# Constructing a Lagrangian whose Euler-Lagrange equation yields a Poisson bracket

I was wondering if it is possible to construct a Lagrangian $$\mathcal{L}$$ such that the first variation of the action $$\mathcal{S}(f, g)$$ in the second argument $$g$$ yields the Poisson bracket of $$f$$ and $$g$$? That is, can we find an $$\mathcal{L}$$ such that

\begin{align} \mathcal{S}(f, g) &= \int_{\Omega^{2}} \mathcal{L}(x, y, f, \nabla f, g, \nabla g) \ dx dy \\ \implies \frac{\delta \mathcal{S}(f, g)}{\delta g} &= \frac{d}{d \epsilon} \left[ \int_{\Omega^{2}} \mathcal{L} (x, y, f, \nabla f, g + \epsilon \phi, \nabla g + \epsilon \nabla \phi) \ dx dy \right]_{\epsilon = 0} \\ &= \int_{\Omega^{2}} \{f, g \} \phi \ dx dy \end{align}

where

$$\{f, g \} := \partial_{x} f \partial_{y} g - \partial_{y} f \partial_{x} g$$

? Note that variation must be with respect to the second argument only. If we consider the Euler-Lagrange equation (it is this particular ELE because we are only computing the variation in $$g$$) for it, we are looking to solve

$$$$\partial_{g} \mathcal{L} - \partial_{x} (\partial_{g_{x}} \mathcal{L}) - \partial_{y} (\partial_{g_{y}} \mathcal{L}) = \{f, g \}$$$$

which I have been unable to do (I'm essentially down to guessing and checking). I know the inverse problem of finding a Lagrangian given a PDE can be quite difficult, so I'm not expecting much, but considering the Poisson bracket is ubiquitous in classical mechanics I thought someone here may know how to get it from the action or possibly provide a reference.

Edit:

I also think I remember reading somewhere that it is difficult to get first derivatives, if not impossible, from minimising an action, as using integration by parts on the gradient of the perturbation $$\nabla \phi$$ necessarily gives a second derivative. For example, assuming the boundary conditions make the evaluated terms vanish, we have

$$\int_{\Omega} \nabla g \cdot \nabla \phi dx = - \int_{\Omega} \phi \nabla^{2} g dx$$

and so we can't ever get a single derivative acting on $$g$$. If this is true then my question is easily answered. However, I am not totally sure of this and I can't find anything online to back up this statement. Could anyone confirm this?

Edit 2 (17/07/23):

• Are you allowing for $L$ to vary differently as a function of $g_x$ and $g_y$? The compressed notation suggests not but sometimes $h(\nabla g)$ can be shorthand for $h(g_x, g_y)$ separately. If you are not, I suspect this is not doable because the left side of the Euler-Lagrange equation is symmetric in the interchange of $g_x\leftrightarrow g_y$ but the right side is not. Jun 24, 2022 at 19:46