| \section{Quantitative purity and sharp structural stability}\label{sec:stability} |
| \begin{lemma}[A two-multiplier switching margin]\label{lem:margin} |
| Suppose $K=\sin\theta\cos\theta+\eta\cos^2\theta-\xi\sin^2\theta$ |
| has $\Delta=1+4\eta\xi>0$. Let $Z$ be its two zeros on |
| $\R/\pi\mathbb Z$, and let $d_Z$ be distance to this set. Then |
| \begin{equation}\label{eq:margin} |
| |K(\theta)|\ge\frac{2\sqrt\Delta}{\pi^2}d_Z(\theta). |
| \end{equation} |
| For the bounds-only calibration $\xi=0$, this is exactly the uniform v1.0 |
| constant $2/\pi^2$, independent of $\eta$. |
| \end{lemma} |
| \begin{proof} |
| Factor the indefinite quadratic as |
| $K=A\sin(\theta-\theta_1)\sin(\theta-\theta_2)$. |
| Let $\sigma\in(0,\pi/2]$ be the shorter distance between its zero lines. |
| At either zero, differentiation gives $|K'|=|A|\sin\sigma=\sqrt\Delta$; |
| the last equality also follows by writing |
| $K=(\eta-\xi)/2+(\eta+\xi)\cos2\theta/2+\sin2\theta/2$. |
| Let $d_1,d_2$ be the distances to the two individual zero lines. |
| The circle triangle inequality gives $d_1+d_2\ge\sigma$, so |
| $d_1d_2\ge\sigma\min(d_1,d_2)/2$. Using |
| $|\sin x|\ge2\dist(x,\pi\mathbb Z)/\pi$ on both factors, |
| \[ |
| |K|\ge\frac{2|A|\sigma}{\pi^2}\min(d_1,d_2) |
| \ge\frac{2\sqrt\Delta}{\pi^2}d_Z, |
| \] |
| because $\sigma\ge\sin\sigma$. |
| \end{proof} |
|
|
| \begin{theorem}[Phase and clock stability from the exact defect]\label{thm:phasestability} |
| For a growing fixed-resource profile with winding $n$, put |
| $d=B-1$, $E=nG_m(S/n)-L(u)$, and $\Delta_n=1+4\eta_n\xi_n$. |
| With the canonical selector $b=b_{\eta_n,\xi_n}$, |
| \begin{align} |
| Q:=\int_{\theta_0}^{\theta_0+n\pi}|\widehat u-b|\dd\theta |
| &\le2\pi B\sqrt{ndE}\,\Delta_n^{-1/4},\label{eq:phaseL1}\\ |
| \inf_{t_0}\int_0^S|u(s)-u^*_{m,n}(s-t_0)|\dd s |
| &\le(1+2nd)\,2\pi B\sqrt{ndE}\,\Delta_n^{-1/4}. |
| \label{eq:timeL1} |
| \end{align} |
| Here $u^*_{m,n}$ denotes all $n$ repeated cells. The same assertions hold |
| for a bounds-only calibrated cell with $\xi=0$, its branch defect, and |
| $\Delta_n=1$. |
| \end{theorem} |
| \begin{proof} |
| Let $e=|\widehat u-b|\le d$. Over $n$ turns the set with $d_Z\le a$ has |
| measure at most $4na$. Layer-cake integration gives |
| \[ |
| \int d_Z e\dd\theta |
| =\int_0^\infty\int_{d_Z>a}e\dd\theta\dd a |
| \ge\int_0^{Q/(4nd)}(Q-4nda)\dd a=\frac{Q^2}{8nd}. |
| \] |
| The denominators of the defect are at most $B^2$. Combining |
| \eqref{eq:resourceDefect} with Lemma~\ref{lem:margin} gives |
| $E\ge\sqrt\Delta_n Q^2/(4\pi^2B^2nd)$, proving the first estimate. |
|
|
| Translate the actual profile to place its expanding angle at the start |
| of the canonical high arc at $s=0$. Both clocks now start at that angle |
| and end at time $S$. Their difference obeys |
| \[ |
| \sup_\theta|s(\theta)-s_*(\theta)| |
| \le\int\left|\frac1{v_{\widehat u}}-\frac1{v_b}\right|\dd\theta |
| \le Q. |
| \] |
| The intermediate binary signal $b(\theta(s))$ differs from the actual |
| profile in temporal $L^1$ by at most $Q$, since $v_{\widehat u}\ge1$. |
| Its ordered switches agree in number and state ordering with the ideal |
| signal; each of the at most $2n$ switch times is displaced by at most $Q$. |
| The symmetric difference of two intervals is bounded by the sum of the |
| absolute displacements of their endpoints. Summing high-state intervals |
| bounds the remaining temporal $L^1$ error by $2ndQ$, proving the result. |
| \end{proof} |
|
|
| \begin{theorem}[Global resource-constrained square-root rigidity]\label{thm:globalstability} |
| Suppose $F=F(R,S,m)>0$. Let $F_n=nG_m(S/n)$ and let $\mathcal O_*$ be |
| the union of all translates of every winning architecture. Define |
| \begin{align} |
| \delta_b&=\min\bigl(\{F\}\cup\{F-F_n:n\in\mathcal N_m\setminus\mathcal W_m\}\bigr)>0,\nonumber\\ |
| C_0&=\max_{n\in\mathcal W_m} |
| \frac{2\pi B(1+2nd)\sqrt{nd}}{S\Delta_n^{1/4}},\qquad |
| C=\max\{C_0,d/\sqrt{\delta_b}\}.\label{eq:resourceC} |
| \end{align} |
| Every admissible measurable profile satisfies |
| \begin{equation}\label{eq:globalstability} |
| \boxed{\dist_1(u,\mathcal O_*):= |
| \inf_{v\in\mathcal O_*}\frac1S\int_0^S|u-v|\dd s |
| \le C\sqrt{F-L(u)}.} |
| \end{equation} |
| Replacing the distance by mean absolute permittivity error is valid after |
| multiplying $C$ by $\varepsilon_+$. |
| \end{theorem} |
| \begin{proof} |
| Set $\delta=F-L$. If $\delta<\delta_b$, then $L>0$ and the actual winding |
| must win; every other branch would force |
| $\delta\ge F-F_n\ge\delta_b$. Apply Theorem~\ref{thm:phasestability}. |
| If $\delta\ge\delta_b$, the normalized $L^1$ distance is at most $d$, |
| which gives the second term in $C$. Finally |
| $|\varepsilon_+/u-\varepsilon_+/v|\le\varepsilon_+|u-v|$ and |
| $\dd t/T=\dd s/S$. |
| \end{proof} |
| The estimate is global for each fixed parameter triple, not uniform as |
| means approach endpoints, as branches approach degeneracy, or as contrast |
| collapses. The explicit constants display those possible deteriorations. |
|
|
| \begin{proposition}[Sharpness while preserving the resource exactly]\label{prop:resourceSharp} |
| The exponent $1/2$ in \eqref{eq:globalstability} cannot generally be improved, |
| even when period, contrast, and mean are all fixed. |
| \end{proposition} |
| \begin{proof} |
| Take $R=2$, $m=2$, $S=3\pi/4$. The only possible growing winding is one, |
| and its unique optimizer has high time $h_0=\pi/4$, low time $\ell_0=\pi/2$, |
| and $F=\log2$. For small $a>0$, use the four states and durations |
| \[ |
| (4,h_0-a),\quad(1,a),\quad(4,a),\quad(1,\ell_0-a). |
| \] |
| Their total high time, low time, period, and mean are unchanged. Exact |
| multiplication of the four segment matrices followed by Taylor expansion gives |
| \[ |
| D(a)=-\frac54+\frac94a^2-\frac{15}{2}a^4+O(a^6),\qquad |
| F-L(u_a)=3a^2+O(a^4). |
| \] |
| The high-state set is $[0,h_0-a]\cup[h_0,h_0+a]$. For sufficiently small |
| $a$, its intersection with any circular interval of length $h_0$ has measure |
| at most $h_0-a$: an interval covering both components loses their intervening |
| gap of length $a$, whereas an interval covering only one component contains |
| at most $h_0-a$. The long complementary gap prevents another circular case. |
| This maximum is attained by $[0,h_0]$. Hence the minimum symmetric difference |
| is $2a$ and |
| \[ |
| \dist_1(u_a,\mathcal O_*)=6a/S. |
| \] |
| For $\alpha>1/2$, its ratio to $(F-L(u_a))^\alpha$ diverges as $a\downarrow0$. |
| The symbolic expansion and independent floating-point checks are supplied |
| in the verification suite. |
| \end{proof} |
|
|
