Conditional Boltzmann entropy inequality and equality form

For a fixed time, write $\mathrm{\ell }=\log (f\mathrm{/}{f}_{\mathrm{r}\mathrm{e}\mathrm{f}})$. By [F3]–[F4], dominated differentiation gives $\frac{dH}{dt}={\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^d}(1+\mathrm{\ell }){\mathrm{\partial }}_{t}fdvdr\mathrm{.}$ Substitute the transport equation. The spatial contribution is $-{\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^d}(1+\mathrm{\ell })v\cdot {\mathrm{\nabla }}_{r}fdvdr=-{\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^d}v\cdot {\mathrm{\nabla }}_r(f\mathrm{\ell })dvdr\mathrm{.}$ For each fixed $v$, the integral of each coordinate derivative over the periodic rectangle is zero by the one-dimensional fundamental theorem of calculus and opposite-face identification; absolute integrability in [A2] permits Fubini [F2]. Thus the transport contribution vanishes.

The weak identity with test $1$ has zero bracket on every elastic collision, so $\int Q(f)dv=0$ at each $r,t$. Consequently the collision contribution from $(1+\mathrm{\ell })Q(f)$ equals that from $\mathrm{\ell }Q(f)$. Apply the weak identity with $\phi =\mathrm{\ell }(r,\cdot ,t)$ at fixed $(r,t)$. Since $f_{\mathrm{r}\mathrm{e}\mathrm{f}}$ is constant, the sum $\mathrm{\ell }(v)+\mathrm{\ell }(w)-\mathrm{\ell }(v^{\mathrm{\prime }})-\mathrm{\ell }(w^{\mathrm{\prime }})$ is $\log (A\mathrm{/}B)$. Hence $\int \mathrm{\ell }(v)Q(f)(v)dv=\frac{1}{4}{\int }_{Q_{\mathrm{e}\mathrm{l}}}(B-A)\log (A\mathrm{/}B)d\nu \mathrm{.}$ The velocity-side function is measurable in $(r,v)$ because $Q(f)={\mathrm{\partial }}_{t}f+v\cdot {\mathrm{\nabla }}_{r}f$ and $f>0$ have the stated regularity. Integration of its positive and negative parts over the sigma-finite Lebesgue velocity factor gives a measurable function of $r$; the weak identity makes that function equal to the collision integral at each $(r,t)$. By [A2] this collision integral is in $L^1(\mathrm{\Omega },dr)$, so integrate the identity in $r$ to obtain the asserted derivative formula. no collision/spatial product measure or interchange of $\nu$ with spatial Lebesgue measure is used.

For positive $A,B$, strict increase of $\log$ [F5] makes $(A-B)\log (A\mathrm{/}B)\ge 0$, with equality exactly when $A=B$. The derivative in step 2.1 is therefore nonpositive. Let $g(r)$ be the nonnegative inner collision integral; it is measurable and integrable by [A2] and step 2.1. If ${\int }_{\mathrm{\Omega }}g(r)dr=0$, then $g(r)=0$ for Lebesgue-almost every $r$. For each such $r$, a nonnegative function with zero $\nu$-integral vanishes $\nu$-almost everywhere: its sets where the value is at least $1\mathrm{/}n$ all have $\nu$-measure zero, and their countable union contains its positive set. This argument holds for an arbitrary measure, with no sigma-finiteness hypothesis. Thus $A=B$ for $\nu$-almost every collision at almost every $r$. The converse follows by substitution in the derivative identity.

Under [A3], fix an $r$ for which collision balance holds $\nu$-almost everywhere. The function $(v,w,v^{\mathrm{\prime }},w^{\mathrm{\prime }})\mapsto A-B$ is continuous. If it were nonzero at an allowed collision, it would remain nonzero on a relative open neighborhood; full support of $\nu$ gives that neighborhood positive measure, contradicting balance. Hence balance holds for every elastic collision. The set of such $r$ has full Lebesgue measure and is dense in $\mathrm{\Omega }$; continuity in $r$ extends the identity to any limit point when needed.

Fix such an $r$ and put $g(v)=\log (f(r,v,t)\mathrm{/}{f}_{\mathrm{r}\mathrm{e}\mathrm{f}})$. For any $a,u\in {\mathbb{R}}^d$ and any orthogonal rotation $R$, the pairs $(a+u,a-u)$ and $(a+Ru,a-Ru)$ have the same total momentum $2a$ and total squared speed $2\mathrm{\mid }a{\mathrm{\mid }}^2+2\mathrm{\mid }u{\mathrm{\mid }}^2$. Step 4.1 therefore gives $g(a+u)+g(a-u)=g(a+Ru)+g(a-Ru)\mathrm{.}$ For unit vectors $\theta ,\eta$, put $u=s\theta$ and $R\theta =\eta$. The second-order Taylor expansion at $a$ gives $g(a+s\theta )+g(a-s\theta )=2g(a)+s^2{\theta }^{\mathrm{\top }}{D}^{2}g(a)\theta +o(s^2)\mathrm{.}$ Equating the expansion for $\eta$, dividing by $s^2$ and sending $s\to 0$ shows that the quadratic form of the symmetric Hessian is constant on the unit sphere. Testing coordinate vectors makes all diagonal entries equal; testing $(e_i+e_j)\mathrm{/}\sqrt{2}$ makes each off-diagonal entry zero. Thus $D^{2}g(a)=\lambda (a)I$.

Since ${\mathrm{\partial }}_{ij}g=0$ for $i\mathrm{\ne }j$, the fundamental theorem of calculus along coordinate segments shows ${\mathrm{\partial }}_{i}g(v)$ depends only on $v_i$. Write it as $G_i(v_i)$. The diagonal Hessian equalities give $G_i^{\mathrm{\prime }}(v_i)=G_j^{\mathrm{\prime }}(v_j)$ for every pair of independent real coordinates $v_i,v_j$. Because $d\ge 2$, varying these coordinates independently forces all these derivatives to be one constant $\lambda$. Integrating each first derivative gives $g(v)=\alpha +b\cdot v+\lambda \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$.

If $\lambda >0$, $e^{g(v)}$ grows quadratically in every sufficiently large direction and is not integrable. If $\lambda =0$, $e^{\alpha +b\cdot v}$ is constant when $b=0$ and grows on a half-line when $b\mathrm{\ne }0$, so it is not integrable over ${\mathbb{R}}^d$. Thus integrability forces $\lambda <0$. Completing the square writes $f(r,v,t)=C(r,t)\exp [-c(r,t)\mathrm{\mid }v-u(r,t){\mathrm{\mid }}^2]$ for $C,c>0$, the Maxwellian velocity form. Conversely, when $g=\alpha +b\cdot v+\lambda \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$, the sum $g(v)+g(w)$ is unchanged by every collision because total momentum and total squared speed are conserved; hence $A=B$ and the production is zero. This equality branch concerns the collision term only; a spatially varying Maxwellian may still evolve under transport.