# Second order equations on manifolds

In my notes, the lecturer considers a smooth vector field $v: TM\to T(TM)$, with $M$ a smooth manifold. Let’s write

$$v(u,e)=((u,e), (a(u,e),b(u,e)).$$

It is said that $v$ is a second order equation if $T\pi_M\circ v=\text{id}$. It implies that $a(u,e)=e$, i.e.

$$v(u,e)=((u,e),(e,b(u,e))).$$

Here, my notes claim that if $c(t)=(u(t),e(t))$ is a curve on $TM$ satisfying the above, then

$$\frac{du}{dt}=e(t), \ \frac{de}{dt}=b(t)$$

This is probably very stupid but I can get why…

#### Solutions Collecting From Web of "Second order equations on manifolds"

The requirement for v(u,e) that a(u,e) = e is called “obeying the canonical flip on the double tangent bundle”; it is precisely the vector fields on the double tangent bundle that satisfy or obey the canonical flip that are called second-order ODEs on a closed, connected, smooth manifold M.

For a second-order ODE on a manifold v(u,e), if c(t) = (u(t),e(t)) is a solution, then u(t) is called a “base curve” on the manifold M and its derivative c(t) is the solution to the second-order ODE; this is one of the reasons that v(u,e) is called a second-order ODE.

For more information on the canonical flip, try Transveral Mappings and Flows by Abraham and Robbin; for more information second-order ODEs on manifolds, try Foundations of Mechanics by Abraham and Marsen