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Finding Equilibrium Points of Nonlinear Systems

  1. Jan 6, 2010 #1

    So I keep making mistakes trying to find all of the equilibrium points of different simple nonlinear systems. These problems aren't difficult, it's just that I keep taking different approaches to finding the equilibrium points.

    Is there a methodological way to know that I have found all of the equilibrium points for a system?

    I have a few examples (below)

    Example 1:

    Example 2:

    Example 3:

    Example 4:

    PS, this isn't homework. This is a component of an exam I need to pass, and I'm just looking for a structured way to approach this type of problem.

    Thanks in advance
  2. jcsd
  3. Jan 7, 2010 #2
    Equilibrium points are points where the derivative of both x and y equals zero.

    So f.e. in this system:


    The equilibrium points satisfy the system of (algebric) equations:


    Which means you have only (0,0) as an equilibrium point.
  4. Jan 7, 2010 #3


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    An "equilibrium solution" is simply a constant solution and so its derivative is 0. At an equilibrium point we must have
    [tex]\dot{x}=x(3-x-2y)= 0[/tex]
    [tex]\dot{y}=y(2-x-y)= 0[/tex]

    x(3- x- 2y)= 0 when x= 0 or when 3- x- 2y= 0.

    If x= 0, then y(2-x-y)= 0 becomes y(2- y)= 0 so either y= 0 or y= 2. So far, two equilibrium points are (0, 0) and (0, 2).

    y(2- x- y)= 0 when y= 0 or 2- x- y= 0.

    If y= 0, then x(3- x- 2y)= 0 becomes x(3- x)= 0 so either x= 0 or x= 3. We already had (0, 0) but now we have (3, 0) as a third equilibrium point.

    If neither x nor y is 0 then we have 3- x- 2y= 0 and 2- x-y= 0. Subtracting the second equation from the first 1- y= 0 or y= 1. With y= 1, both equations become 1- x= 0 so x= 1. The fourth and last equilibrium point is (1, 1).

    Do the same for the rest.
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