An approximate reset uniform over an actual spring-gain interval

Write $a=1+(\lambda-1)c$, $r=(d^2+\lambda^2c^2)^{1/2}$, and $\nu_\pm=(a\pm r)^{1/2}$. A continuously chosen real orthogonal eigenbasis $O$ makes a quarter-turn, since the off-diagonal spring is positive in the interior and the two diagonal endpoint curvatures are exchanged. In its canonical coordinates $Q,P$ set $a_j=\sqrt{\nu_j/2}\,Q_j+iP_j/\sqrt{2\nu_j}$. Direct differentiation of these operators and their adjoints gives the exact equation

If $\Omega=O^TO_u$, its off-diagonal annihilator coefficients are $-\Omega_{ij}c_{ij}$ and its creator coefficients are $-\Omega_{ij}d_{ij}$, with

There is also the local creator coefficient $\nu_{j,u}/(2\nu_j)$; the remaining blocks follow by conjugation on the real interval. In particular $G$ has zero diagonal. No sum-frequency channel has been discarded.

We supply finite analytic bounds for this equation. On the complex distance-$1/100$ neighbourhood of $[0,1]$, $|z(1-z)|\leq.2601$. The real maximum of $S'$ is below ten (its exact value is $65\binom{64}{32}/2^{64}$). Consequently $|S'(z)|<10(1.0404)^{32}<40$, and $|S(z)-S(u)|<.4$ at a nearest real $u$. Thus $|\sin\pi S|,|\cos\pi S|<\cosh(2\pi/5)<2$. The square inside $4r$ differs from one by less than $.002001\cdot4=.008004$. Its analytic square root stays within $.005$ of one, and $\nu_\pm^2$ stay within $.002$ of $5/4,3/4$ respectively. The principal frequency branches are therefore analytic. In particular every gap between distinct entries of $D_0$ exceeds $1/5$ in modulus, $.85<|\nu_j|<1.13$, and

For the latter estimate differentiate $a$ and $r$: their bounds are $.001\pi40/2$ and $.002001\pi40/.995$, respectively, then divide their sum by $2(.85)$. These estimates give row and column sums of $G$ below 200, and hence $\|G\|\leq200$. Entrywise division by the four-channel gaps, with Frobenius norm comparison, gives the safe bound

Put $\varepsilon=T^{-1}$, $N=31$, and seek $W=I+\sum_{k=1}^N\varepsilon^kW_k$ and $D=D_0+\sum_{k=1}^N\varepsilon^kD_k$, with $W_k$ off-diagonal and $D_k$ diagonal. With $W_0=I$, define exactly

Expansion shows cancellation through order $\varepsilon^N$ in $\mathcal R=\varepsilon W'-(D_0+\varepsilon G)W+WD$. The diagonal of $W_{k-1}'$ and of $W_{k-j}D_j$ vanishes, which proves the diagonal choice. In particular $D_1=0$.

Use nested complex widths $(N+1-k)/(100(N+1))$. Each Cauchy derivative costs at most 3200, including the last derivative on the real interval. If $c_k$ bounds $\|W_k\|$ on its assigned domain, then

Induction yields $c_k\leq576000^k/(k+1)^2$: the preceding single term costs at most $4\cdot68000/576000$, and the convolution at most $16\cdot4000/576000$. Their sum is less than one. For the convolution split at its midpoint and use $\sum_{m\geq1}m^{-2}<2$. Every entry of $G$ vanishes to order at least 32 at the real endpoints, because $S'$ does. The recursion loses at most one order at each step, so all $W_k,D_k$, $k\leq31$, vanish there. Thus $W(0)=W(1)=I$ exactly.

Let $q=576000/T\leq.00576<.01$. Then $\|W-I\|\leq q/(1-q)$ and $\|W^{-1}\|<2$. The last derivative and the remaining products of total degree greater than $N$ give

In the dressed equation for $B=W^{-1}A$ the error generator therefore has integrated norm less than $20000q^{31}$. The leading $D_0$ is anti-Hermitian on the real interval. Since $D_1=0$, the full integrated norm of the diagonal correction is at most $200q/(1-q)<6/5$. Including the error keeps the logarithmic growth budget below $5/4$. In Duhamel's formula the diagonal propagation on $[s,1]$ and the perturbed propagation on $[0,s]$ use disjoint subintervals of this one budget. Their product is at most $e^{5/4}<4$, not the product of two independent whole-interval overestimates. The identity endpoint dressing gives $4\cdot20000q^{31}$, proving (12.2).

The diagonal reference sends the initial reset annihilator into the final active mode, so it kills the unused vacuum. On the retained entrance state the four operator norms have squared sum

Cauchy–Schwarz therefore gives final active number norm at most $\ell\sqrt{2M^2+2}$. If $p$ is its nonvacuum probability, then $p\leq\|a_x\Psi_{\rm final}\|^2$. Normalized projection onto the active vacuum changes the wave by $[2(1-\sqrt{1-p})]^{1/2}\leq\sqrt{2p}$. This is at most $2\ell\sqrt{M^2+1}\leq s(M+1)$; the number norm obeys the same larger bound. If the vacuum projection is zero, then $p=1$ and the sharper number estimate forces $s(M+1)\geq\sqrt2$. Any normalized vacuum-product wave is then orthogonal to the true wave and has distance $\sqrt2$, so the conclusion still holds. Thus the statement includes arbitrary finite $M$. The rational power at $T=10^8$ gives the final decimal ceiling.