A one-dimensional obstacle problem and its contact set

By [F7] the principal form is bounded and coercive. Put $s:=\sqrt{1-2\epsilon }$, so $0<s<1$ and $t=1-s\in (0,1)$; from $s^2=1-2\epsilon$ one gets $\epsilon =\frac{1}{2}(1-s^2)=\frac{1}{2}(1-s)(1+s)=\frac{1}{2}t(2-t)=t-\frac{1}{2}{t}^2$, hence $\psi (\pm t)=\epsilon -\frac{1}{2}{t}^2=t-t^2=t(1-t)$ and ${\psi }^{\mathrm{\prime }}(\pm t)=\mp t$.

Define $u(x)=\epsilon -\frac{1}{2}{x}^2$ for $\mathrm{\mid }x\mathrm{\mid }\le t$ and $u(x)=t(1-\mathrm{\mid }x\mathrm{\mid })$ for $t\le \mathrm{\mid }x\mathrm{\mid }\le 1$. At $x=\pm t$ the two formulas agree by step 1.1, and the one-sided derivatives agree as well because the inner derivative is ${\psi }^{\mathrm{\prime }}(x)=-x$ with ${\psi }^{\mathrm{\prime }}(\pm t)=\mp t$ and the outer derivative is $\mp t$; hence $u$ is $C^1$ on $[-1,1]$ by [F5], with $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \max \{t,1\}=1$ and $u(\pm 1)=0$. Therefore $u\in H^1(-1,1)$ with weak derivative $u^{\mathrm{\prime }}$ and $Tu=(0,0)$, so $u\in \ker T=H_0^1(-1,1)$ by [F2]. Finally $u-\psi =0$ on $[-t,t]$, while for $t\le \mathrm{\mid }x\mathrm{\mid }\le 1$ one has $u(x)-\psi (x)=t(1-\mathrm{\mid }x\mathrm{\mid })-\epsilon +\frac{1}{2}{x}^2=\frac{1}{2}(\mathrm{\mid }x\mathrm{\mid }-t{)}^2$ by step 1.1; hence $u\ge \psi$ on $(-1,1)$ with equality exactly on $[-t,t]$, so $u\in K$ and $\{u=\psi \}=[-t,t]$.

The derivative $u^{\mathrm{\prime }}$ equals $t$ on $(-1,-t)$, $-x$ on $(-t,t)$ and $-t$ on $(t,1)$; it is continuous and piecewise affine, hence Lipschitz and absolutely continuous on $[-1,1]$, with $u^{\mathrm{\prime }\mathrm{\prime }}=0$ a.e. on $(-1,-t)\cup (t,1)$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-1$ a.e. on $(-t,t)$. Let $v\in K$ and let $q:=v-u$ be represented by its absolutely continuous representative vanishing at $\pm 1$, which exists by step 2.1 and [F2]. Applying integration by parts [F4] to $F=q$ and $G=u^{\mathrm{\prime }}$ gives ${\int }_{-1}^1{q}^{\mathrm{\prime }}{u}^{\mathrm{\prime }}+{\int }_{-1}^{1}q{u}^{\mathrm{\prime }\mathrm{\prime }}=q(1)u^{\mathrm{\prime }}(1)-q(-1)u^{\mathrm{\prime }}(-1)=0$, that is ${\int }_{-1}^1{u}^{\mathrm{\prime }}{q}^{\mathrm{\prime }}={\int }_{-t}^{t}qdx$.

Since $v\in K$ one has $v\ge \psi$ a.e. on $(-1,1)$ [F1], and $u=\psi$ on $[-t,t]$ by step 2.1, so the representative $q=v-u$ of step 3.1 satisfies $q\ge 0$ a.e. on $(-t,t)$ and ${\int }_{-t}^{t}qdx={\int }_{-t}^t(v-\psi )dx\ge 0$.

For every $v\in K$, [F1] expands $J(v)-J(u)=a(u,q)+\frac{1}{2}a(q,q)={\int }_{-1}^1{u}^{\mathrm{\prime }}{q}^{\mathrm{\prime }}+\frac{1}{2}{\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2$ with $q=v-u$; by steps 3.1 and 4.1 this equals $\frac{1}{2}{\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\int }_{-t}^t(v-\psi )dx\ge 0$. Hence $u$ minimises $J$ on $K$.

If $v\in K$ satisfies $J(v)=J(u)$, then both nonnegative terms in step 5.1 vanish, so ${\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2=0$ and $q^{\mathrm{\prime }}=0$ a.e.; the absolutely continuous representative of $q$ is then constant by [F6], and its endpoint values $Tq=(0,0)$ (it lies in $H_0^1(-1,1)$) force that constant to be $0$. Hence $q=0$ a.e. and $v=u$: the minimiser is unique.

By [F1] the obstacle problem has exactly one minimiser on $K$ and it is the unique solution of the variational inequality; steps 4.1 and 5.1 identify this minimiser with the explicit $u$, so $u$ is the obstacle solution. Step 2.1 gives the contact set $\{u=\psi \}=[-t,t]$ and the noncontact set $\{t<\mathrm{\mid }x\mathrm{\mid }<1\}$, and step 1.1 gives the matching slopes $u^{\mathrm{\prime }}(\pm t)=\mp t={\psi }^{\mathrm{\prime }}(\pm t)$ at the free boundary. The Axiom of Choice enters through the obstacle setting and the trace lemma [F1, F2], and it supplies the Countable and Dependent Choice consumed by the integration by parts [F3, F4]; no further choice principle is used.