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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.

Simplify the real component to $u=(1-x^2-y^2)/D$. Thus $u=0$ is the unit circle with $(1,0)$ excluded from the mapping's domain. Its upper semicircle has $v>0$ and its lower semicircle has $v<0$; the point $(-1,0)$ maps to $u=v=0$. Inside the disk, $u>0$, so the principal arctangent is unambiguous. Setting $\alpha =V$ givesThe Laplace equation follows from (b). Approaching an open upper boundary arc makes the ratio tend to $+\infty$, giving $+V$; the lower arc gives $-V$. This solution is $V(2F-1)$ in terms of the harmonic measure of a semicircle in a disk $F$.

The boundary data jump at $(\pm 1,0)$, so no path-independent continuous boundary value can be imposed at those two junctions. The displayed bounded harmonic function is the solution with the specified limits on the open arcs; those endpoint discontinuities are part of the idealized capacitor geometry.

Set $\Phi =R(r)P(\theta )$ and multiply the Laplace equation by $r^2/(RP)$. Separation of variables givesWith $\xi =\cos \theta$, the angular equation is $[(1-{\xi }^2)P_{\xi }{]}^{\prime }+\lambda P=0$. Regularity at both poles selects the Legendre polynomials $P_n$ and $\lambda =n(n+1)$. The radial Cauchy-Euler equation has solutions $r^n,r^{-n-1}$. Expanding the angular data in this complete orthogonal system therefore giveswith the usual convergence required of a harmonic expansion on the region considered. Angular Legendre functions singular at the poles are excluded.

For $0<a<b$, the outer data have expansion $b^2{\cos }^2\theta =b^2/3+(2b^2/3)P_2(\cos \theta )$, while the inner data are pure $P_0$. Thus only degrees zero and two are needed. A radial interpolation matching each pair of boundary values givesThe first term is a linear combination of $1,r^{-1}$ and takes values $a^2,b^2/3$ at the two radii. The second is a combination of $r^2,r^{-3}$ and takes values zero and $2b^2/3$. These checks establish both boundary conditions; each term is harmonic. The maximum principle for harmonic functions gives uniqueness. This is an example of axisymmetric harmonic interpolation between concentric spheres.

Split the Fourier transform integral at zero:By Fourier inversion applied to this pair, or by the residue theorem, the other transform isFor example, inversion with $\lambda =\mu$ gives $\int e^{ikx}(k^2+{\mu }^2{)}^{-1}dk=(\pi /\mu )e^{-\mu |x|}$; evenness supplies the desired sign convention.

Take the spatial Fourier transform of the Laplace equation. For $y>0$ it gives ${\hat{\psi }}_{yy}-k^2\hat{\psi }=0$. Removing the exponentially growing solution and matching the boundary givesAt $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifiesthe Poisson kernel for the upper half-plane. The convolution theorem therefore gives the Poisson integralThis is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero harmonic function $e^{x^2-y^2}\sin (2xy)=\mathrm{Im}{e}^{(x+iy{)}^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to \infty$, but grows rapidly in $x$ and has no ordinary spatial Fourier transform. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.

For the interval data, direct integration gives the Poisson extension of interval boundary dataAs $y\downarrow 0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi \sim 2ac/(\pi y)$ for fixed $x$ and large $y$.

For separation of variables, write $B=F(r)\Theta (\theta )$. The exterior Laplace equation separates intoWith $\mu =\cos \theta$, angular regularity at the poles selects $\lambda =\ell (\ell +1)$ and the Legendre polynomial $P_{\ell }(\mu )$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }-\ell (\ell +1)F=0$, an Euler-Cauchy equation with solutions $r^{\ell },r^{-\ell -1}$. Decay at infinity excludes the first branch for every $\ell$, including its constant $\ell =0$ case. ThereforeThe expansion uses axisymmetric spherical harmonics. The written scalar boundary-value problem permits the $\ell =0$ term; any additional physical exclusion of that sector must be stated separately.

For separation of variables, let $\Phi =S(\sigma )T(\tau )$. The Laplace equation givesA nonzero $2\pi$-periodic angular solution requires $\lambda =n^2$, $n=0,1,\dots$. For $n\ge 1$, $S$ is a linear combination of $\cos (n\sigma )$ and $\sin (n\sigma )$, while $T^{\prime \prime }-n^{2}T=0$ gives hyperbolic sine and cosine. For $n=0$, periodicity makes $S$ constant, and $T=A_0+B_0\tau$. Negative separation constants would give nonperiodic real exponentials in $\sigma$.

Superposition yieldsCompleteness follows by expanding a harmonic function in a Fourier series in the periodic coordinate: its Fourier coefficients must satisfy exactly these radial equations. As usual, the coefficients are restricted by the convergence required of the series on compact subregions. The bipolar coordinates domain imposes no boundary condition at $\tau =0$ or $\tau =\infty$, so both hyperbolic branches remain available.

Put $\Psi =U(r)V(\varphi )$ in the Laplace equation in polar coordinates. After multiplication by $r^2/(UV)$, separation of variables givesThe angular function is $2\pi$-periodic and single-valued, so its nonconstant modes are $\cos (n\varphi ),\sin (n\varphi )$ with integer $n\ge 1$. Their radial Euler differential equation has solutions $r^n,r^{-n}$; finiteness at the origin excludes $r^{-n}$.

The zero angular mode has constant angular part and radial solutions $1,\log r$. The logarithmic term is again excluded at the origin. Superposition therefore yieldsThis is the regular harmonic expansion inside the disk.

For any interior point $\mathbf{p}$, choose a sufficiently small circle centred on $\mathbf{p}$ and wholly inside the disk. Translating the origin does not change the Laplace equation, so part (b) givesSuppose a real harmonic function, continuous on the closed disk, attains its global maximum $M$ at an interior point. Every value on each such circle is at most $M$, and their average is $M$. By continuity, every value on the circle must therefore equal $M$. Varying its radius shows that a neighbourhood of the point also has value $M$.

Thus the set of points attaining $M$ is open in the disk, and it is closed by continuity. The disk is connected, so this set must be the whole disk. The function is constant and has the same maximum on the boundary. Otherwise the maximum cannot be interior. Compactness of the closed disk guarantees existence of a maximum, proving the maximum principle for harmonic functions:Continuity up to the boundary is the standard hypothesis for this attained-maximum formulation.

For a separated axisymmetric harmonic function $\Psi =R(r)\Theta (\theta )$, multiplying the Laplace equation by $r^2/(R\Theta )$ givesThe two sides depend on different variables, hence both equal a constant. Put $u=\cos \theta$ in the angular equation. Since $\sin \theta {\Theta }^{\prime }=-{\sin }^2\theta d\Theta /du$, it becomes the Legendre differential equationRegularity on both parts of the polar axis restricts $\lambda =\ell (\ell +1)$, with $\ell =0,1,\dots$, and selects the Legendre polynomial $P_{\ell }(u)$. The other angular solution is singular at an endpoint.

The radial Euler differential equation isTrying $R=r^s$ gives $s(s+1)=\ell (\ell +1)$, so $s=\ell$ or $s=-\ell -1$. Superposition of these separated harmonic functions givesThe Legendre polynomials form the angular expansion for regular data with rotational symmetry about the polar axis; the radial domain and the boundary conditions determine which coefficients survive and where the expansion converges.

Substitute $U=R(r)F(\theta )$ into the axisymmetric Laplace equation and multiply by $r^2/(RF)$. The radial and angular terms must be separate constants:ThusWith $x=\cos \theta$, the angular equation becomesRequiring boundedness on the entire polar interval, and using the supplied characterization of bounded Legendre polynomial solutions, gives $\lambda =n(n+1)$ and $F\propto P_n(\cos \theta )$ for $n=0,1,2,\dots$. Trying $R=r^s$ gives $s(s+1)=n(n+1)$, hence $s=n$ or $s=-n-1$.

The globally angular-regular separated modes and their superposition are thereforeCoefficients and convergence are determined by the radial domain and boundary data. A domain containing the origin excludes negative radial powers for a regular solution. If the angular domain excludes a polar axis of the spherical coordinate system and singular angular behavior is allowed, additional Legendre-function branches are possible; the displayed family is the one regular over the complete sphere, as intended by the hint.

A fundamental solution of the Laplace equation satisfies ${\Delta }_{\mathbf{r}}{G}_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ in the sense of distributions, without a boundary condition. Choose it rotationally symmetric about the source. Away from the source, with $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial Laplace equation is $(s{G}_0^{\prime }{)}^{\prime }=0$, giving $G_0=A\log s+B$.

Integrating over a small disc around the source and using the divergence theorem fixes the normalization:ThereforeThe additive constant is arbitrary; adding other entire harmonic functions is also possible if rotational symmetry is not imposed. The positive sign here follows from the convention $\Delta G_0=+\delta$.

Take a smooth variation $u+\epsilon \eta$ with $\eta$ vanishing on the boundary; compactly supported test functions suffice for the interior equation. At an extremum, the first variation is zero:Integration by parts removes the derivatives from $\eta$. The boundary terms vanish, givingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationThis is the classical equation for sufficiently smooth admissible extrema; for weaker regularity the first-variation identity supplies its weak form.

For the squared-gradient integrand, the equation is $2(u_{xx}+u_{yy}+u_{zz})=0$, namely the Laplace equation. Put $u_0=ax+by+cz$ and write any admissible function as $u=u_0+v$, with $v=0$ on the boundary. The energy identity isBy the divergence theorem, the middle integral is ${\int }_{\partial V}v(a,b,c)\cdot ndS=0$; equivalently it vanishes by weak integration by parts for zero-trace functions. The final term is nonnegative and vanishes at $v=0$. Thus this is a global minimum, not merely a stationary value:

The factor ${\eta }_{xx}/(1+{\eta }_x^2{)}^{3/2}$ is the signed curvature of the graph, so the pressure jump is surface tension times curvature, the planar Young–Laplace equation.

Take total velocity potentials ${\Phi }_{-}={\varphi }_{-}$ and ${\Phi }_{+}=Ux+{\varphi }_{+}$. Each disturbance potential satisfies the Laplace equation. The interface moves with the fluid: for $F=y-\eta (x,t)$, the material equation $F_t+v\cdot \nabla F=0$ gives ${\eta }_t+u{\eta }_x=v_y$. Keeping only first-order terms and evaluating at $y=0$ yieldsThe unsteady Bernoulli equation is ${\Phi }_t+|\nabla \Phi {|}^2/2+p/\rho =C(t)$, with any common gravitational contribution canceling across this equal-density interface. Subtracting the basic flow and linearizing gives $p_{-}^{\prime }=-\rho {\varphi }_{-,t}$ and $p_{+}^{\prime }=-\rho ({\varphi }_{+,t}+U{\varphi }_{+,x})$. The linearized pressure jump is $T{\eta }_{xx}$. HenceUse complex modes with real parts understood:The decaying vertical factors solve the Laplace equation in their respective half-planes. The kinematic conditions give $B_{-}=-i\omega A/k$ and $B_{+}=i(\omega -kU)A/k$. Substitution in the dynamic condition givesWhen the discriminant is nonnegative, the phase speeds are $U/2\pm \sqrt{Tk/(2\rho )-U^2/4}$. They are two dispersive capillary-wave branches advected by the mean velocity $U/2$.

When $\boxed{U^2>2Tk/\rho }$, the frequencies are complex conjugates and one mode grows exponentially, with growth rate $k\sqrt{U^2/4-Tk/(2\rho )}$. This is Kelvin-Helmholtz instability. Surface tension stabilizes sufficiently short waves, while for any nonzero shear, sufficiently long waves are unstable in this gravity-free equal-density model. A real cosine with real frequency describes the stable branches; the unstable case is represented by the real part of the corresponding growing complex mode.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

The boundary-value method. In a charge-free region, the electric potential obeys the Laplace equation, ${\nabla }^2\Phi =0$, and $\mathbf{E}=-\nabla \Phi$. Conductors impose equipotential boundaries, with specified voltages or charges, and a far-field condition fixes the remaining behaviour. The Uniqueness of the Dirichlet problem guarantees that a solution meeting the prescribed conductor potentials is the physical one. To see why, the difference $w$ of two candidate solutions is harmonic and vanishes on the boundaries; integration by parts gives $\int |\nabla w{|}^{2}dV=0$, hence no difference in their fields. An exterior problem also needs the appropriate condition at infinity.

Two axisymmetric harmonic functions. For $\Phi =f(r)\cos \theta$ in spherical coordinates, the Laplacian reduces toFor $f=r$, the bracket is $2r-2r=0$. For $f=r^{-2}$, it is $2r^{-2}-2r^{-2}=0$. ThereforeThe second function is singular at the origin, which is allowed here because that point lies inside the excluded conductor. The printed spherical Laplacian has a typo in its azimuthal term: its denominator should be $r^2{\sin }^2\theta$. Both functions are axisymmetric, so that term vanishes and the typo does not affect these calculations.

Potential and field geometry. Set the lower conductor to zero potential and the distant upper plate to $V$. The small boss sees approximately uniform background field $-E_0\hat{\mathbf{z}}$, where $E_0=V/d$. Combining the two harmonic functions to make $\Phi =0$ on $r=a$ gives the conducting hemispherical protrusion in a uniform field,It also vanishes on the surrounding plane $z=0$. The electric field components areEquipotentials crowd near the apex, and field lines meet the conductor normally.

The formula has the correct uniform far field but does not exactly impose a flat upper equipotential at finite $d$. It is the leading approximation justified by $a\ll d$; an exact finite-gap solution would include further corrections.