6.2 Pointcaré Map and Periodic Stability

Nonlinear Stability of Periodic Orbits

If we have a $T$ -periodic orbit $γ (t) = {ϕ^{t} (x_{0}), 0 \leq t \leq T}$ where $T$ is the minimal-period.
We have the distance to the orbit:

d i s t (x, γ) = min_{0 \leq s \leq T} {| x - γ (s) | \geq 0}

Orbitally L-Stable

$\forall ε > 0, \exists δ > 0$

d i s t (y_{0}, γ) < δ ⟹ \forall t \geq 0, d i s t (ϕ^{t} (y_{0}), γ) < ε

If not $L -$ Stable then it is unstable

Orbitally $ω -$ Attracting

$\exists δ > 0$ s.t

d i s t (y_{0}, γ) < δ ⟹ lim_{t \to \infty} d i s t (ϕ^{t} (y_{0}), γ) = 0

Orbitally Asymptotically Stability

if $L -$ stable and $ω -$ attracting

Poincaré Map

Also called the first return map.
Assume we have the system $\dot{x} = f (x)$ for $x \in R^{n}$ . Assume we have $x^{*} \in Σ$ where:

$x^{*}$ satisfying that $ϕ^{T} (x^{*}) = x^{*}$ for some $T > 0$
$Σ \subseteq R^{n}$ is (local) cross-section in dimension $n - 1$
The vector field $f (x)$ interest $Σ$ transversely

For $x \subseteq Σ$ $τ (x) = min {t : ϕ^{t} (x) \in Σ}$ . We call the cross-section $Σ$ a Poincaré section for the flow and the map $P (x) = ϕ^{τ (x) (x)}$ the Poincaré map.

Example: Poincaré Map for Harmonic Oscillator

First consider the most simple example

\begin{aligned} \dot{x} & = y; \\ \dot{y} & = - x . \end{aligned}

Consider the point $(x^{*}, y^{*}) = (1, 0)$ , and the hyperplane

Σ = {(x, y), y = 0} .

Then the Poincaré map for a point near $(1, 0)$ can be calculated in the following way: In polar coordinates, the system reads

\begin{aligned} \dot{r} & = 0; \\ \dot{θ} & = - 1. \end{aligned}

The time to return to $Σ$ for points near $(1, 0)$ is $2 π$ . Moreover $θ (2 π) = - 2 π$ , one has that

x (2 π) = r (2 π) \cos θ (2 π) = r, y (2 π) = r (2 π) \sin θ (2 π) = 0.

Then the Poincaré map for a point near $(1, 0)$ is $P (x, 0) = (r (2 π), 0) = (x, 0)$ , which is the identity map.

Example: Poincaré Map for Anharmonic Oscillator

First, consider the system:

\begin{aligned} \dot{x} & = y + x (1 - x^{2} - y^{2}) \\ \dot{y} & = - x + y (1 - x^{2} - y^{2}) \end{aligned}

Consider the point $(x^{*}, y^{*}) = (1, 0)$ and the cross-section (the line where we measure the return):

Σ = {(x, y), y = 0}

To find the Poincaré map for a point near $(1, 0)$ , we switch to polar coordinates ( $x = r \cos θ, y = r \sin θ$ ). The system simplifies to:

\begin{aligned} \dot{r} & = r (1 - r^{2}) \\ \dot{θ} & = - 1 \end{aligned}

Let $(x, 0) = (r (0), 0) \in Σ$ be a starting point near $(1, 0)$ .

From $\dot{θ} = - 1$ , we know that $θ (t) = - t$ . Therefore, the time $T$ required to make one full rotation and return to the cross-section $Σ$ is $2 π$ (since the angle goes from $0$ to $- 2 π$ ).

So, we need to find the value of the radius $r (t)$ at time $t = 2 π$ .

x (2 π) = r (2 π) \cos (θ (2 π)) = r (2 π)

y (2 π) = r (2 π) \sin (θ (2 π)) = 0

so at $2 π$ we are back on the map.

Step-by-Step Derivation of the Integral

We need to solve the differential equation for $r$ :

\frac{d r}{d t} = r (1 - r^{2})

Separation of Variables

Move all $r$ terms to the left and $d t$ to the right:

\frac{1}{r (1 - r^{2})} d r = d t

Setting up the Definite Integral

We integrate from the start time $t = 0$ to the return time $t = 2 π$ .

At $t = 0$ , the radius is $r (0)$ .
At $t = 2 π$ , the radius is $r (2 π)$ .

\int_{r (0)}^{r (2 π)} \frac{1}{r (1 - r^{2})} d r = \int_{0}^{2 π} d t

The right side is simple: $\int_{0}^{2 π} d t = 2 π$ .

The left side requires Partial Fraction Decomposition.

The Decomposition

We want to break $\frac{1}{r (1 - r^{2})}$ into simpler fractions. Note that $1 - r^{2} = (1 - r) (1 + r)$ .

The decomposition is:

\frac{1}{r (1 - r^{2})} = \frac{A}{r} + \frac{B}{1 - r} + \frac{C}{1 + r}

Solving for the constants ( $A = 1, B = 1 / 2, C = - 1 / 2$ ), we get:

\frac{1}{r (1 - r^{2})} = \frac{1}{r} + \frac{1}{2 (1 - r)} + \frac{1}{2 (1 + r)}

Performing the Integration

Now we integrate term by term:

\int (\frac{1}{r} + \frac{1}{2 (1 - r)} + \frac{1}{2 (1 + r)}) d r

= \ln | r | - \frac{1}{2} \ln | 1 - r | + \frac{1}{2} \ln | 1 + r |

= \ln | r | + \frac{1}{2} \ln | \frac{1 + r}{1 - r} |

The text simplifies the integral result into a combined log form:

\frac{1}{2} \ln (\frac{r^{2}}{1 - r^{2}})

(Checking the derivative of this expression confirms it matches the original integrand).

Evaluating the Bounds

We evaluate $[\frac{1}{2} \ln (\frac{r^{2}}{1 - r^{2}})]$ from $r (0)$ to $r (2 π)$ :

\frac{1}{2} \ln (\frac{r^{2} (2 π)}{1 - r^{2} (2 π)}) - \frac{1}{2} \ln (\frac{r^{2} (0)}{1 - r^{2} (0)}) = 2 π

Multiply the entire equation by 2 to clear the fraction:

\ln (\frac{r^{2} (2 π)}{1 - r^{2} (2 π)}) - \ln (\frac{r^{2} (0)}{1 - r^{2} (0)}) = 4 π

Solving for the Poincaré Map

We want to find the mapping $P (x) = r (2 π)$ based on the input $x = r (0)$ .

Exponentiate both sides to remove the natural logs:

\frac{\frac{r^{2} (2 π)}{1 - r^{2} (2 π)}}{\frac{r^{2} (0)}{1 - r^{2} (0)}} = e^{4 π}

Let $u = r^{2} (2 π)$ (the unknown output squared) and $v = r^{2} (0)$ (the known input squared).

\frac{u}{1 - u} = e^{4 π} (\frac{v}{1 - v})

Now, solve for $u$ :

u = e^{4 π} (\frac{v}{1 - v}) (1 - u)

u (1 + e^{4 π} \frac{v}{1 - v}) = e^{4 π} \frac{v}{1 - v}

After simplifying the algebra to isolate $u$ (which corresponds to $r^{2} (2 π)$ ), and taking the square root, we arrive at the result shown in the image.

As a result:

P ((x, 0)) = r (2 π) = ((1 + e^{- 4 π} (x^{- 2} - 1))^{- 1 / 2}, 0), x > 0

Characteristic Multipliers

Let $x^{*}$ be a point on $γ (t)$ and $Σ$ be a Poincaré section containing $x^{*}$ with the Pincaré map denoted by $P (x) .$ The eigenvalues of the Jacobian $D P (x^{*})$ is called the characteristic multipliers.

if every $| λ_{i} | < 1$ then $γ (t)$ is orbitally asymptotically stable
if there exists $| λ_{i} | > 1$ then $γ (t)$ is not obritally Lyapunov stable.