# On the Properties of a Class of Impulsive Competition Beverton-Holt Equations
## 4 About Extinction Conditions under Small Intrinsic Growth Rates  
Note that Theorem 6 gives extinction conditions for intrinsic growth rates exceeding unity, which is an usual condition in the background literature on Beverton-Holt equations [5; 6; 7; 8; 9; 10; 11; 12; 18; 26; 27; 28; 29; 33]. In this section, the asymptotic extinction is studied for the case of intrinsic growth rates being less than unity. **Remark 6**: _Note that extinction at the right of the \(k-th\) sampling instant of the \(i-th\) species occurs if \(d_{i}(k)=b_{i}(k)x_{i}^{-}(k)\) and it occurs asymptotically if \(\left(d_{i}(k)-b_{i}(k)x_{i}^{-}(k)\right)\to 0\) as \(k\to\infty\). This can happen irrespective if the intrinsic growth rate exceeds unity or is less than unity. The first case, which is the typical situation in the Beverton-Holt equation, might be addressed as a particular case included in Theorem 6. The condition is difficult to achieve in practical cases, since it requires an exact monitoring of the independent consumption through time. In the following and for the case of intrinsic growth rates below unity, this case is excluded from the analysis._To fix ideas, we first discuss the case when just a single species is involved. For this purpose, we establish two preliminary necessary technical results for possible extinction either in finite time or asymptotic extinction, while keeping the solution sequence non-negative and bounded for all time under non-negative initial conditions. **Proposition 5**. : _Let us consider one single-species impulsive Beperton-Holt equation, Equation (1). Let us assume that \(\{\mu(k)\}_{k=0}^{\infty}\subset(0,\,1]\), \(\{b(k)\}_{k=1}^{\infty}\subset(0,\,1]\) and \(\{d(k)\}_{k=1}^{\infty}\subset(-\infty,+\infty)\) Then, for any given \(k\in\mathbb{N}_{0}\), \(x(k)=0\Rightarrow x^{-}(k+1)=0\) and, furthermore, \(x(k+1)=0\) if and only if \(d(k+1)=0\). If \(\{x(k)\}_{k=0}^{\infty}\to 0\), then \(\{d(k)\}_{k=1}^{\infty}\to 0\) and, if, in addition, \(\{b(k)\}_{k=1}^{\infty}\subset\mathbb{R}_{+}\), then \(\{x^{-}(k)\}_{k=0}^{\infty}\to 0\)._  
**Proof. ** One obtains, from (1), that  
\[x(k+1)=\frac{\left[b(k+1)\mu(k)K(k)+d(k+1)\ (1-\mu(k))\right]\ x(k)-d(k+1) K(k)}{K(k)-(1-\mu(k))x(k)}; \tag{77}\]  
\[\forall k\in\mathbb{N}_{0}\]  
Since \(K(k)>0,\ x(k)=0\Rightarrow x^{-}(k+1)=0\) from (1) and \(x(k+1)=0\) if and only if \(d(k+1)=0\) and, if \(b(k+1)>0\), then \(x^{-}(k+1)=\frac{x(k+1)+d(k+1)}{b(k+1)}=0\). Note that, if \(\{x(k)\}_{k=0}^{\infty}\to 0\), then  
\(d(k+1)=\frac{b(k+1)\mu(k)K(k)\ x(k)}{K(k)-(1-\mu(k))x(k)}-x(k+1)\to 0\) as \(k\rightarrow\infty\) and also \(x^{-}(k+1)=\frac{x(k+1)+d(k+1)}{b(k+1)}\to 0\) as \(k\rightarrow\infty\) if \(\{b(k)\}_{k=1}^{\infty}\subset\mathbb{R}_{+}\) since \(\{x(k)\}_{k=0}^{\infty}\to 0\) and \(\{d(k)\}_{k=1}^{\infty}\to 0\). \(\square\)  
**Proposition 6**. : _Let us consider a Beperton-Holt equation for one single species. Let us assume that \(\{\mu(k)\}_{k=0}^{\infty}\subset(0,\,1]\), \(\{b(k)\}_{k=1}^{\infty}\subset(0,\,1]\) and \(\{d(k)\}_{k=1}^{\infty}\subset(-\infty,+\infty)\). Then, for any given \(k\in\mathbb{N}_{0}\), \(x(k+1)\in[0,+\infty)\), with \(x(k)\geq 0\), if and only if_  
\[x(k)\in\left[\min\Bigl{(}0,\ \frac{d(k+1)K(k)}{b(k+1)\mu(k)K(k)+d(k+1)(1- \mu(k))}\Bigr{)},\ \frac{K(k)}{1-\mu(k)}\Bigr{)};\right. \tag{78}\] \[\forall k\in\mathbb{N}_{0}\]  
**Proof. ** Let us consider several cases in (77) with the given assumptions, namely,  
**Case a**: \(x(k+1)\geq 0\), if \(x(k)\in\left[\frac{d(k+1)K(k)}{b(k+1)\mu(k)K(k)+d(k+1)(1-\mu(k))},\ \frac{K(k)}{1-\mu(k)}\right)\). Note that the limits of the admissible domain of \(x(k)\) are mutually compatible, since  
\[\frac{d(k+1)}{b(k+1)\mu(k)K(k)+d(k+1)(1-\mu(k))}\ <\frac{1}{1-\mu(k)}\] \[\Leftrightarrow d(k+1)<d(k+1)+\frac{b(k+1)\mu(k)K(k)}{1-\mu(k)}\]  
which trivially holds, since \(\frac{b(k+1)\mu(k)K(k)}{1-\mu(k)}>0\). Since \(x(k)\) is also requested to be non-negative, then (78) is used. It was proved that (78) implies that \(x(k+1)\geq 0\).