Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 8 (Topics in Kinetic Theory)

Write $z=(x,v)$ and let $\Phi (t,s,z)$ denote the Hamiltonian flow from time $s$ to time $t$. The characteristic curves satisfy Hamilton's equations:The Hamiltonian Liouville equation then reduces along each curve toThe signs and derivative variables here are those in the PDF.

The global characteristic flow for a Hamiltonian with bounded Hessian follows, for example, from $H\in C^2(\mathbb{R}\times {\mathbb{R}}^{2d})$ and, for every finite $T$,Thus the Hamiltonian vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is globally Lipschitz continuous in $z$ on each finite time interval and satisfies $|b(t,z)|\le A_T+L_T|z|$. The Picard-Lindelöf theorem gives local existence and uniqueness, while the Gronwall inequality gives, for example,This excludes finite-time escape. There is a unique Hamiltonian flow for all finite forward and backward times, and $\Phi (s,t)$ is the inverse of $\Phi (t,s)$. These sufficient conditions are deliberately stronger than necessary.

For $S_t(z)=\Phi (t,0,z)$, differentiability with respect to initial data gives the variational equationThe Jacobi determinant derivative formula givesThe mixed derivatives in the Hamiltonian vector field cancel:Consequently $J^{\prime }(t)=0$, and $J(0)=1$ gives preservation of phase space volume:The same argument applies to $\Phi (t,s)$ for every starting time $s$. This is the Liouville theorem in Hamiltonian mechanics. Explicit time dependence of $H$ does not affect the cancellation.

For an autonomous Hamiltonian function, the chain rule and Hamilton's equations giveThus the energy is constant along each characteristic:The time-independent hypothesis is implicit in the displayed expression in this subpart. For the time-dependent Hamiltonian function allowed earlier, the correct Hamiltonian energy balance is insteadFor example $H(t,x,v)=t+|v{|}^2/2$ has a unique global Hamiltonian flow, but its value along a curve increases at unit rate. Thus conservation cannot be claimed for arbitrary time-dependent $H$.

Assume $\omega \ne 0$. With $h=0$, the method of characteristics gives $f(t,\Phi (t,0,z))=f_0(z)$. HenceLet $K=\mathrm{supp}{f}_0$ and $E_{\ast }={\max }_K{H}_{\omega }$; if $f_0=0$, the conclusion is immediate. By the Hamiltonian energy balance, every image point of $K$ remains in the energy sublevelThis is a fixed compact ellipsoid in phase space. Therefore the support bound is uniform in time:This uniform support bound from a coercive conserved energy requires no explicit solution of the curves.

The nonzero-frequency qualification is necessary. At $\omega =0$ the energy does not control $x$, and the equation is free transport equation. Choose a smooth compactly supported $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne 0$. Its transported value at $(x_0+t{v}_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual support is compact, but there is no fixed compact support for all times.

For the isotropic harmonic oscillator flow, Hamilton's equations are $\dot{X}=V$, $\dot{V}=-{\omega }^{2}X$. In dimension three these are three identical uncoupled pairs. For $\omega \ne 0$, put $c_t=\cos (\omega t)$, $s_t=\sin (\omega t)$. The solution from $(x,v)$ at time zero isThe inverse flow is obtained by replacing $t$ by $-t$:Integrating the source along the backward characteristic gives the Duhamel formula for Hamiltonian transport:Indeed $f(t,S_{t}z)=f_0(z)+{\int }_0^{t}h(s,S_{s}z)ds$, and setting $z=S_{-t}(x,v)$ gives the formula. It has the prescribed initial value and differentiation along the characteristic gives the source.

The zero-frequency limit has $s_t/\omega \to t$, $\omega s_t\to 0$, so $S_t(x,v)=(x+tv,v)$ and

For the globally invertible Hamiltonian flow from (a), the general characteristic formula isEvery $\Phi (s,t)$ preserves Lebesgue measure, by (b). Thus composition with it is an isometry of each Lp space. The Minkowski integral inequality givesApplying the Minkowski inequality also in time yields the finite-time Lp bound for Hamiltonian transport:The smoothness assumptions allow the characteristic construction; the norm estimate itself only uses the Lp space data and volume preservation.

For a concrete failure on infinite time, choose $\omega =1$, $f_0=0$, andThis is smooth, time independent and in every finite Lp space; it is invariant under the isotropic harmonic oscillator flow. Therefore the solution grows linearly:This invariant-source secular growth in Hamiltonian transport supplies the counterexample even with zero initial data.

Fix $t,x$. The Cauchy-Schwarz inequality in the integration variable $v_{\ast }$ givesIntegrate first in $v$, then in $x$. The uniform kernel hypothesis bounds the first factor after velocity integration by $C^2$, independently of $t,x$. HenceThe linear Boltzmann collision operator is linear by its integral definition, so this proves it is a bounded linear operator on $L_{x,v}^2$. The estimate is the Hilbert-Schmidt kernel bound applied at each spatial point. The printed real-kernel square is $|k{|}^2$ for a complex kernel. Measurability of the coefficients is understood so that the displayed integrals are defined.

Free transport equation is the measure-preserving shear $(x,v)\mapsto (x-vs,v)$. Its composition operator preserves the $L_{x,v}^2$ norm. Since $a\ge 0$, the damping multiplier has modulus at most one:Consequently the damped free-transport evolution is contractive:This needs no upper bound on $a$. Nonnegative integrals may be interpreted in the extended sense with $e^{-\infty }=0$; local integrability along characteristics additionally gives the usual continuous initial trace.

Denote the damped free-transport evolution from time $s$ to time $t$ byBy the argument in (b), $\Vert U_a(t,s)g{\Vert }_2\le \Vert g{\Vert }_2$. The integral operator is the Boltzmann Volterra operatorThe Minkowski integral inequality and (a) give the useful stronger pointwise boundTherefore the requested estimate isFor strongly measurable bounded $L^2$-valued $f$, these are Bochner integrals; their finite norm bounds establish existence of the integrals.

For fixed $t$, put $M_t={\sup }_{0\le s\le t}\Vert f(s){\Vert }_2$. A Cauchy-Schwarz inequality in time strengthens the organization of the estimate in (c):Start with $\Vert \tau f(t){\Vert }_2^2\le C^2{t}^2{M}_t^2$. If the printed bound holds for $n-1$, thenTaking square roots proves the iterated Cauchy-Schwarz bound for a Volterra operator:There is also a factorial bound for a Volterra iterate, obtained by iterating the unsquared integral estimate in (c):Its factor $t^n/n!$ is the volume of the time-ordered simplex $0<s_n<\cdots <s_1<t$. It is stronger than the printed estimate because ${\prod }_{j=1}^n{j}^2\ge {\prod }_{j=1}^n(2j-1)$.

Work in the Banach space ${\mathcal{B}}_T$ of bounded strongly measurable maps $[0,T]\to L_{x,v}^2$, with norm $\Vert g{\Vert }_{{\mathcal{B}}_T}={\sup }_{0\le t\le T}\Vert g(t){\Vert }_2$. Keeping actual representatives at every time matches the pointwise-in-time mild formulation. The Boltzmann Volterra operator is bounded on this space, and the factorial bound for a Volterra iterate givesThus the Volterra series for the linear Boltzmann equation converges in operator norm for every finite $T$, even when $CT\ge 1$. SetFor its partial sums, $(I-\tau ){\sum }_{n=0}^N{\tau }^{n}F=F-{\tau }^{N+1}F$. The remainder tends to zero by the factorial estimate. Hence $(I-\tau )f=F$, precisely the required characteristic integral equation.

The norm bound in (b) gives an explicit choice of the existence constant:Also $f(0)=f_0$, since every term with $n\ge 1$ vanishes at zero and the damping interval has length zero. This proves existence in the paper's weak, characteristic-integral sense. With merely measurable nonnegative $a$, that sense does not itself require a continuous initial trace; that trace follows under the additional local characteristic-integrability condition described in (b).

Let $f_1,f_2$ have the same initial data and satisfy the characteristic integral equation. Their difference $g=f_1-f_2$ satisfies $g=\tau g$, hence $g={\tau }^{n}g$ for every $n$.

Fix any $T_0<T$ and let $M={\sup }_{0\le s\le T_0}\Vert g(s){\Vert }_2<\infty$. The factorial bound for a Volterra iterate givesThe scalar factor tends to zero, so $g(t)=0$ throughout this interval. Since $T_0<T$ is arbitrary, the weak solution of the linear Boltzmann equation is unique on $[0,T)$. The same argument gives uniqueness at $T$ whenever solutions are defined there by the integral formula.

For the normalized standard Gaussian density,The two terms cancel, so the Ornstein-Uhlenbeck Fokker-Planck equation has stationary density for a Fokker-Planck equationEquivalently its Fokker-Planck probability current $-\nabla \gamma -v\gamma$ vanishes identically. The normalization follows from the Gaussian integral in each coordinate.

For a smooth test function $\phi$, integration by parts gives the moment identityThe assumed decay removes all boundary terms. Apply it to $1$, $v_i$ and $|v{|}^2/2$. Writing $m_i=\int v_{i}f=M{u}_i$ gives the Ornstein-Uhlenbeck moment equationsAssume $M(0)\ne 0$ when using the normalized mean $u$; otherwise $u$ is undefined, while the unnormalized momentum equation still holds. Since $M$ is constant, $u_i^{\prime }=-u_i$. Their solutions areThus mass is conserved, the mean velocity tends to zero, and $E(t)\to d{M}_0/2$. This includes the mean velocity conclusion omitted from the converted TeX.

In the one-dimensional unit-mass setting, put $q=f/\gamma$. Using the convention $0\log 0=0$, the relative entropy isBecause $\int \gamma q=\int f=1=\int \gamma$, we can subtract $q-1$ inside the integral:This is relative entropy nonnegativity from the given scalar inequality. Its integrand vanishes only at $q=1$, so zero relative entropy characterizes $f=\gamma$ almost everywhere. The nonnegativity remains valid with value $+\infty$ when entropy is not finite.

For positive smooth $f$ with the stated decay, integration by parts givesand, using unit mass,For zero values of $f$, these calculations can be made with the positive unit-mass approximation $(f+\epsilon \gamma )/(1+\epsilon )$ and then passed to the limit whenever the quantities are finite. Positive-time solutions also have the usual Gaussian smoothing.

Since $-\log \gamma =v^2/2+\frac{1}{2}\log (2\pi )$, mass conservation givesThe derivative of the first term is $\int {\partial }_t{f}_t\log f_t$, because $\int {\partial }_t{f}_t=0$. The two identities above and the energy equation in (b), with $d=M=1$, yieldNow expand the relative Fisher information:where $\int v{f}^{\prime }=-1$. Thus the entropy dissipation identity for Ornstein-Uhlenbeck flow is

The given relative Fisher information inequality is $I^{\prime }(t)\le -2I(t)$. Multiplying by $e^{2t}$ and integrating gives relative Fisher information decay under Ornstein-Uhlenbeck flow:In particular $I(t)\le I(0)e^{-2t}$ when $I(0)$ is finite; if necessary one starts at a positive time with finite information. It follows that $I(t)\to 0$.

Write $H(t)=H(f_t\mid \gamma )$. It is nonnegative and decreasing, so has a finite limit $H_{\infty }\ge 0$ when $H(0)<\infty$. The printed integrability request concerns the product $H(t)H^{\prime }(t)$. Its sign is nonpositive, and the fundamental theorem of calculus givesPassing to $R\to \infty$ proves time integrability of an entropy-dissipation product:Also ${\int }_0^{\infty }|H^{\prime }|=H(0)-H_{\infty }\le H(0)$. The zero value of $H_{\infty }$ will be used as supplied in (f); positivity and monotonicity alone only establish existence of the limit.

Use the zero entropy limit supplied in this subpart. By the fundamental theorem of calculus and the entropy dissipation identity for Ornstein-Uhlenbeck flow,Apply the relative Fisher information decay estimate starting at time $t$:Thus the requested entropy-dissipation inequality isIt is the Gaussian logarithmic Sobolev inequality along this evolution. Since $H^{\prime }\le -2H$, an integrating factor gives the entropy convergence rate for Ornstein-Uhlenbeck flow:For finite initial entropy, $f_t$ therefore converges to the stationary Gaussian density in relative entropy at rate $2$. If desired, Pinsker's inequality also converts this to the density estimate $\Vert f_t-\gamma {\Vert }_{L^1}\le \sqrt{2H(f_0\mid \gamma )}{e}^{-t}$.