Morphodynamic shallow layer equations featuring bed load and suspended sediment with lattice Boltzmann method

Abstract

Different coupled systems for the shallow water equation, bed elevation, and suspended load equation are proposed until this day. The main differences come from the physical viewpoints, which caused some distinctions in the models. Recently, a coupled shallow water system of equations over an erodible bed has been proposed, in which the water layer, bed morphodynamics, and suspended sediments are interacting with each other. This system possesses a term in the mass conservation equation that couples the water depth and the bed level in the equilibrium distribution function required by lattice Boltzmann method (LBM). In this paper, the main goal is to utilize an advanced LBM to solve this system of equations. Besides solving the bed morphological equation by LBM, another simple and explicit scheme (like LBM) is proposed to investigate the ability of LBM. As the second goal, a practical approach is developed for applying so-called open boundary condition that relaxes the solution onto a prescribed equilibrium flow.

Appendix A

1.1 A.1: Chapman–Enskog analysis of the modified shallow water equation

From Eq. (8), the moments of equilibrium function can be obtained as follows:

$$\begin{aligned} \begin{aligned}&\sum _{\alpha }^{}\, f\,^{\text {eq}}_\alpha = h + z,\\&\sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i \,\, f\,^{\text {eq}}_\alpha = h {\textbf{u}}_i,\\&\sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j\,\, f\,^{\text {eq}}_\alpha = h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij},\\&\sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j {{\textbf{e}}_\alpha }_k \,\, f\,^{\text {eq}}_\alpha = \frac{{\text {e}}\,^2}{3}h\left( {\textbf{u}}_i \delta _{jk} + {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) . \end{aligned} \end{aligned}$$

(24)

Also, mass and momentum of shallow water equation are conservative, so we have

$$\begin{aligned} \sum _{\alpha }^{} f_\alpha = h + z, \quad \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i \, f_\alpha = h {\textbf{u}}_i. \end{aligned}$$

(25)

Using Eqs. (5) and (7) gives the moments of force term as follows:

$$\begin{aligned} \begin{aligned} \sum _{\alpha }^{} q^{shw}_\alpha&= \sum _{\alpha }^{} \frac{1}{6{\text {e}}\,^2} {{\textbf{e}}_\alpha }_i \mathbf {{\mathcal {S}}}\,^{shw}_i = 0,\\ \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i q^{shw}_\alpha&= \sum _{\alpha }^{} \frac{1}{6{\text {e}}\,^2} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j \mathbf {{\mathcal {S}}}\,^{shw}_j = \mathbf {{\mathcal {S}}}\,^{shw}_i,\\ \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j q^{shw}_\alpha&= \sum _{\alpha }^{} \frac{1}{6{\text {e}}\,^2} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j {{\textbf{e}}_\alpha }_k \mathbf {{\mathcal {S}}}\,^{shw}_k = 0. \end{aligned} \end{aligned}$$

(26)

The multi-scale Chapman–Enskog expansion [12, 29, 43] is a tool to show that the lattice Boltzmann equation (6) with conditions (24)–(26) will recover (1a). Also, it helps finding $\tau _f$ and error term too. This analysis was used for classical shallow water equation to create a discretized equilibrium distribution function required in LBM [29, 43].

Knowing $\Delta x = e \Delta t$, helps rewriting (6) as follows:

$$\begin{aligned} \begin{aligned}&f_\alpha ({\textbf{x}} + {\textbf{e}}_\alpha \Delta t, t + \Delta t) - f_\alpha ({\textbf{x}}, t)\\&\quad = -\frac{\Delta t}{\tau _f} \, (f_\alpha ({\textbf{x}}, t)- f_\alpha ^{\text {eq}} ({\textbf{x}}, t)) \\&\qquad + \Delta t q^{shw}_\alpha \left({\textbf{x}} + \frac{1}{2} {\textbf{e}}_\alpha \Delta t,t +\frac{1}{2} \Delta t\right), \end{aligned} \end{aligned}$$

(27)

and using Taylor expansion around arbitrary point $({\textbf{x}},t)$ gives

$$\begin{aligned} \begin{aligned}&\sum _{n=1}^{\infty } \frac{{\Delta t}^n}{n!} (\partial _t + {{\textbf{e}}_\alpha }_i \partial _i)^n f _\alpha ({\textbf{x}},t) \\&\quad = -\frac{\Delta t}{\tau _f} (f_\alpha ({\textbf{x}}, t)- f_\alpha ^{\text {eq}} ({\textbf{x}}, t)) \\&\qquad + \sum _{n=0}^{\infty } \frac{{\Delta t}^{n+1}}{2^n n!} (\partial _t + {{\textbf{e}}_\alpha }_i \partial _i)^n q^{shw}_\alpha ({\textbf{x}},t). \end{aligned} \end{aligned}$$

(28)

In Chapman–Enskog expansion literature, we assume that

$$\begin{aligned} \partial _t = \varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2}, \quad \partial _i = \varepsilon \partial _{i_1}, \quad q^{shw}_\alpha = \varepsilon q^{{(1)}^{shw}}_\alpha , \end{aligned}$$

(29)

and also, we consider the expansion of $f_\alpha$ around $f ^{(0)} _\alpha = f ^{\text {eq}} _\alpha$ as follows:

$$\begin{aligned} f_\alpha = f ^{(0)} _\alpha + \varepsilon f ^{(1)} _\alpha + \varepsilon ^2 f ^{(2)} _\alpha + \cdots , \end{aligned}$$

(30)

where $\varepsilon$ is usually assumed as the order of Knudsen number. Equations (24), (25), and (30), give

$$\begin{aligned} \begin{array}{ll} \sum _{\alpha }^{}\, f\,^{(n)}_\alpha = 0,&\sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i\, f\,^{(n)}_\alpha = 0, \end{array} \end{aligned}$$

(31)

where $n=1,2,3,\ldots$.

Substituting Eqs. (29) and (30) into Eq. (28), results

$$\begin{aligned} \begin{aligned}&\sum _{n=1}^{\infty } \left( \frac{{\Delta t}^n}{n!} (\varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2} + \varepsilon {{\textbf{e}}_\alpha }_i \partial _{i_1})^n \right) \left( \sum _{k=0}^{\infty } {\varepsilon }^k f ^{(k)} _\alpha \right) \\&\quad = -\frac{\Delta t}{\tau _f} \left( \sum _{k=1}^{\infty } {\varepsilon }^k f ^{(k)} _\alpha \right) \\&\qquad + \varepsilon \sum _{n=0}^{\infty } \left( \frac{{\Delta t}^{n+1}}{2^n n!} (\varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2} + \varepsilon {{\textbf{e}}_\alpha }_i \partial _{i_1})^n \right) q^{{(1)}^{shw}}_\alpha , \end{aligned} \end{aligned}$$

(32)

and rounding Eq. (32) up to the second-order yields

$$\begin{aligned} \begin{aligned}&\varepsilon \left( \Delta t {\mathcal {D}} f ^{(0)} _\alpha + \frac{\Delta t}{\tau _f} f ^{(1)} _\alpha - \Delta t q^{{(1)}^{shw}}_\alpha \right) \\&\quad + \varepsilon ^2 \left( \left( \Delta t \partial _{t_2} + \frac{{\Delta t}^2}{2} {\mathcal {D}}\,^2 \right)\, f\,^{(0)}_\alpha + \Delta t {\mathcal {D}} f ^{(1)} _\alpha \right. \\&\left. \quad + \frac{\Delta t}{\tau _f} f ^{(2)} _\alpha - \frac{{\Delta t}^2}{2} {\mathcal {D}} q^{{(1)}^{shw}}_\alpha \right) = O({\varepsilon }^3), \end{aligned} \end{aligned}$$

where ${\mathcal {D}} = \left( \partial _{t_1} + {{\textbf{e}}_\alpha }_i \partial _{i_1} \right)$.

So the first-order of Chapman–Enskog expansion for LBE is as follows:

$$\begin{aligned} \Delta t {\mathcal {D}} f ^{(0)} _\alpha + \frac{\Delta t}{\tau _f} f ^{(1)} _\alpha - \Delta t q^{{(1)}^{shw}}_\alpha , \end{aligned}$$

(33)

and the second-order is

$$\begin{aligned}{} & {} \Delta t \partial _{t_2}\, f\,^{(0)}_\alpha + \frac{{\Delta t}^2}{2} {\mathcal {D}}\,^2\, f\,^{(0)}_\alpha + \Delta t {\mathcal {D}} f ^{(1)} _\alpha \nonumber \\{} & {} \quad + \frac{\Delta t}{\tau _f} f ^{(2)} _\alpha - \frac{{\Delta t}^2}{2} {\mathcal {D}} q^{{(1)}^{shw}}_\alpha . \end{aligned}$$

(34)

For sake of getting the second-order accuracy, we need to assume Eqs. (33) and (34) to be zero. So they can be written as

$$\begin{aligned}{} & {} {\mathcal {D}} f ^{(0)} _\alpha + \frac{1}{\tau _f} f ^{(1)} _\alpha - q^{{(1)}^{shw}}_\alpha =0, \end{aligned}$$

(35)

$$\begin{aligned}{} & {} \partial _{t_2}\, f\,^{(0)}_\alpha + \frac{{\Delta t}}{2} {\mathcal {D}}\,^2\, f\,^{(0)}_\alpha + {\mathcal {D}} f ^{(1)} _\alpha \nonumber \\{} & {} \quad + \frac{1}{\tau _f} f ^{(2)} _\alpha - \frac{{\Delta t}}{2} {\mathcal {D}} q^{{(1)}^{shw}}_\alpha =0. \end{aligned}$$

(36)

Finding $f ^{(1)} _\alpha$ via Eq. (35) yields

$$\begin{aligned} f ^{(1)} _\alpha = -\tau _f \left( {\mathcal {D}} f ^{(0)} _\alpha - q^{{(1)}^{shw}}_\alpha \right) , \end{aligned}$$

(37)

and substituting it in Eq. (36) gives

$$\begin{aligned}{} & {} \partial _{t_2}\, f\,^{(0)}_\alpha - \left( \tau _f - \frac{\Delta t}{2}\right) {\mathcal {D}}\,^2\, f\,^{(0)}_\alpha \nonumber \\{} & {} \quad + \left( \tau _f - \frac{\Delta t}{2}\right) {\mathcal {D}} q^{{(1)}^{shw}}_\alpha + \frac{1}{\tau _f} f ^{(2)} _\alpha = 0. \end{aligned}$$

(38)

Substituting ${\mathcal {D}}$ into Eqs. (35) and (38) and also doing some rearranging give

$$\begin{aligned}&\partial _{t_1}\, f\,^{(0)}_\alpha + \partial _{i_1} {{\textbf{e}}_\alpha }_i\, f\,^{(0)}_\alpha = -\frac{1}{\tau _f} f ^{(1)} _\alpha + q^{{(1)}^{shw}}_\alpha , \end{aligned}$$

(39)

$$\begin{aligned}&\partial _{t_2}\, f\,^{(0)}_\alpha = -\frac{1}{\tau _f} f ^{(2)} _\alpha + \left(\tau _f - \frac{\Delta t}{2}\right) \left[ - \left( \partial _{t_1} q^{{(1)}^{shw}}_\alpha + \partial _{i_1} {{\textbf{e}}_\alpha }_i q^{{(1)}^{shw}}_\alpha \right) \right. \nonumber \\&\quad + \partial _{t_1} \left( \partial _{t_1}\, f\,^{(0)}_\alpha + \partial _{j_1} {{\textbf{e}}_\alpha }_j\, f\,^{(0)}_\alpha \right) \nonumber \\&\quad \left. + \partial _{i_1} \left( \partial _{t_1} {{\textbf{e}}_\alpha }_i\, f\,^{(0)}_\alpha + \partial _{j_1} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j\,\, f\,^{(0)}_\alpha \right) \right] . \end{aligned}$$

(40)

Calculating $\sum _{\alpha }^{} \left( \text {Eq. } (39)\right)$, $\sum _{\alpha }^{} \left( {{\textbf{e}}_\alpha }_i \text {Eq. }(39)\right)$, $\sum _{\alpha }^{} \left( \text {Eq. }(40)\right)$, and $\sum _{\alpha }^{} \left( {{\textbf{e}}_\alpha }_i \text {Eq. }(40)\right)$ using Eqs. (26) and (31) concludes

$$\begin{aligned}&\partial _{t_1} {\mathcal {F}}\,^{(0)}_0 + \partial _{i_1} {\mathcal {F}}\,^{(0)}_{i} = 0, \end{aligned}$$

(41)

$$\begin{aligned}&\partial _{t_1} {\mathcal {F}}\,^{(0)}_i + \partial _{i_1} {\mathcal {F}}\,^{(0)}_{ij} = \frac{1}{\varepsilon }\mathbf {{\mathcal {S}}}\,^{shw}_i, \end{aligned}$$

(42)

$$\begin{aligned}&\partial _{t_2} {\mathcal {F}}\,^{(0)}_0 = 0, \end{aligned}$$

(43)

$$\begin{aligned}&\partial _{t_2} {\mathcal {F}}\,^{(0)}_i = \left( \tau _f - \frac{\Delta t}{2}\right) \partial _{j_1} \left( \partial _{t_1} {\mathcal {F}}\,^{(0)}_{ij} + \partial _{k_1} {\mathcal {F}}\,^{(0)}_{ijk} \right) , \end{aligned}$$

(44)

where

$$\begin{aligned} \begin{aligned} {\mathcal {F}}\,^{(0)}_0&= \sum _{\alpha }^{}\, f\,^{(0)}_\alpha , \quad {\mathcal {F}}\,^{(0)}_i = \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i\, f\,^{(0)}_\alpha , \\ {\mathcal {F}}\,^{(0)}_{ij}&= \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j\, f\,^{(0)}_\alpha , \quad {\mathcal {F}}\,^{(0)}_{ijk} = \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j {{\textbf{e}}_\alpha }_k\, f\,^{(0)}_\alpha . \end{aligned} \end{aligned}$$

From Eq. (24), we obtain

$$\begin{aligned} \begin{aligned}&{\mathcal {F}}\,^{(0)}_0 = h + z,{} & {} {\mathcal {F}}\,^{(0)}_i = h {\textbf{u}}_i,\\&{\mathcal {F}}\,^{(0)}_{ij} = h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij},{} & {} {\mathcal {F}}\,^{(0)}_{ijk} = \frac{{\text {e}}\,^2}{3}h\left( {\textbf{u}}_i \delta _{jk} + {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) , \end{aligned} \end{aligned}$$

and substituting them into Eqs. (41)–(44), gives

$$\begin{aligned}{} & {} \partial _{t_1} ( h + z ) + \partial _{i_1} ( h {\textbf{u}}_i ) = 0, \end{aligned}$$

(45)

$$\begin{aligned}{} & {} \partial _{t_1} ( h {\textbf{u}}_i ) + \partial _{i_1} \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) = \frac{1}{\varepsilon } \mathbf {{\mathcal {S}}}\,^{shw}_i, \end{aligned}$$

(46)

$$\begin{aligned}{} & {} \partial _{t_2} ( h + z ) = 0, \end{aligned}$$

(47)

$$\begin{aligned}{} & {} \partial _{t_2} ( h {\textbf{u}}_i ) = \left(\tau _f - \frac{\Delta t}{2}\right) \partial _{j_1} \left( \partial _{t_1} \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \right. \nonumber \\{} & {} \left. \quad + \partial _{k_1} \left( \frac{{\text {e}}\,^2}{3}h\left( {\textbf{u}}_i \delta _{jk} + {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) \right) \right) . \end{aligned}$$

(48)

To recover Eq. (1a), we should calculate

$$\begin{aligned} \varepsilon \times (\text {Eq. } (45)) + \varepsilon ^2 \times (\text {Eq. } (47)), \end{aligned}$$

and

$$\begin{aligned} \varepsilon \times (\text {Eq. } (46)) + \varepsilon ^2 \times (\text {Eq. } (48)), \end{aligned}$$

which give

$$\begin{aligned}&\partial _t ( h + z ) + \partial _i ( h {\textbf{u}}_i ) = 0, \end{aligned}$$

(49)

$$\begin{aligned}&\partial _t ( h {\textbf{u}}_i ) + \partial _i \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) = -gh\left( \partial _i z + S_f^i \right) \nonumber \\&\quad + \left(\tau _f - \frac{\Delta t}{2}\right) \partial _j \left( \varepsilon \partial _{t_1} \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \right. \nonumber \\&\left. \quad + \partial _k \left( \frac{{\text {e}}\,^2}{3}h \left( {\textbf{u}}_i \delta _{jk} + {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) \right) \right) . \end{aligned}$$

(50)

Equation (1a) has been recovered with Eq. (49) but the second one contains some extra terms comparing with Eq. (50). So we need to do some rearranging on Eq. (50) to recover the second equation of shallow water and also find the error term. So rewriting Eq. (50), gives

$$\begin{aligned} \begin{aligned}&\partial _t ( h {\textbf{u}}_i ) + \partial _i \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \\&\quad = \left( \frac{{\text {e}}\,^2}{3} (\tau _f - \frac{\Delta t}{2}) \right) \partial _j \partial _j \left( h {\textbf{u}}_i \right) \\&\qquad -gh\left( \partial _i z + S_f^i \right) \\&\qquad + \left( \frac{{\text {e}}\,^2}{3} (\tau _f - \frac{\Delta t}{2}) \right) \partial _j \left( \frac{3 \varepsilon }{{\text {e}}\,^2} \partial _{t_1} \left( h {\textbf{u}}_i {\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \right. \\&\left. \qquad + \partial _k \left( h \left( {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) \right) \right) . \end{aligned} \end{aligned}$$

(51)

Let

$$\begin{aligned} \nu _h = \frac{{\text {e}}\,^2}{3} \left( \tau _f - \frac{\Delta t}{2} \right) , \end{aligned}$$

(52)

then Eq. (51) yields

$$\begin{aligned}{} & {} \partial _t ( h {\textbf{u}}_i ) + \partial _i \left( h{\textbf{u}}_i{\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \nonumber \\{} & {} \quad = \nu _h \partial _j \partial _j \left( h {\textbf{u}}_i \right) -gh\left( \partial _i z + S_f^i \right) + \mathbf {{\mathcal {E}}}\,^{shw}, \end{aligned}$$

(53)

where

$$\begin{aligned} \mathbf {{\mathcal {E}}}\,^{shw}= & {} \nu _h \partial _j \left( \frac{3 \varepsilon }{{\text {e}}\,^2} \partial _{t_1} \left( h {\textbf{u}}_i {\textbf{u}}_j + \frac{g}{2} h^2 \delta _{ij} \right) \right. \\{} & {} \left. + \partial _k \left( h \left( {\textbf{u}}_j \delta _{ik} +{\textbf{u}}_k \delta _{ij} \right) \right) \right) , \end{aligned}$$

is the error term. So, Eq. (53) is recovered (1a) with the error term $O(\nu _h)$.

1.2 A.2: Chapman–Enskog analysis of the sediment suspended load

The moments of $g^{\text {eq}}_\alpha$ can be calculated from Eq. (13), which they are as follows:

$$\begin{aligned} \sum _{\alpha }^{} g^{\text {eq}}_\alpha= & {} hc,\quad \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i g^{\text {eq}}_\alpha = h c {\textbf{u}}_i,\quad \nonumber \\ \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j g^{\text {eq}}_\alpha= & {} c_s^2 h c \delta _{ij}. \end{aligned}$$

(54)

Also, the zeroth moment of $g_\alpha$ is as follows:

$$\begin{aligned} \sum _{\alpha }^{} g_\alpha = h c, \end{aligned}$$

(55)

and using Eqs. (5) and (12), we have

$$\begin{aligned} \sum _{\alpha }^{} q^{c}_\alpha = \mathbf {{\mathcal {S}}}\,^{c}, \quad \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i q^{c}_\alpha = 0, \end{aligned}$$

(56)

which they are the moments of $q^{c}_\alpha$. We use multi-scale Chapman–Enskog expansion [12, 43] to show that the second-order lattice Boltzmann formula (11) with assumptions Eqs. (12) and (13) can recover the convection–diffusion Equation (1b). By assuming $\Delta x = e \Delta t$, Eq. (11) can be written as follows:

$$\begin{aligned} \begin{aligned}&g_\alpha ({\textbf{x}} + {\textbf{e}}_\alpha \Delta t, t + \Delta t) - g_\alpha ({\textbf{x}}, t) \\&\quad = -\frac{\Delta t}{\tau _g} (g_\alpha ({\textbf{x}}, t)- g_\alpha ^{\text {eq}} ({\textbf{x}}, t)) \\&\qquad + \Delta t q^{c}_\alpha \left({\textbf{x}} + \frac{1}{2} {\textbf{e}}_\alpha \Delta t,t +\frac{1}{2} \Delta t\right), \end{aligned} \end{aligned}$$

(57)

and using Taylor expansion around $({\textbf{x}},t)$ gives

$$\begin{aligned} \begin{aligned}&\sum _{n=1}^{\infty } \frac{{\Delta t}^n}{n!} (\partial _t + {{\textbf{e}}_\alpha }_i \partial _i)^n g_\alpha ({\textbf{x}},t) \\&\quad = -\frac{\Delta t}{\tau _g} (g_\alpha ({\textbf{x}}, t)- g_\alpha ^{\text {eq}} ({\textbf{x}}, t)) \\&\qquad + \sum _{n=1}^{\infty } \frac{{\Delta t}^{n+1}}{2^n n!} (\partial _t + {{\textbf{e}}_\alpha }_i \partial _i)^n q^{c}_\alpha ({\textbf{x}},t). \end{aligned} \end{aligned}$$

(58)

In Chapman–Enskog expansion literature, we assume that

$$\begin{aligned} \partial _t = \varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2}, \quad \partial _i = \varepsilon \partial _{i_1}, \quad q^{c}_\alpha = \varepsilon q^{{(1)}^{c}}_\alpha , \end{aligned}$$

(59)

and also we consider the expansion of $g_\alpha$ around $g^{(0)} _\alpha = g^{\text {eq}} _\alpha$ as follows:

$$\begin{aligned} g_\alpha = g ^{(0)} _\alpha + \varepsilon g ^{(1)} _\alpha + \varepsilon ^2 g ^{(2)} _\alpha + \cdots , \end{aligned}$$

(60)

where $\varepsilon$ is usually assumed as the order of Knudsen number. Equations (54), (55) and (60), give

$$\begin{aligned} \sum _{\alpha }^{} g^{(n)}_\alpha = 0, \end{aligned}$$

(61)

where $n=1,2,3,\ldots$.

Substituting Eqs. (59) and (60) into Eq. (58), yields

$$\begin{aligned} \begin{aligned}&\sum _{n=1}^{\infty } \left( \frac{{\Delta t}^n}{n!} (\varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2} + \varepsilon {{\textbf{e}}_\alpha }_i \partial _{i_1})^n \right) \left( \sum _{k=0}^{\infty } {\varepsilon }^k g ^{(k)} _\alpha \right) \\&\quad = -\frac{\Delta t}{\tau _g} \left( \sum _{k=1}^{\infty } {\varepsilon }^k g ^{(k)} _\alpha \right) \\&\qquad + \varepsilon \sum _{n=0}^{\infty } \left( \frac{{\Delta t}^{n+1}}{2^n n!} (\varepsilon \partial _{t_1} +\varepsilon ^2 \partial _{t_2} + \varepsilon {{\textbf{e}}_\alpha }_i \partial _{i_1})^n \right) q^{{(1)}^{c}}_\alpha , \end{aligned} \end{aligned}$$

(62)

and with rounding Eq. (62) up to second-order of $\varepsilon ,$ we obtain

$$\begin{aligned} \begin{aligned}&\varepsilon \left( \Delta t {\mathcal {D}} g ^{(0)} _\alpha + \frac{\Delta t}{\tau _g} g ^{(1)} _\alpha - \Delta t q^{{(1)}^{c}}_\alpha \right) \\&\quad + \varepsilon ^2 \left( \left( \Delta t \partial _{t_2} + \frac{{\Delta t}^2}{2} {\mathcal {D}}\,^2 \right) g^{(0)}_\alpha + \Delta t {\mathcal {D}} g ^{(1)} _\alpha \right. \\&\left. \quad + \frac{\Delta t}{\tau _g} g ^{(2)} _\alpha - \frac{{\Delta t}^2}{2} {\mathcal {D}} q^{{(1)}^{c}}_\alpha \right) = O({\varepsilon }^3), \end{aligned} \end{aligned}$$

where ${\mathcal {D}} = \left( \partial _{t_1} + {{\textbf{e}}_\alpha }_i \partial _{i_1} \right)$. So to get the accuracy of second-order, we need to assume

$$\begin{aligned}&{\mathcal {D}} g ^{(0)} _\alpha + \frac{1}{\tau _g} g ^{(1)} _\alpha - q^{{(1)}^{c}}_\alpha =0, \end{aligned}$$

(63)

$$\begin{aligned}&\partial _{t_2} g^{(0)}_\alpha + \frac{\Delta t}{2} {\mathcal {D}}\,^2 g^{(0)}_\alpha + {\mathcal {D}} g ^{(1)} _\alpha + \frac{1}{\tau _g} g ^{(2)} _\alpha - \frac{{\Delta t}}{2} {\mathcal {D}} q^{{(1)}^{c}}_\alpha =0. \end{aligned}$$

(64)

Equations (63) and (64) lead to

$$\begin{aligned}&\partial _{t_1} g^{(0)}_\alpha + \partial _{i_1} {{\textbf{e}}_\alpha }_i g^{(0)}_\alpha = -\frac{1}{\tau _g} g ^{(1)} _\alpha + q^{{(1)}^{c}}_\alpha , \end{aligned}$$

(65)

$$\begin{aligned}&\partial _{t_2} g^{(0)}_\alpha = -\frac{1}{\tau _g} g ^{(2)} _\alpha + \left(\tau _g - \frac{\Delta t}{2}\right) \left[ - \left( \partial _{t_1} q^{{(1)}^{c}}_\alpha + \partial _{i_1} {{\textbf{e}}_\alpha }_i q^{{(1)}^{c}}_\alpha \right) \right. \nonumber \\&\quad ~ + \partial _{t_1} \left( \partial _{t_1} g^{(0)}_\alpha + \partial _{j_1} {{\textbf{e}}_\alpha }_j g^{(0)}_\alpha \right) \nonumber \\&\quad \left. + \partial _{i_1} \left( \partial _{t_1} {{\textbf{e}}_\alpha }_i g^{(0)}_\alpha + \partial _{j_1} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j g^{(0)}_\alpha \right) \right] . \end{aligned}$$

(66)

Calculating $\sum _{\alpha }^{}$ (Eq. (65)) and $\sum _{\alpha }^{}$ (Eq. (66)), with help of Eq. (56) and Eq. (61) gives

$$\begin{aligned}&\partial _{t_1} {\mathcal {G}}\,^{(0)}_0 + \partial _{i_1} {\mathcal {G}}\,^{(0)}_{i} = \frac{1}{\varepsilon } \mathbf {{\mathcal {S}}}\,^{c}, \end{aligned}$$

(67)

$$\begin{aligned}&\partial _{t_2} {\mathcal {G}}\,^{(0)}_0 = \left(\tau _g - \frac{\Delta t}{2}\right) \partial _{i_1} \left( \partial _{t_1} {\mathcal {G}}\,^{(0)}_{i} + \partial _{j_1} {\mathcal {G}}\,^{(0)}_{ij} \right) , \end{aligned}$$

(68)

where

$$\begin{aligned} {\mathcal {G}}\,^{(0)}_0= & {} \sum _{\alpha }^{} g^{(0)}_\alpha ,\quad {\mathcal {G}}\,^{(0)}_i = \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i g^{(0)}_\alpha ,~~~~~\\ {\mathcal {G}}\,^{(0)}_{ij}= & {} \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j g^{(0)}_\alpha , \end{aligned}$$

and from Eq. (54), we obtain

$$\begin{aligned} {\mathcal {G}}\,^{(0)}_0 = h c, \quad {\mathcal {G}}\,^{(0)}_i = h c {\textbf{u}}_i,\quad {\mathcal {G}}\,^{(0)}_{ij} = c_s^2 h c \delta _{ij}, \end{aligned}$$

and substituting them into Eqs. (67) and (68) gives

$$\begin{aligned}{} & {} \partial _{t_1} ( h c ) + \partial _{i_1} ( h c {\textbf{u}}_i ) = \frac{1}{\varepsilon } \mathbf {{\mathcal {S}}}\,^{c}, \end{aligned}$$

(69)

$$\begin{aligned}{} & {} \partial _{t_2} ( hc ) = \left(\tau _g - \frac{\Delta t}{2}\right) \partial _{i_1} \left( \partial _{t_1} (h c {\textbf{u}}_i) + \partial _{j_1} (c_s^2 h c \delta _{ij}) \right) . \end{aligned}$$

(70)

To recover Eq. (1b), we should calculate

$$\begin{aligned} \varepsilon \times \text {Eq. } (69) + \varepsilon ^2 \times \text {Eq. } (70), \end{aligned}$$

which gives

$$\begin{aligned} \partial _t ( h c ) + \partial _i ( h c{\textbf{u}}_i ) = E-D + \partial _i (D_c \partial _i (h c)) +{\mathcal {E}}\,^c, \end{aligned}$$

(71)

where

$$\begin{aligned} D_c = c_s^2 \left( \tau _g - \frac{\Delta t}{2}\right) , \end{aligned}$$

(72)

is the diffusion coefficient and ${\mathcal {E}}\,^c = (\tau _g - \frac{\Delta t}{2}) \partial _i \partial _{t_1} (\varepsilon h c {\textbf{u}}_i)$ is the error term. So Eq. (1b) is recovered from lattice Boltzmann equation.

1.3 A.3: Chapman–Enskog analysis of the morphological bed load

The moments of $l^{\text {eq}}_\alpha$ can be obtained from Eq. (21), which they are as follows:

$$\begin{aligned} \sum _{\alpha }^{} l^{\text {eq}}_\alpha= & {} z,\quad \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i l^{\text {eq}}_\alpha = \xi {\textbf{u}}_i,\quad \nonumber \\ \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i {{\textbf{e}}_\alpha }_j l^{\text {eq}}_\alpha= & {} c_s^2 z \delta _{ij}. \end{aligned}$$

(73)

Also, the zeroth moment of $l_\alpha$ (mass in convection–diffusion is conservative) is as follows:

$$\begin{aligned} \sum _{\alpha }^{} l_\alpha = z, \end{aligned}$$

(74)

and using Eqs. (5) and (20) concludes

$$\begin{aligned} \sum _{\alpha }^{} q^{z}_\alpha = \mathbf {{\mathcal {S}}}\,^{z}, \quad \sum _{\alpha }^{} {{\textbf{e}}_\alpha }_i q^{z}_\alpha = 0, \end{aligned}$$

(75)

which they are the moments of $q^{z}_\alpha$. We don’t mention the Chapman–Enskog analysis here, since it is the same as discussed in 1. But from it, the truncation error is

$$\begin{aligned} {\mathcal {E}}\,^z = \left( \tau _l - \frac{\Delta t}{2}\right) \partial _i \partial _{t_1} (\varepsilon \xi {\textbf{u}}_i). \end{aligned}$$