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Compatible discrete Maxwell constraints and finite algorithms — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Constraint preservation does not imply bounded leapfrog evolution

Statement refuted

The statement refuted is: for every finite cochain datum of the type defined in the item named “Finite cochain Maxwell phase space,” every positive rational step size in the item named “Finite compatible leapfrog evolution,” and every initial state satisfying D2b0=0 and D0Td1/2=0, repeated application of its one-step map has a bounded orbit. Here “bounded orbit” means that the Euclidean norm of the coordinate vector (bn,dn+1/2) has finite supremum over all integers n≥0.

Facts & Assumptions

Given: A finite real cochain complex, its positive-definite mass maps, and the one-step update in Finite compatible leapfrog evolution.

[F1]

Finite cochain Maxwell phase space: the constraints are D2b=0 and D0Td=0, with D1D0=D2D1=0.

[F2]

Finite compatible leapfrog evolution: one step is b+=b−τD1M1−1d followed by d+=d+τD1TM2b+.

[F3]

Exact discrete constraints and semidiscrete and staggered energy: the real one-step map Tτ preserves both constraints for every τ>0, and on the algorithm's rational input domain it is the exact rational update; when M1=M2=I, its σ is the Euclidean operator norm of D1.

[A1]

Define C0=R4, C1=R6, C2=R4, and C3=R, with the displayed standard bases. Take the one-step size τ=2, initial data b0=x=(1,1,0,0)T and d1/2=0, and matrices

D0=(−1100−1010−10010−1100−10100−11),D1=(1−1010010−101001−10010001−11),D2=(−11−11),M1=I6,M2=I4.

Counterexample

1.1givenF1A1algebra

The matrices in [A1] are the signed boundary-incidence maps of one oriented tetrahedron, with edges ordered 01,02,03,12,13,23, faces ordered 012,013,023,123, and cell orientation 0123. Matrix multiplication gives D1D0=04×4 and D2D1=01×6. Also D2x=−1+1=0 and D0Td1/2=04×1, so the initial state satisfies both constraints.

2.1givenA1F2F3step 1.1algebra

Put c:=(−1,1,−1,1)T, so D2=cT. Direct multiplication gives D1D1T=(31−11131−1−11311−113)=4I4−ccT. For every z∈R4, this gives ∥D1Tz∥22=4∥z∥22−(cTz)2≤4∥z∥22. For real coordinate vectors u,v∈Rm, direct expansion gives ∥u∥22∥v∥22−(uTv)2=∑1≤i<j≤m(uivj−ujvi)2≥0, hence ∣uTv∣≤∥u∥2∥v∥2. Applying this with u=D1Tz and v∈R6 gives ∣zTD1v∣≤2∥z∥2∥v∥2. For w=D1v, applying the same identity with u=z and v=w gives ∣zTw∣≤∥w∥2 for unit z; if w≠0 equality is attained at z=w/∥w∥2, while if w=0 both sides are zero. Thus ∥w∥2=max⁡∥z∥2=1∣zTw∣ and ∥D1v∥2≤2∥v∥2 for every v∈R6. Since cTx=0, e:=x/2 is unit and D1D1Te=4e; consequently f:=D1Te/2 is unit and D1f=2e. Thus the operator norm of D1 is exactly 2, attained at f. Here M1=M2=I, so the σ defined in [F3] is 2. For a state b=ue, d=vf, the update in [F2] with τ=2 gives u+=u−4v and v+=v+4u+=4u−15v, hence the invariant two-dimensional subspace has matrix (1−44−15). The initial state has modal coordinates (u,v)=(2,0).

3.1givenstep 2.1algebra

The matrix in step 2.1 has trace −14, determinant 1, and eigenvalues λ−=−7−43 and λ+=−7+43, found by solving λ2+14λ+1=0. Since 6<43<7, λ−<−1 and −1<λ+<0. Direct substitution gives eigenvectors v−=(1,2+3)T and v+=(1,2−3)T; their determinant is −23≠0, so write the initial modal vector as (2,0)T=av−+bv+. The coefficient a is nonzero, since otherwise (2,0)T would be a multiple of v+, whose second coordinate 2−3 is nonzero. The first modal coordinate after n steps is therefore un=aλ−n+bλ+n. If un were bounded, then bλ+n would be bounded because ∣λ+∣<1, so their difference aλ−n would be bounded, contradicting a≠0 and ∣λ−∣>1. Since ∥bn∥2≥∣un∣, the state orbit is unbounded.

4.1givenF1F2F3A1step 1.1step 3.1algebra∎

By [F3], each update preserves both constraints, and step 1.1 sets them to zero initially. Explicitly, D2b+=D2b−τD2D1M1−1d=D2b and D0Td+=D0Td+τ(D1D0)TM2b+=D0Td, so induction gives zero constraints at every iterate. Thus exact constraint preservation holds at every step although this rational finite-complex algorithm has an unbounded orbit for τσ=4>2. This refutes the stated bounded-orbit claim.

Sources