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The inverse function theorem applied to $X_t$ makes $u_0\circ X_t^{-1}$ smooth. Since $u_0(\pm A)=0$, $X_t(\pm A)=\pm A+ct$. Monotonicity of the characteristic flow map and the vanishing of $u_0$ outside $[-A,A]$ implyThus the solution is a smooth function with compact support at each such time. The fact that some interior characteristic speeds differ from $c$ causes no difficulty: they cannot cross the two exterior characteristic curves before $t^{\ast }$.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelöf theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

With $g=0$, pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so $u$ is measurable and $\Vert u{\Vert }_{\infty }\le \Vert u_0{\Vert }_{\infty }$. Choose smooth $u_0^{(k)}$ converging to $u_0$ in $L_{\mathrm{loc}}^1$ with uniformly bounded essential suprema, and definePart a makes each $u^{(k)}$ a classical, hence weak, solution. On every compact subset of spacetime, the $C^1$ change-of-variables formula for the flow and its locally bounded Jacobian determinant give $u^{(k)}\to u$ in $L^1$. Passing to the limit in the weak identity by dominated convergence proves that $u$ is a bounded weak solution with initial datum $u_0$.

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.

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.

The Jacobian determinant of a differentiable characteristic flow map measures its change of volume. The variational equation and the Liouville formula for a fundamental matrix give the displayed formula for a complete smooth field $b$. For kinetic advection with $b=(v,F)$, its divergence is ${\nabla }_v\cdot F$. A zero divergence preserves phase-space Lebesgue measure.

Differentiating a smooth characteristic flow map in its initial point gives this linear equation, with $Y(s)=I$. The Liouville formula for a fundamental matrix then gives $\det Y(t)=\exp ({\int }_s^t\mathrm{div}b(r,X(r))dr)$. Incompressible Hamiltonian flows preserve Lebesgue measure because this divergence vanishes.