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The polar form is $1+i\sqrt{3}=2e^{i\pi /3}$. By De Moivre's theorem,Therefore the real part and imaginary part are

Multiply numerator and denominator by the denominator's complex conjugate:Thus the real part and imaginary part are

The characteristic roots satisfyThey are a nonreal conjugate pair precisely when ${\mu }^{1/2}/9-16{\mu }^{7/6}<0$, or ${\mu }^{2/3}>1/144$. ThereforeIn lake-coupled Röthlisberger drainage instability, above this value the real part is ${\mu }^{1/4}/6>0$, so nonzero linear perturbations are sinusoidal oscillations with an exponentially growing envelope. Their angular frequency is $[4{\mu }^{7/6}-{\mu }^{1/2}/36{]}^{1/2}$.

Physically the feedback represents repeated lake filling and pressure increase, channel enlargement by melting, rapid drainage, pressure reduction and material creep closure. Large drainage events are glacial lake outburst floods, or jökulhlaups. The unbounded amplitude is a conclusion of the linearized equations, not a proof of infinite physical discharge or an indefinitely growing nonlinear solution. Once perturbations are comparable with the equilibrium scales, nonlinear dynamics and constraints such as lake depletion must be retained. At the threshold the two real roots coincide; below it the instability is nonoscillatory.

For the first Hopf boundary of scalar delayed negative feedback, first, if a root has nonnegative real part then $|\lambda +\beta |=\gamma e^{-\mathrm{Re}\lambda }\le \gamma$. For $\beta \ge \gamma$ this is impossible, apart from the potential equality case $\lambda =0$, already excluded. Thus $\beta <\gamma$ is necessary for instability, although not sufficient by itself.

At an imaginary root $\lambda =i\omega$ with $\omega >0$, separate real and imaginary parts:The first crossing uses ${\omega }_c\in (\pi /2,\pi )$ for $\beta >0$, and satisfiesThe period is in scaled time; its physical value is $\tau {\mathcal{T}}_c$. Higher-frequency branches have larger critical gains. For fixed $\beta$, stability at zero delayed gain and absence of any earlier imaginary or zero-root crossing identify this lowest branch as the first stability boundary. Its crossing is transverse: implicit differentiation of $\lambda +\beta +\gamma e^{-\lambda }=0$ gives at the imaginary rootFor very small $\beta$, ${\omega }_c=\pi /2+2\beta /\pi +O({\beta }^2)$, so ${\mathcal{T}}_c\to 4$ and ${\gamma }_c=\pi /2+2\beta /\pi +O({\beta }^2)$. Therefore ${\gamma }_c=\pi /2$ is exact in the limit $\beta =0$ and is the leading approximation for small positive $\beta$. At that limit $\lambda e^{\lambda }=-\gamma$ first has an imaginary pair at $\lambda =\pm i\pi /2$; subsequent pairs occur at $\pi /2+2n\pi$ and cannot be the first loss of stability. A nonlinear Hopf theorem, including its nondegeneracy conditions, is needed to turn the linear imaginary mode into a claim about a particular finite-amplitude periodic branch.

The degree-five row of the wavevector selection rule for equivariant monomials in part (a) has two vectors other than $(1,0,0,0)$. They correspond exactly toThe vector $(1,0,0,0)$ gives only regular fifth-order terms $A_1|A_j{|}^2|A_k{|}^2$. Omitting those as required, and applying the same two square-symmetry generators, the truncated normal form of a dynamical system isHere $e,f$ are real because rotation by $\pi$ acts by complex conjugation. These are equations for steady patterns, so all four complex right-hand sides must vanish; constancy of the two invariant angles alone would only establish a relative equilibrium and could allow translation drift.

Take a nonzero equal-magnitude state, $|A_j|=R>0$, and set $z=R^2$, $S=c_1+c_2+c_3+c_4$, $\rho =\sigma +Sz$. Dividing each equation by its nonzero amplitude gives, in order,Taking imaginary parts, adding and subtracting pairs, yieldsThus, unless $e=f=0$, each invariant phase is zero or $\pi$ modulo $2\pi$. Equality of the real parts further givesThis identifies the equal-amplitude states of an eight-mode square pattern completely. For generic $e\ne f$, the two nonzero phase types, with their amplitude equations, areRepresentatives are respectively $(A_1,A_2,A_3,A_4)=R(1,1,1,1)$ and $R(-1,1,-1,1)$. Spatial translations change individual phases but leave the indicated phase pair fixed. Reflection acts on the invariant angles as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_1,{\chi }_2)$, and interchange of $x,y$ as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_2,-{\chi }_1)$. Thus the two rows are not equivalent under square symmetry and translations. Every positive root of the corresponding polynomial gives an amplitude branch of that type; if there is no positive root, that type is absent at those parameter values.

There are two coefficient degeneracies to include. When $e=f\ne 0$, the mixed pairs $(0,\pi )$ and $(\pi ,0)$ are allowed and are equivalent under square symmetry. They form one additional phase type, with amplitude equation $\boxed{\sigma +Sz=0}$; $iR(1,1,1,1)$ is a representative. When $e=f=0$, all invariant phases are unrestricted, and the same cubic amplitude equation gives a continuous family modulo translations and the finite square-group action. Even if $e+f=0$ but $e\ne f$, only the original two phase types remain: the equality of their amplitude equations does not remove the phase constraints. Finally, the all-zero state is always a separate steady state and has no defined polar phases.

A Hopf crossing of the transverse stability matrix requires zero trace and strictly positive determinant. Since ${\beta }_R<0$, the trace condition gives the positive thresholdThe determinant there isThus the required coefficient condition and the rotating-frame Hopf frequency areThe trace derivative $-2{\beta }_R$ is strictly positive, so with $d,\beta$ fixed the real parts cross zero with nonzero speed. Under this condition the determinant stays positive from $r=0$ to $r_H$; the pure branch first loses linear stability through this Hopf bifurcation. When $Q<Q_{\ast }$ its first instability is instead the steady one of part (a), and $D_H<0$ prevents a Hopf pair at the trace-zero point. Equality gives the already noted double-zero degeneracy.

Generically the new branch is periodic in the symmetry-reduced rotating frame, with a small oscillating odd-mode amplitude and a modulation of the even-mode amplitude. Reconstructing the common temporal phase combines the modulation frequency with the carrier frequency. The original amplitudes therefore describe modulated periodic oscillations, generically a two-frequency quasiperiodic state. If the two frequencies are rationally related, the reconstructed motion can be periodic instead. The spatial pattern no longer has a fixed mixture of the even and odd modes: their relative amplitudes and phase oscillate. As with the steady case, nonlinear nondegeneracy is needed for an ordinary Hopf branch, and its criticality or stability cannot be inferred from the trace and determinant alone.

The spatial reflection $x\mapsto -x$ acts on the two complex amplitudes as $(A,B)\mapsto (A,-B)$, because the first spatial eigenfunction is even and the second is odd. The first equation must therefore be even in $B,\overline{B}$, and the second odd. For a nonresonant double Hopf bifurcation, the temporal phases give independent actions $A\mapsto e^{i{\phi }_1}A$, $B\mapsto e^{i{\phi }_2}B$ in the cubic normal form of a dynamical system. A cubic monomial in the first equation must have net $A$ phase weight one and net $B$ phase weight zero. Only $A|A{|}^2,A|B{|}^2$ satisfy this; the second equation similarly has $B|B{|}^2,B|A{|}^2$. Parametrizing the independent cross-saturation coefficients givesAll coefficients can be complex: their real parts determine amplitude growth and their imaginary parts frequency shifts. These are cubic truncations of a family of reduced vector fields on the four-real-dimensional extended centre manifold, with the two unfolding parameters entering ${\nu }_1,{\nu }_2$.

The nonresonance assumption must be made explicit. Merely having ${\omega }_1\ne {\omega }_2$ does not imply this form: for ${\omega }_1=2{\omega }_2$, the quadratic terms $B^2$ in $\dot{A}$ and $A\overline{B}$ in $\dot{B}$ satisfy both temporal resonance and spatial parity. Thus the initial form is the generic nonresonant double-Hopf form, rather than a statement for every possible unequal frequency pair.

At the one-to-one reflection-symmetric one-to-one Hopf resonance, only the common temporal phase action remains. Cubic terms must then have total phase weight one. Spatial reflection retains, in the first equation, the three terms $A|A{|}^2,A|B{|}^2,B^2\overline{A}$; the second retains $B|B{|}^2,B|A{|}^2,A^2\overline{B}$. Common phase covariance excludes all quadratic terms. The additional resonant terms are consequently $-{\gamma }_1{B}^2\overline{A}$ and $-{\gamma }_2{A}^2\overline{B}$, with independent complex coefficients. This is the reflection-symmetric one-to-one Hopf resonance.

In the specified equal-coefficient specialization, put $r=R^2$. A nonzero pure-$A$ periodic branch exists for $\mathrm{Re}{\nu }_1>0$ and hasIts radial eigenvalue is $-2{\alpha }_{R}r<0$, and its phase is neutral. For a transverse $B$ perturbation the linear equation isIts time dependence can be removed exactly by $B=C{e}^{i\Omega t}$. Using ${\nu }_1-\alpha r=i\Omega$ and ${\nu }_2={\nu }_1-d$ givesWriting $C=u+iv$, $d=d_R+i{d}_I$, $\beta ={\beta }_R+i{\beta }_I$ yields the constant real stability matrixLet $Q=\mathrm{Re}(\beta \overline{d})={\beta }_R{d}_R+{\beta }_I{d}_I$. The transverse characteristic polynomial isBoth transverse eigenvalues have negative real part exactly whenThese hold at $r=0$ because $d_R>0$. Thus a sufficiently small pure branch is transversely stable; it is stable modulo its neutral temporal phase until one of the two inequalities fails. The secondary steady and Hopf thresholds of a resonant pure mode are derived in the two parts below. They refer to accessible points on the pure branch; in a multi-parameter problem $d$ may also vary, so a proposed parameter path must actually cross the corresponding surface.

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.

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.

First use independent fair signs ${\epsilon }_i\in \{-1,1\}$, the Rademacher distribution convention for Bernoulli random variables in this setting. Let $\Omega =\{-1,1{\}}^d$ with uniform probability, write $s(\omega )={\sum }_i{\omega }_i{a}_i$ and $F(\omega )=\Vert s(\omega )\Vert$. We prove the sharp Rademacher second-moment inequality by a complete finite-cube calculation.

For a function on this cube define the coordinate-flip generator on a hypercubewhere ${\omega }^{(i)}$ flips just coordinate $i$. The Walsh characters ${\chi }_A(\omega )={\prod }_{i\in A}{\omega }_i$ form an orthonormal basis: distinct characters have expectation-zero product, and there are $2^d$ of them. Flipping a coordinate in $A$ changes its sign and other flips leave it unchanged, so $L{\chi }_A=-|A|{\chi }_A$. Expand the real function $F$ as ${\sum }_A\hat{F}(A){\chi }_A$. Since $F(-\omega )=F(\omega )$, changing variables $\omega \mapsto -\omega$ shows that $\hat{F}(A)=0$ whenever $|A|$ is odd. ThusThis derives the even-function spectral gap on a hypercube, rather than assuming it.

At each vertex with $s(\omega )\ne 0$, choose a norming functional ${\ell }_{\omega }$ of norm one satisfying ${\ell }_{\omega }(s(\omega ))=F(\omega )$, supplied by the Hahn-Banach theorem. For complex $E$ take real parts in the following inequalities. At a vertex with $s(\omega )=0$, use the zero functional. The supporting inequality for the norm givesBut $Ls=-s$, since every coefficient is a degree-one Walsh character. Summing the preceding inequalities with factor $1/2$ therefore proves $LF(\omega )\ge -F(\omega )$. As $F\ge 0$,Rearranging and taking square roots yieldsNo completeness of $E$ is needed. The constant is sharp: with two identical nonzero scalar coefficients, the absolute sum is zero with probability $1/2$ and twice the coefficient magnitude with probability $1/2$.

If Bernoulli instead means fair $\{0,1\}$ variables $b_i$, write $b_i=(1+{\epsilon }_i)/2$ and $a={\sum }_i{a}_i$. Add an independent fair sign ${\epsilon }_0$ and apply the proved estimate to $Y={\epsilon }_{0}a+{\sum }_i{\epsilon }_i{a}_i$. Conditional on either sign of ${\epsilon }_0$, $\Vert Y\Vert$ has the same distribution as $\Vert a+{\sum }_i{\epsilon }_i{a}_i\Vert$, by symmetry of the sign sum. Hence $\Vert Y{\Vert }_{L^p(E)}=2\Vert {\sum }_i{b}_i{a}_i{\Vert }_{L^p(E)}$ for $p=1,2$, proving the same bound in that convention.

The constant is documented in the primary sharp-constant study.

Fairness is essential. For a general $\{0,1\}$ Bernoulli random variable with success probability $0<t<1/2$, a single nonzero scalar coefficient gives $L^2/L^1=1/\sqrt{t}>\sqrt{2}$. Thus the printed term cannot be read as permitting arbitrary Bernoulli probabilities.

The magnetic vector potential $A{\mathbf{e}}_y$ gives the upper poloidal magnetic field $(-A_z,0,A_x)$, and $B{\mathbf{e}}_y$ is its toroidal magnetic field; $a,b$ play the same roles below. This representation automatically satisfies $\nabla \cdot \mathbf{B}=0$. Diffusion acts in both regions. Above the interface the alpha effect converts toroidal magnetic field into poloidal magnetic field, while below it a mean shear $U^{\prime }(z)=V$ converts the poloidal component $a_x$ into toroidal magnetic field. Separating these source regions still allows an alpha-Omega dynamo through diffusion across the interface.

For electromagnetic matching, keep the closure actually represented by these equations. Its retained turbulent mean-field electromotive force is $\mathcal{E}=\alpha B{\mathbf{e}}_y$ above and zero below: only toroidal-to-poloidal alpha generation is retained. Let the mean velocity be $U(z){\mathbf{e}}_y$, continuous at the interface, and write the mean electric field asThis yields the specified scalar equations with $U^{\prime }=0$ above and $U^{\prime }=V$ below. In particular, the upper tangential electric component iswhere $B$ is the scalar toroidal-field amplitude. The other tangential component is $E_y=-\alpha B-\eta {\nabla }^{2}A=-{\partial }_{t}A$.

Assume equal permeability, no imposed singular surface current and ordinary electromagnetic matching. Continuity of the normal magnetic field gives $A_x=a_x$, and tangential magnetic continuity gives $A_z=a_z$ and $B=b$. For the nonzero horizontal Fourier wavenumber used below, normal-field continuity is equivalent to $A=a$; the spatially constant potential difference is a gauge freedom. Continuity of $E_x$, with continuous $U$ and $A_x$, gives the final condition. Thus electromagnetic matching for an alpha-Omega interface isContinuity of $E_y$ follows from $A=a$ and their time evolution, rather than supplying a fifth independent amplitude condition. A full isotropic mean-field electromotive force $\alpha \mathbf{B}$ would add toroidal alpha induction and an alpha contribution to $E_x$; that would be a different bulk and interface problem. Retaining it only in a boundary condition would be inconsistent with the equations being solved here.

Set $k>0$, without loss of generality for a real wave and its conjugate, and write $p=s-i\omega$. Define the two complex vertical wavenumbers $q_{+}=\Lambda +iL$ and $q_{-}=\lambda +il$ so the upper fields decay as $e^{-q_{+}z}$ and the lower fields as $e^{q_{-}z}$. The original PDF has a lowercase $l$ in the lower phase. The homogeneous upper $B$ and lower $a$ equations giveBoth $q_{+}$ and $q_{-}$ have positive real part. Since their squares agree, they must be equal, rather than opposite. Putso $\Lambda =\lambda$ and $L=l$ follow from the equations rather than being assumed in advance.

Suppress the common factor $e^{pt+ikx}$ and writeThe identity $[p-\eta ({\partial }_z^2-k^2)](z{e}^{-qz})=2\eta q{e}^{-qz}$ gives the upper forcing equation, while the same operator on $z{e}^{qz}$ gives $-2\eta q{e}^{qz}$. HenceThe first and third interface dynamo matching give $a_0=A_0$, $b_0=B_0$. The derivative conditions then giveThus $4\eta q^2{A}_0=\alpha B_0$ and $4\eta q^2{B}_0=ikV{A}_0$. Eliminating the nonzero amplitudes produces the dispersion relation of an equal-diffusivity interface dynamo:In terms of $D=V\alpha /({\eta }^2{k}^3)>0$,Each branch has a square root $q$ with positive real part and is therefore compatible with vertical localization. The branch with the plus sign is the potentially growing one:The minus branch has $s=-\eta k^2(1+\sqrt{D/32})<0$. At $D=32$ the first branch is neutrally oscillatory. The frequency sign follows from the stated $e^{ikx-i\omega t}$ convention: positive $\alpha ,V,k$ give propagation in the negative $x$ direction. Reversing the coupling sign reverses propagation, while the threshold for the conjugate sign depends on the magnitude of the dynamo number.

The equal-diffusivity interface dynamo idealizes a solar configuration in which helical convection regenerates poloidal magnetic field above the tachocline and differential rotation generates strong toroidal magnetic field in or below it. Diffusion communicates the fields between the two regions, closing the feedback loop. This separation is the mechanism emphasized by Parker's interface-wave model. An oscillatory traveling mode gives migrating magnetic patterns and polarity reversals; buoyant toroidal flux can appear as sunspot belts. If $x$ is a local latitude coordinate, the direction of their migration depends on the alpha–shear sign and the coordinate orientation.

The magnetic period of this branch is $2\pi /|\omega |$; polarity reverses after half a period, and activity that depends on field strength rather than sign can repeat on that half-period. This is the qualitative relation between the magnetic solar cycle and its activity cycle. The model does not itself predict the observed period or guarantee equatorward migration: those require actual geometry, transport coefficients and shear signs. Real stars also have diffusivity contrasts, meridional transport, spatially varying regeneration and nonlinear feedback. The kinematic exponential mode describes onset, while saturation and a maintained finite-amplitude cycle require further physics.

Multiply the homogeneous Squire equation by $\bar{\eta }$ and integrate over the channel. The zero boundary values remove the integration by parts boundary term, givingFor real $U$ and $k$, the middle term is purely imaginary. Taking real parts yields the dissipation identity for a homogeneous Squire mode:The strict inequality holds for a nonzero mode with zero endpoint values and positive Reynolds number. Hence all homogeneous Squire modes are temporally stable. In the complete shear-flow perturbation system, the Squire variable can also be forced by the wall-normal velocity; this homogeneous-mode result does not rule out instability of that coupled system or its transient growth.

Take the norm to be the Euclidean norm. The solution is $\mathbf{q}(t)=e^{tA}\mathbf{q}(0)$, so optimal growth is the operator norm $G(t)=\Vert e^{tA}{\Vert }_2$, equivalently the largest singular value of the matrix exponential. Let $s(A)$ be the spectral abscissa and let $w(A)={\lambda }_{\max }((A+A^{\ast })/2)$ be the numerical abscissa. The Euclidean semigroup growth bounds areFor the lower bound, the spectral radius of $e^{tA}$ is $e^{ts(A)}$ and is bounded by its operator norm. For the upper bound, differentiating $\Vert \mathbf{q}{\Vert }_2^2$ gives $2\mathrm{Re}({\mathbf{q}}^{\ast }A\mathbf{q})\le 2w(A)\Vert \mathbf{q}{\Vert }_2^2$, and integration gives the result. A non-normal matrix can have $w(A)>s(A)$, allowing transient amplification even when all its eigenvalues have negative real parts.

For the triangular matrix, $s(M)=-1/R$. The eigenvalues of its Hermitian part are $-3/(2R)\pm \frac{1}{2}\sqrt{1+R^{-2}}$, soThe upper exponent becomes positive when $R>\sqrt{8}$, even though the long-time spectrum is stable.

The two equations are $q_1^{\prime }=-q_1/R$ and $q_2^{\prime }=q_1-2q_2/R$. Solve the first, then use an integrating factor in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ givesThe maximizing initial state is the unit eigenvector of $B^{\ast }B$ belonging to its largest eigenvalue. HereFor $t>0$, defineThe optimal initial state for triangular stable shear is thereforeMultiplication by any nonzero complex scalar gives the same growth ratio; at $t=0$, every nonzero initial state is optimal.