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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 $(v^2-x^2)/2$.

Fix the mixed Fourier transform conventionThe spatial derivative transforms to $2\pi ik\hat{f}$, and multiplication by $v$ transforms to $-(2\pi i{)}^{-1}{\partial }_{\xi }\hat{f}$. Hence the transformed free transport equation isIts characteristic equations for a transport equation give $\dot{\xi }=-k$, so the characteristic ending at $\xi$ at time $t$ began at $\xi +kt$. ConsequentlyThe same sign follows directly by substituting $x=y+tv$ in the Fourier transform of $f_0(x-tv,v)$. No first velocity moment is assumed, so the differential equation may be understood in the sense of tempered distributions; the explicit transform formula is valid pointwise because $f_0$ is integrable.