Integral Geometry of Euler Equations

Abstract

We develop an integral geometry of stationary Euler equations defining some function w on the Grassmannian of affine lines in $\mathbb {R}^3$ 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 $v=0.$

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

$$\begin{aligned} \sum _{j=1}^3\frac{\partial (v^iv^j)}{\partial x_j}+\frac{\partial p}{\partial x_i}=0\quad \hbox {for } i=1,2,3, \end{aligned}$$

(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 $p=p(x),\,x\in \mathbb {R}^3$; it expresses the conservation of fluid’s momentum $v\otimes v +p\delta _{ij}$ and reads in a coordinate free form as follows

$$\begin{aligned} \text{ div }(v\otimes v)+\nabla p=0. \end{aligned}$$

(1.2)

Note that if we add to (1.1) the incompressibility condition

$$\begin{aligned} \text{ div }\,v=0, \end{aligned}$$

(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 Nadirashvili (2014) and also Chae and Constantin (2015). Let us state it explicitly:

Conjecture 1.1

Let $(v;p)\in C_0(\mathbb {R}^3)$ 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-Salas (2012). 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

$$\begin{aligned} \sum _{j=1}^3v^j\frac{\partial v^i}{\partial x_j}+\frac{\partial p}{\partial x_i}=0,\quad i=1,2,3 \end{aligned}$$

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 $\mathbb {R}^3,$ 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 Sharafutdinov (1994) 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 Sharafutdinov (1994) 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 Sharafutdinov (1994) and Nadirashvili et al. (2016).

In what follows we fix a positive scalar product $\langle x,y\rangle , \, x,y\in \mathbb {R}^n$. Let

$$\begin{aligned} T{\mathbb S}^{n-1}=\{(x,\xi )\in {\mathbb R}^n\times {\mathbb R}^n:\,\Vert \xi \Vert =1,\,\langle x,\xi \rangle =0\}\subset {\mathbb R}^n\times ({\mathbb R}^n{\setminus }\{0\}) \end{aligned}$$

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

$$\begin{aligned} (If)(x,\xi )=\sum \limits _{i_1,\dots ,i_h=1}^n\;\int \limits _ {-\infty }^\infty f_{i_1\dots i_h}(x+t\xi )\,\xi _{i_1}\dots \xi _{i_h}\,dt \end{aligned}$$

(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:

$$\begin{aligned} \psi (x,t\xi )= {\mathrm {sgn}}( t) t^{h-1}\psi (x,\xi )\quad (0\ne t\in {\mathbb R}),\quad \psi (x+t\xi ,\xi )=\psi (x,\xi ), \end{aligned}$$

(2.2)

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 $\chi (x,\xi )=\chi (x,\xi /\Vert \xi \Vert )$, to the open subset ${\mathcal Q}\cap \{\xi \ne 0 \}$ of the quadric

$$\begin{aligned} {\mathcal Q}=\{(x,\xi )\in {\mathbb R}^n\times {\mathbb R}^n:\, \langle x,\xi \rangle =0\} \supset T{\mathbb S}^{n-1}. \end{aligned}$$

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

$$\begin{aligned} \psi (x,\xi )=\Vert \xi \Vert ^{h-1}\chi \left( x-\frac{\langle x,\xi \rangle }{\Vert \xi \Vert ^2} \xi ,\frac{\xi }{\Vert \xi \Vert }\right) , \end{aligned}$$

(2.3)

which follows from (2.2); note that $\Big (x-\frac{\langle x,\xi \rangle }{\Vert \xi \Vert ^2}\xi ,\frac{\xi }{\Vert \xi \Vert }\Big )\in T\mathbb {S}^{n-1} \subset {\mathcal Q}$, 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 $I:{\mathcal S}(S^h;{\mathbb R}^n){\longrightarrow }{\mathcal S}(T{\mathbb S}^{n-1})$, 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 Sharafutdinov (1994) as follows.

John’s Conditions. A function $\chi \in {\mathcal S}(T{\mathbb S}^{n-1})\ (n\ge 3)$ belongs to the range of the operator I if and only if the following two conditions hold:

• 1. $\chi (x,-\xi )=(-1)^h\chi (x,\xi );$

• 2. The function $\psi \in C^\infty ({\mathbb R}^n\times ({\mathbb R}^n{\setminus }\{0\}))$ defined by (2.3) satisfies the equations

  $$\begin{aligned} \left( \frac{\partial ^2}{\partial x_{i_1}\partial \xi _{j_1}}-\frac{\partial ^2}{\partial x_{j_1}\partial \xi _{i_1}}\right) \dots \left( \frac{\partial ^2}{\partial x_{i_{h+1}}\partial \xi _{j_{h+1}}}-\frac{\partial ^2}{\partial x_{j_{h+1}}\partial \xi _{i_{h+1}}}\right) \psi =0 \end{aligned}$$

  (2.4)

  written for all indices $1\le i_1,j_1,\ldots ,i_{h+1},j_{h+1}\le n.$

Define the symmetric inner differentiation operator $d_s=\sigma \nabla$ by symmetrization of the covariant differentiation operator $\nabla :C^{\infty } (S^h) {\longrightarrow }C^{\infty }(T^{h+1}),$

$$\begin{aligned} (\nabla u)_{i_1,\ldots i_{h+1}} =u_{i_1\ldots i_h;i_{h+1}}=\frac{{\partial }u_{i_1\ldots i_h}}{{\partial }x_{i_{h+1}}}; \end{aligned}$$

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 Sharafutdinov 1994).

Kernel of the Ray Transform. Let $n\ge 2$ and $h\ge 1$ be integers. For a compactly-supported field $f\in C^{\infty }_0(S^h;\mathbb {R}^n)$ the following statements are equivalent:

• 1. $If = 0;$

• 2. There exists a compactly-supported field $v \in C^{\infty }_0(S^{h-1};\mathbb {R}^n)$ such that its support is contained in the convex hull of the support of f and

  $$\begin{aligned} d_sv = f. \end{aligned}$$

  (2.5)

Note also that an inversion formula for the operator I given by Theorem 2.10.2 in Sharafutdinov (1994) 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 $T\mathbb {S}^2$ over $\mathbb {S}^2$ coincides with the homogeneous space $M=G/(\mathbb {R}\times \mathrm{SO}(2))=G'/(\mathbb {R}\times \mathrm{O}(2))$, where $G=\mathbb {R}^3\rtimes \mathrm{SO}(3)$ is the group of proper rigid motions of $\mathbb {R}^3$, while $G'=G\cdot \{\pm I_3\}$ is the isometry group of $\mathbb {R}^3$. Therefore the operator I for $n=3$ can be written as $I:{\mathcal S}(S^h;{\mathbb {R}}^3){\longrightarrow }{\mathcal S}(M).$

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,

$$\begin{aligned} A=(y_1,y_2,0)+t\alpha =(y_1+\alpha _1t,y_2+\alpha _2t,t), \end{aligned}$$

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:

$$\begin{aligned} k=k(\alpha _1,\alpha _2)=\sqrt{1+\alpha _1^2+ \alpha _2^2}=\sqrt{1+\Vert \alpha \Vert ^2}; \end{aligned}$$

(2.6)

we will use this notation throughout the paper.

Define the diffeomorphism

$$\begin{aligned} \Phi :U\rightarrow {\mathbb R}^4,\quad (x,y,z,\xi )=(x,y,z,\xi _1,\xi _2,\xi _3) \mapsto (y,\alpha )=(y_1,y_2,\alpha _1,\alpha _2), \end{aligned}$$

on the open set $U= T{\mathbb {S}}^2\cap \{ \xi _3>0\}$ by

$$\begin{aligned} y_1=x-\frac{\xi _1}{\xi _3}z,\quad y_2=y-\frac{\xi _2}{\xi _3}z, \quad \alpha _1=\frac{\xi _1}{\xi _3},\quad \alpha _2=\frac{\xi _2}{\xi _3}. \end{aligned}$$

(2.7)

Then $(U,\Phi )$ is a coordinate patch on M; this parametrization was used by F. John in his seminal paper (John 1938).

For a function $\chi \in C^\infty (U)$, we define $\varphi \in C^\infty ({\mathbb R}^4)$ by

$$\begin{aligned} \varphi =k^{h-1}\chi \circ \Phi ^{-1}. \end{aligned}$$

These two functions are expressed through each other by the formulas

$$\begin{aligned}&\chi (x,y,z,\xi )=\xi _3^{h-1}\varphi \left( x-\frac{\xi _1z}{\xi _3} ,y-\frac{\xi _2z}{\xi _3},\frac{\xi _1}{\xi _3},\frac{\xi _2}{\xi _3}\right) ,\nonumber \\&\varphi (y,\alpha ) =k^{h-1}\chi \left( y_1-\frac{\langle y,\alpha \rangle \alpha _1}{k^2},y_2- \frac{\langle y,\alpha \rangle \alpha _2}{k^2}, -\frac{\langle y,\alpha \rangle }{k}, \frac{\alpha _1 }{k},\frac{\alpha _2}{k},\frac{1}{k}\right) .\qquad \qquad \end{aligned}$$

(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

$$\begin{aligned} \varphi =k^{h-1}\chi |_U\circ \Phi ^{-1}\in C^\infty ({\mathbb R}^4). \end{aligned}$$

For a tensor field $f\in {\mathcal S}(S^h;{\mathbb R}^3)$, the function

$$\begin{aligned} \varphi =k^{h-1}(If)|_U\circ \Phi ^{-1}\in C^\infty ({\mathbb R}^4) \end{aligned}$$

(2.9)

is expressed through f by the formula

$$\begin{aligned} \varphi (y,\alpha )=\sum \limits _{i_1,\dots ,i_h=1}^3\;\int \limits _{-\infty }^\infty f_{i_1\dots i_h}(y_1+\alpha _1t,y_2+\alpha _2t,t)\,\alpha _{i_1}\dots \alpha _{i_h}dt, \end{aligned}$$

(2.10)

with $\alpha _3=1$, which easily follows from (2.1).

Let

$$\begin{aligned} L=_\mathrm{def} {\partial ^2\over \partial \alpha _2\partial y_1}- {\partial ^2\over \partial \alpha _1 \partial y_2 } \end{aligned}$$

(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. $\chi (-x,-\xi )=(-1)^h\chi (x,\xi )$;

• 2. The function $\varphi \in C^\infty ({\mathbb R}^4)$ defined by (2.8) solves the equation

  $$\begin{aligned} L^{h+1}\varphi =0. \end{aligned}$$

  (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 $(v,p)\in C^\infty _0(\mathbb {R}^3)$ 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 $\mathbb {R}^3$ and let $\nu _L$ be its unit normal,  then

$$\begin{aligned} \int _{L}\langle v,z\rangle \langle v,\nu _L\rangle \,d\sigma _L =0 \end{aligned}$$

(3.1)

for any $z\in L$ where $d\sigma _L$ 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

$$\begin{aligned} \int _{L}\langle v,z\rangle \langle v,\nu \rangle \,d\sigma _L= & {} \int (z_1v^1+z_2v^2)v^3 \,dx_1dx_2\\= & {} z_1\int v^1v^3 \,dx_1dx_2+z_2\int v^2v^3 \,dx_1dx_2. \end{aligned}$$

Note that by Eq. (1.1) with $i=1$ and $i=2$ there holds

$$\begin{aligned}&\frac{\partial (v^1v^3)}{\partial x_3} =-\frac{\partial (v^1v^2)}{\partial x_2}-\frac{\partial (v^1v^1)}{\partial x_1}-\frac{\partial p}{\partial x_1},\\&\frac{\partial (v^2v^3)}{\partial x_3}=- \frac{\partial (v^2v^1)}{\partial x_1}- \frac{\partial (v^2v^2)}{\partial x_2}-\frac{\partial p}{\partial x_2}, \end{aligned}$$

and thus we get

$$\begin{aligned}&\frac{\partial }{\partial x _3}\left( \int v^1v^3 \,dx_1dx_2\right) =\int \frac{\partial (v^1v^3)}{\partial x_3}\,dx_1dx_2=0,\\&\frac{\partial }{\partial x _3}\left( \int v^2v^3 \,dx_1dx_2\right) =\int \frac{\partial (v^2v^3)}{\partial x_3}\,dx_1dx_2=0. \end{aligned}$$

Therefore, the compactly supported functions

$$\begin{aligned} \int v^1v^3 \,dx_1dx_2\quad \hbox {and}\quad \int v^2v^3 \,dx_1dx_2 \end{aligned}$$

of $x_3$ on $\mathbb {R}$ 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 $\mathbb {R}^2$; note that below we use this notation for vector fields on various planes in $\mathbb {R}^3$. Then let us set

$$\begin{aligned} F= \int _{-\infty }^\infty (-v^2v^3,\,v^1v^3)\,dx_3; \end{aligned}$$

(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 $x^0 = (x^0_1, x^0_2)\in \mathbb {R}^2$, $0\ne \xi = ( \xi _1, \xi _2)\in \mathbb {R}^2$, 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

$$\begin{aligned} \int _{L}\langle v,\tilde{\xi }_a \rangle \langle v,\nu \rangle \,d\sigma _L =0 \end{aligned}$$

and thus we get

$$\begin{aligned} \int _{L}(\xi _1v^1 + \xi _2v^2 + av^3)(-\xi _2v^1 + \xi _1v^2)\,d\sigma _L =0. \end{aligned}$$

Substituting the values $a=0,\,a=1$ and taking the difference we get the equation $IF=0$.

Therefore, by (2.5) we have

$$\begin{aligned} d_sw_0= \nabla w_0=-F \end{aligned}$$

(3.3)

for a unique compactly supported smooth scalar function $w_0=w_0(x)$.

Let us fix for a moment a point $P^0=(x_1^0, x_2^0)\in \mathbb {R}^2$, let $r \subset \Pi _{12}= \mathbb {R}^2$ 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

$$\begin{aligned} \int _{r}\langle e_{r},F\rangle \,ds_{r}=w_0(x_1^0,x_2^0) \end{aligned}$$

(3.4)

for the line element $ds_{r}$ of r. Let $H\subset \mathbb {R}^3$ be a half-plane perpendicular to $\Pi _{12}$ with $\partial H=m(x_1^0, x_2^0,0,0)$, 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

$$\begin{aligned} \int _{H} v^3\langle \nu _{H},v\rangle \,d\sigma _{H}=-\int _{r}\langle e_{r},F\rangle \,ds_{r} \end{aligned}$$

for the area element $d\sigma _{H}$ 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 $\partial H=m(x_1^0, x_2^0,0,0)$. Since the choice of a vertical line in $\mathbb {R}^3$ is arbitrary we see that the following definition is correct:

Definition 3.2

Define

$$\begin{aligned} w= -\int _{H(m)}\langle e_m,v\rangle \langle \nu _{H(m)},v\rangle \,d\sigma _{H(m)} \end{aligned}$$

(3.5)

where H(m) is a half-plane with $\partial H(m)= m$ 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

$$\begin{aligned} w(y_1,y_2,0,0)=w_0(y_1,y_2). \end{aligned}$$

Proof

Indeed, it is sufficient to verify that

$$\begin{aligned} {\partial w\over \partial y_1}(0)=\int _{-\infty }^\infty v^2v^3 \,dx_3,\quad {\partial w\over \partial y_2}(0)=-\int _{-\infty }^\infty v^1v^3 \,dx_3, \end{aligned}$$

(3.6)

which is clear, since

$$\begin{aligned}&w(\delta ,0,0,0)=-\int _{H_{1,\delta }}v^2v^3\,dx_1dx_3 =-\int _{\delta }^\infty dx_1\int _{-\infty }^\infty v^2v^3 \,dx_3,\\&w(0,\mu ,0,0)=\int _{H_{2,\mu }}v^1v^3\,dx_2dx_3 = \int _{\mu }^\infty dx_2\int _{-\infty }^\infty v^1v^3 \,dx_3 \end{aligned}$$

for the half-planes

$$\begin{aligned} H_{1,\delta }=\{(x_1>\delta ,0,x_3)\},\quad H_{2,\mu }=\{(0,x_2>\mu ,x_3)\}. \end{aligned}$$

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

$$\begin{aligned} k_1 =\sqrt{1+\alpha _1^2},\quad k_2 =\sqrt{1+\alpha _2^2}; \end{aligned}$$

(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:

• (i) $H(m)_1$ is parallel to $x_1$-axis, $H(m)_2$ is parallel to $x_2$-axis;

• (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

$$\begin{aligned} \nu _{H(m)_1}= \left( 0,{ 1\over k_2},-{\alpha _2\over k_2}\right) ,\quad \nu _{H(m)_2}= \left( -{1\over k_1},0,{\alpha _1\over k_1} \right) , \end{aligned}$$

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

$$\begin{aligned} e_m=\frac{1}{k}\left( \alpha _1,\alpha _2,1\right) =\left( {\alpha _1\over k},{\alpha _2\over k},{1\over k} \right) \end{aligned}$$

for the positive unit directional vector $e_m$ of m.

Proposition 3.4

Let $d\sigma _i$ 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

$$\begin{aligned} (\mathrm{i})\quad w= & {} -\int _{H(m)_2}\langle e_m,v \rangle \langle \nu _{H(m)_2},v\rangle \,d\sigma _2\nonumber \\= & {} -\int _{-\infty }^\infty \int _{l_2 }^\infty (\langle e_m,v \rangle \langle \nu _{H(m)_2},v\rangle )|_{(l_1,x_2,x_3)} k_1dx_2dx_3\nonumber \\= & {} \int _{-\infty }^\infty \int _{l_2}^\infty \frac{1}{k} ((\alpha _1v^1+\alpha _2v^2+v^3)(v^1-\alpha _1v^3))|_{(l_1,x_2,x_3)}dx_2dx_3\end{aligned}$$

(3.8)

$$\begin{aligned} (\mathrm{ii})\quad w= & {} -\int _{H(m)_1}\langle e_m,v \rangle \langle \nu _{H(m)_1},v\rangle \,d\sigma _1\nonumber \\= & {} -\int _{-\infty }^\infty \int _{l_1}^\infty (\langle e_m,v \rangle \langle \nu _{H(m)_1},v\rangle )|_{(x_1,l_2,x_3)} k_2\,dx_1dx_3\nonumber \\= & {} -\int _{-\infty }^\infty \int _{l_1}^\infty \frac{1}{k}((\alpha _1v^1+\alpha _2v^2+v^3) (v^2- \alpha _2 v^3))|_{(x_1,l_2,x_3)}\,dx_1dx_3.\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} x_2-x_3 \alpha _2-y_2 =0, \end{aligned}$$

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 $d\sigma _{1} = k_2\,dx_1dx_3$. Then one notes that the orthogonal projection $\pi _{13}(m)$ of m on the coordinate plane $\Pi _{13}=\{x_2=0\}$ is given by

$$\begin{aligned} \pi _{13}(m)=\Pi _{13}\cap H_2=\{x_1 =y_1+{\alpha _1x_3}\}, \end{aligned}$$

and thus $H(m)_1$ projects onto

$$\begin{aligned} \{x_1>y_1+{x_3\alpha _1}\}\subset \Pi _{13}, \end{aligned}$$

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 : 

$$\begin{aligned}&w=\int _{-\infty }^\infty \int _{l_2}^\infty (\alpha _2v^1v^2+v^1v^3+\alpha _1(v^1)^2 -\alpha _1(v^3)^2)|_{(l_1,x_2,x_3)}\,dx_2dx_3,\qquad \quad \end{aligned}$$

(3.10)

$$\begin{aligned}&w=\int _{-\infty }^\infty \int _{l_1}^\infty (\alpha _2(v^3)^2- \alpha _1v^1v^2-v^2 v^3-\alpha _2( v^2)^2)|_{(x_1,l_2,x_3)} dx_1dx_3. \end{aligned}$$

(3.11)

This corollary permits to calculate the quantities

$$\begin{aligned} \;{\partial ^m w\over \partial y_1^i\partial y_2^j \partial \alpha _1^k \partial \alpha _2^l}(0),\quad i+j+k+l=m, \end{aligned}$$

for $k+l\le 1$, and in particular, implies the following.

Corollary 3.6

We have

$$\begin{aligned}&{\partial ^2w\over \partial y_2\partial \alpha _1}(0)= \int _{-\infty }^\infty \left( (v^3)^2-(v^1) ^2-x_3{\partial (v^3v^1)\over \partial x_1} \right) \Bigg |_{(0,0,x_3)} \,dx_3,\end{aligned}$$

(3.12)

$$\begin{aligned}&{\partial ^2w\over \partial y_1\partial \alpha _2}(0)=\int _{-\infty }^\infty \left( (v^2)^2 -(v^3) ^2+x_3{\partial (v^3v^2)\over \partial x_2} \right) \Bigg |_{(0,0,x_3)} \,dx_3,\end{aligned}$$

(3.13)

$$\begin{aligned}&{\partial ^2w\over \partial y_1^2}(0)+{\partial ^2w\over \partial y_2^2}(0) =\int _{-\infty }^{\infty }\left( {\partial (v^2v^3)\over \partial x_1}-{\partial \left( v^1v^3\right) \over \partial x_2} \right) \Bigg |_{(0,0,x_3)}\,dx_3. \end{aligned}$$

(3.14)

Proof of (3.12)

From (3.10) we have

$$\begin{aligned} w(0,y_2,\alpha _1,0)=\int _{-\infty }^\infty \int _{y_2}^\infty (v^1v^3+(v^1)^2 \alpha _1 -(v^3)^2\alpha _1)|_{(x_3 \alpha _1,x_2,x_3 )}\,dx_2dx_3, \end{aligned}$$

whence we get

$$\begin{aligned} {\partial w(0,0,\alpha _1,0) \over \partial y_2}=-\int _{-\infty }^\infty (v^1v^3+(v^1)^2 \alpha _1 -(v^3)^2\alpha _1)|_{(\alpha _1x_3, 0,x_3 )}\,dx_3 \end{aligned}$$

and, finally,

$$\begin{aligned} {\partial ^2w\over \partial y_2\partial \alpha _1}(0)=\int _{-\infty }^\infty \left( (v^3)^2 -(v^1)^2 -x_3 {\partial (v^3v^1)\over \partial x_1}\right) |_{(0,0,x_3 )}\,dx_3. \end{aligned}$$

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:

$$\begin{aligned} {\partial ^2w \over \partial y_1\partial \alpha _2}(0)-{\partial ^2w \over \partial y_2\partial \alpha _1}(0) =\int _{-\infty }^\infty (p+(v^1)^2+(v^2)^2-(v^3) ^2 )|_{(0,0,x_3 )} \,dx_3.\nonumber \\ \end{aligned}$$

(3.15)

Indeed, we have $\;{\partial (v^3v^1)\over \partial x_1} +{\partial (v^3v^2) \over \partial x_2} =-{\partial (p +(v^3)^2 )\over \partial x_3}$ by (1.1) and integrating

$$\begin{aligned} \int _{-\infty }^\infty x_3 \left( {\partial (v^3v^1)\over \partial x_1} +{\partial (v^3v^2)\over \partial x_2} \right) \Bigg |_{(0,0,x_3 )} dx_3=-\int _{-\infty }^\infty {x_3 \partial (p +(v^3)^2)\over \partial x_3 } \Bigg |_{(0,0,x_3 )} dx_3 \end{aligned}$$

by parts we get (3.15).

4 Operators P and $\Delta _M$

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

Recall that $M=G/(\mathbb {R}\times \mathrm{SO}(2))=G'/(\mathbb {R}\times \mathrm{O}(2))$ for $G=\mathbb {R}^3\rtimes \mathrm{SO}(3)$ 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

$$\begin{aligned} Pf(m_0)=_\mathrm{def}Lf_g(0), \end{aligned}$$

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 $u\in {\mathbb {R}}\times {\mathrm SO}(2)=St_0$, $St_0<G$ being the stabilizer of the vertical line. It is sufficient to verify the equality separately for $u\in {\mathbb {R}}$ and $u\in {\mathrm SO}(2)$. It is clear for a vertical shift $u\in {\mathbb {R}}$ since L has constant coefficients; for a rotation $u\in {\mathrm SO}(2)$ by angle $\theta$ in the horizontal plane one easily calculates

$$\begin{aligned} F_u(y_1,y_2,\alpha _1,\alpha _2 )= & {} F(y_1\cos \theta -y_2\sin \theta ,y_2\cos \theta +y_1\sin \theta ,\alpha _1\cos \theta \\&-\, \alpha _2\sin \theta ,\alpha _2\cos \theta +\alpha _1\sin \theta ), \end{aligned}$$

and a simple calculation shows the necessary equation, since we get

$$\begin{aligned}&{\partial ^2 F_u(0)\over \partial \alpha _2\partial y_1}= \cos ^2 \theta {\partial ^2 F(0)\over \partial \alpha _2\partial y_1} -\sin ^2 \theta {\partial ^2 F(0)\over \partial \alpha _1\partial y_2}+\cos \theta \sin \theta \left( {\partial ^2 F(0)\over \partial \alpha _1\partial y_1} -{\partial ^2 F(0)\over \partial \alpha _2\partial y_2}\right) ,\\&{\partial ^2 F_u(0)\over \partial \alpha _1\partial y_2} =\cos ^2 \theta {\partial ^2 F(0)\over \partial \alpha _1\partial y_2} -\sin ^2 \theta {\partial ^2 F(0)\over \partial \alpha _2\partial y_1}+\cos \theta \sin \theta \left( {\partial ^2 F(0)\over \partial \alpha _1\partial y_1} -{\partial ^2 F(0)\over \partial \alpha _2\partial y_2}\right) . \end{aligned}$$

The proof is finished.

We can now rewrite (3.15) as follows

$$\begin{aligned} P_0w= & {} \int _{-\infty }^\infty (p+(v^1)^2+(v^2)^2-(v^3)^2)\,dx_3\nonumber \\= & {} \int _{-\infty }^\infty (p+|v|^2-2(v^3)^2)\, dx_3 \end{aligned}$$

(4.1)

for the operator $P_0$ being P evaluated at 0, which implies that

$$\begin{aligned} Pw= & {} \int _m(p+|v|^2-2\langle v,e_m\rangle ^2) \,ds\nonumber \\= & {} \int _m(p+|v|^2-2v\otimes v)\,ds =IQ_0(m) \end{aligned}$$

(4.2)

for the quadratic tensor field $Q_0=(p+|v|^2)\delta _{ij}-2v\otimes v$ 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 $\Delta _M=\Delta _{y_1, y_2}$ acting in tangent planes to $\mathbb {S}^2$; it is defined by the usual formula

$$\begin{aligned} \Delta _Mf(m )=\frac{\partial ^2 f(m )}{\partial {y_1}^2} +\frac{\partial ^2 f(m)}{\partial {y_2}^2} \end{aligned}$$

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 $\Delta _M f(m)$ is determined by the G-invariance condition as for the operator P above, and the rotational symmetry of $\Delta _{y_1, y_2}$ guarantees the correctness of that definition. Note that the operators P, $\Delta _M$ commute and note also that (2.10) implies that for $Q\in C_0^\infty (S^h,\mathbb {R}^3)$ there holds a commutation rule

$$\begin{aligned} I(\Delta Q)=\Delta _M(IQ). \end{aligned}$$

(4.3)

Remark 4.1

The algebra $D_{G'}(M)$ of the $G'$-invariant differential operators on M is freely generated by $\Delta _M$ and $P^2$ as a commutative algebra, see Gonzalez and Helgason (1986).

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

$$\begin{aligned} P=k^2L+\alpha _1\frac{{\partial }}{{\partial }y_2}-\alpha _2\frac{{\partial }}{{\partial }y_1},\quad \Delta _M= k_1^2\frac{ {\partial }^2}{ {\partial }y_1^2}+k_2^2\frac{{\partial }^2}{{\partial }y_2^2} +2\alpha _1\alpha _2 {\partial ^2 \over \partial y_1\partial y_2}. \end{aligned}$$

(4.4)

Now we deduce the principal linear differential equation for w.

Proposition 4.3

We have

$$\begin{aligned} P^2w=-4\Delta _M w. \end{aligned}$$

(4.5)

Proof

We begin with the following simple result.

Lemma 4.4

If $f\in C^{\infty }_0(\mathbb {R}^3)$ 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

$$\begin{aligned} P^2w=PIQ_0= PI((p+|v|^2)\delta _{ij}-2v\otimes v)=-2PI(v\otimes v) \end{aligned}$$

(4.6)

for a compactly supported vector field v solving (1.1). Moreover, we have

$$\begin{aligned} I(v\otimes v)(y_1,y_2,\alpha _1,\alpha _2)=\int _{-\infty }^{\infty }( {(v^3) ^2 +2v^1v^3\alpha _1+2v^2v^3\alpha _2})\,{dx_3 }+O(|\alpha |^2) \end{aligned}$$

and thus by (3.14) we get

$$\begin{aligned} PI(v\otimes v)(0)=P_0I(v\otimes v)=2\int _{-\infty }^{\infty }\left( {\partial \left( v^2v^3\right) \over \partial x_1} -{\partial \left( v^1v^3\right) \over \partial x_2} \right) dx_3 =2\Delta _{M}w(0), \end{aligned}$$

hence $P I(v\otimes v)= 2\Delta _Mw$ everywhere and $P^2w=-4\Delta _Mw$ by (4.6).

Corollary 4.5

The equation

$$\begin{aligned} IQ_0(m)=0,\quad \forall m\in M \end{aligned}$$

(4.7)

implies Conjecture 1.2.

Indeed, if $IQ_0(m)=0$ then $\Delta _M w(m)=-\frac{1}{4} P^2w(m)=-\frac{1}{4} PIQ_0(m)=0$ 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 $\Delta _M$ in terms of the Lie algebra $\mathfrak g$ of G. We have $\mathfrak g= {\mathfrak so}(3)\oplus {\mathfrak r}(3)=\mathbb {R}^3\oplus \mathbb {R}^3$ 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

$$\begin{aligned}{}[(r_1;0),(r_2;0)]=(r_1\times r_2;0),\quad [(0;s_1),(0;s_2)]=0,\quad [(r;0),(0;s)]= (0;r \times s). \end{aligned}$$

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:

$$\begin{aligned} \tilde{\Delta }_M = S_1^2+S_2^2+S_3^2,\quad \tilde{P}= {S_1} {R_1}+ {S_2} {R_2}+ {S_3} {R_3}, \end{aligned}$$

(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 $\tilde{\Delta }_M =\Delta _M$ and $\tilde{P}=P.$

Proof

First, the operators $\tilde{\Delta }_M$ 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 $\tilde{\Delta }_M$ 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:

$$\begin{aligned}{}[ {S_1} {R_1}, {R_3}]= - {S_2} {R_1}- {S_1} {R_2},\quad [ {S_2} {R_2}, {R_3}]= {S_1} {R_2}+ {S_2} {R_1},\quad [ {S_3} {R_3}, {R_3}]= 0. \end{aligned}$$

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

$$\begin{aligned}{}[{S_1} {R_1}, {S_3}]= -{S_1}{S_2},\quad [{S_2}{R_2},{S_3}]={S_1S_2},\quad [S_3R_3, S_3]= 0. \end{aligned}$$

Since any line is ${\mathrm SO}(3)$-conjugate to a vertical one, $\tilde{P}$ is G-invariant. Finally, we have $\tilde{\Delta }_M (m_0)=\Delta _M (m_0),\, \tilde{P}(m_0)=P(m_0)\,$ for $m_0=m(0,0,0,0)$, which finishes the proof, since $\Delta _M$ 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 $t\in \mathbb {R}$, and let $l_t=m(t,0,0,0)$ be a vertical line in the plane $\Pi _{13}$; note that $\Pi _{13}=\bigcup _{t\in \mathbb {R}} l_t$.

Lemma 4.7

For $f\in C^\infty _0(M)$ we have

$$\begin{aligned} \int _\mathbb {R}P^2f(l_t)\,dt =\int _\mathbb {R}S_2^2 R_2^2 f(l_t) \,dt. \end{aligned}$$

(4.9)

Proof

Define the operators A and B by $A= {S_1} {R_1}P,\, B= {S_3} {R_3}P;$ then

$$\begin{aligned} P^2=A+B+ {S_2}{R_2}P=A+B+{S_2R_2}\left( {S_1R_1}+ S_2R_2+ S_3R_3\right) . \end{aligned}$$

Since $Af(l_t)$ is a derivative of a function of t, while B vanishes identically on $\Pi _{13}$, we get that

$$\begin{aligned} \int _\mathbb {R}(Af(l_t)+Bf(l_t))dt=0. \end{aligned}$$

We have also

$$\begin{aligned} {S_2R_2S_1 R_1}(f(l_t))={S_1S_2R_2 R_1}(f(l_t))- {S_3S_2}{R_1}(f(l_t)), \end{aligned}$$

thus the integral of the left-hand side is zero and the same is true for

$$\begin{aligned} {S_2R_2S_3R_3}(f(l_t)) ={S_3S_2R_2R_3}(f(l_t))+{S_1S_2R_3}(f(l_t)); \end{aligned}$$

since ${S_2R_2S_2R_2}=S_2^2R_2^2$ we get the conclusion.

Lemma 4.8

We have

$$\begin{aligned} \int _{\mathbb {R}} R_2^2 w(l_t)\,dt =- 4\int _{\mathbb {R}} w(l_t)\,dt. \end{aligned}$$

(4.10)

Proof

Let us fix a positive constant $c<{\pi \over 2}$, and let $l^\theta _t =({t\over \cos (\theta )},0,\tan (\theta ),0)$ for any $\theta$ with $|\theta |<c$; therefore, $l^\theta _t$ is just the line $l_t$ rotated (in the clockwise direction) through the angle $\theta$ about the origin in the plane $\Pi _{13}$, and for any t we have

$$\begin{aligned} R_2^2w(l^\theta _t)|_{\theta =0}= {\partial ^2\over \partial \theta ^2}w(l^\theta _t)|_{\theta =0}. \end{aligned}$$

Let $e^\theta _1=(\cos \theta ,0,-\sin \theta ),\;e^\theta _3=(\sin \theta ,0,\cos \theta )$ then we have

$$\begin{aligned} w(l^\theta _t)={1\over \cos \theta }\int _0^\infty \int _{l_t}\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle |_{ \left( {t\over \cos \theta }+x_3 \tan \theta ,x_2,x_3\right) }\,dx_3 dx_2 \end{aligned}$$

by (3.8), and if we put $t={x_1\cos \theta }-x_3\sin \theta$ we get that

$$\begin{aligned} \int _{\mathbb {R}} w(l^\theta _t)\,dt= & {} {1\over \cos \theta }\int _0^\infty \int _{\mathbb {R}^2}\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle \,dx_3 dx_2dt\\= & {} \int _{x_2 >0}\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle \,dx_1dx_2 dx_3. \end{aligned}$$

Therefore, since $\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle =v^1v^3\cos 2\theta +((v^1)^2-(v^3)^2){\sin 2\theta \over 2}$ we have

$$\begin{aligned} {\partial ^2\over \partial \theta ^2}\int _{\mathbb {R}} w(l^\theta _t)\,dt= & {} \int _{ x_2>0 }{\partial ^2\over \partial \theta ^2} (\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle ) \,dx_1dx_2 dx_3\\= & {} -{4}\int _{x_2 >0} (\langle v,e^\theta _1\rangle \langle v,e^\theta _3\rangle )\,dx_1dx_2 dx_3 \end{aligned}$$

and evaluating at $\theta =0$ 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

$$\begin{aligned} \int _{\mathbb {R}} P^2 w(l_t)\,dt=\int _{\mathbb {R}} S_2^2 R_2^2 w(l_t)\,dt =-4 \int _{\mathbb {R}} S_2^2w(l_t)\,dt= -4\int _{\mathbb {R}} \Delta _M w(l_t)\,dt.\nonumber \\ \end{aligned}$$

(4.11)

If we define a function $F(x_1,x_2)$ on $\mathbb {R}^2$ by

$$\begin{aligned} F(x_1,x_2)= P^2w(x_1,x_2,0,0)+4\Delta _M w(x_1,x_2,0,0), \end{aligned}$$

then the integral of F over the $x_1$-axis vanishes by (4.11). Changing the coordinate system $x_1,x_2$ in $\mathbb {R}^2$, 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 $h=2$

$$\begin{aligned} P^3\psi +4P\Delta _M \psi =0 \end{aligned}$$

(4.12)

if $\psi =IQ$ for $Q\in C_0^\infty (S^2,\mathbb {R}^3)$. Applying then (4.12) to $\psi =\Delta _M w$ (which can be written as $\Delta _M w=IQ'$ for a certain $Q'$ not given here) we obtain $P^3\Delta _Mw+4P\Delta ^2_M w=\Delta _MP(P^2w+4\Delta _M w)=0$ 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

$$\begin{aligned} P\Delta _Mu =-2\Delta _Mw,\quad \Delta _Mu=IQ_1 \end{aligned}$$

for some $\,Q_1\in C_0^\infty (S^2,\mathbb {R}^3)$ and applying (4.12) to $\psi =\Delta _Mu=IQ_1$ we get

$$\begin{aligned} 0=P^3\Delta _M u+4\Delta _M^2 P u= -2\Delta _M(P^2 w+4\Delta _M w), \end{aligned}$$

and thus we reprove (4.5). We can define u similarly to (3.5) as follows

$$\begin{aligned} u=\int _{H(m)}\mathrm{{dist}}(P,m)(p+\langle \nu _{H(m)},v\rangle ^2) d\sigma _{H(m)}, \end{aligned}$$

where $\mathrm{{dist}}(P,m)$ 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:

$$\begin{aligned} JQ(L)= \int _L\mathrm {tr}(Q_{|L}) \,d\sigma _L \end{aligned}$$

(5.1)

for an affine plane $L\subset \mathbb {R}^3$ and $Q\in C^{\infty }(S^{2};\mathbb {R}^3)$ satisfying

$$\begin{aligned} |Q(x)| \le C(1 +|x|)^{-2-\varepsilon }, \end{aligned}$$

(5.2)

for some $\varepsilon >0,$ where $Q_{|L}$ is the restriction onto L; then we get a bounded linear operator

$$\begin{aligned} J:{\mathcal S}(S^2;{\mathbb R}^3){\longrightarrow }{\mathcal S}(\mathbb {R}P^3) \end{aligned}$$

for the manifold $\mathbb {R}P^3$ parametrizing affine planes $L\subset \mathbb {R}^3.$

Proposition 5.1

We have $JQ_0(L)=0.$

Proof

We have for any affine plane $L\subset \mathbb {R}^3$ that

$$\begin{aligned} JQ_0(L)=2\int _L( p+\langle v, \nu _L\rangle ^2)\, d\sigma _L=0. \end{aligned}$$

Indeed, setting without loss of generality $L=\Pi _{12},\,\nu _L=e_3$ we get that

$$\begin{aligned} {\partial JQ_0\over \partial x_3}(\Pi _{12})= & {} {\partial \over \partial x_3}\left( \int _L( p+\langle v, \nu _L\rangle ^2)\, d\sigma _L\right) \\= & {} \int _{\Pi _{12}}{\partial (p+(v^3)^2)\over \partial x_3}\, dx_1dx_2\\= & {} -\int _ {\Pi _{12}}\left( {\partial (v^1v^3)\over \partial x_1}+{\partial (v^2v^3)\over \partial x_2}\right) \, dx_1dx_2=0. \end{aligned}$$

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 $JQ_0(L)=0, \forall L\in \mathbb {R}P^3$ is equivalent to the following equation for the components $\{q^{ij}\}$ of $Q_0{:}$

$$\begin{aligned} \sum _{ i,j=1}^3 {\partial ^2 q^{ij}\over \partial x_i\partial x_j}=\Delta \text{ tr }\, Q_0, \end{aligned}$$

(5.3)

while $IQ_0(m)=0,\, \forall m\in M$ is equivalent to the following system

$$\begin{aligned} {2\,\partial q^{ij}\over \partial x_i\partial x_j}={\partial ^2q^{ii}\over \partial x_j^2}+ {\partial ^2 q^{jj}\over \partial x_i^2},\quad 1\le i<j\le 3 \end{aligned}$$

(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 $(v,p)\in \mathbb {C}^\infty _0(\mathbb {R}^3)$ 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

$$\begin{aligned} \int _m\langle v,t\rangle \langle v,n\rangle \,ds=0. \end{aligned}$$

(6.1)

Let $L\subset \mathbb {R}^3$ 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 ${{\mathrm{curl}}}_L V =0$ for the curl operator ${{\mathrm{curl}}}_L$ on L which gives

$$\begin{aligned} v_n{{\mathrm{curl}}}_L v_t - \langle v^\perp _t, \nabla _L v_n\rangle =-v_n\omega _n- \langle v^\perp _t, \nabla _L v_n\rangle =0 \end{aligned}$$

(6.2)

for the normal component $\omega _n$ of $\omega ={{\mathrm{curl}}}v$ and the gradient operator $\nabla _L$ on L. We will apply (6.2) to various planes $L \subset \mathbb {R}^3$.

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

$$\begin{aligned} v^3{{\mathrm{curl}}}_L v_t-\langle v^\perp _t, \nabla v^3\rangle =v^3\left( {\partial v^1 \over \partial x_2} -{\partial v^2 \over \partial x_1} \right) -v^1{\partial v^3 \over \partial x_2}+v^2 {\partial v^3 \over \partial x_1} =0. \end{aligned}$$

(6.3)

Let now

$$\begin{aligned} \Omega =_\mathrm{def}\{u\in \mathbb {R}^3|v(u)\ne 0,\,\omega (u)\ne 0\} \end{aligned}$$

and let $D=_\mathrm{def}\mathbb {R}^3{\setminus } \bar{\Omega }$. We can suppose that $\Omega$ is not empty, since otherwise in a neighborhood of a point $x_0$ where the maximum of |v| is attained we have $\Delta v=-{{\mathrm{curl}}}{{\mathrm{curl}}}v+\nabla \text{ div }\, v=0$ 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 $u_0\in \Omega$ that

$$\begin{aligned} v^3\left( {\partial v^1 \over \partial x_2} -{\partial v^2 \over \partial x_1} \right) (u_0) =0\quad \hbox {and therefore}\quad \langle v,\omega \rangle (u_0) =0. \end{aligned}$$

(6.4)

Therefore $\langle v,\omega \rangle =0$ holds everywhere on $\Omega$, thus on $\mathbb {R}^3$ and differentiating this relation in the v-direction we obtain

$$\begin{aligned} \langle v \nabla v,\omega \rangle + \langle v\nabla \omega , v\rangle =0 ; \end{aligned}$$

(6.5)

using the commutation law

$$\begin{aligned} v \nabla \omega = \omega \nabla v \end{aligned}$$

(6.6)

we get then from (6.5) that

$$\begin{aligned} \langle v \nabla v,\omega \rangle + \langle \omega \nabla v, v\rangle =0. \end{aligned}$$

(6.7)

In orthonormal coordinates $x_1,x_2,x_3$ with $x_1$ directed along v and  $x_2$ directed along $\omega$ at $u_0$ we can rewrite (6.7) as follows

$$\begin{aligned} {\partial v^1 \over \partial x_2}(u_0) +{\partial v^2 \over \partial x_1}(u_0) =0, \end{aligned}$$

(6.8)

since $v(u_0)\ne 0$ and $\omega (u_0)\ne 0$. Below we always use that coordinate system.

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

$$\begin{aligned} {\partial v^1\over \partial x_2}(u_0) ={\partial v^2\over \partial x_1}(u_0)\quad \hbox {and therefore}\quad {\partial v^1\over \partial x_2}(u_0) ={\partial v^2 \over \partial x_1}(u_0)=0. \end{aligned}$$

(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

$$\begin{aligned} {\partial v^2 \over \partial x_3} (u_0)=0\quad \hbox {and hence}\quad {\partial v^3 \over \partial x_2}(u_0) =0\quad \hbox {and}\quad {\partial \omega ^3 \over \partial x_1}(u_0)=0. \end{aligned}$$

(6.10)

Then, differentiating (6.4) with respect to $x_1$ and $x_3$ at $u_0$, we get

$$\begin{aligned} {\partial \omega ^1 \over \partial x_1}(u_0)=0\quad \hbox {and}\quad {\partial \omega ^1 \over \partial x_3}(u_0)=0. \end{aligned}$$

(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

$$\begin{aligned} v^1(u_0) \left( {\partial v^2 \over \partial x_2}(u_0)+{\partial v^2\over \partial x_3}(u_0) -{\partial v^3\over \partial x_3} (u_0)-{\partial v^3 \over \partial x_2} (u_0)\right) =0 \end{aligned}$$

and thus

$$\begin{aligned} {\partial v^2 \over \partial x_2}(u_0)+{\partial v^2\over \partial x_3} (u_0)-{\partial v^3\over \partial x_3} (0)-{\partial v^3 \over \partial x_2}(u_0)=0. \end{aligned}$$

Therefore by (6.10) we get also

$$\begin{aligned} {\partial v^2 \over \partial x_2}(u_0) ={\partial v^3 \over \partial x_3}(u_0). \end{aligned}$$

(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

$$\begin{aligned} v^1(u_0)\left( \omega ^2(u_0)+{\partial v^1\over \partial x_3}(u_0)+{\partial v^2\over \partial x_3} (u_0) \right) =0; \end{aligned}$$

therefore, we get by (6.10) that

$$\begin{aligned} \omega ^2(u_0)=-{\partial v^1 \over \partial x_3}(u_0). \end{aligned}$$

(6.13)

If a trajectory $\gamma =\gamma (t)$ of the flow v is parametrized by t,  i.e. $\frac{d\gamma }{d t}=v,$ we have in virtue of (6.6) a differential inequality

$$\begin{aligned} |q'(t)|\le C |q(t)|, \end{aligned}$$

for the function $q(t)=_\mathrm{def } \omega ^2(\gamma (t))$ and a positive constant C. Therefore, if $q(0)\ne 0$ then $q(t)\ne 0$ for any $t\in \mathbb {R}$, thus any trajectory of v does not cross $\partial \Omega =\partial D$ and hence stays either in $\bar{\Omega }$ or in $\bar{D}$.

Using (6.9), we see that |v| is constant on any trajectory $\Gamma$ of the vector field $\omega$ and, conversely, (6.10) and (6.11) imply that $\omega$ has a constant direction on any trajectory $\gamma$ of the flow v and therefore $\gamma \subset \Pi _\gamma$ is a plane curve for an affine plane $\Pi _\gamma$. Set $\xi =\omega /|\omega |$, then we can define the vector $\xi (\gamma )$ for any $\gamma \subset \Omega$ and $\xi (\gamma )\perp \Pi _\gamma$. For $z\in \Omega$ we get

$$\begin{aligned} {\partial \xi \over \partial x_1}(z)=0 \end{aligned}$$

(6.14)

since v(z) is parallel to $x_1$, $\omega (z)$ is parallel to $x_2$ and $\xi (z)=(0,1,0)$; thus (6.11) implies that

$$\begin{aligned} {\partial \xi ^1 \over \partial x_3}(z)=0. \end{aligned}$$

(6.15)

Therefore $\xi$ satisfies the Frobenius integrability condition $\langle \xi ,{{\mathrm{curl}}}\xi \rangle =0$ and hence in a neighborhood of z there exists a smooth function U(x) with $\nabla U\ne 0$ parallel to $\xi$. Moreover, U is a first integral of the flow v since $\partial U/\partial v=0$. 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 $\xi : S\rightarrow \mathbb {S}^2$. Since $\xi$ is constant on the trajectories of v the image $\beta =\xi (S) \subset \mathbb {S}^2$ is a curve or a point. Moreover, $\beta$ is orthogonal to the axis $x_1$ at $\xi (\gamma )$ by (6.14), (6.15) and (6.13) implies that $\gamma$ is not a straight line. Thus we can choose $z'\in \gamma , z'\ne z$ and get that $\gamma$ is orthogonal at $\xi (\gamma )$ to some line not parallel to $x_1$. Therefore $\mathrm{rank}\,\xi (\gamma ) =0,$ hence $\mathrm{rank}\, \xi =0$ on S, $\xi$ is constant on S and thus S is a plane. We see that a neighborhood of z in $\mathbb {R}^3$ is foliated by planes invariant under the flow v.

Denote by $\gamma (s,t)\subset \Omega$ the trajectory of v passing through $z+(0,s,0)$, and let $L_s^z$ be the plane containing $\gamma (s,t)$. Then $L_s^z\perp \omega (z+(0,s,0))$ and the planes $L_s^z$ are invariant under the flow v and foliate a neighborhood of z in $\mathbb {R}^3$. Let $\lambda (s,y)$ for $y\in \Pi _{13}$ be an affine function on $\Pi _{13}$ with the graph $L_s^z$ and let $l_z(y)={\partial \lambda \over \partial s}(0,y)$ then $l_z$ is an affine linear function. Denote by G a connected component of $\Pi _{13}\cap \Omega ,\, z\in G$ then G is invariant under the flow v. We fix some $z'\in G$ and set

$$\begin{aligned} l=l_{z'}, \end{aligned}$$

then $l(z')=1$.

Let $z_1,z_2\in G$, then some neighborhoods of $z_1$ and $z_2$ in $\mathbb {R}^3$ 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

$$\begin{aligned} h(t)={\partial \gamma (s,t)\over \partial s}{|_{s=0}} \end{aligned}$$

then $h(0)=(0,1,0)$ and thus by (6.6) the vector field h is proportional to $\omega$ on $\gamma =\gamma (0,t)$. Since $\omega$ is orthogonal to $\Pi _{13}$ on $\gamma$ we get that $h(y) = (0,l(y),0),$

$$\begin{aligned} {\partial v^2\over \partial x_2}(y)={\partial \ln l\over \partial v}(y) \end{aligned}$$

(6.16)

for any $y\in \gamma$ and l does not vanish on $\gamma$.

It follows by (6.16) and (6.12) that

$$\begin{aligned} {\partial v^3\over \partial x_3}(y) ={\partial \ln l\over \partial v}(y) \end{aligned}$$

(6.17)

and since $\text{ div }\, v=0$ there holds (recall that $x_1$ is directed along v(y))

$$\begin{aligned} {\partial |v|\over \partial x_1}(y)=-2{\partial \ln l\over \partial v}(y) ={\partial \ln |v|\over \partial v}(y); \end{aligned}$$

(6.18)

hence we get that

$$\begin{aligned} |v(y)|= \frac{ C_\gamma }{l^2(y)} \end{aligned}$$

(6.19)

along the trajectory $\gamma$ for a positive constant $C_\gamma$ depending on $\gamma$. 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

$$\begin{aligned} v(y)\ne 0\quad \hbox {for any }y\in \bar{G}\hbox { with }l(y)\ne 0. \end{aligned}$$

(6.20)

Let $z_0\in \bar{G}$ be a point where the function |v| attains its maximum on $\bar{G}$, then $z_0\in \partial G$. Indeed, if it is not the case, we have

$$\begin{aligned} {\partial v^1 \over \partial x_3}(z_0) =0, \end{aligned}$$

and hence $\omega (z_0)=0$ by (6.13) which implies $z_0\in \partial G$.

Let $\gamma _1$ be the trajectory of v starting from $z_1\in \partial G$ with $v(z_1)\ne 0$, then $\omega =0$ on the whole trajectory $\gamma _1$ and $\gamma _1\subset \partial \Omega$. Therefore $\nabla b = v\times \omega =0$ on $\gamma _1$, where $b=p+\frac{1}{2} |v|^2$ is the Bernoulli function (see, e.g., Arnold and Khesin 1998) and we get

$$\begin{aligned} \nabla p=-\frac{1}{2} \nabla |v|^2\quad \hbox {on } \gamma _1. \end{aligned}$$

(6.21)

Let $y\in \gamma _1$ 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 $\omega (y)=0$. Therefore $\langle \nabla |v|^2(y),e\rangle =0$ and (6.21) implies that $\gamma _1$ 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 $\omega (x)\ne 0$ for some $x\in G$ (the direction of $\omega (x)$ does not depend on x), then l is a linear function on $\Pi _{13}$ vanishing on the $x_1$-axis: $l=Cx_3$ for $C\ne 0$. Denote now $D^+_\varepsilon =B_\varepsilon \cap \{ 0< x_3 \}$ and $D^-_\varepsilon =B_ \varepsilon \cap \{0>x_3 \},$ then we have $D^+_\varepsilon \subset G$. Indeed, first note that $(D^+_\varepsilon \cap \partial G)\cup ( D^-_\varepsilon \cap \partial G)=\emptyset$ since otherwise the trajectory $\gamma _1$ starting from $z_1\in (D^+_\varepsilon \cap \partial G)\cup (D^-_\varepsilon \cap \partial G)$ leads to a contradiction since $l|_{\gamma _1}=0$. Moreover, for the trajectory $\alpha _0(t)$ of $v^\perp =(-v^3,v^1)$ starting at 0 we have $\alpha _0(t)\in D^+_\varepsilon$ for a small $t>0$ and $\alpha _0(t)\in D^-_\varepsilon$ for a small $t<0$. Since $\omega \ne 0$ on G we get that |v| strictly decreases along $\alpha _0$ by (6.13) (v(0) being parallel to $x_1$) while $\alpha _0(t)$ stays in G and since |v| attains at 0 its maximum in $\bar{G}$ we get that $D^-_\varepsilon \bigcap G=\emptyset$; therefore, $D^+_\varepsilon \subset G$.

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

• (i) tends to a limit, or

• (ii) tends to a limit cycle $\rho \subset G$, or

• (iii) is closed.

Since (i) contradicts (6.19) and (ii) implies that any trajectory $\gamma _a$ with $s<a<\varepsilon$ tends to $\rho$ which contradicts (6.19) as well, we get that (iii) holds. Moreover, any trajectory starting inside $D^+_\varepsilon$ enters the domain$\{x_3>\delta \}\cap G$ for some fixed $\delta >0$ and the union $A=\bigcup _{0<s<\varepsilon }\gamma _s\subset G$ is a topological annulus.

Note now that $C_{\gamma _s}$ tends to zero for $s \rightarrow 0$ by (6.19). Any trajectory $\alpha$ of $v^\perp$ in A is orthogonal to the trajectories $\gamma _s$ and thus intersects all $\gamma _s$ while |v| strictly decreases along $\alpha$ by (6.13) since $\omega ^2>0$ on G. If $\lambda \in (0,\varepsilon )$ then

$$\begin{aligned} \inf _{\gamma _{\lambda }}|v|>|v(z)| \end{aligned}$$

for a sufficiently small $s\in (0,\lambda )$ and some $z\in \gamma _s$, while the trajectory $\alpha _z$ of $v^\perp$ starting from z intersects $\gamma _{\lambda }$ and |v| strictly decreases along $\alpha _z$ 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 $\mathbb {R}^3$. In this section we suppose that $v\in C_0^\infty (S^1,\mathbb {R}^3)$. We can rewrite (1.2) as follows

$$\begin{aligned} {{\mathrm{curl}}}(\text{ div }(v\otimes v))=0. \end{aligned}$$

(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

$$\begin{aligned} \Psi =\Psi (v)=_{\mathrm {def}}\sigma ({{\mathrm{curl}}}(v\otimes v))=0 \end{aligned}$$

(7.2)

for the symmetrization $\Psi$ of the tensor field ${{\mathrm{curl}}}(v\otimes v),$ i.e.

$$\begin{aligned} {2}\Psi ^{ij}= {\varepsilon _{ilm}}\frac{\partial (v^jv^l)}{\partial x_m}+ {\varepsilon _{jlm}} \frac{\partial ( v^iv^l)}{\partial x_m} , \end{aligned}$$

where $\varepsilon _{ijk}$ 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

$$\begin{aligned} Z=v^1 (v^2, v^3)=(v^1v^2,\,v^1v^3) \end{aligned}$$

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 $m\subset {\Pi _{23}(x_1)},$ a normal $\nu _m$ to m and a line $m'\perp m\subset {\Pi _{23}(x_1)},$ 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

$$\begin{aligned} \;\Psi ^{11}={{\mathrm{curl}}}Z= {{\mathrm{curl}}}\, ^s Z=0, \end{aligned}$$

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 ${{\mathrm{curl}}}(\mathrm {div}(v\otimes v))=0$ then $\sigma ({{\mathrm{curl}}}(v\otimes v))=0.$

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