= Solution
Bounded coefficients and Cauchy-Schwarz give $|B[u,v]|\leq C\|u\|_{H^1}\|v\|_{H^1}$. <Uniform ellipticity> gives $B[u,u]\geq\theta\|Du\|_2^2$ because $c\geq0$. <Poincaré inequality> $\|u\|_2\leq C_P\|Du\|_2$ holds on $H_0^1$ of a bounded domain, so
$$
\boxed{B[u,u]\geq\nu\|u\|_{H^1}^2,\qquad\nu=\theta/(1+C_P^2)>0.}
$$
A <weak solution> is $u\in H_0^1$ satisfying $B[u,\phi]=\int f\phi$ for every $\phi\in H_0^1$. The energy is finite by boundedness of $B$ and the forcing functional, and
$$
I(v)\geq\frac\nu2\|v\|_{H^1}^2-\|f\|_2\|v\|_{H^1}\geq\frac\nu4\|v\|_{H^1}^2-C_f.
$$
It is bounded below and a <minimizing sequence> is bounded in the <Hilbert space> $H_0^1$. Its <closed linear span> is <separable>, so part 1 supplies a <weakly convergent> <subsequence> with limit $u$. Symmetry and nonnegativity of $B$ show
$$
B[v_j,v_j]=B[u,u]+2B[u,v_j-u]+B[v_j-u,v_j-u]\geq B[u,u]+o(1).
$$
The forcing term converges by <weak convergence>. Thus $I(u)\leq\liminf I(v_j)$, proving attainment. <First variation> in every zero-trace direction gives the weak equation. If $u,w$ are two solutions, test their difference equation with $u-w$ and use coercivity to get $u=w$. Hence \b[there is exactly one <weak solution>, and it is the unique energy minimizer]. Indeed $I(v)-I(u)=\tfrac12B[v-u,v-u]$ directly verifies the minimizing property.
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