Solving ODE - Space curve Frame I'm trying to calculate the parallel frame $\{T, U, V\}$ of a space curve $\alpha : I \mapsto \mathbb{R}^3$. It's similar to Frenet frame, except we have instead the projection of torsion $\tau$ on $U$: $\tau_U = 0$
It's described by the three equations,
$$T' = \kappa_U U + \kappa_V V \quad \quad \quad \quad \ \ (1)$$
$$U' = -\kappa_U T = -\langle \kappa,U \rangle \ T \quad \quad (2)$$
$$V' = -\kappa_V T = -\langle \kappa,V \rangle \ T \quad \quad (3)$$
where $\kappa = T'$ is the curvature vector and $\kappa_U, \kappa_V$ are its components.
I know $T$, $\kappa$ and the initial conditions $\{T(0), U(0), V(0)\}$ at $\alpha(0)$ but I don't know how to solve the ODE $(2)$ and $(3)$ in order to get $U$ and $V$. 
Any suggestion or a note on what to search for would be great!
 A: To find $U$ and $V$ first consider the Frenet frame $\{T,N,B\}$ along a unit speed curve (or where it exists). We know this frame explicitly since $N=T'/\|T'\|$ and $B=T\times N$.  Also note that there exists locally a smooth function $\theta\colon I \to \mathbb{R}$ such that
$$U = \cos \theta N + \sin \theta B.$$
Deriving $U$ gives
$$
  U' = -\kappa \cos\theta T + (\theta'+\tau)(-\sin\theta N + \cos\theta B).
$$
We see that $U$ satisfies ODE $(2)$ iff $\theta' = -\tau$. Also the initial value $\theta(0)$ is determined by $U(0)$, so we find $\theta$ from this:
$$
  \theta = -\int \tau\,dt + \mathrm{const.}
$$
The vector field $V$ is determined similarly. Note that the inner product $U\cdot V$ is constant along the curve. So if $U(0)$ and $V(0)$ are orthogonal, we can just take $V=\pm T \times U$.
Remark: You can do the same calculation starting with an arbitrary adapted frame $\{T,N_1,N_2\}$. You will find the condition $\theta' = -N_1' \cdot N_2$. 
You will probably find this article interesting:
There is more than one way to frame a curve. by R.L. Bishop (The American Mathematical Monthly, 1975). 
