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The exact normalizing sum can be expensive on a large loss network. The Erlang fixed point approximation replaces joint resource acceptance by a product of marginal acceptances. Here consider unit requirements $A_{jr}\in \{0,1\}$, positive integer capacities and finitely many fixed routes. Let $B_j$ be resource blocking and $q_j=1-B_j$. A route-$r$ call contributes to the offered traffic at resource $j$ after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is ${\prod }_{j\in r}{q}_j$. This is an independence approximation, not an alternative exact factorization of the stationary law in part (i).

Uniqueness follows from a convex potential for the Erlang fixed point. For a single resource, the carried load of an Erlang loss resource is $m_C(v)=v[1-E(C,v)]$. It increases strictly from zero to $C$ as $v$ increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to $\log v$ gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.

Use $p_j=-\log q_j\ge 0$. Let $v_C(p)$ be the unique offered load with $1-E(C,v_C(p))=e^{-p}$, and set $g_C(p)=v_C(p)e^{-p}=m_C(v_C(p))$, with $g_C(0)=0$. This function is strictly increasing and tends to $C$. DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore $F$ is a strictly convex function. It is also a coercive function on the nonnegative orthant: an unbounded coordinate makes its integral grow asymptotically linearly with positive slope $C_j$. A unique minimizer exists.

If resource $j$ carries some positive offered route, its inward derivative at $p_j=0$ is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by $q_j$ gives precisely $v_j={\sum }_{r:j\in r}{\rho }_r{\prod }_{k\in r\setminus \{j\}}{q}_k$. A resource with no positive offered route uniquely has $p_j=0$, hence $B_j=0$. Thus the minimizer and the fixed point coincide, proving existence and uniqueness of the Erlang fixed point for fixed unit-resource routing. This does not by itself guarantee convergence of every simultaneous substitution algorithm; the uniqueness claim concerns the solution of the equations.

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.

Let $A_{\delta }={\sup }_{K\ge 0}{K}^{\delta }C(K)<\infty$. The call-price decay and moment threshold follows by splitting the preceding integral at one. Since $(S-K{)}_{+}\le S$, for $0<\epsilon <\delta$,For $K\ge 1$, the decay bound givesThe power payoff static call representation therefore yieldsThe $\epsilon =0$ case is the given finite first moment. The strict endpoint matters: a Pareto distribution with $\mathbb{P}(S>x)=x^{-(1+\delta )}$ for $x\ge 1$ has $C(K)=K^{-\delta }/\delta$ for $K\ge 1$, but its moment of order $1+\delta$ is infinite. Thus the stated decay condition does not generally imply the endpoint moment.

Let $u={\delta }_m/H$. The supplied equation has integrating factor $(aH{)}^3$:Consequently the integral linear growth factor in a matter-Lambda universe gives the two independent solutionsThe first is the conventional matter-era decaying mode. A finite lower limit in the second fixes a decaying admixture; adding a multiple of $H$ can choose a pure growing normalization.

During matter domination, write $H=h{a}^{-3/2}$. The integral solution isRemoving the second term by the independent $H$ solution gives the matter-era linear growth factorFor positive cosmological constant at sufficiently late times, $H\to H_{\Lambda }>0$. The integrand approaches $H_{\Lambda }^{-3}{a}^{-3}$, whose tail is integrable. ThereforeIts remaining approach has leading $a^{-2}$ behavior for the usual matter-plus-Lambda background. Growth of structure freezes rather than continuing as $a$. The basis function $H$ also approaches a constant; subtracting a suitable multiple of it isolates a genuinely decaying late-time solution $\propto a^{-2}$. The name “decaying mode” for $H$ refers to its matter-era behavior.

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 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.

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.

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.

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 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.

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.