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.
Exact constant-density compressible viscous shear
Example
Assume Countable Choice. Let , , , and . On with coordinates , set , , and . Then and give a global smooth solution with no external force of where . This is the first sine heat mode of the compatible channel shear evolution at channel height , periodically extended in . It is one exact solution of the compressible viscous system, not a general nonconstant-density existence result. With SI units, are metres, is seconds, are m/s, is kg/m, is Pa, are Pas, is Pa(m/kg), and is m/s.
Example references
Verification
Given: Countable Choice, the torus, constants, constitutive stress, and compressible equations in the Example.
[F1] For channel height , the sine-mode heat evolution has terms and is smooth for compatible odd periodic data (Global compatible no-slip shear evolution).
[A1] Countable Choice is carried from the compatible shear theorem's Fourier-basis input (The Axiom of Countable Choice ()).
[A2] The conservative mass and momentum equations and Newtonian stress tensor displayed in the Example are the mathematical system being checked; no empirical constitutive assertion is made.
Take on . Its odd extension has period and is smooth. Under Countable Choice [A1], the channel Fourier formula [F1] has all sine coefficients zero except , because . The resulting is smooth and -periodic in , so it defines the stated torus field and satisfies .
Since is independent of and is constant, and . Thus the mass equation in [A2] holds.
We have , , and . The symmetric gradient has only and entries, each , so it is trace-free and . Therefore and the bulk-viscosity term is zero. The momentum left side in [A2] is by step 1.1 and , so the momentum equation holds. This verifies the exact solution for all .