# What is a limit cycle?
Isolated, closed trajectories.
1. Not like a center.
2. Centers are closed, but not isolated.
3. Neighboring trajectories are NOT closed. 
Different forms:
1. **Stable** - Trajectories pull onto the limit cycle
2. **Unstable** - Trajectories are repelled by the limit cycle.

**A imit cycle is a explicitly nonlinear phenomenon.**

You can't identify if there is a limit cycle by using linearizing methods.
# How do we find limit cycles?
## How do we rule out a closed loop?
### Dulac's Criterion:
If we have some flow field:
$$ \dot{\vec{x}}= f(\vec x)$$
- If we can find a function $\zeta(x,y)$ such that $\nabla \cdot (\zeta f))$ does not change sign in some region of $R$, then there's no limit cycle in that region. 
- If in some region $R$, $\zeta(x,y)$ s.t :
  $$ \frac{\partial}{\partial x} (\zeta(x,y) f_1(x,y)) + \frac{\partial}{\partial y}(\zeta(x,y) f_2(x,y))$$
  is of constant sign, then there are no closed orbits in R. 
<mark style="background: #FFF3A3A6;">Finding $\zeta$ is tricky.</mark>

Example:
$\dot x = y$
$\dot y = -x -y + x^2 + y^2$

Assume $\zeta(x,y) = 1$
$\partial / \partial x (y) + \partial / \partial y (-x - y +x^2 +y^2) /rightarrow 0 + (-1+2y)$

Assume $\zeta(x,y) = e^{\alpha x}$

$\partial / \partial x (e^{\alpha x} y) + \partial / \partial y (e^{\alpha x} (-x - y +x^2 +y^2))$
$\alpha e^{\alpha x} y + 2 y e^{\alpha x} - e^{\alpha x}$
$e^{\alpha x}((\alpha+2) y -1)$
Now let $\alpha = -2$
$\nabla \cdot (\zeta f) = e^{-2 x}$

Now a special note: These functions can define where limit cycles can't be. If the function doesn't change sign for a subset of R, there can't be a limit cycle contained in that subset. There CAN be a limit cycle that crosses the point the function changes sign.
### Lyapunov Function
Aleksander Lyapunov (Liapunov)
$V(\vec x) = V(x,y) \leftarrow$ a scalar function
$V(\vec x) > 0 \forall \vec x\neq \vec x^*$
$V(\vec x^*) = 0$
$\dot V = \frac{dV}{dt} <0$
$V(\vec x)$ is a positive definite function.
Then the system is stable ISL (in the sense of Lyapunov). 
The system will always asymptotically approach the equilibrium point.
$\frac{dV}{dt} = \frac{dV}{dx} \frac{dx}{dt} + \frac{dV}{dy} \frac {dy}{dt} = \dot x \frac{dV}{dx} + \dot y \frac{dV}{dy}$

Example:
$\dot x = y - x^3$
$\dot y = -x-y^3$
$V(x,y) = c_1 x^2 + c_2 y^2$
$\frac{dV}{dt} = 2 c_1 x \dot x + 2 c_2 y \dot y$
$= 2 c_1 x(y-x^3) + 2c_2 y(-x-y^3)$
Assume $c_1 = c_2$
... $\therefore \frac{dV}{dt} = -2c(x^4+y^4) < 0$
Therefore limit cycles are not possible.