In the Lotka-Volterra system of two competing species:
dx/dt = x(1-0.5x-0.5y)
dy/dy = y(0.25-0.5x)
I have found critical points at (0,0), (1/2,3/2), and (2,0). (0,0) is an unstable point, and (2,0) is a stable point, but I don't know what (1/2,3/2) is. Can you tell me?
Also, what does this system tell us about the interactions between animal X and animal Y?
dx/dt = x(1-0.5x-0.5y)
dy/dy = y(0.25-0.5x)
I have found critical points at (0,0), (1/2,3/2), and (2,0). (0,0) is an unstable point, and (2,0) is a stable point, but I don't know what (1/2,3/2) is. Can you tell me?
Also, what does this system tell us about the interactions between animal X and animal Y?
-
It sounds like you are pretty knowledgeable on the subject, so I won't go into super detail in the answer, but we need to evaluate the Jacobian of the system at the point (1/2, 3/2).
the Jacobian of this system, of course is
J(x,y) = [1 - x - 0.5y -0.5x
-0.5y 0.25 - 0.5x]
(pardon the horrible format of that matrix)
we then evaluate this Jacobian J(0.5,1.5) and get
[-0.25 -0.25
-0.75 0]
a matrix which has eigenvalues of (approximately) 0.32 and -0.58.
Because one of them is negative, and one is positive, this creates a saddle point node, where trajectories are drawn towards the point from the eigenvector associated with the negative eigenvalue, and pushed away from it in the direction associated with the positive one.
All-in-all, this creates an unstable node, where the only way to stay on it is to have been started on it.
As for the interactions between animal x and animal y, graphing the solution space to this might be of assistance in explaining, so PLEASE graph it if you can. But the interaction states that there is a basin of the plane where animal y will simply flourish without bounds while the population of x slowly dies out (given a small enough initial population of animal x.) however, if there is a sufficient population of animal x, they will populate to 2 (units ... hundreds, thousands, ... whatnot) while y dies out quickly.
Hope that helps!
the Jacobian of this system, of course is
J(x,y) = [1 - x - 0.5y -0.5x
-0.5y 0.25 - 0.5x]
(pardon the horrible format of that matrix)
we then evaluate this Jacobian J(0.5,1.5) and get
[-0.25 -0.25
-0.75 0]
a matrix which has eigenvalues of (approximately) 0.32 and -0.58.
Because one of them is negative, and one is positive, this creates a saddle point node, where trajectories are drawn towards the point from the eigenvector associated with the negative eigenvalue, and pushed away from it in the direction associated with the positive one.
All-in-all, this creates an unstable node, where the only way to stay on it is to have been started on it.
As for the interactions between animal x and animal y, graphing the solution space to this might be of assistance in explaining, so PLEASE graph it if you can. But the interaction states that there is a basin of the plane where animal y will simply flourish without bounds while the population of x slowly dies out (given a small enough initial population of animal x.) however, if there is a sufficient population of animal x, they will populate to 2 (units ... hundreds, thousands, ... whatnot) while y dies out quickly.
Hope that helps!
-
You might find this URL helpful. http://www.tiem.utk.edu/~gross/bioed/bea…