Lie symmetry method

Definition

Let $U \subseteq \R^2$ be open and let $\omega : U \to \R$ be smooth; consider the ordinary differential equation $y' = \omega(x, y)$. Let $\Gamma_\varepsilon : (x, y) \mapsto (\hat x, \hat y)$ be a smooth local one-parameter group of maps, with $\Gamma_0 = \id$ and $\Gamma_\varepsilon \circ \Gamma_\delta = \Gamma_{\varepsilon + \delta}$. It is a Lie point symmetry of the equation if it maps graphs of solutions to graphs of solutions. Its infinitesimal generator is the vector field $X = \xi \partial_x + \eta \partial_y$ with $\xi = \partial \hat x / \partial \varepsilon$ and $\eta = \partial \hat y / \partial \varepsilon$ at $\varepsilon = 0$. The group is a symmetry if and only if the linearized symmetry condition holds: $$\eta_x + (\eta_y - \xi_x)\, \omega - \xi_y\, \omega^2 = \xi\, \omega_x + \eta\, \omega_y,$$ where subscripts denote partial derivatives. The Lie symmetry method solves the equation from a known symmetry, near a point where $X \ne 0$, as follows.

1. Find canonical coordinates $(r, s)$, functions of $(x, y)$ with $X r = 0$ and $X s = 1$.
2. In these coordinates the equation takes the form $ds/dr = \Omega(r)$, with no dependence on $s$.
3. Integrate: $s = \int \Omega(r)\, dr + C$, then return to $(x, y)$.