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Let $k$, $l$ be smooth functions from an interval $I$ into $\mathbb R$ and $k>0$. Let's consider system of differential equations

$ t'=k n, $ $ n'=-k t-l b, $ $ b'=ln $ with unknow functions $t,n,b: I\rightarrow \mathbb R^3$.

How to show that scalar products below are zero: $ t\cdot t'=0, $ $ n\cdot n'=0, $ $ b\cdot b'=0. $

Thanks

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    I wish to show that Euclidean norms of functions $t,n,b$ are constant. Then if we assume that for fixed $s_0$ norms of $t(s_0),n(s_0),b(s_0)$ is $1$, we obtain that this norms are equal $1$ at each point.2012-11-30

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Sketch of counterexample. Assume that $k,l$ are constant functions. Then $ \begin{eqnarray*} t'' & = & kn'\\ & = & -k^{2}t-klb, \end{eqnarray*} $ differentiating again, $ \begin{eqnarray*} t''' & = & -k^{2}t'-klb'\\ & = & -k^{2}t'-kl^{2}n\\ & = & -k^{2}t'-l^{2}kn\\ & = & -k^{2}t'-l^{2}t'\\ & = & -(k^{2}+l^{2})t'. \end{eqnarray*} $ This equation can be solved for coordinate functions of $t$. Then $ t_j(x)=c_{1,j}+c_{2,j} e^{\sqrt{k^2+l^2}x}+c_{3,j} e^{-\sqrt{k^2+l^2}x}. $ If I haven't overlooked something, from these you can constructe an explicit counterexample.