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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,

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.

Write $\xi =V_1$, $\eta =\varphi$, and $u_j=d^{j}u/d{x}^j$. On the jet space of a scalar ordinary differential equation, definewhere the total derivative operator isThe order-$N$ prolongation of a vector field is

In the coordinates $(x,u)$, the given action of the special orthogonal group isDifferentiating at $s=0$ gives the infinitesimal generator of a Lie point symmetryso $\xi =u$ and $\eta =-x$. The recursive formula givesandTherefore the second prolongation of the rotation generator for plane graphs is

For ${\Delta }_0=u_{xx}$,which vanishes when ${\Delta }_0=0$. Forwe obtainwhich likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.

Geometrically, $u_{xx}=0$ describes straight lines. The signed curvature of a plane graph isso the second equation says $\kappa =1$ and describes consistently oriented arcs of unit circles. Rotation preserves straight lines, circles, and signed curvature, explaining both invariances.

Write $\xi =V_1$ and $\eta =\varphi$. The vector fieldgenerates a Lie point symmetry of an ordinary differential equation if its local flow maps solution graphs to solution graphs. Its $n$th prolongation of a vector field iswhereandis the total derivative operator. The infinitesimal invariance criterion is

Put $p=u^{\prime }$ and $q=u^{\prime \prime }$. Direct calculation givesandForthe invariance condition isafter imposing $q=p^2/u-u^2$. The coefficient of $p^3$ isIt must vanish identically, soSince $u>0$, integration givesfor functions $F,G$.

Now takeThenand substitution givesThus the infinitesimal invariance criterion holds.

The generated one-parameter transformation is, for $c\ne 0$,while for $c=0$ it is $\tilde{x}=x+d\epsilon$, $\tilde{u}=u$. Combining these transformations gives the two-parameter symmetry groupThis is the affine-scaling symmetry of u double prime equals u prime squared over u minus u squared.