Noether's theorem for first-order systems In a paper I'm reading (https://arxiv.org/abs/1312.6120, equations 8-9) I've seen the following statement - given that $s$ is constant and $a,b$ are some dynamical variables obeying the following equations of motion:
$$\frac{d}{dt}a=b(s-ab), \qquad \frac{d}{dt}b=a(s-ab)\tag{1}$$
then, as this is going down the gradient of the function $$E(a,b)=\frac{1}{2}(s-ab)^2\tag{2}$$ and there's the scaling symmetry $$a\to\lambda a, \qquad b\to\lambda^{-1} b,\tag{3}$$ as per Noether's theorem the quantity $a^2-b^2$ is conserved. While it is easy to manually check that this is true, how is Noether's theorem used here? I haven't managed to formulate a Lagrangian or derive a useful first-order version of the theorem. Is there some generic Noether-analogous result for cases like these?
 A: *

*Yes, OP is right: One can strictly speaking not apply Noether's theorem without an action formulation. (It is of course easy to verify that $a^2-b^2$ indeed is an integral of motion (IOM), as OP already mentioned.)


*Let us for fun try to cook up an action formulation.

*

*First note that we can recast OP's EOMs (1) as Hamilton's equations
$$ \dot{a}~=~\{a,H\}, \qquad \dot{b}~=~\{b,H\} \tag{A}$$
with Hamiltonian
$$H~=~\frac{1}{2}(b^2-a^2)\tag{B}$$
and fundamental Poisson bracket
$$ \{a,b\}~=~s-ab,\tag{C}$$
cf. this Phys.SE post.


*It is no coincidence that the Hamiltonian $H$ is (a function of) OP's IOM, since for a 2D phase space, this is essentially unique.


*The corresponding symplectic 2-form is
$$ \omega~=~\frac{1}{s-ab}\mathrm{d}b\wedge \mathrm{d}a~=~\mathrm{d}\theta,\qquad \theta~=~-\ln|s-ab|\mathrm{d}\ln |a|, \tag{D}$$
cf. e.g this Phys.SE post.


*The corresponding Hamiltonian Lagrangian becomes
$$ L_H~=~ -\frac{1}{a}\ln|s-ab|\dot{a} - \frac{1}{2}(b^2-a^2),\tag{E} $$
cf. e.g this Phys.SE post.


*It is straightforward to check that the corresponding Euler-Lagrange (EL) equations are OP's EOMs (1).


*However, the OP's transformation (3) is not a symmetry of the action (E). This can essentially be traced back to the fact that the Hamiltonian $H$ is not invariant under OP's transformation (3).


*Since $L_H$ has no explicit time dependence, Noether's theorem implies that $H$ is conserved.
