Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 30A / i / Solution

For $u\in H_0^1(\Omega )$, both $u$ and its weak gradient are square integrable. The assumed bounds give $|F(u,x)|\le K(1+|u{|}^2)$, so the potential term is integrable and $E(u)$ is finite. Alsowhich is bounded below and coercive in $H_0^1$ by the Poincaré inequality.

For distinct $u,v$ and $0<t<1$, convexity of $F$ and expansion of the squared gradient giveThe last norm is nonzero: a zero weak gradient makes $u-v$ constant, and its zero trace forces that constant to vanish. Thus $E$ is strictly convex.

Weak lower semicontinuity means that $u_j\rightharpoonup u$ in $H_0^1$ implies $E(u)\le {\mathrm{liminf}}_{j}E(u_j)$. A minimizing sequence is bounded by coercivity. Reflexivity of this Hilbert space gives a weakly convergent subsequence with limit still in $H_0^1$. The assumed weak lower semicontinuity makes that limit a minimizer, and strict convexity makes it unique. Hence $\boxed{E\text{ has exactly one minimizer in }{H}_0^1(\Omega )}$.