Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 2

The Lax-Milgram theorem states that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form, and there is an $\alpha >0$ such thatthen for every bounded linear functional $F\in H^{\prime }$ there is a unique $u\in H$ satisfyingMoreover, $\boxed{\Vert u{\Vert }_H\le {\alpha }^{-1}\Vert F{\Vert }_{H^{\prime }}}$. Symmetry of $B$ is not required.

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

On the product Hilbert space $H=H_0^1(U)\times H^1(U)$ defineandThe Cauchy-Schwarz inequality makes $B$ and $F$ bounded. On the diagonal,The Poincare inequality controls $\Vert u{\Vert }_2$ by $\Vert Du{\Vert }_2$, andThe displayed diagonal value therefore controls the full product $H^1$ norm, so $B$ is coercive. The Lax-Milgram theorem now gives exactly one pair $(u,w)\in H$ satisfying the weak identities. Hence a unique weak solution exists for every $f,g\in L^2(U)$.