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Linearize this system

  1. Mar 31, 2005 #1
    For this system of differential equations

    [tex] \frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
    [tex] \frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})[/tex]
    for all [tex] x,y \geq 0 [/tex]
    now the equilibrium points are
    (x,y) = (0,0), (1000,1088), (3129,0)

    i need to linearize this system such taht i can figure out the eigenvalues of this system and figure out whether this system is a spiral sink, source or whatever

    if i use x = u -1000 and y = v-1088
    i get answers like this
    [tex] \frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000} [/tex]
    [tex] \frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u} [/tex]
    but hte denominator throws things off, doesn't it??
    but if i just accept it like it is and assumethat as u,v approach zero the non linear terms get insignificant (ya ya not proper math language)
    and the matrix becomes
    [tex] \left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right) [/tex]
    is this right so far?? Any help would be appreciated, greatly!
    Last edited: Mar 31, 2005
  2. jcsd
  3. Apr 1, 2005 #2


    User Avatar
    Science Advisor

    The following briefly reviews key features of diff-eq Linearization technique. Steps 1 - 4 at bottom of review section provide guidance for OP to obtain problem solution.

    Remember, linearization is the process of APPROXIMATING the original non-linear diff-eq system with a "Linearized System" about an Equilibrium Point (x0,y0). Thus, let the original non-linear system be given below (assuming no explicit function of "t" on the RHS):

    \left (
    (dx/dt) \\
    \end{array} \right )
    \left (
    f(x,y) \\
    \end{array} \right )

    Then the Linearized System will utilize the Jacobian matrix (in red below) evaluated at an Equilibrium Point (x0,y0):

    \left (
    (dx/dt) \\
    \end{array} \right )
    \color{red} \left (
    \begin{array}{c c}
    f_{x}(x_{0},y_{0}) & f_{y}(x_{0},y_{0}) \\
    g_{x}(x_{0},y_{0}) & g_{y}(x_{0},y_{0})
    \end{array} \right ) \color{black} \cdot
    \left (
    (x - x_{0}) \\
    (y - y_{0})
    \end{array} \right )

    By evaluating the Jacobian Eigenvalues, the Linearized System solutions and their characteristics (sink, spiral, etc.) can then be determined. Hence, the first objective is determining the Jacobian from appropriate Partials:

    [tex] \frac{dx}{dt} = f(x,y) = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
    [tex] \frac{dy}{dt} = g(x,y) = y(-0.5 + \frac{x}{1000+x})[/tex]

    Taking Partials of the above f(x,y) and g(x,y), we get:
    fx(x,y) = (0.8) - (1.6)*x/(3129) - (y)*(1000)/{(x + 1000)^2}
    fy(x,y) = (-x)/(x + 1000)
    gx(x,y) = (y)*(1000)/{(x + 1000)^2}
    gy(x,y) = (-0.5) + (x)/(x + 1000)

    The steps are now the following:
    1) Form Jacobian matrix with above Partials.
    2) Evaluate Jacobian at an Equilibrium Point (x0,y0). ::: <---(by def: f(x0,y0) = g(x0,y0) = 0)
    3) Determine Eigenvalues for the evaluated Jacobian.
    4) Determine and interpret Linearized System Solutions from Eigenvalues.

    Last edited: Apr 2, 2005
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