Henon's trick [1] is notationally easy to follow when using the following notation:

$$ \begin{align} \frac{dx_1}{dt}&=f_1(x_1,\cdots,x_N) \\ &\vdots\\ \frac{dx_N}{dt}&=f_N(x_1,\cdots,x_n) \end{align} $$

For example, dividing through the first $N-1$ equation with the last equation, and taking the inverse of the last equation gives:

$$ \begin{align} \frac{dx_1}{dx_N}&=\frac{f_1}{f_N}(x_1,\cdots,x_N) \\ &\vdots\\ \frac{dx_{N-1}}{dx_N}&=\frac{f_{N-1}}{f_N}(x_1,\cdots,x_n)\\ \frac{dt}{dx_N}&=\frac{1}{f_N(x_1,\cdots,x_n)} \end{align} $$

(where $\frac{f_1}{f_2}(x) = \frac{f_1(x)}{f_2(x)}$) The new system is another ODE, where the integration happens over $x_N$, and not $t$.

Though it's cumbersome, I prefer to write an ODE as

$$ \begin{align} x_1'(t)&=f_1(x_1(t),\cdots,x_N(t)) \\ &\vdots\\ x_n'(t)&=f_N(x_1(t),\cdots,x_n(t)) \end{align} $$ (where there is nothing that resembles a differential)

Pattern matching, after applying Henon's trick we should get $$ \begin{align} \bar{x_1}'(\bar{x_N})&=\frac{f_1}{f_N}(\bar{x_1}(\bar{x_N}),\cdots,\bar{x_{N-1}}(\bar{x_N}),\bar{x_N}) \\ &\vdots\\ \bar{x_{N-1}}(\bar{x_N})&=\frac{f_{N-1}}{f_N}(\bar{x_1}(\bar{x_N}),\cdots,\bar{x_{N-1}}(\bar{x_N}),\bar{x_N})\\ \bar{t}'(\bar{x_N})&=\frac{1}{f_N(\bar{x_1}(\bar{x_N}),\cdots,\bar{x_{N-1}}(\bar{x_N}),\bar{x_N})} \end{align} $$

(Where there are bars, because originally $t$ was an independent variable, and now it became a function of the $x_N$, as did all the other functions that solve the ODE.)

Clearly my "preferred" notation is cumbersome to write and read, but it is explicit and mechanical.

How can I understand the manipulations on the ODEs (which as far as I can tell, in the traditional notation, use differentials) in this more cumbersome notation (and what is a good name for this notation)? I imagined the Inverse Function Theorem and Implicit Function Theorem might be needed, but didn't see how.

[1] Henon, M. On the numerical computation of Poincaré maps http://www.sciencedirect.com/science/article/pii/0167278982900343

  • 1
    $\begingroup$ Yes it is the inverse function theorem: you get an ODE for $t(x_N)$ and then plug that time into the other ODE solutions to get $x_i(x_N)$. $\endgroup$
    – Ian
    Commented Aug 24, 2017 at 20:47
  • 1
    $\begingroup$ This is really what is going on behind the scenes when you write $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ just in general, by the way. $\endgroup$
    – Ian
    Commented Aug 24, 2017 at 20:59


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