> [!NOTE] Definition 1 (Stationary Poimt) > Let $y_{k}=u(y_{k-1}), \quad \text{for all }k\geq 1$where $u:\mathbb{R}\to \mathbb{R}$ is a given function and $y_{0}=\bar{y}$, be a [[First Order Autonomous Recurrence Relation|first order autonomous recurrence relation]]. > > Stationary points are points $y^{*}\in \mathbb{R}$ such that $y_{k}=y^{*}$ for all $k\geq 1$ is a solution. **Note**: If $y_{k}=y^{*}$ is a solution then $y^{*}=u(y^{*}).$ # Properties **Stability**: A stationary point $y^{*}$ is [[Stable Stationary Point of First Order Autonomous Recurrence Relation|stable]] iff whenever the $n$th iteration of $u$ on $y_{0}$ is close to $y^*$ whenever $y_{0}$ is close to $y^{*}.$ By [[Condition for Stability of Stationary Point of First Order Autonomous Recurrence Relation]], $y^{*}$ is stable if $|u'(y^{*})|<1$ and unstable if $|u'(y^{*})|>1.$ **Periodicity**: ... # Applications **Examples**: > [!Example] > Let $u(y)=-\frac{1}{2}y^{3} + \frac{9}{8}y.$ The recurrence relation $y_{k}=u(y_{k-1})$ has three stationary point: $y^{*} =u(y^{*}) \iff 0 = -\frac{1}{2}(y^{*})^{3}+ \frac{1}{8}y^{*} = y^{*}\left( -\frac{1}{2} (y^{*})^{2} + \frac{1}{8} \right) \iff y\in \left\{ -\frac{1}{2},0, \frac{1}{2} \right\}.$As $u'(y)= -\frac{3}{2}y^{2}+ \frac{9}{8},$ we have that $u'\left( -\frac{1}{2} \right)= u'\left( \frac{1}{2} \right) = \frac{3}{4}, \quad u'(0)=\frac{9}{8} $so $y^{*}=\pm \frac{1}{2}$ are stable stationary points whilst $y^{*}=0$ is unstable. > >The figure below shows some solutions for different values of $y_{0}$ and shows that solutions move away from zero even when the starting point is close to $0$ but approach the stationary points $\pm \frac{1}{2}$ when the starting points are close. > >![[Screenshot 2024-04-25 at 20.51.51.png|300]]