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Choose the planar magnetic field convention $\mathbf{B}=\nabla \times (A{\mathbf{e}}_z)$, so $B_s=s^{-1}{A}_{\varphi }$ and $B_{\varphi }=-A_s$. The resistive induction equation, with constant magnetic diffusivity and a suitable time-dependent gauge constant removed, reduces to advection and diffusion of this Cartesian magnetic flux function:For the specified differential rotation, its mth azimuthal Fourier mode obeysThe physical field is the real part of the complex mode. Substitute $a=s^{m}g(t)\exp [-im{\Omega }_{0}t-i{s}^{2}f(t)]$. A direct radial differentiation givesOn the left, rigid rotation cancels, leaving $\dot{g}/g-i{s}^2\dot{f}+im\omega s^2$. Equating the constant and quadratic coefficients yieldsThe initial condition fixes $f(0)=0$ and $g(0)=1$, since the constant C was already factored out. Set ${\mu }^2=4i\eta m\omega$. Then the exact quadratic-shear magnetic flux solution follows by differentiation:Indeed $\dot{f}=m\omega {\mathrm{sech}}^2(\mu t)$ and $4i\eta f^2=m\omega {\tanh }^2(\mu t)$. Also $\dot{g}/g=-(m+1)\mu \tanh (\mu t)=-4i\eta (m+1)f$. For an unspecified initial normalization the second formula has a multiplicative constant $C^{\prime }$, but here $C^{\prime }=1$.

For $\eta >0$ and $\omega >0$, choose $\mu =(1+i)\sqrt{2\eta m\omega }$. The asymptotic condition is $\mathrm{Re}(\mu )t\gg 1$: then $f$ tends to the finite constant $m\omega /\mu$ and $g\sim 2^{m+1}\exp [-(m+1)\mu t]$. At any fixed s, differentiating A to obtain the magnetic field multiplies it by factors involving $m/s$ and $sf$, which have no additional exponential time dependence. Therefore its envelope decays asUsing the paper's magnetic Reynolds number definition gives the precise scalingThus the decay time scales as ${\mathrm{Rm}}^{1/2}/{\Omega }_0$ for positive ${\Omega }_0$ and fixed m. The mechanism is magnetic phase mixing under differential rotation: winding produces progressively fine radial structure on which magnetic diffusion acts efficiently. For negative $\omega$, choose the square root with positive real part, replacing $\omega$ by $|\omega |$ in the decay time. If $\omega =0$, the initial harmonic mode simply rotates, with $f=0$, $g=1$; there is no shear-enhanced decay. The printed positive-root and Reynolds-number expressions implicitly use $\omega >0$.

A field in a circulation plane has components crossing circular streamlines. Magnetic flux freezing carries those lines with fluid elements at different angular speeds: the lines become spirals, their azimuthal components are stretched, and neighboring turns acquire increasingly small separations. This is a nonaxisymmetric winding and phase-mixing problem.

For illustration, in the ideal limit take the initial horizontal field along $x$ and fix a horizontal plane. With $\chi =\varphi -\omega (s,z)t$, the induction equations giveThe first term describes advection around a circle; the term proportional to $t{\partial }_s\omega$ is the stretching by differential rotation. Spatial variation of $\omega$ also produces phase gradients proportional to $t\nabla \omega$. A rigidly rotating region merely turns the horizontal field, whereas a shearing region progressively winds it up.

Even small magnetic diffusivity becomes important once these gradients are large. Locally a phase mode of azimuthal order $m$ has damping rate of order $\eta m^2|\nabla \omega {|}^2{t}^2$, leading to attenuation of order $\exp [-\eta m^2|\nabla \omega {|}^2{t}^3/3]$. The associated magnetic phase mixing under differential rotation time isFor a horizontal uniform seed, $m=1$; if $|\nabla \omega |\sim \Omega /\ell$, this is ${\Omega }^{-1}{R}_m^{1/3}$, much shorter than the ordinary diffusion time at large $R_m$. Resistive smoothing cancels the closely spaced oppositely directed field and, in persistent circulation regions with an imposed surrounding field, drives flux expulsion from the interiors toward their boundaries. The detailed final distribution depends on the geometry and boundary forcing. Winding alone does not supply a self-sustaining poloidal–toroidal regeneration loop, and magnetic feedback can again limit the kinematic process.

The ansatz gives $B_r=A_{\varphi }/r$, $B_{\varphi }=-A_r$, and $B_z=B$. Initially $A=B_{0}r\sin \varphi$ and $B=0$, since $B_r=B_0\cos \varphi$ and $B_{\varphi }=-B_0\sin \varphi$ are a uniform Cartesian field. In ideal magnetohydrodynamics, the transverse flux function is advected, while the axial field is stretched by the axial velocity gradient. More explicitly, the transverse induction equations are equivalent to $({\partial }_t+f{\partial }_{\varphi })A=0$; the axial component of ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$ is $({\partial }_t+f{\partial }_{\varphi })B=B_r{g}^{\prime }$. Substituting the azimuthal mode therefore givesThe initial coefficients are $\hat{A}(r,0)=B_{0}r$ and $\hat{B}(r,0)=0$. Multiplication by $e^{ift}$ integrates both equations, giving $\hat{A}=B_{0}r{e}^{-ift}$ and $\hat{B}=i{B}_{0}t{g}^{\prime }{e}^{-ift}$. Thus the ideal magnetic response to a cylindrical cyclonic event at its end isWith ideal induction these fields remain frozen after the flow is switched off. The EMF calculated below is the active-flow value at $T^{-}$; at $T^{+}$ the velocity itself is zero.

In cylindrical components, $\mathbf{u}\times \mathbf{B}=(rfB+g{A}_r)\hat{\mathbf{r}}+(g{A}_{\varphi }/r)\hat{\mathbf{\varphi }}-f{A}_{\varphi }\hat{\mathbf{z}}$. Put $\psi =\varphi -fT$. Since $A_r=B_0(\sin \psi -rT{f}^{\prime }\cos \psi )$, its Cartesian $x$-component reduces toUse ${\int }_0^{2\pi }\cos \varphi \cos (\varphi -fT)d\varphi =\pi \cos (fT)$ and the area element $rdrd\varphi$. The integrated electromotive response of a finite cyclonic event isWhen the boundary term vanishes, the last two terms combine as $-g{\partial }_r[r^2\sin (fT)]$, giving the useful equivalent integralFor the compactly supported profile, $f{g}^{\prime }-g{f}^{\prime }=0$ on $0<r<a$. Substitution of $v=a^2-r^2$ into the first integral givesAt zero the continuous limiting value is zero, and $\mathcal{E}(T)=-\pi B_0{a}^{6}T/3+O(T^3)$. For $B_0>0$ it begins negative; the nonzero zeros satisfy $\tan (a^{2}T)=a^{2}T$, the first at $a^{2}T\simeq 4.4934$. It then alternates sign, with an asymptotic $1/T$ envelope. The requested sketch uses $x=a^{2}T$ and scales the EMF by $\pi B_0{a}^4$:

This sign alternation is possible even though the flow has fixed-sign kinetic helicity. Indeed, its vorticity components are $(0,-g^{\prime },2f+r{f}^{\prime })$, so $\mathbf{u}\cdot \mathbf{\omega }=2fg+r(g{f}^{\prime }-f{g}^{\prime })$, which reduces to $2f^2\ge 0$ for this profile. The finite-duration response retains winding and phase information; it cannot be assigned a universal sign from the instantaneous helicity alone.

A small magnetic diffusivity replaces the ideal mode equations byDifferential rotation creates radial phase gradients of order $T{f}^{\prime }$, so the resistive damping rate grows like $\eta T^2{f}^{\prime 2}$. Locally, magnetic phase mixing under differential rotation gives an accumulated damping factor of order $e^{-\eta f^{\prime 2}{T}^3/3}$. Thus the ideal result is a good early-time approximation but diffusion eventually limits winding and stretching, smooths the sharp field gradients at $r=a$, and modifies the later EMF oscillations. Small diffusion does not immediately remove all finite-time sign reversals, and a long-time resistive response requires the diffusive boundary-value problem rather than the ideal formula. After the event, the remaining distorted field relaxes by diffusion toward the imposed uniform background.