Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 32C (Integrable Systems)

Write $u_j=u^{(j)}$. The transformation ${\psi }_s:(x,u)\mapsto (\tilde{x},\tilde{u})$ induces a transformation of the $n$th jet space of a scalar ordinary differential equation by differentiating $\tilde{u}$ with respect to $\tilde{x}$ through order $n$. The infinitesimal generator of this induced action is the prolongation of a vector fieldTo determine its coefficients, put ${\eta }_0=\eta$ and use the total derivative operatorDifferentiating transformed derivatives byand then differentiating with respect to the group parameter at $s=0$ gives the recursionThis also proves the displayed form of the prolongation by mathematical induction on the derivative order.

For $V_1={\partial }_x$, one has $\xi =1$, $\eta =0$, and hence $D_x\xi =0$. The recursion from part a gives ${\eta }_j=0$ for every $j$.

For $V_2=x{\partial }_x$, one has $D_x\xi =1$. Starting from ${\eta }_0=0$, induction gives

For $V_3=x^2{\partial }_x$, one has $D_x\xi =2x$. The first coefficients are ${\eta }_1=-2x{u}_1$ and ${\eta }_2=-2u_1-4x{u}_2$, and induction givesThus the three prolongations of the projective vector fields on the line are

Write the Schwarzian derivative asThe first prolonged generator is ${\partial }_x$, and $S$ has no explicit dependence on $x$, so ${\mathrm{pr}}^{(3)}{V}_1(S)=0$. Under ${\mathrm{pr}}^{(3)}{V}_2$, each $u_j$ has weight $-j$. Both terms in $S$ therefore have weight $-2$, giving ${\mathrm{pr}}^{(3)}{V}_2(S)=-2S$. Substitution offor $V_3$ gives ${\mathrm{pr}}^{(3)}{V}_3(S)=-4xS$. Therefore the projective transformation weights of the Schwarzian derivative are

For an equation $\Delta =S-f(x)=0$, the Lie point symmetry of an ordinary differential equation criterion is

When $f=0$, each identity ${\mathrm{pr}}^{(3)}{V}_i(S)=c_{i}S$ vanishes on the equation. The equation $S[u]=0$ has all three infinitesimal symmetries $V_1,V_2,V_3$. They generate the projective transformations of the independent variable.

For $f=1$, the translation generator satisfiesFor $V_2$ and $V_3$, restriction to $S=1$ gives respectively $-2$ and $-4x$, which do not vanish identically. Thus $S[u]=1$ has the symmetry $V_1$ among the three listed generators.

For $f(x)=x^{-2}$,which vanishes when $S=x^{-2}$. The corresponding expressions for $V_1$ and $V_3$ are nonzero on the equation. Thus $S[u]=x^{-2}$ has the dilation symmetry $V_2=x{\partial }_x$ among the three listed generators.