Equilibrium Level

**craig** · December 29th 2009, 05:41 AM

Two species of birds compete with each other according to the equations,

$\frac{dx}{dt} = x-3x^2-3xy$ and $\frac{dy}{dt} = y+\frac{y^2}{2}-\frac{xy}{2}$

where the populations are described by the continuous functions $x(t)$ and $y (t)$ in appropriate units.

Define the equilibrium level and verify that $x = \frac{1}{3}$ , $y = 0$ is an equilibrium level of the above system of equations.

Show that this equilibrium level is unstable.

Firstly, equilibrium level is where $\frac{dx}{dt} = \frac{dy}{dt} = 0$ .

We can easily show that both these equations are equal to zero using the values $(\frac{1}{3},0)$ .

The second part is where I'm stuck however. I'm guessing that an unstable equilibrium level is when $t \to \infty$ , something $\to \infty$ , pretty stuck from here though.

Thanks in advance for the help

**CaptainBlack** · December 29th 2009, 08:29 AM

Originally Posted by craig

Two species of birds compete with each other according to the equations,

$\frac{dx}{dt} = x-3x^2-3xy$ and $\frac{dy}{dt} = y+\frac{y^2}{2}-\frac{xy}{2}$

where the populations are described by the continuous functions $x(t)$ and $y (t)$ in appropriate units.

Define the equilibrium level and verify that $x = \frac{1}{3}$ , $y = 0$ is an equilibrium level of the above system of equations.

Show that this equilibrium level is unstable.

Firstly, equilibrium level is where $\frac{dx}{dt} = \frac{dy}{dt} = 0$ .

We can easily show that both these equations are equal to zero using the values $(\frac{1}{3},0)$ .

The second part is where I'm stuck however. I'm guessing that an unstable equilibrium level is when $t \to \infty$ , something $\to \infty$ , pretty stuck from here though.

Thanks in advance for the help

Linearize about the equilibrium solution and look at the solutions to the linearized problem, if they are decaying then you have a stable equilibrium.

If they grow exponentially then you have an unstable equilibrium.

If the solution is oscillatory then this is nominally stable but you may have to look at a higher order approximation to confirm this.

CB

**shawsend** · December 30th 2009, 06:37 AM

Lemme' tell you something about these. You can't. You just can't (or shouldn't) do these without drawing the phase portrait to see what's happening on the global level. Mathematica 7 has StreamPlot. Also easy to find the other equilibrium points:

Code:

In[37]:=
myEquilPts = {x, y} /. 
   Solve[{x - 3*x^2 - 3*x*y == 0, 
     y + y^2/2 - (x*y)/2 == 0}, {x, y}]
myEqSet = Graphics[{Red, PointSize[
      0.015], Point[myEquilPts]}]; 
Show[{StreamPlot[{x - 3*x^2 - 3*x*y, 
     y + y^2/2 - (x*y)/2}, {x, -3, 3}, 
    {y, -3, 3}], myEqSet}]

Out[37]= {{0, -2}, {0, 0}, {1/3, 0}, {7/6, -(5/6)}}

Yep, they're moving away from (1/3,0). I realize that's not helping with the analytic analysis but I think this "reinforces" such.

**craig** · December 30th 2009, 07:44 AM

Originally Posted by CaptainBlack

Linearize about the equilibrium solution and look at the solutions to the linearized problem, if they are decaying then you have a stable equilibrium.

If they grow exponentially then you have an unstable equilibrium.

If the solution is oscillatory then this is nominally stable but you may have to look at a higher order approximation to confirm this.

CB

Thanks =)

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