= 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 is
$$
\binom{X(t)}{V(t)}=A_t\binom{x_0}{v_0},\qquad
A_t=\begin{pmatrix}\cosh t&\sinh t\\\sinh t&\cosh t\end{pmatrix}.
$$
This is the <hyperbolic characteristic flow for an inverted oscillator>. The addition formulas give $A_sA_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward <characteristic flow map> from the point $(x,v)$ at time $t$ to time $s$ is
$$
S_{s,t}(x,v)=\bigl(x\cosh(t-s)-v\sinh(t-s),\ v\cosh(t-s)-x\sinh(t-s)\bigr).
$$
Along 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 $t$ gives
$$
\boxed{f(t,x,v)=f_0(S_{0,t}(x,v))+\int_0^t h(s,S_{s,t}(x,v))\,ds.}
$$
The 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 $(v^2-x^2)/2$.
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