Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 23H (Analysis of Functions)

This is the boundedness criterion for a bilinear form. Since $B$ is continuous at $(0,0)$, there are $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$. For nonzero $u,v$, apply this toLinearity in each argument then givesThe same inequality is immediate if either vector is zero, so one may take $M=4/(rs)$.

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a bounded linear functional. The Riesz representation theorem supplies a unique vector $Lv$ such thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. The invertibility from lower bounds on an operator and its adjoint therefore shows that $L$ is invertible and $L^{-1}$ is bounded.

Set $v_f=L^{-1}f$. Then $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus the unique vector is $v_f=L^{-1}f$.

The zero-boundary Sobolev space isHere $H^1(\Omega )$ consists of the $L^2$ functions whose first weak derivatives also belong to $L^2$, with normOn a sufficiently regular domain, the closure condition is equivalent to having zero boundary trace.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

Use the Poisson integral to choose a smooth harmonic function $G$ on the unit disk whose boundary value is $g$. Seek $v$ in the form $v=w+G$, which is the standard homogenization of Dirichlet boundary data. Then $w\in H_0^1(\Omega )$ and, since $\Delta G=0$,Part (d) gives a unique such $w$, and henceis the unique weak solution with boundary value $g$.