Find the fixed points of a nonlinear two-dimensional system:
$\dot{x} = \sin y$ $\dot{y} = x - x^3.$
I know that $0 = x(1 - x²) \implies x = 0, 1, -1$. I am not sure what to do after this.
Find the fixed points of a nonlinear two-dimensional system:
$\dot{x} = \sin y$ $\dot{y} = x - x^3.$
I know that $0 = x(1 - x²) \implies x = 0, 1, -1$. I am not sure what to do after this.
As usual for the system of differential equations to find its fixed points you need to solve the equation $ \mathbb f(\mathbb {\tilde x}) = \mathbb 0 $ In your case it looks like
$ \left\{\begin{array}{rcc} \sin y & = & 0 \\ x-x^3 & = & 0 \end{array}\right. \quad \Longrightarrow \left[ \begin{array}{ccl} y & = & \pi k,\ k \in \mathbb Z \\ x & = & \{-1,0,1\} \end{array} \right. $
Using the definition of a fixed point/equilibrium/steady state/etc. we have $x(1 - {x^2})\mathop = \limits^{{\text{set}}} 0 \Rightarrow x = 0, \pm 1$ and $\sin y\mathop = \limits^{{\text{set}}} 0 \Rightarrow y = n\pi, n \in {\mathbb Z}.$ That is all you need to do.