# On the Properties of a Class of Impulsive Competition Beverton-Holt Equations
## 4 About Extinction Conditions under Small Intrinsic Growth Rates
Now, note that the conditions \(\{\mu(k)\}_{k=0}^{\infty}\subset(0,\,1]\), \(\{b(k)\}_{k=1}^{\infty}\subset(0,\,1]\) and \(\{d(k)\}_{k=1}^{\infty}\subset(-\infty,+\infty)\) also imply that the domain of \(x(k)\) is bounded and, from (77), \(x(k+1)\) is also bounded. Thus, Case a fully verifies the statement claim. **Case b**: \(x(k+1)\geq 0\), if \(x(k)\in\left(\frac{K(k)}{1-\mu(k)},\ \frac{d(k+1)K(k)}{b(k+1)\mu(k)K(k)+d(k+1)(1-\mu(k))}\right)\). However, a necessary condition for the above domain of values of \(x(k)\) to be well-posed is that  
\[\frac{1}{1-\mu(k)}<\frac{d(k+1)}{b(k+1)\mu(k)K(k)+d(k+1)(1-\mu(k))}\] \[\Leftrightarrow d(k+1)>d(k+1)+\frac{b(k+1)\mu(k)K(k)}{1-\mu(k)}\]which is a contradiction. Then, Case b does not verify the statement claim. **Case c**: \(x(k+1)\geq 0\), if \(x(k)=\frac{K(k)}{1-\mu(k)}=\frac{d(k+1)K(k)}{b(k+1)\mu(k)K(k)+d(k+1)(1-\mu(k))}\) and the indetermination \(0/0\), resulting from (77), is solvable with a positive limit, as the numerator and denominator converge to zero. However, the identity of both confluent values of \(x(k)\) leads to the following contradiction:  
\[0=d(k+1)-d(k+1)=\frac{b(k+1)\mu(k)K(k)}{1-\mu(k)}>0\]  
so that Case c does not verify the statement claim. As a result, only Case a satisfies the statement claim and the proof is complete. \(\square\)  
The next main result of this section concerned with the case in which one single species relies on the asymptotic convergence of the solution to extinction when the solution to the right of the sampling instants is a non-negative strictly decreasing sequence, that is, \(\{x(k)-x(k+1)\}_{k=0}^{\infty}\subset\mathbf{R}_{+}\) and \(\{x(k)\}_{k=0}^{\infty}(\subset\mathbf{R}_{+})\to 0\). Let us also assume that  
\[ \begin{array}{c}d(k+1)\in\left(\frac{[(b(k+1)\mu(k)-1)K(k)+(1-\mu(k))x(k)] \ x(k)}{K(k)-(1-\mu(k))x(k)},\,\frac{b(k+1)\mu(k)K(k)x(k)}{K(k)-(1-\mu(k))x(k) }\right];\\ \forall k\in N_{0}\end{array}  \tag{79}\]  
Then, \(\{x(k)\}_{k=0}^{\infty}\subset\mathbf{R}_{0+}\), \(\{x^{-}(k)\}_{k=0}^{\infty}\subset\mathbf{R}_{0+}\), \(\{d(k)\}_{k=1}^{\infty}\subset\mathbf{R}\) are bounded, with \(\{x^{-}(k)\}_{k=0}^{\infty}\to 0\) and \(\{x(k)\}_{k=0}^{\infty}\to 0\) being strictly decreasing and \(\{d(k)\}_{k=1}^{\infty}\to 0\). Proof. : Let us consider the Lyapunov sequence candidate \(V(k)=x(k)\); \(\forall k\in\mathbf{N}_{0}\) so that \(\Delta V(k)=V(k+1)-V(k)=x(k+1)-x(k)\). Then, from Proposition 6, \(0<x(k)<\frac{K(k)}{1-\mu(k)}\) for any \(k\in\mathbf{N}_{0}\) implies that \(x(k+1)\geq 0\), so that  
\[ \begin{array}{c}(K(k)-(1-\mu(k))x(k))\Delta V(k)=(K(k)-(1-\mu(k))x(k))\ (V(k+1)-V(k))\\ =[(b(k+1)\mu(k)-1)K(k)+d(k+1)\ (1-\mu(k))+(1-\mu(k))x(k)]\ x(k)-d(k+1)K (k)\\ =[(b(k+1)\mu(k)-1)K(k)+(1-\mu(k))x(k)]\ x(k)-d(k+1)\ [K(k)-(1-\mu(k))x(k)]\leq 0; \\ \forall k\in\mathbf{N}_{0}\end{array}  \tag{80}\]  
and \(\{x(k)\}_{k=0}^{\infty}\subset\mathbf{R}_{0+}\) is bounded and non-increasing, since \(0<x(0)\ <\frac{K(0)}{1-\mu(0)}\) if  
\[d(k+1)\geq\frac{[(b(k+1)\mu(k)-1)K(k)+(1-\mu(k))x(k)]\ x(k)}{K(k)-(1-\mu(k))x(k )};\ \forall k\in\mathbf{N}_{0} \tag{81}\]  
which implies that \(\Delta V(k)\leq 0\); \(\forall k\in\mathbf{N}_{0}\). Since \(K(k)>(1-\mu(k))x(k)\), it follows, from (80), that \(x(k+1)<x(k)\), so that \(\{x(k)\}_{k=0}^{\infty}\) is strictly decreasing, if \(x(k)>0\) and (81) is a strict inequality, \(\Delta V(k)\to 0\) as \(k\to\infty\) and that \(x(k)\to 0\) as \(k\to\infty\). If \(x(k+1)\geq 0\), then  
\[d(k+1)\leq b(k+1)x^{-}(k+1)=\frac{b(k+1)\mu(k)K(k)x(k)}{K(k)-(1-\mu(k))x(k)}; \ \forall k\in\mathbf{N}_{0} \tag{82}\]  
and \(\{x^{-}(k)\}_{k=0}^{\infty}\subset\mathbf{R}_{0+}\) is also bounded, since \(x(0)=x^{-}(0)\) is finite and \(\{d(k)\}_{k=1}^{\infty}\subset(-\infty,+\infty)\). Combining both inequalities, (81), in its strict version, and (82), yields (79). This