Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 2

A locally integrable function $u$ has weak derivative $g_i$ in direction $i$ when $g_i$ is locally integrable andThe test functions have compact support inside the domain, so no boundary values enter this definition. Such $g_i$ is unique almost everywhere and is denoted ${\partial }_{i}u$.

For $1\le p<\infty$, the first-order Sobolev space consists of the Lp space classes whose first weak derivatives are also in that Lp space:This Sobolev norm is equivalent to $\Vert u{\Vert }_p+\Vert \nabla u{\Vert }_p$; either convention is acceptable. When $p=2$, this is the Hilbert space $H^1(B_1)$.

For $\psi \in C_c^{\infty }(B_1)$, the functions $\psi$ and ${\partial }_i\psi$ are bounded. Hence $\psi u$ and $\psi {\partial }_{i}u+u{\partial }_i\psi$ belong to $L^p(B_1)$. For a test function $\varphi$, the classical product rule and the weak derivative identity for $u$, tested with $\psi \varphi$, giveThereforeIn fact $\psi u$ has compact support in $B_1$ and belongs to the corresponding zero-boundary Sobolev space, by mollification within that support.

For arbitrary $\psi \in C^{\infty }(B_1)$, the same identity still proves the local product rule for weak derivatives, but membership in the global Sobolev space can fail. To give a counterexample for every dimension and every $p$ in the question, takeHere $u\in W^{1,p}(B_1)$ and $\psi$ is a smooth function on the open ball, while polar coordinates showThus $\psi u\notin L^p(B_1)$, and the global assertion is false. Boundedness of $\psi$ and its first derivatives would suffice without compact support.

Interpret the paper's unspecified $\Omega$ as containing $B_1$, the domain on which the coefficients are used. Write $A(x)=(a_{ij}(x))$, a real symmetric matrix. The strict ellipticity condition is pointwise positive definiteness:The lower bound may depend on $x$. By continuity, strict ellipticity gives a positive lower bound on each compact subset of $B_1$, but does not require one bound throughout the whole open ball.

With the two-constant convention used in the next part, uniform ellipticity meansBoth constants are independent of $x$. Some terminology uses uniform ellipticity for just the lower bound and states coefficient boundedness separately; here the paper supplies both constants. The spectral theorem for the symmetric matrix gives $|A(x)\xi |\le \Lambda |\xi |$. For example, $A(x)=(1-|x{|}^2)I$ satisfies strict ellipticity in $B_1$ but has no uniform positive lower bound as $|x|\uparrow 1$.

There is a small incompatibility in the printed cutoff function assumptions. No $\phi \in C_c^{\infty }(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla \phi |\le 2$: for a unit vector $e$, choose $t<1$ beyond its support. Continuity gives $\phi (e/2)=1$, while the fundamental theorem of calculus would give $1\le 2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a cutoff function exists, and the requested Caccioppoli inequality is unchanged apart from its constant. We keep $M=\Vert \nabla \phi {\Vert }_{\infty }$ in the calculation, which also yields the formal bound under the printed assumptions.

The weak formulation is ${\int }_{B_1}A\nabla u\cdot \nabla \zeta =0$ for test functions $\zeta$. Since $A$ is bounded, this extends by density of smooth functions in a Sobolev space to every $\zeta \in H_0^1(B_1)$. Part (a)(ii), followed by mollification inside the compact support, makes ${\phi }^2(u-k)$ an admissible choice. The product rule givesPut $X=\int {\phi }^2|\nabla u{|}^2$ and $Y=\int (u-k{)}^2|\nabla \phi {|}^2$. The lower and upper uniform ellipticity bounds, together with Cauchy-Schwarz inequality, implyIf $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\le 4{\Lambda }^2{\lambda }^{-2}Y$. Since $\phi =1$ on $B_{1/2}$,For a fixed valid cutoff function, $M\le C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16{\Lambda }^2/{\lambda }^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.

First repeat the Caccioppoli inequality calculation on two concentric balls. If $h$ is a harmonic function on $B_R$, choose a cutoff function equal to one on $B_r$, supported in $B_R$, and with gradient bounded by $C/(R-r)$. The argument of (ii), with $A=I$ and $k=0$, givesHere $C_{\ast }$ depends at most on the dimension. When $h$ is only known to be smooth inside $B_R$ and in $L^2(B_R)$, the weak formulation and all integrals involving the cutoff function are still legitimate, since its support is strictly inside $B_R$.

Every classical partial derivative of the given smooth harmonic function $u$ is again a harmonic function, since constant-coefficient partial derivatives commute with the Laplacian. Choose indices $i_1,\dots ,i_m$ representing the multi-index $\alpha$, and defineFor each $j\ge 1$, the preceding Caccioppoli inequality yieldsStarting with $u\in L^2(B_1)$, these estimates also establish the integrability needed at each subsequent step. Iterating $m$ times givesThis is the desired interior derivative estimate for a harmonic function, with a constant depending only on $n$ and $m$, uniformly over every multi-index of size $m$.

Set $w=(u-k{)}^{+}$, the positive part of $u-k$. The allowed Sobolev chain rule givesUse the admissible zero-boundary Sobolev space function ${\phi }^{2}w$ in the weak formulation, with a valid cutoff function as explained in (ii). Its gradient is ${\phi }^2\nabla w+2\phi w\nabla \phi$, soIndeed the integrands vanish off $\{u>k\}$, and on that set $\nabla w=\nabla u$. The uniform ellipticity and Cauchy-Schwarz inequality estimates from (ii) now apply to $w$ and giveFixing the cutoff function gives the required constant independent of $k$. This is a superlevel Caccioppoli inequality; it uses the positive part as a test and does not require the truncated function itself to solve the original equation.