The Poincaré map

Poincaré, no beardPoincaré, beard → $\cdots$ → Poincaré, grey beard

Poincaré, Φ(Poincaré), and Φ(Φ(Φ(Poincaré)))

Definition of the Poincaré map

Consider a single differential equation for one variable
\[ \dot x = f(t, x) \tag{1} \]
and assume that the function $f(t,x)$ depends periodically on time with period $T$:
$\displaystyle f(t+T, x) = f(t, x) $ for all $(t, x)\in\R^2$
A typical example is the logistic equation
$\displaystyle \dot x = x( 1 - x) - h(t)$
in which the harvesting rate $h(t)$ depends periodically on time (e.g. because harvesting only occurs during day time). We assume that the function $f(t,x)$ satisfies all the conditions in the existence and uniqueness theorems. Then, for any initial condition $a\in\R$ there will be a solution $x(t)$ of (1) with $x(0)=a$. This solution may not exist all the way from $t=0$ to $t=T$, but if it does, the value $x(T)$ is by definition the value of the Poincaré map at $a$:
Definition. The Poincaré map, return map, or time $T$ map for the differential equation $\dot x=f(t,x)$ is the map $\phi:J\to\R$, given by $\phi(x_0) = x_1$ where $x(t)$ is the solution of the differential equation with $x(0)=x_0$, and where $x_1=x(T)$.

Here $J\subset\R$ is the domain of the Poincaré map, which consists of those $x_0\in\R$ for which the solution $x(t)$ of the differential equation exists for $0\le t\le T$.

Periodic solutions

In a differential equation $\dot x = f(t, x)$ where the right hand side $f(t,x)$ changes periodically in time usually will not have stationary (i.e. time independent) solutions. Instead there can be time periodic solutions.
Definition. A $T$–periodic solution of the differential equation $\dot x = f(t, x)$ is a solution that satisfies $x(t+T) = x(t)$ for all $t\in\R$.
Theorem. The solution $x(t)$ of $\dot x=f(t,x)$ with initial condition $x(0)=a$ is $T$–periodic if and only if $\phi(a)=a$.
Any number $a$ with $\phi(a)=a$ is called a fixed point for the Poincaré map $\phi$.

Proof

If $x(t)$ is a periodic solution, then $x(T)=x(0)$. By definition of the Poincaré map, $x(T) = \phi(x(0)) $. It follows that $x(0)$ satisfies $\phi(x(0)) = x(0)$, i.e. $x(0)$ is a fixed point of the Poincaré map.
Conversely, suppose $x(0)$ is a fixed point of the Poincaré map, i.e. $\phi(x(0)) = x(0)$. The definition of the Poincaré map implies that $x(T) = \phi(x(0)) = x(0)$. Consider the function $\tilde x(t) = x(T+t)$; this function also satisfies the differential equation, because \begin{align*} \frac{d\tilde x(t)}{dt} &= \frac{d x(T+t)}{dt} & &\text{def of }\tilde x(t) \\ &= x'(T+t) & &\text{chain rule}\\ &= f(T+t, x(T+t)) & &\text{$x$ satisfies the diffeq}\\ &=f(T+t, \tilde x(t)) & &\text{def of }\tilde x\\ &= f(t, \tilde x(t)) & &\text{$f(t, x)$ periodic in }t. \end{align*} Since $x(t)$ and $\tilde x(t)$ both are solutions of $x' = f(t, x(t))$ with the same initial condition: \[ \tilde x(0) = x(T+0) = x(T) = x(0) \] the uniqueness theorem implies they are the same. Thus $x(t) = x(T+t)$ for all $t\in\R$.

Properties of the Poincaré map

Iteration

Iteration Theorem.  If a solution $x(t)$ to $\dot x=f(t,x)$ exists for $0\le t\le nT$, for some integer $n$, then \[ x\bigl((k+1)T\bigr) = \phi\bigl(x(kT)\bigr) \qquad (k=0, 1, 2, \ldots, n-1). \] Thus after $n$ time steps of length $T$, \[ x(nT) = \phi\Bigl(\phi\bigl(\cdots\phi(x(0))\cdots\bigr)\Bigr). \]
Definition of the Poincarémap, and proof that $x(nT) = \phi(\cdots\phi(x(0))\cdots)$.

Proof

The function $y(t) = x(kT+t)$ is a solution of (1) because \[ y'(t) =\frac{d x(kT+t)} {dt} = x'(kT+t) = f(kT+t, x(kT+t)) = f(kT+t, y(t)). \] The periodicity condition for the differential equation implies $f(kT+t, y) = f(t,y)$, so we see that \[ y'(t) = f(kT+t, y(t)) = f(t, y(t)), \] which means that $y$ is indeed a solution of the differential equation (1).

By definition of the Poincaré map this implies $\phi(y(0)) = y(T)$, whence $x\bigl((k+1)T\bigr) = \phi\bigl(x(kT)\bigr)$.

The Poincaré map is monotone

Monotone Poincaré map Theorem. Let $\phi:J\to\R$ be the Poincaré map for the differential equation (1), and assume that the domain $J$ is an interval. Then for all $a, b\in J$ one has $a\lt b \implies \phi(a)\lt\phi(b)$.

Proof

We denote the solution of $\dot x = f(t, x)$ with initial value $x(0)=a$ by $x(t, a)$. So, $\phi(a) = x(T,a)$.

If $\phi(b)=\phi(a)$ then $x(T, b)=x(T,a)$, i.e. we have two solutions that coincide at one time ($t=T$). By the uniqueness theorem they must be the same solution, and thus $a=b$.

If $\phi(b)\lt\phi(a)$, then consider the difference \[ y(t) = x(t,b) - x(t,a) \] between the two solutions with initial values $a$ and $b$. We have $y(0) = b-a\gt0$, while \[ y(T) = x(T, b) - x(T, a) = \phi(b)-\phi(a) \lt 0. \] Hence, since $y(t)$ is a continuous function of $t$, there is some time $t_0\in(0,T)$ at which one has $y(t_0)=0$. This means $x(t_0, b)=x(t_0,a)$. But the uniqueness theorem says that two solutions with diffferent initial values can never coincide, so we again have a contradiction.

Since both assumptions $\phi(a) = \phi(b)$ and $\phi(a)\gt \phi(b)$ lead to contradictions with $b\gt a$, the only possible conclusion is that $\phi(b)\gt\phi(a)$.

The iterates are monotone

Assume that the Poincare map is defined on an interval $J\subset\R$, and assume that for some $a\in J$ the iterates \[ x_0=a, \quad x_1=\phi(x_0), \quad x_2=\phi(x_1),\quad \dots,\quad x_n=\phi(x_{n-1}),\quad \dots \] are defined for all $n$ (i.e. every time you compute a new iterate $x_k$ it lies in the domain $J$ of the Poincaré map, so that you can continue the iteration.)
How to find the iterates $a$, $\phi(a)$, $\phi\bigl(\phi(a)\bigr)$, etc. in the graph of the Poincaré-map.
The Monotone Iterates Theorem. The sequence $x_n$ is either strictly increasing, meaning \[ x_0 \lt x_1\lt x_2\lt \dots \lt x_{n-1}\lt x_n\lt \dots \] or strictly decreasing, or else it is constant: $x_0=x_1=\cdots = x_{n-1}=x_n=\cdots$

Proof

We consider three cases: $x_1=x_0$, $x_1\gt x_0$, and $x_1\lt x_0$.
Corollary. If the sequence of iterates $\{x_1, x_2, x_3, \dots\}$ is bounded then $x_\infty = \lim_{n\to\infty} x_n$ exists. The limit $x_\infty$ is a fixed point of the Poincaré map $\phi$.

Proof

If the sequence $\{x_n\}$ is constant, then $a=x_0$ is a fixed point, and the sequence “converges to $a$,” so there is nothing to prove in this case.

If the sequence is either increasing or decreasing then a theorem about real numbers (“every bounded and monotone sequence converges”) implies that $x_\infty = \lim_{n\to\infty} x_n$ exists. We still have to show that the limit $x_\infty$ is a fixed point of $\phi$. This follows from \[ \phi(x_\infty) = \phi\Bigl(\lim_{n\to\infty} x_n\Bigr) =\lim_{n\to\infty} \phi\bigl(x_n\bigr) =\lim_{n\to\infty} x_{n+1} =x_\infty. \]

The derivative of the Poincaré map and the variational equation

If we again denote the solution of $\dot x = f(t, x)$ with $x(0)=a$ by $x(t, a)$, then $x_a =\frac{\pd x} {\pd a}$ satisfies the variational equation, i.e. the equation obtained by differentiating both sides of (1) with respect to $a$: \[ \dot x_a = f_x(t, x(t, a))\, x_a(t,a), \qquad x_a(0, a) = 1. \] Since the Poincaré map is defined to be $\phi(a) = x(T, a)$ it's derivative is given by \[ \phi'(a) = x_a(T, a) \] Thus one can find the derivative of the Poincaré map by solving the variational equation.

Stability of periodic solutions

Suppose $x(t)$ is a $T$–periodic solution of $\dot x = f(t, x)$. Let $y(t)$ be another solution whose initial value $y(0)$ is close but not exactly equal to $x(0)$. If $|y(0)-x(0)|$ is small then on any finite time interval $0\le t\le t_0$ the difference $y(t)-x(t)$ will also be small. But in the limit $t\to\infty$ the difference could grow exponentially (see the uniqueness theorem.)
Definition. A solution $x(t)$, $t\ge 0$ is asymptotically stable if there is an $\varepsilon\gt0$ such that every solution $y(t)$ with $|x(0)-y(0)|\lt\varepsilon$ satisfies \[ \lim_{t\to\infty} |y(t)-x(t)| =0 \]
One can tell if a periodic solution is asymptotically stable by looking at the Poincaré map.
Theorem.  Let $a$ be a fixed point for the Poincaré map, so that $\phi(a) = a$, and so that the solution $x(t)$ of $\dot x = f(t, x)$ with initial value $x(0)=a$ is $T$–periodic.

If $\phi'(a)\lt 1$ then the solution $x(t)$ is asymptotically stable.

If $\phi'(a)\gt 1$ then the solution $x(t)$ is not asymptotically stable.

Proof

We have a solution $x(t)$ with initial value $x(0)=a$. The solution is assumed to be periodic, so we have $x(nT)=a$ for all $n=1,2,3,\dots$    We consider a solution $y(t)$ whose initial value $y(0)$ is close to $a$.

Since $\phi'(a)\lt 1$ we can choose an $r$ with $\phi'(a)\lt r\lt 1$. Since $\phi'$ is continuous, we can find an $\varepsilon\gt0$ such that $\phi'(x)\lt r$ for all $x\in(a-\varepsilon, a+\varepsilon)$. By the Mean Value Theorem we have for any such $x$

\[ |\phi(x) - \phi(a)| \le r |x-a|, \] and, since $\phi(a) = a$, \[ |\phi(x) - a|\le r|x-a|. \] If the initial value $y(0)$ is chosen so that $|y(0)-a| \lt \varepsilon$, then it follows that $|y(T)-a|\lt r\varepsilon$, and in particular, $|y(T)-a|\lt \varepsilon$. We can therefore repeat the argument $n$ times, which leads to \[ |y(nT) - a|\le r|y\bigl((n-1)T\bigr) - a| \le \cdots \le r^n \varepsilon. \] Remember that $0\lt r\lt 1$. This implies that $r^n\to 0$ as $n\to\infty$, and thus we have found that $\lim_{n\to\infty} y(nT) = a$.

So far we have shown that the nearby solution $y(t)$ converges to the solution $x(t)$ along the sequence of times $t=nT$ ($n=1,2,3,\dots$). We can use the exponential separation estimate for to prove convergence for times in between the period multiples $nT$.

Let $f(t, x)$ have Lipschitz constant $L$, and let $mT\le t\lt (m+1)T$. Then we have \[ |x(t) - y(t)| \le e^{L(t-mT)} |x(mT)-y(mT)| \le e^{LT} r^m\varepsilon. \] To finally prove that $\lim_{t\to\infty}(x(t)-y(t)) = 0$, let $\delta\gt 0$ be given, and choose $n$ so large that $e^{LT}r^n\varepsilon \lt \delta$. Then, for any $t\ge nT$ we find the integer $m$ for which $mT \le t\lt (m+1)T$, and conclude \[ |y(t)-x(t)| \le e^{LT}r^m\varepsilon \le e^{LT}r^n \varepsilon \lt \delta. \]

Examples

There are not many differential equations for which you can compute the Poincaré map explicitly. Here are a few of such examples. In each case the procedure for computing the Poincaré map is straightforward, namely, (1) find the general solution of the diffeq, (2) find the solution with initial value $a$, (3) compute the value of that solution at time $t=T$, where $T$ is the specified period. If several choices of $T$ are possible, then one can include $T$ in the notation for the Poincaré map, and write $\phi^T(a)$ instead of just $\phi(a)$.

Linear growth

Consider the differential equation \[ \dot x = x, \] and assume that the period is $T=1$.
On the left, the solution to $\dot x=x$ with initial value $x(0)=a$; the value of the solution at time $t=1$ is, by definition, the Poincaré map applied to $a$. In this case the Poincaré map is a linear function of the initial value, namely $\phi(a) = e\cdot a$. Its graph is on the right.

Computation of the Poincaré map and its fixed points

The solution with initial value $a$ is $x(t) = ae^t$. The time $T=1$ map is given by \[ \phi (a) = e\cdot a. \] The fixed points are the solutions of the equation \[ \phi(a)=a, \text{ i.e. }e a = a. \] Since $e\neq1$ the only solution of $ea=a$ is $a=0$: thus the only fixed point of the Poincaré map is $a=0$. This was to be expected, because such fixed points correspond to periodic solutions $x(t)$ of the differential equation, and of all the solutions $x(t, a) = ae^t$ only the zero solution is periodic.

Stability of the fixed point $a=0$

The Poincaré map is given by $\phi(a) = ea$, i.e. it is linear. Its derivative is given by $\phi'(a) = e$ for any $a$. In particular, at the fixed point $a=0$ we have $\phi'(0)=e$. Since $e\gt 1$ this fixed point is not asymptotically stable.

Explosive growth

Consider $\dot x = x^2$, and choose $T=1$. The general solution is \[ x(t) = \frac{1}{t+C}, \quad\text{or }x(t)=0. \] The solution with initial value $x(0)=a$ is \[ x(t,a) = \frac{a}{1-at}. \] Note that this form of the solution includes the zero solution. The time $T=1$ map is now simply \[ \phi(a) = \frac{a}{1-a}. \] So far, so good, but what is the domain of $\phi$? The domain of $\phi$ consists of all $a\in\R$ for which the solution $x(t, a)$ exists on the whole interval $0\le t\le 1$. Our formula for the solution shows that it becomes singular when $1-at=0$, i.e. when $t=1/a$. If $a\ge1$ then $t=1/a\in[0,1]$, so the solution becomes singular before $t$ reaches $t=1$, and the Poincaré map is not defined. The domain of the Poincaré map $\phi$ therefore consists of all $a$ with $a\lt 1$.

Linear damping, time periodic source

Consider the linear equation where $f(t, x) = -x+2\cos t$, i.e. the equation \[ \dot x = -x + 2\cos t \] This equation is periodic in $t$ with period $T=2\pi$. We can think of this equation as modeling a quantity $x$ which decays exponentially (the “$-x$” term), and is periodically replenished/depleted at a rate $2\cos t$.

Solution of the differential equation

This equation is linear, and it is therefore one of those rare equations where we can compute the general solution by hand. We solve the differential equation (the integrating factor is $e^t$) and after some calculation arrive at the following solution \[ x(t) = e^{-t}\bigl(x_0 - 1\bigr) + \cos t-\sin t. \] This form of the solution shows that for very large values of $t$ the solution is approximately given by \[ x(t) \approx \cos t-\sin t \] no matter which initial value $x_0$ we choose.

Computation of the Poincaré map

The Poincaré map follows from our formula for the solution, namely, for any $x_0$ one defines $\phi(x_0)$ to be the value of the solution at time $T=2\pi$: \[ \phi(x_0) = x(2\pi) = e^{-2\pi}(x_0-1) + 1 = e^{-2\pi}x_0 + 1-e^{-2\pi}. \]