 Solution

Strict positivity throughout phase space is incompatible with compact support. The following calculation uses positive smooth rapidly decaying solutions with all displayed integrals justified; nonnegative cases use the corresponding entropy limits where justified.

Write $B=|{\mathbb{S}}^2|$ and use integration over $(v,v_{\ast },\sigma )$. The two symmetrized weak collision identities areandThey follow from particle interchange and the angular exchange for elastic collisions. They are the weak identities needed here; no invalid fixed-angle Jacobian is used.

The collision invariants $1$, $v$ and $|v{|}^2$ obey ${\varphi }^{\prime }+{\varphi }_{\ast }^{\prime }=\varphi +{\varphi }_{\ast }$. Momentum follows from $v^{\prime }+v_{\ast }^{\prime }=v+v_{\ast }$, and energy from $|v^{\prime }{|}^2+|v_{\ast }^{\prime }{|}^2=|v{|}^2+|v_{\ast }{|}^2$. HenceFor the Boltzmann H functional $H=\int \int f\log fdxdv$, the spatial transport contribution is a boundary divergence and the term $\int Q$ is zero. Set $a=f^{\prime }{f}_{\ast }^{\prime }$, $b=f{f}_{\ast }>0$. The second weak identity with $\varphi =\log f$ gives the Boltzmann H theoremIndeed this integrand with its negative sign is $b(1-X)\log X$ for $X=a/b>0$, which is nonpositive. The kinetic H functional decreases; the physical entropy with opposite sign increases.