Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 3 / a

The zero Dirichlet boundary condition must be part of the definition. Thus $u\in H^1(B)$ is a weak solution precisely when $u\in H_0^1(B)$, equivalently its Sobolev trace vanishes on $\partial B$, andHere ${\chi }_E$ is the indicator function of the measurable set $E$. One may initially require the identity only for test functions $\varphi \in C_c^{\infty }(B)$: both sides are continuous in the Sobolev norm, and the definition of the zero-trace Sobolev space then extends it to all $H_0^1(B)$. This formulation uses first weak derivatives and encodes $-\Delta u={\chi }_E$ without assuming classical second derivatives.

As printed, the answer isThe constant function $u\equiv 1$ has zero gradient and positive Lp norm for every such $q$, including its supremum norm when $q=\infty$.

The missing assumption is a zero Sobolev trace, which is present for the weak solutions in the surrounding parts. With $H_0^1(B)$ in place of $H^1(B)$, the intended answer isThis is the three-dimensional Sobolev inequality at $q=6$, followed by Holder inequality on the bounded ball for smaller $q$; it includes the zero-boundary Poincaré inequality at $q=2$. The subsequent parts use this zero-boundary version.

By the definition in (i), $u$ itself is an admissible test in the weak formulation. HenceThe absolute value makes a proof of positivity unnecessary. Holder inequality with exponents $6$ and $6/5$, followed by the zero-boundary Sobolev inequality, givesTaking square roots of both estimates proves exactly the requested chain:If $\Vert \nabla u{\Vert }_2=0$, the zero-boundary Poincaré inequality gives $u=0$ and the conclusion follows. Otherwise divide the first-to-last estimate by $\Vert \nabla u{\Vert }_2^{1/2}$ and square, obtaining $\Vert \nabla u{\Vert }_2\le C_S|E{|}^{5/6}$. Finally the same Poincaré inequality givesAll constants depend only on the fixed unit ball and the dimension, and are independent of $E$ and $u$. If $|E|=0$, testing the homogeneous equation with $u$ gives the same zero solution directly.