[proofplan] We derive the characteristic ODE system by differentiating two identities: the composition $U(s) = u(X(s))$ and the PDE constraint $F(\nabla u(x), u(x), x) = 0$. The $\dot{U}$ equation follows from a direct chain rule application. The $\dot{G}_i$ equation requires differentiating the PDE identity $F(\tau(x)) = 0$ with respect to $x_i$ to express the second-order term $\sum_j (\partial^2 u / \partial x_i \partial x_j) \cdot (\partial F / \partial a_j)$ in terms of first-order data, then recognising this sum as $\dot{G}_i(s)$ along the characteristic curve. [/proofplan] [step:Derive the $\dot{U}$ equation via the chain rule] Since $U(s) = u(X(s))$ and $u \in C^2(\Omega)$, the chain rule gives \begin{align*} \dot{U}(s) = \sum_{j=1}^n \frac{\partial u}{\partial x_j}(X(s)) \, \dot{X}_j(s) = \sum_{j=1}^n G_j(s) \, \dot{X}_j(s). \end{align*} Substituting the characteristic velocity condition $\dot{X}_j(s) = \frac{\partial F}{\partial a_j}\big|_{T(s)}$: \begin{align*} \dot{U}(s) = \sum_{j=1}^n G_j(s) \, \frac{\partial F}{\partial a_j}\bigg|_{T(s)}. \end{align*} [/step] [step:Differentiate the PDE identity $F(\tau(x)) = 0$ with respect to $x_i$] Define the auxiliary map $\tau: \Omega \to \mathbb{R}^n \times \mathbb{R} \times \mathbb{R}^n$ by $\tau(x) := (\nabla u(x), u(x), x)$, so the [first-order PDE](/page/Imperial%20Theory%20of%20PDEs%3A%20First%20Order%20Equations) reads $F(\tau(x)) = 0$ for all $x \in \Omega$. Since $u \in C^2(\Omega)$ and $F$ is smooth, the composition $F \circ \tau$ is $C^1$. Differentiating the identity $F(\tau(x)) = 0$ with respect to $x_i$ via the chain rule: \begin{align*} 0 = \frac{\partial}{\partial x_i}\bigl[F(\tau(x))\bigr] = \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{\tau(x)} \frac{\partial^2 u}{\partial x_i \partial x_j}(x) + \frac{\partial F}{\partial b}\bigg|_{\tau(x)} \frac{\partial u}{\partial x_i}(x) + \frac{\partial F}{\partial c_i}\bigg|_{\tau(x)}. \end{align*} [guided] The PDE identity $F(\nabla u(x), u(x), x) = 0$ holds for every $x \in \Omega$. Since this is an identity in $x$ (not just at a single point), we may differentiate both sides with respect to $x_i$. The map $\tau(x) = (\nabla u(x), u(x), x)$ has $2n + 1$ component functions of $x$, and $F$ depends on $2n + 1$ variables $(a_1, \ldots, a_n, b, c_1, \ldots, c_n)$. Applying the multivariable chain rule to the composition $x \mapsto F(\tau(x))$: \begin{align*} \frac{\partial}{\partial x_i}\bigl[F(\tau(x))\bigr] &= \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{\tau(x)} \cdot \frac{\partial}{\partial x_i}\bigl[\partial_{x_j} u(x)\bigr] + \frac{\partial F}{\partial b}\bigg|_{\tau(x)} \cdot \frac{\partial u}{\partial x_i}(x) + \sum_{k=1}^n \frac{\partial F}{\partial c_k}\bigg|_{\tau(x)} \cdot \frac{\partial x_k}{\partial x_i}. \end{align*} The third sum simplifies because $\partial x_k / \partial x_i = \delta_{ki}$, so only the $k = i$ term survives. The first sum uses $\frac{\partial}{\partial x_i}[\partial_{x_j} u] = \frac{\partial^2 u}{\partial x_i \partial x_j}$, which exists and is continuous since $u \in C^2(\Omega)$. Setting the result equal to zero: \begin{align*} 0 = \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{\tau(x)} \frac{\partial^2 u}{\partial x_i \partial x_j}(x) + \frac{\partial F}{\partial b}\bigg|_{\tau(x)} \frac{\partial u}{\partial x_i}(x) + \frac{\partial F}{\partial c_i}\bigg|_{\tau(x)}. \end{align*} This identity relates the second derivatives of $u$ to first-order data via $F$, which is the key to eliminating second derivatives from the $\dot{G}_i$ equation. [/guided] [/step] [step:Derive the $\dot{G}_i$ equation by evaluating along $X(s)$] Since $G_i(s) = \partial_{x_i} u(X(s))$ and $u \in C^2(\Omega)$, the chain rule gives \begin{align*} \dot{G}_i(s) = \sum_{j=1}^n \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) \, \dot{X}_j(s) = \sum_{j=1}^n \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) \, \frac{\partial F}{\partial a_j}\bigg|_{T(s)}. \end{align*} Evaluating the differentiated PDE identity from the previous step at $x = X(s)$ and rearranging for the sum involving second derivatives: \begin{align*} \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{T(s)} \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) = -\frac{\partial F}{\partial b}\bigg|_{T(s)} G_i(s) - \frac{\partial F}{\partial c_i}\bigg|_{T(s)}. \end{align*} The left-hand side is exactly $\dot{G}_i(s)$, so \begin{align*} \dot{G}_i(s) = -\frac{\partial F}{\partial c_i}\bigg|_{T(s)} - G_i(s) \, \frac{\partial F}{\partial b}\bigg|_{T(s)}. \end{align*} [guided] The $\dot{G}_i$ equation is the most substantial part of the derivation, because naively $\dot{G}_i$ involves the second derivatives $\partial^2 u / \partial x_i \partial x_j$ evaluated at $X(s)$. These are not part of the characteristic data $(G, U, X)$. The whole point of the [method of characteristics](/page/Method%20of%20Characteristics) is that these second derivatives can be eliminated using the PDE. From the chain rule applied to $G_i(s) = \partial_{x_i} u(X(s))$: \begin{align*} \dot{G}_i(s) = \sum_{j=1}^n \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) \, \dot{X}_j(s). \end{align*} Substituting the characteristic velocity $\dot{X}_j(s) = \frac{\partial F}{\partial a_j}\big|_{T(s)}$: \begin{align*} \dot{G}_i(s) = \sum_{j=1}^n \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) \, \frac{\partial F}{\partial a_j}\bigg|_{T(s)}. \end{align*} Now look at the differentiated PDE identity evaluated at $x = X(s)$. Note that $\tau(X(s)) = (\nabla u(X(s)), u(X(s)), X(s)) = (G(s), U(s), X(s)) = T(s)$. The identity reads \begin{align*} \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{T(s)} \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) + \frac{\partial F}{\partial b}\bigg|_{T(s)} G_i(s) + \frac{\partial F}{\partial c_i}\bigg|_{T(s)} = 0, \end{align*} where we used $\frac{\partial u}{\partial x_i}(X(s)) = G_i(s)$ by definition. Solving for the sum on the left: \begin{align*} \sum_{j=1}^n \frac{\partial F}{\partial a_j}\bigg|_{T(s)} \frac{\partial^2 u}{\partial x_i \partial x_j}(X(s)) = -\frac{\partial F}{\partial b}\bigg|_{T(s)} G_i(s) - \frac{\partial F}{\partial c_i}\bigg|_{T(s)}. \end{align*} The left-hand side is precisely $\dot{G}_i(s)$, and the right-hand side involves only the characteristic data $T(s) = (G(s), U(s), X(s))$. Therefore \begin{align*} \dot{G}_i(s) = -\frac{\partial F}{\partial c_i}\bigg|_{T(s)} - G_i(s) \, \frac{\partial F}{\partial b}\bigg|_{T(s)}. \end{align*} This is the crucial "miracle" of the method of characteristics: the PDE provides exactly the relation needed to close the system -- the second derivatives drop out and the evolution of $(G, U, X)$ depends only on $(G, U, X)$. [/guided] [/step] [step:Assemble the three equations into the characteristic ODE system] Collecting the results, the triple $T(s) = (G(s), U(s), X(s))$ satisfies the autonomous first-order ODE system: \begin{align*} \dot{X}_j(s) &= \frac{\partial F}{\partial a_j}\bigg|_{T(s)}, \\ \dot{G}_i(s) &= -\frac{\partial F}{\partial c_i}\bigg|_{T(s)} - G_i(s) \, \frac{\partial F}{\partial b}\bigg|_{T(s)}, \\ \dot{U}(s) &= \sum_{j=1}^n G_j(s) \, \frac{\partial F}{\partial a_j}\bigg|_{T(s)}, \end{align*} for $i, j = 1, \ldots, n$. The first equation is the defining hypothesis on $X$, and the second and third are derived consequences. The system is closed: the right-hand sides depend only on $T(s)$, not on second derivatives of $u$. [/step]