Two calibrated probe positions give local recovery and two global branches
A fixed smooth pulse design with q_1=c nonzero and q_2=0 has isolated, locally recoverable preparations with identical complete final apparatus records. Both exact initial energies E,H_0 are retained. The two receiver states approach opposite shell points as coupling decreases and differ in both canonical coordinates. Thus the second calibration removes the R28 curve locally but still permits global canonical ambiguity.
R29, 2026-09-12. Use R28’s apparatus and fixed positive preparation box, known T, exact model and the two energies. Choose the fixed pulses below before decreasing coupling. The exact full-record compensator eta_lambda(w) is centred at eta_c, with q_1=c and all other probe coordinates and momenta zero. It is defined on a receiver neighbourhood containing both selected shell points by the uniform R25 construction, for lambda small relative to the fixed box width. Source coverage is in B60.
1. Three regular constraints after compensating the apparatus
Fix a receiver shell point z_bar, to be chosen in section 2, and set Y_lambda=G_lambda(z_bar,eta_c). R28’s smooth divided residuals U_lambda, V_lambda impose q_1=c and H_app=H_0 on eta_lambda(w). Add
\[W_\lambda(w)=q_2(\eta_\lambda(w))/\lambda.\]
The exact pointer drift identity used in R28 gives the smooth zero-coupling limit W_0(w)=-(K/M_2)A_2(w-z_bar), where
\[A_1(w)=\int_0^T t f_1(s_0+v_0t)x_w(t)\,dt,\qquad B_1(w)=\int_0^T f_1(s_0+v_0t)\dot x_w(t)\,dt,\] \[A_2(w)=\int_0^T t f_2(s_0+v_0t)x_w(t)\,dt.\]
The other limits are U_0=-(K/M_1)A_1(w-z_bar) and V_0=-Kc B_1(w-z_bar). Smooth divisibility is joint in lambda and w: each vanishing numerator N equals lambda times the integral of its coupling derivative at theta lambda for 0<=theta<=1. The exact residual system is
\[\mathcal R_\lambda(w)= (H_s(w)-E,U_\lambda(w),V_\lambda(w),W_\lambda(w))=0. \tag{1}\]
All reaction terms are retained in this system.
2. Three pulse rows with a canonical one-dimensional kernel
Consider initially narrow-pulse evaluation rows x(tau), xdot(tau), x(2tau), with tau>0 small and 2tau<T. The first two conditions say that the state at tau is (0,0,Y,Q). Choose Q>0 fixed. Since
\[x(2\tau)=\frac g{2\mu}Y\tau^2+ \frac g{6\mu\nu}Q\tau^3+O(Y\tau^4+Q\tau^5),\]
the coefficient of Y is nonzero for small tau. There is a unique Y with x(2tau)=0, and Y=-Q tau/(3nu)+O(tau cubed). Evolving this state backward from tau gives the common-kernel vector v with
\[v_x=-\frac{gQ}{3\mu\nu}\tau^3+O(\tau^5),\qquad v_P=\frac{5gQ}{6\nu}\tau^2+O(\tau^4). \tag{2}\]
Both are nonzero. The three evaluation rows have rank three: the third restricts the two-dimensional kernel of the first two by its nonzero Y coefficient. Fix such a tau. Positive smooth pulse widths sufficiently small around tau and 2tau preserve rank and a nearby kernel vector with v_x v_P nonzero for the actual rows A_1,B_1,A_2, after dividing each by its positive integral and time factor as appropriate.
Two more disjoint pulses can complete the four ordinary position-signal rows to an invertible R06 signal matrix. The two already chosen position rows are independent, and analytic observability spans all four dimensions on any remaining open interval. Fix all widths once. Force ceilings and cutoff margins are then met by reducing the physical coupling bound.
Let J be the positive receiver energy matrix and set
\[z_{\rm bar}=\sqrt{\frac{2E}{v^T Jv}}v.\]
The three linear rows vanish on z_bar. Their kernel is its span. Therefore at both +z_bar and -z_bar the four rows dH_s,A_1,B_1,A_2 are independent: dH_s evaluated on the corresponding shell vector is 2E.
3. Continue both roots of the exact constrained record system
At lambda=0, (1) has roots +z_bar and -z_bar. Its derivative at each is invertible. Smooth parameter continuation gives two exact roots
\[w_+(\lambda)=z_{\rm bar},\qquad w_-(\lambda)=-z_{\rm bar}+O(\lambda). \tag{3}\]
The first is constant because eta_lambda(z_bar)=eta_c exactly. For the second, precondition the residual derivative by its inverse at (0,-z_bar). On a small fixed receiver ball and coupling interval it differs from identity by at most one half. The residual at the centre is O(lambda), so shrinking the coupling interval gives a self-mapping contraction on this ball. This supplies a uniform branch, not merely formal first-order solutions.
R25’s apparatus estimate is uniform between the two bounded receiver neighbourhoods: ||eta_lambda(w)-eta_c||<=C lambda||w-z_bar||. Hence both branches retain b/2 preparation margins for sufficiently small coupling. Each has q_1=c, q_2=0, H_s=E, H_app=H_0 and exact complete record Y_lambda. Endpoint pulse absence, clock positivity and linear cutoffs persist.
4. Local recovery and global risk have different quantifiers
At each positive fixed small coupling, form the augmented map from all fourteen initial coordinates to the ten final records and four preparation quantities (H_s,q_1,H_app,q_2). Its apparatus-to-record block is invertible by R25. Eliminating that block leaves the derivative of (1), with its last three rows multiplied by lambda. It is invertible at both branches by section 3, since lambda>0.
In fixed units, continuity gives a convex neighbourhood about each branch on which the augmented derivative differs from its value at the centre by less than half its inverse margin. Segment integration gives a positive lower Lipschitz bound there. Restricting to the known preparation values then gives local recovery from the ten records, including local stability. These neighbourhoods and constants may depend on the chosen positive coupling. No uniform limit of their unscaled inverse margins is asserted.
Nevertheless the two neighbourhoods contain distinct preparations with one common record. Every deterministic receiver estimator on the global admitted class therefore satisfies
\[\epsilon_x\ge\tfrac12|(w_+-w_-)_x|, \qquad\epsilon_P\ge\tfrac12|(w_+-w_-)_P|,\] \[\epsilon_x\epsilon_P\ge \tfrac14|(w_+-w_-)_x(w_+-w_-)_P| \longrightarrow |(z_{\rm bar})_x(z_{\rm bar})_P|>0. \tag{4}\]
The limit concerns this lower bound; it has action units. Its constants depend on E,c,b and fixed pulse geometry. This pair yields neither a positive area nor an exact global minimax value. It is not generated by the broken sign symmetry of the full calibrated apparatus: the second compensating apparatus is obtained from the exact inverse record map.
Next R30: calibrate q_3=0 as well. Its divided residual adds A_3=integral t f_3 x. Select it nonzero on the present one-dimensional kernel and test uniform global recovery, rather than only disappearance of these two roots. The resulting four leading linear constraints may make exact receiver energy redundant; establish that on the stated full preparation domain.