Overview

In section 5 of our notes, we discussed an analytic criterion for the stability of a steady-state (or equilibrium) \(x^{\ast}\) corresponding to a one-dimensional autonomous system \(\frac{dx}{dt}=f(x)\). This states that if \(f\) is a continuously differentiable function at \(x^{\ast}\), then

\(f'(x^{\ast}) < 0\) implies \(x^{\ast}\) is stable, while
\(f'(x^{\ast}) > 0\) implies \(x^{\ast}\) is unstable.

In this post, we provide an explanation for the above stated fact.

Linear Stability Analysis

Consider the effect of a small perturbation to an equilibrium. Let \(\xi(t) = x(t) - x^{\ast}\) be a small perturbation away from \(x^{\ast}\). Then by taking the derivative with respect to \(t\) on both sides of the equality, we see that

\[\begin{align} \frac{d}{dt}\xi &= \frac{d}{dt}(x - x^{\ast}) \\ &= \frac{d}{dt}x - \frac{d}{dt}x^{\ast} \\ &= \frac{dx}{dt} - 0 \\ &= f(x) \\ &= f(\xi + x^{\ast}) \end{align}\]

and therefore

\(\frac{d\xi}{dt}=f(\xi + x^{\ast})\)

Now Taylor’s formula leads to

\(f(\xi + x^{\ast}) = f(x^{\ast}) + f'(x^{\ast})\xi + \mathcal{O}(\xi^2)\)

and thus

\(\frac{d\xi}{dt} \approx f'(x^{\ast})\xi\)

where we have used the fact that \(f(x^{\ast})=0\).

Recall that a first-order linear system has the form

\(\frac{dx}{dt}=ax\)

and the general solution to such a linear system is \(x(t)=Ce^{at}\), where \(C\) is a constant. Thus, we see that a solution to a first-order linear system will asymptotically decay if the coefficient \(a\) is negative, but grow if the coefficient \(a\) is positive.

Now, observe that, based on our derivation, the differential equation for a small perturbation \(\xi(t)\) such as we have defined it very nearly satisfies a first-order linear system with coefficient \(f'(x^{\ast})\). So, we expect that the effect of a perturbation will asymptotically decay if \(f'(x^{\ast}) < 0\) but asymptotically increase if \(f'(x^{\ast}) > 0\). This helps to explain our criterion for stability.

It is important to note that our criterion provides no information when \(f'(x^{\ast})=0\). In fact, one can easily construct examples of systems that have a stable equilibrium such that \(f'(x^{\ast})=0\) and also systems with an equilibrium satisfying \(f'(x^{\ast})=0\) that is not stable. It is instructive to quickly draw the phase lines for each of the following systems:

\(\frac{dx}{dt}=-x^2\), and
\(\frac{dx}{dt}=-x^3\).

One can conduct a more careful analysis of stability than what we have done here. However, this requires more advanced mathematics than is appropriate in the context of the Topics in Biomathematics course.