==== TABLE OF CONTENTS ====
Incompressible-Box-Transport-and-Finite-Computation//analysis.tex
Incompressible-Box-Transport-and-Finite-Computation//balanced.tex
Incompressible-Box-Transport-and-Finite-Computation//clocks.tex
Incompressible-Box-Transport-and-Finite-Computation//coding-conventions.tex
Incompressible-Box-Transport-and-Finite-Computation//compact-processors.tex
Incompressible-Box-Transport-and-Finite-Computation//compilers.tex
Incompressible-Box-Transport-and-Finite-Computation//geometry.tex
Incompressible-Box-Transport-and-Finite-Computation//guarded-history.tex
Incompressible-Box-Transport-and-Finite-Computation//initialization.tex
Incompressible-Box-Transport-and-Finite-Computation//introduction.tex
Incompressible-Box-Transport-and-Finite-Computation//main.tex
Incompressible-Box-Transport-and-Finite-Computation//observer-routing.tex
Incompressible-Box-Transport-and-Finite-Computation//persistent-history.tex
Incompressible-Box-Transport-and-Finite-Computation//routing-conventions.tex
Incompressible-Box-Transport-and-Finite-Computation//solid-routing.tex
Incompressible-Box-Transport-and-Finite-Computation//torus-routing.tex
Incompressible-Box-Transport-and-Finite-Computation//references.bib
===========================
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//analysis.tex ====
\section{Analytic classes and changes of physical coordinates}\label{sec:model}
The geometric constructions prescribe a smooth velocity first. This section
records exactly which other solutions are excluded by the energy argument.
Pressure assumptions are kept separate: continuity in a pressure norm, a
norm at each individual time, and an integrable pressure gradient are
different conditions.
Write $\mathcal F_\nu[U]=U_t+(U\cdot\nabla)U-\nu\Delta U$.
All noncompact results below use $\mathbb R^3$; their velocities and forces
have one fixed compact spatial support unless the individual statement
says otherwise. This property concerns the constructed solution, not a
competing solution. On $[0,T]$, write $CH^k=C([0,T];H^k(\mathbb R^3))$
and $C^1L^2=C^1([0,T];L^2(\mathbb R^3))$. Spatial constants in pressure
may depend on time.
\begin{lemma}[The pressure term without a pressure norm]\label{lem:b-gradient}
If $w,g\in L^2(\mathbb R^3;\mathbb R^3)$ satisfy
$\operatorname{div}w=0$ and $\operatorname{curl}g=0$ in distributions, then
$\langle w,g\rangle_{L^2}=0$. In particular, any distributional gradient
$g=\nabla p\in L^2$ has zero pairing with $w$, without an assumption that
$p\in L^2$.
\end{lemma}
\begin{proof}
An $L^2$ field is a tempered distribution. The differential identities,
initially tested on compactly supported smooth functions, extend to
Schwartz tests by multiplying by cutoffs tending to one and using $L^2$
convergence of the tests and their first derivatives. Fourier transformation
gives $\xi\cdot\widehat w=0$ and
$\xi_j\widehat g_k-\xi_k\widehat g_j=0$. These are identities of locally
integrable functions, so they hold almost everywhere. Away from $\xi=0$,
$\widehat g$ is parallel to $\xi$ and $\widehat w$ is perpendicular to it.
Their Hermitian product is zero. The exceptional point has measure zero;
Plancherel proves the assertion.
\end{proof}
\begin{lemma}[Residual realization and separate comparison classes]
\label{lem:b-comparison}\label{lem:p2-realization}
Let $\nu>0$ and let $U$ be a prescribed smooth divergence-free velocity
on $[0,\infty)\times\mathbb R^3$, with $U(0)=0$. On each finite interval,
assume $U\in CH^2\cap C^1L^2$ and $U,\nabla U$ are bounded. Put
$f=\mathcal F_\nu[U]$. Then $(U,0)$ is a solution. Its velocity is unique
in each of the following separately stated comparison classes, with the
same force and initial data:
\begin{enumerate}[label=(\roman*),itemsep=2pt]
\item Smooth classical $v\in CH^2\cap C^1L^2$, with bounded $v$, and a
pressure representative $p\in CH^1$.
\item Smooth classical $v\in CH^2\cap C^1L^2$, with bounded $v,\nabla v$,
and an $H^1$ pressure representative at each time, with no required time
continuity in that pressure norm.
\item Strong solutions with $v\in CH^4\cap C^1H^2$ and distributional
$\nabla p\in CL^2$.
\item Smooth classical $v\in CH^2\cap C^1L^2$, with bounded $v,\nabla v$,
and a pressure representative $p\in CL^2$.
\item Smooth classical $v\in CH^2\cap C^1L^2$, with bounded $v,\nabla v$,
without an additional integrability or normalization condition on pressure.
\end{enumerate}
The pressure gradient is unique. An $L^2$ pressure representative, when
specified, is zero; otherwise pressure is determined up to a function of
time. The reference material flow is a smooth diffeomorphism at every
finite time and exists for every finite forward time.
On a flat torus $\T_L^3$, the same conclusion holds in the classical class
where $v,v_t$, spatial derivatives through order two, and $p,\nabla p$ are
continuous on finite closed cylinders, with periodic mean-zero pressure.
No Sobolev conditions at infinity are involved in that assertion.
\end{lemma}
\begin{proof}
Substitution proves existence of the prescribed solution. Set $w=v-U$.
The difference equation is
\[
w_t+(v\cdot\nabla)w+(w\cdot\nabla)U
=\nu\Delta w-\nabla p,\qquad \operatorname{div}w=0.
\]
The stated velocity regularity implies $w\in C^1L^2\cap CH^2$, so
$\frac d{dt}\|w\|_2^2=2\langle w,w_t\rangle$ at each time. In case
(iii), the stronger assumptions imply these inclusions. They also imply
bounded velocity and spatial derivatives through order two: for $j\le2$,
Cauchy--Schwarz in Fourier space uses the finite integral
$\int_{\mathbb R^3}|\xi|^{2j}(1+|\xi|^2)^{-4}\,d\xi$.
For the transport term, insert a cutoff $\chi_R$ that equals one on the
ball of radius $R$, is supported in the ball of radius $2R$, and satisfies
$|\nabla\chi_R|\le C/R$. Its boundary error is at most
$CR^{-1}\|v\|_\infty\|w\|_2^2$, which tends to zero. Diffusion is
integrated by parts in $H^2$; alternatively its cutoff error is bounded by
$CR^{-1}\|w\|_2\|\nabla w\|_2$. The deformation term is bounded by
$\|\nabla U\|_{\infty,\mathrm{op}}\|w\|_2^2$.
We justify the pressure pairing according to the stated alternatives.
For (i) and (ii), at each fixed time approximate the chosen $H^1$ pressure
by compactly supported smooth functions; distributional incompressibility
and $w\in L^2$ give $\langle\nabla p,w\rangle=0$. Thus no continuity
of $p$ in time is used in (ii). For (iv), integration by parts on the
cutoff gives a pressure boundary term bounded by
$CR^{-1}\|p\|_2\|w\|_2$, also tending to zero. For (iii), apply
Lemma~\ref{lem:b-gradient} directly. Finally, in (v) the equation itself
gives at each time
\[
\nabla p=f-v_t-(v\cdot\nabla)v+\nu\Delta v\in L^2,
\qquad \|(v\cdot\nabla)v\|_2\le\|v\|_\infty\|\nabla v\|_2.
\]
Here $f\in L^2$ follows from the identical estimate for $U$ and its stated
time and space regularity. The smooth pressure gradient is curl free, so
Lemma~\ref{lem:b-gradient} again cancels the term. This does not choose an
$L^2$ representative of the pressure.
In every case the resulting pointwise energy inequality is
\[
\frac12\frac d{dt}\|w\|_2^2+\nu\|\nabla w\|_2^2
\le \|\nabla U\|_{\infty,\mathrm{op}}\|w\|_2^2.
\]
The coefficient is bounded on each finite interval and $w(0)=0$, so an
integrating factor gives $w=0$. The pressure has already disappeared before
time integration; the proof imposes no unlisted temporal pressure norm.
Substitution then gives $\nabla p=0$. A spatially constant $L^2$ function
on $\mathbb R^3$ is zero, which proves the normalization assertion.
On the torus every integration by parts is periodic and has no boundary
term. The specified classical regularity justifies norm differentiation,
diffusion, and pressure cancellation. The same energy inequality gives
uniqueness and the mean-zero normalization fixes pressure. In either domain,
the smooth reference velocity and its bounded spatial derivative give
unique material trajectories; bounded speed prevents finite-time escape.
Solving backwards on a finite interval gives the inverse flow, and smooth
ODE dependence gives its smoothness.
\end{proof}
Cases (i)--(v) identify the precise hypotheses used by the applications;
their presence in a common lemma does not replace a theorem's declared
comparison class by a different one. In particular, a pressure norm is not
silently imposed in a gradient-only application. The proof is the classical
energy comparison for prescribed smooth solutions \cite[Section~18]{Leray1934},
with its noncompact pressure pairings made explicit.
For the flow $\Phi_t$, differentiation of its trajectory equation gives
\[
U(t)=(\partial_t\Phi_t)\circ\Phi_t^{-1},\qquad
\partial_{tt}\Phi_t(a)=
(f-\nabla p+\nu\Delta U)(t,\Phi_t(a)).
\]
These identities describe the material form of the same constructed solution;
they do not add an independent existence claim.
\begin{lemma}[Physical scaling]\label{lem:p2-chart}
For $a,b>0$, $x=x_0+ay$, and $s=bt$, let
$U(t,x)=abV(bt,(x-x_0)/a)$. Then trajectories are transformed by this change
of variables, incompressibility is preserved, and at the prescribed physical
viscosity $\nu$,
\[
\mathcal F_\nu[U](t,x)=ab^2\left[
V_s+(V\cdot\nabla_y)V-\frac{\nu}{a^2b}\Delta_yV
\right](bt,(x-x_0)/a).
\]
\end{lemma}
\begin{proof}
The trajectory equation gives $\dot x=abV$; differentiating in space gives
$\operatorname{div}_xU=b\operatorname{div}_yV$. The three residual terms
scale by $ab^2,ab^2$, and $b/a$, respectively, proving the formula.
\end{proof}
Thus a side-$10$ torus is kept as a physical side-$10$ torus. If a chart
is rescaled, the residual is recomputed with the fixed $\nu$; the equation
is not identified with a differently normalized viscosity.
\begin{proposition}[Projection on a flat torus]\label{prop:projection-l2}
An effective smooth mean-zero force $g$ on $\T_L^3$ has an effective
mean-zero potential $\phi$ satisfying $\Delta\phi=\operatorname{div}g$.
Replacing $(g,p)$ by $(g-\nabla\phi,p-\phi)$ preserves velocity and makes
the force solenoidal. Uniform mixed-derivative bounds, temporal $L^2$
bounds of their spatial suprema, and separately imposed time periodicity,
eventual stationarity, or orderwise decay bounds are preserved.
\end{proposition}
\begin{proof}
In frequencies $\xi_k=2\pi k/L$, set
$\widehat\phi(k)=-i\xi_k\cdot\widehat g(k)/|\xi_k|^2$ for $k\ne0$
and zero for $k=0$. The multiplier from $g$ to $\nabla\phi$ is
$\xi_k\xi_k^{\mathsf T}/|\xi_k|^2$. For a spatial derivative order $r$,
integrate each Fourier coefficient $r+5$ times in a coordinate with largest
$|k_i|$. There are $O(n^2)$ lattice points on the shell $\|k\|_\infty=n$,
so the derivative series and its tail converge absolutely, with
\[
\|\partial_t^h\partial_x^\alpha\nabla\phi\|_\infty
\le C_{\alpha,L}\max_i
\|\partial_t^h\partial_{x_i}^{|\alpha|+5}g\|_\infty.
\]
This bound is pointwise in time, and proves each asserted norm or weighted
time estimate. Derivative bounds give effective Fourier tails; coefficient
integrals can be computed by Riemann sums with derivative error estimates.
The formula gives the Poisson identity, divergence cancellation, and the
pressure sign by substitution. Linearity preserves the time symmetries.
\end{proof}
Projection generally changes pressure. It supplies no compact-support
conclusion on Euclidean space and is distinct from the direct zero-pressure
solenoidal shear construction.
\paragraph{Effective flat profiles.}
For later cutoff formulas put
\begin{equation}\label{eq:p2-step}
\rho(s)=\begin{cases}e^{-1/s},&s>0,\\0,&s\le0,\end{cases}
\qquad \sigma(s)=\frac{\rho(s)}{\rho(s)+\rho(1-s)}.
\end{equation}
Positive-side derivatives of $\rho$ are polynomials in $1/s$ times
$e^{-1/s}$ and extend flatly at zero. The estimate
$v^me^{-v}\le(m+k)!v^{-k}$ gives effective error bounds near the seam,
and $\rho(s)+\rho(1-s)\ge e^{-2}$. Products, translations, rescalings,
and derivatives therefore yield effective cutoffs with all requested
derivative bounds. Periodization uses zero collars; at approximate real
arguments a finite superset of possible translates is evaluated. No real
equality test at a seam or execution of the encoded computation is used.
==== END Incompressible-Box-Transport-and-Finite-Computation//analysis.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//balanced.tex ====
\section{A complete balanced three-stack realization}\label{sec:bb-balanced-three-stack}
In a positional code with base $B$, replacing a prefix of length $r$
by one of length $s$ gives an affine map with slope $B^{r-s}$.
Thus a three-stack instruction has determinant one when it preserves
the sum of the three prefix lengths. A third stack supplies reversible
history under this constraint: we first move a fresh blank from beyond
a work-tape boundary onto the history stack, then replace that blank
by the record.
\begin{theorem}[Balanced three-stack realization]
\label{thm:bb-balanced-three-stack}
For every machine and finite input, there are effective coefficients
$f_0,f_1$ such that $f=f_0+\nu f_1$ on $\R^3$ has an explicit
zero-data solution $(U,0)$. Force and velocity have one compact
spatial support and bounded mixed derivatives of every order, and
are one-periodic after $t=1$. At the fixed label $x_*=(4,0,0)$,
\[
\exists t\ge0:\quad X(t;x_*)\in(-1,2)^3
\quad\Longleftrightarrow\quad\text{the machine halts}.
\]
For a fixed machine its repeated field is independent of the input.
The comparison class is $u\in C H^2\cap C^1L^2$, bounded $u,\nabla u$
on finite intervals, and pressure modulo constants in $C H^1$.
The formula is valid for every fixed real $\nu>0$, with effective
evaluation when $\nu$ is computable and relative evaluation otherwise.
\end{theorem}
\begin{proof}
Normalize the finite machine to one halting state $h$, directing missing
nonhalting instructions to it by unchanged-letter stay moves. This preserves
initial halting as well. Let $\Gamma$ be the work alphabet and
$b\in\Gamma$ its blank. Use three infinite top-first stacks $L,R,J$. A rule replaces three
finite prefixes, leaving their tails untouched. Introduce boundary
$\star$, marked work letters $\bar a$, and controls $A_q,S_q,T_q,D_q$;
only $A_h$ is terminal. For each $q\ne h$, the preparatory rules are
\begin{align}
(A_q;\varepsilon,a,\varepsilon)&\to(S_q;\bar a,\varepsilon,\varepsilon),
\label{eq:bb-balanced-entry}\\
(S_q;\varepsilon,c,\varepsilon)&\to(S_q;c,\varepsilon,\varepsilon),
\label{eq:bb-balanced-out-sweep}\\
(S_q;\varepsilon,\star b,\varepsilon)&\to(T_q;\varepsilon,\star,b),
\label{eq:bb-balanced-blank}\\
(T_q;c,\varepsilon,\varepsilon)&\to(T_q;\varepsilon,c,\varepsilon),
\label{eq:bb-balanced-return-sweep}\\
(T_q;\bar a,\varepsilon,\varepsilon)&\to(D_q;\varepsilon,a,\varepsilon).
\label{eq:bb-balanced-restore}
\end{align}
Here $a,c\in\Gamma$. For $\delta(q,a)=(q',d,\sigma)$ add
\begin{equation}\label{eq:bb-balanced-step-table}
\begin{array}{c|c|c}
\sigma&\text{source}&\text{target}\\\hline
0&(D_q;\varepsilon,a,b)&(A_{q'};\varepsilon,d,\kappa)\\
-1&(D_q;c,a,b)&(A_{q'};\varepsilon,cd,\kappa)\quad(c\in\Gamma)\\
1&(D_q;\varepsilon,ac,b)&(A_{q'};d,c,\kappa)\quad(c\in\Gamma)\\
1&(D_q;\varepsilon,a\star b,b)&(A_{q'};d,b\star,\kappa).
\end{array}
\end{equation}
Every individual row occurrence, including its choice of $c$, receives
a distinct record symbol $\kappa$.
\emph{Full-domain separation and initialized behavior.}
\label{lem:bb-balanced-processor}\label{proof:bb-balanced-processor}
Every rule preserves total prefix length: it is one in the transfer
rows, two in the blank-supply row, and respectively two, three, three,
four in the final rows. Source branches at $A_q$ test the right top;
at $S_q$ the right top separates work letters from the boundary;
at $T_q$ the left top separates work and marked letters. At $D_q$
the right top fixes the work instruction, with the next right symbol
or left top separating its remaining branches. Target branches at
$S_q$ have distinct marked or ordinary left tops; at $T_q$ distinct
boundary or ordinary right tops; at $D_q$ distinct restored $a$;
and at an $A$ control distinct history tops $\kappa$.
This proves separation for arbitrary independent tails.
Initialize
\[
(A_{q_0};b^\infty,v\star b^\infty,b^\infty),\qquad
v=\begin{cases}\omega,&\omega\ne\varepsilon,\\b,&\omega=\varepsilon.
\end{cases}
\]
At a checkpoint, $L$ lists cells strictly left of the head and
$R=a_1\cdots a_m\star b^\infty$, $m\ge1$, lists the current cell
and a finite right block; beyond it the tape is blank. $J$ contains
completed records, newest first. The first rule marks $a_1$ while
moving it left; $m-1$ forward transfers expose $\star$. The blank
row removes one filler blank beyond that boundary and pushes it on
$J$, without changing the infinite blank filler. The reverse transfers
and restoration of $a_1$ restore exactly the original two work stacks.
The final rule replaces the fresh history blank by its record and
performs the work instruction. A left move pops the next cell from
$L$; a right move with $m\ge2$ pops the current cell from $R$;
when $m=1$ the last row exposes a blank and retains a nonempty right
block. Thus the invariant holds and one work step costs
$1+(m-1)+1+(m-1)+1+1=2m+2$ rules. An initially halted input needs
none. The records recover the head displacements as well.
\emph{From balanced words to solid boxes.}
Give the enlarged alphabet $\Sigma$ the odd digits
$1,3,\ldots,2|\Sigma|-1$ in base $B=2|\Sigma|+1$.
For a finite word $v$ put $e(v)=\sum_{j=1}^{|v|}d(v_j)B^{-j}$;
for an infinite stack $\xi$ put
$e(\xi)=\sum_{j\ge1}d(\xi_j)B^{-j}$.
The prefix interval is $I(v)=e(v)+[0,B^{-|v|}]$, with
$I(\varepsilon)=[0,1]$. Every infinite tail satisfies
\[
\frac1{B-1}\le e(\xi)\le\frac{2|\Sigma|-1}{B-1}<1,
\]
so its code lies strictly inside each applicable prefix interval.
Distinct incompatible prefixes have positive gaps, since digits differ
by at least two while the remaining interval has width at most one unit
in the differing place. Place control $s$ at
$o_s=(4i_s,0,0)$, with $i_{A_h}=0$ and distinct positive integers
for the others. The code of $(s;L,R,J)$ is
$o_s+(e(L),e(R),e(J))$.
A rule with input prefixes $\alpha_1,\alpha_2,\alpha_3$
and output prefixes $\alpha'_1,\alpha'_2,\alpha'_3$ gives boxes
\[
Q_i=o_s+\prod_jI(\alpha_j),\qquad
Q_i'=o_{s'}+\prod_jI(\alpha'_j).
\]
Both families are separately disjoint by the preceding full-prefix
separation. In control-relative coordinates their exact maps are
\begin{equation}\label{eq:bb-balanced-affine-rule}
y_j=e(\alpha'_j)+B^{|\alpha_j|-|\alpha'_j|}
(x_j-e(\alpha_j)).
\end{equation}
The identity $e(\alpha\xi)=e(\alpha)+B^{-|\alpha|}e(\xi)$
proves their action on every code. Initialized stacks are eventually
blank, with rational codes
$e(vb^\infty)=e(v)+B^{-|v|}d(b)/(B-1)$.
The affine determinant equals
$B^{\sum|\alpha_j|-\sum|\alpha'_j|}=1$, exactly by balance.
Apply Theorem~\ref{lem:bb-balanced-box-routing} to these full boxes.
\emph{Every intermediate time.}
All half-widths are at most $1/2$, so that lemma uses $R=1$.
Sources have first coordinate at least four, since no rule leaves
$A_h$. Targets with nonhalt control have the same property.
For such a branch the initial expansion and final contraction preserve
first coordinate at least four. During deformation its center has
first coordinate at least $4\Lambda$. Every scaled halt target
center has first coordinate at most $\Lambda$, whereas the endpoints
of each nonhalt parking path have first coordinate at least
$4\Lambda$. Since $\Lambda+4<4\Lambda$, the original segment
cannot meet an obstacle cube around a halt target. Any encountered
obstacle has first coordinate at least $4\Lambda$, so its boundary
detour remains at least $4\Lambda-4$. Subtracting a half-width at
most one still leaves first coordinate greater than two. Hence
\begin{equation}\label{eq:bb-balanced-no-premature-event}
\Phi_\tau(Q_i)\subset\{x_1>2\}\quad(0\le\tau\le1)
\quad\text{for every branch with nonhalt target}.
\end{equation}
This includes delivery after halt targets have already been filled.
Finally load $x_*=(4,0,0)$ along the straight segment to the rational
initial code using the translation part of Lemma~\ref{lem:bb-curl}.
It avoids $(-1,2)^3$ for a nonhalt initial control. Repeat the box
field from time one. Induction at integer times applies the exact
full-box maps; the finite compiler cycles reach a code in $(0,1)^3$
precisely upon halting. Equation~\eqref{eq:bb-balanced-no-premature-event}
excludes every intermediate false visit. The residual proposition
gives the force and uniqueness; compact support gives all finite-slab
Sobolev conditions and uniformly bounded kinetic energy. Every choice
was finite and effective, including the obstacle detours. The input
enters only the first loading path, completing the proof.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//balanced.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//clocks.tex ====
\section{Clock profiles for selected processors}\label{sec:clocks}
Lemma~\ref{ba:log-clock} gives the common logarithmic change of time.
We now apply it to two Euclidean processors with different pressure
classes, and then use a companion's autonomous spatial clock to obtain
stationary or decaying forcing on the torus. In every case the clock
traverses every finite computational time; periodic, stationary and
decaying profiles remain separate choices.
\subsection{Preserving the declared comparison classes}
\label{sec:bb-late-logclock}
The two templates below use different whole-space pressure assumptions.
Slowing their trajectories preserves those declared comparison classes
because the new reference velocity retains the required regularity on
every finite physical-time interval.
\begin{proposition}\label{prop:bb-late-logclock}
The template of Theorem~\ref{thm:bb-direction-shuttle} admits a force
in $L^2([0,\infty)\times\R^3)$ with the same material label and
event. The template of Theorem~\ref{thm:bb-guarded-history} admits
a force in $L^2([0,\infty);H^j(\R^3))$ for every integer $j\ge0$.
Both retain uniform compact support, zero initial velocity, bounded
mixed derivatives, their stated pressure and velocity comparison
classes, and finite effective prescriptions.
\end{proposition}
\begin{proof}\label{proof:bb-late-logclock}
Apply Lemma~\ref{ba:log-clock} to each template, including its loading
interval if present. Both are initially zero, have one compact support,
and have uniformly bounded mixed derivatives. The lemma gives the exact
residual and all mixed bounds; fixed support also gives
$\|f(t)\|_{H^j}\le C_j(1+t)^{-1}$ for every fixed $j$.
On each finite interval the slowed reference field retains every Sobolev
regularity and bounded derivative required by its original comparison
class. Case (i) of Lemma~\ref{lem:b-comparison} therefore applies to the
direction shuttle's continuous $H^1$ pressure representative, and case
(iii) applies to the guarded-history processor's continuous $L^2$
pressure gradient. Both constructed pressures are zero.
The new path is $X(\log(1+t);a_0)$. The clock lemma preserves the event
and the effective prescription. In particular, loaded samples $1+k$
occur at physical times $e^{1+k}-1$, whereas the aligned construction's
samples $k$ occur at $e^k-1$. These times have no finite accumulation.
The new forces have the stated temporal integrability; no periodicity
in physical time is asserted for them.
\end{proof}
\subsection{The input-offset alternative}\label{bc:input-offset-section}
A fixed material label can also be retained by shifting the state charts
by the initial tape code. The complete planar routing and spatial-clock
construction for this choice belongs to
\cite[Theorem~4.1 and Section~5]{OpenAI379StationaryA2}.
The following application fixes its match with the present conventions.
\begin{corollary}[Two profiles of the input-offset processor]
\label{bc:input-offset-event}
On $\T^3$, the fixed label $P=(1/4,1/4,0)$ and strip
$O=\{1/32h_n$. Mark the old head, scan right to $g$,
write the record and install the new frontier at $g+1$, then return to
the mark. All intervening history is nonfrontier, the new frontier
cell is empty, and there is only one mark. Hence the same rows apply
when $r_0=2$. There are $g-h_n-1$ continuing right moves,
$g-h_n$ continuing left moves, and four other instructions, giving
$2(g-h_n)+3$. The last row clears the mark and makes the work
movement; the invariant is restored with frontier $g+1$. This proves
finite completion, indefinite legality for a nonhalting run, and the
stated halt equivalence, including initial halting.
\end{proof}
\subsection{From symbolic cylinders to closed rectangles}
Radix encodings turn a push or pop of a tape letter into an affine map;
this is the generalized-shift viewpoint of
Moore~\cite{Moore1990,Moore1991}. The following calculation adds the
separation of full closed rectangles needed for smooth localization.
\begin{lemma}[Separated rectangle codes]\label{ba:rectangles}
Let a deterministic partial tape table on an alphabet $\Gamma$ of size
$m$ have the two incoming properties of Lemma~\ref{lem:bb-onecell}.
Assign distinct odd digits $e(a)\in\{1,3,\ldots,2m-1\}$ and choose
an integer $B\ge2m+1$. Its local rules admit effective positive diagonal
affine maps of determinant one between finite families of closed planar
rectangles. Sources are pairwise separated, as are targets. Both the
minimal subdivision and the subdivision by both adjacent letters given
below realize every tape in their full symbolic cylinders.
\end{lemma}
\begin{proof}\label{ba:rectangles-proof}
Use Lemma~\ref{lem:bb-onecell} with digits $d(a)=e(a)$. Its digit
hypotheses hold for every $B\ge2m+1$. In the notation
\[
C(a_1a_2\cdots)=\sum_{j\ge1}e(a_j)B^{-j},\quad
h_a(v)=\frac{e(a)+v}{B},\quad
I_{a_1\cdots a_l}=h_{a_1}\cdots h_{a_l}([0,1]),
\]
the left stream starts immediately left of the head and the right stream
starts at the head. Place state $s$ in $k_s+a[0,1]^2$ with a common
rational $a>0$. The minimal branch maps are exactly
\eqref{eq:bb-one-cell-branches}; scaling both charts by $a$ leaves
$\operatorname{diag}(B^{-d},B^d)$ unchanged.
For the additional full subdivision, restrict the right- and stay-move
sources to $x\in I_\gamma$, for every full letter $\gamma$.
Their targets become $I_{\beta\gamma}\times[0,1]$ and
$I_\gamma\times I_\beta$, respectively, for written letter $\beta$.
The already split left-move targets remain
$[0,1]\times I_{\gamma\beta}$. At a fixed target state the
incoming direction is fixed, and the pair $(\beta,\gamma)$
distinguishes the branches in every case. Thus the added subdivision
preserves separate target separation, while source separation follows
by restriction of the shared source rectangles. All these statements
concern the filled closed rectangles and arbitrary symbolic tails.
For completeness the quantitative gaps remain valid for the entire
range $B\ge2m+1$: distinct first-digit intervals have gap at least
$B^{-1}$ and distinct second-digit intervals at least $B^{-2}$.
The source and target families therefore have within-state gaps at
least $aB^{-2}$; different state squares have their own chosen gaps.
Tail values lie between $1/(B-1)$ and $(2m-1)/(B-1)<1$.
An initial constant tail after $l$ digits contributes
$B^{-l}e(a_{\rm tail})/(B-1)$, so initialized codes are rational.
At checkpoints the history records, or the known absolute origin of
the record block, also recover the absolute work-head position.
\end{proof}
\subsection{Conventions for the remaining geometric realizations}\label{bc:section}
\label{bc:frontier-section}
Unless a different table is specified, subsequent routes use the move-first recorder of
Lemma~\ref{lem:bb-movefirst}, with initial frontier two. The controls
$(C_q,J_r,S_r,P_q,T_q)$ there are written $(W_q,P_r,B_r,C_q,D_q)$
here, and its empty/frontier/record symbols $(e,F,r)$ are written
$(E,F,R_r)$. Permuting work, history and mark coordinates in every row
preserves every guard. This identifies the full partial table, its inverse
and the exact checkpoint count $2(2+n-h')+4$, including arbitrary allowed
tapes. A nonterminal branch means one with both endpoint controls
nonterminal; a final branch into a halt can enter the observer.
The full branches are those of Lemma~\ref{lem:bb-onecell}. For an alphabet
of size $m$ we use one of the following explicitly indicated conventions:
\begin{equation}\label{bc:digits}
(B,\{d_a\})=(2m+2,\{1,3,\ldots,2m-1\}),\quad
(2m+1,\{1,3,\ldots,2m-1\}),\quad
(2m,\{0,2,\ldots,2m-2\}).
\end{equation}
All satisfy its digit bound and separation hypotheses. In the last
convention the full blank has digit zero; boundary codes remain in the
closed-rectangle and neighborhood conclusions. Write
$C(a_1a_2\cdots)=\sum_{j\ge1}d_{a_j}B^{-j}$ and
$I_v=\sum_{j=1}^{|v|}d_{v_j}B^{-j}+[0,B^{-|v|}]$.
A constant tail after $k$ positions contributes
$d_{\rm tail}B^{-k}/(B-1)$, so the initial codes are rational.
Guarded three-cell rules use Lemma~\ref{lem:bb-window-rectangles} with
$(a,b)=(1,2)$ after their own full-domain inverse has been proved.
\subsection{Analytic conventions for the geometric variants}\label{bc:analytic-section}
The whole-space baseline in these variants means smooth
$u\in CH^2\cap C^1L^2$, bounded $u,\nabla u$ on finite intervals,
and a pressure representative in $CH^1$. When the competitor-gradient
condition is omitted explicitly, the other conditions remain unchanged.
Both cases fall under Lemma~\ref{lem:b-comparison}(i); separately stated
pressure classes are retained. The torus baseline is its classical torus
clause. Unless a projection is specified, the force is
\begin{equation}\label{bc:residual}
f_U=U_t+(U\cdot\nabla)U-\nu\Delta U,\qquad (u,p)=(U,0).
\end{equation}
Torus projection is the separate choice in
Proposition~\ref{prop:projection-l2}, with changed pressure.
Localized affine motion is supplied by Lemma~\ref{lem:bb-curl} on a
compact time interval, using effective derivative bounds, positive lower
scale bounds and the positive cutoff margins specified in each route.
The potential is \eqref{eq:bb-potential}; the resulting flow fixes every
region disjoint from its support. In particular exponential diagonal
scales and linear reciprocal scales have respectively generators
$\theta'\log D$ and $\diag(l'/l,-l'/l,0)$.
\begin{lemma}[Finite forcing mechanisms]\label{bc:analytic}
A finite family of the preceding pulses, joined with flat time collars,
can be used once for initialization and repeated with period one thereafter.
It gives a smooth effective velocity and residual force with all mixed
derivatives uniformly bounded. In $\R^3$ both have one fixed compact
spatial support, and the velocity has uniformly bounded kinetic energy.
In a torus chart,
periodized curl or planar Hamiltonian pulses have zero spatial mean, as
does their residual force. Initial velocity and pressure are zero.
Uniqueness holds in the comparison classes of Section~\ref{sec:model};
in the whole-space class the competitor's gradient bound can be omitted
if the remaining conditions are retained.
\end{lemma}
\begin{proof}\label{bc:analytic-proof}
Each pulse is a finite expression in rational data, effective smooth
switches, their derivatives, positive diagonal scales and logarithms.
Finite positive separation margins bound every cutoff derivative. A finite
maximum bounds a period and the loading interval, and repetitions preserve
that bound. To evaluate near a join, a coarse enclosure of time gives a
finite superset of possibly active pulses. Summing them avoids an exact
comparison of a computable time with an integer. The flat collars give
smoothness. Compact support gives every required spatial Sobolev bound.
Lemma~\ref{lem:p2-realization} gives the asserted solution and comparison;
on a torus the integral of each curl or Hamiltonian derivative vanishes,
and the convection term is a divergence. Its flow satisfies
$\det D\Phi_t=1$, because differentiating this determinant gives
$(\divg U)(t,\Phi_t)\det D\Phi_t=0$ and its initial value is one.
The velocity and acceleration identities in that lemma consequently apply
to these volume-preserving diffeomorphisms.
\end{proof}
For later use, spatial chart changes are always made at the velocity level.
If $x=o+a y$, set $\widetilde U(t,x)=aU(t,(x-o)/a)$ and then compute
\eqref{bc:residual} at the prescribed physical $\nu$. The time and convection
terms scale by $a$, while the Laplacian term scales by $a^{-1}$.
Thus chart rescaling does not silently change the fixed physical viscosity.
==== END Incompressible-Box-Transport-and-Finite-Computation//coding-conventions.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//compact-processors.tex ====
\section{Compact support and onto slow clocks}\label{ba:euclidean}
A whole-space processor can use arbitrarily separated finite lanes.
We shall construct a periodic cycle with fixed compact support and also
give a separate decaying force with the same material event. We first
establish the time change used for the latter choice. Its clock must tend
to infinity: traversing only a finite amount of machine time would not
preserve the halting event.
\subsection{The logarithmic clock at every derivative order}
\begin{lemma}[Logarithmic clock with full derivative bounds]
\label{ba:log-clock}
Let $W(s,x)$ be a smooth solenoidal field for $s\ge0$, initially zero,
with every mixed derivative bounded and support in one compact set $K$.
For fixed computable $\nu>0$, set
\[
s(t)=\log(1+t),\quad \rho(t)=(1+t)^{-1},\quad
U(t,x)=\rho(t)W(s(t),x),\quad
F=U_t+(U\cdot\nabla)U-\nu\Delta U.
\]
Every derivative $\partial_t^k\partial_x^\alpha U$ and
$\partial_t^k\partial_x^\alpha F$ is uniformly
$O((1+t)^{-1-k})$ and belongs to space-time $L^2$.
Both supports lie in $K$. If $\Xi$ is the material flow of $W$, that
of $U$ is $\Xi(s(t),a)$, so every fixed spatial reachability event is
unchanged. Effective finite formulas for $W$ give effective formulas
for $U,F$ and all derivatives.
\end{lemma}
\begin{proof}\label{ba:log-clock-proof}
Since $s'=\rho$ and $\rho'=-\rho^2$,
\begin{equation}\label{ba:clock-rule}
\partial_t\{\rho^jY(s(t),x)\}
=\rho^{j+1}(\partial_s-j)Y(s(t),x).
\end{equation}
For $P_{k,j}(D)=\prod_{r=0}^{k-1}(D-j-r)$, with $P_{0,j}=1$,
induction yields
\begin{equation}\label{ba:clock-induction}
\partial_t^k\{\rho^jY(s(t),x)\}
=\rho^{j+k}P_{k,j}(\partial_s)Y(s(t),x).
\end{equation}
Spatial differentiation commutes with this identity. Taking $j=1$
proves the velocity bound at every order. The residual is exactly
\begin{equation}\label{ba:clock-residual}
F=\rho^2\{W_s-W+(W\cdot\nabla)W\}(s(t),x)
-\nu\rho\Delta W(s(t),x).
\end{equation}
Every mixed derivative of the braces is bounded by the finite product
rule and the hypotheses. Apply \eqref{ba:clock-induction} with $j=2$
and $j=1$ to obtain orders $\rho^{2+k}$ and $\rho^{1+k}$,
respectively. This gives the force bound, including every spatial
multi-index. In particular no temporal derivative of the convection
term has been omitted.
Supports do not increase under differentiation or multiplication.
For either field $Z$, the squared integral of an indicated derivative
is at most
\[
|K|C_{k,\alpha}^2\int_0^\infty(1+t)^{-2-2k}\,dt
=\frac{|K|C_{k,\alpha}^2}{1+2k}.
\]
The zero datum remains zero, so residual realization applies.
The chain rule shows that $\Xi(s(t),a)$ solves the $U$ trajectory
equation with initial label $a$; uniqueness identifies it with the
actual trajectory. The clock is increasing onto $[0,\infty)$ with
inverse $e^s-1$, so it loses no finite computational time. Finally
logarithm on positive arguments, the rational powers of $1+t$, and
the finite-index evaluation of the template and its derivatives are
effective. This proves all assertions.
\end{proof}
\subsection{Loaded Euclidean processors}
\begin{theorem}[Two full-box processors on Euclidean space]
\label{ba:euclidean-boxes}
For every finite machine and input, and every fixed computable $\nu>0$,
either the move-first recorder of Lemma~\ref{lem:bb-movefirst} or the
six-rule recorder of Lemma~\ref{ba:six-compiler} yields effective smooth forcing on $\mathbb R^3$ with one
compact spatial support, all mixed derivatives bounded, and period one
for $t\ge1$, such that
\[
\exists t\ge0:\ (\Phi_t(0))_1<-1
\quad\Longleftrightarrow\quad\text{halting}.
\]
The zero-data solution is unique in Lemma~\ref{ba:realization}'s class,
has pressure zero and uniformly bounded energy. Separately, the six-rule
processor admits a force square-integrable on all space-time with the
same support and event.
\end{theorem}
\begin{proof}\label{ba:euclidean-boxes-proof}
Write $\Gamma$ for the full alphabet of the recorder being used.
For the six-rule table use initial frontier 2 and base $B=2|\Gamma|+2$.
Give nonhalting controls lanes $3i+[0,1]$ and halting controls lanes
$-3i+[0,1]$, enumerating each group from 1; the second coordinate is
$[0,1]$. The minimal subdivision has gap $B^{-2}$. Use full thickness
$[-1,1]$, heights $10i$, and the transported-cutoff construction in
Section~\ref{sec:ba-transported-localization}, with initial padding $1/(10B^3)$.
Target horizontal padding is at most $1/(10B^2)$, below half the gap;
vertical middle intervals are also separated.
For the move-first table use a single halt lane $-3+[0,1]$, other lanes
$2+2j+[0,1]$, base $B=2|\Gamma|+1$, the minimal subdivision, thickness
$[-1,1]$, and heights $4i$. Use current-box padding $1/(10B^2)$.
The middle-phase separation is
$2(1+1/(10B^2))<4$; the planar gaps handle the other phases.
This localization has a fixed current padding, as distinct from the
transported initial padding in the six-rule version.
The rational initial code $P_*$ is reached from zero along
$\theta(t)P_*$ by a localized translation during $[0,1]$.
Repeating the chosen cycle produces a template $W$ with one compact
support and bounded mixed derivatives. The code identity at times
$1+n$ follows by induction through the local rules; the respective
recorder invariant supplies every next checkpoint in finite positive
time. A halt lies below $-1$ in the first coordinate. In a nonhalting
run all branch endpoints have positive first coordinate, and
\eqref{eq:bb-convex-coordinate} keeps it positive throughout every cycle.
Loading is nonnegative. Lemma~\ref{ba:realization} completes the PDE
argument. Uniform support and boundedness give a uniform energy bound.
The separate square-integrable choice follows from
Lemma~\ref{ba:log-clock}, without claiming periodicity in physical
time for that choice.
\end{proof}
\subsection{Two realizations of a left frontier}
\begin{proposition}[Separate columns or solid-box parking]
\label{prop:bb-mid-stream}\label{prop:bb-mid-parking}
Both left-frontier tables have compact whole-space realizations,
periodic after $t=1$, with fixed label zero and event $X_1(t;0)<-1/2$.
The stopping version can use distinct-height stream functions and
has uniqueness with pointwise-in-time $H^1$ comparison pressure.
The continuing version transports full solid boxes by evacuation,
reshaping and delivery, and represents every local step, including
post-halt idle computation. It has in particular uniqueness with
comparison pressure in $C([0,T];H^1)$.
\end{proposition}
\begin{proof}\label{proof:bb-mid-stream}\label{proof:bb-mid-parking}
Let $\Sigma$ be the full work--history--mark alphabet of the chosen
left-frontier table, including its record symbols and any halting idle
records in the continuing version. Use odd digits in base $B=2|\Sigma|+2$, offset $-2$ for $C_h$ and
$2,4,6,\ldots$ for other controls. The two head-relative fractions
and height zero give a rational initial code $Z_*$. The rightmost
nonempty history cell is the original cell $-2$, an absolute landmark.
For the stopping table, apply the stream-function construction
\eqref{eq:bb-mid-stream-path} at heights $z_i=3i$.
Choose source and target column paddings no greater than $1/2$ and
one third of their respective family gaps. A middle layer padding
$1/2$ keeps distinct layers disjoint. The explicit thickening bound
following that construction proves the neighborhood action.
Write $Z_*=(x_*,y_*,0)$. With a flat switch $\theta$ from zero to one,
load the origin along $\theta(t)Z_*$ using the velocity
$\nabla\times(0,0,H)$, where
$H(t,x,y,z)=\theta'(t)\chi(x,y,z)(x_*y-y_*x)$.
Choose $\chi$ compactly supported in $\R^3$ and equal to one near the
entire segment. The gate $\theta'$ vanishes on the time collars.
All nonhalt source and target lower-left first coordinates are at
least two. Inequality~\eqref{eq:bb-mid-stream-safe} therefore
excludes the observer during each period; loading has nonnegative
first coordinate. Halt codes lie in $(-2,-1)$.
For the continuing table retain its unrestricted first-row history
symbol and its idle work rules. Thicken every rectangle by $[-1,1]$.
Choose $L$ ten larger than the maximum of zero and every source or
target right endpoint. Reserve parking projections
$(L+4i,0)+[-1,1]^2$ and let $\delta$ be the minimum of one and the
gaps in each union of one rectangle family with these reservations.
Use padding $\delta/10$, travel height five, and $7N$ slots: evacuate
all sources by three translations each, reshape all parked boxes by
$\operatorname{diag}(1+\theta(\lambda_i-1),[1+\theta(\lambda_i-1)]^{-1},1)$,
and deliver by three translations each. Horizontal widths stay at
most one, vertical half-width is one, and the reservations miss both
original families. Lemma~\ref{lem:bb-lift-parking} proves exact action
on the whole boxes and their neighborhoods even with source--target
overlap. A first localized translation loads $Z_*$ from the origin.
The sampling identity is now valid for every $k\ge0$, not just until
halting: $\Phi_{1+k}(0)=Z(C_k)$ for the continued table.
For a nonhalting run all used endpoint centers have first coordinate
at least two, parking centers are farther right, and half-widths are
at most $1/2$. Translation, parking strain, and waiting therefore
keep first coordinate positive. A halt checkpoint lies in $(-2,-1)$.
Proposition~\ref{prop:bb-realization} proves the analytic assertions
for both constructions. Their compact support also gives uniformly
bounded kinetic energy.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//compact-processors.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//compilers.tex ====
\section{History records and complete symbolic domains}
\label{sec:bb-frontiers}
The balanced-stack theorem is complete. The remaining realizations
use alternative recorders with tape data in two coordinates.
A motion of boxes is invertible. To use it for an ordinary machine,
we must retain whatever information an instruction erases. We use
finite history records in the sense of reversible computation, but
prove the local inverses below: reversibility of the initialized
execution would not by itself separate the full boxes.
A work instruction is $r=(q,a)\mapsto(p_r,b_r,d_r)$, with
$d_r\in\{-1,0,1\}$. The work tape is indexed by $\Z$, starts with
head zero, and has only finitely many prescribed nonblank cells.
Halting states have no outgoing work instruction. Missing instructions
can be replaced by write-preserving stationary steps to a fresh halt
state. Unless a construction explicitly adds a dummy initial step,
an initially halting state is already a halt checkpoint. When a construction
uses a single halting state $h$, merge the original halting states and
redirect their incoming transitions; this preserves initial halting. In
the recorder tables $Q$ is the work-state set and $\Gamma$ the work
alphabet. Radix bases are chosen from the full cell alphabet, including
all history and mark tracks.
\subsection{A one-cell code and its exact inverse test}
A partial one-cell table sends a control and a scanned letter to a
new control, a written letter, and a displacement. Its two incoming
conditions are: the new control determines the displacement, and the
new control together with the written letter determines the entire
old control and read letter.
\begin{lemma}[From one-cell tables to closed rectangles]
\label{lem:bb-onecell}\label{lem:bb-mid-prefix}
Give an alphabet of size $m$ digits separated by at least two, between
$0$ and $B-2$. Put
\[
E(A)=\sum_{j\ge1}d(A_j)B^{-j},\qquad
I(v)=\sum_{j=1}^{|v|}d(v_j)B^{-j}+[0,B^{-|v|}].
\]
For a head-relative tape $(a_j)_{j\in\Z}$ use the coordinates
\[
(x,y)=\big(E(a_{-1}a_{-2}\cdots),E(a_0a_1\cdots)\big).
\]
After writing at cell zero, a displacement $e$ makes the old cell $e$
the new cell zero.
A deterministic partial one-cell table satisfying the incoming
conditions acts, after separate state translations, by a finite list
of reciprocal affine maps on full closed rectangles. Each source
family and each target family is separately disjoint. For a rule
reading $a$, writing $b$, and moving by $e$, the branches are
\begin{equation}\label{eq:bb-one-cell-branches}
\begin{array}{c|c|c|c}
e&\text{source}&\text{target}&(x',y')\\\hline
1 &[0,1]\times I(a)&I(b)\times[0,1]
&((d(b)+x)/B,By-d(a))\\
0 &[0,1]\times I(a)&[0,1]\times I(b)
&(x,y+(d(b)-d(a))/B)\\
-1&I(c)\times I(a)&[0,1]\times(d(c)+I(b))/B
&(Bx-d(c),[d(c)+(d(b)+By-d(a))/B]/B).
\end{array}
\end{equation}
The last row is split over every letter $c$. Each linear part is
$\operatorname{diag}(B^{-e},B^e)$. With odd digits
$1,3,\ldots,2m-1$ and $B\in\{2m+1,2m+2\}$, every code lies in
the interior of its applicable rectangle, and unscaled within-family
gaps are at least $B^{-2}$.
\end{lemma}
\begin{proof}
The identity $E(aA)=(d(a)+E(A))/B$ gives all three formulas and their
action on the independent tails. At the first differing digit,
incompatible prefix intervals have a positive gap: the digit difference
is at least two while the child interval has length one in that scale.
Thus the source state, read letter, and any left split separate sources.
At one target state the displacement is fixed. For a right or stationary
move the written letter separates the relevant target coordinate;
for a left move the pair $(c,b)$ separates the two-letter prefix.
The incoming inverse condition rules out duplicates. These arguments
concern the complete intervals, including their endpoints.
For odd digits the infinite tail lies between $1/(B-1)$ and
$(2m-1)/(B-1)<1$, proving interiority. The longest target prefix has
length two, giving the asserted gap. Uniformly scaling state squares
by $s$ scales gaps by $s$ and leaves the linear parts unchanged.
\end{proof}
\subsection{Moving the work head before recording}
We first place a frontier to the right of every possible work head.
A temporary mark lets the history excursion return to the correct
cell even though its length grows with the computation.
\begin{lemma}[Move-first recorder]
\label{lem:bb-movefirst}\label{lem:bb-mid-right-compiler}
There is an effective one-cell table satisfying the incoming conditions
whose checkpoint configurations simulate the work machine. After
$n$ work steps its frontier is at $n+2$, its records occupy
$2,\ldots,n+1$, and its temporary marks are all zero. Each nonhalting
checkpoint has a finite positive continuation to the next one, with
no intervening checkpoint.
\end{lemma}
\begin{proof}
Use letters $(c,\ell,j)$ with work letter $c$, history
$\ell\in\{e,F\}\sqcup\{r\}$ and mark $j\in\{0,1\}$.
The controls are $C_q,J_r,S_r,P_p,T_p$. The complete partial table is
\[
\begin{array}{lll}
C_q:(a,\ell,0)&\mapsto J_r:(b_r,\ell,0),d_r&\ell\ne F,\\
J_r:(c,\ell,0)&\mapsto S_r:(c,\ell,1),1&\ell\ne F,\\
S_r:(c,\ell,0)&\mapsto S_r:(c,\ell,0),1&\ell\ne F,\\
S_r:(c,F,0)&\mapsto P_{p_r}:(c,r,0),1,&\\
P_p:(c,e,0)&\mapsto T_p:(c,F,0),-1,&\\
T_p:(c,\ell,0)&\mapsto T_p:(c,\ell,0),-1&\ell\ne F,\\
T_p:(c,\ell,1)&\mapsto C_p:(c,\ell,0),0&\ell\ne F.
\end{array}
\]
Initialize the frontier at two, all other history empty, and all marks
zero. At checkpoint $n$ the head satisfies $|h_n|\le n$.
The first row updates the work tape and moves to $h'\le n+1$, strictly
left of the frontier $f=n+2$. The second row marks $h'$. The forward
scan writes $r$ at $f$, puts a new frontier at $f+1$, and the backward
scan stops exactly at the mark, which it clears. Its length is
$2(f-h')+4$, so it is finite and positive. This proves the invariant
inductively, including the exact work update and halt equivalence.
For the full-domain inverse, $r$ recovers the old work pair into $J_r$;
mark one distinguishes entry into $S_r$ from its mark-zero loop;
a written record determines entry into $P_p$; a written $F$
distinguishes entry into $T_p$ from its loop; and entry into $C_p$
uniquely restores mark one. Every destination has the single incoming
displacement displayed in the table. No invariant on remote cells
was used.
\end{proof}
\subsection{Recording before the work-head displacement}
The next table leaves its mark at the old work head. Its final step
performs the work displacement; storing that displacement in the
checkpoint control keeps incoming moves unambiguous.
\begin{lemma}[Direction-recording scan]
\label{lem:bb-direction}\label{lem:bb-mid-mark-compiler}
Let $D=\{-1,0,1\}$ and let $O$ be either a singleton or $\{0,1\}$.
An effective one-cell table on
$\Gamma\times(\{\bot,F\}\sqcup\mathcal R)\times\{0,1\}\times O$,
where $\mathcal R=\{(q,e,a):q\text{ nonhalting},e\in D,a\in\Gamma\}$,
satisfies the incoming conditions. It preserves the $O$ track in
its physical cells. Starting with frontier at one, its checkpoint
at work step $n$ has frontier $n+1$, records at $1,\ldots,n$, zero
marks, and the correct work configuration. A step from head $i$
takes $5+2(n-i)$ local rules.
\end{lemma}
\begin{proof}\label{proof:bb-mid-mark-compiler}
Use main controls $M(q,e)$, forward controls $G(r)$, and $A(p,d),B(p,d)$
for every work state $p$ and $d\in D$, including pairs not produced by
a work instruction. Copy the $O$ component in each row. In the first
three rows use every $r=(q,e,a)\in\mathcal R$ with
$\delta(q,a)=(p,c,d)$. The final three rows apply independently to
every $(p,d)\in Q\times D$, including pairs that are not outputs of
any work instruction. The variables $v\in\Gamma$ and
$h\in\{\bot,F\}\sqcup\mathcal R$ range over all letters subject to
the displayed restrictions. The whole table is
\[
\begin{array}{lll}
M(q,e):(a,h,0)&\mapsto G(r):(c,h,1),1&h\ne F,\\
G(r):(v,h,0)&\mapsto G(r):(v,h,0),1&h\ne F,\\
G(r):(v,F,0)&\mapsto A(p,d):(v,r,0),1,&\\
A(p,d):(v,\bot,0)&\mapsto B(p,d):(v,F,0),-1,&\\
B(p,d):(v,h,0)&\mapsto B(p,d):(v,h,0),-1&h\ne F,\\
B(p,d):(v,h,1)&\mapsto M(p,d):(v,h,0),d&h\ne F.
\end{array}
\]
Starting at $M(q_0,0)$, the first row writes and marks the old head.
The scan reaches frontier $j=n+1$, stores its record, creates the
next frontier, returns to the unique mark and makes the work move.
The count is $2(j-i)+3=5+2(n-i)$. Since $i\le nh_n$,
the forward scan reaches the frontier, moves it one cell right, and
the return scan reaches the mark. Its neighboring-cell guard holds
because the new frontier is $p_n+1$. The last rule clears the mark
and performs the work displacement. Counting the two traversals and
endpoint actions gives $2(p_n-h_n)+1$; induction gives $|h_n|\le n$.
For injectivity on all relative tapes, into Go the old cursor is
output offset $-1$: a mark there distinguishes entry from the loop,
and $r$ restores the overwritten work letter and old Ready control.
Into Back the old cursor is offset $+1$. An output pointer at $+2$
identifies a log insertion, with record at $+1$ recovering $r$ and
old pair $(P,E)$. A nonpointer at $+2$ identifies the guarded loop.
Into Ready the corresponding Back control uniquely restores the
cleared mark. These are all possibilities, proving the full inverse.
\end{proof}
\begin{lemma}[Closed rectangles for a finite guarded window]
\label{lem:bb-window-rectangles}
Suppose a finite partial injective local tape map reads and writes only
indices $[-a,b-1]$, then shifts the cursor by $e$, where
$-a\le e\le b$. Split its domain by every complete word in that
window. With gapped radix coding the input prefix lengths are
$(a,b)$ and output lengths $(a+e,b-e)$; the affine branch has linear
part $\operatorname{diag}(B^{-e},B^e)$. Both full rectangle families
are separately disjoint.
\end{lemma}
\begin{proof}
After the writes the new left prefix consists of old indices
$e-1,e-2,\ldots,-a$, and the right prefix of $e,e+1,\ldots,b-1$.
The two remaining tails are free and unchanged, so the image is the
whole output cylinder. Distinct source cylinders have disjoint
images by injectivity. If both pairs of output prefixes were
compatible, a common extension would belong to both images; hence
at least one coordinate has incompatible prefixes. The gapped
interval argument then separates the complete closed rectangles.
The prefix lengths give the two reciprocal scales directly.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//compilers.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//geometry.tex ====
\section{Transporting complete boxes}
\label{sec:bb-boxes}
We allow an instruction to change all three side lengths while
preserving their product. The geometric question is concrete:
given finitely many disjoint closed boxes and prescribed determinant-one
affine maps onto another disjoint family, can one carry out every map by
a smooth incompressible motion? The two families may overlap each
other. Consequently a destination cannot be occupied until every source
that obstructs it has been removed.
We first extend an explicitly separated motion to a velocity field.
We then give two ways to produce that motion. Separate horizontal
projections permit a simple lift and transfer. For general solid boxes
we use parking and detours around the other occupied boxes.
Figure~\ref{fig:bb-routing} contrasts the height separation with the
evacuation order that makes overlapping source and target families harmless.
\subsection{An affine motion and its curl extension}
Throughout, a smooth switch is obtained by translating and rescaling
$\sigma$ in \eqref{eq:p2-step}, with its transition strictly inside
its allotted interval. Thus successive motions are stationary on
neighborhoods of their joining times.
\begin{lemma}[A velocity on a neighborhood of each moving box]
\label{lem:bb-curl}
Let $K_i(t)=c_i(t)+D_i(t)K_i^0$, $1\le i\le N$, where $K_i^0$ is
an axis-parallel box, possibly with zero side lengths, and $D_i(t)$ is
positive diagonal with determinant one. Suppose the centers and scales
are smooth and constant near the ends of a compact time interval.
For the effective conclusion, assume computable endpoints for the
initial boxes $K_i^0$, effective evaluation of the centers' and scales'
derivatives, bounds for those derivatives, and positive lower bounds
for every scale on that interval.
If the coordinatewise $2\eta$ enlargements of the moving boxes are
pairwise disjoint and bounded, for a known $\eta>0$, a
compactly supported smooth divergence-free field realizes these motions
on open neighborhoods of the initial boxes. It vanishes near the time
endpoints. When all enlarged boxes lie in an interior torus chart,
the field extends to a mean-zero field on that torus.
The field is effective under the effective-data assumptions above;
smooth existence itself does not require computable data.
\end{lemma}
\begin{proof}
Choose a smooth cutoff $\chi_i$ which is one on the $\eta$
enlargement of $K_i(t)$ and supported in its $2\eta$ enlargement,
using products of the effective switches. Put $y=x-c_i$ and
$L_i=\dot D_iD_i^{-1}$; then $\operatorname{tr}L_i=0$.
Take the curl of
\begin{equation}\label{eq:bb-potential}
\chi_i\{\tfrac12\dot c_i\times y+\tfrac13(L_i y)\times y\}.
\end{equation}
The identity
$\nabla\times(G\times y)=2G-y\operatorname{div}G+(y\cdot\nabla)G$
shows that this curl equals $\dot c_i+L_i y$ on the plateau.
Disjoint supports permit summation over $i$. The path
$c_i(t)+D_i(t)D_i(t_0)^{-1}(x_0-c_i(t_0))$ has exactly that
derivative and remains in its prescribed box. ODE uniqueness proves
the formula. On the compact time interval the normalized matrix norm
has a finite bound $M$, computable under the effective-data assumptions;
an initial perturbation less than
$\eta/(2M)$ remains on the plateau. This proves the neighborhood
claim, also for zero-width initial boxes.
The potential is zero near time endpoints because the centers and
scales are constant there. Its support is bounded. Within a torus
chart extend the potentials by zero; periodic curls have zero integral.
\end{proof}
\begin{proposition}[Fluid realization and a pressure refinement]
\label{prop:bb-realization}
Fix $\nu>0$. A finite collection of the preceding motions, repeated
with period one, or preceded by one finite loading interval, yields a
smooth prescribed velocity $U$ with $U(0)=0$. Suppose its spatial
support is contained in one compact set in $\R^3$, or its potentials
are contained in interior torus charts. Then
\begin{equation}\label{eq:bb-residual}
f=U_t+(U\cdot\nabla)U-\nu\Delta U
\end{equation}
has the same support and time periodicity and bounded mixed derivatives
of every order. The solution is $(U,0)$, with the uniqueness and
material-flow conclusions of Lemma~\ref{lem:p2-realization}.
In $\R^3$ its uniqueness statement also holds when the competing
pressure has an $H^1$ representative at each time, without requiring
continuity into $H^1$, while the other velocity hypotheses of
Section~\ref{sec:model} are retained. On the torus $U$ and $f$ have
zero mean. Under the effective-data assumptions of Lemma~\ref{lem:bb-curl},
the formula is $f_0+\nu f_1$ with effective coefficients;
evaluation is effective for computable $\nu$, and relative to $\nu$
otherwise.
\end{proposition}
\begin{proof}
The residual identity and Lemma~\ref{lem:b-comparison} give existence,
uniqueness and the material flow. In the whole-space pressure refinement
use case (ii): the reference field has compact support and smooth bounded
mixed derivatives, hence belongs to $CH^2\cap C^1L^2$ with bounded
velocity and gradient; the competing velocity and pointwise $H^1$
pressure are exactly that case's hypotheses. No pressure continuity
in time is added. Integrating the residual on a torus uses
$\int U=0$, $\int\Delta U=0$ and
$\int(U\cdot\nabla)U=\int\operatorname{div}(U\otimes U)=0$.
Each derivative of $e^{-1/s}$ for $s>0$ is a polynomial in $1/s$
times $e^{-1/s}$. The estimate
$v^m e^{-v}\le(m+k)!v^{-k}$ supplies effective endpoint error bounds,
and the denominator defining $\sigma$ is at least $e^{-2}$.
Under those effective-data assumptions all positive scales have effective
positive lower bounds. Finite
compositions, differentiation, and the logarithms of positive computable
scales are therefore effective. At an approximate time, a coarse bound
gives a finite superset of possible translates of the zero-extended
unit template. This evaluates the periodic field without deciding
whether time is an integer. None of these operations iterates the
machine execution: the program contains the entire finite instruction
list.
\end{proof}
\subsection{Three geometric schedules}
For later use we describe the exact hypotheses of the schedules.
First suppose $P_i,Q_i\subset\R^2$ are finite families of closed
rectangles, possibly with zero side lengths, separately
pairwise disjoint, with affine maps
\[
F_i(x)=q_i+\operatorname{diag}(\lambda_i,\lambda_i^{-1})(x-p_i),
\qquad \lambda_i>0,
\]
where $p_i,q_i$ are their centers, and require $F_i(P_i)=Q_i$.
Their side lengths have a common finite bound. Source and target rectangles
belonging to different families need not be disjoint. The smooth existence
statements allow arbitrary real rectangle coordinates and scales. For the
effective conclusions, assume rational rectangle endpoints and rational
scales, together with computable chosen heights and vertical half-widths;
then a rational side-length bound and positive gap margins can be computed.
\begin{lemma}[Separated projections and parking]
\label{lem:bb-lift-parking}
These reciprocal maps have the following smooth compactly supported
unit-time realizations, stationary on time collars. They are effective
under the rational-data assumptions just stated.
\begin{enumerate}
\item For the products $P_i\times[z_0-h,z_0+h]$, with common $h\ge0$
and $z_0\in\R$, lift all boxes with their source projections fixed,
transfer them at sufficiently separated private heights, and lower
them with their target projections fixed. The middle motion may use
$g_i(s)=\lambda_i^s$ or $g_i(s)=1+s(\lambda_i-1)$ and matrix
$\operatorname{diag}(g_i,g_i^{-1},1)$, with the center interpolated
linearly in $s$. Each horizontal width stays between its endpoint
widths. In the linear case every point satisfies
\begin{equation}\label{eq:bb-convex-coordinate}
X_1(s)=(1-s)X_1(0)+sX_1(1).
\end{equation}
\item Alternatively, choose disjoint parking rectangles away from both
families. Move every source to its parking site before delivering any
target, using an upward, horizontal, and downward translation. Reshape
at the parking sites, then deliver by the same three translations.
This uses $7N$ motions. The same schedule applies to assigned onto positive
diagonal determinant-one maps between rational solid boxes whose centers
lie in a common horizontal plane and whose source and target horizontal
projections are separately disjoint. Require both horizontal widths during
reshaping to stay inside their reserved parking rectangles and the travel
height above the common plane to exceed twice the largest vertical
half-width.
\end{enumerate}
Both constructions act on neighborhoods of the whole source boxes.
\end{lemma}
\begin{proof}
In the first construction choose padding smaller than a quarter of
the within-family horizontal gaps and choose height differences larger
than twice the padded vertical half-width. During lift and descent,
horizontal separation suffices. During transfer, vertical separation
suffices. The two choices of $g_i$ are positive and monotone between
$1$ and $\lambda_i$; their reciprocals are monotone as well. Thus the
width claim holds. The affine path is
$c_i(s)+\operatorname{diag}(g_i,g_i^{-1},1)(X_0-c_i(0))$.
Its first coordinate gives \eqref{eq:bb-convex-coordinate} exactly for
the linear choice. Lemma~\ref{lem:bb-curl} now applies.
For the second construction choose padding smaller than the gaps in
each union of source projections with reserved parking rectangles, and
of target projections with those rectangles. In the evacuation pass,
vertical motion is separated from uncollected sources and parked boxes.
During delivery it is separated from already filled targets and boxes
still parked. At travel height, the moving box is above every resting
box. A shape change stays within its own reserved projection.
These are the occupied sets at each actual time; source--target
intersection causes no interference because all sources have left
before delivery starts. Lemma~\ref{lem:bb-curl} applies to each motion
with every other box fixed. Finite concatenation gives the required
neighborhood map.
\end{proof}
\begin{figure}[ht]
\centering
\begin{tikzpicture}[x=.75cm,y=.65cm,>=stealth,font=\small]
\draw[gray] (0,0)--(5,0);
\draw[thick,blue!65!black,->] (.5,0)--(.5,2.4)--(4.2,2.4)--(4.2,.1);
\draw[thick,orange!80!black,->] (1.6,0)--(1.6,1.3)--(3.1,1.3)--(3.1,.1);
\node[below] at (1,0) {sources};
\node[below] at (3.7,0) {targets};
\node[above] at (2.5,2.6) {Separate height layers};
\begin{scope}[xshift=5.1cm]
\node[draw,rounded corners,minimum width=2.6cm] (s) at (1.8,2.8) {all source boxes};
\node[draw,rounded corners,minimum width=2.6cm] (p) at (1.8,1.1) {all boxes parked};
\node[draw,rounded corners,minimum width=2.6cm] (t) at (1.8,-.6) {all target boxes};
\draw[->,thick] (s)--(p) node[midway,right] {evacuate};
\draw[->,thick] (p)--(t) node[midway,right] {deliver};
\end{scope}
\end{tikzpicture}
\caption{The two separation mechanisms in
Lemma~\ref{lem:bb-lift-parking}. Left: horizontal projections separate
the lift and descent, and distinct heights separate the transfer;
the second horizontal coordinate is suppressed. Right: a temporal
schedule, not a spatial drawing. Every source is removed before any
target is occupied, so overlap between the two families is harmless.}
\label{fig:bb-routing}
\end{figure}
There is also a useful stream-function implementation of the first
schedule. Suppose $a_i,a_i'$ are the lower left corners of $P_i,Q_i$.
At height $z_i$, put $d_i=a_i'-a_i$, $b_i=\log\lambda_i$ and
$g_i(s)=a_i+sd_i$. On a neighborhood of the swept rectangle use
\begin{equation}\label{eq:bb-mid-stream-path}
\xi_i(s)=g_i(s)+\operatorname{diag}(\lambda_i^s,\lambda_i^{-s})(\xi-a_i).
\end{equation}
The Hamiltonian
\[
H_i=d_{i1}y-d_{i2}x+b_i(x-g_{i1})(y-g_{i2})
\]
generates this motion. Put $\widehat H_i=\chi_iH_i$, where $\chi_i$
is one near the whole sweep and supported in its reserved layer. Use
\[
\dot s(\partial_y\widehat H_i,-\partial_x\widehat H_i,0).
\]
Disjoint height layers separate these cutoffs. Vertical phases use
$\nabla\times(0,x\chi_i,0)$, equal to $(0,0,1)$ on their column
plateaus. If $M=\max_i(1,\lambda_i,\lambda_i^{-1})$ and the column
plateau margins are $\delta,\delta'$, an initial thickening smaller than
$\min(\delta/(4M),\delta'/(4M),1/(8M))$ stays on every plateau.
Thus the affine formula also holds on open neighborhoods. For points of
the original rectangle it has the additional bound
\begin{equation}\label{eq:bb-mid-stream-safe}
X_1(s)\ge\min(a_{i1},a'_{i1}).
\end{equation}
The bound follows because the first relative coordinate in
\eqref{eq:bb-mid-stream-path} is nonnegative.
\subsection{Solid boxes with overlapping projections}
The preceding lift requires separate horizontal projections. We now
remove that restriction; this is needed when all three coordinates
encode independent stacks.
\begin{theorem}[Expansion, obstacle detours, and delivery]
\label{lem:bb-balanced-box-routing}
Let $Q_i=a_i+\prod_{j=1}^3[-h_{ij},h_{ij}]$ and
$Q_i'=b_i+\prod_{j=1}^3[-k_{ij},k_{ij}]$ be rational solid boxes,
with positive half-widths, each family pairwise disjoint. Suppose
$s_{ij}=k_{ij}/h_{ij}$ and $\prod_js_{ij}=1$. An effective compactly
supported smooth solenoidal field, zero outside the time interval
$[1/4,3/4]$, has time-one map
\[
a_i+y\longmapsto b_i+\operatorname{diag}(s_{i1},s_{i2},s_{i3})y
\]
on a neighborhood of each $Q_i$.
\end{theorem}
\begin{proof}\label{proof:bb-balanced-box-routing}
For an empty family take the zero field. Otherwise let
$R=\max(1,h_{ij},k_{ij})$. Choose a rational $\Lambda>10$ so that
centers in each of the lists $(\Lambda a_i)$ and $(\Lambda b_i)$
are separated by more than $20R$ in maximum norm. This is possible
because distinct disjoint boxes have distinct centers. Let
\[
K=\max(4\Lambda,|\Lambda a_{i1}|,|\Lambda b_{i1}|),
\qquad d_i=(K+30Ri,0,0).
\]
The $d_i$ are parking centers, separated from each other and both scaled
families by at least $30R$ in their first coordinate.
First multiply all source centers by a common factor increasing from
$1$ to $\Lambda$, while keeping shapes fixed. Next change all shapes
at those separated centers by factors $s_{ij}^{\theta}$,
$0\le\theta\le1$. Each half-width is the geometric interpolation
$h_{ij}^{1-\theta}k_{ij}^{\theta}\le R$.
Move the boxes one at a time from $\Lambda a_i$ to $d_i$, then,
after all have been parked, from $d_i$ to $\Lambda b_i$.
Finally shrink all target centers by a common factor from $\Lambda$
to $1$, with target shapes fixed.
Here is a finite construction of every single-box path. Around each
stationary center place the closed cube of maximum-norm radius $4R$.
These cubes are disjoint since stationary centers are separated by
more than $20R$, and neither endpoint is in a cube. Start with the
straight segment between the endpoints. Its intersections with the
cubes are rational intervals, found by linear inequalities. Replace
each portion inside a cube by a polygonal path on its boundary: join
the entry point to a vertex of a containing face, use cube edges to a
face containing the exit point, and join to that exit point. The cube
is convex face by face, so these segments remain on its boundary;
no other cube meets them. A tangency needs no replacement. The
resulting rational polygonal path has distance at least $4R$ from
every stationary center. Traverse its segments with flat switches,
stopping to all orders at vertices. Its geometric image is unchanged,
so smoothing the time parameter does not reduce clearance.
For a uniform localization margin, let $g_a$ be the minimum over
$i1$, and terminal codes lie strictly in $(-2,-1)\times(0,1)$.
The infinite tails are in $[\delta,1-\delta]$,
$\delta=1/(D-1)$, so finite-prefix decoding of length $b$ is possible
from error less than $(\delta/2)D^{-b}$ without a boundary ambiguity.
Split full triples $[-1,1]$. By
Lemma~\ref{lem:bb-window-rectangles}, source prefix lengths $(1,2)$
become $(1+d,2-d)$, target cylinders are complete, and both rectangle
families have gaps at least $D^{-3}$. Thicken by $[-1/4,1/4]$,
lift to height $4i$, use linear middle scale
$\mu_i=1+(D^{-d_i}-1)\theta$, and descend. Padding smaller than
$D^{-3}/4$ separates all phases. Repeating the pulse, zero on
$[0,1/12]$ and $[11/12,1]$, gives periodicity from zero without
loading. At a nonhalting branch's lift and descent the first
coordinate is greater than one; during the middle transfer the
particle's third coordinate is exactly $4i$, outside $(-1,1)$.
Thus even a horizontal route passing over the observation box cannot
cause a false visit. The integer samples and the compiler invariant
prove halting equivalence. The record count and pointer site $n+3$
recover absolute indexing at checkpoints.
For uniqueness apply Lemma~\ref{lem:b-comparison}(iii).
The explicit compactly supported smooth field lies in $CH^4\cap C^1H^2$
on every finite interval and has pressure zero. The competing class
in the theorem is precisely case (iii), including its distributional
$CL^2$ gradient. Lemma~\ref{lem:b-gradient} cancels that gradient
against the solenoidal velocity difference without introducing a pressure
$L^2$ norm. Thus the velocity and pressure gradient are unique, while
pressure itself remains free up to a time-dependent spatial constant.
Proposition~\ref{prop:bb-late-logclock} gives the stated
decaying alternative.
\end{proof}
\subsection{A flagged head and two-pass parking}
\label{sec:bb-two-pass-parking}
\begin{theorem}\label{thm:bb-two-pass-parking}
For any machine and finitely specified initial tape, an effective
smooth force on $\R^3$, periodic from time zero, with one compact
spatial support and bounded mixed derivatives, realizes the event
\[
\exists t\ge0:\quad X(t;0)\in(-6,-4)\times(-1,1)\times(-1,1)
\quad\Longleftrightarrow\quad\text{halting}.
\]
The zero-data velocity is unique among classical smooth solutions
with bounded $u,\nabla u$, $u\in C H^2\cap C^1L^2$ and $p\in C L^2$
on every finite interval. The prescribed solution has uniformly
bounded kinetic energy.
\end{theorem}
\begin{proof}\label{proof:bb-two-pass-parking}
Merge terminal states into $q_\dagger$, send missing instructions
there by a stay step, and add a fresh initial state $q_*$, with no
incoming instruction, which writes back and stays while entering
the old initial state. For each normalized nonterminal rule
$r=(q,a)\mapsto(p_r,b_r,d_r)$ use cell letters
\[
(w,h,j)\in\Gamma:=\mathcal A\times
(\{\circ,\star\}\sqcup\{[r]:r\in\mathcal R\})\times\{0,1\},
\]
where $\mathcal A$ is the work alphabet and $\mathcal R$ is the
finite set of normalized instruction labels. Thus $\Gamma$ includes
all three tracks.
The complete table, with $h\ne\star$ wherever $h$ occurs, is
\[
\begin{array}{lll}
M_q:(a,h,1)&\mapsto A_r:(b_r,h,0),d_r,\\
A_r:(w,h,0)&\mapsto R_r:(w,h,1),1,\\
R_r:(w,h,0)&\mapsto R_r:(w,h,0),1,\\
R_r:(w,\star,0)&\mapsto B_{p_r}:(w,[r],0),1,\\
B_q:(w,\circ,0)&\mapsto L_q:(w,\star,0),-1,\\
L_q:(w,h,0)&\mapsto L_q:(w,h,0),-1,\\
L_q:(w,h,1)&\mapsto M_q:(w,h,1),0.
\end{array}
\]
The destination fixes the incoming displacement. The record index
recovers the first row's old work pair; a flag distinguishes the
entries into $R_r$; a record distinguishes the entries into $B_q$;
a written frontier distinguishes entry and loop into $L_q$;
and entry into $M_q$ is the unique cell-preserving last row.
This proves the full one-cell inverse conditions.
Initialize the sole flag at zero and the sole frontier at three,
with empty history elsewhere. At ready step $k$, only the work
head $i$ is flagged, records occupy $3,\ldots,k+2$, and the frontier
is $e_k=k+3$. The work rule clears the flag and moves to
$j=i+d_r\le e_k-2$, which is unflagged even for $d_r=0$.
It is flagged anew; the scan logs at $e_k$, advances the pointer,
returns, and preserves the ready flag. The two loop lengths are
$e_k-j-1$ and $e_k-j$, with five other rules, for a total
$2(e_k-j)+4$. Thus all scans terminate and ready halting is exact.
Use odd digits in base $K=2|\Gamma|+2$, initial fractions $x_*,y_*$,
and state offsets $a_{M_{q_*}}=0$, $a_{M_{q_\dagger}}=-5$, and
$4,8,12,\ldots$ for the others. The code
$(a_s-x_*+x,y-y_*,0)$ starts at zero. Lemma~\ref{lem:bb-onecell}
gives $N$ full rectangle maps with scales $\lambda_i\in\{K^{-1},1,K\}$;
thicken by $[-1/4,1/4]$. The entire terminal block is inside the
observation box because $0-1$. Eventually constant tails are rational even for an
arbitrary finitely specified initial tape, and the pointer and record
count recover its absolute head position at ready checkpoints.
For the required two-pass geometry, let $g$ be the minimum of one
and all positive separating coordinate gaps in either planar family,
and set $\epsilon=g/20$. Choose
\[
C=20+\max(\{0\}\cup\{\text{right endpoints of source and target boxes}\}),
\qquad S_i=(C+4i,0,0).
\]
In a first pass give each source four consecutive slots: lift by ten,
translate above $S_i$, descend, and reshape there by the linear
reciprocal scale. In a second pass give each parked box three slots:
lift by ten, translate above its target, and descend. The total is
$7N$ slots, and every source is gone before a target is filled.
Widths remain at most one. Source gaps protect the first lift,
parking spacing protects strain and parking moves, target gaps protect
delivery, and travel height separates every horizontal transfer from
resting boxes. These inequalities remain strict under $2\epsilon$
padding. Applying Lemma~\ref{lem:bb-curl} in chronological order
fixes every stationary box and gives the exact full-box map to the
moving one. Thus overlapping source and target families cause no
interference.
Repeat the flat-ended template from zero and take its residual.
The integer sampling induction gives the table execution. In a
nonhalting run every endpoint box lies in $x_1>-1$; translations
interpolate corresponding endpoint points, strain occurs far to the
right, and waiting leaves the point fixed. The entire two-pass path
therefore remains in $x_1>-1$. A terminal sample lies inside the
observer. This proves the event for all times.
For the exact pressure class use Lemma~\ref{lem:b-comparison}(iv).
The prescribed compactly supported smooth velocity satisfies the
reference $CH^2\cap C^1L^2$ and bounded-gradient conditions. The
competing velocity has precisely these bounds and a $CL^2$ pressure
representative, as that case requires. Its cutoff argument uses the
pressure itself, with error $CR^{-1}\|p\|_2\|u-U\|_2$, and assumes
no extra integrability of its derivative. The lemma gives velocity
uniqueness; the $L^2$ pressure representative is then zero.
Compact support and
bounded derivatives of the prescribed velocity give its uniform
energy bound and its global material flow.
\end{proof}
\subsection{A two-cell append and its guarded inverse}\label{bc:two-cell-section}
Use Lemma~\ref{lem:bb-window-compiler} with initial frontier $r_0=1$,
renaming Ready, Go and Back as $M,R,L$ and $(E,P)$ as
$(\perp,\star)$. The track order is work, mark, history.
The Ready history coordinate and the marked Back row remain unrestricted;
the unmarked Back row tests precisely the right-neighbor history.
Thus the full table, not only its initialized execution, is identical.
At checkpoint $n$, its frontier is $p_n=1+n>h_n$, where $h_n$ is
the work-head position, and the exact cycle length is $2(1+n-h_n)+1$.
Lemma~\ref{lem:bb-window-rectangles} applies with
$(a,b)=(1,2)$: the complete allowed triple fixes both history writes,
and its full image has prefix lengths $(1+e,2-e)$.
\begin{theorem}[Guarded cylinders in a torus chart]\label{bc:guarded-torus}
On the unit torus, the fixed label $(1/8,0,0)$ and the strip
$\{x:x_1\pmod1\in(-1/4,0)\}$ realize halting by a general mean-zero
force with zero initial velocity and pressure. It is periodic after time
one and all its mixed derivatives are bounded. The torus comparison
class gives uniqueness.
\end{theorem}
\begin{proof}\label{bc:guarded-torus-proof}
Let $a$ be the augmented alphabet size and $k$ the control count in
Lemma~\ref{lem:bb-window-compiler}. Use zero-blank base $B=2a$, and enumerate
nonterminal and terminal states with positive integer indices $j\ge1$
and first-coordinate offsets respectively
$L_0+3j$ and $-L_0-3j$, where $L_0=4k+ka^3+10$.
Refinement by triples gives at most $N=ka^3$ branches. Their source
prefix lengths are $(1,2)$ and image lengths $(1+e,2-e)$, so the
smallest possible grid gap is $B^{-3}$. If $P_i,Q_i$ are lower corners,
the branch map is
$Q_i+\diag(B^{-e_i},B^{e_i})(y-P_i)$.
Choose three flat ramps $\alpha,\beta,\gamma$, active respectively in
$(1/8,1/4)$, $(3/8,5/8)$ and $(3/4,7/8)$. Set
\[
c_i(t)=((1-\beta)P_i+\beta Q_i,\ i(\alpha-\gamma)),
\quad l_i=1-\beta+\beta B^{-e_i}.
\]
The moving rectangle has side lengths $l_i B^{-1}$ and
$l_i^{-1}B^{-2}$. Use a physical padding
$\varepsilon=B^{-3}/4$ around each moving rectangle in three dimensions.
During ascent and descent the planar endpoint gaps separate these
neighborhoods; during the affine phase the unit height gaps separate them.
Use Lemma~\ref{lem:bb-curl} for the paths
$c_i+\diag(l_i,l_i^{-1},1)(y-(P_i,0))$ and sum the disjoint fields.
The first coordinate is an endpoint convex combination even though the
paths are based at lower corners rather than centers. All supports lie
in $[-2L_0,2L_0]^3$, by the definitions of $L_0$ and $N$.
Scale by $\rho=1/(32L_0)$, so the repeated mechanism is supported in
$[-1/16,1/16]^3$. Load the fixed label $(1/8,0,0)$ to the scaled
rational input code by a moving translation with cutoff radius $1/32$;
its support stays in $(-1/4,1/4)^3$. For a nonterminal initial code,
the entire first coordinate during loading is positive, and subsequent
nonterminal branches have first coordinate greater than $1/32$.
Terminal codes have first coordinate less than $-1/32$. Hence the stated
strip detects exactly terminal entry, including initial halting, without
any wraparound artifact. Periodize the compact curl fields and recompute
the force at physical $\nu$. Mean zero, bounded derivatives and comparison
follow from Lemma~\ref{bc:analytic}.
\end{proof}
\subsection{A filled half-line and mean-corrected vertical motion}
\label{bc:halfline-section}
The next recorder starts with an infinite constant history tail rather
than a single frontier symbol. Its guard and its mean-zero vertical
realization both have independent roles.
\begin{lemma}[History at a filled half-line]\label{bc:halfline-table}
A finite guarded table has a full-domain inverse and simulates the original
machine from a tape with a separately constant tail in each direction.
At checkpoint $n$ its nonempty history cells are exactly the absolute
indices below $2+n$.
\end{lemma}
\begin{proof}\label{bc:halfline-table-proof}
Normalize to one halt state $h$. For $i=(q,a)\in I=(Q\setminus\{h\})\times A$,
write $\delta(q,a)=(q_i,b_i,e_i)$. History letters are
$\mathcal H=\{\perp,\#\}\sqcup I$, with
$\mathcal H_+=\{\#\}\sqcup I$. The tracks are work, mark and history;
controls are $\mathsf R_q,\mathsf B_q,\mathsf T_i,\mathsf S_i$.
Write $\ell_j$ for the history at relative cell $j$.
Here is the complete table, with unmentioned cells unchanged:
\begin{equation}\label{bc:halfline-rules}
\begin{array}{c|c|c|l}
\text{input}&\text{output}&\text{move}&\text{condition}\\\hline
\mathsf R_q(a,0,k)&\mathsf T_i(b_i,0,k)&e_i&i=(q,a),\ q\ne h\\
\mathsf T_i(c,0,k)&\mathsf S_i(c,1,k)&1&k\in\mathcal H_+\\
\mathsf S_i(c,0,k)&\mathsf S_i(c,0,k)&1&k\in\mathcal H_+\\
\mathsf S_i(c,0,\perp)&\mathsf B_{q_i}(c,0,i)&0&\ell_{+1}=\perp\\
\mathsf B_q(c,0,k)&\mathsf B_q(c,0,k)&-1&k\in\mathcal H_+\\
\mathsf B_q(c,1,k)&\mathsf R_q(c,0,k)&0&k\in\mathcal H_+.
\end{array}
\end{equation}
Initialize $\mathsf R_{q_0}$ at head zero with marks zero, history $\#$
at every index below two and $\perp$ elsewhere. At checkpoint $n$, the
work head $p_n$ has $|p_n|\le n$ and the boundary is $r_n=2+n$.
After the work move, $p'=p_n+e_i\le n+1-1$, and halt codes $<-5$. Thicken the rectangles of
Lemma~\ref{lem:bb-onecell} by $[-1,1]$, give branch $i$ height $10i$,
and use the linear scale $g_i=1+\theta(\lambda_i-1)$ in the lift
schedule. A padding smaller than one tenth of the minimum of one
and all within-family planar gaps suffices. The middle strain has
potential $(0,0,(\dot g_i/g_i)(x_1-c_{i1})(x_2-c_{i2}))$.
Equation~\eqref{eq:bb-convex-coordinate} excludes every nonhalting
visit to $x_1<-5$. The table and routing are fixed for the machine;
only $x_*,y_*$ change with the input. The passive origin bit remains
at physical cell zero, so no absolute tape information is discarded.
For the second case first merge all halting work states into a single
state $q_h$, redirecting incoming transitions and retaining initial halting.
Delete the passive track from the direction recorder, and now let
$\mathcal A$ denote that full alphabet. Use odd digits with
$B=2|\mathcal A|+2$ and
$\lambda=1/[16(J+1)]$, where $J$ counts the controls. Number them
from zero and set
\[
S_s(x,y)=(A_s+\lambda x,1/4+\lambda y),\qquad
A_s=2\lambda\operatorname{index}(s)+
\begin{cases}5/8,&s=M(q_h,e),\\1/4,&\text{otherwise}.
\end{cases}
\]
The nonhalt squares lie in $[1/4,3/8]\times[1/4,1/4+\lambda]$;
halt squares lie in $[5/8,3/4]\times[1/4,1/4+\lambda]$.
Write $m$ for the number of branches. Use coding height $1/4$,
branch heights $1/2+i/[4(m+1)]$, and
the same linear first scale. One hundredth of the minimum of
$1/16$, $1/[4(m+1)]$, and the source and target planar gaps is
valid padding; absent pair gaps are omitted. All supports lie inside
the cube, and all horizontal widths are at most $\lambda$.
A localized translation with padding $1/1000$ loads the rational
initial point from the fixed label in one unit interval.
In a nonhalting period the center has first coordinate at most $3/8$
and half-width at most $\lambda/2$, hence is below $1/2$; loading
has the same bound. Halting codes lie in the observer. The exact
sampling induction gives one table step per period, with checkpoint
cycle $2(k+1-h_k)+3$. The record block beginning at cell one
recovers absolute indexing at checkpoints. An initial halt is
observed at time one.
\end{proof}
\subsection{A one-time torus loader}
The next placement uses a separate first interval to load the code. Only
the instruction cycle is then repeated, so periodicity begins at time one.
\begin{theorem}[A separate one-unit loader on the torus]
\label{ba:loaded-torus}
For every machine and finite input and every fixed computable $\nu>0$,
there is an effective smooth
mean-zero force, one-periodic for $t\ge1$ with bounded mixed derivatives,
whose unique smooth zero-data solution with zero normalized pressure
satisfies
\[
\exists t\ge0:\ \Phi_t(1/2,1/2,1/4)
\in(1/16,1/4)\times\mathbb T^2
\quad\Longleftrightarrow\quad\text{halting}.
\]
Uniqueness holds in the classical class of Section~\ref{sec:model}.
\end{theorem}
\begin{proof}\label{ba:loaded-torus-proof}
Use the single-halt move-first recorder of
Lemma~\ref{lem:bb-movefirst}, without an entry state. Let
$\Gamma$ be its full alphabet and $N$ its number of controls. Put
$B=2|\Gamma|+1$, $a=1/(16N)$, the halt square at
$(1/8,1/4)+a[0,1]^2$, and the other squares at
$(1/2+2ja,1/4)+a[0,1]^2$, $0\le j\le N-2$.
Use the minimal subdivision, so right and stay moves are not split by
the left letter. Its gaps are at least $a/B^2$.
For $M$ branches choose
\[
h=\frac1{64(M+1)},\quad Z_i=\frac12+\frac{i}{4(M+1)},\quad
\varepsilon=\min\{1/128,h/4,a/(4B^2)\}.
\]
Apply linear-first-scale full-box routing from height $z_0=1/4$.
Middle heights differ by $16h$; source and target gaps handle the
vertical phases. Centers lie in
$[1/8,5/8]\times[1/4,5/16]$, widths are at most $a$, and all padded
boxes lie in $(1/16,15/16)^3$. This is a construction on the specified
unit torus, with no change of physical viscosity.
In the first unit interval translate a box of side $1/64$ from the
fixed label to the rational initial code, using Lemma~\ref{lem:bb-curl}.
The segment and a small padding lie inside the same chart. Repeat
only the instruction cycle afterward. Code induction proves the
checkpoint identities. A halting code has first coordinate in
$[1/8,3/16]$, including an initial halt reached by loading. For a
nonhalting run both centers of every used branch have first coordinate
at least $1/2$, with half-width at most $a/2$; loading has the same
lower center bound. Hence the particle always stays to the right of
$1/4$. Residual realization proves every analytic assertion. The
separate loading field accounts for the qualification $t\ge1$ in the
periodicity statement.
\end{proof}
\subsection{Repeated torus loading branches}\label{ba:torus}
A loader can instead be one branch of every period. To keep its target
separate from all computational targets, place the initial code in a state
square with no incoming instruction. This condition concerns the entire
target family, not only the distinguished execution. We implement the
same startup principle first with full boxes, then with fixed columns.
\begin{theorem}[A loader repeated as part of the processor]
\label{ba:periodic-torus}
For every finite machine and input, and every fixed computable $\nu>0$,
there is an effective smooth mean-zero force on the unit flat torus,
one-periodic from time zero, with all mixed derivatives bounded. Its
zero-data solution has pressure zero, is unique in the classical class
of Section~\ref{sec:model}, and satisfies
\[
\exists t\ge0:\ \Phi_t(1/8,1/8,1/4)
\in\{5/81/(2B)$ from the endpoints
of $[0,1]$, so the new target lies strictly inside the entry square.
There is no other target in that square. Both enlarged branch lists
therefore remain separately separated, with gaps at least $a/B^2$.
The extra source is separated from all state squares.
Number the branches, including loading, by $1\le i\le M$. Set
\[
z_0=\tfrac14,\qquad h=\frac1{16(M+1)},\qquad
Z_i=\tfrac12+\frac{i}{4(M+1)},\qquad
\varepsilon=\frac1{10}\min(h,a/B^2).
\]
Thicken each planar source by $[z_0-h,z_0+h]$. Lift the resulting
full box to center height $Z_i$, move it horizontally, and lower it
to $z_0$. For an instruction with displacement $d_i$, use
$D_i=\operatorname{diag}(B^{-d_iw_2},B^{d_iw_2},1)$;
loading has $d_i=0$. Its center is the planar endpoint interpolation
plus $(Z_i-z_0)(w_1-w_3)e_3$. Adjacent middle heights are $4h$ apart,
while the padded half-thickness is $h+\varepsilon<2h$.
Source and target gaps handle the other phases. Horizontal half-widths
never exceed $a/2$, horizontal centers lie between $1/8$ and $3/4+a$,
and vertical centers lie between $1/4$ and $3/4$. The padded motions
are thus strictly inside the unit cube. Lemma~\ref{lem:bb-curl} gives a
chart field, extended periodically in space and time. Its zero time
collars give zero initial velocity.
The first cycle loads the fixed particle. Each subsequent cycle maps
its exact code to the next local code until the first halt, or forever
in a nonhalting run. Lemma~\ref{lem:bb-movefirst} identifies the finite
checkpoint times with machine steps. The added entry state handles an
original initial halt. A halt code is in the target slab. For a
nonhalting execution every used source and target center, including the
loader, has first coordinate at most $3/8$. Exponential interpolation
keeps the center below that bound and the half-width below $a/2$;
the particle cannot reach $5/8$ at any intermediate time. The final
transition into a halt is not subject to this nonhalting-endpoint bound.
Lemma~\ref{ba:realization} gives the force, its mean, its effective
derivatives, and uniqueness. Repeating the loading field is harmless
to the initialized trajectory and makes the force periodic from zero.
\end{proof}
\paragraph{A fixed-column implementation.}\label{bc:unit-periodic-section}
The same untargeted-entry condition also permits a repeated initializer
using fixed source and target columns. The next construction retains its
explicit column potentials and its fixed label; its middle motion uses
linear interpolation of the first planar scale.
\begin{theorem}[A periodic initializer on the unit torus]
\label{bc:unit-periodic-event}
There is an effective halting realization on $\T^3$ with fixed label
$a_*=(3/8,1/4,1/2)$ and fixed strip $5/8R+2$.
Lemma~\ref{lem:bb-lift-parking} gives the full $7N$-motion schedule.
Repeat this unit pulse from zero. Nonhalting endpoint boxes and all
parking boxes have nonnegative first coordinate. Translations preserve
that bound, and strain takes place in positive parking envelopes.
Therefore a nonhalting path never enters the negative observation box.
A halting code lies in $[-3,-2]\times[0,1]\times\{0\}$ inside it.
The residual proposition finishes the proof. Loading was a repeated
instruction, so no exceptional first-time force is required.
\end{proof}
\begin{remark}[A different observation regime]
For comparison, the distinct velocity-field detector of
\cite[Theorem~1.1(2) and Section~5]{OpenAI379EulerianC} uses
$\R^2\times\T$ and the event
$\int_{\{X_2>0\}\times\T}u_3>1/2$. Its force has globally bounded
mixed derivatives and horizontal support compact on each finite time
interval. The nonnegative scalar velocity has diffusion tails; after
loading its total mass is one. Its cutoff estimate gives loss
$\delta<1/24$, nonhalting mass at most $\delta$ in the observed half-space,
and halting mass above $3/4$. Those conclusions follow from the cited
advection--diffusion proof and have no uniform spatial support assertion.
They are a separate force and observation regime from the material test
proved here.
\end{remark}
==== END Incompressible-Box-Transport-and-Finite-Computation//initialization.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//introduction.tex ====
\section{Introduction}\label{sec:introduction}
A finite instruction can prescribe a volume-preserving affine map on a
box. A fluid realization must do more: it must move every point of that
box, preserve incompressibility outside it, and leave the other instructions
available. Source boxes and destination boxes can overlap. Moreover, a
particle used to report the answer must avoid the observation region
throughout every nonhalting transition. We study these requirements together.
Our geometric result is Theorem~\ref{lem:bb-balanced-box-routing}.
It extends any finite collection of positive diagonal determinant-one maps
between separately disjoint rational solid boxes to the time-one map of
an effective smooth compactly supported incompressible velocity. The maps
hold on neighborhoods of the entire closed boxes. The proof separates
centers, changes shapes, evacuates the sources to storage, and delivers
boxes along paths avoiding everything still occupied. Positive margins make
the construction effective and permit additional observation guards.
The first computational application, Theorem~\ref{thm:bb-balanced-three-stack},
uses all three coordinates for symbolic data. Three stacks retain a tape
and its execution history. Every local instruction preserves the total
length of the prefixes it replaces, so its affine map has determinant one.
A fixed particle, initially in a fluid at rest, enters the fixed cube
$(-1,2)^3$ exactly when the encoded machine halts. The force is smooth,
compactly supported in space and periodic after a one-unit loading interval.
Its prescription uses the finite rule table, not the executed computation.
This application completes the main construction before we vary
initialization, history guards and observers.
The passage from a symbolic instruction to a fluid observation begins
with retained history, which makes the instruction injective on its
entire domain of tapes, not just along the chosen computation. A rule
specifies finite prefixes and leaves their infinite tails free; gapped
radix expansions place the codes in separately disjoint closed source
and target boxes. Prefix replacement extends to an affine map of each
whole box, which smooth incompressible transport then realizes.
An additional geometric bound must exclude unintended visits to
the observer between successive rule times. Finally, the prescribed
velocity determines a body force, and an energy comparison identifies
the resulting solution in the declared class. Each later variant changes
one or more of these steps while retaining this separation of tasks.
The remaining constructions answer more specific questions. Two tape
coordinates can be thickened into solid boxes, or their planar scale changes
can be compensated in the normal direction. A temporary mark, a persistent
mark, an explicit frontier and a boundary between occupied and unused
history give different partial instruction maps. We prove their inverses
on their full domains because this is what separates the target boxes.
A bounded observation box permits horizontal motion above it; a slab needs
a coordinate bound throughout that motion. Storage and obstacle detours
supply further alternatives. Loading once gives eventual periodicity,
whereas placing the initial code at the fixed label or repeating a loading
branch gives periodicity from time zero. Onto slow clocks give separate
decaying forces with derivative norms square-integrable in time.
\subsection{Prior work and the role of forcing}
The computational starting point is Turing's finite machine model
\cite{Turing1936}. Landauer's discussion of retained intermediate information
\cite[Section~3]{Landauer1961} and Bennett's explicit history recording
\cite[Eqs.~(9)--(11)]{Bennett1973} explain how an irreversible computation
can be embedded in an invertible mechanism. We use that principle, but
prove the finite guarded tables and their complete inverses here. No
complexity bound or history-erasure procedure from those works is asserted
for these tables.
Moore's generalized shifts turn symbolic updates into affine operations
on positional expansions \cite{Moore1990,Moore1991}. Our gapped expansions
serve the same purpose. Moore also realizes invertible generalized shifts
by smooth planar diffeomorphisms and three-dimensional flows
\cite[Section~6, Theorem~12 and its corollary]{Moore1991}.
Cardona, Miranda, Peralta-Salas and Presas extend the affine maps on
neighborhoods of separated planar blocks to an area-preserving disk
diffeomorphism, using contraction, transport, expansion and Moser's area
correction \cite[Proposition~5.1]{CardonaMirandaPeraltaSalasPresas2021}.
This is a geometric predecessor of the routing problem here. Our local
curl formulas implement positive diagonal determinant-one motions in
three dimensions directly; the proof supplies effective support and
neighborhood bounds and controls observation at intermediate times.
Computational universality has been established for stationary Euler
flows with adapted geometry \cite{CardonaMirandaPeraltaSalasPresas2021}.
Cardona, Miranda and Peralta-Salas also construct universal Euclidean
Beltrami flows, with infinite energy, and robust simulations of
tape-bounded machines by unforced Navier--Stokes flows on a flat torus
\cite[arXiv version~3, Theorem~1 and Remark~29]{CardonaMirandaPeraltaSalas2023Beltrami}.
Dyhr, Gonz\'alez-Prieto, Miranda and Peralta-Salas obtain stationary
unforced Navier--Stokes universality using an adapted metric and the
Hodge Laplacian
\cite[Theorem~A]{DyhrGonzalezPrietoMirandaPeraltaSalas2026}.
Here the ambient metric is fixed and flat, the viscosity is
prescribed, and the initial velocity is zero. The machine is placed in an
external body force. Given an explicit solenoidal velocity $U$, the identity
$f=U_t+(U\cdot\nabla)U-\nu\Delta U$ supplies that force. The mathematical
work is the effective construction of $U$ and its all-time observation,
followed by uniqueness in the stated comparison class. We do not assert a
regularity theorem for arbitrary fluid data.
The companion \cite{OpenAI379Shear} gives an entry construction using
solenoidal shears, and \cite{OpenAI379Sheets} treats prefix instructions
and other planar or normal-compensated realizations. They explain adjacent
geometric choices; the proofs of the present full-box mechanisms are given
here. The final input-offset application explicitly imports the planar
processor and spatial clock of \cite{OpenAI379StationaryA2}, with the exact
parameters and conclusions stated in Section~\ref{bc:input-offset-section}.
The companion \cite{OpenAI379EulerianC} detects computation from an
integral of the velocity field on $\R^2\times\T$ via an
advection--diffusion estimate. The end of
Section~\ref{sec:bb-initialization} compares that event with the path of
one material particle studied here.
\subsection{Equation, effectiveness and reading path}
Throughout $\nu>0$ is fixed, $\T=\R/\Z$, and the fluid equation on
$\R^3$ or the flat unit torus is
\begin{equation}\label{eq:box-NS}
u_t+(u\cdot\nabla)u=-\nabla p+\nu\Delta u+f,
\qquad \divg u=0,\qquad u(0)=0.
\end{equation}
A material label $a$ has path $\dot\Phi_t(a)=u(t,\Phi_t(a))$,
$\Phi_0(a)=a$. We also write $X(t;a)=\Phi_t(a)$. The smooth constructed
velocities have global material flows. On the torus pressure is normalized
to have mean zero. Whole-space pressure assumptions are specified in
Section~\ref{sec:model} and in the individual statements. Unless explicitly
projected, the force is a general body force and the constructed pressure
is zero.
An effective field is a finite program that evaluates every requested
mixed derivative at computably supplied arguments to any prescribed
rational error and computes the bounds used in the proof. The finite
machine and its finite input determine the coefficients, cutoffs and
routing schedule. Evaluating the field cannot first compute the selected
machine run. All algorithmic statements concern computable $\nu$;
the same formulas are valid for arbitrary positive $\nu$, with evaluation
relative to that parameter. The encoded trajectories are exact real
trajectories. No uniform precision tolerance for an unbounded computation
is asserted.
Section~\ref{sec:model} states the analytic comparison tools.
Section~\ref{sec:bb-boxes} proves the box motions, and
Section~\ref{sec:bb-balanced-three-stack} gives the balanced-stack application.
Sections~\ref{sec:bb-frontiers} and~\ref{ba:codes} establish reusable
recorders and full closed-rectangle coding; Section~\ref{ba:geometry}
supplies the common interpolation formulas. The subsequent constructions
can be read according to the requirement being changed:
\begin{center}
\begin{tabular}{@{}p{.24\textwidth}p{.71\textwidth}@{}}
\toprule
Choice & Where it is developed \\
\midrule
Initialization and time periodicity &
Section~\ref{sec:bb-initialization} compares a separate loader, direct
placement of the initial code, and a loading branch repeated in every
period. An untargeted start square permits the last choice. \\[4pt]
History domains &
Sections~\ref{ba:marked-routes} and~\ref{sec:bb-guarded} distinguish
persistent marks, fresh history and neighboring-cell guards. Their
inverses on arbitrary allowed tapes determine the separated image boxes.
\\[4pt]
Observer geometry &
Section~\ref{ba:ordered-routes} treats bounded boxes, slabs and
half-spaces. Sections~\ref{sec:solid-routing} and~\ref{sec:torus-routing}
develop storage, detours and fixed torus charts with the corresponding
intermediate-time bounds. \\[4pt]
Decay and stationarity &
The logarithmic clock in Section~\ref{ba:euclidean} produces separate
decaying forces. Section~\ref{sec:clocks} treats further profile choices,
including the imported input-offset processor. \\
\bottomrule
\end{tabular}
\end{center}
The five whole-space comparison classes are stated in
Lemma~\ref{lem:b-comparison}. Each alternative fixes its own label,
observer, support, pressure class and time dependence; these are separate
constructions rather than simultaneous guarantees for one force.
==== END Incompressible-Box-Transport-and-Finite-Computation//introduction.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//main.tex ====
\documentclass[11pt]{article}
\ifdefined\pdftrailerid\pdftrailerid{}\fi
\ifdefined\pdfinfoomitdate\pdfinfoomitdate=1\fi
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,booktabs,enumitem,microtype,longtable,array,tikz}
\usepackage[margin=1in]{geometry}
\usetikzlibrary{arrows.meta,positioning}
\usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}\newtheorem{definition}[theorem]{Definition}
\theoremstyle{remark}\newtheorem{remark}[theorem]{Remark}
\newcommand{\R}{\mathbb R}\newcommand{\T}{\mathbb T}\newcommand{\Z}{\mathbb Z}\newcommand{\N}{\mathbb N}\newcommand{\Q}{\mathbb Q}
\DeclareMathOperator{\supp}{supp}\DeclareMathOperator{\divg}{div}\DeclareMathOperator{\diag}{diag}\DeclareMathOperator{\id}{id}
\hypersetup{pdftitle={Incompressible Box Transport and Finite Computation},pdfauthor={OpenAI}}
\title{Incompressible Box Transport and Finite Computation}
\author{OpenAI}
\date{September 27, 2026}
\begin{document}\maketitle
\begin{abstract}
We realize finite positive diagonal affine maps of determinant one by
effective smooth incompressible flows on neighborhoods of entire closed
rational solid boxes. The source and target families are each disjoint, but may
overlap each other. The construction uses localized curls, evacuation to
storage and obstacle detours. A balanced three-stack recorder then assigns
every machine and finite input a smooth Navier--Stokes force with one
compact spatial support, periodic after a loading interval, at any fixed
positive computable viscosity. The fluid starts at rest, and one fixed
particle enters one fixed open cube exactly when the machine halts.
Further constructions give fixed torus charts, periodicity from time zero,
alternative history guards and bounded or slab observers. Onto slow clocks
yield separate decaying forces. Each construction includes an all-time
observation proof, effective derivative bounds and a stated pressure
comparison class.
\end{abstract}
\input{introduction}
\input{analysis}
\input{geometry}
\input{balanced}
\input{compilers}
\input{coding-conventions}
\input{routing-conventions}
\input{initialization}
\input{compact-processors}
\input{persistent-history}
\input{guarded-history}
\input{observer-routing}
\input{solid-routing}
\input{torus-routing}
\input{clocks}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END Incompressible-Box-Transport-and-Finite-Computation//main.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//observer-routing.tex ====
\section{Routing choices dictated by the observer}
\label{ba:ordered-routes}
The preceding processors move every branch by a prescribed affine map.
We now ask which parts of that motion must be controlled between the
integer times. A bounded observation box allows horizontal work above
the observer. A slab observer requires a bound on the horizontal
coordinate throughout the motion. Finally, a half-space observer can be
protected by moving all sources to storage before delivering any target.
These three requirements lead to the constructions below.
\subsection{Four motions above a bounded observer}
\label{ba:four-stage-section}
We first separate ascent, scaling, translation and descent. The two
planar motions occur at a fixed height, so their exclusion from the
bounded observer follows without a horizontal coordinate estimate.
Let $D_i,E_i\subset\mathbb R^2$, $1\le i\le N$, be closed rational
axis-aligned rectangles with positive side lengths. Assume the $D_i$
are pairwise disjoint and the $E_i$ are pairwise disjoint, and let
\[
A_i(b)=c_i'+\operatorname{diag}(\lambda_i,\lambda_i^{-1})(b-c_i),
\qquad \lambda_i\in\mathbb Q_{>0},\qquad A_i(D_i)=E_i,
\]
where $c_i,c_i'$ are their centers. No separation between a source and
a target is assumed. All distances used to choose coordinate padding
below are distances for the maximum norm.
Let $\vartheta=\sigma$ be the effective flat switch in
\eqref{eq:p2-step}. For a closed interval, possibly reduced to one
point, define
\[
\chi_{[a,b],\epsilon}(v)=
\vartheta\!\left(\frac{2(v-a+\epsilon)}\epsilon\right)
\vartheta\!\left(\frac{2(b+\epsilon-v)}\epsilon\right).
\]
For a box $K$, let $\chi_{K,\epsilon}$ be the product over its three
coordinates. It is one on the $\epsilon/2$ enlargement of $K$ and
zero outside its $\epsilon$ enlargement. The effective profile estimates
following \eqref{eq:p2-step} apply to these products and rescalings.
\begin{lemma}[Four-stage realization]\label{ba:four-stage}
The rectangle data determine effectively a smooth divergence-free field
$W(s,x)$ supported in a compact subset of $(0,1)\times\mathbb R^3$.
Its time-one map is $(b,z)\mapsto(A_i(b),z)$ on a neighborhood of
each $D_i\times\{0\}$. For a point of that plate, every planar
motion occurs at height $4i$, while ascent and descent preserve its
source and target planar coordinates, respectively. The scaling can
use either linear interpolation of its positive first factor or linear
interpolation of the logarithm of that factor.
\end{lemma}
\begin{proof}\label{ba:four-stage-proof}
The empty family is realized by zero. Otherwise put $h_i=4i$,
$w_i=c_i'-c_i$, and
\[
\Theta_k(s)=\vartheta(10s-(2k-1)),\qquad 1\le k\le4.
\]
Only the $k$th motion is active on the interval
$[(2k-1)/10,2k/10]$; the four intervals are disjoint. If
$a_{ij},a'_{ij}$ are the source and target half-widths, respectively,
take the four swept boxes
\[
\begin{aligned}
K_{i1}&=D_i\times[0,h_i],\\
K_{i2}&=\{c_i+(u_1,u_2):|u_j|\le\max(a_{ij},a'_{ij})\}
\times\{h_i\},\\
K_{i3}&=\operatorname{box}(E_i-w_i,E_i)\times\{h_i\},\\
K_{i4}&=E_i\times[0,h_i].
\end{aligned}
\]
Here $\operatorname{box}$ denotes the smallest axis-aligned rectangle
containing the two displayed rectangles. Within each of the four
families these boxes are disjoint: source separation handles the first,
target separation the fourth, and distinct heights the other two.
Choose a positive rational $\epsilon<1$ smaller than one quarter of
every pairwise distance within these families. A family with fewer than
two members contributes no restriction. Its $\epsilon$ enlargements
are then still disjoint.
Choose either
\[
\mu_i=1+(\lambda_i-1)\Theta_2
\quad\hbox{or}\quad \mu_i=\lambda_i^{\Theta_2},
\qquad \gamma_i=\dot\mu_i/\mu_i.
\]
The factors $\mu_i$ and $\mu_i^{-1}$ stay between their respective
endpoint values. Use the coordinate potentials
\[
\begin{aligned}
B_{i1}&=(0,h_i x_1,0),&
B_{i2}&=(0,0,(x_1-c_{i1})(x_2-c_{i2})),\\
B_{i3}&=(0,0,w_{i1}x_2-w_{i2}x_1),&
B_{i4}&=(0,-h_i x_1,0).
\end{aligned}
\]
For the logarithmic-scale construction we also retain the alternative
translation potentials, with $e_3=(0,0,1)$,
\[
B_{i1}=\tfrac12(h_ie_3)\times x,\qquad
B_{i3}=\tfrac12(w_{i1},w_{i2},0)\times x,\qquad
B_{i4}=\tfrac12(-h_ie_3)\times x.
\]
Their curls equal the same constant translation vectors. Their localized
fields need not agree away from the plates. Either choice gives
\[
W=\sum_i\left[
\dot\Theta_1\nabla\times(\chi_{K_{i1},\epsilon}B_{i1})
+\gamma_i\nabla\times(\chi_{K_{i2},\epsilon}B_{i2})
+\dot\Theta_3\nabla\times(\chi_{K_{i3},\epsilon}B_{i3})
+\dot\Theta_4\nabla\times(\chi_{K_{i4},\epsilon}B_{i4})\right].
\]
This is a finite sum of spatial curls. Its spatial support is compact,
its time support is contained in $[1/10,8/10]$, and the lower bound
$\mu_i\ge\min(1,\lambda_i)>0$ proves smoothness and effectivity.
For $b\in D_i$, the successive paths are
\[
\begin{array}{c|c}
\text{motion}&\text{position}\\\hline
\text{ascent}&(b,h_i\Theta_1)\\
\text{scaling}&(c_i+\operatorname{diag}(\mu_i,\mu_i^{-1})(b-c_i),h_i)\\
\text{translation}&(A_i(b)-w_i+\Theta_3w_i,h_i)\\
\text{descent}&(A_i(b),h_i(1-\Theta_4)).
\end{array}
\]
Each path stays in its swept box. On a neighborhood of that box the
uncut curls are, in order,
\[
h_ie_3,\qquad (x_1-c_{i1},-(x_2-c_{i2}),0),\qquad
(w_{i1},w_{i2},0),\qquad -h_ie_3.
\]
After multiplication by the time coefficients, these are the prescribed
affine velocities. The motion matrices are positive diagonal of determinant
one, the plateau margin is $\epsilon/2$, and the support padding is
$\epsilon$. The disjoint swept boxes therefore permit the plateau
argument of Lemma~\ref{lem:bb-curl}, with $\eta=\epsilon/2$.
It identifies these particular localized fields with the displayed
paths on the plateaux. The normalized diagonal factors are bounded by
$L_i=\max(1,\lambda_i,\lambda_i^{-1})$, so that lemma gives the
full initial neighborhood of radius $\epsilon/(4L_i)$.
\end{proof}
We use the following precise Euclidean comparison class in the applications:
on each $[0,T]$,
\begin{equation}\label{ba:euclidean-comparison-class}
\begin{gathered}
u\in C([0,T];H^2(\mathbb R^3))\cap C^1([0,T];L^2(\mathbb R^3)),\\
u,\nabla u\in L^\infty([0,T]\times\mathbb R^3),\qquad
p\in C([0,T];H^1(\mathbb R^3)).
\end{gathered}
\end{equation}
The pressure representative is part of this condition; it is zero for
our constructed solution. These comparison velocities and pressures
satisfy case (i) of Lemma~\ref{lem:b-comparison}. We retain the stated
bounded-gradient requirement here, although that case does not need it.
That lemma also gives a smooth material diffeomorphism on every finite
interval. Differentiating its trajectory equation gives
$\dot J=(D_xu)(t,\Phi_t(a))J$, $J(0)=I$, for $J=D_a\Phi_t$.
Consequently
\begin{equation}\label{ba:material-volume}
\frac d{dt}\det J=(\nabla\cdot u)(t,\Phi_t(a))\det J=0,
\qquad \det J=1.
\end{equation}
The same proof applies to periodic lifts on the torus. The residual
formulas below are affine in viscosity, so they also define the asserted
solutions for any real $\nu>0$, with effectivity relative to $\nu$.
\begin{theorem}[Two bounded-box material tests]\label{ba:bounded-box-tests}
Fix a positive computable viscosity $\nu$. A deterministic one-tape
machine and finite input effectively determine each of the following
general body forces on $[0,\infty)\times\mathbb R^3$. Both have zero
initial velocity, a unique global smooth velocity in
\eqref{ba:euclidean-comparison-class}, and constructed pressure zero.
\begin{enumerate}
\item The force and velocity have one fixed compact spatial support, all
mixed derivatives bounded, and period one for $t\ge1$. The particle
from $0$ enters $(-5,-2)\times(-1,2)\times(-1,1)$ exactly when the
machine halts.
\item The force and velocity have one fixed compact spatial support and,
for all $k\ge0$ and spatial multi-indices $\alpha$, satisfy
\begin{equation}\label{ba:eighteen-decay}
\|\partial_t^k\partial_x^\alpha u(t)\|_\infty+
\|\partial_t^k\partial_x^\alpha f(t)\|_\infty
\le C_{k,\alpha}(1+t)^{-1-k}.
\end{equation}
The particle from $(2,0,0)$ enters
$(-1/2,3/2)\times(-1/2,3/2)\times(-1/2,1/2)$ exactly when the
machine halts. In particular $f\in L^2([0,\infty);H^m(\mathbb R^3))$
for every integer $m\ge0$.
\end{enumerate}
The supports and constants may depend on the machine and input. The
labels and observers do not. All force derivatives admit evaluation to
any prescribed rational error by a finite program.
\end{theorem}
\begin{proof}\label{ba:bounded-box-tests-proof}
Normalize to a single halt state $q_H$ and a complete nonhalting table,
as in Section~\ref{ba:codes}. Apply the move-first recorder of
Lemma~\ref{lem:bb-movefirst}. Its history frontier starts at cell $2$;
all other history cells are empty and all mark bits are zero. At
checkpoint $n$, the frontier is $g_n=n+2$, the work head satisfies
$|h_n|\le n$, and the next work instruction ending at $h'$ is completed
in exactly $4+2(g_n-h')$ local steps. Thus every required checkpoint
is reached after finitely many steps. The incoming-direction and
written-symbol inverse properties hold on the whole partial table.
The records recover the successive work instructions and hence the
absolute head position, even though the geometric code is head-relative.
Let $m$ be the full alphabet size. Use odd digits $\beta(a)$ in
$\{1,3,\ldots,2m-1\}$, with base $B=2m+1$ in the first construction
and $B=2m+2$ in the second. Set
\[
C(a_0a_1\cdots)=\sum_{r\ge0}\beta(a_r)B^{-r-1},\quad
P_a(v)=\frac{\beta(a)+v}{B},\quad D_a(v)=Bv-\beta(a),\quad
I_a=P_a([0,1]).
\]
Let $\xi$ encode the left string, nearest cell first, and $\eta$
the string starting at the head. The complete right, stay and left
branch maps are \eqref{eq:bb-one-cell-branches}, with its digit
$d(a)$ equal to $\beta(a)$ and its move $e$ equal to $d$ here.
The left move is split over every full symbol $c$. The shared lemma
proves the full-rectangle endpoint formulas, arbitrary-tail action and
linear part $\operatorname{diag}(B^{-d},B^d)$.
Translate each state square horizontally. In the first construction
give the halt state offset $-4$ and the other states offsets $2+2r$,
$r\ge0$; in the second use halt offset $0$ and other offsets $3r$,
$r\ge1$. Thus the code is $(o_s+\xi,\eta,0)$.
Lemma~\ref{lem:bb-onecell} applies with exactly these odd digits and
either displayed base: its incoming conditions are those supplied by
the move-first recorder, and \eqref{eq:bb-one-cell-branches} is its minimal
left-split branch list. Thus both complete closed rectangle families
are separately disjoint, with within-state gaps at least $B^{-2}$.
This proves the hypotheses of Lemma~\ref{ba:four-stage}; no separation
between the source and target families is required.
Use its linear scale and coordinate potentials for the first construction;
use its logarithmic scale and symmetric translation potentials for the
second. In the latter case the scaling coefficient is exactly
$(\log\lambda_i)\dot\Theta_2$ and the scale matrix is
$\operatorname{diag}(\lambda_i^{\Theta_2},\lambda_i^{-\Theta_2})$.
Denote the resulting step field by $W_{\rm step}$.
The initial code $P_{\rm in}$ is rational. A constant full-symbol
tail of digit $\beta_0$ after $M$ positions contributes
$B^{-M}\beta_0/(B-1)$; all other contributions are finite. For the
appropriate fixed label $a_*$, put $v_0=P_{\rm in}-a_*$ and let
$J_0$ be the smallest box containing $[a_*,P_{\rm in}]$. The loader
\[
W_{\rm in}(s,x)=\frac d{ds}\vartheta(2s-1/2)\,
\nabla\times\left(\chi_{J_0,1}(x)\frac{v_0\times x}{2}\right)
\]
has the exact marked path $a_*+\vartheta(2s-1/2)v_0$, since the
uncut curl is $v_0$ and the entire segment lies in the plateau. Form
\[
W(s,x)=W_{\rm in}(s,x)+\sum_{n\ge1}W_{\rm step}(s-n,x).
\]
The zero time collars make this smooth, initially zero, and periodic for
$s\ge1$, with common compact spatial support and bounded mixed derivatives.
At time $s=1+k$ the marked particle is exactly the code after $k$ local
instructions, through the first halt or indefinitely in a nonhalting run.
This follows by induction from the full-rectangle maps. Each halt code
lies in its stated observation box, including one reached by loading an
initially halted machine. In a nonhalting run, the first loader has
$x_1\ge0$ and later nonhalt codes have $x_1\ge2$; for the second these
bounds are $x_1\ge2$ and $x_1\ge3$. Ascent and descent leave those
planar coordinates unchanged. Every planar motion has height $4i\ge4$,
outside both observers. The intervening stationary intervals add no new
positions. This proves exclusion at every real time.
For the first case take $u=W$ and its residual force. For the second set
\[
s(t)=\log(1+t),\quad \theta(t)=(1+t)^{-1},\quad
u(t,x)=\theta(t)W(s(t),x),
\]
whose force is
\[
f=\theta^2\{-W+\partial_sW+(W\cdot\nabla)W\}(s(t),x)
-\nu\theta\Delta W(s(t),x).
\]
Lemma~\ref{ba:log-clock} applies to this fixed-support, bounded-derivative
template. Its identity
$\partial_t(\theta^jY(s(t)))=\theta^{j+1}(\partial_s-j)Y(s(t))$
proves \eqref{ba:eighteen-decay} for every $k,\alpha$. Fixed support
then gives the stated $L^2_tH^m_x$ conclusion. Its clock is onto, with
inverse $t=e^s-1$, so the material path is the old path composed with
$s(t)$ and the event is unchanged.
In both cases smoothness, fixed support and bounded derivatives on
finite time intervals put the reference velocity in $CH^2\cap C^1L^2$
with bounded velocity and gradient. It is solenoidal and initially zero.
Lemma~\ref{lem:b-comparison}(i) therefore gives the exact solution
$(u,0)$ and uniqueness in \eqref{ba:euclidean-comparison-class}.
The effective profiles following \eqref{eq:p2-step}, applied to the
finite table, rational initial code, finitely many boxes and repeated
template, give all derivative evaluations. A coarse time approximation
encloses finitely many possible active translates, avoiding equality
tests at seams. The prescription never computes the machine's run.
\end{proof}
\subsection{Small plates and a slab observer on the unit torus}
\label{ba:torus-plates-section}
A slab observer has no height restriction, so elevation alone cannot
protect it. We now keep each nonhalting branch inside a narrow range of
first coordinates throughout its motion. The delayed-head recorder also
allows arbitrary finitely supported work data on both sides of the head.
\begin{theorem}[Torus plates and square-integrable forcing]
\label{ba:torus-plates}
Fix a positive computable $\nu$. Every deterministic one-tape machine
with finitely specified tape, blank outside a finite set, effectively
determines a smooth mean-zero general force on
$\mathbb T^3=\mathbb R^3/\mathbb Z^3$, with bounded mixed derivatives
and period one for $t\ge1$. Its zero-data Navier--Stokes solution is
globally smooth and unique among smooth solutions, with pressure normalized
to mean zero and constructed pressure zero. For
$P_*=(1/4,1/2,1/4)$ and $X(t)=\Phi_t(P_*)$, the event
\begin{equation}\label{ba:torus-plate-event}
\exists t\ge0:\quad X_1(t)\pmod1\in(1/2,7/8)
\end{equation}
is equivalent to halting.
A separate effective force has the same zero data, event, uniqueness and
mean-zero property, is supported in one compact sub-box of $(0,1)^3$,
has all mixed derivatives bounded, and satisfies
\begin{equation}\label{ba:cube-rate}
\|\widehat f(t,\cdot)\|_\infty=O((1+t)^{-2/3}).
\end{equation}
In particular $\widehat f\in L^2([0,\infty)\times\mathbb T^3)$.
\end{theorem}
\begin{proof}\label{ba:torus-plates-proof}
Use Lemma~\ref{ba:six-compiler} with the frontier initially at cell $1$.
Its main control $M_{q,e}$ records the last work displacement $e$, initially
zero. In one work step it writes and marks the old head, records
$(q,e,a)$ at the frontier, returns to the mark, and only then makes the
work displacement. All other initial history cells are blank and every
mark is zero. At checkpoint $n$ the frontier is $n+1$, $|h_n|\le n$,
and the next checkpoint takes exactly $2(n+1-h_n)+3$ local steps.
These scans test history and marks, so the proof is unaffected by work
symbols on either side of the head. Main controls with halting $q$ have
no outgoing rules. The full-domain incoming properties are those of
the cited lemma.
Let $S$ be the local state set, $N=|S|$, let $S_h$ be its halting main
states, and enumerate $S$ by $n(s)=0,\ldots,N-1$. Put
\[
\ell=\frac1{16(N+1)},\quad z_0=\frac14,\quad
x_s=\frac18+2\ell n(s)+\frac12\mathbf1_{s\in S_h},\quad
T_s(\xi,\eta)=(x_s+\ell\xi,1/4+\ell\eta).
\]
Nonhalting squares lie in $1/8\le x_1<1/4$ and halting squares in
$5/8\le x_1<3/4$; distinct squares have gap at least $\ell$.
Give the full alphabet, of size $m$, odd digits in base $\beta=2m+1$.
Conjugating \eqref{eq:bb-one-cell-branches} by the source and target charts
gives separately disjoint rectangle families $R_j^0,R_j^1$ with positive
diagonal factors $\alpha_j,\alpha_j^{-1}$,
$\alpha_j\in\{\beta^{-1},1,\beta\}$. Every chart has the same
scale, so this remains the physical linear part.
Lemma~\ref{lem:bb-onecell}, with this uniform chart scaling, gives
within-family coordinate gaps at least $\ell/\beta^2$, including the
two-letter target intervals for left moves.
For $1\le j\le L$, write the centers as $p_j^0,p_j^1$, the source
half-widths as $a_j,b_j$, and the target half-widths as
$\alpha_ja_j,\alpha_j^{-1}b_j$. All endpoint half-widths are at
most $\ell/2$. Choose
\[
h_j=\frac12+\frac{j}{4(L+1)},\qquad
\epsilon=\frac1{100}\min\left(1,\frac\ell{\beta^2},
\frac1{4(L+1)}\right).
\]
With $\theta(s)=\vartheta(2s-1/2)$ and
$\Theta_k(\tau)=\theta(3\tau-k+1)$ for $k=1,2,3$, define
\[
\begin{aligned}
C_j(\tau)&=\big((1-\Theta_2)p_j^0+\Theta_2p_j^1,
z_0+(\Theta_1-\Theta_3)(h_j-z_0)\big),\\
\sigma_j&=1+(\alpha_j-1)\Theta_2,\qquad
\lambda_j=\dot\sigma_j/\sigma_j,\\
(w_{j1},w_{j2},w_{j3})&=(a_j\sigma_j,b_j/\sigma_j,0).
\end{aligned}
\]
These centers and half-widths describe moving plates. Their
$\epsilon$ enlargements are disjoint: the first third uses source
projections, the middle third distinct heights, and the last third
target projections. The padding is at most one hundredth of every
relevant gap. All plates and padding lie in a compact sub-box of
$(0,1)^3$, since their centers interpolate between the stated chart
centers, widths are at most $\ell$, heights lie between $1/4$ and
$3/4$, and $\ell\le1/32$, $\epsilon\le1/100$.
For $w\ge0$ put
\[
\kappa_\epsilon(w,v)=
\vartheta\!\left(\frac{2(w+\epsilon-v)}\epsilon\right)
\vartheta\!\left(\frac{2(w+\epsilon+v)}\epsilon\right),\qquad
\chi_j=\prod_{a=1}^3\kappa_\epsilon(w_{ja},x_a-C_{ja}).
\]
Writing $y=x-C_j$, use
\[
A_j=\tfrac12\dot C_j\times y+(0,0,\lambda_jy_1y_2),
\qquad V(\tau,x)=\sum_j\nabla\times(\chi_jA_j).
\]
Extend the potentials by zero outside the chart. The support remains
away from its faces, so this gives a smooth torus field, zero near the
time endpoints. On the whole plateau of plate $j$ its velocity is
$\dot C_j+(\lambda_jy_1,-\lambda_jy_2,0)$; other summands vanish
there. Hence the path from $(p_j^0+(r,s),z_0)$ is exactly
\begin{equation}\label{ba:plate-path}
C_j(\tau)+(\sigma_j(\tau)r,\sigma_j(\tau)^{-1}s,0),
\qquad |r|\le a_j,\quad |s|\le b_j.
\end{equation}
It ends at the prescribed affine image. The plateau argument of
Lemma~\ref{lem:bb-curl} applies with
$D_j=\operatorname{diag}(\sigma_j,\sigma_j^{-1},1)$,
$\eta=\epsilon/2$, and diagonal bound
$\max(1,\alpha_j,\alpha_j^{-1})$: the required enlarged-box separation
was just proved. It gives the same motion on an initial neighborhood
of radius $\epsilon/[4\max(1,\alpha_j,\alpha_j^{-1})]$, with constant
normal displacement, and therefore includes boundary codes.
The initial code $P_{\rm in}$ is rational, since relative to constant
blank tails only finitely many work and frontier digits differ. Load it
from $P_*$ along $C_*=(1-\theta)P_*+\theta P_{\rm in}$ using
\[
V_0(t,x)=\nabla\times\left[
\prod_{a=1}^3\kappa_{1/64}(0,x_a-C_{*,a}(t))
\frac{\dot C_*(t)\times(x-C_*(t))}{2}\right].
\]
The segment and its collar stay within the chart: its first coordinate
stays between $1/8$ and $3/4$, its second between $1/4$ and
$1/2+\ell$, and its third is $1/4$. Its plateau curl is $\dot C_*$.
Take $U=V_0$ on $[0,1]$ and $U(t)=V(t-n)$ on $[n,n+1]$ for $n\ge1$.
Time collars make this smooth and initially zero. It has fixed compact
chart support, bounded mixed derivatives, and period one after loading.
Define $f=U_t+(U\cdot\nabla)U-\nu\Delta U$ using the stated physical
viscosity. Both $U$ and $f$ have mean zero: $U$ is a curl, the
Laplacian integrates to zero, and
$(U\cdot\nabla)U=\operatorname{div}(U\otimes U)$.
The torus clause of Lemma~\ref{lem:b-comparison} applies to this
smooth, initially zero, solenoidal reference field and gives the unique
smooth solution $(U,0)$ with mean-zero pressure.
Equations~\eqref{eq:bb-one-cell-branches} and \eqref{ba:plate-path} identify
the particle at time $1+k$ with the $k$th local configuration. The
recorder invariant ensures that a nonhalting computation continues
indefinitely. A halt, including an initial halt reached by loading,
puts the particle in $5/8\le X_1<3/4$, proving the event. For a
nonhalting computation the loader has $X_1\le1/4$ and every later
branch has both endpoint centers below $1/4$. Its center is their
convex interpolation and its first half-width is at most $\ell/2$.
Thus $00$; the cutoff
estimates and zero collars allow derivative evaluation at approximate real
arguments without tests for exact boundary equality. Only positive
arguments enter the cube root. The program repeats the entire local
table and never uses a computed execution.
\end{proof}
\subsection{Storage and periodic forcing without a loader}
\label{ba:parking-section}
The next construction absorbs initialization into the positions of the
state rectangles. A zero digit for the full blank symbol places the
initial left stack at zero; an input-dependent second-coordinate translation then
places the entire initial code at one fixed particle. To protect a
half-space observer during every period, we first evacuate all sources
to separate storage rectangles and only afterwards fill the targets.
\begin{lemma}[Zero-blank coding]\label{ba:zero-blank-code}
Let an alphabet of size $m>1$ have digits $0,\ldots,m-1$, with full
blank digit zero. Set $B=m+1$, $\kappa=(m-1)/m$, and
\[
\operatorname{val}(a_0a_1\cdots)=\sum_{r\ge0}a_rB^{-r-1},\qquad
I_a=\frac{a+[0,\kappa]}B,\qquad I_{a,b}=\frac{a+I_b}B.
\]
The code is injective, takes values in $[0,\kappa]$, and its distinct
one-letter and two-letter cylinders have positive rational gaps. For a
deterministic partial table with target-state-determined incoming direction
and incoming rule determined by target state and written symbol, separated
state translates of $[0,\kappa]^2$ yield separately disjoint full source
and target rectangles by the following maps in state-relative coordinates:
\begin{equation}\label{ba:zero-digit-maps}
\begin{array}{c|c|c|c}
d&\text{source}&(\xi',\eta')&\text{target}\\\hline
1&[0,\kappa]\times I_a&((b+\xi)/B,B\eta-a)&I_b\times[0,\kappa]\\
0&[0,\kappa]\times I_a&(\xi,\eta+(b-a)/B)&[0,\kappa]\times I_b\\
-1&I_c\times I_a&(B\xi-c,c/B+(b+B\eta-a)/B^2)&
[0,\kappa]\times I_{c,b}.
\end{array}
\end{equation}
The last row is split over all $c$. Each map has positive diagonal
linear part of determinant one and acts exactly on the full rectangle.
\end{lemma}
\begin{proof}\label{ba:zero-blank-code-proof}
The maximum tail is $(m-1)/(B-1)=\kappa$, and
$(m-1+\kappa)/B=\kappa$. Consecutive first-digit intervals have
gap $(1-\kappa)/B>0$. Within one first-digit interval the second-digit
gaps are this quantity divided by $B$; different first digits already
give a gap. The interval containing a code recovers its first digit,
and $v\mapsto Bv-a$ then exposes the tail. Repetition proves
injectivity, including endpoint codes.
Writing $a$ to $b$ changes the first right digit. A right move pushes
$b$ onto the left string and removes $a$ from the right; a left move
removes $c$ from the left and prepends $c,b$ to the old right tail.
These identities give \eqref{ba:zero-digit-maps}, including its images
and reciprocal slopes. Determinism and digit gaps separate sources by
state, read digit, and $c$ when needed. At one target state the common
incoming direction determines where the written digit is visible: in
$I_b$ on the left or right, or in $I_{c,b}$. The incoming-rule
hypothesis and digit gaps then separate the closed target rectangles.
\end{proof}
\begin{lemma}[Evacuation, storage and delivery]\label{ba:parking}
Suppose $A_i:D_i\to E_i$, $1\le i\le N$, are the rational reciprocal
rectangle maps of Lemma~\ref{ba:four-stage}, with $N\ge1$ and every
$D_i\subset\{x_1<0\}$. There is an effective smooth divergence-free
field on $\mathbb R\times\mathbb R^3$, one-periodic in time, zero near
each integer time, and uniformly compactly supported in space, whose
one-period map is $(b,z)\mapsto(A_i(b),z)$ on a neighborhood of every
$D_i\times\{0\}$. If $E_i\subset\{x_1<0\}$, the entire
one-period path of each point in $D_i\times\{0\}$ remains in that
half-space.
\end{lemma}
\begin{proof}\label{ba:parking-proof}
Let $d_i,b_i$ be the source and target centers, and let $a_{ij},a'_{ij}$
be their half-widths. Put $a^*_{ij}=\max(a_{ij},a'_{ij})$.
Choose $\epsilon$ to be one quarter of the minimum of $1$ and all
pairwise maximum-norm distances within the source and target families;
omit pair entries when $N=1$. Define
\[
\begin{gathered}
x_{\min}=\min\{x_1:(x_1,x_2)\in D_i\cup E_i\text{ for some }i\},\\
R=\max(1,\max_i a^*_{i1}),\qquad J=4(R+1),\qquad
s_i=(x_{\min}-iJ,0).
\end{gathered}
\]
The storage rectangle at $s_i$ with half-widths $a^*_{ij}$ lies in
$x_1<0$. Its first-coordinate projection is separated by more than one
from every source and target and from the other storage projections:
the relevant gaps are at least $J-R>1$ and $J-2R>1$.
We need only a localized translation and a localized reciprocal scaling.
On a time segment $[t_0,t_0+\delta]$, use
$g(t)=\vartheta(3(t-t_0)/\delta-1)$, whose derivative is supported
in the middle third. Using the cutoff $\kappa_\epsilon$ defined in
the proof of Theorem~\ref{ba:torus-plates}, for half-widths $a,b$ write
\[
\chi_{a,b}(y)=\kappa_\epsilon(a,y_1)
\kappa_\epsilon(b,y_2)\kappa_\epsilon(0,y_3).
\]
A plate translation from center $P$ to $Q$ is produced by
\[
C(t)=(1-g)P+gQ,\qquad
\nabla\times\left[
\tfrac12\chi_{a,b}(x-C(t))\dot C(t)\times(x-C(t))\right].
\]
On its plateau the velocity is $\dot C$ and the exact path is
$C(t)+(r,s,0)$. At storage center $P=(s_i,0)$, take
$q(t)=1+(\lambda_i-1)g(t)$ and the field
\[
\nabla\times\left(0,0,
\frac{\dot q}{q}(x_1-P_1)(x_2-P_2)
\chi_{a^*_{i1},a^*_{i2}}(x-P)\right).
\]
It has the exact path $P+(qr,q^{-1}s,0)$. The denominator has the
positive lower bound $\min(1,\lambda_i)$, and the whole path lies in
the storage rectangle, because each half-width stays between its source
and target values.
Partition a period into $7N$ equal segments and take $Z=2$. For each
$i$ in order, use four segments for
\[
(d_i,0)\longrightarrow(d_i,Z)\longrightarrow(s_i,Z)
\longrightarrow(s_i,0),\qquad\text{then scale at }s_i.
\]
Only after all these evacuations, use three segments for each $i$ to
deliver its target shape along
\[
(s_i,0)\longrightarrow(s_i,Z)\longrightarrow(b_i,Z)
\longrightarrow(b_i,0).
\]
Sum these segment fields and extend periodically. The middle-third
time switches give zero neighborhoods of every segment endpoint and
every integer time. The result is a smooth finite sum of curls with
one compact spatial support.
This is the $7N$ schedule of Lemma~\ref{lem:bb-lift-parking}(ii).
Its separation hypotheses have explicit margins here. Unpadded
source-source and target-target gaps are at least $4\epsilon$, so their
$\epsilon$ enlargements leave gap at least $2\epsilon$. Storage
gaps exceed one, leaving more than $1-2\epsilon>0$ after padding.
At horizontal travel height the enlarged moving plate and every
enlarged stationary plate have vertical gap $2-2\epsilon>0$.
Every scaling stays in its reserved storage rectangle. The lemma
therefore gives simultaneous noninterference for evacuation and delivery,
including when a source meets a target.
The localized potentials above have exactly these affine generators on
their plateaux, so the plateau argument of Lemma~\ref{lem:bb-curl}
gives endpoint
$(d_i+(r,s),0)\mapsto(b_i+(\lambda_i r,\lambda_i^{-1}s),0)$,
with normal displacement unchanged. Put
$M=\max_i(1,\lambda_i,\lambda_i^{-1})$ and
$d_*=\min(2\epsilon,1-2\epsilon,2-2\epsilon)>0$.
A neighborhood radius smaller than both $\epsilon/(8M)$ and
$d_*/(4M)$ retains the plateaux and all exclusion margins throughout
the concatenation. This applies the shared proofs to the displayed
potentials and does not identify their fields away from the plateaux.
If $E_i$ is negative, each translation of an individual point joins two
negative first coordinates, and its linear interpolation remains negative.
Scaling occurs inside the negative storage rectangle. The point is
stationary during all other segments. This proves the asserted
whole-period exclusion.
\end{proof}
\begin{theorem}[A periodic force with no loading interval]
\label{ba:periodic-parking}
Fix a positive computable $\nu$. A deterministic one-tape machine and
finite input word in cells $0,\ldots,L-1$, with other cells blank and
initial head zero, effectively determine a general body force on
$\mathbb R^3$ extending smoothly and one-periodically to all real times.
It has uniformly compact spatial support and bounded mixed derivatives.
Its zero-data solution has a unique global smooth velocity in
\eqref{ba:euclidean-comparison-class}, with constructed pressure zero, and
\[
\exists t\ge0:\quad (\Phi_t((-2,0,0)))_1>0
\quad\Longleftrightarrow\quad\text{the machine halts}.
\]
The fixed label is already the initial code; periodicity begins at zero.
\end{theorem}
\begin{proof}\label{ba:periodic-parking-proof}
Insert a fresh nonhalting start state that preserves the read letter,
stays put and enters the original start state. Replace missing
nonhalting instructions by letter-preserving stay instructions to a halt
state. This finite normalization preserves the halting question,
including an initially halted machine, and guarantees a nonhalting
initial control and at least one rule.
Apply the six-row delayed-head recorder of Lemma~\ref{ba:six-compiler}
with initial frontier $1$. Initially every other history cell is blank
and every mark is zero. The main control records $(q,e)$, where $e$
is the preceding displacement, and the history record is $(q,e,a)$.
The old head is marked before the right scan and cleared on return,
immediately before displacement into the next main state. At checkpoint
$k$ the frontier is $k+1$ and the head $i$ satisfies $|i|\le k$;
the next checkpoint uses $2(k+1-i)+3\le4k+5$ local steps. The
incoming-direction and written-symbol properties hold on the full
domain, as required by Lemma~\ref{ba:zero-blank-code}.
Give the full blank symbol digit zero and the other full symbols digits
$1,\ldots,m-1$. Order nonhalting local states with the initial state
first and assign offsets $-2h$, $h=1,2,\ldots$; assign halting main
states offsets $2h$, $h=1,2,\ldots$. Let $\eta_{\rm in}$ be the
initial right-stack value and put $y_*=-\eta_{\rm in}$. Only finitely
many full symbols are initially nonblank, so this is a finite rational
sum. Every cell left of the initial head is fully blank. The code
\[
(x_\sigma+\xi,y_*+\eta,0)
\]
therefore places the initial configuration exactly at $(-2,0,0)$.
The input changes the common second-coordinate translation, while the
observed label and zero initial velocity remain fixed.
Every finitely reached tape still has finite full-symbol support, so its
code is rational and can be decoded by rational digit extraction until
both tails are zero. At a checkpoint the leftmost nonempty history site
has absolute index $1$; its relative position also recovers the head's
absolute index. Thus this coding retains the initialized computation.
Translate the rectangles of \eqref{ba:zero-digit-maps} by the indicated
state offsets and $y_*$. Lemma~\ref{ba:zero-blank-code} gives their
full affine action and separate source and target gaps. Since
$\kappa<1$, every nonhalting state block is strictly negative in the
first coordinate and every halting block strictly positive. Halting
states have no outgoing rules, so all sources are negative. Apply
Lemma~\ref{ba:parking} to obtain its periodic field $v$. Zero tails
can place a code on a rectangle boundary, which is included in that
lemma's full-neighborhood conclusion.
Set
\[
f=f_0+\nu f_1,\qquad
f_0=\partial_tv+(v\cdot\nabla)v,\qquad f_1=-\Delta v.
\]
The field and force are smooth, periodic, uniformly compactly supported,
and have all mixed derivatives bounded. Since $v$ vanishes near integer
times, its initial value is zero. Fixed support and bounded derivatives
give $v\in CH^2\cap C^1L^2$ and bounded $v,\nabla v$ on every finite
interval. Lemma~\ref{lem:b-comparison}(i) therefore supplies the exact
solution $(v,0)$, uniqueness in
\eqref{ba:euclidean-comparison-class}, and its global material flow.
At time zero the marked particle is the initial code. If it is a
nonhalting code at integer $n$, its digits select a source rectangle and
the next period applies exactly that local instruction. Induction
identifies the integer-time codes through the first halt or forever.
Finite completion of every recorder scan identifies the original
machine at the checkpoint integers. A halt therefore reaches a positive
code. If the machine never halts, each used branch has a negative target
rectangle. Lemma~\ref{ba:parking} keeps its marked path negative for
that whole period, ruling out an intermediate false visit.
Finally the force program uses finite normalization, the finite table,
the rational input shift, a finite rectangle list, rational storage and
separation data, $7N$ time segments and explicit cutoffs. All scaling
denominators have known positive lower bounds. The zero-collar and flat
profile estimates following \eqref{eq:p2-step} permit evaluation of the
repeated formula and every derivative from approximate real arguments
without exact seam tests.
Every period executes all branch maps, independently of the marked
particle's current state; the program never computes the input run.
\end{proof}
\subsection{Direction-split marking and a bounded observer}
\label{ba:bounded-box-native}
A stay instruction need not clear and reinstall a persistent mark.
We separate that case in the next table. Its geometric realization
uses a bounded observation box: during horizontal transfers the routing
height itself prevents a false visit.
\begin{lemma}[Direction-split persistent marking]
\label{ba:split-mark-compiler}
In the notation of Lemma~\ref{ba:persistent-compiler}, there is an
effective partial finite table with a persistent checkpoint mark,
pointer at $p_n=n+2$, and both full-domain incoming properties of
Lemma~\ref{ba:rectangles}. A work step ending at $w'$ takes
$2(p_n-w')+4+\mathbf1_{\{d_\tau\ne0\}}$ local steps.
\end{lemma}
\begin{proof}\label{ba:split-mark-compiler-proof}
Use full symbols $[g,e,l]$, with work letter $g$, mark $e\in\{0,1\}$,
and $l\in\{E,P\}\sqcup\{J_\tau\}$. Controls are $A_q$,
$B_\tau$ for moving instructions, $C_\tau,S_\tau$, and $D_q,R_q$.
For $\tau=(q,a)$ writing $b_\tau$, the complete table is
\[
\begin{array}{c|c|c|c|r}
\text{control}&\text{read and guard}&\text{target}&\text{write}&d\\\hline
A_q&[a,1,l],\ d_\tau=0&C_\tau&[b_\tau,1,l]&0\\
A_q&[a,1,l],\ d_\tau\ne0&B_\tau&[b_\tau,0,l]&d_\tau\\
B_\tau&[g,0,l]&C_\tau&[g,1,l]&0\\
C_\tau&[g,1,l],\ l\ne P&S_\tau&[g,1,l]&1\\
S_\tau&[g,0,l],\ l\ne P&S_\tau&[g,0,l]&1\\
S_\tau&[g,0,P]&D_{q^+_\tau}&[g,0,J_\tau]&1\\
D_q&[g,0,E]&R_q&[g,0,P]&-1\\
R_q&[g,0,l],\ l\ne P&R_q&[g,0,l]&-1\\
R_q&[g,1,l],\ l\ne P&A_q&[g,1,l]&0.
\end{array}
\]
The first two rows use nonhalting $q$. In those rows and the
$B_\tau$ row, $l$ is unrestricted, including $P$; no additional
history guard is imposed. Initialize in $A_{q_0}$ at zero, with the
input work tape, a unique mark at zero, pointer at two, and empty
history elsewhere.
At checkpoint $n$, the work move ends at
$w'=w_n+d_\tau\le n+11$. These three exclusions cover every real
time and prove the event equivalence independently of horizontal
excursions in the middle stage. The repeated processor installs every
local branch from the finite table; it does not depend on the executed
run or its outcome.
\end{proof}
\subsection{Separating vertical and planar potentials}
\begin{theorem}[A decaying processor with phasewise planar potentials]
\label{ba:phasewise-flow}
At fixed computable $\nu>0$ there is an effective force on $\mathbb R^3$
with one compact spatial support and all mixed derivatives of time order
$k$ bounded by $C_{k,\alpha}(1+t)^{-1-k}$, whose unique smooth
zero-data solution satisfies
\[
\exists t\ge0:\ (\Phi_t(0))_1<-1/2
\quad\Longleftrightarrow\quad\text{halting}.
\]
It uses the six-rule recorder with initial frontier 1, full subdivision
of the rectangles, and separate vertical-translation and planar
Hamiltonian potentials. The constructed pressure is zero and uniqueness
is in Lemma~\ref{ba:realization}'s whole-space class.
\end{theorem}
\begin{proof}\label{ba:phasewise-flow-proof}
Let $\Gamma$ be the full alphabet of the six-rule table with $r_0=1$.
Use odd digits in base $B=2|\Gamma|+1$ and its fully subdivided maps. Put halt lanes at $-2,-4,\ldots$
and the other lanes at $2,4,\ldots$, each of width one. The second
coordinate lies in $[0,1]$. For the resulting rectangles $P_i,Q_i$,
centers $p_i,q_i$, and first scale $\lambda_i$, choose rational
$0<\varepsilon\le1/4$ smaller than one quarter of all within-source
and within-target maximum-norm distances. Empty pair lists impose no
condition. Thicken by $[-1,1]$ and put $H_i=4i$.
On each third of the cycle use a flat progress function $\vartheta$
from zero to one. Lifting uses $z_i=H_i\vartheta$ and potential
$A_i=(0,z_i'X_1,0)$, whose curl is $(0,0,z_i')$.
At height $H_i$, set $c_i=(1-\vartheta)p_i+\vartheta q_i$ and
$\mu_i=1-\vartheta+\vartheta\lambda_i$. Use the potential
$A_i=(0,0,\psi_i)$, where
\[
\psi_i=c_{i1}'(X_2-c_{i2})-c_{i2}'(X_1-c_{i1})
+\frac{\mu_i'}{\mu_i}(X_1-c_{i1})(X_2-c_{i2}).
\]
Its curl is
$(c_{i1}'+(\mu_i'/\mu_i)(X_1-c_{i1}),
c_{i2}'-(\mu_i'/\mu_i)(X_2-c_{i2}),0)$.
Thus it realizes reciprocal scaling about the interpolated center.
Descending uses the first potential with $z_i=H_i(1-\vartheta)$.
Multiply each potential by its current-box $\varepsilon$ cutoff
before taking the curl. Source gaps, distinct middle heights, and
target gaps separate the three stages. On the whole boxes and a
neighborhood the uncut formulas are exact. The first coordinate is
constant during vertical motion and is the convex interpolation of its
endpoints during the planar stage.
For loading put $\theta_0(s)=\sigma(2s-1/2)$, which is zero near zero
and one near one. Let $P_*$ be the rational initial code. Load it in the
first unit interval by the planar path
$m(s)=\theta_0(s)((P_*)_1,(P_*)_2)$, localized around
$(m(s),0)+[-1,1]^3$, using potential
$(0,0,m_1'(X_2-m_2)-m_2'(X_1-m_1))$.
Repeat the three-phase instruction field afterward to obtain $W$.
Code induction and the delayed-head invariant prove the correspondence
at times $1+n$. Nonhalting codes have first coordinate at least 2,
so loading and the convex-path bound exclude $X_1<-1/2$ for every real
time. A halt code has first coordinate at most $-1$, including one
reached by loading an initially halted machine. Apply
Lemma~\ref{ba:log-clock}; physical code times are $e^{1+n}-1$.
Its support, derivative, event, and uniqueness conclusions give precisely
the claimed separate force choice.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//observer-routing.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//persistent-history.tex ====
\section{Persistent marks and guarded history routes}
\label{ba:marked-routes}
The preceding recorders clear the work-head mark at each checkpoint.
We now retain that mark, or replace the moving history pointer by a
guarded boundary between occupied and unused cells. These changes give
different partial maps on arbitrary tapes, even though their initialized
executions simulate the same machine. We prove each inverse on its
declared domain before using it to separate the geometric branch images.
The use of retained history follows the reversible-computation idea of
Bennett~\cite{Bennett1973}; the local guards and geometric interfaces
are established here.
In this section an input word occupies cells $0,1,\ldots$, the other
work cells are blank, and the initial head is at zero. All constructions
include initial halting. When one halt state $h$ is used, identify the
original halting states with $h$ and redirect incoming instructions;
there are no outgoing halting instructions to preserve. If necessary
adjoin an unreachable halt state.
We write $X(t;a)=\Phi_t(a)$ for the trajectory from $a$.
\subsection{A persistent mark and input-dependent placement}
\label{ba:persistent-periodic}
A force can be periodic from time zero if no separate loading interval
is needed. We achieve this by translating the initial control's coding
square so that the prescribed particle already represents the input.
The translation depends effectively on the finite input. First we give
the recorder that this periodic processor implements.
\begin{lemma}[A recorder with a persistent work-head mark]
\label{ba:persistent-compiler}
Let a machine have work alphabet $\Gamma$, halt state $h$, and
instructions
$\gamma=(q,a)\mapsto(q^+_\gamma,a^+_\gamma,d_\gamma)$,
where $q\ne h$ and $d_\gamma\in\{-1,0,1\}$.
There is an effective finite partial one-head table with these properties.
At the checkpoint after $n$ work steps, it has the correct work tape,
state and head, a unique mark at that head, and a history pointer at
$p_n=n+2$. Each target control determines the incoming displacement;
the target control and written full cell symbol determine the source
control and read full symbol on the entire partial domain. If the next
work-head position is $w'=w_n+d_\gamma$, the next checkpoint is reached
after exactly $2(p_n-w')+4$ local steps.
\end{lemma}
\begin{proof}\label{ba:persistent-compiler-proof}
A full symbol is $(a,e,s)\in\Gamma\times\{0,1\}\times\mathcal H$,
where $\mathcal H=\{E,P\}\sqcup\{r_\gamma\}$ records empty history,
a pointer, or an instruction. All displayed control families are
disjoint. The following is the complete table; free cell indices range
over their finite alphabets:
\[
\begin{array}{c|c|c|c|r}
\text{control}&\text{read and guard}&\text{target}&\text{write}&d\\\hline
A_q&(a,1,s),\ s\ne P&B_\gamma&(a^+_\gamma,0,s)&d_\gamma\\
B_\gamma&(c,0,s),\ s\ne P&R_\gamma&(c,1,s)&1\\
R_\gamma&(c,0,s),\ s\ne P&R_\gamma&(c,0,s)&1\\
R_\gamma&(c,0,P)&C_{q^+_\gamma}&(c,0,r_\gamma)&1\\
C_q&(c,0,E)&L_q&(c,0,P)&-1\\
L_q&(c,0,s),\ s\ne P&L_q&(c,0,s)&-1\\
L_q&(c,1,s),\ s\ne P&A_q&(c,1,s)&0
\end{array}
\]
The first row has $q\ne h$ and $\gamma=(q,a)$; there is no rule from
$A_h$. Initially the actual head and the unique mark are at zero,
the work tape is the input, and history is $E$ except for $P$ at cell 2.
At checkpoint $n$, $|w_n|\le n$, so after the first row the head is at
$w'\le n+10},
\]
maps $S_i$ onto $T_i$, with $c_i^S,c_i^T$ their centers. There is a
smooth nondecreasing switch $\eta:[0,1]\to[0,1]$, zero near 0 and one
near 1, for which either scale
path $r_i=\lambda_i^\eta$ or $r_i=1-\eta+\eta\lambda_i$ yields an
effective smooth compactly supported solenoidal unit-time field, zero
on time collars, realizing $(y,z)\mapsto(F_i(y),z)$ on the whole box
$S_i\times[-1,1]$ and a neighborhood of it. The paths are affine;
each horizontal half-width remains between its endpoint values, and the
horizontal sweep of each original box lies in the coordinate bounding
rectangle of $S_i$ and $T_i$. With the linear choice, a particle's first
coordinate is the convex
interpolation of its endpoint first coordinates.
\end{lemma}
\begin{proof}\label{ba:two-interpolations-proof}
For $N=0$ take the zero field. For $N\ge1$ apply
Lemma~\ref{lem:bb-lift-parking}(i) with vertical half-width one.
Use the shared phase switches with $\eta=w_2$ and set
\[
C_i(s)=\big((1-\eta)c_i^S+\eta c_i^T,\,6i(w_1-w_3)\big),
\qquad D_i(s)=\operatorname{diag}(r_i,r_i^{-1},1).
\]
The affine path is
$G_i(s,y)=C_i(s)+D_i(s)(y-(c_i^S,0))$.
Choose $\epsilon$ to be one quarter of the minimum of 1 and all
pairwise max-norm distances in the two planar lists, omitting empty
pair lists. The $\epsilon$-padded moving boxes are disjoint: source
and target gaps are at least $4\epsilon$ during the vertical phases,
and the middle-phase height spacing is $6>2(1+\epsilon)$.
For later loading formulas, record the centered cutoff explicitly:
\[
\chi(C,L,\epsilon;x)=
\prod_{j=1}^3\sigma\!\left(\frac{L_j+\epsilon+x_j-C_j}{\epsilon/2}\right)
\sigma\!\left(\frac{L_j+\epsilon-x_j+C_j}{\epsilon/2}\right).
\]
Here $L$ is the vector of half-widths; the plateau has padding
$\epsilon/2$ and the support has padding $\epsilon$.
If $L_i(s)$ are the moving half-widths, $Y=x-C_i(s)$ and
$A_i=D_i'D_i^{-1}$, the field supplied by
\eqref{eq:bb-potential} is
\begin{equation}\label{ba:affine-box-field}
V(s,x)=\sum_i\nabla\times\left\{
\chi(C_i,L_i,\epsilon;x)
\left[\tfrac12 C_i'\times Y+\tfrac13(A_iY)\times Y\right]
\right\}.
\end{equation}
Indeed $\det D_i=1$, and the preceding separation verifies
Lemma~\ref{lem:bb-curl} with localization parameter $\epsilon/2$.
Its neighborhood conclusion applies with
$L_* =\max_i\max(1,\lambda_i,\lambda_i^{-1})$ because
$\|D_i(s)\|_\infty\le L_*$ and $D_i(0)=I$.
It therefore gives the stated affine motion on the full boxes and
on their initial max-norm neighborhoods of radius $\epsilon/(4L_*)$.
The paths are stationary for $s\le1/12$ and $s\ge11/12$.
Their finite center ranges and endpoint width bounds give common
compact support. The effective positive lower bound
$r_i\ge\min(1,\lambda_i)$ verifies the scale requirement.
For the linear scale, the first-coordinate identity is
\[
(G_i(s,y))_1=(1-\eta)y_1+\eta(F_i(y_1,y_2))_1.
\]
The second coordinate need not interpolate its own endpoint values.
Instead, if $b_{i2}$ is the source's second half-width, convexity of the
reciprocal function gives
\begin{equation}\label{eq:ba-reciprocal-envelope}
\frac{b_{i2}}{(1-\eta)+\eta\lambda_i}
\le (1-\eta)b_{i2}+\eta\frac{b_{i2}}{\lambda_i}.
\end{equation}
The first half-width is exactly its endpoint interpolation. For the
exponential scale, convexity of the exponential gives the corresponding
upper bounds in both coordinates, since
$\lambda_i^{\pm\eta}\le(1-\eta)+\eta\lambda_i^{\pm1}$.
In either case the center is interpolated linearly and each half-width
is at most the interpolation of its endpoint half-widths. Therefore each
lower edge is at least the interpolation of the endpoint lower edges,
and each upper edge is at most the corresponding interpolation of upper
edges. This proves the claimed bounding rectangle for the entire sweep.
The individual endpoint-width bounds also follow directly from monotonicity
of the positive scales and their reciprocals.
\end{proof}
The same construction applies to $P_i\times[-h,h]$ for any common
$h\ge0$, including plates and rectangles with zero side lengths, under
the hypotheses of Lemma~\ref{lem:bb-lift-parking}(i). Replace $6i$ by
heights $H_i$ whose pairwise gaps exceed $2h+2\varepsilon$, and choose
$\varepsilon$ below one third of every within-family planar gap. The
formula for $C_i$ uses the source and target centers in place of
$c_i^S,c_i^T$; its support padding is $\varepsilon$ and plateau padding
is $\varepsilon/2$. Source projections, distinct heights and target
projections separate the three phases in that order. Empty pair lists
impose no gap restriction, and the empty family gives the zero field.
Arbitrary real data give smooth existence; the effective-data assumptions
of the shared lemma give computable scales, derivative bounds and margins.
The same perturbation argument gives an open neighborhood even when a
side length is zero.
\subsection{Transporting an initial cutoff}
\label{sec:ba-transported-localization}
The transported-initial-cutoff convention is also useful. Write the
affine path as $G_i(t)x=J_i(t)x+g_i(t)$ and localize by
$\psi_i(G_i(t)^{-1}x)$, where $\psi_i$ has support padding
$\varepsilon_0$ around the initial box and equals one on a smaller
positive padding.
If $\lambda_i\in[B^{-1},B]$, its initial padding
$\varepsilon_0$ expands horizontally by at most $B$. Choose
$2B\varepsilon_0$ below every target gap,
$2\varepsilon_0$ below every source gap, and
$\varepsilon_0\le\varepsilon$. These transported supports stay
separated in the same three phases. The affine generator is
$v_i+A_ix$, with $A_i=\dot J_iJ_i^{-1}$ and
$v_i=\dot g_i-A_ig_i$. Its potential is
\[
\tfrac12v_i\times x+\tfrac13(A_ix)\times x.
\]
The identity proved in Lemma~\ref{lem:bb-curl} gives the exact
generator on the transported plateau. Its neighborhood argument then
proves the same full-box endpoint map. The factor $B$ in the margin
choice is essential: transported padding does not stay fixed under
stretching.
\subsection{Independent planar interpolation}
For an observer that needs both planar coordinates to follow their own
endpoint interpolations, the reciprocal path is insufficient. The next
construction interpolates the two factors independently and compensates
the temporary planar area change in the third direction. Here the named
targets may be disjoint containers of the actual affine images; only the
image centers enter the motion.
\begin{lemma}[Independent planar scales and normal compensation]
\label{ba:plate-routing}
Let $P_i,Q_i\subset\R^2$ be finite families of closed rectangles,
separately pairwise disjoint, and let assigned positive diagonal affine
maps $F_i$ satisfy $F_i(P_i)\subseteq Q_i$. Write their diagonal factors as
$\lambda_{i1}^*,\lambda_{i2}^*>0$ with product one.
There is a smooth compactly supported solenoidal unit-time field, zero on
time collars, whose endpoint map is $(y,z)\mapsto(F_i(y),z)$ on a
neighborhood of each entire plate $P_i\times\{0\}$. During the horizontal
phase both planar factors interpolate linearly, while the normal factor
compensates their product. Each planar coordinate of a point on the plate
is its convex endpoint interpolation.
\end{lemma}
\begin{proof}\label{ba:plate-routing-proof}
Use the height phases in Section~\ref{sec:ba-localization}. For source
center $p_i$, end its center path at its actual image $F_i(p_i)$; in the
onto case this is the named target center. Choose separated heights $H_i$
as above and set
\[
c_i=\bigl((1-w_2)p_i+w_2F_i(p_i),\,H_i(w_1-w_3)\bigr),
\]
\[
\lambda_{ij}=1-w_2+w_2\lambda_{ij}^*,\qquad
D_i=\operatorname{diag}\bigl(
\lambda_{i1},\lambda_{i2},(\lambda_{i1}\lambda_{i2})^{-1}\bigr).
\]
The volume-preserving path has logarithmic derivative
\[
T_i=\dot D_iD_i^{-1}=\operatorname{diag}\left(
\frac{\dot\lambda_{i1}}{\lambda_{i1}},
\frac{\dot\lambda_{i2}}{\lambda_{i2}},
-\frac{\dot\lambda_{i1}}{\lambda_{i1}}
-\frac{\dot\lambda_{i2}}{\lambda_{i2}}\right).
\]
This matrix has trace zero, so \eqref{eq:bb-potential} localized
around the moving plate gives the desired field. Its planar motion is
$c_i+(\lambda_{i1}(y_1-p_{i1}),\lambda_{i2}(y_2-p_{i2}),0)$.
Normal displacement changes by the factor
$(\lambda_{i1}\lambda_{i2})^{-1}$. The factors have positive
lower bounds; choose the initial thickness less than their product's
minimum times the normal plateau width. A small enlargement of the
planar rectangle is allowed by the positive collars as well. Thus a
whole thin neighborhood follows the volume-preserving affine motion.
The normal factor returns to one at the endpoint. The final plate image
is contained in $Q_i\times\{0\}$. Source and target-container
gaps protect the vertical phases and distinct heights protect the
middle phase. Linear interpolation of each planar factor gives both
convex-coordinate identities on the plate.
This proves smooth existence for arbitrary real data. Under the effective
input assumptions of Lemma~\ref{lem:bb-curl}, the positive scale bounds and
finite gap margins can be computed and the construction is effective.
\end{proof}
\subsection{Forcing and effective evaluation}
\begin{lemma}[Residual forcing for the finite processors]\label{ba:realization}
Fix computable $\nu>0$. Let $W$ be smooth, solenoidal, initially zero,
with support in one compact subset of $\mathbb R^3$ and every mixed
derivative bounded. Then
\begin{equation}\label{ba:residual}
F_W=W_t+(W\cdot\nabla)W-\nu\Delta W
\end{equation}
has the same support and derivative properties. The zero-data solution
$(W,0)$ is unique among smooth solutions with
$u\in C_tH^2\cap C_t^1L^2$, bounded $u$ on finite intervals, and
$p\in C_tH^1$. The corresponding torus assertion holds in the
classical comparison class of Section~\ref{sec:model}. A sum of curls
in a torus chart gives both $W$ and $F_W$ mean zero. Finite formulas
from the effective-data instances of the preceding constructions, finite
loading, and repetition of one cycle give effective evaluation of every
force derivative.
\end{lemma}
\begin{proof}\label{ba:realization-proof}
Fixed compact support and bounded mixed derivatives imply all reference
velocity hypotheses of Lemma~\ref{lem:b-comparison}. Case (i) gives
exactly the stated whole-space comparison class, without a competing
gradient bound; its torus clause gives the stated classical comparison
class. The same lemma provides the pressure normalization and global
material flow. The product rule applied to \eqref{ba:residual}
proves the force's support and derivative bounds. A chart curl has
zero mean, and integration of the residual uses
$\int\Delta W=0$ and
$\int(W\cdot\nabla)W=\int\operatorname{div}(W\otimes W)=0$.
For these effective-data instances, gate derivatives and seam evaluation are
those of \eqref{eq:p2-step} and Proposition~\ref{prop:bb-realization}.
The explicit affine scales and every reciprocal denominator have positive
computable bounds. Thus differentiating the finite formulas, using a
finite superset of possible period indices, evaluates every derivative
without a machine execution or a real equality test at a seam.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//routing-conventions.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//solid-routing.tex ====
\section{Evacuation, columns and obstacle detours}\label{sec:solid-routing}
The following refinements choose the occupied regions and the storage order
explicitly. Ordered low extraction controls a height observer; simultaneous
lifting controls a first-coordinate observer; expanded centers permit
three-dimensional detours. Fixed columns and recorded pointers give further
ways to keep a whole transition outside a detector. Each proof specifies
which parts of the geometry act on solid neighborhoods and which bounds
are needed only for the coded central plane.
Unless an individual theorem specifies another class, whole-space
comparison means smooth $u\in CH^2\cap C^1L^2$, bounded $u,\nabla u$
on finite intervals, and a pressure representative in $CH^1$, as in
Section~\ref{bc:analytic-section}. The torus comparison class is the
classical class of Lemma~\ref{lem:b-comparison}. The distinct pressure
conditions stated below remain part of their respective results.
\subsection{Ordered extraction below the computation}\label{bc:ordered-section}
\begin{proposition}[Ordered full-box transport]\label{bc:ordered}
Let $D_i,S_i$ be finite families of closed rational axis-aligned boxes in $\R^3$
with positive side lengths, pairwise disjoint within each family.
Suppose the assigned center-to-center affine map $F_i$ satisfies
$F_i(D_i)=S_i$ and has positive diagonal linear part $A_i$ of
determinant one. There is an effective compactly supported solenoidal
unit-time field, zero near the time endpoints, whose time-one map has
this action on every full $D_i$. If both centers of a pair have negative
third coordinate, every point of $D_i$ whose third coordinate equals that
of its center follows a path with negative third coordinate.
\end{proposition}
\begin{proof}\label{bc:ordered-proof}
Choose an integer $L>1$ larger than the absolute values of all endpoint
coordinates. The storage columns have first coordinates $U_i=4(i+1)L$
and common height $Z=-4L$. Write $[\ell_i,r_i]$ for the first-coordinate
interval of $D_i$. Extract sources in decreasing order of $r_i$.
Translate each horizontally right to its column and then vertically to
$Z$, retaining its second coordinate. If an unextracted box meets the
active box in both the second and third projections, disjointness forces
their first intervals to be separated. That box cannot lie to the right:
its right endpoint would then exceed $r_i$, contrary to the extraction
order. It lies strictly to the left of $\ell_i$ and misses the entire
rightward sweep. Stored boxes lie below the original boxes; vertical
moves occur in distinct columns outside them.
At storage height, adjust the second center coordinate to its target
value and apply $A_i^{\theta}$. Each intermediate half-width lies between
its two endpoint values, hence is less than $L$. The columns remain
disjoint throughout these operations. Deliver boxes in increasing order
of the right endpoint of their target first interval: raise a box to its
target height and translate it left into its target. The raising column
is empty. As above, a previously delivered box whose last two intervals
meet the active ones must be separated in the first coordinate. Its
smaller right endpoint puts it wholly to the left of the active target
interval, so the leftward sweep stops before reaching it.
Undelivered boxes remain at storage height. These observations give
positive rational clearances for all swept bounding boxes. Choose cutoff
padding below each clearance and use Lemma~\ref{lem:bb-curl} in successive
flat slots. Induction over the slots proves the full-box assertion.
A point initially in the central third-coordinate plane retains zero
third-coordinate offset during each diagonal deformation. Its height is
interpolated only between the two endpoint heights and $Z$. If those
endpoint heights are negative, every intermediate height is negative.
\end{proof}
The onto hypothesis in the preceding proposition is a useful exact-box
specialization. Its target-container version, including the effective
input conditions, is given in Section~\ref{bc:contained-targets}.
\begin{theorem}[A positive-height observation]\label{bc:positive-height}
On $\R^3$, the fixed label $0$ and observation $\{x_3>0\}$ realize
halting effectively by a compactly supported general force with zero
initial velocity and pressure, bounded mixed derivatives, and period one
after time one. The force has a separate alternative in
$L^2(0,\infty;H^j)$ for every $j$, tending to zero in every spatial
supremum norm at rate $O((1+t)^{-1})$.
\end{theorem}
\begin{proof}\label{bc:positive-height-proof}
Use Lemma~\ref{lem:bb-movefirst} and odd digits in base $2m+2$. Give every
simulator state a distinct positive integer index $i_s$ and put its
third-coordinate center at $z_s=i_s$ for ready halting states and
$z_s=-i_s$ otherwise. Thicken the full rectangles of
Lemma~\ref{lem:bb-onecell} by $[z_s-1/4,z_s+1/4]$. The assigned third
scale is one, so the maps satisfy Proposition~\ref{bc:ordered}.
The rational initial code $Y_0$ is loaded from the origin along the
straight segment, by a localized translation during $[0,1]$. Repeat the
unit-time mechanism after time one. At time $1+k$ the particle is exactly
the code after $k$ defined simulator transitions. If the initial machine
halts, the loading endpoint already has positive height. If it never
halts, loading has nonpositive height and all later transitions have
negative endpoint centers; the proposition excludes a positive height
at every intervening time. Conversely, a halting checkpoint has positive
height. Lemma~\ref{bc:analytic} gives the force and uniqueness, and
Lemma~\ref{ba:log-clock} gives the separate temporal alternative.
\end{proof}
\subsection{Simultaneous lifting with a convex-coordinate guard}
\label{bc:simultaneous-section}
\begin{proposition}[Simultaneous full-box transport]\label{bc:simultaneous}
Suppose finite planar source and target rectangle families $S_i,T_i$
have positive side lengths and endpoints on the grid $B^{-2}\Z^2$,
with integer $B\ge2$, and are separately
disjoint. Require their assigned center-to-center maps to satisfy
$F_i(S_i)=T_i$ and to have linear part $\diag(\lambda_i,\lambda_i^{-1})$ with
$\lambda_i\in\{B^{-1},1,B\}$. There is an effective solenoidal motion
of the full boxes obtained by taking the product with $[-1,1]$.
For every transported point its first coordinate is a convex combination
of its initial and final first coordinates at every time.
\end{proposition}
\begin{proof}\label{bc:simultaneous-proof}
Let $p_i^-,p_i^+$ be the centers in the plane $z=0$, set $h_i=5i$, and
choose a smooth $\Theta$ equal to zero on $(-\infty,1/4]$ and one on
$[3/4,\infty)$. For $0\le t\le1$, put
\[
\theta=\Theta(3t-1),\quad l_i=1-\theta+\theta\lambda_i,
\quad c_i=(1-\theta)p_i^-+\theta p_i^+
+h_i\{\Theta(3t)-\Theta(3t-2)\}e_3.
\]
The desired path is $g_i(t,y)=c_i(t)+\diag(l_i,l_i^{-1},1)(y-p_i^-)$.
Use a source padding $\varepsilon=1/(10B^3)$, transported by this same
affine map, for the cutoff neighborhood. During lifting the planar
source gap is at least $B^{-2}$. During lowering the enlarged target
padding is at most $B\varepsilon20R$ so large that the
expanded source centers $A_i^*=Ha_i$ and target centers $B_i^*=Hb_i$
are separately more than $20R$ apart in the sup norm. Set
\[
Z_0=1+100R+\max_i\{\|A_i^*\|_\infty,\|B_i^*\|_\infty\},
\qquad P_i=(0,0,-Z_0-100Ri).
\]
Each of the unions $\{A_i^*\}\cup\{P_i\}$ and
$\{B_i^*\}\cup\{P_i\}$ has separation greater than $20R$.
The construction has five phases: expand source centers by a factor
from one to $H$ while keeping shapes fixed; move them successively to
$P_i$; change each parked shape by the prescribed positive diagonal
power; move them successively to $B_i^*$; contract target centers to
$b_i$ while keeping the final shapes fixed.
The simultaneous expansion is collision free. For each pair of source
boxes some coordinate has positive gap
$|a_{i,j}-a_{k,j}|-r_{i,j}-r_{k,j}$; multiplication of the center
difference by a factor at least one preserves that gap. The same
argument applies to contraction toward the disjoint targets. Choose
padding less than one quarter of the minimum such gap, and less than
one. During a parked deformation the $j$th half-width is
$r_{i,j}^{1-\theta}w_{i,j}^{\theta}\le\max(r_{i,j},w_{i,j})0$.
Every boundary used around a negative-height obstacle lies below
$-H+4R<0$. Storage heights are negative as well. Expansion, contraction,
and positive diagonal deformation preserve the central-plane condition.
Thus all these paths are negative. Finitely many rational vertices,
endpoint widths, and cutoff paddings give a rational support bound.
\end{proof}
\begin{theorem}[A height test inside one torus chart]\label{bc:expanded-height-event}
On the unit torus, the fixed label $o=(1/2,1/2,1/2)$ and event
\begin{equation}\label{bc:expanded-height-fixed-event}
\{x:x_3\pmod1\in(1/2,1)\}
\end{equation}
realize halting by a smooth general mean-zero force, periodic after time
one, with zero initial velocity and pressure and bounded mixed derivatives.
The repeated mechanism depends on the machine alone; only initialization
depends on its finite input. The solution is unique in the torus class.
\end{theorem}
\begin{proof}\label{bc:expanded-height-event-proof}
Use the frontier recorder, odd base $B=2m+1$, distinct negative integer
heights for nonterminal states and height one for the terminal ready state.
Thicken every branch by a third half-width $1/4$. The reciprocal tape
map is onto its target rectangle; adjoining the unchanged third interval
gives an onto box map with linear part $\diag(B^{-e},B^e,1)$ and
determinant one. All source centers are
negative because terminal controls have no outgoing rules. Apply the
preceding proposition. The logical initial code $Y_0$ is rational and has
a uniform machine-dependent bound as the input varies: its two stack
coordinates lie in $[0,1]$ and its height is fixed by the initial control.
A moving translation loads the logical origin to $Y_0$, with a fixed
source half-width one and padding one. This gives a machine-dependent
rational bound $K$ for the supports of all such loaders and the repeated
mechanism; for example take $K=1+2R+M$ with $M$ an upper bound for all
routing vertices and loading coordinates and $R$ enlarged to cover their
half-widths. Choose $\lambda=1/(4K)$ and use the chart $x=o+\lambda Y$.
Everything remains strictly inside $(1/4,3/4)^3$.
Periodize the resulting compactly supported curl fields, loading during
the first unit interval and repeating thereafter. Compute the residual
at physical viscosity $\nu$, after the chart change. The fields and force
have zero mean and all the bounds in Lemma~\ref{bc:analytic}.
For a nonhalting run, loading has nonpositive logical height and each
later segment remains negative by the proposition. A terminal code has
physical height $1/2+\lambda$, which lies in the event. There is no
chart-boundary crossing that could create a spurious observation.
An initially terminal input is detected at the loading endpoint.
The initialized correspondence is exact; at checkpoints the number of
history records gives $n$, and the frontier's relative position together
with its absolute index $2+n$ also recovers the absolute work-head position.
\end{proof}
\subsection{Fixed columns and separate layers}\label{bc:columns-section}
\begin{proposition}[Column and layer transport]\label{bc:columns}
Let $S_i,T_i$ be finite rational planar rectangle families of positive side
lengths, each separately disjoint, with endpoints on the $B^{-2}$ grid,
where $B\ge2$ is an integer.
Let their assigned center-to-center positive reciprocal diagonal affine
maps satisfy $F_i(S_i)=T_i$. A finite effective
solenoidal field in $\R^3$ realizes all these maps on the rectangles in
$z=0$. Along each tracked path the first coordinate stays between its
initial and final values. The field has compact support and flat time
collars.
\end{proposition}
\begin{proof}\label{bc:columns-proof}
Give rectangle $i$ height $h_i=4i$. A planar cutoff equal to one near
its source and supported within padding $1/(4B^2)$ has support disjoint
from the other source cutoffs. Multiply it by a third-coordinate cutoff
which is one near the entire interval $[0,h_i]$. The curl of the potential
$(0,h_i x,0)$ times these cutoffs equals $(0,0,h_i)$ along that source
column. A common flat schedule lifts all rectangles. The corresponding
target cutoffs give a later descent by $-h_i$.
In the middle phase, let $p_i,q_i$ be the centers and $a_i>0$ the
first-coordinate scale. With phase parameter $\mu\in[0,1]$, set
$c_i=p_i+\mu(q_i-p_i)$, $l_i=1+\mu(a_i-1)$ and
\[
H_i(\mu,x,y)=d_{i,1}y-d_{i,2}x+
\frac{a_i-1}{l_i}(x-c_{i,1})(y-c_{i,2}),
\qquad d_i=q_i-p_i.
\]
On a neighborhood of the desired path, the planar Hamiltonian field
$(\partial_yH_i,-\partial_xH_i)$ generates
$c_i+\diag(l_i,l_i^{-1})(\zeta-p_i)$.
The first coordinate is the convex interpolation of its endpoint values.
The second half-width obeys
\[
\frac{r_{i,2}}{(1-\mu)+\mu a_i}
\le (1-\mu)r_{i,2}+\mu r_{i,2}/a_i,
\]
by convexity of the reciprocal function. Consequently the swept rectangle
lies in the coordinate bounding rectangle of the two endpoints. Localize
$H_i$ around that rectangle in the layer $z=h_i$, with third padding less
than one, and take the curl of $(0,0,H_i)$ times that cutoff. For a
flat middle ramp $\mu=\eta_2(t)$, multiply the resulting phase field by
$\eta_2'(t)$. Layers are disjoint, so the sum has the required action. Lift and descent preserve
planar coordinates. The exact paths and ODE uniqueness establish the
full-rectangle and intermediate-coordinate assertions.
\end{proof}
\subsection{Target containers and prescribed affine images}
\label{bc:contained-targets}
The named targets in Propositions~\ref{bc:ordered}, \ref{bc:simultaneous}
and~\ref{bc:columns} can also be rational \emph{containers} for the actual
images. More precisely, replace the onto condition by
\[
F_i(D_i)\subseteq S_i\quad\text{in Proposition~\ref{bc:ordered}},
\qquad
F_i(S_i)\subseteq T_i\quad\text{in the other two propositions}.
\]
Keep the other geometric hypotheses. For ordered transport retain the
center-to-center condition. For simultaneous and column transport the
actual endpoint center is $F_i(p_i)$, where $p_i$ is the source center;
it need not be the center of the container. The maps have the same positive
diagonal determinant-one linear parts as before, including the specified
three scale choices in Proposition~\ref{bc:simultaneous}.
All transport and path conclusions remain valid. For arbitrary real map
coefficients this is a smooth existence statement. The construction is
effective when those coefficients are computable, and otherwise its
evaluation is relative to them. An empty list again uses the zero field.
Here are the necessary modifications of the proofs. In ordered transport,
write $r_{i,j}$ for a source half-width, $w_{i,j}$ for the corresponding
container half-width, and $a_{i,j}>0$ for the diagonal factor. Containment
gives $a_{i,j}r_{i,j}\le w_{i,j}$, hence
\[
r_{i,j}a_{i,j}^{\theta}
\le\max(r_{i,j},a_{i,j}r_{i,j})
\le\max(r_{i,j},w_{i,j})0$ be the minimum sup-norm separation within the
two center families. Put $K=1+32R/m_0$. Expand all source centers by
factors from one to $K$, leaving their shapes unchanged. Let
$M=\max_i\{|(Kc_i)_1|,|(Kq_i)_1|\}$ and choose parking centers
$P_i=(M+32Ri,0)$. Each union of expanded endpoint centers and parking
centers has separation at least $32R$.
Move expanded sources successively to their parking centers, deform them
there using the linear reciprocal scale, move them successively to the
expanded target centers, then contract those centers to their targets.
For each successive transfer put a square of sup-radius $4R$ around
every stationary center. Replace each nontrivial segment intersection
with an obstacle square by a counterclockwise path along its boundary.
The intersection points and corners are rational; the squares are disjoint,
so the resulting finite polygon stays at distance at least $4R$ from
all stationary centers. Flat schedules on its segments give smooth center
motion. The simultaneous expansion and contraction preserve a positive
coordinate gap for every pair, exactly as in the solid-box argument.
If $g_0,g_1$ are their minimum positive gaps, choose padding
$\varepsilon=\min(g_0,g_1,2R)/4$ for these phases. During individual
transfers use padding below $R$; the stationary half-widths are below $R$.
Thus all required cutoff supports avoid stationary rectangles.
For a moving center $c$ and horizontal scale $l$, the potential
\[
H=\chi\left[\dot c_x(y-c_y)-\dot c_y(x-c_x)
+\frac{\dot l}{l}(x-c_x)(y-c_y)\right]
\]
gives the required motion on the cutoff plateau. The other current
rectangles are fixed. This proves every full planar branch, including
its boundary, by induction through the operations. A one-square loading
translation with half-width $1/4$ takes $(-4,0)$ to the initial rational
code. All these supports lie in a square $[-L,L]^2$, where one can choose
$L=10+S_0+4m_Q$ with $S_0$ a finite bound for routing vertices and cutoff
widths. The same $L$ covers every input of a fixed machine, since initial
codes are uniformly bounded.
The planar detours need not avoid the horizontal observation box.
We instead make them at height three. Choose a switch $\eta$ with
collars and set
$h(s)=3\{\eta(3s)-\eta(3s-2)\}$ on a unit interval. It lifts in the
first third, stays at three in the middle, and lowers in the last third.
Choose $\Gamma$ equal to one near $[-L,L]^2\times[-1,4]$, with compact
support, and $\beta(z)$ compactly supported and equal to one near three.
For any one of the planar fields $w$, zero outside its phase interval,
put
\[
Z(s,x)=h'(s)\nabla\times(0,x_1\Gamma(x),0),\qquad
V_w(s,x)=Z(s,x)+3\beta(x_3)(w(3s-1,x_1,x_2),0).
\]
Along each tracked path the first term is exact vertical translation.
The second term acts only in the middle third, at height three, and runs
the full planar phase. It is divergence free because $w$ is planar
solenoidal and $\beta$ is independent of the planar coordinates.
Thus the trajectory lifts at its exact source, performs every detour
outside the vertical detector interval, and lowers at its exact target.
Use this construction for the loader and for each repeated phase. While
its height can lie in $(-1,1)$, a nonterminal computational particle has
first coordinate at least four; the initial loading source has first
coordinate $-4$. A terminal code lies in $(0,1)^2\times\{0\}$.
These facts prove the all-real-time observation equivalence, including
an initially terminal input.
For the comparison assertion, the constructed field has compact support
and bounded mixed derivatives, so it satisfies the reference hypotheses
of Lemma~\ref{lem:b-comparison}. The displayed competitor class is exactly
case (v) of that lemma. Its proof obtains $\nabla p\in L^2$ from the
equation and cancels the pressure term by Lemma~\ref{lem:b-gradient}; it
does not require an integrable pressure representative. Hence the velocity
is unique, and pressure is determined up to a function of time.
Finally let $U(s)$ denote the eventually periodic field just constructed
and set $\rho(t)=(1+t)^{1/3}-1$, $\widehat U=\rho' U(\rho(t))$.
The new residual is
\[
\widehat f=\rho''U+(\rho')^2\{U_s+(U\cdot\nabla)U\}
-\nu\rho'\Delta U,
\quad \rho'=\tfrac13(1+t)^{-2/3},\quad
\rho''=-\tfrac29(1+t)^{-5/3}.
\]
Uniform derivative bounds and the fixed compact support give the stated
decay and space-time $L^2$ estimate. Chain differentiation also gives all
mixed bounds. The clock is effective on $[0,\infty)$, starts at zero,
is finite at finite time and is onto; hence it preserves precisely the
same material event. The initial zero collar is preserved as well.
This slower force is an alternative to, not a periodic version of, the
first force.
\end{proof}
==== END Incompressible-Box-Transport-and-Finite-Computation//solid-routing.tex ====
==== BEGIN Incompressible-Box-Transport-and-Finite-Computation//torus-routing.tex ====
\section{Torus charts and phasewise localization}\label{sec:torus-routing}
The next constructions keep the whole motion inside a specified torus
chart. A scaled linear interpolation preserves a convex first coordinate;
an exponential interpolation in a fixed chart instead uses endpoint-width
bounds. A phasewise construction then exhibits the vertical and planar
potentials separately, and the final variant retains a permanent origin
track. These choices give different path estimates for their observers.
In every case the force is recomputed at the prescribed physical viscosity
after any spatial rescaling.
\subsection{Physical chart scaling and thickened torus boxes}
\begin{proposition}[Linear scaling into the torus]
\label{prop:bb-mid-scaled-torus}
The direction recorder without its passive track has a unit-torus
realization periodic after $t=1$, with fixed label $(1/8,0,0)$ and
halting event $x_1\in(-1/4,0)\pmod1$. Each period acts on full
solid boxes; the pressure is zero and the force has zero mean.
\end{proposition}
\begin{proof}\label{proof:bb-mid-scaled-torus}
Use the controls of Lemma~\ref{lem:bb-direction}. Let
$\Sigma=\Gamma\times(\{\bot,F\}\sqcup\mathcal R)\times\{0,1\}$
be its full cell alphabet after deleting the singleton passive track.
Let $S$ be the finite set of recorder controls and choose a bijection
$j:S\to\{1,\ldots,|S|\}$.
Use odd digits with $B=2|\Sigma|+1$, and offsets $o_s=-2j(s)$ for halting main
controls and $o_s=2j(s)$ for the others. Number the resulting reciprocal branches
$i=1,\ldots,N$, and write $k_i$ for branch $i$'s first-coordinate scale.
Thicken each reciprocal branch by
$[-1,1]$. Its horizontal family gaps are at least $B^{-2}$.
Lift branch $i$ to $4i$, use the linear first scale
$\kappa_i=1+\theta(k_i-1)$, and descend. Padding smaller than
$(4B^2)^{-1}$ separates all phases: in the middle the half-height
is one and height gaps are four; in the other phases horizontal
gaps suffice. The two horizontal half-widths are at most $1/2$.
Use the true-plateau cutoff of Lemma~\ref{lem:bb-curl}; it supplies
at least the full-box guarantee, together with a neighborhood extension.
The exact first coordinate obeys
\begin{equation}\label{eq:bb-mid-scaled-convex}
X_1(\tau)=(1-\theta)X_1(0)+\theta X_1(1).
\end{equation}
Let $V$ be the period-one extension of the unit-time field just constructed.
All codes, paths and padded supports lie in $(-L,L)^3$ for
$L=10+2|S|+4N$. Put $a=1/(16L)$ and define in the torus patch
$(-1/2,1/2)^3$ the repeated field $U(t,x)=aV(t,x/a)$.
Its support lies in $(-1/16,1/16)^3$ and its curl potential is the
scaled potential $a^2A(t,x/a)$. In particular its mean is zero.
The residual must be recomputed at the physical viscosity:
\begin{equation}\label{eq:bb-mid-viscosity-scale}
\mathcal F_\nu[U](t,x)=a\{V_t+(V\cdot\nabla)V-(\nu/a^2)\Delta V\}(t,x/a).
\end{equation}
This keeps the specified viscosity rather than silently rescaling it.
During the first unit interval load from $(1/8,0,0)$ to the scaled
initial code using a curl translation with a box of half-width $1/32$
and padding $1/32$. Its support lies inside $(-1/4,1/4)^3$.
For a nonhalting run both loading endpoints have positive first
coordinate, and every subsequent pair of coded endpoints is positive.
Equation~\eqref{eq:bb-mid-scaled-convex} preserves that sign throughout.
A halting checkpoint has negative first coordinate in $(-1/16,0)$.
All paths stay in the stated patch, so passage modulo one introduces
no extra visit. Checkpoint times are
$1+\sum_{l0,\\0,&r\le0,\end{cases}
\qquad \Theta(r)=\frac{E(r)}{E(r)+E(1-r)}.
\]
Use the single-halt normalization and the seven-row recorder in
Lemma~\ref{lem:bb-movefirst}, with frontier initially at two and all marks
initially zero. In the present notation its controls are
$S_q,U_r,W_r,C_q,D_q$ (ready, work-move completed, right scan,
frontier extension, left scan), and its full symbols are $(a,g,m)$
with history $g\in\{E,F\}\sqcup\{\widehat r\}$.
The first row writes and moves, the second marks and starts the scan,
and the last clears the mark and stays. These are exactly the seven
rows already proved, after renaming. Thus the new work head $j'$ is
less than $n+2$, a work step takes $2(n+2-j')+4$ local steps, and
both full-domain incoming properties hold.
Let $M$ count the partial controls, let $H=S_h$, and put
\[
\ell=\frac1{100(M+1)},\qquad o_H=(3/4,1/2),\qquad
o_{s_i}=(1/4+2i\ell,1/2)\quad(0\le i0$. Curls have zero mean, and incompressibility
gives zero mean for the residual. Lemma~\ref{ba:realization} supplies
the unique zero-data PDE solution and its global trajectories.
At times $1+k$, induction through the full branch maps gives the exact
local code through the first halt, or for all $k$ in the nonhalting
case. A halt code has first coordinate in
$[3/4,3/4+\ell]\subset(2/3,5/6)$, including a code reached solely by
loading an initially halted machine. For a nonhalting input the loader
stays below $1/2$ in first coordinate. Lift and descent retain the
nonhalting planar endpoint codes. In the middle phase the centers are
convex combinations of nonhalting centers and the first half-width is
at most $\ell/2$, so
\[
X_1\le1/4+2M\ell+\ell/2<1/2.
\]
This excludes the height-independent observer at every intermediate
time. Only the loader depends on the input; the repeated field depends
on the finite machine table.
\end{proof}
\subsection{A permanent origin marker and moving cutoff neighborhoods}
\label{bc:height-separated-section}
\begin{theorem}[A height-separated mechanism with an origin track]
\label{bc:height-separated-event}
On the unit torus, the fixed label $(1/8,3/8,1/4)$ and strip
$2/3