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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.

For $I_1$, the change of variables $t=x^2/\theta$ givesFor $k\ge 1$, use $d(e^{-x^2/\theta })=-(2x/\theta )e^{-x^2/\theta }dx$ and integration by parts:The boundary term vanishes: $x^{k+1}\to 0$ at zero, and the exponential decay dominates the power at infinity. These estimates also establish convergence of the improper integrals used above.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

For spherical symmetry the angular mode is $\ell =0$. Put $u(t,r)=r\psi (t,r)$. Direct differentiation givesso the radial heat equation in three dimensions becomes the ordinary one-dimensional heat equation $u_t=\kappa u_{rr}$. Since $r_{\pm }>0$, the zero endpoint temperatures become $u(t,r_{\pm })=0$.

Write $L=r_{+}-r_{-}$. The two initial sine modes already satisfy these Dirichlet boundary conditions; their squared wave numbers are $(\pi /L{)}^2$ and $(2\pi /L{)}^2$. Each therefore evolves independently with its corresponding exponential decay. Dividing back by $r$ givesAt $t=0$ the initial profile is recovered, and each sine is zero at both endpoints. The first mode decays four times more slowly in its exponential rate than the second. This is the two-mode instance of the spherical-shell heat equation via radial rescaling; uniqueness follows, for example, by applying the heat equation energy identity to the difference of two solutions.

The heart supplies the principal continuing mechanical input to blood circulation; arterial recoil, respiratory movements and the action of skeletal muscles redistribute and supplement that input. They have different sources of energy and should not all be called independent propulsive powers. The pressure gradient remaining after passage through the capillaries is still a consequence of the upstream pump.

Arterial elasticity. During ventricular systole, ejected blood expands the aorta and large arteries, storing elastic potential energy in their walls. During diastole, recoil continues to drive blood into the peripheral vessels while closure of the aortic valve limits return towards the heart. This reduces the intermittency of flow downstream. The recoil is a temporary store of cardiac work, rather than a fresh source of energy. A vessel that is too rigid provides less of this smoothing; a compliant vessel with an open backward outlet would not provide the same useful result.

The skeletal-muscle pump. Contracting skeletal muscles compress adjacent veins. Venous valves make the displacement preferentially towards the heart; when compression ceases, refilling permits another stroke. Walking therefore aids return from dependent limbs. The source of the added work is muscle contraction, and the valves guide it without producing energy. Merely possessing valves cannot make a stationary column circulate.

The respiratory pump. During inspiration, falling intrathoracic pressure and displacement of the abdominal contents favour a pressure gradient from abdominal and peripheral veins towards the thoracic veins and right side of the heart. The respiratory diaphragm thus participates indirectly in venous return as well as pulmonary ventilation. The exact flow varies with posture, vessel collapse and the phase of the respiratory cycle; positive-pressure inflation does not reproduce the same mechanical circumstances as spontaneous inspiration.^[ref]

Venous tone and cardiac filling. Sympathetic contraction of venous smooth muscle can reduce venous storage capacity and transfer blood towards the central circulation. This alters the volume available for filling rather than requiring a wave of venous peristalsis. Cardiac relaxation, recoil and movement of the atrioventricular plane also affect filling pressure. Some traditionally described cardiac suction is therefore a feature of the cardiac cycle and mechanical coupling, not a separate organ pulling blood through every vessel.^[ref]

Gravity and supposed capillary attraction. Gravity assists a downward column but opposes an upward one. In a complete circuit returning to its original height it supplies no net gravitational work; its important practical effects include hydrostatic pressure differences and dependent venous pooling. Ordinary capillary action in an open narrow tube, driven by a liquid meniscus, is not a general pump in a fluid-filled capillary network. Nor does a vague attraction of blood by tissues replace a measurable pressure gradient. Local vasodilation can increase perfusion by lowering vascular resistance, but that change does not manufacture the necessary pressure source.

For a simple mean-flow description, with $Q$ the volume flow rate, $\Delta P$ the driving pressure gradient and $R$ the effective vascular resistance,This relation distinguishes a source of driving pressure from a change of resistance. In a lumped compliant arterial reservoir with vascular compliance $C$ and no inflow between beats, volume balance giveswhere $P_v$ is the downstream pressure. The exponential decay follows from stored volume leaving through the resistance. It explains continued flow during diastole without claiming that the idealization describes every reflected arterial wave.

A physical demonstration can compare a pulsatile pump feeding a narrow outlet through rigid tubing with the same pump feeding it through an elastic reservoir. Collecting the outflow over the pumping and resting phases reveals the reservoir's smoothing action; a side-arm manometer records its falling pressure. Separately compressing a valved flexible tube shows the role of a peripheral muscle pump. Keeping pump stroke and outlet resistance comparable prevents the apparatus from conflating a smoother flow with creation of extra energy.

After three half-lives, $P(T)=P_0/8$. Therefore the minimum initial thermal power isEquality gives the stated terminal power exactly in the idealized exponential decay model. The associated duration is $415.2$ days. No thermal-to-electrical efficiency is needed because both specified powers are thermal.

For a narrow unstable state, the effective Hamiltonian has the complex eigenvalue $m_j-i{\Gamma }_j/2$. With natural units and ${\Gamma }_j=1/{\tau }_j$, its Schrödinger equation givesThe first factor is the ordinary energy phase; squaring the second gives the exponential decay of the survival probability, $e^{-t/{\tau }_j}$. This assumes the approximately constant decay rate regime; exact very early and very late behavior need not be exponential.

Writing the surviving state as $A_S(t)|K_S⟩+A_L(t)|K_L⟩$, invert the two kaon CP eigenstates definitions:Here $t$ is the proper time. For a relativistic beam of fixed Lorentz factor $\gamma$, substitute $t=t_{\mathrm{lab}}/\gamma$; in ordinary units the mass phase is $m_j{c}^{2}t/\hslash$.

A useful differentiation theorem says: if $f(t,x)$ is integrable in $x$, differentiable in $t$ for almost every $x$, and on a neighborhood of each parameter value the derivative is bounded in absolute value by one integrable function of $x$, then the integral is differentiable and its derivative is the integral of ${\partial }_{t}f$. The proof applies dominated convergence to difference quotients, bounded by that same majorant by the mean value theorem.

Fix $t_0>0$ and choose $0<a<t_0<b$. For $t\in [a,b]$,The first part is integrable by part (i). The second is integrable by the permitted hypothesis for $h_b$; alternatively exponential decay dominates its polynomial and logarithmic factors. The theorem therefore applies locally at every $t_0$, giving

An unstable isolated level has survival amplitude $a(t)=\theta (t)e^{-i{E}_{0}t/\hslash }{e}^{-\Gamma t/(2\hslash )}$, because its survival probability falls as $e^{-\Gamma t/\hslash }$. Its energy-domain amplitude is proportional toTaking the modulus squared gives the denominator of the Breit-Wigner resonance cross sectionAt $E-E_0=\pm \Gamma /2$ the resonant factor is half its peak, so $\Gamma$ is the full width at half maximum. exponential decay is the isolated-level approximation; energy-dependent thresholds or overlapping levels need a more general lineshape.

Dating principle. Atmospheric cosmic rays produce secondary neutrons, which can generate carbon-14 through ${}^{14}\mathrm{N}(n,p){}^{14}\mathrm{C}$. Carbon then enters atmospheric carbon dioxide and living organisms. While an organism exchanges carbon with its environment, its carbon-14 to carbon-12 ratio approximately tracks that reservoir. After exchange stops, surviving carbon-14 undergoes beta-minus decay, whereas the stable carbon remains. On the closed-sample, constant-initial-ratio model, radioactive decay givesThus radiocarbon dating infers elapsed time from the loss of an isotope ratio or of radioactive activity per fixed carbon mass. It dates cessation of carbon exchange, which need not equal the date an object was manufactured or buried. Actual calendar dating also requires radiocarbon calibration and control of contamination and reservoir differences.

Mean lifetime. For exponential decay, the half-life obeys $e^{-\lambda T_{1/2}}=1/2$, while the mean lifetime is $\tau =1/\lambda$. HenceUsing $1\mathrm{yr}=365.25\times 86400\mathrm{s}$ gives $\lambda =3.83325\times 10^{-12}{\mathrm{s}}^{-1}$ and $\tau =2.60875\times 10^{11}\mathrm{s}$.