MHB Is a Limit Point Always an Equilibrium in Differential Equations?

  • Thread starter Thread starter simo1
  • Start date Start date
  • Tags Tags
    Equilibrium
simo1
Messages
21
Reaction score
0
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
 
Physics news on Phys.org
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 $\chi$ $\sigma$
 
Last edited:
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
845
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K