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

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Academic year: 2021

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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:

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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)

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