Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 3

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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