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A distributional weak solution of ${\partial }_{t}f+v\cdot {\nabla }_{x}f=g$, with locally integrable $f,g$, becomes ${\partial }_{t}F=G$ after the shear $F(t,x,v)=f(t,x+tv,v)$. For almost every characteristic label, there is an absolutely continuous function representing $F$, and its increments are the integrals of $G$ for every pair of times. Arbitrary modifications on time slices can destroy this every-time identity without changing the distributional solution.

For the Maxwell molecule collision operator and the Fourier transform with exponent $-i\xi \cdot v$, put ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$. The angular exchange for elastic collisions, rotation invariance of spherical surface area and factorization of the two velocity integrals give the displayed formula. This identity converts the collision integral into a spherical average of products at two related frequencies.

The integral uses $0\log 0=0$ when it is defined. For the Boltzmann equation, the Boltzmann H theorem makes it nonincreasing under suitable regularity and integrability. This functional is distinct from the microstate-counting Boltzmann entropy; the physical kinetic entropy has the opposite sign.

For steady equal-pressure flow, mass conservation gives $Aw=A_0{w}_0$ and the kinematic boundary condition gives $(D\sqrt{w}{)}_z=\gamma /(2\mu \sqrt{w})$, where $D=R_e-R_i$. Therefore finite-height closure occurs exactly whenFor positive velocity the integral is increasing. If it diverges, closure is finite; if its limiting value lies below the right-hand threshold, there is no finite closure. Quadratic freely falling profiles have logarithmically divergent integrals, while sufficiently strongly drawn exponentially accelerating profiles can have finite integrals. Thus nozzle speed alone is insufficient to guarantee closure: the axial tension or downstream condition matters.

For an axisymmetric slender viscous thread, the Young–Laplace equation and radial extensional viscous stress give $p=p_e+\gamma /a-\mu w_z$. The axial liquid stress relative to ambient pressure is $3\mu w_z-\gamma /a$. Its cross-sectional integral plus the circumferential surface tension force $2\pi \gamma a$ gives the displayed tensile force. Thus the net capillary contribution is $\pi \gamma a$, and axial force balance with downward gravity is $T_z+\pi \rho g{a}^2=0$.

For a transport equation ${\partial }_{t}f+b(t,z)\cdot {\nabla }_{z}f=h$, solve $\dot{z}=b(t,z)$. The chain rule then gives $d[f(t,z(t))]/dt=h(t,z(t))$. Initial data and the time integral of this source determine the solution along each characteristic curve.

For a real continuous function $f$ on $[0,1]$, let $G_n$ be its secant slope on each length-$2^{-n}$ dyadic cell. Regard $([0,1],\mathcal{B},dt)$ as a probability space and use the filtration of half-open dyadic cells with the endpoint $\{1\}$ as a separate null cell. The mean of the two child slopes equals the parent slope, so $(G_n)$ is a martingale. Its integral gives the linear interpolation of $f$ on that grid. If $f$ is Lipschitz continuous, these slopes are bounded by its Lipschitz constant; the Lp martingale convergence theorem then supplies a bounded integral density for $f$.

For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

Unit-mass solutions of the isotropic Maxwell molecule collision operator with matching first moments obey this comparison. The Bobylev identity, the Fourier bound by one, and $|{\xi }^{+}{|}^2+|{\xi }^{-}{|}^2=|\xi {|}^2$ give a scalar damped differential inequality. Its integral form and the Gronwall inequality yield nonexpansion in the Fourier distance of order two. This does not alone establish strict contraction or equilibrium convergence.

When $H\to H_{\Lambda }>0$, the growth integral has an integrable $a^{-3}$ tail. Its product with $H$ tends to a constant, with leading correction proportional to $a^{-2}$ in a matter-plus-Lambda background. Both displayed basis functions can tend to constants; a linear combination cancels that constant and isolates the truly decaying late-time $a^{-2}$ behavior.

On $z>0$, consider $p_t+a{p}_z=D(z{p}_z{)}_z$ with $a,D>0$, zero scalar flux $ap-Dz{p}_z$ at the endpoints for $t>0$, and unit initial impulse at the origin. Put $r=a/D$ and $\eta =z/(Dt)$. The normalized similarity solution isThe ansatz $p=(Dt{)}^{-1}f(\eta )$ gives $[\eta f^{\prime }+(\eta -r)f{]}^{\prime }=0$. Endpoint decay sets this integrated constant to zero and hence $f\propto {\eta }^r{e}^{-\eta }$; the gamma function normalizes its integral to one. Its scale is $Dt$, so it converges weakly to a unit impulse as $t\downarrow 0$. Its gamma distribution shape is $1+r$, its maximum occurs at $z=at$, and its expected value is $(a+D)t$. For width proportional to $z$, the associated horizontally averaged plume concentration is proportional to $z^{r-1}{e}^{-z/(Dt)}$: its interior maximum is $(a-D)t$ when $a>D$, its supremum occurs at the origin when $a=D$, and it is singular there when $0<a<D$. The physical finite source regularizes this ideal-origin behaviour.

Suppose a nonnegative stopping time $T$ and constants $L>0$, $0<q\le 1$ satisfy $\mathbb{P}(T\le nL+L\mid {\mathcal{F}}_{nL})\ge q$ on $\{T>nL\}$. Take $L$ to be a positive integer in discrete time. ThenOn the surviving event, conditional survival is at most $1-q$. The tower property of conditional expectation therefore givesInduction proves the geometric bound. Partitioning the tail integral into intervals of length $L$ then gives $\mathbb{E}T\le L{\sum }_{n\ge 0}\mathbb{P}(T>nL)\le L/q$; in discrete time, group the tail sum into $L$ successive integer indices instead. This proves finite expected hitting times when every surviving state has a uniform positive chance of escaping within a fixed time. It works in discrete or continuous time without assuming independence of successive survival events.

Write $\alpha =r+\beta$ with an integer $r\ge 0$ and $0<\beta \le 1$, using $r=\alpha -1$ at integer $\alpha$. For $f\in C^{r,\beta }$, subtract its degree-$r$ Taylor polynomial at $x_0$. For $r\ge 1$, the integral Taylor remainder with $f^{(r)}(t)-f^{(r)}(x_0)$ is bounded using the Hölder seminorm; for $r=0$ this is exactly Hölder continuity. The resulting bound is the displayed estimate. At integer order, the convention is $C^{r,1}$; a $C^{r+1}$ function on a compact closed interval also obeys this estimate.

The horizontally integrated plume concentration is the integral of a local concentration $C$ across the plume. It measures scalar amount per unit vertical length and unit span, and its vertical integral is total scalar amount per unit span. Unlike a local concentration, its dimensions include one horizontal length. It must be distinguished from horizontally averaged plume concentration when a plume's width varies with height.

The Hamiltonian $(v^2-x^2)/2$ gives $\dot{x}=v$, $\dot{v}=x$. Its characteristic flow map is multiplication by $A_t$, with $A_t{A}_s=A_{t+s}$ and determinant one. The backward map is $A_{-t}$. This expanding-contracting linear flow preserves phase-space Lebesgue measure and every finite-$p$ integral of a transported density, despite stretching its level sets.