Cutoff elastic-reflection collision measure and gain — Examples

Define ${\mathrm{\Sigma }}_1:=\mathrm{diag}(\alpha ,\beta ,\gamma )$ and ${\mathrm{\Sigma }}_2:=\mathrm{diag}(\beta ,\alpha ,\gamma )$. Set $a_{\ast }:=\min \{\alpha ,\beta ,\gamma \}$ and $b_{\ast }:=\max \{\alpha ,\beta ,\gamma \}$. Since $\alpha ,\beta ,\gamma >0$ and $\alpha \mathrm{\ne }\beta$, one has $0<a_{\ast }<b_{\ast }$ and ${\mathrm{\Sigma }}_1,{\mathrm{\Sigma }}_2\in K_{a_{\ast },b_{\ast }}$. On any flat three-torus with positive periods and on $I=\mathbb{R}$, take $\mathbf{a}=0$ and $Q=0$. Define $F_j(r,v,t)=n_0{\gamma }_{{\mathrm{\Sigma }}_j,0}(v)$ using [F3]. Each $F_j$ is smooth and rapidly decreasing in velocity, is constant in $r,t$, and solves the kinetic equation with zero acceleration and zero source. Gaussian normalization follows from [F4] by scaling each coordinate, and [F5] permits the product integration.

Each density has number density $n_0$ and mean-velocity vector $u=0$ by normalization and oddness. In one coordinate with variance $s^2$, differentiating $x\exp (-x^2\mathrm{/}(2s^2))$ and integrating over $\mathbb{R}$ gives $\int x^2{e}^{-x^2\mathrm{/}(2s^2)}dx=s^2\int e^{-x^2\mathrm{/}(2s^2)}dx$; the boundary term vanishes because the Gaussian decays to zero. Applying this to each coordinate gives covariance matrices ${\mathrm{\Sigma }}_1$ and ${\mathrm{\Sigma }}_2$. Thus both have the same scalar internal energy density $e=m{n}_0(\alpha +\beta +\gamma )\mathrm{/}2$, while ${\mathrm{\Pi }}_1=m{n}_0\mathrm{diag}(\alpha ,\beta ,\gamma )$ and ${\mathrm{\Pi }}_2=m{n}_0\mathrm{diag}(\beta ,\alpha ,\gamma )$. Since $\alpha \mathrm{\ne }\beta$, ${\mathrm{\Pi }}_1\mathrm{\ne }{\mathrm{\Pi }}_2$, contradicting determination of $\mathrm{\Pi }$ by $(n,u,e)$.