Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 2

Let the exterior ball be $B(y,R)$, tangent at $x_0$. Then $|x-y|\ge R$ on $\overline{\Omega }$, with equality only at $x_0$. Choose $\delta =R^{-1}$ and defineIn two dimensions, $\log |x-y|$ is a harmonic function away from $y$, so $\Delta w=0\le 0$ in $\Omega$. Moreover $w(x_0)=0$ and $w>0$ on $\overline{\Omega }\setminus \{x_0\}$. Thus $w$ is a barrier for the Dirichlet problem, and $\boxed{x_0\text{ is a regular boundary point}}$.

The weak maximum principle for elliptic operators here statesfor $u\in C^2(\Omega )\cap C^0(\overline{\Omega })$.

To prove it, suppose the interior maximum exceeds the boundary maximum. Put $q(x)=e^{-\ell x_1}$. Since $c=0$,For sufficiently small $\epsilon >0$, $u_{\epsilon }=u+\epsilon q$ still has an interior maximum. At that point its gradient vanishes and its Hessian matrix is negative semidefinite; ellipticity gives $L{u}_{\epsilon }\le 0$. But $L{u}_{\epsilon }=Lu+\epsilon Lq>0$, a contradiction. This proves the principle.

No. A positive zeroth-order coefficient can overturn the weak maximum principle for elliptic operators. On $\Omega =(0,1)$ takeThen $L\nu =0$, $\nu =0$ on $\partial \Omega$, and $\nu >0$ in $\Omega$. Thus the boundary maximum is zero while the interior maximum is one. This is a counterexample satisfying uniform ellipticity and zero drift.

If $u_1,u_2$ solve the same Dirichlet problem, their difference $v=u_1-u_2$ satisfiesPart 2(ii) applied to $v$ gives $v\le 0$, and applied to $-v$ gives $v\ge 0$. Hence $v=0$ and the solution is unique.

The stated assumptions alone do not imply regularity up to the boundary: continuous boundary data need not have two Hölder derivatives. A standard sufficient set of hypotheses for the Global Schauder estimate iswith $a^{ij},b^i,c\in C^{0,\alpha }(\overline{\Omega })$ and uniform ellipticity. Then the Global Schauder estimate givesand bounds its norm by the forcing, boundary data, and $C^0$ norm.