= Solution
During the positive half-period, write $Y=(A,B)^T$ and $\dot Y=K_+Y$, with $K_+=\begin{pmatrix}0&D^2\\ i&0\end{pmatrix}$. Then $K_+^2=iD^2I=\sigma^2I$. Its two <eigenvalues> are $\pm\sigma$, and corresponding <eigenvectors> are $(\sigma,i)^T$ and $(-\sigma,i)^T$. Since $D>0$, they are independent. Thus \b[the general half-period solution] is
$$
\boxed{Y(t)=P_+\begin{pmatrix}\sigma\\i\end{pmatrix}e^{\sigma t}+Q_+\begin{pmatrix}-\sigma\\i\end{pmatrix}e^{-\sigma t}.}
$$
Using the even and odd terms in the <matrix exponential>, $e^{TK_+}=I\cosh(\sigma T)+(K_+/\sigma)\sinh(\sigma T)$. Because $D^2/\sigma=-i\sigma$, the <fundamental matrix of a linear differential equation> gives
$$
\boxed{M^+=\begin{pmatrix}\cosh\sigma T&-i\sigma\sinh\sigma T\\(i/\sigma)\sinh\sigma T&\cosh\sigma T\end{pmatrix},\qquad Y(T)=M^+Y(0).}
$$
The lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_-=\begin{pmatrix}0&-D^2\\i&0\end{pmatrix}$ and $K_-^2=(\sigma^*)^2I$. The analogous <matrix exponential> has the stated $M^-$, replacing $\sigma$ by its <complex conjugate> $\sigma^*=-i\sigma$. Both matrices satisfy
$$
\boxed{\det M^+=\det M^-=\cosh^2(\sigma T)-\sinh^2(\sigma T)=1,}
$$
with $\sigma^*$ 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 $Y(2nT)=N^nY(0)$. The determinant is therefore $\det N=1$. Let $c_\pm=\cosh(\sigma^{(*)}T)$ and $s_\pm=\sinh(\sigma^{(*)}T)$. Multiplying the two matrices and taking the <trace> gives
$$
\operatorname{tr}N=2c_-c_++\left(\frac{\sigma^*}{\sigma}+\frac\sigma{\sigma^*}\right)s_-s_+=2|\cosh(\sigma T)|^2.
$$
The ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt2DT$, the <hyperbolic-function identity> for $|\cosh(a+ia)|^2$ consequently yields
$$
\boxed{\det N=1,\qquad F\equiv\operatorname{tr}N=\cosh x+\cos x.}
$$
For $x>0$, $F>2$: differentiating gives $F'=\sinh x-\sin x>0$, 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 \b[dominant multiplier and growth rate]
$$
\boxed{\Lambda=\frac{F+\sqrt{F^2-4}}2>1,\qquad
\gamma=\frac{\log\Lambda}{2T}=\frac1{2T}\operatorname{arcosh}\left(\frac{\cosh(\sqrt2DT)+\cos(\sqrt2DT)}2\right).}
$$
The 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 $x\to\infty$, $F=\tfrac12e^x[1+O(e^{-x})]$, $\Lambda=F[1+O(F^{-2})]$, and $\log\Lambda=x-\log2+o(1)$. Therefore
$$
\boxed{\gamma=\frac{\sqrt2DT-\log2}{2T}+o(1).}
$$
In particular, \b[at $T=1$ the rate is asymptotic to $(\sqrt2D-\log2)/2$], as required. The growth calculation uses the product $M^-M^+$, rather than averaging the two generators: the two generators do not commute.
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