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One curve from short to long maturities

Exact Perron growth removes secular error; a moving invariant law has a sharp second-order cost.

A fixed-maturity expansion need not remain accurate at long maturities. In a general chain an omitted term of size $T/m^2$ is order one when $T$ is comparable to $m^2$; in the symmetric two-state case below, symmetry cancels that coefficient and the first secular omission is $T/m^3$. Matched formulas are uniformly accurate for every $T\geq0$, and a moving-projection formula extends the result to any finite irreducible chain.

If the fast generator itself varies, the same formula remains uniform but a moving invariant law generically lowers the stationary-observation order from three to two. Periodic forcing, despite having infinite total variation on the half-line, is also tractable after an exact Floquet renormalization. For unequal two-state switching rates, an additional second-order Floquet drift and an explicit fast initial slip appear.

A positive periodic trial profile also gives a computable a posteriori enclosure of the exact Floquet exponent at any fixed switching speed.

The two-state reduction

Let the regimes switch in either direction at rate $m$, and write their time-dependent pricing forcings as $g_1=\bar g+\delta$ and $g_2=\bar g-\delta$. If $a'=(Q+\operatorname{diag}g)a$ with $a(0)=\mathbf1$, put

Positivity of the pricing system gives $|r|<1$, and the two linear equations reduce exactly to

Suppose $\bar g(t)\to\bar g_\infty$ and $\delta(t)\to\delta_\infty$. The exact principal eigenvalue of the limiting matrix is

The matched formula

For a stationary initial regime, the uniformly matched log amplitude is

The last integral is precisely the transient Green–Kubo correction. For the unscaled generator $Q_0=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$, its quadratic form satisfies $\mathcal K_0((\delta,-\delta),(\delta,-\delta))=\delta^2/2$.

At fixed maturity, expanding $\Lambda$ recovers $\int\bar g+(2m)^{-1}\int\delta^2$. At long maturity, keeping $\Lambda$ exact prevents the omitted $T/m^3,T/m^5,\ldots$ eigenvalue terms from accumulating.

A uniform theorem

Assume $\delta$ is continuously differentiable, $G=\sup_t|\delta(t)|<m$,

Then, simultaneously for every maturity $T\geq0$,

The right-hand side is $O(m^{-2})$ and does not grow with $T$. The average forcing $\bar g$ contributes no error because its transient is retained exactly.

Proof

Set

The value $R(\delta)$ is the stable algebraic root of the Riccati equation. For $z=r-R(\delta)$,

Because $|r|,|R|<1$, the damping is at least $2(m-G)$. Also $|R(x)|\leq |x|/(2m)$ and $|R'(x)|\leq1/(2m)$. Variation of constants therefore gives

The exact correction is $\int\delta r$. Its difference from the composite is the sum of $\int\delta z$ and

Finally $|q_m'(x)|\leq |x|^3/(2m^3)$ for $|x|\leq G$. The two displayed bounds prove the result.

Known starting regimes and the initial layer

If regime 1 has initial probability $p$, put $\eta=2p-1$. The exact log amplitude is $\log s+\log(1+\eta r)$. When $D=\sup|\delta'|<\infty$, define

The second term is the fast initial layer: it makes $r_b(0)=0$. Adding $\log(1+\eta r_b(T))$ to $L_c$ gives another maturity-uniform approximation. Indeed,

Thus its additional log error is at most the right-hand side divided by $1-G/m$. In the Vasicek bond problem $g_i(0)=0$, hence $\delta(0)=0$ and the explicit initial-layer term vanishes; the state correction still approaches the correct eigenvector weight at the far end.

The constant spectral projection

For constant $\bar g$ and $\delta$, let $\omega=\sqrt{m^2+\delta^2}$. The stationary-start solution is

Its slow-mode coefficient is not one:

The corresponding constant $\log((m+\omega)/(2\omega))$ is $O(m^{-2})$. This is why the composite has a uniform $O(m^{-2})$ remainder rather than tending exactly to zero at the far end. For a known starting mixture, the coefficient is multiplied by $1+\eta R(\delta)$.

Vasicek certificate

The certificate uses $\kappa=1$, levels $(8\%,2\%)$, and variances $(0.004,0.001)$. It integrates the exact Riccati system out to $\max(20,m^2)$, checks both known starting regimes, compares the reduction with the original two-component ODE, and separately checks the constant-forcing spectral projection.

The horizons include $T=O(1)$, $T=O(m)$, and $T=O(m^2)$. The scaled error settles, while a fixed-maturity formula with the eigenvalue truncated at first order would eventually accumulate error.

Constant forcing for any irreducible finite chain

The stationary-start result extends cleanly without diagonalizing the generator when the forcing has already reached its limit. Let $Q_0$ be any irreducible finite-state generator, possibly defective, with stationary row vector $\pi$. For a real diagonal matrix $D=\operatorname{diag}g$, set

and let $\Lambda_m$ be its principal eigenvalue and $P_m$ its rank-one Riesz spectral projection. Define the exact slow-mode weight $c_m=\pi P_m\mathbf1$.

This theorem retains the exact eigenvalue but does not even need an explicit projection correction to achieve uniform second order for a stationary start. The projection explains the limiting constant: $\log(\pi e^{A_mT}\mathbf1)-\Lambda_mT\to\log c_m=O(m^{-2})$ as $T\to\infty$.

Proof without diagonalizability

Put $\varepsilon=1/m$ and $B_\varepsilon=Q_0+\varepsilon D$. The zero eigenvalue of $Q_0$ is simple and isolated.

Analytic perturbation of its Riesz projection gives a simple eigenvalue $\lambda_\varepsilon=\Lambda_m/m$ and an analytic projection $P_\varepsilon$, with $P_0=\mathbf1\pi$. The rest of the spectrum stays uniformly separated, so finite-dimensional contour calculus gives

uniformly for small $\varepsilon$. This estimate permits Jordan blocks; no eigenbasis is used.

Both endpoint factors vanish at $\varepsilon=0$:

Since $P_\varepsilon$ commutes with $B_\varepsilon$, the complementary contribution factors as

which is $O(\varepsilon^2e^{-cmT})$. Differentiating $P_\varepsilon^2=P_\varepsilon$ and multiplying by $P_0$ on both sides gives $P_0P'_0P_0=0$, hence $\pi P'_0\mathbf1=0$ and $c_m=1+O(\varepsilon^2)$. Positivity of the irreducible Feynman–Kac semigroup and $c_m\to1$ then convert the relative estimate into the logarithmic bound.

A defective three-state certificate

The certificate uses

with $g=(-0.2,0.3,0.6)$. Every off-diagonal rate is positive.

On the mean-zero space the eigenvalue $-1$ has a nontrivial Jordan block because $(uv^\top)^2=0$ but $uv^\top\ne0$. The certificate evaluates the exact matrix exponential after subtracting $\Lambda_m$, rather than diagonalizing $Q_0$.

The last measured convergence rate is $1.9934$. Independently, the residual after the first Green–Kubo eigenvalue term converges at rate $2.0137$.

Time-dependent forcing for any finite chain

The moving-projection problem also has a clean answer. Let $D(t)$ be a bounded $C^1$ diagonal matrix with $D(0)=0$, $\|D'\|_\infty<\infty$, and $\int_0^\infty\|D'(t)\|dt<\infty$. The last condition implies a limit $D_\infty$.

Set $A_m(t)=mQ_0+D(t)$ and let $\lambda_m(t)$ be its instantaneous Perron eigenvalue. For all sufficiently large $m$, choose positive right and left eigenvectors $r_m(t),\ell_m(t)$ by

The normalization fixes the gauge. Its scalar transport connection is

The connection term is essential for a nonreversible chain: integrating only the instantaneous eigenvalue omits transport of the moving slow eigenspace. The formula is gauge invariant when the matching endpoint factors are retained; $\pi r_m=1$ is simply the most convenient gauge. The factor $\log(\nu r_m(T))$ is the finite-chain version of the two-state known-start layer above.

Why the orders improve

Write $P_m=r_m\ell_m^\top$, $a=\ell_m^\top u_m$, and $z=(I-P_m)u_m$. Then

Uniform spectral separation gives a complementary decay rate $cm$. Analytic resolvent bounds give $P_m'=O(m^{-1}\|D'\|)$, so variation of constants and a standard large-$m$ bootstrap yield

Finally $\ell_m-\pi=O(m^{-1})$ and $\ell_m^\top z=0$. Hence $\pi z=(\pi-\ell_m^\top)z=O(m^{-3})$, proving the stationary rate.

A general $\nu$ does not have that last cancellation and retains $O(m^{-2})$. The assumption $D(0)=0$ makes $u_m(0)=r_m(0)=\mathbf1$ exactly; if the forcing is already regime-dependent at the near endpoint, an additional decaying initial-slip term is required.

A limiting-Perron corollary

The exact moving eigendata can be replaced by the simpler matched formula used on the yield-curve page. In addition to the hypotheses above, assume that $D(t)\to D_\infty$ and

Write $g(t)=\operatorname{diag}D(t)$, $\bar g(t)=\pi g(t)$, $\widetilde g(t)=g(t)-\bar g(t)\mathbf1$, and let $Q_0^\#$ be the group inverse. Define

and let $\Lambda_{m,\infty}$ be the exact principal eigenvalue of $mQ_0+D_\infty$. Then

This is a theorem for every fixed finite irreducible generator, including defective ones; it is not merely a fixed-maturity expansion. It keeps the exact limiting Perron root, so no omitted constant-forcing eigenvalue coefficient can accumulate with maturity.

Proof of the corollary

Uniform analytic perturbation on the bounded range of $D(t)$ gives

where $R_m$ is uniformly Lipschitz in $D$ for all sufficiently large $m$. Subtracting the same expansion at $D_\infty$ leaves an integrable remainder bounded by $Cm^{-2}\|D(t)-D_\infty\|$. In the gauge $\pi r_m=1$, one has $r_m'=O(m^{-1}\|D'\|)$ and $\ell_m-\pi=O(m^{-1})$, hence

Integrating these two estimates and applying the stationary part of the moving-projection theorem proves the first bound. Combining the same estimate for $L_m-C_m$ with its arbitrary-prior part proves the endpoint-corrected bound. The assumptions are important: without an integrable approach to $D_\infty$, a transient second-order coefficient can itself accumulate.

Limiting-Perron certificate

For the defective three-state generator above, take $D(t)=(1-e^{-t})\operatorname{diag}(-0.6,-0.2,-0.4)$. The certificate integrates the normalized exact system through $T=\max(20,m^2)$ and evaluates both transient integrals in $C_m$ analytically.

The final observed orders are $2.0198$ and $1.9472$. Checking the three pure starts controls every fixed mixture because the ratio of two positive mixtures lies between the componentwise ratios. This independently certifies the simpler composite and its endpoint factor, rather than evaluating the moving-eigenvector formula that proves them.

Nonzero near-end forcing: the sharp loss without a slip

The same transport formula remains uniformly useful when $D(0)\ne0$, but its initial slow projection must be retained. Define

These orders are generically sharp for this formula. At $T=0$ the exact logarithm is zero, whereas the two approximations equal $\log a_{m,0}=O(m^{-2})$ and $\log\{a_{m,0}\nu r_m(0)\}=O(m^{-1})$, respectively. A decaying fast-mode slip is therefore necessary to recover the $O(m^{-3})$ stationary and $O(m^{-2})$ specified-start orders through the near endpoint.

Proof of the endpoint-mismatch orders

At time zero put $z_0=\mathbf1-a_{m,0}r_m(0)$. Analytic perturbation and the two normalizations give

The cancellation in $a_{m,0}$ follows by differentiating $\pi r_m=\ell_m^\top r_m=1$ at $m^{-1}=0$. Complementary spectral decay and variation of constants then give

For stationary observation, $\pi z=(\pi-\ell_m^\top)z$ supplies one more factor $m^{-1}$; a general prior has no such cancellation. Taking logarithms, uniformly away from zero for sufficiently large $m$, proves the two bounds.

The frozen initial slip restores the higher orders

The missing boundary layer is explicit. With $A_{m,0}=A_m(0)$ and $\lambda_{m,0}=\lambda_m(0)$, define

and, for any fixed prior $\nu$, set

Because $\ell_m(0)^\top z_0=0$, the frozen semigroup contains only fast modes and $b_{m,0}(T)=O(m^{-1}e^{-cmT})$. At $T=0$ the vector inside braces reconstructs the exact initial condition:

Thus a frozen fast semigroup is enough: no time-ordered exponential is needed in the boundary layer.

Proof of the slip theorem

Scale out the transported slow exponential by writing $\alpha=e^{-L_m}a$ and $y=e^{-L_m}z$. Then $\alpha(0)=a_{m,0}$, $y(0)=z_0$, and

The frozen slip solves $b_{m,0}'=(A_{m,0}-\lambda_{m,0}I)b_{m,0}$. Since $D(T)-D(0)=O(T)$ in the $T=O(m^{-1})$ layer, variation of constants, the complementary spectral gap, and $P_m'=O(m^{-1}\|D'\|)$ give, uniformly in $T$,

For stationary observation there is one further cancellation. Both $y$ and the frozen slip lie in their respective left-eigenvector complements, and $\ell_m(T)-\ell_m(0)=O(m^{-1}\min\{T,1\})$ inside the layer. Hence

Finally $e^{-L_m}\nu u_m=\alpha\nu r_m+\nu y$. Subtracting the slip approximation gives $O(m^{-2})$ for a general prior and $O(m^{-3})$ for $\nu=\pi$; positivity for sufficiently large $m$ converts these relative estimates into the displayed logarithmic bounds.

A moving defective-chain certificate

Using the same defective $Q_0$, the certificate takes $D(t)=(1-e^{-t})\operatorname{diag}(-0.6,-0.2,-0.4)$. It computes $r_m'$ from the bordered differentiated eigenvalue equation rather than differencing eigenvectors, and checks that calculation independently by a centered difference. The exact nonautonomous system is also integrated independently.

The final measured rates are $2.9747$ and $1.9475$, respectively. The test covers $T=O(1),O(m),O(m^2)$ and the far-end limit; reconstructing the original three-component ODE from the normalized calculation agrees below $5\times10^{-12}$.

Endpoint-mismatch certificate

To test the sharp loss, the same defective generator is paired with $D(0)=\operatorname{diag}(-0.25,0.12,-0.02)$ and $D(\infty)=\operatorname{diag}(-0.6,-0.2,-0.4)$. The approximation includes $\log a_{m,0}$ but no fast-mode slip.

Over $T=O(1),O(m),O(m^2)$, the stationary and specified-start residuals have measured orders $1.9957$ and $1.0024$. At $m=64$ they are $6.2040\times10^{-6}$ and $3.2712\times10^{-3}$, with scaled errors $m^2E=0.02541$ and $mE=0.20936$.

Adding the frozen slip restores measured orders $2.9322$ and $1.9478$. At $m=64$ the corrected residuals are $1.2922\times10^{-8}$ and $6.7756\times10^{-6}$, with $m^3E=0.00339$ and $m^2E=0.02775$.

When the fast generator moves

The third-order stationary cancellation above uses more than a spectral gap: it uses the fixed invariant row $\pi$. Let $Q(t)$ now be a $C^1$ family of finite-state generators with a common state space, $Q(t)\mathbf1=0$, invariant rows $\pi(t)$, and uniform exponential ergodicity on the complementary subspace. Assume $Q$ and $D$ are bounded and have finite total variation,

and put $A_m(t)=mQ(t)+D(t)$. Take $D(0)=0$ and fix the gauge row $q=\pi(0)$. For the Perron eigenvectors use

and retain the same connection and transported exponent

Why one order disappears

Although $\ell_m'=\pi'(t)+O(m^{-1})$ is now order one, $Q'(t)\mathbf1=0$ still gives $r_m'=O(m^{-1})$. Hence the full projector derivative $P_m'$ is order one, but its only forcing of the fast component satisfies

The $O(m)$ complementary gap therefore still yields $z/a=O(m^{-2})$. The slow-amplitude feedback is now

because the total variation is finite. This term was $O(m^{-3})$ for fixed $Q$, where $\ell_m'=O(m^{-1})$.

In addition, $q-\ell_m(t)$ is generally order one, so observing with the initial stationary row supplies no extra cancellation. These two changes prove the bound and identify the mechanism behind its sharpness.

A sharp two-state certificate

The certificate takes

It integrates the exact normalized pricing system through $T=\max(20,m^2)$ and computes $r_m'$ from the bordered differentiated eigenvalue equation. A centered difference checks the connection independently. Freezing $Q$ at $Q(0)$ while retaining the same $D(t)$ isolates the effect of the moving invariant law.

The final measured orders are $1.9578$ for the moving generator and $2.9842$ for its fixed-generator control. This is a counterexample to carrying the fixed-generator stationary $O(m^{-3})$ rate over to a moving invariant law.

Periodic forcing: exact Floquet growth plus a bounded phase

Finite total variation is sufficient for the preceding transport estimates, but it is not necessary. Return to the symmetric two-state reduction and suppose that $\bar g,\delta$ are $P$-periodic, with $\delta\in C^2$.

Write $\langle f\rangle=P^{-1}\int_0^P f(t)dt$. For all sufficiently large $m$, the Riccati equation has a unique attracting periodic solution $r_m^*$, and set

The integrand defining $\Phi_{m,1}$ has zero period mean, so the phase is bounded and periodic. The number $\lambda_m$ is the exact principal Floquet exponent: it is equivalently $P^{-1}$ times the logarithm of the Perron eigenvalue of the one-period monodromy matrix.

Proof and the secular term

The interval $(-1,1)$ is invariant for the Riccati flow. If $M=\|\delta\|_\infty$ and $m>M$, its derivative with respect to the initial value contracts over one period by at most $e^{-2(m-M)P}$.

The Poincaré map therefore has one fixed point. Substitution and the same contraction give

uniformly in $t$. For $q=r-r_m^*$, exact subtraction gives

Thus the initial transient contributes only $\int_0^T\delta q\,dt=O(m^{-2})$, uniformly in $T$. On the periodic orbit, subtracting $\lambda_m$ and the derivative of $\Phi_{m,1}$ leaves a periodic, zero-mean function of size $O(m^{-2})$; its integral over any number of complete periods cancels, and the final partial period is also $O(m^{-2})$. This proves the bound.

If $\delta\in C^3$, one further substitution gives

The absent $m^{-2}$ mean follows from $\langle\delta\delta'\rangle=0$. Unless the displayed third-order coefficient vanishes, the first-order exponent accumulates an order-one error at $T\asymp m^3$; retaining the exact Floquet exponent is therefore the periodic analogue of retaining the exact far-end Perron root.

Periodic certificate

The certificate uses period $2\pi$ and

It solves the Riccati Poincaré fixed-point equation, computes the exponent independently from the two-dimensional monodromy matrix, and then starts the exact system from $r(0)=0$.

The measured uniform-error order is $2.0075$, and the Riccati and monodromy exponents agree within $8.9\times10^{-16}$. The limiting drift predicted by the expansion is $0.01473536$, versus $0.01473342$ numerically at $m=64$. This construction is the classical decomposition of a periodic linear system into exponential growth and a periodic factor, originating in Floquet's 1883 paper; here the fast-switching expansion supplies the explicit bounded phase and uniform remainder.

Unequal rates: second-order drift and an explicit slip

The symmetric chain hides a cancellation. Let

and write $v=\pi_1\pi_2$, $\sigma=\pi_2-\pi_1$ and

For $M=\pi_1a_1+\pi_2a_2$ and $r=(a_1-a_2)/M$, the exact reduction is

If $\bar g,\delta$ are smooth and $P$-periodic, the unique attracting periodic orbit satisfies

and its exact principal Floquet exponent therefore has the expansion

The $m^{-2}$ secular coefficient is generically nonzero when $\alpha\ne\beta$. It vanishes for equal rates because then $\sigma=0$; periodicity also gives $\langle\delta\delta'\rangle=0$.

Define the bounded second-order phase

and the frozen initial slip

For the proof, subtract the periodic orbit. The difference $q=r-r_m^*$ obeys exactly

Uniform contraction gives the leading boundary layer $q_{\rm bl}(t)=-\delta(0)(m\rho)^{-1}e^{-m\rho t}$ and, by variation of constants,

Because $\delta(t)-\delta(0)=O(t)$, integrating $v\delta q$ differs from $B_{m,2}$ by $O(m^{-3})$ uniformly in its upper limit. On the periodic orbit, subtracting the exact exponent and the displayed phase leaves a zero-mean periodic function of size $O(m^{-3})$; its primitive is uniformly $O(m^{-3})$. These two estimates prove the theorem.

Unequal-rate certificate

With $\alpha=1.7$, $\beta=0.4$ and the same forcing as above, the certificate compares the reduced Riccati system with an independently integrated two-dimensional monodromy matrix.

The final measured orders are $2.0090$ before the correction and $3.0070$ after it. The numerical second-order drift is $0.00283097$, against the predicted $0.00283589$, and the independent exponents agree within $9.5\times10^{-16}$. This is a concrete asymmetric-chain instance of the rapidly fluctuating Markov-chain expansions developed by Khasminskii, Yin, and Zhang (1996), with the additional Floquet resummation and initial layer needed for a maturity-uniform statement.

Periodic switching rates and the geometric drift

The Floquet construction also survives when the fast generator itself is periodic. Let $\alpha(t),\beta(t)>0$ be smooth and $P$-periodic, and define

The moving stationary average $M=pa_1+(1-p)a_2$ and ratio $r=(a_1-a_2)/M$ satisfy exactly

The $p'$ term is geometric: it comes from differentiating the moving stationary weights. Omitting it is not an innocuous choice of coordinates; its period mean against the first fast corrector can be nonzero and then changes the Floquet exponent at order $m^{-1}$.

Put

The attracting periodic orbit and exact exponent have the expansions

Define the bounded moving-average phase

and its frozen slip

The pricing observation is fixed at the initial stationary prior, not at the moving row $\pi(T)$. Put

The first two terms are the periodic observation phase. The last term is a separate fast correction caused by holding the observation row fixed while $\pi(t)$ starts to move; it is uniformly $O(m^{-2})$ because $c(0)=0$.

An arbitrary fixed initial prior

The stationary initial row is special. For a fixed prior $\nu=(q,1-q)$, set

Proof

Uniform positivity of $\alpha,\beta$ makes the Riccati Poincaré map a contraction for large $m$. Substitution gives the displayed expansion of $r_m^*$. For $q=r-r_m^*$, exact subtraction yields

Its leading initial layer is $q_{\rm bl}=-\delta(0)[m\rho(0)]^{-1}e^{-m\rho(0)t}$. Smoothness of all coefficients gives

Consequently $\int_0^T hq-B^{\rm mov}_{m,2}(T)=O(m^{-3})$ uniformly in $T$. For general $c_\nu=O(1)$, the same layer estimate first gives, uniformly in $T$,

which proves the arbitrary-prior theorem after combining it with the first-order moving-average phase. For $q=p(0)$, $c_\nu(T)=O(T)$ at the near end; expanding one order further gives the observation phase and $C^{\rm obs}_{m,2}$ with a uniform $O(m^{-3})$ remainder.

On the periodic orbit, the exact exponent removes the mean of $\bar g+hr_m^*$, while the two displayed correctors leave a zero-mean periodic remainder of size $O(m^{-3})$. Its primitive is bounded uniformly over arbitrarily many periods.

Moving-generator periodic certificate

The certificate uses the preceding $\bar g,\delta$ and

It evaluates the fixed initial observation $\pi(0)a(T)$ and independently integrates the full two-dimensional monodromy matrix. The first-order Floquet coefficient is $0.02164779$; its geometric part $\langle p'\delta/\rho\rangle=-0.00101852$ would be missed by freezing the stationary weights.

The measured stationary-prior orders are $1.9920$ before the second correction and $3.0111$ after it. For the nonstationary prior $\nu_1=0.86$, omitting the leading observation slip gives order $1.0000$; retaining it restores order $1.9995$. The numerical second-order drift is $0.00203401$, versus $0.00205120$ predicted.

The independent exponents agree within $2.4\times10^{-15}$, while the reduced and full two-component prices agree within $2.8\times10^{-12}$. Time-inhomogeneous rapidly switching chains and their occupation expansions are treated by Yin and Zhang (2007); the theorem here adds the exact Floquet resummation, explicit geometric coefficient, and maturity-uniform initial slip. The arbitrary-start layer is also the finite-dimensional boundary-layer mechanism appearing in the full asymptotic series of Khasminskii, Yin, and Zhang (1996).

Any finite periodic chain: the Floquet projection

The two-state Riccati equation supplies explicit coefficients, but it is not needed for a maturity-uniform theorem. Let $Q(t)$ and the real diagonal forcing $D(t)$ be $C^2$ and $P$-periodic, with $Q(t)$ a uniformly irreducible finite-state generator having a uniform fast spectral gap.

Write $A_m(t)=mQ(t)+D(t)$ and let $U_m(t,s)$ be its evolution family. The positive monodromy $F_m=U_m(P,0)$ has a Perron multiplier $\mu_m$, right and left vectors $v_m,w_m$, and exact Floquet exponent

Define the periodic principal profile, its initial coefficient, and the frozen fast layer by

Thus $r_m(t+P)=r_m(t)$. The profile and exponent are exact Floquet quantities; only the short initial transient is approximated.

The calendar origin is arbitrary

The preceding notation fixes phase zero only for convenience. For a start phase $s\in[0,P]$, let $F_{m,s}=U_m(s+P,s)$ and choose its Perron vectors so that

The exponent has no phase subscript. Indeed, periodicity gives the exact conjugacy

so every calendar phase has the same Floquet multipliers. Define the shifted profile, dual coefficient, complementary datum, and frozen slip by

To prove the statement, apply the finite-chain theorem to the translated coefficients $Q_s(t)=Q(s+t)$ and $D_s(t)=D(s+t)$. Their $C^2$ norms, irreducibility constants, and fast spectral gaps are uniform in $s$ by periodicity and compactness. The exponential-dichotomy and frozen-layer estimates therefore have one constant for every phase.

Conjugacy identifies their exact exponents, while the displayed phase-specific right and left eigenvectors identify the principal coefficient and complementary datum. The fixed horizon is essential for the locally frozen exponential; the exact Floquet complement, rather than a frozen local matrix, supplies the all-maturity exponential dichotomy. This is the origin-invariance implicit in Floquet's original theorem; the uniform fast-layer estimate uses the same singular-perturbation mechanism as the time-inhomogeneous chain expansion of Yin and Zhang (2007).

Shifted-phase certificate

The three-state certificate recomputes the conjugate monodromy, both Perron vectors, and the frozen slip independently at phases $s/P\in\{0,0.13,0.31,0.57,0.83\}$. The table reports the maximum over those phases and over $0\leq T\leq3P$ for the fixed prior $(0.8,0.1,0.1)$, together with the maximum uncorrected error for the phase-specific stationary prior.

The final observed orders are $0.9936$ without the arbitrary-prior slip, $2.0424$ with it, and $2.0078$ for the stationary prior. Across all five phases and speeds, the independently recomputed Floquet exponents differ by at most $3.2\times10^{-15}$. The finite phase and maturity mesh is a numerical order certificate; the preceding compactness argument, not the mesh, supplies the two continuous suprema in the theorem.

A regime-dependent payoff loses one boundary-layer order

The payoff $\mathbf1$ is special: it lies in the nullspace of every fast generator, so its complementary Floquet datum is only $O(m^{-1})$. For a fixed positive payoff vector $x$ with genuinely different regime values, define at phase $s$

Now $z_{m,s}(x)=O(1)$ in general. Under the hypotheses of the phase-uniform theorem, with $\min_i x_i>0$, and for every fixed $H<\infty$, the corresponding frozen-layer approximation satisfies

The order loss follows directly from Duhamel scaling on the fast layer. For $T=O(m^{-1})$, $m\{Q(s+T)-Q(s)\}=O(mT)$, and convolution with two $e^{-cmT}$ fast bounds is $O(m^{-1})$ when it acts on an order-one datum. If $Q$ is fixed, only $D(s+T)-D(s)=O(T)$ remains, giving $O(m^{-2})$.

The complete normalized price is bounded away from zero for large $m$, uniformly in phase and bounded maturity, so the same orders pass from the vector remainder to its logarithm. For the bond payoff $x=\mathbf1$, the additional $z_{m,s}(\mathbf1)=O(m^{-1})$ alignment recovers the stronger theorem above. This is the payoff-sensitive version of the matched boundary layers constructed by Khasminskii, Yin, and Zhang (1996).

The sharp fast-clock profile

The preceding orders have explicit leading coefficients. Put $Q_s=Q(s)$, $\pi_sQ_s=0$, $C_s=I-\mathbf1\pi_s$, and $z_s^0=C_sx$. For $\tau\geq0$ define

Let $E_{m,s,\nu}^x(T)$ denote the signed log error inside the absolute value in the payoff-sensitive theorem. If $Q,D$ are $C^2$, then, locally uniformly in $\tau$ and uniformly in the calendar phase,

Indeed, after multiplying the matched expression by $e^{\lambda_mT}$, the principal Floquet part is exact and the vector error is precisely

On the fast clock its two generators differ first by $rQ'(s)/m$, or by $rD'(s)/m^2$ when $Q$ is fixed. Duhamel's formula and $z_{m,s}(x)=z_s^0+O(m^{-1})$ give the two limits; positivity of $x$ gives the displayed logarithmic denominators.

Moreover $z_s^0=0$ if and only if $x$ is constant, because the kernel of $I-\mathbf1\pi_s$ is exactly $\operatorname{span}\{\mathbf1\}$. Thus the bond-payoff improvement is an exact payoff-space statement, while additional cancellation of either displayed scalar profile is model-, phase-, and observation-specific.

Why the locally frozen slip is not an all-maturity object

The earlier version of this note incorrectly put a supremum over every $T\geq0$ around the frozen-slip estimates. Let $\alpha_{m,s}$ and $P^{\rm loc}_{m,s}$ denote the Perron root and projection of the frozen matrix $A_m(s)$. If

then ordinary finite-dimensional spectral asymptotics give

The exact Floquet complement instead decays under the nonautonomous exponential dichotomy. Thus the locally frozen term eventually grows relative to the exact normalized price; if the displayed coefficient is negative, the proposed positive logarithmic bracket eventually crosses zero.

No constant independent of maturity can then bound its log error. The obstruction can occur even for $x=\mathbf1$: the local Perron leakage is smaller, but a nonzero exponentially amplified coefficient still wins at a logarithmically later maturity.

In the moving three-state certificate at phase zero and $m=64$, $\Delta_{m,0}=0.0551616197852$. At integer periods the frozen bracket for the bond changes sign between $42P=263.893783$ and $43P=270.176968$; for $x=(1.35,0.72,1.08)$ it changes sign between $28P=175.929189$ and $29P=182.212374$. The exact prices remain strictly positive.

This counterexample does not affect the bounded-horizon theorem, the fast-clock profiles, or the all-maturity no-slip Floquet estimates. A genuinely all-maturity matched layer must use the transported nonautonomous complementary bundle (or an explicitly stable surrogate), not the exponential of one locally frozen full matrix.

The exact transported repair uses only one-period data

The stable replacement is explicit. Fix a start phase $s$, let $\Pi_{m,s}=v_{m,s}w_{m,s}^{\top}$, and normalize the monodromy on the principal Floquet scale:

The identities $G_{m,s}\Pi_{m,s}=\Pi_{m,s}G_{m,s}=\Pi_{m,s}$ show that $C_{m,s}$ is the exact complementary monodromy. For $T=kP+\tau$, where $k\in\mathbb N_0$ and $0\leq\tau<P$, define

Indeed, periodicity factors the normalized evolution as $e^{-\lambda_mT}U_m(s+T,s)=e^{-\lambda_m\tau}U_m(s+\tau,s)G_{m,s}^k$, while $G_{m,s}^kx=c_{m,s}(x)v_{m,s}+C_{m,s}^kz_{m,s}(x)$. Positivity and irreducibility make the Perron multiplier simple and give $\rho(C_{m,s})<1$; under the uniform fast-gap hypotheses its powers obey the same uniform complementary decay as the nonautonomous dichotomy. Thus this is an exact all-maturity representation, not a new asymptotic error claim.

It requires the one-period monodromy and only the final within-period propagator, so it avoids both long-horizon integration and the unstable local-Perron leakage of the frozen full matrix. The locally frozen layer remains the simpler analytic approximation on fixed bounded horizons.

At $m=64$ the three-state certificate finds $\rho(C_{m,0})=6.15\times10^{-17}$ and checks the vector identity at six period counts through $80P$ and four non-integer phases with maximum discrepancy $1.74\times10^{-14}$. At the very maturities where the frozen brackets have become negative, the exact transported brackets are $0.999925971$ at $43P$ for the bond and $1.055504121$ at $29P$ for the regime-dependent payoff.

The certificate uses $x=(1.35,0.72,1.08)$ and $\nu=(0.8,0.1,0.1)$. It compares the moving three-state chain with the defective fixed-generator control and maximizes over two periods.

The last-doubling frozen-slip orders are $1.05296$ for moving $Q(t)$ and $2.01769$ for fixed $Q_0$; without the slip, both errors converge to nonzero constants. The calculation uses a positive payoff, so it is an order certificate for the boundary-layer mechanism rather than a test driven by sign cancellation. On $0\leq\tau\leq12$, the maximum discrepancy between the scaled numerical log error and the explicit fast-clock profile falls at rates $0.9950$ and $0.9368$ for moving and fixed generators, respectively; at $m=128$ the discrepancies are $1.02\times10^{-4}$ and $2.49\times10^{-4}$.

Explicit first-order coefficient

The exact Floquet exponent also has an explicit first correction in any number of states. Let $g(t)=\operatorname{diag}D(t)$, $\bar g(t)=\pi(t)g(t)$, and $\widetilde g(t)=g(t)-\bar g(t)\mathbf1$. With $Q(t)^\#$ the group inverse, normalize the first profile correction by $\pi(t)u_1(t)=0$ and set

Expanding the periodic Floquet equation and applying the instantaneous stationary row gives

where $\langle\cdot\rangle$ denotes the period average. The first summand is the dynamic Green–Kubo contribution. The second is a geometric contribution caused by transport of the invariant row; it vanishes for a fixed generator but need not vanish when $Q(t)$ moves.

The identity follows because the next-order solvability condition contains $-\pi u_1'$, while differentiating the gauge $\pi u_1=0$ gives $-\pi u_1'=\pi'u_1$. In two states the formula reduces exactly to the earlier $\langle(v\delta+p')R_1\rangle$. Thus this multi-state coefficient remains distinct from a fixed-parameter Green–Kubo rank calculation: moving stationary weights add a separate geometric term.

Explicit second-order coefficient

If $Q,D$ are $C^3$, the same periodic solvability calculation continues one order. Write the local first drift as

and choose the second profile correction in the gauge $\pi u_2=0$:

The expression in braces is centered under $\pi$ precisely because of the definition of $\gamma_1$. The next solvability condition gives

The apparent $-\gamma_1u_1$ contribution vanishes from the solvability condition because $\pi u_1=0$; it remains indirectly inside $u_2$. Thus the second coefficient has the same dynamic-plus-geometric form as the first, but acts on the transported second profile.

All-order Floquet recursion

The calculation is not limited to two corrections. Put $u_0=\mathbf1$ and $b_0=\bar g$. For $n\geq1$, impose the moving gauge $\pi u_n=0$ and define the local drift

Having constructed $u_0,\ldots,u_n$ and $b_0,\ldots,b_n$, set

The solvability identity $\pi R_n=0$ follows exactly from the definition of $b_n$, the gauges, and $(\pi u_n)'=0$. Consequently, for every fixed $N$, if $Q,D$ are $C^{N+1}$ and periodic and the finite-state generators have a uniform spectral gap,

This is a fixed-order statement: it neither asserts convergence of the infinite series nor a remainder uniform in $N$. Substitution of the truncated periodic profile leaves a centered residual of order $m^{-N}$; the uniform fast exponential dichotomy gains the next factor $m^{-1}$, while period averaging selects $\langle b_N\rangle$.

Induction gives the displayed expansion. The exact Floquet exponent should still be retained in a maturity-uniform price: any finite truncation eventually accumulates secular error as $T$ grows.

An a posteriori finite-speed exponent enclosure

The preceding expansion gives an asymptotic order but not a numerical error constant at a specified $m$. Positivity supplies a separate, nonasymptotic answer. Let $A(t)$ be any continuous $P$-periodic cooperative irreducible matrix, and let $\lambda$ be its principal Floquet exponent.

Choose any positive $C^1$ periodic trial profile $\widehat u(t)$ and any continuous periodic scalar phase $\widehat b(t)$. Put

For the proof, set $q(t)=\exp\{\int_0^t(\widehat b-\widehat\lambda)ds\}\widehat u(t)$, which is positive and periodic. Its residual in the constant-exponent equation is the same componentwise relative residual: $q'-(A-\widehat\lambda I)q=e^{\int(\widehat b-\widehat\lambda)}e$. The inequalities $e\geq\ell\widehat u$ and $e\leq L\widehat u$, followed by the cooperative comparison principle over one period, give

The original Collatz enclosure, applied to the positive monodromy matrix, then gives the displayed bounds. This is the finite-dimensional ODE form of the positive sub- and supersolution characterization used for cooperative periodic-parabolic systems by Antón and López-Gómez (2019); here the one-period comparison proof is self-contained.

A posteriori finite-horizon price enclosure

The same residual controls more than the asymptotic exponent. It gives a direct, computable bracket for the finite-horizon price without first knowing the exact Floquet profile. Let $x>0$ be the terminal vector, let $\nu>0$ be the observation row, and define

Indeed, put $\psi(t)=e^{B(t)}\widehat u(t)$. The residual inequalities imply $e^{-Lt}\psi(t)\leq U(t,0)\psi(0)\leq e^{-\ell t}\psi(t)$ by cooperative comparison. Positivity of $U$ then propagates $\alpha\psi(0)\leq x\leq\beta\psi(0)$ and proves the result.

Unlike the exact-profile theorem, this enclosure is not width-uniform in maturity: a residual of order $m^{-N-1}$ gives a vanishing log-price bracket for $T=o(m^{N+1})$ and a generally nondegenerate bracket at the critical scale. That growth is necessary, not a defect of the proof, because the omitted Floquet exponent is secular.

For the rational degree-16 trial below, take $x=\mathbf1$, $\nu=(0.8,0.1,0.1)$, and $T=2\pi m^4$, an integer $m^4$ periods. The trial profile and its zero-mean phase then return to their initial values. The table applies the exact rational residual and normalization bounds; the middle column is an independent scaled-monodromy calculation.

The bracket remains $O(1)$ at $T\asymp m^4$, exactly as the critical-maturity theorem predicts for a third-order exponent. On every subcritical family $T=o(m^4)$, both the residual contribution and the $O(m^{-1})$ initial normalization vanish. The certificate therefore distinguishes an all-maturity exact-Floquet approximation from a computable truncated approximation with a rigorously stated horizon.

One additional recursion step resolves that same formerly critical window. Retain the exponent through $m^{-4}$ and the trial profile through $m^{-5}$.

The exact rational relative residual is then $O(m^{-5})$, so its contribution at $T=2\pi m^4$ is $O(m^{-1})$ rather than $O(1)$; the initial normalization is also $O(m^{-1})$. The resulting proof-grade bracket is

The larger endpoint has measured order $1.00237$ on the final doubling, and the $m=64$ radius is $28.96$ times smaller than the third-order radius at the same maturity. This is a fixed-order, fixed-window statement: it does not assert convergence as the truncation order grows, nor does it make the fourth-order truncation uniform over all maturities. Its own generally critical window is $T\asymp m^5$.

That next critical window can be resolved in exactly the same fixed-order sense. Retaining the exponent through $m^{-5}$ and the trial profile through $m^{-6}$ makes the exact rational relative residual $O(m^{-6})$.

At $T=2\pi m^5$, both its accumulated contribution and the initial normalization are therefore $O(m^{-1})$. The proof-grade fifth-order bracket is

The rational residual radius has measured order $5.99534$; at $m=64$ it is $4.20531\times10^{-11}$, with $m^6$-scaled radius $2.88987$. The finite-horizon radius has measured order $0.99541$ on the final doubling. This again says nothing about convergence as the order grows: the fifth-order truncation has its own generic critical window at $T\asymp m^6$.

The construction meshes directly with the all-order recursion. For an exponent truncated through $m^{-N}$, use

The extra profile corrector is essential: it makes the residual enclosure have the same $O_N(m^{-N-1})$ order as the exponent remainder. It does not claim convergence uniformly in $N$.

The Fourier certificate need not symmetrize the error. If $\widehat u_i(t)=\sum_k u_{ik}e^{ikt}$ and $e_i(t)=\sum_k e_{ik}e^{ikt}$ are the trigonometric interpolants, set

Whenever $u_i^->0$, ordinary interval division gives $e_i(t)/\widehat u_i(t)\in[e_i^-,e_i^+]/[u_i^-,u_i^+]$ for every phase. Taking the smallest lower endpoint and largest upper endpoint over the components supplies separate computable values of $\ell$ and $L$. This preserves the residual mean and is never weaker than replacing both by the symmetric absolute envelope.

Indeed, each periodic global minimum or maximum of a smooth $r_i$ has derivative zero and lies within $h/2$ of a mesh point. Taylor's theorem expanded at that unknown extremum therefore has no linear term, leaving only $B_i h^2/8$. This improves the earlier generic nearest-node padding from $O(h)$ to $O(h^2)$.

It is fully computable from Fourier coefficients. If $E_{ik}\geq\|e_i^{(k)}\|_\infty$, $U_{ik}\geq\|\widehat u_i^{(k)}\|_\infty$, and $u_i^-\leq\inf_t\widehat u_i(t)$ is positive, quotient differentiation gives

For a trigonometric polynomial, the coefficient $\ell^1$ sums $\sum_k |k|^q|c_k|$ provide the required derivative bounds. This result makes the role of the mesh precise: node values alone are not a certificate, but node values plus the displayed Taylor remainder are.

A proof-grade a posteriori three-state certificate

For the moving nonreversible three-state example, the certificate truncates the exponent through $m^{-3}$ and the trial profile through $m^{-4}$. It retains Fourier modes $|k|\leq16$ and makes their conjugate symmetry exact. The retained binary64 real and imaginary parts define an explicit rational trigonometric trial; no claim about FFT rounding is needed.

The five exact rational Fourier modes of $mQ(t)+D(t)$ are then convolved with that trial using rational arithmetic. For every complex coefficient $z$, its modulus is bounded by $n/10^{40}$, where integer arithmetic chooses the least $n$ satisfying $n^2\geq10^{80}|z|^2$. Thus positivity, residual tails, interval division, and the final Collatz bracket are all outward enclosures.

The zero mode of the explicit phase differs from the formal third-order exponent sum by less than $6\times10^{-17}$ on the displayed range.

The table reports this rational enclosure, with the displayed endpoints rounded outward. The monodromy column is an independent floating-point check, not an ingredient of the proof.

The larger absolute endpoint has measured order $4.00255$, and $m^4$ times it is $0.523691$ at $m=64$; the certified profile lower bound there is $0.997454$. The independent monodromy error lies inside every interval.

The floating phase-mesh refinement remains a useful, slightly sharper diagnostic ($[-2.85382,2.31571]\times10^{-8}$ at $m=64$), but it is not used in the rational certificate. Hence the theorem is exact, the displayed rational bracket is proof-grade for its explicit positive trial polynomial, and the monodromy and phase-mesh comparisons are reproducibility diagnostics with clearly separate scope.

Applying the same exact construction one recursion step later, with the exponent through $m^{-4}$ and profile through $m^{-5}$, gives a residual radius of order $5.00228$. At $m=64$ the exponent-error enclosure is $[-1.06002,1.06463]\times10^{-9}$ and the explicit trial profile remains above $0.9$. This is the certificate used in the shrinking former-critical-window table above.

With the exponent through $m^{-5}$ and profile through $m^{-6}$, the corresponding proof-grade residual enclosure at $m=64$ is $[-4.20531,4.20478]\times10^{-11}$, and the independent monodromy exponent error is $2.65344\times10^{-14}$. This is the certificate used for the $T=2\pi m^5$ table.

The sharp maturity crossover for a truncated exponent

The preceding warning has an exact scale. Write $\beta_n=\langle b_n\rangle$ and, for a fixed $N$,

The proof is immediate from the exact Floquet factorization: the common amplitude cancels, whereas the exponent discrepancy is multiplied by $T_m$. This sharpens the scope distinction between a fixed-order asymptotic exponent and an exact-exponent maturity-uniform price. Whenever a later nonzero coefficient exists, increasing the expansion order moves the secular boundary; only the exact exponent supplies an unconditional all-maturity statement.

Symmetric delayed-scale certificate

The symmetric two-state periodic model above gives a non-generic but important cancellation. Truncating after the $m^{-1}$ term has $\beta_2=0$ because periodicity gives $\langle\delta\delta'\rangle=0$, while

In fact this cancellation persists at every even order. After removing the scalar factor $\exp\{\int_0^t\bar g-mt\}$, the traceless system is

Conjugation by $\sigma_z$ sends $m$ to $-m$, so the monodromy trace is even in $m$. The large-$m$ Floquet branch $\kappa_m\sim m$ is consequently odd as a formal Laurent series.

Since $\lambda_m=\langle\bar g\rangle-m+\kappa_m$, its correction contains only $m^{-1},m^{-3},m^{-5},\ldots$. This is a statement about the fixed-order asymptotic series; it does not assert convergence of that series.

The coefficient after $\beta_3$ is explicit. Substituting $r=\sum_{n\geq1}a_n/(2m)^n$ into $r'=\delta-2mr-\delta r^2$ gives

Periodic integration by parts at the fifth step yields

The first omitted nonzero term is therefore third order, so the critical maturity is $T=m^3$, not $m^2$. Since the nonsecular amplitude is retained exactly, the following values are exactly the log-price errors caused by truncating the exponent after $m^{-1}$:

At $m=64$ the relative coefficient error is $1.32\times10^{-4}$. After subtracting $\beta_3/m^3$, the measured local exponent remainder has order $4.998987$, and $m^5(\lambda_m-\lambda_m^{[3]})=0.0079458603038$, within $1.86\times10^{-4}$ relative error of the analytic $\beta_5$.

Thus truncating through $m^{-3}$ moves the next critical maturity to $T=m^5$ in this symmetric model. This parity conclusion is special to the symmetric two-state system; the moving nonreversible chain below has generic even and odd coefficients.

Critical-scale certificate

On the moving nonreversible three-state example, the recursion gives

The table reports the exact log-price error caused by exponent truncation at its generic critical maturity. Equivalently, it reports $m^{N+1}(\lambda_m-\lambda_m^{[N]})$, which must converge to $\beta_{N+1}$.

At $m=64$ the three critical-scale errors are within $1.76\%$, $1.38\%$, and $0.52\%$ of their predicted limits. This is a log-price statement. Dividing by maturity instead yields the corresponding yield error, which remains the local exponent error and does not accumulate.

Why the orders differ

Perron decomposition of the monodromy gives $\mathbf1=c_mv_m+z_{m,0}$ with $w_m^\top z_{m,0}=0$. Singular perturbation gives $v_m=\mathbf1+O(m^{-1})$, $w_m=\pi(0)+O(m^{-1})$, and $z_{m,0}=O(m^{-1})$. The normalization produces the extra stationary cancellation $\pi(0)z_{m,0}=O(m^{-2})$; an arbitrary observation row does not have it.

The complementary Floquet evolution decays at rate $cm$. On that $t=O(m^{-1})$ layer,

so Duhamel's formula replaces the exact fast evolution by $b_{m,0}$ with a uniform $O(m^{-2})$ error. When $Q$ is fixed, only $D(t)-D(0)=O(t)$ remains and the same calculation gains one further power of $m^{-1}$. This is the periodic finite-chain counterpart of the fixed-horizon matched expansions of Khasminskii, Yin, and Zhang, with the secular growth resummed by Floquet's monodromy exponent.

Three-state certificate

The moving example uses six independently varying transition rates, so its invariant row genuinely moves. The control holds the earlier defective three-state generator fixed while retaining periodic diagonal forcing. Both use an arbitrary prior $(0.8,0.1,0.1)$ in addition to the stationary observation.

The final measured orders are $1.0364$ for the moving arbitrary-prior profile without the slip, $2.0356$ after adding it, and $2.0027$ for the moving stationary-prior profile without a slip. The fixed defective-chain control has order $2.0058$ without the slip and $3.0254$ with it. The principal profile closes after one period within $2.8\times10^{-13}$.

For the moving chain, the first coefficient is $\lambda_1=0.0104925059$: its dynamic part is $0.0107582179$ and its geometric part is $-0.0002657120$. The second solvability coefficient is $\lambda_2=0.0012048119$.

The measured residual orders are $1.9732$ and $3.0197$, respectively. At $m=64$, $m^2\{\lambda_m-\langle\bar g\rangle-\lambda_1/m\}=0.0011836342$; Richardson extrapolation from $m=32,64$ gives $0.0012053936$, within $0.049\%$ of the analytic $\lambda_2$. Including the second coefficient reduces the raw exponent error by a factor of about $56$.

An independent FFT implementation of the all-order recursion reproduces both hand-differentiated coefficients within $2\times10^{-12}$ and gives

The third-corrected residual has measured order $4.0064$. Centering and gauge residuals in every recursive solve are below $2\times10^{-12}$. This verifies the recursion on a genuinely moving, nonreversible three-state generator rather than on the two-state reduction.

What remains beyond this theorem

The fixed-generator theorem assumes a diagonal forcing of bounded total variation and $D(0)=0$ only when the boundary-layer term is omitted. The frozen-slip theorem restores the higher orders for arbitrary fixed near-end forcing. The moving-generator extension requires a uniformly ergodic smooth family and integrable total variation.

The Floquet theorems above separately cover smooth periodic forcing and smooth periodic two-state generators with a uniform rate gap; the exact-profile theorem extends maturity-uniform control to every finite periodic chain and supplies an explicit fixed-order recursion. Rough nonperiodic coefficients with infinite variation, generators that lose their gap, and infinite-state factors still require different estimates. The all-order Floquet expansion is asymptotic at each fixed order; no convergence or order-uniform remainder is claimed.

The fixed-horizon expansion and its initial layers belong to the singular-perturbation framework summarized by Yin and Zhang and originate with Tikhonov's 1952 theorem; the perturbed-semigroup viewpoint goes back to Kurtz. Analytic Riesz projections and the non-diagonalizable perturbation argument follows Kato's perturbation theory; the connection term is the finite-dimensional spectral transport introduced in Kato's original adiabatic theorem.

The exact-eigenvalue resummation here addresses the additional growing-maturity scale. State-dependent fast generators and their boundary layers are treated, at fixed horizons, by Nguyen and Yin (2013); the theorem here shows what changes when the maturity horizon is also required to be uniform.