Is a Limit Point Always an Equilibrium in Differential Equations?

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SUMMARY

In the discussion, it is established that if \( x \) is a solution of the differential equation \( x' = f(x) \) and \( f \) is continuously differentiable, then the limit point \( p \) is an equilibrium point if \( \lim_{t \to \infty} x(t) = p \) and \( f(p) = 0 \). The argument is supported by the definition of limits, showing that for any \( \varepsilon > 0 \), there exists a \( t_0 \) such that for all \( t > t_0 \), \( |x(t) - p| < \varepsilon \). This leads to the conclusion that \( \lim_{t \to \infty} x'(t) = 0 \), confirming \( p \) as an equilibrium.

PREREQUISITES
  • Understanding of differential equations, specifically the form \( x' = f(x) \).
  • Knowledge of limits and continuity in calculus.
  • Familiarity with the concept of equilibrium points in dynamical systems.
  • Basic proficiency in mathematical notation and proofs.
NEXT STEPS
  • Study the properties of continuous functions and their derivatives in the context of differential equations.
  • Learn about stability analysis of equilibrium points in dynamical systems.
  • Explore the application of the Lyapunov method for assessing stability in differential equations.
  • Investigate the implications of limit points in nonlinear differential equations.
USEFUL FOR

Mathematicians, students of differential equations, and researchers in dynamical systems will benefit from this discussion, particularly those interested in equilibrium analysis and stability in mathematical modeling.

simo1
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Say x is a solution of the DE x’=f(x) and f is a continuous derivative on its domain,
if lim┬(t→͚inifinity⁡〖x(t)〗=p then p is equilibriumhow can I show that this is true
 
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simo said:
Say x is a solution of the DE x’=f(x) and f is a continuous derivative on its domain,
if lim┬(t→͚inifinity⁡〖x(t)〗=p then p is equilibriumhow can I show that this is true

I suppose You mean that p is 'equilibrium' if f(p)=0. In this case if $\displaystyle \lim_{t \rightarrow \infty} x(t)= p$ then given $\varepsilon >0$ it exists a $t_{0}$ so that for any $t>t_{0}$ is $|x(t) - p|< \varepsilon$. Consequence of that is that for any $t_{1}> t_{0}$ and $t_{2}>t_{0}$ is $|x(t_{1}) - x(t_{2})| < 2\ \varepsilon$, so that is $\displaystyle \lim_{t \rightarrow \infty} x^{\ '} (t) = 0$... Kind regards $\chi$ $\sigma$
 
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