The decay of $\nabla \mathbf{A}^N-{\mathbf{I}}$

Let us now look in particular at $\nabla \mathbf{A}^N$ for $t > t_0$. We have that $\nabla \mathbf{A}^N$ satisfies the heat equation

$$\begin{array}{c} \displaystyle \frac {\partial }{\partial t}... ...ty) \\ \nabla \mathbf{A}^N(t_0) \quad {\rm given.} \end{array}$$

Therefore also

$$\displaystyle \frac {\partial }{\partial t} (\nabla \mathbf{... ...f {0}}\qquad {\rm in \ } \mbox{\Bbbb R}^3 \times (t_0,\infty)$$

and, especially, at ${\mathbf{x}}={\mathbf {0}}$,

$$\nabla \mathbf{A}^N({\mathbf {0}}, t) - {\mathbf{I}}= \frac ... ...mathbf{A}^N-{\mathbf{I}})({\mathbf{p}},t_0) d{\mathbf{p}}\, .$$ (5.1)

Our first goal will be to show that the function $F$ in Lemma 2.2 satisfies $F(t,s) \to (1,0)$ as $t \to \infty$ uniformly in $s \in [0,2\pi]$. This will complete the proof that $F$ is continuous on $[0,\infty]\times[0,2\pi]$, and that $F(\infty,s) = (1,0)$, so that $F$ is indeed a homotopy.

We have that $\nabla\mathbf{A}^N - {\mathbf{I}}= (\nabla\mathbf{A}^N- \nabla\mathbf{A}^E) + (\nabla\mathbf{A}^E -{\mathbf{I}})$. Using the fact that $\nabla \mathbf{A}^E({\mathbf{p}}, t_0) -{\mathbf{I}}$ has bounded support and $\Vert\nabla (\mathbf{A}^N -\mathbf{A}^E)\Vert _p(t_0) \leq C$, we use the Hölder inequality and end up with

$$\vert\nabla \mathbf{A}^N(0, t) - {\mathbf{I}}\vert \leq \frac{C}{(t-t_0)^a}$$

with some positive power $a$ ($C$ may depend on any constants which appeared above, but is independent of the time).

Since it is clear that $F(\cdot,0)$ is the constant function, the proof of Lemma 2.2 will be complete when we have shown that $F(\cdot,2\pi)$ is homotopic to the function whose winding number is $1$, at least if $\nu$, $R^o-R_i$, and $Z^o-Z_i$ are small enough.

It is clear that the representation of $\nabla\mathbf{A}^E_{2\pi}(\cdot,0)$ has this property. So let us put $s=2\pi$. We need to show that $\nabla\mathbf{A}^E_{2\pi}(\cdot,0) - \nabla\mathbf{A}^N_{2\pi}(\cdot,0)$ is small enough to construct a linear homotopy between the representation of $\mathbf{A}^E_{2\pi}(\cdot,0)$ and $F(\cdot,2\pi)$ that does not pass through $(0,0)$. We have already shown this property for $t \le t_0$ in equation (4.3), at least when $\nu$ is sufficiently small. So all that remains is to show the following result.

Lemma 5.1   There exist $\varepsilon _1$ and $\varepsilon _2 >0$ such that if $\max \{R^o-R_i, Z^o-Z_i\} \leq \varepsilon _1$ and $\nu \leq \varepsilon _2(\varepsilon _1)$ then for $s=2\pi$, $\nabla \mathbf{A}^N - {\mathbf{I}})(0,t) \leq \frac{1}{10}$ for any $t\geq t_0$.

Proof: We denote by $I_1$ the part of integral (5.1) with $(\nabla \mathbf{A}^N-{\mathbf{I}})({\mathbf{p}},t_0)$ replaced by $(\nabla \mathbf{A}^N-\nabla \mathbf{A}^E)({\mathbf{p}},t_0)$, and by $I_2$-$I_5$ the parts of integral (5.1) with $(\nabla \mathbf{A}^N-{\mathbf{I}})({\mathbf{p}},t_0)$ replaced by $(\nabla \mathbf{A}^E-{\mathbf{I}})({\mathbf{p}},t_0)$; namely by $I_2$ the integral over the inner cylinder, by $I_3$ over the cylinder $C(R^o,Z_i)$ without the inner cylinder, by $I_4$ the integral over the outer cylinder $C(R^o,Z^o)$ minus the cylinder $C(R^o,Z_i)$ and finally by $I_5$ over the complement of the outer cylinder.

Evidently, $I_2 = I_5 = 0$ since $s=2\pi$. Let us now consider $I_3$. If we rewrite $\nabla A^E({\mathbf {0}}, t_0)-{\mathbf{I}}$ (in Cartesian components) into the cylindrical coordinates, we get that it is equal to ${\mathbf{M}}_0 + {\mathbf{M}}_1 + {\mathbf{M}}_2$ with

$$\begin{array}{l} {\mathbf{M}}_0 = \left( \begin{array}{ccc} ... ...theta , & 0 \\ 0, & 0, & 0 \end{array} \right)\, . \end{array}$$

The heat kernel is independent of the angle $\vartheta$; after integration over it the matrix ${\mathbf{M}}_2$ disappears and from ${\mathbf{M}}_1$ we are left with integrals of the type

$$\frac{C}{(t-t_0)^{\frac 32} \nu^{\frac 32}} \int_{\index {\b... ...2}{4\nu (t-t_0)}} r^2 \sin [2\pi \alpha (r)] \alpha '(r) dr dz$$

(in some terms, $\sin$ is replaced by $\cos$). Using the standard change of variables and integrating over the $z$ variable we end up with

$$C \int_{\frac{R_i}{\sqrt{\nu(t-t_0)}} \leq u \leq \frac{R^o... ... \sqrt{\nu(t-t_0)})\Big) \alpha '(u\sqrt {\nu(t-t_0)}) d u\, .$$

Now the application of the Taylor theorem on the function ${\rm e}\,^{-\frac{u^2}{4}} u^2$ yields

$${\rm e}\,^{-\frac{u^2}{4}} u^2 = {\rm e}\,^{- \frac{R_i^2}{4... ...i^3}{2}\Big) \Big(u-\frac {R_i} {\sqrt{\nu (t-t_0)}}\Big)\, ,$$

where $\xi \in (\frac {R_i} {\sqrt{\nu (t-t_0)}}, \frac {R^o} {\sqrt{\nu (t-t_0)}})$. Moreover

$$\int_{\frac{R_i}{\sqrt{\nu(t-t_0)}} \leq u \leq \frac{R^o}{... ...t {\nu(t-t_0)})\Big) \alpha '(u\sqrt {\nu(t-t_0)}) d u = 0\, .$$

A similar argument can be applied also on terms coming from ${\mathbf{M}}_0$. Thus we have

$$\begin{array}{c} \displaystyle \vert I_3\vert \leq C{\rm e}... ...-t_0)}} (R^o-R_i)\frac{R^o} {\sqrt{\nu(t-t_0)}}\, . \end{array}$$

We can choose $\alpha (r)$ in such a way that $\alpha '(r) \leq \frac{C}{R^o-R_i}$ and as

$${\rm e}\,^{-\frac{R_i^2}{4\nu (t-t_0)}} \Big(\sqrt{\nu (t-t_0)}\Big)^a \leq C(a, R_i)$$

for any $a \in \mbox{\Bbbb R}$, we finally get

$$\vert I_3\vert \leq C (R^o-R_i)$$

with the constant in particular independent of $\nu$ and $t$. Therefore for the "boundary layer" sufficiently thin, this term can be done arbitrarily small, independently of the viscosity and the time.

Similarly we can estimate $I_4$; here $(\nabla \mathbf{A}^E)_{i,3}^{i=1,2}({\mathbf{p}},t_0)$ are odd functions in $z$ and thus we get zero after the integration of the $z$ variable. For the components $i,j$; $i,j=1,2$ proceed similarly as above and end up with the following integral

$$\int_\Omega {\rm e}\,^{-\frac{u^2+v^2}{4}} u^2 \sin \Bigg(2... ..._0)\Big]}\Bigg) \alpha '\Big(u\sqrt {\nu(t-t_0)} \Big) d u d v$$

with $\Omega =\{(u,v); \frac{R_i}{\sqrt{\nu(t-t_0)}} \leq u \leq \frac{R^o}{\sqrt{\n... ...Z_i}{\sqrt{\nu (t-t_0)}} \leq \vert v\vert \leq \frac{Z^o}{\sqrt{\nu(t-t_0)}}\}$. But now $\beta \neq 1$ and we cannot proceed as above. Nevertheless, we get that the integral above is bounded by

$$C (Z^o-Z_i) (\nu (t-t_0))^a {\rm e}\,^{-\frac{R_i^2 + Z_i^2}{4\nu (t-t_0)}}$$

with the constant independent of $\nu$ and $t$. Thus, if $Z^o-Z_i$ is small, we get that also $I_4$ is small.

Finally,

$$\begin{array}{c} \displaystyle \vert I_1\vert \leq \frac{C}{... ...t _\infty(t_0) \leq \nu C(R^o-R_i, Z^o-Z_i, t_0)\,. \end{array}$$

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Remark 5.1   We have shown that $\mathbf{A}^N$ may have no smooth inverse. However it would be more interesting to provide an example in which it can be shown that $\mathbf{A}^N$ has no inverse at all. Looking at the representation of $\nabla \mathbf{A}^E_s({\mathbf {0}}, t_0)$ it is not difficult to see that $\nabla^2 \mathbf{A}_s^E({\mathbf {0}}, t_0)$ is odd in $x$ and $y$ and therefore, since the same holds also for $\nabla^2 \mathbf{A}_s^N({\mathbf {0}}, t_0)$, we get that $\nabla^2 \mathbf{A}_s^N({\mathbf {0}}, t) = {\mathbf {0}}$ for any $t > t_0$ and any $s \in [0,2\pi]$ and thus $\mathbf{A}^N$ is in fact invertible with a non-smooth inverse.