Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Meshfree kernel approximation and conservative sph — Examples
1 · Prerequisites
- Binary Operations, Monoids, Groups and Subgroups
- Compactness in Metric Spaces
- Completeness, Completion, and Uniform Continuity
- Connectedness
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Continuity, IVT, EVT, and Uniform Continuity
- Countability and Uncountability
- Darboux, L'Hôpital, and Taylor's Theorem
- Determinants of Matrices over a Commutative Ring
- Filters and Ultrafilters
- Finite Counting, Factorials and Binomial Coefficients
- Foundations of the Real Numbers for Analysis
- Fubini and Change of Variables
- Gaussian Elimination, Elementary Matrices and Reduced Row Echelon Form
- Group Actions, Orbits, Stabilisers and Cayley's Theorem
- Limits of Real Functions
- limsup, liminf, and Subsequential Limits
- Linear Independence, Bases and Dimension
- Linear Transformations, Rank-Nullity and Quotient Spaces
- Matrices, the Matrix of a Linear Map, and Change of Basis
- Meshfree kernel approximation and conservative sph
- Metric Spaces
- Mixed Partials, Taylor Formulae, and Extrema
- Monotone Functions, Discontinuities, and Continuity Sets
- Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness
- Order, Zorn's Lemma, and the Axiom of Choice
- Picard-Lindelöf and First-Order Ordinary Differential Equations
- Polynomial Rings, the Division Algorithm and Roots
- Properties of the Integral and the Working FTC
- Relations, Functions, and Quotients
- Rings, Subrings, Integral Domains and Fields
- Rⁿ as a Normed Space; Vector-Valued Functions
- Roots, Rational Powers, and Classical Inequalities
- Sequences and Limits
- Sequences and Series of Functions; Uniform Convergence
- Series: Convergence and the Nonnegative Tests
- Subspaces, Products, and Quotients
- Suprema and Infima
- Symmetric Groups, Cycle Decomposition and the Sign Homomorphism
- The Derivative and the Mean Value Theorems
- The Fundamental Theorems of Calculus
- The Inverse and Implicit Function Theorems
- The Riemann Integral in Rᵐ and Jordan Content
- The Riemann Integral: Definition and Integrability
- The Total Derivative in ℝᵐ → ℝⁿ
- The ZFC Axioms and the Basic Set Constructions
- Topological Spaces and Continuity
- Topology of ℝ
- Vector Spaces, Linear Subspaces, Span and Direct Sums
2 · Summary
The examples test the precise hypotheses and limits of the A-page results.
The first schematic uses coordinates divided by ε. The red segment is the kernel support centered at x=0; the cells are Voronoi intervals. The node at 0 lies in the support but contributes zero to the affine derivative sum, leaving the two off-center terms computed exactly in A disordered fixed-neighbor stencil can fail derivative consistency. The drawing is illustrative; the proof uses the stated coordinates and cell lengths.
\begin{tikzpicture}[x=2.1cm,>=stealth]
\draw[->] (-1.45,0) -- (1.55,0) node[right] {$y/\varepsilon$};
\foreach \a/\b/\lab in {-1.25/-0.25/C_L,-0.25/0.30/C_0,0.30/1.30/C_R} {
\draw[fill=blue!8] (\a,-0.10) rectangle (\b,0.10);
\node[above=3pt] at ({(\a+\b)/2},0.10) {\ensuremath{\lab}};
}
\foreach \x in {-0.5,0,0.6} {
\fill[blue!65!black] (\x,0) circle (2pt);
}
\node[below=4pt] at (-0.5,0) {$-\frac12$};
\node[below=4pt] at (0,0) {$0$};
\node[below=4pt] at (0.6,0) {$\frac35$};
\draw[red,dashed,thick] (-1,0.48) -- (1,0.48);
\draw[red,dashed] (-1,0.43) -- (-1,0.07);
\draw[red,dashed] (1,0.43) -- (1,0.07);
\draw[green!50!black,dotted,thick] (0,-0.43) -- (0,0.65);
\node[red,above] at (0,0.48) {$|x-y|\le\varepsilon$};
\node[green!45!black,below] at (0,-0.43) {$x=0$};
\end{tikzpicture}
The next schematic shows the initial three-particle geometry in A conservative positive-pressure SPH system can close a particle gap, with distances measured in units of ε. Here α is the force magnitude for the 1–2 pair and β is the force magnitude for the 2–3 pair. The 1–3 distance exceeds the kernel radius, while the kernel derivative gives β>2α. The arrows show the pairwise forces; for the gap r=q₂−q₁ the exact calculation is r̈(0)=2α−β<0, as proved in the item.
\begin{tikzpicture}[x=3cm,>=stealth]
\draw[->] (-0.22,0) -- (1.62,0) node[right] {$q/\varepsilon$};
\draw[thick] (0,0) -- (0.9,0) -- (1.3,0);
\fill (0,0) circle (2.5pt) node[below=5pt] {$q_1=0$};
\fill (0.9,0) circle (2.5pt) node[below=5pt] {$q_2=0.9$};
\fill (1.3,0) circle (2.5pt) node[below=5pt] {$q_3=1.3$};
\draw[->,blue,thick] (0,0.28) -- (-0.18,0.28) node[left] {$\alpha$};
\draw[->,blue,thick] (0.9,0.40) -- (1.08,0.40) node[right] {$\alpha$};
\draw[->,red,thick] (0.9,0.68) -- (0.48,0.68) node[left] {$\beta>2\alpha$};
\draw[->,red,thick] (1.3,0.28) -- (1.72,0.28) node[right] {$\beta$};
\draw[<->,gray,dashed] (0,0.98) -- (1.3,0.98) node[midway,above] {$1.3\varepsilon>\varepsilon$};
\end{tikzpicture}
- A normalized kernel loses constant reproduction when truncated at a boundary computes the half-mass boundary bias from a missing kernel footprint.
- Interior constant and affine reproduction by a symmetric kernel verifies exact full-space reproduction of constants and affine fields.
- Two SPH particles receive equal and opposite pressure forces checks the equal-and-opposite central force identity.
- A disordered fixed-neighbor stencil can fail derivative consistency gives a disordered fixed-ratio particle family with a nonvanishing affine derivative error.
- A conservative positive-pressure SPH system can close a particle gap shows a positive-pressure neighbor force can close a gap while semidiscrete invariants remain conserved.
3 · Logical flowchart
4 · Definitions, theorems and proofs
None yet.
5 · Examples, counterexamples and false statements
A normalized kernel loses constant reproduction when truncated at a boundary
Statement refuted
The claim that every normalized symmetric kernel preserves constants after truncation to a domain, even at a boundary point, is false.
Facts & Assumptions
Given: In one dimension define W(s)=(35/32)(1−s²)³ for |s|≤1 and W(s)=0 otherwise, set W_ε(s)=ε^(−1)W(s/ε), take Ω=[0,∞), and define T_εf(x)=∫_0^∞f(y)W_ε(x−y)dy.
Scaled symmetric compactly supported kernel family: an admissible kernel is nonnegative, even, normalized and compactly supported.
Change of variables for an injective map on a compact Jordan set: if T is injective C¹ with invertible derivative on an open set containing compact Jordan K, then ∫_{T(K)}f(y)dy=∫_Kf(T(s))|det DT(s)|ds for Riemann-integrable f.
A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: a continuous real function on a compact interval is Riemann integrable.
Counterexample
The displayed W is nonnegative, even, supported on [−1,1], and C² at ±1 because its value and first two inside derivatives vanish there. [F3] makes it integrable on [−1,1], and its integral is (35/32)·2(1−1+3/5−1/7)=1, so it satisfies [F1].
At x=0, evenness and support reduce T_ε1(0) to ∫_0^εW_ε(y)dy. Apply [F2] on K=[0,1] with T(s)=εs; since W_ε(εs)=ε^(−1)W(s), this equals ∫0^1W(s)ds. By [F1], W is even and normalized, so the half-interval integral is (1/2)∫{−1}^1W(s)ds=1/2.
The truncated value 1/2 differs from the constant 1, so the claimed boundary reproduction fails.
Interior constant and affine reproduction by a symmetric kernel
Example
Let f(y)=a+b·y be affine on R^d. Full-space smoothing by an admissible symmetric kernel reproduces f exactly: S_εf(x)=f(x) for every ε>0 and x∈R^d.
Facts & Assumptions
Given: An admissible kernel W, ε>0, a∈R, and b∈R^d.
Scaled symmetric compactly supported kernel family: S_εf(x)=∫f(x−z)W_ε(z)dz.
Symmetry cancels the first kernel moment: ∫zW_ε(z)dz=0.
Verification
Expand the affine function at x: f(x−z)=a+b·(x−z)=f(x)−b·z.
Substitute this expansion into [F1]. The constant term integrates to f(x) by [F2], and the linear term vanishes by [F3]. Thus S_εf(x)=f(x).
Two SPH particles receive equal and opposite pressure forces
Example
For N=2 in Fixed-width regularized conservative SPH particle system, write F_12 for the contribution of particle 2 to m_1v̇_1 and F_21 for the contribution of particle 1 to m_2v̇_2. These pair contributions are equal and opposite and act along the particle separation line.
Facts & Assumptions
Given: Two positions q_1,q_2 in R^d with positive masses and positive SPH densities.
Fixed-width regularized conservative SPH particle system: F_12=−m_1m_2(A_1+A_2)∇W_ε(q_1−q_2).
SPH pressure forces are equal-opposite and central: F_21=−F_12 and F_12 is parallel to q_1−q_2.
Verification
The two-particle acceleration equations contain precisely F_12 in the first momentum equation and F_21 in the second, by [F1].
Applying [F2] gives F_12+F_21=0 and makes each force a scalar multiple of q_1−q_2; hence the pair exerts no net force or torque.
A disordered fixed-neighbor stencil can fail derivative consistency
Statement refuted
Keeping the same finite neighbor pattern in units of ε while refining ε alone guarantees consistency of a meshfree particle derivative. The following explicit periodic family refutes this claim for an affine field.
Facts & Assumptions
Given: In one dimension use W(s)=(35/32)(1−s²)³ for |s|≤1 and zero otherwise, W_ε(s)=ε^(−1)W(s/ε), and the periodic node set P_ε={ε(4k+r): k∈Z, r∈{0,3/5,2,7/2}} with its nearest-node intervals, bounded by midpoints of consecutive nodes (the Voronoi cells), and cell volumes V_j. The rule is used on Ω=R with periodic node pattern.
Scaled symmetric compactly supported kernel family: W_ε is supported on [−ε,ε] and the particle kernel derivative is ∂_xW_ε(x−y).
Meshfree cell quadrature and fill radius: the particle sum is Q_hg=Σ_jV_jg(x_j), and an integration-compatible rule satisfies ∫Ωg=Σ{j:C_j∩supp(g)≠∅}∫_{C_j}g for each compactly supported continuous g.
Second-order bias for the kernel particle gradient: if f∈C³ has bounded third derivative, then the continuous kernel-gradient target D_εf differs from f′ by at most (m₃/6)ε²||f‴||∞.
A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: a continuous real function on a compact Jordan-measurable set is Riemann integrable there.
For : is integrable on if and only if it is integrable on and on , and then ; with the oriented form for arbitrary : the Riemann integral over a closed interval is the sum of its integrals over any finite partition into adjacent closed subintervals.
Counterexample
The adjacent gaps in one period are 3ε/5, 7ε/5, 3ε/2, and ε/2, so the fill radius is h=3ε/4. The four Voronoi-cell lengths per period are 11ε/20, ε, 29ε/20, and ε; hence every volume is at most 2h. If a cell meets B(x,r), its node lies in B(x,r+h). Nodes are separated by at least ε/2, so at most 1+4(r+h)/ε such nodes can lie in that interval. Each cell volume is at most 29ε/20, and r+h≥3ε/4; therefore the total active volume is at most (29/20)ε+(29/5)(r+h)≤(116/15)(r+h)≤8(r+h). The local-volume bound holds with C_L=8. In particular, the cells at −ε/2 and 3ε/5 are [−5ε/4,−ε/4] and [3ε/10,13ε/10], each of volume ε.
The nearest-node intervals tile R, repeat with period 4ε, and have gaps between their endpoints only at shared cell boundaries. Every compact support meets finitely many intervals. For any compactly supported continuous g, choose a finite consecutive block of cells containing supp(g); [F4] makes g integrable on its compact union, and repeated [F5] splits that integral into the cell integrals. Cells outside the block have zero integral because they do not meet supp(g). Thus the integration-compatibility identity in [F2] holds for every compactly supported continuous g.
On |s|<1, W′(s)=−(105/16)s(1−s²)². For f(y)=y at x=0, set g(y)=y,\partial_xW_\varepsilon(x-y)|_{x=0}. By [F1], g vanishes outside [−ε,ε], and [F2] defines its particle sum as Q_hg=Σ_jV_jg(x_j). The node at 0 contributes zero; only y=−ε/2 and y=3ε/5 contribute, with values −W′(1/2)/2=945/1024 and −(3/5)W′(3/5)=3024/3125. Thus Q_hg=945/1024+3024/3125, independent of ε.
Here f′(0)=1 and f‴=0, so [F3] gives D_εf(0)=1 exactly. Since each summand in the value from step 3.1 exceeds 1/2, that value is greater than 1 for every ε. The derivative error therefore does not tend to zero although h/ε=3/4 remains fixed.
A conservative positive-pressure SPH system can close a particle gap
Statement refuted
In a conservative SPH system with positive pressure, every adjacent particle gap has nonnegative initial relative acceleration. This is false: a third-particle pressure force can close a neighboring gap even while the semidiscrete invariants remain exact.
Facts & Assumptions
Given: Let d=1, N=3, m_i=1, ε>0, q_1=0, q_2=9ε/10, q_3=13ε/10, and v_i=0 initially. Use W(s)=(35/32)(1−s²)³ for |s|≤1 and zero otherwise, the smooth ideal-gas pressure map P(ρ,e)=(γ−1)ρe with γ>1, and e_i=p_0ρ_i(q)/(γ−1) initially, where p_0>0.
Fixed-width regularized conservative SPH particle system: A_i=P(ρ_i,e_i)/ρ_i² and the particle acceleration is the symmetric pressure-gradient sum.
SPH pressure forces are equal-opposite and central: the paired pressure forces are equal-opposite and central.
Linear and angular momentum are conserved by semidiscrete SPH: total linear and angular momenta are constant along solutions.
The matched internal-energy equation conserves total semidiscrete energy: H=Σ_i(|v_i|²/2+e_i) is constant along solutions for unit masses.
Local well-posedness of the regularized finite-particle SPH system: a unique local solution exists from every positive-density state.
Counterexample
The polynomial kernel is nonnegative and even. At s=±1 its value and first two inside derivatives vanish, so extension by zero is C²; its integral is (35/32)·2(1−1+3/5−1/7)=1. Each density is positive because its self term W_ε(0)>0. The chosen internal energies give p_i=(γ−1)ρ_ie_i=p_0ρ_i²>0 and A_i=p_0 at t=0. By [F1] and [F5], a unique local solution exists; pressure positivity persists for a short interval by continuity.
For 0<s<1, |W′(s)|=(105/16)s(1−s²)². Direct substitution gives |W′(2/5)|/|W′(9/10)|=28224/3249>2. Also |q_3−q_1|=13ε/10>ε, so the 1–3 kernel-gradient term vanishes.
Let r(t)=q_2(t)−q_1(t). From [F1] and step 1.2, the accelerations at t=0 are q̈_1=−2p_0ε^(−2)|W′(9/10)| and q̈_2=2p_0ε^(−2)(|W′(9/10)|−|W′(2/5)|). Therefore r̈(0)=2p_0ε^(−2)(2|W′(9/10)|−|W′(2/5)|)<0.
Since ṙ(0)=0 and r̈ is continuous, ṙ(t)<0 and r(t)<r(0) for all sufficiently small t>0. Thus the adjacent gap closes initially under strictly positive pressure.
The exact pair antisymmetry and matched internal-energy identity apply by [F2] and [F4], so total momentum and H are conserved throughout this same local motion. The example therefore refutes the claimed gap-stability implication while retaining those invariants.