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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.

One-dimensional SIPDG jump and penalty calculation

Example

This is a self-contained one-dimensional interval specialization of the jump convention in Jump and average conventions for symmetric interior-penalty DG. Let K−=(0,1/2) and K+=(1/2,1), take unit coefficient, and set hF=1/2 at the interior point face. Define v(x)=x on K− and v(x)=x−1 on K+. Then v(0)=v(1)=0, while the interior jump at x=1/2 is 1 and the average derivative is 1. The face contributions are −{v′}[v]−{v′}[v]+γhF−1[v]2. With penalty parameter γ=1, the bilinear form has value ah(v,v)=1.

Facts & Assumptions

Given: The two-cell interval partition, unit coefficient, counting measure on all point faces, hF=1/2 at each face (the common interval-cell length), the piecewise affine function v, and γ=1. For functions v,w on these cells, define ah(v,w):=∑K∈{K−,K+}∫Kv′w′ dx−∑F∈{0,1/2,1}(qF(v)[w]F+qF(w)[v]F)+γ∑F∈{0,1/2,1}hF−1[v]F[w]F, where q1/2(v)={v′}1/2, q0(v)=−v′(0+), and q1(v)=v′(1−); the endpoint jumps are the one-sided value traces and all face integrals use counting measure.

[F1]

On an interior face the scalar jump is the minus trace minus the plus trace, and the average is their arithmetic mean. (Jump and average conventions for symmetric interior-penalty DG)

Verification

1.1givenF1algebra

The derivative is 1 on both cells, so the broken volume term is ∫01/21 dx+∫1/211 dx=1. At the interior point x=1/2, [F1] gives [v]=1/2−(−1/2)=1 and {v′}=1, while both boundary traces vanish.

2.1givenstep 1.1algebra∎

The interior consistency terms contribute −2{v′}[v]=−2, and the penalty contributes γhF−1[v]2=1⋅2⋅1=2. Adding these to the volume term from step 1.1 gives ah(v,v)=1−2+2=1.

Depends on

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