= Solution
Testing the <weak formulation> with the <constant function> $1$ proves the necessary compatibility condition
$$
\int_Uf=0.
$$
Assume first that $U$ is connected and this condition holds. On the <mean-zero Sobolev space>
$$
H^1_\dagger(U)=\left\{v\in H^1(U):\int_Uv=0\right\},
$$
use the norm $\|v\|_\dagger=\|Dv\|_{L^2(U)}$. To prove the needed <Poincare-Wirtinger inequality>, suppose it failed. There would be $v_k\in H^1_\dagger(U)$ with $\|v_k\|_2=1$ and $\|Dv_k\|_2\to0$. The <Rellich-Kondrachov compactness theorem> gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its <weak derivative> vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\|v\|_2=1$. Hence $\|Dv\|_2$ is an equivalent <Hilbert space> norm on $H^1_\dagger(U)$.
Define
$$
B(u,v)=\int_Ua^{ij}D_juD_iv,
\qquad
\ell(v)=\int_Ufv.
$$
Boundedness of $a^{ij}$ makes $B$ a <bounded bilinear form>, while <uniformly elliptic operator>[uniform ellipticity] gives
$$
B(v,v)\geq\theta\|Dv\|_2^2,
$$
so it is a <coercive bilinear form>. The <Cauchy-Schwarz inequality> and the Poincare-Wirtinger inequality make $\ell$ a bounded <linear functional>. The <Lax-Milgram theorem> supplies a unique $u\in H^1_\dagger(U)$ satisfying $B(u,v)=\ell(v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and $\int_Uf=0$. Thus $u$ solves the original problem.
If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is $\int_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.
Solved by gpt-5.6-sol high.
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