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Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. 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 $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso 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 $u\in H_{†}^1(U)$ 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 ${\int }_{U}f=0$. 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.

The Rellich-Kondrachov compactness theorem says that if $U\subset {\mathbb{R}}^n$ is a bounded Lipschitz domain, thenfor $1\le q<p^{\ast }=np/(n-p)$ when $p<n$. When $p=n$, the embedding is compact into every finite $L^q$, and when $p>n$ it is compact into $C^0(\overline{U})$, hence into every $L^q$.

The boundedness of the domain is essential. Choose a nonzero $\varphi \in C_c^{\infty }(0,1)$ and setTranslation invariance gives $\Vert u_j{\Vert }_{W^{1,1}({\mathbb{R}}_{+})}=\Vert \varphi {\Vert }_{W^{1,1}}$, so after a fixed rescaling these functions lie in the unit ball. Their supports are pairwise disjoint andNo subsequence is Cauchy in $L^1$, so the unit ball is not compact. This is the standard failure of Rellich compactness on an unbounded domain.

Solved by gpt-5.6-sol high.

Let $(u_n)$ be bounded in $H_r^1({\mathbb{R}}^2)$. After passing to a subsequence it converges weakly in $H^1$, while the Rellich-Kondrachov compactness theorem gives strong $L^4$ convergence on every bounded ball. The Radial Sobolev inequality gives uniformlyThe same estimate applies to the weak limit. Choosing $R$ large and then using local compactness proves strong convergence in $L^4({\mathbb{R}}^2)$. Henceis compact.

Solved by gpt-5.6-sol high.

The second estimate in part (a) bounds $({\omega }_m)$ in $H_{\mathrm{per}}^2$. The periodic Poisson equation $-\Delta {\Psi }_m={\omega }_m$ and the supplied curl identity then bound $({\Psi }_m)$ in $H_{\mathrm{per}}^4$ and $(u_m)$ in $H_{\mathrm{per}}^3$. After passing to a subsequence,The Rellich-Kondrachov compactness theorem also gives strong convergence in the corresponding spaces with one fewer derivative. In particular, $u_m\to u$ in $L^{\infty }$ and $\nabla {\omega }_m\to \nabla \omega$ in $L^2$, soPassing to the limit in the Galerkin equations givesThese identities have the claimed Sobolev regularity, and the first holds in $L_{\mathrm{per}}^2$. Finally, the weak lower semicontinuity of the Hilbert norm preserves the estimates

Solved by gpt-5.6-sol high.