Floquet theory ($X(t)=P(t)e^{tB}$)

For a continuous periodic coefficient matrix $A(t+L)=A(t)$, a fundamental matrix $X$ for $X^{\prime }=A(t)X$, normalized by $X(0)=I$, obeys $X(t+L)=X(t)X(L)$. The invertible matrix $X(L)$ has a complex matrix logarithm; set $B=L^{-1}\log X(L)$ and $P(t)=X(t)e^{-tB}$. Since $X(L)=e^{LB}$ commutes with $B$, direct substitution gives $P(t+L)=P(t)$ and proves the factorization. A real logarithm is not required for this complex representation.

The eigenvalues of $X(L)$ are Floquet multipliers. All solutions are bounded for forward time when every multiplier has modulus at most one and those on the unit circle have no nontrivial Jordan blocks. In a real undamped second-order scalar equation with no first-derivative term, the monodromy determinant is one. Distinct conjugate unit-modulus multipliers give bounded solutions; a nontrivial Jordan block at multiplier $1$ or $-1$ gives an unbounded second solution even when one periodic or antiperiodic solution exists.

For a Floquet multiplier $m\ne 0$ of a period-$T$ linear variational equation, a Floquet exponent is any logarithmic rate $\rho$ satisfying the displayed relation. Its real part is uniquely $T^{-1}\log |m|$, while its imaginary part is defined modulo $2\pi /T$. Negative transverse real parts give local attraction to a periodic orbit; the autonomous phase direction has multiplier one and exponent zero. Constant-coefficient amplitude equations supply exponents directly as the eigenvalues of their Jacobian matrix.

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.

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]