Why are the equilibria $z = -2$ and $z = 2$ stable while $z = 0$ is unstable for $dz/dt = 4z - z^3$ ?

Stability is determined by the direction of the vector field, which is given by the sign of

dz/dt

. Near

z = -2

and

z = 2

, the arrows on the phase line point toward these equilibria, meaning nearby trajectories converge to them, making them stable. Near

z = 0

, the arrows point away from the equilibrium, meaning nearby trajectories diverge, making it unstable.

Conditions

The system is a one-dimensional autonomous ODE.
Stability is inferred from the sign of $dz/dt$ on intervals adjacent to the equilibria.

Reasoning, step by step

Determine the sign of $dz/dt = 4z - z^3$ on the intervals created by the equilibria (-∞, -2), (-2, 0), (0, 2), and (2, $\infty$ ).
Observe that $dz/dt > 0$ for $z < -2$ and $dz/dt < 0$ for -2 < $z < 0$ , so arrows point toward $z = -2$ .
Observe that $dz/dt > 0$ for $0 < z < 2$ and $dz/dt < 0$ for $z > 2$ , so arrows point toward $z = 2$ .
Observe that $dz/dt < 0$ for -2 < $z < 0$ and $dz/dt > 0$ for $0 < z < 2$ , so arrows point away from $z = 0$ .
Conclude that $z = -2$ and $z = 2$ are stable, and $z = 0$ is unstable.

Example

The video draws purple arrows on the z-axis. Arrows point right for $z < -2$ and left for -2 < $z < 0$ , converging at -2. Arrows point right for $0 < z < 2$ and left for $z > 2$ , converging at 2. Arrows point left for -2 < $z < 0$ and right for $0 < z < 2$ , diverging from 0.

Common misconceptions

Assuming that a positive derivative always means stability.
Confusing the stability of the equilibrium with the sign of the function value at the equilibrium.
Thinking that trajectories cross equilibria in a 1D autonomous system.

Watch the explanation

YouTubeBifurcations of a differential equation

0:37 – 1:13Watch this moment ↗

Connected concepts

Equilibria and stability →

Why are the equilibria $z = -2$ and $z = 2$ stable while $z = 0$ is unstable for $dz/dt = 4z - z^3$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

Explore next

Related questions

Why are the equilibria z=−2z = -2z=−2 and z=2z = 2z=2 stable while z=0z = 0z=0 is unstable for dz/dt=4z−z3dz/dt = 4z - z^3dz/dt=4z−z3?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

Explore next

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