Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 7 (Mathematical Topics in Kinetic Theory)

Use $(y,w)$ to label the initial point in phase space. The characteristic equations for a transport equation are $\dot{X}=a(V)$ and $\dot{V}=0$, with $X(s)=y$ and $V(s)=w$. Since the velocity remains fixed, the ordinary differential equations integrate directly:This characteristic flow map exists for all real times, even if $a$ grows rapidly at infinity: each trajectory samples just one fixed velocity. Its inverse is $S_{t,s}$, and its Jacobian matrix is the block triangular matrixConsequently its Jacobian determinant is one. The nonzero Jacobian determinant of $a$ is unnecessary for this construction.

The chain rule gives $d[f(t,X(t),V(t))]/dt=0$ along each characteristic curve. Pulling the initial datum back by the inverse characteristic flow map therefore givesDirect differentiation also gives $f_t=-a(v)\cdot {\nabla }_{x}f$, so this is a classical solution of the transport equation. Composition with a smooth function makes $f$ smooth jointly in $(t,x,v)$; nonnegative initial data remain nonnegative.

If $K=\mathrm{supp}{f}_{\mathrm{in}}$, then $\mathrm{supp}f(t)=S_{0,t}(K)$. A continuous image of a compact set is compact. Thus compact support is preserved at every finite time; there is no claim of one fixed compact support for all $t\ge 0$.

Choose $R_0$ with $K\subset \{(y,w):|y|\le R_0,|w|\le R_0\}$ and let $A_0={\max }_{|w|\le R_0}|a(w)|$, which is finite by continuity. Every point transported from $K$ satisfiesHence anycontains the support of a function in the required phase space ball throughout $[0,T]$. Equivalently, the image of the compact set $[0,T]\times K$ under the characteristic flow map is compact. This uniform support control puts the solution in ${\mathcal{C}}_T$ and justifies the following differentiations under the integral.

For two solutions with the same initial datum, let $h=f_1-f_2$. Their uniform compact support on $[0,T]$ puts $h$ and $h_t$ in one compact region of phase space. Smoothness there implies that $E(t)=\int |h(t,x,v){|}^{2}dxdv$ is finite and continuously differentiable, by differentiation under the integral sign. The transport equation and integration by parts in $x$ yieldThere is no boundary contribution because of compact support, and ${\mathrm{div}}_{x}a(v)=0$. Since $E(0)=0$, $h=0$ almost everywhere and then everywhere by continuity. The solution is unique in ${\mathcal{C}}_T$. The same calculation gives Lp conservation for incompressible transport at $p=2$.

For the free transport equation, the mixed Lebesgue norm dispersion estimate isIndeed, use the method of characteristics and then the change of variables $y=x-tv$:Taking the essential supremum over $x$ proves the result. This decay measures the spreading of a velocity average; the total phase space Lp norm is conserved rather than decaying. The datum on the right is the initial datum, not the solution at the same time.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

Fix the standard Jacobian matrix convention $(Da{)}_{ij}={\partial }_{v_j}{a}_i$. With column gradients, the commuting velocity derivative for kinetic transport isIf the symbol $J_a$ is instead defined with derivative indices as rows, the printed expression is this same formula. Under the standard convention it needs a transpose.

To prove the claim, put $D={\partial }_t+a(v)\cdot {\nabla }_x$. Differentiating $Df=0$ in $v_j$ givesThe terms cancel, so each component of $G$ satisfies the transport equation. Multiplication by the smooth coefficient $Da$ and differentiation preserve the same uniform compact support, hence $G\in {\mathcal{C}}_T$ componentwise. At $t=0$, $G(0)={\nabla }_v{f}_{\mathrm{in}}$. The L2 norm calculation in part (d), summed over components, givesOne can also differentiate the explicit solution: the $t(Da{)}^T{\nabla }_{x}f$ term cancels the derivative of $x-ta(v)$, leaving the initial velocity derivative transported along the characteristic curves.

The transpose matters. For the shear $a(v)=(v_1+v_2,v_2)$, $\det Da=1$ but $Da\ne (Da{)}^T$. The printed field satisfies $Dg=(Da-(Da{)}^T){\nabla }_{x}f$, which is nonzero for a generic compactly supported datum. The untransposed formula is false with the standard convention.

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

Take canonical coordinates $q=x_1$ and $p=x_2$. The Hamiltonian system isIn the standard Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$, its Hamiltonian isAn arbitrary function of $t$ can be added without affecting the characteristic curves. The Hamiltonian generally depends on time through $\omega$ and need not be conserved along a trajectory.

Let $S_{s,t}(y)$ be the position at time $t$ of the characteristic curve satisfying $X(s)=y$ and $\dot{X}=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the variational equationThe Liouville formula for a fundamental matrix, or the derivative formula for a determinant, therefore givesBecause $u=(-{\varphi }_{x_2},{\varphi }_{x_1})$, the mixed second derivatives cancel and $\mathrm{div}u=0$. The initial Jacobian determinant is one, soThus this Hamiltonian flow preserves planar Lebesgue measure. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.

The planar vorticity velocity kernel has magnitude $(2\pi |z|{)}^{-1}$. Split its defining integral into $|x-y|<R$ and $|x-y|\ge R$ for any $R>0$. The singular part is absolutely integrable in two dimensions:In particular $R=1$ provesWhen both norms are nonzero, optimization in $R$ also gives $\Vert U[\Omega ]{\Vert }_{\infty }\le \sqrt{2/\pi }\Vert \Omega {\Vert }_1^{1/2}\Vert \Omega {\Vert }_{\infty }^{1/2}$. If either norm is zero, $\Omega =0$ almost everywhere. Absolute convergence supplies a well-defined velocity at every $x$, with these uniform bounds.

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.

Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected log-Lipschitz modulus $\mu$ from part (e). Local existence for the continuous finite-dimensional vector field follows from the Peano existence theorem. Boundedness gives $|X(t)-X(s)|\le \Vert U{\Vert }_{\infty }|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. Solutions exist globally.

For uniqueness, let $X,Y$ have the same initial point and set $\delta (t)=|X(t)-Y(t)|$. This absolutely continuous function satisfies, almost everywhere,This is the Osgood uniqueness criterion, rather than the usual linear Gronwall inequality, because the velocity need not be Lipschitz continuous. To see the argument directly, put $w_{\epsilon }=\delta +\epsilon$. As long as $w_{\epsilon }\le 1$, monotonicity of $\mu$ givesFor $z_{\epsilon }=1-\log w_{\epsilon }$, this becomes $z_{\epsilon }^{\prime }\ge -C{z}_{\epsilon }$. Integration givesFor any fixed finite interval, this upper bound tends uniformly to zero as $\epsilon \downarrow 0$. Taking $\epsilon$ small prevents a first exit through $w_{\epsilon }=1$, so the bound remains valid throughout that interval. Thus $\delta =0$. Backward time has the same proof. Every initial point has exactly one global trajectory.

The decisive fact is ${\int }_0^{1}dr/\mu (r)=\infty$. For different nearby initial points, the same comparison givesuntil the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth diffeomorphism.

For smooth vorticity, parts (b)--(c) and the transport equation give $\omega (t,S_{0,t}(y))={\omega }_{\mathrm{in}}(y)$ and measure preservation. The change of variables formula therefore conserves $\Vert \omega (t){\Vert }_1$ and $\Vert \omega (t){\Vert }_{\infty }$. Parts (d)--(e) give a uniformly bounded velocity and a uniform log-Lipschitz modulus; applying the time-dependent version of part (f) constructs unique global characteristic curves.

For a rough Yudovich characteristic flow, state the needed temporal hypothesis explicitly:The usual weak Euler solution class provides these conditions and the initial time trace. On $[0,T]$, put $A(t)=\Vert \omega (t){\Vert }_1+\Vert \omega (t){\Vert }_{\infty }$. The velocity bounds are $|u(t,x)|\le CA(t)$ and $|u(t,x)-u(t,y)|\le CA(t)\mu (|x-y|)$, with $A$ integrable on this interval.

For completeness, mollify $u$ spatially. The resulting smooth velocities have trajectories and obey the same speed and modulus bounds, up to one common constant. Their speeds have the integrable majorant $CA(t)$, giving uniform boundedness and equicontinuity of trajectories. Arzelà-Ascoli theorem supplies a uniformly convergent subsequence. The modulus estimate controls $u(t,X_{\epsilon })-u(t,X)$, while the spatial mollification error tends to zero with an integrable majorant. Thus the limit solvesReplace $Ct$ by $C{\int }_0^{t}A(s)ds$ in the Osgood uniqueness criterion proof. It gives uniqueness and prevents finite-time escape. The characteristics are globally uniquely defined for each given Euler solution in the usual Yudovich class. This constructs the flow for a fixed solution; it is not by itself the entire uniqueness proof for the nonlinear Euler equation.

Merely saying that both spatial norms are finite separately at each time does not state the local temporal bound or measurability used in this argument. The printed formulation must be interpreted in the usual solution class, or supplemented with these temporal conditions. Also, an initial datum here is a function of $x\in {\mathbb{R}}^2$; the extra time variable in the displayed initial-data space is extraneous.

If $H$ is a finite-dimensional Hilbert space, a choice of orthonormal basis identifies its closed unit ball with a closed bounded ball in a finite-dimensional Euclidean space. The Heine-Borel theorem makes it compact in the norm topology.

Conversely, if $H$ is infinite-dimensional, choose a unit vector $e_1$ and inductively choose $e_{n+1}$ orthogonal to the span of $e_1,\dots ,e_n$. The finite span is a proper closed subspace, so its orthogonal complement contains a nonzero vector that can be normalized. The resulting orthonormal sequence lies in the closed unit ball and satisfiesIt has no norm-convergent subsequence. Sequential compactness is equivalent to compactness in a metric space, hence the closed ball is not compact. This proves the norm-compact unit ball criterion: a Hilbert space has a norm-compact closed unit ball exactly when it is finite-dimensional.

Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.

If $|z|>\Vert L\Vert$, the Neumann series givesThe series converges in operator norm because $\Vert L/z\Vert <1$. HenceThe same argument applied to small perturbations of an invertible operator shows that the resolvent set is open and the spectrum is closed.

Suppose $L=L^{\ast }$ and $\mathrm{Im}z\ne 0$. The quadratic form $⟨Lh,h⟩$ is real, so the Cauchy-Schwarz inequality impliesThus $L-zI$ is bounded below by $|\mathrm{Im}z|$: it is injective and has closed range. Its adjoint operator $L-\bar{z}I$ has the same lower bound and trivial kernel. Sincethe range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\mathrm{Im}z{|}^{-1}$. For a bounded self-adjoint operator, $\Sigma (L)\subset \mathbb{R}$. The closed-range and adjoint steps are essential: merely excluding nonreal eigenvalues would not exclude all nonreal spectral points.

Suppose the compact operator $K$ acts on $H$, and $h_n\rightharpoonup h$. The Uniform boundedness principle makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $K{h}_n\rightharpoonup Kh$: for every $g$, $⟨K{h}_n,g⟩=⟨h_n,K^{\ast }g⟩\to ⟨Kh,g⟩$. Every norm-convergent subsequence of $(K{h}_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that compact operators send weak convergence to norm convergence.

Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a Hilbert space has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the unit ball is relatively sequentially compact and hence relatively compact. Complete continuity and compactness are equivalent on a Hilbert space.

We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

Use the stated Fredholm definition of the essential spectrum, which applies to general bounded operators:For each $z$, part (f) applied to $L-zI$ says that $L-zI$ is Fredholm exactly when $L+K-zI$ is Fredholm. Taking the complement of these equivalent Fredholm conditions givesThis is the general Weyl theorem for bounded compact perturbations. Individual isolated eigenvalues can change; the Fredholm obstruction, and hence this essential spectrum, cannot.

A kinetic model of a dilute gas. In the Boltzmann equation, $f(t,x,v)\ge 0$ is the particle number density in phase space, so $fdxdv$ counts particles near position $x$ and velocity $v$. For unit particle mass in three dimensions, without external forces, the nonlinear equation isThe left side is the free transport equation. The quadratic right side describes binary elastic collisions occurring locally in space. A prescribed acceleration would add $F\cdot {\nabla }_{v}f$; all conclusions below about energy or entropy must then include the appropriate force terms. Assume enough smoothness, velocity decay and finite moments to justify the calculations, and either periodic space or decay that removes spatial boundary fluxes. These are assumptions on a classical solution, not a general existence theorem.

The dilute-gas approximation neglects simultaneous many-body collisions and treats the interaction region as small relative to macroscopic spatial scales. Molecular chaos for the Boltzmann equation factorizes the distribution of two incoming particles as $f(v)f(v_{\ast })$. It applies to incoming collision pairs; it does not assert that an interacting many-particle system stays exactly independent. This approximation closes the equation at the one-particle level and explains its nonlinearity.

Collision geometry and the gain-loss operator. Let $g=v-v_{\ast }$ and $n\in S^2$. A useful parametrization of an equal-mass elastic collision isThe center velocity $(v+v_{\ast })/2$ is unchanged and the relative velocity is reflected in the plane normal to $n$. ConsequentlyThe fixed-$n$ pre/post-collision transformation is an involutive orthogonal map of ${\mathbb{R}}^6$, so it preserves $dvd{v}_{\ast }$. Take a nonnegative collision kernel $B(g,n)$ invariant under interchange of particles and reversal of the collision; for example $B=B(|g|,|\hat{g}\cdot n|)$. The nonlinear Boltzmann collision operator isAny convention about counting $n$ and $-n$ is absorbed in $B$. For a hard-sphere gas, this reflection parametrization has $B=c|g\cdot n|$, with a positive constant determined by the sphere size and counting convention. Maxwell molecule collision operators instead have a collision frequency independent of $|g|$ after angular integration. A Grad angular cutoff requires a finite angular integral; singular kernels need additional cancellation analysis and cannot be handled by every elementary gain-loss argument below.

Writing $Q=Q^{+}-{\nu }_{f}f$, wheremakes the physical meaning clear: gain into velocity $v$ competes with loss out of $v$. Along a free characteristic curve, put $F(t)=f(t,x+tv,v)$. Then $F^{\prime }+{\nu }_f(t,x+tv,v)F=Q^{+}(t,x+tv,v)$. The integrating factor givesFor finite collision integrals, this representation shows preservation of nonnegativity. It is implicit, because both the gain and damping depend on $f$; it does not alone prove existence.

The weak collision identity and conserved quantities. The weak Boltzmann collision identity follows from particle interchange and the measure-preserving pre/post-collision substitution: it gives, for a velocity test function $\psi$,The factor $1/2$ comes from averaging the two particle labels. Equivalently the right side isA collision invariant has zero square-bracketed change in every collision. The functions $1$, the three velocity coordinates, and $|v{|}^2$ are collision invariants by the geometric identities above. Hence the collision term conserves particle number, momentum and kinetic energy locally in position. Multiplying the full Boltzmann equation by these functions and integrating in velocity gives local conservation laws; integrating in space yields constant total mass, momentum and energy when boundary fluxes vanish.

The smooth collision invariants are exactly their linear span, provided the kernel allows all scattering directions. Here is a direct proof. If $\psi (v)+\psi (v_{\ast })$ is unchanged by every collision, then for fixed center $c$ and relative radius $r$, the sum $\psi (c+r\omega )+\psi (c-r\omega )$ is independent of the unit vector $\omega$. Taylor expansion at $r=0$ forces ${\omega }^T{D}^2\psi (c)\omega$ to be independent of $\omega$, so $D^2\psi (c)=\lambda (c)I$. Mixed derivatives vanish. Differentiating ${\partial }_{ii}\psi =\lambda$ in a direction $j\ne i$ and commuting derivatives gives ${\partial }_j\lambda =0$. Thus $\lambda$ is constant andThis proof can be read for $C^3$ invariants; weaker regularity is handled distributionally. The multidimensional scattering hypothesis matters: in one dimension equal-mass elastic collisions only exchange the two velocities, and every function is then a collision invariant.

Entropy dissipation and the H theorem. Define the Boltzmann H functional $H(f)=\int f\log fdxdv$, with $0\log 0=0$. The physical kinetic entropy has the opposite sign, up to constants and the particle-number normalization. Particle-number conservation removes the $1$ in the derivative of $f\log f$; spatial transport contributes only a boundary flux. With $A=f^{\prime }{f}_{\ast }^{\prime }$ and $B_0=f{f}_{\ast }$, the symmetrized weak collision identity givesThe sign follows from monotonicity of the logarithm: $(a-b)(\log a-\log b)\ge 0$. Zeros are treated by the usual extended nonnegative limit. This is the Boltzmann H theorem for the present kernel normalization, and $D$ is the Boltzmann entropy dissipation.

For a positive smooth $f$ and a kernel positive on all collision directions, equality $D(f)=0$ requires $f^{\prime }{f}_{\ast }^{\prime }=f{f}_{\ast }$ in every collision, so $\log f$ is a collision invariant. Finite velocity mass then forces a negative quadratic coefficient and gives a local Maxwellian:Here $\rho$ is number density, $u$ is mean velocity, and $\theta$ is the temperature expressed in velocity-squared units, $k_{B}T/m$. Completing the square proves the formula; conservation of momentum and kinetic energy makes $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$, so $Q(M,M)=0$. Vacuum is another collision equilibrium.