# Using $\delta$-$\varepsilon$ definition to prove stability for autonomous system

I want to prove that an equilibrium point of a simple autonomous system is stable using the $$\delta$$-$$\varepsilon$$ definition. By 'autonomous system' I mean a system that does not depend explicitly on time, that is, $$f(t,x)=f(x)$$.

My criteria for stability (in the Lyapunov sense) is:

$$\forall \varepsilon > 0,\;\exists \delta > 0: \|x(0)\| < \delta \implies \forall t \in \mathbb{R}_+ \; \|x(t)\|<\varepsilon$$

Consider a simple 1D system of the form: $$\dot{x}=-x+x^2$$ which has two equilibrium points, namely $$x_{e,1}=0$$ and $$x_{e,2}=1$$, by solving $$\dot{x}=0$$ .

I want to show that $$x_{e,1}$$ is stable by finding a $$\delta=\delta(\varepsilon)$$ which satisfies my stability criteria. By plotting the system with various initial conditions I know that $$x_{e,1}$$ is also attractive, but where should I go from here? How could I find a right $$\delta(\varepsilon)$$?

• This stability definition makes more sense in higher dimensions, for instance in 2D with the Euclidean metric and an ODE system that has "elliptical" spirals towards the equilibrium point, so that the radius along a trajectory is not monotonous, it oscillates with falling amplitude. Commented Sep 16, 2021 at 8:26

The solution to the initial value problem $$\dot{x}=-x+x^2,\quad x(0)=x_0$$ is $$x(t)=\frac1{1-(1-1/{x_0})e^t}.$$ Its derivative is $$\dot x(t)= \frac{(1-1/x_0)e^t}{(1-(1-1/{x_0})e^t)^2}.$$ Note that the denominator of this fraction can't be equal to zero if $$t\geq 0$$.
Let $$x_0\in(-1,1)\setminus\{0\}$$. We have $$1-\frac1{x_0}<0$$ if $$x_0\in(0,1)$$ and $$1-\frac1{x_0}>0$$ if $$x_0\in(-1,0)$$, thus $$\dot x(t)$$ is positive ($$x(t)$$ is monotonically increasing) if $$x_0\in(-1,0)$$ and $$\dot x(t)$$ is negative ($$x(t)$$ is monotonically decreasing) if $$x_0\in(0,1)$$. Since the solution to the initial value problem cannot cross the equilibrium point $$x=0$$, its norm $$\|x(t)\|$$ decreases monotonically for all $$t\ge 0$$ and $$x_0\in(-1,1)\setminus\{0\}$$. This means that we can choose $$\delta(\varepsilon)= \min(\varepsilon,1)$$ in the stability definition.