H theorem and Maxwellian equality for the regular cutoff model

The bounds in [F3] give constants $c,C>0$ such that $c{e}^{-\mathrm{\mid }v-u{\mathrm{\mid }}^2\mathrm{/}(2a)}\le f_t(v)\le C{e}^{-\mathrm{\mid }v-u{\mathrm{\mid }}^2\mathrm{/}(2b)}$ for all $t,v$. The same upper Gaussian envelope holds for $\mathcal{G}(f_t,f_t)$ because [F3] preserves the mixture class; [F4] therefore bounds $\mathrm{\mid }{\mathrm{\partial }}_t{f}_t\mathrm{\mid }$ by a constant times that envelope. The lower and upper bounds imply $\mathrm{\mid }\log (v_0^3{f}_t(v))\mathrm{\mid }\le C_1(1+\mathrm{\mid }v-u{\mathrm{\mid }}^2)$, after changing the finite constant $C_1$. Thus $f_t\mathrm{\mid }\log (v_0^3{f}_t)\mathrm{\mid }$ is integrable and the entropy derivative integrand has an integrable polynomial-Gaussian majorant on every compact time interval.

For each fixed $v$, the ordinary chain rule gives ${\mathrm{\partial }}_t[f_t\log (v_0^3{f}_t)]=(1+\log (v_0^3{f}_t)){\mathrm{\partial }}_t{f}_t$. The uniform majorant from step 1.1 also bounds the difference quotients by the scalar mean-value theorem. Dominated convergence [F5] therefore gives $H^{\mathrm{\prime }}(t)=\int (1+\log (v_0^3{f}_t)){\mathrm{\partial }}_t{f}_{t}dv$. The mass of $f_t$ is constant by [F4], so $\int {\mathrm{\partial }}_t{f}_{t}dv=0$ and the constant term drops. The integrand in the resulting formula is pointwise continuous in time by [F4] and has the same local integrable majorant from step 1.1; a second application of [F5] shows that $H^{\mathrm{\prime }}(t)$ is continuous.

Fix $t$ and write $\phi (v)=\log (v_0^3{f}_t(v))$. The lower Gaussian bound makes $\phi$ finite. The mean of $f_t$ is still $u$ by the moment conservation in [F4]. Also $A,B$ are bounded above by a constant times $e^{-(\mathrm{\mid }v-u{\mathrm{\mid }}^2+\mathrm{\mid }w-u{\mathrm{\mid }}^2)\mathrm{/}(2b)}$, because the pair reflection fixes the common mean $(u,u)$ and preserves the sum of the two centered squared speeds. The Gaussian lower bound gives $\mathrm{\mid }\log (A\mathrm{/}B)\mathrm{\mid }\le C_2(1+\mathrm{\mid }v-u{\mathrm{\mid }}^2+\mathrm{\mid }w-u{\mathrm{\mid }}^2)$. The same Gaussian upper bound holds for $\mathcal{G}(f_t,f_t)$ by [F3], so $\int \mathrm{\mid }\phi \mathrm{\mid }{f}_{t}dv$ and $\int \mathrm{\mid }\phi \mathrm{\mid }\mathcal{G}(f_t,f_t)dv$ are finite; the pair bound and polynomial bound on $\mathrm{\Delta }\phi$ give $\int \mathrm{\mid }A\mathrm{\Delta }\phi \mathrm{\mid }d\mu$ and $\int \mathrm{\mid }B\mathrm{\Delta }\phi \mathrm{\mid }d\mu$ finite. Thus the four conditions in [F2] hold. By [F2] and [F4], $H^{\mathrm{\prime }}(t)=\frac{\lambda }{4}\int (B-A)\mathrm{\Delta }\phi d\mu \mathrm{.}$ The reference factors cancel in the logarithm: $\mathrm{\Delta }\phi =\log (A\mathrm{/}B)$. Hence $H^{\mathrm{\prime }}(t)=-\frac{\lambda }{4}\int (B-A)\log (B\mathrm{/}A)d\mu$. Since $\log$ is strictly increasing, $(B-A)\log (B\mathrm{/}A)\ge 0$, proving the inequality.

If $H^{\mathrm{\prime }}(t)=0$, the continuous nonnegative integrand in step 3.1 vanishes $\mu$-almost everywhere. The product measure has full support by [F2]; a continuous function nonzero at one point is bounded away from zero on a nonempty open neighborhood of positive measure. Thus $A=B$ for every $(v,w,n)$. Conversely, this pointwise identity makes the production integral zero.

Put $g(v)=\log (v_0^3{f}_t(v))$, a $C^2$ function by [F3]. By the reflection coverage in [F2], for every center $c\in {\mathbb{R}}^3$ and every pair $r,r^{\mathrm{\prime }}\in {\mathbb{R}}^3$ with $\mathrm{\mid }r\mathrm{\mid }=\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }$, the equality in step 4.1 gives $g(c+r)+g(c-r)=g(c+r^{\mathrm{\prime }})+g(c-r^{\mathrm{\prime }})$. For unit vectors $\theta ,\eta$, take $r=s\theta$ and $r^{\mathrm{\prime }}=s\eta$. Taylor expansion at $c$ yields $g(c+s\theta )+g(c-s\theta )=2g(c)+s^2{\theta }^{\mathsf{T}}{D}^{2}g(c)\theta +o(s^2)$, with the same expression for $\eta$. Dividing by $s^2$ and taking $s\to 0$ shows that the quadratic form of $D^{2}g(c)$ is constant on the unit sphere. Testing $e_i$ and $(e_i+e_j)\mathrm{/}\sqrt{2}$ shows $D^{2}g(c)=\alpha (c)I$.

The off-diagonal Hessian entries vanish, so for each $i$, ${\mathrm{\partial }}_{i}g$ depends only on its $i$th coordinate: changing any other coordinate leaves it fixed by the fundamental theorem of calculus along that coordinate segment. Write ${\mathrm{\partial }}_{i}g(v)=G_i(v_i)$. The equal diagonal entries give $G_i^{\mathrm{\prime }}(v_i)=G_j^{\mathrm{\prime }}(v_j)$ for independently variable coordinates $v_i,v_j$; hence every such derivative is one constant $\alpha$. Integrating gives $g(v)=c_0+\beta \cdot v+\frac{\alpha }{2}\mathrm{\mid }v{\mathrm{\mid }}^2$.

Integrability of $f_t=e^g\mathrm{/}{v}_0^3$ forces $\alpha <0$: if $\alpha >0$ the density grows in every sufficiently large direction, and if $\alpha =0$ an exponential of an affine function is nonintegrable on ${\mathbb{R}}^3$. Completing the square writes $f_t=C{e}^{-\mathrm{\mid }v-\bar{u}{\mathrm{\mid }}^2\mathrm{/}(2s^2)}$ with $s^2=-1\mathrm{/}\alpha >0$; Tonelli [F7] and the one-dimensional Gaussian integral [F6] give $\int e^{-\mathrm{\mid }v-\bar{u}{\mathrm{\mid }}^2\mathrm{/}(2s^2)}dv=(2\pi s^2{)}^{3\mathrm{/}2}$, so unit mass fixes the prefactor in the Statement. Conversely, for that Gaussian $\log f_t$ is a constant plus a linear function of $v$ and a scalar multiple of $\mathrm{\mid }v{\mathrm{\mid }}^2$; pair momentum and squared-speed conservation in [F2] give $A=B$ at every collision. Define $\mathrm{sgn}(x)=1,0,-1$ according as $x>0,x=0,x<0$, and set $\phi =\mathrm{sgn}(\mathcal{G}(f_t,f_t)-f_t)$. This test is bounded and measurable; since $\int Ad\mu =\int Bd\mu =1$, both sides of the weak identity are absolutely integrable. Applying [F2] gives $\mathrm{\parallel }\mathcal{G}(f_t,f_t)-f_t{\mathrm{\parallel }}_1=0$. The constant curve satisfies the mild equation because $e^{-\lambda t}+\lambda {\int }_0^t{e}^{-\lambda (t-s)}ds=1$, and uniqueness in [F4] makes this Gaussian stationary.

The density formula in [F3] factors for $aI$ and $bI$ into products of one-dimensional Gaussians; Tonelli [F7] and normalization [F6] give the first-coordinate marginal as their equal mixture. For a centered one-dimensional Gaussian of variance $s^2$, integration of the derivative of $x^{2k-1}{e}^{-x^2\mathrm{/}(2s^2)}$ over $\mathbb{R}$ gives $\int x^{2k}{e}^{-x^2\mathrm{/}(2s^2)}dx=(2k-1)s^2\int x^{2k-2}{e}^{-x^2\mathrm{/}(2s^2)}dx$ for $k=1,2$; the boundary terms vanish by Gaussian decay. Thus the mixture's centered second moment is $(a+b)\mathrm{/}2$ and its fourth moment is $3(a^2+b^2)\mathrm{/}2$. A single Gaussian with the same second moment has fourth moment $3((a+b)\mathrm{/}2{)}^2$, smaller by $3(a-b{)}^2\mathrm{/}4>0$. Thus $f_0$ is not a single Gaussian. By the equality characterization in steps 4.1–7.1, the production at $t=0$ is strictly positive and $H^{\mathrm{\prime }}(0)<0$.