Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 (Analysis of Partial Differential Equations)

Put $Q=Q_r(x)$ and $v_Q=|Q{|}^{-1}{\int }_{Q}v$. The fundamental theorem of calculus along a line segment, followed by averaging, givesFor fixed $t$, use the change of variables formula $w=y+t(z-y)$. Its Jacobian determinant is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a convex set. Moreover, $w\in Q_t$ implies $|w-y|\le \sqrt{n}rt$. Thus Tonelli theorem givesThe diagonal $w=y$ is a null set. The Holder inequality with $p^{\prime }=p/(p-1)$ now applies: $p>n$ implies $(n-1)p^{\prime }<n$, and integration in spherical coordinates bounds the kernel byThis proves the Morrey inequality on a cube, uniformly even when $y$ approaches its boundary:Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain

Let $u_j\in C_c^{\infty }({\mathbb{R}}^n)$ tend to $u$ in the Sobolev space $W^{1,p}$, using density of smooth functions in a Sobolev space. Write $\alpha =1-n/p>0$. Applying the Morrey inequality on a cube to a unit cube centered at $x$, and bounding its average by the Holder inequality, givesfor every smooth $v$. Hence $u_j$ is a Cauchy sequence in the supremum norm and has a bounded continuous limit $u^{\ast }$ by uniform convergence. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^{\ast }$ almost everywhere.

For $x\ne y$, choose $r=4|x-y|$, so that $y\in Q_r(x)$. The second estimate from part (a), followed by passage to the limit, givesThus $u^{\ast }$ is a Hölder continuous function andIt is the unique continuous representative: two continuous functions agreeing almost everywhere agree everywhere.

For $1\le p<n$, take a smooth cutoff function $\chi$ equal to one near zero and setIts weak derivative has size $O(|x{|}^{-a-1})$ near zero, whose $p$th power is integrable because $p(a+1)<n$. Thus $u\in W^{1,p}$, but it is essentially unbounded in every neighborhood of zero and has no continuous representative. The value assigned at zero is irrelevant.

Choose $x$ at which $Du(x)$ is finite and the $p$-mean version of the Lebesgue differentiation theorem holds:These conditions hold almost everywhere. On a cube about $x$, subtract the affine function and put $v(y)=u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. Its weak derivative is $Dv=Du-Du(x)$ and $v(x)=0$. Approximate $u$ as in part (b) and subtract the same affine function from the approximants. Uniform convergence and convergence of their weak derivatives extend part (a)'s estimate to $v$, givingFor $h\ne 0$, choose $r=4|h|$ and $y=x+h$. It follows thatThis is precisely classical Fréchet derivative differentiability, not merely existence of coordinate derivatives. Therefore the differentiability almost everywhere of supercritical Sobolev functions gives

The Lax-Milgram theorem assumes that $H$ is a real Hilbert space and $B:H\times H\to \mathbb{R}$ is a bounded bilinear form and a coercive bilinear form: there are $M,\alpha >0$ such thatThen for every continuous linear functional $\ell \in H^{\prime }$ there is a unique $u\in H$ satisfyingSymmetry of $B$ is not required.

For the proof, the Riesz representation theorem provides a bounded linear operator $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v{)}_H$ and $\ell (v)=(g,v{)}_H$. Coercivity and the Cauchy-Schwarz inequality implyThus $T$ is injective and its range is closed: if $T{u}_j$ converges, the displayed inequality makes $u_j$ a Cauchy sequence; completeness gives $u_j\to u$ and $T{u}_j\to Tu$. If $w$ lies in the orthogonal complement of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha \Vert w{\Vert }^2\le B[w,w]=0$, so $w=0$. A closed subspace with zero orthogonal complement is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.

The principal coefficients satisfy the positive uniformly elliptic operator condition if there is $\theta >0$, independent of $x$, such thatThe smooth coefficients on the compact closure also give a uniform upper bound $\sum a^{ij}{\xi }_i{\xi }_j\le \Theta |\xi {|}^2$. The first-order and zeroth-order coefficients do not enter uniform ellipticity.

Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the Sobolev trace theorem. The weak Robin problem for a uniformly elliptic operator uses all of $H^1(U)$ as the test space, since a Robin boundary condition does not require a zero trace. DefineThe weak formulation isFor a classical solution, integration by parts produces the boundary term $-{\int }_{\partial U}(\sum a^{ij}{u}_{x_i}{\nu }_j)v$. Substituting the Robin boundary condition replaces it by ${\int }_{\partial U}\beta uv$, giving the displayed identity.

Conversely, if $u\in C^2(\overline{U})$ satisfies the identity, tests $v\in C_c^{\infty }(U)$ first show $(L+\lambda )u=f$ in the sense of distributions. Subtracting this interior equality after integration by parts givesEvery smooth boundary function is the trace of a smooth function on $\overline{U}$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood almost everywhere.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives

On the zero-boundary Sobolev space $H_0^1(U)$, the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so $(Lu,v{)}_{L^2}=B[u,v]$ for sufficiently regular $u$ and $v\in H_0^1$.

Formal self-adjointness means that $L$ equals its formal adjoint, or equivalently $(L\varphi ,\chi )=(\varphi ,L\chi )$ for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, $L=L^{\ast }$ is equivalent to $b=0$. It makes $B$ a symmetric bilinear form.

For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention $B[u,u]\ge 0$ does not suffice: $L=-\Delta -{\lambda }_1$ has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.

Formal self-adjointness gives symmetry of $B$, and strict positivity gives positive definiteness, so $B$ is an inner product. Bounded coefficients give $B[u,u]\le C\Vert u{\Vert }_{H^1}^2$. To prove the lower bound, uniform ellipticity givesIf no lower bound by the $H^1$ norm existed, choose $u_j\in H_0^1$ with $\Vert u_j{\Vert }_{H^1}=1$ and $B[u_j,u_j]\to 0$. By weak convergence in a Hilbert space and the Rellich-Kondrachov compactness theorem, a subsequence has $u_j\rightharpoonup u$ in $H_0^1$ and $u_j\to u$ in $L^2$. The principal energy has weak lower semicontinuity: it is the squared $L^2$ norm of the bounded multiplication operator $a^{1/2}Du$. The potential term converges because $c$ is bounded. HenceStrict positivity forces $u=0$. The preceding ellipticity estimate now gives $\Vert D{u}_j{\Vert }_2\to 0$, while $\Vert u_j{\Vert }_2\to 0$, contradicting $\Vert u_j{\Vert }_{H^1}=1$. Thus strict positivity implies coercivity for an elliptic Dirichlet form, andThe new norm is equivalent to the original norm, so the space remains complete.

The assertion about all $H_0^1(U)\cap H^k(U)$ is false as printed in the original PDF. Take $U=(0,\pi )$, $L=-d^2/d{x}^2$, and $u=x(\pi -x)$. This is a nonzero smooth Dirichlet function, but $Lu=2$ and $L^{2}u=0$. In particularAlready at $k=3$, $B[Lu,Lu]$ is not defined on its stipulated domain, because $Lu=2\notin H_0^1$. Extending $B$ to $H^1$ for this example would give zero, not an inner product. Additional boundary conditions are essential.

The corrected spaces are the Sobolev domains of powers of an elliptic Dirichlet operator. Let $A_D$ be the Dirichlet realization of an elliptic operator, and setIn particular $X_1=H_0^1$, $X_2=H^2\cap H_0^1$, and $X_3$ also requires $T(Lu)=0$. The formulas in the question define inner products on these spaces.

Here is the norm-equivalence proof on the corrected domains. The base cases are the $L^2$ inner product at $k=0$ and part (i) at $k=1$. Standard Dirichlet elliptic regularity, together with invertibility from the Lax-Milgram theorem, gives for every integer $r\ge 0$The first estimate applies when $Tu=0$; invertibility absorbs the usual lower-order $L^2$ term. Moreover, $A_D:X_{r+2}\to X_r$ is an isomorphism. To see surjectivity, solve $Lu=f$ with zero Dirichlet trace for $f\in X_r$; regularity gives $u\in H^{r+2}$, and $T(L^{j}u)=T(L^{j-1}f)=0$ for each further required trace. The formulas satisfyInduction therefore givesInjectivity of each iterate of $A_D$ gives positive definiteness. These spaces are $D(A_D^{k/2})$; the spectral characterization of elliptic Dirichlet domains in the next part makes the half-powers precise. On the paper's uncorrected spaces the claim is valid at $k=0,1,2$, but fails in general beyond them.

First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse $S=A_D^{-1}$ exists from $L^2$ into $H_0^1$. Elliptic regularity gives $Sf\in H^2\cap H_0^1$, and the Rellich-Kondrachov compactness theorem makes $S$ a compact operator on $L^2$. Symmetry givesso $S$ is a self-adjoint operator. It is injective, and $(Sf,f)=B[Sf,Sf]>0$ for $f\ne 0$. The spectral theorem for compact Hermitian operators supplies an orthonormal basis $(w_m)$ with $S{w}_m={\mu }_m{w}_m$, ${\mu }_m>0$, and ${\mu }_m\to 0$. Put ${\lambda }_m={\mu }_m^{-1}$. ThenRepeated elliptic regularity makes each $w_m$ smooth up to the boundary.

As with part (ii), the stated characterization omits boundary compatibility. At $k=0$, its right-hand side is finite for every $u\in L^2$ by Parseval identity, whereas its left-hand side would require $u\in H_0^1$. At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on $(0,\pi )$,Although this function belongs to every ordinary $H^k$, the weighted sum diverges at $k=3$.

The correct spectral characterization of elliptic Dirichlet domains iswith $X_k$ defined in part (ii). For finite eigenfunction sums the squared norm $((u,u){)}_k$ is exactly the weighted sum: even orders follow from $L^l{w}_m={\lambda }_m^l{w}_m$, and odd orders also use $B[w_m,w_j]={\lambda }_m{\delta }_{mj}$. These sums are dense in $X_1$: if $v$ is $B$-orthogonal to every $w_m$, then $0=B[v,w_m]={\lambda }_m(v,w_m)$, so $v=0$. They are dense in $X_0$ by construction, and the isomorphisms $A_D:X_{k+2}\to X_k$ propagate density to every $X_k$. Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for $k=1,2$.

Take $\psi \in L^2(U)$, the natural initial-data assumption needed for the requested $L^2$ convergence but not explicitly stated in this subpart. Use the genuine Dirichlet eigensystem from part (b)(iii) and put ${\psi }_m=(\psi ,w_m)$. The spectral construction of a parabolic solution isEach finite sum has zero Dirichlet boundary condition and satisfies $u_t+Lu=0$.

To justify all derivatives, fix $\epsilon >0$. For any integers $k,j\ge 0$, the squared $X_k$ norm of a tail of the $j$th time derivative isuniformly for $t\ge \epsilon$. Use the corrected Sobolev domains of powers of an elliptic Dirichlet operator, not the inaccurate unqualified assertion in the PDF. Their norm equivalence gives convergence in every ordinary $H^k$. The Sobolev embedding theorem then gives convergence of every desired spatial derivative by choosing $k$ large enough. Thus termwise differentiation is justified, $u\in C^{\infty }(U_T)$, and $u(t,\cdot )\in C^{\infty }(\overline{U})$ for every $t>0$. The trace remains zero, and $u_t+Lu=0$ holds pointwise.

Finally Parseval identity givesby the dominated convergence theorem, since each factor tends to zero and is bounded by one. No boundary compatibility of the initial $L^2$ data is required for positive-time smoothness.

For each $t$, letA weak energy solution of a variable-coefficient wave equation is a function $u\in H^1(U_T)$ with zero lateral trace, equivalently $u\in L^2(0,T;H_0^1(U))$, whose time trace satisfies $u(0)={\psi }_0$ in $L^2(U)$ and for whichfor every $v\in H^1(U_T)$ with zero lateral trace and $v(T)=0$. Here $u\in H^1(0,T;L^2(U))$ has a continuous $L^2$ representative, so the displacement trace is well defined.

The velocity condition is encoded by the boundary term in time; an arbitrary space-time $H^1$ function need not have an $L^2$ trace of $u_t$. The equation implies $u_{tt}=f-L(t)u\in L^2(0,T;H^{-1}(U))$, hence $u_t$ has a continuous $H^{-1}$ representative and $u_t(0)={\psi }_1$ in that sense. This is the weak formulation obtained by integration by parts once in time and once in space.

Take the difference of two weak solutions, so $f={\psi }_0={\psi }_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying uniqueness of Sobolev weak solutions of a wave equation.

Fix $s\in (0,T]$ and defineThese are Bochner integrals in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the weak formulation isbecause $u(0)=0$. Symmetry of $a^{ij}$ and integration by parts in time giveFor the lower-order part, integration by parts in space givesBounded coefficients, the Poincaré inequality for $v$, and the Young inequality now implySince $Dv(t)=Dz(s)-Dz(t)$,For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The Gronwall inequality applied to $\Vert u(s){\Vert }_2^2+(\theta /2)\Vert Dz(s){\Vert }_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore

For $F(x,t)=t-\tau (x)$, the conormal to its zero set is $dF=(-D\tau ,1)$. The principal symbol of ${\partial }_t^2+L$ isA characteristic hypersurface requires $p(x,t;dF)=0$ on the surface. Thus $\tau$ satisfies the anisotropic eikonal equationThe first-order and zeroth-order terms do not affect this condition.

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

The all-time conclusion in the original PDF is false. The expanding-flow transformation of a wave equation makes both the correct forward statement and a counterexample transparent. PutFor $D={\partial }_t+\rho {\partial }_{\rho }$, the expanded operator from part (i) is $D^2+D-{\partial }_{\rho }^2$. In the new variables $D=-S{\partial }_S$ and ${\partial }_{\rho }=S{\partial }_X$, henceThus it is the ordinary wave equation on the half-plane $S>0$. By the D'Alembert formula, any $C^2$ solution has the formAt $t=0$, differentiating the zero displacement with respect to $\rho$ and using the zero velocity gives, for $-1<\rho <1$,Subtracting $(\rho -1)$ times the first equation from the second gives $2F^{\prime }(\rho +1)=0$, and then $G^{\prime }(\rho -1)=0$. Thus $F=c$ on $(0,2)$ and $G=-c$ on $(-2,0)$. If $t\ge 0$ and $-1<\rho <1$, both arguments remain in those intervals. Consequently the valid forward result isFor negative time the arguments need not stay in the initial intervals. Take $F(s)=(s-2{)}_{+}^3$ and $G=0$, where $q_{+}=\max \{q,0\}$. Thenis a global $C^2$ solution, has the prescribed zero initial displacement and velocity on $(-1,1)$, but $u(0,-\log 3)=1$. This directly disproves the assertion for $t\in \mathbb{R}$.

The connection with part (i) is directional finite propagation speed. At a future point in the strip, the two characteristics traced back to $t=0$ meet it at ${\rho }_0=e^{-t}(\rho +1)-1$ and ${\rho }_0=e^{-t}(\rho -1)+1$, both in $(-1,1)$. For negative time these intersections can lie outside that interval. The lines $\rho =\pm 1$ are characteristic barriers for this forward domain of dependence, not barriers in both time directions.