# Thread: {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

1. ## {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

I need to find the positive equilibrium point of the system, Evaluate the Jacobi matrix, then determine the type and stability of the equilibrium.

The last three parts I can do (I think!), but we just began these non-linear systems and I need help getting started. Normally, I would set f(x,y)= 1-xy and g(x,y)= x-y^3 equal to zero, giving:
0=1-xy and 0=x-y^3, the latter gives x=y^3, subbing into the former gives y^4=1. I remember from precalc that gives me 4 roots: +/- 1, and +/- i. Now, the problem asks for the positive equilibrium, so I can discard the negative roots. y=1 gives x=1. If y=i, then x= -i. So that pair is not a positive equilibrium. Positive equilibrium at (x,y)=(1,1).

For the Jacobi matrix, I take the partial derivatives of f and g, and put them in their spots, which gives:
| (1-y) (1-x) |
| (1-y^3) (x-3y^2) |

plugging my equilibrium point in for x and y gives:
| 0 0 |
| 0 -2 | and taking this matrix and subtracting a (lambda)I matrix and taking the determinant, then solving for the roots gives lambda=0,2.

And now my brain hurts, because I don't know what to do with a zero root....

2. ## Re: {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

Originally Posted by shane18
I need to find the positive equilibrium point of the system, Evaluate the Jacobi matrix, then determine the type and stability of the equilibrium.

The last three parts I can do (I think!), but we just began these non-linear systems and I need help getting started. Normally, I would set f(x,y)= 1-xy and g(x,y)= x-y^3 equal to zero, giving:
0=1-xy and 0=x-y^3, the latter gives x=y^3, subbing into the former gives y^4=1. I remember from precalc that gives me 4 roots: +/- 1, and +/- i. Now, the problem asks for the positive equilibrium, so I can discard the negative roots. y=1 gives x=1. If y=i, then x= -i. So that pair is not a positive equilibrium. Positive equilibrium at (x,y)=(1,1).

For the Jacobi matrix, I take the partial derivatives of f and g, and put them in their spots, which gives:
| (1-y) (1-x) |
| (1-y^3) (x-3y^2) |

plugging my equilibrium point in for x and y gives:
| 0 0 |
| 0 -2 | and taking this matrix and subtracting a (lambda)I matrix and taking the determinant, then solving for the roots gives lambda=0,2.

And now my brain hurts, because I don't know what to do with a zero root....
J = {{-y,-x},{1,3y2}}

fix that and rework the eigenvalues.

3. ## Re: {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

Thanks! Dumb mistake.

So J={{-y,-x},{1.-3y^2}}

I reworked it and came up with a double root lambda=2. I know if the roots are positive, the phase portrait is an unstable node. Is there a special case for double root? We have never talked about that case...

4. ## Re: {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

Originally Posted by shane18
Thanks! Dumb mistake.

So J={{-y,-x},{1.-3y^2}}

I reworked it and came up with a double root lambda=2. I know if the roots are positive, the phase portrait is an unstable node. Is there a special case for double root? We have never talked about that case...
I guess by root you mean root of the characteristic equation i.e. the eigenvalues. I get a repeated eigenvalue of -2, not 2.
The eigenvectors are {1,1} and {0,0} which is degenerate. Nevertheless some results regarding repeated roots are below.

from here
Repeated Eigenvalues

If the set of eigenvalues for the system has repeated real eigenvalues, then the stability of the critical point depends on whether the eigenvectors associated with the eigenvalues are linearly independent, or orthogonal. This is the case of degeneracy, where more than one eigenvector is associated with an eigenvalue. In general, the determination of the system's behavior requires further analysis. For the case of a fixed point having only two eigenvalues, however, we can provide the following two possible cases. If the two repeated eigenvalues are positive, then the fixed point is an unstable source. If the two repeated eigenvalues are negative, then the fixed point is a stable sink.

You do have a degenerate system at {1,1} but both eigenvalues are negative you can say that the point is an asymptotically stable either proper or improper node, or a spiral point.

You're about at the limit of what I know about linearizing systems now.

5. ## Re: {(dx/dt)=1-xy, (dy/dt)=x-y^3} Systems of Diff Eqns

Got it. Thanks again, romsek. Your help has been well received.