verification: implicit plasticity oracle cases and documentation

Four cases, each against its closed form: uniaxial tension of a J2 cube to a
logarithmic strain of 0.2 (Kirchhoff stress exact; 0.13 % from the explicit
solver's hypoelastic Cauchy stress, the elastic volume change between the two
laws), the Bauschinger effect of linear kinematic hardening on a load path
back to zero strain (exact), Hill's elastic-perfectly plastic thick cylinder
(bore displacement at a plastic radius of 1.5a within 0.009 %, collapse
pressure within 2e-8; the pending case is now active) and a cantilever's
plastic collapse, converged 3.1 % above beam theory's hinge, the bound a
continuum exceeds. docs/fem.md, the literature, ROADMAP (M15), parity and the
oracle page follow.

Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01NmAttHhUZTokgPeEJia8sm

tests/oracle/j2-bauschinger/README.md

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+# Bauschinger effect of linear kinematic hardening
+
+One hex8 cube (E = 200000, ν = 0.3, σ_y = 250, H = 2000, Prager's linear kinematic hardening)
+held by symmetry on three faces, its fourth face pulled to a strain of 0.01 and brought back to
+zero within one nonlinear static step without nonlinear geometry, by the load path
+[[0, 0], [0.5, 1], [1, 0]] in increments of 0.05.
+
+At ε_a the stress is σ_a = σ_y + E_t (ε_a − σ_y/E) = 267.327. The yield surface keeps its size
+and moves with the back stress, so unloading is elastic only until σ_a − 2σ_y and then hardens
+with E_t: back at zero strain σ = σ_a − 2σ_y − E_t (ε_a − 2σ_y/E) = −247.525. Isotropic hardening
+would yield again only at −σ_a and end at −281.835. The equivalent plastic strain accumulates
+(σ_a − σ_y)/H forward and E/(E + H)(ε_a − 2σ_y/E) in reverse, 0.0160891.