So suppose we have an equation of the form

\[\dot y = f(y)\]

with \(y(t) \in \mathbb{R}\), \(f(0) = 0\) and \(f \in C^1(\mathbb{R})\). Let’s assume that \(f\) is of the form such that in some interval \((-a, a)\) the phase space is as in the figure below.

1D phase space

These conditions on \(f\) allow us to show asymptotic stability of the equilibrium at 0.

Proof

Stability. Without loss of generality take \(\varepsilon \lt a\) and denote by \(y(t;c)\) a solution which at \(t=0\) is \(c\). If \(\lvert c \rvert \lt \varepsilon\) then we have that \(\lvert y(t;c) \rvert \leq \lvert c \rvert \lt \varepsilon\) for \(t \geq 0\) because of the sign of \(f\) .

Asymptotic. Let’s consider \(0 \lt c \lt \varepsilon \lt a\) and again consider \(y(t;c)\) as previously defined. By the sign of \(f\) , \(y(t;c)\) is monotonously decreasing (with respect to \(t\)). Therefore, \(\lim_{t \rightarrow \infty } y(t;c) = b\). If \(b=0\) we are done so suppose \(b \gt 0\). But then \(f(b) \lt 0\) so it hasn’t converged which gives a contradiction. Hence, we obtain \(b=0\). Same method works for \(0 \gt c \gt -\varepsilon \gt -a\) and for \(c=0\) it is automatically satisfied.

Remark

This set up will allow you to consider cases for which standard linearization techniques do not apply like \(f(y) = -y^3 + y^4 g(y)\) with \(g \in C^1(\mathbb{R})\). Note that this approach only works in the 1D case. It is a nice exercise to check where it breaks down.