# On the Properties of a Class of Impulsive Competition Beverton-Holt Equations
## 2 The Competition Equation and Its Main Properties
### The Single-Species Impulsive Beverton-Holt Equation
It can be argued that, although it is a common hypothesis to assume that \(\{K(k)\}\subset R_{+}\) and \(\{\mu(k)-1\}\subset R_{+}\), it is possible, under certain extra constraints on the state of sufficient smallness, related to information about the equation parameters, that \(\mu(k)\) could take values under unity. To clarify this concern, let us assume Equation (1) without impulsive effects, that is, \(\{b(k)\}_{k=0}^{\infty}\equiv 1\) and \(\{d(k)\}_{k=0}^{\infty}\equiv 0\), so that \(x(k)=x^{-}(k)\); \(\forall k\in N_{0}\). Thus, for any given \(k\in N_{0}\), if \(\mu(k)\in[0,\,1]\), then \(x(k+1)\geq 0\) if \(0\leq x(k)\leq\frac{K(k)}{1-\mu(k)}\), and \(x(k+1)\)  
Now, let us assume that \(\mu(k)\in(0,\ 1)\) and \(0<x(k)<\frac{K(k)}{1-\mu(k)}\), implying that \(x(k+1)>0\) and finite. Let us also assume that \(\mu(k+1)\in(0,\ 1)\). Then, \(0<x(k+1)<\frac{K(k+1)}{1-\mu(k+1)}\) for \(x(k+2)>0\) and finite. One obtains  
\[x(k+1)=\frac{\mu(k)K(k)x(k)}{K(k)-(1-\mu(k))x(k)}<\frac{K(k+1)}{1-\mu(k+1)}\]  
which implies that  
\[[\mu(k)K(k)(1-\mu(k+1))+K(k+1)(1-\mu(k))]x(k)<K(k)K(k+1)\]  
so that the former non-negativity constraint \(x(k)<\frac{K(k)}{1-\mu(k)}\) is amended by the stronger one  
\[x(k)<\frac{K(k)K(k+1)}{\mu(k)K(k)(1-\mu(k+1))+K(k+1)(1-\mu(k))}<\frac{K(k)}{1- \mu(k)}\]  
And, proceeding recursively in the same way, it follows that \(x(k)\) has a strictly decreasing infinite sequence of bounded non-negative upper-bounds. However, note that, if \(\mu(k+1)=1\), then the upper-bound remains unaltered. Thus, we have the following results. **Lemma 1. **_Let us assume that \(\{d(k)\}_{k=0}^{\infty}\equiv 0\) and \(\{b(k)\}_{k=0}^{\infty}\equiv 1\) (namely, there is no impulsive effect). Let us also assume that, for some finite \(k\in\mathbf{N}_{0}\), \(\{\mu(j)\}_{k_{0}}^{\infty}\subset(0,\ 1)\), \(\{x(j)\}_{k_{0}}^{\infty}\subset\mathbf{R}_{0+}\) is bounded and either \(\{K(k)\}_{k=k_{0}}^{\infty}\) is non-decreasing or \(\{K(k)\}_{j=k_{0}}^{\infty}\subset(0,\ 2)\). Then, \(\exists k_{1}\geq 1\) such that \(x(k)=0\) for \(k\geq k_{0}+k_{1}\) (that is, there is extinction in finite time)._  
**Proof. ** Note that, since \(\{\mu(j)\}_{k_{0}}^{\infty}\subset(0,\ 1)\), \(\{x(k)\}_{k=0}^{\infty}\subset[0,\ \infty)\) is strictly decreasing if and only if  
\[x(k+1)-x(k)=\bigg{(}\frac{\mu(k)K(k)}{K(k)-(1-\mu(k))x(k)}-1\bigg{)}x(k)<0;\ \forall k(\geq k_{0})\in\mathbf{N}_{0}\]  
If \(x(k)=0\) for some finite \(k\), then the result is already proved. Thus, let us assume that \(x(k)>0\); \(\forall k(\geq k_{0})\in\mathbf{N}_{0}\). Note that \(x(k)>0\), if and only if \(x(k)<\min\bigg{(}K(k),\ \frac{K(k)}{1-\mu(k)}\bigg{)}=K(k)\), since \(\frac{\mu(k)K(k)}{K(k)-(1-\mu(k))x(k)}<1\), that is, since \((1-\mu(k))x(k)<(1-\mu(k))K(k)\), leading to \(x(k)<K(k)\); \(\forall k(\geq k_{0})\in\mathbf{N}_{0}\), in order for it to be jointly strictly decreasing with the denominator of the right-hand-side being positive for all \(\forall k(\geq k_{o})\in\mathbf{N}_{0}\). Then, either \((x(k)-K(k))\to 0\) as \(k\to\infty\), or \(x(k)\to 0\) as \(k\to\infty\); furthermore,  
\[x(k)<\min(K(k),\ x(k-1))<\ldots<\inf_{k\geq j\geq 0}K(j);\ \forall k(\geq k_{0})\in \mathbf{N}_{0}\]  
Since \(\{K(k)\}_{j=k_{0}}^{\infty}\) is non-decreasing, then \(\inf_{k_{0}\geq k\geq 0}K(k)=\inf_{\infty\geq k\geq 0}K(k)\geq\inf_{ \infty\geq k\geq k_{0}}K(k)\), so that  
\[x(k)<\!\!\min\!\left(\inf_{k\geq j\geq 0}K(j),\ x(k-1)\right)\geq\inf_{ \infty\geq k\geq k_{0}}K(k)\geq K(k_{0});\ \forall j(\geq k_{0})\in\mathbf{N}_{0}\]  
Therefore, \((x(k)-K(k))\to 0\) as \(k\to\infty\) is not possible, so \(x(k)\to 0\) as \(k\to\infty\). Since \(k_{0}\) is finite and \(\{x(k)\}_{k_{0}}^{\infty}\subset\mathbf{R}_{0+}\) is strictly decreasing and converges asymptotically to zero, then there is some finite \(k_{1}\in\mathbf{N}\) such that \(x(k_{0}+k_{1}-1)<2\) and the species reproduction is not feasible for the next samples, so that extinction happens in finite time \(k_{0}+k_{1}\).