Find a periodic orbit for the differential equation

Question

Let $B=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $P=\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$, and $A=P^{-1}BP$.

Find a periodic orbit for the differential equation $\dot{x}=Bx$

Find a periodic orbit for the differential equation $\dot{x}=Ax$.

So what I did was calculate eigenvalues of matrix $B$ which are $\lambda _1=i$, $\lambda _2=-i$ and its eigenvectors $\begin{pmatrix} i \\ 1\end{pmatrix}$, $\begin{pmatrix} -i \\ 1 \end{pmatrix}$. Then calculated matrix exponential $e^{Bt}=\begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$. But then I don't know how to proceed afterwards. Somehow the period seems to be $2\pi$. I know that the solution is $x(t)=e^{Bt}x_0$ which can be written as $\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}=\begin{pmatrix}\cos t & -\sin t \\ \sin t & \cos t\end{pmatrix}\begin{pmatrix}x_0 \\ y_0\end{pmatrix}$.

How can periodic orbit for $\dot{x}=Bx$ be calculated from this?

I guess same approach can be followed for finding periodic orbit for $\dot{x}=Ax$ by calculating eigenvalues and eigenvectors of $P^{-1}AP$ then finding $e^{At}$. But that calculation is tedius. There must be easier way to find periodic orbit for $\dot{x}=Ax$ from matrix $B$. Any thoughts?

Answer 1

For the equation

$\dot x = Bx, \tag 1$

with

$B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag 2$

we recall the general solution is given by

$x(t) = e^{B(t - t_0)} x(t_0), \tag 3$

as is easily checked: since

$\dfrac{d}{dt}(e^{B(t - t_0)}) = Be^{B(t - t_0)}, \tag 4$

(3) yields

$\dot x(t) = \dfrac{d}{dt}(e^{B(t - t_0)})x(t_0) = Be^{B(t - t_0)}x(t_0) = Bx(t). \tag 5$

We observe that

$x(t + 2\pi) = e^{B(t + 2\pi - t_0)} x(t_0) = e^{(2\pi B) + B(t - t_0)}x(t_0) = e^{2\pi B}e^{B(t - t_0)}x(t_0), \tag 6$

and since

$B^2 = -I, \tag 7$

it follows that

$e^{Bt} = I\cos t + B\sin t, \tag 8$

in parallel with the well-known Euler identity

$e^{it} = \cos t + i \sin t, \tag 9$

which is easily seen via the power series expansion

$e^{it} = \displaystyle \sum_0^\infty \dfrac{(it)^n}{n!} \tag{10}$

and comparing it term-by-term to

$e^{Bt} = \displaystyle \sum_0^\infty \dfrac{(Bt)^n}{n!} \tag{11}$

using (7); from (8),

$e^{2\pi B} = I \cos 2\pi + B\sin 2\pi = I, \tag{12}$

and

$e^{Bt} \ne I \tag{13}$

for

$0 < t < 2\pi, \tag{14}$

which shows that $e^{Bt}$ is of period $2\pi$; thus

$x(t + 2\pi) = e^{2\pi B}e^{B(t - t_0)}x(t_0) = e^{B(t - t_0)}x(t_0) = x(t), \tag{15}$

which shows that every non-trivial orbit of (1) is periodic.

We turn to the equation

$\dot x = Ax. \tag{16}$

Given that

$A = P^{-1}BP, \tag{17}$

with

$P = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}, \tag{18}$

we write (16) in the form

$\dot x = P^{=1}BPx, \tag{19}$

whence

$(Px)' = P\dot x = P(P^{-1}BP)x = (BP)x = B(Px), \tag{20}$

which shows that $Px$ satisfies (1); as such, $Px$ may be written in accord with (3):

$(Px)(t) = e^{B(t - t_0)} (Px)(t_0), \tag{21}$

and then via (8):

$(Px)(t) = (I\cos(t - t_0) + B\sin(t - t_0))(Px)(t_0), \tag{22}$

which we left multiply by $P^{_1}$:

$x(t) = P^{-1}(I \cos(t - t_0) + B \sin(t - t_0)) (Px)(t_0), \tag{23}$

and algebraically re-arrange:

$x(t) = P^{-1} (Px)(t_0) \cos(t - t_0) + P^{-1}B(P x)(t_0)\sin(t - t_0)$ $= (P^{-1} P)x(t_0) \cos(t - t_0) + (P^{-1}BP) x(t_0)\sin(t - t_0)$ $= (I\cos(t - t_0) + A\sin(t - t_0))x(t_0), \tag{24}$

and since via (17) we have

$A^2 = P^{-1}BPP^{-1}BP = P^{-1}B^2P$ $= P^{-1}(-I)P = -P^{-1}IP = -I, \tag{25}$

we may infer exactly as we did in the above discussion of equation (1) that

$e^{A(t - t_0)} = I\cos(t - t_0) + A\sin(t - t_0) \tag{26}$

and so we may write the solution to (16) as

$x(t) = e^{A(t - t_0)} x(t_0). \tag{27}$