The damped, periodically-kicked rotator is one of the simplest dynamical systems that displays chaotic behavior1.
Below we find the equation of motion of the damped, kicked rotator $$\ddot{\phi} + \Gamma\dot{\phi} = F = Kf(\phi)\sum_{n=0}^{\infty}\delta(t – nT) \tag{1}$$ where $n$ is an integer, dots denote time derivatives, $\Gamma$ is the damping constant, and $T$ is the period of time that elapses between successive force impulses or "kicks". Variable $\phi$ is a canonical coordinate denoting the angular position of the rotator at time $t$. The moment of inertia of the rotator has been normalized to unity.
If we make the substitutions $$x=\phi, y=\dot{\phi}, z=t$$ the single, second-order equation of motion of one variable can be written as a system of three first-order, autonomous differential equations
\begin{align} \dot{x} &= y \tag{2a} \\ \dot{y} &=-\Gamma y + Kf(x)\displaystyle\sum_{n=0}^{\infty}\delta(t – nT) \tag{2b} \\ \dot{z} &= 1 \tag{2c}. \end{align}Schuster asserts that the three equation system above $(2a,b,c)$ can be reduced to the two-dimensional map
\begin{align} x_{n+1} &= x_{n} + \frac{1 – e^{-\Gamma T}}{\Gamma} [y_n + K f(x_n)] \tag{s1.18b}\\ y_{n+1} &= e^{-\Gamma T} [y_{n} + K f(x_n)] \tag{s1.18a} \end{align}for the variables $$(x_{n}, y_{n}) \triangleq \lim_{\varepsilon \rightarrow 0}[x(n T – \varepsilon), y(n T – \varepsilon)]$$ by integration1.
If the reader is perhaps unsatisfied with the above statement (I confess it was not at all obvious to me), what follows is intended as proof.
Map transformation of $\dot{y}$¶
We will begin with equation $(2b)$ above. First, we bring the sum to the left to contain $f(x)$ and then substitute the function $x(mT-\varepsilon)$ for $x$. Notice we’re adopting a new dummy indexing variable $m$ since $n$ is used in the integration limits.
Next we will mulitiply both sides of the equation by $e^{\Gamma t’}$ and then integrate by parts, with respect to $t’$, between limits from $nT-\varepsilon$ to $t$ $$ \int_{nT-\varepsilon}^{t}\left(\frac{\mathrm{d}}{\mathrm{d}t’}y(t’)\right) e^{\Gamma t’} dt’ = -\int_{nT-\varepsilon}^{t} \Gamma y(t’) e^{\Gamma t’} dt’ + K \int_{nT-\varepsilon}^{t} \sum_{m=0}^{\infty} f(x(mT-\varepsilon)) \delta(t’-mT) e^{\Gamma t’} dt’ \tag{3}$$
Evaluating the definite integral on the LHS of equation $(3)$ we perform integration by parts selecting $u = e^{\Gamma t’}$ so that $du = \Gamma e^{\Gamma t’} dt’$ leaving $dv = \left(\frac{\mathrm{d}}{\mathrm{d}t’}y(t’)\right) dt’$ so that $v = y(t’)$. Recall that $\int u dv = uv – \int vdu$, so it follows that $$ \int_{nT-\varepsilon}^{t}\left(\frac{\mathrm{d}}{\mathrm{d}t’}y(t’)\right) e^{\Gamma t’} dt’ = e^{\Gamma t’} y(t’)\rvert_{t’=nT-\varepsilon}^{t’=t} – \int_{nT-\varepsilon}^{t}\Gamma y(t’)e^{\Gamma t’}dt’ \tag{4}$$
Notice that the last term on the RHS of $(4)$ exactly matches the first term on the RHS of $(3)$. We can thus subtract these out on writing the equality. We next evaluate the first term on RHS of $(4)$ at the integral limits and then rearrange and divide by $e^{\Gamma t}$. Leaving us with $$ y(t) = e^{\Gamma (nT-\varepsilon – t)} y(nT-\varepsilon) + K \int_{nT-\varepsilon}^{t} \sum_{m=0}^{\infty} f(x(mT-\varepsilon)) \delta(t’-mT) e^{\Gamma (t’-t)} dt’ \tag{5}$$
And finally, turning our attention to the remaining integral term on RHS of equation $(5)$, since $f(x(mT-\varepsilon))$ in no way depends on $t’$, the integration variable, the integral can be expressed $$ K \int_{nT-\varepsilon}^{t} \sum_{m=0}^{\infty} f(x(mT-\varepsilon)) \delta(t’-mT) e^{\Gamma (t’-t)} dt’ = K \sum_{m=0}^{\infty} f(x(mT-\varepsilon)) \int_{nT-\varepsilon}^{t} \delta(t’-mT) e^{\Gamma (t’-t)} dt’ \tag{6}$$
To evaluate the RHS of equation $(6)$ above, we integrate noticing that the integrand contains the Dirac delta function as a factor $$ K \sum_{m=0}^{\infty} f(x(mT-\varepsilon)) \int_{nT-\varepsilon}^{t} \delta(t’-mT) e^{\Gamma (t’-t)} dt’ = K f(x(mT-\varepsilon)) e^{\Gamma(mT-t)} \tag{7}$$ when $m=n$, and 0 when $m \neq n$. Writing out the sum is no longer necessary since only one non-zero term remains. After substituting $n$ for $m$ our complete expression now reads $$y(t) = e^{\Gamma (nT-\varepsilon – t)} y(nT-\varepsilon) + K f(x(nT-\varepsilon)) e^{\Gamma (nT-t)} \tag{8}$$
We can make the substitution $ t \rightarrow (n + 1)T – \varepsilon $ to eliminate $t$ leaving $$y((n + 1)T – \varepsilon) = e^{-\Gamma T} y(nT-\varepsilon) + K f(x(nT-\varepsilon)) e^{-\Gamma (T-\varepsilon)} \tag{9}$$ which, in the limit $ \varepsilon \rightarrow 0 $ becomes $$ y_{n+1} = y_{n} e^{-\Gamma T} + K f(x_{n}) e^{-\Gamma T} \tag{10}$$ using variables $(x_n, y_n)$ defined above. After factoring the RHS $$ y_{n+1} = e^{-\Gamma T} [y_{n} + K f(x_n)] \tag{11}$$ we are left with expression $(1.18a)$ of Schuster.
Map transformation of $\dot{x}$¶
To derive the other equation of the two-equation map system (Schuster, $(1.18b)$), we integrate $ \dot{x} = y $ using equation $(11)$. That is, $$\int_{0}^{T}y_{n+1} dT’ = \int_{0}^{T} e^{-\Gamma T’}[y_n + K f(x_n)] dT’ \tag{12}$$ Moving the bracketed factor outside of the integral, integrating and evaluating the result at the limits of integration we see that $$\int_{0}^{T} e^{-\Gamma T’}[y_n + K f(x_n)] dT’ = [y_n + K f(x_n)] \int_{0}^{T} e^{-\Gamma T’} \tag{13}$$ and $$ [y_n + K f(x_n)] \int_{0}^{T} e^{-\Gamma T’} = [y_n + K f(x_n)] \left(-\frac{1}{\Gamma}\right) e^{-\Gamma T’} \rvert_{T’=0}^{T’=T} \tag{14}$$ leaving us with $$ [y_n + K f(x_n)] \left[ -\frac{1}{\Gamma} e^{-\Gamma T} – \left( – \frac{1}{\Gamma} \right) \right] = \frac{1 – e^{-\Gamma T}}{\Gamma} [y_n + K f(x_n)] \tag{15}$$ after algebraic simplification. Given our original equation $(2a)$ in continuous form, $\dot{x} = y$, we can write its discrete analog as $$ x_{n+1} – x_{n} = \int_{0}^{T} y_{n+1} dT’ \tag{16}$$ which implies the LHS of $(16)$ above equals the RHS of equation $(15)$ $$ x_{n+1} – x_{n} = \frac{1 – e^{-\Gamma T}}{\Gamma} [y_n + K f(x_n)] \tag{17}$$ and therefore $$ x_{n+1} = x_{n} + \frac{1 – e^{-\Gamma T}}{\Gamma} [y_n + K f(x_n)] \tag{18}$$ which is equation $(1.18b)$ of Schuster.
Summary and Significance¶
The complete system of the damped, periodically-kicked rotator after reduction into a two-dimensional map of discrete variables $(x_{n}, y_{n}) \triangleq \lim_{\varepsilon \rightarrow 0} [x(nT – \varepsilon), y(nT – \varepsilon)]$ is
\begin{align} x_{n+1} &= x_{n} + \frac{1 – e^{-\Gamma T}}{\Gamma} [y_n + K f(x_n)] \tag{19a}\\ y_{n+1} &= e^{-\Gamma T} [y_{n} + K f(x_n)] \tag{19b} \end{align}Again, as stated in the text, equations $(19a, b)$ above
reduce the initial set of three-dimensional differential equations to a two-dimensional discrete map, which yields a stroboscopic picture of the variables.2
Schuster goes on to show how the Logistic, Hénon and Chirikov Maps can all be derived from this map system depending upon the particular set of limits imposed on it.
Lastly, by adding a constant torque to the driving force in equations $(19a, b)$ followed by a series of "simplifying substitutions", the one-dimensional Sine Circle Map can be derived3.
References¶
1 Deterministic Chaos: an introduction / Schuster, Heinz Georg – 3rd augm. ed. – Weinheim; New York; Basel; Cambridge; Tokyo: VCH, 1995, p. 17
2 Ibid. p. 18
3 Ibid. p. 156