[proofplan] We restrict the solution $u$ to the characteristic curve $\gamma$ and study the one-variable function $v=u\circ\gamma$. The chain rule converts the derivative of $v$ into $\nabla u(\gamma(s))\cdot\gamma'(s)$, and the characteristic equation $\gamma'=b(\gamma)$ changes this into $b(\gamma(s))\cdot\nabla u(\gamma(s))$. The pointwise transport equation then gives $v'=f\circ\gamma$, and the [fundamental theorem of calculus](/theorems/632) on the interval between $s_0$ and $s_1$ yields the desired identity. [/proofplan]

[step:Compose $u$ with the characteristic curve]Define the function \begin{align*} v:I\to\mathbb{R},\qquad s\mapsto u(\gamma(s)). \end{align*} Since $\gamma\in C^1(I;U)$ and $u\in C^1(U)$, the chain rule for $C^1$ functions along differentiable curves gives $v\in C^1(I)$ and, for every $s\in I$, \begin{align*} v'(s)=\nabla u(\gamma(s))\cdot \gamma'(s). \end{align*}[/step]

[guided]The goal is to turn the [partial differential equation](/page/Partial%20Differential%20Equation) into an [ordinary differential equation](/page/Ordinary%20Differential%20Equation) along the curve. To do that, we introduce the one-variable function \begin{align*} v:I\to\mathbb{R},\qquad s\mapsto u(\gamma(s)). \end{align*} This function is well-defined because $\gamma(s)\in U$ for every $s\in I$ and $u:U\to\mathbb{R}$. We now verify the hypotheses of the chain rule for $C^1$ functions along differentiable curves. The curve $\gamma:I\to U$ is $C^1$ by hypothesis, and the function $u:U\to\mathbb{R}$ is $C^1$ by hypothesis. Therefore the composition $v=u\circ\gamma$ is $C^1$ on $I$, and for each $s\in I$ its derivative is \begin{align*} v'(s)=\nabla u(\gamma(s))\cdot \gamma'(s). \end{align*} This is the key reduction: the left side is a one-dimensional derivative in the parameter $s$, while the right side contains the [directional derivative](/page/Directional%20Derivative) of $u$ in the direction of the velocity vector $\gamma'(s)$.[/guided]

[step:Use the characteristic equation and the transport equation to identify $v'$] Fix $s\in I$. Since $\gamma$ solves the ordinary differential equation classically, \begin{align*} \gamma'(s)=b(\gamma(s)). \end{align*} Substituting this into the chain-rule identity gives \begin{align*} v'(s)=\nabla u(\gamma(s))\cdot b(\gamma(s)). \end{align*} The Euclidean dot product is symmetric, so \begin{align*} v'(s)=b(\gamma(s))\cdot \nabla u(\gamma(s)). \end{align*} Because $u$ satisfies $b\cdot\nabla u=f$ pointwise on $U$ and $\gamma(s)\in U$, we obtain \begin{align*} v'(s)=f(\gamma(s)). \end{align*} Thus $v'=f\circ\gamma$ on $I$. [/step]

[step:Integrate the one-dimensional identity along the parameter interval] The function $f\circ\gamma:I\to\mathbb{R}$ is continuous because $f\in C^0(U)$ and $\gamma\in C^1(I;U)$. Hence $v'\in C^0(I)$ and the fundamental theorem of calculus for continuous derivatives applies on the compact interval with endpoints $s_0$ and $s_1$. If $s_0\leq s_1$, then \begin{align*} v(s_1)-v(s_0)=\int_{s_0}^{s_1} v'(s)\,d\mathcal{L}^1(s). \end{align*} Using $v'=f\circ\gamma$, this becomes \begin{align*} v(s_1)-v(s_0)=\int_{s_0}^{s_1} f(\gamma(s))\,d\mathcal{L}^1(s). \end{align*} If $s_1<s_0$, the same identity follows from the preceding case applied to the ordered pair $(s_1,s_0)$ and the oriented-interval convention \begin{align*} \int_{s_0}^{s_1} f(\gamma(s))\,d\mathcal{L}^1(s)=-\int_{s_1}^{s_0} f(\gamma(s))\,d\mathcal{L}^1(s). \end{align*} Finally, substituting $v(s)=u(\gamma(s))$ gives \begin{align*} u(\gamma(s_1))-u(\gamma(s_0))=\int_{s_0}^{s_1} f(\gamma(s))\,d\mathcal{L}^1(s). \end{align*} This is the claimed characteristic formula. [/step]