Solving Coupled Differential Equations. I have the following differential equations, for modeling predator-prey relationships: dx/dt=Ax-Bxy

curukksm 2022-09-16 Answered
Solving Coupled Differential Equations
I have the following differential equations, for modeling predator-prey relationships:
d x d t = A x B x y
d y d t = C x y D y
Where A, B, C, and D are constants. How could I go about solving this? I've only really worked with basic first order differential equations before, and I've found little to help me figure this out. Any help would be much appreciated.
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Answers (2)

vermieterbx
Answered 2022-09-17 Author has 14 answers
Step 1
Generally for this type of model one would start analysing what happens at equilibrium, i.e. whether there is a fixed point, if they are stable or not by looking at the Jacobian of the system.
Step 2
First find the null-clines setting: 0 = A x B x y = f 1 ( x , y ) and C x y D y = f 2 ( x , y ) = 0 then you get x = 0 , y = A / B for f 1 ( x , y ) and y = 0 , x = D / C for f 2 ( x , y ). Finding the intersection of the null-clines you get two fixed points (0,0) (both population are extinct) and (D/C,A,B) (both populations are non-zero). Then if you look at the Jacobian (matrix of partial derivatives of f 1 ( x , y ) , f 2 ( x , y ) computed in the two fixed points you can understand which is the stable one and where the system ends up.
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Kailey Vargas
Answered 2022-09-18 Author has 2 answers
Step 1
While complete analytic solution is not likely, it's curious that we can reduce this system to a couple of nonlinear first order ODEs for x(t) and y(t).
Let's do this for y only, since the same procedure can be applied to x as well.
First, we write down some useful relationships:
x ˙ = x ( A B y )
C x D = y ˙ y
C x y = y ˙ + D y
Now we differentiate the second equation w.r.t. t:
y ¨ = ( C x D ) y ˙ + C y x ˙ = y ˙ 2 y + ( A B y ) ( y ˙ + D y )
Since the equation doesn't containt t explicitly, we can reduce the order by the usual substitution:
y ˙ = u ( y ) , y ¨ = u u
We obtain:
u u = u 2 y + ( A B y ) ( u + D y )
Now we introduce another function:
u = y v ( y ) , u = v + y v
We get:
y v ( v + y v ) = y v 2 + ( A B y ) ( v + D ) y
Step 2
Simplifying, we obtain:
y v v = ( A B y ) ( v + D )
But this is a separable equation. So:
v d v v + D = ( A y B ) d y
v D ln ( v + D ) = A ln y B y + c 1
Getting back to the original function, we have:
y ˙ y D ln ( y ˙ y + D ) = A ln y B y + c 1
Or:
(1) y ˙ D y ln ( y ˙ + D y ) = ( A D ) y ln y B y 2 + c 1 y
This doesn't look like anything solvable, but it is indeed a 1st order ODE for y only.
We also need to determine c1 from the original system somehow, because the extra constant shouldn't be here.
Step 2
We can actually resolve (1) for the derivative using Lambert W (product logarithm) function. Transforming the equation:
( y ˙ + D y ) D y e y ˙ = y ( A D ) y e B y 2 + c 1 y
y ˙ + D y D y exp ( y ˙ + D y D y ) = e 1 c 1 / D D y A / D exp ( B D y )
This has a solution:
y ˙ + D y D y = W ( e 1 c 1 / D D y A / D exp ( B D y ) )
(2) y ˙ = D y ( 1 + W ( e 1 c 1 / D D y A / D exp ( B D y ) ) )
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