Equilibria Organize Autonomous ODEs

For an autonomous equation $y'=f(y)$, an equilibrium $y^*$ satisfies $f(y^*)=0$. The constant function $y(t)=y^*$ is then a solution.

Build a phase line

Mark every root of $f(y)$ on a vertical line. Between roots, test the sign of $f$. If $f>0$, solutions move toward larger $y$; if $f<0$, they move toward smaller $y$.

Arrows pointing toward an equilibrium indicate local stability. Arrows pointing away indicate instability. Attraction from one side and repulsion from the other gives semistability.

Linear test

When $f'(y^*)<0$, nearby solutions usually decay toward $y^*$. When $f'(y^*)>0$, they move away. If $f'(y^*)=0$, the derivative test is inconclusive and the sign diagram remains useful.

A complete phase-line example

Consider

$y'=y(1-y)(y-2).$

The equilibria are $0$, $1$, and $2$. Test one point in each interval:

• for $y<0$, the derivative is positive;
• for $0<y<1$, it is negative;
• for $1<y<2$, it is positive;
• for $y>2$, it is negative.

Arrows point toward $0$ and $2$, so both are stable. They point away from $1$, so it is unstable. This conclusion needs no explicit solution.

Why the derivative test works

Near $y^*$, linearization gives

$y'\approx f'(y^*)(y-y^*).$

Writing $u=y-y^*$ yields $u'=f'(y^*)u$, hence $u(t)\approx u(0)e^{f'(y^*)t}$. Negative $f'(y^*)$ makes perturbations decay; positive $f'(y^*)$ makes them grow.

Check your understanding

For $y'=y^2$, is $y=0$ stable from both sides?

Show the reasoning

No. For negative initial values, solutions increase toward zero; for positive initial values, they increase away and eventually blow up. The equilibrium is semistable. The derivative test is inconclusive because $f'(0)=0$.