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3.2 Equilibrium EvaluationTo determine the equilibria of System (8) we set both dM dtzero and solve for M and N. Our solutions are:and dN dtequal toEc 0 = (M0 c ,N0 c ) = (0,0), (9)⎧⎪ M ⎨c ∗ = (D(D +δ +γ)(K +R in)(D +γ))−((D +γ) 2 R in µ max ))(D +δ +γ)q(D(D +δ +γ)−(µ max (D +γ)))Ec ∗ = ⎪ ⎩N ∗ c = δ(D(D +δ +γ)(K +R in)−((D +γ)R in µ max ))(D+δ +γ)q(D(D +δ +γ)−(µ max (D +γ)))3.3 Equilibrium Analysis.(10)The equilibria representpopulation sizes thus they must be non-negativevalues.Ec 0 is always feasible because M0 c = 0 and ( N0 c = 0. ) Some algebra is needed toδfind conditions for feasibility of Ec. ∗ Nc ∗ = Mc ∗ thus if Nc ∗ is positiveD +γM ∗ c must be positive becauseδD +γ is positive. If N∗ c is negative M ∗ c must benegativeas well. Thus if N ∗ c is feasible then M ∗ c must also be feasible. Thereforewe will only need to prove that N ∗ c is feasible by proving that it is positive. N∗ cis expressed as a fraction; therefore N ∗ c is feasible if the numerator:and the denominator:δ(D(D +δ +γ)(K +R in )−((D +γ)R in µ max ))(D +δ +γ)q(D(D +δ +γ)−(µ max (D +γ)))arebothpositive(Case1)orbothnegative(Case2). Ifinanycasethenumeratorand denominator are not both positive or both negative then Ec ∗ is not feasible.The two cases of feasibility have the following conditions.(C1) The first condition for Case 1 is a positive numerator. Thuswhich impliesD(D +δ +γ)(K +R in ) > (D+γ)R in µ max ,D(D +δ +γ)D+γ> R inµ maxK +R in. (11)7
The second condition for Case 1 is to have a positive denominator. Thuswhich impliesD(D +δ +γ) > (D +γ)µ max ,D(D +δ +γ)(D +γ)> µ max . (12)Condition (12) implies Condition (11). Therefore if Condition (12) is met thenCase 1 of feasability is obtained.(C2) The first condition for Case 2 is to have a negative numerator. Thuswhich impliesD(D +δ +γ)(K +R in ) < (D+γ)R in µ max ,D(D +δ +γ)D+γ< R inµ maxK +R in. (13)The second condition for Case 2 is to have a negative denominator. Thuswhich impliesD(D +δ +γ) < (D +γ)µ max ,D(D +δ +γ)(D +γ)< µ max . (14)Condition (14) is implied by Condition (13). Therefore if Condition (13) is metCase 2 of feasability is obtained. In both cases Nc ∗ is postive, thus in both casesEc ∗ is feasible.In order to analyze the stability of the equilibria in System (8) we letrhs 1rhs 2= µ maxM(R in −Mq −Nq)K +R in −Mq −Nq= −DN +δM −γN.−δM +γN −DM,(15)We then obtain the partial derivatives, of Equations (15), with respect to Mand N:8
Page 1 and 2: Resting Stages and the PopulationDy
Page 3 and 4: (Baker et al., 2007). It was discov
Page 5 and 6: and therefore it is the summation o
Page 7: dMdt= µ maxRMK +R−DM −δM +γN
Page 11 and 12: So if we are in Case 1 of feasibili
Page 13 and 14: 10.90.8Case 2 of Chemostatmotil cel
Page 15: [9] Brooks, B.W., James, S.V., Vale

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