0

I am learning about omega limit sets, and even though I understand the definition, I am struggling with explicitly determining them.

For example, many of our exercises contain initial value problems and ask for omega limit sets. In the case of the following initial value problem: $$\dot{x} = x^2 +x^3, \; x(0) = -\frac{1}{2},$$

how would we explicitly define the omega limit set here, and how would we calculate it?

EDIT 1:

So I have calculated the equilibrium points: $x_1 = 0, x_2 = -1,$ and it seems that both of these equilibrium points are unstable. But still not sure how does this gie us omega limit set?

variableXYZ
  • 1,053
  • 6
  • 13

2 Answers2

2

Hint: What are the equilibrium points for this differential equation? What does a phase portrait look like?

EDIT: You are correct, the equilibrium points are $0$ and $-1$. A very important fact is that (because of the uniqueness theorem) two solutions can never meet or cross. So since your solution starts between the two constant solutions $0$ and $1$, it has to stay between them at all times in the past and future.

The next thing to notice is that the differential equation says that when $x$ is between $0$ and $-1$, $\dot{x} > 0$. Therefore the solution must be increasing, and must approach $0$ in the limit as $t \to +\infty$ (i.e. any increasing function that is bounded above has a limit, and the slope must be $0$ at that limit, so the limit is an equilibrium point). Similarly, the solution must approach $-1$ as $t \to -\infty$.

Robert Israel
  • 448,999
  • If I calculated correctly the equilibrium points are: $x_1 = 0, ; x_2 = -1.$ But I am not sure how to draw phase portrait? – variableXYZ Mar 07 '24 at 21:42
  • Also I don't understand how does IVP change anything here? – variableXYZ Mar 07 '24 at 22:05
  • The initial condition matters because the limit behavior is different depending on where you are with regards to the equilibria. For example, the solution with $x(0) = 0$ stays constant but if $x(0) < 0$ then the solution goes to $-\infty$. – whpowell96 Mar 08 '24 at 01:15
  • I meant $x(0)<-1$ for the second example – whpowell96 Mar 08 '24 at 01:38
1

Any orbit of a $1$-dimensional smooth vector field must tend either to an equilibrium point or to $\pm\infty$. You also have that any orbit is monotonic increasing or decreasing depending on the sign of $\dot{x}$ at its initial point.

So, the $\omega$-limit for an initial value is the nearest equilibrium point in the positive or negative direction, depending on the sign of $\dot x$, or will be $\pm\infty$ if there are no equilibrium points in that direction.

In your problem, you have two equilibrium points: $-1$ and $0$. Since the value of $\dot x$ in $-\frac{1}{2}$ is $\frac{1}{8}>0$, then its orbit is increasing and must tend to the nearest equilibrium point in the positive direction, which is $0$.

If, for instance, your initial value was $x(0) = 1$, then $\dot x$ is positive, but there are no equilibrium points greater than $1$. In this case, the orbit tends to $+\infty$ and the $\omega$-limit is empty.

Tiago Emilio Siller
  • 1,116