Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 (Analysis of Partial Differential Equations)

Only the second-order terms contribute to the principal symbol. For a covector $\xi =({\xi }_t,{\xi }_r,{\xi }_z)$ it isThe conormal to the hypersurface $\Sigma =\{\varphi =0\}$ is $d\varphi =({\varphi }_t,{\varphi }_r,{\varphi }_z)$. After evaluating the coefficient at the prescribed boundary value of $u_z$, the non-characteristic condition is therefore

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At $x\in \Sigma$, keep the unit normal $N(x)$ fixed and define the $j$th normal derivative byFor $j=1$, the chain rule and $N=\nabla \varphi /|\nabla \varphi |$ give

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WriteDifferentiating the prescribed identity $u(\psi (r,z),r,z)=0$ in its two tangential directions givesThe unit normal is $(1,-{\psi }_r,-{\psi }_z)/\sqrt{D}$, so the second item of Cauchy data becomesConsequentlySubstitution into the principal symbol from part a shows that the graph is non-characteristic exactly where

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The Cauchy-Kovalevskaya theorem says that an order-$k$ scalar quasilinear partial differential equation with real-analytic coefficients has a unique local real-analytic solution near each point of a real-analytic non-characteristic hypersurface, provided the prescribed Cauchy dataare real analytic there. The uniqueness is among local real-analytic solutions agreeing with all of those data.

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For ${\psi }_0=(r^2+z^2)/2$,The prescribed function is $g=\sqrt{D}$, so part c gives $u_z=-z$. Hence the non-characteristic condition reduces toThe Cauchy-Kovalevskaya theorem therefore guarantees a unique local real-analytic solution at exactly those pointsfor which

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Suppose the claimed Poincare inequality with a partial Dirichlet boundary fails. There are $v_k\in V$ withAfter the normalization $w_k=v_k/\Vert v_k{\Vert }_{L^2(U)}$,Thus $(w_k)$ is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrashov compactness theorem and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The Sobolev space with a partial Dirichlet condition $V$ is a closed vector subspace, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a constant function. Its trace vanishes on the positive-measure set ${\Gamma }_1$, so that constant is zero. This contradictsTherefore some $C_P$ satisfiesSince the reverse bound $\Vert \nabla v{\Vert }_2\le \Vert v{\Vert }_{H^1}$ is immediate,The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on ${\Gamma }_1$ is zero.

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Multiply the Poisson equation by $v\in V$ and apply Green's first identity. The Dirichlet boundary condition makes the trace of $v$ vanish on ${\Gamma }_1$, while the Neumann boundary condition makes the boundary flux vanish on ${\Gamma }_2$. Thus the weak formulation is: find $u\in V$ such thatwhere

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First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

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The restricted trace map is continuous, and $V$ is its kernel, so $V$ is a closed vector subspace of the Hilbert space $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient normis equivalent to that norm, so it is complete as well. It comes from the inner productConsequently $(V,\Vert \cdot {\Vert }_V)$ is a Hilbert space.

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On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

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The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

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For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

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Solve the adjoint transport equationbackward with terminal value zero. Along the characteristic flow map $Z_{t,s}(x)=e^{s-t}x$, the required solution isDifferentiation under the integral verifies the equation. If $\psi$ has compact support in $[0,T]\times [-R,R]$, then $\phi$ vanishes for $t>T$ and for $|x|>R$, so $\phi \in C_c^1$ as required.

When the initial datum is zero, inserting this $\phi$ into the weak formulation givesfor every $\psi \in C_c^1$. Thus $u=0$ almost everywhere. The difference of two bounded weak solutions has zero initial datum, so this proves uniqueness.

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SetThe scalar conservation law is $u_t+a(t)f^{\prime }(u)u_x=0$. Its characteristic curve issuing from $\xi$ satisfiesThe Jacobian of the one-dimensional characteristic map isBefore characteristic crossing, differentiation with respect to $\xi$ givesBecause $u_0$ has compact support, $m$ is continuous and vanishes outside a compact set. It therefore attains its minimumby the hypothesis. Since $a(t)>\epsilon$, the function $A$ is strictly increasing and tends to infinity. There is consequently a unique first time $T_{\ast }>0$ satisfyingAt a minimizer of $m$, the numerator $u_0^{\prime }$ is nonzero because $f^{\prime \prime }$ is nonzero, while the denominator tends to zero as $t\uparrow T_{\ast }$. Hence the classical solution has gradient blow-up:

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