# Fixed points of: $\dot{x}=\sin(y) \qquad \dot{y}=\cos(x)$

How can you find the fixed points of this system:

$\dot{x}=\sin(y)\\ \dot{y}=\cos(x)$

Normally I would suggest that you find the points when both functions are equal to 0.

• Free MathJax tip - use \sin instead of sin Oct 14 '14 at 9:40
• No, you're thinking of stationary points. I'm guessing this question is to do with the Banach contraction mapping principle. You would rewrite the coupled DEs in the form $X(t) = \mathcal{L} X(t)$, where $\mathcal{L}$ is an integral operator, then find conditions whereunder it is contractive, i.e. $\|\mathcal{L}^N X_1 - \mathcal{L}^N X_2\|< K \|X_1-X_2\|$, where $N\geq 0$ and $|K|<1$. Under these conditions, repeated iteration of $\mathcal{L}$ on any point in a domain wherein the conditions hold coverges to the unique fixed point of $\mathcal{L}$. So I think you're being asked to solve ... Oct 14 '14 at 9:57
• ..the DE and prove that your solution is unique, Oct 14 '14 at 9:57
• @WetSavannaAnimalakaRodVance; In DE land people tend to use fixed points and stationary points interchangeably. Oct 14 '14 at 13:25
• @MrSlunk Well, one learns something everyday. Thanks. Yuk, though - that's a very confusing jargon for any outsider who has heard of the contraction mapping principle (most generalist mathematicians)! Moreover my understanding, as an outsider, is that "fixed points" (in Banach sense) are highly important to dynamic systems theorists as well as extremal points of DEs, something from your background you would be well qualified to comment on. Oct 14 '14 at 23:30

As you suggested, to solve the fixed points simply set $\dot{x} = \dot{y}= 0$.

So $$\dot{x} = 0 \implies y = n\pi\quad \forall n \in \mathbb{N},$$ and $$\dot{y} = 0 \implies x = k\pi + \frac{\pi}{2} \quad \forall k \in \mathbb{N}.$$ Hence your fixed points are all the pairs $(x,y) = \left(n\pi+\frac{\pi}{2}\pi,k\pi\right)$ for $n,k\in\mathbb{N}.$

Maybe this helps

\begin{eqnarray*} \dot{x} &=&\sin y,\;\dot{y}=\cos x \\ \dot{x}\cos x &=&\dot{y}\sin y=\cos x\sin y \end{eqnarray*} so \begin{eqnarray*} \dot{x}\cos x-\dot{y}\sin y &=&0 \\ \partial _{t}\sin x &=&\dot{x}\cos x,\;\partial _{t}\cos y=-\dot{y}\sin y \\ \partial _{t}\{\sin x+\cos y\} &=&0 \\ \sin x+\cos y &=&C \end{eqnarray*}

• Slick +1. Yes I think this has something to do with it. Why the weird way of asking to solve the DE (find the fixed point) do you think? Perhaps the question is asking for an assertion that your solution must be unique through Picard-Lindelof? Oct 14 '14 at 11:54

No, you're thinking of stationary points. I'm guessing this question is to do with the Banach contraction mapping principle. You would rewrite the coupled DEs in the form:

$$X(t)\stackrel{def}{=}\left(\begin{array}{c}x(t)\\y(t)\end{array}\right) = \int_0^t \left(\begin{array}{c}\sin(y(u))\\\cos(x(u))\end{array}\right)\mathrm{d}u$$

i.e. you write it in the form $X = \mathscr{L} X$, where $\mathscr{L}:C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]\to C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]$ is the integral operator defined above. Then you prove that $\mathscr{L}$ is contractive: i.e. $\exists N\geq0,\,0<K<1$ such that $\|\mathscr{L}^N X_1-\mathscr{L}^N X_2\| \leq K \|X_1-X_2\|$ for all $X_1,\,X_2\in C^1[-t_{max},t_{max}]$. Under these conditions, repeated iteration of $\mathscr{L}$ on any "point" (i.e vector of functions) $X_0\in C^1[-t_{max},t_{max}]$ will converge to the unique fixed point $X_\infty\in C^1[-t_{max},t_{max}]$ such that $\mathscr{L} X_\infty = X_\infty$. In effect, I think you're being asked to work through the proof of the local version of the Picard-Lindelöf Theorem for this differential equation. The inequalities $|\sin(x_1-x_2)| < |x_1-x_2|$ and $|\cos(x_1-x_2)| < |x_1-x_2|$ will help you. Also, use the maximum modulus ($\mathcal{L}^\infty$) norm to define the metric space of functions.

You're probably not being asked to explicitly give a closed form for the fixed point (i.e. the solution of the DE), but simply to prove it exists and is unique by the arguiments shown to you on the Wiki page.