• Aucun résultat trouvé

The perfect mixing paradox and the logistic equation: Verhulst vs. Lotka: reply

N/A
N/A
Protected

Academic year: 2021

Partager "The perfect mixing paradox and the logistic equation: Verhulst vs. Lotka: reply"

Copied!
2
0
0

Texte intégral

(1)

1 Appendix S1 We derive here some asymptotic properties of model (5) but we will use Verhulst’s r-α formulation of the logistic model because it makes mathematical derivation lighter. The usual r-K parameterization is obtained with K= r /α. Firstly, we prove that under perfect mixing (β → ∞) we have the following equality: N1 γ1 = N2 γ2 . For that, we use Eq. (5) to consider the following difference: d dt N1 γ1 − N2 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 γ1 dN1 dt − 1 γ2 dN2 dt = r1 γ1 N1− α1 γ1 N1 2 r2 γ2 N2+ α2 γ2 N2 2β 1 γ1 + 1 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ N1 γ1 −N2 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ . In the limit β → ∞, the last term in the RHS becomes predominant and the equation tends to: d dt N1 γ1 − N2 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ = −β 1 γ1 + 1 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ N1 γ1 −N2 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟. Consequently N1 γ1 − N2 γ2 tends to zero, that is N1 γ1 = N2 γ2 . From this equality, we can also deduce that NT γ1+γ2 = N1 + N2 γ1+γ2 = N1 γ1 γ1 + N2 γ2 γ2 γ1+γ2 = N1 γ1 = N2 γ2 . (A1) Secondly, we calculate the dynamics of total abundance. Equations (5) modified in the r-α formulation give: dNT dt = dN1 dt + dN2 dt = r1N1−α1N1 2+ r 2N2−α2N2 2 and using (A1), we have:

(2)

2 dNT dt = r1N1 γ1 γ1 −α1N1 2 γ1 γ1 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 + r2N2 γ2 γ2 −α2N2 2 γ2 γ2 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 = r

(

1γ1+ r2γ2

)

N1 γ1 − α1γ1 2+α 2γ2 2

(

)

N1 γ1 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 =r1γ1+ r2γ2 γ1+γ2 ≡ r ! "# #$ NT − α 1γ1 2+α 2γ2 2 γ1+γ2

(

)

2 ≡αT ! "# #$ NT 2 We can now return to the r-K parameterization and calculate: KT = r αT =

(

r1γ1+ r2γ2

)

(

γ1+γ2

)

α1γ1 2+α 2γ2 2 , (A2) which can be rewritten as in (6) in order to isolate K1+ K2. Additivity of carrying capacities (KT = K1+ K2) occurs when the numerator of the fraction in (6) is zero. This leads to a second-degree equation in the unknown x=γ1/γ2, which has the following two solutions: γ1 γ2 = K1 K2 , (A3) γ1 γ2 = r2 K2 r1 K1 = α 2 α1 . (A4) Recalling the definitions βi =β/γi, the two occurences of additivity are therefore: β1= β K1 and β2= βK2 , (A5) β1=βα1 and β2 =βα2. (A6)

Références

Documents relatifs

Analogous to the heat-bath algorithm, the reversible Metropolis algorithm also satisfies the detailed-balance condition: At each time step, a randomly chosen sphere i attempts a move

With this wide range of expertise, CMI established a research center on Human Centered Infor- mation and Communication Technology to conduct research in the field of

Setting max search path length at 10, experiments based on various clustering exponent α values on the 5890-peer network pro- duced results shown in Figures 7 (a) and (b).. Given

Institut de Mathématiques de l'Université de Fribourg. Notice sur la loi que la population suit dans son accroissement. Intuition and the logistic equation.. Per capita growth rate as

Finally we would like to stress that, if we agree with Hui’s contention that carrying capacity (understood as C) is not affected by the introduction of an additional mortality factor

Moreover we prove that the existence and construction problems for perfect matchings in bipartite graphs are as difficult as the existence and construction problems for a

As the Twin Condition (respectively the Participation Condition) is not satisfied for continuous aggregation rule on the sphere we say that we have a Twin Paradox (respectively a

a function that satisfies the equation (1.5) except for an expo- nentially small error, then we show that this quasi-solution and the exact solution of equation (1.5) are