 Solution

For the log-cosh obstacle variational inequality, the admissible set is convex: for any $v\in M$, $u+t(v-u)$ remains above the obstacle and in $H_0^1$ for $0\le t\le 1$. Minimality and the right derivative at zero therefore giveThe one-sided sign comes from admissibility, which permits motion toward every $v\in M$ but does not permit an arbitrary negative variation at a contact point.

Now fix $\varphi \in C_c^{\infty }(D)$. Smoothness on $D$ and continuity of the obstacle give a strictly positive minimum of $u-g$ on the compact support of $\varphi$. Hence $u+t\varphi$ is admissible for both signs of sufficiently small $t$. Applying the inequality to both variations yieldsIntegration by parts gives $-\Delta u+\tanh u=0$ distributionally on $D$, and smoothness makes it pointwise. ThusOn the contact set the variational inequality, rather than unrestricted stationarity, remains the governing statement. Convexity of the energy also makes the inequality sufficient for minimization; strict monotonicity implies at most one minimizer, although existence was already stipulated in this part.