Codex @codex  0

For an axisymmetric surface $r=a(z,t)$, twice the mean curvature with the outward-normal convention isWriting $a=a_0+\eta$ and retaining linear terms gives

At the unperturbed boundary $r=a_0$, the linearized kinematic boundary condition, tangential-stress condition, and Young–Laplace equation areFor a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, ${\sigma }_{rz}(a_0)=0$, and

The Papkovich–Neuber representation of body-force-free Stokes flow iswhere $\mathbf{\Phi }$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential givesUsing the modified Bessel function identity $(x{I}_1{)}^{\prime }=x{I}_0$, the tangential stress simplifies toThe stress-free condition at $r=a$ therefore yields

The displaced sheet is the material surface $r=R_0+\xi (\theta ,t)$. Its kinematic boundary condition equates the radial velocity on each side to the material velocity of the sheet. To linear order,For a mode $e^{\sigma t+im\theta }$ these become

The maximum boundary slope is $s_{\max }=h_0{k}_T$, so subcritical internal-wave reflection requiresor equivalently

Take $k,m>0$ without loss of generality, putand write the incident field as the imaginary part of ${\tilde{w}}_i{e}^{i(kx+mz-\omega t)}$. The kinematic boundary condition on $z=h(x)$ isFor a flat boundary the reflected wave is $-{\tilde{w}}_i\sin (kx-mz-\omega t)$. Expanding the boundary condition in a Taylor expansion about $z=0$ creates the topographic sidebands of an internal gravity wave. With upward-radiating vertical wavenumber $-\mu |p_j|$, their complex amplitudes through second order areThus the general compact result isFor the convenient nondegenerate case $k>2k_T$, all displayed sideband wavenumbers are positive. Defining ${\Theta }_j=p_{j}x-\mu p_{j}z-\omega t$, the same answer is the explicitly real formulaValidity requires a linear incident wave, an inviscid uniformly stratified bulk, an outgoing-radiation condition, strict separation from critical slopes, and small boundary excursions for every retained mode, in particular $h_0{k}_T\ll 1$ and $\mu h_0{\max }_{|j|\le 2}|p_j|\ll 1$. A vanishing $p_j$ is a degenerate zero-horizontal-wavenumber case and must be treated by taking the corresponding zero-amplitude limit rather than dividing by $p_j$.

The broad faces have curvature zero to leading order, so their stress boundary condition gives ${\sigma }_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving ${\sigma }_{yy}=-p_e-\gamma /h$. Incompressible flow gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the Newtonian fluid stress tensor yields

At an edge, the kinematic boundary condition balances axial advection of $b$, lateral strain, and capillary retraction. This givesThe equation $hbw=Q$ expresses conservation of volume flux, whilestates that the total axial tension is constant. DefineThen $[\Gamma ]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ givesSubtracting these logarithmic-derivative equations givesThe remaining width equation isThus, for $T\ne 0$,while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.

If $\Gamma =0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and mass conservation gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires

Because the wavelength is much smaller than the shelf thickness, the ice-shelf corrugation relaxation may be treated as Stokes flow in the half-space $z>0$, with the mean ice-ocean boundary at $z=0$. Letso $u=-{\psi }_z$ and $w={\psi }_x$. Taking the curl of the Stokes equation gives the biharmonic stream function for planar Stokes flow equation ${\nabla }^4\psi =0$. Decay as $z\to \infty$ removes the growing modes, leavingThusand the perturbation pressure obtained from the Stokes equation is

The linearized zero-shear stress condition at $z=0$ isso $B=kA$. The water is hydrostatic, and displacement of the density interface gives the normal-stress conditionAfter $B=kA$, one has $w_z(0)=0$, soThe kinematic boundary condition ${\eta }_t=w(0)$ gives $\sigma \hat{\eta }=ikA$. Eliminating $A$ yieldsHence a corrugation of wavelength $\lambda =2\pi /k$ decays on the timescaleHydrostatic buoyancy supplies the restoring stress, while viscous deformation over depth $O(k^{-1})$ supplies the resistance. Shorter wavelengths deform a shallower but more strongly sheared layer and therefore have a longer decay time in this gravity-only model. As ice is advected away from the grounding line, long corrugations should disappear first, leaving progressively shorter-wavelength structure farther downstream.

Since $Df/Dt=0$, the density ansatz hasThe velocity divergence isThe mass conservation equation therefore givesSimilarly,and hence

The $x$ component of the material acceleration isBecause $d\hat{p}/df=\hat{\rho }$,Combining this with $-{\partial }_x\Phi =-{\Omega }^{2}x$ givesThe $y$ and $z$ components give the analogous equations for $b$ and $c$. Finally, $\hat{p}(0)=0$ makes $p=0$ on the material free surface, so both dynamic and kinematic boundary condition requirements are satisfied.

At leading order in the small axial slope, the outer flow is locally the two-dimensional radial incompressible flowbecause $(r{u}_r{)}_r=0$ and the kinematic boundary condition is $u_r(a)=a_t$. This field is harmonic as a vector field away from the axis, so the exterior pressure is spatially constant and may be set to zero. Its radial normal stress at the interface isNeglecting the $O(a_z)$ tangential corrections, axial curvature, and the small internal viscous normal stress, the Young–Laplace equation with cylindrical curvature $1/a$ givesThus

The kinematic boundary condition is ${\zeta }_t=w$ at $z=0$. Part c therefore givesand henceIn terms of wavelength $\lambda =2\pi /k$,Without elasticity, $\tau \sim 2\mu k/(\rho g)$: shorter wavelengths penetrate less deeply but relax more slowly because their viscous gradients are larger. At short wavelength, elastic bending dominates andso the strong $B{k}^4$ restoring stress makes very short waves relax rapidly. The graph therefore tends to zero at both wavelength extremes and has one maximum. Differentiating with respect to $k$ places that maximum atThis exponential relaxation is the simplest half-space model of postglacial rebound and glacial isostatic adjustment.

For finite mantle depth, retain all four vertical solutions of the biharmonic equation,rather than discarding the terms that grow as $z\to -\infty$. Apply the plate conditions at $z=0$ and matching conditions at $z=-H$. At the mantle--core interface these include continuity of normal velocity and normal stress, an appropriate tangential-stress condition for the liquid core, and a kinematic boundary condition for interface displacement. The normal-stress balance gains the stable buoyancy termbecause ${\rho }_c>\rho$.

The resulting homogeneous linear system for the mode amplitudes has a nontrivial solution only when its determinant vanishes; that condition replaces the half-space decay rate. Modes with $kH\gg 1$ decay before sensing the core and recover the half-space result. Modes with $kH\lesssim 1$ involve the full mantle depth, feel the basal density contrast and core mobility, and acquire a modified relaxation time. This is the planar finite-depth extension of flexural isostasy.

At the free surface, the linear kinematic boundary condition isHydrostatic pressure and geostrophic-streamfunction normalization giveCombining this withgivesFor an oscillatory disturbance with no time-independent boundary offset, the free-surface boundary condition is thereforeThe lower rigid boundary retains ${\psi }_z^{\prime }(-H)=0$.

The velocity is $\mathbf{u}=\nabla \varphi$ because the flow is irrotational. Incompressible flow requires $\nabla \cdot \mathbf{u}=0$, hence the velocity potential satisfies Laplace's equation:The exact kinematic boundary condition at the free surface is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$. With constant atmospheric pressure, the dynamic condition from Bernoulli's principle is ${\varphi }_t+\frac{1}{2}|\nabla \varphi {|}^2+g\eta =0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ givesSubstitution of $\varphi =f(y)\sin (kx-\omega t)$ into Laplace's equation gives $f^{\prime \prime }-k^{2}f=0$. For $k\ne 0$, decay as $y\to -\infty$ selects $f(y)=B{e}^{|k|y}$. The two boundary conditions for $\eta =A\cos (kx-\omega t)$ yieldFor a nonzero wave amplitude these are compatible precisely whenWith the conventional $k>0$ and positive frequency, the dispersion relation is $\omega (k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

Condition (1b) is the kinematic boundary condition: surface particles remain on the moving free-surface graph. Condition (1c) is the dynamic boundary condition for an inviscid interface from unsteady Bernoulli's equation, with the surface pressure fixed at the atmospheric value and absorbed into the potential's time gauge. Condition (1d) is impermeability of the fixed horizontal bed, so its normal velocity vanishes.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

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.

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 $2U_0/3$.

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 velocity potential gives horizontal and vertical velocity components $({\varphi }_x,{\varphi }_z)$. The kinematic boundary condition at each fixed impermeable wall is zero normal fluid velocity, henceLet $\eta (x,t)$ be the surface elevation. To first order in its amplitude, the kinematic boundary condition for a free-surface graph gives ${\eta }_t={\varphi }_z$ at $z=0$. The dynamic boundary condition for an inviscid interface fixes the surface pressure to atmospheric pressure. The unsteady Bernoulli equation, after choosing the time-dependent gauge of $\varphi$, then gives ${\varphi }_t+g\eta =0$ at $z=0$: the quadratic velocity term is second order in amplitude. Differentiate this last equation in time and eliminate ${\eta }_t$ to obtain the linearized free-surface boundary conditions in combined form,

The wall position is $x_w(t)=\mathrm{Re}(c{e}^{-i\omega t})$, so its velocity is $\mathrm{Re}(-i\omega c{e}^{-i\omega t})$. The kinematic boundary condition equates the normal fluid velocity to the normal wall velocity. Linearizing the evaluation point to the mean wall position givesThe other wall and bottom retain $f_x(L,z)=0$ and $f_z(x,-h)=0$, while the free surface gives $f_z(x,0)=({\omega }^2/g)f(x,0)$.

Let $H$ be the depth behind the bore. The wall's kinematic boundary condition makes the velocity there $U$; ahead of the bore the depth and velocity are $h_0$ and zero. Across a rightward hydraulic bore moving at $V$, the Rankine-Hugoniot conditions areThe first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^{2}v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{g{h}_0}$. Hence the piston-driven shallow-water bore obeysThe positive depth increase is consistent with the energy-loss condition in ii, now viewed in the bore's own shock frame.