Two-Population Competition for Shared Capacity

Why Model Two Populations Together?

When two cell populations draw on limited space, nutrients, or another shared capacity, the growth of either population can depend on the abundance of both. Modeling each population with its own independent growth equation would miss this cross-effect. A coupled model makes that effect explicit and lets us ask whether the populations can coexist, whether one excludes the other, or whether the eventual outcome depends on their starting abundances.

The model below is a compact phenomenological description: it represents crowding by abundance terms rather than explicitly tracking a resource. Its conclusions depend on fixed growth and competition parameters, so it is most useful when the environment is approximately constant over the time interval of interest.

Definition: In the two-population competition model, $N_1(t)$ and $N_2(t)$ are the abundances of populations 1 and 2 at time $t$, measured in cells. Their dynamics are $$ \frac{dN_1}{dt}=r_1N_1\left(1-\frac{N_1+\alpha_{12}N_2}{K_1}\right), \qquad \frac{dN_2}{dt}=r_2N_2\left(1-\frac{\alpha_{21}N_1+N_2}{K_2}\right). $$ Here $r_1,r_2>0$ are intrinsic per-capita growth rates (per unit time), $K_1,K_2>0$ are the single-population carrying capacities (cells), and $\alpha_{12},\alpha_{21}\geq 0$ are dimensionless competition coefficients. In particular, $\alpha_{12}$ scales the effect of population 2 on population 1; $\alpha_{21}$ scales the reverse effect. They need not be equal.

The bracket in each equation is the population's per-capita growth factor relative to its crowding-adjusted capacity. For population 1, for example, $N_1+\alpha_{12}N_2=K_1$ is its zero-growth condition. One cell of population 2 therefore has the same modeled crowding effect on population 1 as $\alpha_{12}$ cells of population 1. These are relative effects, not necessarily literal cell-for-cell substitutions for a measured physical resource.

The standard assumptions are constant parameters; continuous, deterministic abundance trajectories; no switching between the populations; and crowding that is linear in the two abundances within each growth bracket. The equations preserve nonnegative abundances: if a population is at zero, its derivative is zero, and a trajectory starting positive does not cross through zero. These assumptions make equilibrium and stability analysis meaningful, but they do not establish that the model is appropriate for every cell system.

Equilibria and the Coexistence Calculation

An equilibrium is a pair of abundances at which both derivatives are zero. The equations have the extinction equilibrium $(0,0)$, two single-population equilibria $(K_1,0)$ and $(0,K_2)$, and possibly an equilibrium with both populations present. At a positive interior equilibrium, each bracket must be zero, giving two linear equations:

N_1^*+\alpha_{12}N_2^*=K_1, \qquad \alpha_{21}N_1^*+N_2^*=K_2.

Let $D=1-\alpha_{12}\alpha_{21}$, the determinant of this pair of equations. When $D\neq 0$, solving gives

N_1^*=\frac{K_1-\alpha_{12}K_2}{D}, \qquad N_2^*=\frac{K_2-\alpha_{21}K_1}{D}.

These expressions are candidate equilibrium abundances, not a guarantee of coexistence. Both must be positive to describe a biologically relevant interior equilibrium. In the common case $D>0$, that requires $K_1>\alpha_{12}K_2$ and $K_2>\alpha_{21}K_1$. The boundary equilibria provide a complementary way to assess whether coexistence is possible: ask whether a rare population can increase when introduced into an environment occupied by the other population.

Worked Example: A Stable Coexistence Equilibrium

Suppose, hypothetically, that $K_1=1000$ cells, $K_2=800$ cells, $\alpha_{12}=0.50$, and $\alpha_{21}=0.40$. Let $r_1=0.60$ per day and $r_2=0.40$ per day. First check that the model permits a positive interior equilibrium.

The determinant is $D=1-(0.50)(0.40)=0.80$. The equilibrium abundances are $$ N_1^*=\frac{1000-(0.50)(800)}{0.80} =\frac{600}{0.80}=750\text{ cells}, $$ $$ N_2^*=\frac{800-(0.40)(1000)}{0.80} =\frac{400}{0.80}=500\text{ cells}. $$

Both are positive. Substitution checks the two equilibrium conditions: $750+(0.50)(500)=1000$ cells and $(0.40)(750)+500=800$ cells. Thus neither population is predicted to grow or decline at this pair of abundances.

Invasion and Long-Term Outcomes

At the population 1-only equilibrium $(K_1,0)$, a very small population 2 has approximate per-capita growth rate $r_2(1-\alpha_{21}K_1/K_2)$. Population 2 can invade if this rate is positive, equivalently if $K_2>\alpha_{21}K_1$. Likewise, population 1 can invade the population 2-only equilibrium if $K_1>\alpha_{12}K_2$. These invasion conditions compare the resident's crowding effect with the rare population's capacity; they are not comparisons of the intrinsic growth rates alone.

Theorem: If $r_1,r_2,K_1,K_2>0$, $\alpha_{12},\alpha_{21}\geq0$, and both invasion conditions hold, $$ K_1>\alpha_{12}K_2 \quad\text{and}\quad K_2>\alpha_{21}K_1, $$ then $D>0$, the interior equilibrium has positive abundances, and it is locally asymptotically stable. “Locally asymptotically stable” means that trajectories starting sufficiently close to the equilibrium remain close and converge to it.

A short stability check justifies the final claim. The Jacobian matrix, the matrix of first derivatives of the two right-hand sides, evaluated at a positive interior equilibrium is

J(N_1^*,N_2^*)= \begin{pmatrix} -\dfrac{r_1N_1^*}{K_1} & -\dfrac{r_1\alpha_{12}N_1^*}{K_1}\\ -\dfrac{r_2\alpha_{21}N_2^*}{K_2} & -\dfrac{r_2N_2^*}{K_2} \end{pmatrix}.

Its trace is negative. Its determinant is $$ \det(J)=\frac{r_1r_2N_1^*N_2^*}{K_1K_2}(1-\alpha_{12}\alpha_{21}). $$ Under the theorem's conditions, the first factor is positive and $D>0$, so the determinant is positive. For a two-dimensional linearization, negative trace and positive determinant imply eigenvalues with negative real parts; hence the equilibrium is locally asymptotically stable. This is a local conclusion, not a guarantee about every possible model extension or every parameter regime.

Worked Example: Competitive Exclusion

Consider a hypothetical system with $K_1=1000$ cells, $K_2=700$ cells, $\alpha_{12}=0.80$, and $\alpha_{21}=0.90$. At the population 1-only equilibrium, population 2's per-capita invasion rate is $$ r_2\left(1-\frac{(0.90)(1000)}{700}\right) =r_2(1-1.286)=-0.2857r_2. $$ It is negative, so population 2 cannot increase when rare in this resident environment.

At the population 2-only equilibrium, population 1's per-capita invasion rate is $$ r_1\left(1-\frac{(0.80)(700)}{1000}\right) =r_1(1-0.56)=0.4400r_1. $$ It is positive, so population 1 can invade a population 2 resident. These signs indicate that $(1000,0)$ is locally stable to introduction of population 2, whereas $(0,700)$ is unstable to introduction of population 1. The model therefore predicts population 1 exclusion of population 2 for positive starting abundances in this regime.

If neither population can invade the other when rare, both boundary equilibria can be locally stable. Initial abundances can then determine which population persists. The interior equilibrium, if positive in this case, is typically a saddle that separates the outcomes rather than a stable coexistence state.

Worked Example: Alternative Stable Outcomes

For a hypothetical system, let $K_1=1000$ cells, $K_2=800$ cells, $\alpha_{12}=1.40$, and $\alpha_{21}=0.90$. Population 2's invasion factor at $(K_1,0)$ is $1-(0.90)(1000)/800=-0.1250$. Population 1's invasion factor at $(0,K_2)$ is $1-(1.40)(800)/1000=-0.1200$. Both are negative, so neither population can invade the other when rare.

The interior equilibrium calculation gives $D=1-(1.40)(0.90)=-0.26$, $$ N_1^*=\frac{1000-(1.40)(800)}{-0.26} =\frac{-120}{-0.26}=461.5\text{ cells}, $$ $$ N_2^*=\frac{800-(0.90)(1000)}{-0.26} =\frac{-100}{-0.26}=384.6\text{ cells}. $$ Both values are positive and satisfy the zero-growth equations: $461.5+(1.40)(384.6)\approx1000$ cells and $(0.90)(461.5)+384.6\approx800$ cells. However, the interior equilibrium is a saddle, because the Jacobian determinant has the sign of $D$, which is negative. Thus positive abundances at an interior equilibrium do not by themselves mean stable coexistence. In this parameter regime, the two locally stable boundary outcomes are separated by the unstable interior equilibrium.

Reading the Model Carefully

A reliable analysis separates three questions: whether an interior equilibrium exists, whether it is stable, and whether a boundary population can be invaded. These questions answer different things. In particular, a positive solution to the equilibrium equations answers only the first question. The invasion rates and local stability calculation determine whether that solution describes a stable coexistence outcome.

When diagnosing a fit or simulation, check that abundances remain nonnegative, the two zero-growth conditions hold at any reported equilibrium, and numerical trajectories approach the claimed outcome under the stated initial conditions. A trajectory that appears stationary over a short interval may still be far from equilibrium. Also inspect whether the inferred competition coefficients are plausible and whether the data cover a range of abundances sufficient to distinguish self-limitation from cross-limitation.

The model is simpler than alternatives that explicitly track a shared resource or allow the growth rates and capacities to change with time. Its advantage is a small set of interpretable parameters and direct equilibrium calculations. Its limitations follow from the same simplicity: constant linear cross-effects may not describe resource depletion, spatial structure, time-varying environments, or density-dependent mechanisms that change with abundance. If those features are important, a more mechanistic or flexible model may be needed. The equations describe population-level competition; they do not identify the biological mechanism that produced the competition coefficients.

Common Pitfalls

Swapping the competition coefficients. In the first equation $\alpha_{12}$ is the effect of population 2 on population 1; in the second, $\alpha_{21}$ is the reverse effect. They are not required to be equal.
Stopping after finding positive equilibrium abundances. Check stability as well. A positive interior equilibrium can be a saddle, as in the alternative-outcomes example.
Using intrinsic growth rates to decide the winner. Invasion depends on the resident's crowding and the invader's capacity, as shown by $1-\alpha_{21}K_1/K_2$ and $1-\alpha_{12}K_2/K_1$.
Making an arithmetic error in the determinant division. Evaluate the numerator and divide by $D$ explicitly, then substitute the results back into both equilibrium equations.
Interpreting the model as causal evidence. A fitted competition term describes the trajectory under the model; without additional evidence it does not prove a particular shared-resource mechanism.

Key takeaway: Stable coexistence requires both populations to invade when rare. Calculate the equilibrium, verify both abundances and zero-growth equations, and then use invasion rates and stability—not positivity alone—to interpret the long-term outcome.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

In the model, what effect does $\alpha_{12}$ represent, and must it equal $\alpha_{21}$?
For $K_1=90$ cells, $K_2=60$ cells, $\alpha_{12}=0.40$, and $\alpha_{21}=0.50$, calculate the interior equilibrium and check both equilibrium equations.
If $K_2<\alpha_{21}K_1$, can population 2 invade a population 1-only equilibrium? Explain using its per-capita growth rate.
Why is a positive interior equilibrium not sufficient evidence for stable coexistence?

Solutions. (1) $\alpha_{12}$ scales the effect of population 2 on population 1 in the first equation. It need not equal $\alpha_{21}$, which scales the reverse effect. (2) $D=1-(0.40)(0.50)=0.80$. Thus $N_1^*=[90-(0.40)(60)]/0.80=66/0.80=82.5$ cells and $N_2^*=[60-(0.50)(90)]/0.80=15/0.80=18.75$ cells. Substitution gives $82.5+(0.40)(18.75)=90$ cells and $(0.50)(82.5)+18.75=60$ cells. Both invasion conditions hold: $90-24=66>0$ and $60-45=15>0$; the positive coexistence equilibrium is locally stable. (3) Its invasion rate is $r_2(1-\alpha_{21}K_1/K_2)$, which is negative when $K_2<\alpha_{21}K_1$, so population 2 cannot increase when rare. (4) The equilibrium may be unstable; stability must be checked, for example from the Jacobian determinant and trace.