= Solution
Choose an upper ellipticity bound $\Lambda<\infty$ for the symmetric coefficient matrix, possible because its entries are bounded. Then
$$
\lambda\|Du\|_2^2\leq Q(u,u)\leq\Lambda\|Du\|_2^2.
$$
The <Poincaré inequality> on a bounded open set for zero-boundary functions states $\|u\|_2\leq C_P\|Du\|_2$. Consequently
$$
\frac\lambda{1+C_P^2}\|u\|_{W^{1,2}}^2\leq Q(u,u)\leq\Lambda\|u\|_{W^{1,2}}^2.
$$
The 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, since
$$
|\ell(v)|\leq\|f\|_2\|v\|_2\leq C_P\lambda^{-1/2}\|f\|_2 Q(v,v)^{1/2}.
$$
The 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 $\boxed{\text{the zero-boundary problem has exactly one weak solution}}$.
Back to article page
