Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 (Elliptic Partial Differential Equations)

The operator is strictly elliptic when its symmetric principal matrix $A(x)=(a^{ij}(x))$ is positive definite at every point:It is a uniformly elliptic operator when some $\lambda >0$ satisfiesfor every $x$ and $\xi$. Here the least eigenvalue of the continuous matrix $A(x)$ is a positive continuous function on the compact set ${\overline{B}}_1$. It therefore has a positive minimum, which supplies $\lambda$ and proves uniform ellipticity.

A scaled Interpolation inequality in Holder spaces says that for $u\in C^{l,\alpha }(B_R(x_0))$, $0\le j<l$, and every $\delta >0$,where $C=C(n,l,\alpha )$. Corresponding interpolation between any two Hölder orders follows by applying this estimate to derivatives and rescaling the ball.

One useful form of the Simon absorption lemma is the following. Let $S$ be nonnegative and subadditive on balls contained in $B_R$, and suppose that for every such $B_r$,where $A$ is also subadditive and $\epsilon$ is smaller than a constant depending only on $n$ and $\gamma$. ThenMore generally, an additional controlled remainder on the right remains with a dimensional constant. The proof covers $B_{R/2}$ by finitely many smaller balls, sums by subadditivity, and iterates so that the $S$ term is absorbed.

The Interior Schauder estimate iswhere $C$ depends only on $n$, $\alpha$, the ellipticity constant, and the stated coefficient bounds.

It is enough first to prove the estimate for the Hölder seminorm of $D^{2}u$. The Interpolation inequality in Holder spaces controls $Du$ and $D^{2}u$ by a small multiple of $[D^{2}u{]}_{\alpha }$ plus $\Vert u{\Vert }_0$; the small term is later absorbed. Freezing $a^{ij}$ at the centre of each ball and moving coefficient differences and lower-order terms to the right reduces the local estimate toOnce this is known, the Simon absorption lemma gives the desired estimate on $B_{1/2}$.

For completeness, prove the frozen-coefficient estimate by contradiction. If it failed, choose solutions $u_k$, points $x_k$, and scales ${\rho }_k\downarrow 0$ at which the scale-invariant Hölder quotient is almost maximal. Let $q_k$ be the quadratic Taylor polynomial of $u_k$ at $x_k$ and defineThen $v_k(0)=D{v}_k(0)=D^2{v}_k(0)=0$, the Hessians have uniformly bounded local $C^{0,\alpha }$ seminorms, and the normalization makes their oscillation nonzero on a fixed ball. The localized equation iswhere the coefficient matrices converge locally uniformly to one constant positive-definite matrix and the given normalized error $g_k$ tends locally uniformly to zero.

The Arzela-Ascoli theorem produces a locally $C^2$ convergent subsequence with limit $v$ satisfying a constant-coefficient elliptic equation on ${\mathbb{R}}^n$. A linear rotation and scaling turn it into a harmonic function. The normalization gives $D^{2}v(0)=0$ but a nonconstant Hessian, while maximality of the scaled quotient gives growth at most $C(1+|y{|}^{2+\alpha })$. Applying the Liouville theorem to derivatives shows that every second derivative is constant because $0<\alpha <1$. This contradicts the normalized Hessian oscillation. The frozen estimate follows, and interpolation plus absorption completes the proof.

At $\alpha =0$, bounded Hessians have no uniform modulus of continuity, so the Arzela-Ascoli theorem does not supply the compact $C^2$ limit used in the contradiction. At $\alpha =1$, the blow-up has cubic growth, and the Liouville theorem no longer forces its Hessian to be constant: nonzero harmonic cubic polynomials are possible. These are the two endpoint failures behind the restriction $0<\alpha <1$; ordinary Schauder theory replaces neither endpoint by a general $C^{2,\alpha }$ estimate of the displayed form.

The Lagrangian isThe Euler-Lagrange equation for $X$ isExpanding the divergence and simplifying givesThe equation for $Y$ is

For compactly supported smooth variations $(\varphi ,\psi )$, differentiating at $t=0$ givesThis is the weak formulation. Taking $\psi =0$ or $\varphi =0$ separates its two equations. When $X,Y$ are smooth, integration by parts transfers each derivative from the test function; the fundamental lemma of the calculus of variations then recovers exactly the two pointwise equations in part (a).

Take a minimizing sequence $(X_k,Y_k)$ in the affine Sobolev class $W$. The standing bounds $C^{-1}\le X_k\le C$ make the energy uniformly equivalent toThe fixed boundary values and the Poincare inequality therefore bound the sequence in $H^1(\Omega ;{\mathbb{R}}^2)$. By weak compactness in a reflexive Banach space, a subsequence converges weakly to $(X,Y)$. The assumed weak closedness keeps the limit in $W$, and the assumed weak lower semicontinuity givesThus the direct method in the calculus of variations produces a minimizer. Its first variation vanishes in every compactly supported direction, so part (b) makes it a weak solution of the system.

Let $u\in H_{\mathrm{loc}}^1(\Omega )$ weakly solvewhere $a^{ij},F^i\in C^{0,\alpha }$, $g\in L^p$ for a suitable $p>n/(1-\alpha )$, and $a^{ij}$ is uniformly elliptic. The interior divergence-form Schauder estimate states that for ${\Omega }^{\prime }⋐{\Omega }_1⋐\Omega$,For the homogeneous equation the final two terms vanish.

Put $h=\log X$. The first equation becomeswhile the second remainsThe assumed $C^{0,\alpha }$ regularity and the bounds away from zero make $X^{-2}$ a $C^{0,\alpha }$ uniformly elliptic coefficient. The divergence-form $C^{1,\alpha }$ estimate from part (d) first gives $Y\in C_{\mathrm{loc}}^{1,\alpha }$. Hence the right side of the equation for $h$ is $C^{0,\alpha }$, and the Interior Schauder estimate gives $h$, and therefore $X$, in $C_{\mathrm{loc}}^{2,\alpha }$.

Expanding the $Y$ equation givesThe higher-order Schauder estimates now alternate between the equations for $h$ and $Y$, gaining derivatives at each step. Induction yields $(X,Y)\in C^{\infty }({\Omega }^{\prime };H)$ for every ${\Omega }^{\prime }⋐\Omega$.

The De Giorgi-Nash-Moser theorem gives numbers $\alpha =\alpha (n,\lambda ,\Lambda )\in (0,1)$ and $C=C(n,\lambda ,\Lambda ,\theta )$ such that every weak solution satisfiesfor each fixed $0<\theta <1$. In particular, every such weak solution has a locally Hölder-continuous representative despite the coefficients being only bounded and measurable.

The first variation of the graph area functional isThus the minimal surface equation for a graph isor, in nondivergence form,Differentiate the divergence equation with respect to $x_k$. Then $w=D_{k}u$ satisfieswhere

If $|Du|\le L$, the eigenvalues of $a^{ij}(Du)$ lie between $(1+L^2{)}^{-3/2}$ and $1$, so the differentiated equations are uniformly elliptic with bounded measurable coefficients depending only on $L$. Apply the De Giorgi-Nash-Moser theorem to each $D_{k}u$. Since $|D_{k}u|\le L$, for every $\theta <1$,for some $\alpha =\alpha (n,L)\in (0,1)$. The mapis smooth with bounded derivative on $|p|\le L$. Composition therefore gives

The function $u$ solves the linear nondivergence equationThe coefficient matrix is uniformly elliptic, and part (ii) makes it $C^{0,\alpha }$ on every smaller ball. Solve the corresponding linear Dirichlet problem on $B_r⋐B_1$ with boundary value $u$. Existence gives a $C^{2,\alpha }$ solution, and the weak maximum principle for elliptic operators gives uniqueness, so that solution is $u$. Hence $u\in C_{\mathrm{loc}}^{2,\alpha }(B_1)$.

The local boundedness estimate for uniformly elliptic equations first controls $\Vert u{\Vert }_{C^0(B_r)}$ by $\Vert u{\Vert }_{L^2(B_1)}$. Applying the Interior Schauder estimate and covering $B_{\theta }$ by finitely many balls gives

SetPart (b.iii), applied on each compactly contained ball, gives uniform $C^{2,\alpha }$ bounds for $v_k$. A diagonal use of the Arzela-Ascoli theorem therefore gives a subsequence converging in $C^2(K)$ on every compact $K⋐B_1$ to a $C^2$ function $w$.

The same estimate applied to $u_k$ gives $D{u}_k\to 0$ locally uniformly because $\Vert u_k{\Vert }_2\to 0$. Dividing the minimal-surface equation by $\Vert u_k{\Vert }_2$ shows thatThe coefficients converge locally uniformly to ${\delta }_{ij}$. Passing to the $C^2$ limit yieldsThus $w$ is harmonic and the required normalized subsequence converges to it locally in $C^2$.

Multiply $\Delta u+qu\le 0$ by ${\zeta }^2/u_{\epsilon }$ and integrate. Since $\zeta$ is compactly supported, integration by parts and Young's inequality giveAs $\epsilon \downarrow 0$, $u/u_{\epsilon }$ converges to the indicator of $\{u>0\}$. The dominated convergence theorem therefore gives

Put $v=\log u_{\epsilon }$, so $Dv=Du/u_{\epsilon }$. Choose a cutoff $\eta$ which equals one on $B_{\rho }(z)\cap B_1$, is supported in a comparable ball inside $B_2$, and satisfies $|D\eta |\le C/\rho$. Repeating the calculation from part (i) with ${\eta }^2$ and using $|q|\le \mu$ gives the Caccioppoli inequalityThe same construction works for balls meeting $\partial B_1$ because the cutoff is supported in $B_2$. Hence

For $v=\log u_{\epsilon }$, Cauchy-Schwarz inequality and part (a.ii) implyThe John-Nirenberg inequality therefore supplies $p_0=p_0(n,\mu )>0$ such that, with $v_{B_1}$ denoting the average,Since $e^{p_0(v-v_{B_1})}$ and $e^{-p_0(v-v_{B_1})}$ are each bounded by this integrand,

If $u=0$ on $\Sigma$ and $|\Sigma |>0$, thenPart (b.i) consequently givesLetting $\epsilon \downarrow 0$ and using Fatou lemma yields ${\int }_{B_1}{u}^{p_0}=0$. Since $u\ge 0$ is continuous,

Suppose a nonzero solution $w$ did not change sign. Replacing it by $-w$ if necessary gives $w\ge 0$. The argument of part (b.ii) is local and invariant under translation and scaling: it applies on every ball whose concentric double lies in $B_2$. If the zero set of $w$ had positive measure, a density point and this local result would make $w$ vanish on one ball. Applying the same result successively on overlapping balls would then give $w\equiv 0$ throughout the connected ball $B_2$, a contradiction. Thus the zero set has measure zero in $B_2$. Applying part (a.i) to $u=w$ givesfor every compactly supported $C^1$ function $\zeta$; the omitted zero set has measure zero. But $\inf q>{\lambda }_1$, while the variational characterization of the First Dirichlet eigenvalue provides a $\zeta$ withThis contradiction proves that every nonzero solution takes both positive and negative values.

Yes. Since $|q|\le \mu$ and $u\ge 0$,The operator $\Delta -\mu$ is uniformly elliptic and has nonpositive zeroth-order coefficient. If $u(y)=0$ at an interior point, $u$ attains its nonnegative minimum there, so the strong minimum principle for elliptic operators gives $u\equiv 0$ on the connected ball $B_1$.