Add a limiting capacity and analyze stability

Try

An exponential model grows at a fixed fractional rate regardless of size. To represent a limiting resource in a synthetic model, try y′=ry(1−y/K), with r positive and K positive. Here y is the amount, r has inverse-time units, and K has the same units as y. Begin with K=12 and r=1. Calculate the rates at y=0,3,6,9,12, and15 before drawing a trajectory.

The factor 1−y/K reduces growth as y approaches K. It becomes negative above K, so the same equation can describe decline toward the equilibrium from above.

Predict

Does the largest population have the largest positive growth rate? Predict where the rate is maximal. Then distinguish two quantities: total rate y′ and fractional rate y′/y for positive y. The latter decreases linearly with y, whereas the former is a downward-opening quadratic.

Experiment

Use the following rate trace to draw a phase line and a rate-versus-state graph. The horizontal axis of that graph is state, not time.

State y	Resource factor 1−y/12	Rate y(1−y/12)
0	1	0
3	3/4	9/4
6	1/2	3
9	1/4	9/4
12	0	0
15	−1/4	−15/4

Differentiate the rate with respect to state: r(1−2y/K). It vanishes at K/2, and its derivative is −2r/K, so the maximum rate is rK/4. For r=1,K=12 this is three. Identify the symmetric pair of states with equal rates in the table and explain why equal rates do not imply equal future trajectories.

Understand

The equilibrium solutions are y=0 and y=K. Preserve them before dividing during separation. For a positive initial value y₀ different from K, integration and initial data give y(t)=K/[1+A e^(−rt)], where A=K/y₀−1. One way to derive it is to integrate K/[y(K−y)]=1/y+1/(K−y), obtaining ln∣y/(K−y)∣=rt+C. The second logarithm carries a minus sign after integration.

For K=12,r=1,y₀=3, A=3 and y=12/(1+3e^(−t)). Substitution at zero gives three, and as t increases without bound the exponential tends to zero, giving a limiting value of twelve. The rate is greatest when y=6, which occurs when 3e^(−t)=1, or t=ln 3. The trajectory changes concavity there because y″=r(1−2y/K)y′.

If y₀ exceeds K, A lies between negative one and zero. For forward time the denominator stays positive and increases toward one, so y decreases toward K without crossing it. Thus “logistic growth always increases” is false. Uniqueness for this smooth polynomial rate also prevents a non-equilibrium solution from crossing an equilibrium in finite time.

Negative states are mathematically permitted but may not represent populations. Their behavior should not be used as a biological prediction. Likewise, K is a constant parameter in this model, not proof that a real resource limit stays fixed.

Apply

Change K while holding r fixed and predict the effect on the maximum rate and the equilibrium. Then change r alone and identify what changes in time versus state. Support your explanation with rK/4 and K/2 rather than a vague claim about faster growth.

Propose measurements that could distinguish exponential from logistic behavior. Include observations at more than one state scale and explain why early low-state data may fit both. Complete the practice by checking initial data, equilibria, and domain before interpreting a formula.

Continue

Next, use an integrating factor when time-dependent addition prevents simple separation. Keep the habit of checking an equilibrium directly in the original equation before using a transformed expression.