Five results, with their boundaries.

# 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.

# 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.

# 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.

# N0755 — exact reconstruction in a closed, anchored thermal network **Proved:** the conditional mathematical theorem below. **Not proved:** arbitrary recovery of missing physical measurements, the correctness of a particular thermal model, or novelty. ## Original proposal A physical network could reconstruct missing boundary measurements by enforcing local energy balance. This needs an identifiability condition. Conservation cannot create information about arbitrary unknown losses or multiple unconstrained boundary fluxes. We prove recovery of missing node temperatures when conductances, net powers at unobserved nodes, and sufficient temperature anchors are known. ## Model Let a finite undirected graph have symmetric positive edge conductances k_ij. Partition nodes into measured-temperature anchors B and unobserved nodes U. At every node i in U, assume known net applied thermal power p_i and steady conduction: sum over neighbors j of k_ij(T_i−T_j)=p_i. Any external heat loss at an unobserved node must already be included in p_i or represented by a calibrated connection to a known-temperature node. Conductances are constant and known. No unmodeled advection, radiation, storage, or contact drift is allowed. Let L be the weighted graph Laplacian: L_ii=sum_j k_ij and L_ij=−k_ij for an edge. Then L_UU T_U = p_U−L_UB T_B. A physically unmeasured boundary node can be placed in U only when its net power is independently known; missing temperature and missing heat exchange cannot both be silently assumed recoverable. ## Theorem 1: unique reconstruction If every connected component containing an unobserved node also contains at least one anchor, L_UU is positive definite and T_U=L_UU⁻¹(p_U−L_UB T_B) is the unique solution. ### Proof For any vector x on U, extend it by zero on B, obtaining z. Counting each undirected edge once, xᵀL_UU x = zᵀLz = sum_edges k_ij(z_i−z_j)² ≥ 0. If the sum is zero, each edge connects equal values. Every connected component touching U contains an anchor where z=0. Equality propagates along paths, so z=0 throughout that component and hence x=0. Thus the quadratic form is strictly positive for every nonzero x. A finite positive-definite matrix is invertible, giving existence and uniqueness of the displayed solution. ∎ If a connected component has no anchor, its constant-temperature vector lies in the Laplacian nullspace. Depending on source balance there are either no steady solutions or a family differing by an arbitrary constant. Absolute temperature is not uniquely recoverable there. ## Theorem 2: an anchor-error bound Assume conductances and powers are exact, while every anchor temperature has error at most ε in absolute value. The reconstruction error e obeys L_UU e_U = −L_UB e_B. At each unobserved node, e_i is a conductance-weighted average of neighboring errors. If an unobserved value exceeded ε, a global maximum exceeding ε would have to be shared by all its neighbors; otherwise its weighted average would be smaller. Following a path reaches an anchor, contradicting the anchor bound. Applying the same argument to −e gives max_i∈U |e_i| ≤ ε. Thus harmonic reconstruction does not amplify bounded anchor-temperature error in the maximum norm when the other model inputs are exact. ∎ With power error δp as well, the general Euclidean bound is ||δT_U||₂ ≤ [||δp||₂ + ||L_UB||₂ ||δT_B||₂] / λ_min(L_UU). This follows by subtracting the two linear systems and using the inverse norm of a symmetric positive-definite matrix. Unknown conductance errors require a separate analysis. ## Exact example including an unmeasured boundary Take four nodes in a chain, every conductance equal to 1. Anchor T₀=0. Apply known power 4 at the other end, p₃=4, with p₁=p₂=0. Solving gives T₁=4, T₂=8, T₃=12. The end temperature T₃ is recoverable without measuring it, because its net heating is known and a temperature anchor exists. These are constructed model values in consistent arbitrary units, not measured device data. ## Explicit failures - Without any temperature anchor, adding a constant to all temperatures leaves all balances unchanged. - A single node connected to an anchor with conductance k has T=T_anchor+p/k. Unknown heat input or loss makes its temperature unknown; a small k strongly amplifies power uncertainty. - Knowing only q₁+q₂=P at a boundary does not determine q₁ and q₂ individually. Conservation alone is insufficient. - A small balance residual does not prove the model is right: compensating parameter errors may cancel. Validation needs withheld physical measurements and independently calibrated parameters. The proof supplies exact closure and failure conditions. It does not establish an original research result; related physically informed missing-node reconstruction is already present in the literature.

# N0751 — a differential swelling sensor: exact response and limits **Proved:** the linear two-path model results below. **Not proved:** that a real swelling material obeys this model, chemical specificity, universal detection of arbitrarily fast changes, or novelty. ## Original proposal A mechanical differentiator could detect rapid chemical changes through the mismatch of two swelling pathways. A concrete interpretation is a mechanically read difference of two displacement paths responding to the same concentration c(t), with equal calibrated equilibrium gain g>0 and different positive time constants τ_f<τ_s: τ_f dx_f/dt+x_f=g c(t), τ_s dx_s/dt+x_s=g c(t), y=x_f−x_s. The material, input delivery, mechanical readout, and operating range must be experimentally shown compatible with this approximation. The two paths need equal equilibrium gain, not merely equal dimensions. ## Theorem 1: rejection of a constant concentration For a constant concentration c₀ at equilibrium, both paths equal gc₀, so y=0. For a step c(t)=A from an initially equilibrated zero baseline, y(t)=gA[exp(−t/τ_s)−exp(−t/τ_f)], t≥0. This follows by solving each first-order equation with an integrating factor and subtracting. For A>0, y is positive at every finite t>0, and tends to zero both initially and at long times. Thus a concentration change produces a transient while a constant equilibrated input is rejected. ∎ The peak occurs at t*=τ_f τ_s ln(τ_s/τ_f)/(τ_s−τ_f). Differentiating y and solving dy/dt=0 gives this expression; the derivative changes from positive to negative there, so it is the unique peak. For g=A=1, τ_f=1, τ_s=4, t*=(4/3)ln4 and peak y=(3/4)4^(−1/3)≈0.47247. These are arbitrary-unit model values, not material measurements. ## Theorem 2: the exact frequency band For a steady sinusoid of angular frequency ω, substitution of a complex exponential yields H(iω)=g iω(τ_s−τ_f)/[(1+iωτ_f)(1+iωτ_s)]. Therefore |H|=g |ω|(τ_s−τ_f)/sqrt[(1+ω²τ_f²)(1+ω²τ_s²)]. The response vanishes at ω=0 and tends to zero as |ω|→∞. This device is a band-pass change sensor, not an unlimited-bandwidth differentiator. To find the maximum, set z=ω². Apart from a positive constant, squared gain is F(z)=z/[(1+τ_f²z)(1+τ_s²z)]. Its derivative has the sign of 1−τ_f²τ_s²z², so the unique positive maximum occurs at ω*=1/sqrt(τ_f τ_s), |H(iω*)|=g(τ_s−τ_f)/(τ_s+τ_f). ∎ ## Theorem 3: a controlled derivative approximation Assume the stable causal responses are driven by a bounded history with sufficiently smooth c. Integrating the convolution by parts gives x_τ(t)=g c(t)−g∫₀∞ exp(−s/τ)c′(t−s)ds. Thus y(t)=g∫₀∞ [exp(−s/τ_s)−exp(−s/τ_f)]c′(t−s)ds. If |c″|≤K over the relevant history, subtract g(τ_s−τ_f)c′(t). The mean-value bound |c′(t−s)−c′(t)|≤Ks gives |y−g(τ_s−τ_f)c′(t)| ≤gK∫₀∞ s[exp(−s/τ_s)−exp(−s/τ_f)]ds =gK(τ_s²−τ_f²). The kernel is nonnegative because τ_s>τ_f. This proves a bounded approximation to a scaled derivative for inputs whose derivative changes sufficiently slowly. It is not an approximation for arbitrary sharp inputs; use the exact step or frequency response there. ∎ ## Counterexamples and physical limits - If the time constants are equal, the matched paths produce y=0 for every input: no change signal exists. - If gains differ, a constant concentration leaves y=(g_f−g_s)c₀. Calibration mismatch mimics a change or drift. - Arbitrarily high-frequency changes are attenuated, not amplified. - Temperature sensitivity, hysteresis, diffusion modes, input-delivery lag, changing geometry, and mechanical inertia can invalidate the two-exponential model. - A synthetic transient does not demonstrate chemical specificity or detection at a stated noise level. Both require independent physical calibration and blinded tests. The exact output is therefore predictable only within the specified model. Existing swelling kinetics and linear filtering principles are the foundation; no new physical law or verified original device is claimed.