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.