Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 107 (Elliptic Partial Differential Equations)

The weak maximum principle for elliptic operators says that if $c\le 0$ andthenIndeed, a positive interior maximum has $Du=0$ and $\Delta u\le 0$, so the differential inequality is incompatible with a strict positive maximum. Applying this argument to a standard strictly perturbed function and then letting the perturbation tend to zero handles equality and proves the weak statement.

SetThen $v\ge A$ and, because $1\le e^{x_1+d}\le e^{2d}$ and $c\le 0$,Thus $(\Delta +c)(u-v)=f-(\Delta +c)v\ge 0$ and $u-v\le 0$ on the boundary. The weak maximum principle for elliptic operators gives $u\le v$. Applying the same argument to $-u$ gives $|u|\le v$, hence

If $u_1,u_2$ are two solutions with the same boundary values, put $w=u_1-u_2$. The mean value theorem givesTherefore $\Delta w-qw=0$ with zero boundary data. Apply the weak maximum principle for elliptic operators to $w$ and $-w$ with zeroth-order coefficient $-q\le 0$. It follows that $w=0$, proving uniqueness.

The required estimate follows from one-dimensional barriers on a sufficiently narrow slab. Put $\kappa =\sqrt{\tau }$. On $|x_1|<d_1<\pi /(2\kappa )$, the functionsatisfies $h^{\prime \prime }+\tau h=0$, equals one at $x_1=\pm d_1$, and obeysChoose $\delta (\epsilon ,\tau )$ so thatand so that the analogous solution of $k^{\prime \prime }+\tau k=-1$ with zero endpoint values is bounded by $e^{2d_1}-1$. Comparing $u$ and $-u$ withand using $c\le \tau$ givesThe permitted width is of order ${\tau }^{-1/2}$, so it can be chosen with $\delta (\epsilon ,\tau )\to \infty$ as $\tau \downarrow 0$.

Freeze the lower-order coefficients on a ball and apply the interior Schauder estimate for the Poisson equation to a cutoff of $u$. The terms $b_i{D}_{i}u+cu$ are controlled by the Hölder interpolation inequalityChoosing $\eta$ in terms of the prescribed $\delta$ yields

The standard iteration lemma for nested balls absorbs the first term. The remaining $C^2$ norm is bounded by the interior derivative estimate and interpolation, producingEquivalently, a contradiction-and-rescaling proof would produce a globally Hölder harmonic limit forbidden by the Polynomial-growth Liouville theorem for harmonic functions.

Fix $B_r⋐B_R⋐B$. Local uniform convergence bounds $|u_k{|}_{0;B_R}$ uniformly. Interior Schauder estimates therefore bound $u_k$ uniformly in $C^{2,\alpha }(B_r)$. The Arzela-Ascoli theorem gives a subsequence converging in $C^2(B_r)$. Its limit must be the locally uniform limit $v$, and passing to the limit in $L{u}_k=0$ gives $Lv=0$. A diagonal argument over $r\uparrow 1$ proves

Yes. The local $L^1$-to-$L^{\infty }$ estimate for solutions of a uniformly elliptic equation with bounded Hölder coefficients gives, whenever $B_r⋐B_R$,Thus $(u_k)$ is locally uniformly Cauchy and converges locally uniformly to a continuous representative of the $L^1$ limit $v$. Part (b) then applies, so $v\in C^2(B)$ and $Lv=0$.

No. Let ${\lambda }_1>0$ be the First Dirichlet eigenvalue of $-\Delta$ on $B$, choose a corresponding nonzero eigenfunction $w$, and takeThen $L{u}_k=0$ and every $u_k$ vanishes on $\partial B$, so the boundary values converge uniformly to those of $v=0$. But $|u_k{|}_{0;B}\to \infty$, and hence no subsequence converges uniformly on $B$.

The Alexandrov–Bakelman–Pucci maximum principle and the absence of a zeroth-order term giveThe global Schauder estimate isCombining the two bounds gives $|u{|}_{2,\alpha ;B}\le C|f{|}_{0,\alpha ;B}$.

Use the method of continuity withEvery $L_t$ has the same uniform ellipticity and Hölder coefficient bounds, so the estimates in part (a) hold with one constant independent of $t$. They imply injectivity. The set of $t$ for which $L_t:C_0^{2,\alpha }(B)\to C^{0,\alpha }(B)$ is onto is open by the bounded inverse theorem and a Neumann-series perturbation, and closed by the uniform a priori estimate. It contains $0$ because the Dirichlet Laplacian is bijective. Connectedness of $[0,1]$ therefore gives surjectivity at $t=1$, and $L$ is bijective.

Taylor's theorem and $g(0)=g^{\prime }(0)=0$ giveCombining these with the product estimate in a Hölder space yieldsWritingand using the same product estimate gives

Let $G=L^{-1}$ and use the bound from part (b),On the closed ball ${\mathcal{B}}_{\epsilon }\subset C_0^{2,\alpha }(B)$ defineThe Hölder product estimate and part (i) giveandChoose ${\epsilon }_0$ so that $2M{\epsilon }_0<1/2$, then choose ${\delta }_0$ so that the remaining terms make $T$ preserve ${\mathcal{B}}_{{\epsilon }_0}$ and have contraction constant less than one. The Banach fixed-point theorem gives a unique $u$ in that ball satisfying

The Weak Harnack inequality states that there are $p>0$ and $C=C(n,\lambda ,\Lambda )$ such that every nonnegative weak supersolution satisfiesThe radii may be replaced by any fixed nested pair of balls, with the constant adjusted accordingly.

On a ball $B_{2r}(x_0)⋐B$, bothare nonnegative solutions and hence supersolutions. Combining the Weak Harnack inequality with the local boundedness estimate for subsolutions gives an oscillation decay estimatewhere $\theta$ depends only on $n,\lambda ,\Lambda$. Iteration yields Hölder continuity with some exponent $\mu \in (0,1)$. The local $L^2$-to-$L^{\infty }$ estimate controls the initial oscillation and gives

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the supersolution inequality use the nonnegative test functionwhere $\eta$ is compactly supported. Ellipticity and Young inequality give the Logarithmic Caccioppoli inequalityChoose $\eta =1$ on $B_{\rho }$, supported in $B$, with $|D\eta |\le C(1-\rho {)}^{-1}$. Letting $\epsilon \downarrow 0$ and using Fatou lemma gives

NormalizeThen $0\le v_k\le 1$, and $v_k$ remains a weak supersolution. The ordinary Caccioppoli inequality with a cutoff supported in $B_R$ gives a uniform $W^{1,2}(B_{\rho })$ bound whenever $\rho <R<1$. The Rellich-Kondrachov compactness theorem and a diagonal subsequence therefore give, for every $\rho <1$,The strong convergence preserves nonnegativity. Passing to the limit in the linear supersolution inequality against every nonnegative compactly supported test function shows that $v$ is a weak supersolution in $B$.

Write $u=\phi +w$ with $w\in W_0^{1,2}(B)$. On this space defineBoundedness of the coefficients makes $\mathcal{B}$ bounded, and ellipticity giveswhich is coercive by the Poincare inequality. The functional $z\mapsto -\mathcal{B}(\phi ,z)$ is bounded, so the Lax-Milgram theorem gives a unique $w$ and hence a unique weak solution $u$.

The supplied global De Giorgi-Nash-Moser theorem and the weak maximum principle give $u\in C^{0,\beta }(\overline{B})$ for some $\beta =\beta (n,\gamma ,\Gamma )$. Finally, test the weak equation with $u{\psi }^2$. Ellipticity, the coefficient bound, and Young inequality yieldRearranging gives

The range of $v\in C^0(\overline{B})$ is compact. Continuity and positivity of $\lambda ,\Lambda$ therefore give constantsThe coefficients ${\alpha }_{ij}(x)=a_{ij}(x,v(x))$ meet the hypotheses of part (a), which gives a unique weak solution. Its global De Giorgi–Nash–Moser estimate gives $u\in C^{0,\mu }(\overline{B})$ for some $\mu =\mu (v)\in (0,1)$.

The Leray-Schauder fixed point theorem says that if $X$ is a Banach space and $T:X\to X$ is continuous and compact, and the setis bounded, then $T$ has a fixed point.

For $v\in C^0(\overline{B})$, let $T(v)$ be the solution from part (i). The weak maximum principle givesOn this fixed bounded range of values, $\lambda$ and $\Lambda$ have common positive lower and finite upper bounds. The global De Giorgi-Nash-Moser theorem therefore bounds $T(v)$ uniformly in one Hölder space. The compact embedding $C^{0,\mu }(\overline{B})\hookrightarrow C^0(\overline{B})$ makes $T$ compact. Uniform convergence $v_k\to v$, the continuity of $a_{ij}$ on the relevant compact set, energy bounds, and uniqueness of the limiting linear problem show that $T(v_k)\to T(v)$ uniformly, so $T$ is continuous.

The same maximum-principle bound controls every solution of the Leray–Schauder homotopy after using boundary data $t\phi$. Thus the homotopy set is bounded, and the Leray-Schauder fixed point theorem supplies $u=T(u)$. By definition,and $u\in W^{1,2}(B)\cap C^0(\overline{B})$.