How to prove that limit cycle can exist in 2D curl field? I know limit cycle can definitely exist in a 2D curl field (that is field with no divergence, $\nabla \cdot F = 0$), but how to explicitly construct such an example? Kind of stuck where to start. Any help is appreciated!
 A: A limit cycle cannot exist in a 2D curl vector field. A 2D curl vector field is a hamiltonain vector field and as such it has a first integral, so if you have at least one closed curve trajectory, then there will be a whole smooth foliation of closed curve trajectories in an open neighbourhood of the given closed curve trajectory.
Indeed, let the vector field in question is $$X \, =\, P(x, y)\frac{\partial}{\partial x} \, +\, Q(x, y)\frac{\partial}{\partial y}$$ assume the $\gamma_0$ is a closed curve trajectory tangent to the vector field, i.e. $\gamma_0$ is the trajectory of a periodic solution of  the system of ODEs
$$\frac{dx}{dt} \, =\, P(x, y)$$
$$\frac{dy}{dt} \, =\, Q(x, y)$$
We assume that the vector field $X$ is smooth and well defined in a large enough open domain in the plane that contains $\gamma_0$. Now, $\gamma_0$ is a smooth embedded submanifold of the plane, diffeomorphic to the unit circle, and it bounds a simply connected closed compact domain $K(\gamma_0)$ which is diffeomorphic to the closed unit disc. Let us take a small enough neighbourhood $D(\gamma_0)$ of $K(\gamma_0)$, diffeomorphic to an open disc of the plane. Then the vector field $X$ is smooth and well defined on the this simply connected open domain $D(\gamma_0)$ and $\gamma_0 \, \subset \, D(\gamma_0)$. So from now on all arguments will take place in $D(\gamma_0)$.
Define the smooth differential one form
$$\omega \, = \,  - \,Q(x, y)\,dx \, +\, P(x, y)\, dy$$ and observe that $$\omega(X) \equiv 0$$ Now,
\begin{align}
d\omega \, &=\, d \Big(-Q(x, y)\,dx \, +\, P(x, y)\, dy \,\Big) 
\, =\,   - \, \,dQ(x, y)\wedge dx \, +\, dP(x, y)\wedge dy \\
&= -\, \frac{\partial Q}{\partial y}(x, y)\, dy \wedge dx \, +\, \frac{\partial P}{\partial x}(x, y)\, dx  \wedge dy\\
&= \frac{\partial Q}{\partial y}(x, y)\, dx \wedge dy \, +\, \frac{\partial P}{\partial x}(x, y)\, dx  \wedge dy \\
&= \left( \, \frac{\partial Q}{\partial y}(x, y)\, dx\, + \, \frac{\partial P}{\partial x}(x, y)\right) dx  \wedge dy \\&\equiv 0
\end{align}
Since $D(\gamma_0)$ is simply connected, more precisely it deformation retracts onto a point, there exists a smooth function $f(x, y)$ defined at least on $D(\gamma_0)$ such that
$$df\, =\, \omega$$
Hence,
$$P(x, y) \, =\, \frac{\partial f}{\partial y}(x, y)$$
$$Q(x, y) \, =\, -\, \frac{\partial f}{\partial  x}(x, y)$$
which means that the original vector field can be written as
$$X \, =\, \frac{\partial f}{\partial y}(x, y)\frac{\partial}{\partial x} \, - \, \frac{\partial f}{\partial  x}(x, y) \frac{\partial}{\partial y}$$
and is therefore hamiltonian with first-integral (hamiltonian) $f(x, y)$ so every trajectory in the domain $D(\gamma_0)$ follows a level curve of the function $f$, i.e.
$$f(x, y) = c$$ for some constant $c$.
Since $\gamma_0 \, \subset \, f(x, y) = c_0$, then the trajectories of the vector field $X$ in a small enough tubular neighbourhood of $\gamma_0$ are also closed curve trajectories.
