Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 / 4

Multiply $\Delta u+qu\le 0$ by ${\zeta }^2/u_{\epsilon }$ and integrate. Since $\zeta$ is compactly supported, integration by parts and Young's inequality giveAs $\epsilon \downarrow 0$, $u/u_{\epsilon }$ converges to the indicator of $\{u>0\}$. The dominated convergence theorem therefore gives

Put $v=\log u_{\epsilon }$, so $Dv=Du/u_{\epsilon }$. Choose a cutoff $\eta$ which equals one on $B_{\rho }(z)\cap B_1$, is supported in a comparable ball inside $B_2$, and satisfies $|D\eta |\le C/\rho$. Repeating the calculation from part (i) with ${\eta }^2$ and using $|q|\le \mu$ gives the Caccioppoli inequalityThe same construction works for balls meeting $\partial B_1$ because the cutoff is supported in $B_2$. Hence

For $v=\log u_{\epsilon }$, Cauchy-Schwarz inequality and part (a.ii) implyThe John-Nirenberg inequality therefore supplies $p_0=p_0(n,\mu )>0$ such that, with $v_{B_1}$ denoting the average,Since $e^{p_0(v-v_{B_1})}$ and $e^{-p_0(v-v_{B_1})}$ are each bounded by this integrand,

If $u=0$ on $\Sigma$ and $|\Sigma |>0$, thenPart (b.i) consequently givesLetting $\epsilon \downarrow 0$ and using Fatou lemma yields ${\int }_{B_1}{u}^{p_0}=0$. Since $u\ge 0$ is continuous,

Suppose a nonzero solution $w$ did not change sign. Replacing it by $-w$ if necessary gives $w\ge 0$. The argument of part (b.ii) is local and invariant under translation and scaling: it applies on every ball whose concentric double lies in $B_2$. If the zero set of $w$ had positive measure, a density point and this local result would make $w$ vanish on one ball. Applying the same result successively on overlapping balls would then give $w\equiv 0$ throughout the connected ball $B_2$, a contradiction. Thus the zero set has measure zero in $B_2$. Applying part (a.i) to $u=w$ givesfor every compactly supported $C^1$ function $\zeta$; the omitted zero set has measure zero. But $\inf q>{\lambda }_1$, while the variational characterization of the First Dirichlet eigenvalue provides a $\zeta$ withThis contradiction proves that every nonzero solution takes both positive and negative values.

Yes. Since $|q|\le \mu$ and $u\ge 0$,The operator $\Delta -\mu$ is uniformly elliptic and has nonpositive zeroth-order coefficient. If $u(y)=0$ at an interior point, $u$ attains its nonnegative minimum there, so the strong minimum principle for elliptic operators gives $u\equiv 0$ on the connected ball $B_1$.