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Math Help - Stability of Critical Points of Second Order Equation

  1. #1
    Len
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    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:

    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 y_1 = x, y_2=x'

    therefore

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

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

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

    so 0= y_1^2-1, implies 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
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    Re: Stability of Critical Points of Second Order Equation

    You are.
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  3. #3
    Len
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    Re: Stability of Critical Points of Second Order Equation

    Quote Originally Posted by Danny View Post
    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?
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    Re: Stability of Critical Points of Second Order Equation

    Len, are you in MATH3090 at York University?
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