# Find a periodic orbit for the differential equation

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?

• For the $B$ problem, remember that $e^{P^{-1}AP}= P^{-1}e^AP$ Commented Aug 3, 2020 at 14:39

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

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

• +1 very nice answer Robert. Commented Aug 3, 2020 at 16:33
• Thanks for answering. I had given the solution $x(t)=\begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$ which is same as you had given $x(t)=I \cos t + B \sin t$. And the period is $2\pi$. But what is the periodic orbit of $\dot{x}=Bx$?
– John
Commented Aug 3, 2020 at 16:41
• @John: note my remark after equation (15): every non-trivial orbit is periodic. This may also be seen from you matrix. Feel free to ask if you have more questions. Now I'm on to the equation $\dot x = Ax$. Cheers! Commented Aug 3, 2020 at 17:26