Outline of the Proof of Theorem 1.1

In this section we will give the plan for the proof of Theorem 1.1. The idea of the proof is really quite simple. We will in fact construct a family of divergence free, smooth solutions ${\mathbf{u}}_s$ to (1.8) parameterized by a number $s \in [0,2\pi]$. We will use simple ideas from algebraic topology to show that there exists $s_0 \in [0,2\pi]$ such that $u_{s_0}$ provides an example to prove Theorem 1.1.

In fact all of the solutions we construct will be axisymmetric, indeed, when written in cylindrical coordinates, they have the form: ${\mathbf{u}}_s = (0, u_\theta(r,z,t), 0)$. We will prove that there exists $s_0 \in [0,2\pi]$ and $t_0>0$ such that the Jacobian $\nabla \mathbf{A}^N_{s_0}(0,t_0)$ is non-invertible. To this end we have the following representation result.

Lemma 2.1   For any $t>0$, $\nabla \mathbf{A}^N_s({\mathbf {0}}, t)$ can be uniquely written as

$$\left(\begin{array}{lll} a_s \cos b_s, & -a_s \sin b_s & 0\\... ... \sin b_s, & a_s \cos b_s & 0\\ 0 & 0 & 1 \end{array} \right)$$

for some $a_s(t)\in \mbox{\Bbbb R}^+$ and $b_s(t) \in [0,2\pi)$.

The proof of Theorem 1.1 will proceed as follows. For each $s \in [0,2\pi]$, we will construct ${\mathbf{u}}_s$. The Theorem will be proved if we can show the existence of $s_0 \in [0,2\pi]$ and $t_0>0$ such that $a_{s_0}(t_0) = 0$. We will assume the opposite, and give a proof by contradiction.

We will need some simple facts from algebraic topology. We refer the reader to [8] for more details. Let us consider the collection of continuous functions $[0,\infty] \to \mbox{\Bbbb R}^2-\{(0,0)\}$ which map $0$ and $\infty$ to $(1,0)$. We will say two such functions $f$ and $g$ are homotopic with base point $(1,0)$ (or simply homotopic) if there exists a jointly continuous function $F:[0,\infty] \times [0,2\pi] \to \mbox{\Bbbb R}^2-\{(0,0)\}$ such that $F(\cdot,0) = f$, $F(\cdot,2\pi) = g$ and $F(0,\cdot) = F(\infty,\cdot) = (1,0)$. We will call the function $F$ a homotopy. Clearly being homotopic is an equivalence relation.

It is well known that a constant map $f(t) = (1,0)$, and a map with ``winding number 1'', for example, $g(t) = (\cos(2\pi t/(1+t)),\sin(2\pi t/(1+t)))$ are not homotopic. (Since $\mbox{\Bbbb R}^2-\{(0,0)\}$ is homotopy equivalent to the unit circle, this is basically saying that the fundamental group of the unit circle is non-trivial.)

In order to provide our contradiction we will prove the following result.

Lemma 2.2   If ${\mathbf{u}}_s$ is constructed as described in the next section, with the various parameters chosen appropriately, then the function

$$F(t,s) = (a_s(t) \cos b_s(t), a_s(t) \sin b_s(t))$$

provides a homotopy between the function $f$ and a function homotopic to $g$.