Equilibria, Stability and Logistic Growth

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

The idea

For an autonomous equation $y'=f(y)$ , equilibrium values satisfy $f(y_*)=0$ . The sign of $f$ between its zeros determines whether solutions rise or fall. Nearby arrows toward an equilibrium indicate attraction; arrows away indicate instability.

Visual guide

VISUAL GUIDEPositive logistic solutions approach carrying capacity

Start at 2
Start at 15
Carrying capacity K = 10

For r = 1 and K = 10, solutions below capacity rise while those above fall. The curves begin at 2 and 15. Both approach 10; their slope is determined by current population, not just elapsed time.

Method and assumptions

For a differentiable $f$ , $f'(y_*)<0$ gives local asymptotic stability and $f'(y_*)>0$ gives instability. If the derivative is zero, inspect signs directly. Logistic growth is $y'=ry(1-y/K)$ with $r,K>0$ . For positive initial data, $y=K/(1+A e^{-rt})$ , where $A=(K-y_0)/y_0$ .

Worked example: below capacity

With $r=1,K=10,y_0=2$ , $A=4$ and $y=10/(1+4e^{-t})$ . The solution increases toward $10$ . At $y=5$ , the growth rate is maximal: $ry(1-y/K)=2.5$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For y′=y(1−y), arrows near y=1 point upward below it and downward above it. What is y=1?

A repelling equilibrium

Not an equilibrium

An attracting equilibrium

Hint 1 · Find a starting point

Both nearby directions move toward the same level.

Hint 2 · Take the next step

Check that the right-hand side vanishes at y=1.

Show the reasoning

Answer: An attracting equilibrium

The equilibrium is locally asymptotically stable: nearby positive solutions approach 1.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a degenerate equilibrium

For $y'=y^2$ , the only equilibrium is zero and $f'(0)=0$ . Solutions below zero rise toward it, while those above zero rise away and blow up. Zero is semistable, not asymptotically stable from both sides.

Interpreting the result

Carrying capacity is a modeling assumption, not a universal population law. Negative populations are mathematically possible in some equations but usually outside the model’s intended domain.

Practice

Classify equilibria for $y'=y(1-y)$ .
What happens for positive logistic initial data above $K$ ?
Where is logistic growth fastest?

Show worked solutions

$0$ is unstable and $1$ is asymptotically stable because derivatives are $1$ and $-1$ .
The derivative is negative, so the solution decreases toward $K$ .
At $y=K/2$ , with rate $rK/4$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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