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 $\delta(\varepsilon)$?
Best Answer
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.