# 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!

• When the torsion is null, the curve is planar. – Yves Daoust May 16 at 15:56
• I meant $\tau_U = 0$, the projection of $\tau$ on $U$. Edited it now. – mike May 16 at 16:09
• In the first line, I'm guessing its $T'=\kappa_U U + \kappa_V V$? – Calvin Khor May 16 at 17:34
• Yes :D edited it now too! – Sptmp May 16 at 18:00

## 1 Answer

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).

• Thank you for your answer and the article, I will work through it! – mike May 17 at 21:28