Equilibrium Distribution Function The equilibrium distribution function is defined as:
$$ f_a^{eq}(x)=w_a\rho [1+\frac{3}{c^2}(e_a.v)+\frac{9}{2c^2}(e_a.v)^2-\frac{3}{2c^2}(v.v)] $$
which can be expanded to:
$$ f_0^{eq}=w_0\rho [1-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_1^{eq}=w_1\rho [1+\frac{3}{c^2}v_x+\frac{9}{2c^2}v_x^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_2^{eq}=w_2\rho [1+\frac{3}{c^2}v_y+\frac{9}{2c^2}v_y^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_3^{eq}=w_3\rho [1-\frac{3}{c^2}v_x+\frac{9}{2c^2}v_x^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_4^{eq}=w_4\rho [1-\frac{3}{c^2}v_y+\frac{9}{2c^2}v_y^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_5^{eq}=w_5\rho [1+\frac{3}{c^2}(v_x+v_y)+\frac{9}{2c^2}(v_x+v_y)^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_6^{eq}=w_6\rho [1+\frac{3}{c^2}(-v_x+v_y)+\frac{9}{2c^2}(-v_x+v_y)^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_7^{eq}=w_7\rho [1+\frac{3}{c^2}(-v_x-v_y)+\frac{9}{2c^2}(-v_x-v_y)^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
$$ f_8^{eq}=w_8\rho [1+\frac{3}{c^2}(v_x-v_y)+\frac{9}{2c^2}(v_x-v_y)^2-\frac{3}{2c^2}(v_x^2+v_y^2)] $$
where $w_0$ is $\frac{4}{9}$ , $w_1$ to $w_4$ are $\frac{1}{9}$ , and $w_5$ to $w_8$ are $\frac{1}{36}$ .
The velocities in the equations above can be calculated using the following formulation:
$$ v_x=f_1+f_5+f_8-f_3-f_6-f_7 $$
$$ v_y=f_2+f_5+f_6-f_4-f_7-f_8 $$