 Solution

Choose an upper ellipticity bound $\Lambda <\infty$ for the symmetric coefficient matrix, possible because its entries are bounded. ThenThe Poincaré inequality on a bounded open set for zero-boundary functions states $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$. ConsequentlyThe symmetric positive bilinear form $Q$ is therefore an inner product, and its square-root norm is equivalent to the usual Sobolev norm. Its triangle inequality follows from the Cauchy-Schwarz inequality for that inner product. completeness of $W_0^{1,2}$ in its original norm makes it a Hilbert space in the $Q$ norm as well.

The linear functional $\ell (v)=-\int fv$ is bounded, sinceThe Hilbert-space Riesz representation theorem states that every bounded linear functional is the inner product with a unique element. Applied in the $Q$ inner product it gives a unique $u$ with $Q(u,v)=\ell (v)$ for all $v$, which is precisely the weak-solution identity. Alternatively uniqueness follows directly by testing the difference of two solutions with itself: $Q(u_1-u_2,u_1-u_2)=0$. Thus $\boxed{\text{the zero-boundary problem has exactly one weak solution}}$.