Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 2 / c

The Sobolev trace theorem makes evaluation at zero a continuous linear map $T_0:H^1(0,1)\to \mathbb{R}$. Henceis a closed vector subspace of the Hilbert space $H^1(0,1)$. Every closed vector subspace of a Hilbert space is complete with the restricted inner product, so $H$ is a Hilbert space with the standard $H^1$ inner product.

For $u\in H$, the trace $u(0)$ vanishes. Apply the Hardy inequality on an interval with $p=2$ to obtainThus $u/x\in L^2(0,1)$.

Multiply the differential equation by $v\in H$ and integrate. The Hardy inequality on an interval makes the terms containing $u/x$ and $v/x$ integrable. For smooth $v$, integration by parts givesbecause $v(0)=0$ and the Neumann boundary condition is $u^{\prime }(1)=0$. ThereforeThe density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every $v\in H$.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincare inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.