Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 (Elliptic Partial Differential Equations)

Let $M={\sup }_{\Omega }u$ and suppose $u(y)=M$. The set ${\Omega }_M=\{x\in \Omega :u(x)=M\}$ is closed in $\Omega$ by continuity. If $x\in {\Omega }_M$, choose a ball $B(x,r)⋐\Omega$. The mean value property for harmonic functions assumed in the question givesThe nonnegative continuous function $M-u$ consequently has integral zero and therefore vanishes throughout the ball. Thus ${\Omega }_M$ is also open. Since $\Omega$ is connected and ${\Omega }_M$ is nonempty, ${\Omega }_M=\Omega$, so $u$ is constant. Applying the same argument to $-u$ proves the assertion for an attained infimum.

Fix $B=B(x_0,r)⋐\Omega$. Solvability of the Dirichlet problem on a ball gives a unique harmonic function $h\in C^2(B)\cap C^0(\overline{B})$ with $h=u$ on $\partial B$. Both $u$ and $h$ have the mean value property for harmonic functions, so $v=u-h$ has it as well and vanishes on $\partial B$.

If $v$ were not zero, compactness of $\overline{B}$ would give either a positive maximum or a negative minimum in the interior. Part 1(i) would make $v$ constant, contradicting its zero boundary values. Hence $u=h$ on $B$. Every point lies in such a ball, so

Radiality gives ${\varphi }_{\epsilon }(x-y)={\varphi }_{\epsilon }(y-x)$. Since ${\varphi }_{\epsilon }$ is supported in $B(0,\epsilon )$, the changes of variables $y=x+z$ and then $z=\epsilon r\omega$ in spherical coordinates giveThe restriction $\epsilon <\mathrm{dist}(x,\partial \Omega )$ ensures that every sampled point remains in $\Omega$.

For each $0<r<1$, the spherical mean value property for harmonic functions givesInsert this into part 1(iii). The normalization of the radial mollifier is $d{\omega }_d{\int }_0^1\varphi (r)r^{d-1}dr=1$, henceA convolution with a smooth function of compact support is smooth wherever it is defined. Every point admits such an $\epsilon$, so $u\in C^{\infty }(\Omega )$.

Fix $x\in {\Omega }^{\prime }$ and $0<\rho <R({\Omega }^{\prime })$. Differentiate the ball mean-value formula and apply the divergence theorem:Since $|{\nu }_j|\le 1$ and $|\partial B(x,\rho )|=d{\omega }_d{\rho }^{d-1}$,Letting $\rho \uparrow R({\Omega }^{\prime })$ and taking both maxima proves

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.

Here $w$ is the harmonic replacement of $u$ in $B_r=B(x_0,r)$, so $v=u-w\in H_0^1(B_r)$. Subtract the weak equations and test with $v$:Because $B_r\subset B(0,1)$, $|x_1|\le 1$. The Sobolev inequality $\Vert v{\Vert }_6\le C\Vert \nabla v{\Vert }_2$ and the Holder inequality giveAfter division and squaring,

On the finite-measure ball, the Holder inequality givesSquaring and using $|B_r|={\omega }_3{r}^3$ yieldsSince $2=1+2\alpha$ for $\alpha =1/2$, this is (6), with $\boxed{C_3={\omega }_3^{2/3}}$.

The Poincare-Wirtinger inequality on a ball and the assumed estimate (8) implyThis is exactly the stated Campanato space criterion with a constant $M$ independent of the ball. Therefore

For $w\in X_{\delta }$, hypothesis (2) gives $F(w)\in C^{0,\alpha }(\overline{\Omega })$. Thus $f-F(w)$ is Hölder continuous and the classical Dirichlet problem for the Poisson equation has a unique solution $u=T(w)\in C^{2,\alpha }(\overline{\Omega })$. The Global Schauder estimate givesThe maximum estimate for the Poisson problem also givesThese estimates prove that $T$ is well defined and give the requested norm control.

For $w\in X_{\delta }$, hypothesis (2) givesCombining the two estimates in part 4(i) with (5) yieldsChoose $0<\delta <\min \{1,(2C{)}^{-1}\}$, so that $C{\delta }^2\le \delta /2$, and then choose $\epsilon \le \delta /(2C)$. Uniformly for $w\in X_{\delta }$ we obtain $\Vert T(w){\Vert }_{C^{2,\alpha }}\le \delta$. Hence

As printed, the hypotheses admit no function $F$. Take the constant function $g\equiv z$ in (2). Its Hölder seminorm vanishes, soBut (1) simultaneously requires $F(z)\ge |z|$. For every $0<|z|<1$, this would give $|z|\le z^2$, which is false. Therefore no $F\in C^{\infty }(\mathbb{R})$ satisfies (1) and (2), and the contraction question is vacuous under the stated assumptions. This contradiction is present in the official paper, rather than arising from the text conversion.