[proofplan] The proof is the characteristic reduction of the transport equation to a scalar ordinary differential equation. We first verify that the composition $v(t)=u(t,X(t))$ is differentiable along the characteristic curve. Then the chain rule expresses $v'$ as the sum of the time derivative of $u$ and the spatial directional derivative of $u$ in the velocity direction $X'(t)$. The characteristic equation replaces $X'(t)$ by $b(t,X(t))$, and the PDE then gives the asserted inhomogeneous linear ODE. The initial condition follows from $X(0)=x_0$ and $u(0,\cdot)=u_0$. [/proofplan] [step:Differentiate the solution along the characteristic curve] The map \begin{align*} \Gamma: [0,\tau] \to [0,T] \times U, \quad t \mapsto (t,X(t)) \end{align*} is $C^1$ because $X \in C^1([0,\tau];U)$. Since $u \in C^1([0,T]\times U)$, the composition $v=u\circ \Gamma$ belongs to $C^1([0,\tau])$. Applying the chain rule for $C^1$ maps to $u \circ \Gamma$ gives, for every $t \in (0,\tau)$, \begin{align*} v'(t) = \partial_t u(t,X(t)) + \nabla u(t,X(t)) \cdot X'(t). \end{align*} (citing a result not yet in the wiki: chain rule for $C^1$ maps) [guided] Define the path in space-time by \begin{align*} \Gamma: [0,\tau] \to [0,T] \times U, \quad t \mapsto (t,X(t)). \end{align*} This is the right object because $u$ is a function of both time and space, while $v$ is obtained by restricting $u$ to the one-dimensional curve $t \mapsto (t,X(t))$ in space-time. The curve $\Gamma$ is $C^1$: its first component is the identity map $t \mapsto t$, and its second component is the assumed $C^1$ curve $X: [0,\tau]\to U$. Since $u \in C^1([0,T]\times U)$, the composition \begin{align*} v = u \circ \Gamma \end{align*} is a $C^1$ function from $[0,\tau]$ to $\mathbb{R}$ by the chain rule for $C^1$ maps. The derivative of $\Gamma$ at $t$ is the pair consisting of the time velocity $1$ and the spatial velocity $X'(t)$. Therefore the chain rule gives \begin{align*} v'(t) = \partial_t u(t,X(t)) \cdot 1 + \nabla u(t,X(t)) \cdot X'(t). \end{align*} Since multiplication by $1$ does not change the first term, this becomes \begin{align*} v'(t) = \partial_t u(t,X(t)) + \nabla u(t,X(t)) \cdot X'(t). \end{align*} This is the exact point where the PDE begins to reduce to an ODE: the only remaining spatial information is the directional derivative of $u$ in the direction of the characteristic velocity. (citing a result not yet in the wiki: chain rule for $C^1$ maps) [/guided] [/step] [step:Use the characteristic equation to replace the spatial velocity] The characteristic equation states that \begin{align*} X'(t) = b(t,X(t)) \end{align*} for every $t \in (0,\tau)$. Substituting this into the derivative formula gives \begin{align*} v'(t) = \partial_t u(t,X(t)) + b(t,X(t)) \cdot \nabla u(t,X(t)). \end{align*} [/step] [step:Substitute the transport equation along the curve] For each $t \in (0,\tau)$, the point $(t,X(t))$ lies in $(0,T)\times U$, so the transport equation applies at that point: \begin{align*} \partial_t u(t,X(t)) + b(t,X(t)) \cdot \nabla u(t,X(t)) + c(t,X(t))u(t,X(t)) = f(t,X(t)). \end{align*} Using $v(t)=u(t,X(t))$ and the expression for $v'(t)$ from the previous step, we obtain \begin{align*} v'(t) + c(t,X(t))v(t) = f(t,X(t)). \end{align*} Thus $v$ satisfies the asserted scalar inhomogeneous linear ordinary differential equation on $(0,\tau)$. [/step] [step:Evaluate the initial value at the starting point of the characteristic] Using the definition of $v$, the initial condition for the characteristic, and the initial condition for $u$, we compute \begin{align*} v(0) = u(0,X(0)). \end{align*} Since $X(0)=x_0$, this becomes \begin{align*} v(0) = u(0,x_0). \end{align*} Since $u(0,x)=u_0(x)$ for every $x \in U$, evaluating at $x_0 \in U$ gives \begin{align*} v(0) = u_0(x_0). \end{align*} This is the required initial condition, and the proof is complete. [/step]