# H0199 — witness-patch layout: a complete minimax theorem

**Status:** the theorem below is proved under explicit assumptions. The original physical hypothesis is not proved. No mathematical originality or new physical discovery is claimed.

## Original claim and the bounded claim we can establish

The original claim says that interleaving inert reference patches estimates spatial deposition bias better than putting all references at the edge. That is not true for every field, noise model, or estimator.

Here is an exact replacement: for a field whose spatial variation is bounded by a known Lipschitz constant, the smallest achievable worst-case reconstruction error equals the coverage radius of the reference locations times that constant, plus the maximum measurement error. A particular nine-point interior layout has a strictly smaller worst-case error than any nine-point boundary-only layout on a square.

## Assumptions and definitions

Let Ω be a nonempty compact subset of Euclidean space, with distance d. Let S={s₁,…,sₙ} be fixed reference positions, n≥1. We observe

    yᵢ = f(sᵢ) + ηᵢ,       |ηᵢ| ≤ ε.

Assume f is L-Lipschitz: |f(x)−f(z)|≤L d(x,z), with L≥0. Noise is bounded adversarially; no independence or Gaussian assumption is made. Reference patches do not perturb f. Their readings have already been calibrated into the same units as f. We are reconstructing f, not yet separating chemical uptake from a physically altered deposition field.

Define distance to the reference set and its coverage radius:

    d(x,S) = minᵢ d(x,sᵢ),       h(S) = maxₓ∈Ω d(x,S).

For any reconstruction rule A mapping the vector y to a function on Ω, define

    R(A,S) = sup over admissible f and η of maxₓ |A(y)(x)−f(x)|.

## Theorem 1

    inf over all reconstruction rules A of R(A,S) = L h(S) + ε.

### Upper bound

At x, choose a nearest reference sⱼ and estimate f(x) by yⱼ. Then

    |yⱼ−f(x)| ≤ |f(sⱼ)−f(x)|+|ηⱼ|
                ≤ L d(x,S)+ε ≤ Lh(S)+ε.

Thus an available estimator attains risk at most Lh(S)+ε.

### Matching lower bound

The distance function d(·,S) is 1-Lipschitz: the triangle inequality gives d(x,S)≤d(x,z)+d(z,S), and reversing x,z gives the absolute bound.

Choose any constant M and define

    f₊(x)=M+ε+L d(x,S),       f₋(x)=M−ε−L d(x,S).

Both functions are L-Lipschitz. At every reference they equal M+ε and M−ε respectively. With noise −ε for f₊ and +ε for f₋, both produce the identical observation vector y=(M,…,M).

Compactness provides x* attaining h(S). The two possible true values at x* differ by 2(ε+Lh(S)). Any estimate a based on the common observations must therefore satisfy

    max(|a−f₊(x*)|, |a−f₋(x*)|) ≥ ε+Lh(S).

Otherwise the triangle inequality would make the separation smaller than 2(ε+Lh(S)), a contradiction. Consequently every reconstruction rule has worst-case error at least ε+Lh(S). Together with the upper bound this proves equality. If nonnegative deposition is required, choose M≥ε+Lh(S); both adversarial fields remain nonnegative. ∎

## Theorem 2: an explicit equal-count square comparison

Take Ω=[0,1]² and nine reference positions. For any boundary-only layout, the center (1/2,1/2) is at least 1/2 from every reference, so h(edge)≥1/2.

For a 3×3 interior grid with each coordinate in {1/6,1/2,5/6}, every coordinate is within 1/6 of a grid coordinate. Hence distance to the nearest grid point is at most √2/6. Square corners attain that distance, so h(grid)=√2/6 exactly.

Therefore

    R*(grid)=L√2/6+ε,
    R*(edge)≥L/2+ε.

For L>0, the grid has strictly smaller worst-case risk. For ε=0, its risk is at most √2/3≈0.4714 of the boundary-layout lower bound. This is a worst-case statement, not a 52.86% improvement on every physical deposition field. For L=0 the two minimax risks are both ε; strict superiority disappears. ∎

## What this does not prove

- It does not prove superiority on every individual field. For a constant noiseless field both layouts reconstruct exactly. A known affine field can be reconstructed from suitable boundary measurements as well.
- It does not compare all statistical estimators under independent random noise or an integrated mean-square objective.
- References have finite area in real devices. The proof treats calibrated point observations and ignores shadowing, adsorption, chemical perturbation, and footprint-dependent averaging.
- The supplied earlier polynomial simulation and this theorem use different estimators and function classes. One cannot use the theorem as a proof of that simulation's exact performance percentages.

## Physical result still needed

Establish the relevant field smoothness, verify nonperturbing references, measure their error, use equal reference count and area, and compare against independent full-field ground truth. See experiments/H0199.md. This is the bridge from the mathematical result to the original research claim.


# H0346 — perforated hydrogel skin: a complete transport-model proof

**Status:** the stated lumped transport result is proved. Its applicability to a real swelling gel remains an experimental question. The calculation is not a new transport law.

## Original claim and the ambiguity to remove

The original claim says sparse perforations reduce swelling delay by more than their fractional open area. Define that precisely as

    D(φ) = 1 − t_half(φ)/t_half(0) > φ,

where φ is the perforated fraction of skin area, 0<φ<1, and t_half is time to half of equilibrium uptake. This comparison is a defined dimensionless criterion, not a general meaning of “more than expected.” It is not a claim to exceed a correctly specified area-weighted permeability model.

## Model assumptions

Normalize total surface area to one. Let k_s>0 be conductance of intact skin per unit area, k_h≥k_s conductance through a hole, and b>0 bulk conductance behind the skin. Skin and hole pathways act in parallel; the resulting surface layer and bulk act in series. Their conductances are linear, constant, and independent of swelling.

Let S>0 be a fixed storage coefficient. The interior is represented by one well-mixed potential, and the bath potential is a fixed step. Perforating the skin leaves equilibrium capacity S unchanged. There are no access/spreading resistances, inter-hole interactions, cracks, changing modulus, nonlinear osmotic effects, or additional exposed edges.

These are explicit model restrictions. Real poroelastic gels need not behave as a single well-mixed storage element.

## Theorem 1: exact half-time and monotonicity

Parallel conductance is

    g(φ)=(1−φ)k_s+φk_h=k_s+φd,       d=k_h−k_s≥0.

Series addition gives

    G(φ)=1/(1/b+1/g(φ)).

If z is normalized uptake, conservation is S dz/dt=G(φ)(1−z), z(0)=0. Separating variables gives

    z(t)=1−exp(−G(φ)t/S),
    t_half(φ)=S ln(2)[1/b+1/g(φ)].

Differentiation yields

    dt_half/dφ=−S ln(2)d/[k_s+φd]²≤0.

For d>0, half-time strictly decreases with perforated area. For d=0 it is unchanged. ∎

## Theorem 2: exact condition for delay reduction exceeding open fraction

Direct algebra gives

    D(φ)= b φd / [(k_s+φd)(b+k_s)].

For φ>0, all denominators are positive, so

    D(φ)>φ
    iff b d > (k_s+φd)(b+k_s)
    iff d[b−φ(b+k_s)] > k_s(b+k_s).

This last inequality is a necessary and sufficient condition. Equivalently, the effect requires

    φ < b/(b+k_s)

and

    d > k_s(b+k_s)/[b−φ(b+k_s)].

If b−φ(b+k_s)≤0, the claimed inequality is impossible. This establishes both the predicted regime and its failure regime without a numerical fit. ∎

## An exact example

Use consistent arbitrary conductance units and set

    b=1, k_s=1/100, k_h=1, φ=1/100.

Then g=199/10000, and

    D=9900/20099≈0.492562,
    t_half(φ)/t_half(0)=10199/20099≈0.507438.

Thus a 1% opening fraction produces about a 49.26% half-time reduction within this model. This is a constructed mathematical example, not a measured gel result or a predicted performance specification.

## Counterexamples and limits

1. If k_h=k_s, perforations provide no transport contrast and D=0.
2. For fixed k_s,k_h,φ, letting b→0 gives D→0. If bulk transport is the bottleneck, opening the skin has arbitrarily little effect.
3. If k_h<k_s, outside the theorem's beneficial-hole assumption, the derivative changes sign and holes can slow uptake in this same algebraic model.
4. As b→∞, D approaches φd/(k_s+φd). The gain remains explained by ordinary parallel conductance; it is not evidence for extra physics.
5. As φ→0, D/φ approaches b d/[k_s(b+k_s)]. A large initial gain is possible when intact skin is a strong resistance and hole conductance is high. For a finite φ, the exact condition above still controls the result.

## What remains unproved

The proof does not establish that a selected gel has a dense transport-limiting skin, that holes remain open and non-damaging, or that its full uptake curve follows a single exponential. It also does not show an effect beyond an area-weighted transport prediction: that prediction is precisely the mechanism used here.

Measure skin and bulk transport independently, keep equilibrium uptake fixed, and test the complete uptake curves on new specimens. If one parameter set cannot predict intact and perforated specimens, reject this lumped explanation before claiming an enhanced physical effect. See experiments/H0346.md.


# H0836 — compliant buffering: complete conditional proofs and counterexamples

**Status:** the circuit-model results below are proved. Reduced physical sand breakthrough is not established. No new buffering principle is claimed.

## Model and assumptions

Consider a fluid-storage node with constant positive compliance C and a linear downstream hydraulic resistance R>0. Pressure p is measured relative to the downstream outlet; inflow is u(t), outflow is y(t)=p(t)/R. Conservation gives

    C dp/dt = u−p/R,
    τ dy/dt + y = u,             τ=RC>0.

The baseline is subtracted and initially equilibrated. Assume no inertia, resonance, cavitation, leakage, hysteresis, or deformation-dependent resistance. Comparing the rigid limit uses the same R and specified inlet waveform, with C=0 giving y=u. These assumptions are not yet a model of particle transport.

## Theorem 1: exact attenuation

Multiplying the differential equation by exp(t/τ) and integrating gives the zero-state solution

    y(t)=∫₀∞ k(s)u(t−s)ds,
    k(s)=exp(−s/τ)/τ for s≥0.

For a nonzero-frequency sinusoid of amplitude A, inserting a complex exponential gives transfer function

    H(iω)=1/(1+iωτ),
    output amplitude = A/√(1+(ωτ)²) < A.

The DC gain is 1; this model does not remove the mean flow.

For a positive rectangular pulse with A>0, u=A on [0,T] and zero otherwise,

    y=A(1−exp(−t/τ))                     for 0≤t≤T,
    y=A(1−exp(−T/τ))exp(−(t−T)/τ)       for t≥T.

The first branch increases and the second decreases, so

    peak(y)=A(1−exp(−T/τ))<A.

Integrating the differential equation over the whole pulse and recovery gives ∫y dt=∫u dt=AT. A lower peak is accompanied by a longer tail, not the destruction of fluid volume. ∎

## Theorem 2: a sufficient particle-release closure

For nonnegative bounded inflow pulses about a zero-flow baseline, suppose, additionally, that release mass per unit time is Φ(flow), where Φ is convex, nonnegative, memoryless, Φ(0)=0, and both full-event integrals exist. This is an assumption requiring measurement, not a proven law for sand.

Because k is nonnegative and integrates to 1, Jensen's inequality gives

    Φ(y(t)) = Φ(∫k(s)u(t−s)ds)
            ≤ ∫k(s)Φ(u(t−s))ds.

Integrating over all real time and changing order of the nonnegative integrals yields

    ∫Φ(y(t))dt ≤ ∫Φ(u(t))dt.

Thus, under this closure and a matched inlet waveform, buffering cannot increase total modeled release. An example is Φ(q)=a max(q−qcrit,0), a≥0. Strict reduction requires further conditions; Φ(q)=aq produces exactly equal total release because volume is conserved. ∎

## Counterexample 1: a lower peak can release more material

The original claim does not specify convexity. Let release be constant whenever flow exceeds a threshold:

    Φ(q)=1 if q>0.1, otherwise 0.

Take A=T=τ=1. Rigid flow exceeds the threshold for exactly one time unit. Buffered flow crosses it during the rise at

    t↑=−ln(0.9),

and during decay at

    t↓=1+ln((1−exp(−1))/0.1).

Its modeled release duration is t↓−t↑≈2.73855>1, although its peak is only 1−exp(−1)≈0.6321. This explicitly refutes the inference that a lower peak alone guarantees less integrated particle release. The counterexample closure is deliberately nonconvex and is not asserted to describe real sand.

## Counterexample 2: equal inlet waveform is not equal hydraulic work

For this energy comparison use zero initial and final flow and pressure, not an unspecified nonzero operating baseline. Pump work is W=∫p u dt. Multiplying conservation by p and integrating over a complete return to equilibrium gives

    W = R∫y²dt + (C/2)[p²]initial to final = R∫y²dt.

For the rectangular pulse,

    Wbuffer=R A²[T−τ(1−exp(−T/τ))],
    Wrigid=R A²T.

Consequently the same inlet pulse does not supply the same work. With R=T=τ=1, compare a rigid amplitude 1 with a buffered amplitude exp(1/2). Both have work 1, but the buffered peak is

    exp(1/2)−exp(−1/2)=2sinh(1/2)>1.

The strict inequality follows from sinh(x)>x for x>0. This equal-work example does not also match total injected volume; it demonstrates a limitation of energy matching alone, not a refutation of every jointly controlled physical experiment.

## Exact conclusion

Linear compliant storage attenuates matched-waveform fluctuations. A specific convex release closure then bounds integrated modeled release. Neither attenuation nor that closure establishes the original sand-breakthrough claim under all the original matching conditions. Experiments must establish actual compliance, resistance, flow and pressure histories, energy, volume, and release kinetics.
