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When link delays are separable and nondecreasing, Wardrop equilibria minimize the Beckmann potentialover the feasible route-flow polytope. Its derivative with respect to $h_r$ is $c_r(h)$. Since the integrals are convex, the first-order inequality for a global minimum is exactly the Wardrop equilibrium inequality in part (a). Alternatively, the Karush-Kuhn-Tucker conditions equate used-route derivatives to the multiplier for their demand constraint, with larger derivatives on unused routes. Continuity gives existence on the compact feasible set. Strictly increasing link delays give unique link flows, although different route decompositions can still give route-flow nonuniqueness at a Wardrop equilibrium.

The system objective is instead total travel time,For differentiable delays its marginal cost is ${\ell }_j(y_j)+y_j{\ell }_j^{\prime }(y_j)$, not simply ${\ell }_j(y_j)$. Thus an equilibrium optimizing the Beckmann potential generally fails to minimize total delay. A marginal external cost toll $y_j{\ell }_j^{\prime }(y_j)$ makes travelers face the social marginal cost and implements a system optimum under the appropriate convexity assumptions.

These formulations clarify the modeling distinction: nonatomic users optimize their own paths; a planner optimizes total delay. For nonseparable or nonmonotone costs, the variational description can remain applicable while the Beckmann potential argument, existence or uniqueness needs new hypotheses.

In the slotted ALOHA model, a station retains one packet until it successfully transmits. Conditional on $(S_t,N_t)=(S,N)$, take fresh independent trials with a Bernoulli distribution of parameter $1/S$ for the $N$ backlogged stations. Let arrivals be independent across slots and independent of these trials, with Poisson distribution of mean $\nu$. New arrivals join the next slot's backlog. An idle slot serves nobody, exactly one attempt serves one packet, and a collision serves nobody. Consequently the backlog changes by arrivals minus the success indicator.

Under these independence assumptions, $(S_t,N_t)$ is a time-homogeneous Markov chain: the feedback probabilities and the independent arrival law depend only on its current state. Merely specifying Poisson arrival marginals would not suffice. For example, take $S_t=1$, $a=b=c=0$, $N_0=0$, and arrivals $Y_0=H$, $Y_1=K$, $Y_2=H$ for independent random variables $H,K$ with Poisson distributions. Histories with $(N_1,N_2)=(0,2)$ and $(2,2)$ both have current backlog two, but the next backlog is respectively two and four. The usual model therefore includes fresh arrivals as an assumption.

Write $p_0=(1-1/S{)}^N$ and $p_1=(N/S)(1-1/S{)}^{N-1}$. Set $p_1=0$ for $N=0$, and use their exact binomial distribution probabilities at $S=1$. Away from the clipping boundary, the conditional drifts areNear $S=1$, the latter must use $\max (1,S+d)-S$ for each update $d$, rather than $d$ itself.

When both coordinates are large with $N/S\to \kappa$, the Poisson limit theorem approximates the attempt count, which has a binomial distribution, by a Poisson distribution of mean $\kappa$. Thus $p_0\to e^{-\kappa }$ and $p_1\to \kappa e^{-\kappa }$. Rescaling state by $K$ and slot time by $K$ motivates the fluid approximation of slotted ALOHAThis is an interior fluid approximation of slotted ALOHA, not the exact conditional drift at a clipped boundary.

Here is one explicit set of sufficient conditions, independent of the arrival rate within the subcritical range:Put $c_0=2/(e-2)$. ThenHence $g$ is negative below one and positive above one: the controller decreases the attempt denominator when offered contention is too small, and increases it when contention is too large.

Use the ratio time change for a homogeneous fluid model, $du/dt=1/s$. The equations becomeFor $\kappa \ge 1$,since $g(\kappa )\ge 0$ and $A(1+c_0)>1$. Also $h(1)=\nu -e^{-1}<0$. Therefore $h$ is negative on $[1,\infty )$, and by continuity on $[\bar{\kappa },\infty )$ for some $\bar{\kappa }<1$. Its value at zero is $\nu \ge 0$, so nonnegative backlog is preserved.

The ratio remains bounded by $K=\max (\kappa (0),1)$ and eventually enters $[0,\bar{\kappa }]$: on the compact interval $[\bar{\kappa },K]$, its derivative is bounded above by a strictly negative number. Once inside it cannot cross upward. The bounded ratio and smooth coefficients make the transformed equations exist for all $u\ge 0$. In this region $g(\kappa )\le g(\bar{\kappa })<0$, so $s(u)$ decreases at least exponentially in $u$. The original time satisfies $t(u)={\int }_0^{u}s(v)dv$ and has a finite limit $T$; boundedness of $\kappa$ gives $n(u)=\kappa (u)s(u)\to 0$ as well. Thus every nonnegative interior fluid trajectory drains to the origin in finite fluid time. At the origin the ratio equation is undefined; the usual stopped fluid trajectory is held there afterward. A state with $s=0<n$ enters the interior under the continuous limiting boundary drift $\dot{s}=c>0$; the same argument then applies.

This proof supplies sufficient conditions, not a characterization of every stabilizing triplet. It also does not assert the positive recurrent Markov chain property of the original stochastic chain solely from the heuristic ODE.

If $\nu >e^{-1}$, the ALOHA throughput bound givesNo choice of the three feedback increments can make this fluid backlog drain, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$. For the displayed controller, the supercritical fluid trajectory is global and grows instead of draining. At $\nu =e^{-1}$ the ray $n=s>0$ consists of stationary fluid states, explaining why the strict load inequality matters.

Use the usual essential two-person Nash bargaining problem: $F\subset {\mathbb{R}}^2$ is a compact convex set, $d\in F$ is the disagreement point, and some $u\in F$ satisfies $u_i>d_i$ for both players. The Nash bargaining solution isThe positive maximum exists by compactness and essentiality. On positive gains, maximizing this Nash product is equivalent to maximizing $\log (u_1-d_1)+\log (u_2-d_2)$, a strictly concave function. Convexity then gives a unique maximizer. The essentiality and compactness hypotheses matter: for example, with $F=[0,1]\times \{0\}$ and $d=0$, the product is zero everywhere and its argmax alone is not a single-valued definition.

The rule satisfies all four axioms. Pareto efficiency: a feasible vector dominating the chosen vector with at least one strict improvement would increase its positive Nash product. Bargaining symmetry: if the problem is unchanged by swapping players, uniqueness makes the answer unchanged, so the two payoffs agree. Positive affine invariance in bargaining: for $v_i=a_i{u}_i+b_i$ with $a_i>0$, gains transform to $a_i(u_i-d_i)$ and the product is multiplied by the positive constant $a_1{a}_2$, preserving its maximizer. Bargaining independence of irrelevant alternatives: if $G\subseteq F$ is another admissible feasible set containing the chosen vector and the same disagreement point, that vector remains the unique product maximizer over $G$.

To prove characterization, let $f$ be any feasible single-valued rule satisfying these axioms, and let $u^{\ast }=N(F,d)$. Normalize payoffs by the positive affine transformationThe transformed set $H$ has disagreement point zero and product maximizer $e=(1,1)$. For any $z\in H$, convexity puts $e+t(z-e)$ in $H$; for sufficiently small $t>0$ both gains remain positive. The one-sided derivative of the product at its maximum is therefore nonpositive:Thus $H\subseteq \{z:z_1+z_2\le 2\}$. By compactness choose $M\ge 0$ so every coordinate of every $z\in H$ is at least $-M$. The supporting triangle for Nash bargaining isIt contains $H$, is compact, convex, symmetric and essential, and contains disagreement zero. Symmetry forces $f(T_M,0)$ onto the diagonal; Pareto efficiency then forces it to be $e=(1,1)$. Since $e\in H\subseteq T_M$, bargaining independence of irrelevant alternatives gives $f(H,0)=e$. Undoing the normalization by positive affine invariance in bargaining gives $f(F,d)=u^{\ast }$. Hence the four axioms uniquely characterize the Nash bargaining solution on this domain.

A security level payoff maximizes what a player guarantees against the opponent. Let $p$ be the row player's probability of the first action. Its guaranteed payoff isThe decreasing and increasing terms cross at $p=1/5$, attaining $d_1=12/5$; moving either way lowers the smaller term. For the column player, any mixture has zero payoff against the first row, while its second-row payoff is nonnegative. Therefore $d_2=0$ andThe feasible set is the convex hull of the four joint pure-action payoff vectors. Its upper Pareto frontier connects $(4,0)$ to $(3,2)$ to $(2,3)$. Write the row payoff as $u$ and column payoff as $v$. On the first Pareto frontier segment, $v=8-2u$ for $3\le u\le 4$. The Nash product isa concave quadratic with derivative $64/5-4u$, maximized at $u=16/5$, $v=8/5$, giving product $32/25$. On the portion of the other Pareto frontier segment satisfying bargaining individual rationality, $12/5\le u\le 3$ and $v=5-u$. Its product derivative $37/5-2u$ is positive throughout, so its largest product is $6/5$ at $(3,2)$, smaller than $32/25$. All dominated points can be discarded by Pareto efficiency. ConsequentlyThis payoff is implemented by a correlated payoff lottery choosing the payoff $(4,0)$ with probability $1/5$ and $(3,2)$ with probability $4/5$. The convex hull permits such lotteries over joint outcomes; it is not restricted to independent mixed strategies.

Transposing the second player's matrix leaves the row player's security level payoff at $12/5$. The column player's second action now yields payoffs $2$ and $3$, while its first yields zero for either row. Its security level payoff is therefore $2$, guaranteed by the second action; against the first row no mixture can guarantee more. Thus $d=(12/5,2)$.

The relevant upper Pareto frontier joins the payoff vectors $(4,2)$ and $(2,3)$, so $v=4-u/2$. On the segment satisfying bargaining individual rationality $12/5\le u\le 4$, the Nash product becomesIts derivative is $16/5-u$ and its second derivative is $-1$. HenceThe implementing correlated payoff lottery chooses $(4,2)$ with probability $3/5$ and $(2,3)$ with probability $2/5$. Both players' gains are strictly positive: $4/5$ and $2/5$. The new disagreement point must be recomputed after transposition; reusing the previous column security payoff would solve a different Nash bargaining problem.

Differentiate the linearized cosmological continuity equation, remembering the derivative of $1/a$:Insert the linear cosmological Euler equation for the peculiar velocity, givingThus the linear matter perturbation growth equation isThe factor $2H$ includes one expansion term from continuity and one from momentum dilution. This equation uses sub-Hubble, pressureless linear perturbations and the stated matter-only source for the gravitational potential.

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le \Vert g{\Vert }_{\infty }$. Its approximate identity property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is continuous. This bounded solution satisfies the required Gaussian-growth bound.

For heat-equation uniqueness under Gaussian growth, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0{e}^{a|x{|}^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le \tau <1/(4b)$. The positive comparison solutionsatisfies $({\Phi }_b{)}_t=\Delta {\Phi }_b$. For any $\epsilon >0$, its faster spatial growth makes $|w|\le \epsilon {\Phi }_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The heat equation maximum principle applied to $\pm w-\epsilon {\Phi }_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\epsilon \downarrow 0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.

At time $T$, this is Gaussian filtering with covariance $2TI$, hence $\sigma =\sqrt{2T}$ per coordinate. With the angular-frequency Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, the Gaussian filtering Fourier multiplier isFor merely bounded $g$, this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency $k$, the multiplier is $e^{-4{\pi }^{2}T|k{|}^2}$.

For the printed Perona-Malik equation, retain the supplied conductance $c(s)=s{e}^{-s^2/(2{\lambda }^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2{\lambda }^2)}$. Then $u_t=F^{\prime }(u_x)u_{xx}$, withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt $|u_x|=0$ and $\sqrt{2}\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2{\lambda }^2)}$ would give threshold $\lambda$, but that is a different equation.

The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf $u(x_0)={\mathrm{mean}}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the limit give $\Delta u(x_0)=0$. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.

For the area median over the disk, the disk-median curvature expansion, at a point where $|\nabla u|\ne 0$, isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.

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.

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.

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.