Milestone 03

Author

Andreas Greiner, Lars Pastewka

Collision operator

Introduction

The Boltzmann transport equation (BTE) was given in the lecture:

\[\begin{equation} \frac{\partial f\left(\mathbf{r},\mathbf{v},t\right)}{\partial t}+\mathbf{v}\nabla_{\mathbf{r}} f\left(\mathbf{r},\mathbf{v},t\right) +\mathbf{a}\nabla_{\mathbf{v}} f\left(\mathbf{r},\mathbf{v},t\right)=C(f). \end{equation}\]

The l.h.s. of this equation is called the streaming term. The collision term on the r.h.s. represents the interactions between particles and usually is a complicated two particle scattering integral. We approximate this term by a relaxation time approximation. This approximation implies that the distribution function \(f\left(\mathbf{r},\mathbf{v},t\right)\) locally relaxes to an equilibrium distribution \(f^\mathrm{eq}\left(\mathbf{r},\mathbf{v},t\right)\). Viewing the streaming term as the total time derivative of the distribution function we can write the BTE as \[ \frac{d}{dt} f\left(\mathbf{r},\mathbf{v},t\right) = -\frac{f\left(\mathbf{r},\mathbf{v},t\right)- f^\mathrm{eq}\left(\mathbf{r},\mathbf{v},t\right)}{\tau} \tag{1}\] where \(\tau\) is the relaxation time constant.

What does this relaxation process look like? For the sake of clarity we imagine for a moment the distribution function to be independent of \(\mathbf{r}(t)\) and \(\mathbf{v}(t)\). Let us take the discrete form to read \[ f_i(\mathbf{r}+\mathbf{c}_i\Delta t,t+\Delta t) = f_i(\mathbf{r},t)+\omega\left(f_i^{eq}(\mathbf{r},t)-f_i(\mathbf{r},t)\right) \tag{2}\] The remaining question is how does \(f_i^{eq}(\mathbf{r})\) look like. Observe that the equilibrium is a local one, i.e. it depends on local variables. As given in the lecture these are the local density \(\rho(\mathbf{r})\) and the local average velocity \(\mathbf{u}(\mathbf{r})\). To this end you first have to calculate these two local quantities

\[ \rho(\mathbf{r})=\sum_i f_i \tag{3}\]

\[ \mathbf{u}(\mathbf{r})= \frac{1}{\rho(\mathbf{r})}\sum_i \mathbf{c}_i f_i(\mathbf{r}) \tag{4}\]

And the equilibrium distribution function is given as \[ f_i^\mathrm{eq}(\rho,\mathbf{u})=w_i\rho(\mathbf{r}) \left[1 +3\mathbf{c}_i\cdot\mathbf{u} +\frac{9}{2}\left(\mathbf{c}_i\cdot\mathbf{u}\right)^2 -\frac{3}{2}|\mathbf{u}|^2 \right] \tag{5}\] where \(w_i\) for the D2Q9 lattice is as follows: \[ \mathbf{w} = \left(\dfrac{4}{9}, \dfrac{1}{9}, \dfrac{1}{9}, \dfrac{1}{9}, \dfrac{1}{9}, \dfrac{1}{36}, \dfrac{1}{36}, \dfrac{1}{36}, \dfrac{1}{36}\right) \tag{6}\]

Tasks

So how do you run dynamic simulations in practice?

  1. After the streaming step recalculate \(\rho\) from Equation 3.
  2. Calculate \(\mathbf{u}\) from Equation 4.
  3. Compute the new \(f_i(\mathbf{r},t)\) using Equation 2. What does \(\omega\) stand for? Discretize Equation 1 using finite differences to arrive at Equation 2.

Let us run two tests:

  • Create a uniform density on your grid and set the density to a slightly higher value at the center. What happens?
  • Choose an initial distribution of \(\rho(\mathbf{r})\) and \(\mathbf{u}(\mathbf{r})\) at \(t=0\). Observe what happens dynamically as well as in the long time limit \(t\rightarrow\infty\).

Notes

  • Choose \(\omega\) with care, i.e. \(0<\omega<2\).
  • Choose \(0<\rho<1\) and \(|\mathbf{u}|<0.1\).
  • You need to set \(\rho\) and \(\mathbf{u}\) as initial conditions before the first streaming step.