With $R=0$, substitute $h=X(x)T(t)$ into the dimensional Boussinesq equation for an unconfined aquifer:For a positive draining profile, take $\lambda >0$. Normalize $T(t^{\ast })=1$, so $X$ has the dimensions of height. The temporal ordinary differential equation integrates toThe spatial ordinary differential equation and boundary conditions areHere $\lambda$ has units of inverse length with this normalization. The scale freedom between $X$ and $T$ is fixed by the chosen value $T(t^{\ast })=1$; the amplitude of $X$, and hence $\lambda$, must be matched to the available water.

This separable aquifer drawdown is exact if the inherited height profile has the separated spatial shape. For a general rainfall-generated initial condition, it describes the late drought regime; it cannot reproduce an arbitrary profile at the instant rainfall stops merely by choosing one temporal constant.

Return to dimensional variables. The river at base level imposes the drained boundary condition, while the right-hand end is impermeable:As long as the right-hand end remains wet, the latter is $h_x(L,t)=0$. The initial condition for the drought is the height profile inherited at $t=t^{\ast }$; stopping rainfall changes the forcing but does not instantaneously remove stored water.

Introduce the positive outlet constantFor a draining similarity profile, $J$ is finite and positive. A regular linear profile would have $f{f}^{\prime }\to 0$ and $(f{f}^{\prime }{)}^{\prime }\ge 0$ at the origin, incompatible with the rain-fed aquifer similarity solution $(f{f}^{\prime }{)}^{\prime }(0)=-1$. Instead there is a square-root drainage boundary. At leading order, $(f^2{)}^{\prime }=2f{f}^{\prime }\to 2J$, givingMore explicitly, the rain-fed aquifer similarity solution implies $(f{f}^{\prime }{)}^{\prime }=-1+O(\sqrt{\eta })$, hence $f{f}^{\prime }=J-\eta +O({\eta }^{3/2})$ and $f^2=2J\eta -{\eta }^2+O({\eta }^{5/2})$. The height vanishes but its derivative diverges; their product remains finite.

Use $t_d,x_d,h_d$ temporarily for dimensional quantities. The nondimensionalization in part (a) givesTaking the drainage flux to be positive out of the left-hand end, $Q_0=-q(0,t_d)$, givesThe printed expression in terms of $f(0)$ and $f^{\prime }(0)$ must be understood as this limit; a literal product of their separate endpoint values is undefined. This is discharge per unit transverse width. Very close to the river, the divergent slope marks the limit of the Dupuit approximation, while the outer solution still supplies the leading discharge.

The condition $f(0)=0$ means the river maintains the water table at the impermeable base level: $h(0,t)=0$. The liquid is exposed to atmospheric pressure there. It does not imply zero drainage, because $h_x$ may be singular at the outlet.

Far from the river, lateral drainage has not significantly affected storage. Rainfall simply fills pore volume, so the dimensional height is $h\sim Rt/\varphi$. In the dimensionless variables this is $h\sim t$, giving $f(\infty )=1$.

Although the physical aquifer is finite, these boundary conditions describe its early similarity solution before the drainage-affected region reaches $x=L$. In dimensional variables its horizontal extent is of order $\sqrt{KR}t/\varphi$, so the semi-infinite approximation requires $t\ll T=\varphi L/\sqrt{KR}$.

Use the dimensionless equation from part (a). For the similarity ansatz $h=t^{a}f(\eta )$, $\eta =x/t^b$, its three terms have time factorsA nontrivial similarity solution retaining storage, lateral drainage and nonzero rainfall requires the exponents to agree:Unlike a source-free diffusion equation, the nonzero prescribed rainfall fixes an amplitude-versus-time scale directly, so no additional conserved quantity or boundary condition is required to determine these exponents. Degenerate spatially uniform solutions are an exception to this all-terms balance. The rain-fed aquifer similarity solution satisfiesIf the dimensional equation is used directly without scaling out coefficients, its similarity profile instead obeys $\varphi (f-\eta f^{\prime })=K(f{f}^{\prime }{)}^{\prime }+R$. The normalization above is the one in which the subsequent condition $f(\infty )=1$ applies.

Set $x=L\tilde{x}$, $h=H\tilde{h}$, and $t=T\tilde{t}$. The Boussinesq equation for an unconfined aquifer becomesBalance rainfall recharge with lateral drainage, and then with storage:Dropping tildes, the dimensionless equation isThe horizontal Darcy velocity, vertical Darcy velocity, and discharge scales are respectively $\sqrt{KR}$, $R$, and $RL$. Thus the vertical thickness scale is set by rainfall balanced against hydraulic conductivity over the finite drainage length $L$, rather than by porosity. The latter determines the storage time $T$. The Dupuit approximation requires $H/L=\sqrt{R/K}\ll 1$ for these scales. The initially dry initial condition is $h(x,0)=0$.

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Let $H\ll L$ be the characteristic saturated thickness and let $\delta =H/L$. Scale pressure by $\rho gH$. The horizontal part of Darcy's law gives the scaleBy incompressibility, $w\sim \delta u\sim K{\delta }^2$. The vertical part of Darcy's law isIts left side is smaller than the scale $K$ by ${\delta }^2$. Therefore $p_z+\rho g=O(\rho g{\delta }^2)$, and the leading pressure is hydrostatic pressure:Here the water table has pressure $p_a$. This is the Dupuit approximation, valid away from the small region near an outlet where the surface slope need not remain small.

Take $z$ vertically upward and the porous skeleton stationary. Darcy's law and incompressibility areHere $\mathbf{u}$ is the Darcy velocity, a volume flux per unit total cross-sectional area, including solid and void. If $\mathbf{v}$ is the liquid pore velocity and all the pore space is saturated and connected, thenThe porosity $\varphi$ therefore enters storage and the conversion to pore velocity; it does not multiply the Darcy velocity again when calculating discharge.

Define $A=2\cot \alpha /3+8\mathrm{Re}/15$. The linearization about $h=1$ givesThe Galilean transformation $\xi =x-2t$, $\tau =t$ has ${\partial }_t={\partial }_{\tau }-2{\partial }_{\xi }$ and ${\partial }_x={\partial }_{\xi }$. It removes the leading advection, leavingFor a normal mode proportional to $e^{\sigma \tau +ik\xi }$, the second and fourth derivatives contribute $-k^2$ and $k^4$. The dispersion relation is thereforeFor $A>0$ and $S>0$, the unstable wavenumbers satisfy $0<|k|<\sqrt{A/S}$. In the laboratory coordinates, the temporal exponent is $\sigma -2ik$, so this linear stability calculation predicts growth together with downslope transport at dimensionless speed $2$.

For an inverted falling film, keep $z$ directed from the solid into the liquid. The downslope component of gravity is unchanged, but its normal component is now positive. Thus the hydrostatic pressure changes from $(h-z)\cot \alpha$ to $-(h-z)\cot \alpha$. The resulting Benney equation isOnly the normal-gravity contribution changes sign; the surface tension and inertial contributions retain their signs.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

The two wall boundary conditions areDefine the capillary parameter for a falling filmRetain $S=O(1)$ in the long-wave approximation. Thus surface tension is retained in the leading normal stress balance even though geometric slope corrections are small. For $h_x=O(1)$ in the rescaled coordinates, the dimensional surface slope is $\epsilon h_x$.

The three free surface boundary conditions, accurate through $O(\epsilon )$, areThe first is the kinematic boundary condition; the second is the zero tangential traction condition. For the third, the dimensional liquid-side pressure is, to the required order,Division by $\rho g{h}_0\sin \alpha$ gives $p=-3S{h}_{xx}/2+\epsilon w_z$ in dimensionless variables. Subtracting $\bar{p}(h)=(1-h)\cot \alpha$ gives the stated perturbation condition. The additional slope-times-shear contribution to normal stress vanishes at this order because the tangential traction is zero. The omitted geometric and tangential-stress corrections are $O({\epsilon }^2)$ under this ordering.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

All daughter particles are massless, so the parent mass $M$ supplies the only kinematic scale at tree level. Each vertex has two derivatives, making its scattering amplitude scale as ${\lambda }_i{M}^2$. For the four-$b$ vertex, symmetrizing over the derivative-bearing legs yields a sum of pairwise four-momentum products, which is proportional to $M^2$ by four-momentum conservation.

For $n$ massless daughters, the Lorentz invariance Lorentz-invariant phase-space measure scales as $M^{2n-4}$: each $d^{3}q/(2E)$ contributes two powers and the four-dimensional Dirac delta removes four powers. Including the parent factor $1/(2M)$ gives $\Gamma \sim |{\lambda }_i{|}^2{M}^{2n-1}$. ThusThe unspecified factors are dimensionless angular integrals and identical-particle factors. Each result has mass dimension one, as a decay width must in natural units.

Counting the mass dimensions of each interaction operator givesIndeed $\Phi (\partial a{)}^2$ and $\Phi F^2$ have dimension five, whereas $\Phi b^2(\partial b{)}^2$ has dimension seven. These are interactions of an effective field theory with dimensionful couplings.

The direct one-vertex decay channels are $\Phi \to aa$, $\Phi \to bbbb$ and $\Phi \to \gamma \gamma$. Each leading scattering amplitude is linear in its corresponding coupling, and a decay width is quadratic in the amplitude. Therefore, holding the parent mass fixed,There are no tree-level interference terms between these different final states.

An $A\to B$ hop can occur only while the charge occupies $A$. Hence its stationary number per unit time isThe reverse count is $R_{B\to A}=f_3{p}_B^{\mathrm{st}}$, which equals this result. There is therefore no signed stationary current along that edge. The individual hopping traffic need not equal the contact-arrival traffic; their ratio is $R_{A\to B}/R_{\mathrm{in}}=f_2/f_4$. A charge can make several internal visits before leaving, so neither conservation nor the single-charge constraint forces those two one-way rates to coincide.

Condition on the molecule being empty. Its instantaneous incoming-charge jump rate is $f_1$, so the unconditional stationary number of arrivals per unit time isThis stationary jump traffic at a single contact is a unidirectional jump count, not a signed net current. Stationarity also gives $R_{\mathrm{out}}=f_4{p}_A^{\mathrm{st}}=R_{\mathrm{in}}$. Multiplying their difference by the charge gives zero net stationary electrical current through the molecule's single contact, even though individual arrival and departure events continue.

In the magnetic-dominated case the suitable Lorentz boost instead hasThe hypothesis gives $|w|<c$. The electromagnetic field invariants now have $E^2-c^2{B}^2<0$, allowing a purely magnetic inertial frame. The initial velocity is again $-w\hat{\mathbf{z}}$, perpendicular to ${\mathbf{B}}^{\prime }$. Since a magnetic field does no work, the electron's relativistic energy and speed in this inertial frame remain constant. Its trajectory is a circle in the $x^{\prime }{z}^{\prime }$ plane with Lorentz factor ${\gamma }_w$ and cyclotron frequency $\Omega =e{B}^{\prime }/({\gamma }_{w}m)$.

For $E,B>0$, set $\theta =\Omega t^{\prime }$ and $\rho =w/\Omega$. A parametrization with the correct initial velocity and Lorentz force isTransforming the world line back to the laboratory givesThe laboratory trajectory is a stretched cycloid, with cusps at $\theta =2\pi n$ where the electron is momentarily at rest. It first accelerates toward $-x$, then curves toward $+z$, returning to its original $x$ coordinate each cycle while advancing in $z$. Its mean drift is $\Delta z/\Delta t=w=E/B$, the E-cross-B drift. Its laboratory kinetic energy oscillates as the electric field does positive and then negative work; it does not grow indefinitely. The strict inequalities in the two branches exclude the null case $|E/B|=c$, where neither field can be removed by a finite Lorentz boost.