[proofplan] The first characteristic equation is immediate after replacing $z(s)$ by its definition $u(x(s))$. The second characteristic equation follows by differentiating the composed function $z = u \circ x$. The derivative is computed from the differentiability of $u$ and $x$, and the PDE then converts the resulting directional derivative into $b(x(s),z(s))$. [/proofplan] [step:Rewrite the prescribed equation for $x$ using the graph variable $z$] Let $I := (s_0,s_1)$ denote the domain of the curve $x$. Fix $s \in I$. Since $z(s) = u(x(s))$, the assumed differential equation for $x$ gives \begin{align*} \dot{x}(s) = a(x(s),u(x(s))). \end{align*} Replacing $u(x(s))$ by $z(s)$ gives \begin{align*} \dot{x}(s) = a(x(s),z(s)). \end{align*} Thus the first equation in the characteristic system holds for every $s \in I$. [/step] [step:Differentiate the graph variable along the base curve] Fix $s \in I$. Because $u \in C^1(U)$ and $x \in C^1(I;U)$, the composition $z = u \circ x$ is differentiable at $s$. By differentiability of $u$ at $x(s)$ and differentiability of $x$ at $s$, its derivative is \begin{align*} \dot{z}(s) = \nabla u(x(s)) \cdot \dot{x}(s). \end{align*} Substituting the assumed equation for $\dot{x}(s)$ gives \begin{align*} \dot{z}(s) = \nabla u(x(s)) \cdot a(x(s),u(x(s))). \end{align*} [guided] We need to prove that the height coordinate $z(s)$ evolves according to the second characteristic equation. Since $z$ was defined by following the graph of $u$ along the curve $x$, namely $z(s) = u(x(s))$, its derivative must measure how $u$ changes in the direction of motion of $x$. Fix $s \in I$. The assumptions $u \in C^1(U)$ and $x \in C^1(I;U)$ imply that $u$ is differentiable at $x(s)$ and that $x$ is differentiable at $s$. Therefore the derivative of the composition $z = u \circ x$ is computed by the directional derivative of $u$ in the velocity direction $\dot{x}(s)$: \begin{align*} \dot{z}(s) = \nabla u(x(s)) \cdot \dot{x}(s). \end{align*} The base curve $x$ is assumed to solve \begin{align*} \dot{x}(s) = a(x(s),u(x(s))). \end{align*} Substituting this expression for $\dot{x}(s)$ into the derivative of $z$ gives \begin{align*} \dot{z}(s) = \nabla u(x(s)) \cdot a(x(s),u(x(s))). \end{align*} This is the exact point where the motion of the base curve and the graph relation $z = u \circ x$ combine: the velocity of the lifted curve in the vertical direction is the directional derivative of $u$ in the characteristic direction. [/guided] [/step] [step:Use the first-order equation to identify the vertical characteristic equation] The hypothesis on $u$ applies at the point $x(s) \in U$, so \begin{align*} a(x(s),u(x(s))) \cdot \nabla u(x(s)) = b(x(s),u(x(s))). \end{align*} Since the Euclidean dot product is symmetric, the expression obtained for $\dot{z}(s)$ becomes \begin{align*} \dot{z}(s) = b(x(s),u(x(s))). \end{align*} Using again $z(s) = u(x(s))$, we obtain \begin{align*} \dot{z}(s) = b(x(s),z(s)). \end{align*} Together with the first characteristic equation proved above, this shows that $\gamma(s) = (x(s),z(s))$ solves the characteristic system on $I$. [/step]