A one-dimensional obstacle with an explicit free boundary

The function $\psi$ is continuous and piecewise affine, hence belongs to $H^1(0,1)$; its endpoint values are $0$ and $-1\mathrm{/}2$, so [F1] applies. The displayed $u$ is continuous at $a_0$, has derivative $0$ from both sides there, and has endpoint values $u(0)=u(1)=0$. By [F2], its endpoint trace is zero and therefore $u\in H_0^1(0,1)$. On $[0,a_0]$, $u=\psi =0$; on $(a_0,1]$, $u\ge 0>\psi$. Thus $u\in K$, and equality $u=\psi$ holds exactly on $[0,a_0]$.

On $(a_0,1)$, direct differentiation gives $u^{\mathrm{\prime }\mathrm{\prime }}(x)=4-6x$; on $(0,a_0)$, $u^{\mathrm{\prime }\mathrm{\prime }}=0$. The derivative $u^{\mathrm{\prime }}$ is continuous at $a_0$, so its distributional derivative is the piecewise function just computed, with no point mass at $a_0$. For $z=v-u\in H_0^1(0,1)$, [F6] supplies a sequence $z_m\in C_c^{\mathrm{\infty }}(0,1)$ converging in $H^1$. Piecewise integration by parts for each $z_m$ has cancelling interface terms because $u^{\mathrm{\prime }}$ is continuous; passing to the limit in the $L^2$ pairings gives ${\int }_0^1{u}^{\mathrm{\prime }}{z}^{\mathrm{\prime }}=-{\int }_0^1{u}^{\mathrm{\prime }\mathrm{\prime }}z$. Since $f=-1$ on the first interval and $f=-u^{\mathrm{\prime }\mathrm{\prime }}$ on the second, this yields $a(u,v-u)-F(v-u)={\int }_0^{a_0}(v-u)dx={\int }_0^{a_0}vdx\ge 0$, where the last inequality uses $v\ge \psi =0$ almost everywhere on $(0,a_0)$. Hence $u$ solves the variational inequality. The same integration by parts against compactly supported tests gives reaction density $-u^{\mathrm{\prime }\mathrm{\prime }}-f=1$ on $(0,a_0)$ and $0$ on $(a_0,1)$; no point mass occurs at $a_0$ because $u^{\mathrm{\prime }}$ is continuous.

The function $f$ is bounded on both subintervals, hence belongs to $L^2(0,1)$. The form is symmetric and bounded since $\mathrm{\mid }a(w,z)\mathrm{\mid }\le \mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\mathrm{\parallel }{z}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$ by [F4]. By [F3], $a(w,w)=\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\ge c_P^{-2}\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2$ for some $c_P>0$. Thus the form is bounded, symmetric and coercive in the sense of [F5]. The load is bounded because $\mathrm{\mid }F(z)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_2\le c_P\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$ by [F3]–[F4]. If $\tilde{u}\in K$ is another solution, test its inequality with $u$ and the one for $u$ with $\tilde{u}$, then add; symmetry gives $a(u-\tilde{u},u-\tilde{u})\le 0$. Coercivity forces $u=\tilde{u}$.