Version 3 · earlier version; the current one is v6 · External review, round 2: major errors

# Formation of the anchored made field: place-dependent and uniform parts

- **Subproject:** 08-ilang-propagation
- **Package:** 09-anchor-formation
- **Version:** v3
- **Mode:** external regeneration
- **Date:** 2026-10-26

## Changes from previous version

- Step 1, Result 1 (I4): only the connected equal-time correlations ($w$ among them) are those of $\Omega$; $\Gamma$ and $\Sigma$ are static response coefficients of the contract and are no longer called quantities of the evolving state.
- Step 5, new (9.17) (I2): bounds on the free field uniform in $\lambda$, along the axis (folds of condition (d), question note (b)) and in every direction (rank-one points, low-label Airy front); finite propagation by a contour shift.
- Step 6, new (9.18) (I2, I3): the limit $\lambda\to\infty$ in order (O1) is replaced by the $\lambda$-mean; the persistent oscillation of the localized mode is kept; the anchor kernel is integrable only for $\kappa_a\ne\kappa_h/\gamma_3$, and the threshold is excluded from the image statements; an image bound uniform in $\lambda$ is added.
- Step 7, (9.11) (I2): the positive formation statement now rests on (9.17), (9.18), uniformly in every later $\lambda$; the negative one on the $\lambda$-mean (an exact identity). The far-anchor statement is made in the order distances to $h\to\infty$ at fixed $\rho$, then $\rho\to\infty$.
- Step 7, shell, (9.12), Result 6 (I1, I2): restricted to radial $\Phi$ (question note (a)); the reference does not vanish for $a>2$; uniformity in $\lambda$ added; the marginal member $\theta_a=\varepsilon/(1+\varepsilon)$ is left undecided.
- Step 8, (9.13) (I2): uniformity from the exact gradient on the line; the marginal $\varepsilon=\frac12$ is left undecided.
- Step 10, (9.16), Result 10 (I4, I5): the three statements are stated as properties of the description (question note (c)); the carry-over of the weighing to the evolving state is removed; the grip conditions of the weighing and $\kappa_a\ne\kappa_h/\gamma_3$ are included.
- Open issues, Code: updated; `check_formation.py` also prints the end of the window, the $\lambda$-mean of $\Delta_3^\perp$ and the range of $\Delta_1$ across the image front; every output under `code/v3/output/` comes from the v3 run.

## Response to verification

- I1: Accepted and fixed in Step 7 (*Shell*), (9.12) and Result 6: the result is restricted to radial $\Phi$, as the question's note (a) allows. The mean-value property then applies, and for $a>2$ the weighted reference is $\propto1-\theta_a$ with $\theta_a<2/a<1$, so it does not vanish.
- I2: Accepted and fixed in Steps 5–8, with the new bounds (9.17), (9.18). The degenerate points are treated: along the axis the stationary points are folds only (the point $(\frac\pi2,\frac\pi2,\frac\pi2)$ is not stationary for $\mathbf e_1$); in other directions, which enter only through the image, the rank-one points set the dispersive exponent $7/6$. The low-label front and the localized mode are treated as well. The negative statements now rest on exact $\lambda$-means. The code results are stated as finite-window checks.
- I3: Accepted and fixed in Step 6, (9.18): the pointwise limit is replaced by the $\lambda$-mean, and the oscillation $\psi_b(x)\psi_b(x_s)\cos(\omega_b\lambda)/\nu_b$ is kept. The threshold $\kappa_a=\kappa_h/\gamma_3$, where the kernel is not integrable, is excluded from the image statements.
- I4: Accepted and fixed in Step 1 and Step 10: (9.16) now states that the three properties are available in one order, the mark and the weighing for lowest eigenvectors and the formation for the evolution (note (c)). It no longer claims a weighing in the evolving state.
- I5: Accepted and fixed in Step 10 and Result 10: square grips with a common $\hat\beta$, the controlled regime, order $\mu^2$, leading order in the grip, and (5.10) / (5.26) for the lowest-eigenvector form.

## Setup and assumptions

**Field, anchor, source, start** (assumptions, M8, M9, A11). The setting of question.md: oscillator cells at the places of a ring ($D=1$) or a cubic torus ($D=3$) of side $L$, contract $C_M$ plus the anchor $\frac{\kappa_a}2q_h^2$ plus the source term $g_sq_{x_s}$ of a held object, listed at every $\lambda\ge0$ (A9 restricted to $\lambda\ge0$); at $\lambda=0$ the state is the lowest eigenvector $\Omega$ of the contract without the source term. As $M\to\infty$ the contract becomes $H_a+g_sq_{x_s}$ with $H_a=\frac12\sum_xp_x^2+\frac12\sum_{x,y}q_x(K_a)_{xy}q_y$, $K_a=\kappa_h\Delta+\kappa_aP_h$. $K_a>0$, because the kernel of $\Delta$ is spanned by the uniform vector $\mathbf 1$ and $\langle\mathbf 1,P_h\mathbf 1\rangle=1$. So $\Omega$ is nondegenerate, and the contract with the source has a lowest eigenvector.

**Inputs and notation.** I use (4.10)–(4.14) and (4.19) of 04-formation@v4, and (5.8), (5.9), (5.11), (5.15), (5.18) and the weighing of 05-anchor@v3, as quoted in question.md, with $g,r,\gamma_\nu,R,G_a,c_\phi,\gamma_3,\Gamma,w,\Sigma$ defined there. The labels are $\mathbf k\in\frac{2\pi}L\mathbb Z_L^D$, with $\Omega_{\mathbf k}^2=\hat K_0(\mathbf k)=2\kappa_h\sum_i(1-\cos k_i)$ and $c_0=\sqrt{\kappa_h}$. $\tilde\gamma_\nu:=\gamma_\nu-1/(L^D\nu^2)$ is the part of $\gamma_\nu$ from the labels $\mathbf k\ne0$ ($\tilde\gamma_\nu\to g$ as $\nu\to0$), and $\gamma,\tilde\gamma,R$ are continued to complex $\nu=s$ off the spectrum. Normal modes: $K_a\psi_n=\nu_n\psi_n$, $\nu_n=\omega_n^2$, $\nu_0<\nu_1\le\dots$. Distances: $\rho=\lvert x-x_s\rvert$, $d_s=\lvert x_s-h\rvert$, $d_x=\lvert x-h\rvert$ are Euclidean separations of lattice places, as in 04-formation and 05-anchor. Their relation to the distances of 03-ilang-space is the coupling map used there, and it is not needed here. Ratios: $w=c_0\lambda/\rho$, $a=d_s/\rho$, $b=d_x/\rho$. $\theta$ is the unit step. Units: with $[\lambda]=E^{-1}$ ($E$ a unit of cost), $[q]=E^{-1/2}$, $[\kappa_h]=[\kappa_a]=E^2$, $[g_s]=E^{3/2}$, $[\delta a]=[g]=[c_\phi]=E^{-2}$, and $[c_0]=E$ (places per unit $\lambda$).

**Made field** (question.md): $\delta a(x,\lambda)=\partial_{g_s}\langle q_x\rangle_\lambda\vert_{g_s=0}$; $\delta a_{\rm st}$ is the same in the lowest eigenvector with the source, and $u:=\delta a-\delta a_{\rm st}$.

**Criteria.**
- *Along an axis:* (4.11) applied to the place-dependent part $\delta a^\perp$ of Step 3.
  - $D=3$: $\Delta_3^\perp(\rho,\lambda):=\delta a^\perp(x_s+\rho\mathbf e_1,\lambda)/\delta a^\perp_{\rm st}(x_s+\rho\mathbf e_1)-1$. The reference value is $\delta a^\perp_{\rm st}=g(x-h)-g(x-x_s)$, the static part with zero place average. With $L\to\infty$ first it is the place difference $\delta a_{\rm st}(x)-\delta a_{\rm st}(y)$ with $\lvert y\rvert\to\infty$.
  - $D=1$: $\Delta_1$ of (4.11), with $\nabla$ taken along the ring from the source toward the anchor. Place differences remove every uniform part.
- *On a shell:* (4.19) with $\delta a^\perp$ in place of $\delta a$ and $\varphi_\ell(x)=\Phi((x-x_s)/\ell)$, for weights $\Phi$ that depend only on $\lvert y\rvert$ (question note (a)).
- *Uniform part* $\bar U$ of Step 3 (criterion stated before use): it has reached its static value to accuracy $\varepsilon\in(0,1)$ from $\lambda_f$ on if

$$
\lvert\bar U(\lambda)-\bar U_{\rm st}\rvert\le\varepsilon\,\lvert\bar U_{\rm st}\rvert\qquad\text{for every }\lambda\ge\lambda_f .
\tag{9.1}
$$

**Excluded members** (cancellations).
- $x_s=h$: by (5.8), $\delta a_{\rm st}\equiv-1/\kappa_a$, so there is no place-dependent part.
- $D=3$: places with $g(x-h)=g(x-x_s)$ (in the scaling limit $b=1$), where the reference value vanishes.
- $D=1$: places beyond the source as seen from the anchor, or beyond the anchor, where the $L\to\infty$ limit of the static gradient vanishes ((5.11) on the line).
- Places within $O(1)$ of $x_s$ or $h$, where the place-dependent part is not dominated by its long-range part. The limits $\rho,d_s,d_x\to\infty$ remove them.
- Shell: $a=d_s/\ell\le2$, where the shell reaches $h$. For $a>2$ the weighted reference does not vanish (Step 7).
- $D=3$, $\kappa_a=\kappa_h/\gamma_3$ (threshold of the localized mode), for the statements on the image before its $\lambda$-mean (Step 6).

**Orders.** $g_s\to0$ (the means are exactly linear in $g_s$, Step 1) and $M\to\infty$ come first. Then either
- (O1): $L\to\infty$ before every limit in $\lambda$; up to exponentially small terms this is the same as $\lambda$ below the wrap time with $\rho,d_s,d_x\ll L$ (Step 5); or
- (O2): $\lambda$ grows with $L$.

## Derivation

### Step 1. Means at $M\to\infty$

Fix $L$, $g_s$ and a bounded interval of $\lambda$.
- On every vector with finitely many levels, the compressed contract with the source acts as $H_a+g_sq_{x_s}$ once $M$ exceeds those levels by two.
- $H_a+g_sq_{x_s}$ is essentially self-adjoint on these vectors. So the evolutions converge strongly, uniformly in $\lambda$ on bounded intervals (Trotter–Kato), and the lowest eigenvectors converge, since the lowest eigenvalue is isolated and simple.
- The level moments of the states stay bounded, so $\langle q_x\rangle_\lambda$ converges.

This is how $M\to\infty$ first is used: at fixed $L,g_s,\lambda$. All later limits act on the limit.

In the limit, (1.9) gives the Heisenberg equations $\mathrm dq_x/\mathrm d\lambda=p_x$ and $\mathrm dp_x/\mathrm d\lambda=-(K_aq)_x-g_s\delta_{xx_s}$. They are linear. Hence:
- The means $\bar q_x=\langle q_x\rangle_\lambda$ obey $\ddot{\bar q}=-K_a\bar q-g_se_{x_s}$ with $\bar q(0)=\dot{\bar q}(0)=0$; the initial data vanish because $\Omega$ is even under $(q,p)\to(-q,-p)$.
- $q-\bar q$ obeys the source-free equations, under which $\Omega$ is stationary. So the state at $\lambda$ is $\Omega$ displaced by $(\bar q,\bar p)$, and every connected equal-time correlation, for example $\langle q_x^2\rangle_\lambda-\langle q_x\rangle_\lambda^2=w(x)$, is that of $\Omega$ at every $\lambda$. $\Gamma$ and $\Sigma$ of 05-anchor are static response coefficients of the contract (changes of lowest eigenvectors under added terms), not correlations of the evolving state.
- $\bar q$ is exactly linear in $g_s$.

With the normal modes,

$$
\delta a(x,\lambda)=-\Bigl[\frac{1-\cos(\sqrt{K_a}\,\lambda)}{K_a}\Bigr]_{xx_s},\qquad u(x,\lambda)=\Bigl[\frac{\cos(\sqrt{K_a}\,\lambda)}{K_a}\Bigr]_{xx_s},\qquad \delta a_{\rm st}(x)=-G_a(x,x_s).
\tag{9.2}
$$

The last identity holds because the lowest eigenvector with the source is $\Omega$ displaced to $\bar q=-g_sK_a^{-1}e_{x_s}$, so $g_s\delta a_{\rm st}=\phi_a$ of (5.11). $u$ is the anchored analogue of (4.10). Its Laplace transform, for ${\rm Re}\,s>0$, is

$$
\widehat{\delta a}(x,s)=\int_0^\infty e^{-s\lambda}\delta a(x,\lambda)\,\mathrm d\lambda=-\frac1s\Bigl[\frac1{s^2+K_a}\Bigr]_{xx_s}=-\frac{R_{xx_s}(s)}{s},
\tag{9.3}
$$

with $R$ of (5.9) continued from $\nu>0$ to $\nu=s$.

### Step 2. The image at the anchor

The anchor acts on the means as the force $-\kappa_a\bar q_h$ on the cell $h$. Define the *image strength* $\sigma_h(\lambda):=-\kappa_a\delta a(h,\lambda)$. Let $S_0(y,\lambda)=[\sin(\sqrt{\kappa_h\Delta}\,\lambda)/\sqrt{\kappa_h\Delta}]_{y0}$ be the free response and $\delta a_0(y,\lambda)=-\int_0^\lambda S_0(y,\lambda')\,\mathrm d\lambda'$ the free made field (no anchor). On the torus the label $\mathbf k=0$ contributes $\lambda/L^D$ to $S_0$ and $-\lambda^2/2L^D$ to $\delta a_0$. Duhamel's formula for the free equation with the two forces $-g_se_{x_s}$ and $g_s\sigma_he_h$ gives

$$
\delta a(x,\lambda)=\delta a_0(x-x_s,\lambda)+\int_0^\lambda S_0(x-h,\lambda-\lambda')\,\sigma_h(\lambda')\,\mathrm d\lambda',\qquad
\sigma_h(\lambda)=-\kappa_a\,\delta a_0(h-x_s,\lambda)-\kappa_a\!\int_0^\lambda\! S_0(0,\lambda-\lambda')\,\sigma_h(\lambda')\,\mathrm d\lambda' .
\tag{9.4}
$$

This is a Volterra equation for $\sigma_h$, with $\hat\sigma_h(s)=\kappa_a\gamma_s(x_s-h)/\{s[1+\kappa_a\gamma_s(0)]\}$. Since $\hat S_0(y,s)=\gamma_s(y)$, the Laplace transform of (9.4) is (9.3) with (5.9). So the made field is the free field of the source plus the free field of an image at $h$, whose strength $\sigma_h(\lambda)$ the anchor sets. By (5.8) the static image strength is $\kappa_aG_a(h,x_s)=1$. The code check (i) confirms (9.4) against the exact (9.2).

### Step 3. Uniform part and place-dependent part

Sum the equation of the means over all places; since $\sum_x(\Delta v)_x=0$, $\frac{\mathrm d^2}{\mathrm d\lambda^2}\sum_x\bar q_x=-g_s[1-\sigma_h(\lambda)]$: the place sum is driven by the net force of source and image. The uniform part is the place average, i.e. the component along the zero mode $\mathbf 1$ of $\kappa_h\Delta$; the place-dependent part is the rest:

$$
\bar U(\lambda):=\frac1{L^D}\sum_x\delta a(x,\lambda)=-\frac1{L^D}\int_0^\lambda(\lambda-\lambda')\,\bigl[1-\sigma_h(\lambda')\bigr]\,\mathrm d\lambda',\qquad \delta a^\perp:=\delta a-\bar U .
\tag{9.5}
$$

Since $\sum_xg(x)=0$, the place average of (5.11) is its uniform shift. So the static values are exactly the two parts of (5.11):

$$
\bar U_{\rm st}=-\Bigl[\frac1{\kappa_a}+\frac{r(x_s,h)}2\Bigr],\qquad \delta a^\perp_{\rm st}(x)=g(x-h)-g(x-x_s)=-\tfrac12\bigl[r(x,h)-r(x,x_s)\bigr].
\tag{9.6}
$$

Place differences of $\delta a$ equal those of $\delta a^\perp$. To transform (9.5), use $\hat\sigma_h$ of Step 2 and $\gamma_s=1/(L^Ds^2)+\tilde\gamma_s$; the label $\mathbf k=0$ cancels in the numerator:

$$
\hat{\bar U}(s)=-\frac1s\;\frac{1+\kappa_a\bigl[\tilde\gamma_s(0)-\tilde\gamma_s(x_s-h)\bigr]}{\kappa_a+L^Ds^2\bigl[1+\kappa_a\tilde\gamma_s(0)\bigr]} .
\tag{9.7}
$$

The lowest normal mode of $K_a$ is not used for the separation. It is not uniform: it carries an image at $h$ ((9.14)). On the ring at large $L$ it is not separated from the other modes either (Step 9).

### Step 4. Finite $L$: mean value and recurrence

(9.2) is a finite sum of terms $\cos(\omega_n\lambda)$ with $\omega_n>0$. Hence, for every $L$, on both graphs and at every place,

$$
\lim_{T\to\infty}\frac1T\int_0^T\delta a(x,\lambda)\,\mathrm d\lambda=\delta a_{\rm st}(x),\qquad
\forall\,\lambda_f,\eta>0\ \ \exists\,\lambda>\lambda_f:\ \max_x\lvert\delta a(x,\lambda)\rvert<\eta ,
\tag{9.8}
$$

and the same holds for $\bar U$ and $\delta a^\perp$. The second statement follows from Dirichlet's simultaneous approximation theorem: there are arbitrarily large $\lambda$ at which every $\omega_n\lambda$ lies within any given distance of $2\pi\mathbb Z$, and there (9.2) vanishes up to that distance. At such $\lambda$:
- $\Delta_3^\perp=\Delta_1=-1$;
- the left side of (4.19) equals the sum on its right side without the factor $\varepsilon$;
- $\lvert\bar U-\bar U_{\rm st}\rvert=\lvert\bar U_{\rm st}\rvert$.

So at fixed $L$ neither part meets (4.11), (4.19) or (9.1) for any $\varepsilon<1$, and the static values are reached only as $\lambda$-means. Formation in the sense of (4.11) and (4.19) therefore needs order (O1).

### Step 5. Order (O1): finite propagation, speeds and the free field

*Finite propagation.* $(1-\cos\sqrt X\lambda)/X=\sum_{n\ge1}(-1)^{n+1}\lambda^{2n}X^{n-1}/(2n)!$, and $(K_a^{n-1})_{xy}=0$ when $n-1$ is below the graph distance $\lvert x-y\rvert_1$.
- So the torus field at fixed places and the field of the same equations on $\mathbb Z^D$ (anchor at $h$) differ only by terms that wind around the torus. These are bounded by $\sum_{n>\ell}\lambda^{2n}\lVert K_a\rVert^{n-1}/(2n)!$, with $\ell$ of order $L$.
- They stay exponentially small up to the wrap time $\simeq(L-\rho)/c_0$. The reason is that $\sin(\lambda\Omega)/\Omega$ and $(1-\cos\lambda\Omega)/\Omega^2$ are entire functions of $\Omega^2_{\mathbf k}$, hence of $\mathbf k$. Shifting the contour by $\mathbf k\to\mathbf k+is\hat y$ with small $s>0$ gives the factor $e^{-s(\lvert y\rvert-c_0\lambda)+O(s^3\lambda)}$, since $\lvert{\rm Im}\,\Omega\rvert\le c_0s+O(s^3)$. So every free kernel is $O(e^{-c_\delta\lvert y\rvert})$ for $c_0\lambda\le(1-\delta)\lvert y\rvert$.
- By (9.5), $\lvert\bar U(\lambda)\rvert\le\lambda^2\sup\lvert1-\sigma_h\rvert/(2L^D)$.

Hence in order (O1), at every fixed $\lambda$, $\bar U\to0$ and $\delta a^\perp\to\delta a^\infty$, the field on $\mathbb Z^D$.

*Speeds.* $\lvert\nabla_{\mathbf k}\Omega_{\mathbf k}\rvert^2=\kappa_h\sum_i\sin^2k_i/\sum_i4\sin^2(k_i/2)\le c_0^2$, with equality only as $\mathbf k\to0$. So the largest slope of the normal-mode frequency is its slope $c_0$ at small labels, and the front of the free field is at $c_0\lambda=\rho$, up to an Airy layer of width $O(\rho^{1/3})$. The anchor is a rank-one term. It leaves the continuous spectrum $[0,4D\kappa_h]$ of $\kappa_h\Delta$ on $\mathbb Z^D$ unchanged and adds at most one normal mode $\psi_b$ above it, localized at $h$, with $\nu_b=\omega_b^2$ solving $1+\kappa_a\gamma_{i\omega_b}(0)=0$. It exists for $D=1$ at every $\kappa_a$, and for $D=3$ iff $\kappa_a>\kappa_h/\gamma_3$, since $\gamma(0)=-\gamma_3/\kappa_h$ at the top of the band (shift $\mathbf k\to\mathbf k+(\pi,\pi,\pi)$). Its share of $\delta a(x,\cdot)$ is $-\psi_b(x)\psi_b(x_s)(1-\cos\omega_b\lambda)/\nu_b=O(e^{-(d_x+d_s)/\xi})$.

*Free field.* On $\mathbb Z^D$, $\delta a_0(y,\lambda)=-\int\frac{\mathrm d^Dk}{(2\pi)^D}e^{i\mathbf k\cdot y}\,\frac{1-\cos\Omega_{\mathbf k}\lambda}{\Omega_{\mathbf k}^2}$.
- $D=3$: the static part is $-g_\infty(y)=-\frac1{4\pi\kappa_h\rho}+O(\rho^{-3})$ (input). In the oscillating part put $\mathbf k=\mathbf q/\rho$.
  - Small $\lvert\mathbf k\rvert$ gives $\frac1{2\pi^2\kappa_h\rho}\int_0^\infty\frac{\sin q}q\cos(qw)\,\mathrm dq=\frac{\theta(1-w)}{4\pi\kappa_h\rho}$: the angular average of $e^{iq_1}$ is $\sin q/q$, and the last step is Dirichlet's integral.
  - The rest of the zone is a smooth oscillatory integral, of order at most $\rho^{-4/3}$ along $\mathbf e_1$. The order $\rho^{-4/3}$ occurs at the fold caustics $w=1+\sqrt2,\sqrt2+\sqrt3$ of condition (d) of 04-formation, which covers these phases for the nearest-neighbour contract at $m_h\to0$ (question note (b)). In other directions the order is at most $\lvert y\rvert^{-7/6}$, by (9.17) below. The angular average and the step are the same in every direction.
  - The cubic term of $\Omega_{\mathbf k}$ only smooths the step over $\lvert w-1\rvert=O(\rho^{-2/3})$.
- $D=1$: the same steps with the factor $e^{ik}-1$ give the static gradient $\frac1{2\kappa_h}$ (Dirichlet kernel) and the oscillating gradient $-\frac{\theta(1-w)}{2\kappa_h}$, with remainder $O(\rho^{-1/2})$.

Hence, as $\rho\to\infty$ at fixed $w\ne1$,

$$
\rho\,\delta a_0(\rho\mathbf e_1,\lambda)\to-\frac{\theta(w-1)}{4\pi\kappa_h}\quad(D=3),\qquad \nabla\delta a_0(\rho,\lambda)\to\frac{\theta(w-1)}{2\kappa_h}\quad(D=1).
\tag{9.9}
$$

*Uniform bounds on the free field ($D=3$).* (4.11) and (4.19) ask for every $\lambda\ge\lambda_f$, so limits at fixed $w$ do not suffice. Let $u_0:=\delta a_0+g_\infty$ be the oscillating part, and fix $\delta>0$. Split the labels with a smooth cutoff $\chi_0(\mathbf k)$, equal to 1 near $\mathbf k=0$ and supported where $\lvert\nabla\Omega\rvert>c_0/(1+\delta/2)$.
- *Low labels.* In spherical coordinates $\mathrm d^3k/\Omega^2=c_0^{-2}(1+O(k^2))\,\mathrm dk\,\mathrm d\sigma$ is smooth, and $\Omega=c_0k-c_0\alpha(\hat k)k^3+O(k^5)$ with $\alpha=\sum_i\hat k_i^4/24\in[\frac1{72},\frac1{24}]$.
  - For $c_0\lambda\ge(1+\delta)\lvert y\rvert$ the radial derivative of the phase $k\lvert y\rvert\cos\vartheta+\lambda\Omega$ is at least of order $\lambda$. Two integrations by parts in $k$ leave a purely imaginary boundary term at $k=0$, plus $O(\lambda^{-2})$. Since $u_0$ is the real part of this integral, its low-label part is $O(\lambda^{-2})$. The same steps give $O(\lambda^{-2})$ for the low-label part of $S_0$, whose amplitude vanishes at $k=0$.
  - Across the front, uniform Airy asymptotics in $(\lvert y\rvert-c_0\lambda)/\lambda^{1/3}$ apply ($\alpha$ is bounded away from 0). For $\delta a_0$ they give the step of height $1/(4\pi\kappa_h\lvert y\rvert)$, smoothed over $O(\lvert y\rvert^{1/3})$. For $S_0$ they give a layer of height $O(\lvert y\rvert^{-4/3})$ whose oscillating tail has $\int\lvert S_0\rvert\,\mathrm d\lambda=O(\lvert y\rvert^{-1/2})$.
- *High labels* (amplitudes vanishing near $\mathbf k=0$).
  - In every direction, the dispersive bound of the lattice wave equation on $\mathbb Z^3$ applies: $C(1+\lambda)^{-7/6}$, uniformly in $y$. Its exponent is set by the points with all $k_i=\pm\frac\pi2$, where the Hessian of $\Omega$ has rank one.
  - Along $\mathbf e_1$, the stationary points of $\rho k_1\pm\lambda\Omega$ lie on the lines $k_2,k_3\in\{0,\pi\}$. There $\partial_2^2\Omega,\partial_3^2\Omega=\kappa_h\cos k_{2,3}/\Omega\ne0$, and $\partial_1^2\Omega$ vanishes only at the maxima of $\kappa_h\sin k_1/\Omega$ on the lines with one or two labels $\pi$. These are folds, at $w=1+\sqrt2$ and $\sqrt2+\sqrt3$ (condition (d), note (b)). As $v=\rho/\lambda\to0$ the stationary points tend to the corners $\mathbf k\in\{0,\pi\}^3\setminus\{0\}$, which are nondegenerate. Stationary phase uniform in $v\in[0,c_0/(1+\delta)]$ then gives $O(\lambda^{-4/3})$.
- Ahead of the front all kernels are exponentially small (contour shift above).

Together, for $y\ne0$, with $C$ independent of $\lambda$ and of the direction:

$$
\sup_{c_0\lambda\ge(1+\delta)\rho}\rho\,\lvert u_0(\rho\mathbf e_1,\lambda)\rvert\le C_\delta\,\rho^{-1/3},\qquad
\sup_{\lambda\ge0}\lvert\delta a_0(y,\lambda)\rvert\le\frac C{\lvert y\rvert},\qquad
\int_0^\infty\lvert S_0(y,\lambda)\rvert\,\mathrm d\lambda\le\frac C{\lvert y\rvert^{1/6}} .
\tag{9.17}
$$

The last bound is $\int_{(1-\delta)\lvert y\rvert/c_0}^\infty C\lambda^{-7/6}\,\mathrm d\lambda$, plus the Airy tail and the low-label $O(\lambda^{-2})$.

### Step 6. Order (O1): the image strength

By Step 2, $\hat\sigma_h(s)=\frac{\gamma_s(x_s-h)}s\,\chi(s)$, with $\chi(s)=\kappa_a/[1+\kappa_a\gamma_s(0)]$ on $\mathbb Z^D$. The first factor is the transform of $-\delta a_0(x_s-h,\cdot)$, which by (9.9) is a step at $c_0\lambda=d_s$.
- $D=3$: $\chi$ is the transform of $\kappa_a\delta(\lambda)+K(\lambda)$, plus, for $\kappa_a>\kappa_h/\gamma_3$, the undamped term of the localized mode, whose share of $\sigma_h$ is $O(e^{-d_s/\xi})$.
  - For $\kappa_a\ne\kappa_h/\gamma_3$, $1+\kappa_a\gamma_s(0)$ has no zero on the band. Its singularities there (band edges, van Hove points) are of square-root type, so $K=O(\lambda^{-3/2})$ is integrable.
  - At $\kappa_a=\kappa_h/\gamma_3$ the zero sits at the top edge, and $K$ decays only as $\lambda^{-1/2}$.
  - The total weight is $\chi(0)=\kappa_a/(1+\kappa_a\gamma_3/\kappa_h)=1/c_\phi$.
  - By (9.17), $A(y,\lambda):=-\delta a_0(y,\lambda)$ is bounded by $C/\lvert y\rvert$, and by (9.9) in the direction of $x_s-h$ it tends to $g_\infty(x_s-h)\theta(c_0\lambda-d_s)$ at fixed $c_0\lambda/d_s\ne1$. Dominated convergence in $\sigma_h=\chi*A(h-x_s,\cdot)$ gives (9.10) for $\kappa_a\ne\kappa_h/\gamma_3$.
- $D=1$: $\gamma_s(y)=\gamma_s(0)\zeta_s^{\lvert y\rvert}$ with $\gamma_s(0)=[s\sqrt{s^2+4\kappa_h}]^{-1}\to\infty$. So $\hat\sigma_h=\frac{\zeta_s^{d_s}}s\,\frac{\kappa_a\gamma_s(0)}{1+\kappa_a\gamma_s(0)}$, and the second factor is $1-2c_0s/\kappa_a+\dots$, a relaxation over $\lambda\sim2c_0/\kappa_a$.

Hence, as $d_s\to\infty$, on the scale $\lambda\propto d_s$,

$$
\sigma_h(\lambda)\to\beta\,\theta(c_0\lambda-d_s),\qquad \beta=\frac{g_\infty(x_s-h)}{c_\phi}=\frac{1+O(d_s^{-2})}{4\pi\kappa_hc_\phi d_s}\quad(D=3,\ \kappa_a\ne\kappa_h/\gamma_3),\qquad \beta=1\quad(D=1).
\tag{9.10}
$$

By (9.4) the image adds $-\beta\,\delta a_0(x-h,\lambda-d_s/c_0)$, a free field from $h$ that arrives at $x$ at $c_0\lambda=d_s+d_x$.

*$D=3$: $\lambda$-mean and image bound in order (O1).* Let $\delta a^\infty=-[(1-\cos\sqrt{K_a}\lambda)K_a^{-1}]_{xx_s}$ be the field on $\mathbb Z^3$ with the anchor (Step 5).
- *$\lambda$-mean.* The spectral measure of $K_a$ on $\mathbb Z^3$ has no atom at 0, and the mean of $\cos\omega\lambda$ vanishes for $\omega>0$, the localized mode included. So the $\lambda$-mean of $\delta a^\infty$ is $-(K_a^{-1})_{xx_s}$. By (5.9) on $\mathbb Z^3$ at $\nu\to0$, this is the first part of (9.18), for every $\kappa_a$.
- *Pointwise limit.* The continuous-spectrum part of $\delta a^\infty-(\text{mean})$ tends to 0 as $\lambda\to\infty$ (Riemann–Lebesgue). For $\kappa_a>\kappa_h/\gamma_3$ the localized mode adds the persistent oscillation $\psi_b(x)\psi_b(x_s)\cos(\omega_b\lambda)/\nu_b$, of size $O(e^{-(d_x+d_s)/\xi})$. So a pointwise limit exists only without this mode.
- *Image bound.* By (9.4), $I:=\delta a^\infty-\delta a_0(x-x_s,\cdot)=S_0(x-h)*\sigma_h$ and $\sigma_h=\chi*A(h-x_s)$, where $*$ is the convolution on $[0,\lambda]$. Since $A(y,0)=0$ and $\partial_\lambda A=S_0$, also $I=A(x-h)*\chi*S_0(h-x_s)$. Hence $\lvert I\rvert\le\sup\lvert A(x-h,\cdot)\rvert\,(\kappa_a+\lVert K\rVert_1)\int\lvert S_0(h-x_s,\cdot)\rvert$, plus the share of the localized mode. With (9.17) this is the second part of (9.18):

$$
\lim_{T\to\infty}\frac1T\int_0^T\delta a^\infty(x,\lambda)\,\mathrm d\lambda=-g_\infty(x-x_s)+\beta\,g_\infty(x-h),\qquad
\sup_{\lambda\ge0}\lvert I(x,\lambda)\rvert\le\frac C{d_x\,d_s^{1/6}}+O\bigl(e^{-(d_x+d_s)/\xi}\bigr)\ \ (\kappa_a\ne\kappa_h/\gamma_3).
\tag{9.18}
$$

The $L\to\infty$ limit of (5.11) is instead $-[c_\phi-g_\infty(x_s-h)]-g_\infty(x-x_s)+g_\infty(x-h)$. So in (O1) the uniform part and an image of strength $1-\beta$ are missing even from the $\lambda$-mean: they never form. The image of strength 1 in (5.11) expresses the balance of forces on the finite torus ($\sigma_h^{\rm st}=1$); on $\mathbb Z^3$ the anchor takes only the share $\beta$ of the force of the source.

### Step 7. Formation on the cubic torus

Take order (O1), then $\rho\to\infty$ along $\mathbf e_1$ at fixed $w$, $a$ and $b\ne1$. In (O1), $\delta a^\perp=\delta a^\infty$ at every $\lambda$ (Step 5), so

$$\Delta_3^\perp=\frac{-g_\infty(x-h)+u_0(\rho\mathbf e_1,\lambda)+I(x,\lambda)}{g_\infty(x-h)-g_\infty(\rho\mathbf e_1)} .$$

- By (9.6), $\rho\,\delta a^\perp_{\rm st}\to(\frac1b-1)/4\pi\kappa_h$, and $-g_\infty(x-h)/\delta a^\perp_{\rm st}\to1/(b-1)$.
- By (9.9), $\rho\,u_0\to\theta(1-w)/4\pi\kappa_h$. By (9.10) and (9.4), the image contributes $O(\beta\rho/d_x)=O(1/\rho)$ at fixed $w$ ($\kappa_a\ne\kappa_h/\gamma_3$).

So

$$
\lim\Delta_3^\perp(\rho,\lambda)=\begin{cases}-1,&w<1,\\[2pt] \dfrac1{b-1},&w>1,\end{cases}\qquad
\lim_{\rho\to\infty}\frac{c_0\lambda_{\rm form}(\rho;\varepsilon)}\rho=1\ \ \text{if } b>1+\frac1\varepsilon,\qquad \lambda_{\rm form}(\rho;\varepsilon)=\infty\ \ \text{for all large }\rho\ \text{if } \lvert b-1\rvert<\frac1\varepsilon .
\tag{9.11}
$$

The profile holds at fixed $w\ne1$ for $\kappa_a\ne\kappa_h/\gamma_3$. The two statements on $\lambda_{\rm form}$ need every later $\lambda$:
- *Middle* ($\kappa_a\ne\kappa_h/\gamma_3$). By (9.17) and (9.18), uniformly in $c_0\lambda\ge(1+\delta)\rho$, $\lvert\Delta_3^\perp+g_\infty(x-h)/\delta a^\perp_{\rm st}\rvert\le C_\delta\rho^{-1/3}+C(a\rho)^{-1/6}/\lvert1-b\rvert$, which tends to 0. At $w=1-\delta$, $\Delta_3^\perp\to-1$. So for $\frac1{b-1}<\varepsilon$ and large $\rho$, $(1-\delta)\rho<c_0\lambda_{\rm form}\le(1+\delta)\rho$. As in (4.14), $\lambda_{\rm form}-\rho/c_0=O(\rho^{1/3})$ once the caustic wakes and the image term are below $\varepsilon-\frac1{b-1}$.
- *Right* (every $\kappa_a$). If $\lvert\Delta_3^\perp\rvert\le\varepsilon$ for all $\lambda\ge\lambda_f$, its $\lambda$-mean obeys the same bound. By (9.18) that mean is $(\beta-1)g_\infty(x-h)/\delta a^\perp_{\rm st}$, which tends to $\frac1{b-1}$ since $\beta=O(1/\rho)$. So $\lambda_{\rm form}=\infty$ for large $\rho$ when $\lvert b-1\rvert<1/\varepsilon$. This rests on an exact identity of the mean, not on a limit at fixed $w$.

The marginal member $b=1+1/\varepsilon$ is not decided by (9.11).
- *Far from the anchor* ($d_s,d_x\to\infty$ at fixed $\rho$, then $\rho\to\infty$; $\kappa_a\ne\kappa_h/\gamma_3$). By (9.18), $\rho\sup_\lambda\lvert I\rvert\to0$, and the reference tends to $-g_\infty(\rho\mathbf e_1)$. So $\Delta_3^\perp\to-u_0(\rho\mathbf e_1,\lambda)/g_\infty(\rho\mathbf e_1)$ uniformly in $\lambda$. By (9.17) and (9.9), the $\limsup$ and $\liminf$ over $d_s,d_x$ of $c_0\lambda_{\rm form}/\rho$ both tend to 1 as $\rho\to\infty$, for every $\varepsilon\in(0,1)$. The place-dependent part forms within a $\lambda$ linear in $\rho$. The formation speed is $c_0$, equal to the front speed and to the largest slope of the normal-mode frequency, and independent of $\kappa_a$. The middle statement of (9.11) gives the same for $b\to\infty$ after $\rho\to\infty$.
- *Distances to $h$ comparable with $\rho$:* it does not form to accuracy below $1/\lvert b-1\rvert$, because the missing image of (5.11) is a fixed fraction of the reference value.

The code check (ii) (`check_formation.py`, cubic part) is a finite-window check: $c_0\lambda\le40$, i.e. $w\le40/\rho$. It tests the profile, not the bound for every later $\lambda$, which rests on (9.17) and (9.18). It finds, at $\rho=3$ to $16$, $\Delta_3^\perp=-1$ before the front and a later plateau equal to $1/(b-1)$ (from $0.04$ to $2.0$), to within $0.006$ at $\rho\ge12$, with wakes decaying about as $\rho^{-0.4}$. The $\lambda$-mean $(\beta-1)g_\infty(x-h)/\delta a^\perp_{\rm st}$, computed by the script, is $0.74$ to $1.98$ for the near anchor at $\rho\ge6$. So by the argument of the right statement, these members do not form at $\varepsilon=0.5$ at any later $\lambda$.

*Shell* ($\Phi$ depending only on $\lvert y\rvert$, question note (a)). Take order (O1), then $\ell\to\infty$ at fixed $w=c_0\lambda/\ell$ and $a=d_s/\ell>2$.
- *Static.* Write $z=(x-x_s)/\ell$ and $H=(h-x_s)/\ell$, with $\lvert H\rvert=a>2$. Since $\Phi$ is radial, the shell sum of $1/\lvert x-h\rvert$ is $\ell^2\int\Phi(z)\lvert z-H\rvert^{-1}\mathrm d^3z\,(1+o(1))=\ell^2\int\Phi/a\,(1+o(1))$. This is the mean-value property of the harmonic function $\lvert z-H\rvert^{-1}$ on the spheres $\lvert z\rvert\le2$, which do not reach $H$. With $\theta_a:=\int\Phi/(a\int\Phi\lvert z\rvert^{-1})\in(\frac1a,\frac2a)$, the reference sum is $-\frac{\ell^2}{4\pi\kappa_h}\int\Phi\lvert z\rvert^{-1}(1-\theta_a)(1+o(1))$. It does not vanish, since $\theta_a<2/a<1$.
- *Free part.* By Poisson summation, $\sum_x\varphi_\ell(x)u_0(x-x_s,\lambda)=\int\ell^3\hat\Phi(\ell\mathbf k)\cos(\lambda\Omega)\Omega^{-2}\frac{\mathrm d^3k}{(2\pi)^3}+O(\ell^{-\infty})$. On compact ranges of $w$ it tends to the continuum value $\frac{\ell^2}{4\pi\kappa_h}\int_{\lvert z\rvert>w}\Phi\lvert z\rvert^{-1}$. For large $w$, note that $\hat\Phi$ is radial. Along each ray the amplitude $k^2\hat\Phi(\ell k)/\Omega^2$ is even in $k$ and the phase $\lambda\Omega$ is odd, so the radial integral runs over the whole line with no boundary term. Its phase derivative is at least $c_0\lambda/2$ where $\hat\Phi(\ell\mathbf k)$ is not negligible, which gives $O(\ell^2w^{-N})$ uniformly in $\ell$.
- *Image.* By (9.18), $\lvert\sum_x\varphi_\ell I\rvert=O(\ell^3\cdot\ell^{-7/6})$, negligible against $\ell^2$ ($\kappa_a\ne\kappa_h/\gamma_3$).
- Define $\Psi(w):=\int_{\lvert z\rvert>w}\Phi\lvert z\rvert^{-1}/\int\Phi\lvert z\rvert^{-1}$: continuous and nonincreasing, equal to $1$ for $w\le1$ and to $0$ for $w\ge2$. Then, uniformly in $w\ge0$, the left side of (4.19) divided by the sum on its right side (without $\varepsilon$) tends to $\lvert\Psi(w)-\theta_a\rvert/(1-\theta_a)$. By (9.18), the $\lambda$-mean of the same ratio tends to $\theta_a/(1-\theta_a)$, for every $\kappa_a$.

So, with $w_\varepsilon:=\min\{w:\Psi(w)\le\theta_a+\varepsilon(1-\theta_a)\}\in[1,2]$,

$$
\text{(4.19) for }\delta a^\perp:\quad\lim_{\ell\to\infty}\frac{c_0\lambda_{\rm form}}\ell=w_\varepsilon\ \ \text{if}\ \theta_a<\frac\varepsilon{1+\varepsilon},\qquad \lambda_{\rm form}=\infty\ \text{for all large }\ell\ \ \text{if}\ \theta_a>\frac\varepsilon{1+\varepsilon} .
\tag{9.12}
$$

The first statement assumes $\kappa_a\ne\kappa_h/\gamma_3$ and that $\Psi$ is strictly decreasing at $w_\varepsilon$; the second holds for every $\kappa_a$, by the $\lambda$-mean argument. The marginal member $\theta_a=\varepsilon/(1+\varepsilon)$ is not decided. As $a\to\infty$, the field forms in the mean by $c_0\lambda\le2\ell$ for every $\varepsilon$. For $a<(1+\varepsilon)/\varepsilon$ it never does, since $\theta_a>1/a$.

*Order (O2).* By (9.8) neither criterion holds at fixed $L$. Beyond the wrap time, the slow mode of (9.14) adds to $u$ the image $-(1-\beta_L)g(x-h)\cos(\lambda/T_U)$. This is an oscillation of relative size $(1-\beta_L)\lvert g(x-h)\rvert/\lvert\delta a^\perp_{\rm st}(x)\rvert$, which tends to $1/\lvert b-1\rvert$.

### Step 8. Formation on the ring (control)

Take order (O1), then $\rho\to\infty$ at fixed $w$ and $a=d_s/\rho>1$, with the observation place between source and anchor and the gradient toward the anchor.
- The reference is the $L\to\infty$ limit of $\nabla\delta a_{\rm st}$, which is $1/\kappa_h$ between source and anchor by (5.11). On the finite ring it is $(L-d_s)/(\kappa_hL)$.
- By (9.10), $\beta=1$. By (9.9), the image $-\delta a_0(x-h,\lambda-d_s/c_0)$ has the gradient $\frac1{2\kappa_h}\theta(c_0\lambda-d_s-d_x)$ toward $h$, and $d_s+d_x=(2a-1)\rho$.

So

$$
\lim\Delta_1(\rho,\lambda)=\begin{cases}-1,&w<1,\\-\tfrac12,&1<w<2a-1,\\0,&w>2a-1,\end{cases}\qquad
\lim_{\rho\to\infty}\frac{c_0\lambda_{\rm form}(\rho;\varepsilon)}\rho=\begin{cases}2a-1,&\varepsilon<\tfrac12,\\1,&\varepsilon>\tfrac12.\end{cases}
\tag{9.13}
$$

The free half of the gradient forms at the front, at speed $c_0$. The other half is the image, which arrives along the path $d_s+d_x$ at speed $c_0$.

*Every later $\lambda$.* On the line, $\gamma_s(y)=\gamma_s(0)\zeta_s^{\lvert y\rvert}$, and (9.3) with (5.9) gives exactly

$$\nabla\delta a^\infty(\rho,\lambda)=F(\rho,\lambda)+F(m,\lambda)-(K_1*F(m,\cdot))(\lambda)+O(e^{-m/\xi}),\qquad m=2d_s-\rho-1 .$$

- Here $F(n,\cdot)$, with transform $-\gamma_s(0)(\zeta_s-1)\zeta_s^n/s$, is the free gradient at distance $n$.
- $K_1$ has the transform $1/(1+\kappa_a\gamma_s(0))$. It is integrable: $1+\kappa_a\gamma_s(0)$ has no zero on the band, and $\gamma_s(0)$ diverges as an inverse square root at both band edges. And $\int K_1=0$, since $\gamma_s(0)\to\infty$ as $s\to0$.
- The last term is the localized mode.

The stationary points of $kn\pm\lambda\Omega$ with $\Omega'=c_0\cos(k/2)\le c_0/(1+\delta)$ have $\Omega''=-\frac{c_0}2\sin(k/2)\ne0$: there are no caustics in $D=1$ (note (b)). So, uniformly in $c_0\lambda\ge(1+\delta)n$, $F(n,\lambda)=\frac1{2\kappa_h}+O(\lambda^{-1/2})$; and $F$ is exponentially small for $c_0\lambda\le(1-\delta)n$. Across its front, $\kappa_hF\simeq\frac12\int_X^\infty{\rm Ai}(t)\,\mathrm dt$, with $X$ the distance to the front in units of the Airy width $O(n^{1/3})$. So $\kappa_hF$ stays in $[0,0.64]$ up to $o(1)$, since $0<\int_X^\infty{\rm Ai}\le1.28$. Splitting the convolution at a large fixed $\lambda'$ gives $K_1*F(m,\cdot)\to0$ uniformly. Hence, uniformly:
- $\Delta_1\to-\frac12$ on $1+\delta\le w\le(2a-1)(1-\delta)$;
- $\Delta_1\to0$ on $w\ge(2a-1)(1+\delta)$;
- $\Delta_1\in[-\frac12,0.14]+o(1)$ across the image front.

This gives both limits of $\lambda_{\rm form}$ in (9.13); the marginal $\varepsilon=\frac12$ is not decided.
- For $\varepsilon<\frac12$ and the anchor far away ($a\to\infty$), the place-dependent part does not form within any $\lambda$ linear in $\rho$. This uses only the lower bound $\liminf c_0\lambda_{\rm form}/\rho\ge2a-1$. This is the control of item 2.
- For comparable distances it forms within a $\lambda$ linear in $\rho$, with slope $(2a-1)/c_0>1/c_0$.
- In order (O2) the field varies on the scale $L/c_0$ and does not settle (Step 9 and (9.8)).

The code check (ii) (`check_formation.py`, ring part) is a finite-window check ($w\le4$ for $a=2$). It finds the plateaus $-0.500$ and $0.000$, and $c_0\lambda_{\rm form}/\rho\to3.02$ and $1.02$ at $\rho=160$ for $a=2$. Across the image front, $\Delta_1$ lies in $[-0.515,0.117]$ at $\rho=160$, and $\max_{w>1.4}\lvert\Delta_1\rvert$ falls from $0.577$ to $0.530$ between $\rho=20$ and $160$. With $d_s=400$, formation at $\varepsilon=0.25$ occurs only at $c_0\lambda=2d_s-\rho+3$.

### Step 9. The uniform part

*Order (O1).* By Step 5, $\bar U(\lambda)\to0$ at every $\lambda$. Meanwhile $\bar U_{\rm st}\to-[c_\phi-g_\infty(x_s-h)]$ for $D=3$ and $\bar U_{\rm st}\to-[\frac1{\kappa_a}+\frac{d_s}{2\kappa_h}]$ for $D=1$ at fixed $d_s$. So the uniform part never reaches its static value.

*Cubic torus, order (O2): the modes that enter $\bar U$.*
- A normal mode with $\psi_n(h)\ne0$ satisfies $(\kappa_h\Delta-\nu_n)\psi_n=-\kappa_a\psi_n(h)e_h$. So $\psi_n\propto v_n:=(\kappa_h\Delta-\nu_n)^{-1}e_h$, and $\nu_n$ solves $1+\kappa_a\gamma_{i\sqrt\nu}(0)=0$, i.e. $\frac1{L^D\nu}=\frac1{\kappa_a}+\tilde\gamma_{i\sqrt\nu}(0)$.
- Summing the mode equation over places gives $\sum_x\psi_n=\kappa_a\psi_n(h)/\nu_n$. Modes with $\psi_n(h)=0$ are orthogonal to $\mathbf 1$ and do not enter $\bar U$.
- By interlacing, the lowest root lies below the lowest nonzero eigenvalue $\mu_1=4\kappa_h\sin^2(\pi/L)$ of $\kappa_h\Delta$.

*The slow mode.* For $D=3$, $L^{-3}\sum_{\mathbf k\ne0}\hat K_0^{-2}=O(L)$, so $\tilde\gamma_{i\omega}(x)=g(x)+O(\omega^2L)$. With $\nu_0=O(L^{-3})$ this gives, with $c_\phi^{(L)}:=\frac1{\kappa_a}+g(0)\to c_\phi$:
- $\nu_0^{-1}=L^3c_\phi^{(L)}[1+O(L^{-2})]$;
- $v_0(x)=-[c_\phi^{(L)}-g(x-h)]+O(L^{-2})$;
- $\lVert v_0\rVert^2=L^3c_\phi^{(L)2}[1+O(L^{-2})]$.

*The other modes.* By Cauchy–Schwarz, their share of $\bar U$ is at most $\frac2{L^3}\bigl[\sum_{n\ge1}\langle\mathbf1,\psi_n\rangle^2/\nu_n\bigr]^{1/2}\Gamma(x_s)^{1/2}$.
- By (5.8), $\sum_{n\ge0}\langle\mathbf1,\psi_n\rangle^2/\nu_n=\langle\mathbf1,K_a^{-1}\mathbf1\rangle=L^6c^{(L)}_\phi$, and the mode $n=0$ carries $L^6c^{(L)}_\phi[1+O(L^{-2})]$ of it.
- $\Gamma(x_s)=\frac1{\kappa_a}+r(x_s,h)\le2\lvert\bar U_{\rm st}\rvert$, and $c^{(L)}_\phi\le2\lvert\bar U_{\rm st}\rvert$ for $x_s\ne h$ and large $L$. The last bound uses $g(0)\to\gamma_3/\kappa_h$, $6\gamma_3<2$, and $g(0)-g(x)\ge g(0)-g(\mathbf e_1)=(1-L^{-3})/(6\kappa_h)$ for $x\ne0$: $g$ lies below the mean of its neighbours away from $0$, so by the maximum principle its largest value off $0$ is at a neighbour of $0$.
- So this share is $O(L^{-1})\lvert\bar U_{\rm st}\rvert$.

Hence, uniformly in $\lambda\ge0$,

$$
T_U:=\nu_0^{-1/2}=\sqrt{L^3c^{(L)}_\phi}\,\bigl[1+O(L^{-2})\bigr],\qquad
\bar U(\lambda)=\bar U_{\rm st}\bigl[1-\cos(\lambda/T_U)\bigr]+O(L^{-1})\lvert\bar U_{\rm st}\rvert,\qquad
\frac{\psi_0(x)\psi_0(x_s)}{\nu_0}=\frac{\bigl[c^{(L)}_\phi-g(x-h)\bigr]\bigl[c^{(L)}_\phi-g(x_s-h)\bigr]}{c^{(L)}_\phi}\bigl[1+O(L^{-2})\bigr].
\tag{9.14}
$$

What (9.14) says:
- The uniform part oscillates between $0$ and $2\bar U_{\rm st}$ on the scale $T_U\propto L^{3/2}$.
- $T_U$ depends on the anchor only through $c_\phi=\frac1{\kappa_a}+\frac{\gamma_3}{\kappa_h}$, and not on $d_s$; $d_s$ enters $\bar U_{\rm st}$ and the delay $d_s/c_0\le L/c_0$.
- $T_U/(L/c_0)=c_0(Lc^{(L)}_\phi)^{1/2}\to\infty$, so the uniform part moves over many wrap times.
- The place-dependent part of the slow term is the image $-(1-\beta_L)g(x-h)$, with $\beta_L:=g(x_s-h)/c^{(L)}_\phi$; it is used in Step 7.

The same argument holds on the ring for a weak anchor, $\kappa_aL\ll\kappa_h$, using $L^{-1}\sum_{k\ne0}\hat K_0^{-2}=L^3/(720\kappa_h^2)$ and $g(0)=(L^2-1)/(12\kappa_hL)$. There $T_U\simeq(L/\kappa_a)^{1/2}$, with corrections $O(\kappa_aL/\kappa_h)$. The code check (iii) confirms (9.14) on both graphs.

*Ring, order (O2), $\kappa_aL\gg\kappa_h$.* Let $L\to\infty$ at fixed $\tau=c_0\lambda/L$ and $\alpha=d_s/L$.
- On the scale $L$ the chain is a string of tension $\kappa_h$.
- Matching at $h$ gives the anchor the reflection amplitude $-\kappa_a/(\kappa_a-2i\kappa_hk)$ for a wave of label $k$. This tends to $-1$ for $k\ll\kappa_a/\kappa_h$, so the anchor acts as a fixed point, and the ring is a string of length $L$ fixed at both ends.
- Its modes are $\sqrt{2/L}\sin(n\pi y/L)$, with $y$ the distance from $h$ along the ring and $\omega_n=n\pi c_0/L$. Their place averages are $\sqrt{2/L}\,[1-(-1)^n]/(n\pi)$.

So

$$
\bar U(\lambda)=-\frac{4L}{\pi^3\kappa_h}\sum_{n\ {\rm odd}}\frac{\sin(n\pi\alpha)\,(1-\cos n\pi\tau)}{n^3}+O(L^0)=-\frac L{2\kappa_h}\Bigl[f(\alpha)-\tfrac12f(\alpha+\tau)-\tfrac12f(\alpha-\tau)\Bigr]+O(L^0),
\tag{9.15}
$$

where $f$ is the odd $2$-periodic extension of $x(1-x)$; the second form follows from $\sum_{n\,{\rm odd}}\sin(n\pi x)/n^3=\frac{\pi^3}8x(1-x)$ on $[0,1]$. The $O(L^0)$ term (finite $\kappa_a$, dispersion) grows at most linearly in $\tau$. What (9.15) says:
- $\bar U=-\lambda^2/(2L)$ until the signals reach the anchor.
- $\bar U$ is periodic in $\lambda$ with period $2L/c_0$, and oscillates between $0$ ($\tau$ even) and $2\bar U_{\rm st}$ ($\tau$ odd).
- Its mean is $-L\alpha(1-\alpha)/(2\kappa_h)=\bar U_{\rm st}+O(1)$.
- Its scale is the wrap time, independent of $\kappa_a$ at leading order.

The code check (iii) confirms (9.15).

So the uniform part reaches its static value in the sense of (9.1) in no order of the limits: not in (O1), where it stays $0$; not in (O2), where it oscillates with amplitude $\lvert\bar U_{\rm st}\rvert$; and not at fixed $L$, by (9.8). Its $\lambda$-mean equals $\bar U_{\rm st}$ at every $L$.

### Step 10. All three together (short argument)

Only the cubic torus carries a long-range mark, so item 4 concerns $D=3$. As the question's note (c) states, the three are properties of the stated description:
- (5.15) in its order (a) and the weighing of 05-anchor are properties of the lowest eigenvectors of the contracts with the sources and the bodies, on the finite torus, followed by $L\to\infty$;
- (9.11) is a property of the evolution from $\Omega$ in order (O1).

They are available together in the order

$$
g_s\to0,\ M\to\infty;\qquad L\to\infty;\qquad d/\ell\to\infty;\qquad \ell\to\infty\ \ (w\ \text{fixed}),
\tag{9.16}
$$

where $d$ is the smallest distance of the places involved to $h$, and $\ell$ stands for every other distance ($\rho$, $\lvert\mathbf r\rvert$, $R$; the comparison vector $\delta$ and the placements of family (A) stay fixed or grow with $\ell$). Here $\kappa_a\ne\kappa_h/\gamma_3$.

In this order:
- (5.15) gives the separation dependence $-g_1g_2/(4\pi\kappa_h\lvert\mathbf r\rvert)$ (its order (a), $\lvert\mathbf r\rvert/D_X\to0$, with $L\to\infty$ first and all distances $\ge\ell\gg1$). It comes from the source term $g(x_1-x_2)$ alone.
- Step 7, far from the anchor ($d\to\infty$ at fixed $\rho$, then $\rho\to\infty$), gives formation of the place-dependent part within a $\lambda$ linear in the distance, at speed $c_0$, for every $\varepsilon\in(0,1)$.
- The weighing of 05-anchor holds within (A), or within (B) with $R/d\to0$ at fixed $c$, under its own conditions. In its controlled regime, at order $\mu^2$ and leading order in the grip, $\rho_{\rm mark}=1$ and $\rho_{{\rm bond},i}=m_{\mathbf r}$ for every level and both branches iff the grips are square with a common $\hat\beta$. The same holds in the lowest eigenvector of the stated contract iff in addition (5.10) holds for one body ((5.26) of 05-anchor for a pair). For other grips the weighing fails, as quoted.

So, for square grips with a common $\hat\beta$ (and (5.10), (5.26) for the lowest-eigenvector form), the long-range mark, the formation within a $\lambda$ linear in the distance and the weighing are available together in (9.16), at leading order in $\ell/d$.

This does not say that a weighing is realized in the evolving state (note (c): not asked). That would need a dynamical observable. The evolving state is $\Omega$ displaced by the means (Step 1); its $\lambda$-mean in (O1) lacks the uniform part and the image of strength $1-\beta$ of $\phi_a$ ((9.18)); and its total cost is conserved, so it is not the lowest eigenvalue with the source.

The order cannot be relaxed for the places of the formation. If the source or an observation place is at a distance from $h$ comparable to $\rho$, then $b\le a+1$ stays bounded, and (9.11) gives $\lambda_{\rm form}=\infty$ for every $\varepsilon<1/\lvert b-1\rvert$. In order (O2), formation in the sense of (4.11) and (4.19) fails by (9.8). In no order does the state reach the lowest eigenvector of the contract with the source: the evolution from $\Omega$ does not relax (A11), and the static values appear only as $\lambda$-means.

## Result

Class for all results: the nearest-neighbour oscillator cells without pinning, with the anchor $\frac{\kappa_a}2q_h^2$, every $\kappa_h,\kappa_a>0$, every $h$ and every $x_s\ne h$, on the ring and on the cubic torus, at first order in $g_s$ (exact for the means). The order is $M\to\infty$ first; each item states its remaining order. The excluded members are those of "Setup".

1. **(9.2), (9.3), (9.4)** (formula; every $L$, both graphs, every $\lambda\ge0$). The made field is $\delta a=-[(1-\cos\sqrt{K_a}\lambda)K_a^{-1}]_{xx_s}$, with $\delta a_{\rm st}=-G_a(x,x_s)$ and Laplace transform $-R_{xx_s}(s)/s$. Equivalently, it is the free field of the source plus the free field of an image at $h$, whose strength $\sigma_h=-\kappa_a\delta a(h,\cdot)$ solves the Volterra equation (9.4); the static image strength is 1. The state at $\lambda$ is $\Omega$ displaced by the means, so its connected equal-time correlations ($w$ among them) are those of $\Omega$.
2. **(9.5), (9.6), (9.7)** (formula; every $L$, both graphs). The uniform part is the place average $\bar U=-L^{-D}\int_0^\lambda(\lambda-\lambda')[1-\sigma_h]\,\mathrm d\lambda'$, driven by the net force of source and image, with Laplace transform (9.7). The place-dependent part is $\delta a^\perp=\delta a-\bar U$. Their static values are exactly the uniform shift and the place-dependent part of (5.11).
3. **(9.8)** (theorem; every finite $L$, both graphs). The $\lambda$-mean of $\delta a$ (and of $\bar U$, $\delta a^\perp$) is the static value. For every $\lambda_f$ there are $\lambda>\lambda_f$ at which $\delta a$ is arbitrarily close to $0$. So at fixed $L$ neither part meets (4.11), (4.19) or (9.1) for any $\varepsilon<1$.
4. **(9.9), (9.10), (9.17), (9.18)** (formula; order (O1), then the stated distance $\to\infty$ at fixed $w$; (9.17) and (9.18): $D=3$, order (O1), bounds uniform in $\lambda$ and the $\lambda$-mean).
   - The free field forms at the front: $\rho\,\delta a_0\to-\theta(w-1)/(4\pi\kappa_h)$ ($D=3$) and $\nabla\delta a_0\to\theta(w-1)/(2\kappa_h)$ ($D=1$). The front speed equals the largest slope $c_0$ of the normal-mode frequency.
   - The image strength tends to $\beta\,\theta(c_0\lambda-d_s)$, with $\beta=g_\infty(x_s-h)/c_\phi$ for $D=3$ ($\kappa_a\ne\kappa_h/\gamma_3$) and $\beta=1$ for $D=1$.
   - Uniform bounds (9.17): $\rho\lvert u_0(\rho\mathbf e_1,\lambda)\rvert\le C_\delta\rho^{-1/3}$ for $c_0\lambda\ge(1+\delta)\rho$; $\lvert\delta a_0(y,\lambda)\rvert\le C/\lvert y\rvert$ for all $\lambda$; $\int_0^\infty\lvert S_0(y,\lambda)\rvert\,\mathrm d\lambda\le C\lvert y\rvert^{-1/6}$.
   - (9.18): for every $\kappa_a$, the $\lambda$-mean of the field on $\mathbb Z^3$ is $-g_\infty(x-x_s)+\beta g_\infty(x-h)$. For $\kappa_a\ne\kappa_h/\gamma_3$, the image term obeys $\sup_\lambda\lvert I\rvert\le C/(d_xd_s^{1/6})+O(e^{-(d_x+d_s)/\xi})$. For $\kappa_a>\kappa_h/\gamma_3$, a localized mode adds a persistent oscillation of size $O(e^{-(d_x+d_s)/\xi})$.
   - In order (O1) the uniform part and, for $D=3$, an image of strength $1-\beta$ never form: they are missing from the $\lambda$-mean.
5. **(9.11)** (criterion; cubic torus, order (O1), then $\rho\to\infty$ along $\mathbf e_1$ at fixed $w$, $a$, $b\ne1$).
   - For $\kappa_a\ne\kappa_h/\gamma_3$: $\Delta_3^\perp\to-1$ for $w<1$ and $\to1/(b-1)$ for $w>1$. If $b>1+1/\varepsilon$, then $c_0\lambda_{\rm form}/\rho\to1$, uniformly controlled in every later $\lambda$: formation within a $\lambda$ linear in the distance, at speed $c_0$ (equal to the front speed and the largest slope, independent of $\kappa_a$).
   - For every $\kappa_a$: if $\lvert b-1\rvert<1/\varepsilon$, then $\lambda_{\rm form}=\infty$ for all large $\rho$ (by the $\lambda$-mean).
   - The marginal $b=1+1/\varepsilon$ is not decided.
   - Far from the anchor ($d_s,d_x\to\infty$ at fixed $\rho$, then $\rho\to\infty$; $\kappa_a\ne\kappa_h/\gamma_3$): $c_0\lambda_{\rm form}/\rho\to1$ for every $\varepsilon\in(0,1)$.
6. **(9.12)** (criterion; cubic torus, radial $\Phi$, order (O1), then $\ell\to\infty$ at fixed $w$ and $a=d_s/\ell>2$, where the weighted reference $\propto1-\theta_a$ does not vanish).
   - If $\theta_a<\varepsilon/(1+\varepsilon)$, $\kappa_a\ne\kappa_h/\gamma_3$, and $\Psi$ is strictly decreasing at $w_\varepsilon$: (4.19) for $\delta a^\perp$ holds with $c_0\lambda_{\rm form}/\ell\to w_\varepsilon\in[1,2]$.
   - If $\theta_a>\varepsilon/(1+\varepsilon)$ (every $\kappa_a$): $\lambda_{\rm form}=\infty$ for all large $\ell$.
   - So the field forms in the mean by $c_0\lambda\le2\ell$ as $a\to\infty$, and never forms for $a<(1+\varepsilon)/\varepsilon$.
7. **(9.13)** (negative, control; ring, order (O1), then $\rho\to\infty$ at fixed $w$ and $a=d_s/\rho>1$, observation between source and anchor). $\Delta_1\to-1$, $-\frac12$, $0$ on $w<1$, $1<w<2a-1$, $w>2a-1$, uniformly for $w$ at distance $\ge\delta$ from $1$ and $2a-1$. Across the image front, $\Delta_1\in[-\frac12,0.14]+o(1)$. For $\varepsilon<\frac12$, $c_0\lambda_{\rm form}/\rho\to2a-1$; for $\varepsilon>\frac12$, it tends to $1$. With the anchor far away ($a\to\infty$) and $\varepsilon<\frac12$, the place-dependent part does not form within any $\lambda$ linear in $\rho$.
8. **(9.14)** (formula; cubic torus, every $\lambda\ge0$, as $L\to\infty$; and the ring with $\kappa_aL\ll\kappa_h$, with corrections $O(\kappa_aL/\kappa_h)$ and $T_U\simeq(L/\kappa_a)^{1/2}$). $\bar U=\bar U_{\rm st}[1-\cos(\lambda/T_U)]+O(L^{-1})\lvert\bar U_{\rm st}\rvert$, with $T_U=\sqrt{L^3c^{(L)}_\phi}\,[1+O(L^{-2})]$, which depends on $\kappa_a$ only through $c_\phi$ and not on $d_s$. The slow mode carries the oscillating image $-(1-\beta_L)g(x-h)$.
9. **(9.15)** (formula; ring, $L\to\infty$ at fixed $c_0\lambda/L$ and $d_s/L$). $\bar U$ is that of a string fixed at both ends at the anchor: periodic with period $2L/c_0$, between $0$ and $2\bar U_{\rm st}$, with corrections $O(L^0)$ growing at most linearly in $c_0\lambda/L$. Together with items 3 and 4 (negative): the uniform part reaches its static value (9.1) in no order of the limits, on both graphs; it reaches it only as a $\lambda$-mean.
10. **(9.16)** (theorem, short argument; cubic torus, $\kappa_a\ne\kappa_h/\gamma_3$, grips square with a common $\hat\beta$). In the order $L\to\infty$, then $d/\ell\to\infty$, then $\ell\to\infty$ at fixed $w$, three properties of the description are available together at leading order in $\ell/d$:
    - the long-range mark (5.15)(a) and the weighing of 05-anchor within (A), or within (B) with $R/d\to0$ at fixed $c$, as properties of lowest eigenvectors. The weighing holds in its controlled regime, at order $\mu^2$ and leading order in the grip; for the lowest eigenvector of the stated contract it holds iff in addition (5.10) holds ((5.26) of 05-anchor for a pair);
    - the formation of the place-dependent part within a $\lambda$ linear in the distance at speed $c_0$, for every $\varepsilon\in(0,1)$, as a property of the evolution from $\Omega$.

    A weighing realized in the evolving state is not claimed. Formation fails for small $\varepsilon$ when the source or an observation place is at a distance from $h$ comparable to $\rho$, and it fails in order (O2).

## Consistency checks

1. **Dimensions.** $[K_a^{-1}]=E^{-2}=[\delta a]$ in (9.2). $\beta=g/c_\phi$ is dimensionless, and so are (9.11), (9.13). $[T_U]=[(L^3c_\phi)^{1/2}]=E^{-1}=[\lambda]$, and the ring scale $L/c_0$ also has dimension $E^{-1}$. $[L/\kappa_h]=E^{-2}=[\bar U]$ in (9.15). Passed.
2. **Comparison with (4.12)–(4.13) at small $m_h$.** For $b\to\infty$, (9.11) gives $\Delta_3^\perp\to-\theta(1-w)$. This is (4.12) at $\mu=0$, since $e^0-1-0=0$ for $w>1$, and it is the $\mu\to0$ limit of the bound (4.13). Far from the anchor, the unpinned anchored field forms like the pinned field at $\mu\to0$. Passed.
3. **Limit $\kappa_a\to\infty$** (a fixed cell). $c_\phi\to\gamma_3/\kappa_h$, so $T_U\to(L^3\gamma_3/\kappa_h)^{1/2}$ stays finite. $\beta\to g_\infty(x_s-h)/g_\infty(0)<1$: even a fixed cell takes only part of the force on $\mathbb Z^3$. For $D=1$ the anchor time $2c_0/\kappa_a\to0$ and the reflection amplitude is exactly $-1$. (9.6) goes to $\bar U_{\rm st}=-r(x_s,h)/2$, which is (5.11) at $1/\kappa_a=0$. (9.11), (9.13) and (9.15) are unchanged at leading order. Passed.

## Open issues

- The bounds (9.17), (9.18) are stated without their constants. (9.17) uses two standard results: the dispersive bound $\lambda^{-7/6}$ of the lattice wave equation on $\mathbb Z^3$, and uniform Airy asymptotics of the low-label front. At $\rho\le16$ the code shows wakes of $0.30$ to $0.16$, decaying about as $\rho^{-0.4}$. So at these distances, formation at $\varepsilon\lesssim0.3$ is delayed beyond $c_0\lambda\approx\rho$, and the code's formation times are finite-window values.
- At the threshold $\kappa_a=\kappa_h/\gamma_3$ ($D=3$) the anchor kernel decays only as $\lambda^{-1/2}$. The image statements (9.10), (9.18) (second part), the profile and positive parts of (9.11) and (9.12), and (9.16) are not derived there.
- The shell criterion is treated only for radial $\Phi$ (question note (a)).
- The error terms of (9.14) are upper bounds. The code finds deviations far below them.
- In order (O2) on the cubic torus, the part of the field from the modes other than the slow one is controlled beyond the wrap time only by (9.8). No criterion restricted to a window of $\lambda$ is studied.
- Item 4 uses the weighing of 05-anchor as quoted in words. Whether a weighing is realized in the evolving state, through a dynamical observable, is not studied (question note (c)).
- The ring with $\kappa_aL$ comparable to $\kappa_h$ is not treated. (9.14) and (9.15) are its two limits.

## Code

All files are under `code/v3/`. Each script runs from the project root with `.venv/Scripts/python.exe` and writes to `code/v3/output/`. "Exact solution" means the solution of the linear equations of the means at $M\to\infty$ (Step 1), computed from the eigen-decomposition of $K_a$ or by fourth-order Runge–Kutta. $\kappa_h=1$ ($c_0=1$).

- `check_made_field.py` (check; item 1). On the ring at $L=64,96$ and the cubic torus at $L=8,12$, with $\kappa_a=1$, the anchor at two distances from the source and three places each, it compares the image representation (9.4) with the exact (9.2) up to $\lambda=3L$. It also compares the force formula (9.5) with the exact place average, and checks the static split (9.6) and the static image strength 1. Output: `code/v3/output/made_field_check.txt`. The largest relative deviation is $7.7\times10^{-6}$ (trapezoid rule); the static split agrees to $10^{-12}$.
- `check_formation.py` (check; item 2). Outputs: `code/v3/output/formation_check.txt`, `code/v3/output/formation_Delta.csv`.
  - *Cubic torus.* $L=64,96$; Runge–Kutta with $\mathrm d\lambda=0.05$ up to $\lambda=40$, below the wrap time. Anchor far (antipode, $d_s=55,83$) and near ($h=x_s-8\mathbf e_1$); $\rho=3,4,6,8,12,16$ along $\mathbf e_1$. Reference: $g_\infty(x-h)-g_\infty(x-x_s)$, with $g_\infty$ from the Bessel-integral representation ($g_\infty(0)=0.2527310$).
  - *Order (O1) emulated.* The field at the probe places is the same at both sizes to $2\times10^{-7}$ (finite propagation), and the time-step error is $9\times10^{-6}$. The finite-$L$ place average equals $-\lambda^2/2L^3$ to $0.5\%$; it vanishes in order (O1), so $\Delta_3^\perp=\delta a/{\rm ref}-1$.
  - *Before the front:* $\Delta_3^\perp=-1.000$ for $w<0.6$ in all cases.
  - *Plateau* (mean over $1.6<w\le3$): it follows the finite-$\rho$ prediction $-g_\infty(x-h)/{\rm ref}$, i.e. $1/(b-1)$, over the range $0.04$ to $2.0$. Deviations: $\le0.08$ at $\rho=3$, $\le0.025$ at $\rho=6$, $\le0.014$ at $\rho=8$, $\le0.006$ at $\rho=12,16$.
  - *Wakes:* $\max_{w>1.3}\lvert\Delta_3^\perp-{\rm pred}\rvert$ falls from $0.30$ to $0.16$ (in units of the free field) between $\rho=3$ and $16$, about as $\rho^{-0.4}$, consistent with the $O(\rho^{-1/3})$ caustic wakes of Step 5.
  - *$\lambda$-mean in order (O1)* (computed, not simulated): $(\beta-1)g_\infty(x-h)/{\rm ref}$ with $\beta=g_\infty(x_s-h)/c_\phi$. For the far anchor it equals the finite-$\rho$ plateau prediction to $0.001$. For the near anchor it is $0.354,0.483,0.736,0.986,1.484,1.981$ for $\rho=3,\dots,16$, against $1/(b-1)=0.375$ to $2.000$.
  - *Formation times* (finite-window values; the window ends at $w_{\max}=40/\rho=13.3$ to $2.5$): at $\varepsilon=0.5$ (far anchor, $L=96$), $c_0\lambda_{\rm form}/\rho=1.00,1.05,1.04,1.03,1.02,1.23$ for $\rho=3,\dots,16$. ($\lambda_{\rm form}$ is the first $\lambda$ from which $\lvert\Delta_3^\perp\rvert\le\varepsilon$ up to the end of the window $\lambda=40$.) Where $1/(b-1)>0.5$ (near anchor at $\rho\ge6$, $1/(b-1)=0.75$ to $2.0$), there is no formation within the window, as (9.11) predicts. Since the $\lambda$-mean exceeds $0.5$ there, there is none at any later $\lambda$ either. The marginal members $1/(b-1)=0.5=\varepsilon$ ($b=3$) are not decided by (9.11); there the finite-$\rho$ corrections decide. For the near anchor at $\rho=4$ (plateau $0.496$, finite-$\rho$ prediction $0.487$), $c_0\lambda_{\rm form}/\rho=9.91$ at both sizes, just before the end of the window at $w=10$. For the far anchor at $L=64$, $\rho=16$ (plateau $0.501$, finite-$\rho$ prediction $0.499$), there is no formation within the window ($w\le2.5$). At $\varepsilon=0.2,0.3$ the wakes, which still exceed $\varepsilon-1/(b-1)$ at these $\rho$, delay formation to $c_0\lambda/\rho\approx1.6$–$4.2$ or beyond the window.
  - *Ring.* $L=1024,2048$, exact.
    - (a) $a=2$, $\rho=20,40,80,160$: the plateau is $-0.500$ (to $0.004$) on $1.4<w<2.6$ and $0.000$ (to $0.007$) on $w>3.4$. $c_0\lambda_{\rm form}/\rho=3.08,3.06,3.03,3.02$ at $\varepsilon=0.25$ (prediction 3) and $1.34,1.20,1.12,1.02$ at $\varepsilon=0.6$ (prediction 1), identical at both sizes. $\max_{w>1.4}\lvert\Delta_1\rvert=0.577,0.561,0.543,0.530$. Across the image front ($2.6<w<3.4$), $\Delta_1$ lies in $[-0.544,0.069]$ at $\rho=20$ and in $[-0.515,0.117]$ at $\rho=160$, approaching the range $[-\frac12,0.14]$ of Step 8.
    - (b) $d_s=400$, $\rho=10,20,40,80$: $\Delta_1$ has mean $-0.500$ on $1.4\rho<c_0\lambda<0.9(2d_s-\rho)$, with largest deviation $0.11$ to $0.045$. Formation at $\varepsilon=0.25$ occurs at $c_0\lambda=2d_s-\rho+3$, i.e. $c_0\lambda_{\rm form}/\rho=79,39,19,9$.
- `check_uniform.py` (check; item 3). On the cubic torus at $L=8,12,16,20$ with $\kappa_a=1$ (Runge–Kutta, $\mathrm d\lambda=0.05$, over $2.2\pi T_U$), plus $L=12$ with a farther anchor and with $\kappa_a=4$, it compares $\bar U$ with (9.14). The relative deviations are $3.2\times10^{-4}$, $1.5\times10^{-4}$, $8.3\times10^{-5}$, $5.5\times10^{-5}$ for $L=8,12,16,20$, $1.3\times10^{-4}$ for the farther anchor and $9.6\times10^{-4}$ for $\kappa_a=4$. Over all six runs, $\nu_0^{-1/2}/\sqrt{L^3c^{(L)}_\phi}$ lies between $1.00001$ and $1.00017$ (from a sparse eigen-solver): $1.00006$ to $1.00001$ for $L=8$ to $20$, $1.00003$ for the farther anchor, $1.00017$ for $\kappa_a=4$. The first minimum lies at $\lambda/(\pi\sqrt{L^3c^{(L)}_\phi})=1.000\pm0.0002$, where $\bar U/\bar U_{\rm st}=2.000$. The place-dependent part of the slow term matches $-(1-\beta_L)g(x-h)$ to $0.03/L^2$ (relative) in the five runs with $\kappa_a=1$ and to $0.05/L^2$ for $\kappa_a=4$. On the ring at $L=128,256,512$ ($d_s=L/4$, exact), it compares $\bar U$ with (9.15). The deviation stays at about 3 in absolute value (relative $30/L$ for $c_0\lambda\le2L$). The first minimum lies at $c_0\lambda/L=1+4.1/L$ with $\bar U/\bar U_{\rm st}=2.000$, and $\bar U(2L/c_0)/\bar U_{\rm st}=0.019$, $0.005$, $0.0013$. On the weak-anchor ring ($\kappa_aL=0.128,0.032$), (9.14) holds to $10^{-4}$. Outputs: `code/v3/output/uniform_check.txt`, `code/v3/output/uniform_U.csv`.

## Methods used

- Heisenberg equations of a quadratic contract; displaced lowest eigenvector; Trotter–Kato convergence of compressions
- Normal modes, rank-one (Sherman–Morrison) updates, secular equation, interlacing
- Duhamel formula, Volterra equation, Laplace transform
- Dirichlet's simultaneous approximation (recurrence), mean values of quasi-periodic functions
- Finite propagation of power series; stationary phase; lattice Green functions; Dirichlet integral
- Mean-value property of harmonic functions; continuum string limit with a reflection amplitude
- Contour shift of entire integrands; stationary phase uniform in a parameter, folds, Airy asymptotics; dispersive bound of the lattice wave equation; Poisson summation
- $\lambda$-means of spectral integrals, Riemann–Lebesgue lemma, convolution bounds (Young)
- Cauchy–Schwarz bounds
- Numerics: eigen-decomposition, fourth-order Runge–Kutta, trapezoid rule for Volterra equations

Author: Ivan Honis · ivan.honis@ndot.io · License: CC BY 4.0 · Human–AI note