A compact stock-preserving baker module

In the first octant, $t=\tan\theta\in[0,1]$ gives

Symmetry implies $1\leq R_p\leq\sqrt2$ and $|R'_p|\leq R_p$ globally. Thus $K\geq1$ gives radial invertibility, and the inverse function theorem gives smoothness off the origin. Near the origin the chart is ordinary polar coordinates and $H_*=\pi|q|^2$, so there is no origin singularity. Flat joins in $\chi$ make the intermediate patch smooth. Since $R\leq R_p$ and $R_p|\cos\theta|,R_p|\sin\theta|\leq1$, the level stays in its axis-intercept square.

Let $Z(r)=\int_0^{2\pi}w(r,\theta)\,d\theta$. Differentiating (2.4) gives $A_r=rZ$. The chart Jacobian is $\det(Q_r,Q_\theta)=rRK=rw$. Since the Hamiltonian is constant on each level, its equations are $\dot r=0$ and $\dot\theta=A_r/(rw)=Z/w$. The lifted angular clock

therefore satisfies $\dot{\mathcal T}=1$. Fourfold symmetry gives $\mathcal T(r,\theta+\pi/2)=\mathcal T(r,\theta)+1/4$, proving the exact endpoint and period claims. Each path remains on its level, which proves confinement. For $r\geq1/2$ the enclosed area is the full rounded-square area $A_p r^2$: changing its inner foliation subtracts no area offset.

First the whole inner square $[-1+2h,1-2h]^2$ is contained in the unchanged region $r<1-h$. In the outer homogeneous region, its rounded-square radius is at most $2^{1/p}(1-2h)$, and

Points whose homogeneous radius is below $1/2$ lie inside the outer half-radius curve, so their actual nested radius is also below $1/2<1-h$. This covers the patched center as well.

The first quantile Hamiltonian is $\mathcal H_g(u,v)=H_c(2u-1,2v-1)/4$. Its time $1/4$ map is $(u,v)\mapsto(1-v,u)$ on $[h,1-h]^2$. The divisor four is the coordinate Jacobian. The second Hamiltonian is the sum of two disjoint terms

Each support is strictly inside its own horizontal half-square, with a positive gap between them. The divisor eight is its coordinate Jacobian, and the negative sign gives clockwise motion. On their unchanged regions the maps are respectively $(x,y)\mapsto(2y,(1-x)/2)$ and $(x,y)\mapsto(2y-1,1-x/2)$.

The global quarter-turn sends $L_h$ into the lower unchanged region and $R_h$ into the upper one. In normalized coordinates their first coordinate is $1-2v$, bounded by $1-2h$ in absolute value, and their second is $4u-1$ or $4u-3$, bounded by $1-4h$. Their composed maps are exactly $(2u,v/2)$ and $(2u-1,(v+1)/2)$. Every intermediate point follows a confined complete level; ordinary rigid rotation of square corners, which would leave the square, has not been used.

For each of two consecutive stages of duration $1/2$, use the nonnegative rate $g(\tau)=\Theta'(2\tau)/2$ in its local stage time. It integrates to $1/4$, has maximum at most one, and is flat at both ends. Multiplying the corresponding Hamiltonian by this rate preserves its autonomous endpoint by time reparameterization. Smooth compact Hamiltonian fields have complete diffeomorphic area-preserving flows and smooth joins. The two rectangles have combined area $(1-4h)(1-2h)$, proving (2.6).

The inverse flow is realized by time reversal of the compact Hamiltonian path. The baker is a bijection almost everywhere with inverse (2.8); direct substitution verifies both branches. Area preservation maps the two original rectangles to $E_h$ with unchanged total area.

On the fixed central annulus, $r\chi'\leq4$ and $r^2|\chi''|\leq800$. With $R_p-1<1/2$, the chart quantities satisfy

For example $rK_r=2rR_r+r^2R_{rr}$ and $rw_r=(rR_r)K+R(rK_r)$. These bounds involve only the first angular derivative of $R_p$ and the fixed radial step jets.

The inverse-radius gradient has radial and angular components $1/K$ and $-R_\theta/(RK)$, so $|\nabla r|\leq\sqrt2<2$. For the clock (2.5),

The two quotient terms in $\mathcal T_r$ each cost at most $616$; one can take $\theta\in[0,2\pi]$. Other angular lifts add an integer to $\mathcal T$ and have the same derivatives. The last bound follows from $\nabla\mathcal T=\mathcal T_r\nabla r+ \mathcal T_\theta\nabla\theta$ and $|\nabla\theta|\leq1/r$.

Regard $Q$ now as the inverse chart in $(r,\mathcal T)$. Since $|Q_\theta|\leq(9/4)r$,

using $\pi<22/7$. It follows that $|Q_r|_{\mathcal T}<100000$ and $|Q_{\mathcal T}|<75r$. The full uncut flow is $Q(r,\mathcal T+s)$; differentiation therefore gives $100000\cdot2+75r(2465/r)=384875$. The inverse uses $-s$ and the same estimate. The circular origin supplies its continuous smooth extension.

For the cut flow replace $\mathcal T+s$ by $\mathcal T+s\psi_h(r)$. Its extra derivative is at most $75r(1/4)(4/h)2\leq150/h$ on its support. Outside the support the map is the identity. This proves (2.10).

The global square conjugation has condition number one; the half-square conjugations have condition number two. On a good completed stage, the map is exactly an affine quarter-turn, so it costs only one or two, respectively. At most one incomplete stage incurs the uncut bound. Thus any prefix or inverse prefix of one module costs less than $800000$ on that tube. Since $\|DC\|_{\rm op}\leq\phi(0)<2/5$, conversion to $\ell^1$ gives $\sqrt2\cdot800000\cdot\phi(0)<480000<10^6$. The inverse-label estimate involves the forward CDF derivative. The set $E_{2h}$ is at least $h/2$ in $\ell^1$ distance from the complement of $E_h$. The stipulated $h/8$ drift therefore keeps the virtual entrance label in $E_h$, where the preceding good-level proof applies. The corresponding forward version uses $L_{2h}\cup R_{2h}$. The excluded sets have the stated areas, by the same rectangle calculation. Using cut derivatives for both stages instead gives $2D_h^2$.

Finally $A_r=rZ\leq(21\pi/2)r$ and $|\nabla r|<2$, so $|J\nabla H_*|\leq21\pi r<66$ on the compact support. An inverse linear coordinate conjugation costs at most $1/2$, giving quantile speed at most $33$ per unit rate. Every support quantile is at least $h/8$ from zero and one. The function $I(u)=\phi(\Phi^{-1}(u))$ is concave, since $I''=-1/I<0$, and symmetric. Its chords imply $I(u)\geq2\phi(0)\min(u,1-u)$. Hence $\|DC^{-1}\|\leq4\sqrt{2\pi}/h<12/h$ on the support. The physical standardized speed is below $33\cdot12/h=396/h<400/h$.