For an equation $\dot{x}=f(x)$, recall that a fixed point $\bar{x}$ is a point such that $f(\bar{x})=0$; it corresponds to a constant or stationary solution $x(t)=\bar{x}$ that is obviously defined for all $t\in\mathbb{R}$.

Consider again the logistic equation $\dot{x}=bx-dx^2$ for $x\geq 0$; it has two fixed points, namely $0$ and $b/d$. Intuitively, $0$ is an *unstable fixed point* since it ‘repulses’ solutions that start from some $x_0\in (0,b/d)$, while $b/d$ is a *stable fixed point* since it ‘attracts’ all solutions starting from $x_0$.

Time to reach a fixed point

Before studying the notion of stability, we answer a basic question that we raised in the first chapter: how long does it take to reach a fixed point $\bar{x}$ ? By the uniqueness of solutions asserted in the fundamental theorem, the answer is: an infinite amount of time ! Indeed, let $x(t)=\bar{x}$, $t\in\mathbb{R}$. Now suppose that $\tilde{x}(t)$ is a solution that starts out of the fixed point or any other fixed point and is such that there exists some finite time $t^*$ such that $\tilde{x}(t^*)=\bar{x}=x(t^*)$. By the fundamental theorem, one must have $x(t)=\tilde{x}(t)$ for all $t$ in some interval $(-\tau,\tau)$ about $t^*$. But we obtain a contradiction since $x(t)$ and $\tilde{x}(t)$ are distinct functions. Thus $t^*=\pm\infty$.

Derivative test for stability of fixed points

In order to describe the behavior of solutions near a fixed point we introduce the process of *linearization*. If $\bar{x}$ is a fixed point of the equation $\dot{x}=f(x)$ so that $f(\bar{x})=0$, we make the change of variable $u(t)=x(t)-\bar{x}$, representing the deviation of the solution from the constant solution. Substitution gives

$$ \dot{u}(t)=f(\bar{x}+u(t)) $$

and an application of Taylor’s theorem gives$$ \dot{u}(t)=f(\bar{x})+f’(\bar{x})\thinspace u(t)+\frac{f’(\xi)}{2}\thinspace (u(t))^2 $$

for some $\xi$ between $\bar{x}$ and $\bar{x}+u(t)$. Since $f(\bar{x})=0$ and setting $g(u)=\frac{f’(\xi)}{2} u^2$, we can rewrite the previous equation in the equivalent form$$ \dot{u}=f’(\bar{x})\thinspace u+g(u). $$

The function $g$ is small in the sense that $g(u)/u\to 0$ as $u\to 0$. The*linearization*of the differential equation at the fixed point $\bar{x}$ is defined to be the linear differential equation

$$ \dot{v}=f’(\bar{x})\thinspace v $$

that is obtained by neglecting the higher order term $g(u)$ in $\dot{u}=f’(\bar{x})\thinspace u+g(u)$. This linear equation was the first we met in this ebook and it is easy to solve: the solution is $v(0)\ \exp\big(\thinspace f’(\bar{x})\thinspace t\big)$: if $f’(\bar{x})>0$, the solution grows exponentially, as $t\to+\infty$; if $f’(\bar{x})<0$, it decays exponentially; if $f’(\bar{x})=0$, it is constantly equal to $v(0)$.We now define two notions of stability. Informally, we say that a fixed point is *stable* if starting close (enough) guarantees that you stay close. We say that it is *asymptotically stable* if all sufficiently small initial deviations produce small excursions that eventually return to the fixed point.

Formally this means the following.

We say that $\bar{x}$ is *stable* if, for any (arbitrarily small) $\epsilon>0$, there exists a $\delta>0$ (depending on $\epsilon$) such that, for all initial conditions $x(0)=x_0$ satisfying $|x_0-\bar{x}|<\delta$, we have $|x(t)-\bar{x}|<\epsilon$ for all $t>0$.

We say that $\bar{x}$ is *asymptotically stable* if it is stable and if there exists a $\rho>0$ such that

$$ \lim_{t\to+\infty} |x(t)-\bar{x}|=0 $$

for all $x_0$ satisfying $|x_0-\bar{x}|<\rho$.The obvious question is: is a given fixed point stable or unstable? The following theorem answers this question in a way that shouldn’t surprise you.

**Theorem.** *Suppose that $\bar{x}$ is a fixed point and that $f’(\bar{x})\neq 0$. Then it is asymptotically stable if $f’(\bar{x})<0$, and unstable if $f’(\bar{x})>0$.
*

Coming back to the logistic equation $\dot{x}=bx-dx^2$ ($b,d>0$), it is clear that $0$ is unstable and $b/d$ is asymptotically stable. This is of course confirmed by the theorem: since $f’(0)=b>0$ and $f’(b/d)=-b<0$.

In most examples, we are able to ascertain the *local stability* of fixed *by graphical inspection*, hence the previous theorem seems only to confirm what is obvious from a graphical study. But it says more: we now have a measure of how stable is a fixed point by looking at the magnitude of $f’(\bar{x})$. This magnitude plays the role of an exponential growth or decay rate. Its reciprocal $1/|f’(\bar{x})|$ is a characteristic time scale determining the time required for $x(t)$ to vary significantly in the neighborhood of $\bar{x}$.

It is not surprising that the theorem does not allow to conclude in the case where $f’(\bar{x})=0$; we have to look for the next nonzero order term in Taylor’s expansion.

The derivative test prefigures what we can do for a higher-dimensional system: the qualitative behavior near fixed points should be predicted by looking at the linear approximation of the system.