[Paper Review] Laminates supported on cubes

Gabriella Svantnerné Sebestyén, László Székelyhidi|arXiv (Cornell University)|Apr 24, 2015

Structural Analysis and Optimization13 references5 citations

TL;DR

This paper investigates the relationship between rank-one convexity and quasiconvexity in the space of 2×2 matrices, demonstrating that all homogeneous gradient Young measures arising from periodic deformations with three rank-one directions in the 2D case are laminates. The key result is that for any choice of vectors in R², the associated probability measure on the vertices of a rank-one cube is always a laminate, supporting the conjecture that rank-one convexity implies quasiconvexity in the 2×2 setting.

ABSTRACT

In this paper we study the relationship between rank-one convexity and quasiconvexity in the space of 2x2 matrices. We show that a certain procedure for constructing homogeneous gradient Young measures from periodic deformations, that arises from V.~\v Sverák's celebrated counterexample in higher dimensions, always yields laminates in the 2x2 case.

Motivation & Objective

• To determine whether rank-one convexity implies quasiconvexity in the 2×2 matrix setting, a long-standing open problem in the calculus of variations.
• To analyze the structure of homogeneous gradient Young measures generated by periodic deformations with multiple rank-one directions.
• To investigate whether the measure arising from three-phase periodic deformations on a 2D cube is a laminate, which would imply quasiconvexity for all rank-one convex functions.

Proposed method

• Constructs periodic test functions with three distinct rank-one directions in R², using sawtooth functions with phase shifts to generate piecewise-constant gradients.
• Derives the resulting probability measure ν on the 8 vertices of a 3D rank-one cube (hypercube) in R²×², computing the weights ν_ε via volume fractions in the unit torus.
• Analyzes the barycenter and support of ν, showing it is a probability measure with zero mean.
• Uses geometric decomposition techniques to split the measure along rank-one segments, constructing new laminates by convex combinations.
• Applies symmetry and harmonic-arithmetic mean inequalities to prove that the resulting laminate ν satisfies ν(−X₀) ≥ 3ν(−Xₖ) for all k=1,2,3.
• Considers two cases based on the determinant of the matrix X₂, using different splitting strategies to maintain symmetry and barycenter zero.

Experimental results

Research questions

• RQ1 Does every rank-one convex function on 2×2 matrices satisfy the quasiconvexity inequality when tested with three-phase periodic deformations?
• RQ2 Are all homogeneous gradient Young measures arising from such deformations in the 2×2 case necessarily laminates?
• RQ3 Can the measure supported on the vertices of a 3D rank-one cube in R²×² be decomposed into a laminate through rank-one connections, regardless of the choice of vectors?
• RQ4 What is the minimal ratio ν(−X₀)/ν(−Xₖ) for symmetric laminates constructed from three-directional deformations?
• RQ5 Does the structure of the measure depend on the determinant sign of the matrix X₂, and if so, how does this affect the decomposition?

Key findings

• All homogeneous gradient Young measures generated by three-directional periodic deformations in the 2×2 case are laminates, regardless of the choice of vectors in R².
• The symmetric laminate constructed via convex combination satisfies ν(−X₀) ≥ 3ν(−Xₖ) for all k=1,2,3, with equality when a=b=c.
• In the case where det X₂ > 0, the minimal ratio ν′(−X₀)/ν′(−Xₖ) exceeds 4, confirming a stronger lower bound than in the symmetric case.
• The measure ν is symmetric and supported on 6 points: X₀, X₁, X₂, X₃, −X₀, −X₁, −X₂, −X₃, depending on the decomposition.
• The construction is robust across different configurations: when det X₂ ≤ 0, the laminate is built using three symmetric components; when det X₂ > 0, a modified splitting along [X₁, −X₂] is used.
• The harmonic-arithmetic mean inequality is used to prove that the minimal value of ν(−X₀)/ν(−Xₖ) is 3, achieved when a=b=c.