## Iteration of a nonlinear map

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Let $T\colon R^n\to R^n$ be a linear map. If we want to study the behavior of $T^kx$ for some $x\in R^n$ as integer $k$ grows, we usually look at the eigen structure of $T$.

Now let $S\colon R^n\to R^n$ be a linear map plus a nonlinear perturbation. And I want to study the behavior of $S^kx$ for some $x\in R^n$ as integer $k$ grows. I am wondering if there exists a theory that discusses this kind of problem.

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## 2 Answers

Yes, there is such a theory, and you already gave yourself the answer in the tag! For particular shape of your map $S$, i.e. a linear hyperbolic map (meaning: the spectrum is disjoint from the unit circle) $T$ plus a small (in a sense to be precised) Lipschitz perturbation , the Hartman-Grobman tells you that $S$ is conjugated with $T$ by a Hölder continuous homeomorphism. The proof is very elementary and follows from the contraction principle; see e.g. M.Shub's book, Global stability of Dynamical Systems.

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Suppose there is a linear $T$ with $\| Sx - Tx \| = o(\|x\|)$ as $x \to 0$, and all eigenvalues of $T$ have absolute value $< 1$. Then there exist positive integer $N$ and positive reals $\delta$ and $\epsilon$ such that for $\|x\| < \delta$, $\|S^N x\| \le (1 - \epsilon) \|x\|$. It follows that for $\|x\|$ sufficiently small, $S^k x \to 0$ as $k \to \infty$. Was that the sort of result you were looking for?

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The nonlinear map $S$ that I am interested in is equal to a linear map $T$ plus a polynomial in $x$. I want to know for what kind of $x$ $S^kx$ will converge as $k$ grows and where it will converge to. – silkrain Apr 17 2011 at 16:04
The general theory (and in particular the Hartman-Grobman theorem, as mentioned by Pietro) can tell you about what happens in a neighbourhood of 0.  So if $T$ has $k$ eigenvalues (counted by algebraic multiplicity) with absolute value $< 1$ and none with absolute value $=1$, there will be a $k$-dimensional manifold near 0 on which the iterates will converge to 0.  If there are fixed points other than 0, the linearizations around those fixed points will tell you about convergence to them. – Robert Israel Apr 17 2011 at 18:09
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