Viability Analysis of Fisheries Management on Hermaphrodite Population

Abstract

We study the viability domains of bio-economic constraints for fishing model of hermaphrodite population, displaying three stages, juvenile, female and male. The dynamic of this model is subject to two constraints: an ecological constraint ensuring the stock perennity, and an economic constraint ensuring a minimum revenue for fishermen. Using viability kernel, we find out a viability domain which simultaneously guarantees a minimum stock level and a minimum income for fleets.

Appendices

Appendix1: Proof of Lemma (2)

Let’s consider a solution $(n_1(t), n_2(t), n_3(t))$ of (1) such that $n_2 + n_3 \ge \alpha _1$.

In the phase plane $(n_2, n_3)$, we have above the nullcline $\mathcal{N}_3^{min},\;\dot{n}_3\ge 0$ and below $\bar{\mathcal{N}}_2^{min},\;\dot{n}_3\le 0$ (see Fig. 4). We shall distinguish two cases:

If $n_2$ is decreasing, then it is necessarily above $n'_2$ (because if the solution is in the phase plane $(n_1, n_2)$, it is to the right of $\mathcal{L}_1$ and below $n'_2$, then it is below $\mathcal{N}_2^{min}$, thus, $n_2$ is increasing).

If the solution is above $\mathcal{N}_3^{min}$, then $n_2$ is decreasing and $n_3$ is increasing, the trajectory will intersect $\mathcal{N}_3^{min}$ while checking the constraint. Subsequently, after intersection, both $\dot{n}_2$ and $\dot{n}_3$ are negative and thus $(n_2(t),n_3(t))$ will decrease until reaching the line $n_2 = n'_2$. However, in parallel, in the other phase plane $(n_1, n_2)$, the solution will be below $\mathcal{N}_2^{min}$, thus $n_2$ becomes increasing and to the right of $\mathcal{L}_1$. In this case, the trajectory will increase and move away from the constraint (the trajectory crosses the nullcline $\mathcal{N}_2^{min}$ before the line $n_2 = n'_2$).

A same proof can be done if $n_2$ is increasing.

Appendix 2: Proof of Lemma (3)

• 1. If $n''_2 > n'_2$ then $n''_3 > n'_3$ (see Fig. 5a). Let’s consider a solution $(n_1(t), n_2(t), n_3(t))$ of (1) such that (8) and (9) are fulfilled. If $n_2$ is increasing, then the trajectory stay viable and will move away from the two constraints since (8) and (9) are always verified. If $n_2$ is decreasing then after a certain time, the trajectory will cross the nullcline $\mathcal{N}_3^{min}$. Accordingly, we have both $n_2$ and $n_3$ are decreasing. In parallel, in the other phase plane $(n_1, n_2)$, the trajectory will intersect $\mathcal{N}_2^{min}$ before reaching $n''_2$. Thus $n_2$ becomes increasing, the trajectory will go away from the constraints and it will be viable.

• 2. If $n''_2 < n'_2$ then the viability domain which satisfies (8) and (9) is the domain $\mathcal{V}_1 \cap \mathcal{D}_2$. Same proof as the first case (see Fig. 5b).

Fig. 5

Viability domain (shaded part) in the two phase planes $(n_2, n_3)$ and $(n_1, n_2)$. a Case where $n''_2 > n'_2$. b Case where $n''_2 < n'_2$

Appendix 3: Proof of Theorem (2)

• 1. If $n''_2 > n'_2$ and $n'''_3 > n''_3$ ie $n'''_2 > n''_2$, then $\mathcal {V} = \mathcal{V}_3 \cap \mathcal{D}_2$.

  Let’s consider a solution $(n_1(t), n_2(t), n_3(t))$ of (1) such that (8)–(10) are verified. If it is above $\mathcal{N}_3^{min}$, then $n_3$ is increasing. In the two cases ($n_2$ is increasing or decreasing), the trajectory is viable and moves away from all constraints (see Fig. 6a).

  If the solution is below $\mathcal{N}_3^{min}$, then $n_3$ is decreasing: consequently, if $n_2$ is increasing and located above $\mathcal{L}_3$, The trajectory will go away from all constraints. If $n_2$ is decreasing, then in the phase plane $(n_1, n_2)$, the trajectory will intersect $\mathcal{N}_2^{min}$ before reaching the line $n_2 = n'''_2$. Below $\mathcal{N}_2^{min}$, $n_2$ becomes increasing and the trajectory will increase and moves away from all constraints.

• 2. If $n''_2 > n'_2$ and $n'''_3 < n''_3$ ie $n'''_2 < n''_2$, then $\mathcal {V} = \mathcal{V}_2 \cap \mathcal{D}_3$. Same proof as the previous case (we replace $n''_2$ by $n'''_2$). Figure 7a explains the situation.

• 3. If $n''_2 < n'_2$ and $n'''_3 > n''_3$ and $n'''_3 < n'_3$,    then $\mathcal {V} = \mathcal{V}_1 \cap \mathcal{D}_2 \cap \mathcal{D}_3.$ Analogous proof as the first case. Figure (7b explains the situation.

• 4. If $n''_2 < n'_2$ and $n'''_3 > n''_3$ and $n'''_3 > n'_3$,    then $\mathcal {V} = \mathcal{V}_3 \cap \mathcal{D}_2$. Analogous proof as the first case.

Fig. 6

Viability domain (shaded part) in the two phase planes $(n_2, n_3)$ and $(n_1, n_2)$. a Case where $n''_2 > n'_2$ and $n'''_3 > n''_3$. b Case where $n''_2 > n'_2$ and $n'''_3 < n''_3$

Fig. 7

Viability domain (shaded part) in the two phase planes $(n_2, n_3)$ and $(n_1, n_2)$. a Case where $n''_2 > n'_2$ and $n'''_3 < n''_3$. b Case where $n''_2 < n'_2$ and $n'''_3 > n''_3$ and $n'''_3 < n'_3$