# Is there a “natural” embedding of the Henon map into a continuous flow?

I was wondering about this. Now, I'm a big fan of "continuous iteration", and I was curious about this problem. What I was wondering was whether or not there exists a continuous dynamical system which embeds the Henon map, even if over a larger space such as $\mathbb{R}^3$ or $\mathbb{C}^2$ than $\mathbb{R}^2$. I suspect, perhaps, $\mathbb{C}^2$ might be better, but finding an embedding for $\mathbb{R}^3$ would be interesting, too. Extra points if it's "natural" in some sense, and analytic.

As an example, consider the simple discrete dynamical system given by

$$x_{n+1} = y_n$$ $$y_{n+1} = x_n$$.

Suppose we want to extend this to a continuous flow, that is, parameterized by a real number instead of a natural number, within $\mathbb{C}^2$. We can do so via

$$\mathbf{\phi}(t, \mathbf{v}_0) = \begin{bmatrix} \frac{1 + (-1)^t}{2} & \frac{1 - (-1)^t}{2} \\ \frac{1 - (-1)^t}{2} & \frac{1 + (-1)^t}{2} \end{bmatrix} \mathbf{v_0}$$

where $(-1)^t = e^{i \pi t}$ and $\mathbf{v}_0 = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}$. This can be obtained by taking the Frechet derivative (Jacobian, essentially, in this case) and the matrix power by diagonalization.

Visually, it is a rotation in the 4-real-dimensional space.

This dynamical system can also be embedded in a flow in $\mathbb{R}^3$: add a $z$-coordinate and then consider rotation about the line $y = x$, $z = 0$.

Now, given an analytic function $f$ with a fixed point at 0, if we take the Taylor series about the fixed point, and iterate, we can find that the coefficients of the iterates can be described by polynomials in exponentials of the iteration count, with exponential base $f'(0)$. This gives what is called the "regular iteration" about the fixed point. It can also be obtained by solving the "Schroder equation"

$$\Psi(f(z)) = K \Psi(z)$$

for the unknown function $\Psi$, with $f$ having a fixed point at 0. Then, given the condition of analyticity of $f$ and its having a fixed point at $0$, there is a natural analytic solution for $\Psi$. We can then obtain the regular iteration by $f^t(z)_{\mathrm{reg}} = \Psi^{-1}(K^t \Psi(z))$.

We can also obtain a regular iteration about fixed points not at $0$ by a simple "fixed-point translation". Let $L$ be the fixed point. Then $g(x) = f(x + L) - L$ has the fixed point at $0$. We can iterate this, and then translate back to get the regular iterates of $f$ about $L$.

However, this method seems most developed for the case of functions of a single real or complex variable. What I'm wondering about is whether or not there's a generalization to the $\mathbb{C}^2$ case, and the specific application to the Henon map. It'd be even better if there were also a way to calculate the resulting dynamical system on a computer. I wonder whether or not the famous "attractor" would carry over to this new, continuous-time, higher-dimensional version.

Failing that, is there some other method to obtain a (not necessarily unique) embedding into a continuous flow, one preferably smooth and even analytic in the time parameter as well as the space parameters?

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im looking for the answer to the same question. but,could you please explain how you extend the discrete case to a continuous flow using a jacobian/Fre'chet derivative?(need to learn how its done) –  Sunny Marella Apr 14 '13 at 14:24
In the case above, the function can be described entirely by the action of the Jacobian matrix upon the vector of variables: $f(\mathbf{v}) = \mathbf{J}\mathbf{v}$. Then $f^n(\mathbf{v}) = \mathbf{J}^n \mathbf{v}$ (note that, for example, $(f \circ f)(\mathbf{v}) = \mathbf{J} \mathbf{J} \mathbf{v}$, etc.). We can compute $\mathbf{J}^n$ by writing $\mathbf{J}$ as $\mathbf{J} = \mathbf{V}^{-1} \mathbf{D} \mathbf{V}$ where $\mathbf{D}$ is diagonal and $\mathbf{V}$ is the matrix of eigenvectors. Then $\mathbf{J}^n = \mathbf{V}^{-1} \mathbf{D}^n \mathbf{V}$, (...) –  mike4ty4 Apr 14 '13 at 23:45
(...) which can be easily extended to non-integer $n$ to get the continuous flow $\phi(t, \mathbf{v}) = f^t(\mathbf{v}) = \mathbf{J}^t \mathbf{v}$. –  mike4ty4 Apr 14 '13 at 23:46
I am sure that this cannot happen in general. For the area-preserving Henon map for example cannot happen. Check the paper "Borel summation and splitting of separatrices for the Henon map" by Gelfreich and Sauzin. –  tst May 9 '13 at 22:37