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A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

The hydrostatic buoyancy anomaly is $b^{\prime }=f_0{\psi }_z$, and its leading adiabatic evolution is $D_g{b}^{\prime }/Dt+N_0^{2}w=0$. The small vertical velocity advects the background stratification; vertical advection of the perturbation buoyancy is higher order. ConsequentlyThe impermeability condition at a stationary lower boundary $z=\alpha y$ is $w=\alpha v$. Using $v_g={\psi }_x$ gives the topographic buoyancy boundary condition for quasi-geostrophic flowFor application at the reference plane, the actual small-height condition is $|\alpha |L/D\ll 1$. Slopes scaled by $D/L$ are of Rossby-number order, so this topographic contribution is retained alongside the small ageostrophic vertical velocity. The dimensional slope should not be compared to the Rossby number without also specifying the aspect ratio.

The basic quasi-geostrophic streamfunction is $\overline{\psi }=-\Lambda zy$. It gives $U=\Lambda z$, basic buoyancy gradient ${\overline{b}}_y=-f_0\Lambda$, and spatially constant interior quasi-geostrophic potential vorticity. Linearizing its conservation law therefore givesThe printed equation omits the shear-advection term. This cannot be omitted for general disturbances: an interior anomaly proportional to $\cos [k(x-\Lambda zt)]$ is materially conserved but not independent of time at a fixed location. The intended zero-interior-PV Eady model modes do have $q^{\prime }=0$ and satisfy both expressions; the ensuing mode calculation is valid in that subspace.

At either boundary, linearizing the geostrophic material derivative includes $v^{\prime }{\overline{\psi }}_{zy}=-\Lambda {\psi }_x^{\prime }$. LetThe two boundary equations areThese are exactly the respective bottom and top slope corrections to the rigid-boundary buoyancy condition for quasi-geostrophic waves.

Assume periodicity horizontally and first consider $q^{\prime }=0$. Multiply its defining equation by ${\psi }_x^{\prime }$ and integrate over the layer. The two horizontal Laplacian terms vanish by integration by parts; the remaining volume term after one vertical integration is an $x$-derivative. ThereforeFor nonzero $A_0,A_D$, the boundary equations giveThe top advection term integrates to zero. Subtracting proves conservation of the boundary pseudomomentum of a sloping Eady layer:The original PDF uses squared vertical derivatives and the upper slope $\gamma$ in the first denominator; these details are corrupted in the converted TeX. For arbitrary nonzero interior PV the same calculation instead gives $d\mathcal{P}/dt=(f_{0}s{)}^{-1}\int {\psi }_x^{\prime }{q}^{\prime }dV$, which need not vanish. For example, at an instant take ${\psi }^{\prime }=\cos kx+(z/D{)}^2\sin kx$, independent of $y$; the zonal average of ${\psi }_x^{\prime }{q}^{\prime }$ is $-ks/D^2$. Thus the boundary-only conservation statement also needs the zero-interior-PV qualification.

For an exponentially growing smooth normal mode, this qualification is automatic: the interior equation is $ik(\Lambda z-c)\hat{q}=0$, and a phase speed with nonzero imaginary part cannot equal the real basic velocity. Therefore $\hat{q}=0$. If $A_0{A}_D<0$, the conserved boundary quadratic form is definite, so a nonzero exponentially growing boundary amplitude is impossible. Instability consequently requiresfor $\Lambda \ne 0$. When one $A$ vanishes, the corresponding boundary derivative is simply advected and must vanish for a growing mode. The other boundary then yields a real single-edge-wave speed; if both vanish, the zero-interior-PV Neumann problem has no nonzero growing mode. These degenerate cases are neutral and should not be handled by dividing by zero.

For equal slopes, put $\zeta =z/D$, $C=c/(\Lambda D)$, $M=N_0|k|D/|f_0|$, and $a=N_0^2\alpha /(f_0\Lambda )-1$. The normal-mode interior equation is ${\Phi }_{\zeta \zeta }-M^2\Phi =0$ for the zero-interior-PV modes. ThusThe coefficient equations areSetting their determinant to zero, without dividing by $C$ or $a$, givesso the Eady model with parallel sloping boundaries hasFor positive $k,f_0$, $M$ is the printed $\mu$, and $a$ is the printed $\tilde{\alpha }$. The expression is even in the sign of that vertical scale. For $a\ge 0$ the radicand is at least $1/4$, so both speeds are real and these modes have no exponential growth. For equal slopes the earlier necessary condition is merely $a^2>0$, which is satisfied in many stable cases and is not sufficient. At $a=0$ the speeds are $0$ and $\Lambda D$; the invariant with nonzero denominators cannot be used, but the determinant still correctly gives neutral modes. No claim excluding transient amplification is needed.

For a harmonic function $u$, put $g=u_x-i{u}_y$. Its Wirtinger derivative isso the Cauchy-Riemann equations make $g$ holomorphic. In the simply connected domain it has a single-valued primitiveThe integral is path independent: local holomorphic primitives make its value invariant under path homotopies, and being simply connected makes any two paths with the given endpoints homotopic, as in Question 3.

Since $F^{\prime }=g$, the real part satisfies $(\mathrm{Re}F{)}_x=u_x$ and $(\mathrm{Re}F{)}_y=\mathrm{Re}(ig)=u_y$. Hence $\mathrm{Re}F-u$ is constant on the connected domain. Adding a real constant corrects it, producing a single-valued holomorphic $f$ with $\mathrm{Re}f=u$. The imaginary part is a harmonic conjugate and is unique up to an additive real constant. This is the global harmonic conjugate criterion by vanishing periods in the simply connected case.

An explicit disk map of a slit half disk makes the two boundary approaches visible without assuming an extension theorem for the boundary. We construct the map as a succession of bijections and then determine its limits.

First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the square root whose argument lies in $(0,\pi )$. The removed segment on the imaginary axis becomes $[-1,0)$, so the image of $D$ is preciselyThe Möbius transformation $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the unit disc, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an automorphism of the unit disk after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued square root $u=\sqrt{v}$ and maps that slit disc bijectively ontoAll these maps have nonzero derivatives on their respective domains, since their potentially problematic zero or branch point is on the removed slit.

Next the Joukowski map $J(u)=u+1/u$ is a conformal bijection from $U$ onto the lower half-plane. For $u\in U$,If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the lower half-plane and consider $u^2-qu+1=0$. Neither root lies on the unit circle, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\mathrm{Im}u>0$. Thus it lies in $U$. Also $J^{\prime }(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the lower half-plane bijectively onto the unit disc: the inequality $|q+i|<|q-i|$ is exactly $\mathrm{Im}q<0$, and its inverse is $q=i(\zeta +1)/(\zeta -1)$. Consequentlyis a conformal bijection of $D$ onto the disc.

At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero real part. Since $\mathrm{Im}(z^2)=2\mathrm{Re}z\mathrm{Im}z$ and the real-coefficient Möbius map $w\mapsto v$ preserves the sign of the imaginary part, right-side approach gives $\arg v\to 0$, while left-side approach gives $\arg v\to 2\pi$. HenceThe two limits of $F$ are thereforeThey have modulus one and are distinct, since $L>0$ and the Möbius map $C$ is injective on the extended real line. The two-sided cluster set at an interior slit point is a geometric consequence of opening the slit into two boundary intervals.

For the given map $f$, the composition $B=f\circ F^{-1}$ is an automorphism of the unit disk. To see its precise boundary behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta )=(\zeta -a)/(1-\overline{a}\zeta )$. This gives a disc automorphism fixing zero. Applying Schwarz lemma to it and its inverse shows it is a rotation. ThusIts denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous bijection of the unit circle. Put $\alpha =B({\gamma }_{+})$ and $\beta =B({\gamma }_{-})$. These are distinct points of that circle, independent of the approaching sequence.

For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore exactly $\{\alpha \}$, $\{\beta \}$ or $\{\alpha ,\beta \}$, according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm {\epsilon }_n$ with positive ${\epsilon }_n\downarrow 0$.

For the unheaded continuation, set $w=\log z$ in the food equation. The two square roots are $w=\pm i\pi /4$, so exponentiating givesBoth values have $|z-1{|}^2=2-\sqrt{2}<1$ and hence lie strictly inside the disk. Both also have the indicated principal logarithms. Allowing the multivalued complex logarithm produces no additional points: the only permitted values of $w$ are still the two square roots already used.

For drink, the PDF squares only the denominator. Put $z=x+iy$. Its numerator is $(3x+iy)/2$ and its unsquared denominator is $(x+3iy)/2$, which is nonzero unless $z=0$. Multiplying the equation by this denominator squared and equating the real part and imaginary part yieldsthat is $x^2-9y^2=2x$ and $y(1-9x)=0$. If $y=0$, the real equation gives $x=0$ or $x=2$, and zero is excluded. If $y\ne 0$, then $x=1/9$, but the real equation would require $9y^2=-17/81$, impossible for real $y$. ThusIt lies on the allowed disk boundary, and direct substitution gives $3/1^2=3$. The denominator-only square matters: squaring the entire quotient would define a different locus.

Count oscillator polarizations at fixed center-of-mass momentum. In a critical closed bosonic string, each chiral sector has 24 transverse oscillator species. A species at frequency $m$ can be occupied $0,1,2,\dots$ times, so its contribution is $(1-q^m{)}^{-1}$. The full chiral generating function is thereforeThe two sectors are independent, while closed-string level matching restricts their oscillator numbers to the same $N$. Thus the number of states at the closed-string mass level is $g_N=d_N^2$, not $d_{2N}$.

Put $q=e^{-\beta }$, with $\beta >0$. The Dedekind eta function obeysThe transformed argument $2\pi i/\beta$ has large imaginary part, so its product approaches one and $\eta (2\pi i/\beta )\sim e^{-{\pi }^2/(6\beta )}$. HenceThis establishes the singular exponential as well as its power prefactor. The corrections to the transformed eta product are exponentially small in $1/\beta$.

The Cauchy coefficient formula, followed by $q=e^{-\beta }$, gives a steepest-descent integral with leading formLet $n=N-1$. The saddle of $S(\beta )=n\beta +4{\pi }^2/\beta$ isAlong the vertical steepest-descent direction $\beta ={\beta }_{\ast }+it$, the quadratic part is $-S^{\prime \prime }({\beta }_{\ast })t^2/2$. Thus the saddle-point approximation gives the Gaussian prefactorThe point $q=1$ supplies the largest exponential contribution. At another fixed root of unity of order $k>1$, the corresponding modular singularity gives a smaller exponential $e^{4\pi \sqrt{n}/k}$, so it does not change this leading asymptotic. Squaring the chiral result gives the large-level degeneracy of a closed bosonic stringReplacing $N-1$ by $N$ gives the equivalent usual leading asymptotic. The power $-27/2$ matters: the two-sector degeneracy is not just an unspecified exponential.

Since $M=2\sqrt{(N-1)/{\alpha }^{\prime }}$, the exponential is $e^{{\beta }_{H}M}$ withHere Boltzmann's constant is one. Coarse-graining the levels into a density per unit rest mass introduces the Jacobian $dN/dM={\alpha }^{\prime }M/2$. ConsequentlyThis is distinct from the degeneracy per level, whose power in $M$ is $-27$. The internal-state contribution to a canonical partition function behaves at large mass asIt converges for $\beta >{\beta }_H$ and diverges for $\beta <{\beta }_H$. At $\beta ={\beta }_H$, this particular internal rest-mass integral converges because of the power $M^{-26}$; one should not discard the prefactor and claim divergence there without specifying the full thermodynamic ensemble. For example, including continuous momenta in 25 spatial dimensions multiplies the high-mass integrand by a factor proportional to $M^{25/2}$, still leaving convergence of this ideal one-string integral at the endpoint, although sufficiently high derivatives are singular. Multi-string effects, volume and interactions affect the detailed limiting thermodynamics.

The leading microcanonical entropy grows as $S(E)\sim {\beta }_{H}E$, with logarithmic corrections, so increasingly large energies are stored in increasingly excited, long strings rather than in an arbitrarily hot ordinary gas. The Hagedorn temperature is the boundary of canonical convergence from above in inverse temperature; extrapolating the free-string canonical description to $T>T_H$ fails. This counting does not settle the interacting phase above that scale. Also, the bosonic ground-state tachyon is a separate vacuum instability; the high-level asymptotic is meaningful as a formal spectrum calculation without asserting a stable bosonic thermal vacuum.

Write $r=\sqrt{\mu /\delta }$. At first order the amplitude-phase perturbation is $A=r+\epsilon (A_1+ir\theta )+O({\epsilon }^2)$. Expanding the cubic term in the focusing nonlinear Schrodinger equation gives $A|A{|}^2=r^3+\epsilon (3r^2{A}_1+i{r}^3\theta )+O({\epsilon }^2)$. The constant terms cancel because $\delta r^2=\mu$. Separating real and imaginary parts yieldsFor the Fourier normal modes, these equations give $-i\Omega {\hat{A}}_1=-\chi r{K}^2\hat{\theta }$ and $-ir\Omega \hat{\theta }+(2\mu -\chi K^2){\hat{A}}_1=0$. Their determinant vanishes exactly whenThe focusing nonlinear Schrodinger modulation dispersion is negative for $0<K^2<2\mu /\chi$. One of the two frequencies then has positive imaginary part and grows exponentially: the constant state has modulational instability. The growth rate is $\sqrt{\chi K^2(2\mu -\chi K^2)}$, maximized at $K^2=\mu /\chi$ with value $\mu$. Modes beyond the unstable band have real frequencies. The zero-wavenumber phase mode is neutral, with possible linear phase drift from a uniform amplitude perturbation.

Let $V=U-c$ and suppose a growing normal mode has $\mathrm{Im}c=c_i>0$, taking $\alpha >0$ without loss of generality. Since $V$ never vanishes on the real channel, choose a continuous square-root branch and write $\varphi =V^{1/2}H$. Direct differentiation givesSubstitution into the Taylor–Goldstein equation and division by $V^{1/2}$ givesThus the required transformation isThe impermeable walls give $\varphi (\pm 1)=0$, hence $H(\pm 1)=0$. Multiply by $\overline{H}$, integrate across the channel, and use integration by parts. With the boundary term zero, the resulting power-transformed Taylor–Goldstein energy identity isBoth $U^{\prime \prime }$ and $J$ are real. Since $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part, divided by $-c_i$, isIf $J\ge U^{\prime 2}/4$ everywhere, the integrand is nonnegative and the first two terms have positive integral for any nonzero mode. This contradicts the identity. Therefore instability requiresThis is the necessary instability condition of the Miles–Howard theorem. Where $U^{\prime }\ne 0$, it says that the gradient Richardson number $J/U^{\prime 2}$ must be below $1/4$ somewhere. It is not a sufficient condition for instability, and the conclusion concerns exponential normal-mode growth, not exclusion of the algebraic lift-up effect in part (a). The algebra uses the given nondimensional $J$; in dimensional notation its counterpart is the squared buoyancy frequency.

Write the complex number as $z=x+iy$. Since $iz=-y+ix$, the modulus of the complex exponential givesThis equals one precisely when $y=0$. The quadratic formula shows that the imaginary part of either root is $\pm \sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:Both endpoints satisfy the condition. The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the unit circle.

Write the system as ${\mathbf{x}}^{\prime }=M\mathbf{x}$, with $M=\left(\begin{array}{cc}2& 5\\ -1& -2\end{array}\right)$. Its characteristic polynomial is ${\lambda }^2+1$, so its eigenvalues are $i$ and $-i$. An eigenvector for $i$ is $v=(5,-2+i{)}^{\mathsf{T}}$. Thus $e^{it}v$ is a complex solution, and its real part and imaginary part give real solutions:Their initial vectors are $(5,-2{)}^{\mathsf{T}}$ and $(0,1{)}^{\mathsf{T}}$, with determinant $5\ne 0$. Consequently they are linearly independent and span the two-dimensional solution space.

The odd-potential branch uses $s=\tanh (kL)$, whereas the even-potential branch uses $1/s$. The imaginary parts are identical becauseConsequently the equality of top-hat jet parity growth rates givesHere $\gamma =k{c}_i$ is the positive temporal growth rate. The real phase velocities differ: $c_{r,\mathrm{odd}}=\frac{V}{2}(1-e^{-2kL})$ and $c_{r,\mathrm{even}}=\frac{V}{2}(1+e^{-2kL})$. Thus equal amplification does not imply equal propagation. As $kL\to \infty$, both approach the isolated-vortex sheet phase velocity $V/2$ and growth rate $k|V|/2$.

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.

Choose $a=1$. The power-transformed Taylor–Goldstein energy identity reduces toThe right side is real. For an unstable mode, $c_i>0$, so taking the imaginary part givesConsequently the phase velocity is a weighted average:The weights are nonnegative and not identically zero. For a nonconstant shear profile, equality at an extremum would force $q$ to vanish on an interval where $U$ differs from that extremum; uniqueness for the regular ordinary differential equation would then give the zero mode. This is the phase-speed bound for unstable stratified shear modes.

Choose $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and let $Q=N^2-(U^{\prime }{)}^2/4$. The result isBecause $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part givesIf $4N^2>(U^{\prime }{)}^2$ everywhere, the integral is strictly positive for every nonzero eigenfunction, so a mode with $c_i\ne 0$ is impossible. Thus the flow is neutrally stable to these inviscid normal modes. This is the Miles–Howard theorem, expressed as a lower bound $\mathrm{Ri}>1/4$ on the gradient Richardson number. It excludes exponential modal growth; it does not by itself exclude transient growth.

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

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