Version 1 · earlier version; the current one is v6 · Review of v1

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

- **Subproject:** 08-ilang-propagation
- **Package:** 09-anchor-formation
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-26

## 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 $\delta a$.
- *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.

**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 correlation, in particular $w,\Gamma,\Sigma$ of 05-anchor, is that of $\Omega$ at every $\lambda$.
- $\bar q$ 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 $L$.
- By stationary phase (there is no real stationary point beyond the largest group speed $c_0$), they stay exponentially small up to the wrap time $\simeq(L-\rho)/c_0$.
- 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 above it, localized at $h$: for $D=1$ at every $\kappa_a$, for $D=3$ above a threshold of $\kappa_a$. Its share of $\delta a(x,\cdot)$ is $O(e^{-(d_x+d_s)/\xi})$.

*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 $\rho^{-4/3}$ by stationary phase ($\rho^{-4/3}$ at fold caustics, assumed as in condition (d) of 04-formation).
  - 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 $w\ne1$,

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

### 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 $\kappa_a\delta(\lambda)$ plus an integrable kernel and the localized mode of Step 5. The kernel's singularities on the imaginary axis are band edges and van Hove points. Its total weight is $\chi(0)=\kappa_a/(1+\kappa_a\gamma_3/\kappa_h)=1/c_\phi$.
- $D=1$: $\gamma_s(y)=\gamma_s(0)\zeta_s^{\lvert y\rvert}$ 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 $\lambda\propto d_s$,

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

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 $c_0\lambda=d_s+d_x$.

For $D=3$, the limit $\lambda\to\infty$ in order (O1) is therefore the static field on $\mathbb Z^3$, $-g_\infty(x-x_s)+\beta g_\infty(x-h)$. 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 $1-\beta$ 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 $b\ne1$.
- By (9.4), (9.9) and (9.10), $\rho\,\delta a^\perp\to-\theta(w-1)/4\pi\kappa_h$; the image contributes only $O(\beta\rho/d_x)=O(1/\rho)$.
- By (9.6), $\rho\,\delta a^\perp_{\rm st}\to(\frac1b-1)/4\pi\kappa_h$.

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 middle statement holds because $\lvert\Delta_3^\perp\rvert\to1$ on $w<1$ and $\to\frac1{b-1}<\varepsilon$ uniformly on $w\ge1+\delta$, where the corrections of Step 5 are uniform. As in (4.14), $\lambda_{\rm form}-\rho/c_0=O(\rho^{1/3})$ once the caustic wakes are below $\varepsilon-\frac1{b-1}$.
- *Far from the anchor* ($b\to\infty$ after $\rho\to\infty$, which forces $a\ge b-1\to\infty$; or $d_s,d_x\to\infty$ at fixed $\rho$ first, with the same result): the place-dependent part forms within a $\lambda$ linear in $\rho$ for every $\varepsilon\in(0,1)$. The formation speed is $c_0$, equal to the front speed and to the largest slope of the normal-mode frequency, and independent of $\kappa_a$.
- *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) 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 $\rho^{-0.4}$.

*Shell.* Take order (O1), then $\ell\to\infty$ at fixed $w=c_0\lambda/\ell$ and $a=d_s/\ell>2$.
- In the continuum limit of the sums, $\delta a^\perp\to-\theta(c_0\lambda-\lvert y\rvert)/(4\pi\kappa_h\lvert y\rvert)$, with $y=x-x_s$.
- The shell sum of $1/\lvert x-h\rvert$ is $\ell^3\int\Phi/d_s$, by the mean-value property of $1/\lvert x-h\rvert$ on spheres around $x_s$ that do not reach $h$.
- Define $\Psi(w):=\int_{\lvert z\rvert>w}\Phi\lvert z\rvert^{-1}/\int\Phi\lvert z\rvert^{-1}$ (equal to $1$ for $w\le1$ and to $0$ for $w\ge2$) and $\theta_a:=\int\Phi/(a\int\Phi\lvert z\rvert^{-1})\in(\frac1a,\frac2a)$. The left side of (4.19) divided by the sum on its right side (without $\varepsilon$) tends to $\lvert\Psi(w)-\theta_a\rvert/(1-\theta_a)$.

So

$$
\text{(4.19) for }\delta a^\perp\ \text{holds from } c_0\lambda=w_\varepsilon\ell\ \text{on}\iff\theta_a\le\frac\varepsilon{1+\varepsilon},\qquad w_\varepsilon=\min\bigl\{w:\Psi(w)\le\theta_a+\varepsilon(1-\theta_a)\bigr\}\in[1,2] .
\tag{9.12}
$$

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 $c_0$.
- 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 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) finds the plateaus $-0.500$ and $0.000$, and $c_0\lambda_{\rm form}/\rho\to3.02$ and $1.02$ at $\rho=160$ for $a=2$. 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 $D=3$. (5.15) in its order (a) and the weighing of 05-anchor are statements about the lowest eigenvector on the finite torus, followed by $L\to\infty$. (9.11) is a statement about the evolution in order (O1). They are compatible 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 $\ell$).

In this order:
- (5.15) gives the separation dependence $-g_1g_2/(4\pi\kappa_h\lvert\mathbf r\rvert)$ (its order (a), $\lvert\mathbf r\rvert/D_X\to0$). It comes from the source term $g(x_1-x_2)$ alone.
- (9.11) with $b\to\infty$ gives formation of the place-dependent part at speed $c_0$ for every $\varepsilon$.
- The weighing of 05-anchor holds within (A), or within (B) with $R/d\to0$.

The weighing rests on place differences of $\phi_a$, through $W^{(1)}$ and $W^{\rm mark}$ of (5.18), and on $w,\Gamma,\Sigma$, which belong to $\Omega$ and are unchanged by the evolution (Step 1). The parts of the static made field that do not form in (O1) change these only slightly:
- The missing image $(1-\beta)g(x-h)$ changes a place difference over $\delta$ by $O(\lvert\delta\rvert/d^2)$, against at least $c\lvert\delta\rvert/(4\pi\kappa_hR^2)$ for the formed source term.
- The uniform part enters differences of $W^{\rm mark}$ only through $\bar U_{\rm st}[\Gamma(x)-\Gamma(x')]$, with $\Gamma(x)-\Gamma(x')=r(x,h)-r(x',h)=O(\lvert\delta\rvert/d^2)$.
- Both are relative $O(R^2/d^2)\to0$.

So in (9.16) the long-range mark, the formation within a $\lambda$ linear in the distance and the weighing hold together, at leading order in $\ell/d$.

No such order exists with $d$ comparable to $\ell$: then (9.11) fails for $\varepsilon<1/\lvert b-1\rvert$, (5.15) changes to its orders (b) and (c), and at the mirror placement no grip weighs. 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 $w,\Gamma,\Sigma$ are static.
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)** (formula; order (O1), then the stated distance $\to\infty$ at fixed $w$). 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 image strength tends to $\beta\,\theta(c_0\lambda-d_s)$, with $\beta=g_\infty(x_s-h)/c_\phi$ for $D=3$ and $\beta=1$ for $D=1$. In order (O1) the uniform part and, for $D=3$, an image of strength $1-\beta$ never form. The front speed equals the largest slope $c_0$ of the normal-mode frequency.
5. **(9.11)** (criterion; cubic torus, order (O1), then $\rho\to\infty$ along $\mathbf e_1$ at fixed $w$, $a$, $b\ne1$). $\Delta_3^\perp\to-1$ for $w<1$ and $\to1/(b-1)$ for $w>1$. If $b>1+1/\varepsilon$, then $c_0\lambda_{\rm form}/\rho\to1$: 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 $\varepsilon\in(0,1)$ once $b\to\infty$. If $\lvert b-1\rvert<1/\varepsilon$, there is no formation.
6. **(9.12)** (criterion; cubic torus, order (O1), then $\ell\to\infty$ at fixed $w$ and $a=d_s/\ell>2$). (4.19) for $\delta a^\perp$ holds from $c_0\lambda=w_\varepsilon\ell$ on, with $w_\varepsilon\in[1,2]$, iff $\theta_a\le\varepsilon/(1+\varepsilon)$. 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 $w<1$, $1<w<2a-1$, $w>2a-1$. 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). In the order $L\to\infty$, then $d/\ell\to\infty$, then $\ell\to\infty$ at fixed $w$, three statements hold together at leading order in $\ell/d$: the long-range mark (5.15)(a), the formation of the place-dependent part within a $\lambda$ linear in the distance at speed $c_0$, and the weighing of 05-anchor within (A), or within (B) with $R/d\to0$. No order exists with $d$ comparable to $\ell$, nor 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 lattice corrections in (9.11) and (9.13), near $w=1$ and near $w=2a-1$, and the estimate $\lambda_{\rm form}-\rho/c_0=O(\rho^{1/3})$ rest on the stationary-phase structure of Step 5. For $m_h=0$, fold caustics are assumed as in condition (d) of 04-formation, not proven. 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 $c_0\lambda\approx\rho$.
- 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. The carry-over in Step 10 assumes that the weighing rests only on place differences of $\phi_a$ and on $w,\Gamma,\Sigma$.
- 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/v1/`. Each script runs from the project root with `.venv/Scripts/python.exe` and writes to `code/v1/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/v1/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/v1/output/formation_check.txt`, `code/v1/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.
  - *Formation times:* 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 $\rho=3,\dots,16$. Where $1/(b-1)\ge0.5$ (near anchor at $\rho\ge4$; far anchor at $L=64$, $\rho=16$, $b=3$), there is no formation, as (9.11) predicts. 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.
    - (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 $L=8,12,16,20$, and $9.6\times10^{-4}$ for $\kappa_a=4$. $\nu_0^{-1/2}/T_U=1.00006$ to $1.00001$ (from a sparse eigen-solver), and the first minimum lies at $\lambda/(\pi T_U)=1.000\pm0.0002$, 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). 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/v1/output/uniform_check.txt`, `code/v1/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
- 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