One-dimesional system non-existence fixed points

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In a one-dimensional dynamic system defined by x' = f(x) = rx - x^3, fixed points are found by solving f(x) = 0, yielding x* = {0, -sqrt(r), +sqrt(r)}. For r < 0, the solutions -sqrt(r) and +sqrt(r) become imaginary, indicating no real fixed points exist. The absence of real fixed points suggests that the system lacks stable positions. An example of a system with only imaginary fixed points is x' = x^2 + 1, which further illustrates this concept. Thus, systems with only imaginary fixed points do not exhibit stability.
lahanadar
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Homework Statement


First things first, this is not a HW but a coursework question. I try to understand a concept.

Assume we have a one-dimensional dynamic system with:

x'=f(x)=rx-x^3

Homework Equations


Fixed points are simply calculated by setting f(x)=0.

The Attempt at a Solution


If I compute f(x)=0:

f(x)=x(r-x^2)=0 and so x*={0, -sqrt(r), +sqrt(r)}

If r<0, then -sqrt(r), +sqrt(r) becomes obsolete since they become imaginary.

What if I only come up with only imaginary fixed points for another system? How would the system behave in terms of stability?
 
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lahanadar said:
What if I only come up with only imaginary fixed points for another system? How would the system behave in terms of stability?
Simple: there would be no stable position.
x'=1 is probably the easiest example of such a system. x'=x2+1 if you want imaginary solutions.
 

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