Are All One-Dimensional Vector Fields Gradient Systems?

[xibeisiber]

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!

 

[pasmith]

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.

 

[xibeisiber]

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...

 

[pasmith]

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
\dot x = -\frac{dV}{dx}
for x \in \mathbb{R} and some V(x). A vector field on the line is essentially just a function f: \mathbb{R} \to \mathbb{R} which is at least continuous.

So, given any continuous function f: \mathbb{R} \to \mathbb{R}, can we always find V(x) such that
<br /> \frac{dV}{dx} = -f(x)?<br />

 

[xibeisiber]

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
\dot x = -\frac{dV}{dx}
for x \in \mathbb{R} and some V(x). A vector field on the line is essentially just a function f: \mathbb{R} \to \mathbb{R} which is at least continuous.

So, given any continuous function f: \mathbb{R} \to \mathbb{R}, can we always find V(x) such that
<br /> \frac{dV}{dx} = -f(x)?<br />

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.

 

[xibeisiber]

em,I feel like I understand that "on the line" means one dimension.
I wasted too much time on such a simple question!