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The convex-obstacle nodal set is conforming

Statement

Under Conforming nodal P1 Signorini contact discretization, Kh is a nonempty closed convex subset of the finite-dimensional space Vh and Kh⊆K. If v∈K∩H2(Ω;R2)∩C(Ω‾;R2), its global P1 nodal interpolant Ihv is well defined, belongs to Vh, and lies in Kh.

Facts & Assumptions

Given: A boundary-conforming matching triangular mesh, the continuous P1 space Vh, a continuous convex obstacle ψ≤0 on the straight segment ΓC, and the nodal set Kh of Conforming nodal P1 Signorini contact discretization.

[F1]

The contact set is K={v∈V:TCv⋅n≥ψ a.e. on ΓC} and contains zero when ψ≤0. (The vector Signorini contact set and variational inequality, The Signorini contact set is nonempty, closed, and convex)

[F2]

A matching mesh has nondegenerate triangles with shared facets equal to whole facets, and conforming P1 traces are affine on each facet. (Matching affine simplicial meshes and shape regularity, Broken and conforming polynomial spaces on a matching mesh)

[F3]

Ihψ is the affine interpolant of the endpoint values on each contact edge, and Kh imposes vh(a)⋅n≥Ihψ(a) at every contact vertex. (Conforming nodal P1 Signorini contact discretization)

[F4]

If an H1 function has a representative extending continuously to each closed collar, TD and TC agree with its boundary restriction to the corresponding segments. In particular, this holds for H1∩C(Ω‾) functions. (Partial-trace displacement space on a polygon, The Lp trace operator on a bounded C1 domain)

[F5]

Convexity on an interval means ψ((1−t)a+tb)≤(1−t)ψ(a)+tψ(b) for every a,b in the interval and t∈[0,1]. (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval)

Proof

technique · direct

Given: The hypotheses above, with Vh the matching continuous piecewise-affine vector space.

1.1F2F3givenalgebra

The set Vh is finite dimensional. By [F3], Kh is the intersection of finitely many closed affine halfspaces in Vh, so it is closed and convex. The zero field is in Vh, and 0≥Ihψ(a)=ψ(a) at every contact vertex since ψ≤0; hence 0∈Kh and Kh is nonempty.

1.2F1F2F3F4F5givenalgebra

Fix a contact edge e=[a,b]. For vh∈Kh, the function qh:=vh⋅n−Ihψ is affine on e and is nonnegative at a,b by [F3]. Thus qh((1−t)a+tb)=(1−t)qh(a)+tqh(b)≥0 throughout e. Convexity [F5] gives Ihψ((1−t)a+tb)≥ψ((1−t)a+tb), so vh⋅n≥ψ on every contact edge. Since vh∈C(Ω‾), [F4] identifies its contact trace with this restriction. Therefore vh∈K and Kh⊆K.

2.1F1F2F3F4givenconstructalgebra∎

Let v∈K∩H2∩C(Ω‾). Its continuous gap TCv⋅n−ψ is nonnegative almost everywhere by [F1], hence everywhere: if it were negative at one point, continuity would make it negative on a subsegment of positive length. Define Ihv cell by cell as the affine polynomial matching v at the three vertices. On a shared edge the two affine traces agree at both endpoints, so they agree along the whole edge. For every test function compactly supported in Ω, integration by parts on each triangle cancels the internal-edge boundary terms because these traces agree; the remaining piecewise constant gradients are in L2(Ω). Hence Ihv∈H1(Ω;R2) and is globally continuous. Its boundary values on ΓD are zero: v∈V means TDv=0 almost everywhere, and [F4] identifies this trace with the continuous boundary restriction, which therefore vanishes everywhere on the segment. Since the mesh resolves ΓD, the affine boundary traces of Ihv vanish there, so Ihv∈Vh. At every contact node, Ihv(a)=v(a) and [F4] identifies the contact trace with the continuous restriction; the continuous gap is nonnegative, so the nodal constraints hold and Ihv∈Kh.

Depends on

Used by

Cited to discharge well-definedness by Conforming nodal P1 Signorini contact discretization.

Dependency tree · two levels

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