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The obstacle admissible set can be empty when trace and obstacle are incompatible

Statement refuted

Refuted: that after omitting the trace-compatibility hypothesis $T ψ \leq 0$ from The closed convex obstacle set and the obstacle variational inequality, the set $K_{ψ} = {v \in H_{0}^{1} (Ω; R) : v \geq ψ a.e.}$ is automatically nonempty for every real $ψ \in H^{1} (Ω)$ . The definition itself retains that hypothesis and has the admissible element $ψ^{+}$ ; the assertion refuted here concerns arbitrary obstacle data without boundary compatibility.

Assume the Axiom of Choice (The Axiom of Choice), inherited from the trace suppliers. The witness is the interval $I = (- 1, 1)$ with the incompatible obstacle datum $ψ \equiv 1$ (Endpoint trace commutes with Sobolev truncation on an interval). If $v \in K$ , the unique absolutely continuous representative $v^{*}$ satisfies $v^{*} \geq 1$ at every point of $[- 1, 1]$ : a strict violation at one point would, by continuity, persist on an interval of positive measure, contradicting $v \geq 1$ almost everywhere. But $v \in H_{0}^{1} (I)$ has zero endpoint trace, $T v = (0, 0)$ , so $v^{*} (- 1) = v^{*} (1) = 0$ , a contradiction. Hence $K = \emptyset$ , and boundary compatibility ( $T ψ \leq 0$ in the sense of the definition) is necessary for nonemptiness. [The obstacle $ψ \equiv 1$ satisfies $T ψ = (1, 1) \leq̸ (0, 0)$ .]

Facts & Assumptions

Given: The interval $I = (- 1, 1)$ , the incompatible obstacle datum $ψ \equiv 1$ , and the admissible set $K = {v \in H_{0}^{1} (I) : v \geq 1 a.e.}$ .

[F1]

Endpoint trace commutes with Sobolev truncation on an interval: for $I = (- 1, 1)$ and $p = 2$ , $T$ is well defined and linear with $\ker T = H_{0}^{1} (I)$ ; a class $v \in H_{0}^{1} (I)$ therefore has endpoint trace $T v = (0, 0)$ .

[F2]

One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives: every $v \in H^{1} (I)$ has a unique continuous absolutely continuous representative $v^{*}$ on $[- 1, 1]$ , and the endpoint trace is $T v = (v^{*} (- 1), v^{*} (1))$ .

[F3]

The closed convex obstacle set and the obstacle variational inequality: the definition requires $T ψ \leq 0$ before defining its admissible set. Here the same set formula is used for arbitrary real $H^{1}$ data, without imposing that compatibility condition; $ψ \equiv 1$ is outside the definition's permitted obstacles.

Counterexample

technique · direct

Given: The interval $I = (- 1, 1)$ , the incompatible obstacle datum $ψ \equiv 1$ and the set $K$ above; suppose, towards the case analysis, that $v \in K$ .

1.1givenF2

Since $v \geq 1$ almost everywhere and $v^{*}$ is continuous with $v^{*} = v$ almost everywhere [F2], the representative $v^{*}$ satisfies $v^{*} (x) \geq 1$ for every $x \in [- 1, 1]$ : if $v^{*} (x_{0}) < 1$ at some point, then by continuity $v^{*} < 1$ on the intersection of $I$ with a sufficiently small interval about $x_{0}$ , which has positive measure even when $x_{0}$ is an endpoint, contradicting $v \geq 1$ a.e.

2.1step 1.1F1F2

On the other hand $v \in H_{0}^{1} (I) = \ker T$ by the definition of $K$ and [F1, F3], and the endpoint trace of the class is read from its absolutely continuous representative, so $T v = (v^{*} (- 1), v^{*} (1)) = (0, 0)$ [F2]. This contradicts step 1.1, which gives $v^{*} (- 1) \geq 1$ and $v^{*} (1) \geq 1$ .

3.1step 2.1givenF1F3∎

More generally, if $ψ \in H^{1} (I)$ and $v \in H_{0}^{1} (I)$ obeys $v \geq ψ$ a.e., then $(ψ - v)^{+} = 0$ as a class, so [F1] gives $(T ψ - T v)^{+} = T ((ψ - v)^{+}) = 0$ . Since $T v = 0$ , necessarily $T ψ \leq (0, 0)$ . No $v$ satisfies both requirements of membership in $K$ , so $K = \emptyset$ . Since $T ψ = (1, 1) \leq̸ (0, 0)$ , this is exactly the failure of the boundary compatibility hypothesis of [F3]; hence that hypothesis (equivalently $ψ^{+} \in H_{0}^{1} (I)$ ) is necessary for the admissible set to be nonempty.

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Sources

John Andersson, The Obstacle Problem, KTH lecture notes, 16 December 2015 (complete 52-page notes) (standard reference, not scraped)