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I am studying the generalized $n$-species competitive Lotka-Volterra system where populations of species $i$ are defined by the standard differential equation:

$$ \dot x_i = f_i(\mathbf{x}) := x_i \left( 1 - \sum_j a_{ij}x_j \right) $$

where all $a_{ij} \geq 0$ and $a_{ii}=1$. I know that any asymptotic behavior can be observed in general. However, I am wondering if there exist some constraints on the competition (or interaction) matrix $\mathbf{A}$ — with elements $a_{ij}$ — such that only a single species survives. More formally, do there exist characteristics of $\mathbf{A}$ such that there is only one fixed point $\mathbf{x}^\star_k$ wherein $x_k = 1$ and $\forall j\neq k, x_j=0$?

I have already observed that if $\mathbf{A}$ is a lower triangular matrix with all elements $1$ we obtain that only the dominant species $x_1$ will survive. We can also always reorder/relabel any system such that the species with index $1$ is the dominant one. It will the the only one to survive as we have in the fixed point $\mathbf{x}^\star$: $$ f_1(\mathbf{x}^\star) = x_1^\star(1-x_1^\star) = 0, \;\text{thus}\; x_1^\star = 1, $$ as we are interested in systems with $x_i(0) > 0$. This is just logistic growth of species $1$. For subsequent species $x_2$ we obtain $$ f_2(\mathbf{x}^\star) = x_2^\star(1-x_2^\star-x_1^\star) = 0, \;\text{thus}\; x_2^\star = 0, $$ and the same holds for any other $i>2$.

Is this the only way that we can make only a single (dominant) species survive if all initial abundances are positive, $x_i>0$? Or are there perhaps more constraints we can place on the competition matrix $\mathbf{A}$ (or its elements $a_{ij}$) that ensure only a single species survives in the end?

Johannes Nauta
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    Rough idea: it is well known that lotka--volterra system is equivalent to a replicator equation with some matrix $\bar A$. For this replicator equation is it sufficient to have a dominant strategy so that all other strategies will go extinct, in terms of lotka-volterra system this will be equivalent to the survival of only one species. – Artem Jun 28 '23 at 19:11
  • The competition (or interaction) matrix is the matrix $\mathbf{A}$ with elements $a_{ij}$ that determine the competition rate as the influence of species $j$ on species $i$. – Johannes Nauta Jun 28 '23 at 21:40
  • In the fixed point $dx^\star_i/dt=0$. When $\mathbf{A}$ is a lower triangular matrix with all non-zero elements equal to $1$, we have that $a_{11}=1$. Thus $dx_1/dt = x_1(1-x_1)$. Then, in the fixed point the abundance of species $1$ is given by $x_1^\star$, which solves the equation $x_1^\star(1-x_1^\star) = 0$, thus $x_1^\star=1$ (or $0$, but we are interested in systems with $x_i(0)>0$). The notation $\frac{dx_1}{dt}|_{\mathbf{x}^\star}$ is just the time derivative of $x_1$ evaluated at the fixed point $\mathbf{x}^\star = (x_1^\star, x_2^\star, \ldots, x_n^\star)$. – Johannes Nauta Jun 29 '23 at 09:27
  • Ok, I have updated the question to reflect this. – Johannes Nauta Jun 29 '23 at 09:37
  • The definition of $f_i$ clashes with the one on Wikipedia – Rodrigo de Azevedo Jun 29 '23 at 09:56

2 Answers2

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The simplest criteria I know that are sufficient to ensure that only a single species survives in any Lotka-Volterra system are that:

  • the interaction matrix $(-A)$ is Volterra-Lyapunov stable, which means there is some diagonal matrix $D$ with positive diagonal for which $DA+A^TD$ is positive definite; and
  • some single-species equilibrium is linearly (asymptotically) stable.

The Volterra-Lyapunov stability condition generally ensures that all solutions of the system approach some unique globally attracting equilibrium state, according to results in the 1998 book of Hofbauer & Sigmund, Evolutionary Games and Population Dynamics.

The linear stability condition is easy to check for a single-species state, as it just means $a_{kk}>0$ for all $k$ in this case. In other words, it means the diagonal elements of $A$ are positive (whether or not the system is competitive, with all nonnegative matrix entries). The reason is that linear asymptotic stability implies the state is a local attractor; but the system has a unique global attractor due to the Volterra-Lyapunov stability property.

Without the Volterra-Lyapunov stability condition, it turns out that competitive Lotka-Volterra systems can admit quite large numbers of linearly stable steady states, at least $3^{N/3}$ for certain examples. Won Eui Hong and I published a paper about this recently, also available on the arXiv.

One has to be careful doing as @Artem suggests and carrying over stability results from replicator systems to Lotka-Volterra systems. The known mappings do not ensure equivalence in every case. Moreover, the famous evolutionarily stable state criterion for replicator systems neither implies nor is implied by linear asymptotic stability in the corresponding Lotka-Volterra system, in general. This is discussed in section 6 in our paper. For single-species dominance in particular, though, I haven't thought about it.

Bob Pego
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  • Thank you for the suggestions. I will have to look a bit deeper into the statements in order to check for a scheme that ensures only a single species survives. Please also check the answer I just posted. – Johannes Nauta Jul 02 '23 at 07:43
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I have found some work by M.L. Zeeman (1995) that pertains precisely to the question I was asking. More specifically, in the paper it is proven that for an $n$-species competitive Lotka-Volterra model the following algebraic criteria on the parameters guarantee that all but one of the species is driven to extinction. For a system given by $$ \dot{x}_i = x_i\left(b_i- \sum_{j}^n a_{ij}x_j\right) $$ the criteria are $$ \begin{align} \frac{b_j}{a_{jj}} < \frac{b_i}{a_{ij}} \quad \forall i<j, \quad \text{and} \quad \frac{b_j}{a_{jj}} > \frac{b_i}{a_{ij}} \quad \forall i>j. \end{align} $$ Which gives the fixed point $R_1 = (b_1/a_{11}, 0, 0, \ldots, 0)$ to be globally attracting on $\mathbb{R}^n_+$; i.e. the state with only positive abundance of species $x_1$ (after reordering if needed) will be globally attracting, thus only species $1$ will not go extinct.

Johannes Nauta
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