# Thread: alpha and omega limit sets

1. ## alpha and omega limit sets

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)

2. Originally Posted by Borkborkmath
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

$x(t)=\sin(t)$

The definition of an omega limit point is if there exists a sequence that goes to infinity such that

$a_n \to \infty \implies x(a_n)=r, r \in \mathbb{R}$

Since there exists such a sequence for any

$r \in [-1,1]$

This is the omega set.

The alpha set is the same except the limit must go to negative infinity.

for the next part let

$y(t) = x'(t) \implies y'(t)=x''(t)$

Now use the ODE to write this as a first order system.

3. 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.
$\alpha$-limit set = {sin(t) : t $\in$ [-2pi, 0]}
$\omega$-limit set = {sin(t) : t $\in$ [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))

4. Originally Posted by Borkborkmath
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.
$/alpha$-limit set = {sin(t) : t $/in$ [-2pi, 0]}
$/omega$-limit set = {sin(t) : t $/in$ [0, 2pi]}
Unless, what you wrote and what I wrote mean the same thing.
sin(t) and cos(t) take on all values between -1 and 1.

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))
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.