# On the Properties of a Class of Impulsive Competition Beverton-Holt Equations
## 5 Numerical Experiments  
This Section aims at illustrating, through numerical simulation examples, some of the theoretical results discussed in the previous Sections. In this way, this Section is organized into a number of examples concerned with the Theorems and results stated in Sections 2-4. **Example 2**. : _Let us consider the system (1) parameterized by \(b(k)=1\), \(d(k)=0\), \(K(k)=300\) and \(\mu(k)=0.6+5\times 0.6^{k}\) with \(x(0)=100\). The values of \(\mu(k)\) with iteration along with the evolution of \(x(k)\) are displayed in Figure 1. Thus, Lemma 1 holds, since the growth rate converges to 0.6 and \(\mu(k)\in(0,1)\) for all \(k\geq 5\). It is observed, in Figure 1, how the population extinguishes in finite time, since \(x(j)=0\) for \(j>30\) according to Lemma 1. On the other hand, when \(\mu(k)=1+0.62^{k}\), we have that \(\mu(k)\) converges to unity (in fact \(\mu(k)\) is forced to be exactly unity from \(j=10\) onwards) and we are in the condition of applying Lemma 2. Therefore, Figure 2 shows how the population remains strictly positive for all discrete time (and even converges to a positive constant value) as Lemma 2 claims._  
**Example 3**. : _Let us consider the system (1) parameterized by the constant values \(b(k)=0.9\), \(d(k)=0.1\), \(K(k)=300\) and \(\mu(k)=2.4\) with \(x(0)=100\). The evolution of \(x(k)\) is displayed in Figure 3. It can be seen, in this figure, that the population converges to the equilibrium point given numerically by 248.385. This value is in accordance with the calculations in Theorem 1(i), since \(\mu(k)=2.4>max(1,1/b(k))=1.11\) and \(d=0.1\) satisfies the constraint_  
\[0.1=d \in\left[-x_{1},\ \frac{\left(\sqrt{\mu}b-1\right)^{2}k}{\mu-1} \right]\cup\left[\frac{\left(\sqrt{\mu}b+1\right)^{2}k}{\mu-1},\ \infty\right)\] \[=\left[-0.0863,47.2741\right]\cup\ \left[1307,\ \infty\right).\]  
Figure 1: Evolution of \(\mu(k)\) and \(x(k)\) with iteration in Example 2 when \(\mu(k)\) converges to a limit be-low unity. Figure 3: Evolution of \(x(k)\) for constant parameters in Example 3, including harvesting and external consumption. The population converges to a positive equilibrium point. Figure 2: Evolution of \(\mu(k)\) and \(x(k)\) with iteration in Example 2 when \(\mu(k)\) converges to unity in finite time. Thus, Theorem 1(i) holds. Furthermore, Theorem 1(i) provides two equilibrium points given by \(x_{1}=0.0863\) and \(x_{2}=248.385\). Moreover, Figure 3 also depicts the left value of the population \(x^{-}(k)\). It satisfies at the equilibrium \(276.09=x^{-}(k)=\frac{x_{2}+d}{b}=\frac{248.385+0.1}{0.6}=276.09\), as Theorem 1(i) states. If we now consider the case when \(b(k)=1\) and \(d=0\), i.e., there is no harvesting nor independent consumption, then the left and right values of the population are the same, as Figure 4 shows, while the equilibrium point is in accordance with Theorem 1(ii), since \(x_{2}=\frac{(\mu b-1)K}{\mu-1}=300\), as it can be readily seen in Figure 4. In addition, when \(b=\mu^{-1}\) (and \(d=0\)), the trajectory of the population is depicted in Figure 5. It can be observed, in this figure, that the population ends up extinguishing, as predicted by Theorem 1(ii). **Example 4. **_Let us consider the multi-species Beverton-Holt Equations (2) and (3) parameterized by the constant values \(b_{i}(k)=1\), \(d_{i}(k)=0\), \(K_{1}(k)=300\), \(K_{2}(k)=200\) and \(K_{3}(k)=100\) and \(\mu_{1}(k)=2.4\), \(\mu_{2}(k)=2\) and \(\mu_{3}(k)=1.8\) with \(x_{1}(0)=100\), \(x_{2}(0)=50\) and \(x_{3}(0)=30\). The coupling matrix is given by_  
\[C=\left( \begin{array}{ccc}1&0.4&0.3\\ 0.25&1&0.4\\ 0.5&0.35&1\end{array} \right)\]  
Figure 4: Evolution of \(x(k)\) for constant parameters in Example 3, while no harvesting nor independent consumption is considered. Figure 5: Evolution of \(x(k)\) for constant parameters in Example 3 and \(b=\mu^{-1}\) with \(d=0\). These parameters correspond to the case when the species 1 dominates the other two.