Wikipedia:Reference desk/Archives/Mathematics/2019 February 11

= February 11 =

Turning an iterative equation into a differential equation
I have something of the form

$$x(t+1)=f(x(t))$$,

and would like to convert this into a differential equation, e.g.

$$\dot{x}(t)=g(x(t))$$.

How would I do this?--Leon (talk) 08:17, 11 February 2019 (UTC)


 * You could derive an approximately equivalent differential equation as follows:


 * $$\dot{x}(t) \approx x(t+1) - x(t) = f(x(t)) - x(t)$$


 * In effect, the difference equation is then a numerical solution to the differential equation with a step size of 1. Note, however, that solutions to the difference equation and the differential equation may have quite different behaviours. Gandalf61 (talk) 10:33, 11 February 2019 (UTC)
 * They do! Is there a procedure for finding differential equations that satisfy such relations?--Leon (talk) 11:52, 11 February 2019 (UTC)


 * You may be interested in Delay differential equation (which seems to indicate that this isn't possible). –Deacon Vorbis (carbon &bull; videos) 12:35, 11 February 2019 (UTC)


 * Suppose $$f=\operatorname{id}$$; then your first equation becomes $$x(t+1)=x(t)$$ and it is satisfied by any periodic function with period 1. IMHO no function $$g$$ will give the second equation $$\dot x = g(x)$$ a potential of generating all possible periodic functions $$x(t)$$ (with restriction of 1 being a period). Consider $$x(t)$$ being a sine or a Dirichlet's function
 * $$x(t)=\begin{cases}1&\mathrm{if\ } t\in\mathbb Q\\0&\mathrm{otherwise}\end{cases}$$
 * --CiaPan (talk) 14:17, 11 February 2019 (UTC)


 * This is the fractional iteration problem, solutions are not unique, but you can try to find solutions with nice analytical properties. The simplest way to get to solutions is to find fixed points of $$f(x)$$. If $$x^{*}$$ is a fixed point of $$f(x)$$ and $$f'(x^{*})$$ is larger than zero, then taking $$x(0)$$very close to $$x^{*}$$ and applying the function $$f$$ $$k$$ times leads to $$x(k)\approx x^{*} + \left(x(0) - x^{*}\right)f'(x^{*})^{k}$$, and you can then extend this exponential relationship to real $$k$$. This then defines $$x(t)$$ for real $$t$$ even if $$x(0)$$ is not close to a fixed point, as long as you can get arbitrarily close to a fixed point by iterating using either $$f(x)$$ or its inverse. Count Iblis (talk) 22:30, 11 February 2019 (UTC)


 * See also Abel equation. Count Iblis (talk) 22:38, 11 February 2019 (UTC)