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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 $\omega >0$.