The Poincaré map
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 $T$,
\[
x(nT) = \phi\Bigl(\phi\bigl(\cdots\phi(x(0))\cdots\bigr)\Bigr).
\]
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.)
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 $x_1\lt
x_0$.
- If $x_1=x_0$, then we have found, by definition of $x_0=a$ and $x_1=
\phi(x_0)$, that $\phi(a) = a$. In other words, $a$ is a fixed point for
$\phi$. In this case we have $x_n=a$ for all $n$ (proof by induction: we know
$x_0=a$; if $x_{n-1}=a$, then $x_n= \phi(x_{n-1}) = \phi(a) = a$).
- If $x_1\gt x_0$, (i.e. $\phi(a)\gt a$), then we have $ x_n\gt x_{n-1}$
for all $n$. This follows by induction: by assumption, $x_n\gt x_{n-1}$ is
true for $n=1$. If $x_n\gt x_{n-1}$ then it follows from the fact that $\phi$
is monotone, and the definition of the $x_i$ that
\[
x_n\gt x_{n-1}\implies
\phi(x_n) \gt \phi(x_{n-1}) \implies x_{n+1} \gt x_n.
\]
- The case where $x_1\lt x_0$ is similar to the previous case and leads to the
conclusion that the sequence $x_n$ is strictly decreasing.
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 $T=1$.
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
$T=2\pi$:
\[
\phi(x_0) = x(2\pi) = e^{-2\pi}(x_0-1) + 1 = e^{-2\pi}x_0 + 1-e^{-2\pi}.
\]
→
→
$\cdots$
→
