I have found a theorem but I did not understand the proof. I'm looking for a clarification of the proof or a different proof.

Let $f_1, f_2, f_3$ be the three components of a curve in $R^3$ parameterized by $t$. Let $W$ be the Wronskian matrix of $f$. whose $(i,j)$ entry is $f_j^{(i-1)}$.

Theorem: If $|W'| = 0$ then $f$ is a planar curve i.e. it satisfies $a_1 f_1 + a_2 f_2 + a_3 f_3 + a_0 = 0$ for some constant $a_i$s.

The proof goes something like this:
We can find $b_i$ which are ordinary functions of $t$ such that $W' b = 0$, that is, $\sum b_i f_i' = 0, \sum b_i f_i'' = 0, \sum b_i f_i''' = 0$.
Although the proof does not explain why such functions $b_i$ exist, I imagine I could watch the null space of $W'$ evolve smoothly with time, and select an evolving vector $b$ living inside that nullspace.
By takng the derivative of the equations, we can arrive at $\sum b_i'f_i' = 0, \sum b_i f_i'' = 0$. Then the proof says something I don't understand: "The $b$'s are proportional respectively to the cofactors of the elements of the last row in the determinant of $W'$. The same is true of the $b'$'s"
Then we get $\frac{b_1'}{b_1} = \frac{b_2'}{b_2} =\frac{b_3'}{b_3} = \phi(t)$ and then $b_i = a_i e^\phi$ so $\sum a_i e^\phi f_i' = 0$ so $e^\phi\sum a_i f_i' = 0$ and then we remove the exponential and integrate to get the final result.


$W'=W(f'), f=(f_1,f_2,f_3)$

If a set of analytic functions has a wronskian of $0$ then the functions are linearly dependent, since the $0^{th}$ to $n-1^{th}$ derivatives of each function can be considered as components of a vector, and checking if the determinant is $0$ is a standard test of linear dependence. If they are linearly dependent, there will be one coefficient for each function, $n$ in total, which is roughly why the derivatives up to $n-1$ contain enough information to determine linear dependence/independence.

$W'=0$ implies that there exist $a_1,a_2,a_3$ such that $a_1f_1'+a_2f_2'+a_3f_3'=0$. Integrating gives the required result.

  • $\begingroup$ I think you mixed up dependent and independent at least once. But yes this seems to be correct although you additionally require that the functions be analytic. But maybe it's possible to reduce that assumption? $\endgroup$ – Mark May 17 '13 at 19:20
  • $\begingroup$ Thanks for spotting the mistakes. Edited. $\endgroup$ – Angela Richardson May 17 '13 at 19:22

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