[proofplan] We derive the prolongation formula by induction on the derivative order $k$. The base case ($k = 1$) is established by applying the contact condition $d\tilde{u} = \tilde{u}_{\tilde{x}}\, d\tilde{x}$ (i.e., the chain rule) to the infinitesimally transformed variables and extracting the $O(\varepsilon)$ coefficient. The inductive step uses the higher-order contact condition $d\tilde{u}^{(k)} = \tilde{u}^{(k+1)}\, d\tilde{x}$ together with the same extraction procedure, yielding the recursion $\eta_{k+1} = D_x \eta_k - u^{(k+1)} D_x \xi$. [/proofplan]

[step:Set up the infinitesimal transformation and define the prolonged coefficients] Let $V = \xi(x, u)\, \partial/\partial x + \eta(x, u)\, \partial/\partial u$ be a vector field on the $(x, u)$-space, and let $g^\varepsilon$ be its one-parameter group. By the [Vector Field Generates A One-Parameter Group](/theorems/1349) theorem, the transformed variables are \begin{align*} \tilde{x} &= x + \varepsilon\, \xi(x, u) + O(\varepsilon^2), \\ \tilde{u} &= u + \varepsilon\, \eta(x, u) + O(\varepsilon^2). \end{align*} The $n$-th prolongation $\mathrm{pr}^{(n)} g^\varepsilon$ acts on the jet space $(x, u, u^{(1)}, \ldots, u^{(n)})$ by extending the transformation to all derivatives up to order $n$. Since the transformation is near the identity, the prolonged action has the form \begin{align*} \tilde{u}^{(k)} = u^{(k)} + \varepsilon\, \eta_k + O(\varepsilon^2), \qquad k = 0, 1, \ldots, n, \end{align*} where $\eta_0 := \eta(x, u)$ and the coefficients $\eta_k$ are to be determined. The prolonged vector field is then \begin{align*} \mathrm{pr}^{(n)} V = \xi\, \frac{\partial}{\partial x} + \eta\, \frac{\partial}{\partial u} + \sum_{k=1}^n \eta_k\, \frac{\partial}{\partial u^{(k)}}. \end{align*} Our task is to derive the recursion for $\eta_k$. [/step]

[step:Derive $\eta_1 = D_x \eta - u^{(1)} D_x \xi$ from the first-order contact condition] The derivative $\tilde{u}^{(1)} = d\tilde{u}/d\tilde{x}$ is defined by the contact condition \begin{align*} d\tilde{u} = \tilde{u}^{(1)}\, d\tilde{x}, \end{align*} which is the statement that $\tilde{u}$ is a function of $\tilde{x}$ (the chain rule). We compute each side to $O(\varepsilon)$. **Left-hand side.** Differentiating $\tilde{u} = u + \varepsilon\, \eta(x, u) + O(\varepsilon^2)$ with respect to $x$ (the original independent variable): \begin{align*} \frac{d\tilde{u}}{dx} = u_x + \varepsilon\, D_x \eta + O(\varepsilon^2), \end{align*} where $D_x$ denotes the total derivative with respect to $x$: \begin{align*} D_x \eta = \frac{\partial \eta}{\partial x} + u_x\, \frac{\partial \eta}{\partial u}. \end{align*} So $d\tilde{u} = (u_x + \varepsilon\, D_x \eta)\, dx + O(\varepsilon^2)$. **Right-hand side.** Similarly, $d\tilde{x} = (1 + \varepsilon\, D_x \xi)\, dx + O(\varepsilon^2)$ where \begin{align*} D_x \xi = \frac{\partial \xi}{\partial x} + u_x\, \frac{\partial \xi}{\partial u}. \end{align*} Therefore the contact condition $d\tilde{u} = \tilde{u}^{(1)}\, d\tilde{x}$ becomes \begin{align*} (u_x + \varepsilon\, D_x \eta)\, dx = \tilde{u}^{(1)} (1 + \varepsilon\, D_x \xi)\, dx. \end{align*} Dividing both sides by $dx$ and solving for $\tilde{u}^{(1)}$: \begin{align*} \tilde{u}^{(1)} = \frac{u_x + \varepsilon\, D_x \eta}{1 + \varepsilon\, D_x \xi}. \end{align*} Expanding the denominator using $(1 + \varepsilon\, D_x \xi)^{-1} = 1 - \varepsilon\, D_x \xi + O(\varepsilon^2)$: \begin{align*} \tilde{u}^{(1)} &= (u_x + \varepsilon\, D_x \eta)(1 - \varepsilon\, D_x \xi + O(\varepsilon^2)) \\ &= u_x + \varepsilon(D_x \eta - u_x\, D_x \xi) + O(\varepsilon^2). \end{align*} Comparing with $\tilde{u}^{(1)} = u^{(1)} + \varepsilon\, \eta_1 + O(\varepsilon^2)$ and recalling $u^{(1)} = u_x$, we read off \begin{align*} \eta_1 = D_x \eta - u^{(1)} D_x \xi. \end{align*} [/step]

[step:Prove the recursion $\eta_{k+1} = D_x \eta_k - u^{(k+1)} D_x \xi$ by induction using the higher-order contact condition] We proceed by induction on $k$. The base case $k = 0$ (giving $\eta_1$) was established in the previous step. Assume the formula holds for $\eta_k$, so that $\tilde{u}^{(k)} = u^{(k)} + \varepsilon\, \eta_k + O(\varepsilon^2)$. The higher-order contact condition at order $k$ states \begin{align*} d\tilde{u}^{(k)} = \tilde{u}^{(k+1)}\, d\tilde{x}. \end{align*} This is the requirement that $\tilde{u}^{(k+1)} = d\tilde{u}^{(k)}/d\tilde{x}$, i.e., each prolonged derivative is obtained by differentiating the previous one with respect to the transformed independent variable. **Left-hand side.** Differentiating $\tilde{u}^{(k)} = u^{(k)} + \varepsilon\, \eta_k + O(\varepsilon^2)$ with respect to $x$: \begin{align*} \frac{d\tilde{u}^{(k)}}{dx} = u^{(k+1)} + \varepsilon\, D_x \eta_k + O(\varepsilon^2), \end{align*} where $D_x \eta_k$ is the total $x$-derivative of $\eta_k$ (which depends on $x$, $u$, $u^{(1)}, \ldots, u^{(k+1)}$ through the chain rule). **Right-hand side.** As before, $d\tilde{x}/dx = 1 + \varepsilon\, D_x \xi + O(\varepsilon^2)$. Substituting into the contact condition and dividing by $dx$: \begin{align*} u^{(k+1)} + \varepsilon\, D_x \eta_k = \tilde{u}^{(k+1)} (1 + \varepsilon\, D_x \xi) + O(\varepsilon^2). \end{align*} Solving for $\tilde{u}^{(k+1)}$ by expanding the inverse: \begin{align*} \tilde{u}^{(k+1)} &= (u^{(k+1)} + \varepsilon\, D_x \eta_k)(1 - \varepsilon\, D_x \xi) + O(\varepsilon^2) \\ &= u^{(k+1)} + \varepsilon(D_x \eta_k - u^{(k+1)} D_x \xi) + O(\varepsilon^2). \end{align*} Comparing with $\tilde{u}^{(k+1)} = u^{(k+1)} + \varepsilon\, \eta_{k+1} + O(\varepsilon^2)$, we obtain \begin{align*} \eta_{k+1} = D_x \eta_k - u^{(k+1)} D_x \xi, \end{align*} completing the induction. [/step]

[step:Assemble the prolongation formula]Combining the results of the previous steps, the $n$-th prolongation of $V = \xi\, \partial/\partial x + \eta\, \partial/\partial u$ is \begin{align*} \mathrm{pr}^{(n)} V = \xi\, \frac{\partial}{\partial x} + \eta\, \frac{\partial}{\partial u} + \sum_{k=1}^n \eta_k\, \frac{\partial}{\partial u^{(k)}}, \end{align*} where $\eta_0 = \eta(x, u)$ and the coefficients satisfy the recursion \begin{align*} \eta_{k+1} = D_x \eta_k - u^{(k+1)} D_x \xi, \qquad k = 0, 1, \ldots, n-1. \end{align*} This determines each $\eta_k$ explicitly in terms of $\xi$, $\eta$, and the jet variables $u^{(1)}, \ldots, u^{(k)}$. For instance, the first two coefficients are \begin{align*} \eta_1 &= D_x \eta - u^{(1)} D_x \xi, \\ \eta_2 &= D_x \eta_1 - u^{(2)} D_x \xi. \end{align*}[/step]

[guided]The recursion has a clean structure: at each order, the new coefficient $\eta_{k+1}$ is obtained by applying the total derivative $D_x$ to the previous coefficient $\eta_k$ and subtracting a correction term $u^{(k+1)} D_x \xi$ that accounts for the transformation of the independent variable. This correction is the same at every order -- only the derivative $u^{(k+1)}$ that multiplies $D_x \xi$ changes. The formula becomes more transparent if one considers the special case $\xi = 0$ (the transformation does not move $x$). Then $D_x \xi = 0$ and the recursion simplifies to $\eta_{k+1} = D_x \eta_k$, meaning each prolongation coefficient is just the total $x$-derivative of the previous one. This is natural: if $x$ is unchanged, then $\tilde{u}^{(k)} = (d/dx)^k \tilde{u}$, and the infinitesimal change in the $k$-th derivative is simply the $k$-th total derivative of the infinitesimal change in $u$. The $-u^{(k+1)} D_x \xi$ term accounts for the additional complication when the independent variable is also being transformed.[/guided]