# Hyperbolic equilibrium point

In the study of dynamical systems, a hyperbolic equilibrium point or hyperbolic fixed point is a fixed point that does not have any center manifolds. Near a hyperbolic point the orbits of a two-dimensional, non-dissipative system resemble hyperbolas. This fails to hold in general. Strogatz notes that "hyperbolic is an unfortunate name—it sounds like it should mean 'saddle point'—but it has become standard."[1] Several properties hold about a neighborhood of a hyperbolic point, notably[2]

## Maps

If ${\displaystyle T\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n}}$ is a C1 map and p is a fixed point then p is said to be a hyperbolic fixed point when the Jacobian matrix ${\displaystyle \operatorname {D} T(p)}$ has no eigenvalues on the complex unit circle.

One example of a map whose only fixed point is hyperbolic is Arnold's cat map:

${\displaystyle {\begin{bmatrix}x_{n+1}\\y_{n+1}\end{bmatrix}}={\begin{bmatrix}1&1\\1&2\end{bmatrix}}{\begin{bmatrix}x_{n}\\y_{n}\end{bmatrix}}}$

Since the eigenvalues are given by

${\displaystyle \lambda _{1}={\frac {3+{\sqrt {5}}}{2}}}$
${\displaystyle \lambda _{2}={\frac {3-{\sqrt {5}}}{2}}}$

We know that the Lyapunov exponents are:

${\displaystyle \lambda _{1}={\frac {\ln(3+{\sqrt {5}})}{2}}>1}$
${\displaystyle \lambda _{2}={\frac {\ln(3-{\sqrt {5}})}{2}}<1}$

Therefore it is a saddle point.

## Flows

Let ${\displaystyle F\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n}}$ be a C1 vector field with a critical point p, i.e., F(p) = 0, and let J denote the Jacobian matrix of F at p. If the matrix J has no eigenvalues with zero real parts then p is called hyperbolic. Hyperbolic fixed points may also be called hyperbolic critical points or elementary critical points.[3]

The Hartman–Grobman theorem states that the orbit structure of a dynamical system in a neighbourhood of a hyperbolic equilibrium point is topologically equivalent to the orbit structure of the linearized dynamical system.

### Example

Consider the nonlinear system

{\displaystyle {\begin{aligned}{\frac {dx}{dt}}&=y,\\[5pt]{\frac {dy}{dt}}&=-x-x^{3}-\alpha y,~\alpha \neq 0\end{aligned}}}

(0, 0) is the only equilibrium point. The Jacobian matrix of the linearization at the equilibrium point is

${\displaystyle J(0,0)=\left[{\begin{array}{rr}0&1\\-1&-\alpha \end{array}}\right].}$

The eigenvalues of this matrix are ${\displaystyle {\frac {-\alpha \pm {\sqrt {\alpha ^{2}-4}}}{2}}}$. For all values of α ≠ 0, the eigenvalues have non-zero real part. Thus, this equilibrium point is a hyperbolic equilibrium point. The linearized system will behave similar to the non-linear system near (0, 0). When α = 0, the system has a nonhyperbolic equilibrium at (0, 0).