A Discrete Nine-Velocity Model of the Boltzmann Equation: Solution in the Form of Wild Sum and Applications to Simulating Incompressible Flows

Abstract

A discrete kinetic nine-velocity model of the Boltzmann equation on a plane is considered. In the limit of small free path and low bulk velocities, this model describes flows of viscous incompressible fluids. The complete discretization of the model over the time and spatial variables, which is, in particular, required for the numerical solution, is carried out using the truncated Wild sum. It is shown that the scheme has the second order of accuracy. As an example of the application of the proposed method, numerical solutions of two benchmark problems are obtained—Taylor–Green vortices and flow in a cavity with a moving boundary. The simulation results are compared with the solutions obtained on the basis of the classical nine-velocity lattice Boltzmann model.

APPENDIX

1.1 THE CHAPMAN–ENSKOG EXPANSION

The Chapman–Enskog expansion for scheme (27)–(29) is carried out similarly to how this is done in the LBE theory (see [1, 28]). First, we expand the left-hand side of (27) in the Taylor series

$${{f}_{i}}(t + \Delta t,{\mathbf{r}} + {{{\mathbf{c}}}_{{\mathbf{i}}}}\Delta t) = {{f}_{i}}(t,{\mathbf{r}}) + \Delta t{{D}_{i}}{{f}_{i}}(t,{\mathbf{r}}) + \frac{1}{2}\Delta tD_{i}^{2}{{f}_{i}}(t,{\mathbf{r}}) + o(\Delta {{t}^{2}}),$$

where

$${{D}_{i}} = \frac{\partial }{{\partial t}} + {{{\mathbf{c}}}_{i}} \cdot \frac{\partial }{{\partial {\mathbf{r}}}}.$$

As in the standard approach used in the LBE theory (see [1, 28]), we expand the distribution function ${{f}_{i}}$ in a series in the small parameter $\epsilon$ (Knudsen number) in a neighborhood of the local equilibrium

$${{f}_{i}} = f_{i}^{{eq}} + \epsilon {{f}_{{1,i}}} + {{\epsilon }^{2}}{{f}_{{2,i}}} + \; \ldots ;$$

we also assume that there is a number of time scales (see [1, 28]) and the derivatives with respect to the spatial variables and time are written as (see [1, 28])

$$\frac{\partial }{{\partial t}} = \epsilon \frac{\partial }{{\partial {{t}_{1}}}} + {{\epsilon }^{2}}\frac{\partial }{{\partial {{t}_{2}}}} + \; \ldots ,\quad \frac{\partial }{{\partial {\mathbf{r}}}} = \epsilon \frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}}.$$

Substitute the expansions of the derivatives and of the DF in the parameter $\epsilon$ into Eq. (27) and collect the terms of equal orders in $\epsilon$. The terms of order $O(1)$ are identically equal to zero. We have the equations

$$O(\epsilon )\,:\quad \Delta t{{D}_{{1,i}}}f_{i}^{{eq}} = - {{\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)}^{2}}{{f}_{{1,i}}} + \tau {{e}^{{ - \frac{{\Delta t}}{\tau }}}}\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right){{L}_{i}}({{{\mathbf{f}}}_{1}}),$$

(33)

where

$${{D}_{{1,i}}} = \left( {\frac{\partial }{{\partial {{t}_{1}}}} + {{{\mathbf{c}}}_{i}} \cdot \frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}}} \right)$$

and ${{L}_{i}}$ is the collision operator for Eqs. (4)–(12) linearized with respect to the local Maxwellian. Since we only consider the flows with small Mach numbers, we my use the absolutely Maxwellian distribution instead of the locally Maxwellian distribution; the former distribution is specified by the weights ${{\rho }_{0}}{{w}_{i}}$. In this case, the operator ${{L}_{i}}({{{\mathbf{f}}}_{1}})$ has the form

$${{L}_{i}}({{{\mathbf{f}}}_{1}}) = {{\rho }_{0}}\sum\limits_{jkl = 1}^N {A_{{kl}}^{{ij}}} ({{w}_{k}}{{f}_{{1,l}}} - {{w}_{i}}{{f}_{{1,j}}} + {{f}_{{1,k}}}{{w}_{l}} - {{f}_{{1,i}}}{{w}_{j}}).$$

By multiplying Eqs. (33) by ${{{\mathbf{c}}}_{i}}$ and summing over $i$, we obtain equations of momentum variation for inviscid medium (the Euler equation)

$$\frac{\partial }{{\partial {{t}_{1}}}}{{\rho }_{0}}{\mathbf{u}} + \frac{\partial }{{\partial {{{\mathbf{r}}}_{{\mathbf{1}}}}}} \cdot ({{\rho }_{0}}{\mathbf{uu}} + {{\rho }_{0}}{{\theta }_{0}}) = 0.$$

(34)

For the terms of order $O({{\epsilon }^{2}})$, we obtain

$$O({{\varepsilon }^{2}})\,:\quad \Delta t\frac{\partial }{{\partial {{t}_{2}}}}f_{i}^{{eq}} + \Delta t{{D}_{{1,i}}}{{f}_{{1,i}}} + \frac{{\Delta {{t}^{2}}}}{2}D_{{1,i}}^{2}f_{i}^{{eq}} = \; \ldots ,$$

(35)

where the ellipsis denotes the second-order terms in $\epsilon$ that appear on the right-hand side; their explicit form is not required here. Now multiply Eqs. (35) by ${{{\mathbf{c}}}_{i}}$ and sum them over $i$ to obtain

$$\frac{\partial }{{\partial {{t}_{2}}}}{{\rho }_{0}}{\mathbf{u}} + \frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}} \cdot {{{\mathbf{P}}}_{1}} + \frac{{\Delta t}}{2}\frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}} \cdot \sum\limits_i {{{{\mathbf{c}}}_{i}}} {{{\mathbf{c}}}_{i}}{{D}_{{1,i}}}f_{i}^{{eq}} = 0,$$

(36)

where ${{{\mathbf{P}}}_{1}} = \sum\nolimits_i {{{{\mathbf{c}}}_{i}}} {{{\mathbf{c}}}_{i}}{{f}_{{1,i}}}$. Sum Eqs. (34) and (36) to obtain

$$\left( {\epsilon \frac{\partial }{{\partial {{t}_{1}}}} + {{\epsilon }^{2}}\frac{\partial }{{\partial {{t}_{2}}}}} \right){{\rho }_{0}}{\mathbf{u}} + \frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}}\left( {\epsilon ({{\rho }_{0}}{\mathbf{uu}} + {{\rho }_{0}}{{\theta }_{0}}) + {{\epsilon }^{2}}{{{\mathbf{P}}}_{1}} + {{\epsilon }^{2}}\frac{{\Delta t}}{2}\sum\limits_i {{{{\mathbf{c}}}_{i}}} {{{\mathbf{c}}}_{i}}{{D}_{{1,i}}}f_{i}^{{eq}}} \right) = 0;$$

now return to the original variables $t$ and ${\mathbf{r}}$ to obtain (up to small terms $O\left( {{{\epsilon }^{3}}} \right)$) the equation

$$\frac{\partial }{{\partial t}}({{\rho }_{0}}{\mathbf{u}}) + \frac{\partial }{{\partial {\mathbf{r}}}} \cdot \left( {{{\rho }_{0}}{\mathbf{uu}} + {{\rho }_{0}}{{\theta }_{0}} + \epsilon {{{\mathbf{P}}}_{1}} + \epsilon \frac{{\Delta t}}{2}\sum\limits_i {{{{\mathbf{c}}}_{i}}} {{{\mathbf{c}}}_{i}}{{D}_{{1,i}}}f_{i}^{{eq}}} \right) = 0,$$

(37)

where the third and the fourth terms in the second term in (37) are responsible for viscous corrections to the Euler equation, which should be found. To find ${{{\mathbf{P}}}_{1}}$, we need expressions for ${{f}_{{1,i}}}$. The total derivative of the locally Maxwellian distribution in (33) in the case of small Mach numbers up to orders $O\left( {{{{\text{M}}}^{3}}} \right)$ and incompressible flows is (see [23])

$${{\rho }_{0}}{{w}_{i}}\frac{{{{{\mathbf{c}}}_{i}}{{{\mathbf{c}}}_{i}}}}{{{{\theta }_{0}}}}:\frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}}{\mathbf{u}};$$

recall that “:” is the tensor convolution operator. Nonequilibrium DFs ${{f}_{{1,i}}}$ should be sought in the form (see [23, 29])

$${{f}_{{1,i}}} = {{a}_{i}}\frac{{{{{\mathbf{c}}}_{i}}{{{\mathbf{c}}}_{i}}}}{{{{\theta }_{0}}}}:\frac{\partial }{{\partial {{{\mathbf{r}}}_{1}}}}{\mathbf{u}},$$

where the coefficients ${{a}_{i}}$ are the same for the indices $i$ corresponding to the velocities ${{{\mathbf{c}}}_{i}}$ for which the kinetic energy of particles is the same; it is clear that there are three different coefficients ${{a}_{i}}$. From Eq. (33), we obtain for the coefficient ${{a}_{1}}$, which corresponds to the velocities $\pm 1$, $\pm 2$,

$${{a}_{1}} = - \frac{{\frac{1}{9}{{\rho }_{0}}\Delta t}}{{{{{\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)}}^{2}} + \frac{4}{9}\tau {{\rho }_{0}}{{e}^{{ - \frac{{\Delta t}}{\tau }}}}\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)\alpha }};$$

similarly, for the coefficient ${{a}_{3}}$, which corresponds to the velocities $\pm 3, \pm 4$, we have

$${{a}_{3}} = - \frac{{\frac{1}{{12}}{{\rho }_{0}}\Delta t}}{{{{{\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)}}^{2}} + \frac{1}{9}\tau {{\rho }_{0}}{{e}^{{ - \frac{{\Delta t}}{\tau }}}}\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)(\gamma + 4\beta + 4\lambda )}};$$

the coefficient ${{a}_{0}}$ corresponding to the zero velocity is zero. For convenience, we assume below that ${{\rho }_{0}} = 1$.

As has been shown above, the viscous corrections to the stress tensor ${{{\mathbf{P}}}_{1}}$ depend on the nonequilibrium DFs ${{f}_{{1,i}}}$; i.e., ${{{\mathbf{P}}}_{1}} = \sum\nolimits_i {{{f}_{{1,i}}}} {{{\mathbf{c}}}_{i}}{{{\mathbf{c}}}_{i}}$. Consider in more detail the viscous terms

$$\epsilon \frac{\partial }{{\partial {\mathbf{r}}}} \cdot {{{\mathbf{P}}}_{1}} = \epsilon \sum\limits_\sigma {\frac{\partial }{{\partial {{r}_{\sigma }}}}} {{P}_{{1,\eta \sigma }}} = \epsilon \sum\limits_\sigma {\frac{\partial }{{\partial {{r}_{\sigma }}}}} {\kern 1pt} \sum\limits_i \,{{f}_{{1,i}}}{{c}_{{i\eta }}}{{c}_{{i\sigma }}} = \theta _{0}^{{ - 1}}\sum\limits_{\sigma \kappa w} {\sum\limits_i {{{a}_{i}}} } {{c}_{{i\eta }}}{{c}_{{i\sigma }}}{{c}_{{i\kappa }}}{{c}_{{iw}}}\frac{{{{\partial }^{2}}}}{{\partial {{r}_{\kappa }}\partial {{r}_{\sigma }}}}{{u}_{w}},$$

where $\sigma$, $\eta$, $\kappa$, and $w$ are equal to $x$ or $y$, and we also denote ${{r}_{x}} = x$ and ${{r}_{y}} = y$. Recall that the vectors ${{{\mathbf{c}}}_{i}}$ have the form $({{e}_{{ix}}},{{e}_{{iy}}})c = ({{e}_{{ix}}},{{e}_{{iy}}})\sqrt {3{{\theta }_{0}}}$, where ${{e}_{{ix}}}$ and ${{e}_{{iy}}}$ take the values $0$, $\pm 1$. Hence, we have, for example, that

$$\begin{gathered} \epsilon \frac{\partial }{{\partial x}}{{P}_{{1,xx}}} + \epsilon \frac{\partial }{{\partial y}}{{P}_{{1,xy}}} = 2(9{{\theta }_{0}}){{a}_{1}}\frac{{{{\partial }^{2}}{{u}_{x}}}}{{\partial {{x}^{2}}}} + 4(9{{\theta }_{0}}){{a}_{3}}\left( {\frac{{{{\partial }^{2}}}}{{\partial x\partial y}}{{u}_{y}} + \frac{{{{\partial }^{2}}}}{{\partial {{y}^{2}}}}{{u}_{x}}} \right) \\ = \;2(9{{\theta }_{0}}){{a}_{1}}\frac{{{{\partial }^{2}}{{u}_{x}}}}{{\partial {{x}^{2}}}} + 4(9{{\theta }_{0}}){{a}_{3}}\left( {\frac{{{{\partial }^{2}}}}{{\partial x\partial y}}{{u}_{y}} + \frac{{{{\partial }^{2}}}}{{\partial {{y}^{2}}}}{{u}_{x}}} \right) - 4(9{{\theta }_{0}}){{a}_{3}}\frac{\partial }{{\partial x}}\left( {\frac{\partial }{{\partial x}}{{u}_{x}} + \frac{\partial }{{\partial y}}{{u}_{y}}} \right) \\ = \;9{{\theta }_{0}}(2{{a}_{1}} - 4{{a}_{3}})\frac{{{{\partial }^{2}}}}{{\partial {{x}^{2}}}}{{u}_{x}} + 4(9{{\theta }_{0}}){{a}_{3}}\frac{{{{\partial }^{2}}}}{{\partial {{y}^{2}}}}{{u}_{x}}; \\ \end{gathered} $$

the last expression implies that the viscous terms have the same form as in the Navier–Stokes equations if

$${{a}_{1}} = 4{{a}_{3}},$$

which is equivalent to (17); however, the viscosity of scheme (27)–(29) is no longer given by formula (18).

The terms $\epsilon \frac{{\Delta t}}{2}\frac{\partial }{{\partial {\mathbf{r}}}}\sum\nolimits_i {{{{\mathbf{c}}}_{i}}} {{{\mathbf{c}}}_{i}}{{D}_{{i,1}}}f_{i}^{{eq}}$ were considered in the previous work [23], and they are equal to ${{\rho }_{0}}{{\theta }_{0}}\frac{{\Delta t}}{2}\left( {\frac{{{{\partial }^{2}}}}{{\partial {{x}^{2}}}} + \frac{{{{\partial }^{2}}}}{{\partial {{y}^{2}}}}} \right){\mathbf{u}}$.

Thus, the viscosity is given by the formula

$$\nu = \frac{{{{\theta }_{0}}\Delta t}}{{{{{\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)}}^{2}} + \frac{4}{9}\tau {{\rho }_{0}}{{e}^{{ - \frac{{\Delta t}}{\tau }}}}\left( {1 - {{e}^{{ - \frac{{\Delta t}}{\tau }}}}} \right)\alpha }} - \frac{{{{\theta }_{0}}\Delta t}}{2},\quad \tau = \nu {\text{/}}{{\theta }_{0}},$$

and additionally the following condition must be fulfilled:

$$4\alpha = \gamma + 4\beta + 4\lambda .$$