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Derivation of Multicomponent Lattice Boltzmann Equations by Introducing a Nonequilibrium Distribution Function into the Maxwell Iteration Based on the Convective Scaling

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Abstract

This study firstly proposes a simple recursive method for deriving the macroscale equations from lattice Boltzmann equations. Similar to the Maxwell iteration based on the convective scaling, this method is used to expand the lattice Boltzmann (LB) equations with the time step \(\delta _{t}\). It is characterised by the incorporation of a nonequilibrium distribution function not appearing in the Maxwell iteration to considerably reduce the mathematical manipulations required. Next, we define the kinetic equations of a multicomponent (i.e. N-component) system based on a model using the Maxwell velocity distribution law for the equilibrium distribution function appearing in the cross-collision terms. Then, using this simple recursive method, we derive the generalized Stefan–Maxwell equation, which is the macroscale governing equation of a multicomponent system while ensuring the mass conservation. In short, our objective is to firstly define the kinetic equations of a multi-component system having a clear physical interpretation and then formulate the LB equations of any N-component system deductively.

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Correspondence to Keiichi Yamamoto.

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Communicated by Shin-ichi Sasa.

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Appendices

Appendix A

Appendix A proves that coefficients of the same order must be equal on both sides for (6) to hold true for any \(\delta _{t}\). The zero-th order terms of (6) are gathered on the left-hand side and all other terms are gathered on the right-hand side. The absolute value is taken on both sides, and the right-hand side is evaluated with an inequality sign.

$$\begin{aligned}&\left| -\frac{1}{\varphi }\left( {{f}_{i}^{\left( 0 \right) }}-{{f}_{i}^{eq}} \right) \right| \nonumber \\&\quad \le \left| \left( {\hat{D}}_{i} {{f}_{i}^{\left( 0 \right) }}+\frac{1}{\varphi }{{f}_{i}^{\left( 1 \right) }} \right) \right| {{\delta }_{t}}+\left| \left( {\hat{D}}_{i} {{f}_{i}^{\left( 1 \right) }}+\frac{1}{2}{{{{\hat{D}}_{i}}}^{2}}{{f}_{i}^{\left( 0 \right) }}+\frac{1}{\varphi }{{f}_{i}^{\left( 2 \right) }} \right) \right| \delta _{t}^{2}+\cdots \nonumber \\&\quad \le Max\left( \left| \left( {\hat{D}}_{i} {{f}_{i}^{\left( 0 \right) }}+\frac{1}{\varphi }{{f}_{i}^{\left( 1 \right) }} \right) \right| ,\left| \left( {\hat{D}}_{i} {{f}_{i}^{\left( 1 \right) }}+\frac{1}{2}{{{{\hat{D}}_{i}}}^{2}}{{f}_{i}^{\left( 0 \right) }}+\frac{1}{\varphi }{{f}_{i}^{\left( 2 \right) }} \right) \right| \cdots \right) \frac{{{\delta }_{t}}}{1-{{\delta }_{t}}}\nonumber \\ \end{aligned}$$
(80)

Here, when \(\delta _{t} \rightarrow 0\), the right-hand side converges to 0. Therefore, the necessary conditions for (6) to hold true for any \(\delta _{t}\) can be obtained as:

$$\begin{aligned} {{f}_{i}^{\left( 0 \right) }}-{{f}_{i}^{eq}}=0 \end{aligned}$$
(81)

In addition, (81) is substituted in (6), both sides are divided by \(\delta _{t}\), and an evaluation similar to (80) is performed to obtain (82).

$$\begin{aligned} {\hat{D}}_{i} f_{i}^{\left( 0 \right) }=-\frac{1}{\varphi }f_{i}^{\left( 1 \right) } \end{aligned}$$
(82)

The proof is completed by repeating this operation in the same manner.

Appendix B

Appendix B shows the detailed derivation of the mass conservation law, the Euler equation and the Navier–Stokes equation from (21) to (29).

First, the mass conservation law is derived. Concerning (8), the sum is taken with i as below.

$$\begin{aligned} \mathop {\sum }\limits _{i}{{{{{\hat{D}}}}_{i}}f_{i}^{(0)}}=-\frac{1}{\varphi }\mathop {\sum }\limits _{i}{f_{i}^{(1)}}=0 \end{aligned}$$
(83)

The lef-hand side of (83) is transformed as:

$$\begin{aligned} \mathop {\sum }\limits _{i}{{{{{\hat{D}}}}_{i}}f_{i}^{(0)}}&=\mathop {\sum }\limits _{i}{\left( \frac{\partial }{\partial t}+{{e}_{i,\beta }}\frac{\partial }{\partial {{x}^{\beta }}} \right) f_{i}^{(0)}} \nonumber \\&=\frac{\partial }{\partial t}\mathop {\sum }\limits _{i}{f_{i}^{(0)}}+\frac{\partial }{\partial {{x}^{\beta }}}\mathop {\sum }\limits _{i}{{{e}_{i,\beta }}f_{i}^{(0)}} \end{aligned}$$
(84)

(84) is substituted into (83), the mass conservation law (85) is derived.

$$\begin{aligned} \frac{\partial \rho }{\partial t}+\frac{\partial \rho {{u}_{\beta }}}{\partial {{x}^{\beta }}}=0 \end{aligned}$$
(85)

Next, the Euler equation is derived. Concerning (8), both sides of (8) are multiplied by the discrete velocity and the sum is taken with i as below.

$$\begin{aligned} \mathop {\sum }\limits _{i}{{{{{\hat{D}}}}_{i}}f_{i}^{(0)}{{e}_{i,\alpha }}}=-\frac{1}{\varphi }\mathop {\sum }\limits _{i}{f_{i}^{(1)}{{e}_{i,\alpha }}}=0 \end{aligned}$$
(86)

The left-hand side of (86) is transformed as:

$$\begin{aligned} \mathop {\sum }\limits _{i}{{{{{\hat{D}}}}_{i}}f_{i}^{(0)}{{e}_{i,\alpha }}}&=\mathop {\sum }\limits _{i}{\left( \frac{\partial }{\partial t}+{{e}_{i,\beta }}\frac{\partial }{\partial {{x}^{\beta }}} \right) f_{i}^{(0)}{{e}_{i,\alpha }}} \nonumber \\&=\frac{\partial }{\partial t}\mathop {\sum }\limits _{i}{{{e}_{i,\alpha }}f_{i}^{(0)}}+\frac{\partial }{\partial {{x}^{\beta }}}\mathop {\sum }\limits _{i}{{{e}_{i,\alpha }}{{e}_{i,\beta }}f_{i}^{(0)}} \nonumber \\&=\frac{\partial \rho {{u}_{\alpha }}}{\partial t}+\frac{\partial }{\partial {{x}^{\beta }}}\left( \rho {{u}_{\alpha }}{{u}_{\beta }}+p{{\delta }_{\alpha \beta }} \right) \end{aligned}$$
(87)

(87) is substituted into (86), and using (85) the Euler equation (88) is derived.

$$\begin{aligned} \frac{\partial {{u}_{\alpha }}}{\partial t}+{{u}_{\beta }}\frac{\partial {{u}_{\alpha }}}{\partial {{x}^{\beta }}}=-\frac{1}{\rho }\frac{\partial p}{\partial {{x}^{\alpha }}} \end{aligned}$$
(88)

Finally, the Navier–Stokes equation is derived. As the derivation process up to (21) have been shown, we begin the calculation from that equation below.

$$\begin{aligned}&\frac{\partial \left( \rho {{u}_{\alpha }} \right) }{\partial t}+\frac{\partial \left( \rho {{u}_{\alpha }}{{u}_{\beta }}+p{{\delta }_{\alpha \beta }} \right) }{\partial {{x}^{\beta }}} \nonumber \\&\quad ={{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\mathop {\sum }\limits _{i}{\left( {{e}_{i,\gamma }}\frac{\partial }{\partial {{x}^{\gamma }}}+\frac{\partial }{\partial t} \right) {{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}} \nonumber \\&\quad ={{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \left( \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial t}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}}+\frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \right) \nonumber \\&\quad ={{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \left( \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial t}\left( \rho {{u}_{\alpha }}{{u}_{\beta }}+P{{\delta }_{\alpha \beta }} \right) +\frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \right) \nonumber \\&\quad ={{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( \frac{\partial \rho }{\partial t}{{u}_{\alpha }}{{u}_{\beta }}+\rho \frac{\partial {{u}_{\alpha }}}{\partial t}{{u}_{\beta }}+\rho {{u}_{\alpha }}\frac{\partial {{u}_{\beta }}}{\partial t}+\frac{\partial P}{\partial t}{{\delta }_{\alpha \beta }} \right) \nonumber \\&\qquad +{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \end{aligned}$$
(89)

Here, the Euler equation (88) is substituted into the above, the right-hand side is further transformed as:

$$\begin{aligned}= & {} {{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \left( \frac{\partial }{\partial {{x}^{\beta }}}\left( -\frac{\partial \rho {{u}_{\gamma }}}{\partial {{x}^{\gamma }}}{{u}_{\alpha }}{{u}_{\beta }}+\frac{\partial P}{\partial t}{{\delta }_{\alpha \beta }} \right) \right) \\&+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \left( \frac{\partial }{\partial {{x}^{\beta }}}\left( \rho \left( -\frac{1}{\rho }\frac{\partial P}{\partial {{x}^{\alpha }}}-{{u}_{\gamma }}\frac{\partial {{u}_{\alpha }}}{\partial {{x}^{\gamma }}} \right) {{u}_{\beta }}+\rho {{u}_{\alpha }}\left( -\frac{1}{\rho }\frac{\partial P}{\partial {{x}^{\beta }}}-{{u}_{\gamma }}\frac{\partial {{u}_{\beta }}}{\partial {{x}^{\gamma }}} \right) \right) \right) \\&+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \\= & {} -{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial \rho {{u}_{\gamma }}}{\partial {{x}^{\gamma }}}{{u}_{\alpha }}{{u}_{\beta }}+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial P}{\partial t}{{\delta }_{\alpha \beta }} \\&+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( -\frac{\partial P}{\partial {{x}^{\alpha }}}{{u}_{\beta }}-\frac{\partial {{u}_{\alpha }}}{\partial {{x}^{\gamma }}}{{u}_{\beta }}\rho {{u}_{\gamma }} \right) +{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( -\frac{\partial P}{\partial {{x}^{\beta }}}{{u}_{\alpha }}-{{u}_{\alpha }}\frac{\partial {{u}_{\beta }}}{\partial {{x}^{\gamma }}}\rho {{u}_{\gamma }} \right) \\&+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \\= & {} {{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( -\frac{\partial P}{\partial {{x}^{\alpha }}}{{u}_{\beta }}-\frac{\partial P}{\partial {{x}^{\beta }}}{{u}_{\alpha }}+\frac{\partial P}{\partial t}{{\delta }_{\alpha \beta }}-\frac{\partial \rho {{u}_{\gamma }}{{u}_{\alpha }}{{u}_{\beta }}}{\partial {{x}^{\gamma }}}+\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{f}^{(0)}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}} \right) \end{aligned}$$

To continue the calculations, it is necessary to use the equation of state (23) and the equilibrium distribution function, so we take the equilibrium distribution function to be the conventional expression [26] :

$$\begin{aligned} f_{i}^{(0)}={{A}_{\sigma }}\left\{ 1+{{B}_{\sigma }}\left( {{e}_{i,\mu }}{{u}_{\mu }} \right) +{{C}_{\sigma }}{{\left( {{e}_{i,\mu }}{{u}_{\mu }} \right) }^{2}}+{{D}_{\sigma }}{{u}_{\mu }}{{u}_{\mu }} \right\} ,\sigma =1,2 \end{aligned}$$
(90)

where, \(\sigma \) represents the type of the lattice velocity vector.

(23) and (90) are substituted into the right-hand side of (89) and the time derivative is rewritten as the space derivative using the mass conservation law (85), the right-hand side of the equation is transformed as:

$$\begin{aligned}= & {} {{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( -{{C}_{s}}^{2}\frac{\partial \rho }{\partial {{x}^{\alpha }}}{{u}_{\beta }}-{{C}_{s}}^{2}\frac{\partial \rho }{\partial {{x}^{\beta }}}{{u}_{\alpha }}-{{C}_{s}}^{2}\frac{\partial \rho {{u}_{\gamma }}}{\partial {{x}^{\gamma }}}{{\delta }_{\alpha \beta }} \right) \nonumber \\&+{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( -\frac{\partial \rho {{u}_{\gamma }}{{u}_{\alpha }}{{u}_{\beta }}}{\partial {{x}^{\gamma }}}+\frac{\partial }{\partial {{x}^{\gamma }}}\mathop {\sum }\limits _{i}{{{A}_{\sigma }}{{B}_{\sigma }}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}{{e}_{i,\mu }}{{u}_{\mu }}} \right) \end{aligned}$$
(91)

Here, the last term tensor in (91), is calculated about the D2Q9, for example, as follows:

$$\begin{aligned}&\frac{\partial }{\partial {{x}^{\gamma }}}{{u}_{\mu }}\mathop {\sum }\limits _{i}{{{A}_{\sigma }}{{B}_{\sigma }}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}{{e}_{i,\mu }}} \nonumber \\&\quad =\frac{\partial }{\partial {{x}^{\gamma }}}{{u}_{\mu }}\mathop {\sum }\limits _{i=1}^{4}{{{A}_{1}}{{B}_{1}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}{{e}_{i,\mu }}}+\frac{\partial }{\partial {{x}^{\gamma }}}{{u}_{\mu }}\mathop {\sum }\limits _{5}^{8}{{{A}_{2}}{{B}_{2}}{{e}_{i,\alpha }}{{e}_{i,\beta }}{{e}_{i,\gamma }}{{e}_{i,\mu }}} \nonumber \\&\quad =\frac{\partial }{\partial {{x}^{\gamma }}}\frac{2{{c}^{4}}}{3}{{A}_{1}}{{B}_{1}}{{u}_{\mu }}\left( {{\delta }_{\alpha \beta }}{{\delta }_{\gamma \mu }}+{{\delta }_{\alpha \gamma }}{{\delta }_{\beta \mu }}+{{\delta }_{\alpha \mu }}{{\delta }_{\beta \gamma }} \right) \nonumber \\&\qquad +\frac{\partial }{\partial {{x}^{\gamma }}}\frac{4{{c}^{4}}}{3}{{A}_{2}}{{B}_{2}}{{u}_{\mu }}\left( {{\delta }_{\alpha \beta }}{{\delta }_{\gamma \mu }}+{{\delta }_{\alpha \gamma }}{{\delta }_{\beta \mu }}+{{\delta }_{\alpha \mu }}{{\delta }_{\beta \gamma }} \right) \nonumber \\&\quad =\frac{\partial }{\partial {{x}^{\gamma }}}\left( \frac{2{{c}^{4}}}{3}{{A}_{1}}{{B}_{1}}+\frac{4{{c}^{4}}}{3}{{A}_{2}}{{B}_{2}} \right) {{u}_{\mu }}\left( {{\delta }_{\alpha \beta }}{{\delta }_{\gamma \mu }}+{{\delta }_{\alpha \gamma }}{{\delta }_{\beta \mu }}+{{\delta }_{\alpha \mu }}{{\delta }_{\beta \gamma }} \right) \nonumber \\&\quad =\frac{\partial }{\partial {{x}^{\gamma }}}\left( \frac{2{{c}^{4}}}{3}{{A}_{1}}{{B}_{1}}+\frac{4{{c}^{4}}}{3}{{A}_{2}}{{B}_{2}} \right) \left( {{u}_{_{\gamma }}}{{\delta }_{\alpha \beta }}+{{u}_{\beta }}{{\delta }_{\alpha \gamma }}+{{u}_{\alpha }}{{\delta }_{\beta \gamma }} \right) \end{aligned}$$
(92)

By setting \(\frac{2{{c}^{4}}}{3}{{A}_{1}}{{B}_{1}}+\frac{4{{c}^{4}}}{3}{{A}_{2}}{{B}_{2}} =a \rho \), (92) is calculated as below:

$$\begin{aligned}= & {} a\frac{\partial \rho {{u}_{_{\gamma }}}}{\partial {{x}^{\gamma }}}{{\delta }_{\alpha \beta }}+a\frac{\partial \rho {{u}_{\beta }}}{\partial {{x}^{\alpha }}}+a\frac{\partial \rho {{u}_{\alpha }}}{\partial {{x}^{\beta }}} \nonumber \\= & {} a\frac{\partial \rho {{u}_{_{\gamma }}}}{\partial {{x}^{\gamma }}}{{\delta }_{\alpha \beta }}+a\rho \frac{\partial {{u}_{\beta }}}{\partial {{x}^{\alpha }}}+a\frac{\partial \rho }{\partial {{x}^{\alpha }}}{{u}_{\beta }}+a\rho \frac{\partial {{u}_{\alpha }}}{\partial {{x}^{\beta }}}+a\frac{\partial \rho }{\partial {{x}^{\beta }}}{{u}_{\alpha }} \end{aligned}$$
(93)

Substituting this calculated tensor into (91) yields,

$$\begin{aligned}= & {} {{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\left( \left( a-{{C}_{s}}^{2} \right) \frac{\partial \rho }{\partial {{x}^{\alpha }}}{{u}_{\beta }}+\left( a-{{C}_{s}}^{2} \right) \frac{\partial \rho }{\partial {{x}^{\beta }}}{{u}_{\alpha }}+\left( a-{{C}_{s}}^{2} \right) \frac{\partial \rho {{u}_{\gamma }}}{\partial {{x}^{\gamma }}}{{\delta }_{\alpha \beta }} \right) \nonumber \\&- {{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\frac{\partial \rho {{u}_{\gamma }}{{u}_{\alpha }}{{u}_{\beta }}}{\partial {{x}^{\gamma }}} \nonumber \\&+a{{\delta }_{t}}\left( \varphi -\frac{1}{2} \right) \frac{\partial }{\partial {{x}^{\beta }}}\rho \left( \frac{\partial {{u}_{\alpha }}}{\partial {{x}^{\beta }}}+\frac{\partial {{u}_{\beta }}}{\partial {{x}^{\alpha }}} \right) \end{aligned}$$
(94)

Then, by setting \(a=C_{s}^{2}\), we confirm that the the Navier–Stokes equation (28) can be derived.

By the way, about the D2Q9 equilibrium distribution functions (25), we can confirm \(a=C_{s}^{2}\) directly. From the equilibrium distribution functions (25), \({{A}_{1}}{{B}_{1}}\), \({{A}_{2}}{{B}_{2}}\) are defined by (95), (96).

$$\begin{aligned} {{A}_{1}}{{B}_{1}}= & {} \frac{\rho }{9C_{s}^{2}} \end{aligned}$$
(95)
$$\begin{aligned} {{A}_{2}}{{B}_{2}}= & {} \frac{\rho }{36C_{s}^{2}} \end{aligned}$$
(96)

From these relations, \(a=C_{s}^{2}\) is confirmed as below.

$$\begin{aligned} \frac{2{{c}^{4}}}{3}{{A}_{1}}{{B}_{1}}+\frac{4{{c}^{4}}}{3}{{A}_{2}}{{B}_{2}}=C_{s}^{2}\rho =a \rho \end{aligned}$$
(97)

As a matter of fact, (97) is a necessary condition in order to construct the equilibrium distribution functions [26].

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Yamamoto, K., Seta, T. Derivation of Multicomponent Lattice Boltzmann Equations by Introducing a Nonequilibrium Distribution Function into the Maxwell Iteration Based on the Convective Scaling. J Stat Phys 182, 4 (2021). https://doi.org/10.1007/s10955-020-02686-x

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