Differentiable Flows

March 26th, 2005 ~ Posted in: Mathematics

Recall the problem I had earlier of trying to define a non-integral number of compositions of a function. Specifically, if h \, : \, [0,1] \rightarrow [0,1] is an increasing function that maps onto, what meaning can we assign to h^{(t)}, (the composition of h with itself t times, where t \in \R)?

I presented some ideas I had come up with earlier this week, and was surprised to see that they were nowhere near the mark.

A little background on the problem: Dr. J. said one of his friends had posed it to him, and he couldn’t figure it out, even over a period of years. Everytime that friend saw him, he would ask him if he had figured it out yet, so eventually Dr. J. asked someone for a hint. Apparently, the hint was enough of an aid for him to figure the problem out.

He passed the hint along to us, really more of a definition— define h^{(t)} to be

h^{(t)}(x) = \phi^{-1}(a^t \phi(x)),

where \phi\,:\, [0,1] \rightarrow [0,\infty) is increasing.

You can check that this works out to give you the properties that you would expect: h^{(t)} \circ h^{(s)} = h^{(t+s)}, h^{(0)} = j (the identity), etc. But how do you find the particular \phi, a, given h? And are they unique?

It took me a while to get over the outlandishness of this definition, but I can see that it is plausible– a^t \phi(x) is an increasing function from [0,1], and \phi^{-1} is an increasing function to [0,1], so their composition determines some increasing function from [0,1] onto [0,1). Only, is 1 in the range of this function?

Apparently the theory this is concerned with is that of ‘differentiable flows’, and is useful in engineering.

There is one complication, but it’s pretty simple to overcome. The problem is this, as the definition lies, h cannot be just any increasing function. Look at the definition for a while, and see if you can figure out why not (it has to do with a).

The problem is, by the definition, \phi(h(x)) = a \phi(x) and \phi is increasing, so if h(x)&lt; x , a&lt;1, while if h(x)>x, a>1. So if h crosses the identity function, or even touches it (then a=1), the definition doesn’t make sense. But this can be overcome by noticing that if you partition [0,1] into a maximal partition a_1 = 0 < a_2 < \cdots < a_n = 1 where a_2, \ldots, a_{n-1} are the fixed points of h and call h_j the restriction of h to [a_j, a_{j+1}], then each h_j maps from its domain back onto its domain (because of h is increasing). So if you let t_j \, : \, [0,1] \rightarrow [a_j, a_{j+1}] and s_j \, : \, [a_j, a_{j+1}] \rightarrow [0,1], then s_j \circ h_j \circ t_j \, :\, [0,1] \rightarrow [0,1], so the original definition applies, and

\displaystyle (s_j \circ h_j \circ t_j)^{(t)} = \phi_j^{-1}(a_j^t \phi_j(x))),

and we can define

\displaystyle h_j^{(t)} = s_j^{-1} \circ (s_j \circ h_j \circ t_j)^{(t)} \circ t_j^{-1} = s_j^{-1}(\phi_j^{-1}(a_j^t \phi(t_j^{-1}(x)))) \chi_{[a_j, a_{j+1})},

so that

\displaystyle h^{(t)}(x) = \sum s_j^{-1}(\phi_j^{-1}(a_j^t \phi(t_j^{-1}(x)))) \chi_{[a_j, a_{j+1})} .

Once again, there is a problem here at the fixed points \{a_j\}. I don’t think it would be cheating too much to just stipulate that h^{(t)}(a_j)=a_j.

Mathematics ~

This entry was posted on Saturday, March 26th, 2005 at 12:50 pm and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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