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Equilibrium points

  1. Feb 3, 2009 #1
    [tex]V(x,y) = x^{2}+y^{2} - z; [/tex]

    I need to find the equilibrium points of this guy. I can do this in my sleep for two dimensions but I'm not getting anything useful with this particular function.

    What would the graph look like for V(x,y) = c, c is arbitrary constant. It's a cone, I know, but I'm not sure how to graph it out.

    Any suggestions?

    If anybody cares, this is what the plot looks like in matlab. Looks like the EP is (0,0,0).

    Last edited: Feb 3, 2009
  2. jcsd
  3. Feb 4, 2009 #2
    What is z? Is it a constant, or did you mean V(x, y, z)?
  4. Feb 4, 2009 #3
    I meant V(x,y,z)
  5. Feb 4, 2009 #4


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    What's your definition of equilibrium point? I've only heard it used in the context of differential equations.

    With respect to graphing, notice that for each fixed z=constant plane, you just the the equation for a circle centered at the origin with radius sqrt(c+z), so you expect it to look like a cone, but with a parabolic shape
  6. Feb 4, 2009 #5
    Yes, this is part of a dynamical systems problem. It's really frustrating when I get stuck on the Calculus parts.

    Here, the equilibrium point (or critical point) is defined as V(x,y,z) = 0.
  7. Feb 4, 2009 #6
    I'm not sure what the problem is, you have a function of 3 variables, the set of equil. points will be [tex] z = x^2 + y^2 [/tex]
  8. Feb 5, 2009 #7


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    If V is a potential energy function, then the equilibrium points are at the minimum (stable equilibrium) and maximum (unstable equilibrium) values of V(x,y,z). The minimum and maximum values of a function of three variables occur where the gradient,
    [tex]\nabla V= 2x\vec{i}+ 2y\vec{j}- \vec{k}= \vec{0}[/tex]
    Since the [itex]\vec{k}[/itex] component is never 0, there are NO equilibrium points.
  9. Feb 5, 2009 #8
    That's what I thought too. I don't know how the phase portrait would look with no EQ PTS. I guess the solutions just pass through [somewhere].
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