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Phase portrait of nonlinear system of differential equations

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Describe the phase portrait of the nonlinear system x' = x^2, y' = y^2
    Also, find the equilibrium points and describe the behaviour of the associated linearized system.

    3. The attempt at a solution

    We have an equilibrium point at (0,0).
    The associated linearized system is x' = 0, y' = 0. The phase portrait for this consists of lines of equilibria along x = 0, and y = 0.

    For the nonlinear system, I have found solutions x(t) = -1/t and y(t) = -1/t. I don't know what these solutions mean in terms of a phase portrait. Nor can I express the solutions in terms of constants x_0 and y_0.
     
  2. jcsd
  3. Nov 8, 2009 #2

    Mark44

    Staff: Mentor

    It looks like you have left off the constant of integration when you found solutions x(t) and y(t). Both of your differential equations are separable.

    dx/dt = x2 ==> dx/x2 = dt ==> [itex]\int dx/x^2 = \int dt[/itex]
    ==> -1/x = t + C1 ==> x = -1/(t + C1)

    Similarly, y = -1/(t + C2)
    You should be able to determine the constant from your initial conditions.

    In the special case where C1 = C2 = 0, the trajectories follow the line y = x. If t > 0, the solution points approach the origin along the part of the line in the third quadrant. If t < 0, the solution points approach the origin along the part of the line in the first quadrant. Different initial conditions will generate different trajectories, but I believe all of them will be straight lines pointing into the origin.

    Does that make sense? It has been a lot of years since I studied dynamical systems, so I might be a little off base on some of this.
     
  4. Nov 8, 2009 #3
    Thanks :) It makes sense.
     
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