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Equilibria in Nonlinear Systems

  1. Nov 8, 2014 #1
    So I'm reading the Example on page 161 of Differential Equations, Dynamical Systems and an Introduction to Chaos by Hirsh, Smale, and Devaney.

    I'm not understanding everything.

    So given the system

    [itex]x' = x + y^2 [/itex]
    [itex]y' = -y[/itex]

    we see this is non-linear. I get it that near the origin, y^2 tends to zero "fast". So we can consider the system

    x' = x

    y' = -y

    I see that the solution to this system is [itex]X = c_1 e^t (1,0) + c_2e^{-t}(0,1) = (c_1e^{t},c_2e^{-2}) [/itex]and we have that the y-axis is the stable line and the x-axis is the unstable line.

    Here's where I'm unsure of where things are going on since we can solve [itex]y' = -y \rightarrow y = y_0e^{-t}[/itex], we solve [itex]x' = x \rightarrow x = x_0e^t[/itex]

    Since we have the solution to the homogenous eig buation and we'll say that the particular solution [itex] x_p(t) = Ce^{-2t}, x_p'(t) = -2Ce^{-2t},[/itex] we'll get that [itex]C = -\dfrac{1}{3}y_0^2[/itex], so we'll get that [itex]x(t) = ce^t - \dfrac{1}{3}y_0^2e^{-2t}[/itex]

    Not seeing how they got that
    [itex]x(t) = \left( x_0 + \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]

    [itex]y(t) = y_0e^{-t} [/itex]

    why isn't it [itex]x(t) = \left( x_0 - \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]?

    This seems like a big deal because we're getting ready to do a change of coordinates.
     
  2. jcsd
  3. Nov 9, 2014 #2

    pasmith

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    Homework Helper

    You have [itex]x(t) = ce^t - \frac13 y_0^2 e^{-2t}[/itex] and you need [itex]x(0) = x_0[/itex]. Hence [itex]c - \frac13 y_0^2 = x_0[/itex] so that [itex]c = x_ 0 + \frac13 y_0^2[/itex].
     
  4. Nov 10, 2014 #3
    Thanks so much! I know we want to do that in order to make the linearization work correct?
     
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