Stability of an Equilibrium¶

Proposition¶

Let $x^*$ be an equilibrium of $x'(t)=f\big(x(t)\big)$ $f (x (t))$ , where $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$ $\frac{d x}{d t} (t)$ .

$x^*$ is stable if solutions that start close to $x^*$ stay bounded near it: for every $\varepsilon>0$ there is $\delta>0$ such that $|x(0)-x^*|<\delta$ implies $|x(t)-x^*|<\varepsilon$ for all $t\geq0$ $0$ .
$x^*$ is (locally) asymptotically stable if it is stable and, in addition, $x(t)\to x^*$ as $t\to\infty$ for every $x(0)$ close enough to $x^*$ .
$x^*$ is unstable if it is not stable.

Linearization test. If $f$ is continuously differentiable near $x^*$ , then

$f'(x^*)<0$ implies that $x^*$ is locally asymptotically stable;
$f'(x^*)>0$ implies that $x^*$ is unstable;
$f'(x^*)=0$ gives no conclusion.

In Math V, stable means that solutions stay bounded (near the equilibrium); if, in addition, every nearby solution converges to the equilibrium, it is asymptotically stable.

Beyond the course

Many textbooks call the $\varepsilon$ - $\delta$ notion above Lyapunov stability. Math V does not use that name; it simply says stable.

Interpretation¶

Near $x^*$ , the Taylor polynomial of degree 1 gives $f(x)\approx f'(x^*)(x-x^*)$ $x^{*})$ , so the deviation $y=x-x^*$ approximately obeys $y'=f'(x^*)\,y$ $f^{'} (x^{*}) y$ , with solution $y(0)\mathrm{e}^{f'(x^*)t}$ . A negative derivative makes small deviations decay; a positive one makes them grow. Graphically: $x^*$ is asymptotically stable when $f$ crosses zero from positive to negative.

The graph of f(x) = x(1 − x) is drawn with arrows on the x-axis that point away from the unstable equilibrium 0 and towards the asymptotically stable equilibrium 1.

The figure shows the logistic example below, $x'=f(x)=x(1-x)$ $x)$ . The sign of $f$ gives the direction of motion on the $x$ -axis: to the right where $f>0$ $0$ , to the left where $f<0$ $0$ . At $x^*=0$ the graph crosses zero upwards, $f'(0)=1>0$ $0$ , and nearby solutions move away; at $x^*=1$ it crosses downwards, $f'(1)=-1<0$ $0$ , and nearby solutions move towards it from both sides.

Choose $f$ and drag $x(0)$ close to each equilibrium: solutions return when $f'(x^*)<0$ and move away when $f'(x^*)>0$ $0$ . For $-x^3$ , $x^3$ and $x^2$ the test is inconclusive, as $f'(0)=0$ $0$ , yet the three outcomes differ.

The continuous-time criterion is a sign condition, $f'(x^*)<0$ $0$ . The discrete-time criterion is a size condition: $|a|<1$ for the linear equation $x_{t+1}=ax_t+b$ (see stability of a scalar difference equation), and, by the same linearization, $|g'(x^*)|<1$ for a fixed point of $x_{t+1}=g(x_t)$ with $g$ continuously differentiable. Mixing the two is a common error.

Stability in the test is local: it says nothing about initial values far from $x^*$ . Asymptotic stability also requires the limit at infinity $\lim_{t\to\infty}x(t)=x^*$ $x^{*}$ ; bounded motion near $x^*$ is not enough.

Key Results¶

For $x'=ax+b$ with $a\neq0$ $0$ , $x(t)-x^*=\mathrm{e}^{at}\big(x(0)-x^*\big)$ $x^{*})$ : the equilibrium $x^*=-b/a$ is globally asymptotically stable if $a<0$ and unstable if $a>0$ $0$ .
For $x'=0$ every point is an equilibrium that is stable but not asymptotically stable.
When $f'(x^*)=0$ $0$ , all three outcomes occur: $x'=-x^3$ (asymptotically stable), $x'=x^3$ (unstable) and $x'=x^2$ (unstable, although solutions starting below $0$ approach it).
For linear systems, the analogue is that all eigenvalues must have negative real part; see linear vector ODE and stability of higher-order ODEs.

Examples¶

Logistic equation. $x'=x(1-x)$ $x)$ , so $f(x)=x-x^2$ and $f'(x)=1-2x$ $2 x$ . At $x^*=0$ $0$ , $f'(0)=1>0$ $0$ : unstable. At $x^*=1$ $1$ , $f'(1)=-1<0$ $0$ : locally asymptotically stable. Near $1$ , deviations shrink roughly like $\mathrm{e}^{-t}$ .

Linear with forcing. $x'=-0.5x+3$ has $x^*=6$ and $a=-0.5<0$ $0$ , so $x^*$ is asymptotically stable. From $x(0)=10$ $10$ : $x(t)=6+4\mathrm{e}^{-0.5t}$ and $x(2)=6+4\mathrm{e}^{-1}\approx7.47$ $7.47$ .

Inconclusive test. For $x'=-x^3$ $- x^{3}$ , $f'(0)=0$ $0$ . Yet $x\,x'=-x^4<0$ for $x\neq0$ $0$ , so $|x(t)|$ decreases, and indeed $x(t)\to0$ $0$ : the equilibrium is asymptotically stable although the test is silent.

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 3

Aliases: stable equilibrium, asymptotic stability, asymptotically stable equilibrium Keywords: f'(x*)<0, unstable equilibrium, linearization test, local stability, Lyapunov stability Symbols: delta, epsilon, infinity

Last updated: 14 October 2026 08:21