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The maximal asymptotic growth rate of a finite-dimensional periodic linear differential equation is $P^{-1}\log \rho (N)$, where $P$ is the coefficient period and $N$ the monodromy matrix of a periodic linear system. The spectral radius selects the dominant Floquet multiplier. Generic initial data attain this rate; data lying entirely in other invariant subspaces can grow or decay at different rates.

A Parker dynamo wave is a traveling or oscillatory mode of a mean-field dynamo coupling poloidal and toroidal magnetic-field components through an alpha effect and shear. A reduced complex-amplitude model can represent fluctuating alpha by a time-dependent coupling. Its periodically switched growth must be calculated from the monodromy matrix of a periodic linear system, rather than by averaging noncommuting generators.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

For $\dot{A}=D^{2}f(t)B$ and $\dot{B}=iA$, with $f$ alternating between $+1$ and $-1$ for intervals of length $T$, the monodromy matrix of a periodic linear system has determinant one and trace $\cosh (\sqrt{2}DT)+\cos (\sqrt{2}DT)>2$. Its reciprocal Floquet multipliers give positive dominant Floquet growth rate for $D,T>0$, despite the zero mean of $f$. An exceptional initial state in the stable eigenspace decays.