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The Floquet growth rate measures the exponential trend of a periodic linear differential equation over many coefficient periods. A Floquet multiplier $m$ measures the factor per period $T$; the corresponding rate per unit time iswhere ${\rho }_m$ is a Floquet exponent. Taking the real part removes the ambiguity in the complex logarithm.

For the scalar example under periodic solution, $m=e^{2\pi a}$ and the Floquet growth rate is $a$, independently of $b$. Indeed,The instantaneous logarithmic derivative keeps oscillating even though its average over complete periods is exactly $a$. Sampling at the same phase in each period removes the modulation altogether: $y(nT)=m^{n}y(0)$.

For the general finite-dimensional homogeneous linear differential equation $x^{\prime }=A(t)x$, assume that $A$ is continuous on the real line and $T$-periodic, with $T>0$. Normalize its fundamental matrix by $\Phi (0)=I$ and set $M=\Phi (T)$. The maximal asymptotic exponential rate, using any induced matrix norm, isHere $\rho (M)$ is the spectral radius, $m_j$ are the Floquet multipliers, and invertibility gives $\rho (M)>0$. This concerns growth of the solution operator; it does not say that every initial condition attains the maximum.^[ref]

To prove the formula, write $t=nT+s$ with $0\le s<T$. Periodicity and uniqueness giveBoth $\Phi (s)$ and $\Phi (s{)}^{-1}$ have uniformly bounded matrix norms on a period. Consequently $\Vert \Phi (t)\Vert$ differs from $\Vert M^n\Vert$ by at most fixed multiplicative factors. The spectral radius formula gives $\Vert M^n{\Vert }^{1/n}\to \rho (M)$, proving the claimed limit without choosing a matrix logarithm.

For a particular nonzero initial condition $v$, decompose $v$ into the generalized eigenspaces of $M$ over the complex numbers. The rate of $\Phi (t)v$ is the largest $T^{-1}\log |m_j|$ among components that are nonzero. Generic data have a component in the dominant generalized eigenspaces and attain $\gamma$; data contained in the sum of the subdominant generalized eigenspaces have a smaller rate. For example, with $A=\mathrm{diag}(0.2,-0.3)$, the maximal rate is $0.2$, while $v=(0,1)$ decays at rate $-0.3$. These invariant subspaces describe initial data at phase zero. Their transported versions $E(t)=\Phi (t)E(0)$ can vary with time and satisfy $E(t+T)=E(t)$.

A negative Floquet growth rate allows temporary amplification. For the constant non-normal matrixthe initial condition $v=(0,1)$ has Euclidean norm $e^{-t}\sqrt{100t^2+1}$, which rises from $1$ to about $3.697$ at $t=1$ before eventually decaying. Its exponential rate is still $-1$: the Jordan block contributes a polynomial factor, whose logarithm divided by $t$ tends to zero. At a unit-modulus Floquet multiplier, such polynomial growth can even be unbounded with zero exponential rate; $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ gives $e^{tA}v=(t,1)$. Thus zero Floquet growth rate alone proves neither boundedness nor a periodic solution.

A Floquet multiplier is an eigenvalue of a one-period linear evolution map. For a homogeneous linear differential equation with continuous coefficients of period $T>0$, this map is the monodromy matrix of a periodic linear system $N$. If $Nv=mv$ for a nonzero eigenvector $v$, the associated solution of a differential equation with $Y(0)=v$ satisfies $Y(t+T)=mY(t)$. Its exponential rate is $T^{-1}\log |m|$, the corresponding Floquet growth rate. A Floquet multiplier equal to one gives a $T$-periodic solution in that eigenvector direction; a nontrivial Jordan block at a unit-modulus Floquet multiplier can still produce polynomial growth.

For a nonlinear periodic orbit, the Floquet multipliers are eigenvalues of the one-period map of its variational equation. A nonconstant periodic orbit of a smooth autonomous differential equation has a Floquet multiplier equal to one in its tangent direction: differentiating the solution of a differential equation with respect to its starting phase gives a periodic solution of the variational equation. The derivative of a transverse Poincaré map removes this phase direction and retains the transverse Floquet multipliers. Transverse Floquet multipliers strictly inside the unit circle give local attraction, while one outside gives instability; a transverse Floquet multiplier on the unit circle requires further analysis. For a planar periodic orbit, the transverse Floquet multiplier is given by the divergence test for a planar periodic orbit.

For the homogeneous linear differential equation $\dot{Y}=K(t)Y$ with continuous coefficient matrix of period $P>0$, the monodromy matrix of a periodic linear system is the one-period map from $Y(0)$ to $Y(P)$. If $X$ is any fundamental matrix,Its eigenvalues are the Floquet multipliers. Multipliers are nonzero, since a fundamental matrix is invertible. Changing the starting phase conjugates $N$, so the Floquet multipliers and Jordan blocks are independent of that choice.^[ref]

This turns a continuous-time problem into repeated matrix multiplication:If $Nv=mv$, the associated solution of a differential equation obeys $Y(t+P)=mY(t)$ at every time, not merely at the sampled times. In particular, $Nv=v$ gives a nonzero $P$-periodic solution, and every nonzero $P$-periodic solution arises this way. If $Nv=-v$, the state changes sign after one period and repeats after two. A unit-modulus Floquet multiplier different from $1$ need not give repetition after one period, and a generalized eigenvector in a nontrivial Jordan block need not even give bounded motion.

For the scalar example under periodic solution, $N$ is the one-by-one matrix $(e^{2\pi a})$. Numerically solving a matrix problem uses the same idea: integrate one initial value problem for each column of $X(0)=I$ through one period, then compute the eigenvalues of $X(P)$. This is useful when an explicit solution of a differential equation is unavailable: one period determines the Floquet growth rate over arbitrarily many periods.

For piecewise constant coefficients, which are also covered by the one-period construction, $N$ is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. For example, take period $P=2$ and alternateSince $K_1^2=K_2^2=0$,Replacing the coefficient by its mean would instead give $e^{K_1+K_2}=\left(\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right)$ and the incorrect rate $0.5$. The matrices do not commute. The scalar mean-coefficient formula therefore cannot simply be transferred to matrix systems. Nevertheless the Liouville formula for a fundamental matrix gives the correct determinant even in the noncommuting case:In the example, this is $1=m_{+}{m}_{-}$; it fixes the product of the Floquet multipliers, rather than either multiplier separately.

In Sur les équations différentielles linéaires à coefficients périodiques (1883), Gaston Floquet calls a function satisfying $f(t+P)=mf(t)$ periodic of the second kind, and distinguishes the genuinely periodic function obtained when $m=1$. This historical terminology captures precisely the difference between repetition of the coefficient and repetition of the state.^[ref]

For $r>0$, the polar coordinates satisfyA positive constant radius must therefore be one. The nonconstant periodic solutions arewith least period $2\pi$ and counterclockwise orientation. The origin is also a stationary solution, hence periodic in the degenerate sense of a constant function; it is a point rather than a nontrivial closed orbit. There are no other constant-radius solutions.

For $r_0>0$, put $v=r^2$. Then $\dot{v}=2v(1-v)$, a logistic differential equation. Solving this separable differential equation givesThe denominator is positive for every forward time. When $0<r_0<1$, the radius increases monotonically to one; when $r_0>1$, it decreases monotonically to one; and when $r_0=1$ it stays there. Thus every nonzero orbit approaches the unit circle while continuing to rotate. More precisely, its distance from the phase-matched circular solution tends to zero; its Cartesian coordinates do not converge to a single point. For $r_0=0$, the solution stays at the origin.

The origin is the only equilibrium point. Its Jacobian matrix iswhose eigenvalues are $1\pm i$. It is an unstable focus. The unit circle is an attracting limit cycle, rather than a curve of equilibria: on it the vector field remains nonzero. Linearizing the radial equation at $r=1$ gives $\dot{\delta r}=-2\delta r+O(\delta r^2)$, confirming radial attraction. All other radii are strictly monotone, so there are no additional nontrivial periodic orbits.

The second-order equation is equivalent to the planar system after setting $y=\dot{x}-ax+b{x}^3$. Any nonconstant periodic solution has a maximum $M$ and minimum $m$. At an extremum $\ddot{x}=x^3-x$, so $M>1$ and $m<-1$ are impossible. Equality at either endpoint would give one of the equilibrium points, and uniqueness would force a constant solution. Therefore a nonconstant periodic orbit lies in $-1<x<1$.

If $b/a<1$, throughout this striphas the sign of $a$ whenever $x\ne 0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus there are no nonconstant periodic solutions. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.

A periodic solution is a solution of a differential equation which is a periodic function: for some $T>0$, $x(t+T)=x(t)$ at every time. This demands repetition of the state, rather than merely repetition of the coefficients in the differential equation. A nonconstant periodic solution of an autonomous differential equation traces a periodic orbit; constant equilibrium points are also periodic functions, so they must be excluded when asserting absence of periodic orbits.

For an explicit example, consider the initial value problemIts coefficient has period $T=2\pi$. Taking a logarithmic derivative and then applying integration givesDirect differentiation verifies both the differential equation and the initial condition. After one coefficient period,Thus the repeating coefficient generally produces a repeating shape with a changed amplitude. With $b=1$, the three cases $a=-0.2,0,+0.2$ have one-period factors approximately $0.2846,1,3.5136$, respectively. Only $a=0$ gives a periodic solution with this initial condition. For $a<0$ the positive solution of a differential equation tends to zero, and for $a>0$ it tends to infinity; neither can have any positive period. The zero solution of a differential equation, obtained with a different initial condition, is periodic for every $a$.

The exponential function separates two roles:For $b=1$, even the decaying case $a=-0.2$ initially rises, because $y^{\prime }(0)=0.8>0$. The growing case $a=0.2$ has intervals of decline, because $a+\cos t$ becomes negative. This is the distinction between the instantaneous logarithmic derivative $y^{\prime }/y$ and the Floquet growth rate measured over many periods.

More generally, for a real continuous periodic function $q$ of period $T$, the scalar linear differential equation $y^{\prime }=q(t)y$ hasA nonzero solution of a differential equation is $T$-periodic exactly when $m=1$, or equivalently when the coefficient has zero integral over a period. This scalar calculation is the simplest example of a Floquet multiplier.^[ref]