Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 (C4.3, A14690W1)

Original solutions to all three questions of the functional analytic methods for partial differential equations course of the University of Oxford Part C examination, held on 2 June 2023. Source: released parent PDF, bundle pages 339–342; paper pages 1–4. The native PDF confirms three substantive printing issues: the cutoff function bound in 2(b)(ii), the missing boundary restriction in 3(a)(ii), and the Sobolev space used for the asserted convergence in 3(b)(iv). Each is addressed explicitly at the relevant leaf.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^3)$ with $\int \rho =1$, and put ${\rho }_{\epsilon }(x)={\epsilon }^{-3}\rho (x/\epsilon )$. Sobolev approximation by mollifiers in the Sobolev space under consideration statesIn particular, it gives density of smooth functions in a Sobolev space. We prove convergence of both the function and all its first weak derivatives.

The usual smoothing properties of convolution give $u_{\epsilon }\in C^{\infty }$. For a test function $\varphi \in C_c^{\infty }({\mathbb{R}}^3)$ and $i\in \{1,2,3\}$, Fubini's theorem and the definition of a weak derivative giveThe integrals are absolutely convergent: the compact supports of ${\rho }_{\epsilon }$ and $\varphi$ restrict $y$ to a compact set, on which $u$ and ${\partial }_{i}u$ are integrable by Holder inequality. ConsequentlyThe approximate identity theorem in the Lp space $L^6$ applies separately to $u$ and ${\partial }_{i}u$. ThusEach convolution belongs to $L^6$ by Young's convolution inequality. With the componentwise Sobolev norm, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent Sobolev norm gives the same convergence.

Here Morrey's inequality has Hölder exponent $1-3/6=1/2$. Every $u\in W^{1,6}({\mathbb{R}}^3)$ has a unique continuous representative $u^{\ast }$ such thatWe use the Hölder norm $\Vert f{\Vert }_{C^{0,1/2}}=\Vert f{\Vert }_{\infty }+[f{]}_{C^{0,1/2}}$, where the Hölder seminorm is ${\sup }_{x\ne y}|f(x)-f(y)|/|x-y{|}^{1/2}$. The second bound is the homogeneous form of Morrey's inequality; the first includes control of the values themselves.

Let $u_j={\rho }_{1/j}\ast u$ be the mollifications from (i). They converge to $u$ in the Sobolev norm. Apply the assumed smooth version of Morrey's inequality to the differences $u_j-u_{\ell }$:Thus $(u_j)$ is a Cauchy sequence in the complete Hölder space $C^{0,1/2}({\mathbb{R}}^3)$ and converges there to a function $w$. In particular the convergence is uniform. On every bounded set $K$ of finite volume,Exhausting ${\mathbb{R}}^3$ by balls proves $w=u$ almost everywhere. Set $u^{\ast }=w$. Taking limits in the smooth Hölder norm bound gives the first boxed estimate. For any fixed $x\ne y$, taking limits ingives the second boxed estimate after taking the supremum. Finally, two continuous representatives equal almost everywhere must coincide everywhere, since a nonzero continuous difference would remain nonzero on an open ball of positive volume.

On the unbounded domain ${\mathbb{R}}^3$, translations already prevent compactness in every indicated Hölder space. Choose a bump function $\eta \in C_c^{\infty }(B_1)$ with $\eta (0)=1$, put $e_1=(1,0,0)$, and defineA change of variables in the Lebesgue integrals defining the Sobolev norm gives $\Vert u_m{\Vert }_{W^{1,6}({\mathbb{R}}^3)}=\Vert \eta {\Vert }_{W^{1,6}({\mathbb{R}}^3)}$, so the sequence is bounded. Each $u_m$ is a smooth function of compact support and belongs to every Hölder space $C^{0,\gamma }({\mathbb{R}}^3)$, $0<\gamma \le 1$: its bounded gradient gives the Lipschitz estimate for distances at most one, and its bounded values control larger distances.

For $m\ne \ell$, the supports are disjoint, and at $x=3m{e}_1$ we have $u_m(x)=1$ and $u_{\ell }(x)=0$. ThereforeNo subsequence can be a Cauchy sequence in any of these Hölder norms, hence none converges. This uses the translation mechanism for noncompactness of a Sobolev embedding by translation; concentration at the critical Hölder exponent alone would not rule out convergence at every smaller Hölder exponent.

Write $r=|x|$, $u(x)=\bar{u}(r)$, and let $g(r)$ be the radial profile of the radial derivative $\nabla u(x)\cdot x/|x|$, which may be assumed to be a radial function here. The spherical coordinate system givesSince $1\le r^2\le 4$, both $\bar{u}$ and $g$ belong to the Lp space $L^6(1,2)$. It remains to identify the weak derivative of the radial profile directly.

For $\varphi \in C_c^{\infty }(1,2)$, each component of $\varphi (r)x/r$ is a test function on $A$. Sum the three defining weak derivative identities, using the stated divergence formula:Given an arbitrary test function $\zeta \in C_c^{\infty }(1,2)$, choose $\varphi (r)=\zeta (r)/r^2$. Dividing by $4\pi$ gives $\int g\zeta =-\int \bar{u}{\zeta }^{\prime }$, exactly the definition of the weak derivative. We have provedThis radial Sobolev profile on an annulus estimate follows from the two weighted integrals above and equivalence of the usual Sobolev norms.

Use the continuous representative of $\bar{u}$ furnished by the one-dimensional Sobolev embedding theorem, and define $u^{\ast }(x)=\bar{u}(|x|)$ on $A$. The one-dimensional Sobolev representative is absolutely continuous; Holder inequality applied to its fundamental theorem of calculus givesAlso $\Vert \bar{u}{\Vert }_{\infty }\le \Vert \bar{u}{\Vert }_6+\Vert {\bar{u}}^{\prime }{\Vert }_6$: select a point where $|\bar{u}|$ is no larger than its $L^6$ norm on the unit-length interval and apply the same fundamental theorem of calculus. Thus the one-dimensional Hölder norm is controlled by $\Vert \bar{u}{\Vert }_{W^{1,6}(1,2)}$.

The two Hölder seminorms must be compared using the correct distances. By the reverse triangle inequality,Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two Hölder seminorms. Their supremum norms are equal as well. Part (i) now givesThe representative agrees with $u$ almost everywhere, and the assignment is linear and injective on the radial Sobolev space. This proves the requested continuous embedding. The inner radius being positive makes the radial profile Sobolev norm comparable to its weighted radial version.

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.

The zero Dirichlet boundary condition must be part of the definition. Thus $u\in H^1(B)$ is a weak solution precisely when $u\in H_0^1(B)$, equivalently its Sobolev trace vanishes on $\partial B$, andHere ${\chi }_E$ is the indicator function of the measurable set $E$. One may initially require the identity only for test functions $\varphi \in C_c^{\infty }(B)$: both sides are continuous in the Sobolev norm, and the definition of the zero-trace Sobolev space then extends it to all $H_0^1(B)$. This formulation uses first weak derivatives and encodes $-\Delta u={\chi }_E$ without assuming classical second derivatives.

As printed, the answer isThe constant function $u\equiv 1$ has zero gradient and positive Lp norm for every such $q$, including its supremum norm when $q=\infty$.

The missing assumption is a zero Sobolev trace, which is present for the weak solutions in the surrounding parts. With $H_0^1(B)$ in place of $H^1(B)$, the intended answer isThis is the three-dimensional Sobolev inequality at $q=6$, followed by Holder inequality on the bounded ball for smaller $q$; it includes the zero-boundary Poincaré inequality at $q=2$. The subsequent parts use this zero-boundary version.

By the definition in (i), $u$ itself is an admissible test in the weak formulation. HenceThe absolute value makes a proof of positivity unnecessary. Holder inequality with exponents $6$ and $6/5$, followed by the zero-boundary Sobolev inequality, givesTaking square roots of both estimates proves exactly the requested chain:If $\Vert \nabla u{\Vert }_2=0$, the zero-boundary Poincaré inequality gives $u=0$ and the conclusion follows. Otherwise divide the first-to-last estimate by $\Vert \nabla u{\Vert }_2^{1/2}$ and square, obtaining $\Vert \nabla u{\Vert }_2\le C_S|E{|}^{5/6}$. Finally the same Poincaré inequality givesAll constants depend only on the fixed unit ball and the dimension, and are independent of $E$ and $u$. If $|E|=0$, testing the homogeneous equation with $u$ gives the same zero solution directly.

Let $w_m=|E_m{|}^{5/6}{u}_m$. By linearity of the weak formulation, $w_m$ is the zero-boundary weak solution with forcing ${\chi }_{E_m}$. The estimate in (a)(iii) therefore givesThe change of variables $x=r_{m}y$ gives $|E_m|=r_m^3|E|>0$, so the division is legitimate. Since $H^1(B)$ is a Hilbert space, weak sequential compactness in a Hilbert space now gives a subsequence and some $u\in H^1(B)$ such thatIn fact $u\in H_0^1(B)$: the Sobolev trace operator is continuous and linear, so its zero kernel is preserved by weak convergence.

The Lp norm of the forcing at the critical exponent is constant:To prove weak convergence in $L^{6/5}(B)$, the duality of Lp spaces says that we must test against an arbitrary $g\in L^6(B)$. Holder inequality, restricted to the shrinking measurable set $E_m$, givesThe last step is absolute continuity of the Lebesgue integral, since $|E_m|=r_m^3|E|\to 0$ and $|g{|}^6$ is integrable. ThusTesting only bounded or smooth functions would not by itself establish this weak convergence; the restricted Holder inequality handles every member of the full dual space.

Take the weakly convergent subsequence in (i), with limit $u\in H_0^1(B)$. For each $\varphi \in H_0^1(B)$, the weak formulation givesThe left side tends to $\int \nabla u\cdot \nabla \varphi$ by weak convergence in $H^1$. The zero-boundary Sobolev inequality gives $\varphi \in L^6(B)$, so (ii) makes the right side tend to zero. Taking $\varphi =u$ in the limiting identity yields ${\int }_B|\nabla u{|}^2=0$. The zero-boundary Poincaré inequality then gives $u=0$.

The same argument applies to a weakly convergent subsequence of any subsequence of $(u_m)$, because all remain bounded as in (i). To see why this proves convergence of the full sequence, suppose a continuous linear functional $F$ on $H^1(B)$ did not satisfy $F(u_m)\to 0$. Some subsequence would have $|F(u_{m_j})|\ge \epsilon >0$. Its boundedness gives a further weakly convergent subsequence, whose limit has just been proved to be zero; then $F$ tends to zero along it, a contradiction. ThereforeUniqueness of a possible limit alone is not enough without the weak sequential compactness in a Hilbert space used to extract a further subsequence.

The zero extension of H01 makes $v_m$ an element of $H^1({\mathbb{R}}^3)$. The chain rule and change of variables give the exact scaling identitiesIndeed $\nabla v_m(x)=r_m^{3/2}\nabla u_m(r_{m}x)$ inside the expanded ball, and there is no boundary jump because the Sobolev trace is zero.

First take the printed convergence assertion as an assumption, as the question requests. It would imply $\Vert \nabla v_m{\Vert }_2\to \Vert \nabla v{\Vert }_2>0$. The strict inequality follows because an $H^1({\mathbb{R}}^3)$ function with zero weak gradient is constant almost everywhere, and the only constant in $L^2({\mathbb{R}}^3)$ is zero; this contradicts the asserted positivity of $v$. If $u_m$ converged strongly in $H^1(B)$, its limit would have to be its weak limit zero from (iii), forcing $\Vert \nabla u_m{\Vert }_2\to 0$. The first scaling identity contradicts this. Thus the intended answer is

There is nevertheless a substantive printing error: the asserted strong convergence of $v_m$ in the full $H^1({\mathbb{R}}^3)$ is false for these positive sources. Here is a check that does not rely on any assertion about the limiting profile. Put $R_m=r_m^{-1}$ and $c=|E{|}^{-5/6}$. Rescaling the weak formulation givesThe spherical mean of a Poisson solution outside its source provides an exact obstruction. The total source is $M=c|E|=|E{|}^{1/6}>0$. Let $a_m(s)=(4\pi {)}^{-1}{\int }_{S^2}{v}_m(s\omega )d\omega$ be its spherical mean. On $1<s<R_m$ the function is a harmonic function. The radial flux identity, obtained by testing against radial test functions or by the divergence theorem, gives $-4\pi s^2{a}_m^{\prime }(s)=M$, while the zero boundary value gives $a_m(R_m)=0$. HenceCauchy-Schwarz inequality on each sphere, followed by spherical coordinates, now yieldsThus the printed sequence is not even bounded in the full $H^1$ norm. The result on Poisson convergence on expanding balls in a homogeneous Sobolev space gives the appropriate corrected convergence in the homogeneous Sobolev space ${\dot{H}}^1({\mathbb{R}}^3)$, which controls the gradient in $L^2$ and the function in $L^6$, without requiring it to belong to $L^2$.

For completeness, the desired nonconvergence of $u_m$ has a direct proof that needs neither the incorrect claim nor its correction. Choose a nonnegative nonzero test function $\eta \in C_c^{\infty }(E)$, which exists since $E$ is a domain containing the origin. The rescaled weak formulation and Cauchy-Schwarz inequality giveConsequently, for every $m$,Together with (iii), this rigorously proves failure of strong convergence for the actual sequence defined in the paper.