Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
How 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.

Unique solvability of an implicit θ heat step

Statement

For every finite two-dimensional rectangular Dirichlet grid, Δt>0, θ∈[0,1], and interior right-hand side (including the known source and boundary contributions), the θ step of Fully discrete θ heat finite-difference scheme has a unique interior solution. Its implicit matrix is I+θΔtAh, where Ah is the zero-boundary operator.

Facts & Assumptions

Given: A finite rectangular interior grid, Δt>0, θ∈[0,1], and the θ update of Fully discrete θ heat finite-difference scheme.

[F1]

Fully discrete θ heat finite-difference scheme writes the Dirichlet step with unknown terms (I+θΔtAh)Un+1; all source and prescribed boundary rows are known right-hand-side terms.

[F2]

Symmetry and positivity of the rectangular Dirichlet heat operator: ⟨Ahv,v⟩h≥0 for every zero-boundary interior vector v.

Proof

1.1F1given

If θ=0, the implicit matrix in [F1] is the identity, so for every right-hand side there is exactly one solution, namely that right-hand side.

1.2F1F2givenalgebra

Suppose θ>0 and (I+θΔtAh)v=0. Taking the weighted inner product with v gives ∥v∥h2+θΔt⟨Ahv,v⟩h=0. By [F2] the second term is nonnegative, so ∥v∥h2=0 and v=0.

2.1step 1.2algebra∎

The interior vector space is finite dimensional and the matrix is square. Its zero kernel from step 1.2 means its rank equals the dimension by rank-nullity; hence its range is the whole space. Thus every right-hand side has exactly one solution.

Remarks

This is finite-dimensional solvability only; it makes no condition-number or mesh-uniform iterative-solver claim. No choice axiom is used.

Depends on

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources