The French Newbie and the Lotka-Voltera Crazy Idea

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I am back to talk with you and to have criticisms on ideas crossing my mind!

I am currently working on a disease that is striking french vineyards really hard. I am currently using a model that is working pretty well which is this one.

$S′= −qS(pI+ϵ)^2+k_1I+k_4G$

$I′= qG(pI+ϵ)^2−k_1I−k_3I$

$G′= −qG(pI+ϵ)^2−k_4G+τ_{−1}(D−S−I−G)$

However, I do want to replace the constant p, which stands for pesticides, by a variable varying in time. Basically, p can take every value between 0 and 1, 0 represents a state in which the pesticide is eradicating all the studied insects and 1 represents a state in which the substance is ineffective.

So I had the idea to model p like this:

$p = \dfrac{(p_{O} + p_{NO})}{2}$

to integrate the fact that both organic pesticides and synthetic pesticides are being used. With O standing for organic and NO for non-organic.

$p = \dfrac{\dfrac{Area_O}{Area_T}*e_O + \dfrac{Area_{NO}}{Area_T}*e_{NO}}{2}$

with $e_O$ the effectiveness of the non-organic pesticide and $e_{NO}$ the effectiveness for non-organic one.

So, my first question is : Does it make sense to model the general effectiveness like this ?

$E = 1 - p$

with $p$ the expression above

Furthermore, I don't want $Area_O$ or $Area_{NO}$ to be constants. So, I tried to figure out a way to make them varying in a dependent way. Here come the Competitive Lotka–Volterra equations.

I tried to imagine my two types of areas, O and NO, as two species competing for a ressource which is the soil. This ressource is finite so that $A_T = A_O + A_{NO} = 80$.

Then I first came up with this system of ODE with:

$A'_O = \alpha A_O - \beta A_{NO}$

$A'_{NO} = \beta A_{NO} - \alpha A_{O}$

And therefore :

$A'_O = \beta A_{NO} - \alpha A_{NO}$ + $A_T$ ( $\alpha$ - $\beta$ )$

$A'_{NO} = \alpha A_{NO} - \beta A_{O}$ + $A_T$ ( $\beta$ - $\alpha$ )$

I used ODE45 to simulate it using $\alpha = 0.09$ and $\beta = 0.001$. It gave me this :

enter image description here

Of course I am not happy with it because, even if there is the symmetry that I need, it doesn't have a logistic form. In fact, I would like the blue curve to be convergent toward $80$ and the orange one to be convergent toward $0$.

Therefore, I tried to look for the competitive Lotka–Volterra equation, which integrate the logistic dimension, but I am not sure if it is the right path to take.

What do you think about that ? Do you have suggestions on these ideas or criticisms ?

Thanks you all,

Have a good day or a good night,

Hugo

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I'll use a different notation for the variables, just to avoid typing subscripts and Greek letters.

If $X$ is organic and $Y$ is nonorganic and if $X+Y=80$ then you can get a single differential equation for either one, say $X$.

If the original two-variable model is written as

$X'= mX-nY$ and $Y'=-mX+nY$

then the single model for $X$ is $X'=mX -n (80-X) = (m+n)X-80n$. The solution should look logistic. The variable $X$ tends to the equilibrium value $X= \frac{80n}{m+n}$ and by a similar argument $y$ approaches the complementary equilibrium value.

P.S. I think the difficulties you had can be traced to algebra errors, since the signs of some of your terms seem backwards.