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Nodal feasibility can miss a nonconvex obstacle between vertices

Statement

Let $Ω = (0, 1)^{2}$ , $Γ_{D} = {0} \times [1 / 4, 3 / 4]$ , $Γ_{C} = {1} \times [1 / 4, 3 / 4]$ , and $n = (1, 0)$ . Use the grid $x \in {0, 1 / 2, 1}$ , $y \in {0, 1 / 4, 3 / 4, 1}$ and split each rectangular cell by a diagonal. This is a matching triangular mesh on which both segments are single boundary edges. Parameterize $Γ_{C}$ by normalized arclength $s = 2 (y - 1 / 4) \in [0, 1]$ , set $ψ (s) = - 1 + 4 s (1 - s)$ , and let $v_{h} (x, y) = (- x, 0)$ . The obstacle is continuous, nonpositive and concave, and $v_{h}$ satisfies both endpoint nodal inequalities, but $v_{h} \notin K$ because its contact gap is negative at the midpoint.

Facts & Assumptions

Given: The nodal scheme Conforming nodal P1 Signorini contact discretization except that its convexity hypothesis on $ψ$ is removed for this counterexample.

[F1]

The nodal constraints require $v_{h} (a) \cdot n \geq I_{h} ψ (a)$ at each contact vertex; the continuum feasible set requires $T_{C} v_{h} \cdot n \geq ψ$ almost everywhere on the whole contact segment. (Conforming nodal P1 Signorini contact discretization, The vector Signorini contact set and variational inequality)

[F2]

A real function is concave on an interval when its negative is convex there, with convexity defined by the chord inequality. (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval)

Proof

1.1F1F2givenalgebra

The polynomial $ψ (s) = - 1 + 4 s (1 - s)$ has endpoint values $- 1$ and midpoint value $0$ , and $ψ (s) = - (2 s - 1)^{2}$ shows its maximum is $0$ . For $s_{0}, s_{1} \in [0, 1]$ and $t \in [0, 1]$ , direct expansion gives $ψ ((1 - t) s_{0} + t s_{1}) - [(1 - t) ψ (s_{0}) + t ψ (s_{1})] = 4 t (1 - t) (s_{1} - s_{0})^{2} \geq 0$ , with strict inequality when $0 < t < 1$ and $s_{0} \neq s_{1}$ . Thus [F2] says $ψ$ is strictly concave, not convex, and the convex-obstacle premise is absent.

2.1F1givenalgebra∎

The vector field $v_{h} (x, y) = (- x, 0)$ is affine on each triangle and is zero on $Γ_{D}$ , hence belongs to $V_{h}$ . On the two contact vertices its normal value is $- 1$ , equal to the endpoint obstacle values, so both nodal inequalities hold. Along all of $Γ_{C}$ , $v_{h} \cdot n = - 1$ , while at $s = 1 / 2$ the obstacle equals $0$ ; continuity gives an interval around that midpoint on which $- 1 < ψ (s)$ . Therefore the continuum contact inequality fails on a set of positive arclength and $v_{h} \notin K$ .

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Sources

Richard S. Falk, Error Estimates for the Approximation of a Class of Variational Inequalities, complete 10-page paper (standard reference, not scraped)