Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2008 / 4

One-dimensional Sobolev embedding gives a continuous representative for each $v\in H_0^1(0,l)$, with zero endpoint values, so the pointwise obstacle condition is meaningful. The positive part $g_{+}=\max \{g,0\}$ is Lipschitz and vanishes near both endpoints because $g(0),g(l)<0$. It therefore belongs to $H_0^1$ and satisfies $g_{+}\ge g$, proving that $M$ is nonempty.

If $v,w\in M$ and $t\in [0,1]$, then $(1-t)v+tw\ge g$, so $M$ is a convex set. If $v_n\to v$ in $H_0^1$, the estimate $\Vert v_n-v{\Vert }_{\infty }\le \sqrt{l}\Vert v_n^{\prime }-v^{\prime }{\Vert }_2$ gives uniform convergence. Passing to the limit in $v_n(x)\ge g(x)$ shows $v\in M$. Thus the admissible set is nonempty, closed and convex.

For any $v\in M$, the segment $u+t(v-u)$ stays in $M$ for $0\le t\le 1$. Since $u$ is a minimizer, the right derivative of its energy at $t=0$ is nonnegative. The first variation therefore gives the variational inequalityThe factor two comes from the quadratic gradient term. Only a one-sided inequality follows because a variation pointing below the obstacle need not be admissible.

The noncontact set $D=\{x\in (0,l):u(x)>g(x)\}$ is open. For $\phi \in C_0^{\infty }(D)$, the positive gap above the obstacle has a positive minimum on the support of $\phi$. Thus both $u+t\phi$ and $u-t\phi$ are admissible for sufficiently small $t>0$. Applying the variational inequality to both signs yieldsHence the Euler-Lagrange equation on the noncontact set isIt holds classically under the assumed smoothness. On the contact set the obstacle prevents the negative variations used to obtain this equality, so a variational inequality remains the appropriate global description.

Let $u,w\in M$ both solve the variational inequality. Test the inequality for $u$ with $w$ and the one for $w$ with $u$, then add. Writing $d=u-w$ givessince cosine is Lipschitz with constant one. Since $d(0)=0$, the fundamental theorem of calculus and Cauchy-Schwarz inequality give $|d(x)|\le \sqrt{l}\Vert d^{\prime }{\Vert }_2$; integrating gives $\Vert d{\Vert }_2^2\le l^2\Vert d^{\prime }{\Vert }_2^2$. ThereforeFor $\boxed{0<\gamma <2c_0/l^2}$, the coefficient is positive, so $d^{\prime }=0$ and the zero boundary value gives $d=0$. This proves uniqueness with an explicit sufficient smallness condition. More generally any Poincaré inequality constant $C_P$ gives the condition $\gamma C_P^2<2c_0$; optimizing that constant is unnecessary for the requested conclusion. The underlying principle is uniqueness for a strongly monotone variational inequality.