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For $v\in H_0^2(-1,1)$, integrate its weak derivative from the left endpoint, where $v=v^{\prime }=0$. The Cauchy-Schwarz inequality gives $\Vert v^{\prime }{\Vert }_2\le 2\Vert v^{\prime \prime }{\Vert }_2$ and $\Vert v{\Vert }_2\le 2\Vert v^{\prime }{\Vert }_2$. Adding the three squared derivative norms gives the displayed bound. The constant is sufficient, rather than asserted sharp. Thus the second-derivative norm is equivalent to the full Sobolev norm on this clamped space.

For a conformal bijection $f:D\to \tilde{D}$ and smooth $u,v$ of compact support on $\tilde{D}$, $\boxed{(u\circ f,v\circ f{)}_{\nabla ,D}=(u,v{)}_{\nabla ,\tilde{D}}}$. The Jacobian matrix of $f$ is a rotation times $|f^{\prime }|$. Consequently the chain rule introduces $|f^{\prime }{|}^2$ into the gradient pairing, while the change of variables formula introduces precisely the same factor into area. They cancel. The identity extends by completion to the Dirichlet energy spaces; it does not preserve the $L^2$ term of an inhomogeneous Sobolev norm.

For arbitrary real $s,t$, the convolution of $u\in H^s(\mathbb{R})$ and $v\in H^t(\mathbb{R})$ has a canonical definition as a tempered distribution. With $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$, write $A=⟨\xi {⟩}^s\hat{u}$ and $B=⟨\xi {⟩}^t\hat{v}$, both in the L2 space. Then $AB\in L^1$ by Cauchy-Schwarz inequality, andThus the product defines a tempered distribution, and the displayed inverse Fourier transform defines $u\ast v$. This extends ordinary convolution of Schwartz functions by the convolution theorem and density in the Sobolev norm.

The result need not belong to any Sobolev space, even when both inputs belong to every Sobolev space. For $\hat{u}=\hat{v}={\xi }^{-1/3}{1}_{(0,1)}(\xi )$, each weighted squared input is integrable, but the squared product is ${\xi }^{-4/3}{1}_{(0,1)}(\xi )$, which is not locally integrable at zero. Every weight $(1+{\xi }^2{)}^r$ is bounded above and below there, proving the failure for all real $r$.

On the unit interval, the weak form is $a(u,v)={\int }_0^{1}fv$, with $a(u,v)={\int }_0^1(u^{\prime }{v}^{\prime }+xuv)$ on the zero-boundary Sobolev space. The nonnegative potential makes this form coercive in the derivative norm, and the Poincaré inequality makes that norm equivalent to the Sobolev norm. The Lax-Milgram theorem gives a unique solution. A conforming Ritz method minimizes the corresponding strictly convex quadratic energy and produces a symmetric positive-definite tridiagonal matrix for piecewise-linear hats.

For $u_{tt}-\Delta u+u^3=0$ in three spatial dimensions, conserved positive wave energy $E$ controls $\Vert \nabla u{\Vert }_2$. The Sobolev inequality gives $\Vert u^2\nabla u{\Vert }_2\lesssim \Vert \nabla u{\Vert }_2^2\Vert D^{2}u{\Vert }_2\lesssim E\Vert D^{2}u{\Vert }_2$. Differentiated wave energy estimates and the Gronwall inequality then give $\Vert u(t){\Vert }_{{\dot{H}}^2}+\Vert u_t(t){\Vert }_{{\dot{H}}^1}\le CD{e}^{CEt}$, where $D$ bounds the corresponding initial Sobolev norms.

On $I=(-a,a)$, a first-order Sobolev space function has a continuous representative and obeysBound the average by the Cauchy-Schwarz inequality and average the identity $u(x)-u(y)={\int }_y^x{u}^{\prime }$. This also gives $[u{]}_{C^{0,1/2}}\le \Vert u^{\prime }{\Vert }_2$. A unit constant with the standard Sobolev norm cannot be asserted on arbitrary interval lengths: $u=1$ already disproves it when $2a<1$.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

Solutions to all three questions of the finite element methods for partial differential equations course of the University of Oxford Part C examination, sat on 6 June 2023. Source: original released PDF, bundle pages 325–328; paper pages 1–4. The original numbering is retained; question statements are not reproduced.

All function spaces and bilinear forms below are real. Write $(p,q{)}_0={\int }_{\Omega }pq$ for the L2 inner product, and $\Vert p{\Vert }_1^2=\Vert p{\Vert }_0^2+\Vert \nabla p{\Vert }_0^2$ for the first-order Sobolev norm; the notation applies componentwise to vector fields. We use the Poincaré inequality explicitly allowed by the paper. On an arbitrary unbounded Lipschitz domain that inequality would need an additional assumption. The convergence assertions use the usual full conforming finite element spaces, or vector subspaces containing the indicated interpolants; merely being a vector subspace of the stated piecewise polynomial functions is insufficient. Part 1(d) explains the intended meaning of positivity for its nonsymmetric matrix.

Use the Sobolev space with a partial Dirichlet conditionwhere $T$ is the Sobolev trace operator. The restricted Sobolev trace operator is continuous into $L^2({\Gamma }_D)$; its kernel is consequently a closed vector subspace of $H^1(\Omega )$. Thus $V$, with the first-order Sobolev norm, is a Hilbert space. The endpoint convention for ${\Gamma }_D$ and ${\Gamma }_N$ does not affect this definition: finitely many boundary points have zero boundary length. It would be inappropriate to impose pointwise boundary values on arbitrary $H^1$ functions in two dimensions.

For a smooth solution, Green's first identity givesThe boundary term vanishes because $Tv=0$ on ${\Gamma }_D$ and the normal derivative is zero on ${\Gamma }_N$. This suggests the weak formulationHere $a$ is exactly the first-order Sobolev inner product. In particular the Cauchy-Schwarz inequality givesThus $a$ is a bounded bilinear form and a coercive bilinear form, both with constant one. The Lax-Milgram theorem proves existence and uniqueness, together withThe positive reaction term already controls the entire Sobolev norm, so no Poincaré inequality is needed here. The zero Neumann boundary condition is the natural part of the weak formulation; it is not an additional restriction on the test space. For this weak solution, $-\Delta u+u=f$ holds as a distributional identity, and the zero normal derivative on ${\Gamma }_N$ is understood through the corresponding Green's first identity.

The finite-dimensional vector subspace $V_h\subset V$ is closed, so it is a Hilbert space with the inherited Sobolev norm. The bounded bilinear form, coercive bilinear form, and bounded linear functional estimates from part (b) restrict to $V_h$. Applying the Lax-Milgram theorem proves

For any $v_h\in V_h$, put $d_h=v_h-u_h\in V_h$. Expansion using symmetry of the bilinear form givesThus $u_h$ minimizes $J$ on $V_h$, uniquely by the coercive bilinear form estimate. Similarly, the continuous weak formulation gives the exact identityCombining the two identities provesMore precisely, $J(v_h)-J(u_h)=\frac{1}{2}\Vert v_h-u_h{\Vert }_a^2$ and $J(u_h)-J(u)=\frac{1}{2}\Vert u_h-u{\Vert }_a^2$, with the energy norm justified next. These identities connect the energy excess directly to the squared error.

Use the real zero-trace Sobolev space $H_0^1(\Omega )$ with norm $\Vert \nabla u{\Vert }_2$. The Poincaré inequality on the bounded smooth domain gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$, making this an equivalent Sobolev norm. The weak formulation is to find $u\in H_0^1(\Omega )$ with $u\ge 0$ almost everywhere andThe zero Dirichlet boundary condition is encoded in the Sobolev trace. Formally, the identity follows by multiplying the partial differential equation by a test function and using integration by parts; it then extends to all $v\in H_0^1(\Omega )$ by density.

To check all terms, boundedness of $\Omega$ gives finite Lebesgue measure $m=|\Omega |$. The Cauchy-Schwarz inequality gives $u\in L^1(\Omega )$ andThus the right side is a bounded linear functional of $v$, since its absolute value is at most $\Vert 1+\sqrt{u}{\Vert }_2{C}_P\Vert \nabla v{\Vert }_2$. The left side is finite by the Cauchy-Schwarz inequality. These estimates work for every stated $n\ge 2$.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.

The distributional derivative is defined by $⟨u^{\prime },\varphi ⟩=-⟨u,{\varphi }^{\prime }⟩$. Since differentiation under the integral sign gives $(\hat{\varphi }{)}^{\prime }=\widehat{-i\xi \varphi }$, we obtainThis proves the Fourier transform of a derivative identity $\boxed{\widehat{u^{\prime }}=i\xi \hat{u}}$ for all tempered distributions.

If $u,u^{\prime }\in H^s(\mathbb{R})$, their Fourier transforms are represented by locally square-integrable functions, and the identity identifies $\widehat{u^{\prime }}$ with $i\xi \hat{u}$ almost everywhere. Hence the Sobolev norms satisfyso $\boxed{u\in H^{s+1}(\mathbb{R}),\Vert u{\Vert }_{H^{s+1}}^2=\Vert u{\Vert }_{H^s}^2+\Vert u^{\prime }{\Vert }_{H^s}^2}$ for every real $s$.

For real $s$, the Sobolev space $H^s(\mathbb{R})$ consists of those tempered distributions $u$ whose Fourier transform is represented by a locally square-integrable function and for whichThis is the Sobolev norm used here; inserting the fixed factor $(2\pi {)}^{-1}$ gives an equivalent convention.

For the constant function $c$, Fourier inversion at zero givesA nonzero Dirac delta distribution cannot be represented by a locally square-integrable function. Indeed, if $G$ represented $2\pi c{\delta }_0$, choose a test function $\rho$ with $\rho (0)=1$, and put ${\rho }_{\epsilon }(\xi )=\rho (\xi /\epsilon )$. On a fixed compact interval containing their supports, Cauchy-Schwarz inequality givesThus $c=0$. Every Sobolev space requires local square integrability of the Fourier transform, regardless of how negative $s$ is. ThereforeThe zero constant function belongs to every Sobolev space.

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.

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.

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.

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.

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.