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\begin{document}
\title
{An Explicit Example of a Fast Dynamo \\
Generated by a Steady-state Solution of Euler's Equation}
\author{C.~Chicone}
\address{ Department of Mathematics\\
University of Missouri\\ Columbia, MO 65211}
\thanks{
The first author's research was supported by
the National Science
Foundation under the grant DMS-9303767;
the second author was
supported by the National Science
Foundation under the grant DMS-9400518}
\author{Y.~Latushkin}
\keywords {Kinematic dynamo, geodesic flow}
\email{carmen@chicone.cs.missouri.edu\\
mathyl@mizzou1.missouri.edu}
\subjclass {76W05, 58F99, 58G25}
\maketitle
\begin{abstract}
We give elementary and explicit arguments to show that the geodesic
flow on the unit tangent bundle of a two dimensional Riemannian
manifold with constant negative curvature provides an example of a
``fast'' dynamo for the magnetic kinematic dynamo equation.
\end{abstract}
\section{Introduction}
Consider a steady state solution $\bv$ of Euler's equation for
an ideal fluid with pressure $p$ on a three dimensional space
$X$:
\begin{equation}\lb{Euler}
\dfrac{d\bv}{dt}+(\bv,\bigtriangledown)\bv=-\grad p,
\quad \div \bv=0.
\end{equation}
The kinematic dynamo equation for the induction of a magnetic field $\bu$
by the fluid velocity field $\bv$
is given by
\begin{equation}\lb{Kge}
\dot\bu = \eps \triangle \bu + \curl{(\bv\times \bu)},
\quad \div \bu=0,
\end{equation}
where $\eps=R_m^{-1}$, and $R_m$ is the magnetic Reynolds number.
Let $\calD_{\eps}$ denote the linear operator defined by
\eqref{Kge} in $L_2(X,\mu)$ for a volume $\mu$ on $X$
invariant with respect to
the flow $\varphi^t$ induced by $\bv$:
\begin{equation}\lb{defA}
\calD_{\eps}\bu:=\eps\triangle \bu+ \curl{(\bv\times \bu)}.
\end{equation}
Also, note that $\calD_\eps$ depends on the choice of the
volume and the metric on $X$.
Define the spectral bound for $\calD_\eps$ by
$s_{\eps}: = \sup\{\Re\lambda :\lambda \in \sigma(\calD_{\eps})\}$.
If $s_{\eps} > 0$ for all sufficiently small positive $\eps$, then
$\bv$ is called a kinematic dynamo.
The dynamo is called ``fast'' if, in addition,
$\limsup_{\eps\to 0} s_{\eps} > 0$.
There exist several examples of fast dynamos, see
\cite{AZRS,BC,FO,Sow} and the literature cited therein. In particular,
by a deep result of M.~M.~Vishik~\cite{Vishik},
every Anosov flow with smooth foliations
on a compact three dimensional manifold produces a fast dynamo.
The proof of this result uses the full range of
hyperbolic theory together with
the theory of asymptotic expansions for pseudo differential operators.
These general methods do not seem to
provide explicit formulas for eigenvalues of $\calD_\eps$.
In the present note we give an elementary and
explicit example of an Anosov flow that is a steady state solution of
Euler's equation and that produces fast dynamo action. In fact,
let $M$ denote a smooth two dimensional compact Riemannian manifold without
boundary that has constant negative Gauss curvature $k$ and let
$X$ denote the total space of its unit tangent bundle.
The set $X$ is a compact three dimensional manifold.
For the appropriate natural choice of
Riemannian metric (Sasaki metric) and volume element on $X$,
the geodesic vector field $\bv$, the vector field on
$X$ that generates the geodesic flow,
is a steady state solution of Euler's equation with
$\grad p=0$. This vector field produces the fast dynamo. In fact,
we will prove the following theorem:
\begin{thm} \lb{main}
If $\bv$ generates the geodesic flow on
the total space of the unit tangent bundle
$X$ of a two dimensional compact Riemannian manifold $M$ with
constant negative curvature $k$ and if $\calD_{\eps}$ denotes the
associated dynamo operator
defined with respect to the Sasaki metric and volume
on $X$,
then, for each magnetic Reynolds number
$R_m>\sqrt{-k}$, the dynamo operator $\calD_{\eps}$
has the positive eigenvalue
\[
\lambda_{\eps}:=\dfrac{1}{2} \left[-\eps(1+k^2) +
\sqrt{\eps^2(1-k^2)^2-4k}\, \right],
\]
and
$
\limsup_{\eps \to 0} s_{\eps}\ge
\lim_{\eps \to 0} \lambda_{\eps}=\sqrt{-k} > 0
$.
In particular,
$\bv$ is a steady state solution of Euler's equation
that induces a fast kinematic dynamo.
\end{thm}
The proof of the theorem uses only elementary formulas from vector
calculus and certain commutation relations for Lie brackets obtained by
L.~Green in \cite{Green0}, see also the related papers
\cite{BChE,Ch0,Ch,ChE,Green}.
\section{Proof of the Theorem}
%\setcounter{equation}{0}
We recall the construction of the parallelization of $X$.
Let $g$ be the Riemannian metric on $M$
and let $T_mM$
denote the tangent plane at $m\in M$
with the scalar product $g_m(\cdot,\cdot)$
induced by $g$.
The total space of the unit tangent bundle of $M$ is the set $X$,
$$
X:=\{(m,\bw) : m\in M, \, \bw\in T_mM, \; g_m(\bw,\bw) = 1\}.
$$
This set has the natural structure of a
three dimensional compact manifold.
For $\gamma: \bbR\to M$, a curve in $M$, we let
$\dot{\gamma}(t)\in \calT_{\gamma(t)}M$
denote the tangent vector to $\gamma$ at $\gamma(t)$.
If $x=(m,\bw)\in X$, then there is a unique geodesic
$\gamma:\bbR\to M$,
such that $\gamma(0)=m$ and $\dot{\gamma}(0)=\bw$.
The geodesic flow $\varphi^t$ in $X$ is defined as follows:
If $x=(m,\bw) \in X$ and
$\gamma$ denotes the geodesic starting at $m$ with tangent vector $\bw$,
then $\varphi^tx=(\gamma(t), \dot{\gamma}(t)).$
The geodesic vector field $\bv$ is defined to be the generator of
the flow $\varphi^t$. We note that $\bv$ has no
zeros on $X$.
There are two additional
nonvanishing vector fields on $X$, that we denote by $\by$ and $\bz$, such
that for each $x\in X$ the set $\{\bv(x),\by(x),\bz(x)\}$
is a basis for the tangent space $T_xX$.
These vector fields are defined as follows.
Since $M$ is orientable, there is a well-defined notion of
positive rotation through a right angle.
That is, there is a well-defined map
$(m,\bw) \mapsto (m,\bw^{\perp})$
such that $g_m(m,\bw^{\perp})=0$ and
$g_m(\bw^{\perp},\bw^{\perp})=g_m(\bw,\bw)$.
We define $\by$ as the generator of the flow
$(m,\bw)\mapsto \varphi^t (m,\bw^{\perp})$ on $X$.
Also, let $r_m(t)\bw=e^{it} \bw$ give the
unit speed rotation of the unit circle in the positive sense relative
to the orientation in $T_mM$. Define $\bz$ as the generator of the
fiber rotation flow on $X$ given by $(m,\bw) \mapsto (m,r_m(t)\bw)$.
We use the notation $[\bu,\bv]=\bu\bv-\bv\bu$ for the
Lie bracket of the vector fields $\bu,\bv$ on $X$ and we let
$k$ denote the lift of the Gauss curvature of $M$ to $X$.
The following bracket relations, as reported in \cite{Green0}, are valid:
\begin{equation}\lb{brackets}
[\bv,\by]=k\bz, \quad [\bv,\bz]=-\by, \quad [\by,\bz] =\bv.
\end{equation}
There is a unique Riemannian metric on $X$ defined by declaring,
for each $x\in X$,
the set $\{\bv(x), \by(x), \bz(x)\}$ to be an orthonormal basis
for $T_x X$. We will denote the induced scalar product in $T_xX$
by $<\cdot,\cdot>$. It can be shown that
this Riemannian metric coincides with the Sasaki metric, see
\cite{Ch} for a definition, but this fact will not be needed here.
Let
$\mu$ denote the volume element on $X$ chosen so that
$\mu(\bv,\by,\bz)=1$
and let $\omega_\bu(\bxi)=<\bxi,\bu>$.
Using the standard (see, e.g., \cite{Marsden,ArnoldCM}) notations
$\d$ and $\i$ for the exterior and the interior
derivative,
we recall that
the maps $\bu\mapsto \omega_\bu$ and
$\bu\mapsto \i_\bu\mu$ are isomorphisms and that
the operations of
vector analysis on $X$ correspond to the exterior differential
calculus applied to the target differential forms.
For example,
the operators $\curl$ and $\div$, are defined by
\begin{equation}\lb{stform}
\d\omega_\bu=\i_{\curl \bu}\mu,\quad \d\i_{\bu}\mu = \div{\bu}\,\mu.
\end{equation}
\begin{lem}\lb{l1}
For the vector fields $\bv,\by,\bz$ on $X$ one has:
\begin{xalignat}{3}\lb{curlid}
\curl\bv & =-\bv,& \curl\by &=-\by,&\curl\bz &=-k\bz, \\
\div\bv & =0, & \div\by &=0, &\div \bz &=0. \notag
\end{xalignat}
\end{lem}
\begin{pf}
Let $\L_{\bxi}$ denote Lie differentiation in the direction $\bxi$
and recall the formula
(see, e.g., \cite[n$^\circ$6.4.11(ii)]{Marsden})
\begin{equation}\lb{domega}
\d\omega_{\bu}(\bxi,\ba)=\L_{\bxi} (\omega_u(\ba))
-\L_{\ba}(\omega_\bu(\bxi))-\omega_\bu([\bxi,\ba]).
\end{equation}
Since $\{\bv,\by,\bz\}$ is an orthonormal basis, an
application of this formula gives:
\begin{equation}\notag
\d\omega_\bv(\by,\bz)
=\L_{\by}(\omega_\bv(\bz))- \L_\bz(\omega_\bv(\by))-\omega_\bv([\by,\bz])
=-\omega_\bv(\bv)=-\|\bv\|^2=-1. \notag
\end{equation}
On the other hand, by \eqref{stform},
$\d\omega_\bv(\by,\bz)=\i_{\curl\bv}\mu(\by,\bz)=\mu (\curl{\bv},\by,\bz)$,
and, as a result, $\mu(\curl{\bv},\by,\bz)=-1$.
We conclude from the last formula that $\curl\bv=-\bv$.
Indeed, since $\bv,\by,\bz$ are of unit length and
mutually orthogonal with
respect to the scalar product $<\cdot,\cdot>$,
a similar application
of formula~\eqref{domega} yields:
$\d\omega_\bv(\bv,\bz)=0$ and
$\d\omega_\bv(\bv,\by)=0$. There are functions
$\alpha$, $\beta$ and $\delta$ such that
$\curl\bv=\alpha \bv+\beta \by+\delta \bz$.
Using the identities
\begin{align}\notag
-1& =\d\omega_\bv(\by,\bz)=\mu(\curl\bv,\by,\bz)
=\alpha\mu(\bv,\by,\bz)+\beta\mu(\by,\by,\bz)+
\delta\mu (\bz,\by,\bz),\\ \notag
0& =\d\omega_\bv(\bv,\bz)=\mu(\curl\bv,\bv,\bz)
=\alpha\mu(\bv,\bv,\bz)+\beta\mu(\by,\bv,\bz)+
\delta\mu (\bz,\bv,\bz),\\ \notag
0& =\d\omega_\bv(\bv,\by)
=\mu(\curl\bv,\bv,\by)=\alpha\mu(\bv,\bv,\by)
+\beta\mu(\by,\bv,\by) +\delta\mu(\bz,\bv,\by) \notag
\end{align}
we have $\alpha=-1$, $\beta=0$ and $\delta=0$, as required.
To prove $\div \bv=0$, we use \eqref{stform} and the formula
$\curl\bv=-\bv$:
\[
\div\bv\,\mu = \d\i_{\bv}\mu=\d\i_{-\curl\bv} \mu =
\d\d\omega_\bv=0.
\]
The remaining formulas in \eqref{curlid} are proved similarly.
For example, to prove
$\curl\bz=-k\bz$ we compute
$$
\d\omega_\bz(\bv,\by) =\L_\bv(\omega_\bz(\by)) -
\L_\by(\omega_\bz(\bv))-\omega_\bz([\bv,\by]) =- k \|\bz\|^2=-k.
$$
If $\curl\bz=\alpha \bv+\beta \by + \delta \bz$, then,
using the identities
\begin{align}\notag
-k& = \d\omega_\bz(\bv,\by)=\mu (\curl\bz,\bv,\by) =
\alpha\mu(\bv,\bv,\by) + \beta\mu(\by,\bv,\by)
+ \delta\mu (\bz,\bv,\by),\\ \notag
0& =\d\omega_\bz(\bz,\by)=\mu(\curl\bz,\bz,\by)
=\alpha\mu(\bv,\bz,\by)+\beta\mu(\by,\bz,\by) +
\delta\mu (\bz,\bz,\by),\\ \notag
0 & =\d\omega_\bz (\bz,\bv) = \mu (\curl\bz,\bz,\bv)
=\alpha\mu (\bv,\bz,\bv) + \beta\mu (\by,\bz,\bv) +
\delta\mu (\bz,\bz,\bv). \notag
\end{align}
we conclude $\alpha =0$, $\beta=0$ and $\delta =-k$.
\end{pf}
A coordinate free interpretation of the
Euler equation \eqref{Euler} that depends on the Riemannian
metric and Riemannian volume on $X$,
namely
\begin{equation}\lb{ManifoldEuler}
\dfrac{d\bv}{dt}-\bv\times\curl\bv+\grad\dfrac{\|\bv\|^2}{2}=-\grad
p, \quad \div \bv=0,
\end{equation}
is obtained from the identity
$$
(\bxi,\bigtriangledown)\bxi=\grad\dfrac{\bxi^2}{2} -\bxi\times\curl\bxi.
$$
Since, for the geodesic vector field, we have
$\|\bv\|^2=1$, $\div\bv=0$ and $\curl \bv=-\bv$, we see that
$\bv$ is a solution of \eqref{ManifoldEuler} with $\grad p=0$.
The Laplacian in the kinematic dynamo equation \eqref{Kge} must also
be defined on $X$. Using standard formulas
(see e.g., \cite[p.~605]{Marsden}),
the Laplace-Beltrami-deRham operator $\Delta$ that appears in \eqref{Kge}
may be defined as a differential operator on a divergence free vector
field $\bxi$ by
$\Delta \bxi =-\curl\curl \bxi$.
This operator depends on the choice of the
volume and the Riemannian metric. However, the operator
defined in this manner on Euclidean three dimensional
space with its usual volume agrees with the definition of
$\Delta$ as componentwise application of the usual Laplacian.
At any rate, using this definition and Lemma~\ref{l1},
we have
\begin{equation}\lb{delta}
\Delta \bv=-\bv, \quad \Delta \by=-\by, \quad \Delta \bz=-k^2\bz.
\end{equation}
If $\div\bxi=0$ and $\div \ba=0$, then a computation shows
$\curl{(\bxi\times \ba)} =-[\bxi,\ba]$. Thus,
the kinematic dynamo operator \eqref{defA}
is given by $\calD_{\eps}\bu=\eps\Delta \bu-[\bv,\bu]$.
Using this representation together with the formulas
\eqref{delta} and \eqref{brackets}, we compute
\[
\calD_{\eps}\by=-k\bz-\eps \by,\quad
\calD_{\eps} \bz=\by-\eps k^2 \bz.
\]
Observe that $\calD_{\eps}$ preserves the two
dimensional subspace
$$\calL:=\{a\by+b\bz: a\in \bbR,\; b\in\bbR\}$$
of divergence free vector fields in $L_2(X,\mu)$ spanned by
$\by$ and $\bz$ and that
the operator
$\calD_{\eps}$ on $\calL$
is represented by the matrix
\[
D=
\left[
\begin{array}{ccc}
-\eps & 1 \\
-k & -\eps k^2
\end{array}
\right].
\]
Each eigenvalue $\lambda_\eps$ of $D$ is an
eigenvalues of $\calD_{\eps}$. Hence, the spectral bound
$s_\eps$ of $\calD_\eps$ must satisfy the estimate
$$
s_{\eps} \ge \lambda_{\eps}:=
\dfrac{1}{2} \left[
-\eps(1+k^2)+\sqrt{\eps^2(1-k^2)^2-4k}\,
\right].
$$
A corresponding eigenfunction $\bu_\eps$ for $\calD_\eps$ is given by
$\bu_\eps=\by+(\eps+\lambda_\eps)\bz$. If, in
particular, $k=-1$, then $\lambda_\eps=-\eps+\sqrt{-k}$ and
$\bu_\eps=\by+\sqrt{-k}\bz$.
If $k<0$ and $\eps\in (0,1/\sqrt{-k})$, then
$\lambda_{\eps} > 0$, and $\bv$ induces a kinematic dynamo.
Moreover, since
\begin{equation}\lb{ineq}
\limsup_{\eps\to 0} s_{\eps} \ge \sqrt{-k} > 0,
\end{equation}
the dynamo action is fast.\qed
It can be shown that $\sigma(\calD_0)$ on $L_2(X,\mu)$
lies between the vertical lines:
$\Re\lambda=\pm\sqrt{-k}$. Then, in
accordance with \cite{CLM}, the spectrum of $\calD_0$ on the space of
divergence free vector fields is the strip $\{\lambda: |\text{ Re
}\lambda |\le\sqrt{-k}\}$. By a result in \cite{Vishik} (see also a
discussion in \cite{CLM}) one has $\limsup_{\eps\to 0} s_{\eps} \le
s_0$ for {\it any}, not necessarily geodesic, flow. Formula
\eqref{ineq} confirms a conjecture from \cite{CLM} that
$\limsup_{\eps\to 0} s_{\eps} = s_0$ for any flow.
For the case of negative nonconstant curvature
$k(x)\le k_0<0$ and small enough $\eps>0$ one can construct an
explicit $\bu_\eps\in L_2(X,\mu)$ with $\|\bu_\eps\|=1$
such that
\[\text{ Re}<\calD_\eps\bu_\eps\;,\; \bu_\eps >_{L_2}>0,
\quad \text{ and }\quad \limsup_{\eps\to 0}
\text{ Re}<\calD_\eps\bu_\eps\;,\; \bu_\eps >_{L_2}>0,\]
where $<\cdot\; ,\;\cdot>_{L_2}$ denotes
the scalar product in $L_2(X,\mu)$, induced by $<\cdot\; ,\;\cdot>$.
\newpage
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\end{document}