Suppose $\gamma \colon [0,\alpha] \to\mathbb{R}^{3}$ is a smooth unit-speed curve. Let $(E,V,W)$ be a $\gamma$-adapted orthonormal frame along $\gamma$ (by $\gamma$-adapted I mean $E=\dot{\gamma}$).
It follows that there exist smooth functions $\kappa_{1}, \kappa_{2}, \kappa_{3} \colon I \to \mathbb{R}$, such that
$$
\begin{pmatrix}
\dot{E}\\
\dot{V}\\
\dot{W}
\end{pmatrix}
=
\begin{pmatrix}
0 & \kappa_{1} & \kappa_{2}\\
-\kappa_{1} & 0 & \kappa_{3}\\
-\kappa_{2} & -\kappa_{3} & 0
\end{pmatrix}
\begin{pmatrix}
E\\
V\\
W
\end{pmatrix}.
$$
Conversely, for any choice of $(f_{1},f_{2},f_{3}) \in C^{\infty}(I,\mathbb{R}^{3})$ and suitable initial condition $(v,w)$, the system
$$
\begin{cases}
\dot{V}=-f_{1}E + f_{3}W\\
\dot{W}=-f_{2}E-f_{3}V\\
V(0)=v,\:W(0)=w
\end{cases}
$$
has unique global solution $(\overline{V},\overline{W})$.
Question: If I keep the initial condition fixed, what condition should $(f_{1},f_{2},f_{3})$ satisfy so that $(E,\overline{V},\overline{W})$ is a $\gamma$-adapted orthonormal frame along $\gamma$?
Best Answer
I am not completely sure whether I understand your question well, but if you just require $E=\dot\gamma$, then the freedom in finding an adapted orthonormal frame is given by a smooth function $I\to O(2)$. Explicitly, if you fix one adapted orthonormal frame $E,V,W$, then you can write as $\bar V(t)=a(t)V(t)+b(t)W(t)$, $\bar W(t)=c(t)V(t)+d(t)W(t)$ (which exhausts all vectors pependiculat to $E(t)$, which is fixed). Then the condition that $V(t)$ and $W(t)$ are orthonormal is exactly that $\begin{pmatrix} a(t) & b(t)\\ c(t) & d(t)\end{pmatrix}\in O(2)$. (What you are writing up corresponds to the infinitesmial verstion of this condition and does not take into account that $E$ has to be fixed.
However, the notion of an adapted orthonormal frame is not the usual one. What one usually assume in this situation is that $\ddot\gamma(t)$ (which automatically is perpendicular to $\dot\gamma(t)$ is non-zero for all $t$. Then one gets a unique adapted orthonormal frame by putting $V(t):=\frac{ \ddot\gamma(t)}{|\ddot\gamma(t)|}$ and then $W(t)$ to be $E(t)\times V(t)$ (i.e. the unique vector such that $\{E(t),V(t),W(t)\}$ is a positively oriented orthonormal basis for each $t$.
Edit (in view of the comments and the modification of the question): I would not believe that the question has a simple answer. For example, if $\gamma$ is a straight line, then your must indeed have $f_1=f_2=0$ but this does not work for any other curve. The simplest way to solve the problem (at least for curves with $\ddot\gamma$ nowhere vanishing) is "backwards": Take the special adapted frame that I have described above. For this the $\kappa_i$ can be explicitly described via the curvature and the torsion of the curve. Now any other frame is described by a curve with values in $O(2)$ as described above. (If you take your frame to be positively oriented, then basically this curve can be described by one parmeter - the angle of rotation.) Inserting this will give you the $\kappa_i$ as built up from the curvature and torsion of $\gamma$. I am not sure how difficult it will be to interpret the result.