Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 9 (Elliptic PDEs)

The Weak Harnack inequality is an estimate for a nonnegative weak supersolution. Write $F=(f^1,\dots ,f^n)$, let $B_{2R}(x_0)\subset \subset \Omega$, and assume $u\ge 0$ almost everywhere on this ball. There are $p>0$ and $C<\infty$, depending only on the dimension, the uniform ellipticity bounds, $q$, and the scaled norms of the lower-order coefficients, such thatHere $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$ is the integral average. The weak supersolution inequality uses the sign convention $Lu\le \mathrm{div}F+g$. For instance, the coefficient dependence can be expressed usingThus one may use the same $p,C$ on all sufficiently small balls when the global coefficient norms and uniform ellipticity bounds are fixed. The nonnegativity condition is essential to this formulation; a signed weak supersolution may first be shifted, with the resulting change in its forcing included. As usual, supremum and infimum statements for functions in a Sobolev space mean essential suprema and essential infima. The standard multidimensional statement uses $n\ge 2$, so $q/2>1$. In dimension one the analogous formulation requires $q\ge 2$; the printed restriction $q>n$ alone would allow $q/2<1$, which does not suffice. To see the obstruction, smooth the nonnegative capped function $\min \{1,|x|/\epsilon \}$ by a mollifier of width $\tau$. Its positive second derivative is a bump of mass $2/\epsilon$ and has $L^{q/2}$ size of order ${\epsilon }^{-1}{\tau }^{2/q-1}$. For fixed $\epsilon$ and $1<q<2$, this size and the minimum both tend to zero as $\tau \downarrow 0$, whereas the average on a fixed larger interval stays positive. Thus the Weak Harnack inequality cannot have the displayed uniform forcing bound in that range. Shrinking $\epsilon$ and choosing $\tau$ still smaller likewise defeats a uniform Hölder estimate based on that norm alone.

Let $M={\mathrm{esssup}}_{\Omega }u$. The standard local boundedness of weak elliptic subsolutions gives a finite upper bound on every relatively compact ball. In particular, the assumed equality of essential suprema makes $M$ finite. Equivalently, when using the maximum principle for an upper-bounded weak subsolution, this preliminary fact is already part of the hypothesis.

Set $v=M-u$. Because the divergence-form elliptic operator here annihilates constants, $v\ge 0$ and $Lv=-Lu\le 0$. The hypothesis says ${\mathrm{essinf}}_{B}v=0$. Cover the compact closure of $B$ by finitely many balls $D_j$ whose doubled balls lie in $\Omega$. At least one $D_j$ has essential infimum zero: otherwise the minimum of their finitely many positive lower bounds would be a positive lower bound on $B$. The Weak Harnack inequality with zero forcing now givesConsequently $v=0$ almost everywhere on $D_j$. If another admissible ball overlaps $D_j$ in an open set, its essential infimum of $v$ is also zero, and the same argument makes $v$ vanish there. Since a domain is connected, a chain of overlapping interior balls reaches every point of $\Omega$. This propagation of zeros by the weak Harnack inequality proves $u=M$ almost everywhere on $\Omega$, which is the claimed strong maximum principle. Notice why the absence of the $b,d$ terms matters: it is precisely what makes $L(M-u)=-Lu$.

Put $\alpha =1-n/q>0$ andFirst suppose $A<\infty$. For an interior ball $B_R$, with $R\le 1$, write $M_R={\mathrm{esssup}}_{B_R}u$, $m_R={\mathrm{essinf}}_{B_R}u$, and $\omega (R)=M_R-m_R$. Constant shifts of a divergence-form elliptic operator change its forcing. The two nonnegative weak supersolutions $v=M_R-u$ and $w=u-m_R$ satisfy, respectively,Their Weak Harnack inequalities on $B_{R/2}$ therefore have forcing errors bounded by $CN{R}^{\alpha }$. Indeed $|M_R|,|m_R|\le A$, the new divergence forcing is bounded by $\Vert F{\Vert }_q+A\Vert b{\Vert }_q$, and the scalar forcing by $\Vert g{\Vert }_{q/2}+A\Vert d{\Vert }_{q/2}$; also $R^{2\alpha }\le R^{\alpha }$.

On $B_{R/2}$ we have $v+w=\omega (R)$. Even if $p<1$, the elementary inequality $(s+t{)}^p\le C_p(s^p+t^p)$ impliesThe sum of the two essential infima is $\omega (R)-\omega (R/2)$. Enlarging a uniform constant $C_0>1$ giveswhere $\theta =1-C_0^{-1}\in (0,1)$. This is an oscillation decay estimate; inhomogeneous oscillation decay implies Hölder continuity. ChooseIterating on concentric balls, starting at a fixed admissible radius $R_{\ast }\le 1$, yieldsThe strict choice of $\mu$ bounds the geometric series, including the case in which the two original decay rates coincide. Monotonicity of oscillation extends this bound to intermediate radii.

Take $R_{\ast }$ uniformly smaller than the distance of ${\Omega }^{\prime }$ from the boundary of $\Omega$. The vanishing essential oscillation produces a continuous representative of $u$. A ball centred at either of two sufficiently close points of ${\Omega }^{\prime }$ contains both, so its oscillation bounds their difference by $CN|x-y{|}^{\mu }$. For separated points use $|u(x)-u(y)|\le 2A$. Thus the Hölder seminorm satisfiesThe constants depend only on the indicated domain separation, dimension, coefficient norms, uniform ellipticity bounds and $q$. If the global essential supremum is infinite, the displayed global estimate is vacuous; local boundedness of weak elliptic subsolutions on slightly larger compact subsets still makes the same argument prove interior Hölder continuity.

For a bounded domain, the nondivergence-form elliptic operator obeys the weak maximum principle: $Lu\ge 0$ and $c\le 0$ implyHere $u^{+}=\max \{u,0\}$. Boundedness of $\Omega$ is needed for this formulation: it is not explicitly imposed in this question. On an unbounded domain one needs an appropriate condition at infinity instead. Without either condition the claim is false, as $u(x)=x_n$ and $L=\Delta$ on the half-space $\{x_n>0\}$ show.

For the bounded-domain proof choose $\gamma >0$ so large that $\lambda {\gamma }^2-\Vert b^1{\Vert }_{\infty }\gamma -\Vert c{\Vert }_{\infty }>0$. The positive exponential $h(x)=e^{\gamma x_1}$ then satisfiesLet $u_{\epsilon }=u+\epsilon h$, $M={\sup }_{\partial \Omega }{u}^{+}$ and $H={\sup }_{\overline{\Omega }}h<\infty$. If $u_{\epsilon }$ had a value greater than $M+\epsilon H$, it would attain a positive maximum at an interior point. There its gradient vanishes and its Hessian matrix is negative semidefinite. The symmetric part of $(a^{ij})$ is positive-definite by uniform ellipticity, so $a^{ij}{D}_{ij}{u}_{\epsilon }\le 0$; also $c{u}_{\epsilon }\le 0$. Hence $L{u}_{\epsilon }\le 0$, contradicting $L{u}_{\epsilon }=Lu+\epsilon Lh>0$. Thus $u_{\epsilon }\le M+\epsilon H$. Letting $\epsilon \downarrow 0$ proves the weak maximum principle, without requiring continuity of the coefficients.

For example, the unbounded-domain version follows by exhausting the domain with bounded truncations if ${\mathrm{limsup}}_{|x|\to \infty ,x\in \Omega }u(x)\le M$: the added spherical boundaries then have supremum at most $M+o(1)$.

In the bounded-domain setting of the weak maximum principle, let $u_1,u_2$ be two classical solutions with the same forcing and boundary values. Their difference $z=u_1-u_2$ satisfies $Lz=0$ and $z=0$ on the boundary. Apply the weak maximum principle first to $z$ and then to $-z$:Thus $u_1=u_2$, proving uniqueness of the Dirichlet problem. The same proof works on an unbounded domain if the difference and its negative satisfy the required condition at infinity. As literally printed without either restriction, uniqueness need not hold: on $\{x_n>0\}$ the two harmonic functions $0$ and $x_n$ have the same zero boundary values.

On the unit ball take the uniformly elliptic operator $L=\Delta -2n$ and the smooth function $u(x)=-1-|x{|}^2$. The zeroth-order coefficient is negative andNeverthelessThis is consistent with the weak maximum principle, whose upper bound here is ${\sup }_{\partial B_1}{u}^{+}=0$. At a negative interior maximum the term $cu$ can be positive; this is why a negative boundary maximum cannot replace its positive part. If $c=0$, subtracting the boundary maximum preserves the differential inequality and gives the stronger bound. That subtraction is unavailable for general $c\le 0$.

Use a Lipschitz continuous cutoff function $\eta$ which equals one on $B_R(x_0)$, vanishes outside $B_{2R}(x_0)$, takes values in $[0,1]$, and has $|\nabla \eta |\le 2/R$. Its gradient is supported in the annulus $A_R=B_{2R}(x_0)\setminus B_R(x_0)$. The function ${\eta }^2(u-c)$ is an admissible test function in the zero-boundary Sobolev space for the weak solution, by approximation with smooth compactly supported test functions. If ${\Lambda }_0$ bounds the operator norm of $(a^{ij})$, we can take ${\Lambda }_0=n\Lambda$. The weak equation and uniform ellipticity giveYoung inequality bounds the right side byAfter absorption, this annular Caccioppoli inequality isThe dimension is fixed in the notation $C(\lambda ,\Lambda )$ of the question; with the entrywise coefficient bound its dependence on $n$ is absorbed there. If the larger ball merely lies in $B$ without its closure being compactly contained, approximation from smaller cutoff functions gives the same admissible test function and estimate. Crucially, the right side uses only the annulus, where the cutoff function varies. The constant $c$ is arbitrary because constants have zero gradient.

Write $E(t)={\int }_{B_t(x_0)}|\nabla u{|}^2$. For $n\ge 2$, choose $c$ in the annular Caccioppoli inequality to be the integral average of $u$ over $A_R$. The supplied Poincare inequality on an annulus givesMoving the $E(R)$ term to the left is the hole-filling argument:Consequently $E(2^{-k}{r}_0)\le {\theta }^{k}E(r_0)$. Put $\beta =-{\log }_2\theta >0$ and choose $\mu =\min \{\beta /2,1/2\}\in (0,1)$. For $2^{-k-1}{r}_0<r\le 2^{-k}{r}_0$, monotonicity givesDyadic endpoints can be assigned to either adjacent interval. Since $E(r_0)\le {\int }_B|\nabla u{|}^2$, the requested dyadic energy decay isBoth constants depend only on the dimension and uniform ellipticity bounds, not on $u,x_0,r,r_0$.

There is a dimensional detail in the printed hint: an annulus is disconnected in dimension one, so that Poincaré inequality with a single average is false there. The conclusion still holds. In one dimension the weak equation gives $a(x)u^{\prime }(x)=J$ almost everywhere for a constant flux for a one-dimensional divergence-form equation $J$. Since $\lambda \le a\le \Lambda$,This implies the required estimate with, for example, $\mu =1/2$ and $K=(\Lambda /\lambda {)}^2$. If $J=0$, the estimate is immediate. Thus the proof also covers dimension one without using the inapplicable hint.

We use a blow-up compactness proof of an interior Schauder estimate, with Taylor normalization in elliptic blow-up arguments. Fix $n,\mu ,\delta$ and suppose no suitable constant exists. There would be solutions $u_j$ with forcing $f_j$ for whichChoose distinct $x_j,y_j\in B_{1/2}$ such that, for $r_j=|y_j-x_j|$,Then $S_j/A_j<2/\delta$, $N_j/A_j<2/j$, andIn particular $r_j\to 0$. Subtract the quadratic Taylor polynomial of $u_j$ at $x_j$ and rescale:Its domain $\{z:x_j+r_{j}z\in B_1\}$ exhausts ${\mathbb{R}}^n$. The normalization givesTaylor theorem, or integration along line segments from zero, consequently bounds $D^2{v}_j,\nabla v_j,v_j$ on every fixed ball by constants times $|z{|}^{\mu },|z{|}^{1+\mu },|z{|}^{2+\mu }$, respectively. The Hessian matrices are equicontinuous. The Arzelà-Ascoli theorem and a diagonal subsequence therefore give $v_j\to v$ in $C^2$ on compact subsets of ${\mathbb{R}}^n$.

The rescaled equation ison each compact set. Thus $v$ is harmonic and is smooth. Every entry of its Hessian matrix is harmonic and has a global Hölder seminorm at most $2/\delta$. The allowed Liouville lemma for globally Hölder harmonic functions makes each entry constant. Since $D^{2}v(0)=0$, all these constants are zero.

On the other hand, $e_j=(y_j-x_j)/r_j$ is a unit vector and $|D^2{v}_j(e_j)|=1$ by construction. A further subsequence has $e_j\to e$; the local $C^2$ convergence implies $|D^{2}v(e)|=1$, a contradiction. Hence the desired constant exists, andThe global Hölder seminorm on the limiting Hessian matrix, rather than a bound on its magnitude throughout space, is what makes this Liouville argument work.

We prove compactness of zero-boundary Poisson solutions. Here the printed $\partial \Omega$ must mean $\partial B_1$; otherwise the boundary condition refers to an unspecified set. Use this intended boundary condition throughout. Let $f=\Delta u$. A maximum-principle barrier gives a uniform bound on $u$: with $\psi (x)=(1-|x{|}^2)/(2n)$, we have $\Delta \psi =-1$ and $\psi =0$ on $\partial B_1$. Since $|f|\le 1$,The weak maximum principle gives $-\psi \le u\le \psi$, so $\Vert u{\Vert }_{\infty }\le 1/(2n)$.

Standard elliptic regularity on the smooth ball first gives $u\in C^{2,\mu }(\overline{B_1})$. The global Schauder estimate then givesThus the functions and all their derivatives through order two are uniformly bounded and equicontinuous on the closed ball. The Arzelà-Ascoli theorem gives a subsequence converging uniformly along with all these derivatives. Integration along line segments identifies the limiting derivatives, so the convergence is in $C^2(\overline{B_1})$.

It remains to keep the limit in the set, rather than merely proving relative compactness. If $u_j\to u$ in $C^2$, then the zero boundary values pass to the limit and $f_j=\Delta u_j\to f=\Delta u$ uniformly. For each pair of distinct points,Taking suprema gives $[f{]}_{\mu }\le {\mathrm{liminf}}_j[f_j{]}_{\mu }$, while $\Vert f_j{\Vert }_{\infty }\to \Vert f{\Vert }_{\infty }$. Hence $\Vert f{\Vert }_{C^{0,\mu }}\le 1$. The limit belongs to the set, proving sequential compactness in $C^2(\overline{B_1})$. The fixed positive Hölder exponent supplies the compactness; a bound on the supremum norm of $\Delta u$ alone would not supply this argument.

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.

Choose continuous boundary values equal to $\sin x$ on the lower side and zero on the other three sides. These values agree at all four corners, so they define a continuous function on the boundary of the rectangle. Any solution would satisfy the zero vertical-side conditions of part (a), whose Fourier sine series forces its lower boundary trace to be zero. This contradicts the prescribed value $\sin x$ for $0<x<\pi$. Hence these continuous Dirichlet data admit no solution in $C^0(\overline{R})\cap C^2(R)$.

The missing hypothesis is uniform ellipticity up to the boundary. The principal coefficient matrix and its quadratic form areAlthough this is positive-definite at every interior point, no single positive uniform ellipticity constant works on $R$: taking $\xi =(0,1)$ requires $y^2\ge \lambda$ for arbitrarily small positive $y$. The normal second-derivative coefficient vanishes on the lower side. This degenerate ellipticity allows the equation and boundedness to force a boundary trace, instead of allowing arbitrary continuous Dirichlet data. The coefficient matrix is positive-definite at every interior point, but lacks the uniform lower bound used by the usual bounded-domain existence theorem.

With only $u\in L^2(\Omega )$ available, use a distributional weak solution: for every test function $\eta \in C_c^{\infty }(\Omega )$ requireAll terms are meaningful because the smooth coefficients are bounded on the compact support of the test function and $u$ is locally integrable. This is exactly $\Delta u={\sum }_i{D}_i(b^{i}u)+cu+f$ as an identity of distributions. No first weak derivative of $u$ is assumed in this definition, and no boundary condition is being imposed.

Fix $x$ with $\mathrm{dist}(x,\partial \Omega )>\sigma$ and use ${\eta }_x(y)={\varphi }_{\sigma }(x-y)$ as the test function in part (a). Its support lies compactly inside $\Omega$. Differentiating the convolution under the integral is allowed, and $D_{y_i}{\eta }_x=-D_{x_i}{\varphi }_{\sigma }(x-y)$ while ${\Delta }_y{\eta }_x={\Delta }_x{\varphi }_{\sigma }(x-y)$. ConsequentlyThusat every stated interior point. This is the convolution derivative identity. The convolutions are local, so no integrability of the smooth coefficients all the way to the boundary is needed. In particular, $(b^{i}u{)}_{\sigma }$ is the convolution of the product; it must not be replaced by $b^i{u}_{\sigma }$ for a variable coefficient.

Choose nested interior open sets $U\subset \subset V\subset \subset W\subset \subset \Omega$. For small $\sigma$, part (b) gives a smooth equation on $V$ of the formThe divergence-forcing interior H1 estimate isOne can obtain this estimate directly by testing the smooth equation against ${\eta }^2{u}_{\sigma }$ and applying Young inequality, with a cutoff function equal to one on $U$ and supported in $V$. The approximate identity and the $L^2$ convolution bound give, uniformly in $\sigma$,with analogous bounds for $(cu{)}_{\sigma }$ and $f_{\sigma }$. The coefficients need only be bounded on $W$. Therefore $u_{\sigma }$ is bounded in $H^1(U)$. It converges to $u$ in $L^2(U)$, and weak sequential compactness in a Hilbert space in this Sobolev space gives $u\in H^1(U)$. Since $U$ was arbitrary, $u\in H_{\mathrm{loc}}^1(\Omega )$.

Now expand the distributional derivative:Its right side belongs to $L_{\mathrm{loc}}^2$, so the interior $H^2$ elliptic regularity estimate gives $u\in H_{\mathrm{loc}}^2$. More generally, if $u\in H_{\mathrm{loc}}^m$ for an integer $m\ge 1$, multiplication by the smooth coefficients puts the right side in $H_{\mathrm{loc}}^{m-1}$. Interior elliptic regularity then gives $u\in H_{\mathrm{loc}}^{m+1}$. This elliptic regularity bootstrap proves $u\in H_{\mathrm{loc}}^k$ for every integer $k$.

For any nonnegative integer $\ell$, choose $k>\ell +n/2$. The Sobolev embedding theorem on compact interior subsets gives $u\in C_{\mathrm{loc}}^{\ell }$. Taking all $\ell$ proves $u\in C^{\infty }(\Omega )$, for its smooth representative. The first gain from $L^2$ to $H^1$ is the step supplied by the mollified equation; assuming $H^1$ in the initial definition would miss that step.