Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 / 3

Let $B$ and $F$ satisfy the hypotheses of the Lax-Milgram theorem. By the Riesz representation theorem, there are a bounded linear operator $T:H\to H$ and $z\in H$ such thatCoercivity and the Cauchy-Schwarz inequality implyso $\Vert Tu{\Vert }_H\ge \alpha \Vert u{\Vert }_H$. Thus $T$ is injective and its range is closed. If $y$ is orthogonal to its range, then $B(x,y)=0$ for every $x$; taking $x=y$ and using coercivity gives $y=0$. The range is therefore dense as well as closed, hence all of $H$. There is a unique $u=T^{-1}z$, and it satisfies $B(u,v)=F(v)$ for all $v$. The lower bound also gives $\Vert u{\Vert }_H\le \Vert F\Vert /\alpha$.

Set $\mathcal{H}=H_0^1(U{)}^2$. A weak solution is a vector $\varphi \in \mathcal{H}$ such thatwhere $D\varphi :Dv={\sum }_{j=1}^{2}D{\varphi }_j\cdot D{v}_j$ and the natural data space is $F\in H^{-1}(U{)}^2=(H_0^1(U{)}^2{)}^{\ast }$.

If $\varphi \in C^2(\overline{U}{)}^2$, take $v\in C_c^{\infty }(U{)}^2$ and integrate the gradient term by parts. The weak identity becomesThe fundamental lemma of the calculus of variations gives $-\Delta \varphi +A\varphi =F$ pointwise. Membership in $H_0^1(U{)}^2$, together with continuity up to the boundary, says that the boundary trace is zero, so the Dirichlet condition also holds classically.

For compactly supported smooth vector fields, two integrations by parts show that $-\Delta$ is formally self-adjoint. Pointwise,ThereforeIt follows that $L=L^{†}$ exactly when $A=A^T$, that is, when $A$ is a symmetric matrix at every point.

Use the bilinear formon $H_0^1(U{)}^2$. Smoothness on the compact set $\overline{U}$ makes $A$ bounded, so $B$ is bounded. Positive semidefiniteness givesThe Poincare inequality makes the right side coercive for the $H_0^1(U{)}^2$ norm. Hence the Lax-Milgram theorem gives a unique weak solution for everyIn particular, every $F\in L^2(U{)}^2$ is admissible.