Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 13 (Elliptic Partial Differential Equations)

Write ${\omega }_n=|B_1|$. If $u$ is a $C^2$ harmonic function and $\overline{B_r(x)}\subset \Omega$, its two mean value properties for harmonic functions areTo prove the spherical property, express the normalized spherical mean on the fixed unit sphere:Differentiation and the divergence theorem giveSince continuity gives $M(r)\to u(x)$ as $r\downarrow 0$, the spherical mean equals $u(x)$ for every admissible radius. Integrate this identity in radial coordinates:This proves the ball mean property as well. For $n=1$, the sphere consists of the two endpoints and the same argument is the fundamental theorem of calculus.

The strong maximum principle for harmonic functions states that a $C^2$ harmonic function on a connected domain which attains its supremum at an interior point is constant. Applying it to $-u$ gives the corresponding assertion for an attained infimum.

Let $M=u(x_0)={\sup }_{\Omega }u$. The ball mean value property for harmonic functions gives, for every sufficiently small ball about $x_0$,The integrand is continuous and nonnegative. If it were positive at any point, it would be bounded below by a positive number on a smaller ball, contradicting the zero integral. Therefore $u=M$ on the whole ball. The set $A=\{x\in \Omega :u(x)=M\}$ is relatively closed by continuity and relatively open by this same argument at each of its points. It is nonempty. Connected space forces $A=\Omega$, proving the principle. If the open set is disconnected, constancy is asserted on the relevant connected component.

First suppose $u$ is a classical harmonic function. At a local maximum of $u-v$, its Hessian matrix is negative semidefinite, so $0=\Delta u(x_0)\le \Delta v(x_0)$. At a local minimum the Hessian inequality reverses and gives $\Delta v(x_0)\le 0$. Thus both test conditions hold.

Conversely, suppose only that $u$ is continuous and satisfies the two conditions. We prove the stronger conclusion that these conditions themselves imply $u\in C^{\infty }$ and $\Delta u=0$. Use the standard Dirichlet existence theorem on a ball: every continuous function on its boundary has a harmonic extension $h\in C^{\infty }(B_R(a))\cap C(\overline{B_R(a)})$ with exactly those boundary values. This follows, for example, from the Poisson integral. Apply it to $u$ on an arbitrary relatively compact ball, obtaining a harmonic replacement $h$.

If $u-h$ is positive somewhere, choose $\epsilon >0$ small enough that $u-h-\epsilon (R^2-|x-a{|}^2)$ is still positive there. It vanishes on the boundary and therefore has a positive maximum at an interior point. The upper testhas $\Delta v=-2n\epsilon <0$, contradicting the maximum test condition. Similarly, if $u-h$ is negative somewhere, $v=h-\epsilon (R^2-|x-a{|}^2)$ yields an interior negative minimum of $u-v$ for small $\epsilon$, but $\Delta v=2n\epsilon >0$, contradicting the minimum condition. Hence $u=h$ throughout the ball.

If the quantifier requires $v\in C^2(\Omega )$ rather than a local test, multiply the displayed smooth $v$ by a smooth cutoff supported within the ball and equal to one near the touching point, extending by zero outside. The resulting global test has the same germ and Laplacian there, so the argument remains valid. Since every point lies in such a ball, $\boxed{u\text{ is a smooth harmonic function throughout }\Omega }$. This is the continuous viscosity harmonic functions are classical characterization.

The space $W_0^{1,2}(\Omega )$ is the closure of $C_c^{\infty }(\Omega )$ in the Sobolev norm; membership imposes the homogeneous Dirichlet condition in the weak sense, without assuming that the boundary is smooth. A weak solution is an element of that space satisfyingRepeated indices are summed. The minus sign comes from integration by parts in the divergence equation. Boundedness of the coefficients, Hölder's inequality and density extend this identity to every $\varphi \in W_0^{1,2}(\Omega )$. Symmetry permits interchanging the derivative indices in the left side.

Choose an upper ellipticity bound $\Lambda <\infty$ for the symmetric coefficient matrix, possible because its entries are bounded. ThenThe Poincaré inequality on a bounded open set for zero-boundary functions states $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$. ConsequentlyThe symmetric positive bilinear form $Q$ is therefore an inner product, and its square-root norm is equivalent to the usual Sobolev norm. Its triangle inequality follows from the Cauchy-Schwarz inequality for that inner product. completeness of $W_0^{1,2}$ in its original norm makes it a Hilbert space in the $Q$ norm as well.

The linear functional $\ell (v)=-\int fv$ is bounded, sinceThe Hilbert-space Riesz representation theorem states that every bounded linear functional is the inner product with a unique element. Applied in the $Q$ inner product it gives a unique $u$ with $Q(u,v)=\ell (v)$ for all $v$, which is precisely the weak-solution identity. Alternatively uniqueness follows directly by testing the difference of two solutions with itself: $Q(u_1-u_2,u_1-u_2)=0$. Thus $\boxed{\text{the zero-boundary problem has exactly one weak solution}}$.

The required functional, with the sign appropriate to a positive divergence operator, isIt is differentiable on $W_0^{1,2}$, with $D\mathcal{F}(u)[v]=Q(u,v)+\int fv$. A critical point consequently satisfies the weak equation. We now prove existence by the direct method in the calculus of variations, rather than invoke the representation theorem again.

The ellipticity and Poincare estimates implyThus the infimum is finite and any minimizing sequence is bounded in the Sobolev norm. A bounded sequence in a Hilbert space has a weakly convergent subsequence; write $u_j\rightharpoonup u$ in $W_0^{1,2}$. The equivalent $Q$ norm has the same continuous linear functionals and hence the same weak convergence. A Hilbert norm is Weak lower semicontinuity, giving $Q(u,u)\le {\mathrm{liminf}}_{j}Q(u_j,u_j)$. The term $\int f{u}_j$ converges to $\int fu$ because it is a bounded linear functional. ThereforeThe limit attains the infimum. Differentiating $t\mapsto \mathcal{F}(u+tv)$ at its minimum, for every $v$, gives $D\mathcal{F}(u)[v]=0$. This produces the desired weak solution. Strict positivity of $Q$ also makes the functional strictly convex, consistent with uniqueness.

For every $v\in L^2(\Omega )$, the existence result in Question 2 applies to the identity coefficient matrix: $a_{ij}={\delta }_{ij}$ is bounded, symmetric and uniformly elliptic. It gives a unique $u\in W_0^{1,2}(\Omega )$ satisfyingThus every input has exactly one output and $\boxed{T\text{ is well defined}}$. The weak Dirichlet theorem used here requires only a bounded open set; no smoothness of its boundary is needed.

If $u_1=T{v}_1$ and $u_2=T{v}_2$, linearity of the weak equation makes $\alpha u_1+\beta u_2$ solve the equation with forcing $\alpha v_1+\beta v_2$. Uniqueness therefore gives $T(\alpha v_1+\beta v_2)=\alpha T{v}_1+\beta T{v}_2$.

Test the equation for $u=Tv$ with $u$ itself. The Poincaré inequality yieldsIf the gradient norm is nonzero, divide by it; if it is zero, Poincare gives $u=0$ and the estimate is automatic. In either caseHence $T:L^2(\Omega )\to W_0^{1,2}(\Omega )$ is a bounded linear operator. Notice that the energy identity has a minus sign, so $⟨Tv,v⟩=-\Vert D(Tv){\Vert }_2^2\le 0$. Testing the equations for $Tv$ and $Tz$ against each other gives $⟨Tv,z⟩=-\int D(Tv)\cdot D(Tz)=⟨v,Tz⟩$. Thus the bounded Dirichlet inverse is a negative self-adjoint operator on $L^2$.

For finite $p>n$, chooseSince $n\ge 3$ and $p>n$, we have $2<q<2^{\ast }=2n/(n-2)$. The zero-boundary versions of the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem say that $W_0^{1,2}(\Omega )$ embeds continuously into $L^{2^{\ast }}$ and compactly into $L^q$ for $q<2^{\ast }$ on a bounded open set. These versions hold on arbitrary bounded open sets by zero extension; a boundary extension operator is not needed.

If $(v_j)$ is bounded in $L^2$, then $(T{v}_j)$ is bounded in $W_0^{1,2}$ by the preceding part. A subsequence consequently converges in $L^q$ to some $z$. Hölder's inequality givesMultiplication by $w$ is bounded from $L^q$ to $L^2$, and the same estimate proves that $K$ is bounded and well defined. Every bounded input sequence has a subsequence whose images converge in $L^2$, so $\boxed{K\text{ is compact}}$. If $p=\infty$ is included, take $q=2$ and use the same argument. This is the compact L2 potential perturbation of the Dirichlet Laplacian mechanism.

First, multiplication by $w$ maps $W_0^{1,2}$ into $L^2$ by the Hölder and Sobolev estimates just given. Thus a weak solution $u$ of the perturbed equation has $v=\Delta u=f+wu\in L^2$. Uniqueness for the unperturbed Dirichlet problem gives $u=Tv$. Its equation becomesConversely, if this equation has a solution $v\in L^2$, then $u=Tv$ belongs to $W_0^{1,2}$ and satisfies $\Delta u=v=f+wTv=f+wu$ weakly. These constructions are inverse to one another.

The Fredholm alternative for a compact operator states that, on a Banach space, $I-K$ for compact $K$ is bijective if and only if its kernel is zero; in that case its inverse is bounded. Apply it on $L^2$. Kernels correspond under $T$: a nonzero $v$ with $v=wTv$ gives a nonzero $u=Tv$, since $\Delta Tv=v$; a nonzero homogeneous solution $u$ gives $v=wu$, which cannot be zero because a zero-boundary harmonic weak solution is zero. Therefore triviality of the homogeneous PDE kernel is equivalent to triviality of $\ker (I-K)$ and hence to unique solvability for every $f\in L^2$. The reverse implication also follows immediately by taking $f=0$. No sign assumption on $w$ is required.

With the convention $Lu\ge 0$ for a classical subsolution, the weak maximum principle for elliptic operators isHere $L$ is uniformly elliptic with bounded drift, has no zeroth-order term, and the domain is bounded. Equivalently, $Lu\le 0$ gives ${\inf }_{\Omega }u\ge {\min }_{\partial \Omega }u$.

One can see why measurable coefficients suffice for the pointwise classical statement by adding $\epsilon e^{\kappa x_1}$. Choose $\kappa$ large enough that $\lambda \kappa >\Vert b_1{\Vert }_{\infty }$; then $L(e^{\kappa x_1})=e^{\kappa x_1}({\kappa }^2{a}_{11}+\kappa b_1)>0$. The perturbed subsolution cannot attain a maximum inside: its gradient would be zero and its Hessian negative semidefinite, forcing its elliptic second-order term to be nonpositive. Its maximum is therefore on the boundary. Letting $\epsilon \downarrow 0$ proves the stated estimate. Only the symmetric part of $a$ acts on the Hessian, and ellipticity gives its positive definiteness.

Put $F={\sup }_{\Omega }{f}^{+}\ge 0$ and $m={\inf }_{\partial \Omega }u$, both finite by continuity on the compact closure. Set $z=Fv+m-u$. Constants are annihilated by $L$, soOn the boundary, $z\le m-u\le 0$, using $v\le 0$ and $F\ge 0$. The weak maximum principle for elliptic operators therefore gives $z\le 0$ inside as well. Rearrangement provesThe nonnegative multiplier is essential when multiplying the barrier inequality.

Apply the preceding result to $L=\Delta$, $u=w$ and $f=\Delta w$. Define the barrierThen $\Delta v=1$. Since $0\in \Omega$ and $d=\mathrm{diam}\Omega$, $|x|=|x-0|\le d$ on $\overline{\Omega }$, so $v\le 0$ on the boundary. All hypotheses of the barrier comparison hold, givingThe assumption that the origin belongs to the domain is what validates this particular choice of centered quadratic barrier.

Recall that $g_i\in L_{\mathrm{loc}}^1(\Omega )$ is the weak derivative $D_{i}u$ if $u\in L_{\mathrm{loc}}^1(\Omega )$ and $\int u{D}_i\varphi =-\int g_i\varphi$ for all $\varphi \in C_c^{\infty }(\Omega )$. A global weak derivative restricts to every open neighborhood simply by taking test functions supported there.

Conversely, suppose local weak derivatives exist near each point. local integrability of $u$ follows by a finite cover of each compact subset. On overlapping neighborhoods the two candidates for $D_{i}u$ have identical integrals against every smooth compactly supported test function. The fundamental lemma for locally integrable functions therefore makes them equal almost everywhere. Choose a countable local cover, possible in Euclidean space; the countably many exceptional overlap sets can all be discarded. The derivatives consequently patch to measurable $g_i\in L_{\mathrm{loc}}^1(\Omega )$.

For a test function $\varphi$, take a smooth partition of unity $({\chi }_j)$ subordinate to the local cover and locally finite on its support. Only finitely many terms occur there. Because ${\sum }_j{\chi }_j=1$ near the support,The derivatives of the partition sum to zero, explaining the first equality. Thus the patched functions are the global weak derivatives. This proves both directions of locality of weak differentiability.

By locality of weak differentiability, it is enough to work near the exceptional point; translate it to zero. Let $g_i$ denote the classical derivative off zero, assigning any value at zero. For $\varphi \in C_c^{\infty }(\Omega )$ choose a smooth cutoff ${\chi }_{\epsilon }$ which is zero on $B_{\epsilon }$, one outside $B_{2\epsilon }$, satisfies $|D{\chi }_{\epsilon }|\le C/\epsilon$, and has its transition region contained in $\Omega$. The product $\varphi {\chi }_{\epsilon }$ avoids the singular point, so classical integration by parts givesThe middle term has absolute value at mostwhen $n\ge 2$. The other terms tend to $\int u{D}_i\varphi$ and $-\int g_i\varphi$, since $u$ is locally integrable and $g_i\in L_{\mathrm{loc}}^1$. Therefore $\boxed{D_{i}u=g_i\text{ weakly on all of }\Omega }$. Its value at the single point is irrelevant. This is bounded isolated-point removability for weak first derivatives.

For $n=1$, take $u(x)=0$ for $x<0$ and $u(x)=1$ for $x>0$ on $(-1,1)$, with any value at zero. It is bounded and $C^1$ off zero, with classical derivative zero there. Butso its distributional derivative is a Dirac delta. If a locally integrable weak derivative existed, it would be zero off zero and hence zero almost everywhere, contradicting this identity. The step function is the required counterexample; the cutoff error in dimension one need not vanish.

Work first on $B=B_1$, and let $m=|B{|}^{-1}{\int }_{B}u$. Put $Z=\{u=0\}$. Since $|Z|\ge \theta |B|$,Also ${\int }_B{u}^2={\int }_B|u-m{|}^2+|B|m^2$, soThe Poincare-Wirtinger inequality now bounds the last integral by $C_n{\int }_B|Du{|}^2$. Here is a direct proof on the ball. For smooth $u$, the variance identity and the fundamental theorem along line segments giveFor $0\le t\le 1/2$, fix $y$ and change variables $z=(1-t)x+ty$. Convexity keeps $z$ inside $B$, and the inverse Jacobian is at most $2^n$. Thus the double integral of this gradient term is at most $2^n|B|{\int }_B|Du{|}^2$. For $1/2\le t\le 1$, fix $x$ instead and obtain the same bound. Integration over $t$ proves ${\int }_B|u-m{|}^2\le 2^{n+1}{\int }_B|Du{|}^2$. Density of smooth functions in $W^{1,2}(B)$ passes the estimate to all Sobolev functions.

For a ball centered at $a$ with radius $R$, set $v(y)=u(a+Ry)$. Its zero set has at least the same volume fraction $\theta$, while ${\int }_{B_1}{v}^2=R^{-n}{\int }_{B_R}{u}^2$ and ${\int }_{B_1}|Dv{|}^2=R^{2-n}{\int }_{B_R}|Du{|}^2$. ThereforeThis establishes the Poincare inequality with a positive-measure zero set, with an explicit admissible constant. If $\theta =1$, the function is already zero almost everywhere.

For smooth $u$, the fundamental theorem along the segment in direction $e_k$ gives, for either sign of $h$,The entire segment lies in $\Omega$ under the distance hypothesis. The Cauchy-Schwarz inequality on $[0,1]$, followed by integration and translation, yieldsFor general $u\in W^{1,2}(\Omega )$, extend $u$ and its weak gradient separately by zero as $L^2$ functions on ${\mathbb{R}}^n$, and mollify $u$ with radius $\epsilon$. At points farther than $\epsilon$ from the boundary the mollified gradient equals the convolution of the zero-extended gradient, because the convolution kernel is supported wholly inside $\Omega$. Choose $\epsilon <\mathrm{dist}({\Omega }^{\prime },\partial \Omega )-|h|$. All the segments then lie in this region, and the preceding calculation, followed by the $L^2$ contraction bound for convolution, gives the same right side $\Vert D_{k}u{\Vert }_{L^2(\Omega )}$. The mollifications converge locally in $L^2$, as do their fixed-$h$ translated differences, so passage to the limit preserves the estimate. HenceThe argument proves the sharper estimate with $D_{k}u$ on the right. It is the basic Sobolev difference quotient bound.

Put $\delta =\mathrm{dist}({\Omega }^{\prime },\partial \Omega )>0$. Choose a smooth cutoff $0\le \eta \le 1$, equal to one on ${\Omega }^{\prime }$, supported within its $\delta /2$ neighborhood, with $|D\eta |\le C_n/\delta$. Such a cutoff is obtained by smoothing a distance cutoff; its support is relatively compact in $\Omega$. Let $0<|h|<\delta /8$, fix a coordinate $k$, and put $z={\Delta }_k^{h}u$. It lies in $W^{1,2}$ on the cutoff neighborhood because both translated copies of $u$ do. Extend ${\eta }^{2}z$ by zero, and use the test functionThe weak equation is $\int Du\cdot D\varphi =-\int f\varphi$. The discrete integration-by-parts identity $\int a{\Delta }_k^{-h}b=-\int ({\Delta }_k^{h}a)b$, valid by translation with these supports, turns it intoWrite $X=\Vert \eta Dz{\Vert }_2$, $U=\Vert z{\Vert }_{L^2(\mathrm{supp}\eta )}$, $A=\Vert D\eta {\Vert }_{\infty }$ and $F=\Vert f{\Vert }_2$. Expanding the left side gives $X^2$ plus a cross term of absolute value at most $2AUX$. Applying part (a) on the whole space to the zero extension of ${\eta }^{2}z$ bounds the right side byConsequentlywhere the last step is the elementary Young inequality. Part (a), on a slightly larger cutoff neighborhood if needed, gives $U\le \Vert D_{k}u{\Vert }_{L^2(\Omega )}$. We conclude thatuniformly as $h\to 0$.

To obtain actual second weak derivatives, choose $h_j\to 0$. For each $i,k$, the bounded sequence ${\Delta }_k^{h_j}{D}_{i}u$ has a weakly convergent subsequence in $L^2({\Omega }^{\prime })$, say to $g_{ik}$. For every smooth compactly supported $\psi$ in ${\Omega }^{\prime }$,Thus $g_{ik}=D_k{D}_{i}u$ is the second weak derivative. weak lower semicontinuity gives the same uniform bound for these derivatives. Adding the already known zeroth and first derivative norms and summing over the finitely many coordinates provesAll derivative existence and estimates have been established by the difference-quotient argument; no second derivatives were assumed in choosing the test. This proves the Interior second-derivative estimate for the Poisson equation in the stated setting.