# 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
Since the given implicit constraint has to be checked for \(d_{i}\
eq 0\), it becomes identical to \(1\geq\frac{(\lambda_{i}-\mu_{i}+1)^{2}d_{i}}{4(\mu_{i}-1)K_{i}}\), or equivalently identical to  
\[d_{i}\leq\overline{d}_{i}=\frac{4(\mu_{i}-1)K_{i}}{(\lambda_{i}-\mu_{i}+1)^{2}}= \frac{4(\mu_{i}-1)K_{i}}{\lambda_{i}^{2}+(\mu_{i}-1)^{2}-2(\mu_{i}-1)\lambda_{i}}\]Note that \((\mu_{i}-1)\lambda_{i}=\frac{\mu_{i}-1}{d_{i}}K_{i}(b_{i}\mu_{i}-1)\), so that, under the given necessary conditions, \(sgn[(\mu_{i}-1)\lambda_{i}]=sign[K_{i}(b_{i}\mu_{i}-1)]=sgn(b_{i}\mu_{i}-1)\), which is \(+1\) if \(b_{i}>\mu_{i}^{-1}\) and \(-1\) if \(b_{i}<\mu_{i}^{-1}\). Thus, if \(b_{i}\geq\mu_{i}^{-1}\), then \(d_{i}\leq\overline{d}_{i}\) if \(d_{i}\leq\frac{4(\mu_{i}-1)K_{i}}{\lambda_{i}^{2}+(\mu_{i}-1)^{2}}\), which is, in turn, guaranteed if \(d_{i}\leq\frac{4K_{i}}{\mu_{i}-1}\). As a result, the sufficient condition (b) is guaranteed if \(d_{i}\in\left(\frac{K_{i}(b_{i}\mu_{i}-1)}{\mu_{i}-1},\ \frac{4K_{i}}{\mu_{i}-1}\right]\), if \(b_{i}\in\left(\mu_{i}^{-1},\ 5\mu_{i}^{-1}\right]\). On the other hand, if \(b_{i}<\mu_{i}^{-1}\), then \(\overline{d}_{i}=\frac{4(\mu_{i}-1)K_{i}}{\lambda_{i}^{2}+|\mu_{i}-1|(|\mu_{i }-1|+2|\lambda_{i}|)}\leq\frac{4K_{i}}{\mu_{i}-1}\) if, furthermore, \(\mu_{i}>1\), implying again that the condition (b) is guaranteed if \(d_{i}\in\left(\frac{K_{i}(b_{i}\mu_{i}-1)}{\mu_{i}-1},\ \frac{4K_{i}}{\mu_{i}-1}\right]\) and \(\overline{d}_{i}=-\frac{4|\mu_{i}-1|K_{i}}{\lambda_{i}^{2}+|\mu_{i}-1|(|\mu_{ i}-1|+2|\lambda_{i}|)}\geq-\frac{4K_{i}}{1-\mu_{i}}=\frac{4K_{i}}{\mu_{i}-1}\) if \(\mu_{i}<1\). Note that, in both,  
_Case \(2\)_ (\(g(\Delta_{i})=0\)_has complex conjugate zeros_\(\Delta_{i1,\ i2}\)). This condition holds if and only if \(B_{i}^{2}<C_{i}\) (which also trivially requires that \(C_{i}>0\)), which, in view of (23), requires that \((\mu_{i}-1)d_{i}<0\), which holds if and only if \(sgn(d_{i})=sgn(\mu_{i}-1)\). Since \(g(0)=C_{i}>0\) and \(\Delta_{i1,\ i2}\) are complex conjugate zeros, \(g(\Delta_{i})>0\) in all its definition domain, that is, if \(\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}\in[0,\ \infty)\). Property (ii) follows from Cases 1 and 2. \(\square\)  
Note, from Equation (20), that both equilibrium points of the \(i-\)th species are zero if \(d_{i}=c_{ji}=0\); \(\forall j(\
eq i)\in\overline{n}\) and \(b_{i}=\mu_{i}^{-1};\forall i\in\overline{n}\). Excluding this trivial case, also note from Equation (20) that, provided that the radianci is non-negative, a necessary condition for at least one of the equilibrium points to be non-negative is that either \(\hat{b}_{i}>0\) or \(\hat{b}_{i}<0\) and \(\hat{c}_{i}\geq 0\). The first case implies that  
\[d_{i}+\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}<\frac{K_{i}(b_{i}\mu_{i}-1)}{\mu_{i}-1}\]  
The second case implies that \(\mu_{i}<1\) and  
\[d_{i}+\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}>\frac{K_{i}(b_{i}\mu_{i}-1)}{\mu_{i}-1};\]  
\[d_{i}\Big{(}K_{i}-(1-\mu_{i})\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}\Big{)}\geq 0\]  
while the second constraint above is fulfilled with \(d_{i}\geq 0\) and \(\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}\leq\frac{K_{1}}{1-\mu_{i}}\), or \(d_{i}\leq 0\) and \(\sum_{j(\
eq i)=1}^{n}c_{ji}x_{j}\geq\frac{K_{1}}{1-\mu_{i}}\). **Example 1** (_Positive equilibrium points for \(n=2\) and \(n\geq 3\) with zero independent consumptions_). : _The allocation and positivity of the equilibrium points is easy to deal with in an analytical way from a linear algebraic system in the absence of independent consumptions at the equilibrium state, as it is now discussed. If \(n=2\) and \(d_{1}=d_{2}=0\), then_  
\[x_{i}=\frac{\mu_{i}b_{i}K_{i}x_{i}}{K_{i}+(\mu_{i}-1)\big{(}x_{i}+c_{ji}x_{j} \big{)}}\ \text{for}\ i,j(\
eq i)=1,2 \tag{24}\]  
_which leads, for \(x_{i}>0\) and for \(i=1,2\), denoted also as \(x=(x_{1}\,x_{2})^{T}\succ\succ 0\), to_  
\[x_{i}+c_{ji}x_{j}=\frac{K_{i}(b_{i}\mu_{i}-1)}{\mu_{i}-1}\ \text{for}\ i,j(\
eq i)=1,2 \tag{25}\]leading to_  
\[