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The linearized perturbation obeys $\dot{\eta }=D\mathbf{f}(\mathbf{X}(t))\eta$, so its fundamental matrix satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that $\dot{\mathbf{X}}(t)=\mathbf{f}(\mathbf{X}(t))$ is a solution of the variational equation. For a nonconstant periodic orbit this tangent solution is nonzero and periodic, giving one multiplier equal to $1$.

The Liouville formula for a fundamental matrix gives $(\det \Phi {)}^{\prime }=(\mathrm{tr}D\mathbf{f})\det \Phi$. Since $\det \Phi (0)=1$, the product of the two multipliers is $\exp ({\int }_0^T\nabla \cdot \mathbf{f}(\mathbf{X}(t))dt)$. The other multiplier is therefore exactly this exponential.

Use the nearly conserved oscillator energy $E=({\dot{x}}^2+x^2)/2$. The differential equation givesFor a leading-order orbit $x=A\cos \theta$, $\dot{x}=-A\sin \theta$, the cycle averages are $⟨{\dot{x}}^4⟩=3A^4/8$, $⟨x^3\dot{x}⟩=0$, $⟨x^4{\dot{x}}^2⟩=A^6/16$, and $⟨{\dot{x}}^2⟩=A^2/2$. The energy balance method therefore givesThe two nonzero simple roots give the leading limit cycles $x\simeq \cos (t+{\theta }_0)$ and $x\simeq \sqrt{2}\cos (t+{\theta }_0)$. Below amplitude $1$ the energy increases, between $1$ and $\sqrt{2}$ it decreases, and above $\sqrt{2}$ it increases. Hence the amplitude-$1$ cycle is stable and the amplitude-$\sqrt{2}$ cycle unstable for positive small $\epsilon$.

The linearized averaged radial exponents are $-\epsilon /2$ and $\epsilon$, respectively, consistent with nontrivial Floquet multipliers $e^{-\pi \epsilon +O({\epsilon }^2)}$ and $e^{2\pi \epsilon +O({\epsilon }^2)}$. The nonzero derivatives at these averaged zeros also justify persistence of nearby cycles for sufficiently small $\epsilon$.

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.

Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The Floquet matrix is the one-period matrix $F(E)$ defined byEquivalently, it advances the Cauchy data $(\psi ,{\psi }^{\prime }{)}^T$ by one period. The Wronskian of two solutions of the stationary Schrodinger equation is constant between the delta functions, and the derivative jump at each delta function also has determinant one. ConsequentlyFor a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its Floquet multipliers ${\rho }_{\pm }$ satisfyBy Bloch theorem, an energy is in the interior of an allowed energy band when ${\rho }_{\pm }=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\mathrm{tr}F$ is real, this is equivalent to

For the attractive delta-comb Kronig-Penney model, integrating the Schrodinger equation across a lattice point givesAt negative energy write $E=-{\hslash }^2{\mu }^2/(2m)$ and use the local basis $e^{\pm \mu (x-na)}$. Free propagation through one cell followed by the derivative jump givesThusThe band condition is thereforeUsing $(\cosh u-1)/\sinh u=\tanh (u/2)$ and $(\cosh u+1)/\sinh u=\coth (u/2)$ gives the requested inequality

At positive energy write $E={\hslash }^2{k}^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix issoHence the positive-energy bands are exactly the values of $k>0$ for whichFinally, the zero-energy limit has $\frac{1}{2}\mathrm{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore zero energy lies in a forbidden gap.

Linearization about $X(t)$ givesOne Floquet multiplier of $\Phi (T)$ is $1$, corresponding to a phase shift along the orbit. Liouville's formula givesIn the plane this determinant is the other, transverse multiplier. The periodic orbit is transversely asymptotically stable when the displayed quantity is less than one, equivalently when the divergence integral is negative, and unstable when it is greater than one.

Let $\dot{z}=F(z)$ be a $C^1$ planar system on a simply connected domain $D$. The Bendixson-Dulac criterion states that if some $B\in C^1(D)$ makeshave one sign throughout $D$ and not vanish identically on any open subset, then $D$ contains no periodic orbit.

Indeed, if a periodic orbit $\Gamma$ bounded a region $\Omega \subset D$, then $F$ and hence $BF$ would be tangent to $\Gamma$. Its normal flux would therefore vanish. The divergence theorem would givecontradicting the sign hypothesis. Taking $B=1$ gives the usual Bendixson divergence criterion.

The stability version of the divergence test concerns an existing periodic orbit $\gamma$ of period $T$. Liouville's formula for the variational equation shows that its nontrivial Floquet multiplier isThus the divergence test for a planar periodic orbit says that $\gamma$ is asymptotically stable if the integral is negative and unstable if it is positive.

The nontrivial Floquet multiplier is the factor by which a transverse perturbation changes after one period. Liouville's formula makes its logarithm the integral of the divergence. Here $\nabla \cdot f=\mu (1-3a{x}^2-y^2)$, yielding the stated formula.

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.