Exact equilibrium approach in the BGK model

Assume Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$: every $\mathbb{N}$-indexed family of nonempty sets has a choice function. Let $v_{\ast }=1\mathrm{m}{\mathrm{s}}^{-1}$ and ${\omega }_3(v)={\sum }_{j=0}^3(\mathrm{\mid }v\mathrm{\mid }\mathrm{/}{v}_{\ast }{)}^j$. Let $f_0:{\mathbb{R}}^3\to [0,\mathrm{\infty })$ be a measurable velocity number density, with $f_0$ in ${\mathrm{s}}^3{\mathrm{m}}^{-6}$ and $f_0{d}^{3}v$ measured in ${\mathrm{m}}^{-3}$, satisfying $\int {\omega }_3{f}_0{d}^{3}v<\mathrm{\infty }$, number density $n_0=\int f_0{d}^{3}v>0$, positive centered second moment, and finite reference-relative entropy $\int f_0\mathrm{\mid }\log (f_0\mathrm{/}{f}_{\mathrm{r}\mathrm{e}\mathrm{f}})\mathrm{\mid }{d}^{3}v$, where $f_{\mathrm{r}\mathrm{e}\mathrm{f}}>0$ is constant in the same units as $f_0$. Take the primitive particle mass $m>0$ in kilograms and Boltzmann constant $k_B>0$ in joules per kelvin. Define $u_0=n_0^{-1}\int v{f}_0{d}^{3}v$ and $T_0=m(3k_B{n}_0{)}^{-1}\int \mathrm{\mid }v-u_0{\mathrm{\mid }}^2{f}_0{d}^{3}v>0$. For any nonnegative velocity density $g$ with finite ${\omega }_3$ moment, number density $n[g]=\int g{d}^{3}v>0$, and positive centered second moment, define $u[g]=n[g{]}^{-1}\int vg{d}^{3}v$, $T[g]=m(3k_{B}n[g]{)}^{-1}\int \mathrm{\mid }v-u[g]{\mathrm{\mid }}^{2}g{d}^{3}v$, and its normalized moment-matched Maxwellian

$M[g](v)=n[g]{\left(\frac{m}{2\pi k_{B}T[g]}\right)}^{3\mathrm{/}2}\exp (-\frac{m\mathrm{\mid }v-u[g]{\mathrm{\mid }}^2}{2k_{B}T[g]})\mathrm{.}$

Write $M=M[f_0]$; it has the same number, momentum, and kinetic-energy moments as $f_0$. For $\tau >0$ in seconds and dimensionless $\epsilon >0$, consider the homogeneous equation

${\mathrm{\partial }}_t{f}^{\epsilon }(v,t)=\frac{M[f^{\epsilon }](v,t)-f^{\epsilon }(v,t)}{\epsilon \tau },f^{\epsilon }(v,0)=f_0(v),t\ge 0.$

There is a nonnegative solution $f^{\epsilon }\in C^1([0,\mathrm{\infty });L^1({\omega }_3{d}^{3}v))$ given by

$f^{\epsilon }(v,t)=M(v)+e^{-t\mathrm{/}(\epsilon \tau )}(f_0(v)-M(v))\mathrm{.}$

It solves the equation in $L^1({\omega }_3{d}^{3}v)$ and preserves the number, momentum, and kinetic-energy moments of $f_0$. It is also the unique solution in the following pointwise class: a jointly measurable $\tilde{f}:{\mathbb{R}}^3\times [0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with finite ${\omega }_3$-weighted velocity integral at every time and initial trace $\tilde{f}(v,0)=f_0(v)$ for almost every $v$, whose velocity sections are continuous on $[0,\mathrm{\infty })$ and differentiable on $(0,\mathrm{\infty })$ for almost every $v$. Require a common full-measure set of velocities on which the equation holds for every $t>0$, and, on each compact time interval, an integrable ${\omega }_3$-weighted majorant for ${\mathrm{\partial }}_t\tilde{f}$ independent of $t$. Also require the density and temperature of $\tilde{f}$ to remain positive. These envelopes permit differentiation of the three moment integrals and justify the conservation argument in the proof.

For each $T_{\ast }>0$, with the dimensionless fixed velocity weight $\mathrm{\parallel }g{\mathrm{\parallel }}_{1,3}={\int }_{{\mathbb{R}}^3}(1+\mathrm{\mid }v\mathrm{/}{v}_{\ast }{\mathrm{\mid }}^3)\mathrm{\mid }g(v)\mathrm{\mid }{d}^{3}v$, which has units ${\mathrm{m}}^{-3}$ for velocity densities $g$, it satisfies

$\mathrm{\parallel }{f}^{\epsilon }(t)-M{\mathrm{\parallel }}_{1,3}=e^{-t\mathrm{/}(\epsilon \tau )}\mathrm{\parallel }{f}_0-M{\mathrm{\parallel }}_{1,3},{\int }_0^{T_{\ast }}\mathrm{\parallel }{f}^{\epsilon }(t)-M{\mathrm{\parallel }}_{1,3}dt\le \epsilon \tau \mathrm{\parallel }{f}_0-M{\mathrm{\parallel }}_{1,3}\mathrm{.}$

Writing $H(g)={\int }_{{\mathbb{R}}^3}g\log (g\mathrm{/}{f}_{\mathrm{r}\mathrm{e}\mathrm{f}})d^{3}v$ with $0\log 0:=0$, which has units ${\mathrm{m}}^{-3}$ for velocity densities $g$, the entropy gap obeys

$0\le H(f^{\epsilon }(t))-H(M)\le e^{-t\mathrm{/}(\epsilon \tau )}(H(f_0)-H(M))\mathrm{.}$

Thus for each fixed $t>0$, $f^{\epsilon }(t)\to M$ in this weighted $L^1$ norm as $\epsilon \to 0$, while $f^{\epsilon }(0)=f_0$; if $f_0$ differs from $M$ on a set of positive measure, the initial mismatch is nonzero. If $f_0=M$ almost everywhere, the formula gives the stationary equilibrium with zero norm and entropy gaps. This is an exact homogeneous BGK model prediction and does not assert an inhomogeneous Boltzmann-to-Euler limit or an observed relaxation.

Countable Choice inherited from the kinetic measure and Gaussian-moment conventions

François Golse and C. David Levermore, Fluid Dynamics from Kinetic Equations, Lecture 2, “The BGK Operator”

David Tong, Lectures on Kinetic Theory, §2.5.1 (relaxation-time approximation and its non-exactness qualification)