Construction of the Fluid Flow

We will consider the following vector field in cylindrical coordinates:

$${\mathbf{u}}_s = {\mathbf{u}}= (0, u_\theta(r,z,t), 0)$$ (4.1)

with $u_\theta(r,z,t) = \alpha (r) \beta (\vert z\vert) \gamma _s(t) r$, where

$$\begin{array}{l} \begin{array}{rlllrlllll} \alpha (r) &= 0 &... ...0^{t_0} \gamma _s(\tau) d \tau = s \in [0,2\pi]\, . \end{array}$$

The vector field ${\mathbf{u}}$ from (4.1) is divergence free and smooth (in Cartesian coordinates $(x,y,z)$). Evidently, there exist ${\mathbf{f}}^E$ and ${\mathbf{f}}^N$, smooth axially symmetric vector fields such that ${\mathbf{u}}$ satisfies (with constant pressure) the Euler equations and the Navier-Stokes equations, respectively.

In the cylinder $\vert z\vert\leq Z_i$, $r\leq R_i$ it corresponds to the rotation by the angle $s$ during the time interval $[0,t_0]$ and outside of the cylinder $\vert z\vert\leq Z^o$, $r\leq R^o$ the fluid does not move at all.

Let us start by analyzing $\mathbf{A}^E$. This is actually quite easy to compute explicitly. Writing the input vector in cylindrical coordinates, and the output in Cartesian coordinates, we have

$$\mathbf{A}^E(r,z,\vartheta , t_0) = \Big(r \cos [\vartheta -... ...r \sin[\vartheta - s \alpha (r) \beta (\vert z\vert) ], z\Big)$$

that is,

$$\begin{array}{lcr} \mathbf{A}^E(x,y,z,t_0)& = &\Big( x \cos[... ...qrt{x^2+ y^2}) \beta (\vert z\vert) ] , z \Big)\, . \end{array}$$

Inside the inner cylinder we have

$$\nabla \mathbf{A}^E = \left( \begin{array}{ccc} \cos s, & \s... ...-\sin s, & \cos s, & 0 \\ 0, & 0, & 1 \end{array} \right)\, ;$$

outside the outer cylinder

$$\nabla \mathbf{A}^E = \left( \begin{array}{ccc} 1, & 0, & 0 \\ 0, & 1, & 0 \\ 0, & 0, & 1 \end{array} \right)\, ;$$

for $\vert z\vert\leq Z_i$, $R_i \leq r\leq R^o$

$$\nabla \mathbf{A}^E = \left( \begin{array}{ccc} \cos [s \alp... ..., & 0,& 1 \end{array} \right) + {\mathbf{M}}_1 +{\mathbf{M}}_2$$

with

$$\begin{array}{l} {\mathbf{M}}_1 = \left( \begin{array}{ccc} ... ...a '(r), & 0 \\ 0, & 0, & 0 \end{array} \right)\, ; \end{array}$$

for $Z_i\leq \vert z\vert\leq Z^o$, $r\leq R_i$

$$\nabla \mathbf{A}^E = \left( \begin{array}{ccc} 1, & 0, & -s... ...] \beta '(\vert z\vert)\\ 0, & 0, & 1 \end{array} \right)\, ;$$

and finally for $Z_i\leq \vert z\vert\leq Z^o$, $R_i \leq r\leq R^o$ we get a combination of the last two cases. We will use the structure of $\nabla \mathbf{A}^E$ later.

Let us now look at the difference between $\mathbf{A}^N$ and $\mathbf{A}^E$, our goal being inequality (4.3) below. We have

$$\begin{array}{c} \displaystyle \frac {\partial }{\partial t}... ...mathbf{A}^{E})({\mathbf{x}}, 0) = {\mathbf {0}}\, . \end{array}$$

Taking the spatial gradient we get

$$\frac {\partial }{\partial t} [\nabla (\mathbf{A}^N - \mathb... ... - (\nabla (\mathbf{A}^N-\mathbf{A}^E))\nabla {\mathbf{u}}\, .$$ (4.2)

Now, since

$$\sup _{t \in [0,t_0]}\Vert\nabla ^3 \mathbf{A}^N\Vert _p \leq C (\Vert\nabla {\mathbf{u}}\Vert _{k,p})$$

for some $k$ sufficiently large, we have, after testing equation (4.2) by $\vert\nabla (\mathbf{A}^N-\mathbf{A}^E)\vert^{p-2}\nabla (\mathbf{A}^N - \mathbf{A}^E)$

$$\frac{d}{dt} \Vert\nabla (\mathbf{A}^N-\mathbf{A}^E)\Vert _p... ...rt _\infty \Vert\nabla (\mathbf{A}^N-\mathbf{A}^E)\Vert _p\, .$$

Thus, as $\nabla (\mathbf{A}^N-\mathbf{A}^E)({\mathbf{x}},0) = {\mathbf {0}}$, we get

$$\sup _{t \in [0,t_0]} \Vert\nabla (\mathbf{A}^N-\mathbf{A}^E)\Vert _p \leq \nu C(\Vert\nabla {\mathbf{u}}\Vert _{k,2}, t_0)$$ (4.3)

for all $p\in (1,\infty]$.