One-dimesional system non-existence fixed points

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SUMMARY

The discussion focuses on the analysis of fixed points in a one-dimensional dynamic system defined by the equation x' = f(x) = rx - x^3. The fixed points are determined by solving f(x) = 0, yielding x* = {0, -sqrt(r), +sqrt(r)}. For values of r less than zero, the solutions -sqrt(r) and +sqrt(r) become imaginary, indicating the absence of real fixed points. The absence of real fixed points implies that the system lacks stable positions, as illustrated by the example x' = x^2 + 1, which also results in imaginary solutions.

PREREQUISITES
  • Understanding of one-dimensional dynamic systems
  • Familiarity with fixed point theory
  • Knowledge of stability analysis in dynamical systems
  • Basic algebra for solving polynomial equations
NEXT STEPS
  • Study the stability criteria for fixed points in dynamical systems
  • Explore the implications of imaginary fixed points on system behavior
  • Learn about bifurcation theory in one-dimensional systems
  • Investigate other examples of dynamic systems with complex fixed points
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This discussion is beneficial for students and researchers in mathematics, particularly those studying dynamical systems, stability analysis, and fixed point theory.

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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