Trace Determinant Plane

Somefantastik

Hello,

I have a Jacobian matrix

$$A = \left( \begin{array}{cc} -2x+1-a_{1}y & -a_{1}x \\ -a_{2}ry & r-2ry-ra_{2}x \end{array} \right)\$$

where

$$a_{1} > 1 \ ; \ a_{2} > 1 \ ; \ and \ r>0$$

the trace is

$$Tr(A) = 1-2x-a_{1}y+r(1-y-a_{2}x)-ry = x(-2-ra_{2}) +y(-a_{1}-2r)+1+r$$;

and the determinant is

$$det(A) = r(1-2y-a_{2}x-2x+4yx+2a_{2}x^{2}-a_{1}y+2y^{2}a_{1})$$;

I need to show that Tr(A)>0 or det(A)<0. Neither one of them seem to be working out for me. Can anyone point me in the right direction? The point of this exercise is showing an equilibrium point is unstable.

Update: I can see from matlab that the determinant must be zero as the phase portrait shows a saddle point.

Last edited:
Related Differential Equations News on Phys.org

"Trace Determinant Plane"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving