Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 4

Use the genuine planar Givens rotationIn the second component the cosine multiplies $v_j$: the repeated $v_i$ in the printed formula is an error. With that printed expression, at $\theta =0$ the pair becomes $(v_i,v_i)$, which does not preserve length or measure. The rotation-based claims require the corrected expression. Also take $N\ge 2$, since the normalization by $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ is undefined for $N=1$.

Let $C_N=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ and $U_{ij,\theta }F=F\circ R_{ij,\theta }$. The change of variables formula and determinant one give $\Vert U_{ij,\theta }F{\Vert }_2=\Vert F{\Vert }_2$. Thus each is a unitary operator, with adjoint $U_{ij,-\theta }$. The Kac collision operator is the averageThe Minkowski integral inequality gives $\Vert QF{\Vert }_2\le \Vert F{\Vert }_2$, so $Q$ is bounded. For the Hilbert space inner product, integration and the angular change $\theta \mapsto -\theta$ giveHence $Q=Q^{\ast }$ and $\Vert Q\Vert \le 1$. In fact a nonzero radial Gaussian function is fixed by every rotation, showing $\Vert Q\Vert =1$. Angular averages can be understood as strong Bochner integrals; continuity of rotations in $L^2$ follows first for smooth compactly supported functions, then by density.

For each rotation, unitarity givesAveraging and using self-adjointness of $Q$ yields the Dirichlet form of the Kac collision operatorThe velocity integral is necessary: its omission from the printed right-hand side would leave a function of $\mathbf{v}$ rather than a scalar. This identity applies to every $L^2$ function, with complex modulus when necessary, and is nonnegative.

If $(I-Q)F=0$, every nonnegative angular integral is zero, so $\Vert U_{ij,\theta }F-F{\Vert }_2=0$ for almost every angle. Strong continuity in angle extends equality to every angle. The coordinate-plane Givens rotations generate the special orthogonal group $SO(N)$, hence $F$ is invariant in $L^2$ under every element of this group. To identify its shape rigorously despite almost-everywhere representatives, average over the normalized Haar measure of $SO(N)$. This averaging leaves $F$ unchanged, while transitivity of the rotation group on each sphere makes the average a radial function. Thus $F(\mathbf{v})=\varphi (|\mathbf{v}|)$ almost everywhere.

Conversely, every radial function is fixed by every coordinate-plane rotation, and so by $Q$. ThereforeThis is the radial kernel of the Kac collision operator. The rotation correction is essential to this conclusion: with the literal printed map, even $F(\mathbf{v})=e^{-|\mathbf{v}{|}^2}$ in dimension two is not fixed. At $\mathbf{v}=(0,1)$ its printed-map angular average is the average of $e^{-{\sin }^2\theta }$, strictly greater than its value $e^{-1}$.

Write $F=F_N$ and integrate the Kac master equation over $v_2,\dots ,v_N$. For a pair $i,j\ge 2$, the rotation acts only on integrated variables. Its unit Jacobian determinant makes the integrated gain identical to the integrated loss, so all those pairs cancel.

The only remaining pairs are $(1,j)$, $j=2,\dots ,N$. For such a pair, first integrate over every variable except $v_1$ and $v_j$. This yields the corresponding two-coordinate marginal distribution evaluated at the rotated pair. Permutation symmetry of $F_N$ makes all $N-1$ resulting integrals identical to the one for $(1,2)$. The coefficient isConsequently the Kac marginal evolution equation isThe time argument has been suppressed on the right. The loss is consistent with normalization, since $\int {\Pi }_2(v_1,v_2)d{v}_2={\Pi }_1(v_1)$. Under the printed definition $k<N$, this use of ${\Pi }_2$ requires $N\ge 3$. For $N=2$ the same formula holds with the natural extension ${\Pi }_N(F_N)=F_N$.

This identity is exact and generally unclosed. Replacing the two-coordinate marginal distribution by the product of one-coordinate marginals would produce the quadratic collision equation associated with Kac chaos. Permutation symmetry alone does not imply that product approximation.

Put $u=F_N/{\gamma }_N$, where the Gaussian density is strictly positive. Extend $u\log u$ continuously at zero by $0\log 0=0$. Both densities have integral one, so the relative entropy can be written asThe bracket is nonnegative and vanishes only at $u=1$: its derivative for $u>0$ is $\log u$, with a unique minimum at one. HenceThe inequality holds also for infinite entropy. Its negative integrand part is integrable, since $u\log u\ge -1/e$ and ${\gamma }_N$ has integral one, so the extended-value integral is well defined. This is relative entropy in Kac's model; no differentiation is needed for nonnegativity.

Use the correct Gaussian entropy decompositionThe last coefficient is $1/2$, as follows from $-\log {\gamma }_N=N\log (2\pi )/2+|\mathbf{v}{|}^2/2$; the printed hint omits it. Under the allowed differentiability and integrability assumptions, the supplied collision invariants conserve mass and energy. Differentiating therefore giveswhere the extra derivative term $\int {\partial }_{t}F$ vanishes by mass conservation. Thus the Kac entropy production isFor one pair, call the double integral $A_{ij}$. The measure-preserving substitution $(\mathbf{v},\theta )\mapsto (R_{ij,\theta }\mathbf{v},-\theta )$ interchanges $F$ and $F\circ R_{ij,\theta }$, with angles taken modulo $2\pi$. Averaging the original and substituted expressions givesSince $N/(4\pi C_N)=1/[2\pi (N-1)]$, the desired Kac entropy dissipation formula isFor positive values $a,b$, $(a-b)(\log a-\log b)\ge 0$ because the logarithm is increasing. At two zeros use value zero; at one zero and one positive value use the nonnegative extended value $+\infty$. One may first use positive densities and then regularize by $(F+\epsilon {\gamma }_N)/(1+\epsilon )$; rotation invariance of ${\gamma }_N$ preserves the formula and permits the usual limit at zeros under the stated assumptions.

Thus relative entropy is nonincreasing along the evolution. When the dissipation is finite, zero dissipation means pairwise rotation invariance and hence radiality, by the preceding kernel argument. Radial normalized densities other than ${\gamma }_N$ can be stationary with positive relative entropy: vanishing dissipation is not a claim that the unique stationary density is Gaussian.