Research ContributionArnold Mathematical Journal

Received: 23 May 2017 / Revised: 24 July 2017 / Accepted: 29 July 2017

Integral Geometry of Euler Equations
Aix Marseille Université, CNRS, Centrale Marseille, I2M UMR 7373 13453 Marseille France nnicolas@yandex.ru, Aix Marseille Université, CNRS, Centrale Marseille, I2M UMR 7373 13453 Marseille France IITP RAS 19 B. Karetnyi Moscow Russia serge.vladuts@univ-amu.fr
###### Abstract
We develop an integral geometry of stationary Euler equations defining some function $w$ on the Grassmannian of affine lines in $\R^3$ depending on a putative compactly supported solution $(v;p)$ of the system and deduce some linear differential equations for $w$. We conjecture that $w=0$ everywhere and prove that this conjecture implies that $v=0.$
###### Keywords
Euler equations, Integral geometry, Tensortomography
76B03, 35J61

## 1. Introduction

In the present paper we introduce and develop a version of integral geometry for the steady Euler system.

More precisely, the system which we consider is as follows

 \begin{eqnarray}\label{eqE} \sum_{j=1}^3\frac{\pa (v^iv^j)}{\pa x_j}+\frac{\pa p}{\pa x_i}=0\quad\hbox{for } i=1,2,3, \end{eqnarray} (1.1)
for an unknown smooth vector field $v=v(x)=(v^1(x),v^2(x),v^3(x))$ and an unknown smooth scalar function $p=p(x),\,x\in \R^3$; it expresses the conservation of fluid's momentum $v\otimes v +p\de_{ij}$ and reads in a coordinate free form as follows
 \begin{eqnarray}\label{Ed} \div (v\otimes v)+\nabla p=0. \end{eqnarray} (1.2)
Note that if we add to (1.1) the incompressibility condition
 \begin{eqnarray}\label{div} \div \,v=0, \end{eqnarray} (1.3)
the system (1.2) and (1.3) describes a steady state flow of the ideal fluid.

A long-standing folklore conjecture states that a smooth compactly supported solution of (1.2) and (1.3) should be identically zero, and this is known for Beltrami flows; see [Nadirashvili2014] and also [Chae and Constantin2015]. Let us state it explicitly:

###### Conjecture 1.1.
Let $(v;p)\in C_0(\R^3)$ be a solution of (1.2) (1.3) . Then $v=0,\,p=0.$

Note, however, that there do exist nontrivial Beltrami flows slowly decaying at infinity; see [Enciso and Peralta-Salas2012]. Note also that nontrivial compactly supported solutions of system (1.1) exist, e.g., any spherically symmetric vector field $v$ is a solution of (1.1) for a suitable pressure $p$, but we do not know whether the system \begin{eqnarray*} \sum_{j=1}^3v^j\frac{\pa v^i}{\pa x_j}+\frac{\pa p}{\pa x_i}=0,\quad i=1,2,3 \end{eqnarray*} which is equivalent to (1.1) and to (1.2) under the additional condition (1.3) , admits a non-trivial compactly supported solution not satisfying (1.3) .

Our idea is to separate the study of system (1.1) from the study of Eq. (1.3) using an integral geometry technique which permits to encode (almost) all information about a compactly supported solution $v$ of (1.1) in a single scalar function $w$ defined on the Grassmannian $M$ of affine lines in space. We conjecture that this function $w$ is in fact identically zero, which can readily be verified for any spherically symmetric solution. If it were the case, then adding the incompressibility condition (1.3) to the condition $w=0$ we were able to obtain that $v=0$ everywhere. In order to proof that $w=0$ we deduce a linear differential equation for $w$ on $M$. Unfortunately, this equation is not elliptic and admits non-trivial solutions, and therefore, the purported annulation of $w$ remains conjectural, despite some additional integral geometry arguments going in the same direction.

More concretely, below we characterize the kernel of (1.1) in terms of integral transforms of the quadratic forms $v^iv^j$. Thus, given any smooth compactly supported solution $(v;p)$ of (1.1) , we define a smooth function $w$ on the Grassmannian $M$ classifying lines in space using the $X$-ray transforms of $v^iv^j$ and then derive a linear differential equation for $w$. Using a Radon plane transform of $w$ we deduce one more linear homogeneous differential equation which suggests that $w=0$ everywhere. However, we are not able to justify this claim rigorously and we formulate it as a conjecture; we show that assuming the conjecture and (1.3) one can deduce Conjecture 1.1. Therefore, we put forth

###### Conjecture 1.2.
Let $w$ be the function on the Grassmannian $G_{1,3}$ of affine lines in $\R^3,$ defined below in Sect. 3, which depends on a compactly supported solution $(v;p)$ of (1.1) . Then $w=0$ everywhere.

Note, in particular, that this conjecture holds for any spherically symmetric compactly supported vector field $v.$

Note also that it is possible to prove that the general solution of Eq. (4.5) depends on two arbitrary (smooth compactly supported) functions (V.A. Sharafutdinov, personal communication) and therefore to prove Conjecture 1.2 one needs to use some non-linear argument, e.g. using the fact that the tensor $v\otimes v$ is of rank 1 as opposed to the general symmetric tensor of rank 3 which also gives a solution of (4.5) by constructing the corresponding function $w$; thus, we can say that we have (at least) two more independent conditions on $w$. Unfortunately, we not know how to use this rank conditions to restrict possible functions $w$'s. However, there do exist some natural linear conditions for a suitable tensor which are sufficient to prove Conjecture 1.2, for instance, system (5.4) which is partially supported by Eq. (5.3) .

The rest of the paper is organized as follows: in Sect. 2 we recall some definitions and results from [Sharafutdinov1994] concerning the $X$-ray transform of symmetric tensor fields. In Sect. 3 we define and study a smooth function $w\in C^{\infty}(M)$ which depends on a smooth compactly supported solution $(v,p)$ of (1.1) . Section 4 contains a description of two invariant order 2 differential operators on $C^\infty(M)$ and a differential equation for $w$ in terms of those operators. In Sect. 5 we define a plane Radon transform for quadratic tensor fields, prove that it vanishes for some quadratic tensor field connected with $w$ and explain why this annulation partially confirms Conjecture 1.2. Finally, in Sect. 6 we deduce Conjecture 1.1 assuming Conjecture 1.2 together with (1.3) .

## 2. Tensor $X$-ray Transform

We use throughout our paper the integral geometry of tensor fields developed in [Sharafutdinov1994] and discussed in [Nadirashvili et al.2016] in its three-dimensional form. Let us give some of its points in our simple situation. For details see [Sharafutdinov1994] and [Nadirashvili et al.2016].

In what follows we fix a positive scalar product $\langle x,y\rangle, \, x,y\in \R^n$. Let \begin{eqnarray*} T{\mathbb S}^{n-1}=\{(x,\xi)\in{\mathbb R}^n\times{\mathbb R}^n:\,\|\xi\|=1,\,\langle x,\xi\rangle =0\}\subset{\mathbb R}^n\times({\mathbb R}^n{\setminus}\{0\}) \end{eqnarray*} be the tangent bundle of ${\mathbb S}^{n-1}\subset{\mathbb R}^n$.

Given a continuous rank $h$ symmetric tensor field $f$ on ${\mathbb R}^n$, the X-ray transform of $f$ is defined for $(x ,\xi)\in{\mathbb R}^n\times ({\mathbb R}^n{\setminus}\{0\})$ by

 \begin{eqnarray} (If)(x,\xi)=\sum\limits_{i_1,\dots,i_h=1}^n\;\int\limits_ {-\infty}^\infty f_{i_1\dots i_h}(x+t\xi)\,\xi_{i_1}\dots\xi_{i_h}\,dt \label{eq0} \end{eqnarray} (2.1)
under the assumption that $f$ decays at infinity so that the integral converges.

We denote by ${\mathcal S} (S^h;{\mathbb R}^n)$ the space of symmetric degree $h$ tensor fields with all components lying in the Schwartz space, and denote by ${\mathcal S}(T{\mathbb S}^{n-1})$ the Schwartz space on $T{\mathbb S}^{n-1}$. Below we consider only tensors from ${\mathcal S} (S^h;{\mathbb R}^n)$ and functions from ${\mathcal S}(T{\mathbb S}^{n-1})$. For such $f\in {\mathcal S} (S^h;{\mathbb R}^n)$ we get a $C^\infty$-smooth function $\psi(x,\xi)=(If)(x,\xi)$ on ${\mathbb R}^n\times({\mathbb R}^n{\setminus}\{0\})$ satisfying the following conditions:

which mean that $(If)(x,\xi)$ actually depends only on the line passing through the point $x$ in direction $\xi$, and we parameterize the manifold of oriented lines in ${\mathbb R}^n$ by $T{\mathbb S}^{n-1}$. For $\chi(x,\xi)\in {\mathcal S}(T{\mathbb S}^{n-1})$ we can extend $\chi$ by homogeneity, setting $\chi(x,\xi)=\chi(x,\xi/\|\xi\|)$, to the open subset ${\mathcal Q}\cap \{\xi\ne 0 \}$ of the quadric \begin{eqnarray*} {\mathcal Q}=\{(x,\xi)\in {\mathbb R}^n\times{\mathbb R}^n:\, \langle x,\xi\rangle =0\} \supset T{\mathbb S}^{n-1}. \end{eqnarray*}

Conversely, for a tensor field $f\in{\mathcal S}(S^h;{\mathbb R}^n)$, the restriction $\chi=\psi|_{T{\mathbb S}^{n-1}}$ of the function $\psi=If$ to the manifold $T{\mathbb S}^{n-1}$ belongs to ${\mathcal S}(T{\mathbb S}^{n-1} )$. Moreover, the function $\psi$ is uniquely recovered from $\chi$ by the formula

 \begin{eqnarray} \psi(x,\xi)=\|\xi\|^{h-1}\chi\left(x-\frac{\langle x,\xi\rangle}{\|\xi\|^2} \xi,\frac{\xi}{\|\xi\|}\right), \label{1sps} \end{eqnarray} (2.3)
which follows from (2.2); note that $\Big(x-\frac{\langle x,\xi\rangle}{\|\xi\|^2}\xi,\frac{\xi}{\|\xi\|}\Big)\in T\S^{n-1} \subset {\mathcal Q}$, and thus the right-hand side of (2.3) is correctly defined. Therefore, the X-ray transform can be considered as a linear continuous operator $I\colon {\mathcal S}(S^h;{\mathbb R}^n)\lra{\mathcal S}(T{\mathbb S}^{n-1})$, and now we are going to describe its image and kernel.

The image of the operator $I$ is described by Theorem 2.10.1 in [Sharafutdinov1994] as follows. John's Conditions. A function $\chi\in{\mathcal S}(T{\mathbb S}^{n-1})\ (n\geq3)$ belongs to the range of the operator $I$ if and only if the following two conditions hold :

1. 1. $\chi(x,-\xi)=(-1)^h\chi(x,\xi);$
2. 2. The function $\psi\in C^\infty({\mathbb R}^n\times({\mathbb R}^n{\setminus}\{0\}))$ defined by (2.3) satisfies the equations
 \begin{eqnarray} \left(\frac{\partial^2}{\partial x_{i_1}\partial\xi_{j_1}}-\frac{\partial^2}{\partial x_{j_1}\partial\xi_{i_1}}\right)\dots \left(\frac{\partial^2}{\partial x_{i_{h+1}}\partial\xi_{j_{h+1}}}-\frac{\partial^2}{\partial x_{j_{h+1}}\partial\xi_{i_{h+1}}}\right)\psi=0 \label{JE} \end{eqnarray} (2.4)
written for all indices $1\leq i_1,j_1,\ldots,i_{h+1},j_{h+1}\leq n.$

Define the symmetric inner differentiation operator $d_s=\si \nabla$ by symmetrization of the covariant differentiation operator $\nabla\colon C^{\infty} (S^h) \lra C^{\infty}(T^{h+1}),$ \begin{eqnarray*} (\nabla u)_{i_1,\ldots i_{h+1}} =u_{i_1\ldots i_h;i_{h+1}}=\frac{\pt u_{i_1\ldots i_h}}{\pt x_{i_{h+1}}}; \end{eqnarray*} it does not depend on the choice of a coordinate system.

The kernel of the operator $I$ is given by (Theorem 2.2.1, (1),(2) in [Sharafutdinov1994]). Kernel of the Ray Transform. Let $n\ge 2$ and $h\ge 1$ be integers. For a compactly-supported field $f\in C^{\infty}_0(S^h;\R^n)$ the following statements are equivalent :

1. 1. $If = 0;$
2. 2. There exists a compactly-supported field $v \in C^{\infty}_0(S^{h-1};\R^n)$ such that its support is contained in the convex hull of the support of $f$ and
 \begin{eqnarray}\label{dv} d_sv = f. \end{eqnarray} (2.5)

Note also that an inversion formula for the operator $I$ given by Theorem 2.10.2 in [Sharafutdinov1994] implies that it is injective on the subspace of divergence-free ($=$solenoidal) tensor fields. The 3-Dimensional Case For $n=3$ one notes that the tangent bundle $T\S^2$ over $\S^2$ coincides with the homogeneous space $M=G/(\R\times{\rm SO}(2))=G'/(\R\times {\rm O}(2))$, where $G=\R^3\rtimes{\rm SO}(3)$ is the group of proper rigid motions of $\R^3$, while $G'=G\cdot\{\pm I_3\}$ is the isometry group of $\R^3$. Therefore the operator $I$ for $n=3$ can be written as $I\colon{\mathcal S}(S^h;{\R}^3)\lra{\mathcal S}(M).$

Let us define coordinates on the open subset $M_{nh}$ of $M$ consisting of non-horizontal affine lines. Namely, $m=m\left(y_1,y_2,\alpha_1,\alpha_2 \right)$ is given by a parametric equation for a current point $A$ on $m$, \begin{eqnarray*} A=(y_1,y_2,0)+t\alpha =(y_1+\alpha_1t,y_2+\alpha_2t,t), \end{eqnarray*} where $t$ grows in the positive direction of $m$ and thus the vector $\alpha=(\alpha_1,\alpha_2,1)$ defines the positive direction of $m$.

We can now rewrite the above general formulas using the coordinates $(y_1,y_2,\alpha_1,\alpha_2)$. First we fix the following notation:

 \begin{eqnarray}\label{eqk} k=k(\alpha_1,\alpha_2)=\sqrt{1+\alpha_1^2+ \alpha_2^2}=\sqrt{1+\|\alpha\|^2}; \end{eqnarray} (2.6)
we will use this notation throughout the paper.

Define the diffeomorphism \begin{eqnarray*} \Phi\colon U\rightarrow{\mathbb R}^4,\quad(x,y,z,\xi)=(x,y,z,\xi_1,\xi_2,\xi_3) \mapsto(y,\alpha)=(y_1,y_2,\alpha_1,\alpha_2), \end{eqnarray*} on the open set $U= T{\S}^2\cap \{ \xi_3> 0\}$ by

Then $(U,\Phi)$ is a coordinate patch on $M$; this parametrization was used by F. John in his seminal paper ([John1938]).

For a function $\chi\in C^\infty(U)$, we define $\varphi\in C^\infty({\mathbb R}^4)$ by \begin{eqnarray*} \varphi=k^{h-1}\chi\circ\Phi^{-1}. \end{eqnarray*} These two functions are expressed through each other by the formulas

 \begin{eqnarray} &&\chi(x,y,z,\xi)=\xi_3^{h-1}\varphi\left(x-\frac{\xi_1z}{\xi_3} ,y-\frac{\xi_2z}{\xi_3},\frac{\xi_1}{\xi_3},\frac{\xi_2}{\xi_3}\right),\nonumber\\ &&\label{phi} \varphi(y,\alpha) =k^{h-1}\chi\left(y_1-\frac{\langle y,\alpha\rangle\alpha_1} {k^2},y_2- \frac{\langle y,\alpha\rangle\alpha_2}{k^2}, -\frac{\langle y,\alpha \rangle }{k}, \frac{\alpha_1 }{k},\frac{\alpha_2}{k},\frac{1}{k}\right).\qquad\qquad \end{eqnarray} (2.8)

If a function $\chi\in C^\infty(T{\mathbb S}^2)$ satisfies $\chi(-x,-\xi)=(-1)^h\chi(x,\xi)$, then it is uniquely determined by \begin{eqnarray*} \varphi=k^{h-1}\chi|_U\circ\Phi^{-1}\in C^\infty({\mathbb R}^4). \end{eqnarray*} For a tensor field $f\in{\mathcal S}(S^h;{\mathbb R}^3)$, the function

 \begin{eqnarray}\label{phik} \varphi=k^{h-1}(If)|_U\circ\Phi^{-1}\in C^\infty({\mathbb R}^4) \end{eqnarray} (2.9)
is expressed through $f$ by the formula
 \begin{eqnarray}\label{eq1.4} \varphi(y,\alpha)=\sum\limits_{i_1,\dots,i_h=1}^3\;\int\limits_{-\infty}^\infty f_{i_1\dots i_h}(y_1+\alpha_1t,y_2+\alpha_2t,t)\,\alpha_{i_1}\dots\alpha_{i_h}dt, \end{eqnarray} (2.10)
with $\alpha_3=1$, which easily follows from (2.1).

Let

 \begin{eqnarray} \label{eql} L=_{\rm def} {\partial^2\over \partial \alpha_2\partial y_1}- {\partial^2\over \partial \alpha_1 \partial y_2 } \end{eqnarray} (2.11)
be the John operator. The main result of [Nadirashvili et al.2016] says that for $n=3$, a function $\chi\in{\mathcal S}(T{\mathbb S}^2)$ belongs to the range of the operator $I$ for a given $h\ge 0$ if and only if the following two conditions hold:
1. 1. $\chi(-x,-\xi)=(-1)^h\chi(x,\xi)$;
2. 2. The function $\varphi\in C^\infty({\mathbb R}^4)$ defined by (2.8) solves the equation
 \begin{eqnarray} L^{h+1}\varphi=0. \label{eq1.5} \end{eqnarray} (2.12)
Thus ${h^2+5h+6\over 2}$ Eq. (2.4) for $n=3$ are equivalent to Eq. (2.12) .

## 3. Function $w$

In what follows we fix a compactly supported smooth solution $(v,p)\in C^\infty_0(\R^3)$ of system and define a function $w\in C^\infty_0 (M)$ using the following result.

###### Lemma 3.1.
Let $L$ be an affine plane in $\R^3$ and let $\nu_L$ be its unit normal$,$ then
 \begin{eqnarray}\label{eq3.2} \int_{L}\langle v,z\rangle\langle v,\nu_L\rangle \,d\sg_L =0 \end{eqnarray} (3.1)
for any $z\in L$ where $d\sg_L$ is the area element on $L$.

###### Proof.
We can assume without loss of generality that $L=\{(x_1,x_2,0)\}$ and $\nu_L =(0,0,1)=e_3,\, z=z_1e_1+z_2e_2$ for $e_1=(1,0,0),\,e_2=(0,1,0),\, e_3=(0,0,1)$. Then we have \begin{eqnarray*} \int_{L}\langle v,z\rangle\langle v,\nu\rangle \,d\sg_L &=&\int (z_1v^1+z_2v^2)v^3 \,dx_1dx_2\\ &=&z_1\int v^1v^3 \,dx_1dx_2+z_2\int v^2v^3 \,dx_1dx_2. \end{eqnarray*}

Note that by Eq. (1.1) with $i=1$ and $i=2$ there holds \begin{eqnarray*} &&\frac{\pa (v^1v^3)}{\pa x_3} =-\frac{\pa (v^1v^2)}{\pa x_2}-\frac{\pa(v^1v^1)}{\pa x_1}-\frac{\pa p}{\pa x_1},\\ &&\frac{\pa (v^2v^3)}{\pa x_3}=- \frac{\pa (v^2v^1)}{\pa x_1}- \frac{\pa(v^2v^2)}{\pa x_2}-\frac{\pa p}{\pa x_2}, \end{eqnarray*} and thus we get \begin{eqnarray*} &&\frac{\pa }{\pa x _3}\left(\int v^1v^3 \,dx_1dx_2\right)=\int \frac{\pa (v^1v^3)}{\pa x_3}\,dx_1dx_2=0,\\ &&\frac{\pa }{\pa x _3}\left(\int v^2v^3 \,dx_1dx_2\right)=\int \frac{\pa (v^2v^3)}{\pa x_3}\,dx_1dx_2=0. \end{eqnarray*} Therefore, the compactly supported functions \begin{eqnarray*} \int v^1v^3 \,dx_1dx_2\quad\hbox{and}\quad \int v^2v^3 \,dx_1dx_2 \end{eqnarray*} of $x_3$ on $\R$ are constant and thus vanish everywhere which finishes the proof.

For any fixed value of $x_3$, we define the vector field $v^\perp v^3=(-v^2v^3,\,v^1v^3)$ on the plane $(x_1,x_2,x_3)$ with coordinates $\{x_1,x_2\}$ depending on $x_3$ as on a parameter, where $u^\perp=(-u_2, u_1)$ for a vector field $u=(u_1, u_2)$ on $\R^2$; note that below we use this notation for vector fields on various planes in $\R^3$. Then let us set

 \begin{eqnarray}\label{n7} F= \int_{-\infty}^\infty(-v^2v^3,\,v^1v^3)\,dx_3; \end{eqnarray} (3.2)
note that $F$ is a compactly supported vector field on the plane $\Pi_{12}=\{(x_1,x_2,0)\}$ with coordinates $\{x_1,x_2\}$ and (3.1) implies that $IF=0$. Indeed, choose $x^0 = (x^0_1, x^0_2)\in \R^2$, $0\ne \xi= ( \xi_1, \xi_2)\in \R^2$, and let $L$ be the 2-plane through the point $x^0$ parallel to the vectors $\xi$ and $(0, 0, 1).$ Then the vector $\nu= (- \xi_1, \xi_2,0)$ is orthogonal to $L$ and the vector $\tilde \xi_a = (- \xi_1, \xi_2,a)$ is parallel to $L$ for every $a.$ By (3.1) we have \begin{eqnarray*} \int_{L}\langle v,\tilde \xi_a \rangle\langle v,\nu\rangle \,d\sg_L =0 \end{eqnarray*} and thus we get \begin{eqnarray*} \int_{L}(\xi_1v^1 + \xi_2v^2 + av^3)(-\xi_2v^1 + \xi_1v^2)\,d\sg_L =0. \end{eqnarray*} Substituting the values $a=0,\,a=1$ and taking the difference we get the equation $IF=0$.

Therefore, by (2.5) we have

 \begin{eqnarray}\label{eq3.4} d_sw_0= \nabla w_0=-F \end{eqnarray} (3.3)
for a unique compactly supported smooth scalar function $w_0=w_0(x)$.

Let us fix for a moment a point $P^0=(x_1^0, x_2^0)\in \R^2$, let $r \subset\Pi_{12}= \R^2$ be a ray emanating from $P^0$ and let $e_r$ be a unit directional vector of $r$, then in virtue of (3.3) we have

 \begin{eqnarray}\label{st} \int_{r}\langle e_{r},F\rangle \,ds_{r}=w_0(x_1^0,x_2^0) \end{eqnarray} (3.4)
for the line element $ds_{r}$ of $r$. Let $H\subset \R^3$ be a half-plane perpendicular to $\Pi_{12}$ with $\pa H=m(x_1^0, x_2^0,0,0)$, where $m(x_1^0,x_2^0,0,0)$ is the vertical line passing through the point $(x^0_1,x^0_2,0) \in \Pi_{12}$; therefore, $H$ orthogonally projects onto some ray $r$ emanating from $P_0$. Let us consider the integral \begin{eqnarray*} \int_{H} v^3\langle\nu_{H},v\rangle \,d\si_{H}=-\int_{r}\langle e_{r},F\rangle \,ds_{r} \end{eqnarray*} for the area element $d\sg_{H}$ of $H$ and a suitable unit normal $\nu_{H}$ to $H$, then by (3.4) it does not depend on $H$ for a fixed point $P^0=(x_1^0, x_2^0)$ and a fixed line $\pa H=m(x_1^0, x_2^0,0,0)$. Since the choice of a vertical line in $\R^3$ is arbitrary we see that the following definition is correct:
###### Definition 3.2.
Define
 \begin{eqnarray}\label{dw} w= -\int_{H(m)}\langle e_m,v\rangle \langle\nu_{H(m)},v\rangle \,d\si_{H(m)} \end{eqnarray} (3.5)
where $H(m)$ is a half-plane with $\pa H(m)= m$ and $\nu_{H(m)}$ is the unit normal to $H(m)$ such that the basis $(e_m, \nu_{m},\nu_{H(m)})$ is positively oriented for the interior unit normal $\nu_{m}$ to $m$ lying in $H(m)$.

Therefore, $w$ is a compactly supported smooth function on $M$ and it can be written as $w=w(y_1,y_2,\alpha_1,\alpha_2)$ on $M_{nh}$; moreover, we get

###### Lemma 3.3.
We have \begin{eqnarray*} w(y_1,y_2,0,0)=w_0(y_1,y_2). \end{eqnarray*}

###### Proof.
Indeed, it is sufficient to verify that
 \begin{eqnarray}\label{P0} {\pa w\over \pa y_1}(0)=\int_{-\infty}^\infty v^2v^3 \,dx_3,\quad {\pa w\over \pa y_2}(0)=-\int_{-\infty}^\infty v^1v^3 \,dx_3, \end{eqnarray} (3.6)
which is clear, since \begin{eqnarray*} &&w(\de,0,0,0)=-\int_{H_{1,\de}}v^2v^3\,dx_1dx_3 =-\int_{\de}^\infty dx_1\int_{-\infty}^\infty v^2v^3 \,dx_3,\\ && w(0,\mu,0,0)=\int_{H_{2,\mu}}v^1v^3\,dx_2dx_3 = \int_{\mu}^\infty dx_2\int_{-\infty}^\infty v^1v^3 \,dx_3 \end{eqnarray*} for the half-planes \begin{eqnarray*} H_{1,\de}=\{(x_1> \de,0,x_3)\},\quad H_{2,\mu}=\{(0,x_2> \mu,x_3)\}. \end{eqnarray*}

Now we give two explicit formulas for $w$ which use two specific choices of $H(m)$. We begin by putting

 \begin{eqnarray} k_1 =\sqrt{1+\alpha_1^2},\quad k_2 =\sqrt{1+\alpha_2^2}; \end{eqnarray} (3.7)
recall also that $k=\sqrt{1+\alpha_1^2+\alpha_2^2}.$

Given a line $m\in M_{nh},$ let $H(m)_1$ and $H(m)_2$ be the half-planes with the border-line $m$ which are determined by the following conditions:

1. (i) $H(m)_1$ is parallel to $x_1$-axis, $H(m)_2$ is parallel to $x_2$-axis;
2. (ii) $\langle\nu_i,e_i\rangle> 0,\;i=1,2,$
for $e_1=(1,0,0),\,e_2=(0,1,0)$ and the internal normals $\nu_i\in H(m)_i$, $i=1,2$.

We have then \begin{eqnarray*} \nu_{H(m)_1}= \left(0,{ 1\over k_2},-{\alpha_2\over k_2}\right),\quad \nu_{H(m)_2}= \left(-{1\over k_1},0,{\alpha_1\over k_1} \right), \end{eqnarray*} and the plane $H(m)_1$ forms angle $\beta_1$ with the coordinate plane $\Pi_{13}=\{x_2=0\}$ where $\cos\beta_1={1/k_2}$, while the plane $H(m)_2$ forms angle $\beta_2$ with the coordinate plane $\Pi_{23}=\{x_1=0\}$, $\cos\beta_2={1/ k_1}$. Note also that we have \begin{eqnarray*} e_m=\frac 1k\left(\alpha_1,\alpha_2,1\right) =\left({\alpha_1\over k},{\alpha_2\over k},{1\over k} \right) \end{eqnarray*} for the positive unit directional vector $e_m$ of $m$.

###### Proposition 3.4.
Let $d\si_i$ be the surface area element on $H(m)_i$ and let $l_i=y_i+x_3\alpha_i$ for $i=1,2.$ Then in the introduced notation we have
 \begin{align}\label{eqi} ({\rm i})\quad w&= -\int_{H(m)_2}\langle e_m,v \rangle\langle\nu_{H(m)_2},v\rangle \,d\si_2\nonumber\\ &=-\int_{-\infty}^\infty\int_{l_2 }^\infty (\langle e_m,v \rangle\langle\nu_{H(m)_2},v\rangle)|_{(l_1,x_2,x_3)} k_1dx_2dx_3\nonumber\\ &= \int_{-\infty}^\infty\int_{l_2}^\infty \frac 1k ((\alpha_1v^1+\alpha_2v^2+v^3)(v^1-\alpha_1v^3))|_{(l_1,x_2,x_3)}dx_2dx_3 \end{align} (3.8)
 \begin{align} \label{eqii} ({\rm ii})\quad w&= -\int_{H(m)_1}\langle e_m,v \rangle\langle\nu_{H(m)_1},v\rangle \,d\si_1\nonumber\\ &=-\int_{-\infty}^\infty\int_{l_1}^\infty(\langle e_m,v \rangle\langle\nu_{H(m)_1},v\rangle)|_{(x_1,l_2,x_3)} k_2\,dx_1dx_3\nonumber\\ &= -\int_{-\infty}^\infty\int_{l_1}^\infty \frac1k((\alpha_1v^1+\alpha_2v^2+v^3) (v^2- \alpha_2 v^3))|_{(x_1,l_2,x_3)}\,dx_1dx_3.\nonumber\\ \end{align} (3.9)

###### Proof.
This is an elementary calculation which we give only for $H(m)_1$, since the case of $H(m)_2$ is completely similar; note only that the choice of normals $\nu_{H(m)_1}$ and $\nu_{H(m)_2}$ comes from the orientation condition. Let us fix the values of $y_1$, $y_2$, $\alpha_1$, and $\alpha_2$, and let $H_i\supset H(m)_i$ be an affine plane containing $H(m)_i$ for $i=1,2$. Then an equation of $H_1$ is of the form $ax_2+bx_3+c=0$, and therefore $c=-ay_2$. Since $e_m\in {\bar H_1}$ for the vector plane ${\bar H_1}$ parallel to $H_1$, we get $a \alpha_2+b=0$, and we can choose $a=1$, $b= - \alpha_2$, so the equation takes the form \begin{eqnarray*} x_2-x_3 \alpha_2-y_2 =0, \end{eqnarray*} and therefore $x_2=x_3\alpha_2+y_2$ on the half-plane $H(m)_1$. Since $\cos\beta_1 = {\cos \arctan \alpha_2}= {1 \over k_2}$, we see that $d\si_{1} = k_2\,dx_1dx_3$. Then one notes that the orthogonal projection $\pi_{13}(m)$ of $m$ on the coordinate plane $\Pi_{13}=\{x_2=0\}$ is given by \begin{eqnarray*} \pi_{13}(m)=\Pi_{13}\cap H_2=\{x_1 =y_1+{\alpha_1x_3}\}, \end{eqnarray*} and thus $H(m)_1$ projects onto \begin{eqnarray*} \{x_1> y_1+{x_3\alpha_1}\}\subset \Pi_{13}, \end{eqnarray*} since $\langle \nu_2,e_2\rangle> 0$, which completes the proof.

The formulas (3.8) and (3.9) are somewhat cumbersome and use below only the following simple consequence.

###### Corollary 3.5.
In the first order of $(\alpha_1,\alpha_2),$ ignoring terms with total $deg_\alpha\ge 2,$ we have the following expressions$:$
 $$\label{wI} w=\int_{-\infty}^\infty\int_{l_2}^\infty (\alpha_2v^1v^2+v^1v^3+\alpha_1(v^1)^2 -\alpha_1(v^3)^2)|_{(l_1,x_2,x_3)}\,dx_2dx_3,$$ (3.10)
 $$\label{wII} w=\int_{-\infty}^\infty\int_{l_1}^\infty(\alpha_2(v^3)^2- \alpha_1v^1v^2-v^2 v^3-\alpha_2( v^2)^2)|_{(x_1,l_2,x_3)} dx_1dx_3.$$ (3.11)

This corollary permits to calculate the quantities \begin{eqnarray*} \;{\pa^m w\over \pa y_1^i\pa y_2^j \pa \alpha_1^k \pa \alpha_2^l}(0),\quad i+j+k+l=m, \end{eqnarray*} for $k+l\le 1$, and in particular, implies the following.

###### Corollary 3.6.
We have
 $$\label{w12} {\pa^2w\over\pa y_2\pa\alpha_1}(0)= \int_{-\infty}^\infty\left((v^3)^2-(v^1) ^2-x_3{\pa (v^3v^1)\over\pa x_1} \right)\Bigg|_{(0,0,x_3)} \,dx_3,$$ (3.12)
 $$\label{w21} {\pa^2w\over\pa y_1\pa \alpha_2}(0)=\int_{-\infty}^\infty \left((v^2)^2 -(v^3) ^2+x_3{\pa (v^3v^2)\over\pa x_2} \right)\Bigg|_{(0,0,x_3)} \,dx_3,$$ (3.13)
 $$\label{de} {\pa^2w\over\pa y_1^2}(0)+{\pa^2w\over\pa y_2^2}(0) =\int_{-\infty}^{\infty}\left({\pa (v^2v^3)\over\pa x_1}-{\pa \left(v^1v^3\right) \over\pa x_2} \right)\Bigg|_{(0,0,x_3)}\,dx_3.$$ (3.14)

###### Proof of (3.12).
From (3.10) we have \begin{eqnarray*} w(0,y_2,\alpha_1,0)=\int_{-\infty}^\infty\int_{y_2}^\infty(v^1v^3+(v^1)^2 \alpha_1 -(v^3)^2\alpha_1)|_{(x_3 \alpha_1,x_2,x_3 )}\,dx_2dx_3, \end{eqnarray*} whence we get \begin{eqnarray*} {\pa w(0,0,\alpha_1,0) \over \pa y_2}=-\int_{-\infty}^\infty(v^1v^3+(v^1)^2 \alpha_1 -(v^3)^2\alpha_1)|_{(\alpha_1x_3, 0,x_3 )}\,dx_3 \end{eqnarray*} and, finally, \begin{eqnarray*} {\pa^2w\over\pa y_2\pa \alpha_1}(0)=\int_{-\infty}^\infty\left((v^3)^2 -(v^1)^2 -x_3 {\pa (v^3v^1)\over\pa x_1}\right)|_{(0,0,x_3 )}\,dx_3. \end{eqnarray*} The proof of (3.13) is completely similar and that of (3.14) is even simpler.

Taking then the difference of (3.12) and (3.13) we get the following formula:

 \begin{eqnarray}\label{eqL} {\pa^2w \over\pa y_1\pa \alpha_2}(0)-{\pa^2w \over\pa y_2\pa \alpha_1}(0) =\int_{-\infty}^\infty(p+(v^1)^2+(v^2)^2-(v^3) ^2 )|_{(0,0,x_3 )} \,dx_3.\nonumber\\ \end{eqnarray} (3.15)

Indeed, we have $\;{\pa (v^3v^1)\over\pa x_1} +{\pa (v^3v^2) \over\pa x_2} =-{\pa (p +(v^3)^2 )\over\pa x_3}$ by (1.1) and integrating \begin{eqnarray*} \int_{-\infty}^\infty x_3 \left({\pa(v^3v^1)\over\pa x_1} +{\pa (v^3v^2)\over\pa x_2} \right)\Bigg|_{(0,0,x_3 )} dx_3=-\int_{-\infty}^\infty {x_3 \pa (p +(v^3)^2)\over\pa x_3 } \Bigg|_{(0,0,x_3 )} dx_3 \end{eqnarray*} by parts we get (3.15) .

## 4. Operators $P$ and $\De_M$

Let us define first an order 2 differential operator $P$ on the space $C^{2}(M).$

Recall that $M=G/(\R\times{\rm SO}(2))=G'/(\R\times {\rm O}(2))$ for $G=\R^3\rtimes{\rm SO}(3)$ and $G'=G\cdot\{\pm I_3\}.$

###### Definition 4.1.
Let $f\in C^{2}(M),\,m_0\in M$, and let $g(m_0)=0=(0,0,0,0)$ for $g\in G$. Then \begin{eqnarray*} Pf(m_0)=_{\rm def}Lf_g(0), \end{eqnarray*} where $f_g(m)=f(g^{-1}(m))$ for any $m\in M$ and $L$ is defined by (2.11) .
###### Lemma 4.2.
This definition is correct.

###### Proof.
We must verify that $Lf_g(0)=Lf_h(0)$ for any $g,h\in G$ such that $g(m_0)=h(m_0)=(0)$.

We put $u=g^{-1}h,\,F=f_u$, and thus we have to verify that $LF(0)=LF_u(0)$ for $u\in {\R}\times {\mathrm SO}(2)=St_0$, $St_0< G$ being the stabilizer of the vertical line. It is sufficient to verify the equality separately for $u\in {\R}$ and $u\in {\mathrm SO}(2)$. It is clear for a vertical shift $u\in {\R}$ since $L$ has constant coefficients; for a rotation $u\in {\mathrm SO}(2)$ by angle $\theta$ in the horizontal plane one easily calculates \begin{eqnarray*} F_u(y_1,y_2,\alpha_1,\alpha_2 )&=&F(y_1\cos \theta-y_2\sin \theta,y_2\cos \theta+y_1\sin \theta,\alpha_1\cos \theta\\ &&-\, \alpha_2\sin \theta,\alpha_2\cos \theta+\alpha_1\sin \theta), \end{eqnarray*} and a simple calculation shows the necessary equation, since we get \begin{eqnarray*} &&{\partial^2 F_u(0)\over \partial \alpha_2\partial y_1}= \cos^2 \theta {\partial^2 F(0)\over \partial \alpha_2\partial y_1} -\sin^2 \theta{\partial^2 F(0)\over \partial \alpha_1\partial y_2}+\cos \theta \sin \theta\left({\partial^2 F(0)\over \partial \alpha_1\partial y_1} -{\partial^2 F(0)\over \partial \alpha_2\partial y_2}\right),\\ &&{\partial^2 F_u(0)\over \partial \alpha_1\partial y_2} =\cos^2 \theta {\partial^2 F(0)\over \partial \alpha_1\partial y_2} -\sin^2 \theta {\partial^2 F(0)\over \partial \alpha_2\partial y_1}+\cos \theta \sin \theta\left({\partial^2 F(0)\over \partial \alpha_1\partial y_1} -{\partial^2 F(0)\over \partial \alpha_2\partial y_2}\right). \end{eqnarray*} The proof is finished.

We can now rewrite (3.15) as follows

 \begin{eqnarray}\label{eqP} P_0w&=&\int_{-\infty}^\infty(p+(v^1)^2+(v^2)^2-(v^3)^2)\,dx_3\nonumber\\ &=&\int_{-\infty}^\infty(p+|v|^2-2(v^3)^2)\, dx_3 \end{eqnarray} (4.1)
for the operator $P_0$ being $P$ evaluated at 0, which implies that
 \begin{eqnarray}\label{Psps} Pw&=&\int_m(p+|v|^2-2\langle v,e_m\rangle^2) \,ds\nonumber\\ &=&\int_m(p+|v|^2-2v\otimes v)\,ds =IQ_0(m) \end{eqnarray} (4.2)
for the quadratic tensor field $Q_0=(p+|v|^2)\de_{ij}-2v\otimes v$ and any $m\in M$, since $P$ is $G$-invariant; therefore $Pw=IQ_0$ as functions on $M$.

We will also use the fiber-wise Laplacian $\De_M=\De_{y_1, y_2}$ acting in tangent planes to $\S^2$; it is defined by the usual formula \begin{eqnarray*} \De_Mf(m )=\frac{\pa^2 f(m )}{\pa{y_1}^2} +\frac {\pa^2 f(m)}{\pa{y_2}^2} \end{eqnarray*} for $f\in C^2(M)$ and a vertical line $m=m(y_1,y_2,0,0)$. For any $m\in M$ the value $\De_M f(m)$ is determined by the $G$-invariance condition as for the operator $P$ above, and the rotational symmetry of $\De_{y_1, y_2}$ guarantees the correctness of that definition. Note that the operators $P$, $\De_M$ commute and note also that (2.10) implies that for $Q\in C_0^\infty(S^h,\R^3)$ there holds a commutation rule

 \begin{eqnarray}\label{dd} I(\De Q)=\De_M(IQ). \end{eqnarray} (4.3)
###### Remark 4.1.
The algebra $D_{G'}(M)$ of the $G'$-invariant differential operators on $M$ is freely generated by $\De_M$ and $P^2$ as a commutative algebra, see [Gonzalez and Helgason1986].

One can also give explicit formulas for $P$ and $\De_M$ in our coordinates, namely,

 \begin{eqnarray}\label{pe} P=k^2L+\alpha_1\frac{\pt}{\pt y_2}-\alpha_2\frac{\pt}{\pt y_1},\quad\De_M= k_1^2\frac{ \pt^2}{ \pt y_1^2}+k_2^2\frac{\pt^2}{\pt y_2^2} +2\alpha_1\alpha_2 {\partial^2 \over \partial y_1\partial y_2}. \end{eqnarray} (4.4)

Now we deduce the principal linear differential equation for $w$.

###### Proposition 4.3.
We have
 \begin{eqnarray}\label{p4} P^2w=-4\De_M w. \end{eqnarray} (4.5)

###### Proof.
We begin with the following simple result.

###### Lemma 4.4.
If $f\in C^{\infty}_0(\R^3)$ is a scalar function then $P(If)=0.$

Indeed, since $P$ is $G$-invariant, it is sufficient to verify the equation at a single point $0\in M$ which follows from (2.12) with $h=0$.

Lemma 4.4 implies by (4.2) that

 \begin{eqnarray}\label{pwq} P^2w=PIQ_0= PI((p+|v|^2)\de_{ij}-2v\otimes v)=-2PI(v\otimes v) \end{eqnarray} (4.6)
for a compactly supported vector field $v$ solving (1.1) . Moreover, we have \begin{eqnarray*} I(v\otimes v)(y_1,y_2,\alpha_1,\alpha_2)=\int_{-\infty}^{\infty}( {(v^3) ^2 +2v^1v^3\alpha_1+2v^2v^3\alpha_2})\,{dx_3 }+O(|\alpha|^2) \end{eqnarray*} and thus by (3.14) we get \begin{eqnarray*} PI(v\otimes v)(0)=P_0I(v\otimes v)=2\int_{-\infty}^{\infty}\left({\pa \left(v^2v^3\right)\over\pa x_1} -{\pa \left(v^1v^3\right)\over\pa x_2} \right)dx_3 =2\De_{M}w(0), \end{eqnarray*} hence $P I(v\otimes v)= 2\De_Mw$ everywhere and $P^2w=-4\De_Mw$ by (4.6) .
###### Corollary 4.5.
The equation
 \begin{eqnarray}\label{iq0} IQ_0(m)=0,\quad\forall m\in M \end{eqnarray} (4.7)
implies Conjecture 1.2.

Indeed, if $IQ_0(m)=0$ then $\De_M w(m)=-\frac14 P^2w(m)=-\frac14 PIQ_0(m)=0$ and thus $w=0$, since $w$ is compactly supported. Invariant definitions and the second proof of (4.5) Now let us give a description of $P$ and $\De_M$ in terms of the Lie algebra $\mathfrak g$ of $G$. We have $\mathfrak g= {\mathfrak so}(3)\oplus {\mathfrak r}(3)=\R^3\oplus\R^3$ as vector spaces, where ${\mathfrak r}(3)$ is 3-dimensional and abelian. Thus, we can write any $g\in\mathfrak g$ as $g=(r;s) \in{\mathfrak so}(3)\oplus {\mathfrak r}(3)$, and the commutators in $\mathfrak g$ are given by \begin{eqnarray*} [(r_1;0),(r_2;0)] {\!}= {\!}(r_1\times r_2;0),\quad [(0;s_1),(0;s_2)]=0,\quad [(r;0),(0;s)] {\!}= {\!} (0;r \times s). \end{eqnarray*}

Let $(R_1,R_2,R_3)$ be the standard basis of ${\mathfrak so}(3)$, and $(S_1,S_ 2, S_3)$ be that of ${\mathfrak r}(3)$. Consider the following operators on $M$:

 \begin{eqnarray}\label{0p1} \tilde \De_M = S_1^2+S_2^2+S_3^2,\quad\tilde P= {S_1} {R_1}+ {S_2} {R_2}+ {S_3} {R_3}, \end{eqnarray} (4.8)
where we denote simply by $g$ the action on $M$ of an element $g\in U (\mathfrak g)$ of the universal enveloping algebra $U(\mathfrak g)$; therefore, $S_i$ acts as the infinitesimal shift in the $x_i$-direction, and $R_i$ as the infinitesimal rotation about $x_i$-axis.
###### Proposition 4.6.
We have $\tilde \De_M =\De_M$ and $\tilde P=P.$

###### Proof.
First, the operators $\tilde\De_M$ and $\tilde P$ are $G$-invariant. Indeed, it follows from the rotational invariance of the quadratic form $x_1^2+x_2^2+x_3^2$ that $\tilde \De_M$ is rotationally invariant; for translations, the same follows from the commutation rule ${S_i}{S_j} ={S_j}{S_i}$ for $i,j=1,2,3$.

To prove the invariance of $\tilde P$ under the $x_3$-axis rotation we verify that $\tilde P$ and ${R_3}$ commute which can be shown as follows: \begin{eqnarray*} [ {S_1} {R_1}, {R_3}]= - {S_2} {R_1}- {S_1} {R_2},\quad [ {S_2} {R_2}, {R_3}]= {S_1} {R_2}+ {S_2} {R_1},\quad [ {S_3} {R_3}, {R_3}]= 0. \end{eqnarray*} Similarly we get its invariance under the $x_1$- and $x_2$-axis rotations and thus its ${\mathrm{SO}}(3)$-invariance, while its $x_3$-translations invariance follows from \begin{eqnarray*} [{S_1} {R_1}, {S_3}]= -{S_1}{S_2},\quad [{S_2}{R_2},{S_3}]={S_1S_2},\quad [S_3R_3, S_3]= 0. \end{eqnarray*} Since any line is ${\mathrm SO}(3)$-conjugate to a vertical one, $\tilde P$ is $G$-invariant. Finally, we have $\tilde \De_M (m_0)=\De_M (m_0),\, \tilde P(m_0)=P(m_0)\,$ for $m_0=m(0,0,0,0)$, which finishes the proof, since $\De_M$ and $P$ are both $G$-invariant. Indeed, e.g., in $\tilde P(m_0)$ the terms $S_1R_1+S_2R_2$ give $L(0)$, while $S_3R_3$ vanishes since $m_0$ is invariant under both $S_3$ and $R_3.$

###### Second Proof of Proposition 4.3.
Let $t\in \R$, and let $l_t=m(t,0,0,0)$ be a vertical line in the plane $\Pi_{13}$; note that $\Pi_{13}=\bigcup_{t\in \R } l_t$.

###### Lemma 4.7.
For $f\in C^\infty_0(M)$ we have
 \begin{eqnarray} \int_\R P^2f(l_t)\,dt =\int_\R S_2^2 R_2^2 f(l_t) \,dt.\label{l2f} \end{eqnarray} (4.9)

###### Proof.
Define the operators $A$ and $B$ by $A= {S_1} {R_1}P,\, B= {S_3} {R_3}P;$ then \begin{eqnarray*} P^2=A+B+ {S_2}{R_2}P=A+B+{S_2R_2}\left({S_1R_1}+ S_2R_2+ S_3R_3\right). \end{eqnarray*} Since $Af(l_t)$ is a derivative of a function of $t$, while $B$ vanishes identically on $\Pi_{13}$, we get that \begin{eqnarray*} \int_\R(Af(l_t)+Bf(l_t))dt=0. \end{eqnarray*} We have also \begin{eqnarray*} {S_2R_2S_1 R_1}(f(l_t))={S_1S_2R_2 R_1}(f(l_t))- {S_3S_2}{R_1}(f(l_t)), \end{eqnarray*} thus the integral of the left-hand side is zero and the same is true for \begin{eqnarray*} {S_2R_2S_3R_3}(f(l_t)) ={S_3S_2R_2R_3}(f(l_t))+{S_1S_2R_3}(f(l_t)); \end{eqnarray*} since ${S_2R_2S_2R_2}=S_2^2R_2^2$ we get the conclusion.

###### Lemma 4.8.
We have
 \begin{eqnarray} \int_{\R} R_2^2 w(l_t)\,dt =- 4\int_{\R} w(l_t)\,dt. \label{l3f} \end{eqnarray} (4.10)

###### Proof.
Let us fix a positive constant $c< {\pi\over 2}$, and let $l^\th_t =({t\over\cos(\th)},0,\tan(\th),0)$ for any $\th$ with $|\th|< c$; therefore, $l^\th_t$ is just the line $l_t$ rotated (in the clockwise direction) through the angle $\th$ about the origin in the plane $\Pi_{13}$, and for any $t$ we have \begin{eqnarray*} R_2^2w(l^\th_t)|_{\th=0}= {\pa^2\over \pa \th^2}w(l^\th_t)|_{\th=0}. \end{eqnarray*} Let $e^\th_1=(\cos \th,0,-\sin\th),\;e^\th_3=(\sin\th,0,\cos \th)$ then we have \begin{eqnarray*} w(l^\th_t)={1\over\cos\th}\int_0^\infty\int_{l_t}\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle |_{ \left({t\over \cos \th}+x_3 \tan\th,x_2,x_3\right)}\,dx_3 dx_2 \end{eqnarray*} by (3.8) , and if we put $t={x_1\cos \th}-x_3\sin\th$ we get that \begin{eqnarray*} \int_{\R} w(l^\th_t)\,dt&=&{1\over \cos \th}\int_0^\infty \int_{\R^2}\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle \,dx_3 dx_2dt\\ &=& \int_{x_2 > 0}\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle \,dx_1dx_2 dx_3. \end{eqnarray*}

Therefore, since $\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle=v^1v^3\cos 2\th+((v^1)^2-(v^3)^2){\sin 2\th\over 2}$ we have \begin{eqnarray*} {\pa^2\over \pa \th^2}\int_{\R} w(l^\th_t)\,dt&=&\int_{ x_2 > 0 }{\pa^2\over \pa \th^2} (\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle) \,dx_1dx_2 dx_3\\ &=& -{4}\int_{x_2 > 0} (\langle v,e^\th_1\rangle \langle v,e^\th_3\rangle)\,dx_1dx_2 dx_3 \end{eqnarray*} and evaluating at $\th=0$ we get a proof of Lemma 4.8.

We can finish now our second proof of (4.5) . Indeed, (4.9) and (4.10) imply that

 \begin{eqnarray}\label{4l} \int_{\R} P^2 w(l_t)\,dt=\int_{\R} S_2^2 R_2^2 w(l_t)\,dt =-4 \int_{\R} S_2^2w(l_t)\,dt= -4\int_{\R} \De_M w(l_t)\,dt.\nonumber\\ \end{eqnarray} (4.11)

If we define a function $F(x_1,x_2)$ on $\R^2$ by \begin{eqnarray*} F(x_1,x_2)= P^2w(x_1,x_2,0,0)+4\De_M w(x_1,x_2,0,0), \end{eqnarray*} then the integral of $F$ over the $x_1$-axis vanishes by (4.11) . Changing the coordinate system $x_1,x_2$ in $\R^2$, we get the same for the integral of $F$ over any line in the plane $\{ x_1,x_2\}$. Thus $F=0$ by Radon's theorem, and we get the conclusion.

###### Remark 4.2.
One can compare (4.5) with results that can be deduced from (2.12) for $h=2$. A simple direct calculation using (4.4) gives for $h=2$
 \begin{eqnarray}\label{pde} P^3\psi+4P\De_M \psi=0 \end{eqnarray} (4.12)
if $\psi=IQ$ for $Q\in C_0^\infty (S^2,\R^3)$. Applying then (4.12) to $\psi=\De_M w$ (which can be written as $\De_M w=IQ'$ for a certain $Q'$ not given here) we obtain $P^3\De_Mw+4P\De^2_M w=\Delta_MP(P^2w+4\Delta_M w)=0$ and thus $P(P^2w+4\Delta_M w)=0$ which is much weaker than (4.5) since the kernel of $P$ is enormous.

However, it is possible to construct a function $u\in C_0^\infty(M)$ verifying \begin{eqnarray*} P\De_Mu =-2\De_Mw,\quad \De_Mu=IQ_1 \end{eqnarray*} for some $\,Q_1\in C_0^\infty(S^2,\R^3)$ and applying (4.12) to $\psi=\De_Mu=IQ_1$ we get \begin{eqnarray*} 0=P^3\De_M u+4\De_M^2 P u= -2\De_M(P^2 w+4\De_M w), \end{eqnarray*} and thus we reprove (4.5) . We can define $u$ similarly to (3.5) as follows \begin{eqnarray*} u=\int_{H(m)}{\rm{dist}}(P,m)(p+\langle \nu_{H(m)},v\rangle^2) d\si_{H(m)}, \end{eqnarray*} where ${\rm{dist}}(P,m)$ is the distance from a point $P\in H(m)$ to $m$.

## 5. A Radon Plane Transform

Let us define a Radon tensor plane transform $J$ as follows:

 \begin{eqnarray} \label{pt} JQ(L)= \int_L\mathrm{tr}(Q_{|L}) \,d\si_L \end{eqnarray} (5.1)
for an affine plane $L\ss\R^3$ and $Q\in C^{\infty}(S^{2};\R^3)$ satisfying
 \begin{eqnarray} \label{est2} |Q(x)| \le C(1 +|x|)^{-2-\varepsilon}, \end{eqnarray} (5.2)
for some $\varepsilon> 0,$ where $Q_{|L}$ is the restriction onto $L$; then we get a bounded linear operator \begin{eqnarray*} J\colon {\mathcal S}(S^2;{\mathbb R}^3)\lra{\mathcal S}(\R P^3) \end{eqnarray*} for the manifold $\R P^3$ parametrizing affine planes $L\ss\R^3.$
###### Proposition 5.1.
We have $JQ_0(L)=0.$

###### Proof.
We have for any affine plane $L\ss\R^3$ that \begin{eqnarray*} JQ_0(L)=2\int_L( p+\langle v, \nu_L\rangle^2)\, d\sg_L=0. \end{eqnarray*} Indeed, setting without loss of generality $L=\Pi_{12},\,\nu_L=e_3$ we get that \begin{eqnarray*} {\pa JQ_0\over \pa x_3}(\Pi_{12})&=&{\pa \over \pa x_3}\left(\int_L( p+\langle v, \nu_L\rangle^2)\, d\sg_L\right)\\ &=&\int_{\Pi_{12}}{\pa(p+(v^3)^2)\over\pa x_3}\, dx_1dx_2\\ &=&-\int_ {\Pi _{12}}\left({\pa(v^1v^3)\over\pa x_1}+{\pa(v^2v^3)\over\pa x_2}\right) \, dx_1dx_2=0. \end{eqnarray*} Therefore, $JQ_0(\Pi_{12})$ does not depend on $x_3$ and hence equals 0.

Let us explain in what Proposition 5.1 partially confirms (4.7) and thus Conjecture 1.2. One can verify that the condition $JQ_0(L)=0, \forall L\in\R P^3$ is equivalent to the following equation for the components $\{q^{ij}\}$ of $Q_0{:}$

 \begin{eqnarray}\label{jq0} \sum_{ i,j=1}^3 {\pa^2 q^{ij}\over\pa x_i\pa x_j}=\De\tr\, Q_0, \end{eqnarray} (5.3)
while $IQ_0(m)=0,\, \forall m\in M$ is equivalent to the following system
 \begin{eqnarray}\label{jq} {2\,\pa q^{ij}\over\pa x_i\pa x_j}={\pa^2q^{ii}\over\pa x_j^2}+ {\pa^2 q^{jj}\over\pa x_i^2},\quad 1\le i< j\le 3 \end{eqnarray} (5.4)
of three equations and their sum gives (5.3) .

## 6. A Uniqueness Theorem

Now we can deduce Conjecture 1.1 from Conjecture 1.2.

###### Theorem 6.1.
Let $(v,p)\in\C^\infty_0(\R^3)$ be a solution of (1.1) (1.3) and let the corresponding function $w$ vanish everywhere on $M$ then $(v;p)=0$ everywhere.

###### Proof.
For $m\in M$ denote by $t$ a vector parallel to $m$ and by $n$ a vector perpendicular to $m,$ then the equality $w=0$ implies that
 \begin{eqnarray}\label{int} \int_m\langle v,t\rangle\langle v,n\rangle \,ds=0. \end{eqnarray} (6.1)

Let $L\ss \R^3$ be an affine plane; let $v_n=\langle\nu_L,v\rangle$ and $v_t=v-v_n\nu_L$ be its components normal to $L$ and tangent to $L$, respectively. Then by (6.1) we have $IV=0$ for the vector field $V=v_n v_t$ on $L$, and hence $\curl_L V =0$ for the curl operator $\curl_L$ on $L$ which gives

 \begin{eqnarray}\label{eqr} v_n\curl_L v_t - \langle v^\perp_t, \nabla_L v_n\rangle=-v_n\om_n- \langle v^\perp_t, \nabla_L v_n\rangle=0 \end{eqnarray} (6.2)
for the normal component $\om_n$ of $\om=\curl v$ and the gradient operator $\nabla_L$ on $L$. We will apply (6.2) to various planes $L \ss \R^3$.

First take $L=\Pi_{12}$, then $v_t=(v^1,v^2,0),\,v_n=v^3,\,V=v^3v_t$ and we get

 \begin{eqnarray} \label{eq01} v^3\curl_L v_t-\langle v^\perp_t, \nabla v^3\rangle=v^3\left({\pa v^1 \over \pa x_2} -{\pa v^2 \over \pa x_1} \right)-v^1{\pa v^3 \over \pa x_2}+v^2 {\pa v^3 \over \pa x_1} =0. \end{eqnarray} (6.3)
Let now \begin{eqnarray*} \Om=_{\rm def}\{u\in\R^3|v(u)\ne 0,\,\om(u)\neq 0\} \end{eqnarray*} and let $D=_{\rm def}\R^3{\setminus} \bar \Om$. We can suppose that $\Om$ is not empty, since otherwise in a neighborhood of a point $x_0$ where the maximum of $|v|$ is attained we have $\De v=-\curl\curl v+\nabla \div\, v=0$ and thus $v$ is harmonic which contradicts the maximum principle for harmonic fields. It follows then that $v=0$ in this neighborhood and thus everywhere.

In orthonormal coordinates with $v^1(u_0)=v^2(u_0) =0$ we get for $u_0\in \Om$ that

 \begin{eqnarray}\label{eqc5} v^3\left({\pa v^1 \over \pa x_2} -{\pa v^2 \over \pa x_1} \right)(u_0) =0\quad \hbox{and therefore}\quad \langle v,\om\rangle(u_0) =0. \end{eqnarray} (6.4)
Therefore $\langle v,\om\rangle=0$ holds everywhere on $\Om$, thus on $\R^3$ and differentiating this relation in the $v$-direction we obtain
 \begin{eqnarray}\label{eq40} \langle v \nabla v,\om\rangle+ \langle v\nabla \om, v\rangle=0 ; \end{eqnarray} (6.5)
using the commutation law
 \begin{eqnarray}\label{eq41} v \nabla \om= \om\nabla v \end{eqnarray} (6.6)
we get then from (6.5) that
 \begin{eqnarray}\label{eq42} \langle v \nabla v,\om\rangle+ \langle\om \nabla v, v\rangle=0. \end{eqnarray} (6.7)
In orthonormal coordinates $x_1,x_2,x_3$ with $x_1$ directed along $v$ and $x_2$ directed along $\om$ at $u_0$ we can rewrite (6.7) as follows
 \begin{eqnarray}\label{eq43} {\pa v^1 \over \pa x_2}(u_0) +{\pa v^2 \over \pa x_1}(u_0) =0, \end{eqnarray} (6.8)
since $v(u_0)\neq 0$ and $\om(u_0)\ne0$. Below we always use that coordinate system.

Moreover, since the vector $\om$ is directed along $x_2$, we get

 \begin{eqnarray}\label{eq45} {\pa v^1\over \pa x_2}(u_0) ={\pa v^2\over\pa x_1}(u_0)\quad \hbox{and therefore}\quad{\pa v^1\over \pa x_2}(u_0) ={\pa v^2 \over \pa x_1}(u_0)=0. \end{eqnarray} (6.9)

Let then $L=\Pi_{13}$, and thus $v_t=(v^1,0,v^3),\,v_n=v^2,\,V=v^2v_t$. Since $v_n(u_0)=0$, we get from (6.2) and (6.6) that

Then, differentiating (6.4) with respect to $x_1$ and $x_3$ at $u_0$, we get
 \begin{eqnarray}\label{eq48} {\pa \om^1 \over \pa x_1}(u_0)=0\quad \hbox{and}\quad {\pa \om^1 \over \pa x_3}(u_0)=0. \end{eqnarray} (6.11)

Now we take $L=\{x_2+x_3=0\}$, therefore $v_n={v^2+v^3\over \sqrt 2},\nu_L=\left(0, {1\over \sqrt 2},{1\over \sqrt 2} \right)$, $v_t=\left(v^1, {v^2-v^3\over 2},{v^3-v^2\over 2} \right)$ and $V={v^2+v^3\over \sqrt 2} \left(v^1, {v^2-v^3\over \sqrt 2} \right)_{\mathcal{B}}$ in the orthonormal basis ${\mathcal{B}}=\left\{e_1'=(1,0,0), e_2'= \left(0, {1\over \sqrt 2},-{1\over \sqrt 2} \right)\right\}$. Since $v_n(u_0)=0$ and the vector $v_t^\perp(u_0)$ is directed along $e'_1$, we get from (6.2) that \begin{eqnarray*} v^1(u_0) \left ({\pa v^2 \over \pa x_2}(u_0)+{\pa v^2\over\pa x_3}(u_0) -{\pa v^3\over\pa x_3} (u_0)-{\pa v^3 \over \pa x_2} (u_0)\right)=0 \end{eqnarray*} and thus \begin{eqnarray*} {\pa v^2 \over \pa x_2}(u_0)+{\pa v^2\over\pa x_3} (u_0)-{\pa v^3\over\pa x_3} (0)-{\pa v^3 \over \pa x_2}(u_0)=0. \end{eqnarray*} Therefore by (6.10) we get also

 \begin{eqnarray}\label{eq49} {\pa v^2 \over \pa x_2}(u_0) ={\pa v^3 \over \pa x_3}(u_0). \end{eqnarray} (6.12)

For $L=\{x_1+x_2=0\}$ we then have $v_n={v^1+v^2\over\sqrt 2}$, $\nu_L=\left( {1\over \sqrt 2},{1\over \sqrt 2},0 \right),$ $v_t=\left( {v^1-v^2\over 2},{v^2-v^1\over 2},v^3\right)$ and thus $V={v^1+v^2\over\sqrt 2}\left({v^1-v^2\over\sqrt 2},v^3\right)_{\mathcal{B'}}$ in the plane basis ${\mathcal{B'}}=\left\{\left({1\over \sqrt 2},-{1\over \sqrt 2},0 \right), (0,0,1)\right\}$. Since the vector $v_t^\perp(u_0)$ is directed along $(0,0,1)$ we get from (6.2) and (6.9) that \begin{eqnarray*} v^1(u_0)\left(\om^2(u_0)+{\pa v^1\over\pa x_3}(u_0)+{\pa v^2\over\pa x_3} (u_0) \right)=0; \end{eqnarray*} therefore, we get by (6.10) that

 \begin{eqnarray}\label{eq51} \om^2(u_0)=-{\pa v^1 \over \pa x_3}(u_0). \end{eqnarray} (6.13)

If a trajectory $\ga=\ga(t)$ of the flow $v$ is parametrized by $t,$ i.e. $\frac{d\ga}{d t}=v,$ we have in virtue of (6.6) a differential inequality \begin{eqnarray*} |q'(t)|\leq C |q(t)|, \end{eqnarray*} for the function $q(t)=_{\rm def } \om^2(\gamma(t))$ and a positive constant $C$. Therefore, if $q(0)\ne0$ then $q(t)\ne0$ for any $t\in \R$, thus any trajectory of $v$ does not cross $\pa\Om =\pa D$ and hence stays either in $\bar\Om$ or in $\bar D$.

Using (6.9) , we see that $|v|$ is constant on any trajectory $\Ga$ of the vector field $\om$ and, conversely, (6.10) and (6.11) imply that $\om$ has a constant direction on any trajectory $\ga$ of the flow $v$ and therefore $\ga\ss \Pi_\ga$ is a plane curve for an affine plane $\Pi_\ga$. Set $\xi =\om/|\om|$, then we can define the vector $\xi(\ga)$ for any $\ga \ss \Om$ and $\xi(\ga)\perp \Pi_\ga$. For $z\in \Om$ we get

 \begin{eqnarray}\label{eq82} {\pa \xi \over\pa x_1}(z)=0 \end{eqnarray} (6.14)
since $v(z)$ is parallel to $x_1$, $\om(z)$ is parallel to $x_2$ and $\xi(z)=(0,1,0)$; thus (6.11) implies that
 \begin{eqnarray}\label{eq80} {\pa \xi^1 \over\pa x_3}(z)=0. \end{eqnarray} (6.15)
Therefore $\xi$ satisfies the Frobenius integrability condition $\langle\xi,\curl\xi \rangle =0$ and hence in a neighborhood of $z$ there exists a smooth function $U(x)$ with $\nabla U\neq 0$ parallel to $\xi$. Moreover, $U$ is a first integral of the flow $v$ since $\pa U/\pa v=0$. Let then $S$ be a level surface of $U$ containing $z$, then $S$ is foliated by the trajectories of $v$ and the vector field $\xi$ defines the Gauss map $\xi : S\to \S^2$. Since $\xi$ is constant on the trajectories of $v$ the image $\be=\xi(S) \ss \S^2$ is a curve or a point. Moreover, $\be$ is orthogonal to the axis $x_1$ at $\xi (\ga)$ by (6.14) , (6.15) and (6.13) implies that $\ga$ is not a straight line. Thus we can choose $z'\in \ga, z'\ne z$ and get that $\ga$ is orthogonal at $\xi (\ga)$ to some line not parallel to $x_1$. Therefore $\rank\,\xi (\ga) =0,$ hence $\rank\, \xi =0$ on $S$, $\xi$ is constant on $S$ and thus $S$ is a plane. We see that a neighborhood of $z$ in $\R^3$ is foliated by planes invariant under the flow $v$.

Denote by $\ga(s,t)\ss \Om$ the trajectory of $v$ passing through $z+(0,s,0)$, and let $L_s^z$ be the plane containing $\ga(s,t)$. Then $L_s^z\perp\om (z+(0,s,0))$ and the planes $L_s^z$ are invariant under the flow $v$ and foliate a neighborhood of $z$ in $\R^3$. Let $\la (s,y)$ for $y\in \Pi_{13}$ be an affine function on $\Pi_{13}$ with the graph $L_s^z$ and let $l_z(y)={\pa \la\over \pa s}(0,y)$ then $l_z$ is an affine linear function. Denote by $G$ a connected component of $\Pi_{13}\cap \Om,\, z\in G$ then $G$ is invariant under the flow $v$. We fix some $z'\in G$ and set \begin{eqnarray*} l=l_{z'}, \end{eqnarray*} then $l(z')=1$.

Let $z_1,z_2\in G$, then some neighborhoods of $z_1$ and $z_2$ in $\R^3$ are foliated by the same set of planes invariant under the flow $v$ and thus the sets of planes $L^{z_1}_s$ and $L^{z_2}_s$ are the same after a reparametrization. Therefore $l$ does not vanish in $G$ and we have $l_{z_1}=l_{z_2}/l_{z_2}(z_1)$. Set now \begin{eqnarray*} h(t)={\pa \ga(s,t)\over\pa s}{|_{s=0}} \end{eqnarray*} then $h(0)=(0,1,0)$ and thus by (6.6) the vector field $h$ is proportional to $\om$ on $\ga=\ga(0,t)$. Since $\om$ is orthogonal to $\Pi_{13}$ on $\ga$ we get that $h(y) = (0,l(y),0),$

 \begin{eqnarray}\label{eq52a} {\pa v^2\over\pa x_2}(y)={\pa \ln l\over\pa v}(y) \end{eqnarray} (6.16)
for any $y\in \ga$ and $l$ does not vanish on $\ga$.

It follows by (6.16) and (6.12) that

 \begin{eqnarray}\label{d51} {\pa v^3\over\pa x_3}(y) ={\pa \ln l\over\pa v}(y) \end{eqnarray} (6.17)
and since $\div\, v=0$ there holds (recall that $x_1$ is directed along $v(y)$)
 \begin{eqnarray}\label{d52} {\pa |v|\over\pa x_1}(y)=-2{\pa \ln l\over\pa v}(y) ={\pa \ln |v|\over\pa v}(y); \end{eqnarray} (6.18)
hence we get that
 \begin{eqnarray}\label{eq521} |v(y)|= \frac{ C_\gamma }{l^2(y)} \end{eqnarray} (6.19)
along the trajectory $\ga$ for a positive constant $C_\gamma$ depending on $\ga$. Note also that equations (6.16) (6.19) hold for any trajectory of $v$ in $G$ and hence by continuity in $\bar G$ outside the zero locus of $l$; in particular, we see that
 \begin{eqnarray} \label{ln0} v(y)\ne 0\quad\hbox{for any }y\in\bar G\hbox{ with }l(y)\ne 0. \end{eqnarray} (6.20)

Let $z_0\in \bar G$ be a point where the function $|v|$ attains its maximum on $\bar G$, then $z_0\in \pa G$. Indeed, if it is not the case, we have \begin{eqnarray*} {\pa v^1 \over \pa x_3}(z_0) =0, \end{eqnarray*} and hence $\om(z_0)=0$ by (6.13) which implies $z_0\in \pa G$.

Let $\ga_1$ be the trajectory of $v$ starting from $z_1\in \pa G$ with $v(z_1)\neq 0$, then $\om=0$ on the whole trajectory $\ga_1$ and $\ga_1\subset \pa \Om$. Therefore $\nabla b = v\times \om=0$ on $\ga_1$, where $b=p+\frac12 |v|^2$ is the Bernoulli function (see, e.g., [Arnold and Khesin1998]) and we get

 \begin{eqnarray}\label{eq52} \nabla p=-\frac12 \nabla |v|^2\quad \hbox{on } \ga_1. \end{eqnarray} (6.21)
Let $y\in \ga_1$ and let $e$ be a unit vector in $\Pi_{13}$ orthogonal to $v(y)$, then $\langle v_e(y),v(y)\rangle =0$ for $v_e= (\nabla_e v^1,\nabla_e v^3)$ by (6.13) since $\om(y)=0$. Therefore $\langle \nabla |v|^2(y),e\rangle =0$ and (6.21) implies that $\ga_1$ is a straight line interval $I$ which is finite since $v$ has a compact support, $v$ vanishes at its end points and thus $l|_I=0$ by (6.20) .

Let now $z_0=0$ and continue to assume that $x_1$ is directed along $v(0)\ne 0$ and $x_2$ is directed along $\om(x)\ne 0$ for some $x\in G$ (the direction of $\om(x)$ does not depend on $x$), then $l$ is a linear function on $\Pi_{13}$ vanishing on the $x_1$-axis: $l=Cx_3$ for $C\neq 0$. Denote now $D^+_\epsilon =B_\epsilon\cap\{ 0< x_3 \}$ and $D^-_\epsilon=B_ \epsilon\cap\{0> x_3 \},$ then we have $D^+_\epsilon\ss G$. Indeed, first note that $(D^+_\epsilon\cap\pa G)\cup( D^-_\epsilon\cap \pa G)=\emptyset$ since otherwise the trajectory $\ga_1$ starting from $z_1\in (D^+_\epsilon\cap \pa G)\cup (D^-_\epsilon\cap \pa G)$ leads to a contradiction since $l|_{\ga_1}=0$. Moreover, for the trajectory $\al_0(t)$ of $v^\perp =(-v^3,v^1)$ starting at 0 we have $\al_0(t)\in D^+_\epsilon$ for a small $t> 0$ and $\al_0(t)\in D^-_\epsilon$ for a small $t< 0$. Since $\om\ne 0$ on $G$ we get that $|v|$ strictly decreases along $\al_0$ by (6.13) ($v(0)$ being parallel to $x_1$) while $\al_0(t)$ stays in $G$ and since $|v|$ attains at $0$ its maximum in $\bar G$ we get that $D^-_\epsilon\bigcap G=\emptyset$; therefore, $D^+_\epsilon\ss G$.

Furthermore, any trajectory $\ga_s$ of $v$ starting from the point $(0,0,s)\in D^+_ \epsilon$ with $0< s< \epsilon$ and a sufficiently small $\epsilon > 0$ is closed. Indeed, by (6.19) we may assume that $C_{\ga_s}$ strictly increases as a function of $s\in(0,\epsilon)$ and thus $\ga_s$ with $s\in(0,\epsilon)$ intersects the interval $(0,(0,\epsilon))$ only once. By the Poincaré–Bendixson theorem we get that $\ga_s$ either

1. (i) tends to a limit, or
2. (ii) tends to a limit cycle $\rho\ss G$, or
3. (iii) is closed.
Since (i) contradicts (6.19) and (ii) implies that any trajectory $\ga_a$ with $s< a< \epsilon$ tends to $\rho$ which contradicts (6.19) as well, we get that (iii) holds. Moreover, any trajectory starting inside $D^+_\epsilon$ enters the domain$\{x_3> \de\}\cap G$ for some fixed $\de > 0$ and the union $A=\bigcup_{0< s< \epsilon}\ga_s\ss G$ is a topological annulus.

Note now that $C_{\ga_s}$ tends to zero for $s \to 0$ by (6.19) . Any trajectory $\al$ of $v^\perp$ in $A$ is orthogonal to the trajectories $\ga_s$ and thus intersects all $\ga_s$ while $|v|$ strictly decreases along $\al$ by (6.13) since $\om^2> 0$ on $G$. If $\lambda\in (0,\epsilon)$ then \begin{eqnarray*} \inf_{\ga_{\lambda}}|v|> |v(z)| \end{eqnarray*} for a sufficiently small $s\in (0,\la)$ and some $z\in \gamma_s$, while the trajectory $\al_z$ of $v^\perp$ starting from $z$ intersects $\ga_{\lambda}$ and $|v|$ strictly decreases along $\al_z$ which gives a contradiction and thus finishes the proof.

## 7. Vector Analysis' Framework

Let us briefly discuss Conjectures 1.1 and 1.2 in terms of the vector analysis for compactly supported tensor fields in $\R^3$. In this section we suppose that $v\in C_0^\infty(S^1,\R^3)$. We can rewrite (1.2) as follows

 \begin{eqnarray}\label{Ec} \curl(\div (v\otimes v))=0. \end{eqnarray} (7.1)
###### Proposition 7.1.
If (7.1) holds and the corresponding function $w\in C_0^\infty(M)$ is everywhere zero on $M$ then
 \begin{eqnarray}\label{sc} \Psi=\Psi(v)=_{\mathrm{def}}\si(\curl(v\otimes v))=0 \end{eqnarray} (7.2)
for the symmetrization $\Psi$ of the tensor field $\curl(v\otimes v),$ i.e. \begin{eqnarray*} {2}\Psi^{ij}= {\epsilon_{ilm}}\frac{\pa(v^jv^l)}{\pa x_m}+ {\epsilon_{jlm}} \frac{\pa ( v^iv^l)} {\pa x_m} , \end{eqnarray*} where $\epsilon_{ijk}$ is the standard permutation $($pseudo-$)$tensor$,$ giving the sign of the permutation $(ijk)$ of $(123)$ and the summation convention applies.

###### Proof.
For any fixed value of $x_1$, we define the vector field \begin{eqnarray*} Z=v^1 (v^2, v^3)=(v^1v^2,\,v^1v^3) \end{eqnarray*} on the vertical plane ${\Pi_{23}(x_1)} =\{x_1,x_2,x_3\}$ with coordinates $\{x_2,x_3\}$ depending on $x_1$ as on a parameter.

We have then $IZ(m')=w_{\nu_m}(m)=\langle \nabla w, \nu_m\rangle$ for a line $m\ss {\Pi_{23}(x_1)},$ a normal $\nu_m$ to $m$ and a line $m'\perp m\ss {\Pi_{23}(x_1)},$ thus $IZ=0$ and hence the solenoidal component $^sZ$ equals zero, where $Z=^sZ+\,^pZ$ is the Helmholtz decomposition of the vector field $Z$. Therefore we have \begin{eqnarray*} \;\Psi^{11}=\curl Z= \curl\, ^s Z=0, \end{eqnarray*} thus $\Psi^{ii}=0$ for $\,i=1,2,3$ and rotating the coordinate system in each plane ${\{ x_i,x_j\}}$ through the angle $\frac{\pi}{4}$ we get $\Psi^{ij}=0$ for all $\,1\le i \le j\le 3.$

Moreover, the proof of Theorem 6.1 shows that the conditions $\Psi(v)=0$ and $\mathrm{div}\,v=0$ imply $v=0$.

Therefore Conjectures 1.1 and 1.2 follow from

###### Conjecture 7.2.
If $\curl(\mathrm{div}(v\otimes v))=0$ then $\si(\curl(v\otimes v))=0.$

Another equivalent statement can be formulated as follows

###### Conjecture 7.3.
If $I({\mathrm{div}}(v\otimes v))=0$ then $PI(v\otimes v)=0.$

One can also ask whether the condition $\curl(\mathrm{div}(v\otimes v))=0$ implies that $v$ is spherically symmetric, which would grant Conjectures 1.1, 1.2, 7.2 and 7.3.

###### Acknowledgements.
We would like to thank A. Enciso, A. Jollivet and V. Sharafutdinov for their aid at different stages of our work.

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