# On the Properties of a Class of Impulsive Competition Beverton-Holt Equations
## 2 The Competition Equation and Its Main Properties
### Non-Negativity and Boundedness of the Solutions
** One obtains that (2) is equivalent to  
\[x_{i}^{-}(k+1)=\frac{\mu_{i}x_{i}(k)}{1+\frac{\mu_{i}-1}{K_{i}}\sum_{j=1}^{n} c_{ji}x_{j}(k)}=\frac{\mu_{i}x_{i}(k)}{1+\frac{\mu_{i}-1}{K_{i}(k)}x_{i}(k)}; \forall k\in\mathbf{N}_{0} \tag{7}\]  
where the equivalent \(i-th\) carrying capacity \(K_{i\epsilon}\) is given by  
\[0\leq K_{i\epsilon}(k)=\frac{K_{i}x_{i}(k)}{\sum_{j=1}^{n}c_{ji}x_{j}(k)}= \frac{K_{i}}{1+\sum_{j(\
eq i)=1}^{n}c_{ji}\frac{x_{j}(k)}{x_{i}(k)}}\leq K_{ i};\;\forall k\in\mathbf{N}_{0} \tag{8}\]  
since \(c_{ii}=1\) and \(c_{ij}\geq 0\); \(\forall j(\
eq i)\in\overline{n}\). By combining (7) and (8), one obtains  
\[ \begin{split}& x_{i}^{-}(k+1)=\frac{\mu_{i}x_{i}(k)}{1+\frac{\mu_{i }-1}{K_{i}(k)}x_{i}(k)}\leq\frac{\mu_{i}x_{i}(k)}{1+\frac{\mu_{i}-1}{K_{i}}x_{ i}(k)}=\frac{\mu_{i}K_{i}x_{i}(k)}{K_{i}+(\mu_{i}-1)x_{i}(k)};\\ &\forall i\in\overline{n},\forall k\in\mathbf{N}_{0}\end{split}  \tag{9}\]  
If Assumption 1 holds, \(\{b_{i}(k)\}_{0}^{\infty}\equiv 1\) and \(\{d_{i}(k)\}_{0}^{\infty}\equiv 0\), then \(x_{i}^{-}(k)=x_{i}(k)\); \(\forall i\in\overline{n}\), \(\forall k\in\mathbf{N}_{0}\). Let us define \(y_{i}(k)=1/\left(x_{i}^{-}(k)\right)\); \(\forall i\in\overline{n}\), \(\forall k\in\mathbf{N}_{0}\). One obtains, from recursive calculations in (8) using (7),  
\[ \begin{split} 0<\frac{\mu_{i}-1}{\mu_{i}K_{i}}\leq\frac{\mu_{i}-1}{ \mu_{i}\max\limits_{0\leq\zeta<\infty}K_{i\epsilon}(k)}\leq\frac{\mu_{i}-1}{ \mu_{i}}\sum_{j=0}^{k}\frac{\mu_{j}^{-k}}{K_{i\epsilon}(0)}\\ \leq\frac{1}{x_{i}(k+1)}=y_{i}(k+1)=y_{i}^{-}(k+1)=\mu_{i}^{-1} y_{i}(k)+\frac{\mu_{i}-1}{\mu_{i}K_{i\epsilon}(k)}\\ =\mu_{i}^{-(k+1)}y_{i}(0)+\frac{\mu_{i}-1}{\mu_{i}}\sum_{j=0}^{ \mu_{j}^{-k}}\frac{\mu_{j}^{-k}}{K_{i}(j)}\\ \leq y_{i}(0)+\frac{\mu_{i}-1}{\mu_{i}\min\limits_{0\leq j\leq 1}K_{i \epsilon}(j)}\sum_{j=0}^{k}\mu_{i}^{-k}\\ \leq\frac{1}{x_{i}(0)}+\frac{\mu_{i}-1}{\mu_{i}\min\limits_{0\leq j \leq 1}K_{i\epsilon}(j)}\sum_{j=0}^{\infty}\mu_{i}^{j-k}\\ \leq\frac{1}{x_{i}(0)}+\frac{\mu_{i}-1}{\mu_{i}\min\limits_{0\leq j \leq 1}K_{i\epsilon}(j)}\frac{1-\mu_{i}^{-(k+1)}}{1-\mu_{i}^{-1}}\\ \leq\frac{1}{x_{i}(0)}+\frac{1}{\min\limits_{0\leq k<\infty}K_{ i\epsilon}(k)};\\ \forall i\in\overline{n},\forall k\in\mathbf{N}_{0}\end{split}  \tag{10}\]  
since \(K_{i\epsilon}(k)\leq K_{i}<\infty\). Since \(\infty>x_{i}(0)>0\) and \(\frac{1}{x_{i}(k+1)}>0\); \(\forall i\in\overline{n}\), \(\forall k\in\mathbf{N}_{0}\), then \(\{x_{i}(k)\}_{0}^{\infty}\) is bounded; \(\forall i\in\overline{n}\). Note that \(\{x_{i}(k)\}_{0}^{\infty}\) is a non-negative sequence for any given \(i\in\overline{n}\) by construction and, furthermore, a positive sequence, unless \(\liminf\limits_{k\to\infty}\min\limits_{0\leq k<\infty}K_{i\epsilon}(k)=0\), which would eventually imply, from (10), that \(\{1/x_{i}(k)\}_{0}^{\infty}\to\infty\), so that \(\{x_{i}(k)\}_{0}^{\infty}\to 0\). Since it has been already proved (Property (i)) that \(\{x_{i}(k)\}_{0}^{\infty}\) is bounded for any \(j\in\overline{n}\)then \(\left\{x_{j}(k)/x_{i}(k)\right\}_{0}^{\infty}\rightarrow\infty\), which would translate into \(\liminf\limits_{k\rightarrow\infty}\min\limits_{0\leq k<\infty}K_{ie}(k)=0\) from (8), if and only if \(\left\{x_{i}(k)\right\}_{0}^{\infty}\to 0\). Property (i) has been proved. From Assumptions 1-2, with non-negative independent consumption, one directly obtains that \(\frac{x_{i}(k+1)}{x_{i}(k)}\leq b_{i}(k+1)\mu_{i}\), which leads to the first part of Property (ii). If Assumptions 1 and 2 hold with non-necessarily non-negative independent consumption sequence, then \(\left\{x_{i}(k)\right\}_{k=0}^{\infty}\to 0\) if \(\prod_{k=0}^{\infty}[\mu_{i}b_{i}(k)-d_{i}(k+1)]=0\), which proves the second part of Property (ii). Property (iii) is proved by contradiction arguments. Let us assume that Assumptions 1 and 2 hold and \(\left\{x_{i}(k)\right\}_{k=0}^{\infty}\) is unbounded.