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In the positive-support exponential family $q(\theta )=\int p(x)e^{-\theta x}dx$, a proper natural conjugate prior with zero endpoint density values has mean parameter strictly inside the essential convex support $(a,b)$. Updating that mean by a positive-weight average with supported observations keeps it inside $(a,b)$. At finite parameter endpoints, prior vanishing forces $q\to \infty$, so the increased posterior power $q^{-k-n}$ still vanishes. At positive infinity, positive base-measure mass below a point $d$ smaller than the updated mean bounds $q(\theta )$ below by $C{e}^{-d\theta }$, giving exponential decay of the posterior kernel. Mass above the updated mean gives the analogous bound at negative infinity. Thus the updated density is proper and has the boundary condition required to integrate its score.

At the first half-temperature time, the solution of Newton's law of cooling givesSince $\alpha >2$, this time is positive and finite. The instantaneous temperature increase supplies the new initial condition $T(t_1+)=\alpha T_0/2+\beta$. Restart the same first-order linear differential equation at $t_1$:This is Newton cooling with an instantaneous temperature jump. The time translation in the exponential decay is essential: its prefactor is the excess immediately after the jump.

With a positive cooling rate $k$, Newton's law of cooling gives $T^{\prime }=-k(T-T_0)$. The temperature excess $U=T-T_0$ satisfies the first-order linear differential equation $U^{\prime }=-kU$. Applying the initial condition to its exponential decay solution givesThe temperature approaches the ambient value $T_0$, rather than zero. The intended cooling situation has $T_0>0$ and $k>0$.

Impose the target temperature on the post-jump exponential decay:Solving for the jump and eliminating $t_1$ using $e^{k{t}_1}=2(\alpha -1)/(\alpha -2)$ givesThis is the unique required temperature increase, and it satisfies the stipulated lower bound. Indeed $t_2>t_1$ impliesThus no additional restriction on the prescribed time is needed.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

Declare an edge of the planar dual graph open exactly when its crossed primal edge is closed. This dual bond percolation has parameter $q=1-p<1/2=p_c$, using the Harris-Kesten theorem. We use the standard exponential tail of subcritical cluster size: for some $K,a>0$, uniformly in the dual vertex $v$,This is decay of the number of vertices in the percolation cluster, not merely its radius.

On $\{|C|=n\}$, the geometric fact allowed in the paper supplies a simple dual cycle in a graph surrounding $C$, with length $\ell \ge \alpha \sqrt{n}$. Every crossed edge is in the edge boundary of $C$ and is therefore closed, so the dual cycle is open. Its $\ell$ distinct vertices lie in one dual percolation cluster of size at least $\ell$.

A dual cycle in a graph of length $\ell$ surrounding the origin has all its vertices within sup-norm distance $\ell +1$ of the origin: its coordinate spans are at most $\ell$, and the origin lies between each pair of extreme coordinates. There are at most $K_0(\ell +1{)}^2$ possible dual vertices there. The union bound over lengths and possible vertices yieldsfor some $K_1,b>0$, since a polynomial factor can be absorbed into a slower exponential decay.

To remove the constant prefactor for all $n\ge 1$, first choose $n_0$ so that the bound is at most $e^{-(b/2)\sqrt{n}}$ for $n\ge n_0$. For the remaining finitely many $n$, use ${\mathbb{P}}_p(|C|=n)\le 1-\theta (p)<1$ and chooseFor $1\le n<n_0$, the chosen $c_2$ ensures $e^{-c_2\sqrt{n}}\ge 1-\theta (p)$.

For a steady flow independent of $y$, geostrophic balance givesHence $\zeta =v_x=(g/f){\eta }^{\prime \prime }$, and the definition of $q$ becomesEquivalently,With the Rossby deformation radius $R=\sqrt{gh}/f$, the bounded solution for the stated potential-vorticity step isContinuity of $\eta$ and $v=(g/f){\eta }^{\prime }$ at $x=0$ givesThereforeand

When $q_1>q_2$, the surface height increases smoothly from $-h{q}_1/f$ to $-h{q}_2/f$, while $v$ is a positive eastward jet with its maximum at the discontinuity and exponential decay on the Rossby-radius scale:

The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

WriteThe quasi-steady model is the coupled pairFor a thin shell, $T_s$ is close to $T_m$. A first-order asymptotic expansion givesand thereforeTo leading order $F_c\sim F_{\ast }$, so an initially freezing shell grows linearly:

For a thick shell the conductive correction in the surface balance is small. The radiative equilibrium temperature and its first correction areThus, with $A=a\log (T_m/T_e)$,When $F_w$ is negligible over an intermediate range, this ordinary differential equation gives the square-root growth lawRetaining $F_w$, its implicit solution isConsequently a sketch of $h(t)$ starts approximately linearly, crosses to square-root growth, and approaches $h_{\infty }$ with exponential decay of $h_{\infty }-h$. A shell placed above $h_{\infty }$ instead thins because the basal oceanic heat flux exceeds the conductive loss.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life

The time constant $\tau >0$ in exponential decay is the time over which the decaying quantity is multiplied by $e^{-1}$. A response approaching equilibrium as $q(t)=q_{\infty }+(q(0)-q_{\infty })e^{-t/\tau }$ has the same time constant. For a resistor-inductor circuit it is $L/R$.