For the ideal magnetic response to a cylindrical cyclonic event, the plane-integrated $x$-component of $\mathbf{u}\times \mathbf{B}$ at duration $t$ isThis follows from $(\mathbf{u}\times \mathbf{B}{)}_x=(rf{B}_z+g{A}_r)\cos \varphi -(g/r)A_{\varphi }\sin \varphi$ and azimuthal integration. If boundary terms vanish, integration by parts gives the equivalent form $\pi B_0{\int }_0^{\infty }{r}^2{g}^{\prime }[\sin (ft)+tf\cos (ft)]dr$. For $f=g=a^2-r^2$ on $r<a$, zero outside, the response isThe value at zero is its continuous limit zero. It is initially $-\pi B_0{a}^{6}t/3$, but changes sign repeatedly at the nonzero roots of $\tan x=x$, with oscillatory envelope of order $t^{-1}$. The flow's kinetic helicity density is $2f^2\ge 0$, showing that the sign of a finite-duration ideal response is not fixed by the instantaneous helicity sign.

Turbulent magnetic pumping is an effective transport of the mean magnetic field produced by correlations in inhomogeneous or anisotropic turbulence. Its contribution to the mean-field electromotive force is ${\mathcal{E}}_{\mathrm{pump}}=\mathbf{\gamma }\times \overline{\mathbf{B}}$. In the mean induction equation this enters as $\nabla \times (\mathbf{\gamma }\times \overline{\mathbf{B}})$, the same algebraic form as advection by a velocity $\mathbf{\gamma }$. Thus it supplements the mean material velocity without requiring an equal mean mass flow. Downward pumping can help retain or couple fields in a solar dynamo.

Suppose a steady solenoidal velocity satisfies $\Delta \mathbf{u}=-\mathbf{u}$, and spatial averaging permits integration by parts without boundary terms. For a uniform test magnetic field $\mathbf{B}$, the small-magnetic Reynolds number expansion gives ${\mathbf{b}}_1=(\mathbf{B}\cdot \nabla )\mathbf{u}$ and $\Delta {\mathbf{b}}_2=-\nabla \times (\mathbf{u}\times {\mathbf{b}}_1)$. Using the self-adjointness of the Laplacian inside the average,The second equality concerns the averaged mean-field electromotive force; it does not assert that the quadratic forcing itself is monochromatic.

// Target: fluid-mechanics.bigb

In the residual unsatisfied formula, choose a Boolean literal contained in the largest number $a$ of clauses and make it true. Its opposite occurs in at most $a$ clauses, since it too was an eligible literal. The step permanently satisfies $a$ clauses; only clauses containing the opposite can become empty, so it loses at most $a$ clauses. Delete satisfied clauses and false literals and repeat. No satisfied clause or newly empty clause is counted twice. Summing gives total lost nonempty clauses at most total satisfied clauses $G$. All initially nonempty clauses are eventually in one of these classes, so $G\ge N_{+}/2\ge \mathrm{OPT}/2$. Initial empty clauses can never be satisfied and are excluded from $N_{+}$. This proves the approximation algorithm guarantee even with tautological clauses or repeated literals.

Write $\psi =\sqrt{\rho }{e}^{i\theta }$ away from zeros, and ${\mathbf{v}}_s=\nabla \theta$. Its Madelung equations give ${\rho }_t+\nabla \cdot (\rho {\mathbf{v}}_s)=0$ and ${\theta }_t+|{\mathbf{v}}_s{|}^2/2=1-\rho +{\nabla }^2\sqrt{\rho }/(2\sqrt{\rho })-\zeta D_n\rho$. For an incompressible normal fluid, $D_n\rho =-\nabla \cdot [\rho ({\mathbf{v}}_s-{\mathbf{v}}_n)]$. Taking a gradient proves the displayed equation when each space-time derivative of the slowly varying amplitudes is $O(ϵ)$: the quantum potential force is $O({ϵ}^3)$, while the bulk term is $O({ϵ}^2)$.

With the normal fluid at rest, linearize about the unit-amplitude plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-k^{2}t/2)}$. For amplitude and phase perturbations proportional to $e^{i\mathbf{l}\cdot \mathbf{x}+\sigma t}$, the continuity equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})R=l^2\Theta /2$, and the phase equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})\Theta =-(2+l^2/2+2\zeta \sigma )R$. Their determinant is the displayed dispersion relation. The damping contains the laboratory exponent $\sigma$, not the Doppler-shifted exponent, because the reference flow is the normal component.

For an integrable density $Z=dQ/dP$, $Z\ge 0$ and ${\mathbb{E}}_{P}Z=1$, its density process is $Z_t={\mathbb{E}}_P[Z\mid {\mathcal{F}}_t]$. If $Z_0=1$, then $P$ and $Q$ agree on ${\mathcal{F}}_0$, so every ${\mathcal{F}}_0$-measurable random variable integrable under $P$ remains integrable under $Q$. This covers the normalized stochastic exponential densities used in the Brownian Girsanov theorem.

Without that initial normalization the assertion can fail. On the positive integers with the full sigma-algebra already known at time zero, take $P(n)=2^{-n}$ and $Q(n)=c/n^2$, where $c=({\sum }_{n\ge 1}{n}^{-2}{)}^{-1}$. The constant-in-time process $X_t=n$ is a martingale under $P$, but ${\mathbb{E}}_Q{X}_0=c{\sum }_{n\ge 1}1/n=\infty$. The density process $Z_t=Q(n)/P(n)$ is constant in time with random $Z_0$ and mean one. It is not the Doléans-Dade exponential of a constant process, which would start at one and remain one. Thus drift-corrected zero-starting increments and integrability of an added initial variable are separate issues.

This linear functional extracts one coefficient of a B-spline expansion using local polynomial data. With ${\psi }_i(x)={\prod }_{\ell =1}^{k-1}(x-t_{i+\ell })/(k-1)!$ and a point $\xi$ inside a knot cell in the support, it is ${\sum }_{r=0}^{k-1}(-1{)}^r{s}^{(r)}(\xi ){\psi }_i^{(k-1-r)}(\xi )$. The local polynomial identity for the Marsden dual functional gives the desired coefficient. Its bounded restriction to the local spline space can be extended to continuous data on $[t_i,t_{i+k}]$ by the Hahn-Banach theorem, with the same bound. Such extensions supply spline quasi-interpolation coefficients without requiring derivatives of the data.

For a nonvanishing superconducting order parameter $\psi =\sqrt{n_s}{e}^{i\theta }$, $j=(q{n}_s/m)(\hslash \nabla \theta -qA)$. Single-valuedness gives phase winding $2\pi N$ around a closed curve, hence the fluxoid and flux quantization in a superconductor identity above. Magnetic flux alone is $Nh/q$ when the current integral vanishes. If the entire spanning surface is in a region where $D\psi =0$, the field vanishes there and the winding is zero; nonzero quantized flux requires a hole or a vortex core where that condition does not hold.

On a connected closed oriented surface, a magnetic Anosov flow with $\int f{\Omega }_a=0$ preserves a smooth contact form only when $f=0$. To see the obstruction, take $d\eta =f{\Omega }_a$ and $\sigma =\alpha -{\pi }^{\ast }\eta$. An invariant contact form can be normalized to evaluate to one on $F$; its differential is a constant multiple of ${\iota }_F\mu$, since smooth invariant functions are constant. Averaging closed one-forms forces that multiple to be $-1$, so the contact form differs from $\sigma$ by a closed form. Its fibre-average decomposition gives $Fh=(\eta -\xi )(v)-af$. The degree-zero/one kernel theorem gives $af=0$ and $d\eta =d\xi =aK{\Omega }_a$, forcing $f=0$ whether $a$ vanishes or not. For $f=0$, the canonical form $\alpha$ is invariant and contact.

In a stationary M/M/1 queue with independent Poisson arrivals of rate $\lambda$ and exponential service times of rate $\mu >\lambda$, the departure counting process is a Poisson process of rate $\lambda$. Moreover, the departures before a fixed time $t$ are independent of the queue population $N_t$. This independence concerns the past departures: future departures depend on whether the queue is currently empty.

The proof uses time reversal. The stationary Geometric distribution satisfies Detailed balance, ${\pi }_n\lambda ={\pi }_{n+1}\mu$, so the stationary queue population is a Reversible Markov chain. Departures in forward time are arrivals in reversed time. In the reversed chain the upward rate is again the constant $\lambda$, so those arrivals are an independent Poisson process, including independence of their future increments from the present queue population. Reversing time gives both assertions. Consequently, feeding the output of a stationary M/M/1 queue into another queue supplies Poisson input to that second queue. The claim depends on equilibrium; a queue started empty has no immediate departures even though arrivals have positive rate.

With packet arrival rate lambda, independent exponential payloads of mean one and byte-service rate C, let $\rho =\lambda /C<1$. The number N of unfinished packets has law $(1-\rho ){\rho }^n$. Conditional on N=n, the remaining payloads are n independent unit-mean exponentials, including the in-service residual by memorylessness. Summing their Laplace transforms gives $\mathbb{E}{e}^{-sW}=(1-\rho )(1+s)/(1+s-\rho )$. Thus W has mass $1-\rho$ at zero and, conditional on positivity, is exponential with rate $1-\rho$. The probability of at least r headers is instead ${\rho }^r$ for integer r.

For $L=\dot{F}{F}^{-1}$, the product rule proves the displayed lower-convected derivative identity. Under a superposed rotation $Q(t)$, ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\dot{Q}{Q}^T$. The two rotation terms in ${\dot{\tau }}^{\ast }$ cancel those in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$, proving that the lower-convected rate is an objective time derivative.

Positive numbers $a,b,c$ are the areas of the three coordinate projections of a volume-one Euclidean body exactly when $abc\ge 1$. Necessity is the three-dimensional Loomis--Whitney inequality. For sufficiency take box lengths $x=\sqrt{ac/b}$, $y=\sqrt{ab/c}$ and $z=\sqrt{bc/a}$, and retain the union of the three slabs with first, second or third normalized coordinate at most $t$. All three projections are full rectangles, while volume is $\sqrt{abc}[1-(1-t{)}^3]$. Choosing $t=1-(1-1/\sqrt{abc}{)}^{1/3}$ gives volume one. The slabs have positive thickness, so the construction uses a body with interior rather than zero-volume additions.

Each nonnegative integrable $f_j$ depends on all coordinates except the $j$th. In dimension three, Cauchy-Schwarz inequality first in the third coordinate and then in the remaining pair proves the displayed estimate. Induction integrates the last coordinate by Hölder's inequality on the first $d-1$ factors, then applies Hölder's inequality to the remaining factor and the dimension-$(d-1)$ product. Taking indicators of coordinate projections of a Euclidean body gives the volume version of the Loomis--Whitney inequality.

In isotropic incompressible three-dimensional turbulence, $3S_2(r)/4=\int E(k)H(kr)dk$, with $H(x)=1+3\cos x/x^2-3\sin x/x^3$. It is smooth at zero, where $H=x^2/10+O(x^4)$, and tends to one at large argument. A sharp wavenumber cutoff omits the finite large-scale-gradient contribution.

A smooth velocity varying on a scale larger than $r$ contributes $\Delta U_L\simeq r{\partial }_L{U}_L$ to a velocity increment. Higher powers produce mixed moments with smaller-scale increments. longitudinal structure functions therefore attenuate, but do not exactly filter out, large-scale strain and intensity modulation.

A velocity increment is the difference of a velocity field between two positions. Its longitudinal component is the projection onto their separation direction. Moments of increments define longitudinal structure functions; the Kolmogorov 1941 theory estimates a typical inertial-range magnitude as $(ϵr{)}^{1/3}$.

Invert the two-dimensional Fourier transform in the long-path projection of a magnetic correlation, then perform the angular integration in polar coordinates. The angular integral is $2\pi J_0(kr)$, a Bessel function of the first kind, giving the displayed Hankel transform with the stated Fourier-Hankel normalization for an isotropic spectrum. The result recovers the scalar tensor coefficient $H$, whose shell-integrated magnetic energy spectrum has an additional factor $k^2$.

Orient a degree-$(d+1)$ regular tree towards one end, and let height $g$ decrease along the unique parent edge and increase along the $d$ child edges. For $w_{\rho }(A)={\sum }_{x\in A}{\rho }^{g(x)}$, suppressing the occupied-target constraint only increases births, so its infinitesimal generator satisfies $L{w}_{\rho }\le [-1+\lambda ({\rho }^{-1}+d\rho )]w_{\rho }$. For $d>1$, choose $\rho =d^{-1/2}$. If $\lambda <1/(2\sqrt{d})$, the expected weight and the infection probability at each fixed site decay exponentially. The expected number of recoveries there equals its expected total occupation time, which is finite. Each infection episode ends almost surely at a recovery, so infinitely late infection is impossible. For $d=1$, take $\rho <1$ sufficiently close to one for each $\lambda <1/2$.