Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 / 1

WriteAs a differential operator of order two, its principal symbol isIf $\nu =({\nu }_t,{\nu }_x,{\nu }_y)$ is a unit normal to the real-analytic hypersurface $\Sigma$, the non-characteristic hypersurface condition is thereforeUnder this condition the equation can be solved for the second derivative normal to $\Sigma$. The Cauchy-Kovalevskaya theorem then gives a unique real-analytic solution in a neighbourhood of $\Sigma$ for the prescribed analytic Cauchy data $u{|}_{\Sigma }$ and ${\partial }_{\nu }u{|}_{\Sigma }$.

The Lax-Milgram theorem says that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form satisfyingfor some $\alpha >0$, and $F\in H^{\ast }$, then there is a unique $u\in H$ such that $B(u,v)=F(v)$ for every $v\in H$.

For this problem take $H=H_0^1(U)$ andThe Cauchy-Schwarz inequality and the continuous embedding $H_0^1(U)\hookrightarrow L^2(U)$ make both maps bounded. For smooth zero-boundary functions, integration by parts givesdensity extends this identity to $H_0^1(U)$. Consequentlyso $B$ is coercive. Lax–Milgram supplies the unique weak solution.

The Sobolev fundamental theorem of calculus on lines gives, for almost every $x$,Apply the Minkowski integral inequality and translation invariance of Lebesgue measure:The restriction on $h$ ensures that every translated copy used above lies in $U$.

Choose $h_j\to 0$. The assumed bound and the weak subsequence of a bounded Hilbert-space sequence result give a subsequence for whichFor every test function $\phi \in C_c^{\infty }(V)$, a change of variables yields the difference-quotient integration-by-parts identityThe right side converges to $-{\int }_{V}u{D}_i\phi$, while the left side converges to ${\int }_V{g}_i\phi$. Thus $g_i$ is the $i$th weak derivative of $u$. This holds for every $i$, so

The Rellich-Kondrachov compactness theorem states in particular that for a bounded open set $U\subset {\mathbb{R}}^n$ with smooth boundary, the embeddingis compact. More generally it is compact into $L^q(U)$ for $1\le q<2n/(n-2)$ when $n\ge 3$, for every finite $q$ when $n=2$, and into $C^0(\overline{U})$ in dimension one.

Boundedness is essential. Choose a nonzero $\phi \in C_c^{\infty }({\mathbb{R}}^n)$ and set $u_k(x)=\phi (x-k{e}_1)$ with pairwise disjoint supports. Their $H^1({\mathbb{R}}^n)$ norms are equal, whilefor $j\ne k$. Thus no subsequence converges in $L^2$.