Boltzmann entropy code

[Test]

1. Calculating Q

1.1 1d illustration

Goal: $$ Q_{k}=\sum_{l=-n}^{n} \hat{f}{l} \hat{f}{k-l} \hat{\beta}(l, k-l),\quad k = -n,\dots,n $$

Method	description
Classical way	$f$ 标准顺序存储，$\hat{\beta}$ 按$(l,k)$二维数组存储
duplicate $f$	$f$ 按2倍的fftw3顺序存储，$\hat{\beta}$ 按$(l,k)$二维数组存储
duplicate $f$ and rearrange $B$	$f$ 按2倍的fftw3顺序存储，$\hat{\beta}$ 按$(l,k-l)$二维数组存储

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3d example

1d数组+行或列存储	i->j（f[n-j]）	j->i（f[i]）+2层	$3^3$倍f
n=$25^3$	0.33	0.32	0.24
n=$27^3$	0.53(4.82)	0.51(4.81)	0.38(4.56)
n=$29^3$	0.81	0.77	0.58
n=$31^3$	1.29	1.15	0.86
n=$33^3$	1.77	1.75	1.24
n=$35^3$	2.50	2.43	1.72
	100%	>96%	70%

2. Evaluation of B

[G. Dimarco and L. Pareschi 2014] Section 5.1 $$ \hat{Q}{k}=\sum{l=-N }^{N} \hat{f}{l} \hat{f}{k-l} \hat{\beta}(l, k-l), \quad k=-n, \ldots, n $$ where the Boltzmann kernel modes $\hat{\beta}(l, m)=\hat{B}(l, m)-\hat{B}(m, m)$ are now given by $$ \hat{B}(l, m)=\int_{\mathcal{B}{0}(2 \lambda \pi)} \int{\mathbb{S}^{2}}|q| \sigma(|q|, \cos \theta) e^{-i\left(l \cdot q^{+}+m \cdot q^{-}\right)} d \omega d q $$ In the VHS case, $|q| \sigma(|q|, \cos \theta)=C_{\alpha}|q|^{\alpha}$ where we can chose $C_{\alpha}=\left((4 \pi)^{2}(2 \lambda \pi)^{3+\alpha}\right)^{-1}$. Furthermore, the integral can be reduces to a one-dimensional integral (Pareschi and Russo 2000b) $$ \hat{B}(l, m)=\int_{0}^{1} r^{2+\alpha} \operatorname{Sinc}(\xi r) \operatorname{Sinc}(\eta r) d r=F_{\alpha}(\xi, \eta) $$ where $\xi=|l+m| \lambda \pi, \eta=|l-m| \lambda \pi$, $\lambda = 2/(3+\sqrt{2})$. We report the expressions for the case of Maxwell molecules $\alpha = 0$ and hard spheres $\alpha = 1$ $$ \begin{aligned} F_{0}(\xi, \eta) &=\frac{p \sin (q)-q \sin (p)}{2 \xi \eta p q} \ F_{1}(\xi, \eta) &=\frac{q \sin (q)+\cos (q)}{2 \xi \eta q^{2}}-\frac{p \sin (p)+\cos (p)}{2 \xi \eta p^{2}}-\frac{2}{p^{2} q^{2}} \end{aligned} $$ where $p=(\xi+\eta), q=(\xi-\eta)$.

Special cases $$ \begin{aligned} F_0(\xi,0) &= \frac{-\xi \cos (\xi) + \sin (\xi)}{\xi^3}\ F_0(0,0) &= \frac{1}{3}\ F_0(\xi,\xi) &= \frac{\xi-\cos(\xi)\sin(\xi)}{2\xi^3}\ F_1(\xi,0) &= \frac{-2-(-2+\xi^2)\cos(\xi) + 2\xi \sin (\xi)}{\xi^4}\ F_1(0,0) &= \frac{1}{4}\ F_1(\xi,\xi) &= -\frac{-1-2\xi^2+\cos(2\xi) + 2\xi \sin (2\xi)}{8\xi^4} \end{aligned} $$

3. Runge--Kutta method

1. initial guess
2. fftw3 interface
3. 1rk-4
4. nofiltering: for smooth case, it enjoys higher accuracy
5. change to array for faster evaluation

4. Experiments

Remarks

1. $J_{max} = N/2 + c, c < N/2$

 ./testBKW3D 17 0.000855 0.5 490

• entropy tu
• 比较误差 dt 取非常小，
• 不加entropy fix

Setup

Forward Euler	Detail
$T_\mathrm{END}$	0.08
$\Delta t_{fe}=\frac{1-\exp (-\beta \Delta t_\mathrm{cfe})}{\beta }$	0.0007