A small note on a result in McDuff-Salamon
Originally posted: February 9th, 2026
I would like to dedicate this short blog post to filling in a few details which are glossed-over in the discussion of the Lagrangian-Hamiltonian correspondence in the first chapter of McDuff and Salamon's book on symplectic topology.
It is straightforward to demonstrate that a path \(x(t)\) is a critical point of the action functional corresponding to Lagrangian \(L\) if and only if it is a solution to the Euler-Lagrange equation,
\begin{equation} \frac{d}{dt} \frac{d L}{d v_i}(t, x(t), \dot{x}(t)) - \frac{d L}{d x_i}(t, x(t), \dot{x}(t)) = 0 \end{equation}over all \(i = 1, \dots, n\). Note that
\begin{equation} \frac{d}{dt} \frac{d L}{d v_j}(t, x(t), \dot{x}(t)) = \frac{d}{ds} \frac{d L}{d v_j}(s, x(t), \dot{x}(t)) \biggr\rvert_{s = t} + \sum_{i} \left[ \dot{x}(t) \frac{d^2 L}{d x_i d v_j}(t, x(t), \dot{x}(t)) + \ddot{x}(t) \frac{d^2 L}{d v_i d v_j}(t, x(t), \dot{x}(t))\right] \end{equation}In the case that we have the Legendre condition on \(L\), as explained in the book, then the above is a system of second-order differential equations for \(x(t)\). As per usual, we may convert a system of second-order differential equations into a larger first-order system. Usually, we do this by setting \(y = \dot{x}\), however, in this case, we can be somewhat more tactful and utilize the Legendre transform, where we define the functions \(y_k(t, x, v) = \frac{dL}{dv_k}(t, x, v)\), and we use the notation \(y_k(t) = y_k(t, x(t), \dot{x}(t)) = \frac{dL}{dv_k}(t, x(t), \dot{x}(t))\) for some trajectory \(x(t)\). It then follows that \(x(t)\) being a solution to the original Euler-Lagrange equation is equivalent to the condition that
\begin{equation} \label{eq:1} \dot{y_k}(t) = \frac{d}{dt} \frac{dL}{dv_k}(t, x(t), \dot{x}(t)) = \frac{dL}{dx_k}(t, x(t), \dot{x}(t)) \end{equation}Suppose that the Legendre condition holds at some point \((t_0, x_0, v_0)\) with \(y_0 = \frac{dL}{dv}(t_0, x_0, v_0)\), then it follows from implicit function theorem that in a neighbourhood of this point there are unique smooth functions \(G_k(t, x, y)\) such that \(y - \frac{dL}{dv}(t, x, G(t, x, y)) = 0\) in a neighbourhood \(U\) of \((t_0, x_0, y_0)\). We can use the notation \(v_k = G_k(t, x, y)\) as shorthand. It follows that on \(U\), we can define a Hamiltonian \(H : U \rightarrow \mathbb{R}\) as
\begin{equation} H(t, x, y) = \sum_{j = 1}^{n} y_j G_j(t, x, y) - L(t, x, G(t, x, y)) = \sum_{k = 1}^{n} y_j v_j - L \end{equation}Note that
\begin{equation} \label{eq:g} \frac{dH}{dy_j}(t, x, y) = G_j(t, x, y) + \sum_{k = 1}^{n} \frac{dG_k}{dy_j} \left[ y_k - \frac{dL}{dv_k}(t, x, G(t, x, y)) \right] = G_j(t, x, y) \end{equation}and
\begin{equation} \label{eq:l} \frac{dH}{dx_j}(t, x, y) = \sum_{k = 1}^{n} \frac{dG_k}{dx_j} \left[y_k - \frac{dL}{dv_k}(t, x, G(t, x, y))\right] - \frac{dL}{dx_j} = - \frac{dL}{dx_j}(t, x, G(t, x, y)) \end{equation}on \(U\). It follows immediately that if \(x(t)\) is a solution to the Euler-Lagrange equation, and the Legendre condition is satisfied at \((t_0, x(t_0), \dot{x}(t_0))\), then in a neighbourhood \(U\) of \((t_0, x(t_0), y(t_0))\), the Hamiltonian will be defined. We can choose \(t\) sufficiently close to \(t_0\) (i.e. in \((t_0 - \varepsilon, t_0 + \varepsilon)\)) so that \((t, x(t), y(t)) \in U\), and we will have
\begin{equation} \label{eq:2} \frac{dH}{dy_j}(t, x(t), y(t)) = G_j(t, x(t), y(t)) \ \ \ \ \text{and} \ \ \ \ \frac{dH}{dx_j}(t, x(t), y(t)) = -\frac{dL}{dx_j}(t, x(t), G(t, x(t), y(t))) \end{equation}By definition, \(y(t) = \frac{dL}{dv}(t, x(t), \dot{x}(t))\), so both \(t \mapsto G(t, x(t), \dot{x}(t))\) and \(t \mapsto \dot{x}(t)\) are smooth functions on \((t_0 - \varepsilon, t_0 + \varepsilon)\) such that \(y(t) - \frac{dL}{dv}(t, x(t), v(t)) = 0\) when plugged-in as \(v(t)\). However, the Legendre condition and implicit function theorem imply that such a \(v(t)\) in a neighbourhood of \(t_0\) is unique, so on interval \((t_0 - \delta, t_0 + \delta)\), we have \(\dot{x}(t) = G(t, x(t), y(t))\), and using Eq. \eqref{eq:1}, we have
\begin{equation} \label{eq:3} \frac{dH}{dy_j}(t, x(t), y(t)) = \dot{x}_j(t) \ \ \ \ \text{and} \ \ \ \ \frac{dH}{dx_j}(t, x(t), y(t)) = -\dot{y}_j(t) \end{equation}Thus, in summary, we have shown that if \(x(t)\) is a solution to the Euler-Lagrange equation, and the Legendre condition is satisfied at \((t_0, x(t_0), \dot{x}(t_0))\), then there exists an interval around \(t_0\) such that \(x(t)\) and \(y(t)\) satisfy the above differential equations, which we refer to as the Hamilton equations.
Conversely, suppose the Legendre condition of the Lagrangian \(L\) is satisfied at point \((t_0, x_0, v_0)\). We define \(G_j(t, x, y)\) as before, so that \(y - \frac{dL}{dv}(t, x, G(t, x, y)) = 0\) on \(U\) about \((t_0, x_0, y_0)\) with \(y_0 = \frac{dL}{dv}(t_0, x_0, v_0)\). Suppose the functions \(x(t)\) and \(y(t)\) with \(x(t_0) = x_0\) and \(y(t_0) = y_0\) satisfy the Hamilton equations on some interval about \(t_0\). Note that Eq. \eqref{eq:g} and Eq. \eqref{eq:3} imply \(G(t, x(t), y(t)) = \dot{x}(t)\) so \(y(t) = \frac{dL}{dv}(t, x(t), \dot{x}(t))\), and we have
\begin{align} 0 = \dot{y}_j(t) - \frac{d}{dt} \frac{dL}{dv_j}(t, x(t), \dot{x}(t)) &= -\frac{dH}{dx_j}(t, x(t), y(t)) - \frac{d}{dt} \frac{dL}{dv_j}(t, x(t), \dot{x}(t)) \\ &= \frac{dL}{dx_j}(t, x(t), \dot{x}(t)) - \frac{d}{dt} \frac{dL}{dv_j}(t, x(t), \dot{x}(t)) \end{align}where the final equality uses Eq. \eqref{eq:l}. This is precisely the Euler-Lagrange equation. Hence, we have proved the following claim:
Claim. If \(L\) is a Lagrangian which satisfies the Legendre condition at some point \((t_0, x_0, v_0)\), then trajectory \(x(t)\) with \(x(t_0) = x_0\) and \(\dot{x}(t_0) = v_0\) satisfies the Euler-Lagrange equation on some open interval around \(t_0\) if and only if there exists some \(y(t)\) also defined on an interval around \(t_0\) with \(y(t_0) = y_0 = \frac{dL}{dv}(t_0, x_0, v_0)\) such that \(x(t)\) and \(y(t)\) satisfy the Hamilton equations, for the Hamiltonian \(H\) defined on some open set around \((t_0, x_0, y_0)\). Moreover, when such a \(y\) exists, it is always the case that \(y(t) = \frac{dL}{dv}(t, x(t), \dot{x}(t))\).