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Take $z$ upwards and write $\zeta =z+Wt$, so that the unperturbed interface is $\zeta =0$, with carbon dioxide above and water below. Use the sharp-interface, incompressible flow version of Darcy's law, with constant porosity and permeability of a porous medium, no capillary pressure, and effectively unbounded layers. The distinction between Darcy flux and interface speed matters: the base flux is $q_z=-\varphi W$, since the printed $W$ is the speed of the interface. Thus the base pressure gradients satisfyThe mass density ordering is stabilizing, whereas injection of the less viscous fluid into the more viscous fluid produces the Saffman–Taylor instability.

For linear stability analysis, displace the interface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$. Taking the divergence of Darcy's law shows that the perturbation pressure satisfies the Laplace equation. Decay away from the interface gives, for $a_{\alpha }=|\alpha |>0$,The kinematic boundary condition and continuity of normal Darcy flux giveMeanwhile pressure continuity, evaluated on the displaced interface, givesEliminating $A,B$ yields the buoyancy-modified Darcy fingering dispersion relation:Every nonzero wavenumber decays when $W<W_c$, grows when $W>W_c$, and is neutral at equality, whereThe zero wavenumber is a uniform displacement and is neutral. Without a short-scale regularization such as surface tension, this model has no fastest-growing finite wavenumber: on the unstable side its growth rate increases without bound as $|\alpha |$ increases. Finite layer depths would change the dispersion relation; none are specified here.

Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a year of $365.25$ days, the data giveThe numerical paragraph supplies no porosity, so it cannot determine a unique numerical interface speed. If $W$ were instead intended to mean Darcy velocity, the numerical critical value would be $103.2\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$, and the interface would move at $W/\varphi$. These two velocity conventions must not be conflated.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

Take $k>0$; reflection in $x$ supplies the result for negative wavenumber. Since each region has uniform base velocity, the perturbation is an incompressible, irrotational flow with a velocity potential satisfying the Laplace equation. Its decaying outer forms areFor a varicose mode of a planar jet, reflection symmetry makes the inner potential proportional to $\cosh (ky)$; for a sinuous mode of a planar jet, it is proportional to $\sinh (ky)$.

At the upper interface, the linearized kinematic boundary condition isThe first equation gives ${\varphi }_{+}(h)=-\sigma \eta /k$. The second givesThe linearized Unsteady Bernoulli equation gives $p^{\prime }=-\rho ({\partial }_t+U{\partial }_x)\varphi$. Applying pressure continuity therefore produces the equal-density top-hat planar-jet dispersion relationSolving the quadratic equation givesEquivalently,The plus sign is the unstable Kelvin-Helmholtz instability branch. In particular, the two modes have exactly the same growth rate,Its limiting forms areThus neither type grows faster in this equal-density idealization, although their phase speeds differ. The unbounded large-$k$ growth is the familiar short-wave pathology of an inviscid vortex sheet.

For an axisymmetric separated solution of the Laplace equation whose angular factor is the Legendre polynomial $P_n(\cos \theta )$, the radial ordinary differential equation has the two powers $r^n$ and $r^{-n-1}$. Thusfor constants $A,B$.

Write $x=\cos \theta$. Since $3\cos (2\theta )=6x^2-3=4P_2(x)-P_0(x)$, only the spherical harmonics of degrees zero and two occur. The degree-zero radial coefficient satisfying its two Dirichlet boundary conditions is $1-2/r$. For degree two, write $C{r}^2+D{r}^{-3}$; thenso $C=-4/31$ and $D=128/31$. By linearity and uniqueness of this boundary value problem,Direct substitution at $r=1,2$ verifies both boundary values.

The Wirtinger derivative $u_z=(u_x-i{u}_y)/2$ is holomorphic, since ${\partial }_{\bar{z}}{u}_z=\Delta u/4=0$ by the Laplace equation. The conformal map $w=z^2$ takes the quadrant to the complex upper half-plane. Integrating the prescribed tangential boundary derivatives, choose a common corner value $c$ and define the real boundary functionThe decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its Poisson integral converges. A holomorphic function whose real part is that Poisson integral is supplied by the regularized Schwarz integral formula:For $u_{\ast }(z)=\mathrm{Re}W(z^2)$, the chain rule gives $(u_{\ast }{)}_z=z{W}^{\prime }(z^2)$. Apply integration by parts to obtainSubstituting $s=r^2$ and $s=-r^2$ gives the requested particular integral representation:Both integrals converge absolutely for an interior point of the quadrant. The Sokhotski–Plemelj formula gives $\mathrm{Re}(u_{\ast }{)}_z(x)=g_2(x)/2$ on the horizontal edge and $\mathrm{Im}(u_{\ast }{)}_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a holomorphic function $H$ satisfyingEquivalently, write $H(z)=ih(z^2)/z$, where $h$ is holomorphic in the complex upper half-plane with real boundary values on the nonzero real axis. An antiderivative of $2H$ has a harmonic real part with zero tangential boundary derivatives. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular harmonic function contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the Poisson integral representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

The far-field composition $-{\varphi }_B$ has chemical potential zero. Local curved-interface coexistence imposes $\mu (R)={\mu }_s=\gamma /({\varphi }_{B}R)$. The decaying spherically symmetric solution of the exterior Laplace equation isWith constant mobility, its outward current isThe linearized exterior composition is $\varphi (r)=-{\varphi }_B+\mu (r)/\alpha$, consistent with the surface shift in the previous part.

For diffusion-controlled droplet growth, the interior is uniform to leading order and has no diffusive current. Replacing exterior material by interior material requires the composition jump $2{\varphi }_B$, so conservation gives $2{\varphi }_B\dot{R}=-J_r(R)$. HenceThe negative sign represents evaporation. The total outward current is $4\pi R^2{J}_r(R)=4\pi M\gamma /{\varphi }_B$, independent of $R$ in this leading quasi-static capillary approximation.

The flat interface is an equilibrium reservoir with $\mu =0$, while the droplet has the constant surface value ${\mu }_s=\gamma /({\varphi }_{B}R)$. The exterior chemical potential satisfies the Laplace equation, tends to zero far away, and obeys these two surface values. These are exactly the electrostatic boundary conditions at a conductor for a sphere held at potential ${\mu }_s$ above a grounded infinite plane.

Introduce an arbitrary permittivity $ϵ$. In the electrostatic problem, $Q=-ϵ{\int }_{S_R}{\partial }_n\mu dS=C(R,h){\mu }_s$. Therefore the diffusion-capacitance analogy gives the outward currentFor an isolated sphere, its capacitance is $C(R,\infty )=4\pi ϵR$, so $I(R,\infty )=4\pi MR{\mu }_s=4\pi M\gamma /{\varphi }_B$. Dividing givesThus sphere-plane capacitance determines the evaporation enhancement. The assumed spherical shape gives the corresponding law $\dot{R}=-M\gamma c(R/h)/(2{\varphi }_B^2{R}^2)$ for droplet evaporation near a planar reservoir. The local surface current is generally nonuniform, even though its total is described by a single capacitance.

Expand $g$ in its Fourier sine seriesBy separation of variables, the decaying solution of Laplace equation in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin (k_{n}y)$. Continuity at $x=0$ and integration of the equation across the Dirac delta function require the normal-derivative jumpThe jump of $C_n{e}^{-k_n|x|}$ is $-2k_n{C}_n$, so

For $g(y)=\delta (y-c)$, its Fourier coefficients areThusTranslation in $x$ gives the Dirichlet Green function for an infinite strip with source $(b,c)$:

For a velocity potential, $\mathbf{u}=\nabla \varphi$. The incompressible flow condition $\nabla \cdot \mathbf{u}=0$ therefore becomes Laplace equation ${\nabla }^2\varphi =0$. The Laplace equation in polar coordinates givesso positivity of $\gamma$ and $\lambda$ requiresThe velocity components areThe no-penetration boundary condition on both walls says $u_{\theta }=0$ at $\theta =0$ and $\theta =\alpha$, hence $\lambda \alpha =n\pi$. If $n\ge 2$, $\sin (n\pi \theta /\alpha )$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and

Equation (1) is Laplace equation for the velocity potential. It expresses incompressibility, $\nabla \cdot \mathbf{u}=0$, together with $\mathbf{u}=\nabla \varphi$. Equation (2) is the linearized dynamic free-surface condition: the unsteady Bernoulli equation and constant atmospheric pressure balance the potential acceleration against the gravitational restoring term $g\zeta$. Equation (3), with the printed $c$ understood as $\zeta$, is the kinematic free-surface condition: the surface moves with the vertical fluid velocity. Equations (4) and (5) are no-penetration boundary conditions at the rigid bottom and cylindrical sidewall, respectively.

Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the Fourier transform in $x$, the Laplace equation becomesDecay as $y\to \infty$ selectsPart (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernelFor $x,y>0$ this givesThe odd extension enforces $u(0,y)=0$, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

Let ${\varphi }_1,{\varphi }_2$ be two solutions with the same Neumann boundary condition, and set $u={\varphi }_1-{\varphi }_2$. ThenMultiplying by $u$ and applying Green's first identity givesThe integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the modified Helmholtz equation is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the Laplace equation with prescribed normal derivative are unique only up to an additive constant, so

For irrotational flow, $\mathbf{u}=\nabla \varphi$. Incompressibility gives the Laplace equationThe rigid side walls impose no penetration:and decay in the infinitely deep fluid requires $\nabla \varphi \to 0$ as $z\to -\infty$.

Linearizing the kinematic condition from part (a) about $z=0$ givesThe Unsteady Bernoulli equation, evaluated at the surface and linearized about hydrostatic equilibrium, gives the dynamic conditionThese are the linearized free-surface boundary conditions. Initial values for $\eta$ and $\varphi$ specify the transient; below we also record the initially quiescent solution.

The spherically symmetric Laplace equation with $\tilde{\varphi }(R)=\delta (R)$ and $\tilde{\varphi }(\infty )=\epsilon$ has solutionThusConservation at the moving interface, whose composition jump is $2{\varphi }_B$, gives $2{\varphi }_B\dot{R}=-J_r(R^{+})$ and hence

Since $\delta (R)=C/R$, the right side is proportional to $\epsilon /R-C/R^2$. It is negative for $R<R^{\ast }$, zero at $R^{\ast }=C/\epsilon$, positive for $R>R^{\ast }$, and approaches zero from above for large $R$. Thus $R^{\ast }$ is the unstable critical nucleus: smaller droplets dissolve, while larger droplets grow.

Write the velocity potentials in the upper, middle and lower domains as ${\varphi }_{+}$, ${\varphi }_0$ and ${\varphi }_{-}$. Irrotational flow and incompressible flow imply the Laplace equationThe far-field boundary conditions are

Every shear layer is a material interface. The exact kinematic boundary condition at $y=h+\eta$ isand at $y=-h-\eta$ it isHere each equation is evaluated on the indicated moving interface.

Finally, pressure continuity is the dynamic boundary condition for an inviscid interface. The Unsteady Bernoulli equation givesafter absorbing spatially constant functions of time into the potentials.

Write the base-state gradients asThe base heat flux and Stefan condition relations are

Under the quasi-stationary approximation, and with advective heat transport neglected, each temperature perturbation satisfies the Laplace equation. The decaying normal-mode solutions areKeeping the displaced ice interface at $T_m$ omits the curvature correction described by the Gibbs--Thomson relation.
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives

Put $r=k_a/k_w$. Since $G_a=G_w/r$, the temperature condition in the limits $r\ll 1$ and $\alpha h\ll 1$ givesAlthough $r$ is small, the product $rC$ is therefore order one and cannot be discarded. The flux condition then givesThis is why the stated asymptotic warning matters.

The perturbed Stefan condition at $x=0$ isUsing $B\sim {\eta }_2{G}_w$ and $-k_w{G}_w={\rho }_i{L}_{f}V$ yieldsSubstitution of the interface-amplitude ratio from part i produces the thin-film icicle-ripple instability dispersion relation

Set $b=\Gamma h{\alpha }^3$. Rationalizing this complex number givesso its growth rate and imaginary part areDifferentiating the growth rate with respect to the wavenumber shows that its positive stationary point satisfiesIt is the unique global maximum, and henceAt this wavenumber $b=1/\sqrt{5}$. With the convention $e^{i\alpha z+\sigma t}$, a constant phase travels with phase velocityThereforeBecause $z$ increases downward, the negative sign means that the icicle ripples migrate upward with speed $\sqrt{5}V/6$.

The scalar Cauchy-Kovalevskaya theorem says that a partial differential equation solved for its highest derivative normal to a real-analytic non-characteristic hypersurface, with real-analytic coefficients and Cauchy data, has a unique local real-analytic solution. In coordinates, an equationhas such a solution near the origin when $F$ and the prescribed values of ${\partial }_t^j\varphi (0,x)$ for $0\le j<m$ are real analytic.

Choose a real-analytic primitive $F_0$ of $f$ near zero and apply the theorem to the scalar Laplace equationThe line $t=0$ is non-characteristic because the coefficient of ${\varphi }_{tt}$ is one. DefineThen $u(0,x)=f(x)$ and $v(0,x)=g(x)$, while equality of mixed derivatives and the Laplace equation giveThus $(u,v)$ is the required local real-analytic solution.

Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

The boundary Schauder estimate iswhere $C$ depends only on the dimension, $\alpha$, the ellipticity constants, and the coefficient Hölder norms.

Subtract a $C^{2,\alpha }$ extension of $\phi$ to reduce to a function $v$ with zero data on the flat boundary; this changes the forcing by a controlled $C^{0,\alpha }$ term. The key local estimate is that for every $\delta >0$,To prove it, argue by contradiction. A failing normalized sequence has points $x_k,y_k$ at which the second-derivative Hölder quotient stays nonzero. Set $r_k=|x_k-y_k|$, subtract the appropriate second-order Taylor polynomial, and rescale space by $r_k$ and the functions by $r_k^{2+\alpha }[D^2{v}_k{]}_{\alpha }$. Necessarily $r_k\to 0$.

If the rescaled distance to the flat boundary tends to infinity, the domains converge to all of ${\mathbb{R}}^n$; otherwise they converge to a half-space. Coefficient compactness freezes the principal matrix, while the normalized right-hand sides converge locally uniformly to zero. A linear change of variables turns every limiting equation into the Laplace equation. In the whole-space case the polynomial-growth Liouville theorem for harmonic functions makes the limit a polynomial of degree at most two. In the half-space case the zero boundary data permit odd reflection across the flat boundary, after which the same theorem applies. The subtracted Taylor normalization forces that polynomial to vanish to second order, contradicting the nonzero limiting Hölder oscillation.

Apply the local estimate on all interior balls and boundary half-balls. The interior and boundary forms of the Simon absorption lemma absorb the term multiplied by $\delta$, while the Hölder interpolation inequality absorbs the remaining lower $C^2$ norm into the top seminorm and the $C^0$ norm. Restoring $\phi$ yields the displayed estimate.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.