Finding Equilibrium Points of V(x,y) = x^2+y^2 - z

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    Equilibrium Points
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Discussion Overview

The discussion revolves around finding the equilibrium points of the function V(x,y,z) = x² + y² - z. Participants explore the nature of the function, its graphical representation, and the definition of equilibrium points in the context of dynamical systems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in finding equilibrium points for the function V(x,y,z) and seeks suggestions for graphing it.
  • Another participant questions whether z is a constant or if V should be considered as a function of three variables.
  • A participant clarifies that V is indeed a function of three variables.
  • There is a discussion about the definition of equilibrium points, with one participant noting it is typically used in the context of differential equations.
  • One participant explains that for fixed z, the graph represents circles centered at the origin, suggesting a conical shape with a parabolic profile.
  • A participant defines equilibrium points as where V(x,y,z) = 0, indicating a connection to dynamical systems.
  • Another participant proposes that the equilibrium points can be described by the equation z = x² + y², suggesting a relationship between the variables.
  • One participant argues that if V is a potential energy function, equilibrium points correspond to the minimum and maximum values, which occur where the gradient is zero, concluding that there are no equilibrium points since the k-component of the gradient cannot be zero.
  • A later reply expresses confusion about the implications of having no equilibrium points for the phase portrait, suggesting that solutions may pass through unspecified locations.

Areas of Agreement / Disagreement

Participants express differing views on the existence and definition of equilibrium points, with some suggesting that they can be found while others argue that none exist based on the gradient analysis. The discussion remains unresolved regarding the implications of these differing perspectives.

Contextual Notes

There are limitations in the definitions and assumptions regarding equilibrium points, particularly in the context of potential energy functions and their graphical representations. The discussion also highlights the dependence on the interpretation of the variables involved.

Somefantastik
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[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).

cone.jpg
 
Last edited:
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What is z? Is it a constant, or did you mean V(x, y, z)?
 
I meant V(x,y,z)
 
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
 
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.
 
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]
 
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.
 
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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