y' = y^3 - y stable points

Question

So I have this differential equation: $$y' = y^3 - y$$ And I need to find out which of the points (0, 1, -1) are stable.

So here we go:

$$y = \pm \frac{1}{\sqrt{e^{2x+c}+1}}$$ (solution to differential equation without y=0)

Assume that $y = y^*(x, y_0^*)$ is a solution which satisfies $y(x_0) = y_0$ then: $$y_0^* = \pm \frac{1}{\sqrt{e^{2x_0+c}+1}}$$ $$c = \frac{1-y_0^*{^2}}{y_0^*{^2}}e^{-2x_0}$$

$$y^*(x, y_0^*) = \pm\frac{y_0^*{^2}}{(1-y_0^*{^2})e^{2(x-x_0)}+y_0^*{^2}}$$

Firstly, let examine point 0 using Lyapunov's stability definition:

$$\forall \varepsilon \gt 0 : \exists \delta(\varepsilon) \gt 0 : |y(x_0) - y^*(x, y_0^*)| \lt \varepsilon, |y_0 - y_0^*| \lt \delta$$

$$|0-y^*(x, y_0^*)| = \frac{y_0^*{^2}}{|(1-y_0^*{^2})e^{2(x-x_0)}+y_0^*{^2}|} \le \frac{y_0^*{^2}}{|1-y_0^*{^2}|}e^{-2(x-x_0)} \le \frac{y_0^*{^2}}{|1-y_0^*{^2}|} \lt \varepsilon, 1-y_0^*{^2} \gt 0$$ And $$\pm\lim_{x\to \infty}\frac{y_0^*{^2}}{(1-y_0^*{^2})e^{2(x-x_0)}+y_0^*{^2}} = 0$$ so it means that point 0 is asymptotical stable.

Now, let examine points 1 and -1: $$|\pm 1-y^*(x, y_0^*)| = \left\lbrace\frac{|1-y_0^*{^2}|e^{2(x-x_0)}}{|(1-y_0^*{^2})e^{2(x-x_0)}+y_0^*{^2}|} \le 1 \lt \varepsilon; \frac{|1-y_0^*{^2}|e^{2(x-x_0)}+2y_0^*{^2}}{|(1-y_0^*{^2})e^{2(x-x_0)}+y_0^*{^2}|} \le 1+\frac{2y_0^*{^2}}{|1-y_0^*{^2}|}e^{-2(x-x_0)} \le 1+\frac{2y_0^*{^2}}{|1-y_0^*{^2}|} \lt \varepsilon \right.$$ That means that in general $$1+\frac{y_0^*{^2}}{|1-y_0^*{^2}|} \lt \varepsilon, 1-y_0^*{^2} \gt 0$$

is that mean that 1 and -1 are stable points (but not asymptotical)? Or where am I doing it wrong? And what about my assumption $$1-y_0^*{^2} \gt 0$$

Answer 1

Is this really how you are supposed to solve the question? Let me suggest instead to determine the sign of $y^3-y$, depending on $y$, and to deduce that the phase diagram of this differential equation is: $$\large{-\!\!\!-\!\!\!-\!\!-\!\!\!\longleftarrow\!\!\!-\!\!-\!\!\!\!\!\stackrel{-1}{\bullet}\!\!\!\!\!-\!\!-\!\!\!\longrightarrow\!\!\!-\!\!\!-\!\!\!-\!\!\!\!\stackrel{0}{\bullet}\!\!\!-\!\!\!-\!\!\!-\!\!\!\longleftarrow\!\!\!-\!\!-\!\!\!\!\!\stackrel{+1}{\bullet}\!\!\!\!\!-\!\!-\!\!\!\longrightarrow\!\!\!-\!\!\!-\!\!\!-\!\!\!-}$$ Thus, there is exactly one stable fixed point, which is $__$.

Answer 2

Let me show you a way I was taught $$ y = y_0 + \delta y $$ where $y_0 = \lbrace{0,\pm1\rbrace}$ Linearise the original equation given by $$ y' = F(y) $$ as $$ \delta y' = \frac{\partial F(y)}{\partial y}\lvert_{y_{0}}\delta y = \left(3y_0^2-1\right)\delta y $$ solutions for $\delta y$ i.e the perturbations $$ \delta y = \delta y_0 \mathrm{e}^{\left(3y_0^2-1\right)t} $$ it is clear that for $y_0 =\pm 1$ the solution is $$ \delta y = \delta y_0 \mathrm{e}^{2t}\implies \text{unstable for t>0 i.e. $|\delta t|\rightarrow \infty$ as $t\rightarrow \infty$} $$ and for $y_0$ $$ \delta y = \delta y_0 \mathrm{e}^{-t}\implies \text{stable for t>0 } $$