How to prove that limit cycle can exist in 2D curl field?
Answer & Explanation
Wernbergbo9d
Beginner2022-04-07Added 12 answers
Step 1 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 assume the is a closed curve trajectory tangent to the vector field, i.e. is the trajectory of a periodic solution of the system of ODEs We assume that the vector field X is smooth and well defined in a large enough open domain in the plane that contains . Now, is a smooth embedded submanifold of the plane, diffeomorphic to the unit circle, and it bounds a simply connected closed compact domain which is diffeomorphic to the closed unit disc. Let us take a small enough neighbourhood of , 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 and So from now on all arguments will take place in Define the smooth differential one form and observe that Now, Since is simply connected, more precisely it deformation retracts onto a point, there exists a smooth function defined at least on such that Hence, which means that the original vector field can be written as and is therefore hamiltonian with first-integral (hamiltonian) f(x,y) so every trajectory in the domain follows a level curve of the function f, i.e. for some constant c. Since , then the trajectories of the vector field X in a small enough tubular neighbourhood of are also closed curve trajectories.