Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 107 (Elliptic Partial Differential Equations)

The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

For the bounded-domain correction, let $\mathcal{S}$ be the family of Perron subfunctions for the Poisson equation $v\in C^0(\overline{\Omega })$ with $v\le \phi$ on $\partial \Omega$. The subfunction condition is ball comparison: whenever $\overline{B}\subset \Omega$ and a classical solution $H$ of $\Delta H=f$ on $B$ satisfies $v\le H$ on $\partial B$, it also satisfies $v\le H$ throughout $B$. For a $C^2$ function this is the sign condition $\Delta v\ge f$. A superfunction reverses both inequalities.

The Perron solution is the pointwise upper envelope of the subfunctions:This is the Perron method for the Dirichlet problem with the Poisson equation replacing the homogeneous Laplace equation. The power barrier for an exterior cone supplies a global member of $\mathcal{S}$ and a global superfunction, as constructed below. The permitted subfunction-superfunction comparison therefore makes this envelope finite. The standard interior Perron theorem, available in the question, gives $u\in C^2(\Omega )$ and $\Delta u=f$; neither result needs to be proved in this part.

Fix $x_0\in \partial \Omega$, $\epsilon >0$, and the power barrier for an exterior cone $g$ from part (i). Write $M=\Vert f{\Vert }_{\infty }$. By continuity of the boundary data, there is $\rho >0$ such that $|\phi (x)-\phi (x_0)|<\epsilon$ on $\partial \Omega \cap B_{\rho }(x_0)$. On the rest of the boundary, $-g\ge m{\rho }^{\nu }$. The boundary is compact, so $\phi$ is bounded. Choose $A>0$ large enough that $A\delta \ge M$ andThen the Perron subfunction for the Poisson equation and its superfunction counterpartsatisfy $\Delta v_{-}\ge M\ge f$, $\Delta v_{+}\le -M\le f$, and $v_{-}\le \phi \le v_{+}$ on the whole boundary. The near-boundary ordering uses the choice of $\rho$; the remaining ordering uses the choice of $A$. In particular the Perron family is nonempty and bounded above.

The Perron method for the Dirichlet problem and the permitted comparison give $v_{-}\le u\le v_{+}$. Since $g(x)\to 0$ as $x\to x_0$, we obtainLetting $\epsilon \downarrow 0$ provesThe interior Perron theorem gives continuity inside $\Omega$, and the displayed limit together with continuity of $\phi$ proves the continuous extension $u\in C^0(\overline{\Omega })$ with the required boundary trace. Thus every boundary point is a regular boundary point.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.

Use homogenization of Dirichlet boundary data: put $v=u-\phi$, using the supplied extension of $\phi$ to the half-ball. Then $v=0$ on the flat face and $\Delta v=f-\Delta \phi$. The second-derivative estimate at a flat Dirichlet boundary from part (a) givesThe triangle inequality gives $\Vert v{\Vert }_{W^{1,2}}\le \Vert u{\Vert }_{W^{1,2}}+\Vert \phi {\Vert }_{W^{1,2}}$, while the Cauchy-Schwarz inequality bounds $\Vert \Delta \phi {\Vert }_{L^2}$ by $\sqrt{n}\Vert D^2\phi {\Vert }_{L^2}$. Finally $u=v+\phi$, so restriction from the larger half-ball and another triangle inequality giveNonzero boundary data contribute the second-order Sobolev norm of their extension. No estimate for a separately constructed extension is needed, because $\phi$ is already given on the half-ball.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

Fix nested relatively compact subsets of $\Omega$ and let $M$ bound the gradient on the larger one. The fundamental theorem of calculus along a line segment gives $|w_h|\le M$ on the smaller one. Part (i) gives coefficient bounds independent of small $h$, so the De Giorgi-Nash-Moser theorem gives uniform interior Hölder continuity estimates for these weak solutions. Since $Du$ is continuous, the difference quotients $w_h$ converge locally uniformly to $D_{\ell }u$. Passing to the limit in the Hölder seminorm estimate, and repeating for every coordinate, yieldsThe first derivatives are locally Hölder continuous. Here the initially obtained exponent depends on $n$ and the local gradient bound. This is the interior interpretation of the printed regularity conclusion; a uniform global estimate does not follow from $u\in C^1(\Omega )$ alone.

Assume $u\in W^{1,2}(B_2(0))$ is nonnegative almost everywhere, and that the real coefficient matrix $A=(a^{ij})$ is measurable and satisfies, for almost every $x$ and every $\xi \in {\mathbb{R}}^n$,The operator norm bound controls all coefficients, including any antisymmetric part. For a symmetric matrix, these assumptions are exactly the usual lower and upper quadratic-form bounds. The weak solution condition isThen the Harnack inequality for uniformly elliptic divergence-form equations isA nonnegative weak solution has its supremum controlled by its infimum on the smaller ball. The De Giorgi-Nash-Moser theorem supplies a Hölder continuous function representative, for which ordinary supremum and infimum can replace the essential supremum and essential infimum. Nonnegativity and the larger ball are essential hypotheses; there is no differentiability requirement on the coefficients.

There is no need to assume that the original zeroth-order coefficient is nonpositive. Set $c_{+}=\max (c,0)$ and $\tilde{c}=\min (c,0)$, and use the uniformly elliptic operatorBecause $u\ge 0$ and $f\le 0$,Now $\tilde{c}\le 0$, so the strong minimum principle for elliptic operators applies. If $u$ vanished at an interior point of the connected domain, this principle would force $u$ to vanish identically. The given nontriviality therefore provesEvery nontrivial nonnegative solution is strictly positive inside. This uses the nondivergence-form strong minimum principle for elliptic operators; the divergence-form Harnack inequality for uniformly elliptic divergence-form equations in part (a) is not being applied to this operator.

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzelà-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.

Multiply the Dirichlet Laplacian eigenfunction equation by $u$ and use integration by parts. The zero Dirichlet boundary condition removes the boundary term and givesThe denominator is positive because the eigenfunction is nonzero. The numerator is also positive: if it were zero, the given Sobolev inequality for $u\in W_0^{1,2}(\Omega )$ would force $u=0$. Therefore the Dirichlet Laplacian eigenvalue has the positive Rayleigh quotientEvery nonzero Dirichlet eigenfunction in this problem has a positive eigenvalue.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

Put $p_j=2{\kappa }^j$, where $\kappa =n/(n-2)>1$. Since $p_j^2/(p_j-1)\le 2p_j$, part (ii), followed by a $p_j$-th root, gives the Moser iteration stepAfter $N$ steps,The geometric series and its differentiated form giveIn particular the exact convergent product isThe printed hint's equality to a single power of $2\kappa$ is not correct in general; its claimed finiteness is correct and the displayed expression supplies the correction.

On a finite-measure domain, Lp norms converge to the supremum norm for a bounded continuous function. Indeed, the upper bound is $\Vert u{\Vert }_p\le |\Omega {|}^{1/p}\Vert u{\Vert }_{\infty }$; for every $a<\Vert u{\Vert }_{\infty }$, the set $\{|u|>a\}$ has positive measure, giving $\Vert u{\Vert }_p\ge a|\{|u|>a\}{|}^{1/p}$. Let $N\to \infty$ in the iteration and use part (i), which gives $\lambda >0$. The Dirichlet eigenfunction supremum estimate isThe eigenvalue exponent is exactly $n/4$. The constant depends only on $n$: the supplied zero-boundary Sobolev inequality has a dimension-only constant, and every product factor above depends only on $\kappa$.