Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 (Topics in Kinetic Theory)

The characteristic equations for a transport equation are $\dot{X}=V$ and $\dot{V}=X$, so $\ddot{X}=X$. For an initial point $(x_0,v_0)$ their solution isThis is the hyperbolic characteristic flow for an inverted oscillator. The addition formulas give $A_s{A}_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward characteristic flow map from the point $(x,v)$ at time $t$ to time $s$ isAlong this characteristic curve, the chain rule changes the transport equation into $d[f(s,S_{s,t}(x,v))]/ds=h(s,S_{s,t}(x,v))$. Integrating from zero to $t$ givesThe assumed $C^1$ regularity makes this a classical solution: on every compact set, the integrand and its needed derivatives are continuous, so differentiation under the finite-time integral is justified. At $t=0$ it has the required initial value, and the characteristic calculation verifies the equation. Conversely every classical solution must satisfy the same integrated identity, proving uniqueness. This is the Duhamel formula for Hamiltonian transport, with Hamiltonian $(v^2-x^2)/2$.

The derivative of the backward characteristic flow map isThus the change of variables formula preserves phase-space Lebesgue measure. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,Taking the $p$th root proves $\Vert f_t{\Vert }_p=\Vert f_0{\Vert }_p$. For $0<p<1$ this is a quasi-norm, and the argument still works because it uses only a change of variables, not the triangle inequality. The identity also holds in the extended sense when an integral is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the essential supremum, so the same conclusion holds for $p=\infty$.

Taking the essential supremum in the characteristic solution gives the sharper estimateIndeed, the bijective characteristic flow map preserves each spatial-velocity essential supremum. If $H_t={\mathrm{esssup}}_{0\le s\le t,x,v\in \mathbb{R}}|h(s,x,v)|$, this proves $\Vert f_t{\Vert }_{\infty }\le \Vert f_0{\Vert }_{\infty }+t{H}_t$. A time-independent source has $H_t=\Vert h{\Vert }_{L^{\infty }({\mathbb{R}}^2)}$, which is the displayed form. For a time-dependent source, the same symbol must mean a bound uniform over the elapsed time interval; its norm at the final time alone need not bound the accumulated forcing.

The bound is sharp. Take $f_0=1$ and $h=1$, giving $f(t,x,v)=1+t$ and equality for every $t\ge 0$. Both functions are smooth and bounded, although their finite-$p$ integrals over the whole plane are infinite.

Choose $f_0(x,v)=e^{-(x^2+v^2)}$ and $h(t,x,v)=x^2+v^2$. The initial Gaussian function is smooth and belongs to every finite Lp space, and is also bounded. The source is smooth and nonzero. The accumulated source along the backward characteristic curve isIts quadratic-form eigenvalues are $(e^{2t}-1)/2$ and $(1-e^{-2t})/2$. Both are positive for $t>0$, soConsequently $\Vert f_t{\Vert }_p=\infty$ for every $t>0$ and every finite $p\ge 1$; it is unbounded, so its $L^{\infty }$ norm is infinite as well. This spatially nonintegrable forcing in transport example avoids any ambiguity about whether the last endpoint is included in “all $p$”.

For each $p\ge 1$ with $f_0\in L^p$, a sufficient condition is $h\in L^1([0,T];L^p({\mathbb{R}}^2))$ on every finite interval. The Minkowski integral inequality and measure preservation give the finite-time Lp bound for Hamiltonian transportFor all finite $p$ simultaneously, impose, for example, $h\in L_{\mathrm{loc}}^1([0,\infty );L^1\cap L^{\infty })$. The elementary bound $\Vert h_s{\Vert }_p\le \Vert h_s{\Vert }_1^{1/p}\Vert h_s{\Vert }_{\infty }^{1-1/p}$ and the Holder inequality in time show that its $L^p$ norms are locally integrable for every $1\le p<\infty$. If a bounded initial value is also required, the same assumption controls the $p=\infty$ endpoint by the previous part.

A concrete stronger condition, compatible with a nonzero $C^1$ source, is that on each finite time interval the source has a common compact support in $(x,v)$. Its continuity then makes it bounded on that compact cylinder, so all these integrability conditions hold. A nonzero smooth source compactly supported in phase space supplies examples.

Use the unit-period circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, with total measure one. The free characteristic flow map givesTranslations in $x$ preserve its periodic measure. The Tonelli theorem and the integral triangle inequality yieldThe Fubini's theorem therefore applies also to signed data, and the same translation givesThis holds for positive or negative time. The printed $L^1(\mathbb{R})$ in this subpart is a domain typo: the spatial variable is periodic, so the correct space is $L^1(\mathbb{T})$. For example, the smooth initial value $f_0(x,v)=e^{-v^2}$ gives the constant density $\sqrt{\pi }$, whose periodic extension is not integrable on the real line.

Fix the mixed Fourier transform conventionThe spatial derivative transforms to $2\pi ik\hat{f}$, and multiplication by $v$ transforms to $-(2\pi i{)}^{-1}{\partial }_{\xi }\hat{f}$. Hence the transformed free transport equation isIts characteristic equations for a transport equation give $\dot{\xi }=-k$, so the characteristic ending at $\xi$ at time $t$ began at $\xi +kt$. ConsequentlyThe same sign follows directly by substituting $x=y+tv$ in the Fourier transform of $f_0(x-tv,v)$. No first velocity moment is assumed, so the differential equation may be understood in the sense of tempered distributions; the explicit transform formula is valid pointwise because $f_0$ is integrable.

The Fourier coefficient of the velocity-integrated density is its mixed transform at zero velocity frequency. ThusThe constant ${\rho }_{\infty }$ contributes only to the zero Fourier mode, where it equals ${\hat{f}}_0(0,0)$. ThereforeIt is important to remove the zero mode, which is conserved rather than mixed away.

Interpret the given velocity regularity in the stated Sobolev space sense. Repeated integration by parts, or the Fourier transform of a derivative in distributions, givesThe usual one-dimensional one-dimensional Sobolev representative or a smooth approximation justifies this identity without imposing extra decay of classical derivatives at specific boundary points. The transform of an integrable function has absolute value at most its $L^1$ norm. ThereforeSetting $\xi =kt$ and taking the supremum gives the required uniform weighted bound. The $k=0$ term on the left is zero; there is no division by that frequency in this argument.

Take the positive integer $n\ge 1$ and put $m=n+1\ge 2$. From the preceding estimate and the absent zero Fourier mode, for $t\ne 0$,Here $\zeta$ is the Riemann zeta function, and its displayed series is finite because $m>1$. One may replace the last derivative norm by the given full mixed Sobolev norm to obtain the requested constant depending only on $n$ and $f_0$.

An absolutely summable sequence of Fourier coefficients gives a uniformly and absolutely convergent Fourier series, with supremum bounded by the sum of their absolute values. Its sum agrees almost everywhere with $r_t$ by uniqueness of Fourier coefficients for integrable periodic functions. HenceThis is the uniform phase-mixing bound from velocity derivatives: uniform convergence for the continuous representative of the density, with rate $O(|t{|}^{-n-1})$. No uniform decay of the full phase-space distribution is asserted. The free-transport phase mixing acts by shifting nonzero spatial modes to large velocity frequency. If a convention allows $0\in \mathbb{N}$, that endpoint needs separate assumptions or an argument: the harmonic series in this proof would diverge.

Let $E=L^1({\mathbb{R}}_x^d\times {\mathbb{R}}_v^d)$, and denote the free-transport semigroup by $(U_{t}g)(x,v)=g(x-tv,v)$. For fixed $v$, the spatial translation has unit Jacobian determinant, so the Tonelli theorem givesThus the given free term is an isometry on $E$. The operators form a strongly continuous semigroup: continuity first holds for smooth compactly supported functions by dominated convergence, and density plus the isometry extends it to every $L^1$ function.

The Fubini's theorem and integral triangle inequality giveDefine the normalized velocity-reset collision operator using the normalized velocity-reset projection $Pg(x,v)=\rho (g)(x)M(v)$ and $B=P-I$. Since $M\ge 0$ and $\int M=1$,Also $P^2=P$; the collision gain replaces the velocity distribution by $M$ while preserving the spatial mass. In Bochner integral notation the printed, undamped source operator isUsing the transport isometry and the integral triangle inequality provesThese estimates hold for measurable, locally time-bounded $E$-valued functions. If the displayed supremum is infinite, the numerical bound is interpreted in the extended sense; the construction below works in a space where it is finite.

Fix $T$ and set $A_T={\sup }_{0\le s\le T}\Vert f(s){\Vert }_1$. The preceding integral estimate gives the result for one iterate. If for some $n\ge 0$ the bound $\Vert {\tau }^{n}f(s){\Vert }_1\le A_T{2}^n{s}^n/n!$ holds for $s\le T$, thenTaking $T=t$ yields exactlyThe case $n=0$ uses the identity operator. This factorial bound for a Volterra iterate comes from time ordering, so no commutation between free transport and the collision projection is assumed.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

For $0\le s\le t$, sum the same absolutely convergent Neumann series estimate:ConsequentlyThis coarse bound is sufficient for the requested locally uniform control and the uniqueness argument. It is not claimed to be the sharp dissipative estimate for the collision model.

If $f$ and $\tilde{f}$ are two weak solutions with the same initial value and the required local time bound, their difference $w$ satisfies $w=\tau w$. By linearity this implies $w={\tau }^{n}w$ for every $n$. On $[0,T]$ put $A_T={\sup }_{s\le T}\Vert w(s){\Vert }_1<\infty$. The factorial bound for a Volterra iterate now givesFor fixed $T$, the factor tends to zero; its successive-term ratio is $2T/(n+1)$. Therefore $w(t)=0$ as an $L^1$ element at every time on this interval. Since $T$ is arbitrary, the locally time-bounded $L^1$ weak solution is unique globally. This proof uses precisely the additional condition requested, rather than assuming arbitrary pointwise-in-time integrability alone supplies a finite uniform bound.

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.