Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 29D (Integrable Systems)

The flow of the vector field is obtained by solvingDefine $g^{\epsilon }(x,u)$ to be this solution at parameter $\epsilon$. Uniqueness implies $g^0=\mathrm{id}$ and $g^{\epsilon +\delta }=g^{\epsilon }\circ g^{\delta }$ wherever both sides are defined: shifting the solution parameter gives the same initial-value problem. In particular $g^{-\epsilon }$ is the local inverse. This constructs the one-parameter group generated by the vector field, generally as a local group if the flow is not complete.

A Lie point symmetry is a one-parameter transformation of $(x,u)$ which maps the graph of a function solving the differential equation to another such graph of a function, wherever the transformed independent variable is invertible along the graph of a function. Its induced transformation on derivatives is the prolongation. Infinitesimally the criterion is ${\mathrm{pr}}^{(n)}V(\Delta )=0$ on $\Delta =0$. The derivative coefficients can be computed recursively from ${\eta }^{(0)}=\eta$ by ${\eta }^{(j)}=D_x{\eta }^{(j-1)}-u^{(j)}{D}_x\xi$. This expresses the requirement that the transformed derivatives still satisfy the equation.

A homogeneous linear differential operator satisfies $L(e^{\epsilon }u)=e^{\epsilon }L(u)$, so the scaling $x\mapsto x$, $u\mapsto e^{\epsilon }u$ preserves every homogeneous linear equation. Its infinitesimal generator is $V=u{\partial }_u$.

On a patch with $u\ne 0$, take $s=\log |u|$, $t=x$. Then $V(s)=1$, $V(t)=0$. Set $w=ds/dt=u^{\prime }/u$. Since $u^{\prime \prime }/u=w^{\prime }+w^2$, division by $u$ reduces the second-order equation to the Riccati equationRecover a nonzero solution by $u=C\exp (\int wdt)$, allowing the sign in $C$. The zero solution is separate, and intervals containing zeros of other solutions require separate coordinate patches. The order reduction occurs because the transformed equation depends on $s^{\prime }$ but not on $s$ itself.