Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 (Elliptic Partial Differential Equations)

Fix $x\in \Omega$ and $R<\mathrm{dist}(x,\partial \Omega )$. The mean value property for harmonic functions isTo prove the spherical identity, translate $x$ to the origin and put $F(r)=|S^{n-1}{|}^{-1}{\int }_{S^{n-1}}u(r\theta )d{S}_{\theta }$. The divergence theorem givesHence $F(r)={\lim }_{s\downarrow 0}F(s)=u(0)$. Integrating the spherical identity in polar coordinates gives the ball identity.

Choose $0<r<\mathrm{dist}(x,\partial \Omega )$. Differentiating the ball mean-value formula with respect to its centre and then using the divergence theorem givesConsequently the interior derivative estimate for a harmonic function yieldsThe constant may depend on $x$ and its distance from the boundary, but it is independent of $u$.

Take a nonnegative radial mollifier ${\eta }_{\epsilon }$ supported in $B_{\epsilon }(0)$, with $\epsilon <\mathrm{dist}(x,\partial \Omega )$. Writing the convolution in polar coordinates and applying the spherical mean value property for harmonic functions at every radius shows directly thatThe convolution is smooth, so $u$ agrees locally with a smooth function. Since $x$ was arbitrary, $u\in C^{\infty }(\Omega )$.

Choose a cutoff function $\eta$ supported in $B_{\rho }(x)$, equal to one on $B_{{\rho }^{\prime }}(x)$, and satisfying $|D\eta |\le (\rho -{\rho }^{\prime }{)}^{-1}$. Use ${\eta }^{2}u$ as a test function in the weak formulation:The Cauchy-Schwarz inequality followed by Young inequality givesand henceThis is the Caccioppoli inequality. The numerical constant is convention-dependent and is normally absorbed into the displayed estimate; replacing the radii by fixed intermediate radii gives the stated form with one universal constant.

On a relatively compact ball, mollification commutes with the Laplacian, so $u_{\epsilon }=u\ast {\eta }_{\epsilon }$ is a smooth harmonic function. Repeated interior derivative estimates, together with the Caccioppoli inequality, bound every derivative of $u_{\epsilon }$ on a smaller ball by the local $L^2$ norm of $u$, uniformly as $\epsilon \downarrow 0$. The Arzela-Ascoli theorem and a diagonal argument give a smooth local limit, while mollification gives $u_{\epsilon }\to u$ in $L_{\mathrm{loc}}^2$. Thus the limit equals $u$ almost everywhere. After choosing this smooth representative, $u$ is harmonic pointwise. This is the Weyl lemma for an $H^1$ weak solution.

For $u\in L_{\mathrm{loc}}^2(\Omega )$, the correct formulation is the distributional identityThus $\Delta u=0$ as a distribution. The Weyl lemma applies already to locally integrable distributions, so $u$ agrees almost everywhere with a smooth harmonic function.

Uniform ellipticity means that there is $\lambda >0$ such that the symmetric part of the principal coefficient matrix satisfiesfor every $x\in \Omega$ and $\xi \in {\mathbb{R}}^n$. This is the defining coercive bound for a uniformly elliptic operator.

Assume $c\le 0$, the coefficients are bounded, and $u$ is bounded above. Choose $\kappa$ so large that the bounded positive function $q(y)=e^{\kappa y}$ satisfiesAlso set $\psi (x)=\log (1+|x{|}^2)$. Boundedness of the coefficients, $x\cdot D\psi \le 2$, and $c\le 0$ give a global upper bound $L\psi \le C$.

For $\epsilon >0$ and $0<\delta <\epsilon m/C$, the functiontends to $-\infty$ as $|x|\to \infty$ and satisfies $Lv>0$. If $v$ exceeded both zero and its values on $y=\pm 1$, it would attain a positive interior maximum. At that point $Dv=0$ and $D^{2}v\le 0$, whence $Lv\le cv\le 0$, a contradiction. Letting $\delta \downarrow 0$ and then $\epsilon \downarrow 0$ provesThis is the weak maximum principle for elliptic operators on the slab. The boundedness or a comparable growth condition is necessary because the domain is unbounded.

The weak maximum principle for elliptic operators fails without a condition at infinity. The functionis harmonic on the upper half-space, continuous on its closure, and vanishes on the boundary, but it is positive and unbounded in the interior.

For $n\ge 3$, the radial functionis harmonic on ${\mathbb{R}}^n\setminus \overline{B_1(0)}$, vanishes on the unit sphere, and is positive in the domain. It therefore violates the weak maximum principle for elliptic operators. In two dimensions the corresponding counterexample is $u(x)=\log |x|$, since the fundamental solution of the Laplace equation changes from a power to a logarithm.

TakeIts principal symbol is ${\xi }^4$, so it is elliptic, and $a_4{a}_0=-1<0$. Yet $u(x)=\sin x$ satisfies $Lu=0$ on $(0,\pi )$ and vanishes at both boundary points while remaining positive inside. Thus second-order ellipticity is essential to the usual weak maximum principle for elliptic operators.

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the weak subsolution inequality use the admissible truncations approximating ${\eta }^2{u}^{\alpha -1}$. Uniform ellipticity, the coefficient bound, Cauchy-Schwarz inequality, and Young inequality giveand thereforeApply the Sobolev embedding theorem to $w=\eta u^{\alpha /2}$. The preceding estimate yields, for concentric balls $B_r\subset B_R\subset B_1$,Starting with $\alpha =p>1$, taking ${\alpha }_k=p{\sigma }^k$, and choosing radii decreasing to $1/2$, the product of constants converges because ${\sum }_k{\alpha }_k^{-1}<\infty$. Letting $k\to \infty$ provesThis exponent-raising argument is Moser iteration.

Use the Weak Harnack inequality: for some $q>1$ and every nonnegative weak supersolution,where $C$ and $q$ depend only on $n,\lambda ,\Lambda$. A weak solution is both a subsolution and a supersolution. Applying part (i), after rescaling from $B_1$ to $B_{1/2}$, and then the weak Harnack inequality givesThis is the Harnack inequality for uniformly elliptic divergence-form equations.

For a compactly supported variation $v+t\varphi$, differentiation of the area functional at $t=0$ gives the first variationAfter integration by parts, this is the minimal surface equation for a graphFor $v_R(x)=R^{-1}v(Rx)$ one has $D{v}_R(x)=Dv(Rx)$. The equation is invariant under this scaling, so $v_R$ solves it on $B_1$ for every $R>0$.

Differentiate the equation for $v_R$ with respect to $x_k$. The derivative $w=D_k{v}_R$ is a weak solution ofwhereThe eigenvalue in the direction of $p$ is $(1+|p{|}^2{)}^{-3/2}$ and every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. Thus a uniform bound on $|Dv|$ makes this a uniformly elliptic operator with constants independent of $R$.

Suppose $|Dv|\le M$. The coefficient matrices in part (ii) then have uniform ellipticity constants depending only on $M$. Applying the Harnack inequality for uniformly elliptic divergence-form equations to the nonnegative solutions $M+D_k{v}_R$ and $M-D_k{v}_R$ gives a scale-independent oscillation contractionScaling back,For fixed $r$, iterate this estimate with $R=4^{m}r$ and use the global bound $|D_{k}v|\le M$ to obtain ${\mathrm{osc}}_{B_r}{D}_{k}v\le 2M{\theta }^m\to 0$. Every partial derivative is therefore constant, so $v(x)=a\cdot x+b$ is an affine function. This is a bounded-gradient Bernstein theorem for entire minimal graphs.

Choose smooth boundary data ${\phi }_j$ converging uniformly to $\phi$, and let $v_j$ be their harmonic extensions. The maximum principle for harmonic functions givesso $v_j$ converges uniformly on $\overline{B}$ to a continuous function $v$ with boundary value $\phi$. Interior derivative estimates make the convergence smooth on compact subsets of $B$, hence $v$ is harmonic there. Uniqueness follows by applying the maximum principle to the difference of two solutions.

Let $B⋐\Omega$ and let $h$ be the harmonic function whose boundary values are $\max (u_1,u_2)$. The harmonic extensions $(u_i{)}_B$ satisfy $(u_i{)}_B\le h$ by the maximum principle for harmonic functions, while subharmonicity gives $u_i\le (u_i{)}_B$ in $B$. Hence $\max (u_1,u_2)\le h$, proving that the maximum of two subharmonic functions is subharmonic.

The difference $u_1-u_2$ is subharmonic. If $M={\max }_{\overline{\Omega }}(u_1-u_2)>0$, its maximum set lies in $\Omega$. At any point of this set, comparison with the harmonic replacement on a small ball and the Strong maximum principle for harmonic functions show that the whole ball belongs to the maximum set. The set is therefore both open and closed in the connected domain $\Omega$, so it is all of $\Omega$, contradicting the boundary inequality. Thus $u_1\le u_2$ throughout $\Omega$. This is the comparison principle for subharmonic and superharmonic functions.

The constant $m={\min }_{\partial \Omega }\phi$ is subharmonic and lies below the boundary data, so the Perron family is nonempty. The constant $M={\max }_{\partial \Omega }\phi$ is superharmonic. Part (ii) gives $u\le M$ for every member $u$ of the family, while the member $m$ gives the lower bound. Henceso the pointwise supremum is finite and well-defined.

Fix $B⋐\Omega$ and $x_0\in B$. Choose $u_j$ in the Perron family with $u_j(x_0)\uparrow \overline{u}(x_0)$, replace successive terms by finite maxima using part (i), and take their harmonic lifts on $B$. The lifts remain in the family, are increasing, and are uniformly bounded. Interior estimates and the Arzela-Ascoli theorem give a harmonic limit $h$ on $B$ with $h\le \overline{u}$ and $h(x_0)=\overline{u}(x_0)$.

If $h(y)<\overline{u}(y)$ somewhere in $B$, take another family member larger than $h(y)$ and repeat the maximum-and-lift construction. Its harmonic limit $H$ satisfies $H\ge h$ and $H(x_0)=h(x_0)$. The strong minimum principle for elliptic operators forces $H=h$, contradicting the strict inequality at $y$. Thus $h=\overline{u}$ on $B$. Since $B$ was arbitrary, $\overline{u}$ is smooth and harmonic in $\Omega$. This is the Perron method for the Dirichlet problem.

Fix $\epsilon >0$ and choose a boundary neighbourhood $U$ of $z$ on which $|\phi -\phi (z)|<\epsilon$. The positive continuous barrier $w$ has a positive minimum $m$ on the compact set $\partial \Omega \setminus U$. Choosing $A$ large enough makeson all of $\partial \Omega$. The left function is subharmonic and belongs to the Perron family; the right function is superharmonic and dominates every family member by part (ii). ThereforeAs $x\to z$, continuity gives $w(x)\to 0$. Letting $\epsilon \downarrow 0$ proves $\overline{u}(x)\to \phi (z)$. Such a $w$ is a barrier for the Dirichlet problem, and $z$ is a regular boundary point.

Every boundary point of $D$ satisfies the exterior sphere condition. For a point on the inner spherical boundary, use a smaller ball inside the removed ball and tangent at that point; for a point on the cube, use a ball in a supporting exterior half-space. If the exterior ball has centre $y$ and radius $r$, a local positive harmonic barrier isfor $n\ge 3$, while in two dimensions use $w(x)=\log (|x-y|/r)$. Adding a sufficiently large positive multiple of a global superharmonic function extends the local barrier across the bounded domain. Hence every boundary point is regular by part (v), and the Perron method for the Dirichlet problem produces a harmonic function attaining the prescribed continuous boundary data. The maximum principle for harmonic functions gives uniqueness.