scalar transport equation (temperature)
Moment-method also used to impose boundary conditions on θ.
Only one unknown at the walls
θwall = i gi = i gi (heated walls)
∂Nθ = i ciNgi = 0 (other walls)

70. Streamfunction and Temperature plots
Contours of flow fields for convection in a square cavity. From
left to right, Ra = 1000, Ra = 10000, Ra = 1000000
(Work of Rebecca Allen [PhD (2015), Allen and Reis (2016)])

71. Nusselt numbers
Nu = x y qx dxdy, where qx is the heat flux
Have looked at other Nu too - same level of accuracy observed.
Ra Study Nu
103 Present 1.1178
de Vahl Davis (1983) 1.118
106
Present 8.8249
Le Quere (1991) 8.8252
de Vahl Davis (1983) 8.800
108
Present 30.23339
Le Quere (1991) 30.225
All fields show second order convergence
(Work of Rebecca Allen [PhD (2015), Allen and Reis (2016)])

72. Some further remarks and light analysis

73. Analysis of existing methods
The grouping of moments at a boundary allows us to examine
other methods.
Moments Unknowns
ρ, ρuy , Πyy f2 + f5 + f6
ρux , Πxy , Qxyy f5 − f6
Πxx , Qxxy , Rxxyy f5 + f6
The (on-node) bounce-back condition can be translated into
ρuy = 0, , Qxxy = 0, Qxyy = 0

74. Moments Unknowns
ρ, ρuy , Πyy f2 + f5 + f6
ρux , Πxy , Qxyy f5 − f6
Πxx , Qxxy , Rxxyy f5 + f6
The Zou and He conditions at a wall can be translated into
ρux = ρuwall, , ρuy = 0, Qxxy = 0.
Note that both impose a condition on the "energy moment" but
NOT on the viscous stress tensor!

75. Moments Unknowns
ρ, ρuy , Πyy f2 + f5 + f6
ρux , Πxy , Qxyy f5 − f6
Πxx , Qxxy , Rxxyy f5 + f6
The diffuse reflection conditions at a wall can be translated into
ρuy = 0, Qxxy =
1
3
+ u2
wall Πyy −Rxxyy , Qxyy = −Πxy +uwallΠyy
These are difficult to interpret physically. No stress condition.
Large errors.

76. Analytic solution of the LBE
Recall that for flows satisfying
∂
∂x
=
∂
∂t
= 0, FFF = (ρG, 0)
with walls located at j = 1 and j = n, the LBE velocity field
satisfies

77. Poiseuille flow
uj+1vj+1 − uj−1vj−1
2
= ν uj+1 + uj−1 − 2uj + G,
This is the second order finite–difference form of the
incompressible Navier–Stokes equations with a constant body
force:
∂(uv)
∂y
= ν
∂2u
∂y2
+ G

78. Solution of the difference equation
uj+1vj+1 − uj−1vj−1
2
= ν uj+1 + uj−1 − 2uj + G
We can show ρ is constant and vj = 0
The solution to this second order difference equation is
uj =
4Uc
(n − 1)2
(j − 1)(n − j) + Us, j = 1, 2, . . . , n
where Uc = H2G/8ν is the centre-line velocity and H = (n − 1)
is the channel height.

79. Numerical slip for bounce–back
If we use bounce–back boundary conditions, we find the
numerical slip to be [He et al. (1997)]
Us =
48ν2 − 1
n2
Uc

80. Notes
This slip error is second order in grid spacing
It is large for very viscous flows
The point where the velocity vanishes only asymptotically
coincides with the midpoint between gridpoints
It can be eliminated using TRT [Ginzbourg and Adler (1994)]
Ideally, we’d like to use TRT/MRT for improving stability and not
fixing errors due to boundary treatments
It is NOT Knudsen slip!!
Us = 0 for Moment-Method

81. Imposing physical slip boundary conditions via
moments

82. Flow in a microchannel
No slip Slip flow Transition Molecular
Kn 10−3 10−3 Kn 10−1 10−1 Kn 10 Kn 10
In shear flow the LBE reduces to a linear second–order
recurrence relation =⇒ linear or parabolic profiles at all Kn
But we can capture flow in the bulk from with slip conditions

83. Maxwell–Navier boundary condition
Wall boundary conditions:
uslip = σKnH∂y u|wall , σ = (2 − σa)/σa.
These can be expressed in terms of moments [Reis and Dellar
(2012)]:
f2 = f1 + f3 + f4 + 2 f7 + f8 − P − ρu2
slip ,
f5 = −f1 − f8 + (P + ρu2
slip + ρuslip)/2,
f6 = −f3 − f7 + (P + ρu2
slip − ρuslip)/2,
and since Πxy |wall =
2τ ¯Πxy |wall
(2τ+∆t) = µ∂y u|wall,
uslip = −
6 −f1 + f3 + 2f7 − 2f8
ρ(2τ + 1 + 6KnH)
.

84. Flow in a microchannel: asymptotic solution
We consider a viscous fluid in a channel with an aspect ratio
δ = L/H 1.
The relevant dimensionless numbers are
Re =
ρoUoH
µ
, Ma =
Uo
√
γRT
, Kn =
πγ
2
Ma
Re
An expansion in δ yields the leading–order solution
u(x, y) = −
Re
8Ma2
p 1 − 4y2
+ 4σ
Kn
p
v(x, y) =
2Re
8pMa2
1
2
(p2
) 1 −
4
3
y2
+ 4σKnp
P (x) = (6Kn)2
+ (1 + 12Kn)x + θ(θ + 12Kn)(1 − x) − 6Kn

85. Flow in a microchannel: Kn = 0.1

86. Convergence

87. Dipole wall collision with slip
With the following initial conditions, two counter–rotating
vortices are self–propelled to the right (F = 0):
u0 = −
1
2
|ω| (y − y1) exp (−r1/r0)2
+
1
2
|ω| (y − y2) exp (−r2/r0)2
,
v0 =
1
2
|ω| (x − x1) exp (−r1/r0)2
−
1
2
|ω| (x − x2) exp (−r2/r0)2
.

88. Dipole wall collision with slip
The velocity at the walls is the Navier-slip condition
u|| = sL
∂u||
∂n
From left: t = 0.5, 0.6, 0.7, 1. Re = 2500, sL = 0.002.

89. Energy and Enstrophy: Re = 1252
(Work of Seemaa Mohammed [PhD thesis (2018)])
[Sutherland et al. (2013)]

90. Summary
Most lattice Boltzmann models impose boundary conditions on
the populations (fi),
but we are usually only interesting in hydrodynamic variables.
We can impose the same conditions on the LBM and the target
equations via moment constraints.
This gives a very general methodology for imposing a variety of
hydrodynamic boundary conditions at grid points precisely

91. Summary
Don’t fall into the trap of thinking one method is always “better"
than another!!
Moment-based approach is general, simple, and local
Shown to be very accurate
But is not as stable nor efficient as bounce-back at high Re
numbers
Should be used with MRT/TRT (so should all LBMs?)
Currently lacks geometric flexibility and requires further,
deeper, analysis.

92. References
S. Bennet, PhD Thesis, University of Cambridge (2010)
T. Reis and P.J. Dellar, Moment-based formulation of
Navier-Maxwell slip boundary conditions for lattice Boltzmann
simulations of rarefied flow in microchannels, Phys. Fluids 24
(2012), 112001
R. Allen and T. Reis, Moment-based conditions for lattice
Boltzmann simulations of natural convection in cavities, Prog.
Comp. Fluid Dyn.: An Int. J 16 (2016), 1-4
R. Allen, PhD Thesis, KAUST (2015)

93. References
A. Hantsch, T. Reis, and U. Gross, Moment method boundary
conditions for multiphase lattice Boltzmann simulations of
partially-wetted walls, J. Comp. Multiphase Flow 7 (2015), 1-4
A. Hantsch, PhD Thesis, TU Freiberg (2013)
S. Mohammed and T. Reis, Using the lid-driven cavity flow to
validate moment-based boundary conditions for the lattice
Boltzmann equation, Arch. Mech. Eng, 64 (2017), 57-74
S. Mohammed, D. Graham, and T. Reis, Assessing
moment-based boundary conditions for the lattice Boltzmann
equation: A study of dipole-wall collisions, Comput. Fluids 176
(2018) 79-96

94. References
I. Ginzbourg and P. Adler, Boundary flow condition analysis for
the three-dimensional lattice Boltzmann model, J. Physique II 4
(1994), 191-214
X. He, Q. Zou, and L-S Luo, Analytic solutions and analysis on
non-slip boundary conditions for lattice Boltzmann BGK model
J. Comp. Phys 87 (1997), 115-136
M. Junk and Z. Yang, Analysis of lattice Boltzmann boundary
conditions Proc. Appl. Math. Mech., 3 (2003), 76?79.
F. Verhaeghe, L-S Luo, and B. Blanpain, Lattice Boltzmann
modeling of microchannel flow in slip flow regime, J. Comp.
Phys, 228 (2009), 146-157
Q. Zou and X. He, On pressure and velocity boundary
conditions for lattice Boltzmann BGK model, Phys Fluids (1997)

95. References
J.E Broadwell, Study of rarefied shear flow by the discrete
velocity method, J. Fluid. Mech, 19 (1964), 401-414
R. Gatignol, Kinetic theory boundary conditions for discrete
velocity gases, Phys. Fluids, 20 (1977)
S. Ansumali and IV Karlin, Kinetic boundary conditions in the
lattice Boltzmann model, Phys. Rev. E, 66 (2002), 026311
T. Reis, Burnett order stress and spatially-dependent boundary
conditions for the lattice Boltzmann method, Comm. Comp.
Phys (2019, in press)

96. References
I. Krastins, A. Kao, K. Pericleous, and T. Reis, 3D
Moment-based boundary conditions for the lattice Boltzmann
equation, Int. J. Num. Meth Fluids (2019, submitted)
I. Krastins Parallel lattice Boltzmann method for convection in
dendritic solidification , PhD thesis (2018)

97. Dipole wall collision
With the following initial conditions, two counter–rotating
vortices are self–propelled to the right (F = 0):
u0 = −
1
2
|ω| (y − y1) exp (−r1/r0)2
+
1
2
|ω| (y − y2) exp (−r2/r0)2
,
v0 =
1
2
|ω| (x − x1) exp (−r1/r0)2
−
1
2
|ω| (x − x2) exp (−r2/r0)2
.

98. Dipole wall collision: Snapshots at Re = 2500
(a) t = 0 (b) t = 0.2 (c) t = 0.3
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]

99. (a) t = 0.4 (b) t = 0.49 (c) t = 0.617
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]

100. (a) t = 0.8 (b) t = 0.1 (c) t = 0.1.5
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]

101. At different Re
From left: Re = 625, 1250, 2500, 5000 at t = 1
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]

102. Energy and Enstrophy
E(t) =
1
2
1
−1
1
−1
|u2
|dxdy, Ω(t) =
1
2
1
−1
1
−1
|ω2
|dxdy
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]

103. Energy and Enstrophy: The numbers
current work Clercx and Bruneau
Re t E(t)LB Ω(t)(LB) E(t)FD E(t)SM Ω(t)(FD) Ω(t) (SM)
625
0.25 1.501 472.480 1.502 1.5022 472.7 472.6
0.50 1.013 382.915 1.013 1.0130 380.4 380.6
0.75 0.767 256.511 0.767 0.7673 255.0 255.2
1250
0.25 1.719 614.309 1.721 1.7209 615.0 615.0
0.50 1.312 613.574 1.313 1.3132 611.3 611.9
0.75 1.061 487.081 1.061 1.0613 484.4 484.7
2500
0.25 1.848 727.210 1.851 1.8509 727.8 728.2
0.50 1.540 919.228 1.541 1.5416 916.6 920.5
0.75 1.325 811.549 1.326 1.3262 805.5 808.1
5000
0.25 1.919 823.026 1.923 1.9225 822.8 823.1
0.50 1.690 13334.348 1.692 1.6924 1328 1340
0.75 1.496 1552.926 1.495 1.4980 1659 1517
[Clercx and Bruneau (2006)]

104. Energy and Enstrophy: Convergence
Work of Seemaa Mohammed [PhD thesis (2018)],
[Mohammed, Graham, TR (2018)]