Csdrhrt

Suppose $gamma colon [0,alpha] tomathbb{R}^{3}$ 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 $gamma$?

I am not completely sure whether I understand your question well, but if you just require $E=dotgamma$, then the freedom in finding an adapted orthonormal frame is given by a smooth function $Ito O(2)$. 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 $ddotgamma(t)$ (which automatically is perpendicular to $dotgamma(t)$ is non-zero for all $t$. Then one gets a unique adapted orthonormal frame by putting $V(t):=frac{ ddotgamma(t)}{|ddotgamma(t)|}$ 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 $ddotgamma$ 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.