# Conjugate point criteria on the area-preserving diffeomorphism group

Stephen C. Preston

###### Abstract

This paper answers some questions about conjugate points along the geodesics
corresponding to steady 2D Euler flows, posed by a paper of Drivas-Misiołek-
Shi-Yoneda. We present a new sufficient criterion for the existence of
conjugate points, which improves on the criterion of Misiołek. It applies in
any rotational cell of a steady 2D Euler flow, and in case of rotational
symmetry it captures all known conjugate points. We give a general
construction of the surfaces that admit a steady fluid with given area form,
velocity profile, and vorticity profile, and from this we show how to detect
conjugate points in a single rotational cell of a steady flow. When the
velocity profile has a local extremum, the criterion becomes particularly
simple. Several examples are provided, and in an appendix we use the Misiołek
criterion to give some new examples of conjugate points along Kolmogorov flows
on the torus.

###### Contents

  1. 1 Introduction
  2. 2 Background
  3. 3 Polar coordinates in a steady fluid cell
  4. 4 Proofs of Theorem 1 and Corollary 2
  5. 5 Proofs of Corollaries 3, 4, and 5
  6. 6 Proof of Theorem 6
  7. 7 Examples
  8. A Appendix: The M-criterion for Kolmogorov flows on $\mathbb{T}^{2}$

##  1 Introduction

Suppose $M$ is a two-dimensional compact surface, possibly curved and possibly
with boundary, with a Riemannian metric inducing an area form $\mu$ and a
Poisson bracket $\\{\cdot,\cdot\\}$. If $f\colon M\to\mathbb{R}$ satisfies the
PDE

$\\{f,\Delta f\\}=0,$ (1)

then $f$ generates a steady solution of the 2D Euler equation for ideal
fluids, via $U=\operatorname{sgrad}f$. The velocity field $U$ in turn
generates a geodesic $\eta$ defined by

$\frac{\partial\eta}{\partial t}(t,p)=U(\eta(t,p)),\qquad\eta(0,p)=p,$ (2)

and $\eta(t)$ a curve in the group of area-preserving diffeomorphisms
$\mathcal{D}_{\mu}(M)$ with $\eta(0)=\mathrm{id}$. In this paper we are
interested in the question of whether there are conjugate points along this
geodesic, i.e., whether $\eta(T)$ is conjugate to $\eta(0)$ for some $T>0$.
This is closely related to the question of whether the geodesic is minimizing
between its endpoints, and is also related to the question of the stability of
particle trajectories.

Misiołek [12] was the first to find conjugate points in $\mathcal{D}_{\mu}(M)$
for the case $M=S^{2}$, along a rigid rotation (along with conjugate points on
the $3$-ball and the $3$-sphere), taking advantage of the fact that $S^{2}$,
$S^{3}$, and the spheres making up the $3$-ball all have conjugate points
along them. A more difficult problem is to find conjugate points along a fluid
geodesic when $M$ itself does not have conjugate points; this was solved soon
after by Misiołek as well [13], who found them along the geodesic which comes
from stream function $f(x,y)=\cos{6x}\cos{2y}$ on the torus $\mathbb{T}^{2}$.
The method he used in this paper generalizes to give a technique for finding
conjugate points along any steady flow, and is based on the well-known
criterion in Riemannian geometry [5]: if for some $T>0$ there is a vector
field $\mathbf{Y}\colon[0,T]\to\mathcal{D}_{\mu}(M)$ with
$\mathbf{Y}(0)=\mathbf{Y}(T)=0$ such that the index form

$I(\mathbf{Y},\mathbf{Y})=\int_{0}^{T}\left\langle\frac{D\mathbf{Y}}{dt},\frac{D\mathbf{Y}}{dt}\right\rangle+\langle\mathbf{R}(\mathbf{Y},\dot{\eta})\dot{\eta},\mathbf{Y}\rangle\,dt$
(3)

is negative, then there is a Jacobi field $J(t)$ vanishing at both $t=0$ and
$t=\tau$ for some $\tau<T$, and hence $\eta(\tau)$ is conjugate to $\eta(0)$.

By considering variation fields of the form $\mathbf{Y}(t)=Y(t)\circ\eta(t)$
with $Y(t)=\sin{(\tfrac{t}{T})}W$, where $W$ is a time-independent vector in
$T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$ (i.e., a divergence-free vector field on
$M$), Misiołek showed that negativity of the index form
$I(\mathbf{Y},\mathbf{Y})$ for sufficiently large $T$ is implied by positivity
of the following quantity:

$MC(W,W)=\int_{M}\langle\nabla_{[U,W]}U+\nabla_{U}[U,W],U\rangle\,d\mu.$ (4)

Given a particular steady solution $U$ of the 2D Euler equation (for example
$U=\operatorname{sgrad}f$ as above), finding a single divergence-free $W$
which makes $MC(W,W)>0$ is sufficient to prove that there are eventually
conjugate points along the geodesic $\eta$ which is the flow of $U$. This is
far easier than actually solving the Jacobi equation to find the conjugate
points explicitly. The quantity $MC(W,W)$ is called “Misiołek curvature” by
Tauchi and Yoneda (see [23]–[22]) and the condition $MC(W,W)>0$ called the
M-criterion (see [21]). This criterion has been successfully applied to find
conjugate points along many steady flows, and even along nonsteady Rossby-
Haurwitz waves by Benn [3].

Nonetheless it is quite far from being a complete characterization. One
drawback is that the test fields $Y$ are not necessarily related in any
obvious way to the given steady field $U$; in fact the easiest way to apply
the M-criterion is to guess the form of $Y$ based on some parameters and solve
for them numerically. A bigger drawback is that although the criterion applies
for certain rotationally symmetric surfaces like spheres with a bulge around
the equator, it does not apply to rigid rotation on the standard sphere, which
is the only case where we know all conjugate points and Jacobi fields
explicitly. (See Example 1 below for details.) We can compute that
$MC(W,W)\leq 0$ for every divergence-free field $W$, yet we can show by direct
construction that the rotation $\eta(\tau)$ is conjugate to the identity for
$\tau=\frac{n(n+1)\pi}{k}$ whenever $k,n\in\mathbb{N}$ with $k\leq n$.
Essentially the difference is that instead of choosing
$Y(t)=\sin{(\tfrac{t}{T})}\,W$, we choose
$Y(t)=\sin{(\tfrac{t}{T})}\,\eta(\alpha t)_{*}W$, where we push-forward the
velocity field by the Lagrangian flow for some constant $\alpha$.

It turns out that incorporating this additional drift term gives a much
stronger condition, which captures many more conjugate points along two-
dimensional steady flows. Moreover we obtain a simpler criterion which only
depends on a single function. The idea is that we can work things out fairly
well in the rotationally symmetric case, and the Jacobi field that generates
the earliest conjugate point will have a $\theta$-dependence of the form
$\cos{\theta}$; higher frequencies generate conjugate points happening later
in time (but the frequency needs to be nonzero to generate any conjugate
points). Even those steady flows which are not rotationally symmetric still
generically have closed trajectories, and split $M$ into cells where they
either rotate clockwise or counterclockwise. We will construct a variation
field supported in one of these cells, with a low-frequency oscillation term
and a time-dependent drift.

Now let us make this precise. We suppose $U$ is a steady solution of the Euler
equation with $U=\operatorname{sgrad}f$ for a stream function $f$. Suppose $f$
has a nondegenerate local minimum at a point $p_{0}\in M$ (i.e., its Hessian
is positive definite), and that $f$ has no other critical points and remains
negative in the interior of a cell $C$, while $f=0$ on the boundary $\partial
C$. (A stream function is only determined up to a constant, so this is no loss
of generality.) The basic examples we have in mind are a function $f$ that
depends solely on the radius on a rotationally symmetric manifold, and the
Kolmogorov flows generated by Laplacian eigenfunctions
$f(x,y)=-\cos{mx}\cos{ny}$ on the $2$-torus. Within the cell $C$, the flow
$\eta$ of $U$ consists of counterclockwise rotation along level sets of $f$
with finite period depending only on the value of $f$.

Figure 1: Level sets of the function $f(x,y)=\sin{x}\sin{y}$, the simplest
Kolmogorov flow on the torus. Red curves are negative values of $f$, while
blue curves are positive values. The fluid rotates in each of the four cells
independently, while the corner points are all hyperbolic fixed points.

We will show in Lemma 10 that in this situation, we can define a polar
coordinate system $(r,\theta)$ with $r\in(0,R)$ and $\theta\in[0,2\pi)$ such
that the stream function is given by $f=F(r)$, the steady velocity field is
given by $U=u(r)\,\partial_{\theta}$, and the area form is given by
$\mu=\varphi(r)\,dr\wedge d\theta$, for some functions $F$, $u$, and $\varphi$
satisfying $u(r)=F^{\prime}(r)/\varphi(r)$. Here the metric components are
given by

$ds^{2}=g_{11}(r,\theta)\,dr^{2}+2g_{12}(r,\theta)\,drd\theta+g_{22}(r,\theta)\,d\theta^{2},$

with no simplifications except that

$g_{11}g_{22}-g_{12}^{2}=\varphi(r)^{2}\qquad\text{and}\qquad\frac{\partial}{\partial\theta}\big{(}u(r)g_{12}(r,\theta)\big{)}=\frac{\partial}{\partial
r}\big{(}u(r)g_{22}(r,\theta)\big{)}-\varphi(r)\omega(r).$

In the case where the speed $\lvert U\rvert$ is constant along trajectories,
the manifold must be rotationally symmetric already, and these polar
coordinates will reduce to the standard ones.

We prove in Lemma 12 that the functions $F(r)$, $\varphi(r)$, and $G(r)$ may
be specified arbitrarily (subject to the right behavior at the origin $r=0$
and the requirements that $F^{\prime}$, $G$, and $\varphi$ are all positive
for $r>0$) together with a single arbitrary but sufficiently small function
$\zeta(r,\theta)$, and any such choice will generate a steady flow in a cell
on some curved surface. When the function $\zeta$ is identically zero, we
obtain rotationally symmetric examples. This principle can be used to make up
more interesting examples, especially since it can be quite difficult to
compute these quantities if the function and manifold are already given.

The main theorem of this paper is the following criterion for conjugate points
along the geodesic $\eta$.

######  Theorem 1.

Suppose we are given a solution $f$ of the steady Euler equation (1), with a
unique critical point in a cell $C$ which is a local minimum, with $f=0$ on
$\partial C$. Construct a polar coordinate chart $(r,\theta)$ as in Lemma 10.
Define $E,G\colon[0,R]\to[0,\infty)$ by the formulas

$E(r)=\frac{1}{2\pi}\int_{0}^{2\pi}g_{11}(r,\theta)\,d\theta,\qquad
G(r)=\frac{1}{2\pi}\int_{0}^{2\pi}g_{22}(r,\theta)\,d\theta.$ (5)

Write $\Delta f=\omega(r)$ for the vorticity.

If for some $\alpha\in\mathbb{R}$ and some $\xi\colon[0,R]\to\mathbb{R}$ such
that $\xi(0)=\xi(R)=0$, we have

$I=\int_{0}^{R}\frac{G(r)}{\varphi(r)}\left(\frac{d}{dr}\Big{[}\big{(}\alpha-u(r))\xi(r)\Big{]}\right)^{2}+\frac{E(r)}{\varphi(r)}\big{(}\alpha-u(r)\big{)}^{2}\xi(r)^{2}-\omega^{\prime}(r)\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\,dr<0,$
(6)

then $\eta(\tau)$ is conjugate to $\eta(0)=\mathrm{id}$ for some $\tau>0$.

The case where $f$ has a local maximum is identical and leads to a clockwise
flow; we have chosen this case only because it gives the more standard polar
coordinates.

Having chosen $\xi$, the quantity $I$ in (6) is a quadratic function
$I=A\alpha^{2}+2B\alpha+C$ of $\alpha$, with $A>0$, which can be made negative
for some $\alpha$ if and only if $B^{2}-AC>0$.

If we assume that $u^{\prime}(r)\neq 0$ in the cell (i.e., the velocity is
both positive and monotone), we obtain the following alternative forms of
Theorem 1. The condition (8) is sometimes easier than (6), while the condition
(10) is typically harder than either, but the latter is useful to compare to
previous results on the sign of the curvature of $\mathcal{D}_{\mu}(M)$, as
well as for concluding that the index is always positive in some situations
(see Section 6).

######  Corollary 2.

Suppose we have the same situation as in Theorem 1. Define
$v\colon[0,R]\to\mathbb{R}$ by

$v(r)=\frac{G^{\prime}(r)u(r)}{2\varphi(r)}.$ (7)

Then the index $I$ appearing in (6) can be written in the form

$I=\int_{0}^{R}\Big{[}\frac{\big{(}\alpha-u(r)\big{)}^{2}}{\varphi(r)}\big{(}G(r)\xi^{\prime}(r)^{2}+E(r)\xi(r)^{2}\big{)}-2v^{\prime}(r)\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\Big{]}\,dr.$
(8)

If in addition $u^{\prime}(r)$ is nowhere zero on $(0,R)$, define

$Q(r)=\frac{v^{\prime}(r)}{u^{\prime}(r)}.$ (9)

Then the index (8) may also be written in the form

$I=\int_{0}^{R}\big{(}\alpha-u(r)\big{)}^{2}\left[\frac{G(r)}{\varphi(r)}\Big{(}\xi^{\prime}(r)-\frac{Q(r)\varphi(r)}{G(r)}\,\xi(r)\Big{)}^{2}\\\
-\frac{\varphi(r)}{G(r)}\,M(r)\xi(r)^{2}\right]\,dr,$ (10)

where

$M(r):=\frac{G(r)}{\varphi(r)}\,Q^{\prime}(r)+Q(r)^{2}-\frac{E(r)G(r)}{\varphi(r)^{2}}.$
(11)

The condition on the function $M$ defined by (11) that

$M(r)>0\text{ for some $r\in(0,R)$}$ (12)

is sufficient to have some positive sectional curvature of
$\mathcal{D}_{\mu}(M)$ in a section containing $U$; see [16]. Existence of
such positive curvature is necessary but not sufficient for the existence of
conjugate points; Corollary 2 shows that one needs (12) to hold not merely at
one point but on a sufficiently large interval $(a,b)\subset(0,R)$.

If the manifold is rotationally symmetric, i.e., there is a polar coordinate
system with $ds^{2}=dr^{2}+\varphi(r)^{2}\,d\theta^{2}$ for some function
$\varphi(r)$ already, then any purely radial function $f=F(r)$ will satisfy
(1), and the polar coordinates we construct in Lemma 10 will reduce to those.
In that case we will have $E(r)=1$ and $G(r)=\varphi(r)^{2}$. Furthermore in
that case we do not need to assume $f$ has only one critical point in the cell
$C$, since the only purpose of that assumption is to construct the $\theta$
coordinate manually; instead we can work on the entire manifold $M$. This
situation occurs if and only if the speed $\lvert U\rvert$ is constant along
trajectories, as we will see in Remark 1.

We now demonstrate several consequences of Theorem 1. We begin with
“isochronal flows” on a rotationally symmetric manifold, as defined by Drivas
et al. [6], or in other words Killing fields of the rotationally symmetric
metric.

######  Corollary 3.

Suppose that the steady Euler velocity field is given by
$U=\partial_{\theta}$, i.e., $u(r)\equiv 1$, so that the field is isochronal
(i.e., all orbits have the same period $2\pi$).

Then there is a conjugate point along $\eta$ if and only if
$\frac{d}{dr}(G^{\prime}(r)/\varphi(r))$ is not identically zero on $[0,R]$.
In the rotationally symmetric case where $E(r)=1$ and $G(r)=\varphi(r)^{2}$,
this reduces to the condition that the Gaussian curvature is not identically
zero on $M$.

In particular Corollary 3 captures the conjugate points we know for rotations
on the $2$-sphere, as well as for any ellipsoid (see [21]). The only
rotationally symmetric situations in which we will not have a conjugate point
are the flat torus with $\varphi(r)\equiv 1$ and rigid rotations in a disc
with $\varphi(r)=r$. It is already known that the curvature operator is
nonnegative in the first case by Misiołek [12] and in the second case by the
author [16], so we could not have gotten conjugate points anyway.

Above we have considered the cases where $u^{\prime}$ is identically zero and
where $u^{\prime}$ is nowhere zero. It is also interesting to consider the
case where $u^{\prime}(r_{0})$ is zero at one point (while assuming for
nondegeneracy that $u^{\prime\prime}(r_{0})\neq 0$). In [16] we saw (in the
rotationally symmetric case) that this was sufficient to find some positive
sectional curvature in a plane containing $U$. To obtain a conjugate point one
needs a little more.

######  Corollary 4.

Consider again the situation of Theorem 1. Suppose that the velocity function
$u(r)$ has an isolated critical point at $r_{0}\in(0,R)$, with
$u^{\prime}(r_{0})=0$ and $u^{\prime\prime}(r_{0})\neq 0$. If

$-\frac{u(r_{0})}{\varphi(r_{0})u^{\prime\prime}(r_{0})}\frac{d}{dr}\Big{|}_{r=r_{0}}\left(\frac{G^{\prime}(r)}{2\varphi(r)}\right)>\frac{9}{16},$
(13)

then there is eventually a conjugate point along the geodesic $\eta$. In the
rotationally symmetric case where $G(r)=\varphi(r)^{2}$ and
$\varphi^{\prime\prime}(r)=-\kappa(r)\varphi(r)$ in terms of the Gaussian
curvature $\kappa$, condition (13) reduces to

$\frac{\kappa(r_{0})u(r_{0})}{u^{\prime\prime}(r_{0})}>\frac{9}{16}.$

As the reader might guess, the trick here is to choose $\alpha$ in the index
(8) to be equal to the critical value $u(r_{0})$, which makes
$(\alpha-u(r))\approx-\tfrac{1}{2}u^{\prime\prime}(r_{0})(r-r_{0})^{2}$. For
$r$ close to $r_{0}$, the first two positive terms will thus look like
$(r-r_{0})^{4}$ while the last one looks like $(r-r_{0})^{2}$, and thus it can
be made to dominate if the test function $\xi$ is supported in a small
neighborhood of $r_{0}$.

The velocity function $u(r)$ typically has a local maximum or minimum at
$r=0$, so we can get a similar criterion to Corollary 4 at $r=0$ by using the
same trick. Things work out differently however since $E(r)=O(1)$,
$\varphi(r)=O(r)$, and $G(r)=O(r^{2})$ near $r=0$.

######  Corollary 5.

Consider again the situation of Theorem 1. Suppose $u(0)\neq 0$,
$u^{\prime}(0)=0$, and $u^{\prime\prime}(0)\neq 0$. If

$3E(0)+12G^{\prime\prime}(0)<\frac{2u(0)}{u^{\prime\prime}(0)}\left(\frac{\phi^{\prime\prime\prime}(0)}{\phi^{\prime}(0)}-G^{iv}(0)\right).$
(14)

then there is eventually a conjugate point along the geodesic $\eta$. In the
rotationally symmetric case where $G(r)=\varphi(r)^{2}$ and $E(r)=1$, this
reduces to

$\frac{\kappa(0)u(0)}{u^{\prime\prime}(0)}>\frac{9}{4}.$ (15)

Finally we will use the results above to answer the three open questions posed
by Drivas, Misiołek, Shi, and Yoneda in [6].

  1. 1.

For what values of $(m,n)$ does the Kolmogorov stream function
$f(x,y)=-\cos{mx}\cos{ny}$ for $n$ and $m\geq n$ positive integers have a
conjugate point along its geodesic? This had been established by Misiołek in
[13] for the case $m=6$ and $n=2$, and extended in [6] to a condition
equivalent to $n\geq 2$ and $m\geq\sqrt{3}(n+2/n)$. Theorem 1 does not produce
conjugate points along Kolmogorov flows since we will show in Section 6 that
(10) is never satisfied. However we show that $(m,1)$ for $m\geq 2$, along
with $(2,2)$, $(3,2)$, and $(3,3)$ all have conjugate points along their flows
in the Appendix. We conjecture that all values of $(m,n)$ produce conjugate
points except possibly $(1,1)$, but at this point we cannot prove the full
result.

  2. 2.

Can any Arnold stable solution on a surface $M$ have conjugate points? Yes.
For simplicity take $u(r)=1$, so that $F^{\prime}(r)=\varphi(r)$ and
$\omega(r)=2\varphi^{\prime}(r)$. Then the ratio
$\omega^{\prime}(r)/F^{\prime}(r)$ is given by $-2\kappa(r)$, in terms of the
Gaussian curvature $\kappa$. On a disc with strictly negative curvature, the
Arnold stability condition is satisfied, and Corollary 3 yields a conjugate
point along the geodesic of rigid rotations. One could construct many other
examples using the criterion of Theorem 1. We give the details for a
nonisochronal flow in Section 7.

  3. 3.

Are vortices with constant strength on elliptical domains free of conjugate
points? Yes. In fact this is easy from the general index formula, which shows
that if the vorticity is constant, then the third term in the index formula
disappears (and what remains must be positive). See Proposition 1 below.

######  Theorem 6.

Suppose $m$ and $n$ are both positive integers and
$M=\mathbb{T}^{2}=(\mathbb{R}/2\pi\mathbb{Z})^{2}$ with the usual flat metric
$ds^{2}=dx^{2}+dy^{2}$. Let $f(x,y)=-\cos{mx}\cos{ny}$ be a Kolmogorov stream
function, satisfying $\Delta f=-(m^{2}+n^{2})f$ and thus (1). Consider the
cell $C$ defined by $\lvert x\rvert\leq\frac{\pi}{2m}$ and $\lvert
y\rvert\leq\frac{\pi}{2n}$, and define the radial coordinate $r$ by
$r=\sqrt{1-f^{2}}$. Then for every nontrivial radial function $\xi$ with
$\xi(0)=\xi(1)=0$, the index $I$ appearing in Theorem 1 is positive. Hence we
cannot detect conjugate points by considering variations supported in a single
cell.

Finally we discuss the three-dimensional case. Here it is much simpler to find
conjugate points: instead of constructing a function $g(t,r,\theta)$ with
$Y=\operatorname{sgrad}g$ a divergence-free vector field on $M$ and trying to
show that $I(Y,Y)<0$, it is sufficient to find a vector field
$Y(t,\eta(t,p_{0}))$ along a _single_ Lagrangian trajectory
$t\mapsto\eta(t,p_{0})$ such that the pointwise index form

$I_{p_{0}}(Y,Y)=\int_{0}^{T}\langle
Z(t)_{\eta(t,p_{0})},Z(t)_{\eta(t,p_{0})}\rangle+dU^{\flat}\big{(}Y(t)_{\eta(t,p_{0})},Z(t)_{\eta(t,p_{0})}\big{)}\,dt<0,\qquad
Z(t)=\frac{DY}{dt}-\nabla_{Y(t)}U.$

Given such a field along one trajectory, we can build a variation field
supported in a small neighborhood of it which will make the full index form
(3) negative on $\mathcal{D}_{\mu}(M)$. See [15] for the criterion and some
examples, along with [20] for an improved criterion that captures the
locations of these conjugate points. Examples in the axisymmetric case were
given in [24]. The fact that variations are easy to find in three dimensions
is a reflection of the fact that the Riemannian exponential map is not
Fredholm, which allows conjugate points to cluster along geodesics and have
infinite order; see [8]. The fact that we can easily find an infinite
dimensional space of variations (all of which have disjoint supports) means
that there are infinitely many conjugate points on a finite interval; the fact
that the exponential map is Fredholm in two dimensions prevents this (see
Misiołek [14]) and means that the energy-reducing variations must have
overlapping support and cannot be selected in a small neighborhood of any
point or trajectory. The only exceptions are Corollaries 4 and 5, where the
supports of energy-reducing variations are in a small neighborhood of the
fastest or slowest trajectory.

Here is a summary of the paper’s contents. In Section 2 we provide full
details on conjugate points along 2D fluid flows, including the simplification
of the index form in two dimensions along either steady or nonsteady flows. We
also describe the explicit Jacobi fields and conjugate points along rigid
rotations of the $2$-sphere, which inspires the main construction of the
paper. In Section 3 we describe the general construction of polar coordinates
in a cell. We also prove Lemma 12 which shows that any stream function $F(r)$,
area form coefficient $\varphi(r)$, and vorticity function $\omega(r)$ of the
radial coordinate can appear on some curved surface, and provides a
parameterization of all such surfaces. In Section 4 we prove our main Theorem
1 by considering simple explicit variations and the long-time limit. We also
prove Corollary 2 by showing how to derive formulas (72) and (10) from (6). In
Section 5 we prove Corollary 3 giving the necessary and sufficient condition
for an isochronal flow to have a conjugate point. We also prove Corollaries 4
and 5, which are closely related and give the only local criteria we have
along 2D flows. In Section 6 we work out the polar coordinate representation
for the Kolmogorov flows in terms of Jacobian elliptic functions, and prove
Theorem 6 which shows that the index form is positive. In Section 7 we work
out the details of four examples of nontrivial flows where the new criteria of
this paper are effective. Finally in the Appendix we use Misiołek’s criterion
from [13] to find additional Kolmogorov flows with conjugate points.

## Acknowledgements:

The author conducted this reesarch while partially funded by Simons Foundation
Collaboration Grant 318969. The author thanks Martin Bauer, Theo Drivas, and
Gerard Misiołek for helpful discussions, as well as the organizers of the
Workshop on Small Scale Dynamics in Fluid Motion at the Simons Center for
Geometry and Physics, where the work was completed.

##  2 Background

We will give a brief introduction to the geometry of volumorphisms and the
Euler equation as a geodesic on this group. For more details in the smooth
case, see Arnold-Khesin [2], along with the paper of Ebin-Marsden [7] where
the Sobolev manifold structure was discussed and Misiołek [12] where the basic
properties of the Jacobi equation, conjugate points, and the index form were
demonstrated. For what we need, we can assume everything is $C^{\infty}$,
though the rigorous approach requires only that the diffeomorphisms and
velocity fields are in $H^{s}$ for $s>\dim{M}/2+1$, and everything we do will
make sense in that context as well.

Suppose $M$ is a Riemannian manifold, possibly with boundary $\partial M$. The
Euler equation for a time -dependent vector field $U$, divergence-free and
tangent to the boundary, is given by

$\frac{\partial U}{\partial t}+U\cdot\nabla U=-\nabla p,\qquad U(0)=U_{0}.$
(16)

Here $p$ is a real-valued function determined up to a constant (so that
$\nabla p$ is uniquely determined) by

$\Delta p=-\operatorname{div}{(U\cdot\nabla U)},\qquad\langle\nabla
p,\nu\rangle=-\langle U\cdot\nabla U,\nu\rangle.$

We can write this more simply in terms of the projection $P$:

$P(X)=X+\nabla f,\qquad\Delta f=-\operatorname{div}{X},\quad\langle\nabla
f,\nu\rangle=-\langle X,\nu\rangle.$ (17)

For any vector field $X$ on $M$, we see that $P(X)$ is divergence-free and
tangent to $\partial M$. Hence equation (16) becomes

$\frac{\partial U}{\partial t}+P(\nabla_{U}U)=0,\qquad U(0)=U_{0}.$ (18)

The flow of $U$ is a volume-preserving diffeomorphism $\gamma(t)$ given by

$\frac{\partial\gamma}{\partial
t}(t,x)=U\big{(}t,\gamma(t,x)\big{)},\qquad\gamma(0,x)=x,$ (19)

and in terms of $\gamma$ the Euler equation becomes the second-order equation

$\frac{\partial^{2}\gamma}{\partial t^{2}}=-\nabla
p\circ\gamma,\qquad\gamma(0)=\mathrm{id},\quad\dot{\gamma}(0)=U_{0}.$ (20)

The map $\eta$ satisfies, for each fixed time, the volume-preserving condition
$\operatorname{Jac}(\eta)\equiv 1$, and maps the boundary of $M$ to itself by
the conditions on $U$. Thus it can be viewed as a curve in the infinite-
dimensional space $\mathcal{D}_{\mu}(M)$ of volume-preserving diffeomorphisms,
or volumorphisms. Under composition, this is a group. We now consider the
manifold structure on it.

Elements of the tangent space $T_{\eta}\mathcal{D}_{\mu}(M)$ at a volumorphism
$\eta\in\mathcal{D}_{\mu}(M)$ are given by right-translations
$\mathbf{Y}=Y\circ\eta$, where $Y\in T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$ is a
divergence-free vector field tangent to $\partial M$. The kinetic energy
defines a Riemannian metric on $\mathcal{D}_{\mu}(M)$ by the formula

$\langle\mathbf{Y},\mathbf{Y}\rangle_{\eta}=\int_{M}\langle
Y\circ\eta,Y\circ\eta\rangle\,d\mu=\int_{M}\langle Y,Y\rangle\,d\mu,$

the last formula representing right-invariance of the metric and following
from the change of variables formula and $\operatorname{Jac}(\eta)\equiv 1$.
Along a curve $\eta(t)$, a time-dependent vector field $\mathbf{Y}(t)$ along
$\eta(t)$ is similarly given by $\mathbf{Y}(t)=Y(t)\circ\eta(t)$ for each $t$.
Right-invariance of the Riemannian metric leads to right-invariance of the
covariant derivative and curvature. The covariant derivative of $\mathbf{Y}$
along $\eta$ is given by

$\frac{D\mathbf{Y}}{dt}=\left(\frac{\partial Y}{\partial
t}+P(\nabla_{V(t)}Y(t))\right)\circ\eta(t),$ (21)

where $V$ is the right-translated tangent vector of $\eta$, i.e.,
$\frac{d\eta}{dt}=V\circ\eta$, and $\nabla_{V}Y$ is the covariant derivative
computed pointwise in $M$.

The Riemannian curvature tensor along the curve $\eta$ is given by

$\mathbf{R}(\mathbf{Y},\dot{\eta})\dot{\eta}=\Big{(}P(\nabla_{Y}P(\nabla_{V}V)-\nabla_{V}P(\nabla_{Y}V)-\nabla_{[Y,V]}V)\Big{)}\circ\eta,$
(22)

again using the right-translations $\dot{\eta}=V\circ\eta$ and
$\mathbf{Y}=Y\circ\eta$. Its importance comes from the equation in variations:
if $\eta(t,\delta)$ is a family of geodesics depending on a parameter
$\delta$, with $\eta(t,0)=\gamma(t)$, then the variation field

$\mathbf{J}(t)=\frac{\partial\eta}{\partial\delta}\Big{|}_{\delta=0}$

is called a Jacobi field, and satisfies the equation

$\frac{D^{2}\mathbf{J}}{dt^{2}}+\mathbf{R}(\mathbf{J},\dot{\gamma})\dot{\gamma}=0.$
(23)

This is the linearization of the geodesic equation. If there is a time $T>0$
such that $\mathbf{J}(0)=0$ and $\mathbf{J}(T)=0$, then we say that there is a
monoconjugate point along $\gamma$. The length of the geodesic is
infinitesimally minimizing up to the first monoconjugate point, and ceases to
be minimizing after that. Jacobi fields vanishing at two points represent
stable perturbations of the particle paths, which return (to first order) to
where they would have been without the perturbation. By right-invariance, if
$\mathbf{J}(t)=J(t)\circ\eta(t)$ for $J(t)\in
T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$, and $\dot{\gamma}(t)=U(t)\circ\eta(t)$,
then the Jacobi equation (23) may be rewritten as an equation on the Lie
algebra:

$\left(\frac{\partial}{\partial
t}+\nabla_{U(t)}\right)^{2}J(t)+\mathbf{R}\big{(}J(t),U(t)\big{)}U(t)=0.$ (24)

The fact that the geodesic equation (20) splits into the two decoupled
equations (16) is a consequence of the right-invariance of the metric. It has
the further consequence that the Jacobi equation (23) splits into decoupled
equations: the system is given by

$\frac{\partial J}{\partial t}+\nabla_{U}J-\nabla_{J}U=K,\qquad\frac{\partial
K}{\partial t}+P(\nabla_{U}K+\nabla_{K}U)=0,$ (25)

where we view the first equation as defining $K$. The second equation is the
linearized Euler equation, which is used to study perturbations of the
velocity field itself without looking at the perturbed particle paths. Solving
the Jacobi equation in the form (24) is rather complicated due to the fact
that the curvature operator (22) is difficult to compute explicitly, due to
the projections which require solution of the nonlocal equation (17). Even the
decoupled version (25) is difficult to find explicit solutions for in general
due to those same projections. See [17] for general discussion on this point
and some simple cases. The following example is sketched in [18], but we do
not know if the full solution has ever appeared in print.

######  Example 1.

On a simply-connected surface $M$, we may write $J=\operatorname{sgrad}{g}$
and $K=\operatorname{sgrad}{h}$, for some time-dependent functions $g$ and $h$
on $M$. Equations (25) then become the system

$\frac{\partial g}{\partial t}+\\{f,g\\}=h,\qquad\frac{\partial\Delta
h}{\partial t}+\\{f,\Delta h\\}+\\{h,\Delta f\\}=0,$ (26)

where $U=\operatorname{sgrad}f$. See [17] for details.

Now suppose $M$ is the round sphere, with spherical polar coordinates giving
the metric $ds^{2}=dr^{2}+\sin^{2}{r}d\theta$ and stream function
$f=-\cos{r}$. Then the steady Euler velocity field is given by
$U=\partial_{\theta}$, and the vorticity is given by $\Delta f=2f$. The system
(26) simplifies in this case to the system

$\frac{\partial g}{\partial t}+\frac{\partial
g}{\partial\theta}=h,\qquad\frac{\partial\Delta h}{\partial
t}+\frac{\partial\Delta h}{\partial\theta}+2\,\frac{\partial
h}{\partial\theta}=0.$ (27)

If $P^{k}_{n}$ denotes the Legendre polynomial for positive integers $k$ and
$n$, then the general solution of (27) with $g(0,r,\theta)=0$ is given by

$g(t,r,\theta)=\sum_{n=1}^{\infty}c_{n}tP_{n}(\cos{r})\\\
+\sum_{n=1}^{\infty}\sum_{k=1}^{n}\sin{\left(\frac{kt}{n(n+1)}\right)}\,P^{k}_{n}(\cos{r})\left(a_{nk}\cos{k\left(\theta-t+\frac{t}{n(n+1)}\right)}+b_{nk}\sin{k\left(\theta-t+\frac{t}{n(n+1)}\right)}\right),$

for arbitrary coefficients $a_{nk}$, $b_{nk}$, $c_{n}$. This can be checked by
plugging in, based on the fact that

$\Delta\big{(}P^{k}_{n}(\cos{r})\cos{k\theta}\big{)}=-n(n+1)P^{k}_{n}(\cos{r})\cos{k\theta}.$

Setting $c_{n}=0$ since those Jacobi fields grow linearly in time, we find
that conjugate points happen at times $\tau=\frac{\pi n(n+1)}{k}$ for positive
integers $k,n$ with $k\leq n$. In addition the basic Jacobi field stream
functions are all of the form

$g(t,r,\theta)=\sin{\left(\frac{t}{T}\right)}\xi(r)\cos{k(\theta-\alpha t)}$

for $\alpha=1-\frac{1}{n(n+1)}$. In particular $\alpha$ can _never_ be zero.
See Figure 2.

$t=0$ $t=\pi$ $t=2\pi$ $t=3\pi$

Figure 2: The transported Jacobi field stream function $g(t,r,\theta+t)$,
dragged by the uniform rotation, when $k=2$ and $n=2$, given by
$g(t,r,\theta+t)=\sin{\frac{t}{3}}\sin^{2}{r}\sin{(2\theta+\frac{t}{3})}$. As
$t$ goes from $0$ to $\tau=3\pi$, we get a conjugate point as $g$ expands and
shrinks, while rotating a quarter-turn around the sphere due to the drift.

An easier way to detect conjugate points, without explicitly finding the
Jacobi fields, is to use the index form. Multiplying the Jacobi equation (23)
by $\mathbf{J}(t)$ and integrating by parts from $t=0$ to $t=T$ gives the
formula

$I(\mathbf{J},\mathbf{J})=\int_{0}^{T}\left\langle\frac{D\mathbf{J}}{dt},\frac{D\mathbf{J}}{dt}\right\rangle_{L^{2}(M)}-\langle\mathbf{R}(\mathbf{J},\dot{\gamma})\dot{\gamma},\mathbf{J}\rangle_{L^{2}(M)}\,dt=0$

if $\mathbf{J}$ is a Jacobi field vanishing at $t=0$ and $t=T$. This index
makes sense for any vector field along $\gamma$, not just a Jacobi field, and
it can be shown that we have $I(\mathbf{Y},\mathbf{Y})=0$ for some
$\mathbf{Y}$ satisfying $\mathbf{Y}(0)=\mathbf{Y}(T)=0$ if and only if
$\mathbf{Y}$ is a Jacobi field. Furthermore if $I(\mathbf{Y},\mathbf{Y})<0$
for some such field $\mathbf{Y}$, then there is a Jacobi field $\mathbf{J}$
along $\gamma$ vanishing at $t=0$ and for $t=\tau$ for some $\tau<T$.
Establishing negativity of the index form for some vector field $\mathbf{Y}$
is thus an effective way to show there must be a conjugate point without
actually having to find it.

Since the metric, covariant derivative, and Riemann curvature tensor are all
right-invariant, the index form is as well, and we can write (using the
formulas (21) and (22))

$I(Y,Y)=\int_{0}^{T}\int_{M}\left\langle\frac{\partial Y}{\partial
t}+P(\nabla_{U}Y),\frac{\partial Y}{\partial t}+P(\nabla_{U}Y)\right\rangle\\\
-\langle
P\big{(}\nabla_{Y}P(\nabla_{U}U)\big{)}-P\big{(}\nabla_{U}P(\nabla_{Y}U)\big{)}+P(\nabla_{[U,Y]}U),Y\rangle\,d\mu\,dt.$
(28)

In fact it turns out this formula simplifies quite a lot, and in fact the
divergence-free projections $P$ do not need to be computed at all. The
following version of the formula is similar to one that appears in the
author’s work [15]. (Note that we do this slightly differently from Misiołek
[13] who first assumes that $U$ is a steady solution of the Euler equation.)

######  Lemma 7.

Suppose $U$ solves the Euler equation (18) on $[0,T]$. For a time-dependent
family of divergence-free vector fields $Y(t)$ on $M$, tangent to the boundary
of $M$, and vanishing at $t=0$ and $t=T$, the index form (28) takes the form

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z(t),Z(t)\rangle+dU(t)^{\flat}\big{(}Y(t),Z(t)\big{)}\,d\mu\,dt,\qquad
Z(t):=\dot{Y}(t)+[U(t),Y(t)].$ (29)

###### Proof.

First express $\tfrac{\partial Y}{\partial t}=Z-[U,Y]$ where $Z$ is defined in
terms of $Y$ by (29). Then the covariant derivative of $Y$ is

$\frac{\partial Y}{\partial
t}+P(\nabla_{U}Y)=Z+P(\nabla_{U}Y)-[U,Y]=Z+P(\nabla_{Y}U),$ (30)

using the fact that

$P(\nabla_{U}Y)-P(\nabla_{Y}U)=P(\nabla_{U}Y-\nabla_{Y}U)=P([U,Y])=[U,Y],$

since the Lie bracket of two vector fields in
$T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$ is also in it by the Lie group property.
Observe that $Z(t)$ must also be in $T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$.

Using formula (30), the index form becomes

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z+P(\nabla_{Y}U),Z+P(\nabla_{Y}U)\rangle-\langle
P\big{(}\nabla_{Y}P(\nabla_{U}U)\big{)},Y\rangle\\\ +\langle
P\big{(}\nabla_{U}P(\nabla_{Y}U)\big{)},Y\rangle-\langle
P(\nabla_{[U,Y]}U),Y\rangle\,d\mu\,dt.$ (31)

To simplify further, we use the fact that if $V$, $W$, and $X$ are any fields
in $T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$, then we have

$\int_{M}\langle
P(\nabla_{V}W),X\rangle\,d\mu=\int_{M}\langle\nabla_{V}W,X\rangle\,d\mu,$ (32)

since $P$ is the $L^{2}$ orthogonal projection onto
$T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$ and $X$ is already in that subspace.
Thus the index (31) becomes

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z,Z\rangle+2\langle\nabla_{Y}U,Z\rangle+\langle
P(\nabla_{Y}U),\nabla_{Y}U\rangle-\langle\nabla_{Y}P(\nabla_{U}U),Y\rangle\\\
+\langle\nabla_{U}P(\nabla_{Y}U),Y\rangle-\langle\nabla_{[U,Y]}U,Y\rangle\,d\mu\,dt.$
(33)

We next use integration by parts, based on the fact that $\int_{M}\langle
X,\nabla f\rangle\,d\mu=0$ for any function $f$ and any element $X\in
T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$. This implies that if $V,W,X\in
T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$, then

$\begin{split}\int_{M}\langle\nabla_{V}W,X\rangle\,d\mu&=\int_{M}\langle
V,\nabla\langle W,X\rangle\rangle\,d\mu-\int_{M}\langle
W,\nabla_{V}X\rangle\,d\mu\\\ &=-\int_{M}\langle
W,\nabla_{V}X\rangle\,d\mu.\end{split}$ (34)

Using this in (33) gives

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z,Z\rangle+2\langle\nabla_{Y}U,Z\rangle+\langle
P(\nabla_{Y}U),\nabla_{Y}U\rangle\\\ +\langle
P(\nabla_{U}U),\nabla_{Y}Y\rangle-\langle
P(\nabla_{Y}U),\nabla_{U}Y\rangle-\langle\nabla_{[U,Y]}U,Y\rangle\,d\mu\,dt.$
(35)

Note that we can combine the two terms containing $P(\nabla_{Y}U)$ to get

$\int_{M}\langle P(\nabla_{Y}U),\nabla_{Y}U\rangle-\langle
P(\nabla_{Y}U),\nabla_{U}Y\rangle\,d\mu=-\int_{M}\langle
P(\nabla_{Y}U),[U,Y]\rangle\,d\mu=-\int_{M}\langle\nabla_{Y}U,[U,Y]\rangle,$

using (32) again. In addition the fact that $U$ solves the Euler equation
means that $P(\nabla_{U}U)=-\tfrac{\partial U}{\partial t}$. With these two
facts in mind, the index form (35) now becomes

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z,Z\rangle+2\langle\nabla_{Y}U,Z\rangle-\langle\nabla_{Y}U,[U,Y]\rangle-\langle\tfrac{\partial
U}{\partial t},\nabla_{Y}Y\rangle-\langle\nabla_{[U,Y]}U,Y\rangle\,d\mu\,dt.$
(36)

Integrating by parts in time, using that $Y(0)=Y(T)=0$, gives for the $U_{t}$
term that

$\displaystyle-\int_{0}^{T}\int_{M}\langle\tfrac{\partial U}{\partial
t},\nabla_{Y}Y\rangle\,d\mu\,dt$ $\displaystyle=\int_{0}^{T}\int_{M}\langle
U,\nabla_{\tfrac{\partial Y}{\partial t}}Y\rangle+\langle
U,\nabla_{Y}\tfrac{\partial Y}{\partial t}\rangle\,d\mu\,dt$
$\displaystyle=\int_{0}^{T}\int_{M}\langle U,\nabla_{Z-[U,Y]}Y\rangle+\langle
U,\nabla_{Y}(Z-[U,Y])\,d\mu\,dt$
$\displaystyle=\int_{0}^{T}\int_{M}\langle\nabla_{[U,Y]}U,Y\rangle-\langle\nabla_{Z}U,Y\rangle+\langle\nabla_{Y}U,[U,Y]\rangle-\langle\nabla_{Y}U,Z\rangle\,d\mu\,dt.$

Inserting this into (36) gives, after some cancellations, that

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle
Z,Z\rangle+\langle\nabla_{Y}U,Z\rangle-\langle\nabla_{Z}U,Y\rangle\,d\mu\,dt.$

This becomes (29) upon realizing that Cartan’s formula gives

$dU^{\flat}(Y,Z)=\langle\nabla_{Y}U,Z\rangle-\langle\nabla_{Z}U,Y\rangle.$
(37)

∎

In the special case where $U$ is independent of time, Misiołek [13] obtained a
simple criterion for existence of conjugate points, which we reproduce below.

######  Theorem 8 (Misiołek).

Suppose $U$ is a steady (time-independent) solution of the Euler equation
(16). If there is a $W\in T_{\mathrm{id}}\mathcal{D}_{\mu}(M)$ such that

$MC(W,W):=\frac{\langle\nabla_{[U,W]}U+\nabla_{U}[U,W],W\rangle_{L^{2}(M)}}{\langle
W,W\rangle_{L^{2}(M)}}>0,$ (38)

then there is eventually a conjugate point along the geodesic $\gamma$ which
is the flow of $U$, i.e., for some $\tau>0$ there is a Jacobi field
$\mathbf{J}(t)$ along $\gamma$ such that $\mathbf{J}(0)=\mathbf{J}(\tau)=0$.

###### Proof.

For $T>0$ we take a test field of the form

$Y(t,x)=\sin{\left(\frac{t}{T}\right)}\,W(x),\qquad\text{where }W\in
T_{\mathrm{id}}\mathcal{D}_{\mu}(M).$ (39)

Obviously $Y(0)=Y(\pi T)=0$. Then $Z$ from (29) becomes

$Z(t)=\frac{1}{T}\cos{\left(\frac{t}{T}\right)}\,W+\sin{\left(\frac{t}{T}\right)}\,[U,W].$

Performing the time integration first, we easily get

$\int_{0}^{\pi T}\langle Z,Z\rangle+dU^{\flat}(Y,Z)\,dt=\frac{\pi
T}{2}\Big{(}\langle[U,W],[U,W]\rangle+dU^{\flat}\big{(}W,[U,W]\big{)}+\frac{1}{T^{2}}\langle
W,W\rangle\Big{)}.$

Thus we compute in the limit as $T\to\infty$ that

$\lim_{T\to\infty}\frac{2}{\pi
T}I(Y,Y)=\int_{M}\langle[U,W],[U,W]\rangle+dU^{\flat}\big{(}W,[U,W]\big{)}\,d\mu.$
(40)

If the right side is negative for some $W$, then for sufficiently large $T$
the index form $I(Y,Y)$ will be negative, and hence there will be a conjugate
point location occurring at some $\tau<\pi T$.

The right side of (40) can be written using the Cartan formula (37) again to
get

$\displaystyle\int_{M}\langle[U,W],[U,W]\rangle+dU^{\flat}\big{(}W,[U,W]\big{)}\,d\mu$
$\displaystyle=\int_{M}\langle[U,W],[U,W]\rangle+\langle\nabla_{W}U,[U,W]\rangle-\langle\nabla_{[U,W]}U,W\rangle\,d\mu$
$\displaystyle=\int_{M}\langle\nabla_{U}W,[U,W]\rangle-\langle\nabla_{[U,W]}U,W\rangle\,d\mu$
$\displaystyle=-\langle W,W\rangle_{L^{2}(M)}MC(W,W),$

using the integration-by-parts formula (34) in the last line. Thus negativity
of (40) is equivalent to positivity of $MC(W,W)$. ∎

The expression $MC(W,W)$ is called the _Misiołek curvature_ of $W$, and
Theorem 8 is called the M-criterion in [21]–[21].

Now let us specify to the two-dimensional case, not necessarily assuming that
$U$ is steady (though in our examples it will be). To make things simpler, we
assume that either $M$ is simply-connected or that at least the support of $Y$
is in a simply-connected cell (which will happen later when
$M=\mathbb{T}^{2}$). In this case a divergence-free vector field $Y$ can be
expressed as the skew-gradient of a stream function $g$, defined in terms of
the area form $\mu$ so that

$\mu(X,\operatorname{sgrad}g)=dg(X)\text{ for every vector field $X$ on $M$.}$

Since $Y$ must be tangent to the boundary of $M$, we can show that $g$ must be
constant on the boundary of $M$, and we can choose this constant to be zero.
Lemma 7 then becomes the following.

######  Corollary 9.

Suppose $M$ is a simply-connected surface (or portion of a larger surface),
and that $U=\operatorname{sgrad}f$ solves the Euler equation (18) on $[0,T]$.
For a time-dependent family of functions $g(t)$ on $M$, vanishing on the
boundary of $M$ and at $t=0$ and $t=T$, the index form (29) for
$Y=\operatorname{sgrad}g$ becomes

$I(Y,Y)=\int_{0}^{T}\int_{M}\langle\nabla h,\nabla h\rangle+\Delta
f\\{g,h\\}\,d\mu\,dt,\qquad h:=\frac{\partial g}{\partial t}+\\{f,g\\}.$ (41)

###### Proof.

See [17] for details of these computations. Expressing
$U=\operatorname{sgrad}f$, the vorticity $2$-form becomes $dU^{\flat}=\Delta
f\,\mu$ in terms of the usual Laplacian on functions. With
$Y=\operatorname{sgrad}g$ and $Z=\operatorname{sgrad}h$, the formula (29)
relating $Y$ and $Z$ can be written in terms of the stream functions as

$g_{t}+\\{f,g\\}=h$ (42)

where the Poisson bracket is defined in terms of the area form $\mu$ by the
formula

$df\wedge dg=\\{f,g\\}\,\mu.$

This follows from the fact that
$[\operatorname{sgrad}g,\operatorname{sgrad}g]=\operatorname{sgrad}\\{g,h\\}$.

We then have

$dU^{\flat}(\operatorname{sgrad}g,\operatorname{sgrad}h)\,\mu=\Delta
f\,\mu(\operatorname{sgrad}g,\operatorname{sgrad}h)\,\mu=(\Delta f)dg\wedge
dh=(\Delta f)\\{g,h\\}\,\mu,$

which gives the second term. The first term comes from the fact that if $\mu$
comes from a Riemannian metric, then
$\langle\operatorname{sgrad}h,\operatorname{sgrad}h\rangle=\lvert\nabla
h\rvert^{2}$. ∎

Corollary 9 makes it easy to answer the third question posed by Drivas et al.
[6]: if the velocity field has constant vorticity, there can be no conjugate
points. A special case of this is the rigid rotation of a disc by
$U=\partial_{\theta}$, where the vorticity function is
$\operatorname{curl}{U}=2$, for which the absence of conjugate points (or
indeed any positive curvature) is known [16].

######  Proposition 1.

If $M$ is simply connected and $U=\operatorname{sgrad}f$ where $\Delta f$ is
constant on $M$ and $f$ is constant on $\partial M$, then $U$ is a steady
solution of the Euler equation on $M$, and there are no conjugate points along
the geodesic $\gamma$ which is the flow of $U$.

###### Proof.

The curl of the Euler equation (16) is given by

$\frac{\partial\Delta f}{\partial t}+\\{f,\Delta f\\}=0,$

so that if $\Delta f$ is constant in both space and time, it solves this
equation. Note that $M$ must have a boundary in order for any nonconstant
function to have constant nonzero Laplacian, since on a compact manifold we
have $\int_{M}\Delta f\,d\mu=0$.

Writing $Y=\operatorname{sgrad}g$ and $Z=\operatorname{sgrad}h$ as in
Corollary 9, we must have that $g$ is constant on the boundary. Since
$h=g_{t}+\\{f,g\\}$ and $f$ is also constant on the boundary, we know that
$df$ and $dg$ are collinear, so that $df\wedge dg=0$ and thus $\\{f,g\\}=0$.
So $h$ is also constant on the boundary.

The second term in (41) is now given by

$\int_{M}\Delta f\\{g,h\\}\,d\mu=\Delta f\int_{M}dg\wedge dh=\Delta
f\int_{\partial M}gdh,$

and this is zero since $h$ is constant on $\partial M$ for each time $t$.
Therefore only the positive-definite term remains in (41), so $I(Y,Y)>0$ for
every nontrivial field $Y$, and there can be no conjugate points along
$\gamma$. ∎

If $\omega=\Delta f$ is piecewise constant instead of constant on all of $M$,
the situation changes. Consider for example solutions of the Euler equation
with $\omega=\operatorname{curl}{U}$ in $L^{\infty}(M)$. These exist as weak
solutions of the Euler equation by a famous result of Yudovich [25]. These are
not in $H^{s}$ for $s>2$, and thus the results of Ebin-Marsden [7] and [12] do
not apply; in particular there may not be a geodesic which integrates the
velocity field, and the Jacobi equation may not be a reasonable ODE along the
corresponding weak geodesic. Nonetheless one might expect to be able to make
sense of a vorticity function that is very close to piecewise-constant on
patches in a limiting sense if the index form still makes sense there. For
this purpose if we suppose that $\omega$ is given by a nonzero constant
$\omega_{i}$ on each simply connected open set $O_{i}\subset M$ and zero
otherwise, then the second term of the index form (41) would become

$\int_{M}\Delta f\\{g,h\\}\,d\mu=\sum_{i}\omega_{i}\int_{O_{i}}dg\wedge
dh=-\sum_{i}\omega_{i}\int_{\partial O_{i}}h\,dg,$

and it is conceivable that this can be made negative for some choice of $g$.
For example one could consider a Kirchoff ellipse, where there is a single
patch of nonzero vorticity in an ellipse, which must then rotate at a constant
speed. (See for example Majda-Bertozzi [11] for details.) This would take us
too far afield here, but would be very interesting to study.

##  3 Polar coordinates in a steady fluid cell

We make the following assumptions: there is an open cell $C\subset M$ for
which the closure $\overline{C}$ is compact and simply connected, and in which
a steady solution $U$ of the Euler equation $P(\nabla_{U}U)=0$ is expressed by
$U=\operatorname{sgrad}f$. We assume that $f$ has a local minimum on $C$ at
some unique point $p\in C$, that $\nabla^{2}f$ is nondegenerate at $p$, and
that $f=0$ on $\partial C$, and that $f$ has no critical points in
$\overline{C}$ except for $p$ and points in $\partial C$. (In particular $f$
is a Morse function on $C$ and is always negative.) This is no loss of
generality: adding a constant to $f$ does not change the velocity field $U$,
and flipping the sign of $f$ would merely reverse the direction of the flow.
The two main examples we have in mind are rotationally invariant functions
$f(r)$ on rotationally symmetric surfaces with metric
$ds^{2}=dr^{2}+\varphi(r)^{2}\,d\theta^{2}$, and the function
$f(x,y)=-\cos{mx}\cos{ny}$ on the flat torus $\mathbb{T}^{2}$, where in the
latter case the cell is
$C=(-\tfrac{\pi}{2m},\tfrac{\pi}{2m})\times(-\tfrac{\pi}{2n},\tfrac{\pi}{2n})$.

Since there are no other critical points, all level sets of $f$ except the
singleton $\\{p\\}$ and the boundary $\partial C$ are one-dimensional compact
manifolds, and hence are circles. The velocity field $U$ is nowhere zero on
these level sets, since $\lvert U\rvert=\lvert\nabla f\rvert$ is nowhere zero.
By our choice of sign of $f$, these flows are all counterclockwise. Let
$\gamma_{t}$ be the flow of $U$ defined by (2). Choose a curve
$c\colon[0,R]\to M$ such that $c(0)=p$ and $c^{\prime}(s)\neq 0$ for all
$s\in[0,R]$. Then

######  Lemma 10.

Suppose $f\colon M\to\mathbb{R}$ satisfies the steady Euler equation
$\\{f,\Delta f\\}=0$, and let $U=\operatorname{sgrad}f$. Suppose $f$ has a
nondegenerate local minimum at $p_{0}\in M$, and no other critical points in a
simply connected region $C$ with compact closure $\overline{C}$, with $f=0$ on
$\partial C$. Then the level sets of $f$ contained in $C$ are all
diffeomorphic to circles, and there is a polar coordinate chart $(r,\theta)$
for $0\leq r\leq R$ and $\theta\in[0,2\pi)$ for $C$ +with $r=0$ at $p_{0}$,
$r=R$ on $\partial C$ such that $f(p)=F(r)$ for some
$F\colon[0,R]\to\mathbb{R}$, with the area form given by

$\mu=\varphi(r)\,dr\wedge d\theta$ (43)

for some function $\varphi\colon[0,R]\to\mathbb{R}_{+}$ satisfying

$\varphi(0)=0\quad\text{and}\quad\varphi^{\prime}(0)\neq 0$ (44)

and the velocity field satisfying

$U=\frac{F^{\prime}(r)}{\varphi(r)}\,\frac{\partial}{\partial\theta}.$ (45)

###### Proof.

On $C\backslash\\{p_{0}\\}$, we have that $df$ is nonzero, so the implicit
function theorem says that the level set $f^{-1}\\{k\\}$ is a smooth one-
dimensional closed submanifold of $M$, and so it must be diffeomorphic to a
circle. Construct a curve $c\colon[0,R]\to M$ such that $c(0)=p_{0}$ with
$c^{\prime}(0)\neq 0$, and chosen so that $c$ intersects all level sets of $f$
orthogonally; we can do this by solving the differential equation
$c^{\prime}(s)=\beta(s)\frac{\nabla f(c(s))}{\lvert\nabla f(c(s))\rvert}$ for
$s>0$, for any positive function $\beta$, since $\nabla f$ is nowhere zero
except at $p_{0}$.

For each $s\in(0,R)$, the vector field $U$ is tangent to the level set
$\\{p\in C\,|\,f(p)=f(c(s))\\}$, and points counterclockwise along it. It is
nowhere zero since $\nabla f$ is nowhere zero on $C\backslash\\{p_{0}\\}$, and
its flow $\gamma$ defined by

$\frac{\partial\gamma}{\partial t}(t,p)=U\big{(}\gamma(t,p)\big{)}$

satisfies $f(\gamma(t,p))=f(p)$ for all $t\in\mathbb{R}$. For each $s\in(0,R)$
there is a smallest positive number $P(s)$ such that $\gamma(P(s),c(s))=c(s)$,
which is the period. Because $U$ is steady, $\gamma$ is a one-parameter group,
which implies $\gamma(t+P(s),c(s))=\gamma(t,c(s))$ for all $t\in\mathbb{R}$.

Now define polar coordinates $\Phi\colon(0,R)\to[0,2\pi)\to M$ by

$\Phi(r,\theta)=\gamma\big{(}\frac{P(r)\theta}{2\pi},c(r)\big{)}.$

Then $\Phi$ maps $(0,R)\times[0,2\pi)$ bijectively onto
$C\backslash\\{p_{0}\\}$, and extends $2\pi$-periodically for all
$\theta\in\mathbb{R}$, while sending the origin $r=0$ to the local minimum
$p_{0}$. We have $f(\Phi(r,\theta))=f(c(r))$, so define
$F\colon[0,R)\to\mathbb{R}$ by $F(r)=f(c(r))$. Since the flow of $U$ satisfies

$\gamma\big{(}t,\Phi(r,\theta)\big{)}=\gamma\Big{(}t,\gamma\big{(}\frac{P(r)\theta}{2\pi},c(r)\big{)}\Big{)}=\gamma\Big{(}t+\frac{P(r)\theta}{2\pi},c(r)\Big{)}=\Phi\Big{(}r,P(r)\big{(}\theta+\frac{2\pi
t}{P(r)}\big{)}\Big{)},$

we see that the flow is given in polar coordinates by
$(t,r,\theta)\mapsto(r,\theta+\tfrac{2\pi t}{P(r)})$. Thus we conclude that
the velocity field $U$ must be given in coordinates by

$U=\frac{2\pi}{P(r)}\,\frac{\partial}{\partial\theta}.$ (46)

The area form in $(r,\theta)$ coordinates is given by

$\mu=\varphi(r,\theta)\,dr\wedge d\theta$

for some function $\varphi$, and the divergence of $U$ in these coordinates is
given by

$\operatorname{div}{U}=\frac{1}{\varphi(r,\theta)}\,\frac{\partial}{\partial\theta}\Big{(}\frac{2\pi\varphi(r,\theta)}{P(r)}\Big{)}=\frac{2\pi
P(r)}{\varphi(r,\theta)}\,\frac{\partial\varphi(r,\theta)}{\partial\theta}.$

But since we know $U$ is divergence-free, we conclude that $\varphi$ depends
only on $r$, which is (43).

Recall that $U$ is the skew-gradient of $f$, which is given in $(r,\theta)$
coordinates by $f(\Phi(r,\theta))=F(r)$. The skew-gradient depends only on the
area form (not the metric) and is given by (45) for a purely radial function
with radial area form.

The local behavior at $r=0$ comes from working in different coordinates that
make sense there. Since the metric is locally Euclidean and $f$ has a
nondegenerate critical point at $p_{0}$, we can choose coordinates $(x,y)$
such that $ds^{2}=dx^{2}+dy^{2}+O(x^{2}+y^{2})$, rotated if necessary so that

$f(x,y)=f(p_{0})+\frac{1}{2}ax^{2}+\frac{1}{2}by^{2}+O((x^{2}+y^{2})^{3/2}$

for some $a>0$ and $b>0$. The area form in these coordinates is $\mu=dx\wedge
dy+O(x^{2}+y^{2})$, and the skew-gradient of $f$ is given by

$U=-\frac{\partial f}{\partial y}\,\frac{\partial}{\partial x}+\frac{\partial
f}{\partial x}\,\frac{\partial}{\partial
y}=-by\,\partial_{x}+ax\,\partial_{y}+O(x^{2}+y^{2}),$

so that the flow is given near $p_{0}$ by

$\gamma(t,x,y)=\Big{(}x\cos{(\sqrt{ab}t)}-\sqrt{\tfrac{b}{a}}y\sin{(\sqrt{ab}t)},\sqrt{\tfrac{a}{b}}x\sin{(\sqrt{ab}t)}y\cos{(\sqrt{ab}t)}\Big{)}+O(x^{2}+y^{2}).$

In particular the limit of the periods defined above is
$\displaystyle\lim_{r\to 0}P(r)=\frac{2\pi}{\sqrt{ab}}$. In addition on the
overlap of the coordinate charts, if $(x,y)=\big{(}c_{1}(s),c_{2}(s)\big{)}$
defines the radial curve $c(s)$ in locally Euclidean coordinates, then the
coordinates are related by

$\displaystyle(x,y)$
$\displaystyle=\Phi(r,\theta)=\gamma\Big{(}\frac{P(r)\theta}{2\pi},c(r)\Big{)}=\Big{(}c_{1}(r)\cos{(\sqrt{ab}P(r)\theta)/2\pi}-\sqrt{\tfrac{b}{a}}c_{2}(r)\sin{(\sqrt{ab}P(r)\theta/2\pi)},$
$\displaystyle\qquad\qquad\sqrt{\tfrac{a}{b}}c_{1}(r)\sin{(\sqrt{ab}P(r)\theta/2\pi)}c_{2}(t)\cos{(\sqrt{ab}P(r)\theta/2\pi)}\Big{)}+O(r^{2})$
$\displaystyle=\Big{(}rc_{1}^{\prime}(0)\cos{\theta}-\sqrt{\tfrac{b}{a}}rc_{2}^{\prime}(0)\sin{\theta},\sqrt{\tfrac{a}{b}}rc_{1}^{\prime}(0)\sin{\theta}+rc_{2}^{\prime}(0)\cos{\theta}\Big{)}+O(r^{2}).$

Since by assumption $c^{\prime}(0)\neq 0$, this gives a standard polar
coordinate transformation near the origin, and so all the functions involved
behave exactly as they would in standard polar coordinates, for example (44).
∎

With $\omega=\Delta f$, the vorticity satisfies $\\{f,\omega\\}=0$ in the
simply connected cell $C$. Since $f$ is strictly increasing from $p_{0}$ to
$\partial C$, we can express the vorticity as a function of $f$, via $\Delta
f=\Omega\circ f$. In our polar coordinates we can simply write
$\omega=\omega(r)$. We now see how to compute this in terms of the metric
components.

######  Corollary 11.

In the coordinates $(r,\theta)$ given by Lemma 10, suppose the metric is given
by

$ds^{2}=g_{11}(r,\theta)\,dr^{2}+2g_{12}(r,\theta)\,dr\,d\theta+g_{22}(r,\theta)\,d\theta^{2}.$

Then the vorticity $\omega=\Delta f$ is given by the equation

$\omega(r)=\frac{1}{\varphi(r)}\,\frac{d}{dr}\big{(}G(r)u(r)\big{)},\qquad\text{where
}G(r)=\frac{1}{2\pi}\int_{0}^{2\pi}g_{22}(r,\theta)\,d\theta.$ (47)

Equivalently we may write

$G(r)=\frac{1}{u(r)}\int_{0}^{r}\varphi(s)\omega(s)\,ds.$ (48)

In addition the components satisfy

$g_{11}(r,\theta)g_{22}(r,\theta)-g_{12}(r,\theta)^{2}=\varphi(r)^{2}$ (49)

and

$\frac{\partial
g_{12}(r,\theta)}{\partial\theta}=\frac{1}{u(r)}\,\frac{\partial}{\partial
r}\Big{(}u(r)\big{[}g_{22}(r,\theta)-G(r)\big{]}\Big{)}.$ (50)

###### Proof.

Recall that the velocity field is given by
$U=u(r)\,\frac{\partial}{\partial\theta}$. The vorticity function $\omega$ is
given by $dU^{\flat}=\omega\,\mu,$ and so the first thing we do is compute the
lift of $U$ to a $1$-form:

$U^{\flat}=u(r)g_{12}(r,\theta)\,dr+u(r)g_{22}(r,\theta)\,d\theta.$

Then

$dU^{\flat}=\left(\frac{\partial}{\partial
r}\big{(}u(r)g_{22}(r,\theta)\big{)}-\frac{\partial}{\partial\theta}\big{(}u(r)g_{12}(r,\theta)\big{)}\right)\,dr\wedge
d\theta.$ (51)

We conclude that

$\frac{\partial}{\partial
r}\big{(}u(r)g_{22}(r,\theta)\big{)}-u(r)\,\frac{\partial
g_{12}(r,\theta)}{\partial\theta}=\omega(r)\varphi(r).$ (52)

Now the metric components $g_{12}$ and $g_{22}$ are $2\pi$-periodic functions
of $\theta$, and we may integrate around this formula around the circle to
obtain (47).

Note that subtracting the averaged formula (47) from the full vorticity
formula (52) gives the formula (50). Finally the formula (49) follows from the
usual formula for the Riemannian area form in terms of the metric components,
and the formula (43) that we already know. The formula (48) comes from solving
(47) for $G$ and using the fact that $G(0)=0$. ∎

######  Remark 1.

Observe that $\lvert U\rvert^{2}=u(r)^{2}g_{22}(r,\theta)$. Thus $\lvert
U\rvert^{2}$ is constant along trajectories if and only if $g_{22}(r,\theta)$
is independent of $\theta$. If that is the case, then $g_{22}(r,\theta)=G(r)$
as in (47), and this means that $g_{12}(r,\theta)$ is constant. It is
therefore determined by its values when $\theta=0$, i.e., along the curve $c$.
If we choose $c$ to be parallel to the gradient of $f$ (i.e., perpendicular to
$U$), then $g_{12}(r,\theta)\equiv 0$, and the metric becomes diagonal.
Furthermore since $g_{22}(r,\theta)=G(r)$ and
$g_{11}(r,\theta)G(r)=\varphi(r)$, we also conclude that
$g_{11}(r,\theta)=E(r)$, so the metric is
$ds^{2}=E(r)\,dr^{2}+G(r)\,d\theta^{2}$. We may rescale $r$ so that $E\equiv
1$ (corresponding to choosing $c$ to be unit speed), which means
$G(r)=\varphi(r)^{2}$, and obtain the standard polar coordinate metric

$ds^{2}=dr^{2}+\varphi(r)^{2}\,d\theta^{2}.$

We have thus shown the standard result that $\lvert U\rvert$ is constant on
trajectories if and only if $U$ is rotationally symmetric on a rotationally
symmetric manifold.

If $\lvert U\rvert$ is not constant on trajectories, then there is no
particular reason to demand that $c$ be orthogonal to level sets or that it be
unit speed, since those properties will not be preserved as $c$ is rotated
around the cell $C$ by the flow.

Now we consider how much flexibility there is to construct steady fluid flows
with given functions $F(r)$, $\varphi(r)$, $E(r)$, and $G(r)$. (Note that
these functions also determine $u(r)$ and $\omega(r)$ by formulas (45) and
(47).

######  Lemma 12.

Suppose $F(r)$, $\varphi(r)/r$, and $G(r)$ are given smooth functions of
$r\in[0,R)$ that extend to smooth even functions on $(-R,R)$. (In particular
$\varphi(r)$ is a smooth odd function of $r$.) Assume that $F^{\prime}(r)$,
$\varphi(r)$, and $G(r)$ are all positive for $r\in(0,R)$, with $G(0)=0$.
Suppose $\zeta(r,\theta)$ is a smooth function on the disc $r\in[0,R]$,
$\theta\in S^{1}$, i.e., it can be expressed as

$\zeta(r,\theta)=\sum_{n=0}^{\infty}r^{n}a_{n}(r)\cos{n\theta}+\sum_{n=1}^{\infty}r^{n}b_{n}(r)\sin{n\theta},$

where each $a_{n}$ and $b_{n}$ extends to a smooth even function of $r$.

Then there is a topological disc $M$ with Riemannian metric

$ds^{2}=g_{11}(r,\theta)\,dr^{2}+2g_{12}(r,\theta)\,dr\,d\theta+g_{22}(r,\theta)\,d\theta^{2},$

where

$\begin{split}g_{11}(r,\theta)&=\frac{\varphi(r)^{2}}{G(r)}\,\frac{1+\frac{1}{F^{\prime}(r)^{2}}\,\big{(}\frac{\partial\zeta}{\partial
r}(r,\theta)^{2}\big{)}}{1+\frac{\varphi(r)}{F^{\prime}(r)G(r)}\,\frac{\partial\zeta}{\partial\theta}(r,\theta)}\\\
g_{12}(r,\theta)&=\frac{\varphi(r)}{F^{\prime}(r)}\,\frac{\partial\zeta}{\partial
r}(r,\theta)\\\
g_{22}(r,\theta)&=G(r)+\frac{\varphi(r)}{F^{\prime}(r)}\,\frac{\partial\zeta}{\partial\theta}(r,\theta),\end{split}$
(53)

such that the area form is $dA=\varphi(r)\,dr\wedge d\theta$, and such that
the velocity field $U=\frac{F^{\prime}(r)}{\varphi(r)}\,\partial_{\theta}$ is
a steady solution of the 2D Euler equation on $M$ with vorticity given by
(47).

Conversely all discs with steady flows with stream function strictly
increasing from a single minimum must arise in this way.

###### Proof.

It is clear that

$g_{11}g_{22}-g_{12}^{2}=\varphi(r)^{2}$ (54)

by the definitions. Thus the area form is $dA=\varphi(r)\,dr\wedge d\theta$,
and the divergence of $U=u(r)\,\partial_{\theta}$ is indeed zero. In
particular if $U$ is the skew-gradient of $F$, with
$u(r)=F^{\prime}(r)/\varphi(r)$.

It remains to show that the curl of $U$ is a function of $r$ alone. We have
already computed in (51) that the curl of $U=u(r)\,\partial_{\theta}$ in a
general metric is a function given by

$\operatorname{curl}{U}=\frac{1}{\varphi(r)}\left(\frac{\partial}{\partial
r}\big{(}u(r)g_{22}(r,\theta)\big{)}-\frac{\partial}{\partial\theta}\big{(}u(r)g_{12}(r,\theta)\big{)}\right),$

which in this case reduces to

$\displaystyle\operatorname{curl}{U}$
$\displaystyle=\frac{1}{\varphi(r)}\left(\frac{\partial}{\partial
r}\Big{(}\frac{F^{\prime}(r)G(r)}{\varphi(r)}+\frac{\partial\zeta}{\partial\theta}\Big{)}-\frac{\partial}{\partial\theta}\Big{(}\frac{\partial\zeta}{\partial
r}\Big{)}\right)$
$\displaystyle=\frac{1}{\varphi(r)}\,\frac{d}{dr}\left(\frac{F^{\prime}(r)G(r)}{\varphi(r)}\right).$

We conclude that $U\cdot\nabla\operatorname{curl}{U}=0$, so indeed $U$ is a
steady solution of the Euler equation.

The conditions at $r=0$ are required in order to get smoothness of the metric,
since $\varphi(r)\,dr$ must be a smooth $1$-form on $M$, and
$g_{22}(0,\theta)$ must be zero. We then get smooth metric coefficients since
$u(r)=F^{\prime}(r)/\varphi(r)$ is a smooth, nowhere-zero function on $[0,R)$.

The converse simply comes from (52), after writing the metric as
$g_{22}(r,\theta)=G(r)+\psi(r,\theta)$, where $\psi(r,\theta)$ has mean zero
for each fixed $r$. Then we have that

$\frac{\partial}{\partial
r}\big{(}u(r)\psi(r,\theta)\big{)}=u(r)\,\frac{\partial
g_{12}(r,\theta)}{\partial\theta}.$

We conclude from the Poincaré Lemma since the disc is simply connected that
there must be a smooth function $\zeta$ such that
$u(r)\psi(r,\theta)=\frac{\partial}{\partial\theta}\zeta(r,\theta)$ and
$u(r)g_{12}(r,\theta)=\frac{\partial}{\partial r}\zeta(r,\theta)$. Finally the
formula for $g_{11}$ comes from solving the equation (49) for $g_{11}$ in
terms of $\varphi$ and the other components. ∎

##  4 Proofs of Theorem 1 and Corollary 2

We consider a variation field supported on the cell $C$ as in the previous
section, of the form

$Y(t,r,\theta)=\sin{(\tfrac{t}{T})}\operatorname{sgrad}\big{(}\xi(r)\cos{(\theta-\alpha
t)}\big{)},$

on the time interval $[0,\pi T]$. Performing the time integral first in the
index formula (41) turns out to make things much easier due to the following
Lemma.

######  Lemma 13.

Suppose $\alpha\in\mathbb{R}$ is nonzero. For any $\theta\in\mathbb{R}$ and
any vectors $p,q,r,s\in\mathbb{R}^{n}$ we have the following limits:

$\displaystyle\lim_{T\to\infty}\frac{4}{\pi T}\int_{0}^{\pi
T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\,dt$ $\displaystyle=2$ (55)
$\displaystyle\lim_{T\to\infty}\frac{4}{\pi T}\int_{0}^{\pi
T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\cos^{2}{(\theta-\alpha t)}\,dt$
$\displaystyle=1$ (56) $\displaystyle\lim_{T\to\infty}\frac{4}{\pi
T}\int_{0}^{\pi T}\Big{(}\sin{\big{(}\frac{t}{T}\big{)}}\sin{(\theta-\alpha
t)}\,p+\frac{1}{T}\cos{\big{(}\frac{t}{T}\big{)}}\cos{(\theta-\alpha
t)}\,q\Big{)}^{2}\,dt$ $\displaystyle=\lvert p\rvert^{2},$ (57)
$\displaystyle\lim_{T\to\infty}\frac{4}{\pi T}\int_{0}^{\pi
T}\Big{(}\sin{\big{(}\frac{t}{T}\big{)}}\cos{(\theta-\alpha
t)}\,p-\frac{1}{T}\cos{\big{(}\frac{t}{T}\big{)}\sin{(\theta-\alpha
t)}\,q}\Big{)}^{2}\,dt$ $\displaystyle=\lvert p\rvert^{2},$ (58)
$\begin{split}&\lim_{T\to\infty}\frac{4}{\pi
T}\int_{0}^{T}\Big{\langle}\sin{\big{(}\frac{t}{T}\big{)}}\sin{(\theta-\alpha
t)}p+\frac{1}{T}\cos{\big{(}\frac{t}{T}\big{)}}\cos{(\theta-\alpha t)}q,\\\
&\qquad\qquad\qquad\sin{\big{(}\frac{t}{T}\big{)}}\cos{(\theta-\alpha
t)}r-\frac{1}{T}\,\cos{\big{(}\tfrac{t}{T}\big{)}}\sin{(\theta-\alpha
t)}s\Big{\rangle}\,dt=0.\end{split}$ (59)

###### Proof.

These are easy to evaluate from explicit formulas. For (55) we have

$\frac{4}{\pi T}\int_{0}^{\pi T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\,dt=2$

even without taking the limit. For (56) with $\alpha\neq 0$, we get

$\frac{4}{\pi T}\int_{0}^{\pi
T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\cos^{2}{(\theta-\alpha
t)}\,dt=1-\frac{\sin{(\pi T\alpha)}\cos{(2\theta-\pi T\alpha)}}{\pi
T\alpha(T^{2}\alpha^{2}-1)},$

and the limit of this is obviously $1$ as $T\to\infty$.

For the next two limits, note that the terms involving $\langle p,q\rangle$
and $\lvert q\rvert^{2}$ have terms $\frac{1}{T}$ and $\frac{1}{T^{2}}$
respectively multiplying them, so they are of obviously smaller order and
disappear in the limit; thus (57) follows from (56), while (58) follows since
it’s the same as (57) with $\theta$ shifted by $\pi/2$.

For the last limit, by the same reasoning (ignoring the terms that have
$\frac{1}{T}$), the limit is equal to

$\displaystyle\langle p,r\rangle\lim_{T\to\infty}\frac{4}{\pi T}\int_{0}^{\pi
T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\sin{(\theta-\alpha
t)}\cos{(\theta-\alpha t)}\,dt$ $\displaystyle=-\langle
p,r\rangle\lim_{T\to\infty}\frac{\sin{(\pi T\alpha)}\sin{(2\theta-\pi
T\alpha)}}{8\alpha(\alpha^{2}T^{2}-1)}=0.$

∎

Note that if $\alpha=0$, instead of (56) we get

$\frac{4}{\pi T}\int_{0}^{\pi
T}\sin^{2}{\big{(}\frac{t}{T}\big{)}}\cos^{2}{\theta}\,dt=2\cos^{2}{\theta}=1+\cos{2\theta},$

so it really does matter that $\alpha\neq 0$ in the Lemma above. If $\alpha$
is close to zero, we may need to choose $T$ very large in order to actually
get close to this limit. Similarly if $\alpha=0$ then (59) becomes
$\sin{2\theta}\langle p,r\rangle$ rather than zero. We will eventually be
integrating with respect to $\theta$, so the resulting terms probably
integrate to zero anyway, but it is less obvious.

###### Proof of Theorem 1.

For $T>0$ and $\alpha\in\mathbb{R}$, with $\xi\colon[0,R]\to\mathbb{R}$
satisfying the conditions $\xi(0)=0$, $\xi(R)=0$, and $\xi^{\prime}(0)\neq 0$,
let

$g(t,r,\theta)=\sin{(\tfrac{t}{T})}\xi(r)\cos{(\theta-\alpha t)}.$ (60)

Note that the conditions on $\xi$ at $r=0$ are necessary and sufficient to
make $g$ continuous and differentiable at $p_{0}$, since near $p_{0}$ we will
have

$g(t,r,\theta)\approx\sin{(\tfrac{t}{T})}\xi^{\prime}(0)r\cos{(\theta-\alpha
t)}\approx\sin{(\tfrac{t}{T})}\xi^{\prime}(0)(x\cos{\alpha t}+y\sin{\alpha
t})$

in terms of the locally Euclidean coordinates discussed in Lemma 10.

We compute that $h$ defined by (42) is given by

$\begin{split}h(t,r,\theta)&=\frac{\partial g}{\partial
t}(t,r,\theta)+u(r)\,\frac{\partial g}{\partial\theta}(t,r,\theta)\\\
&=\frac{1}{T}\,\cot{(\tfrac{t}{T})}g(t,r,\theta)+\big{(}\alpha-u(r)\big{)}\sin{(\tfrac{t}{T})}\xi(r)\sin{(\theta-\alpha
t)}.\end{split}$ (61)

In particular we get

$\begin{split}\\{g,h\\}&=\frac{1}{\varphi}\,\left(\frac{\partial g}{\partial
r}\,\frac{\partial h}{\partial\theta}-\frac{\partial
g}{\partial\theta}\,\frac{\partial h}{\partial r}\right)\\\
&=\frac{\sin^{2}{(\tfrac{t}{T})}\xi(r)}{\varphi(r)}\left(\xi(r)u^{\prime}(r)\cos^{2}{(\theta-\alpha
t)}+\frac{d}{dr}\big{(}(\alpha-u(r))\xi(r)\big{)}\right).\end{split}$ (62)

We also compute from (61) that

$\nabla
h=\left(\frac{1}{T}\,\cos{\big{(}\frac{t}{T}\big{)}}\cos{(\theta-\alpha
t)}\xi^{\prime}(r)+\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\sin{\big{(}\frac{t}{T}\big{)}}\sin{(\theta-\alpha
t)}\right)\nabla r\\\
+\left(\big{(}\alpha-u(r)\big{)}\sin{\big{(}\frac{t}{T}\big{)}}\xi(r)\cos{(\theta-\alpha
t)}-\frac{1}{T}\,\cos{\big{(}\tfrac{t}{T}\big{)}}\xi(r)\sin{(\theta-\alpha
t)}\right)\nabla\theta.$ (63)

Referring to (41), we compute (without integrating in space) using formulas
(57) and (58) from Lemma 13 that

$\begin{split}&\lim_{T\to\infty}\frac{4}{\pi T}\int_{0}^{T}\lvert\nabla
h(t,r,\theta)\rvert^{2}+\omega(r)\\{g(t,r,\theta),h(t,r,\theta)\\}\,dt\\\
&\qquad\qquad=\left\lvert\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\right\rvert^{2}\lvert\nabla
r\rvert^{2}+\left\lvert\big{(}\alpha-u(r)\big{)}\xi(r)\right\rvert^{2}\,\lvert\nabla\theta\rvert^{2}.\end{split}$
(64)

The term that would involve $\langle\nabla r,\nabla\theta\rangle$ has zero
limit by formula (59).

Similarly using (55) and (56), we see from equation (62) that

$\begin{split}\lim_{T\to\infty}\frac{4}{\pi
T}\omega(r)\\{g(t,r,\theta),h(t,r,\theta)\\}&=\frac{\xi(r)}{\varphi(r)}\left(\xi(r)u^{\prime}(r)+2\frac{d}{dr}\big{(}(\alpha-u(r))\xi(r)\big{)}\right)\\\
&=\frac{1}{\varphi(r)}\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\Big{)}.\end{split}$
(65)

Combining (64) and (65), we see that the index form from (41) will satisfy

$\lim_{T\to\infty}\frac{4}{\pi
T}I(Y,Y)=\int_{M}\left\lvert\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\right\rvert^{2}\lvert\nabla
r\rvert^{2}+\left\lvert\big{(}\alpha-u(r)\big{)}\xi(r)\right\rvert^{2}\,\lvert\nabla\theta\rvert^{2}+\frac{1}{\varphi(r)}\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\Big{)}\,d\mu.$
(66)

Observing that the gradients of the coordinate functions in a general
Riemannian metric are given by

$\nabla
r=\frac{g_{22}\partial_{r}-g_{12}\partial_{\theta}}{g_{11}g_{22}-g_{12}^{2}}\qquad\text{and}\qquad\nabla\theta=\frac{-g_{12}\partial_{r}+g_{22}\partial_{\theta}}{g_{11}g_{22}-g_{12}^{2}},$

formula (49) shows that

$\lvert\nabla
r\rvert^{2}=\frac{g_{22}(r,\theta)}{\varphi(r)^{2}}\qquad\text{and}\qquad\lvert\nabla\theta\rvert^{2}=\frac{g_{11}(r,\theta)}{\varphi(r)^{2}}.$

Inserting these formulas into (66), and using the formula (43) for the area
element $d\mu$ gives

$\lim_{T\to\infty}\frac{4}{\pi
T}I(Y,Y)=\int_{0}^{R}\int_{0}^{2\pi}\frac{1}{\varphi(r)}\left\lvert\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\right\rvert^{2}g_{22}(r,\theta)+\frac{1}{\varphi(r)}\left\lvert\big{(}\alpha-u(r)\big{)}\xi(r)\right\rvert^{2}\,g_{11}(r,\theta)\\\
+\omega(r)\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\Big{)}\,d\theta\,dr.$

The only terms depending on $\theta$ are $g_{11}(r,\theta)$ and
$g_{22}(r,\theta)$, and performing the $\theta$ integration and using the
definitions (5) of $E(r)$ and $G(r)$, we obtain

$\lim_{T\to\infty}\frac{4I(Y,Y)}{\pi
T}=2\pi\int_{0}^{R}\frac{G(r)}{\varphi(r)}\left\lvert\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\right\rvert^{2}+\frac{E(r)}{\varphi(r)}\left\lvert\big{(}\alpha-u(r)\big{)}\xi(r)\right\rvert^{2}+\omega(r)\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\Big{)}\,dr.$

Finally we integrate the last term by parts in $r$ and use the fact that
$\xi(0)=\xi(R)=0$ to obtain

$\lim_{T\to\infty}\frac{2I(Y,Y)}{\pi^{2}T}=\int_{0}^{R}\frac{G(r)}{\varphi(r)}\left\lvert\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}\xi(r)\Big{)}\right\rvert^{2}+\frac{E(r)}{\varphi(r)}\left\lvert\big{(}\alpha-u(r)\big{)}\xi(r)\right\rvert^{2}-\omega^{\prime}(r)\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\,dr.$

This limit is the quantity $I$ appearing in (6); if it is negative for some
$\alpha\neq 0$, then for sufficiently large $T$ we can make the actual index
$I(Y,Y)$ negative on the interval $[0,T]$, and so there is a conjugate point
occurring at some time $\tau<T$. ∎

###### Proof of Corollary 2.

The easiest way to proceed is to simply compute the difference between $I_{1}$
given by (6) and $I_{2}$ given by (8), and show that it is zero. We get

$I_{1}-I_{2}=\int_{0}^{R}\frac{G(r)}{\varphi(r)}\Big{(}u^{\prime}(r)^{2}\xi(r)^{2}-2u^{\prime}(r)\big{(}\alpha-u(r)\big{)}\xi(r)\xi^{\prime}(r)\Big{)}-\big{(}\alpha-u(r)\big{)}\xi(r)^{2}\,\frac{d}{dr}\big{(}\omega(r)-2v(r)\big{)}\,dr.$

Now note that by formula (47) and (7), we have
$\omega(r)-2v(r)=\frac{G(r)u^{\prime}(r)}{\varphi(r)}$, so that we get

$I_{1}-I_{2}=\int_{0}^{R}\Big{[}-\frac{G(r)u^{\prime}(r)}{\varphi(r)}\,\frac{d}{dr}\Big{(}(\alpha-u(r))\xi(r)^{2}\big{)}-(\alpha-u(r))\xi(r)^{2}\,\frac{d}{dr}\Big{(}\frac{G(r)u^{\prime}(r)}{\varphi(r)}\Big{)}\Big{]}\,dr,$

which is obviously the integral of a total derivative, and the function
vanishes at the endpoints since $\xi(0)=\xi(R)=0$.

Similarly to prove this is equivalent to (10), we show that $I_{2}-I_{3}$
simplifies to zero.

$\displaystyle I_{2}-I_{3}$
$\displaystyle=\int_{0}^{R}2\big{(}\alpha-u(r)\big{)}^{2}Q(r)\xi(r)\xi^{\prime}(r)+\big{(}\alpha-u(r)\big{)}^{2}Q^{\prime}(r)\xi(r)^{2}-2Q(r)\big{(}\alpha-u(r)\big{)}u^{\prime}(r)\xi(r)^{2}\,dr$
$\displaystyle=\int_{0}^{R}\frac{d}{dr}\Big{(}\big{(}\alpha-u(r)\big{)}^{2}Q(r)\xi(r)^{2}\Big{)}\,dr,$

and again this vanishes since $\xi(0)=\xi(R)=0$. ∎

##  5 Proofs of Corollaries 3, 4, and 5

###### Proof of Corollary 3.

We assume that $u(r)=1$. Then formula (47) implies that the vorticity is given
by $\omega(r)=G^{\prime}(r)/\varphi(r)$, and thus its derivative is given by

$\omega^{\prime}(r)=\frac{d}{dr}\left(\frac{G^{\prime}(r)}{\varphi(r)}\right).$
(67)

The condition (6) for existence of conjugate points then takes the form (using
$\beta=\alpha-1$ for simplicity)

$I=\int_{0}^{R}\frac{\beta^{2}}{\varphi(r)}\big{[}G(r)\xi^{\prime}(r)^{2}+E(r)\xi(r)^{2}\big{]}-\beta\frac{d}{dr}\left(\frac{G^{\prime}(r)}{\varphi(r)}\right)\xi(r)^{2}\,dr<0.$
(68)

For any fixed $\xi$, this takes the form $I(\beta)=\beta^{2}A-\beta B$ for
some positive number $A$ and some number
$B=\int_{0}^{R}\frac{d}{dr}\big{(}\frac{G^{\prime}}{\varphi}\big{)}\xi^{2}\,dr$,
which reaches its minimum at

$I\left(\frac{B}{2A}\right)=-\frac{B^{2}}{2A}.$

Hence as long as we can find a function $\xi$ that makes $B$ nonzero, we can
make the index $I$ negative and obtain a conjugate point by Theorem 1. The
only way $B$ must be zero for every $\xi$ is if
$\frac{d}{dr}(G^{\prime}(r)/\varphi(r))$ is identically zero, i.e., if
$G^{\prime}(r)=c\varphi(r)$ for some $c\in\mathbb{R}$.

In particular in the rotationally symmetric case we have
$G(r)=\varphi(r)^{2}$, so that
$G^{\prime}(r)/\varphi(r)=2\varphi^{\prime}(r)$, and the only way this is
constant is if $\varphi^{\prime\prime}(r)=0$. Since
$\varphi^{\prime\prime}(r)=-\kappa(r)\varphi(r)$ in terms of the Gaussian
curvature, and $\varphi(r)$ is nonzero on $(0,R)$, the only way $B$ would be
identically zero in the rotationally symmetric case is if the curvature is
identically zero. ∎

The only cases in which we do not get conjugate points are rotational flows on
flat spaces: e.g., rotation of a disc or annulus with $\varphi(r)=r$, or on a
flat torus with $\varphi(r)\equiv 1$. It is interesting that either positive
curvature or negative curvature of the underlying manifold $M$ could both
contribute to conjugate points on $\mathcal{D}_{\mu}(M)$.

###### Proof of Corollary 4.

We use Corollary 2, in particular the formula (8). Choose $\alpha=u(r_{0})$ in
formula (8). For a small $\varepsilon>0$, we consider functions $\xi$ having
support in $(r_{0}-\varepsilon,r_{0}+\varepsilon)$, which is possible since
$r_{0}$ is neither $0$ nor $R$.

Since $u^{\prime}(r_{0})=0$ and $u^{\prime\prime}(r_{0})\neq 0$, we have

$u(r)-\alpha=\tfrac{1}{2}u^{\prime\prime}(r_{0})(r-r_{0})^{2}+O(\varepsilon^{3}).$

Hence the condition (8) becomes

$I:=\int_{r_{0}-\varepsilon}^{r_{0}+\varepsilon}\frac{\big{[}\tfrac{1}{2}u^{\prime\prime}(r_{0})(r-r_{0})^{2}+O(\varepsilon^{3})\big{]}^{2}}{\varphi(r_{0})+O(\varepsilon)}\Big{(}\big{[}G(r_{0})+O(\varepsilon)\big{]}\xi^{\prime}(r)^{2}+\big{[}E(r_{0})+O(\varepsilon)\big{]}\xi(r)^{2}\Big{)}\\\
+2\big{[}v^{\prime}(r_{0})+O(\varepsilon)\big{]}\big{[}\tfrac{1}{2}u^{\prime\prime}(r_{0})(r-r_{0})^{2}+O(\varepsilon^{3})\big{]}\xi(r)^{2}\,dr<0.$

Changing variables to $s=\frac{r-r_{0}}{\varepsilon}$ and writing
$\xi(r)=\zeta((r-r_{0})/\varepsilon)$, this condition becomes

$I=\varepsilon^{3}\int_{-1}^{1}\tfrac{1}{4}u^{\prime\prime}(r_{0})^{2}s^{4}\varphi(r_{0})\zeta^{\prime}(s)^{2}-u(r_{0})K(r_{0})\varphi(r_{0})u^{\prime\prime}(r_{0})s^{2}\zeta(s)^{2}\,ds+O(\varepsilon^{4})<0.$

Clearly since $\varphi(r_{0})>0$, for sufficiently small $\varepsilon>0$, we
can make $I$ negative as long as for some function $\zeta$ supported in
$[-1,1]$, we have

$\tfrac{1}{4}u^{\prime\prime}(r_{0})^{2}\int_{-1}^{1}s^{4}\zeta^{\prime}(s)^{2}\,ds-K(r_{0})u(r_{0})u^{\prime\prime}(r_{0})\int_{-1}^{1}s^{2}\zeta(s)^{2}\,ds<0.$
(69)

The inequality

$\int_{0}^{1}s^{4}\zeta^{\prime}(s)^{2}\,ds\geq\frac{9}{4}\int_{0}^{1}s^{2}\zeta(s)^{2}\,ds\qquad\text{if
$\zeta(1)=0$}$

is a special case of a general weighted inequality appearing in [19]; this
case is easily proved by computing

$\displaystyle 0$
$\displaystyle\leq\int_{0}^{1}\big{(}s^{2}\zeta^{\prime}(s)+\tfrac{3}{2}s\zeta(s)\big{)}^{2}\,ds$
$\displaystyle=\int_{0}^{1}s^{4}\zeta^{\prime}(s)^{2}\,ds+3\int_{0}^{1}s^{3}\zeta(s)\zeta^{\prime}(s)\,ds+\tfrac{9}{4}\int_{0}^{1}s^{2}\zeta(s)^{2}\,ds$
$\displaystyle=\int_{0}^{1}s^{4}\zeta^{\prime}(s)^{2}\,ds-\frac{9}{4}\int_{0}^{1}s^{2}\zeta(s)^{2}\,ds,$

after integrating the middle term by parts. The constant is optimal, as seen
by using the test function $\zeta(s)=(1-s^{2})^{2}\lvert
s\rvert^{-3/2+\delta}$ for $\delta>0$, which gives

$\frac{\int_{-1}^{1}s^{4}\zeta^{\prime}(s)^{2}\,ds}{\int_{-1}^{1}s^{2}\zeta(s)^{2}\,ds}=\frac{9}{4}+\frac{4\delta}{3}+\frac{\delta^{2}}{3}.$
(70)

Thus choosing a smooth $\zeta$ which is close in $H^{1}$ to such a near-
optimal test function, with the ratio in (70) smaller than
$\tfrac{9}{4}+\varepsilon$, we find that the condition (69) is equivalent to

$\tfrac{1}{4}u^{\prime\prime}(r_{0})^{2}(\tfrac{9}{4}+\varepsilon)-K(r_{0})u(r_{0})u^{\prime\prime}(r_{0})<0.$
(71)

Thus condition (71) holds for some $\varepsilon>0$ if and only if the
condition (69) holds. ∎

###### Proof of Corollary 5.

As in the proof of Corollary 4, we set $\alpha=u(0)$. For a small
$\varepsilon>0$, we consider odd functions $\xi$ of $r$ having support in
$(-\varepsilon,\varepsilon)$.

Since $u^{\prime}(0)=0$ and $u^{\prime\prime}(0)\neq 0$, we have

$u(r)-\alpha=\tfrac{1}{2}u^{\prime\prime}(0)r^{2}+O(\varepsilon^{3}).$

We also recall that by Lemma 10, the functions $E$, $\varphi$, and $G$ behave
like those in standard Euclidean coordinates. That is,
$E(r)=E(0)+O(\varepsilon)$,
$\varphi(r)=\varphi^{\prime}(0)r+O(\varepsilon^{2})$, and
$G(r)=\tfrac{1}{2}G^{\prime\prime}(0)r^{2}+O(\varepsilon^{3})$, where all the
lowest-order coefficients are positive. The formula (7) gives

$v^{\prime}(r)=\frac{d}{dr}\left(\frac{G^{\prime}(r)u(r)}{2\varphi(r)}\right).$

Now $G^{\prime}(r)$ and $\varphi(r)$ are both odd functions of $r$, while
$u(r)$ is an even function of $r$; thus $v$ is an even function of $r$, and we
have $v^{\prime}(r)=v^{\prime\prime}(0)r+O(\varepsilon^{2})$.

The condition (8) becomes

$I:=\int_{0}^{\varepsilon}\frac{\big{[}\tfrac{1}{2}u^{\prime\prime}(0)r^{2}+O(\varepsilon^{3})\big{]}^{2}}{\varphi^{\prime}(0)r+O(\varepsilon^{2})}\Big{(}\big{[}\tfrac{1}{2}G^{\prime\prime}(0)r^{2}+O(\varepsilon^{3})\big{]}\xi^{\prime}(r)^{2}+\big{[}E(0)+O(\varepsilon)\big{]}\xi(r)^{2}\Big{)}\\\
+2\big{[}rv^{\prime\prime}(0)+O(\varepsilon^{2})\big{]}\big{[}\tfrac{1}{2}u^{\prime\prime}(0)r^{2}+O(\varepsilon^{3})\big{]}\xi(r)^{2}\,dr<0.$
(72)

So we just need to compute $v^{\prime\prime}(0)$. Using L’Hopital’s rule and
the fact that all odd derivatives of $G$ and $u$ are zero at $r=0$, while all
even derivatives of $\varphi(r)$ are zero at $r=0$, we get

$v^{\prime\prime}(0)=\frac{u(0)\varphi^{\prime}(0)G^{iv}(0)-u(0)\varphi^{\prime\prime\prime}(0)G^{\prime\prime}(0)+3u^{\prime\prime}(0)\varphi^{\prime}(0)G^{\prime\prime}(0)}{6\varphi^{\prime}(0)^{2}}.$
(73)

Again in the rotationally symmetric case, we have $G(r)=\varphi(r)^{2}$, so
that we get the simplification

$v^{\prime\prime}(0)=u(0)\varphi^{\prime\prime\prime}(0)+u^{\prime\prime}(0)\varphi^{\prime}(0)=\big{(}-\kappa(0)u(0)+u^{\prime\prime}(0)\big{)}\varphi^{\prime}(0).$
(74)

As in the proof of Corollary 4, we change variables to
$s=\frac{r}{\varepsilon}$ and write $\xi(r)=\zeta(r/\varepsilon)$. Then the
index (72) becomes

$I=\varepsilon^{3}\int_{0}^{1}\bigg{[}\frac{u^{\prime\prime}(0)^{2}G^{\prime\prime}(0)}{8\varphi^{\prime}(0)}s^{5}\zeta^{\prime}(s)^{2}+\Big{(}\frac{u^{\prime\prime}(0)^{2}E(0)}{4\varphi^{\prime}(0)}+v^{\prime\prime}(0)u^{\prime\prime}(0)\Big{)}s^{3}\zeta(s)^{2}\bigg{]}\,ds+O(\varepsilon^{4})<0.$

For sufficiently small $\varepsilon>0$, we can make $I$ negative as long as
for some function $\zeta$ supported in $[0,1]$, extending to a differentiable
odd function on $[-1,1]$ we have

$\frac{G^{\prime\prime}(0)u^{\prime\prime}(0)^{2}}{8\varphi^{\prime}(0)}\int_{0}^{1}s^{5}\zeta^{\prime}(s)^{2}\,ds+\Big{(}\frac{u^{\prime\prime}(0)^{2}E(0)}{4\varphi^{\prime}(0)}+v^{\prime\prime}(0)u^{\prime\prime}(0)\Big{)}\int_{0}^{1}s^{3}\zeta(s)^{2}\,ds<0.$
(75)

Using the same strategy as above, we prove that

$\int_{0}^{1}s^{5}\zeta^{\prime}(s)^{2}\,ds\geq
4\int_{0}^{1}s^{3}\zeta(s)^{2}\,ds\qquad\text{if $\zeta(1)=0$}$

by computing

$0\leq\int_{0}^{1}\big{(}s^{3}\zeta^{\prime}(s)+2s\zeta(s)\big{)}^{2}\,ds,$

again integrating the middle term by parts. Using the test function
$\zeta(s)=s^{-2+\delta}(1-s)$ for $\delta>0$ gives

$\frac{\int_{0}^{1}s^{5}\zeta^{\prime}(s)^{2}\,ds}{\int_{0}^{1}s^{3}\zeta(s)^{2}\,ds}=\delta^{2}+\delta+4,$
(76)

so again we can get as close as we want by approximating this function.

Thus as in the proof of Corollary 4, the sufficient condition is

$\frac{G^{\prime\prime}(0)u^{\prime\prime}(0)^{2}}{2\varphi^{\prime}(0)}+\frac{u^{\prime\prime}(0)^{2}E(0)}{4\varphi^{\prime}(0)}+v^{\prime\prime}(0)u^{\prime\prime}(0)<0.$
(77)

Using formula (74), we see that condition (77) holds for some $\varepsilon>0$
if and only if the condition (14) holds.

In the rotationally symmetric case we have $G(r)=\varphi(r)^{2}$, along with
$E(0)=1$ and $\varphi^{\prime}(0)=1$. Using
$\phi(r)=\phi^{\prime}(0)r+\frac{1}{6}\phi^{\prime\prime\prime}(0)r^{3}+\cdots$,
we get

$G(r)=\phi^{\prime}(0)^{2}r^{2}+\tfrac{1}{3}\phi^{\prime}(0)\phi^{\prime\prime\prime}(0)r^{4}+\cdots,$

so that $G^{\prime\prime}(0)=2\phi^{\prime}(0)^{2}$ and
$G^{iv}(0)=8\phi^{\prime}(0)\phi^{\prime\prime\prime}(0)=-8\kappa(0)\phi^{\prime}(0)^{2}$.
Plugging these into (14), we get the simplification (15). ∎

##  6 Proof of Theorem 6

In this section we study the Kolmogorov flows, with stream function on the
torus of the form $f(x,y)=-\cos{mx}\cos{ny}$ with $m$ and $n$ both natural
numbers. We compute the flow explicitly in terms of Jacobi elliptic functions
and find explicit formulas for all the terms appearing in the formula (10). It
is interesting that the index form is essentially the same for all of them,
regardless of $m$ and $n$. This becomes more obvious if we change the
coordinates.

######  Lemma 14.

Suppose $m$ and $n$ are positive integers, and
$f\colon\mathbb{T}^{2}\to\mathbb{R}$ is given by $f(x,y)=-\cos{mx}\cos{ny}$.
Define coordinates $(X,Y)$ by $X=\sin{mx}$ and $Y=\sin{ny}$. Then in these
coordinates the metric becomes

$ds^{2}=\frac{dX^{2}}{m^{2}(1-X^{2})}+\frac{dY^{2}}{n^{2}(1-Y^{2})},$ (78)

with area form given by

$\mu=\frac{dX\wedge dY}{mn\sqrt{(1-X^{2})(1-Y^{2})}},$ (79)

and the function $f$ is given by

$f(u,v)=-\sqrt{(1-X^{2})(1-Y^{2})}.$ (80)

The cell $C$ in Lemma 10 is given in $(X,Y)$ coordinates by the square
$-1<X<1$ and $-1<Y<1$.

###### Proof.

The critical points of $f$ happen when both $\cos{mx}\sin{ny}$ and
$\sin{mx}\cos{ny}$ are zero, and these conditions imply that either
$x=\frac{j\pi}{m}$ and $y=\frac{k\pi}{n}$, or $x=\frac{(j+1/2)\pi}{m}$ and
$y=\frac{(k+1/2)\pi}{n}$, for integers $j$ and $k$. The origin is the former
type and has $f_{xx}(0,0)=m^{2}$, $f_{xy}(0,0)=0$, and $f_{yy}(0,0)=n^{2}$.
Points of the latter type form the corners of a rectangular grid in the level
set $f=0$, and the component of the set containing $(0,0)$ and having $f<0$ is
given by $\lvert x\rvert<\frac{\pi}{2m}$ and $\lvert y\rvert<\frac{\pi}{2n}$.
This becomes the square $\lvert X\rvert<1$ and $\lvert Y\rvert<1$ under the
coordinate change, which is bijective on $C$.

Everything else is easy: since $x=\frac{1}{m}\arcsin{X}$ and
$y=\frac{1}{n}\arcsin{Y}$, we get that the metric $ds^{2}=dx^{2}+dy^{2}$
becomes (78). Thus the area form is (80) by the standard definition of area
form on a manifold (or by computing the Jacobian directly). Clearly the
function $f$ is given by (80) on the cell $C$, where $f$ must be negative. ∎

In the $(X,Y)$ coordinates, the velocity field $U=\operatorname{sgrad}f$ is
given by

$U=\frac{1}{\varphi(X,Y)}\left(-\frac{\partial f}{\partial
Y}\,\frac{\partial}{\partial X}+\frac{\partial f}{\partial
X}\,\frac{\partial}{\partial
Y}\right)=-mnY(1-X^{2})\,\partial_{X}+mnX(1-Y^{2})\,\partial_{Y}.$ (81)

To construct the polar coordinates as in Lemma 10, we need to find the flow,
which requires the use of Jacobi elliptic functions.

######  Lemma 15.

For the velocity field $U$ given by (81), the polar coordinates defined by
Lemma 10 can be chosen as

$X=\frac{r\operatorname{cn}(\tau,r)}{\operatorname{dn}(\tau,r)},\qquad
Y=r\operatorname{sn}(\tau,r),\qquad\text{where
}\tau=\frac{2\mathbf{K}(r)\theta}{\pi}.$ (82)

Here $\operatorname{sn}$, $\operatorname{cn}$, and $\operatorname{dn}$ are the
usual Jacobi elliptic functions, while $\mathbf{K}(r)$ is the complete
elliptic integral

$\mathbf{K}(r)=\int_{0}^{\pi/2}\frac{d\phi}{\sqrt{1-r^{2}\sin^{2}{\phi}}}.$

###### Proof.

We choose the curve $c$ in Lemma 10 to be $c(s)=(s,0)$ in the $(X,Y)$
coordinates. Then the flow $\gamma$ is obtained by solving the system

$\frac{dX}{dt}=-mnY(1-X^{2}),\qquad\frac{dY}{dt}=mnX(1-Y^{2}),\qquad
X(0)=s,\qquad Y(0)=0.$ (83)

Using the basic derivative formulas for the Jacobi elliptic functions (see [4]
or [9] for these and other formulas)

$\frac{d}{d\tau}\operatorname{sn}(\tau,r)=\operatorname{cn}(\tau,r)\operatorname{dn}(\tau,r),\qquad\frac{d}{d\tau}\operatorname{cn}(\tau,r)=-\operatorname{sn}(\tau,r)\operatorname{dn}(\tau,r),\qquad\frac{d}{d\tau}\operatorname{dn}(\tau,r)=-r^{2}\operatorname{sn}(\tau,r)\operatorname{cn}(\tau,r)$

together with the initial values

$\operatorname{sn}(0,r)=0,\qquad\operatorname{cn}(0,r)=1,\qquad\operatorname{dn}(0,r)=1$

and the basic identities

$\operatorname{cn}(\tau,r)^{2}+\operatorname{sn}(\tau,r)^{2}=1,\qquad
r^{2}\operatorname{sn}(\tau,r)^{2}+\operatorname{dn}(\tau,r)^{2}=1,\qquad\operatorname{dn}(\tau,r)^{2}-r^{2}\operatorname{cn}(\tau,r)^{2}=1-r^{2},$
(84)

we can compute that

$X(t,s)=\frac{s\operatorname{cn}(mnt,s)}{\operatorname{dn}(mnt,s)},\qquad
Y(t,s)=s\operatorname{sn}(mnt,s)$ (85)

satisfy the system (83).

We have the following identities (analogous to the
$(\theta\mapsto\pi/2-\theta)$ shift identities for standard trigonometric
functions):

$\operatorname{sn}(\tau+\mathbf{K}(r),r)=\frac{\operatorname{cn}(\tau,r)}{\operatorname{dn}(\tau,r)}\qquad\text{and}\qquad\frac{\operatorname{cn}(\tau+\mathbf{K}(r),r)}{\operatorname{dn}(\tau+\mathbf{K}(r),r)}=-\operatorname{sn}(\tau,r),$
(86)

which imply that the period of the flow (85) is
$P(r)=\frac{4\mathbf{K}(r)}{mn}$. Thus the polar coordinate transformation is
(82). ∎

Now we compute the velocity field, metric components, and area form in these
polar coordinates.

######  Lemma 16.

In polar coordinates defined by Lemma 15, the velocity field is given by

$U=\frac{\pi mn}{2\mathbf{K}(r)}\,\frac{\partial}{\partial\theta}$ (87)

while the area form is given by

$\mu=\varphi(r)\,dr\wedge d\theta,\qquad\text{where
}\varphi(r)=\frac{2r\mathbf{K}(r)}{\pi mn\sqrt{1-r^{2}}}.$ (88)

The averages of the metric components from (5) are given by

$G(r)=\frac{4(m^{2}+n^{2})r(1-r^{2})\mathbf{K}(r)\mathbf{K}^{\prime}(r)}{m^{2}n^{2}\pi^{2}}$
(89)

and

$E(r)=\frac{(m^{2}+n^{2})J(r)}{m^{2}n^{2}(1-r^{2})^{2}\mathbf{K}(r)},$ (90)

where $J$ is defined by

$J(r)=\int_{0}^{\mathbf{K}(r)}\big{[}\operatorname{sn}(\tau,r)\operatorname{dn}(\tau,r)-\operatorname{cn}(\tau,r)Z(\tau,r)\big{]}\,d\tau.$
(91)

###### Proof.

We first recall that the function is given in $(X,Y)$ coordinates by (80), and
since our radial curve is $X=r$, $Y=0$, we have $f(X,Y)=f(r,0)$ for all
$(X,Y)$, so that

$(1-X^{2})(1-Y^{2})=1-r^{2}\qquad\text{or}\qquad
r^{2}=X^{2}+Y^{2}-X^{2}Y^{2}.$ (92)

Hence in particular we have

$F(r)=-\sqrt{1-r^{2}}\text{ for }0\leq r\leq 1.$ (93)

Formula (87) follows from formula (46), since the period is
$P(r)=\frac{4\mathbf{K}(r)}{mn}$. Finally formula (45) gives a quick way to
compute the area element $\varphi(r)$, which is

$\varphi(r)=\frac{F^{\prime}(r)}{u(r)}=\frac{2r\mathbf{K}(r)}{\pi
mn\sqrt{1-r^{2}}}.$

Of course we could also compute these quantities directly using the coordinate
transformation (82) and the usual transformation formulas.

Next we need to compute the metric averages $E(r)$ and $G(r)$ from (5). The
easiest way to compute $G(r)$ is to use formula (47). Since

$\omega(r)\varphi(r)=-(m^{2}+n^{2})F(r)\varphi(r),$

formulas (93), (87), and (88) imply that

$\frac{\pi
mn}{2}\frac{d}{dr}\Big{(}\frac{G(r)}{\mathbf{K}(r)}\Big{)}=(m^{2}+n^{2})\frac{2r\mathbf{K}(r)}{\pi
mn},$

and integrating gives

$G(r)=\frac{4(m^{2}+n^{2})}{\pi^{2}m^{2}n^{2}}\,\mathbf{K}(r)\big{[}\mathbf{E}(r)-(1-r^{2})\mathbf{K}(r)\big{]},$

which is equivalent to (89).

Computing $E(r)$ is much more difficult, but it becomes somewhat easier if we
first observe that by the shift formulas (86), the oddness of
$\operatorname{sn}$ and evenness of $\operatorname{cn}$ and
$\operatorname{dn}$, and the definitions (82) of $u$ and $v$, we have the
complementary formula

$X(r,\tfrac{\pi}{2}-\theta)=Y(r,\theta).$ (94)

Also using (86) we can see that integrating any function involving only even
powers of $X$, $X_{\theta}$, or $X_{r}$ around the full $[0,2\pi]$ interval is
the same as integrating it from $0$ to $\pi/2$ and multiplying by $4$. Finally
integrating any even power of $X$ or its derivatives from $0$ to $\pi/2$ is
the same as integrating the corresponding power of a derivative of $Y$
backwards from $\pi/2$ to $0$, which gives the same answer by the obvious
change of variables. Hence we can reduce everything to integrals of
$Y(r,\theta)$.

We first compute

$g_{11}(r,\theta)=\frac{1}{m^{2}(1-X(r,\theta)^{2})}\,\Big{(}\frac{\partial
X}{\partial
r}(r,\theta)\Big{)}^{2}+\frac{1}{n^{2}(1-Y(r,\theta)^{2})}\,\Big{(}\frac{\partial
Y}{\partial r(r,\theta)}\Big{)}^{2}.$

Then using the symmetry described above, we see that

$\begin{split}E(r)&=\frac{1}{2\pi}\int_{0}^{2\pi}g_{11}(r,\theta)\,d\theta\\\
&=\frac{2}{\pi}\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right)\int_{0}^{\pi/2}\frac{1}{(1-Y(r,\theta)^{2}}\,\Big{(}\frac{\partial
Y}{\partial r}(r,\theta)^{2}\Big{)}\,d\theta.\end{split}$ (95)

Computing the radial derivative of $Y(r,\theta)$, we get (using formula 710.51
of [4] or (3.10.9) of [10])

$\displaystyle\frac{1}{\sqrt{1-Y(r,\theta)^{2}}}\frac{\partial
Y(r,\theta)}{\partial r}$
$\displaystyle=\frac{1}{(1-r^{2})\operatorname{dn}(\tau,r)}\,\Big{(}\operatorname{sn}(\tau,r)\operatorname{dn}^{2}(\tau,r)^{2}-\operatorname{dn}(\tau,r)\operatorname{cn}(\tau,r)\operatorname{zn}(\tau,r)\Big{)}$
$\displaystyle=\frac{\operatorname{sn}(\tau,r)\operatorname{dn}^{2}(\tau,r)^{2}-\operatorname{dn}(\tau,r)\operatorname{cn}(\tau,r)\operatorname{zn}(\tau,r)}{1-r^{2}},$

where $\operatorname{zn}$ is the Jacobi Zeta function defined by

$\frac{\partial}{\partial\tau}\operatorname{zn}(\tau,r)=\operatorname{dn}(\tau,r)^{2}-\frac{\mathbf{E}(r)\tau}{\mathbf{K}(r)},$

so that (by the known integral for $\operatorname{dn}^{2}$) we get
$\operatorname{zn}(0,r)=0$ and $\operatorname{zn}(\mathbf{K}(r),r)=0$.

Thus we obtain from (95) that

$\displaystyle E(r)$ $\displaystyle=\frac{2(m^{2}+n^{2})}{\pi
m^{2}n^{2}(1-r^{2})^{2}}\int_{0}^{\pi/2}\big{[}\operatorname{sn}(\tau,r)\operatorname{dn}(\tau,r)-\operatorname{cn}(\tau,r)\operatorname{zn}(\tau,r)\big{]}^{2}\,d\theta$
$\displaystyle=\frac{(m^{2}+n^{2})}{m^{2}n^{2}(1-r^{2})^{2}\mathbf{K}(r)}\int_{0}^{\mathbf{K}(r)}\big{[}\operatorname{sn}(\tau,r)\operatorname{dn}(\tau,r)-\operatorname{cn}(\tau,r)\operatorname{zn}(\tau,r)\big{]}^{2}\,d\tau,$

which is equation (90). ∎

Unfortunately we are not able to get a more explicit form of the function
$J(r)$ from (91). However this ends up not mattering, since we only need a
lower bound for $E$, as in the next proof.

###### Proof of Theorem 6.

First we observe the dependence on the pair $(m,n)$ of positive integers. Let
us denote by $\varphi_{mn}(r)$, $u_{mn}(r)$, $G_{mn}(r)$, and $E_{mn}(r)$ the
formulas (88), (87), (89), and (90), with their explicit dependence on
$(m,n)$. Then we obviously have:

$\displaystyle\varphi_{mn}(r)$
$\displaystyle=\frac{1}{mn}\,\varphi_{11}(r),\qquad$ $\displaystyle\qquad
u_{mn}(r)$ $\displaystyle=mnu_{11}(r)$ $\displaystyle G_{mn}(r)$
$\displaystyle=\frac{m^{2}+n^{2}}{m^{2}n^{2}}\,G_{11}(r),$ $\displaystyle
E_{mn}(r)$ $\displaystyle=\frac{m^{2}+n^{2}}{m^{2}n^{2}}\,E_{11}(r).$

Similarly if we compute $v_{mn}$ using (7), we get

$v_{mn}=\frac{G_{mn}^{\prime}(r)u_{mn}(r)}{2\varphi_{mn}(r)}=(m^{2}+n^{2})\,\frac{G_{11}^{\prime}(r)u_{11}(r)}{2\varphi_{11}(r)}=(m^{2}+n^{2})v_{11}(r),$

so that using (9) we get

$Q_{mn}(r)=\frac{m^{2}+n^{2}}{mn}\,Q_{11}(r).$

We thus find that in the quantity $M_{mn}(r)$ defined by (12), all the terms
have exactly the same dependence on $(m,n)$, with

$M_{mn}(r)=\frac{(m^{2}+n^{2})^{2}}{m^{2}n^{2}}\,M_{11}(r).$

We conclude that when seeking variation fields supported in a single cell of
the flow, the value of $(m,n)$ doesn’t matter: every flow has energy-reducing
variations in a cell if and only if the $(1,1)$ flow does.

Now we will show that the $(1,1)$ flow _does not_. So from now on we set
$m=n=1$ in all the formulas for $u$, $\varphi$, $E$, and $G$. For this purpose
it is easier, since we don’t have an explicit formula for $E(r)$, to estimate
from below using Cauchy-Schwarz: we have

$\displaystyle E(r)G(r)$
$\displaystyle=\frac{1}{2\pi^{2}}\left(\int_{0}^{2\pi}g_{11}(r,\theta)\,d\theta\right)\left(\int_{0}^{2\pi}g_{22}(r,\theta)\,d\theta\right)$
$\displaystyle\geq\frac{1}{2\pi^{2}}\left(\int_{0}^{2\pi}\sqrt{g_{11}(r,\theta)g_{22}(r,\theta)}\,d\theta\right)^{2}$
$\displaystyle\geq\frac{1}{2\pi^{2}}\left(\int_{0}^{2\pi}\varphi(r)\,d\theta\right)^{2}$
$\displaystyle=\varphi(r)^{2}.$

We conclude that

$M_{11}(r)=\frac{G(r)Q^{\prime}(r)}{\varphi(r)}+Q(r)^{2}-\frac{E(r)G(r)}{\varphi(r)^{2}}\leq\overline{M}(r):=\frac{G(r)Q^{\prime}(r)}{\varphi(r)}+Q(r)^{2}-1.$
(96)

For the quantity $\overline{M}(r)$ defined by the right side, we have explicit
formulas. We compute $v_{11}(r)$ from (7), using (89), (87), and (88), as
follows:

$v_{11}(r)=\frac{G^{\prime}(r)u(r)}{2\varphi(r)}=\sqrt{1-r^{2}}\left(1+\frac{(1-r^{2})\mathbf{K}^{\prime}(r)^{2}}{\mathbf{K}(r)^{2}}\right).$

Then we compute from (9) that

$Q_{11}(r)=\frac{v_{11}^{\prime}(r)}{u_{11}^{\prime}(r)}=\frac{2}{\pi
r\sqrt{1-r^{2}}}\Big{(}\frac{r^{2}\mathbf{K}(r)^{2}}{\mathbf{K}^{\prime}(r)}-2r(1-r^{2})\mathbf{K}(r)+(2-3r^{2})(1-r^{2})\mathbf{K}^{\prime}(r)+\frac{r(1-r^{2})^{2}\mathbf{K}^{\prime}(r)^{2}}{\mathbf{K}(r)}\Big{)}.$

With these we can compute for example that for small $r$ we have

$\overline{M}(r)=-\frac{1}{2}\,r^{2}-\frac{35}{64}\,r^{4}\mathrm{O}(r^{8}),$

while for $r\approx 1$ we have

$\overline{M}(r)=-\frac{1}{\pi^{2}}\Big{(}\frac{2}{1-r}+4\big{[}\ln{(1-r)}+\tfrac{15}{4}-3\ln{2}\big{]}^{2}+O(1).$

The graph plotted in Figure 3 shows that $\overline{M}(r)$ is always negative
for $r\in(0,1)$, and thus so is $M(r)$. Now examining the index formula (10)
for variations $\xi$ supported in $[0,R]$, we see that it is positive-definite
since $M(r)\leq 0$, and thus no such variations can detect conjugate points. ∎

Figure 3: A plot of the simplified upper bound $\overline{M}(r)$ from (96),
for the quantity appearing in (10). Negativity of this function everywhere
implies that the index form is positive definite on variations supported in a
single cell of any Kolmogorov flow on the torus.

Obviously there _are_ conjugate points along Kolmogorov flows, as found by
Misiołek [13] and Drivas et al. [6], so it must be that the variation fields
_cannot_ be supported in a single cell. To use the computations in this paper
to prove a general theorem on existence of conjugate points along Kolmogorov
flows, one would need to patch together variation fields in different cells
which are _not_ assumed to vanish on the boundary. The tricky bit here is that
what seem like very well-behaved stream functions on the torus look rather bad
in our polar coordinates, since the polar coordinates themselves degenerate on
the cell boundaries. This difficulty seems quite surmountable, but would take
us too far afield here, so we leave it for future work.

We thus find that for Kolmogorov flows on the flat torus, the method proposed
in this paper does not improve upon the original M-criterion for detecting
conjugate points. So instead we will apply that method in the Appendix to find
a few more examples.

##  7 Examples

In this section we will work out some details for explicit examples where the
criteria of Theorem 1 and its corollaries produce conjugate points.

First we give an example of Arnold-stable and nonisochronal flows with
conjugate points, answering a question posed by Drivas et al. [6]. Of course
the isochronal flow on the same constant curvature hyperbolic disc will also
be an Arnold-stable flow with conjugate points.

######  Example 2.

Consider the hyperbolic disc with metric
$ds^{2}=dr^{2}+\sinh^{2}{r}\,d\theta^{2}$ for $r\in[0,R]$, for $R=\ln{\rho}$
with $\rho>1$. Choose $F(r)=\frac{1}{2}\cosh^{2}{r}$, so that $u(r)=\cosh{r}$
and

$\omega(r)=\frac{1}{\varphi(r)}\,\frac{d}{dr}\big{(}\phi(r)^{2}u(r)\big{)}=3\cosh^{2}{r}-1=6F(r)-1.$

Then $\frac{\omega^{\prime}(r)}{F^{\prime}(r)}=6$, which implies that $U$
satisfies the nonlinear Arnold stability criterion for steady 2D Euler flows.
The function $v$ from (7) is given by $v(r)=\cosh^{2}{r}$.

Choose test function $\xi(r)=\sinh{r}\sinh{(R-r)}$. Then the index given by
(8) can be computed explicitly and gives

$I(\alpha,\rho)=\frac{\left(\rho-1\right)^{4}}{3360\rho^{5}}\,\Big{(}84\rho^{2}\left(\rho^{2}+8\rho+1\right)\alpha^{2}-105\rho(\rho+1)^{4}\alpha\\\
+23\rho^{6}+92\rho^{5}+181\rho^{4}+248\rho^{3}+181\rho^{2}+92\rho+23\Big{)}.$

This is minimized at

$\alpha=\frac{5(\rho+1)^{4}}{8\rho(\rho^{2}+8\rho+1)},$

and for this $\alpha$ we obtain

$I=-\frac{(157\rho^{4}+412\rho^{3}+366\rho^{2}+412\rho+157)(\rho-1)^{8}}{53760\rho^{5}(\rho^{2}+8\rho+1)},$

which is obviously negative, so we get a conjugate point regardless of the
radius $R=\ln{\rho}$.

Here we give an example of how to use the criterion from Corollary 2 to select
an appropriate test function $\xi$.

######  Example 3.

Consider the standard $2$-sphere with $R=\pi$ and $\varphi(r)=\sin{r}$. Set
$u(r)=\frac{7}{4}+4\cos{r}+\cos^{2}{r}$, which has $u^{\prime}(r)<0$ for
$r\in(0,\pi)$. We compute that

$Q(r):=\frac{1}{u^{\prime}(r)}\frac{d}{dr}\big{(}\phi^{\prime}(r)u(r)\big{)}=\frac{7+32\cos{r}+12\cos^{2}{r}}{16+8\cos{r}},$

which leads to $M$ defined by (12) being given by

$M(r)=\frac{3}{64}\,\frac{80\cos^{4}{r}+384\cos^{3}{r}+496\cos^{2}{r}-64\cos{r}-221}{(2+\cos{r})^{2}}.$

This function $S$ is positive on $[0,0.3\pi]$ and again on $[0.83\pi,\pi]$ as
shown in Figure 4. Choose

$\xi(r)=\begin{cases}\sin{4r}&0\leq r\leq\frac{\pi}{4},\\\
0&\text{otherwise.}\end{cases}$

Then the index form (6) is given by

$I(\alpha)=\int_{0}^{\pi/4}\big{[}u(r)-\alpha\big{]}^{2}\Big{[}\sin{r}\Big{(}\xi^{\prime}(r)-\frac{Q(r)}{\varphi(r)}\xi(r)\Big{)}^{2}-\frac{M(r)}{\sin{r}}\,\xi(r)^{2}\Big{]}\,dr,$
(97)

and we can smooth $\xi$ out to a $C^{\infty}$ function at the expense of an
arbitrarily small increase in $I$. As a function of $\alpha$, the index $I$ is
given by

$I(\alpha)\approx 3.549+\alpha^{2}-38.57\alpha+104.2,$

so that in particular $I(5.431)\approx-0.5635$; see the graph in Figure 4.
Thus $\eta(T)$ is eventually conjugate to the identity for sufficiently large
$T$.

Figure 4: On the left, the function $S=\varphi Q^{\prime}+Q^{2}-1$, for $u$
defined on the round $2$-sphere by $u(r)=\frac{7}{4}+4\cos{r}+\cos^{2}{r}$. In
particular it is positive on $[0,\frac{\pi}{4}]$, which is the support of the
perturbation $\xi$. On the right, the index form $I$ defined by (97),
considered for the function $\xi(r)=\sin{4r}$ on $[0,\tfrac{\pi}{4}]$ as a
function of $\alpha$.

Next we give an example applying the local criterion of Corollary 4, in the
case where $u$ has a local minimum in the interior.

######  Example 4.

If $\varphi(r)=\sin{r}$ and $u(r)=\frac{9}{8}-\sqrt{2}\cos(r)+\cos^{2}(r)$,
then $u^{\prime}(\tfrac{\pi}{4})=0$ and
$\dfrac{u(\tfrac{\pi}{4})}{u^{\prime\prime}(\tfrac{\pi}{4})}=\frac{5}{8}>\frac{9}{16}$.
We compute that

$\omega(r)=-3\sqrt{2}\,\cos^{2}{r}+4\cos^{3}{r}+\sqrt{2}+\tfrac{1}{4}\cos{r}.$

If we choose $\epsilon=\delta=\frac{1}{10}$ in the notation of the proof of
Corollary 4, and set

$\xi(r)=\frac{\frac{\pi}{4}+\tfrac{1}{10}-r}{\lvert
r-\frac{\pi}{4}\rvert^{\frac{7}{5}}}\qquad\text{for
$r\in[\tfrac{\pi}{4},\tfrac{\pi}{4}+\tfrac{1}{10}$,}$

and zero otherwise, then the integrand in (6) can be expressed in a series as

$\varphi(r)\left(\frac{d}{dr}\Big{[}\big{(}\alpha-u(r))\xi(r)\Big{]}\right)^{2}+\frac{1}{\varphi(r)}\big{(}\alpha-u(r)\big{)}^{2}\xi(r)^{2}-\omega^{\prime}(r)\big{(}\alpha-u(r)\big{)}\xi(r)^{2}=\\\
-\frac{7\sqrt{2}}{40000(r-\frac{\pi}{4})^{\frac{4}{5}}}+\frac{51\sqrt{2}\,\left(r-\frac{\pi}{4}\right)^{\frac{1}{5}}}{10000}+\cdots,$

and this is integrable on $[\tfrac{\pi}{4},\tfrac{\pi}{4}+\frac{1}{10}]$,
giving a negative value.

The variation field here is supported in a very small neighborhood of the
circle $r=\tfrac{\pi}{4}$ on the sphere but approaches infinity as
$r\to\tfrac{\pi}{4}$. Setting $\tilde{\xi}(r)$ to be the even extension of
$\xi(r)$ around $r=\tfrac{\pi}{2}$, with
$\tilde{\xi}(r)=\xi(\tfrac{\pi}{2}+\beta)$ for $\lvert
r-\tfrac{\pi}{2}\rvert\leq\beta$ for an even smaller $\beta$, will make this
function finite and piecewise $C^{1}$, while still yielding a negative
integral for (6).

Finally we present an example of applying Corollary 5, using also Lemma 12 to
demonstrate how this all can work in the non-rotationally symmetric case.

######  Example 5.

Suppose $\varphi(r)=r$, $G(r)=r^{2}-\frac{r^{4}}{8}$, and
$u(r)=5+\tfrac{1}{2}r^{2}$. Let $\eta(r,\theta)=r^{6}\cos{2\theta}$. By
formulas (53), we have

$\begin{split}g_{11}&=\frac{1152r^{8}\cos^{2}{2\theta}+8(r^{2}+10)^{2}}{(r^{2}+10)(80-r^{4}-32r^{2}\sin{2\theta}-2r^{2})}\\\
g_{12}&=\frac{12r^{5}\cos{2\theta}}{10+r^{2}}\\\
g_{22}&=\frac{r^{2}(80-r^{4}-32r^{2}\sin{2\theta}-2r^{2})}{8(10+r^{2})}.\end{split}$
(98)

We easily verify that $g_{11}(0,\theta)=1$ so that $E(0)=1$, that
$g_{11}g_{22}-g_{12}^{2}=r^{2}$, and that the vorticity of $U$ is given by
$\omega=10r-\tfrac{1}{2}r^{3}-\tfrac{3}{8}r^{5}$.

Since $E(0)=1$, $G^{\prime\prime}(0)=2$, $G^{iv}(0)=-3$, $u(0)=5$,
$u^{\prime\prime}(0)=1$, $\phi^{\prime}(0)=1$, and
$\phi^{\prime\prime\prime}(0)=0$, the condition (14) becomes

$3+24<\frac{10}{1}\cdot\big{(}0-(-3)\big{)},$

which is of course satisfied. Hence there is a conjugate point along this
geodesic, and the variation test field is supported in an arbitrarily small
neighborhood of the origin.

To get a sense of the geometry of this very nonflat surface, we plot the
sectional curvature on the disc of radius $R=1$, as computed in Maple; in
particular it is positive near the origin but takes on both signs as we move
away.

Figure 5: The sectional curvature for the metric given by the components (98)
on the disc $r\leq 1$. The steady flow velocity profile has a local minimum at
the origin, and satisfies the condition of Corollary 5 to have an energy-
reducing variation supported near the origin, but the surface itself does not
have any rotational symmetry.

##  Appendix A Appendix: The M-criterion for Kolmogorov flows on
$\mathbb{T}^{2}$

As noted in Section 6, the criterion of Theorem 1 is not able to detect
conjugate points along Kolmogorov flows on $\mathbb{T}^{2}$, where the stream
functions are given by $f(x,y)=-\cos{mx}\cos{ny}$ for $m,n\in\mathbb{N}$.
(Because of the symmetry between $x$ and $y$, there is no loss of generality
in assuming $m\geq n$, which we will do throughout.)

In this case the Misiołek criterion becomes, using Corollary 9 with a stream
function of the form $g(t,x,y)=\sin{\tfrac{t}{T}}\zeta(x,y)$, that

$\int_{\mathbb{T}^{2}}\lvert\nabla\phi\rvert^{2}-(m^{2}+n^{2})\phi^{2}\,dx\,dy<0,\qquad\text{where
}\phi=\\{f,\zeta\\}$ (99)

for some function $\zeta\colon\mathbb{T}^{2}\to\mathbb{R}$.

The result of Drivas, Misiołek, Shi, and Yoneda [6] can be stated as follows.

######  Theorem 17 (DMSY).

If $(m,n)\in\mathbb{N}^{2}$ with $n\geq 2$ and $m>\frac{3n^{2}+6}{\sqrt{3}n}$,
then the test function $\zeta(x,y)=\cos{(mx+y)}\cos{ny}$ gives a negative
value in (99) on the stream function $f(x,y)=-\cos{mx}\cos{ny}$. Hence the
corresponding geodesic eventually has a conjugate point.

In this appendix, we will fill out some of the diagram of Kolmogorov flows
known to eventually have conjugate points, as shown in Figure 6.

Figure 6: Here we display the values of $(m,n)$ with $m\geq n$ for which the
Kolmogorov flow $f(x,y)=-\cos{mx}\cos{ny}$ is known to have conjugate points.
The diamond shapes in red,enclosed in the region $n\geq 2$,
$m>\frac{3n^{2}+6}{\sqrt{3}n}$, are those found by Drivas et al. [6]. The red
X marks when $n=0$ are known to have no conjugate points. The green circles
are the new ones found here, while all unfilled circles are still unknown.

######  Theorem 18.

The geodesics corresponding to Kolmogorov flows $f(x,y)=-\cos{mx}\cos{ny}$
eventually have conjugate points if $n=1$ and $m\geq 2$. In addition the pairs
$(2,2)$, $(3,2)$, and $(3,3)$ also give geodesics with eventual conjugate
points.

###### Proof.

All we have to choose a test function $\zeta$ to make the M-criterion (99)
satisfied in each case.

For $f(x,y)=-\cos{mx}\cos{y}$, choose

$\zeta_{m1}(x,y)=\cos{x}(4+\cos{2y}).$

Then

$\phi:=\\{f,g\\}=2m\sin{mx}\cos{y}\sin{2y}\cos{x}+\cos{mx}\sin{x}\sin{y}(4+\cos{2y}),$

and the index is

$\iota(\zeta,\zeta)=\iint_{\mathbb{T}^{2}}\lvert\nabla\phi\rvert^{2}-(m^{2}+1)\phi^{2}\,dA=\frac{\pi^{2}(29-12m^{2})}{4},$

which is negative for all integers $m\geq 2$.

For $f(x,y)=-\cos{2x}\cos{2y}$, we choose

$\zeta_{22}(x,y)=235(\cos{x}+\cos{y})-27(\cos{3x}+\cos{3y})-9(\cos{5x}+\cos{5y})\\\
-5(\cos{4x}\cos{5y}+\cos{5x}\cos{4y})-10(\cos{3x}\cos{4y}+\cos{4x}\cos{3y}).$
(100)

We then compute that if $\phi=\\{f,\zeta\\}$, then the index (99) is

$\iota(\zeta,\zeta)=\iint_{\mathbb{T}^{2}}\lvert\nabla\phi\rvert^{2}-8\phi^{2}\,dA=-854\pi^{2}.$

Finally for $f(x,y)=-\cos{3x}\cos{2y}$, choose

$\zeta_{32}(x,y)=(1320\cos{6x}+235\cos{12x})\cos{y}+(1000+1060\cos{6x}+190\cos{12x})\cos{3y}\\\
+(550+650\cos{6x}+125\cos{12x})\cos{5y}+(195+270\cos{6x}+63\cos{12x})\cos{7y}\\\
+(50+73\cos{6y}+22\cos{12y})\cos{9y}+(10+10\cos{6x}+3\cos{12x})\cos{11y}.$
(101)

Then we obtain

$\iota(\zeta,\zeta)=\iint_{\mathbb{T}^{2}}\lvert\nabla\phi\rvert^{2}-13\phi^{2}\,dA=-\frac{764865\pi^{2}}{8}.$

For $f(x,y)=-\cos{3x}\cos{3y}$, choose

$\zeta_{33}(x,y)=1000\cos{x}-18\cos{5x}\cos{6y}-42\cos{5x}-20\cos{7x}-11\cos{7x}\cos{6y}.$

Then the index is

$\iota(\zeta,\zeta)=\iint_{\mathbb{T}^{2}}\lvert\nabla\phi\rvert^{2}-18\phi^{2}\,dA=-51939\pi^{2}.$

∎

Once the test functions are guessed, verifying that the index is negative is
straightforward. We can find them by searching in the space of Fourier
coefficients, but especially as seen from the $(3,2)$ case, this starts to
require many modes beyond those we start with. We have scaled the variation
fields so that all the coefficients are integers (which allows us to compute
the integrals exactly rather than numerically). The coefficients are quite
sensitive: most of the ones in the $(3,2)$ case cannot be changed to a
different integer without making the index positive. All computations here
were done in Maple 2021. In Figure 7 we plot the variation fields whose
formulas are given in the proof of Theorem 18.

Figure 7: Variation fields which eventually reduce the energy along specified
Kolmogorov flows determined by parameters $(m,n)$. Top row: $(2,1)$ and
$(2,2)$. Bottom row: $(3,2)$ and $(3,3)$.

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