Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 3 / a / iv / Solution

Work in the mean-zero Sobolev spaceThe integral is a continuous functional on $H^1(U)$, so $H$ is closed. The Neumann-Poincare inequality shows that $\Vert v{\Vert }_a=\Vert \nabla v{\Vert }_2$ is equivalent to the usual $H^1$ norm on $H$, making $a$ a complete inner product there.

For $f\in L^2(U)$ the functional $L(v)={\int }_{U}fv$ satisfiesThe Riesz representation theorem, or the Lax-Milgram theorem, gives a unique $u\in H$ with $a(u,v)=L(v)$ for all $v\in H$. To recover every $H^1(U)$ test, write $v=(v-v_U)+v_U$. The constant contributes zero to $a$ and contributes $v_U{\int }_{U}f=0$ to $L$. Thus the same equality holds for all $v\in H^1(U)$.

A weak solution exists whenever ${\int }_{U}f=0$; fixing its average to zero makes it unique. Moreover $\Vert \nabla u{\Vert }_2\le \sqrt{C_P}\Vert f{\Vert }_2$ and the Neumann-Poincare inequality also controls $\Vert u{\Vert }_2$.