Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 (Topics in Kinetic Theory)

The characteristic equations for a transport equation on phase space areA sufficient global hypothesis is that $F$ is continuous in time, locally Lipschitz continuous in $(x,v)$ uniformly on compact time intervals, and, for every finite $T$, satisfies $|F(t,x,v)|\le C_T(1+|x|+|v|)$ for $|t|\le T$. These hypotheses give unique characteristic curves for all real times if the field is defined for all real times. It suffices for forward existence to impose them on nonnegative time. One may strengthen the spatial hypothesis to $C^1$ when differentiating the characteristic flow map below. Local Lipschitz continuity alone guarantees only local existence; the linear-growth bound prevents finite-time escape.

Put $z=(X,V)$ and $b(t,z)=(V,F(t,X,V))$. On a small closed time interval and a closed ball around the initial point, let $M$ bound $b$ and let $K$ be its spatial Lipschitz constant. The mapsends the corresponding closed ball of continuous paths into itself when the time length times $M$ is at most the ball radius. If the time length times $K$ is less than one, it is a contraction in the supremum norm. The Banach fixed-point theorem gives a local solution and uniqueness; differentiating its integral equation gives the ordinary differential equation. Overlapping local solutions agree by this uniqueness.

For continuation, the growth assumption gives, on any bounded time interval,for forward time, with the analogous reversed-time bound. The Gronwall inequality bounds the trajectory on that interval. A finite terminal time is impossible: within the resulting compact ball the field is bounded, so the path has a limit at the endpoint, and the local construction restarts there. This proves the global characteristic flow under linear growth. No differentiability of the flow with respect to its initial point is needed for this existence and uniqueness proof.

For this force, $\dot{V}=V$, hence $V(t)=e^{t}v$ and $X(t)=x+(e^t-1)v$. Invert the characteristic flow map: the initial velocity for an endpoint $(x,v)$ is $e^{-t}v$, and the initial position is $x-(1-e^{-t})v$. Constancy along each characteristic curve givesThere is no amplitude factor here: this is scalar advection, not the conservative continuity equation for a compressible density.

Differentiating the explicit characteristic flow map gives the block triangular Jacobian matrixIts Jacobian determinant is the product of its diagonal block determinants. Therefore

Let $A(t)=D_{(x,v)}(X,V)$ along one characteristic curve. Differentiating the characteristic equation gives $\dot{A}=D_{z}b(t,X,V)A$, $A(0)=I_6$, with $b=(V,F)$. The Liouville formula for a fundamental matrix then gives the phase-space flow JacobianIndeed $\mathrm{tr}{D}_{z}b={\mathrm{div}}_{x,v}(v,F)={\nabla }_v\cdot F$, since $v$ is independent of $x$. Differentiation proves $\dot{J}=({\nabla }_v\cdot F)J$. If that divergence vanishes, $J(0)=1$ gives $\boxed{J\equiv 1}$ and the flow preserves Lebesgue measure on phase space.

The finite-$p$ assertion needs the zero-divergence hypothesis from the preceding part, which is not repeated in this part's printed hypotheses. The norm is on ${\mathbb{R}}_x^3\times {\mathbb{R}}_v^3$, and finite conservation requires $f_0$ to belong to the corresponding Lp space in addition to its smoothness.

Writing ${\Phi }_t$ for the characteristic flow map, the general solution is $f_t=f_0\circ {\Phi }_t^{-1}$. A change of variables givesThus the Lp conservation for incompressible transport isWithout that condition the requested conclusion is false: the force in part (c), together with any nonzero smooth integrable initial datum given by a Gaussian function, gives $\Vert f_t{\Vert }_p=e^{3t/p}\Vert f_0{\Vert }_p$. The essential supremum is still conserved by a complete invertible flow, because composition does not change the range of values.

For a linear spatial operator, let $U(t,s)$ denote the formal homogeneous evolution from time $s$ to time $t$, satisfying ${\partial }_{t}U(t,s)h+D(t)U(t,s)h=0$ and $U(s,s)h=h$. This also allows time-dependent coefficients in the spatial operator. The Duhamel principle statesDifferentiation of the integral contributes $g(t)$ at its upper endpoint and $-D(t)$ times the integral, while the initial value is $f_0$. If $D$ is time-independent, one writes $U(t,s)=S(t-s)=e^{-(t-s)D}$. The principle uses linearity; an arbitrary nonlinear differential operator would not justify this superposition formula. No analytic construction of $U$ or boundary conditions is required for this formal statement.

The damped free-transport evolution is $S(t)h(x,v)=e^{-t}h(x-tv,v)$. Substituting it into the Duhamel principle givesThe spatial shifts all follow the same backward characteristic curve ending at $(x,v)$ at time $t$; the velocity is constant.

Separate the transported initial datum from the gain term. Defineand the Boltzmann Volterra operatorThe Duhamel principle makes the linear Boltzmann equation equivalent, whenever the integrals and solution are legitimate, toThe gain samples velocities $v_{\ast }$ at the backward spatial position of the outgoing velocity $v$; replacing that spatial shift by one depending on $v_{\ast }$ would give a different equation.

Fix an arbitrary finite $T$ and use the Banach space of bounded continuous functions $C_b([0,T]\times {\mathbb{R}}^{2d})$ with the supremum norm. Set $A=\Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1$. For $f$ in this space, the velocity integralis bounded by $\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. If $(t_n,x_n)\to (t,x)$, the integrands converge pointwise and are bounded by the integrable function $\Vert f{\Vert }_{\infty }{k}_2$. The dominated convergence theorem proves continuity. Thus the rank-one gain on bounded continuous functions, $Kf=k_1{H}_f$, is continuous and bounded. The same theorem applied to the time integral shows that $\tau$ maps this space into itself, including at $t=0$.

Time ordering gives the factorial bound for a Volterra iterate:Consequently the Volterra series for the linear Boltzmann equationconverges uniformly on the whole finite time slab, so its sum is continuous and bounded and satisfies the fixed point equation. Tracking the damping factors gives the sharper pointwise bound $\Vert f(t){\Vert }_{\infty }\le e^{(A-1)t}\Vert f_0{\Vert }_{\infty }$. This proves existence for every finite $T$, without requiring $AT<1$ or uniform continuity of $f_0$. Boundedness on finite slabs does not assert a uniform bound over infinite time.

Let $h_0=f_{1,0}-f_{2,0}\ge 0$. The damped free-transport evolution preserves nonnegativity, and a nonnegative kernel makes $\tau$ preserve it as well. Every term in ${\sum }_{n\ge 0}{\tau }^{n}F(h_0)$ is therefore nonnegative. By linearity, the difference $f_1-f_2$ solves the fixed point equation with initial value $h_0$. The assumed uniqueness identifies it with this nonnegative sum on every finite interval. Hence the order preservation for the linear Boltzmann equation givesContinuity upgrades pointwise limits in the series to the stated continuous solution; no pointwise differentiability or minimum principle is needed.

Use the weighted Hilbert structure of velocity-reset relaxation, with inner product $⟨f,g⟩=\int f\overline{g}/M$ and norm $\Vert f{\Vert }_H$. The Cauchy-Schwarz inequality givesThus $\rho (f)$ is an absolutely convergent integral and a bounded linear functional. Since $\Vert M{\Vert }_H^2=\int M=1$, the normalized velocity-reset collision operator obeysThis establishes boundedness. The subsequent orthogonal decomposition improves the operator norm to exactly one.

For the normalized velocity-reset collision operator,This bounded everywhere-defined operator is self-adjoint. Put $h=f/M$. Expanding the square with the probability measure $M(v)dv$ givesIt follows thatReplacing the difference by $(f(v_{\ast })M(v)-f(v)M(v_{\ast }))/(M(v)M(v_{\ast }))$ gives the other printed integral expression. For real functions the modulus squares are ordinary squares; the modulus version also proves the complex-space statement.

The weighted inner product satisfies $⟨f,M⟩=\int f=\rho (f)$. ThereforeThe identity operator and $L$ are linear, bounded and self-adjoint, hence so is $\Pi$. Directly, $\Vert \Pi f{\Vert }_H=|\rho (f)|\le \Vert f{\Vert }_H$; equality holds at $f=M$. Thus $\Vert \Pi \Vert =1$.

Since $\rho (M)=1$, ${\Pi }^{2}f=\rho (f)\rho (M)M=\Pi f$. Also $Lf=0$ exactly when $f=\rho (f)M$, which is exactly the range of $\Pi$. ThusA bounded self-adjoint idempotent is an orthogonal projection: if $u=\Pi u$ and $w=(I-\Pi )w$, then $⟨u,w⟩=⟨u,\Pi w⟩=0$. Here the orthogonal complement is precisely the zero-mass subspace $\rho (f)=0$.

Using $L=\Pi -I$ and the orthogonal projection identity ${\Pi }^{\ast }\Pi =\Pi$, we have $⟨\Pi f,f⟩=\Vert \Pi f{\Vert }_H^2$. The orthogonal decomposition $f=\Pi f+(I-\Pi )f$ then givesThe sign is negative, as in the PDF; the converted TeX loses it at the last equality. This establishes the exact spectral gap of normalized velocity relaxation: on the zero-mass subspace $L=-I$, with gap one rather than the weaker half-gap estimate.

The orthogonal projection satisfies $\Pi L=\Pi (\Pi -I)=0$, hence $d\Pi f_t/dt=0$ and $\Pi f_t=\Pi f_0=\rho (f_0)M$. Set $h_t=f_t-\Pi f_0$. Then $\Pi h_t=0$ and ${\dot{h}}_t=L{h}_t$. Crucially, applying the energy identity of part (e) to $h_t$ givesSolving this scalar equation and taking square roots provesThis argument uses (e) explicitly, rather than bypassing the requested energy method with an explicit solution formula.

Nonnegativity and unit mass give, by the triangle inequality for the Fourier transform, $|\hat{f}(\xi )|\le \int f=1=\hat{f}(0)$. The finite second moment implies a finite first absolute moment, so differentiating under the Fourier integral is justified andThe exponent convention has no $2\pi$, so no such factor occurs in this derivative.

If either vector vanishes, both arguments are zero. Otherwise choose a rotation matrix $R$ sending $x/|x|$ to $y/|y|$. The surface measure on a sphere is rotation invariant. Changing variables $\sigma =R\omega$ givesThus the two spherical integrals are equal. The property extends to complex continuous $\varphi$ by treating real and imaginary parts separately, as needed for the Fourier exponential.

The printed hint with unchanged $\sigma$ is not a valid change of variables: at fixed $\sigma$, the outgoing velocities forget the direction of $v-v_{\ast }$. A correct proof uses the angular exchange for elastic collisions.

Set $C=(v+v_{\ast })/2$ and $v-v_{\ast }=r\omega$, where $r>0$ and $\omega \in {\mathbb{S}}^2$. Then $dvd{v}_{\ast }=dC{r}^{2}drd\omega$, while $v^{\prime }=C+r\sigma /2$ and $v_{\ast }^{\prime }=C-r\sigma /2$. The gain integral becomesExchange the two independently integrated angular variables $\omega$ and $\sigma$. The measure is unchanged, and reverting to $v=C+r\omega /2$, $v_{\ast }=C-r\omega /2$ givesThe measure-zero set $r=0$ causes no difficulty. This proves the requested identity without invoking the false fixed-$\sigma$ Jacobian assertion.

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. The surface measure on a sphere here is two-dimensional surface area, not the restriction of ambient three-dimensional Lebesgue measure, which would give the sphere measure zero.

Expansion of the two squares givesLet $D=d(f,g)$ be the Fourier distance of order two, with the supremum taken over $\xi \ne 0$. Both Fourier transforms have modulus at most one. Add and subtract $\hat{g}({\xi }^{+})\hat{f}({\xi }^{-})$, then use the triangle inequality:Dividing by $|\xi {|}^2$ proves the bound by $D$, and splitting the last expression into its two weighted terms gives the requested intermediate inequality. If one of ${\xi }^{\pm }$ vanishes, its unweighted difference is zero by equal mass; its weighted term is interpreted as zero, avoiding a $0/0$ quotient.

Equal mass and first moment also explain finiteness of this distance: subtract the constant and linear Taylor terms in the Fourier integral and use $|e^{-ia}-1+ia|\le a^2/2$. This bounds the difference by $|\xi {|}^2\int |v{|}^2(f+g)/2$. No direction-independent extension of the quotient at zero is required.

The reference to part (f) within this part is a printed self-reference; the needed product estimate is part (e). Since both masses are one, subtracting the Bobylev identities gives, for $\xi \ne 0$,The normalized angular average and part (e) giveFor completeness, the Duhamel principle for this scalar equation yields $d(t)\le e^{-t}d(0)+{\int }_0^t{e}^{-(t-s)}d(s)ds$. Apply the Gronwall inequality to $e^{t}d(t)$ to obtain the Fourier nonexpansion for Maxwell molecules, $\boxed{d(f_t,g_t)\le d(f_0,g_0)}$. This estimate is nonexpansion; by itself it does not prove strict decay or convergence to a specified equilibrium.

For every test function $\varphi \in C_c^{\infty }({\mathbb{R}}_t\times {\mathbb{R}}_x^3\times {\mathbb{R}}_v^3)$, the distributional weak solution satisfiesThere is no initial-time boundary term because the test functions are on the whole time line. In divergence notation the flux in the seven coordinates is $(f,vf,0)$, whose divergence is ${\partial }_{t}f+v\cdot {\nabla }_{x}f$. For a general $N$-dimensional divergence the introductory flux must be vector-valued; the scalar codomain printed for $A$ is dimensionally inconsistent when $N>1$. The transport identity above has the correct flux dimensions.

The shear $(t,x,v)\mapsto (t,x+tv,v)$ is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus $F,G$ are locally integrable functions. On a compact time interval $I$ and compact label set $K\subset {\mathbb{R}}^6$, the integral ${\int }_K{\int }_I|G|dtdxdv$ is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that $G(\cdot ,x,v)\in L_{\mathrm{loc}}^1(\mathbb{R})$ for almost every $(x,v)$.

To transform the weak equation, choose $\varphi (t,x,v)=\rho (x-tv,v)\psi (t)$, where $\rho \in C_c^{\infty }({\mathbb{R}}^6)$ and $\psi \in C_c^{\infty }(\mathbb{R})$. Including a velocity cutoff in $\rho$ ensures the required compact support. The transport derivative of this test function is $\rho (x-tv,v){\psi }^{\prime }(t)$. Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which ${\partial }_{t}F=G$ as one-dimensional distributions.

For each remaining label, subtract the primitive $H(t)={\int }_0^{t}G(s)ds$. The distributional derivative of $F-H$ is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, $\tilde{F}(t)=C+H(t)$. The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.

Interpret $L$ as the velocity operator appearing in the equation and in the stated quadratic-form hypothesis. Let $E(t)=\int \int |f(t,x,v){|}^{2}dxdv$. Smoothness turns the distributional equation into the usual equation. Multiply it by $f$ and integrate; for complex functions use the real part of the inner product. Spatial integration by parts gives $\int v\cdot {\nabla }_x(f^2)=0$, while the coercivity hypothesis at each $x$ givesThus $e^{2\delta t}E(t)$ is nonincreasing. The coercive energy estimate for kinetic transport isNo division by $E$ is required, so the zero solution causes no exception. Compact spatial support, or sufficient decay to eliminate the boundary flux, is enough; compact support in time is unnecessary.

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.

A local Maxwellian can be written as $\mathcal{M}=\exp (a+b\cdot v+c|v{|}^2)$, with $c<0$ and coefficients depending on $(t,x)$. Its logarithm is a linear combination of collision invariants, so ${\mathcal{M}}^{\prime }{\mathcal{M}}_{\ast }^{\prime }=\mathcal{M}{\mathcal{M}}_{\ast }$ and $Q(\mathcal{M},\mathcal{M})=0$ at each spatial point.

Comparing powers of $v$ in $({\partial }_t+v\cdot {\nabla }_x)\log \mathcal{M}=0$ givesHere $\mathrm{sym}\nabla b$ is the symmetric part of a matrix formed by the coefficients of the quadratic polynomial. For constants $A,\alpha ,\beta >0$, choose $a=\log A-\alpha |x{|}^2$, $b=2\alpha tx$ and $c=-\beta -\alpha t^2$. All four coefficient equations hold. The resulting expanding Gaussian solution of the Boltzmann equation isDirect verification is particularly simple: $x-tv$ and $v$ are constant along free characteristic curves, so the transport derivative is zero. At fixed $(t,x)$ its logarithm is the displayed Maxwellian quadratic, so the collision term is also zero. Its spatial dependence is nonconstant, it is positive and rapidly decaying, and its Gaussian velocity integral is finite because $\beta +\alpha t^2>0$. It is not a compactly supported example, nor does this final part require one.