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The Signorini contact set is nonempty, closed, and convex

Statement

Under The vector Signorini contact set and variational inequality, K is nonempty, closed, and convex in the Hilbert space V.

Facts & Assumptions

Given: The polygon and partial-trace Hilbert space V, a continuous obstacle ψ≤0, a unit normal n, and K={v∈V:TCv⋅n≥ψ a.e. on ΓC}.

[A1]

The Axiom of Choice is inherited through the trace map and the Hilbert-space interface for V. (The Axiom of Choice, The partial-trace displacement space is a Hilbert space)

[F1]

The contact trace TC:V→L2(ΓC;R2) is bounded and linear. (The partial-trace displacement space is a Hilbert space)

[F2]

The contact set is the preimage of the nonnegative cone under v↦TCv⋅n−ψ. (The vector Signorini contact set and variational inequality)

[F3]

The contact segment has finite length and its continuous obstacle belongs to L2(ΓC). (Partial-trace displacement space on a polygon, The vector Signorini contact set and variational inequality)

Proof

technique · direct

Given: V, TC, n, and ψ≤0 as above.

1.1A1F1F2givenalgebra

The zero field belongs to V, and TC0⋅n=0≥ψ almost everywhere, so 0∈K. If v,w∈K and 0≤t≤1, linearity of TC gives TC(tv+(1−t)w)⋅n−ψ=t(TCv⋅n−ψ)+(1−t)(TCw⋅n−ψ)≥0 almost everywhere, so tv+(1−t)w∈K.

2.1A1F1F2F3algebrastep 1.1∎

If gm∈L+2(ΓC) and gm→g in L2, then g−:=max⁡{−g,0}≤∣g−gm∣ almost everywhere for each m, hence ∥g−∥2≤∥g−gm∥2→0 and g≥0 almost everywhere. Thus the cone is closed; it is convex by pointwise convexity. By [F1] and [F3], the affine map v↦TCv⋅n−ψ is continuous into L2(ΓC), so its preimage of that closed convex cone is closed in V. This proves the claim.

Depends on

Used by

Cited to discharge well-definedness by The vector Signorini contact set and variational inequality.

Dependency tree · two levels

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Sources