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No slip on the two fixed radial walls gives $g(\pm \alpha )=0$. The outward flux through a circular arc is ${\int }_{-\alpha }^{\alpha }{u}_{r}rd\theta ={\int }_{-\alpha }^{\alpha }gd\theta$, so source strength $Q$ requires that integral to equal $Q$. Set $g=QG$ and $ϵ=Q/\nu$. ThenAt leading order $G_0^{(3)}+4G_0^{\prime }=0$, whose general solution is a constant plus sine and cosine of $2\theta$. The symmetric source-driven branch has no sine term; for generic $\alpha$ this also follows directly by subtracting its two wall conditions. ThusIndeed its integral is $C(\sin 2\alpha -2\alpha \cos 2\alpha )=1$. At isolated angles where an odd zero-flux homogeneous mode satisfies both wall conditions, symmetry selects the stated branch.

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

Suppress hats temporarily. Mass conservation between any two cross-sections, together with impermeability of both walls, makes their fluxes equal. Where the wall is differentiable, the same result follows locally by differentiating the flux with a variable upper limit and using continuity givesBoth the upper-limit term and the boundary velocities vanish by the wall conditions. Thus the flux is constant. Since $p=p(x)$, integrate $u_{zz}=p^{\prime }(x)$ twice and impose $u(0)=1$, $u(h)=0$:Integrating across the gap gives $Q=h/2-p^{\prime }{h}^3/12$. Restoring hats,The endpoint conditions are the specified equal reservoir pressures after subtracting $p_0$ and dividing by the pressure scale.

The flux average for a translating lubrication gap follows by integrating the pressure gradient between the endpoints:If $2Q$ were smaller than the minimum height, the integrand would be strictly positive everywhere, contradicting zero total pressure change. If it were larger than the maximum, the integrand would be strictly negative. ThereforeEquivalently,is a weighted average of the height with strictly positive weight ${\hat{h}}^{-3}$. This also identifies the constant flux uniquely.

Put $v=u-V$, the velocity in the shock frame. The mass jump condition gives a common mass flux $j=\rho v$. Subtracting $V$ times the mass balance from the momentum balance givesFor energy, transform the conserved energy to the shock frame by subtracting $V$ times momentum and adding $V^2/2$ times mass. The resulting energy flux isFor a genuine shock, $j\ne 0$; dividing its continuous energy flux by the same nonzero $j$ on both sides givesThe second term is the ideal-gas specific enthalpy. A contact discontinuity has zero relative mass flux and does not in general require equal specific enthalpies; the division here is appropriate for the shock specified in the question.

For elastic scattering by a real short-range scalar potential, write $C=m/(2\pi {\hslash }^2)$, ${\mathbf{k}}_i=\mathbf{k}$ and $|{\mathbf{k}}_f|=k$. In the far field, $|\mathbf{r}-{\mathbf{r}}^{\prime }|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(r^{-1})$ and ${\mathbf{k}}_f=k\hat{\mathbf{r}}$. The outgoing spherical-wave coefficient isSubstitution of the incident plane wave inside this integral givesThe incoming and outgoing flux factors cancel because their speeds agree. Here $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ is a wavevector transfer; the actual momentum transfer is $\hslash \mathbf{q}$. This distinction preserves dimensions. The approximation requires the scattered correction to the interior wave to be small, and does not in general apply to an unscreened long-range potential.

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

The negative-kinetic-energy branch is interpreted as an antiparticle channel. In the overlap $m{c}^2<E<V_0-m{c}^2$, particles can escape to the left and oppositely charged antiparticles to the right. This interval exists only when $V_0>2m{c}^2$. The external electrostatic source supplies the energy for pair creation; the reflected particle flux can exceed the incident one because the outgoing state need not contain only the incident particle.

The conserved quantity is signed charge current, not a positive one-particle probability in both regions. Negative $\mathcal{T}$ denotes rightward transport of opposite charge, so it is consistent with $\mathcal{R}>1$ and $\mathcal{R}+\mathcal{T}=1$. A quantized charged field with particle and antiparticle creation operators gives the appropriate interpretation of this Klein paradox. This bosonic result must not be replaced by a fermionic reflection law or interpreted as a failure of charge conservation.^[ref]

A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by temperature alone. A cavity with a small aperture approximates this ideal. Photon number is not conserved in thermal equilibrium, so its chemical potential is zero. Combining the Bose occupation with the two-polarization electromagnetic mode density gives the spectral energy densityChanging variable to $x=\hslash \omega /(k_{B}T)$ shows directly why the total energy density is proportional to $T^4$:Three powers of T arise from the number of thermally populated modes and one from their characteristic energy. Isotropic radiation incident on a surface has outward hemispherical flux $cu/4$, givingThe factor 1/4 is the hemispherical angular average; multiplying energy density by c alone would overcount. The spectrum's frequency scale grows with T. Classical equipartition over all modes would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency modes. Real materials can have wavelength-dependent emissivity, and a body's net radiative exchange also includes absorbed environmental radiation.

The term $\alpha {\nabla }^{2}q$ is cell diffusion arising from random movement. Chemotaxis up the concentration gradient gives the cell flux $\beta q\nabla c$, whose negative divergence is the transport contribution $-\beta \nabla \cdot (q\nabla c)$. The reaction term $q(1-q)$ gives growth governed by the logistic differential equation with dimensionless carrying capacity one.

The chemoattractant itself diffuses, is produced by cells at rate $\kappa \eta q$, and decays at rate $\kappa c$. The homogeneous Neumann boundary conditions suppress the normal diffusive flux of each quantity. They also suppress the normal chemotactic cell flux because the normal derivative of $c$ vanishes. Thus the rectangular boundary is impermeable, and the spatially integrated cell count changes through cell proliferation rather than transport across that boundary.

Define ${\chi }_1$ to be one in phase 1 and zero in phase 2, including its dependence on both transverse coordinates even though the averaged fields depend only on $x,t$. With no interphase mass transfer, its distributional phase-mass equation is ${\partial }_t(\rho {\chi }_1)+\nabla \cdot (\rho \mathbf{v}{\chi }_1)=0$. At a pipe cross section $\Sigma$ of area $A$, setThus ${\rho }_1$ is an intrinsic phase density and $u_1$ is the density-weighted mean axial velocity. This weighting is necessary when density varies within the phase; it becomes the ordinary intrinsic mean for an incompressible phase of constant density.

Integrate over the cross section. The transverse divergence contributes no boundary mass flux at the impermeable pipe wall, and the material interface has no mass leakage. The section area is constant, soFor phase 2 use ${\chi }_2=1-{\chi }_1$ and fraction $1-\alpha$, with the analogous intrinsic density and density-weighted velocity:The intrinsic quantities are defined where their phase fraction is nonzero. These are the intrinsic phase averages in a two-fluid pipe model.

For constant phase density, let $u_1$ be the intrinsic mean axial speed and defineThe profile coefficient measures the momentum flux relative to a uniform velocity profile with the same mean. Cauchy-Schwarz gives $D_1\ge 1$ where the mean is nonzero, with equality for a uniform phase profile. At zero mean it is safer to retain the second moment itself rather than divide by $u_1^2$.

In the stated common-pressure, no-extra-force closure, the averaged conservative momentum equation isFor constant $D_1$, expand its derivatives and use ${\alpha }_t+u_1{\alpha }_x+\alpha (u_1{)}_x=0$. The terms combine asThereforeThis is the phase momentum profile coefficient correction. It accounts for the second moment of velocity, not merely a rescaling of the convective speed. A variable coefficient would also produce ${\rho }_1\alpha u_1^2(D_1{)}_x$.

Stretch $z=\delta Z$ and write $f=\delta h$, $w=\delta W$, with horizontal velocity and time left order one. The longitudinal viscous term is ${\delta }^{-2}{u}_{ZZ}$, so set $P={\delta }^{2}p$. For $Re{\delta }^2\ll 1$, the leading bulk equations are $P_Z=0$, $u_{ZZ}=P_x$ and $u_x+W_Z=0$.

The normal-stress condition gives $P=-{\delta }^3\hat{C}{h}_{xx}$ at leading order, and the tangential condition gives $u_Z(h)=0$. HenceNo slip at the bed, $u(0)=0$, yields $u=P_x(Z^2/2-hZ)$. Averaging through the depth givesThe kinematic condition becomes $h_t+u(h)h_x-W(h)=-\widehat{ϵ}/\delta$ at leading order. Integration of continuity from the impermeable bed then gives

A nonzero contact speed or evaporative flux is singular in the strictly no-slip sharp-contact-line model with mobility proportional to $h^3$. A finite nonzero limiting $\overline{u}$ would require $h_{xxx}=3\overline{u}/(C{h}^2)$, which is singular for a linearly vanishing height and generally incompatible with a prescribed finite slope at the exact line. Thus a dynamic finite-contact-angle condition needs microscopic regularization, such as slip or a precursor film. In the slow-drying limit a fixed contact angle can instead be used for the leading quasistatic outer profile; it is not an exact regular finite-rate contact-line solution.

At leading quasistatic order $h_{xxx}=0$, so the symmetric profile satisfying the imposed outer angle isIts cross-sectional volume is $V={\int }_{-s}^{s}hdx=2\theta s^2/3$. The moving endpoints contribute no boundary-height term, and the end flux vanishes, giving $V_t=-2ϵs$. ConsequentlyThese are leading slow-time results up to extinction; the approximation need not remain uniform at the final vanishing scale.

To recover the small velocity, differentiate the quasistatic profile using $s_t=-3ϵ/(2\theta )$:Symmetry makes the flux zero at the centre, so $h\overline{u}=-ϵx(s^2-x^2)/(4s^2)$. Division by the height givesThe flow is inward on both sides, suggesting transport of suspended matter toward a central deposit rather than a perimeter ring. Coffee commonly leaves the opposite, ring-like pattern. A possible missing ingredient is contact-line pinning: holding the perimeter fixed makes evaporation require outward replenishment, and stronger evaporation near the edge reinforces that transport. Receding constant-angle and pinned contact lines therefore have qualitatively different deposit mechanisms.

For comparison, the same uniform-evaporation parabolic model with fixed $s$ gives ${\theta }_t=-3ϵ/s$ and $\overline{u}=ϵx/(s\theta )$, an outward velocity. This comparison isolates the boundary assumption even before adding edge-enhanced evaporation.

The receding solution is the quasistatic evaporation of a receding parabolic drop.

A per-capita mortality rate $a$ means that the death flux from a biological population of size $I$ is $aI$. With constant rate and no other transfers, $\dot{I}=-aI$ gives exponential survival $I(t)=I(0)e^{-at}$. In an epidemic model this can specify mortality from the infective compartment rather than mortality of the entire population.

With constant lateral supply $M$, a discharge law of the displayed form gives equilibrium area $[(m+1)(Q_{\mathrm{in}}+Ms)/c{]}^{1/(m+1)}$ and characteristic speed $c{A}^m$. Under a Chezy law, a wetted-perimeter relation $P_b\propto A^q$ gives $m=(1-q)/2$. Small area perturbations are transported downstream with amplitude inversely proportional to this characteristic speed; finite compressive disturbances can still form shocks because the flux is convex.

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.

For flux $F$ and effective dust opacity $\kappa$, radiative acceleration is $\kappa F/c$. Comparing it with the gravitational acceleration gives an Eddington luminosity condition appropriate to dusty gas. In an optically thick starburst galaxy, feedback can limit the star formation rate per area and drive a galactic outflow, while the gas supply limits the duration. A surface-density limit is not a universal ceiling on total star formation rate: the star-forming area matters. Radiation-supported starburst disk calculations give characteristic optically thick infrared fluxes near $10^{13}{L}_{\odot }{\mathrm{kpc}}^{-2}$.

For a constant-opacity, gas-pressure-supported radiative disk surface at $z=H$, imposing $p=T=0$ does not impose $F=0$. If the outgoing surface flux is finite and positive, dividing hydrostatic balance by the temperature gradient gives $dp/dT\sim 16\sigma {\Omega }^{2}H{T}^3/(3\kappa F_s)$. Consequently $p\propto T^4$, $\rho \propto T^3$, and the diffusion equation gives $T\propto H-z$. Thus the mass density also vanishes regularly while the surface radiates a finite flux. This ideal boundary replaces the finite-temperature photospheric transition; its accuracy is not controlled solely by the disk aspect ratio.

// Target: astrophysics.bigb