| \section{The fixed-period, fixed-resource sharp theorem}\label{sec:resource} |
| Let $R>1$, $S>0$, $1<m<R^2$, and put |
| \begin{equation}\label{eq:duty} |
| p=\frac{m-1}{R^2-1}\in(0,1). |
| \end{equation} |
| For a proposed one-turn duration $\tau$, define |
| \begin{align} |
| h=p\tau,\qquad \ell=(1-p)\tau,\qquad |
| D_m(\tau)&=\cos(Rp\tau)\cos((1-p)\tau)\nonumber\\ |
| &\quad-\frac{R+R^{-1}}2\sin(Rp\tau)\sin((1-p)\tau). |
| \label{eq:resourceD} |
| \end{align} |
| On $0<\tau<\tau_{\rm cap}:=\min\{\pi/(Rp),\pi/(1-p)\}$ set |
| \[ |
| t(\tau)=\tan(Rp\tau/2)\tan((1-p)\tau/2). |
| \] |
| This function is continuous and strictly increasing from zero to infinity. |
| Define the unique two roots |
| \begin{equation}\label{eq:resourceedges} |
| t(\tau_-)=R^{-1},\qquad t(\tau_+)=R, |
| \end{equation} |
| and, for $\tau\in(\tau_-,\tau_+)$, define |
| \begin{equation}\label{eq:resourceG} |
| G_m(\tau)=\arcosh[-D_m(\tau)]>0. |
| \end{equation} |
| These definitions require only monotone scalar root finding and elementary |
| functions. The finite winding set is |
| \begin{equation}\label{eq:resourceN} |
| \mathcal N_m(R,S)=\{n\in\N: S/\tau_+<n<S/\tau_-\}. |
| \end{equation} |
|
|
| \begin{theorem}[Global resource-constrained extremality and equality]\label{thm:resource} |
| Over every measurable $S$-periodic profile with $1\le u\le R^2$ and |
| $S^{-1}\int_0^S u=m$, |
| \begin{equation}\label{eq:resourceF} |
| \boxed{\sup_u L(u)=F(R,S,m):= |
| \max\bigl(\{0\}\cup\{nG_m(S/n):n\in\mathcal N_m(R,S)\}\bigr).} |
| \end{equation} |
| If $\mathcal N_m$ is nonempty, let $\mathcal W_m$ be the set of all maximizing |
| integers. Every equality profile, and only those profiles, is up to a temporal |
| translation and null-set changes $n$ repetitions of |
| \begin{equation}\label{eq:resourcecell} |
| u^*_{m,n}(s)= |
| \begin{cases} |
| R^2,&0\le s<pS/n,\\ |
| 1,&pS/n\le s<S/n, |
| \end{cases}\qquad n\in\mathcal W_m. |
| \end{equation} |
| If $\mathcal N_m$ is empty, every admissible profile has $L(u)=0$ and no |
| binary-purity implication is asserted. The endpoint means $m=1,R^2$ force |
| constant profiles and have zero growth. Collapsed contrast $R=1$ is the |
| same constant-oscillator case. |
| \end{theorem} |
|
|
| \begin{proof} |
| First consider any growing profile and its expanding eigenline, with winding |
| $n$. Average the exact variables $z$ over its $n$ turns, identifying each |
| turn with the projective circle. The average is in $[0,1]$ and has |
| \[ |
| \int za=\pi-S/n,\qquad \int zb=mS/n-\pi, |
| \qquad \int zc_0=L(u)/n>0. |
| \] |
| By Lemma~\ref{lem:boundary}, this moment is interior. Lemma~\ref{lem:normal} |
| therefore gives a unique binary phase selector with a high-state arc and a |
| low-state arc, each connected and shorter than one full projective turn. |
| Its time profile is one high segment of duration $h$ followed by one low |
| segment of duration $\ell$, up to translation. Time and resource impose |
| $h+\ell=S/n$ and $R^2h+\ell=mS/n$, hence precisely |
| $h=pS/n$ and $\ell=(1-p)S/n$. The associated phase gain is at least |
| $L(u)/n>0$. Its transfer matrix is $E(1,\ell)E(R,h)$, whose half trace is |
| \eqref{eq:resourceD}; its expanding multiplier is negative, because the |
| phase advances by $\pi$. Thus $D_m(S/n)<-1$, and its gain equals $G_m(S/n)$. |
|
|
| To identify the admissible duration interval explicitly, set |
| $x=\tan(Rp\tau/2)$ and $y=\tan((1-p)\tau/2)$. Substitution of the |
| half-angle identities gives |
| \begin{equation}\label{eq:factoredD} |
| D_m(\tau)+1=\frac{2(xy-R)(xy-R^{-1})}{(1+x^2)(1+y^2)}. |
| \end{equation} |
| For the binary selector above, $Rh<\pi$ and $\ell<\pi$, since each angular |
| arc is shorter than $\pi$. Equation~\eqref{eq:factoredD} now proves |
| $\tau_-<S/n<\tau_+$, so $n\in\mathcal N_m$ and |
| $L(u)\le nG_m(S/n)\le F$. |
|
|
| Conversely, for any $n\in\mathcal N_m$, the binary cell |
| \eqref{eq:resourcecell} has $D_m<-1$ and positive gain $G_m$. Each segment |
| advances any lifted angle by less than $\pi$, because its duration is less |
| than its own full projective-turn time. The expanding eigenline of the |
| negative eigenvalue has a positive odd multiple of $\pi$ as its angle |
| advance, and the total advance is less than $2\pi$. It is therefore exactly |
| $\pi$. This actual expanding trajectory has the interior moment above. |
| The linear-program maximizer at that moment is also a binary cell with |
| these same durations, so the given cell is the canonical maximizer up to |
| translation. Repetition attains $nG_m$. This proves the maximum and its |
| attainment without assuming a layered admissible class. |
|
|
| To classify equality among arbitrary profiles, integrate |
| \eqref{eq:twoPointDefect} for the dual of this one-turn moment. Both the time |
| and resource multipliers cancel, giving Theorem~\ref{thm:resourceDefect} |
| below. Equality forces the selected endpoint at every angle except its two |
| zeros. The phase-to-time clock then fixes every residence time and every |
| turn; no independently variable timing remains. This gives precisely |
| \eqref{eq:resourcecell}, with every tied winning integer retained. |
| If no candidate exists, the first part rules out every growing profile. |
| The endpoint-mean statements follow by integrating $u-1\ge0$ or $R^2-u\ge0$. |
| \end{proof} |
|
|
| \begin{theorem}[Resource defect and unique dual parameters]\label{thm:resourceDefect} |
| For any admissible growing profile with winding $n$, the one-turn moment |
| $(\pi-S/n,mS/n-\pi)$ has a unique finite pair $(\eta_n,\xi_n)$. |
| Writing $b=b_{\eta_n,\xi_n}$ and $\Delta_n=1+4\eta_n\xi_n>0$, |
| \begin{equation}\label{eq:resourceDefect} |
| \boxed{nG_m(S/n)-L(u)= |
| \int_{\theta_0}^{\theta_0+n\pi} |
| \frac{|K_{\eta_n,\xi_n}(\theta)|\,|\widehat u(\theta)-b(\theta)|} |
| {v_{\widehat u}(\theta)v_b(\theta)}\dd\theta.} |
| \end{equation} |
| The same dual is the one associated with the expanding eigenline of the |
| binary cell of durations $pS/n$ and $(1-p)S/n$. |
| \end{theorem} |
| \begin{proof} |
| Existence and uniqueness follow from Lemma~\ref{lem:normal}. The integral |
| of the maximum of $H_{\eta_n,\xi_n}$ over one turn is |
| $G_m(S/n)-\eta_n S/n-\xi_n mS/n$. The actual integral over $n$ turns is |
| $L(u)-\eta_nS-\xi_nmS$. Their difference gives |
| \eqref{eq:resourceDefect} exactly. |
| \end{proof} |
|
|
| \subsection{Constructive phase and dual recovery} |
| Let $\tau=S/n$ and $x_0$ be an expanding eigenvector of |
| $P=E(1,(1-p)\tau)E(R,p\tau)$. Set |
| $x_1=E(R,p\tau)x_0$ and choose lifted angles $\theta_0,\theta_1$ in the |
| positive high-state arc. Solve the two linear equations |
| \begin{equation}\label{eq:dualrecover} |
| \begin{pmatrix} |
| \cos^2\theta_0&-\sin^2\theta_0\\ |
| \cos^2\theta_1&-\sin^2\theta_1 |
| \end{pmatrix} |
| \binom{\eta_n}{\xi_n} |
| =-\binom{\sin\theta_0\cos\theta_0}{\sin\theta_1\cos\theta_1}. |
| \end{equation} |
| This system is nonsingular. Indeed the low state has zero radial gain, |
| whereas on the high arc |
| \[ |
| G_m(\tau)=\tfrac12\log\frac{v_B(\theta_0)}{v_B(\theta_1)}>0. |
| \] |
| Thus $\cos^2\theta_0\ne\cos^2\theta_1$, which is exactly the determinant |
| condition in \eqref{eq:dualrecover}. This supplies the dual, the switch |
| phases, and the defect weight without nonlinear multiplier fitting. |
|
|
| \subsection{Symmetries and physical interpretation} |
| Temporal translation conjugates the monodromy and preserves its spectral |
| radius. Reversal also preserves the scalar discriminant: if |
| $J_0=\diag(1,-1)$, the reversed monodromy is $J_0M^{-1}J_0$ and |
| $\tr M^{-1}=\tr M$ for determinant one. Reversal of the canonical two-block |
| period lies in its translation orbit. With fixed, unequal labeled material |
| bounds and fixed $m$, exchange of the material values is not an additional |
| symmetry. There is no unspecified gauge freedom in the fixed state convention. |
|
|
| In physical time, the optimizer uses $\varepsilon_-$ for $pT/n$ and |
| $\varepsilon_+$ for $(1-p)T/n$, repeated $n$ times. Therefore its high-$u$ |
| (low-permittivity) duty cycle is forced by the resource, not fitted from a |
| spectral observation. The optimal physical amplitude rate is $F(R,S,m)/T$. |
| The weak-$*$ argument proving the angular maximum also excludes a need for |
| chattering or singular intermediate states at positive equality. |
|
|
