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Help! One problem of nonlinear dynamics

  1. Dec 5, 2012 #1
    Show that all vector fields on the line are gradient systems.

    This is exercise 7.2.4 in the book "Nonlinear Dynamics and Chaos" by Steven H.Strogatz

    Thanks very much!
     
  2. jcsd
  3. Dec 5, 2012 #2

    pasmith

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    Homework Helper

    Re-read the definitions Strogatz gives of "gradient system" and "vector field", and then answer the following:

    What's a gradient system on the line?

    What's a vector field on the line?

    If you can answer those questions, then the proof should be straightforward.
     
  4. Dec 5, 2012 #3
    Thank you for your reply!
    I tried to prove that ∂f/∂y=∂g/∂x,as shown in the attachment,but failed...
     

    Attached Files:

  5. Dec 5, 2012 #4

    pasmith

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    You're barking up the wrong tree, I'm afraid.

    "The line" means the real numbers. A http://planetmath.org/GradientSystem.html [Broken] on the line is a system in which
    [tex]\dot x = -\frac{dV}{dx}[/tex]
    for [itex]x \in \mathbb{R}[/itex] and some V(x). A vector field on the line is essentially just a function [itex]f: \mathbb{R} \to \mathbb{R}[/itex] which is at least continuous.

    So, given any continuous function [itex]f: \mathbb{R} \to \mathbb{R}[/itex], can we always find V(x) such that
    [tex]
    \frac{dV}{dx} = -f(x)?
    [/tex]
     
    Last edited by a moderator: May 6, 2017
  6. Dec 6, 2012 #5
    But the system I consider is 2D,the V(x,y) is not always there for all f(x,y) and g(x,y).So there must be anything else I haven't realised.
     
  7. Dec 6, 2012 #6
    em,I feel like I understand that "on the line" means one dimension.:surprised
    I wasted too much time on such a simple question!!!!!!!:cry:
     
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