Integral geometry of Euler equations
aa r X i v : . [ m a t h . A P ] A p r Integral Geometry of Euler Equations
Nikolai Nadirashvili ∗ , Serge Vl˘adut¸ † Abstract.
We develop an integral geometry of stationary Euler equationsdefining some function w on the Grassmannian of affine lines in R dependingon a putative compactly supported solution ( v ; p ) of the system and deducesome linear differential equations for w . We conjecture that w = 0 everywhereand prove that this conjecture implies that v = 0 . AMS 2000 Classification: 76B03; 35J61
In the present paper we introduce and develop a version of integral geometryfor the steady Euler system.More precisely, the system which we consider is as follows X j =1 ∂ ( v i v j ) ∂x j + ∂p∂x i = 0 for i = 1 , , , (1.1)for an unknown vector field v = v ( x ) = ( v ( x ) , v ( x ) , v ( x )) and an unknownscalar function p = p ( x ) , x ∈ R ; it expresses the conservation of fluid’s mo-mentum v ⊗ v + pδ ij and reads in a coordinate free form as followsdiv( v ⊗ v ) + ∇ p = 0 . (1.2)Note that if we add to (1.1) the incompressibility conditiondiv v = 0 , (1.3)the system (1.1)–(1.3) describes a steady state flow of the ideal fluid.A long-standing folklore conjecture states that a smooth compactly sup-ported solution of (1.1)–(1.3) should be identically zero, and this is known forBeltrami flows; see [N] and also [CC]. Let us state it explicitly: Conjecture 1.1
Let ( v ; p ) ∈ C ( R ) be a solution of (1.1) – (1.3) . Then v = 0 , p = 0 . ∗ Aix Marseille Universit´e, CNRS, Centrale Marseille, I2M UMR 7373, 13453, Marseille,France, [email protected] † Aix Marseille Universit´e, CNRS, Centrale Marseille, I2M UMR 7373, 13453, Marseille,France and IITP RAS, 19 B. Karetnyi, Moscow, Russia, [email protected] v is a solution of(1.1) for a suitable pressure p , but we do not know whether the system X j =1 v j ∂v i ∂x j + ∂p∂x i = 0 , i = 1 , , v i v j . More precisely, given any smooth compactly sup-ported solution ( v ; p ) of (1.1), we define a smooth function w on the Grassman-nian M classifying lines in space using the X -ray transforms of v i v j and thenderive a linear differential equation for w . Using a Radon plane transform of w we deduce one more linear homogemenous differential equation which suggeststhat w = 0 everywhere. However, we are not able to deduce this fact and for-mulate it as a conjecture; we show that assuming the conjecture and (1.3) onecan deduce Conjecture 1.1. Therefore, we put forth Conjecture 1.2.
Let w be the function on the Grassmannian G , of affinelines in R , defined below in section 3, which depends on a compactly supportedsolution ( v ; p ) of (1.1) . Then w = 0 everywhere. Note, in paricular, that this conjecture holds for any spherically symmetriccompactly supported vector field v. The rest of the paper is organized as follows: in Section 2 we recall somedefinitions and results from [S] concerning the X -ray transform of symmetrictensor fields. In Section 3 we define and study a smooth function w ∈ C ∞ ( M )which depends on a smooth compactly supported solution ( v, p ) of (1.1). Section4 contains a description of two invariant order 2 differential operators on C ∞ ( M )and a differential equation for w in terms of those operators. In Section 5 wedefine a plane Radon transform for quadratic tensor fields, prove that it vanishesand explain why this partially confirms Conjecture 1.2. Finally, in Section 6 wededuce Conjecture 1.1 assuming Conjecture 1.2 together with (1.3). Acknowledgement . We would like to thank A. Enciso, A. Jollivet and V.Sharafutdinov for their aid at different stages of our work. X -ray Transform We use throughout our paper the integral geometry of tensor fields developedin [S] and discussed in [NSV] in its three-dimensional form. Let us give some ofits points in our simple situation. For details see [S] and [NSV].In what follows we fix a positive scalar product h x, y i , x, y ∈ R n . Let T S n − = { ( x, ξ ) ∈ R n × R n : k ξ k = 1 , h x, ξ i = 0 } ⊂ R n × ( R n \ { } )2e the tangent bundle of S n − ⊂ R n .Given a continuous rank h symmetric tensor field f on R n , the X-ray trans-form of f is defined for ( x, ξ ) ∈ R n × ( R n \ { } ) by( If )( x, ξ ) = n X i ,...,i h =1 ∞ Z −∞ f i ...i h ( x + tξ ) ξ i . . . ξ i h dt (2.1)under the assumption that f decays at infinity so that the integral converges.We denote by S ( S h ; R n ) the space of symmetric degree h tensor fields with allcomponents lying in the Schwartz space, and denote by S ( T S n − ) the Schwartzspace on T S n − . Below we consider only tensors from S ( S h ; R n ) and functionsfrom S ( T S n − ). For such f ∈ S ( S h ; R n ) we get a C ∞ -smooth function ψ ( x, ξ ) =( If )( x, ξ ) on R n × ( R n \ { } ) satisfying the following conditions: ψ ( x, tξ ) = sgn( t ) t h − ψ ( x, ξ ) (0 = t ∈ R ) , ψ ( x + tξ, ξ ) = ψ ( x, ξ ) , (2.2)which mean that ( If )( x, ξ ) actually depends only on the line passing throughthe point x in direction ξ , and we parameterize the manifold of oriented lines in R n by T S n − . For χ ( x, ξ ) ∈ S ( T S n − ) we can extend χ by homogeneity, setting χ ( x, ξ ) = χ ( x, ξ/ k ξ k ), to the open subset Q ∩ { ξ = 0 } of the quadric Q = { ( x, ξ ) ∈ R n × R n : h x, ξ i = 0 } ⊃ T S n − . Conversely, for a tensor field f ∈ S ( S h ; R n ), the restriction χ = ψ | T S n − ofthe function ψ = If to the manifold T S n − belongs to S ( T S n − ). Moreover,the function ψ is uniquely recovered from χ by the formula ψ ( x, ξ ) = k ξ k h − χ (cid:16) x − h x, ξ ik ξ k ξ, ξ k ξ k (cid:17) , (2.3)which follows from (2.2); note that (cid:16) x − h x,ξ ik ξ k ξ, ξ k ξ k (cid:17) ∈ T S n − ⊂ Q , and thusthe right-hand side of (2.3) is correctly defined. Therefore, the X-ray transformcan be considered as a linear continuous operator I : S ( S h ; R n ) −→S ( T S n − ),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 [S] as follows. John’s Conditions.
A function χ ∈ S ( T S n − ) ( n ≥ belongs to the rangeof the operator I if and only if the following two conditions hold: (1) χ ( x, − ξ ) = ( − h χ ( x, ξ );(2) The function ψ ∈ C ∞ (cid:0) R n × ( R n \ { } ) (cid:1) defined by (2.3) satisfies theequations (cid:16) ∂ ∂x i ∂ξ j − ∂ ∂x j ∂ξ i (cid:17) . . . (cid:16) ∂ ∂x i h +1 ∂ξ j h +1 − ∂ ∂x j h +1 ∂ξ i h +1 (cid:17) ψ = 0 (2.4)3 ritten for all indices ≤ i , j , . . . , i h +1 , j h +1 ≤ n . Define the symmetric inner differentiation operator d s = σ ∇ by symmetriza-tion of the covariant differentiation operator ∇ : C ∞ ( S h ) −→ C ∞ ( T h +1 ) , ( ∇ u ) i ,...i h +1 = u i ...i h ; i h +1 = ∂u i ...i h ∂x i h +1 ;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 [S]). Kernel of the ray transform.
Let n ≥ and h ≥ be integers. For acompactly-supported field f ∈ C ∞ ( S h ; R n ) the following statements are equiva-lent: (1) If = 0;(2) There exists a compactly-supported field v ∈ C ∞ ( S h − ; R n ) such that itssupport is contained in the convex hull of the support of f and d s v = f. (2.5)Note also that an inversion formula for the operator I given by Theo-rem 2.10.2 in [S] 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 over S coincides withthe homogeneous space M = G/ ( R × SO(2)) = G ′ / ( R × O(2)), where G = R ⋊ SO(3) is the group of proper rigid motions of R , while G ′ = G · {± I } isthe isometry group of R . Therefore the operator I for n = 3 can be written as I : S ( S h ; R ) −→S ( M )Let us define coordinates on the open subset M nh of M consisting of non-horizontal affine lines. Namely, m = m ( y , y , α , α ) is given by a parametricequation for a current point A on m , A = ( y , y ,
0) + tα = ( y + α t, y + α t, t ) , where t grows in the positive direction of m and thus the vector α = ( α , α , m .We can now rewrite the above general formulas using the coordinates ( y , y , α , α ).First we fix the following notation: k = k ( α , α ) = q α + α = p k α k ; (2.6)we will use this notation throughout the paper.Define the diffeomorphismΦ : U → R , ( x, y, z, ξ ) = ( x, y, z, ξ , ξ , ξ ) ( y, α ) = ( y , y , α , α ) ,
4n the open set U = T S ∩ { ξ > } by y = x − ξ ξ z, y = y − ξ ξ z, α = ξ ξ , α = ξ ξ . (2.7)Then ( U, Φ) is a coordinate patch on M ; this parametrization was used by F.John in his seminal paper [J].For a function χ ∈ C ∞ ( U ), we define ϕ ∈ C ∞ ( R ) by ϕ = k h − χ ◦ Φ − . These two functions are expressed through each other by the formulas χ ( x, y, z, ξ ) = ξ h − ϕ (cid:18) x − ξ zξ , y − ξ zξ , ξ ξ , ξ ξ (cid:19) ,ϕ ( y, α ) = k h − χ (cid:18) y − h y, α i α k , y − h y, α i α k , − h y, α i k , α k , α k , k (cid:19) . (2.8)If a function χ ∈ C ∞ ( T S ) satisfies χ ( − x, − ξ ) = ( − h χ ( x, ξ ), then it isuniquely determined by ϕ = k h − χ | U ◦ Φ − ∈ C ∞ ( R ) . For a tensor field f ∈ S ( S h ; R ), the function ϕ = k h − ( If ) | U ◦ Φ − ∈ C ∞ ( R ) (2.9)is expressed through f by the formula ϕ ( y, α ) = X i ,...,i h =1 ∞ Z −∞ f i ...i h ( y + α t, y + α t, t ) α i . . . α i h dt, (2.10)with α = 1, which easily follows from (2.1).Let L = def ∂ ∂α ∂y − ∂ ∂α ∂y (2.11)be the John operator. The main result of [NSV] says that for n = 3, a function χ ∈ S ( T S ) belongs to the range of the operator I for a given h ≥ χ ( − x, − ξ ) = ( − h χ ( x, ξ );(2) The function ϕ ∈ C ∞ ( R ) defined by (2.8) solves the equation L h +1 ϕ = 0 . (2.12)Thus h +5 h +62 equations (2.4) for n = 3 are equivalent to equation (2.12).5 Function w In what follows we fix a compactly supported smooth solution ( v, p ) ∈ C ∞ ( R )of system and define a function w ∈ C ∞ ( M ) using the following result. Lemma 3.1
Let L be an affine plane in R and let ν L be its unit normal, then Z L h v, z ih v, ν L i dσ L = 0 (3.1) for any z ∈ L where dσ L is the area element on L .Proof. We can assume without loss of generality that L = { ( x , x , } and ν L = (0 , ,
1) = e , z = z e + z e for e = (1 , , , e = (0 , , , e = (0 , , Z L h v, z ih v, ν i dσ L = Z ( z v + z v ) v dx dx == z Z v v dx dx + z Z v v dx dx . Note that by equation (1.1) with i = 1 and i = 2 there holds ∂ ( v v ) ∂x = − ∂ ( v v ) ∂x − ∂ ( v v ) ∂x − ∂p∂x ,∂ ( v v ) ∂x = − ∂ ( v v ) ∂x − ∂ ( v v ) ∂x − ∂p∂x , and thus we get ∂∂x (cid:16) Z v v dx dx (cid:17) = Z ∂ ( v v ) ∂x dx dx = 0 ,∂∂x (cid:16) Z v v dx dx (cid:17) = Z ∂ ( v v ) ∂x dx dx = 0 . Therefore, the compactly supported functions Z v v dx dx and Z v v dx dx of x on R are constant and thus vanish everywhere which finishes the proof.For any fixed value of x , we define the vector field v ⊥ v = ( − v v , v v )on the plane ( x , x , x ) with coordinates { x , x } depending on x as on aparameter, where u ⊥ = ( − u , u ) for a vector field u = ( u , u ) on R ; notethat below we use this notation for vector fields on various planes in R . Thenlet us set F = Z ∞−∞ ( − v v , v v ) dx ; (3.2)6ote that F is a compactly supported vector field on the plane Π = { ( x , x , } with coordinates { x , x } and (3.1) implies that IF = 0. Indeed, choose x = ( x , x ) ∈ R , 0 = ξ = ( ξ , ξ ) ∈ R , and let L be the 2-plane through thepoint x parallel to the vectors ξ and (0 , , . Then the vector ν = ( − ξ , ξ , L and the vector e ξ a = ( − ξ , ξ , a ) is parallel to L for every a. By (3.1) we have Z L h v, e ξ a ih v, ν i dσ L = 0and thus we get Z L ( ξ v + ξ v + av )( − ξ v + ξ v ) dσ L = 0 . Substituting the values a = 0 , a = 1 and taking the difference we get theequation IF = 0.Therefore, by (2.5) we have d s w = ∇ w = − F (3.3)for a unique compactly supported smooth scalar function w = w ( x ).Let us fix for a moment a point P = ( x , x ) ∈ R , let r ⊂ Π = R bea ray emanating from P and let e r be a unit directional vector of r , then invirtue of (3.3) we have Z r h e r , F i ds r = w ( x , x ) (3.4)for the line element ds r of r . Let H ⊂ R be a half-plane perpendicular toΠ with ∂H = m ( x , x , , m ( x , x , ,
0) is the vertical line passingthrough the point ( x , x , ∈ Π ; therefore, H orthogonally projects ontosome ray r emanating from P . Let us consider the integral Z H v h ν H , v i dσ H = − Z r h e r , F i ds r for the area element dσ H of H and a suitable unit normal ν H to H , then by(3.4) it does not depend on H for a fixed point P = ( x , x ) and a fixed line ∂H = m ( x , x , , R is arbitrary we seethat the following definition is correct: Definition 3.2
Define w = − Z H ( m ) h e m , v ih ν H ( m ) , v i dσ H ( m ) (3.5) where H ( m ) is a half-plane with ∂H ( m ) = m and ν H ( m ) is the unit normal to H ( m ) such that the basis (cid:0) e m , ν m , ν H ( m ) (cid:1) is positively oriented for the interiorunit normal ν m to m lying in H ( m ) . w is a compactly supported smooth function on M and it can bewritten as w = w ( y , y , α , α ) on M nh ; moreover, we get Lemma 3.3
We have w ( y , y , ,
0) = w ( y , y ) . Proof.
Indeed, it is sufficient to verify that ∂w∂y (0) = Z ∞−∞ v v dx , ∂w∂y (0) = − Z ∞−∞ v v dx , (3.6)which is clear, since w ( δ, , ,
0) = − Z H ,δ v v dx dx = − Z ∞ δ dx Z ∞−∞ v v dx ,w (0 , µ, ,
0) = Z H ,µ v v dx dx = Z ∞ µ dx Z ∞−∞ v v dx for the half-planes H ,δ = { ( x > δ, , x ) } , H ,µ = { (0 , x > µ, x ) } . Now we give two explicit formulas for w which use two specific choicesof H ( m ). We begin by putting k = q α , k = q α ; (3.7)recall also that k = p α + α . Given a line m ∈ M nh , let H ( m ) and H ( m ) be the half-planes with theborder-line m which are determined by the following conditions:( i ) H ( m ) is parallel to x -axis , H ( m ) is parallel to x -axis;( ii ) h ν i , e i i > , i = 1 , , for e = (1 , , , e = (0 , ,
0) and the internal normals ν i ∈ H ( m ) i , i = 1 , ν H ( m ) = (cid:18) , k , − α k (cid:19) , ν H ( m ) = (cid:18) − k , , α k (cid:19) , and the plane H ( m ) forms angle β with the coordinate plane Π = { x = 0 } where cos β = 1 /k , while the plane H ( m ) forms angle β with the coordinateplane Π = { x = 0 } , cos β = 1 /k . Note also that we have e m = 1 k ( α , α ,
1) = (cid:18) α k , α k , k (cid:19) for the positive unit directional vector e m of m .8 roposition 3.4 Let dσ i be the surface area element on H ( m ) i and let l i = y i + x α i for i = 1 , . Then in the introduced notation we have ( i ) w = − Z H ( m ) h e m , v ih ν H ( m ) , v i dσ == − Z ∞−∞ Z ∞ l (cid:0) h e m , v ih ν H ( m ) , v i (cid:1) | ( l ,x ,x ) k dx dx == Z ∞−∞ Z ∞ l k (cid:0) ( α v + α v + v )( v − α v ) (cid:1) | ( l ,x ,x ) dx dx (3.8)( ii ) w = − Z H ( m ) h e m , v ih ν H ( m ) , v i dσ == − Z ∞−∞ Z ∞ l (cid:0) h e m , v ih ν H ( m ) , v i (cid:1) | ( x ,l ,x ) k dx dx == − Z ∞−∞ Z ∞ l k (cid:0) ( α v + α v + v )( v − α v ) (cid:1) | ( x ,l ,x ) dx dx . (3.9) Proof.
This is an elementary calculation which we give only for H ( m ) , sincethe case of H ( m ) is completely similar; note only that the choice of normals ν H ( m ) and ν H ( m ) comes from the orientation condition. Let us fix the valuesof y , y , α , and α , and let H i ⊃ H ( m ) i be an affine plane containing H ( m ) i for i = 1 ,
2. Then an equation of H is of the form ax + bx + c = 0, andtherefore c = − ay . Since e m ∈ ¯ H for the vector plane ¯ H parallel to H , weget aα + b = 0, and we can choose a = 1, b = − α , so the equation takes theform x − x α − y = 0 , and therefore x = x α + y on the half-plane H ( m ) . Since cos β = cos arctan α = k , we see that dσ = k dx dx . Then one notes that the orthogonal projection π ( m ) of m on the coordinate plane Π = { x = 0 } is given by π ( m ) = Π ∩ H = { x = y + α x } , and thus H ( m ) projects onto { x > y + x α } ⊂ Π , since h ν , e i >
0, which completes the proof.The formulas (3.8)–(3.9) are somewhat cumbersome and use below only thefollowing simple consequence.
Corollary 3.5
In the first order of ( α , α ) , ignoring terms with total deg α ≥ ,we have the following expressions: w = Z ∞−∞ Z ∞ l (cid:16) α v v + v v + α (cid:0) v (cid:1) − α (cid:0) v (cid:1) (cid:17) | ( l ,x ,x ) dx dx , (3.10) w = Z ∞−∞ Z ∞ l (cid:16) α (cid:0) v (cid:1) − α v v − v v − α (cid:0) v (cid:1) (cid:17) | ( x ,l ,x ) dx dx . (3.11)9his corollary permits to calculate the quantities ∂ m w∂y i ∂y j ∂α k ∂α l (0) , i + j + k + l = m, for k + l ≤
1, and in particular, implies the following.
Corollary 3.6
We have ∂ w∂y ∂α (0) = Z ∞−∞ ( v ) − ( v ) − x ∂ (cid:0) v v (cid:1) ∂x ! | (0 , ,x ) dx , (3.12) ∂ w∂y ∂α (0) = Z ∞−∞ ( v ) − ( v ) + x ∂ (cid:0) v v (cid:1) ∂x ! | (0 , ,x ) dx , (3.13) ∂ w∂y (0) + ∂ w∂y (0) = Z ∞−∞ ∂ (cid:0) v v (cid:1) ∂x − ∂ (cid:0) v v (cid:1) ∂x ! | (0 , ,x ) dx . (3.14) Proof of (3.12) . From (3.10) we have w (0 , y , α ,
0) = Z ∞−∞ Z ∞ y (cid:0) v v + ( v ) α − ( v ) α (cid:1) | ( x α ,x ,x ) dx dx , whence we get ∂w (0 , , α , ∂y = − Z ∞−∞ (cid:0) v v + ( v ) α − ( v ) α (cid:1) | ( α x , ,x ) dx and, finally, ∂ w∂y ∂α (0) = Z ∞−∞ ( v ) − ( v ) − x ∂ (cid:0) v v (cid:1) ∂x ! | (0 , ,x ) dx . 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: ∂ w∂y ∂α (0) − ∂ w∂y ∂α (0) = Z ∞−∞ (cid:0) p + ( v ) + ( v ) − ( v ) (cid:1) | (0 , ,x ) dx . (3.15)Indeed, we have ∂ ( v v ) ∂x + ∂ ( v v ) ∂x = − ∂ ( p +( v ) ) ∂x by (1.1) and integrating Z ∞−∞ x ∂ (cid:0) v v (cid:1) ∂x + ∂ (cid:0) v v (cid:1) ∂x ! | (0 , ,x ) dx = − Z ∞−∞ x ∂ (cid:0) p + ( v ) (cid:1) ∂x | (0 , ,x ) dx by parts we get (3.15). 10 Operators P and ∆ M Let us define first an order 2 differential operator P on the space C ( M ) . Recall that M = G/ ( R × SO(2)) = G ′ / ( R × O(2)) for G = R ⋊ SO(3) and G ′ = G · {± I } . Definition 4.1
Let f ∈ C ( M ) , m ∈ M , and let g ( m ) = 0 = (0 , , , for g ∈ G . Then P f ( m ) = def Lf g (0) , where f g ( m ) = f ( g − ( m )) for any m ∈ 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 ∈ G such that g ( m ) = h ( m ) = (0).We put u = g − h, F = f u , and thus we have to verify that LF (0) = LF u (0)for u ∈ R × S O (2) = St , St < G being the stabilizer of the vertical line.It is sufficient to verify the equality separately for u ∈ R and u ∈ S O (2). It isclear for a vertical shift u ∈ R since L has constant coefficients; for a rotation u ∈ S O (2) by angle θ in the horizontal plane one easily calculates F u ( y , y , α , α ) == F ( y cos θ − y sin θ, y cos θ + y sin θ, α cos θ − α sin θ, α cos θ + α sin θ ) , and a simple calculation shows the necessary equation, since we get ∂ F u (0) ∂α ∂y = cos θ ∂ F (0) ∂α ∂y − sin θ ∂ F (0) ∂α ∂y + cos θ sin θ (cid:18) ∂ F (0) ∂α ∂y − ∂ F (0) ∂α ∂y (cid:19) ,∂ F u (0) ∂α ∂y = cos θ ∂ F (0) ∂α ∂y − sin θ ∂ F (0) ∂α ∂y + cos θ sin θ (cid:18) ∂ F (0) ∂α ∂y − ∂ F (0) ∂α ∂y (cid:19) . The proof is finished.We can now rewrite (3.15) as follows P w = Z ∞−∞ (cid:0) p + ( v ) + ( v ) − ( v ) (cid:1) dx = Z ∞−∞ (cid:0) p + | v | − v ) (cid:1) dx (4.1)for the operator P being P evaluated at 0, which implies that P w = Z m (cid:0) p + | v | − h v, e m i (cid:1) ds = Z m (cid:0) p + | v | − v ⊗ v (cid:1) ds = IQ ( m ) (4.2)for the quadratic tensor field Q = (cid:0) p + | v | (cid:1) δ ij − v ⊗ v and any m ∈ M , since P is G -invariant; therefore P w = IQ as functions on M .11e will also use the fiber-wise Laplacian ∆ M = ∆ y ,y acting in tangentplanes to S ; it is defined by the usual formula∆ M f ( m ) = ∂ f ( m ) ∂y + ∂ f ( m ) ∂y for f ∈ C ( M ) and a vertical line m = m ( y , y , , m ∈ M thevalue ∆ M f ( m ) is determined by the G -invariance condition as for the operator P above, and the rotational symmetry of ∆ y ,y guarantees the correctness ofthat definition. Note that the operators P , ∆ M commute and note also that(2.10) implies that for Q ∈ C ∞ ( S h , R ) there holds a commutation rule I (∆ Q ) = ∆ M ( IQ ) . (4.3) Remark 4.1.
The algebra D G ′ ( M ) of the G ′ -invariant differential operatorson M is freely generated by ∆ M and P as a commutative algebra, see [GH].One can also give explicit formulas for P and ∆ M in our coordinates, namely, P = k L + α ∂∂y − α ∂∂y , ∆ M = k ∂ ∂y + k ∂ ∂y + 2 α α ∂ ∂y ∂y . (4.4)Now we deduce the principal linear differential equation for w . Proposition 4.3
We have P w = − M w. (4.5) Proof.
We begin with the following simple result.
Lemma 4.4 If f ∈ C ∞ ( R ) is a scalar function then P ( If ) = 0 . Indeed, since P is G -invariant, it is sufficient to verify the equation at asingle point 0 ∈ M which follows from (2.12) with h = 0.Lemma 4.4 implies by (4.2) that P w = P IQ = P I (cid:0)(cid:0) p + | v | (cid:1) δ ij − v ⊗ v (cid:1) = − P I ( v ⊗ v ) (4.6)for a compactly supported vector field v solving (1.1). Moreover, we have I ( v ⊗ v )( y , y , α , α ) = Z ∞−∞ (cid:0) ( v ) + 2 v v α + 2 v v α (cid:1) dx + O ( | α | )and thus by (3.14) we get P I ( v ⊗ v )(0) = P I ( v ⊗ v ) = 2 Z ∞−∞ ∂ (cid:0) v v (cid:1) ∂x − ∂ (cid:0) v v (cid:1) ∂x ! dx = 2∆ M w (0) , hence P I ( v ⊗ v ) = 2∆ M w everywhere and P w = − M w by (4.6).12 orollary 4.5 The equation IQ ( m ) = 0 , ∀ m ∈ M (4.7) implies Conjecture 1.2. Ideed, if IQ ( m ) = 0 then ∆ M w ( m ) = − P w ( m ) = − P IQ ( m ) = 0 andthus 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 ∆ M in terms of the Lie algebra g of G .We have g = s o (3) ⊕ r (3) = R ⊕ R as vector spaces, where r (3) is 3-dimensionaland abelian. Thus, we can write any g ∈ g as g = ( r ; s ) ∈ s o (3) ⊕ r (3), and thecommutators in g are given by[( r ; 0) , ( r ; 0)] = ( r × r ; 0) , [(0; s ) , (0; s )] = 0 , [( r ; 0) , (0; s )] = (0; r × s ) . Let ( R , R , R ) be the standard basis of s o (3), and ( S , S , S ) be that of r (3). Consider the following operators on M : e ∆ M = S + S + S , e P = S R + S R + S R , (4.8)where we denote simply by g the action on M of an element g ∈ U ( g ) of theuniversal enveloping algebra U ( g ); therefore, S i acts as the infinitesimal shift inthe x i -direction, and R i as the infinitesimal rotation about x i -axis. Proposition 4.6
We have e ∆ M = ∆ M and e P = P .Proof. First, the operators e ∆ M and e P are G -invariant. Indeed, it followsfrom the rotational invariance of the quadratic form x + x + x that e ∆ M isrotationally invariant; for translations, the same follows from the commutationrule S i S j = S j S i for i, j = 1 , , e P under the x -axis rotation we verify that e P and R commute which can be shown as follows:[ S R , R ] = − S R − S R , [ S R , R ] = S R + S R , [ S R , R ] = 0 . Similarly we get its invariance under the x − and x − axis rotations and thusits S O (3)-invariance, while its x -translations invariance follows from[ S R , S ] = − S S , [ S R , S ] = S S , [ S R , S ] = 0 . Since any line is S O (3)-conjugate to a vertical one, e P is G -invariant. Finally,we have e ∆ M ( m ) = ∆ M ( m ) , e P ( m ) = P ( m ) for m = m (0 , , , M and P are both G -invariant. Indeed, e.g., in e P ( m )the terms S R + S R give L (0), while S R vanishes since m is invariantunder both S and R . Second proof of Proposition 4.3.
Let t ∈ R , and let l t = m ( t, , ,
0) be avertical line in the plane Π ; note that Π = S t ∈ R l t .13 emma 4.7 For f ∈ C ∞ ( M ) we have Z R P f ( l t ) dt = Z R S R f ( l t ) dt. (4.9) Proof.
Define the operators A and B by A = S R P, B = S R P ; then P = A + B + S R P = A + B + S R ( S R + S R + S R ) . Since Af ( l t ) is a derivative of a function of t , while B vanishes identically onΠ , we get that Z R ( Af ( l t ) + Bf ( l t )) dt = 0 . We have also S R S R ( f ( l t )) = S S R R ( f ( l t )) − S S R ( f ( l t )) , thus the integral of the left-hand side is zero and the same is true for S R S R ( f ( l t )) = S S R R ( f ( l t )) + S S R ( f ( l t ));since S R S R = S R we get the conclusion. Lemma 4.8
We have Z R R w ( l t ) dt = − Z R w ( l t ) dt. (4.10) Proof.
Let us fix a positive constant c < π , and let l θt = (cid:16) t cos( θ ) , , tan( θ ) , (cid:17) for any θ with | θ | < c ; therefore, l θt is just the line l t rotated (in the clockwisedirection) through the angle θ about the origin in the plane Π , and for any t we have R w ( l θt ) | θ =0 = ∂ ∂θ w ( l θt ) | θ =0 . Let e θ = (cos θ, , − sin θ ) , e θ = (sin θ, , cos θ ) then we have w ( l θt ) = 1cos θ Z ∞ Z l t h v, e θ ih v, e θ i| ( t cos θ + x tan θ,x ,x ) dx dx by (3.8), and if we put t = x cos θ − x sin θ we get that Z R w ( l θt ) dt = 1cos θ Z ∞ Z R h v, e θ ih v, e θ i dx dx dt = Z x > h v, e θ ih v, e θ i dx dx dx . Therefore, since h v, e θ ih v, e θ i = v v cos 2 θ + (( v ) − ( v ) ) sin 2 θ we have ∂ ∂θ Z R w ( l θt ) dt = Z x > ∂ ∂θ ( h v, e θ ih v, e θ i ) dx dx dx = − Z x > ( h v, e θ ih v, e θ i ) dx dx dx θ = 0 we get a proof of Lemma 4.8.We can finish now our second proof of (4.5). Indeed, (4.9)–(4.10) imply that Z R P w ( l t ) dt = Z R S R w ( l t ) dt = − Z R S w ( l t ) dt = − Z R ∆ M w ( l t ) dt. (4.11)If we define a function F ( x , x ) on R by F ( x , x ) = P w ( x , x , ,
0) + 4∆ M w ( x , x , , , then the integral of F over the x -axis vanishes by (4.11). Changing the coordi-nate system x , x in R , we get the same for the integral of F over any line inthe plane { x , x } . 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 P ψ + 4 P ∆ M ψ = 0 (4.12)if ψ = IQ for Q ∈ C ∞ ( S , R ). Applying then (4.12) to ψ = ∆ M w (whichcan be written as ∆ M w = IQ ′ for a certain Q ′ not given here) we obtain P ∆ M w + 4 P ∆ M w = ∆ M P ( P w + 4∆ M w ) = 0 and thus P ( P w + 4∆ M w ) = 0which is much weaker than (4.5) since the kernel of P is enormous.However, it is possible to construct a function u ∈ C ∞ ( M ) verifying P ∆ M u = − M w, ∆ M u = IQ for some Q ∈ C ∞ ( S , R ) and applying (4.12) to ψ = ∆ M u = IQ we get0 = P ∆ M u + 4∆ M P u = − M (cid:0) P w + 4∆ M w (cid:1) , and thus we reprove (4.5). We can define u similarly to (3.5) as follows u = Z H ( m ) dist( P, m ) (cid:0) p + h ν H ( m ) , v i (cid:1) dσ H ( m ) , where dist( P, m ) is the distance from a point P ∈ H ( m ) to m . Let us define a Radon tensor plane transform J as follows: JQ ( L ) = Z L tr( Q | L ) dσ L (5.1)for an affine plane L ⊂ R and Q ∈ C ∞ ( S ; R ) satisfying | Q ( x ) | ≤ C (1 + | x | ) − − ε , (5.2)15or some ε > , where Q | L is the restrictio onto L ; definition is correct and weget a bounded linear operator J : S ( S ; R ) −→S ( R P )for the manifold R P of affine planes L ⊂ R . Proposition 5.1
We have JQ ( L ) = 0 .Proof. We have for any affine plane L ⊂ R that JQ ( L ) = 2 Z L (cid:0) p + h v, ν L i (cid:1) dσ L = 0 . Indeed, setting without loss of generality L = Π , ν L = e we get that ∂JQ ∂x (Π ) = ∂∂x (cid:16) Z L (cid:0) p + h v, ν L i (cid:1) dσ L (cid:17) == Z Π ∂ (cid:0) p + ( v ) (cid:1) ∂x dx dx = − Z Π (cid:18) ∂ ( v v ) ∂x + ∂ ( v v ) ∂x (cid:19) dx dx = 0 . Therefore, JQ (Π ) does not depend on x and hence equals 0.Let us explain in what Proposition 5.1 partially confirms (4.7) and thusConjecture 1.2. One can verify that the condition JQ ( L ) = 0 , ∀ L ∈ R P isequivalent to the following equation for the components { q ij } of Q : X i,j =1 ∂ q ij ∂x i ∂x j = ∆tr Q , (5.3)while IQ ( m ) = 0 , ∀ m ∈ M is equivalent to the following system2 ∂q ij ∂x i ∂x j = ∂ q ii ∂x j + ∂ q jj ∂x i , ≤ i < j ≤ Now we can deduce Conjecture 1.1 from Conjecture 1.2.
Theorem 6.1
Let ( v, p ) ∈ C ∞ ( R ) be a solution of (1.1) – (1.3) and let thecorresponding function w vanish everywhere on M then ( v ; p ) = 0 everywhere.Proof. For m ∈ M denote by t a vector parallel to m and by n a vectorperpendicular to m, then the equality w = 0 implies that Z m h v, t ih v, n i ds = 0 . (6.1)16et L ⊂ R be an affine plane; let v n = h ν L , v i and v t = v − v n ν L be itscomponents 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 curloperator curl L on L which gives v n curl L v t − h v ⊥ t , ∇ L v n i = − v n ω n − h v ⊥ t , ∇ L v n i = 0 (6.2)for the normal component ω n of ω = curl v and the gradient operator ∇ L on L .We will apply (6.2) to various planes L ⊂ R .First take L = Π , then v t = ( v , v , , v n = v , V = v v t and we get v curl L v t − h v ⊥ t , ∇ v i = v (cid:18) ∂v ∂x − ∂v ∂x (cid:19) − v ∂v ∂x + v ∂v ∂x = 0 . (6.3)Let now Ω = def { u ∈ R | v ( u ) = 0 , ω ( u ) = 0 } and let D = def R \ ¯Ω. We can suppose that Ω is not empty, since otherwisein a neighborhood of a point x where the maximum of | v | is attained we have∆ v = − curl curl v + ∇ div v = 0 and thus v is harmonic which contradictsthe maximum principle for harmonic fields. It follows then that v = 0 in thisneighborhood and thus everywhere.In orthonormal coordinates with v ( u ) = v ( u ) = 0 we get for u ∈ Ω that v (cid:18) ∂v ∂x − ∂v ∂x (cid:19) ( u ) = 0 and therefore h v, ω i ( u ) = 0 . (6.4)Therefore h v, ω i = 0 holds everywhere on Ω, thus on R and differentiating thisrelation in the v -direction we obtain h v ∇ v, ω i + h v ∇ ω, v i = 0; (6.5)using the commutation law v ∇ ω = ω ∇ v (6.6)we get then from (6.5) that h v ∇ v, ω i + h ω ∇ v, v i = 0 . (6.7)In orthonormal coordinates x , x , x with x directed along v and x directedalong ω at u we can rewrite (6.7) as follows ∂v ∂x ( u ) + ∂v ∂x ( u ) = 0 , (6.8)since v ( u ) = 0 and ω ( u ) = 0. Below we always use that coordinate system.Moreover, since the vector ω is directed along x , we get ∂v ∂x ( u ) = ∂v ∂x ( u ) and therefore ∂v ∂x ( u ) = ∂v ∂x ( u ) = 0 . (6.9)17et then L = Π , and thus v t = ( v , , v ) , v n = v , V = v v t . Since v n ( u ) = 0, we get from (6.2) and (6.6) that ∂v ∂x ( u ) = 0 and hence ∂v ∂x ( u ) = 0 and ∂ω ∂x ( u ) = 0 . (6.10)Then, differentiating (6.4) with respect to x and x at u , we get ∂ω ∂x ( u ) = 0 and ∂ω ∂x ( u ) = 0 . (6.11)Now we take L = { x + x = 0 } , therefore v n = v + v √ , ν L = (cid:16) , √ , √ (cid:17) , v t = (cid:16) v , v − v , v − v (cid:17) and V = v + v √ (cid:16) v , v − v √ (cid:17) B in the orthonormal ba-sis B = n e ′ = (1 , , , e ′ = (cid:16) , √ , − √ (cid:17)o . Since v n ( u ) = 0 and the vector v ⊥ t ( u ) is directed along e ′ , we get from (6.2) that v ( u ) (cid:18) ∂v ∂x ( u ) + ∂v ∂x ( u ) − ∂v ∂x ( u ) − ∂v ∂x ( u ) (cid:19) = 0and thus ∂v ∂x ( u ) + ∂v ∂x ( u ) − ∂v ∂x (0) − ∂v ∂x ( u ) = 0 . Therefore by (6.10) we get also ∂v ∂x ( u ) = ∂v ∂x ( u ) . (6.12)For L = { x + x = 0 } we then have v n = v + v √ , ν L = (cid:16) √ , √ , (cid:17) ,v t = (cid:16) v − v , v − v , v (cid:17) and thus V = v + v √ (cid:16) v − v √ , v (cid:17) B ′ in the plane ba-sis B ′ = n(cid:16) √ , − √ , (cid:17) , (0 , , o . Since the vector v ⊥ t ( u ) is directed along(0 , ,
1) we get from (6.2) and (6.9) that v ( u ) (cid:18) ω ( u ) + ∂v ∂x ( u ) + ∂v ∂x ( u ) (cid:19) = 0;therefore, we get by (6.10) that ω ( u ) = − ∂v ∂x ( u ) . (6.13)If a trajectory γ = γ ( t ) of the flow v is parametrized by t, i.e. dγdt = v, wehave in virtue of (6.6) a differential inequality | q ′ ( t ) | ≤ C | q ( t ) | , q ( t ) = def ω ( γ ( t )) and a positive constant C . Therefore, if q (0) = 0 then q ( t ) = 0 for any t ∈ R , thus any trajectory of v does not cross ∂ Ω = ∂D and hence stays either in ¯Ω or in ¯ D .Using (6.9), we see that | v | is constant on any trajectory Γ of the vector field ω and, conversely, (6.10) and (6.11) imply that ω has a constant direction onany trajectory γ of the flow v and therefore γ ⊂ Π γ is a plane curve for an affineplane Π γ . Set ξ = ω/ | ω | , then we can define the vector ξ ( γ ) for any γ ⊂ Ω and ξ ( γ ) ⊥ Π γ . For z ∈ Ω we get ∂ξ∂x ( z ) = 0 (6.14)since v ( z ) is parallel to x , ω ( z ) is parallel to x and ξ ( z ) = (0 , , ∂ξ ∂x ( z ) = 0 . (6.15)Therefore ξ satisfies the Frobenius integrability condition h ξ, curl ξ i = 0 andhence in a neighborhood of z there exists a smooth function U ( x ) with ∇ U = 0parallel to ξ . Moreover, U is a first integral of the flow v since ∂U/∂v = 0. Letthen S be a level surface of U containing z , then S is foliated by the trajectoriesof v and the vector field ξ defines the Gauss map ξ : S → S . Since ξ is constanton the trajectories of v the image β = ξ ( S ) ⊂ S is a curve or a point. Moreover, β is orthogonal to the axis x at ξ ( γ ) by (6.14)–(6.15) and (6.13) implies that γ is not a straight line. Thus we can choose z ′ ∈ γ, z ′ = z and get that γ isorthogonal at ξ ( γ ) to some line not parallel to x . Therefore rank ξ ( γ ) = 0 , hence rank ξ = 0 on S , ξ is constant on S and thus S is a plane. We see that aneighborhood of z in R is foliated by planes invariant under the flow v .Denote by γ ( s, t ) ⊂ Ω the trajectory of v passing through z + (0 , s, L zs be the plane containing γ ( s, t ). Then L zs ⊥ ω ( z + (0 , s, L zs are invariant under the flow v and foliate a neighborhood of z in R . Let λ ( s, y ) for y ∈ Π be an affine function on Π with the graph L zs and let l z ( y ) = ∂λ∂s (0 , y ) then l z is an affine linear function. Denote by G a connectedcomponent of Π ∩ Ω , z ∈ G then G is invariant under the flow v . We fix some z ′ ∈ G and set l = l z ′ , then l ( z ′ ) = 1.Let z , z ∈ G , then some neighborhoods of z and z in R are foliated bythe same set of planes invariant under the flow v and thus the sets of planes L z s and L z s are the same after a reparametrization. Therefore l does not vanish in G and we have l z = l z /l z ( z ). Set now h ( t ) = ∂γ ( s, t ) ∂s | s =0 then h (0) = (0 , ,
0) and thus by (6.6) the vector field h is proportional to ω on γ = γ (0 , t ). Since ω is orthogonal to Π on γ we get that h ( y ) = (0 , l ( y ) , ,∂v ∂x ( y ) = ∂ ln l∂v ( y ) (6.16)19or any y ∈ γ and l does not vanish on γ .It follows by (6.16) and (6.12) that ∂v ∂x ( y ) = ∂ ln l∂v ( y ) (6.17)and since div v = 0 there holds (recall that x is directed along v ( y )) ∂ | v | ∂x ( y ) = − ∂ ln l∂v ( y ) = ∂ ln | v | ∂v ( y ); (6.18)hence we get that | v ( y ) | = C γ l ( y ) (6.19)along the trajectory γ for a positive constant C γ depending on γ . Note alsothat equations (6.16)–(6.19) hold for any trajectory of v in G and hence bycontinuity in ¯ G outside the zero locus of l ; in particular, we see that v ( y ) = 0 for any y ∈ ¯ G with l ( y ) = 0 . (6.20)Let z ∈ ¯ G be a point where the function | v | attains its maximum on ¯ G ,then z ∈ ∂G . Indeed, if it is not the case, we have ∂v ∂x ( z ) = 0 , and hence ω ( z ) = 0 by (6.13) which implies z ∈ ∂G .Let γ be the trajectory of v starting from z ∈ ∂G with v ( z ) = 0, then ω = 0 on the whole trajectory γ and γ ⊂ ∂ Ω. Therefore ∇ b = v × ω = 0 on γ , where b = p + | v | is the Bernoulli function (see, e.g., [AK]) and we get ∇ p = − ∇| v | on γ . (6.21)Let y ∈ γ and let e be a unit vector in Π orthogonal to v ( y ), then h v e ( y ) , v ( y ) i =0 for v e = ( ∇ e v , ∇ e v ) by (6.13) since ω ( y ) = 0. Therefore h∇| v | ( y ) , e i = 0and (6.21) implies that γ is a straight line interval I which is finite since v hasa compact support, v vanishes at its end points and thus l | I = 0 by (6.20).Let now z = 0 and continue to assume that x is directed along v (0) = 0and x is directed along ω ( x ) = 0 for some x ∈ G (the direction of ω ( x ) doesnot depend on x ), then l is a linear function on Π vanishing on the x -axis: l = Cx for C = 0. Denote now D + ε = B ε ∩ { < x } and D − ε = B ε ∩ { > x } , then we have D + ε ⊂ G . Indeed, first note that ( D + ε ∩ ∂G ) ∪ ( D − ε ∩ ∂G ) = ∅ sinceotherwise the trajectory γ starting from z ∈ ( D + ε ∩ ∂G ) ∪ ( D − ε ∩ ∂G ) leads to acontradiction since l | γ = 0. Moreover, for the trajectory α ( t ) of v ⊥ = ( − v , v )starting at 0 we have α ( t ) ∈ D + ε for a small t > α ( t ) ∈ D − ε for a small t <
0. Since ω = 0 on G we get that | v | strictly decreases along α by (6.13)( v (0) being parallel to x ) while α ( t ) stays in G and since | v | attains at 0 itsmaximum in ¯ G we get that D − ε T G = ∅ ; therefore, D + ε ⊂ G .20urthermore, any trajectory γ s of v starting from the point (0 , , s ) ∈ D + ε with 0 < s < ε and a sufficiently small ε > C γ s strictly increases as a function of s ∈ (0 , ε ) and thus γ s with s ∈ (0 , ε ) intersects the interval (0 , (0 , ε )) only once. By the Poincar´e-Bendixsontheorem we get that γ s either( i ) tends to a limit, or( ii ) tends to a limit cycle ρ ⊂ G , or( iii ) is closed.Since ( i ) contradicts (6.19) and ( ii ) implies that any trajectory γ a with s < a < ε tends to ρ which contradicts (6.19) as well, we get that ( iii ) holds. Moreover,any trajectory starting inside D + ε enters the domain { x > δ } ∩ G for some fixed δ > A = S G . If λ ∈ (0 , ε ) theninf γ λ | v | > | v ( z ) | for a sufficiently small s ∈ (0 , λ ) and some z ∈ γ s , while the trajectory α z of v ⊥ starting from z intersects γ λ and | v | strictly decreases along α z which gives acontradiction and thus finishes the proof. Let us briefly discuss Conjectures 1.1. and 1.2 in terms of the vector analysisfor compactly supported tensor fields in R . In this section we suppose that v ∈ C ∞ ( S , R ). We can rewrite (1.2) as followscurl(div( v ⊗ v )) = 0 . (7.1) Proposition 7.1 If (7.1) holds and the corresponding function w ∈ C ∞ ( M ) iseverywhere zero on M then Ψ = Ψ( v ) = def σ (curl( v ⊗ v )) = 0 (7.2) for the symmetrization Ψ of the tensor field curl( v ⊗ v ) , i.e. ij = ε ilm ∂ ( v j v l ) ∂x m + ε jlm ∂ ( v i v l ) ∂x m , where ε ijk is the standard permutation (pseudo-)tensor, giving the sign of thepermutation ( ijk ) of (123) and the summation convention applies.Proof. For any fixed value of x , we define the vector field Z = v ( v , v ) = ( v v , v v )21n the vertical plane Π ( x ) = { x , x , x } with coordinates { x , x } dependingon x as on a parameter.We have then IZ ( m ′ ) = w ν m ( m ) = h∇ w, ν m i for a line m ⊂ Π ( x ) , anormal ν m to m and a line m ′ ⊥ m ⊂ Π ( x ) , thus IZ = 0 and hence thesolenoidal component s Z equals zero, where Z = s Z + p Z is the Helmholtzdecomposition of the vector field Z . Therefore we haveΨ = curl Z = curl s Z = 0 , thus Ψ ii = 0 for i = 1 , , { x i , x j } through the angle π we get Ψ ij = 0 for all 1 ≤ i ≤ j ≤ . Moreover, the proof of Theorem 6.1 shows that the conditions Ψ( v ) = 0 anddiv v = 0 imply v = 0.Therefore Conjectures 1.1 and 1.2 follow from Conjecture 7.1. If curl(div( v ⊗ v )) = 0 then σ (curl( v ⊗ v )) = 0 . Another equivalent statement can be formulated as follows
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