Consider the solution to x''= -x with the initial conditions x(0) = 0, x'(0) = 1. Find the alpha and omega limit sets for this solution.
(Then rewrite as a 2x2 system)
If you solve the ODE you get
$\displaystyle x(t)=\sin(t)$
The definition of an omega limit point is if there exists a sequence that goes to infinity such that
$\displaystyle a_n \to \infty \implies x(a_n)=r, r \in \mathbb{R}$
Since there exists such a sequence for any
$\displaystyle r \in [-1,1]$
This is the omega set.
The alpha set is the same except the limit must go to negative infinity.
for the next part let
$\displaystyle y(t) = x'(t) \implies y'(t)=x''(t)$
Now use the ODE to write this as a first order system.
Ty for your fast response.
I also got the x(t) = sin(t) after staring at the problem for a little.
But, I think that the way you wrote the limit set and what I got are different.
$\displaystyle \alpha$-limit set = {sin(t) : t $\displaystyle \in$ [-2pi, 0]}
$\displaystyle \omega$-limit set = {sin(t) : t $\displaystyle \in$ [0, 2pi]}
Unless, what you wrote and what I wrote mean the same thing.
Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
But would it be:
(cos(t) sint(t)
-sin(t) cos(t))
sin(t) and cos(t) take on all values between -1 and 1.
There is only one equation given: x''= -x. As TheEmptySet suggested, let y= x' so your equation becomes y'= -x and you have the system of equations x'= y, and y'= -x.Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
But would it be:
(cos(t) sint(t)
-sin(t) cos(t))