Are All One-Dimensional Vector Fields Gradient Systems?

In summary, the conversation discusses exercise 7.2.4 in the book "Nonlinear Dynamics and Chaos" by Steven H. Strogatz, in which the task is to show that all vector fields on the line are gradient systems. The definitions of gradient system and vector field on the line are reviewed, and the question is asked whether it is always possible to find a function V(x) such that the gradient is equal to the given vector field. The conversation ends with the realization that the system being considered is 2D, and there may be other factors that have not been considered.
  • #1
xibeisiber
8
0
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!
 
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  • #2
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.
 
  • #3
pasmith said:
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.

Thank you for your reply!
I tried to prove that ∂f/∂y=∂g/∂x,as shown in the attachment,but failed...
 

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  • #4
You're barking up the wrong tree, I'm afraid.

"The line" means the real numbers. A http://planetmath.org/GradientSystem.html 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]
 
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  • #5
pasmith said:
You're barking up the wrong tree, I'm afraid.

"The line" means the real numbers. A http://planetmath.org/GradientSystem.html 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]
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 realized.
 
  • #6
em,I feel like I understand that "on the line" means one dimension.
I wasted too much time on such a simple question!:cry:
 

1. What is nonlinear dynamics?

Nonlinear dynamics is a branch of science that studies systems that do not follow a linear relationship between cause and effect. These systems can exhibit complex and unpredictable behavior, making them difficult to analyze and understand.

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