# Stability of Critical Points of Second Order Equation

• Dec 6th 2011, 11:57 PM
Len
Stability of Critical Points of Second Order Equation
Find the critical points of the following second-order equation (by transferring into a system of first order differential equations) and determine the stability of these critical points:

$\displaystyle x''+x*cos(x'- \frac{\pi}{2})=x^2-1$

My start at a solution: transferring into a system of first order differential equations

let $\displaystyle y_1 = x, y_2=x'$

therefore

$\displaystyle y_1'=x'=y_2$
$\displaystyle y_2'= x''= x^2-1 -x*cos(x'- \frac{\pi}{2})$$\displaystyle = y_1^2-1 -y_1*cos(y_2- \frac{\pi}{2})$

So to get the critical points I can set $\displaystyle y_1'= y_2' = 0$ This implies$\displaystyle y_2= 0$

so $\displaystyle 0= y_1^2-1 -y_1*cos(0- \frac{\pi}{2})$, where $\displaystyle cos(- \frac{\pi}{2})=0$

so $\displaystyle 0= y_1^2-1$, implies $\displaystyle y_1=+-1$

Then my critical points are

(1,0) and (-1,0)

At this point I would take the Jacobian at the two critical points and solve for the eigenvalues. The eigenvalues would then tell me about the stability? I will do this, I just want to confirm I am on the right track first.

Thanks so much,

Len
• Dec 7th 2011, 04:29 AM
Jester
Re: Stability of Critical Points of Second Order Equation
You are.
• Dec 7th 2011, 11:25 AM
Len
Re: Stability of Critical Points of Second Order Equation
Quote:

Originally Posted by Danny
You are.

Thank you,

So for the point (1,0) I got eigenvalues -2 and 1. This would mean the point is unstable and a saddle.

and for (-1,0) I got eigenvalues 1/2 + (sqrt(7)/2)i and 1/2 - (sqrt(7)/2)i

So it has two positive real parts. therefore it is unstable.

Is there anything else I can say about the stability and the critical points?
• Dec 7th 2011, 12:24 PM
djiang12
Re: Stability of Critical Points of Second Order Equation
Len, are you in MATH3090 at York University?