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It performs a trial division test for a prime number. An integer below $2$ is rejected; otherwise the function searches for a divisor from $2$ through $\lfloor \sqrt{n}\rfloor$. A composite number has a prime factor no larger than its square root, so finding no such divisor proves primality. For example, mystery2(97) returns True, while mystery2(91) returns False, since $91=7\cdot 13$.

This describes the intended mathematical test. For extremely large integers, the floating-point computation n**0.5 can round or overflow; an implementation intended for arbitrary-size integers should use an exact integer square root.

Expanding the squared Euclidean norm and applying the Cauchy-Schwarz inequality givesBoth sides are nonnegative, so taking square roots proves the triangle inequality, $\boxed{|a+b|\le |a|+|b|}$. For nonzero vectors, equality requires a positive real proportionality factor.

Put $\mu =|b{|}^2>0$ and $\lambda =a\cdot b$. Expanding the nonnegative squared Euclidean norm of $c=\mu a-\lambda b$ givesDividing by $\mu$ and taking nonnegative square roots proves the Cauchy-Schwarz inequality:Equality holds exactly when $c=0$, equivalently when the nonzero vectors $a,b$ are linearly dependent.

Since $(1+i{)}^2=2i$, the two square roots are $1+i$ and $-1-i$. The principal square root is $1+i$, whose real part isIf the notation is meant to include both square-root values, their real parts are $+1$ and $-1$. Stating the principal-branch convention removes that ambiguity.

Expand the sine of the complex number using hyperbolic functions:Its squared complex modulus isTaking nonnegative square roots proves $\boxed{|\sin z|\ge |\sin x|}$. Equality holds exactly when $y=0$.

To invert the complex cosine, let $z=\cos w$ and put $u=e^{iw}$. Since the exponential function is nonzero, $u\ne 0$, and Euler's formula gives $2z=u+u^{-1}$. Multiplying by $u$ yields the quadratic equation $u^2-2zu+1=0$. ThusChoosing either square root of $1-z^2$ merely exchanges the two signs. The two values have product one, so neither vanishes; taking a complex logarithm is legitimate.

For each choice of sign, all solutions of $e^{iw}=u$ satisfy $iw=\mathrm{Log}u+2\pi in$. Therefore the full inverse relation isConversely, exponentiating any displayed value recovers one root of the quadratic equation, and hence $(e^{iw}+e^{-iw})/2=z$, proving completeness. With a multivalued complex logarithm, the $2\pi n$ term is already included and this is the source's compact formula for inverse cosine.

The principal inverse cosine requires compatible choices of logarithm branch, square root and sign. Both signs cannot represent that same principal function indiscriminately: at $z=0$, the two principal-log values are $\pi /2$ and $-\pi /2$, whereas the principal $\arccos 0$ is $\pi /2$. Thus the printed $\pm$ is correct for the inverse relation, not a pair of interchangeable principal-value formulas.

Integrate the equation without dividing by $y$:This holds on a connected interval excluding $x=-1$. The real solutions are differentiable branches of this implicit curve. Where the radicand is positive, $y=\pm \sqrt{2x-2\log |x+1|+C}$, with a locally fixed sign.

A zero of $y$ can belong to a classical solution only at $x=0$, since the original equation then requires $x/(x+1)=0$. For $C=0$ near zero, $2x-2\log (1+x)=x^2+O(x^3)$; the smooth continuations have slopes $+1$ or $-1$ and switch between square-root signs across zero. Keeping the principal nonnegative square root across zero instead gives a cusp. Thus the implicit expression requires a smooth branch choice, not an unrestricted global choice of one square-root sign.

Rationalize the difference of square roots before taking the limit. For positive $x$, divide the resulting numerator and denominator by $x^2$:Each factor now has a finite nonzero denominator limit, giving

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

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 ${\epsilon }_n\downarrow 0$.

Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

A useful strong form of the Hensel lemma is the following. Let $F$ be a complete discretely valued field, with valuation ring $\mathcal{O}$, normalized valuation $v$, and $P\in \mathcal{O}[T]$. If $a_0\in \mathcal{O}$ satisfiesthen there is a unique root $a\in \mathcal{O}$ in the ball $v(a-a_0)>v(P^{\prime }(a_0))$. When $P(a_0)=0$ the assertion is immediate; otherwise the condition includes $P^{\prime }(a_0)\ne 0$.

Here is a proof by Newton iteration over a valued field. Write $c=v(P^{\prime }(a_0))$ and $t_0=v(P(a_0))>2c$, and setSuppose $v(P^{\prime }(a_j))=c$ and $t_j=v(P(a_j))>2c$. The correction has valuation $t_j-c>c\ge 0$, so $a_{j+1}\in \mathcal{O}$. Taylor expansion over $\mathcal{O}$ cancels the linear term and givesThus the derivative still has valuation $c$, and $t_{j+1}-2c\ge 2(t_j-2c)$. Unless an exact root is reached earlier, the corrections tend to zero with exponentially increasing valuations. Completeness gives $a_j\to a$, and continuity gives $P(a)=0$ and $v(a-a_0)=t_0-c>c$. For uniqueness, if $a,b$ are in this ball, polynomial expansion around $a_0$ givesThe second factor has valuation exactly $c$ and is nonzero, so two roots in the ball must be equal. In particular, a simple residue root lifts uniquely: $v(P(a_0))\ge 1$ and $v(P^{\prime }(a_0))=0$ satisfy the hypothesis.

For the square-class group of a p-adic field, write every element as $p^{r}u$ with $u\in {\mathbb{Z}}_p^{\times }$. Multiplication by a square changes $r$ by an even integer, so valuation parity gives one independent factor $C_2$.

For odd $p$, a unit $u$ is a square if and only if its residue is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency applies the simple-root Hensel lemma to $T^2-u$, whose derivative is a unit at a nonzero residue root. The cyclic group ${\mathbb{F}}_p^{\times }$ has square subgroup of index two. If $u_0$ is a unit with nonsquare residue, thenFor $p=2$, every odd square is one modulo eight. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $P(T)=T^2-u$ and $a_0=1$. Then $v_2(P(1))\ge 3>2v_2(P^{\prime }(1))=2$, so the strong Hensel lemma gives a square root. Thus unit square classes are exactly their residues $1,3,5,7$ modulo eight. The classes $-1$ and $5$ generate this group of order four independently, and valuation parity supplies the class $2$. Hence

For conjugate exponents $p,q\in [1,\infty ]$, $1/p+1/q=1$, Hölder's inequality states that $\int |fg|\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their essential suprema.

Here is an explicit finite-exponent proof of the remaining Generalized Holder inequality. Put $q_i=p_i/r$, so ${\sum }_{i}1/q_i=1$, and normalize $h_i=|f_i{|}^r/\Vert f_i{\Vert }_{p_i}^r$. If one norm is zero the product vanishes almost everywhere. Otherwise $\int h_i^{q_i}=1$. The weighted arithmetic-geometric mean inequality, applied to $h_i^{q_i}$ with weights $1/q_i$, givesIntegrating gives $\int {\prod }_i{h}_i\le 1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root provesIn particular the product belongs to $L^r$.

For the projection estimate, use the three-dimensional functional Loomis-Whitney inequalityThis permitted three-dimensional result also follows directly by Cauchy-Schwarz inequality: integrate the square root of $g_1{g}_2$ in $x_3$, obtaining the square root of the product of their marginal integrals, and then apply Cauchy-Schwarz inequality in $(x_1,x_2)$ against $\sqrt{g_3}$.

Let $f_j$ be any nonnegative integrable function of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First Hölder's inequality with three equal exponents yieldsThe fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further Hölder's inequality with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore givesThe marginal-integral identifications use Tonelli's theorem. Now take $f_j=1_{K_j}$. The projections are compact and hence measurable with finite measure. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $1_K(x)\le {\prod }_j{1}_{K_j}({\hat{x}}_j{)}^{1/3}$. ConsequentlyThis includes zero-measure projections and is sharp for axis-parallel rectangular boxes.

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.