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Consider the two species competition model given by

$$ \frac{da}{dt }= [λ_1 a /(a+K1)] - r_{ab}\cdot ab - da, \ \ \ \ \ \ \ \ \ \ (1)$$

$$\frac{db}{dt }= [λ_2 b *(1-b/K2)] - r_{ba}\cdot ab , \ \ \ \ \ \ \ t>0,\ \ \ \ \ \ \ \ (2)$$

for two interacting species denoted a=a(t) and b=b(t), with initial conditions a=a0 and b=b0 at t=0. Here λ1, λ2, K1,K2, r_(ab), r_(ba) and d are all positive parameters. (a) Describe the biological meaning of each term in the two equations.

=> A series expansion of 1/(a+K1), gives

1/(a+K1) ≈ (K1 -a)/ K1 ^2 + O (a^2)

Now,

da/dt = [λ1 a * (a+K1)/ K1^2] - r_(ab) ab - da,

λ1 a represents the exponential growth of population

da represents the exponential decay of population

λ1 is the growth rate

d is the decay rate

what does r_(ab) ab represent?

The first term of RHS equation 1: [λ1 a * (a+K1)/ K1^2] represents logistic growth at a rate λ1 with carrying capacity K1.

db/dt = [λ2 b *(1-b/K2)] - r_(ba) ab ,

λ2 b represents the exponential growth of population

λ2 is the growth rate

what does r_(ba) ab represent?

The first term of RHS equation 2: [λ2 b *(1-b/K2)] represents logistic growth at a rate λ2 with carrying capacity K2.

Kindly please check my answer.

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  • $\begingroup$ Try to improve your question formatting like this $\endgroup$ – Semsem Mar 15 '14 at 3:51
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A typical Interspecific competition between two species a and b can be represented by the following equations:

$$\frac{da}{dt} = \lambda_{1}a [1-\frac{a}{k}]$$

This is the logistic population growth model in the absence of competing species b.

Now , $$\frac{da}{dt} = \lambda_{1}a [1-\frac{a}{k_{1}} - \frac{\beta_{12}b}{k_{1}}]$$ is the logistic population growth model for a under the presence of species b. Now we are going to add one more term to the model to represent the decay by exponential decay model.

Exponential decay model is $$\frac{da}{dt} = -\lambda_{d}a$$

Add this term to the original equation, we get $$\frac{da}{dt} = \lambda_{1}a [1-\frac{a}{k_{1}} - \frac{\beta_{12}b}{k_{1}}] -\lambda_{d}a$$

Now to define each of the term in the above model (ODE).

$\lambda_{1}$ = logistic population growth rate.

$k_1$ - Carrying capacity of species a

**$\beta_{12}b$ - can be thought of as the decrease in growth rate of species "a" due to the presence of species "b"

$\lambda_d$ = exponential decay rate.

Similarly,

$$\frac{db}{dt} = \lambda_{2}b [1-\frac{b}{k_{2}} - \frac{\beta_{21}a}{k_{2}}] $$

Now to define each of the term in the above model (ODE).

$\lambda_{2}$ = logistic population growth rate of b.

$k_2$ - Carrying capacity of species b

**$\beta_{21}a$ - can be thought of as the decrease in growth rate of species "b" due to the presence of species "a".

This is typically the interspecific competition model. In your original equation, you have $\beta_{12}$ to be $r_{ab}$

Hope this answers your question.

Good luck

Satish

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  • $\begingroup$ Is this MODEL: da/dt = λ1 a[1- (a/k1) - (r_ab ab)/k1] -da is equal (same) to da/dt = λ1 a / (a+k1) - r_ab ab -da ? $\endgroup$ – Manjushree Mar 15 '14 at 14:59
  • $\begingroup$ It is a variation of it. I am hoping that with some algebraic manipulation, you could derive yours from mine or vice versa. What I have given you is the usual model that is used in physical sciences and competitive model amongst products in marketing. Let me try to derive from the other when I have free time and let you know $\endgroup$ – Satish Ramanathan Mar 15 '14 at 15:06
  • $\begingroup$ The McClaurin's series for $$\frac{1}{a+k_{1}} = \frac{1}{k_{1}} - \frac{a}{k_{1}^{2}} + O(a^2) = \frac{1}{k_{1}}[1-\frac{a}{k_{1}}]$$. I am suspecting that you have missed a $k_1$ in the numerator somewhere. If you had it, the ODE conforms to my model. $\endgroup$ – Satish Ramanathan Mar 15 '14 at 15:59
  • $\begingroup$ sorry satish, I don't think I have missed k1 in the numerator while expanding series. $\endgroup$ – Manjushree Mar 15 '14 at 16:19
  • $\begingroup$ I am talking about missing a $k_1$ in the original ODE for "a". Your series is correct, I don't dispute it. $\endgroup$ – Satish Ramanathan Mar 15 '14 at 16:20

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