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Math Help - Stability of a nonlinear system

  1. #1
    Senior Member bkarpuz's Avatar
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    Stability of a nonlinear system

    Dear MHF members,

    I have the following problem.
    Problem. Find all the singular points of the system
    \dot{x}=xy+12
    \dot{y}=x^{2}+y^{2}-25.
    Investigate the stability and determine the type of each singular point. \rule{0.2cm}{0.2cm}

    I know that the critical points of the system are (4,-3), (3,-4), (-4,3) and (-3,4). My problem is determining the stability properties of these points. Please let me know if my idea below is correct. For instance for the point critical point (4,-3), I linearize the equation at (4,-3) and get
    \dot{x}=-3 (x - 4) + 4 (y + 3)
    \dot{y}=8 (x - 4) - 6 (y + 3)
    This is possible since the remainders tend to zero, i.e.,
    \lim_{(x,y)\to(4,-3)}\frac{(x-4) (y+3)}{\sqrt{x^{2}+y^{2}}}=0
    and
    \lim_{(x,y)\to(4,-3)}\frac{25 + x(x-8) + y (y+6)}{\sqrt{x^{2}+y^{2}}}=0.
    Since the linearized equations eigenvalues \lambda_{1,2}=(-9\pm\sqrt{137})/2 are of opposite sign, we see that the critical point (4,-3) is a saddle point and is therefore unstable. I will repeat the same steps above for each critical point.

    Thanks a lot.
    bkarpuz
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Since the linearized equations eigenvalues \lambda_{1,2}=(-9\pm\sqrt{137})/2 are of opposite sign, we see that the critical point (4,-3) is a saddle point and is therefore unstable. I will repeat the same steps above for each critical point.

    Right, but its is sufficient to compute

    A=J_v(4,-3)=\begin{bmatrix}{\dfrac{{\partial v_1}}{{\partial x}}(4,-3)}&{\dfrac{{\partial v_1}}{{\partial y}}(4,-3)}\\{\dfrac{{\partial v_2}}{{\partial x}}(4,-3)}&{\dfrac{{\partial v_2}}{{\partial y}}(4,-3)}\end{bmatrix}=\begin{bmatrix}{-3}&{\;\;4}\\{\;\;8}&{-6}\end{bmatrix}

    whose eigenvalues are certainly

    \lambda=\dfrac{-9\pm \sqrt{137}}{2}

    (saddle point)
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