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An affine shear field passes the frictionless contact patch test

Statement

On the unit square with ΓD={0}×[1/4,3/4], ΓC={1}×[1/4,3/4], and n=(1,0), use any matching boundary-conforming affine triangular mesh that resolves both segments as unions of boundary edges, and take obstacle ψ=0. Let u(x,y)=(0,x) and define the load F(v)=a(u,v) on V. Then u is the unique continuous solution and, on every such mesh, the same field is the unique nodal P1 solution. Its contact gap is identically zero.

Facts & Assumptions

Given: The elasticity and Signorini conventions of Small-strain elasticity with a frictionless contact gap and The vector Signorini contact set and variational inequality, with the square and boundary segments above.

[A1]

Countable Choice is used by the imported continuous and discrete Stampacchia theorems. (The Axiom of Countable Choice (ACω))

[A2]

The Axiom of Choice is inherited by the trace and partial-boundary Korn interfaces in the continuous well-posedness theorem. (The Axiom of Choice, Continuous frictionless elasticity contact is well posed)

[F1]

V consists of H1 vector fields with zero Dirichlet trace, and a is the isotropic symmetric-strain elasticity form. (Small-strain elasticity with a frictionless contact gap)

[F2]

The feasible set is K={v∈V:TCv⋅n≥ψ} and the nodal scheme defines finite-dimensional Vh and a nonempty closed convex set Kh⊆K. (The vector Signorini contact set and variational inequality, Conforming nodal P1 Signorini contact discretization, Broken and conforming polynomial spaces on a matching mesh, Every finite-dimensional normed space is Banach, The convex-obstacle nodal set is conforming)

[F3]
[F4]

The continuous VI has a unique solution. (Continuous frictionless elasticity contact is well posed)

[F5]

On a Hilbert space, a nonempty closed convex set and bounded coercive form yield a unique VI solution under Countable Choice. (Stampacchia's variational inequality)

Proof

1.1F1F2givenalgebra

The affine field u lies in H1(Ω;R2) and vanishes on the left side, so u∈V. On the right side u⋅n=0=ψ, hence u∈K and has zero gap. Every affine function restricts to a continuous piecewise-affine function on the given matching mesh, so u∈Vh; its contact nodal inequalities hold with equality, so u∈Kh.

2.1A1A2F2F3F4F5step 1.1algebra∎

The gradient is ∇u=(0010), so div⁡u=0 and ε(u)=(01/21/20). By [F3], F(v)=a(u,v) is bounded. For every v∈K, a(u,v−u)−F(v−u)=0, so u solves the continuous VI; uniqueness follows from [F4]. For every vh∈Kh⊆K, the same identity holds, so u solves the discrete VI. The inherited H1 inner product makes finite-dimensional Vh a Hilbert space by [F2]; the restricted form is bounded and coercive by [F3], and [F5] gives uniqueness.

Depends on

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