Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 105 / 2

Use the interior $\Omega =\{1/2<|x|<2\}$ for the Sobolev space notation; the closed shell in the PDF specifies its boundary. Put $A(x)=\mathrm{diag}(1,1,g(|x|))$. A weak solution has zero boundary trace in the sense of the Sobolev trace theorem, equivalently $u\in H_0^1(\Omega )$, and satisfiesThe minus sign is necessary because the operator in the paper is positive divergence, not its negative. One can first test with $C_c^{\infty }$ functions and then use density in the zero-boundary Sobolev space. Smoothness of $g$ makes its restriction to the compact radial interval bounded, so both sides define continuous functionals on this space. Merely belonging to $H^1$ without the zero trace would omit the boundary condition.

Positivity and continuity of $g$ give $\mu ={\min }_{[1/2,2]}g>0$ and a finite maximum $G$. Hence, for every $x$ in the shell and every $\xi \in {\mathbb{R}}^3$,This is the uniformly elliptic operator condition. Positivity at separate points without a compactness/continuity lower bound would not alone supply a common ellipticity constant.

Test the weak formulation with $u$ itself. Uniform ellipticity and the Cauchy-Schwarz inequality giveThe Poincaré inequality for the bounded shell with zero trace gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$. If the gradient is nonzero, divide to obtain $\Vert \nabla u{\Vert }_2\le C_P{\lambda }^{-1}\Vert f{\Vert }_2$; if it is zero the estimate is immediate and the zero trace forces $u=0$. ThusThe constant depends on the shell and $g$, not on $u$ or $f$. Subtracting two solutions with the same forcing gives a zero-forcing solution and hence proves uniqueness. This is a coercive energy argument, not an appeal to real analytic Cauchy solvability.

A fixed smooth choice $g\equiv 0$ already gives a counterexample. Select nonzero smooth functions $\varphi (x_1,x_2)$ and $\chi (x_3)$ with product support compactly contained in a small box inside the shell. LetThen $u_k$ has zero boundary trace and is a classical, hence weak, solution. The $L^2$ norms of $f_k$ are bounded independently of $k$. On the other hand,The squared norm of $\chi \cos (k{x}_3)$ tends to $\frac{1}{2}\Vert \chi {\Vert }_2^2$, by the oscillatory integral identity for ${\cos }^2$. The other term has bounded norm, so $\Vert u_k{\Vert }_{H^1}$ grows at least proportionally to $k$. ThereforeThis failure of a coercive Dirichlet estimate under degeneracy loses control in the third direction; it does not rely on varying $g$ with the sequence.

For the spherical coordinates in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin \theta$. Their metric tensor is therefore $\mathrm{diag}(1,r^2,r^2{\sin }^2\theta )$ and the Jacobian determinant is $r^2\sin \theta$. Applying the divergence formula to the gradient givesFor example, the formula follows directly from $\Delta u=|\det h{|}^{-1/2}{\partial }_i(|\det h{|}^{1/2}{h}^{ij}{\partial }_{j}u)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the Laplacian is a smooth geometric operator there as well, described by other charts. The coordinate volume form integrates over $1/2<r<2$, $0<\theta <\pi$, $0\le \varphi <2\pi$, with the usual periodic identification.

Here is a direct use of the spherical representation that also avoids estimating separate coordinate derivatives at the poles. Put $A={\Delta }_{S^2}$, $T={\partial }_r^2+2r^{-1}{\partial }_r$, and write $d\omega$ for unit-sphere area. Then $f=Tu+r^{-2}Au$. Expanding its squared norm with the physical measure givesOn each boundary sphere $u=0$, so every angular derivative, including $Au$ and ${\nabla }_{S}u$, is zero there. Radial and angular integration by parts consequently giveIndeed the $u_{rr}$ term gives $\int |{\nabla }_S{u}_r{|}^2$ with boundary remainder $[\int u_{r}Au]=0$, while the $2u_r/r$ term is $-\int r^{-1}{\partial }_r|{\nabla }_{S}u{|}^2=-\int r^{-2}|{\nabla }_{S}u{|}^2$, its boundary remainder also being zero. Thuswhere the last step uses the preceding $H^1$ energy estimate and $r\ge 1/2$. All unmarked integrals in these two identities are with $drd\omega$.

To control every angular second derivative, the spherical Hessian identity isIt follows by integrating the derivative-commutation identity ${\nabla }^a{\nabla }_a{\nabla }_{b}w={\nabla }_{b}Aw+{\mathrm{Ric}}_b{}^c{\nabla }_{c}w$ and using $\mathrm{Ric}=h$ on the unit sphere. There is no sphere boundary. Hence the preceding bound controls ${\nabla }_S^{2}u$, ${\nabla }_S{u}_r$, and $u_{rr}=Tu-2u_r/r$ in their appropriate weighted norms.

In an orthonormal polar frame the Cartesian Hessian matrix has componentsand $D^{2}u(e_A,e_B)=r^{-2}({\nabla }_S^{2}u{)}_{AB}+r^{-1}{u}_r{\delta }_{AB}$. Since $r$ is bounded above and below, these bounds plus the first-order estimate proveThis is the direct spherical-shell H2 estimate. Intrinsic angular integration combines the suggested angular tests, with no spurious boundary at the polar coordinate singularities.

The coefficients in the spherical Laplacian are independent of $\varphi$, so $[\Delta ,{\partial }_{\varphi }]=0$. Also ${\partial }_{\varphi }$ is tangent to both boundary spheres; differentiating the zero boundary trace gives another zero trace. Therefore $w={\partial }_{\varphi }u$ satisfies $\Delta w={\partial }_{\varphi }f$ and the same homogeneous Dirichlet boundary conditions.

In Cartesian coordinates ${\partial }_{\varphi }=x_1{\partial }_2-x_2{\partial }_1$, a globally smooth rotation field even where angular coordinates are singular. Permuting coordinate axes gives $R_{12}$, $R_{23}$ and $R_{31}$. For each, the rotational commutator for the Laplacian follows directly from $\Delta (x_i{\partial }_{j}u)=x_i{\partial }_j\Delta u+2{\partial }_i{\partial }_{j}u$; antisymmetrizing cancels the extra terms. ThusIf $f$ is radial, all these right-hand sides vanish. Uniqueness from the energy estimate makes every $R_{ij}u$ zero. At each point of a sphere the rotation fields span its tangent plane, so all tangential derivatives of $u$ vanish. Each sphere is connected, giving $\boxed{u(x)=U(|x|)}$. This proves radiality without assuming it as a separation-of-variables ansatz.

The Sobolev embedding theorem $H^2(\Omega )\hookrightarrow L^{\infty }(\Omega )$ in dimension three, combined with the preceding estimate, givesOne direct justification of this embedding is a bounded Sobolev extension operator into $H^2({\mathbb{R}}^3)$, followed by Fourier inversion and the Cauchy-Schwarz inequality: ${\int }_{{\mathbb{R}}^3}(1+|\xi {|}^2{)}^{-2}d\xi <\infty$ bounds the inverse Fourier integral by the $H^2$ norm.

For radial forcing, the preceding part makes $u=U(r)$ and $f=F(r)$. Set $a=1/2$, $b=2$ and $H(r)={\int }_a^r{s}^{2}F(s)ds$. The radial equation gives $r^2{U}^{\prime }(r)=C_0+H(r)$. Both zero boundary values implyMoreover $\Vert H{\Vert }_{\infty }\le ({\int }_a^b{s}^{2}ds{)}^{1/2}({\int }_a^b{s}^2|F{|}^{2}ds{)}^{1/2}\le C\Vert f{\Vert }_2$. Since $r\ge a$, this proves the radial Poisson gradient estimate:

For a nonradial counterexample choose $x_0$ in the shell and a nonconstant $\phi \in C_c^{\infty }(B(0,1))$. For sufficiently small $\epsilon$, set $u_{\epsilon }(x)={\epsilon }^{1/2}\phi ((x-x_0)/\epsilon )$ and $f_{\epsilon }=\Delta u_{\epsilon }$. These are smooth with support away from the boundary. Scaling givesThus the failure of L2-to-Linfinity Poisson gradient bounds in three dimensions is explicit, despite the valid $H^2$ and value-supremum estimates.