# 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
Then, the following equilibrium constraints hold:_  
\[y_{i}^{-}\leq\frac{b_{i}}{\mu_{i}b_{i}-1}\Bigg{[}\frac{\mu_{i}-1}{K_{i}}+\frac {\mu_{i}-1}{K_{i}}\sum_{(j,i)=1}^{n}c_{ji}\Bigg{(}\frac{b_{i}-d_{ij}y_{i}^{-}+d _{i}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-}\big{)}}\Bigg{)}+\frac{1}{b_{i}-d_{i}y_{i }^{-}}\Bigg{]} \tag{32}\]  
\[y_{i}=\frac{b_{i}-d_{ij}y_{i}^{-}+d_{i}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-}\big{)} }y_{i}^{-}\leq\frac{b_{i}-d_{ij}y_{i}^{-}+d_{i}}{\big{(}b_{i}-d_{ij}y_{i}^{-} \big{)}(y_{i}b_{i}-1)} \tag{33}\]  
If \(d_{i}\leq 0\) with \(|d_{i}|\leq(1-b_{i})x_{i}^{-}\), then the following further upper-bounds for (32) and (33) are obtained:  
\[y_{i}^{-}\leq\frac{b_{i}}{\mu_{i}b_{i}-1}\bigg{[}\frac{\mu_{i}-1}{K_{i}}+\frac {\mu_{i}-1}{K_{i}}\sum_{j(j,i)=1}^{n}c_{ji}\bigg{(}\frac{(1-|d_{i}|)b_{i}^{-1}} {b_{i}+|d_{i}|y_{i}^{-}}\bigg{)}+\frac{1}{b_{i}+|d_{i}|y_{i}^{-}}\bigg{]}\]  
\[\leq\frac{b_{i}}{\mu_{i}b_{i}-1}\bigg{[}\frac{\mu_{i}-1}{K_{i}}+\frac{\mu_{i}-1 }{K_{i}}\sum_{j(j,i)=1}^{n}\frac{1}{b_{i}}\Big{(}c_{ji}+\frac{1}{b_{i}}\Big{)}\]  
**Proof. ** So that one obtains, from (2)-(3) the following equilibrium constraints, since \(c_{ii}=1\); \(\forall i\in\overline{n}\),  
\[y_{i}^{-}=\frac{K_{i}+\sum_{j=1}^{n}c_{ji}(\mu_{i}-1)x_{j}}{\mu_{i}K_{i}x_{i}} \tag{34}\]  
\[=\mu_{i}^{-1}y_{i}+\frac{\mu_{i}-1}{\mu_{i}K_{i}}+\frac{\mu_{i}-1}{\mu_{i}K_{ i}}\sum_{j(j,i)=1}^{n}c_{ji}y_{i}y_{j}^{-1}\]  
\[y_{i}=\frac{1}{b_{i}x_{i}^{-}-d_{i}}=\frac{y_{i}^{-}}{b_{i}-d_{i}y_{i}^{-}}=b_ {i}^{-1}y_{i}^{-}+Y_{i} \tag{35}\]  
where  
\[Y_{i}=\Bigg{(}\frac{1}{b_{i}-d_{i}y_{i}^{-}}-\frac{1}{b_{i}}\Bigg{)}y_{i}^{-} =\frac{d_{i}y_{i}^{-}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-}\big{)}} \tag{36}\]  
Then, from (36) into (35), one obtains  
\[y_{i}=\Bigg{(}b_{i}^{-1}+\frac{d_{i}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-}\big{)} }\Bigg{)}y_{i}^{-}=\frac{b_{i}-d_{i}y_{i}^{-}+d_{i}}{b_{i}\big{(}b_{i}-d_{i}y_ {i}^{-}\big{)}}y_{i}^{-} \tag{37}\]  
Which, replaced in (34), yields  
\[ \begin{gathered}\bigg{(}1-\mu_{i}^{-1}\bigg{(}b_{i}^{-1}+\frac{d_ {i}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-}\big{)}}\bigg{)}\bigg{)}y_{i}^{-}=\frac{ \mu_{i}-1}{\mu_{i}K_{i}}+\frac{\mu_{i}-1}{\mu_{i}K_{i}}\sum_{j(j,i)=1}^{n}c_{ji} y_{i}y_{j}^{-1}\\ =\frac{\mu_{i}-1}{\mu_{i}K_{i}}+\frac{\mu_{i}-1}{\mu_{i}K_{i}}\sum_{j (j,i)=1}^{n}c_{ji}y_{j}^{-1}\bigg{(}b_{i}^{-1}+\frac{d_{i}}{b_{i}\big{(}b_{i}-d _{j}y_{i}^{-}\big{)}}\bigg{)}y_{i}^{-}\end{gathered}  \tag{38}\]  
and then  
\[\bigg{(}1-\mu_{i}^{-1}\bigg{(}b_{i}^{-1}+\frac{d_{i}}{b_{i}\big{(}b_{i}-d_{i}y_ {i}^{-}\big{)}}\bigg{)}-\frac{\mu_{i}-1}{\mu_{i}K_{i}}\sum_{j(j,i)=1}^{n}c_{ji} y_{j}^{-1}\bigg{(}b_{i}^{-1}+\frac{d_{i}}{b_{i}\big{(}b_{i}-d_{i}y_{i}^{-} \big{)}}\bigg{)}\bigg{)}y_{i}^{-}=\frac{\mu_{i}-1}{\mu_{i}K_{i}} \tag{39}\]  
Note, from (3), that \(x_{i}^{-}\geq max(0,\,d_{i}/b_{i})\) (equivalently, \(y_{i}^{-}\leq b_{i}/d_{i}\)) in order for the \(i-th\) equilibrium point to be non-negative to the left of the sampling instants, provided that it is non-negative to the right of such sampling instants. Therefore, the last twoadditive terms on the left-hand-side of the above equation are necessarily negative under that stipulation. Let us assume that the \(i-th\) species is dominant so that \(x_{i}\geq\underset{1\leq j\ll(\
eq i\leq n)}{max}x_{j}\), so that \(y_{i}/y_{j}\leq 1;\forall j(\
eq i)\in\overline{n}\). Then,  
\[ \begin{split}&\left(1-\mu_{i}^{-1}\left(b_{i}^{-1}+\frac{d_{i}}{b_{ i}\left(b_{i}-d_{i}y_{i}^{-1}\right)}\right)\right)y_{i}^{-}\\ &\leq\frac{\mu_{i}-1}{\mu_{i}K_{i}}+\frac{\mu_{i}-1}{\mu_{i}K_{i} }\Sigma_{j(\
eq i)=1}^{n}\,c_{ji}\left(b_{i}^{-1}+\frac{d_{i}}{b_{i}\left(b_{i} -d_{i}y_{i}^{-}\right)}\right)\end{split}  \tag{40}\]  
and, since \((d_{i}/b_{i})y_{i}^{-}\leq 1\),  
\[