Computational Physics

Exercises 463

To proceed, we must work out the tensors(^0 )and (^0 ). In order to do this, we
note that, on the hexagonal grid, the following relations hold:
∑

i

eiαeiβ= 3 δαβ; (14.55a)

∑

i

eiαeiβeiγeiη=

3

4

(δαβδγη+δαγδβη+δαηδβγ), (14.55b)

whereas similar moments containing an odd number ofeiαvanish. Together with
(14.32), this yields
(αβ^0 )=ρuiαuiβ+Pδαβ, (14.56)

wherePis found using the ideal gas law relation with the kinetic energy as in
Section 14.2. Furthermore

(αβ γ^0 ) =ρ

c^2

3

(uαδβγ+uβδαγ+uγδαβ). (14.57)

Substituting these equations into(14.54), we obtain

∂tjα+∂β(αβ^0 )=

(

τ−

1

2

)

t

[

∂t∂β(ρuαuβ+Pδαβ)+ρ

c^2

3

(∂β∂βuα+ 2 ∂α∂βuβ)

]

+O(t^2 ). (14.58)

In vector notation, this reads

∂tj+∇·(^0 )=

(

τ−

1

2

)

t

[

∇

(

∂P

∂t

)

+ρ∇·

(

∂uu

∂t

)

+ρ

c^2

3

∇^2 u

]

+O(t^2 ).

(14.59)

To obtain the last form, we have neglected the term∇(∇·u), which is small in the
incompressible limit as can be seen from the continuity equation. Furthermore, in
this limit, the time derivatives on the right hand side are of the order ofM^2 – hence
we are left with
∂tj+ρ∇(uu)=∇p+ρν∇^2 u, (14.60)

where we have neglected corrections beyond t^2 and tM^2 and where we have
used (14.36).
This is the Navier–Stokes equation in the incompressible limit for which∇·u=

0. This derivation does not depend on the fact that we have used the hexagonal
   lattice – it yields the same result for the d2q9 model.

Exercises

14.1 Derive from the Navier–Stokes equation that the flow in a two-dimensional pipe
forms a parabolic velocity profile. Check that the curvature (i.e. the second