Autonomous Equations and Phase-Line Stability

Definition With $t$ as the independent variable and $y' = dy/dt$, a first-order equation is autonomous if it has the form

where the rate $f$ depends on the state $y$ alone and not on $t$. Throughout this lesson $f$ is continuously differentiable on the real line, so every initial-value problem has exactly one solution on its maximal interval. A number $y_*$ with $f(y_*) = 0$ is an equilibrium, and the constant function $y(t) = y_*$ for all $t$ is then a solution, called an equilibrium solution. Conversely, a constant solution $y(t) = c$ has $0 = y' = f(c)$, so the equilibrium solutions are exactly the constant solutions. Because $f$ does not involve $t$, if $y(t)$ is a solution then so is $y(t - t_0)$ for every constant $t_0$, and starting the clock at $t = 0$ loses nothing.

Definition The phase line of $y' = f(y)$ is the $y$-axis with every equilibrium marked and, on each open interval between neighboring equilibria, an arrow toward larger $y$ where $f > 0$ and toward smaller $y$ where $f < 0$. The axis carries states, not times, and the arrows show only direction, not speed. Since $f$ is continuously differentiable, two different solutions never share a point, so no solution can cross or touch an equilibrium solution. A solution that starts in one of these open intervals therefore stays in it, moves strictly in the direction of the arrow for as long as it exists, and never reaches an equilibrium at a finite time. If an equilibrium lies at the head of the arrow, the solution exists for all $t \ge 0$ and tends to that equilibrium as $t \to \infty$. If the interval has no end in that direction, the solution is unbounded in that direction, and it may become infinite at a finite time.

Definition An equilibrium $y_*$ of $y' = f(y)$ is stable if for every $\varepsilon > 0$ there is a $\delta > 0$ such that every solution with $\lvert y(0) - y_* \rvert < \delta$ is defined for all $t \ge 0$ and satisfies $\lvert y(t) - y_* \rvert < \varepsilon$ for all $t \ge 0$. An equilibrium that is not stable is unstable. Then for some $\varepsilon > 0$ there are starts as close to $y_*$ as we like whose solutions later leave the interval $(y_* - \varepsilon, y_* + \varepsilon)$. When only part of the axis is meaningful, such as $y \ge 0$ for a population, stability relative to that domain uses the same definition with the starts restricted to the domain, and the scope has to be stated.

Definition An equilibrium $y_*$ is asymptotically stable if it is stable and there is an $\eta > 0$ such that every solution with $\lvert y(0) - y_* \rvert < \eta$ satisfies $y(t) \to y_*$ as $t \to \infty$. The set of all starts whose solutions tend to $y_*$ is its basin of attraction. On a phase line, an equilibrium with arrows pointing toward it from both sides is asymptotically stable.

Definition Let $y_*$ be an equilibrium of $y' = f(y)$, with $f$ continuously differentiable on an open interval containing $y_*$. If $f'(y_*) < 0$, then $y_*$ is asymptotically stable. If $f'(y_*) > 0$, then $y_*$ is unstable. If $f'(y_*) = 0$, the test gives no conclusion. The reason is the mean value theorem. Since $f(y_*) = 0$, we have $f(y) = f'(c)(y - y_*)$ for some $c$ between $y_*$ and $y$, and when $f'(y_*) \ne 0$ continuity keeps $f'(c)$ of the same sign as $f'(y_*)$ for $y$ near $y_*$. So the arrows next to $y_*$ point toward it when $f'(y_*) < 0$ and away from it when $f'(y_*) > 0$.

Definition An equilibrium $y_*$ of $y' = f(y)$ is semistable if, for some $\delta > 0$, the rate $f$ has the same strict sign on $(y_* - \delta, y_*)$ and on $(y_*, y_* + \delta)$. Both arrows next to $y_*$ then point the same way. If $f > 0$ on both sides, solutions just below $y_*$ rise toward it and solutions just above rise away from it, and if $f < 0$ on both sides the roles are reversed. A semistable equilibrium is unstable on the full line, although relative to its attracting side alone it is asymptotically stable.