Predator-Prey Equations: Modification of Lotka-Volterra

[Kinsbutt]

Homework Statement

Consider the system

x'=x(1-\sigma x-0.5y), y'=y(-0.75+0.25x)
where σ > 0.

[STRIKE]a) Find all of the critical points...[/STRIKE] (DONE)
b) Determine the type and stability property of each critical point. Find the value σ(1) < 1/3 where the nature of the critical point in the interior of the first quadrant changes. Describe the change that takes place in this critical point as σ passes through σ(1).

Homework Equations

<br /> J(x,y)= \left(<br /> \begin{array}{cc}<br /> F_{x} &amp; F_{y} \\<br /> G_{x} &amp; G_{y} \\<br /> \end{array}<br /> \right)<br />

<br /> 0=\left|<br /> \begin{array}{aa}<br /> F_{x}(x,y) -\lambda &amp; F_{y}(x,y) \\<br /> G_{x}(x,y) &amp; G_{y}(x,y) - \lambda \\<br /> \end{array} \right|<br />

The Attempt at a Solution

So, I found the roots easily via the solutions to the first two equations given in the problem.
(0,0), (1/σ, 0), and (3, 2-6σ).

When linearizing the equations using the Jacobian/partials, we get
<br /> J(x,y)= \left(<br /> \begin{array}{cc}<br /> 1-2\sigma x - 0.5y &amp; -0.5x \\<br /> 0.25y &amp; -0.75+0.25x \\<br /> \end{array}<br /> \right)<br />
Which seems to work out OK for the (0,0) critical point - it gives two eigenvalues with opposite signs, which is a saddle point, which agrees with the solution in the back of the book.

However, when moving to the second critical point, (1/σ, 0), I'm stuck. We end up with:

0=\lambda^{2} - \lambda (1- \frac{1}{4 \sigma})+ \frac{1}{4 \sigma}
\Rightarrow \lambda = \frac{\frac{1-4\sigma}{4 \sigma}\pm \sqrt{\frac{4\sigma -1}{4 \sigma}-\sigma}}{2}
(Might be a calculation error here) \Rightarrow \frac{1-4\sigma}{4\sigma}\pm i \frac{2\sigma-1}{2\sigma}

...which would give a spiral point, regardless of the value of σ. The solution for the second critical point is a saddle point for σ < 1/3, and an asymptotically stable node for σ > 1/3.

I expect my problem has somewhere to do with the local linearization of the critical points, but I have no idea where I'd begin to fix it.

If anyone has the 9th edition of the Boyce and DiPrima "Elementary Diff Eq and Boundary Value Problems", this is problem 11 from Ch. 9.5.

 

[D H]

However, when moving to the second critical point, (1/σ, 0), I'm stuck. We end up with:

0=\lambda^{2} - \lambda (1- \frac{1}{4 \sigma})+ \frac{1}{4 \sigma}

How did you get that characteristic equation? (Show your work.)

 

[Kinsbutt]

How did you get that characteristic equation? (Show your work.)

From:

<br /> 0=\left|<br /> \begin{array}{aa}<br /> F_{x}(x,y) -\lambda &amp; F_{y}(x,y) \\<br /> G_{x}(x,y) &amp; G_{y}(x,y) - \lambda \\<br /> \end{array} \right|<br />

So we have, for (1/σ,0):

<br /> J(1/ \sigma, 0)= \left(<br /> \begin{array}{cc}<br /> 1-2\sigma (\frac{1}{\sigma}) - 0.5(0) - \lambda &amp; -0.5 (\frac{1}{\sigma}) \\<br /> 0.25(0) &amp; -0.75+0.25(\frac{1}{\sigma}) - \lambda \\<br /> \end{array}<br /> \right)<br />

<br /> J(1/ \sigma, 0)= \left(<br /> \begin{array}{cc}<br /> 1-2 &amp; -0.5/ \sigma \\<br /> 0 &amp; -0.75+0.25/ \sigma \\<br /> \end{array}<br /> \right)<br />

So we can get the characteristic equation from:
det(J(x,y)-I\lambda)=0

Which gives:

<br /> J(1/ \sigma, 0)= \left|<br /> \begin{array}{cc}<br /> -1-\lambda &amp; -0.5/ \sigma \\<br /> 0 &amp; -0.75+0.25/ \sigma-\lambda \\<br /> \end{array}<br /> \right|<br />

aaand I find my error. Looks like in the 3 or 4 times I redid the calculation I forgot the -0.75 term. Wasn't until I was copying/pasting LaTex that I now see it's supposed to be there. Whoops.

Out of completeness:

0=(-1-\lambda)(-3/4+\frac{1}{4\sigma}-\lambda)
0=\lambda^2+\lambda(7/4-\frac{1}{4\sigma})+(3/4-\frac{1}{4\sigma})<br />

After checking that (7/4-1/4\sigma)^{2}-4(3/4-1/4\sigma)=\Delta&gt;0 for all σ, we can immediately see that at σ=1/3, the roots change sign as c (in terms of ax^2+bx+c) changes from positive to negative. Since for σ < 1/3, c is negative, the roots are opposite signs; therefore, saddle point. For σ > 1/3, c is positive, and b is also positive; thus the roots are both negative, and we get an asymptotically stable node.

Idiotic calculation error. Sorry 'bout that, and thanks.

 

[D H]

aaand I find my error.

Glad I could help!

 

[sara_87]

I found this post useful.
Just one question:
this is a 2-d problem where the eigen-values of the jacobian were used to determine the type of linear stability of the steady states.
Can such a technique be used for a 3-dimensional system?

If so, do we require ALL 3 eigenvalues to be of the same sign for assymptotic stability?
if 2 are positive and 1 is negative, does that also mean its a saddle point?

Thank you very much in advance,
and please let me know if I should have posted this is a new thread.