= 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. Also
$$
E(u)\ge\tfrac12\|\nabla u\|_2^2-K|\Omega|,
$$
which 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 give
$$
E(tu+(1-t)v)\le tE(u)+(1-t)E(v)-\tfrac12t(1-t)\|\nabla(u-v)\|_2^2.
$$
The last norm is nonzero: a zero weak gradient makes $u-v$ constant, and its zero <trace> forces that constant to vanish. Thus \b[$E$ is strictly convex].
<Weak lower semicontinuity> means that $u_j\rightharpoonup u$ in $H_0^1$ implies $E(u)\le\liminf_jE(u_j)$. 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 \b[$\boxed{E\text{ has exactly one minimizer in }H_0^1(\Omega)}$].
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