Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 2 / b / ii / Solution

There is a small incompatibility in the printed cutoff function assumptions. No $\phi \in C_c^{\infty }(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla \phi |\le 2$: for a unit vector $e$, choose $t<1$ beyond its support. Continuity gives $\phi (e/2)=1$, while the fundamental theorem of calculus would give $1\le 2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a cutoff function exists, and the requested Caccioppoli inequality is unchanged apart from its constant. We keep $M=\Vert \nabla \phi {\Vert }_{\infty }$ in the calculation, which also yields the formal bound under the printed assumptions.

The weak formulation is ${\int }_{B_1}A\nabla u\cdot \nabla \zeta =0$ for test functions $\zeta$. Since $A$ is bounded, this extends by density of smooth functions in a Sobolev space to every $\zeta \in H_0^1(B_1)$. Part (a)(ii), followed by mollification inside the compact support, makes ${\phi }^2(u-k)$ an admissible choice. The product rule givesPut $X=\int {\phi }^2|\nabla u{|}^2$ and $Y=\int (u-k{)}^2|\nabla \phi {|}^2$. The lower and upper uniform ellipticity bounds, together with Cauchy-Schwarz inequality, implyIf $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\le 4{\Lambda }^2{\lambda }^{-2}Y$. Since $\phi =1$ on $B_{1/2}$,For a fixed valid cutoff function, $M\le C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16{\Lambda }^2/{\lambda }^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.