We are given the following equations for the rate of change of each population.
u' = u(1-u-av/(u+d))
v' = bv(1-v/u)
To examine the behavior of this system, we must first find the steady states of the system of equations. Note that u and v are positive, u can't be 0.
To find the steady states we must solve for when u' and v' both equal 0. Solving for v'=0, we find that v=0, or v=u.
We will examine these steady states one at a time, beginning with the fixed point with v=0. When v=0, u=1 or u=1. The fixed point is (1,0). We rejected u=0, because when that occurs, v' has a zero in the denominator. To determine stability, it is often useful to find the eigenvalues and use them to determine the behavior around these points. To determine the eigenvalues, we must compute the jacobian and substitute the values of u and v into it.
Solving for the eigenvalues of this matrix, we get lambda=-1, or b. As b is always positive, we have one positive and one negative eigenvalue, which indicates that this is a saddle point. It is therefore unstable.
To find the other fixed point, we replaced v with u and solved u' for u, we get that u is:
As this equation being the value of u and v makes it very difficult to find the eigenvalues, we will re-express out equations as follows.
u' = u*k1(u,v)
v' = v*k2(u,v)
The jacobian would now be
( k1+u(dk1/du) , u(dk1/dv) )
( v(dk2/du) , k2+v(dk2/dv) )
At the fixed point, k1=k2=0. The jacobian would then be:
( u(dk1/du) , u(dk1/dv) )
( v(dk2/du) , v(dk2/dv) )
After substituting in u=v, we get the following jacobian.
The determinant is
The trace is
a,b,d,and v are all defined to be positive. The determinant will always be positive. Therefore both eigenvalues have the same sign, if the eigenvalues are real. This gives us three scenarios for the stability at this point. When the trace < 0, trace = 0, and trace > 0. Let us now examine what occurs when the trace is negative. To do this we will find what values of b cause the trace to be 0. We will then choose a and d and find b such that b satisfies the condition we are looking for. b1 is the value of b which make the trace 0. The trace will be negative if b > b1, and positive if b < b1.
Substituting in v in terms of a and d yields
Putting in a=2, d=.4, we find that b=.2 will give a negative trace. This is the graph with these parameters
We then chose a=5, d=.2, b=.5
As we can see from these plots, if the trace is negative, the fixed point where u=v is stable. We expected this result, as a negative trace and a positive determinant, will give eigenvalues with the real part negative, indicating a stable point as solutions decay to that point. We can also see in these graphs that (1,0) is indeed a saddle point as we predicted earlier.
Parameters wih the trace positive include a=2, d=.4, b=.01.
We can see from these graphs that there is a periodic solution when the trace is greater than 0. Graph 1 shows it moves in to the stable periodic solution if it is outside of it, Graph 2 shows it moves out to it if it is inside.
One set of parameters at which trace = 0 is a=5, d=.2, and b=.1345160394. Here are two graphs showing the solutions when this occurs.
We see that this acts much like the trace > 0 solution, except this curve is not as flat moving towards the periodic solution curve.
Matt Shaw, firstname.lastname@example.org