Analytical solutions of the lattice Boltzmann BGK model

10 Pages·0.1 MB·English

by  Qisu Zou

Analytical solutions of the lattice Boltzmann BGK model
Qisu Zou1,2, Shuling Hou1,3, Gary D. Doolen1,
1Center for Nonlinear Studies and Theoretical Division, Los Alamos National Laboratory,
Los Alamos, NM87545
5
9 2Department of Mathematics, Kansas State University,Manhattan, KS 66506.
9 3Department of Mechanical Engineering, Kansas StateUniversity,Manhattan, KS 66506
1
n Abstract
a
J Analytical solutions of the two dimensional triangular and square lattice Boltzmann

3 BGK models have been obtained for the plain Poiseuille flow and the plain Couette flow.
1
The analytical solutions are written in terms of the characteristic velocity of the flow,

1 the single relaxation time τ and the lattice spacing. The analytic solutions are the exact
v representation of these two flows without any approximation.
4
0
0 1 Introduction
1
0
5 Since the appearance of lattice gas automata (LGA) and its later derivative, the lattice
9 Boltzmann equation method (LBE) as alternative computational methods to study trans-
/
s portphenomena,someanalyticalsolutionshavebeenobtainedforthesemethodsfornonuni-
a
g form flows in 2D or 3D models [1, 2, 3, 4]. They are based on linearized Boltzmann models,
- and they have employed approximations. In [1, 2, 3], the solution around a global equi-
p
m librium with constant density and isotropic velocity (zero velocity) was considered. In [4],
the first order and second order of deviation of distribution function from equilibrium were
o
c assumed to take a certain form in terms of flow quantities, and the coefficients in this form
:
v were obtained using a Chapman-Enskog procedure. These analytic results provided insight
i for applications of the methods. For example, the analytical results allow one to calculate
X
viscosity from given collision rules and to estimate and to improve bounce-back boundary
r
a conditions. They are valuable in enhancing our understanding of the method. Neverthe-
less, analytical solutions for real flows with boundaries like the Poiseuille flow, which is
represented exactly by a second-order finite-difference scheme on a uniform mesh, have not
been obtained previously for LGA or LBE. One reason may be that the boundary condi-
tions used in LGA and LBE are not exact. For example, bounce-back or a combination of
bounce-back and specular reflection [2] for modeling the non-slip boundary condition are
only approximate. The effective non-slip boundary is inside the bounce-back row [2, 4].
Recently, Noble et al. [5] have proposed a boundary condition for the lattice Boltzmann
BGK model (LBGK) on a triangular lattice. When this boundarycondition was applied to
plain Poiseuille flow, the steady state solution of the distribution function gave a parabolic
velocity profile up to machine accuracy. The result suggests the existence of an analytical
solution to LBGK, which is an exact representation of the Poiseuille flow. This analytical
solution is derived in this paper together with an analytical solution for the plain Couette
flow.
1
2 Analytical Solutions of the triangular lattice LBGK model
Firstletusconsiderthelattice Boltzmann modelon atriangular lattice. For achannelflow,
a triangular lattice is constructed as shown in Fig. 1. There are two types of particles on
each node of an FHP model: rest particles (type 0) with e = 0 and moving particles (type
0
1 to 6) with unit velocity e = (cos((i 1)π/6),sin((i 1)π/6), i = 1, ,6 along 6 direc-
i
− − ···
tions. Consider the particle distribution functions f (x,t), which represent the probability
i
of finding a particle at node x = (x,y) and time t with velocity e . The lattice Boltzmann
i
BGK model proposed in [6, 7] is the equation for the evolution of f :
i
1
(0)
f (x+δe ,t+δ) f (x,t) = [f (x,t) f (x,t)], i= 0, ,6 (1)
i i − i −τ i − i ···
(0)
wheref (x,t)is theequilibriumdistributionoftheparticle oftypeiat x,t, therighthand
i
side represents the collision term and τ is the single relaxation time which controls the rate
of approach to equilibrium. The density per node, ρ, and the macroscopic flow velocity, u,
are defined in terms of the particle distribution function by
6 6
Xfi = ρ, Xfiei = ρu. (2)
i=0 i=1
Theequilibriumdistributionfunctionsdependonlyonlocaldensityandvelocity. Asuitable
equilibrium distribution for the FHP model can be chosen in the following form [7]:
f(0) = d ρu2 =αρ ρu2,
0 0− −
1 1
f(0) =d+ ρ[(e u)+2(e u)2 u u], i = 1, ,6, (3)
i 3 i· i· − 2 · ···
where α is an adjustable parameter, d = ρ−d0, and f(0) = ρ, f(0)e = ρu. Note that
6 Pi i Pi i i
Eq. (1) is written in physical units with the value of the lattice link being δ, Using unit
speed for particles (with some physical time unit), a time step has the value of δ as well. A
Chapman-Enskog procedure can be applied to Eq. (1) to derive the macroscopic equations
of the model. They are given by: the continuity equation (with an error term O(δ2) being
omitted):
∂ρ
+ (ρu) = 0, (4)
∂t ∇·
and the momentum equation (with terms of O(δ2) and O(δu3) being omitted):
∂ (ρu )+∂ (ρu u ) = ∂ (c2ρ)+∂ (2νρS ), (5)
t α β α β α s β αβ
−
where the Einstein summation convention is used, S = 1(∂ u +∂ u ) is the strain-rate
αβ 2 α β β α
1 α
tensor, the pressure is given by p = c2ρ, where c is the speed of sound with c2 = − ,
s s s 2
2τ 1
and ν = − δ, with ν being the the kinematic viscosity. The form of the error terms
8
and the derivation of these equations can be found in [8, 9]. The macroscopic equations of
LBGK represent the incompressible Navier-Stokes equations in the limit as δ 0, ρ ρ
0
→ →
(a constant ) and the Mach number approaches zero,
2
The plain Poiseuille flow in a channel with width 2L and velocity u = (u,v) is given by:
y2 ∂p ∂p
u= u (1 ), v = 0, = G, = 0, (6)
0 − L2 ∂x − ∂y
where G is a constant related to the characteristic velocity u by
0
G = 2ρνu /L2, (7)
0
and the flow density ρ is a constant. This is an exact solution of the incompressible Navier-
Stokes equations:
u = 0,
∇·
∂ (ρu )+∂ (ρu u )= ∂ p+µ∂ u , (8)
t α β α β α ββ α
−
where µ = ρν. Without loss of generality, we assume that L = 1 (a simple scaling of
′ ′
y = y/L makes y [ 1,1]). To approximate Poiseuille flow using the lattice Boltzmann
∈ −
model,itisconvenienttoreplacetheconstantgradientbyabodyforcegsothatρg = p.
−∇
The momentum equation of N-S equations with a body force is:
∂ (ρu )+∂ (ρu u ) = ∂ p+ρg +µ∂ u . (9)
t α β α β α α ββ α
−
The Poiseuille flow can be generated with a body force g with ρg = G,g = 0, where the
x y
pressure is held constant. An LBGK which incorporates the body force is a modification of
Eq. (1) given by:
1
(0)
f (x+δe ,t+δ) f (x,t) = [f (x,t) f (x,t)]+δh , (10)
i i − i −τ i − i i
where the h are chosen as:
i
h = 0, h = +G/4, i= 1,2,6, h = G/4, i= 3,4,5, (11)
0 i i
−
so that
Xhi = 0, Xhiei = ρg, Xhieiαeiβ = 0.
i i i
Now suppose there exists a solution f (x,t) of Eq. (10) and it exactly represents the
i
Poiseuille flow. We then expect the following properties:
1. f (x,t) is steady (independent of t).
i
2. f (x,t) is independent of x, hence is only a function of y, denoted by f (y).
i i
3. f (y) = f ( y), and f (y)= f ( y) from the symmetry of the flow.
2 6 3 5
− −
4. f (y) = ρ (constant).
Pi i
5. f (y)e = ρu(y) , where u(y) = u (1 y2) (remember L = 1).
Pi i ix 0
−
6. f (y)e = 0 .
Pi i iy
According to Eq. (3), the equilibrium distributions are given by
f(0) = d ρu2, (u= u(y))
0 0−
ρ ρ ρ ρ
f(0) = d+ u+ u2, f(0) = d u+ u2,
1 3 2 4 − 3 2
ρ ρ
(0) (0)
f = d+ u, f = d u,
2 6 3 − 6
ρ ρ
(0) (0)
f = d u, f = d+ u. (12)
5 − 6 6 6
3
Using properties 1,2 and Eq. (10) for i = 0 gives:
1
(0)
f (y) = f (y) (f (y) f (y)),
0 0 − τ 0 − 0
hence
f (y) = f(0)(y) = d ρu2. (13)
0 0 0−
Eq. (10) for i= 1 gives:
1
(0)
f (y)= f (y) (f (y) f (y))+δG/4,
1 1 − τ 1 − 1
hence
(0)
f (y)= f (y)+τδG/4. (14)
1 1
Similarly
(0)
f (y)= f (y) τδG/4. (15)
4 4 −
It is seen that f ,f ,f are functions of y4, y2 through dependence of u and u2. To find the
0 1 2
remaining f (y), we note that f(0),i = 2,3,5,6 have no u2 term and thus are functions of
i i
y2 only, so the following form is suggested:
f (y) = a +b y+c y2, i = 2,3,5,6, (16)
i i i i
where the twelve unknown coefficients a ,b ,c depend on flow quantities, τ, dy, but not on
i i i
y. Using property 3, we obtain
a +b y+c y2 = a b y+c y2, a +b y+c y2 = a b y+c y2,
2 2 2 6 6 6 3 3 3 5 5 5
− −
which should be true for any y, thus:
a = a , b = b , c = c
6 2 6 2 6 2
−
a = a , b = b , c = c . (17)
5 3 5 3 5 3
−
Similarly, using property 6, we find:
2b y+2b y = 0, hence b = b . (18)
2 3 3 2
−
Property 4 with information obtained gives:
2(a +a )+2(c +c )y2+f (y)+f (y)+f (y)= ρ, (19)
2 3 2 3 0 1 4
On using the expressions for f ,f ,f given in Eq. (13 - 15), we find
0 1 4
1
a +a = (ρ d 2d) = 4d (on using the expression of d), c +c = 0; (20)
2 3 0 3 2
2 − −
which gives
c = c , a = 2d a . (21)
3 2 3 2
− −
Similarly, property 5 with information obtained yields:
1 1
a = ρu +d τδG/4; c = ρu . (22)
2 0 2 0
6 − −6
4
At this point, only b remains unknown. Using Eq. (10) for i = 2, we have
2
1
(0)
f (y+dy) = f (y) (f (y) f (y))+δG/4, (23)
2 2 − τ 2 − 2
where dy is the vertical spacing between two lattice rows and dy = (√3/2)δ. On using the
(0)
expression for f , we can obtain:
2
1 1 1
c [y2+2ydy+(dy)2]+b y+b dy+a = (1 )(c y2+b y+a )+ [d+ ρu (1 y2)]+δG/4.
2 2 2 2 2 2 2 0
−τ τ 6 −
(24)
The balance of terms linear in y yields:
1
b = 2τc dy = τρu dy, (25)
2 2 0
− 3
and fortunately the equations for the coefficients of y2 and y0 are both satisfied. It is easy
to check that the evolution equations for f ,f ,f are all satisfied with the choice of a ,b ,c
3 5 6 i i i
obtained so far.
Putting these results all together, we find that the quantities
f = d ρu2,
0 0
−
ρ ρ
f = d+ u+ u2+τδG/4,
1
3 2
ρ ρ
f = d u+ u2 τδG/4,
4
− 3 2 −
1 1 1
f = ρu y2+ τρu ydy+ ρu +d τδG/4,
2 0 0 0
−6 3 6 −
1 1 1
f = + ρu y2 τρu ydy ρu +d+τδG/4,
3 0 0 0
6 − 3 − 6
1 1 1
f = + ρu y2+ τρu ydy ρu +d+τδG/4,
5 0 0 0
6 3 − 6
1 1 1
f = ρu y2 τρu ydy+ ρu +d τδG/4, (26)
6 0 0 0
−6 − 3 6 −
all satisfy properties 1-6 and that they together with the equilibrium distribution given in
Eq. (12) satisfy the LBGK equation, Eq. (9). Hence it is an exact representation of the
Poiseuille flow.
Next, letusseetowhatboundaryconditionthesolutioninEq.(26)corresponds. Taking
the bottom boundary with y = 1,u = 0, we have
−
f = d , f = d+τδG/4, f = d τδG/4
0 0 1 4
−
1 1
f = τρu dy+d τδG/4, f = + τρu dy+d+τδG/4
2 0 3 0
−3 − 3
1 1
f = τρu dy+d+τδG/4, f = + τρu dy+d τδG/4. (27)
5 0 6 0
−3 3 −
It is seen that on the bottom, after the collision and forcing, f = f 2τδG/4, f =
2 5 3
−
f + 2τδG/4. Hence, if a bounce-back boundary condition f = f ,f = f , on f ,f is
6 2 5 3 6 2 3
applied at the bottom to replace the collision and forcing step, the error is of order δ.
5
This shows that the bounce-back boundary condition is first-order accurate. This has been
confirmed in computations [9, 10, 11].
To obtain the steady-state analytical solution in the LBGK simulations, the boundary
condition should be suitably chosen for the simulation. The boundary condition proposed
by Nobel et al. [5] is a suitable choice. If we are looking at a node B on the bottom, after
streaming, f and f are empty at the node B since no particle is coming from outside.
2 3
Then Eq. (2) with u = v = 0 are used to determine ρ,f ,f . Then the normal collision
2 3
with force as given in Eq. (10) is applied to f on the boundaries. Suppose that initially,
i
(0)
we use uniform density ρ and zero velocity through the flow field; then we compute f
0 i
(0)
and set f = f through the field. Since there are no pressure gradients, it is natural
i i
that the density at each node remains constant ρ (confirmed by simulations). Therefore
0
Eq.(2)canbeusedtofindtheuniquef ,f withthecorrectdensity andvelocity, henceitis
2 3
consistent with the evolution of f ,f in the analytical solution. Simulation results indicate
2 3
that thenumericalsolution with thisboundarycondition approaches theanalytical solution
as t .
→ ∞
Next, let us consider a plain Couette flow, where the flow between two parallel plates
(corresponding to y = 0 and y = 1) are driven by the constantly moving top plate with
velocity u . In this case, the solution is given by:
0
u = u y, 0 y 1, v = 0, p =0, (28)
0
≤ ≤ ∇
with ρ a constant and with no body force. So the LBGK model Eq. (1) is used. Using a
similar procedure, we find the analytical solution of Eq. (1) representing the Couette flow:
ρ ρ ρ ρ
f = d ρu2, f = d+ u+ u2, f = d u+ u2,
0 0 1 4
− 3 2 − 3 2
1 1 1 1
f = + ρu y+d τρu dy, f = ρu y+d+ τρu dy,
2 0 0 3 0 0
6 − 6 −6 6
1 1 1 1
f = ρu y+d+ τρu dy, f = + ρu y+d τρu dy. (29)
5 0 0 6 0 0
−6 6 6 − 6
We note that these analytical solutions are valid for any u , τ and dy.
0
3 Analytical Solutions of the square lattice LBGK model
The square lattice Boltzmann BGK model is proven more robustthan the triangular model
in numerical simulations [9, 12], it is important and interesting to find analytical solutions
for it. The square lattice Boltzmann BGK model uses 3 types of particles. Particles of
type 1 move along the x axis or the y axis with speed e = (cos(π(i 1)/2),sin(π(i
i
− −
1)/2),i = 1,2,3,4, and particles of type 2 move along the diagonal directions with speed
e = √2(cos(π(i 4 1)/2),sin(π(i 4 1)/2),i = 5,6,7,8. Rest particles of type 0
i − − 2 − − 2
with e = 0 (speed zero) are also allowed at each node. Each node is connected to its 8
0
nearest neighbors by 8 links of length δ (in physical units) or √2δ as shown in Fig. 2. The
single-particle distribution function f (x,t) again satisfies the LBGK model Eq. (1) (with
i
i = 0, ,8). The density ρ and the macroscopic velocoty u are still defined in Eq. (2).
···
For the square lattice, the equilibrium distribution can be chosen in the following form for
6
particles of each type (the model d2q9 [7]):
4 3
(0)
f = ρ[1 u u],
0 9 − 2 ·
1 9 3
f(0) = ρ[1+3(e u)+ (e u)2 u u], i= 1,2,3,4
i 9 i· 2 i· − 2 ·
1 9 3
f(0) = ρ[1+3(e u)+ (e u)2 u u], i= 5,6,7,8, (30)
i 36 i· 2 i· − 2 ·
(0) (0)
with XXfσi = ρ, XXfσi eσi = ρu. The macroscopic equations of the model are the
σ i σ i
2τ 1
same as given in Eqs. (4,5) with c2 = 1/3, and ν = − δ. To incorporate a body force
s 6
in the model to model Poinseuille flow, Eq. (10) ( i = 0, ,8) is used, with h chosen in
i
···
the following way [13]:
1 1
h = 0, h = G, h = 0, h = G, h = 0,
0 1 2 3 4
3 −3
1 1
h = h = G, h = h = G, (31)
5 8 6 7
12 −12
To derive an analytical solution of Eq. (10) for the square lattce, we note that the six
properties in Section 2 still applies except that property 3 is replaced by:
3. f (y)= f ( y), f (y) = f ( y) and f (y) = f ( y) from the symmetry of the flow.
2 4 5 8 6 7
− − −
Using a similar procedure as in Section 2, we can find that
4 3
f (y) = f(0)(y)= ρ(1 u2), (u =u (1 y2)), (32)
0 0 9 − 2 0 −
1 2
f (y) = ρ(1+3u+3u2)+ τνρu δ. (33)
1 0
9 3
1 2
f (y) = ρ(1 3u+3u2) τνρu δ. (34)
3 0
9 − − 3
and
f (y) = a +b y+c y2+d y3+e y4, i= 2,4,5,6,7,8, (35)
i i i i i i
with
a = a , b = b , c = c , d = d , e = e ,
4 2 4 2 4 2 4 2 4 2
− − −
a = a , b = b , c = c , d = d , e = e ,
8 5 8 5 8 5 8 5 8 5
− − −
a = a , b = b , c = c , d = d , e = e , (36)
7 6 7 6 7 6 7 6 7 6
− − −
and
7 2 1 1 1 1
a = 4τ4ρu2δ4 +6τ3ρu2δ4 τ2ρu2δ4+ τ2ρu2δ2 + τρu2δ4 τρu2δ2+ ρ ρu2,
2 − 0 0 − 3 0 3 0 6 0 − 3 0 9 − 6 0
2 2
b = 4τ3ρu2δ3 4τ2ρu2δ3 + τρu2δ3 τρu2δ,
2 0 − 0 3 0 − 3 0
1 2 1
c = 2τ2ρu2δ2+τρu2δ2 + ρu2, d = τρu2δ, e = ρu2, (37)
2 − 0 0 3 0 2 3 0 2 −6 0
7
7 1 1 1
a = 2τ4ρu2δ4 3τ3ρu2δ4+ τ2ρu2δ4 τ2ρu2δ2 τ2ρu δ2 τρu2δ4 +
5 0 − 0 6 0 − 3 0 − 6 0 − 12 0
1 1 1 1 1 1
τρu2δ2 + τρu δ2+ ρ+ ρu2+ ρu + τνρu δ,
6 0 12 0 36 12 0 12 0 6 0
1 1 1
b = 2τ3ρu2δ3 +2τ2ρu2δ3 τρu2δ3+ τρu2δ+ τρu δ,
5 − 0 0 − 3 0 3 0 6 0
1 1 1 1 1
c = τ2ρu2δ2 τρu2δ2 ρu2 ρu , d = τρu2δ, e = ρu2, (38)
5 0 − 2 0 − 6 0− 12 0 5 −3 0 5 12 0
and
7 1 1 1
a = 2τ4ρu2δ4 3τ3ρu2δ4+ τ2ρu2δ4 τ2ρu2δ2 + τ2ρu δ2 τρu2δ4 +
6 0 − 0 6 0 − 3 0 6 0 − 12 0
1 1 1 1 1 1
τρu2δ2 τρu δ2+ ρ+ ρu2 ρu τνρu δ,
6 0 − 12 0 36 12 0− 12 0− 6 0
1 1 1
b = 2τ3ρu2δ3 +2τ2ρu2δ3 τρu2δ3+ τρu2δ τρu δ,
6 − 0 0 − 3 0 3 0 − 6 0
1 1 1 1 1
c = τ2ρu2δ2 τρu2δ2 ρu2+ ρu , d = τρu2δ, e = ρu2. (39)
6 0 − 2 0 − 6 0 12 0 6 −3 0 6 12 0
Eqs. (32,33,34) together with Eqs. (35,36,37,38,39) completely specify the analytical solu-
tion, which is a solution of Eq. (10) and it exactly represents the Poiseuille flow.
Next, letusseetowhatboundaryconditionthisanalytical solution corresponds. Taking
the bottom boundary with y = 1,u= 0, we find the relation of f after the collision and
σi
−
forcing:
4 2
f f = τνρu δ, f f = 2δ3(4τ3ρu2 4τ2ρu2+ τρu2),
1− 3 3 0 2− 4 − 0− 0 3 0
2 1 2
f f = τ2ρu δ2+ τρu2δ2+4τ3ρu2δ3 4τ2ρu2δ3 + τρu2δ3,
5− 7 −9 0 9 0 0 − 0 3 0
2 1 2
f f =+ τ2ρu δ2 τρu2δ2+4τ3ρu2δ3 4τ2ρu2δ3 + τρu2δ3, (40)
6− 8 9 0 − 9 0 0 − 0 3 0
if a bounce-back boundary condition in which f exchanges with f , f = f , f = f , f =
1 3 2 4 5 7 6
f , is applied at the bottom to replace the collision and forcing step, the error introduced to
8
f and f is of order δ. This shows that the bounce-back boundary condition is first-order
1 3
accurate. This has been confirmed in computations [9, 10, 11]. To obtain the steady-state
analytical solution derived in this paper in LBGK simulations, the boundary condition
should be suitably chosen. No numerical simulation on a square lattice Boltzmann BGK
model has obtained an exact solution for the Poiseuille flow so far. Of course, if we use the
analytical solution derived as theinitial condition andprovidethe rightform of distribution
functionsontheboundaries,wewillbeableto obtaintheanalytical solutioninasimulation
(confirmed in simulations). Specification of the analytical solution on boundary does not
provide a boundary condition of general purpose. Nevertheless, the analytical solution will
give some guidance in developing better boundary conditions of general purpose for the
model.
The analytical solution for plain Couette flow are given by
4 3 1 1
f = ρ(1 u2y2), f = ρ(1+3u y+3u2y2), f = ρ(1 3u y+3u2y2),
0 9 − 2 0 1 9 0 0 3 9 − 0 0
8
1 1 1 1 1
f = τ2ρu2δ2+ τρu2δ2+ ρ+ τρu2δy ρu2y2,
2 −3 0 6 0 9 3 0 − 6 0
1 1 1 1 1
f = τ2ρu2δ2+ τρu2δ2+ ρ τρu2δy ρu2y2,
4 −3 0 6 0 9 − 3 0 − 6 0
1 1 1 1 1 1 1
f = τ2ρu2δ2 τρu2δ2 τρu δ+ ρ+( τρu2δ+ ρu )y+ ρu2y2,
5 6 0 − 12 0 − 12 0 36 −6 0 12 0 12 0
1 1 1 1 1 1 1
f = τ2ρu2δ2 τρu2δ2+ τρu δ+ ρ+( τρu2δ ρu )y+ ρu2y2,
6 6 0 − 12 0 12 0 36 −6 0 − 12 0 12 0
1 1 1 1 1 1 1
f = τ2ρu2δ2 τρu2δ2 τρu δ+ ρ+(+ τρu2δ ρu )y+ ρu2y2,
7 6 0 − 12 0 − 12 0 36 6 0 − 12 0 12 0
1 1 1 1 1 1 1
f = τ2ρu2δ2 τρu2δ2+ τρu δ+ ρ+(+ τρu2δ+ ρu )y+ ρu2y2. (41)
8 6 0 − 12 0 12 0 36 6 0 12 0 12 0
For the Couette flow, the top boundary is a moving one, the analytical solution given here
will give a guidance in developing a suitable boundary condition for moving boundaries.
We note that these analytical solutions are valid for any u , τ and δ. They will enhance
0
our understanding of the method and will give guidance in applications.
Acknowledgments
WewouldliketothankS.Chen,D.Noble,G.McNamara, D.d’Humi`eres,D.Levermore,
Y.H. Qian for helpful discussions. Q.Z. would like to thank S. Chen for helping to arrange
for his visit to the Los Alamos National Lab.
9
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Figure captions
Figure 1. The geometry of the plain channel flow.
Figure 2. Schematic of a square lattice.
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