Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 1 / a / Solution

The characteristic equations for a transport equation are $\dot{X}=V$ and $\dot{V}=X$, so $\ddot{X}=X$. For an initial point $(x_0,v_0)$ their solution isThis is the hyperbolic characteristic flow for an inverted oscillator. The addition formulas give $A_s{A}_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward characteristic flow map from the point $(x,v)$ at time $t$ to time $s$ isAlong this characteristic curve, the chain rule changes the transport equation into $d[f(s,S_{s,t}(x,v))]/ds=h(s,S_{s,t}(x,v))$. Integrating from zero to $t$ givesThe assumed $C^1$ regularity makes this a classical solution: on every compact set, the integrand and its needed derivatives are continuous, so differentiation under the finite-time integral is justified. At $t=0$ it has the required initial value, and the characteristic calculation verifies the equation. Conversely every classical solution must satisfy the same integrated identity, proving uniqueness. This is the Duhamel formula for Hamiltonian transport, with Hamiltonian $(v^2-x^2)/2$.