Resting Stages and the Population Dynamics of Harmful Algae in ...

So if we are in Case 1 of feasibility then Condition (12) is met which impliesCondition (19) is also met. Therefore in Case 1 the determinant of the Jacobianmatrix (18) must be positive. The trace of the Jacobian matrix (18) is−2D−δ + R inµ maxK +R in−γ.Thus the condition for the trace to be negative is2D+δ +γ > R inµ maxK +R in. (20)Using the fact thatD(D+δ +γ)2D+δ +γ > ,(D +γ)Condition (11), and the transitivity property we know that if we are in Case1 then Condition (20) has been met. Therefore we can say that in Case 1 offeasibility the trace of the Jacobian matrix (18) is negative. Furthermore wecan say that Ec 0 is stable in Case 1.The Jacobianmatrix (17) is evaluated at Ec ∗ for the chemostat model. Againwe want to find conditions for a positive determinant and a negative trace. Thecondition for a positive determinant isD(D +γ) . (23)K +R in D+γSo for Case 1 of feasibility the equilibrium Ec ∗ is unstable, because Condition(11) is directly contradicted by Condition (23), meaning that the determinant isnotpositive andthe traceis notnegative. In Case1offeasibility the equilibriumEc 0 is stable and the equilibrium E∗ c is unstable. This is shown in Figure 2.10