4 Periodic orbits

Floquet Theory

Suppose \(u(t)\) is a periodic solution of \(\dot{x}=f(x)\), so \(u(t+T)=u(t)\) for all \(t \in \mathbb{R}\). What is the linearisation of the equation about this solution? Note that since it is a solution \(u\) satisfies \(\dot{u}=f(u)\). Set \(x(t) = u(t) + v(t)\) where \(|v| \ll 1\). Then \begin{align*} \dot{x} = \dot{u} + \dot{v} = f(u+v) = f(u) + \mathrm{D} f(u) v + O(|v|^2). \end{align*} i.e. \( \dot{v} = \mathrm{D} f(u) v \) is the linearisation of the equation about \(u\), where \( \mathrm{D} f(u)\) is the \(n \times n\) Jacobian matrix evaluated at \(u\). This system of ODEs is linear (in \(v\)), but its coefficient is periodic, different from the constant coefficient ODEs we considered earlier. The Floquet theory deals with the structure of such system of ODEs \[ \dot{v}= A(t)v, \quad A(t+T) = A(t). \]

Example 4.4 ( One dimensional case) The one-dimensional case is elementary and somewhat misleading if we are thinking about its generalization to higher dimensions. Nonetheless it is worth considering as an exercise. We have \[\dot{v}=a(t) v , \qquad a(t+T) = a(t),\] where both \(a(t)\) and \(v(t)\) are scalar. This ODE is separable, and the solution is given by \[v(t)=v_0\exp\left({\int^t_0 \; a(s) \; ds}\right).\] To find the instruction of the solution, we look at \(t=nT\) first, that is \(v(nT) = v_0 \exp(\int_0^{nT} a(s) \; ds)\). Then because \(a\) is periodic, \begin{align*} \int_0^{nT} a(s) \; ds &= \int_0^{T} a(s) \; ds+ \cdots + \int_{(n-1)T}^{nT} a(s) \; ds\\ &= n \int_0^{T} a(s) \; ds \qquad \text{ as } \qquad a(s+T) = a(s) \end{align*} This implies that \(v(nT)=v_0 \Phi^n\) where the constant \(\Phi = \exp\big({\int_0^{T} a(s) \; ds}\big)\). Therefore, \(|v(nT)|\) is increasing/decreasing (i.e. unstable or stable) according to whether \(|\Phi|\) is greater than or less than one. If we write \begin{equation}\label{eq:per1d} v(t) = w(t) \Phi^{t/T} = w(t)\exp\left(\frac{t}{T}\int_0^T a(s)ds\right), \end{equation} then \(w(t)\) is a periodic. In fact, we have \[ v(t+T) = v(t)\exp\left(\int_t^{t+T} a(s)ds\right) =v(t)\exp\left(\int_0^T a(s)ds\right) =w(t)\exp\left(\frac{t+T}{T}\int_0^T a(s)ds\right). \] Comparing with the expression \eqref{eq:per1d} (at \(t+T\) instead of \(t\)), then \[ v(t+T) = w(t+T)\exp\left(\frac{t+T}{T}\int_0^T a(s)ds\right) = w(t)\exp\left(\frac{t+T}{T}\int_0^T a(s)ds\right), \] that is \(w(t)=w(t+T)\) and \(w\) is periodic with period \(T\). Therefore, in general the solution \(v\) is not periodic, unless \(\Phi = 1\) or \(\int_0^T a(s)ds \).

The nice solution in terms of exponentials does not work in \(\mathbb{R}^n\): even though the solution to the system \(\dot{x} = A(t)x\) with initial condition \(v(0)=v_0\) can be written as \(v(t) = \Phi(t)v_0\), the matrix \(\Phi(t)\) depends on the coefficient \(A(t)\) in a much more complicated way than that in one dimension. Nevertheless, such a matrix \(\Phi(t)\) exists, and plays a similar role of as in one dimension.

Example 4.5 (A one-way coupled linear PDE with periodic coefficient) Consider the following two systems \[ (1)\qquad \frac{d}{dt} \begin{pmatrix} x \cr y \end{pmatrix} =\begin{pmatrix} -1 & 0 \cr 1 & \sin t \end{pmatrix} \begin{pmatrix} x \cr y \end{pmatrix};\qquad (2)\qquad \frac{d}{dt} \begin{pmatrix} x \cr y \end{pmatrix} =\begin{pmatrix} -1 & 0 \cr \sin t & 1 \end{pmatrix} \begin{pmatrix} x \cr y \end{pmatrix}, \] where the coefficient \(A(t)\) is periodic with period \(2\pi\). In both cases, \(x\) is governed by the same equation \(\dot{x} = -x\) and the general solution is given by \(x(t) = x_0e^{-t}\). For system (1), \(y\) is governed by \( \dot{y} = x + y\sin t\), whose solution is given by \[ y(t) = x_0e^{-\cos t}\int_0^t e^{-\tau+\cos\tau}d\tau +y_0e^{1-\cos t}. \] For system (2), \(y\) is governed by \(\dot{y}= x\sin t + y\), \[ y(t) = -\frac{e^{-t}}{5}(\cos t + 2\sin t)x_0 + \left(y_0+\frac{x_0}{5}\right)e^{t}. \]
Exercise. Are there any periodic solution with period \(2\pi\) for either system in the previous example with particular initial condition?

The Structure of linear ODEs with periodic coefficients. Let \(v_k\) be the solution to the system \[ \dot{v} = A(t)v,\qquad A(t+T) = A(t) \] with the initial condition \(v_k(0)=e_k\), the \(k\)-th canonical basis in \(\mathbb{R}^n\) (the vector with \(1\) at \(k\)-th entry, and \(0\) otherwise). Because of the linearity, any solution with initial condition \(v(0)=(\alpha_1,\cdots,\alpha_n)^t=\alpha_1e_1+\cdots+\alpha_ne_n\) is \[ v(t) = \sum_{k=1}^n \alpha_n v_k(t) = \Phi(t)v(0), \] where \(\Phi(t)\) is the so called fundamental matrix with \(v_k\) at \(k\)-th column, i.e, \[ \Phi(t) = \begin{bmatrix} v_1(t) & v_2(t) & \cdots & v_n(t) \end{bmatrix}. \] In other words, \(\Phi(t)\) is the solution to the matrix ODEs \(\dot{\Phi}=A(t)\Phi\) starting with the identity matrix as initial condition. We can get from the solutions in Example 4.5 that the fundamental matrices for both systems are \[ \Phi(t) = \begin{pmatrix} e^{-t} & 0 \cr e^{-\cos t}\int_0^t e^{-\tau+\cos\tau}d\tau & e^{1-\cos t} \end{pmatrix} \] and \[ \Phi(t) = \begin{pmatrix} e^{-t} & 0 \cr \frac{e^{-t}}{5}\big( e^t - e^{-t}\cos t\big) - \frac{2}{5}e^{-t}\sin t& e^{t} \end{pmatrix} \] respectively.

There is no need to start with \(v_k(0)=e_k\) as the initial conditions to construct \(\Phi(t)\). We can start with any \(n\)-linearly independent solutions \(\tilde{v}_1,\tilde{v}_2,\cdots,\tilde{v}_n\) of the ODE \(\dot{x}=A(t)x\). If we define \(\widetilde{\Phi}(t) = [\tilde{v}_1\ \tilde{v}_2\ \cdots \tilde{v}_n]\), then \(\Phi(t) = \widetilde{\Phi}(t)\widetilde{\Phi}(0)^{-1}\) is the desired fundamental matrix.
The fundamental matrix \(\Phi(t,s)\) exists for general linear ODEs \(\dot{x}=Ax\), where \(A\) is any general matrix (no need to be periodic).

Now we can see how the periodicity of the coefficient matrix \(A\) appears in the solution. In general, for non-autonomous ODEs (as the ones with periodic coefficients considered in this section), if \(v(t)\) is a solution, \(v(t+s)\) does not have to be a solution. But for periodic solution, if we differentiate both sides of \(\tilde{v}(t) = v(t+T)\), which is in general different from \(v(t)\), then \[ \frac{d}{dt} \tilde{v}(t) = \dot{v}(t+T) = A(t+T)v(t+T) = A(t)\tilde{v}(t). \] That is \(\tilde{v}(t)=v(t+T)\) is also a solution. Then from the facts that \[ \tilde{v}(t) = \Phi(t)\tilde{v}(0)=\Phi(t)v(T)=\Phi(t)\Phi(T)v(0) \] and \(v(t+T)=\Phi(t+T)v(0)\) for any \(v(0)\), we get \(\Phi(t+T)=\Phi(t)B\) and \(\Phi(nT) = B^n\) for any integer \(n\), where \(B\equiv\Phi(T)\) is called the monodromy matrix of the system. Similar to the one dimensional case, if there is a constant matrix \(H\) such \(B = \exp(TH)\), then \(\Phi(t)\exp(-tH)\) is periodic, or \[ \Phi(t) = P(t)\exp(tH), \] the fundamental matrix \(\Phi(t)\) is the product of a periodic matrix \(P(t)\) and a matrix exponential \(\exp(tH)\). The stability of periodic solutions is reduced to the eigenvalues of the monodromy matrix \(B=\Phi(T)=\exp(TH)\): if all eigenvalues have modulus less than unit, the periodic solutions are locally stable; otherwise if there is one eigenvalue with modulus greater than unit, then the periodic solutions are unstable.

Example 4.6 (Another linear ODEs with periodic coefficients) If \[A(t) = \begin{pmatrix} -1 + \frac{1}{2} \cos^2 t & 1 - \frac{1}{2} \cos t \sin t \\ -1 - \frac{1}{2} \cos t \sin t & -1 + \frac{1}{2} \sin^2 t \end{pmatrix}\] with period \(\pi\) or \begin{align*} \dot{u} & = (-1 + \tfrac{1}{2} \cos^2 t) u + (1 - \tfrac{1}{2} \cos t \sin t)v\\ \dot{v} & = (-1 - \tfrac{1}{2} \cos t \sin t) u + (-1 + \tfrac{1}{2} \sin^2 t)v \end{align*} Then \[\begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} e^{-\tfrac{1}{2}t}, \qquad \qquad \begin{pmatrix} \sin t \\ \cos t \end{pmatrix} e^{-t}\] are solutions with initial conditions \(e_1\) and \(e_2\). Therefore \[ \Phi(t) = \begin{pmatrix} e^{-\frac{t}{2}}\cos t & e^{-t}\sin t \cr -e^{-\frac{t}{2}}\sin t & e^{-t} \cos t \end{pmatrix} \] and \[ B = \Phi(\pi)= \begin{pmatrix} e^{-\frac{1}{2}\pi} & 0 \\ 0 & e^{-\pi} \end{pmatrix}\] and hence the origin is stable.

For linear ODEs with periodic coefficients from linearisation, more information is available about the eigenvalues.

Theorem (Special value w.r.t perturbed periodic solutions) Let \(\phi(t)\) be a periodic solution of the autonomous system \(\dot{x} =f(x)\), and \(\dot{v} = A(t)v\) with \(A(t)=Df(\phi(t))\) is the linearisation around \(\phi(t)\). Then the monodromy matrix \(B\) corresponding to \(A(t)\) always have eigenvalue 1.

Proof. Let \(v = \dot{\phi}\), then taking derivative of the equations for \(\phi\), \(v = f(\phi(t))\), we have \[ \dot{v} = Df(\phi(t))\dot{\phi}(t) = A(t)v. \] That is the derivative \(v=\dot{\phi}\) satisfies the linearised ODEs and is periodic with period \(T\) (the same period as \(\phi\)). Therefore, \[ v(0) = v(T) = \Phi(T)v(0) = Bv(0), \] and \(1\) is an eigenvalue of \(B\) with eigenvector \(v(0)=\dot{\phi}(0)\).

Another result is to generalise the solution \(x(t) = x_0\exp(\int_0^T a(s)ds)\) of the scalar ODE \(\dot{x}=a(t)x\) into higher dimension.

Theorem 4.3 (Evolution of the determinant) If \(\Phi(t)\) is a non-singular matrix that satisfies the system of ODEs \(\dot{x} = A(t)x\) (the matrix \(A\) does not have to be periodic and \(\Phi(0)\) does not have to be the identity matrix), then \[ \mbox{ det} \Phi(t) = \exp\left(\int_s^t \mbox{ tr}A(s)ds\right) \mbox{ det} \Phi(s). \]

Sketched proof. We can actually show the equivalent versions \[ \frac{d}{dt} \mbox{ det} \Phi(t) = \mbox{ tr} A(t) \mbox{ det} \Phi(t). \] Without loss of generality, we can assume \(s=0\) and \(\Phi(0)=I\). Then \(\Phi(t) = I + tA(0)+O(t^2)\) when \(t\) is small. Then \[ \mbox{ det} \Phi(t) = \mbox{ det} \big( I + tA(0)+O(t^2)\big) = 1 + t \mbox{ tr}A(0) + O(t^2). \] Therefore, \(\left.\frac{d}{dt} \mbox{ det} \Phi(t)\right|_{t=0}= \mbox{tr}A(0) \).

This theorem is a more general fact: if the points evolve under the ODE \(\dot{x} = f(x,t)\), then the rate of change of the local volume element is \(\mbox{div} f(x,t)\), the divergence of the vector field \(f(x,t)\). The volume (or area in two dimension) does not change if and only if the divergence of the vector field is zero. In the special case of linear ODEs with \(f(x,t) = A(t)x\), we recover the above result since \(\mbox{div} f(x,t) =\mbox{tr} A(t)\).

Now we can look at the stability. The eigenvalues of the monodromy matrix are denoted as \(\rho_1,\rho_2,\cdots,\rho_n\), also called characteristic multipliers. Their logarithms divided by \(T\) are called characteristic exponents, i.e. \(\rho_k = \exp(\mu_k T)\). Therefore \[ \rho_1\rho_2\cdots \rho_n = \exp(\mu_1 T)\exp(\mu_2 T) \cdots \exp(\mu_n T) = \mbox{det}(B)= \exp\left(\int_0^T \mbox{tr} A(s)ds\right). \]

If the ODEs \(\dot{v}=A(t)v\) is two dimensional, and are derived from periodic solutions of \[ \dot{x}_1 = f_1(x_1,x_2),\qquad \dot{x}_2 = f_2(x_1,x_2). \] Then we have one characteristic multiplier \(\rho_1=1\) (from Theorem 4.3) and the other one \[ \rho_2 = \exp\left( \int_0^T \mbox{tr} A(s)ds\right) =\exp\left( \int_0^T \left(\frac{\partial f_1}{\partial x_1} +\frac{\partial f_2}{\partial x_2}\right)ds\right). \] Therefore, the stability is determined by the sign of the integral of the divergence of the vector field \((f_1,f_2)\) along the periodic solution. If the periodic solution is known, the above integral could be evaluated in some special cases, leading to conclusions about the stability of the periodic solution.

Example 4.7 Consider the ODEs \[ \dot{x} = x-y-x(x^2+y^2),\qquad \dot{y} = x+y-y(x^2+y^2). \] In polar coordinates \(x(t) = r(t)\cos \theta(t), y(t) = r(t)\sin \theta(t)\), the equations become \[ \dot{r} = r(1-r^2),\qquad \dot{\theta} = 1. \] There is a periodic solution \(r(t) = 1\) with period \(T=2\pi\). The stability of this periodic solution is easy in the polar coordinates, or using the criteria above. We have \[ A = \frac{\partial}{\partial x} \left( x-y-x(x^2+y^2)\right) + \frac{\partial }{\partial y}\left( x+y-y(x^2+y^2)\right) =2-4(x^2+y^2). \] Therefore, on the periodic orbits \(r=1\) (or \(x^2+y^2=1\)), \(A(t)\) is the constant \(-2<0\) or \(\rho_2 = \exp(-2T)=\exp(-4\pi)<1\). The periodic solution is stable.