source_sections/04_resource.tex

\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.