Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 (Analysis of Partial Differential Equations)

Write $u_Q=|Q{|}^{-1}{\int }_{Q}u$. Since $u-u_Q$ has arithmetic mean zero,and the pairwise-difference identity givesFirst take $u$ smooth. Join $x$ to $y$ by changing one coordinate at a time and apply the Cauchy-Schwarz inequality:For a one-dimensional slice $v$, the fundamental theorem of calculus yieldsIntegrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\Vert D_{i}u{\Vert }_{L^2(Q)}^2$. HenceThe density of smooth functions in a Sobolev space extends the estimate to every $u\in H^1({\mathbb{R}}^n)$. Thus

With the standard Sobolev space inner product, $u_i\rightharpoonup u$ weakly in $H^1(U)$ meansfor every $v\in H^1(U)$. Equivalently, every bounded linear functional on $H^1(U)$ takes convergent values on the sequence.

The Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives

Suppose the claimed Poincare-Wirtinger inequality were false. There would be $u_k\in H^1(U)$ such that, after settingwe have $(v_k{)}_U=0$, $\Vert v_k{\Vert }_2=1$, and $\Vert D{v}_k{\Vert }_2\to 0$. The sequence is bounded in $H^1(U)$, so part 1(b)(ii) supplies a subsequence converging strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $v$.

The weak gradient of $v$ is zero. Because $U$ is connected, $v$ is a constant function; its mean is zero, so $v=0$. Strong convergence would then give $\Vert v_k{\Vert }_2\to 0$, contradicting $\Vert v_k{\Vert }_2=1$. Therefore

If the Poincare inequality with a boundary trace failed, after normalization there would be $u_k\in H^1(U)$ withAs in part 1(c)(i), a subsequence converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to a constant function $u$. The Sobolev trace theorem is a bounded linear map, so the traces converge weakly while their norms tend to zero; hence the trace of $u$ is zero. A constant with zero trace is zero, contradicting $\Vert u{\Vert }_2=1$. Consequently

The Lax-Milgram theorem states that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form, and there is an $\alpha >0$ such thatthen for every bounded linear functional $F\in H^{\prime }$ there is a unique $u\in H$ satisfyingMoreover, $\boxed{\Vert u{\Vert }_H\le {\alpha }^{-1}\Vert F{\Vert }_{H^{\prime }}}$. Symmetry of $B$ is not required.

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

On the product Hilbert space $H=H_0^1(U)\times H^1(U)$ defineandThe Cauchy-Schwarz inequality makes $B$ and $F$ bounded. On the diagonal,The Poincare inequality controls $\Vert u{\Vert }_2$ by $\Vert Du{\Vert }_2$, andThe displayed diagonal value therefore controls the full product $H^1$ norm, so $B$ is coercive. The Lax-Milgram theorem now gives exactly one pair $(u,w)\in H$ satisfying the weak identities. Hence a unique weak solution exists for every $f,g\in L^2(U)$.

The function $f$ is real analytic at $y$ if there is a neighborhood of $y$ on which its multivariable Taylor seriesconverges to $f(x)$. Here $\alpha$ is a multi-index; equivalently, $f$ agrees locally with a convergent real power series centred at $y$.

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

One prescribes analytic Cauchy data: the value of $u$ and one first derivative transverse to $\Sigma$, for examplewhere $\varphi$ and $\psi$ are real analytic on $\Sigma$. Being a non-characteristic hypersurface allows the equation to solve for the second derivative in the transverse direction. The Cauchy-Kovalevskaya theorem then gives one and only one local real analytic solution near $p$.

Expanding the equation givesso its principal symbol isIf a characteristic curve is locally a graph $y=y(x)$, its conormal is proportional to $(-y^{\prime },1)$. The characteristic equation is thereforeOn each region separated by $y=\pm 1$, separation of variables givesThus all the characteristic curves areThe inner curves approach the horizontal characteristics $y=\pm 1$ as $x\to \pm \infty$; the outer hyperbolic-cotangent branches have vertical asymptotes and also approach $y=\pm 1$. This describes the requested sketch.

The initial line $x=0$ has conormal $(1,0)$, and $p(y,1,0)=1$, so it is a non-characteristic hypersurface at every point. Since the coefficients and prescribed data are real analytic, the Cauchy-Kovalevskaya theorem gives a unique real analytic solution in a neighborhood of each point of $\{x=0\}$, hence in a neighborhood of that line.

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give

Assume $0<\epsilon <1$ and setThe travel-time coordinate for a one-dimensional variable-speed wave equationsends the initial interval $I_{\epsilon }=(-a,a)$ to $(-S_{\epsilon },S_{\epsilon })$. By the characteristic curves for speed one minus y squared, the two characteristic coordinates are $s-x$ and $s+x$. The finite propagation speed and uniqueness theorem for hyperbolic partial differential equations therefore give the maximal characteristic diamondEquivalently,In the $(x,s)$ plane this is a diamond with vertices $(0,\pm S_{\epsilon })$ and $(\pm S_{\epsilon },0)$; transforming back bends its four sides into the characteristic curves found in part 3(c). Beyond any one of those sides, a point's backward characteristics meet $x=0$ outside $I_{\epsilon }$, where no Cauchy data were prescribed, so uniqueness cannot be extended farther.

Extend $u$ by zero to ${\mathbb{R}}^n$; its compact support inside the ball makes the extension smooth. The Fourier transform of a derivative givesThe assumption says that $L_0$ is a uniformly elliptic operator, soMoreover,The Plancherel theorem therefore yieldsAll integrands vanish outside the original support where appropriate, so

Write $Lu=L_{0}u+(a^{ij}-A^{ij})D_{ij}u$, with repeated indices summed. The triangle inequality and the Cauchy-Schwarz inequality over the $n^2$ coefficient pairs giveChooseThen the strict coefficient bound in the question implies

The continuous coefficients are uniformly continuous on a compact neighborhood of $\overline{W}$. For every $x_0\in \overline{W}$, choose a ball $B_r(x_0)⋐U$ small enough thatfor all $i,j$. Part 4(b), with the frozen symmetric matrix $A=(a^{ij}(x_0))$, then applies on this ball.

Choose a finite collection of these balls and a smooth partition of unity $({\eta }_k)$ that sums to one near $\overline{W}$, with each ${\eta }_k$ supported in its corresponding ball. Applying part 4(b) to ${\eta }_{k}u$ givesSince $(a^{ij})$ is symmetric, the Leibniz rule gives the commutator formulaThe coefficients and the finitely many derivatives of the cutoff functions are bounded, soFinally $u={\sum }_k{\eta }_{k}u$ near $\overline{W}$. Summing the finite set of local estimates provesThis is the coefficient-freezing interior second-derivative estimate.

Fix $W⋐V⋐U$. The standard local regularization of the maximal graph norm domain supplies smooth compactly supported approximants $u_k$ on $V$ such thatApply part 4(c), with $V$ as the outer domain, to $u_k-u_m$. It givesThus $(u_k)$ is Cauchy in $H^2(W)$. Its $H^1$ limit is $u$, so $u\in H^2(W)$. Since $W⋐U$ was arbitrary, the definition of a Local Sobolev space givesThis is the Interior H2 regularity for continuous nondivergence coefficients.