Rigidity estimate for two incompatible wells

Rigidity and gamma convergence for solid‐solid phase transitions with SO(2) invariance

“…The above construction shows that, if at least one rank-one connection is present, no global rigidity result can be expected without assuming some bound on the surface energy. A different picture arises for the case without rank-one connections (i.e., when rank(Q A − B) = 1 has no solutions for Q ∈ SO(2)); see Chaudhuri and Müller [10]. We consider the case that ∇u is close to K and that the distributional second gradient ∇ 2 u is small.…”

Simple proof of two-well rigidity

“…Building on the methods developed in [5] Chaudhuri and Müller in [3] obtained the corresponding rigidity estimate for the case of two strongly incompatible wells K = SO(n)A 1 ∪ SO(n)A 2 . An important ingredient in the proof of Chaudhuri and Müller is the result of Matos in [6] that under certain conditions on the matrices A 1 and A 2 the exact solutions of the inclusion problem ∇u ∈ K are solutions of a certain strongly elliptic system.…”

A Two Well Liouville Theorem

“…Proof of Step 1 4 . This follows by self adjointness of H and Cauchy Schwartz inequality, let ψ denote the clockwise normal to ψ…”

“…One of the main reasons for working on the grid is that when two lines (say 4 such that they only intersect on "good subsquares" function u on the boundary of the diamond will be L 1 close (with error δ 1 8 say) to a fixed rotation. Since there are so many good subsquares finding these four lines is just a matter of careful counting.…”

“…Using methods of [9] (for the non-linear functional) Conti, Schweizer proved a strong generalisation of Theorem 1.1, their strategy was to use hypotheses (4) and (5) …”