Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 33A (Integrable Systems)

Writing ${\psi }^s(x,u)=(\tilde{x}(s;x,u),\tilde{u}(s;x,u))$, the infinitesimal generator of a Lie point symmetry is

On the jet space of a scalar ordinary differential equation, the $n$th prolongation of a vector field iswhere $D_x$ is the total derivative operator. In particular,

Let $\Delta (x,u,u^{\prime })=u^{\prime }-F(x,u)$. Applying the first prolongation givesOn the solution manifold $\Delta =0$, one has $u^{\prime }=F$. The infinitesimal symmetry criterion is therefore exactly

Put $q=u/x^2$, so $F=xG(q)$ andSubstitution into the Lie-symmetry determining equation for a first-order ordinary differential equation and separation with respect to the arbitrary function $G$ givesThese equations imply $\xi =Cx$ and $\eta =2Cu$. Apart from the zero generator, the common symmetry for arbitrary $G$ is therefore generated byIts one-parameter group is the scaling symmetrywhich leaves $u/x^2$ invariant. Special choices of $G$ can have additional symmetries, but this is the complete symmetry valid for an arbitrary smooth $G$.