Garrett M et al., 2013 doesn’t provide insights on how to ‘explain’ following:
What does make up the carrying capacity? (while paper claims students explain that carrying capacity is based on space and resource limitations, but it’s connection is not coming from the given lesson plan - rather it could be intution or prior knowledge.)
This can be potentially due to higher requirement of math integration.
How it’s connected (GenAI Conversation)
why does the population growth curve become oscillating when growth rate becomes more than 1… like 2.5 or something?
Like why the population goes down after reaching certain point. And why does that not happen with growth rate lower than 1?
The oscillations occur because the equation being used is a discrete-time model with an inherent time lag, meaning population adjustments are not instantaneous. In the discrete logistic equation:
N(t+1)=N(t)+rN(t)(1−KN(t))
the number of individuals born or added in the next step is determined entirely by the conditions at the current step t.
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1. Why the Population Drops After Reaching a High Point
When the population is below the carrying capacity (N(t)<K), the limiting factor (1−KN(t)) is positive, so the population grows.
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The Overshoot: If the growth rate r is high, the population adds a massive surge of new individuals in a single time step. Because the growth calculation only “looks” at the resources that were available at time t, it does not slow down gradually during the interval. It leaps right past the carrying capacity, landing at N(t+1)>K.
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The Crash: At this new step, because N>K, the term (1−KN) flips and becomes negative. When multiplied into the equation, the growth component becomes a large deduction. Biologically, the population has severely exceeded available resources, resulting in food shortages, competition, and spiked mortality, causing the population to plummet back down.
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2. Why This Does Not Happen When r<1
When r<1, the population applies the “brakes” gently enough that it never overshoots K.
Mathematically, analyzing the behavior near the carrying capacity K reveals why: If N(t) is close to K, the deviation from equilibrium xt=N(t)−K evolves from one step to the next approximately as:
xt+1≈(1−r)xt
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When 0<r<1, the multiplier (1−r) is positive and strictly less than 1 (between 0 and 1).
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Because it is positive, the deviation does not change sign—meaning if the population starts below K, it stays below K on every single step.
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Because it is less than 1, the gap between the population and K steadily shrinks toward zero. The population approaches K smoothly from below, producing the classic S-shaped (sigmoidal) curve.
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3. What Happens as r Increases Above 1
As r increases, (1−r) becomes negative, which causes the population to flip back and forth across K:
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1<r<2 (Damped Oscillations): Here, −1<(1−r)<0. The negative sign causes the population to overshoot K, then undershoot K, alternating sides each generation. However, because the magnitude ∣1−r∣<1, each bounce is smaller than the last, eventually dampening down and stabilizing exactly at K.
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2<r≤2.45 (Stable 2-Point Oscillations): Once r>2, the equilibrium point K becomes unstable because ∣1−r∣>1. The overshoot is so extreme that the resulting crash is equally drastic. The population enters a perpetual loop bouncing back and forth between two fixed values (one high overshoot, one low crash).
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r>2.5 (Period-Doubling & Chaos): As r climbs further, the cycle doubles into a 4-point cycle, then 8-point, quickly devolving into deterministic chaos. In this state, population sizes fluctuate wildly from generation to generation without ever settling into a repeating cycle.
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where is this coming from?
This equation comes from linear stability analysis—specifically, looking at what happens to a tiny deviation from the equilibrium point K using a linear approximation (or first-order Taylor expansion).
Step-by-Step Derivation
Start with the discrete logistic equation:
N(t+1)=N(t)+rN(t)(1−KN(t))
1. Define the deviation (perturbation)
Let xt be the difference between the population and the carrying capacity at time t:
xt=N(t)−K⟹N(t)=K+xt
Similarly, at the next step:
N(t+1)=K+xt+1
2. Substitute into the growth equation
Substitute N(t)=K+xt and N(t+1)=K+xt+1 into the discrete logistic model:
K+xt+1=(K+xt)+r(K+xt)(1−KK+xt)
3. Simplify the term inside the brackets
Simplify the limiting term (1−KK+xt):
1−KK+xt=1−(1+Kxt)=−Kxt
4. Substitute back and expand
Now insert that back into the equation:
K+xt+1=K+xt+r(K+xt)(−Kxt)
Subtract K from both sides:
xt+1=xt−Kr(K+xt)xt
Expand the right-hand fraction:
Kr(K+xt)xt=KrKxt+rxt2=rxt+Krxt2
Substituting this back gives:
xt+1=xt−(rxt+Krxt2)=(1−r)xt−Krxt2
5. Drop the higher-order term
When the population is very close to carrying capacity, the deviation xt is tiny compared to K (∣xt∣≪K).
Because squaring a small number makes it much smaller (for example, if xt/K=0.01, then (xt/K)2=0.0001), the quadratic term −Krxt2 is negligible compared to the linear term. Ignoring this higher-order term leaves:
xt+1≈(1−r)xt
Alternative View via Calculus (Derivative at the Fixed Point)
If you view the update rule as a function f(N):
f(N)=N+rN−KrN2
Take the first derivative with respect to N:
f′(N)=1+r−K2rN
Evaluate this derivative at the carrying capacity equilibrium N∗=K:
f′(K)=1+r−K2rK=1+r−2r=1−r
By Taylor expansion around K:
xt+1≈f′(K)⋅xt=(1−r)xt
The factor (1−r) represents the eigenvalue (or slope) of the map at the fixed point. Its magnitude and sign dictate whether deviations shrink, alternate, or explode.