A new flocking model through body attitude coordination
AA new flocking model through body attitudecoordination
Pierre Degond (1) , Amic Frouvelle (2) , and Sara Merino-Aceituno (3) (1)(3)
Department of Mathematics, Imperial College London, London SW7 2AZ, UK (1) [email protected] (3) [email protected] (2)
CEREMADE, UMR CNRS 7534, Universit´e de Paris-Dauphine, PSL Research University,Place du Mar´echal De Lattre De Tassigny, 75775 PARIS CEDEX 16 - FRANCE,[email protected]
October 10, 2018
Abstract
We present a new model for multi-agent dynamics where each agent is describedby its position and body attitude: agents travel at a constant speed in a given direc-tion and their body can rotate around it adopting different configurations. In thismanner, the body attitude is described by three orthonormal axes giving an elementin SO (3) (rotation matrix). Agents try to coordinate their body attitudes with the onesof their neighbours. In the present paper, we give the Individual Based Model (par-ticle model) for this dynamics and derive its corresponding kinetic and macroscopicequations.The work presented here is inspired by the Vicsek model and its study in [24]. Thisis a new model where collective motion is reached through body attitude coordina-tion. Key words:
Body attitude coordination; collective motion; Vicsek model; Gener-alized Collision Invariant; Rotation group.
AMS Subject Classification:
Contents a r X i v : . [ m a t h - ph ] M a y Discussion of the main result: the Self-Organized Hydrodynamics for body at-titude coordination (SOHB) 53 Modelling: the Individual Based Model and its mean-field limit 7 SO (3) . . . . . . . . . . . . . . . . . . . 144.3 Equilibrium solutions and Fokker-Planck formulation . . . . . . . . . . . . . 164.4 Generalized Collision Invariants . . . . . . . . . . . . . . . . . . . . . . . . . 194.4.1 Definition and existence of GCI . . . . . . . . . . . . . . . . . . . . . . 194.4.2 The non-constant GCIs . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.5 The macroscopic limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 SO (3) In this paper we model collective motion where individuals or agents are described bytheir position and body attitude. The body attitude is given by three orthonormal axis;one of the axes describes the direction in which the agent moves at a constant speed; theother two axis indicate the relative position of the body with respect to this direction.Agents try to coordinate their body attitude with those of near neighbours (see Figure 1).Here we present an Individual Based Model (particle model) for body attitude coordina-tion and derive the corresponding macroscopic equations from the associated mean-fieldequation, which we refer to as the Self-Organized Hydrodynamics for body attitude coor-dination (SOHB), by reference to the Self-Organized Hydrodynamics (SOH) derived fromthe Vicsek dynamics (see [24] and discussion below).There exists already a variety of models for collective behaviour depending on thetype of interaction between agents. However, to the best of our knowledge, this is thefirst model that takes into account interactions based on body attitude coordination. Thishas applications in the study of collective motion of animals such as birds and fish andit is a stepping stone to model more complex agents formed by articulated bodies (corp-ora) [13, 14]. In this section we present related results in the literature and the structureof the document.There exists a vast literature on collective behaviour. In particular, here we deal withthe case of self-propelled particles which is ubiquitous in nature. It includes, among2 a) Birds with coordinated body atti-tude. Three orthonormal axis describethe body attitude: the green arrow in-dicates the direction of movement; theblue and red ones indicate the positionof the body with respect to this direc-tion. (b) Birds with no coordinated body at-titude.(c) Dolphins moving in the same direction but with dif-ferent body attitude. In this example one can see that thebody attitude coordination model gives more informationthan the Vicsek model (which only describes the directionof movement).
Figure 1: Examples of body attitude coordination/dis-coordination and the use of therotation matrix representation. These images are in public domain (released under Creative Commons CC0 by pixabay.com). N agents labelledby k = 1 , . . . , N the positions and body attitudes ( X k , A k ) ∈ R × SO (3) over time are givenby the Stochastic Differential Equations (16)-(17). In Section 3.2 we give the corresponding(formal) mean-field limit (Prop. 3.4) for the evolution of the empirical measure when thenumber of agents N → ∞ . 4he last part concerns the derivation of the macroscopic equations (Theorem 4.16) forthe total density of the particles ρ = ρ ( t, x ) and the matrix of the mean body attitude Λ =Λ( t, x ) . To obtain these equations we first study the rescaled mean-field equation (Eq.(22) in Section 4.1), which is, at leading order, a Fokker-Planck equation. We determineits equilibria, which are given by a von Mises distribution on SO (3) (Eq. (25), Section 4.3).Finally in Section 4.4 we obtain the Generalized Collision Invariants (Prop. 4.14), whichare the main tool to be able to derive the macroscopic equations in Section 4.5. The main result of this paper is Theorem 4.16 which gives the following macroscopicequations for the density of agents ρ = ρ ( t, x ) and the matrix of the mean body atti-tude Λ = Λ( t, x ) ∈ SO (3) (i.e., the Self-Organized Hydrodynamics for body attitudecoordination (SOHB)): ∂ t ρ + c ∇ x · ( ρ Λ e ) = 0 , (1) ρ (cid:16) ∂ t Λ + c (cid:0) (Λ e ) · ∇ x (cid:1) Λ (cid:17) + (cid:2) (Λ e ) × (cid:0) c ∇ x ρ + c ρ r x (Λ) (cid:1) + c ρ δ x (Λ)Λ e (cid:3) × Λ = 0 . (2)In the equations above c , c , c and c are explicit constants which depend on the parame-ters of the model (namely the rate of coordination and the level of noise). The expressionsof the constants c , c and c depend on the Generalized Collision Invariant mentionedin the introduction (and computed thanks to Prop. 4.14). The constant c is obtained as a“consistency” relation (Lemma 4.8). In (2), the operation [ · ] × transforms a vector v in anantisymmetric matrix such that [ v ] × u = v × u for any vector u (see (8) for the exact defi-nition). The scalar δ x (Λ) and the vector r x (Λ) are first order differential operators intrinsicto the dynamics : if Λ( x ) = exp ([ b ( x )] × ) Λ( x ) with b smooth around x and b ( x ) = 0 ,then δ x (Λ)( x ) = ∇ x · b ( x ) | x = x and r x (Λ)( x ) = ∇ x × b ( x ) | x = x , where ∇ x × is the curl operator. These operators are well-defined as long as Λ is smooth:as we will see in the next section, we can always express a rotation matrix as exp ([ b ] × ) for some vector b ∈ R , and this function b (cid:55)→ exp ([ b ] × ) is a local diffeomorphism be-tween a neighborhood of ∈ R and the identity of SO (3) . This gives a unique smoothrepresentation of b in the neighborhood of when x is in the neighborhood of x sincethen Λ( x )Λ( x ) − is in the neighborhood of Id .We express Eq. (2) in terms of the basis vectors { Ω = Λ e , u = Λ e , v = Λ e } and wewrite Λ = Λ(Ω , u , v ) . System (1)-(2) can be expressed as an evolution system for ρ and the5asis { Ω , u , v } as follows: ∂ t ρ + c ∇ x · ( ρ Ω) = 0 , (3) ρD t Ω + P Ω ⊥ ( c ∇ x ρ + c ρ r ) = 0 , (4) ρD t u − u · ( c ∇ x ρ + c ρ r ) Ω + c ρ δ v = 0 , (5) ρD t v − v · ( c ∇ x ρ + c ρ r ) Ω − c ρ δ u = 0 , (6)where D t := ∂ t + c (Ω · ∇ x ) , δ = δ x (Λ(Ω , u , v )) and r = r x (Λ(Ω , u , v )) . The operator P Ω ⊥ denotes the projection on the orthogonal of Ω . We easily see here that these equationspreserve the constraints | Ω | = | u | = | v | = 1 and Ω · u = Ω · v = u · v = 0 . The expressionsof δ and r are: δ = [(Ω · ∇ x ) u ] · v + [( u · ∇ x ) v ] · Ω + [( v · ∇ x )Ω] · u , r = ( ∇ x · Ω)Ω + ( ∇ x · u ) u + ( ∇ x · v ) v . Eq. (1) (or equivalently Eq. (3)) is the continuity equation for ρ and ensures massconservation. The convection velocity is given by c Λ e = c Ω and Ω gives the directionof motion. Eq. (2) (or equivalently Eqs. (4)-(6)) gives the evolution of Λ . We remark thatevery term in Eq. (2) belongs to the tangent space at Λ in SO (3) ; this is true for the firstterm since ( ∂ t + c (Λ e ) · ∇ x ) is a differential operator and it also holds for the second termbecause it is the product of an antisymmetric matrix with Λ (see Prop. A.2). Alternately,this means that (Ω( t ) , u ( t ) , v ( t )) is a direct orthonormal basis as soon as (Ω(0) , u (0) , v (0)) .The term corresponding to c in (2) gives the influence of ∇ x ρ (pressure gradient) onthe body attitude Λ . It has the effect of rotating the body around the vector directedby (Λ e ) × ∇ x ρ at an angular speed given by c ρ (cid:107) (Λ e ) × ∇ x ρ (cid:107) , so as to align Ω with −∇ x ρ .Indeed the solution of the differential equation d Λ dt + γ [ n ] × Λ = 0 , when n is a constantunit vector and γ a constant scalar, is given by Λ( t ) = exp( − γ t [ n ] × )Λ , and exp( − γ t [ n ] × ) is the rotation of axis n and angle − γ t (see (10), it is called Rodrigues’ formula). Sincewe will see that c is positive the influence of this term consists of relaxing the directionof displacement Λ e towards ∇ x ρ . Alternately, we can see from (4) that Ω turns in theopposite direction to ∇ x ρ , showing that the ∇ x ρ term has the same effect as a pressuregradient in classical hydrodynamics. We note that the pressure gradient has also theeffect of rotating the whole body frame (see influence of ∇ x ρ on u and v ) just to keepthe frame orthonormal. Similarly to what happens with the ∇ x ρ term in Eq. (2), theterm containing c ρ r in Eq. (4) has the effect of relaxing the direction of displacement Ω towards − r (we will indeed see that c is positive). Finally, the last terms of Eqs. (5)-(6)have the effect of rotating the vectors u and v around Ω along the flow driven by D t atangular speed c δ .If we forget the term proportional to r in (4), System (3)-(4) is decoupled from (5)-(6),and is an autonomous system for ρ and Ω , which coincides with the Self-Organized Hy-drodynamic (SOH) model. The SOH model provides the fluid description of a particlesystem obeying the Vicsek dynamics [24]. As already discussed in [24], the SOH model6ears analogies with the compressible Euler equations, where (3) is obviously the massconservation equation and (4) is akin to the momentum conservation equation, wheremomentum transport ρD t Ω is balanced by a pressure force − P Ω ⊥ ∇ x ρ . There are howevermajor differences. The first one is the presence of the projection operation P Ω ⊥ which isthere to preserve the constraint | Ω | = 1 . Indeed, while the velocity in the Euler equationsis an arbitrary vector, the quantity Ω in the SOH model is a velocity orientation and isnormalized to . The second one is that the convection speed c in the convection opera-tor D t is a priori different from the mass convection speed c appearing in the continuityequation. This difference is a signature of the lack of Galilean invariance of the system,which is a common feature of all dry active matter models.The major novelty of the present model, which can be referred to as the Self-OrganizedHydrodynamic model with Body coordination (or SOHB) is that the transport of the di-rection of motion Ω involves the influence of another quantity specific to the body orien-tation dynamics, namely the vector r . The overall dynamics tends to align the velocityorientation Ω , not opposite to the density gradient ∇ x ρ but opposite to a composite vec-tor ( c ∇ x ρ + c ρ r ) . The vector r is the rotational of a vector b locally attached to the frame(namely the unit vector of the local rotation axis multiplied by the local angle of rotationaround this axis). This vector gives rise to an effective pressure force which adds up tothe usual pressure gradient. It would be interesting to design specific solutions where thiseffective pressure force has a demonstrable effect on the velocity orientation dynamics.In addition to this effective force, spatial inhomogeneities of the body attitude alsohave the effect of inducing a proper rotation of the frame about the direction of motion.This proper rotation is also driven by spatial inhomogeneities of the vector b introducedabove, but are now proportional to its divergence. The body attitude is given by a rotation matrix. Therefore, we work on the Riemannianmanifold SO (3) (Special Orthogonal Group), which is formed by the subset of matrices A such that AA T = Id and det( A ) = 1 , where Id stands for the identity matrix.In this document M indicates the set of square matrices of dimension ; A is the setof antisymmetric matrices of dimension ; S is the set of symmetric matrices of dimen-sion . Typically we will denote by A, Λ matrices in SO (3) and by P matrices in A . Boldsymbols n , v , e indicate vectors.We will often use the so-called Euler axis-angle parameters to represent an elementin SO (3) : to A ∈ SO (3) there is associated an angle θ ∈ [0 , π ] and a vector n ∈ S sothat A = A ( θ, n ) corresponds to the anticlockwise rotation of angle θ around the vector n .It is easy to see that tr( A ) = 1 + 2 cos θ (7)7for instance expressing A in an orthonormal basis with n ), so the angle θ is uniquelydefined as arccos( (tr( A ) − . Notice that n is uniquely defined whenever θ ∈ (0 , π ) (if θ = 0 then n can be any vector in S and if θ = π then the direction of n is uniquelydefined but not its orientation). For a given vector u , we introduce the antisymmetricmatrix [ u ] × , where [ · ] × is the linear operator from R to A given by [ u ] × := − u u u − u − u u , (8)so that for any vectors u , v ∈ R , we have [ u ] × v = u × v . In this framework, we have thefollowing representation for A ∈ SO (3) (called Rodrigues’ formula): A = A ( θ, n ) = Id + sin θ [ n ] × + (1 − cos θ )[ n ] × (9) = exp( θ [ n ] × ) . (10)We also have n × ( n × v ) = ( n · v ) n − ( n · n ) v , therefore when n is a unit vector, we have : [ n ] × = n ⊗ n − Id , (11)where the tensor product a ⊗ b is the matrix defined by ( a ⊗ b ) u = ( u · b ) a for any u ∈ R .Finally, SO (3) has a natural Riemannian metric (see [38]) induced by the following innerproduct in the set of square matrices of dimension : A · B = 12 tr( A T B ) = 12 (cid:88) i,j A ij B ij . (12)This normalization gives us that for any vectors u , v ∈ R , we have that [ u ] × · [ v ] × = ( u · v ) . (13)Moreover, the geodesic distance on SO (3) between Id and a rotation of angle θ ∈ [0 , π ] isexactly given by θ (the geodesic between Id and A is exactly t ∈ [0 , θ ] (cid:55)→ exp( t [ n ] × ) ). SeeAppendix A for some properties of SO (3) used throughout this work.Seeing SO (3) as a Riemannian manifold, we will use the following notations: T A is thetangent space in SO (3) at A ∈ SO (3) ; P T A denotes the orthogonal projection onto T A ; theoperators ∇ A , ∇ A · are the gradient and divergence in SO (3) , respectively. These operatorsare computed in Section 4.2 in the Euler axis-angle coordinates. Consider N agents labelled by k = 1 , . . . , N with positions X k ( t ) ∈ R and associatedmatrices (body attitudes) A k ( t ) ∈ SO (3) . For each k , the three unit vectors representingthe frame correspond to the vectors of the matrix A k ( t ) when written as a matrix in the8anonical basis ( e , e , e ) of R . In particular, the direction of displacement of the agentis given by its first vector A k ( t ) e . Evolution of the positions.
Agents move in the direction of the first axis with constantspeed v d X k ( t ) dt = v A k ( t ) e . Evolution of the body attitude matrix.
Agents try to coordinate their body attitude withthose of their neighbours. So we are facing two different problems from a modellingviewpoint, namely to define the target body attitude, and to define the way agents relaxtheir own attitude towards this “average” attitude.As for the Vicsek model [24], we consider a kernel of influence K = K ( x ) ≥ anddefine the matrix M k ( t ) := 1 N N (cid:88) i =1 K ( | X i ( t ) − X k ( t ) | ) A i ( t ) . (14)This matrix corresponds to the averaged body attitude of the agents inside the zone ofinfluence corresponding to agent k . Now M k ( t ) / ∈ SO (3) , so we need to orthogonalize andremove the dilations, in order to construct a target attitude in SO (3) . We will see that thepolar decomposition of M k ( t ) is a good choice in the sense that it minimizes a weightedsum of the squared distances to the attitudes of the neighbours. We also refer to [40] forsome complements on averaging in SO (3) .We give next the definition of polar decomposition: Lemma 3.1 (Polar decomposition of a square matrix, [33, Section 4.2.10]) . Given a matrix M ∈ M , if det( M ) (cid:54) = 0 then there exists a unique orthogonal matrix A (givenby A = M ( √ M T M ) − ) and a unique symmetric positive definite matrix S such that M = AS . Proposition 3.2.
Suppose that the matrix M k ( t ) has positive determinant. Then the followingassertions are equivalent:(i) The matrix A minimizes the quantity N (cid:80) Ni =1 K ( | X i ( t ) − X k ( t ) | ) (cid:107) A i ( t ) − A (cid:107) among theelements of SO (3) .(ii) The matrix A is the element of SO (3) which maximizes the quantity A · M k ( t ) .(iii) The matrix A is the polar decomposition of M k ( t ) .Proof. We get the equivalence between the first two assertions by expanding: (cid:107) A i ( t ) − A (cid:107) = 12 [tr( A i ( t ) T A i ( t )) + tr( A T A )] − A · A i ( t ) = 3 − A · A i ( t ) , since A and A i ( t ) are both orthogonal matrices. So minimizing the weighted sum of thesquares distances amounts to maximizing inner product of A and the weighted sum M k of the matrices A i given by (14). 9herefore if det M k > , and A is the polar decomposition of M k , we immediately getthat det A > , hence A ∈ SO (3) . We know that S can be diagonalized in an orthogonalbasis : S = P T DP with P T P = Id and D is a diagonal matrix with positive diagonalelements λ , λ , λ . Now if B ∈ SO (3) maximizes tr( B T M k ) among all matrices in SO (3) ,then it maximizes tr( B T AP T DP ) = tr( P B T AP T D ) . So the matrix ¯ B = P B T AP T maxi-mizes tr( ¯ BD ) = λ ¯ b + λ ¯ b + λ ¯ b among the elements of SO (3) (the map B (cid:55)→ P B T AP T is a one-to-one correspondence between SO (3) and itself). But since ¯ B is an orthogonalmatrix, all its column vectors are unit vectors, and so b ii (cid:54) , with equality for i = 1 , and if and only if ¯ B = Id , that is to say P B T AP T = Id , which is exactly B = A .We denote by PD ( M k ) ∈ O (3) the corresponding orthogonal matrix coming from thePolar Decomposition of M k .We now have two choices for the evolution of A k . We can use the second point ofProposition 3.2 and follow the gradient of the function to maximize : dA k ( t ) dt = ν ∇ A ( M k · A ) | A = A k = νP T Ak M k , (15)(see (54) for the last computation, P T Ak is the projection on the tangent space, this way thesolution of the equation stays in SO (3) ).Or we can directly relax to the polar decomposition PD ( M k ) , in the same manner: dA k ( t ) dt = νP T Ak ( PD ( M k )) . We can actually see that the trajectory of this last equation, when PD ( M k ) belongsto SO (3) and does not depend on t , is exactly following a geodesic (see Prop. A.4). There-fore in this paper we will focus on this type of coordination. The positive coefficient ν gives the intensity of coordination, in the following we will assume that it is a functionof the distance between A k and PD ( M k ) (the angle of the rotation A Tk PD ( M k ) ), which isequivalent to say that ν depends on A k · PD ( M k ) . Remark 3.3.
Some comments:1. One could have used the Gram-Schmidt orthogonalization instead of the Polar Decomposi-tion, but it depends on the order in which the vector basis is taken (for instance if we startwith e , it would define the first vector as the average of all the directions of displacement, in-dependently of how the other vectors of the body attitudes of the individuals are distributed).The Polar Decomposition gives a more canonical way to do this.2. We expect that the orthogonal matrix coming from the Polar Decomposition of M k belongsin fact to SO (3) . Firstly, we notice that O (3) is formed by two disconnected compo-nents: SO (3) and the other component formed by the matrices with determinant -1. Weassume that the motion of the agents is smooth enough so that the average M k stays ‘close’to SO (3) and that, in particular, det( M k ) > . simple example is when we only average two different matrices A and A of SO (3) .We then have M = ( A + A ) . If we write A A T = exp( θ [ n ] × ) thanks to Rodrigues’formula (10) and we define A = A exp( θ [ n ] × ) , we get that A = A exp( θ [ n ] × ) andso M = A (exp( θ [ n ] × ) + exp( − θ [ n ] × )) = A (cos θ Id + (1 − cos θ ) n ⊗ n ) , thanks toRodrigues’ formula (9) and to (11) . Since the matrix S = cos θ Id + (1 − cos θ ) n ⊗ n isa positive-definite symmetric matrix as soon as θ ∈ [0 , π ) , we have that det( M ) > . Thepolar decomposition of M is then A , which is the midpoint of the geodesic joining A to A (which corresponds to the curve t ∈ [0 , θ ] (cid:55)→ A exp( t [ n ] × ) ).As soon as we average more than two matrices, there exist cases for which det( M ) < : forinstance if we take A = − − , A = − − , A = − − , we have M = ( A + A + A ) = − Id .Noise term. Agents make errors when trying to coordinate their body attitude with thatof their neighbours. This is represented in the equation of A k by a noise term: √ DdW kt where D > and W kt = (cid:16) W k,i,jt (cid:17) i,j =1 , , are independent Gaussian distributions (Brown-ian motion).From all these considerations, we obtain the IBM d X k ( t ) = v A k ( t ) e dt, (16) dA k ( t ) = P T Ak ◦ (cid:104) ν ( PD ( M k ) · A k ) PD ( M k ) dt + 2 √ DdW kt (cid:105) , (17)where the Stochastic Differential Equation is in Stratonovich sense (see [32]). The projec-tion P T Ak and the fact that we consider the SDE in Stratonovich sense ensures that the solu-tion A k ( t ) stays in SO (3) . The normalization constant √ D ensures that the diffusion co-efficient is exactly D : the law p of the underlying process given by dA k = 2 √ DP T Ak ◦ dW kt satisfies ∂ t p = D ∆ A p where ∆ A = ∇ A · ∇ A is the Laplace-Beltrami operator on SO (3) .Notice the factor √ D instead of the usual √ D which is encountered when consideringdiffusion process on manifolds isometrically embedded in the euclidean space R n , be-cause we are here considering SO (3) embedded in M (isomorphic to R ), but with themetric (12), which corresponds to the canonical metric of R divided by a factor . Werefer to the book [37] for more insight on such stochastic processes on manifolds. We assume that the kernel of influence K is Lipschitz, bounded, with the following prop-erties: K = K ( | x | ) ≥ , (cid:90) R K ( | x | ) dx = 1 , (cid:90) R | x | K ( | x | ) dx < ∞ . (18)11n [7] the mean-field limit is proven for the Vicsek model. Using the techniques there it isstraightforward to see that for M ( x, t ) := 1 N N (cid:88) i =1 K ( X i − x ) A i the law f N = f N ( x, A, t ) of the empirical measure associated to the Stratonovich Stochas-tic Differential Equation (SDE): d X k ( t ) = v A k ( t ) e dt, (19) dA k ( t ) = P T Ak ◦ (cid:104) ν ( M ( X k , t ) · A k ) M ( X k , t ) dt + 2 √ DdW kt (cid:105) , (20)converges weakly f N → f as N → ∞ . The limit satisfies the kinetic equation: ∂ t f + v A e · ∇ x f = D ∆ A f − ∇ A · ( F [ f ] f ) , with F [ f ] := ν ( M f · A ) P T A ( M f ) , M f = (cid:90) R × SO (3) K ( x − x (cid:48) ) f ( x (cid:48) , A (cid:48) , t ) A (cid:48) dA (cid:48) dx (cid:48) . The equations we are dealing with (16)-(17), since we consider the Polar Decomposi-tion of the averaged body attitude M k , are slightly different from (19)-(20), which wouldcorrespond to the modelling point of view of Eq. (15). As a consequence, the correspond-ing coefficient of the SDE is not Lipschitz any more and the known results for existenceof solutions and mean-field limit (see [43, Theorem 1.4]) fail. More precisely, the prob-lem arises when dealing with matrices with determinant zero; the orthogonal matrix ofthe Polar Decomposition is not uniquely defined for matrices with determinant zero and,otherwise, PD ( M k ) = M k ( (cid:112) M Tk M k ) − (Lemma 3.1).A complete proof of the previous results in the case of Eq. (16)-(17) would involveproving that solutions to the equations stay away from the singular case det( M k ) = 0 .This is an assumption that we make on the Individual Based Model (see the second pointof Remark 3.3). This kind of analysis has been done for the Vicsek model (explained inthe introduction) in [28] where the authors prove global well-posedness for the kineticequation in the spatially homogeneous case.In our case one expects the following to hold: Proposition 3.4 (Formal) . When the number of agents in (16) - (17) N → ∞ , its correspondingempirical distribution f N ( x, A, t ) = 1 N N (cid:88) k =1 δ ( X k ( t ) ,A k ( t )) onverges weakly to f = f ( x, A, t ) , ( x, A, t ) ∈ R × SO (3) × [0 , ∞ ) satisfying ∂ t f + v A e · ∇ x f = D ∆ A f − ∇ A · ( f F [ f ]) , (21) F [ f ] := νP T A ( ¯ M [ f ]) , ¯ M [ f ] = PD ( M [ f ]) , M [ f ]( x, t ) := (cid:90) R × SO (3) K ( x − x (cid:48) ) f ( x (cid:48) , A (cid:48) , t ) A (cid:48) dA (cid:48) dx (cid:48) , where PD ( M [ f ]) corresponds to the orthogonal matrix obtained on the Polar Decomposition of M [ f ] (see Lemma 3.1); and ν = ν ( ¯ M [ f ] · A ) . The goal of this section will be to derive the macroscopic equations (Theorem 4.16). Fromnow on, we consider the kinetic equation given in (21).
We express the kinetic Eq. (21) in dimensionless variables. Let ν be the typical interac-tion frequency scale so that ν ( ¯ A · A ) = ν ν (cid:48) ( ¯ A · A ) with ν (cid:48) ( ¯ A · A ) = O (1) . We introducealso the typical time and space scales t , x such that t = ν − and x = v t ; the asso-ciated variables will be t (cid:48) = t/t and x (cid:48) = x/x . Consider the dimensionless diffusioncoefficient d = D/ν and the rescaled influence kernel K (cid:48) ( | x (cid:48) | ) = K ( x | x (cid:48) | ) . Skipping theprimes we get ∂ t f + A e · ∇ x f = d ∆ A f − ∇ A · ( f F [ f ]) ,F [ f ] := ν ( ¯ M [ f ] · A ) P T A ( ¯ M [ f ]) , ¯ M [ f ] = PD ( M [ f ]) , M [ f ]( x, t ) := (cid:90) R × SO (3) K ( x − x (cid:48) ) f ( x (cid:48) , A (cid:48) , t ) A (cid:48) dA (cid:48) dx (cid:48) . Here d , ν and K are assumed to be of order 1. Remark 4.1.
Notice in particular that before and after scaling the ratio νD = ν (cid:48) d remains the same. Now, to carry out the macroscopic limit we rescale the space and time variables bysetting ˜ t = εt , ˜ x = εx to obtain (skipping the tildes): ∂ t f ε + A e · ∇ x f ε = 1 ε ( d ∆ A f ε − ∇ A · ( f ε F ε [ f ε ])) ,F ε [ f ] := ν ( ¯ M ε [ f ] · A ) P T A ( ¯ M ε [ f ε ]) , ¯ M ε [ f ] = PD ( M ε [ f ]) , M ε [ f ]( x, t ) := (cid:90) R × SO (3) K (cid:18) x − x (cid:48) ε (cid:19) f ( x (cid:48) , A (cid:48) , t ) A (cid:48) dA (cid:48) dx (cid:48) . emma 4.2. Assuming that f is sufficiently smooth (with bounded derivatives), we have theexpansion ¯ M ε [ f ]( x, t ) = Λ[ f ]( x, t ) + O ( ε ) , where Λ[ f ]( x, t ) = PD ( λ [ f ]) and λ [ f ] = (cid:90) SO (3) A (cid:48) f ( x, A (cid:48) , t ) dA (cid:48) . Proof.
This is obtained by performing the change of variable x (cid:48) = x + εξ in the definitionof M ε [ f ] and using a Taylor expansion of f ( x + εξ, A (cid:48) , t ) with respect to ε . We use that K isisotropic and with bounded second moment by assumption (see Eq. (18)).From the lemma, we rewrite ∂ t f ε + A e · ∇ x f ε = 1 ε Q ( f ε ) + O ( ε ) , (22) F [ f ] := ν (Λ[ f ] · A ) P T A (Λ[ f ]) , Λ[ f ] = PD ( λ [ f ]) , λ [ f ]( x, t ) := (cid:90) SO (3) f ( x, A (cid:48) , t ) A (cid:48) dA (cid:48) ,Q ( f ) := d ∆ A f − ∇ A · ( f F [ f ]) . Λ[ f ] , Q ( f ) and F [ f ] are non-linear operators of f , which only acts on the attitudevariable A . SO (3) In the sequel we will use the volume form, the gradient and divergence in SO (3) ex-pressed in the Euler axis-angle coordinates ( θ, n ) (explained at the beginning of Section 3).In this section we give their explicit forms; the proofs are in appendix A. Proposition 4.3 (The gradient in SO (3) ) . Let f : SO (3) → R be a smooth scalar function.If ¯ f ( θ, n ) = f ( A ( θ, n )) is the expression of f in the Euler axis-angle coordinates by Rodrigues’formula (9) , we have then ∇ A f = ∂ θ ¯ f A [ n ] × + 12 sin( θ/ A (cid:16) cos( θ/ (cid:2) ∇ n ¯ f (cid:3) × + sin( θ/ (cid:2) n × ∇ n ¯ f (cid:3) × (cid:17) , (23) where A = A ( θ, n ) and ∇ n is the gradient on the sphere S . The volume form in SO (3) is left invariant (it is the Haar measure), due to the fact thatthe inner product in M is also left invariant: A · B = tr( A T B ) = Λ A · Λ B when Λ ∈ SO (3) .We give its expression in the Euler axis-angle coordinates ( θ, n ) : Lemma 4.4 (Decomposition of the volume form in SO (3) ) . If ¯ f ( θ, n ) = f ( A ( θ, n )) is theexpression of f in the Euler axis-angle coordinates by Rodrigues’ formula (9) , we have (cid:90) SO (3) f ( A ) dA = (cid:90) π W ( θ ) (cid:90) S ¯ f ( θ, n ) d n dθ, here dn is the Lebesgue measure on the sphere S , normalized to be a probability measure, and W ( θ ) = 2 π sin ( θ/ . (24)We have seen in Prop. 4.3 that the gradient is decomposed in the basis { A [ n ] × , A (cid:2) ∇ n ¯ f (cid:3) × , A (cid:2) n × ∇ n ¯ f (cid:3) × } , which are three orthogonal vectors of T A (by Prop. A.2).More generally if B ∈ T A for A = A ( θ, n ) ∈ SO (3) , then B is of the form AH with H antisymmetric, so H = [ u ] × for some u ∈ R . Decomposing u on n and its orthogonal, weget that there exists v ⊥ n and b ∈ R such that B = bA [ n ] × + A [ v ( θ, n )] × . Expressing B inthis form, we compute the divergence in SO (3) . Proposition 4.5 (The divergence in SO (3) ) . Consider B : SO → T ( SO (3)) a smooth function(so that B ( A ) ∈ T A for all A ∈ SO (3) ), and suppose that B ( A ( θ, n )) = b ( θ, n ) A [ n ] × + A [ v ( θ, n )] × for some smooth function b and smooth vector function v such that v ( θ, n ) ⊥ n . Then ∇ A · B = 1sin ( θ/ ∂ θ (cid:0) sin ( θ/ b ( θ, n ) (cid:1) + 12 sin( θ/ ∇ n · (cid:16) v ( θ, n ) cos( θ/
2) + ( v ( θ, n ) × n ) sin( θ/ (cid:17) . Now we can compute the Laplacian in SO (3) : Corollary 4.6.
The Laplacian in SO (3) can be expressed as ∆ A f = 1sin ( θ/ ∂ θ (cid:16) sin ( θ/ ∂ θ ˜ f (cid:17) + 14 sin ( θ/
2) ∆ n ˜ f , where ∆ n is the Laplacian on the sphere S and f ( A ) = f ( A ( θ, n )) = ˜ f ( θ, n ) .Proof. Let B ( θ, n ) := ∇ A f ( A ( θ, n )) ∈ T A . Then, using the notations of Prop. 4.5 and theresult of Prop. 4.3, we have that b = ∂ θ ˜ f , v = 12 sin( θ/ (cid:0) cos( θ/ ∇ n ˜ f + sin( θ/ n × ∇ n ˜ f ) (cid:1) , from here we just need to apply Prop. 4.5 knowing that ( n × ∇ n ˜ f ) × n = ∇ n ˜ f since ∇ n ˜ f is orthogonal to n . 15 .3 Equilibrium solutions and Fokker-Planck formulation We define a generalization of the von-Mises distributions on SO (3) by M Λ ( A ) = 1 Z exp (cid:18) σ ( A · Λ) d (cid:19) , (cid:90) SO (3) M Λ ( A ) dA = 1 , Λ ∈ SO (3) , (25)where Z = Z ( ν, d ) is a normalizing constant and σ = σ ( µ ) is such that ( d/dµ ) σ = ν ( µ ) .Observe that Z < ∞ is independent of Λ since the volume form on SO (3) is left-invariant.Therefore we have Z = (cid:90) SO (3) exp( d − σ ( A · Λ)) dA = (cid:90) SO (3) exp( d − σ (Λ T A · Id)) dA = (cid:90) SO (3) exp( d − σ ( A · Id)) dA, and we also obtain that M Λ ( A ) is actually M Id (Λ T A ) .We are now ready to describe the properties of Q in terms of these generalized von-Mises distributions. Lemma 4.7 (Properties of Q ) . The following holds:i) The operator Q can be written as Q ( f ) = d ∇ A · (cid:20) M Λ[ f ] ∇ A (cid:18) fM Λ[ f ] (cid:19)(cid:21) and we have H ( f ) := (cid:90) SO (3) Q ( f ) fM Λ[ f ] dA = − d (cid:90) SO (3) M Λ[ f ] (cid:12)(cid:12)(cid:12)(cid:12) ∇ A (cid:18) fM Λ[ f ] (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dA ≤ . (26) ii) The equilibria, i.e., the functions f = f ( x, A, t ) such that Q ( f ) = 0 form a -dimensionalmanifold E given by E = { ρM Λ ( A ) | ρ > , Λ ∈ SO (3) } , where ρ is the total mass while Λ is mean body attitude of ρM Λ ( A ) , i.e., ρ = (cid:90) SO (3) ρM Λ ( A ) dA, Λ = Λ[ ρM Λ ] . Furthermore, H ( f ) = 0 iff f = ρM Λ for arbitrary ρ ∈ R + and Λ ∈ SO (3) . To prove Lemma 4.7 we require the following one, which is of independent interestand for which we introduce the following notation: for any scalar function g : (0 , π ) → R and a given integrable scalar function h : (0 , π ) → R which remains positive (or negative)on (0 , π ) , we define (cid:104) g ( θ ) (cid:105) h ( θ ) := (cid:90) π g ( θ ) h ( θ ) (cid:82) π h ( θ (cid:48) ) dθ (cid:48) dθ. (27)16 emma 4.8 (Consistency relation for the ‘flux’) . λ [ M Λ ] = c Λ where c ∈ (0 , is equal to c = (cid:104) + cos θ (cid:105) m ( θ ) sin ( θ/ (28) for m ( θ ) = exp( d − σ ( + cos θ )) . (29) Proof.
Using the fact that the measure on SO (3) is left invariant, we obtain λ [ M Λ ] = 1 Z (cid:90) SO (3) A exp( d − σ (( A · Λ ))) dA = Λ Z (cid:90) SO (3) Λ T A exp( d − σ ( tr(Λ T A ))) dA = Λ Z (cid:90) SO (3) B exp( d − σ ( tr( B ))) dB. We now write B = Id + sin θ [ n ] × + (1 − cos θ )[ n ] × thanks to Rodrigues’ formula (9).Therefore, using Lemma 4.4, we get λ [ M Λ ] = Λ (cid:82) SO (3) B exp( d − σ ( tr( B ))) dB (cid:82) SO (3) exp( d − σ ( tr( B ))) dB = Λ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ )) (cid:0) (cid:82) S (Id + sin θ [ n ] × + (1 − cos θ )[ n ] × ) d n (cid:1) dθ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ )) dθ . Next, we see that since the function n (cid:55)→ [ n ] × is odd, we have (cid:82) S [ n ] × d n = 0 . We also have(see (11)) that [ n ] × = n ⊗ n − Id . Since we know that (cid:82) S n ⊗ n d n = Id (by invariance byrotation), it is easy to see that the integral in S has to be proportional to Id , the coefficientis given by computing the trace), we get that λ [ M Λ ] = Λ (cid:82) SO (3) B exp( d − σ ( tr( B ))) dB (cid:82) SO (3) exp( d − σ ( tr( B ))) dB = Λ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ ))(Id + (1 − cos θ )( − dθ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ )) dθ = (cid:82) π ( + cos θ ) sin ( θ/
2) exp( d − σ ( + cos θ )) dθ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ )) dθ Λ = c Λ , which gives the formula (28) for c .It remains to prove that c ∈ (0 , . We have that c is the average of ( + cos θ ) forthe probability measure on (0 , π ) proportional to sin ( θ/
2) exp( d − σ ( + cos θ )) . Since we17ave ( + cos θ ) ≤ with equality only for θ = 0 , we immediately get that c < . Toprove the positivity, we remark that the function in the exponent θ (cid:55)→ d − σ ( + cos θ ) is strictly decreasing for θ ∈ (0 , π ) (since ν > is the derivative of σ ), so we obtainthat σ ( + cos θ ) > σ ( + cos π ) = σ (0) for θ ∈ (0 , π ) . Therefore, for θ ∈ (0 , π ) , ( + cos θ ) exp( d − σ ( + cos θ )) > ( + cos θ ) exp( d − σ (0)) , since + cos θ > . When θ ∈ ( π , π ) , we have exactly the same inequality above since wehave + cos θ < . Therefore we get c > (cid:82) π ( + cos θ ) sin ( θ/
2) exp( d − σ (0)) dθ (cid:82) π sin ( θ/
2) exp( d − σ ( + cos θ )) dθ = 0 , since (cid:82) π ( + cos θ ) sin ( θ/ dθ = (cid:82) π ( + cos θ )( − cos θ ) dθ = π − (cid:82) π cos θdθ = 0 . Proof of Lemma 4.7.
We follow the structure of the analogous proof in [24]:i) To prove the first identity we have that (see expression (54)) ∇ A (cid:0) ln M Λ[ f ] (cid:1) = d − ∇ A ( σ ( A · Λ[ f ]))= d − ν ( A · Λ[ f ]) P T A (Λ[ f ])= d − F [ f ] and so d ∇ A · (cid:20) M Λ[ f ] ∇ A (cid:18) fM Λ[ f ] (cid:19)(cid:21) = d ∇ A · (cid:2) ∇ A f − f ∇ A (cid:0) ln( M Λ[ f ] ) (cid:1)(cid:3) = d ∇ A f − ∇ A · ( f F [ f ]) . Inequality (26) follows from this last expression and the Stokes theorem in SO (3) .ii) From the inequality (26) we have that if Q ( f ) = 0 , then fM Λ[ f ] is a constant that wedenote by ρ (which is positive since f and M Λ[ f ] are positive). Conversely, if f = ρM Λ then λ [ ρM Λ ] = (cid:90) SO (3) ρM Λ ( A ) A dA = ρc Λ by Lemma 4.8. Now, by uniqueness of the Polar Decomposition and since ρc Id is asymmetric positive-definite matrix, we have that Λ[ ρM Λ ] = Λ .Let us describe the behavior of these equilibrium distributions for small and largenoise intensities. We have that for any function g , the average (cid:104) g ( + cos θ ) (cid:105) m ( θ ) sin ( θ/ isthe average of g ( A · Λ) with respect to the probability measure M Λ (by left-invariance, thisis independent of Λ ). 18ne can actually check that the probability measure M Λ on SO (3) converges in distri-bution to the uniform measure when d → ∞ (by Taylor expansion) and it converges to aDirac delta at matrix Λ when d → (this can be seen for M Id thanks to the decompositionof the volume form and the Laplace method, since the maximum of σ ( + cos θ ) is reachedonly at θ = 0 which corresponds to the identity matrix, and we then get the result forany Λ since M Λ ( A ) = M Id (Λ T A ) ). So for small diffusion, at equilibrium, agents tend toadopt the same body attitude close to Λ .With these asymptotic considerations, we have in particular the behaviour of c : c −→ d →∞ and c −→ d → . To obtain the macroscopic equation, we start by looking for the conserved quantities ofthe kinetic equation: we want to find the functions ψ = ψ ( A ) such that (cid:90) SO (3) Q ( f ) ψ dA = 0 for all f. By Lemma 4.7, this can be rewritten as − (cid:90) SO (3) M Λ [ f ] ∇ A (cid:18) fM Λ [ f ] (cid:19) · ∇ A ψ dA. This happens if ∇ A ψ ∈ T ⊥ A which holds true only if ∇ A ψ = 0 , implying that ψ is constant.Consequently, our model has only one conserved quantity: the total mass. Howeverthe equilibria is -dimensional (by Lemma 4.7). To obtain the macroscopic equations for Λ ,a priori we would need 3 more conserved quantities. This problem is sorted out by usingGeneralized Collision Invariants (GCI) a concept first introduced in [24]. Define the operator Q ( f, Λ ) := ∇ A · (cid:18) M Λ ∇ A (cid:18) fM Λ (cid:19)(cid:19) , notice in particular that Q ( f ) = Q ( f, Λ[ f ]) . Using this operator we define: 19 efinition 4.9 (Generalised Collision Invariant) . For a given Λ ∈ SO (3) we say that a real-valued function ψ : SO (3) → R is a Generalized Collision Invariant associated to Λ , or forshort ψ ∈ GCI (Λ ) , if (cid:90) SO (3) Q ( f, Λ ) ψ dA = 0 for all f s.t P T Λ0 ( λ [ f ]) = 0 . In particular, the result that we will use is : ψ ∈ GCI (Λ[ f ]) = ⇒ (cid:90) SO (3) Q ( f ) ψ dA = 0 . (30)Indeed, since Λ[ f ] is the polar decomposition of λ [ f ] , we have λ [ f ] = Λ[ f ] S , with S a sym-metric matrix. Therefore (see Proposition A.2), we get that λ [ f ] belongs to the orthogonalof T Λ[ f ] , so the definition 4.9 and the fact that Q ( f ) = Q ( f, Λ[ f ]) gives us the property (30).The definition 4.9 is equivalent to the following: Proposition 4.10.
We have that ψ ∈ GCI (Λ ) if and only ifthere exists B ∈ T Λ such that ∇ A · ( M Λ ∇ A ψ ) = B · A M Λ . (31) Proof of Prop. 4.10.
We denote L the linear operator Q ( · , Λ ) , and L ∗ its adjoint. We havethe following sequence of equivalences, starting from Def. 4.9: ψ ∈ GCI (Λ ) ⇔ (cid:90) SO (3) ψ L ( f ) dA = 0 , for all f such that P T Λ0 ( λ [ f ]) = 0 ⇔ (cid:90) SO (3) L ∗ ( ψ ) f dA = 0 , for all f such that (cid:90) SO (3) Af ( A ) dA ∈ ( T Λ ) ⊥ ⇔ (cid:90) SO (3) L ∗ ( ψ ) f dA = 0 , for all f s.t. ∀ B ∈ T Λ , (cid:90) SO (3) ( B · A ) f ( A ) dA = 0 ⇔ (cid:90) SO (3) L ∗ ( ψ ) f dA = 0 , for all f ∈ F ⊥ Λ ⇔ L ∗ ( ψ ) ∈ (cid:0) F ⊥ Λ (cid:1) ⊥ , where F Λ := { g : SO (3) → R , with g ( A ) = ( B · A ) , for some B ∈ T Λ } , and F ⊥ Λ is the space orthogonal to F Λ in L . F Λ is a vector space in L isomorphic to T Λ and ( F ⊥ Λ ) ⊥ = F Λ since F Λ is closed (finite dimensional). Therefore we get ψ ∈ GCI (Λ ) ⇔ L ∗ ( ψ ) ∈ F Λ ⇔ there exists B ∈ T Λ such that L ∗ ( ψ )( A ) = B · A, which ends the proof since the expression of the adjoint is L ∗ ( ψ ) = M Λ0 ∇ A · ( M Λ ∇ A ψ ) .We prove the existence and uniqueness of the solution ψ satisfying Eq. (31) in thefollowing: 20 roposition 4.11 (Existence of the GCI) . For a given B ∈ T Λ fixed, there exists a unique (up toa constant) ψ B ∈ H ( SO (3)) , satisfying the relation (31) .Proof of Prop. 4.11. We would like to apply the Lax-Milgram theorem to prove the exis-tence of ψ in an appropriate functional space. For this, we rewrite the relation (31) weakly a ( ψ, ϕ ) := (cid:90) SO (3) M Λ ∇ A ψ · ∇ A ϕ dA = (cid:90) SO (3) B · P T Λ0 ( A ) M Λ ϕ dA =: b ( ϕ ) . (32)Our goal is to prove that there exists a unique ψ ∈ H ( SO (3)) such that a ( ψ, ϕ ) = b ( ϕ ) forall ϕ ∈ H ( SO (3)) .To begin with we apply the Lax-Milgram theorem on the space H ( SO (3)) := (cid:26) ϕ ∈ H | (cid:90) SO (3) ϕ dA = 0 (cid:27) . In this space the H -norm and the H semi-norm are equivalent thanks to the Poincar´einequality, i.e., there exists C > such that (cid:90) SO (3) |∇ A ϕ | dA ≥ C (cid:90) SO (3) | ϕ | dA for some C > , for all ϕ ∈ H ( SO (3)) . Notice that the Poincar´e inequality holds in SO (3) because it is compact Riemannian man-ifold [12]. This gives us the coercivity estimate to apply the Lax-Milgram theorem. Hence,there exists a unique ψ ∈ H ( SO (3)) s.t a ( ψ, ϕ ) = b ( ϕ ) for all ϕ ∈ H ( SO (3)) .Now, define for a given ϕ ∈ H ( SO (3)) , ϕ := ϕ − (cid:82) SO (3) ϕ dA ∈ H ( SO (3)) . It holdsthat a ( ψ, ϕ ) = a ( ψ, ϕ ) and b ( ϕ ) = b ( ϕ ) since b (1) = 0 given that it has antisymmetric integrand. Hence, we obtain that thereexists a unique ψ ∈ H ( SO (3)) such that a ( ψ, ϕ ) = b ( ϕ ) for all ϕ ∈ H ( SO (3)) . Suppose next, that there exists another solution ¯ ψ ∈ H ( SO (3)) to this problem, thenthe difference Ψ = ψ − ¯ ψ satisfies: a (Ψ , ϕ ) = (cid:90) SO (3) M Λ ∇ A Ψ · ∇ A ϕ dA for all ϕ ∈ H ( SO (3)) . Take in particular ϕ = Ψ , then (cid:90) SO (3) M Λ |∇ A Ψ | dA = 0 . Hence,
Ψ = c for some constant c , so all solutions are of the form ψ + c where ψ is theunique solution satisfying (cid:82) SO (3) ψ dA = 0 . 21y writing that B ∈ T Λ if and only if there exists P ∈ A , B = Λ P, (33)with A the set of antisymmetric matrices, we deduce the: Corollary 4.12.
For a given Λ ∈ SO (3) , the set of Generalized Collision Invariants associatedto Λ are GCI (Λ ) = span { , ∪ P ∈A ψ Λ P } (where A is the set of antisymmetric matrices) with ψ Λ P the unique solution in H ( SO (3)) of a ( ψ Λ P , ϕ ) = b P ( ϕ ) for all ϕ ∈ H ( SO (3)) , where a and b P are defined by (32) with B substituted by Λ P . Notice that since the mapping P (cid:55)→ ψ Λ P is linear and injective from A (of dimension )to H ( SO (3)) , the vector space GCI (Λ ) is of dimension . From now on, we omit the subscript on Λ , and we are interested in a simpler expressionfor ψ Λ P . Rewriting expression (31) using (33), for any given P ∈ A we want to find ψ suchthat ∇ A · ( M Λ ∇ A ψ ) = (Λ P ) · A M Λ = P · (Λ T A ) M Λ , P ∈ A . (34) Proposition 4.13.
Let P ∈ A and ψ be the solution of (34) belonging to H ( SO (3)) . If wedenote ¯ ψ ( B ) := ψ (Λ B ) , then ¯ ψ is the unique solution in H ( SO (3)) of: ∇ B · (cid:0) M Id ( B ) ∇ B ¯ ψ (cid:1) = P · BM Id ( B ) . (35) Proof.
Let ψ ( A ) = ¯ ψ (Λ T A ) . Consider A ( ε ) a differentiable curve in SO (3) with A (0) = A, dA ( ε ) dε (cid:12)(cid:12)(cid:12)(cid:12) ε =0 = δ A ∈ T A . Then, by definition lim ε → ψ ( A ( ε )) − ψ ( A ) ε = ∇ A ψ ( A ) · δ A , and therefore we have that lim ε → ¯ ψ (Λ T A ( ε )) − ¯ ψ (Λ T A ) ε = ∇ B ¯ ψ (Λ T A ) · Λ T δ A since Λ T A (0) = Λ T A, ddε Λ T A ( ε ) (cid:12)(cid:12)(cid:12)(cid:12) ε =0 = Λ T δ A .
22e conclude that ∇ A ψ ( A ) · δ A = ∇ B ¯ ψ (Λ T A ) · Λ T δ A . Now we check that
12 tr (cid:16) ( ∇ A ψ ( A )) T δ A (cid:17) = 12 tr (cid:16)(cid:0) ∇ B ¯ ψ (Λ T A ) (cid:1) T Λ T δ A (cid:17) = 12 tr (cid:16)(cid:0) Λ ∇ B ¯ ψ (Λ T A ) (cid:1) T δ A (cid:17) , implying (since this is true for any δ A ∈ T A ) that ∇ A ψ ( A ) = Λ ∇ B ¯ ψ (Λ T A ) . Now to deal with the divergence term, we consider the variational formulation. Con-sider ϕ ∈ H ( SO (3)) , then our equation is equivalent to − (cid:90) SO (3) M Λ ( A ) ∇ A ψ ( A ) · ∇ A ϕ ( A ) dA = (cid:90) SO (3) P · (Λ T A ) M Λ ( A ) ϕ ( A ) dA for all ϕ ∈ H ( SO (3)) . The left hand side can be written as: − (cid:90) SO (3) M Id ( B )(Λ T A ) (cid:0) Λ ∇ B ¯ ψ (Λ T A ) (cid:1) · (cid:0) Λ ∇ B ¯ ϕ (Λ T A ) (cid:1) dA = − (cid:90) SO (3) M Id ( B ) ∇ B ¯ ψ ( B ) · ∇ B ¯ ϕ ( B ) dB ; and the right hand side is equal to (cid:90) SO (3) P · B M Id ( B ) ¯ ϕ ( B ) dB, where we define analogously ¯ ϕ ( B ) = ϕ (Λ B ) . This concludes the proof.Therefore it is enough to find the solution to (35). Inspired by [24] we make the ansatz: ¯ ψ ( B ) = P · B ¯ ψ ( tr( B )) for some scalar function ¯ ψ . Proposition 4.14 (Non-constant GCI) . Let P ∈ A , then the unique solution ¯ ψ ∈ H ( SO (3)) of (35) is given by ¯ ψ ( B ) = P · B ¯ ψ ( tr( B )) , (36) where ¯ ψ is constructed as follows: Let (cid:101) ψ : R → R be the unique solution to ( θ/ ∂ θ (cid:16) sin ( θ/ m ( θ ) ∂ θ (cid:16) sin θ (cid:101) ψ (cid:17)(cid:17) − m ( θ ) sin θ ( θ/ (cid:101) ψ = sin θ m ( θ ) , (37)23 here m ( θ ) = M Id ( B ) = exp( d − σ ( + cos θ )) /Z . Then (cid:101) ψ ( θ ) = ¯ ψ (cid:18)
12 tr( B ) (cid:19) (38) by the relation tr( B ) = + cos θ . (cid:101) ψ is π -periodic, even and negative (by the maximum princi-ple).Going back to the GCI ψ ( A ) , we can write it as ψ ( A ) = P · (Λ T A ) ¯ ψ (Λ · A ) . (39) Proof of Prop. 4.14.
Suppose that the solution is given by expression (36). We check that (cid:101) ψ given by Eq. (38) satisfies Eq. (37) using the gradient and divergence in SO (3) computedin Prop. 4.3 and 4.5. First notice that P is antisymmetric, thus if we write Rodrigues’formula (9) for B ( θ, n ) , the symmetric part of B ( θ, n ) gives no contribution when com-puting P · B and we get ¯ ψ ( B ) = P · B ¯ ψ ( tr( B )) = sin θ (cid:101) ψ ( θ ) P · [ n ] × = sin θ (cid:101) ψ ( θ )( p · n ) , where the vector p is such that P = [ p ] × and this leads to ∇ B · (cid:0) M Id ( B ) ∇ B ¯ ψ (cid:1) = 1sin ( θ/ ∂ θ (cid:16) sin ( θ/ m ( θ ) ∂ θ (cid:16) sin θ (cid:101) ψ ( θ ) (cid:17) (cid:17) ( p · n )+ m ( θ ) sin θ ( θ/ (cid:101) ψ ( θ )∆ n ( p · n ) . Using that the Laplacian in the sphere has the property ∆ n ( p · n ) = − p · n ) ( p · n corresponds to the first spherical harmonic), we conclude that expression (37) is satisfied.In the computation we used the same procedure as for the proof of the expression of theLaplacian in SO (3) (Corollary 4.6), but (using the same notations) we have taken b ( θ, n ) = m ( θ ) ∂ θ (sin θ (cid:101) ψ ( θ ))( p · n ) .To conclude the proof we just need to check that (cid:101) ψ exists and corresponds to a func-tion ¯ ψ in H ( SO (3)) . Using the expression of the volume form, since (cid:82) S p · n d n = 0 ,we get that if ψ is smooth, we have (cid:82) SO (3) ¯ ψ ( A ) dA = 0 , and using the expression of thegradient, we get that (cid:90) SO (3) |∇ ¯ ψ ( A ) | dA = 2 π (cid:90) π sin ( θ/ | ∂ θ (sin θ (cid:101) ψ ( θ )) | dθ (cid:90) S | p · n | d n + 2 π (cid:90) π | sin θ (cid:101) ψ ( θ ) | dθ (cid:90) S |∇ n ( p · n ) | d n . Therefore by density of smooth functions in H ( SO (3)) , we get that ¯ ψ ∈ H ( SO (3)) ifand only if (cid:101) ψ ∈ H , where H := (cid:26) ψ | (cid:90) (0 ,π ) ψ sin θ dθ < ∞ , (cid:90) (0 ,π ) | ∂ θ (sin θψ ( θ )) | sin ( θ/ dθ < ∞ (cid:27) . (cid:107) ψ (cid:107) H = (cid:90) (0 ,π ) ψ sin θ dθ + (cid:90) (0 ,π ) | ∂ θ (sin θψ ( θ )) | sin ( θ/ dθ. Now, Eq. (37) written in weak form in H and tested against any φ ∈ H reads a ( (cid:101) ψ , φ ) := − (cid:90) (0 ,π ) m ( θ ) (cid:20) sin ( θ/ ∂ θ (sin θ (cid:101) ψ ( θ )) ∂ θ (sin θφ ( θ )) dθ + 12 sin θ (cid:101) ψ ( θ ) φ ( θ ) (cid:21) dθ = (cid:90) (0 ,π ) sin θ sin ( θ/ m ( θ ) φ dθ =: b ( φ ) . It holds for some c, c (cid:48) , c (cid:48)(cid:48) > that: | a ( ψ, φ ) | ≤ c (cid:107) ψ (cid:107) H (cid:107) φ (cid:107) H since m = m ( θ ) is bounded; andalso | a ( ψ, ψ ) | ≥ c (cid:48) (cid:107) ψ (cid:107) H since there exists m > such that m ( θ ) > m for all θ ∈ [0 , π ] ;finally, we also have that | b ( φ ) | ≤ c (cid:48)(cid:48) (cid:107) φ (cid:107) H . Therefore, by the Lax-Milgram theorem, thereexists a (unique) solution (cid:101) ψ ∈ H to (37), which corresponds to a (unique) ¯ ψ in H ( SO (3)) . In this section we investigate the hydrodynamic limit. To state the theorem we first givethe definitions of the first order operators δ x and r x . For a smooth function Λ from R to SO (3) , and for x ∈ R , we define the following matrix D x (Λ) such that for any w ∈ R ,we have ( w · ∇ x )Λ = [ D x (Λ) w ] × Λ . (40)Notice that this first-order differential equation D x is well-defined as a matrix; for a givenvector w , the matrix ( w · ∇ x )Λ is in T Λ and thanks to Prop. A.3, it is of the form P Λ , with P an antisymmetric matrix. Therefore there exists a vector D x (Λ)( w ) ∈ R depending on w such that P = [ D x (Λ)( w )] × . The function w (cid:55)→ D x (Λ)( w ) is linear from R to R , so D x (Λ) can be identified as a matrix.We now define the first order operators δ x (scalar) and r x (vector), by δ x (Λ) = tr (cid:0) D x (Λ) (cid:1) and [ r x (Λ)] × = D x (Λ) − D x (Λ) T . (41)We first give an invariance property which allows for a simple expression for theseoperators. Proposition 4.15.
The operators D x , δ x and r x are right invariant in the following sense: if A isa fixed matrix in SO (3) and Λ : R → SO (3) a smooth function, we have D x (Λ A ) = D x (Λ) , δ x (Λ A ) = δ x (Λ) and r x (Λ A ) = r x (Λ) . Consequently, in the neighborhood of x ∈ R , we can write Λ( x ) = exp ([ b ( x )] × ) Λ( x ) where b is a smooth function from a neighborhood of x into R such that b ( x ) = 0 , and we have (cid:0) D x (Λ) (cid:1) ij ( x ) = ∂ j b i ( x ) , nd therefore δ x (Λ)( x ) = ( ∇ x · b ) ( x ) , and r Λ ( x ) = ( ∇ x × b ) ( x ) . Proof.
For any w ∈ R , we have, since A is constant: [ D x (Λ A ) w ] × Λ A = w · ∇ x (Λ A ) = ( w · ∇ x Λ) A = [ D x (Λ) w ] × Λ A. This proves that D x (Λ A ) = D x (Λ) , and by (41), the same is obviously true for δ x and r x .We now write, in the neighborhood of x , that Λ( x ) = exp([ b ( x )] × )Λ( x ) , with b smooth in the neighborhood of x and b ( x ) = 0 . Then we have D x (Λ) = D x (cid:0) exp([ b ] × ) (cid:1) .We perform a Taylor expansion around x of exp([ b ] × ) : exp([ b ( x )] × ) = Id + [ b ( x )] × + M ( x ) , where M ( x ) is of order 2 in the coordinates b , b , b , (since b is smooth in the neighbor-hood of x and b ( x ) = 0 ), therefore ∂ M ( x ) = ∂ M ( x ) = ∂ M ( x ) = 0 . We then get, since exp([ b ( x )]) = Id , that [ D x (cid:0) exp([ b ] × ) (cid:1) ( x ) w ] × = w · ∇ x (cid:0) exp([ b ] × ) (cid:1) ( x ) = (cid:2) ( w · ∇ x b )( x ) (cid:3) × , and therefore D x (Λ)( x ) w = D x (exp([ b ] × ))( x ) w = ( w · ∇ x b )( x ) . Taking w = e j , weget D x (Λ)( x ) e j = ∂ j b ( x ) , and thus (cid:0) D x (Λ)( x ) (cid:1) ij = e i · D x (Λ)( x ) e j = ∂ j b i . The formulafor δ x (Λ) follows from (41), since ∇ x · b = (cid:80) i ∂ i b i . Finally by the definition of [ · ] × (see (8)),we get [ ∇ x × b ] × = ∂ b − ∂ b ∂ b − ∂ b ∂ b − ∂ b ∂ b − ∂ b ∂ b − ∂ b ∂ b − ∂ b , so from (41) we obtain ( ∇ x × b )( x ) = r x (Λ)( x ) .We are now ready to state the main theorem of our paper (see Section 2 for a discussionon this result). Theorem 4.16 ((Formal) macroscopic limit) . When ε → in the kinetic Eq. (22) it holds(formally) that f ε → f = f ( x, A, t ) = ρM Λ ( A ) , Λ = Λ( t, x ) ∈ SO (3) , ρ = ρ ( t, x ) ≥ . Moreover, if this convergence is strong enough and the functions Λ and ρ are smooth enough, theysatisfy the following first-order system of partial differential equations: ∂ t ρ + ∇ x · (cid:0) c ρ Λ e (cid:1) = 0 , (42) ρ (cid:16) ∂ t Λ + c (cid:0) (Λ e ) · ∇ x (cid:1) Λ (cid:17) + (cid:2) (Λ e ) × (cid:0) c ∇ x ρ + c ρ r x (Λ) (cid:1) + c ρ δ x (Λ)Λ e (cid:3) × Λ = 0 , (43)26 here c = c ( ν, d ) = (cid:104) + cos θ (cid:105) m ( θ ) sin ( θ/ is the constant given in (28) and c = (cid:104) θ (cid:105) (cid:101) m ( θ ) sin ( θ/ ,c = d (cid:104) ν ( + cos θ ) − (cid:105) (cid:101) m ( θ ) sin ( θ/ ,c = (cid:104) − cos θ (cid:105) (cid:101) m ( θ ) sin ( θ/ , where the notation (cid:104)·(cid:105) (cid:101) m ( θ ) sin ( θ/ is defined in (27) . The function (cid:101) m : (0 , π ) → (0 , + ∞ ) is givenby (cid:101) m ( θ ) := ν ( + cos θ ) sin θ m ( θ ) (cid:101) ψ ( θ ) , (44) where m ( θ ) = exp( d − σ ( + cos θ )) is the same as in (29) and (cid:101) ψ is the solution of Eq. (37) .Proof. Suppose that f ε → f as ε → , then using (22) we get Q ( f ε ) = O ( ε ) , which formallyyields Q ( f ) = 0 and by Lemma 4.7 we have that f = f ( x, A, t ) = ρM Λ ( A ) , with Λ = Λ( t, x ) ∈ SO (3) , ρ = ρ ( t, x ) ≥ . Using the conservation of mass (integrating (22) on SO (3) ), we have that ∂ t ρ ε + ∇ x · j [ f ε ] = O ( ε ) , where ρ ε ( t, x ) := (cid:90) SO (3) f ε ( x, A, t ) dA, j [ f ε ] := (cid:90) SO (3) A e f ε dA, and in the limit (formally) ρ ε → ρ,j [ f ε ] → ρ (cid:90) SO (3) A e M Λ ( A ) dA = ρλ [ M Λ ] e = ρc Λ e , thanks to Lemma 4.8. This gives us the continuity equation (42) for ρ .Now, we want to obtain the equation for Λ . We write Λ ε = Λ[ f ε ] , and we take P ∈ A a given antisymmetric matrix. We consider the non-constant GCI associated to Λ ε andcorresponding to P in (39): ψ ε ( A ) = P · ((Λ ε ) T A ) ¯ ψ (Λ ε · A ) . Since we have ψ ε ∈ GCI (Λ[ f ε ]) ,we obtain, thanks to the main property (30) of the GCI, that (cid:90) SO (3) Q ( f ε ) ψ ε dA = 0 . Multiplying (22) by ψ ε , integrating w.r.t. A on SO (3) and using the expression of ψ ε asstated above, we obtain (cid:90) SO (3) (cid:0) ∂ t f ε + A e · ∇ x f ε + O ( ε ) (cid:1) P · (cid:0) (Λ ε ) T A (cid:1) ¯ ψ (Λ ε · A ) dA = 0 . f ε → f is sufficiently strong, we get in the limit (cid:90) SO (3) (cid:0) ∂ t ( ρM Λ ) + A e · ∇ x ( ρM Λ ) (cid:1)(cid:0) P · Λ T A (cid:1) ¯ ψ (Λ · A ) dA = 0 . (45)Since (45) is true for any P ∈ A , the matrix (cid:90) SO (3) (cid:0) ∂ t ( ρM Λ ) + A e · ∇ x ( ρM Λ ) (cid:1) ¯ ψ (Λ · A ) Λ T A dA = 0 . is orthogonal to all antisymmetric matrices. Therefore, it must be a symmetric matrix,meaning that we have X := (cid:90) SO (3) (cid:0) ∂ t ( ρM Λ ) + A e · ∇ x ( ρM Λ ) (cid:1) ¯ ψ (Λ · A ) (Λ T A − A T Λ) dA = 0 . (46)We have with the definition of M Λ in (25) that ∂ t ( ρM Λ ) = M Λ ( ∂ t ρ + d − ν (Λ · A ) ρ ( A · ∂ t Λ)) , ( A e · ∇ x )( ρM Λ ) = M Λ (cid:0) A e · ∇ x ρ + d − ν (Λ · A ) ρ ( A · ( A e · ∇ x )Λ) (cid:1) . Inserting the two previous expressions into (46), we compute separately each componentof X defined by: X := (cid:90) SO (3) ∂ t ρM Λ ¯ ψ (Λ · A ) (Λ T A − A T Λ) dA,X := (cid:90) SO (3) d − ν (Λ · A ) ρ ( A · ∂ t Λ) M Λ ¯ ψ (Λ · A ) (Λ T A − A T Λ) dA,X := (cid:90) SO (3) A e · ∇ x ρ M Λ ¯ ψ (Λ · A ) (Λ T A − A T Λ) dA,X := (cid:90) SO (3) d − ν (Λ · A ) ρ ( A · ( A e · ∇ x )Λ) M Λ ¯ ψ (Λ · A ) (Λ T A − A T Λ) dA, so X = X + X + X + X .For the first term we have (changing variables B = Λ T A ): X = ∂ t ρ (cid:90) SO (3) M Id ( B ) ¯ ψ (Id · B ) ( B − B T ) dB = 0 since both M Id ( B ) and ¯ ψ (Id · B ) are invariant by the change B (cid:55)→ B T .For the term X we make the change of variables B = Λ T A and compute X = ρ (cid:90) SO (3) d − ν (Id · B )(Λ B · ∂ t Λ) M Id ( B ) ¯ ψ (Id · B )( B − B T ) dB = 2 d − ρπZ (cid:90) (0 ,π ) × S (cid:0) Λ (cid:0) Id + sin θ [ n ] × + (1 − cos θ )[ n ] × (cid:1)(cid:1) · ∂ t Λsin ( θ/ ν ( + cos θ ) m ( θ ) (cid:101) ψ ( θ ) 2 sin θ [ n ] × dθd n , dB = π sin ( θ/ dθd n (seeLemma 4.4) and that writing B = B ( θ, n ) = Id + sin θ [ n ] × + (1 − cos θ )[ n ] × thanks toRodrigues’ formula (9), we have B − B T = 2 sin θ [ n ] × . Removing odd terms with respectto the change n (cid:55)→ − n , we obtain X = 4 d − ρπZ (cid:90) (0 ,π ) × S ν ( + cos θ ) sin θ m ( θ ) (cid:101) ψ ( θ ) sin ( θ/ n ] × · ∂ t Λ) [ n ] × dθd n . Now since ∂ t Λ ∈ T Λ , we have Λ T ∂ t Λ ∈ A (antisymmetric, see Prop. A.2), and so Λ T ∂ t Λ = [ λ t λ t λ t ] × for some vector λ t λ t λ t . Therefore (Λ[ n ] × ) · ∂ t Λ = [ n ] × · (Λ T ∂ t Λ) = [ n ] × · [ λλλ t ] × = ( n · λ t λ t λ t ) . So using the definition (44) of (cid:101) m ( θ ) , we get X = 4 d − ρπZ (cid:90) (0 ,π ) × S (cid:101) m ( θ ) sin ( θ/ n · λ t λ t λ t ) [ n ] × dθd n = 4 d − ρπZ (cid:20)(cid:90) (0 ,π ) × S (cid:101) m ( θ ) sin ( θ/ n · λ t λ t λ t ) n dθd n (cid:21) × = 4 d − ρ πZ (cid:18)(cid:90) π (cid:101) m ( θ ) sin ( θ/ dθ (cid:19) [ λ t λ t λ t ] × , because the mapping w (cid:55)→ [ w ] × is linear, and (cid:82) S n ⊗ n d n = Id .Denote by C := 4 d − πZ (cid:18)(cid:90) π (cid:101) m ( θ ) sin ( θ/ dθ (cid:19) , then we conclude that X = C ρ Λ T ∂ t Λ . Now, for the term X we compute the following, starting again by the change of vari-29bles B = Λ T A : X = (cid:90) SO (3) (Λ B e · ∇ x ρ ) M Id ( B ) ¯ ψ ( Id · B ) ( B − B T ) dB = 4 πZ (cid:90) (0 ,π ) × S m ( θ ) (cid:101) ψ ( θ ) sin θ sin ( θ/ (cid:0) Λ (cid:0) Id + sin θ [ n ] × + (1 − cos θ )[ n ] × (cid:1) e · ∇ x ρ (cid:1) [ n ] × dθd n = 4 πZ (cid:90) (0 ,π ) × S m ( θ ) (cid:101) ψ ( θ ) sin θ sin ( θ/
2) (Λ[ n ] × e · ∇ x ρ ) [ n ] × dθd n = 4 πZ (cid:20)(cid:90) (0 ,π ) × S (cid:101) m ( θ ) ν ( + cos θ ) sin ( θ/ (cid:0) n · ( e × Λ T ∇ x ρ ) (cid:1) n dθd n (cid:21) × = 43 πZ (cid:18)(cid:90) π (cid:101) m ( θ ) ν ( + cos θ ) sin ( θ/ dθ (cid:19) [ e × Λ T ∇ x ρ ] × , where we used similar considerations as for X , as well as that Λ[ n ] × e · ∇ x ρ = [ n ] × e · (Λ T ∇ x ρ ) = ( n × e ) · (Λ T ∇ x ρ ) = n · ( e × Λ T ∇ x ρ ) . Denote by C := 43 πZ (cid:18)(cid:90) π (cid:101) m ( θ ) ν ( + cos θ ) sin ( θ/ dθ (cid:19) , then X = C [ e × Λ T ∇ x ρ ] x . We now compute X in the same way, with the change of variables B = Λ T A : X = ρd − (cid:90) SO (3) (cid:0) ν (Id · B )(Λ B · (Λ B e · ∇ x )Λ) (cid:1) M Id ( B ) ¯ ψ (Id · B ) ( B − B T ) dB . We now use the definition of D x (Λ) given in (40) to get X = ρd − (cid:90) SO (3) (cid:0) ν (Id · B )(Λ B · ([ D x (Λ)Λ B e ] × Λ) (cid:1) M Id ( B )( B − B T ) ¯ ψ (Id · B ) dB . Using the fact that Λ T [ w ] × = [Λ T w ] × Λ T for all w ∈ R , we have Λ B · ([ D x (Λ)Λ B e ] × Λ) = B · [Λ T D x (Λ)Λ B e ] × . To simplify the notations, we denote L = Λ T D x (Λ)Λ . Since the symmetric part of B doesnot contribute to the scalar product B · [ LB e ] × , we get Λ B · ([ D x (Λ)Λ B e ] × Λ) = B · [ LB e ] × = sin θ [ n ] × · [ LB e ] × = sin θ n · LB e , X = 4 ρd − πZ (cid:90) π (cid:101) m ( θ ) sin ( θ/ (cid:20)(cid:90) S (cid:16) n · (cid:0) L (Id + sin θ [ n ] × + (1 − cos θ )[ n ] × ) e (cid:1)(cid:17) n d n (cid:21) × dθ, and we have to know the value of y ( θ ) := (cid:90) S (cid:16) n · (cid:0) L (Id + sin θ [ n ] × + (1 − cos θ )[ n ] × ) e (cid:1)(cid:17) n d n = (cid:90) S (cid:18) n · (cid:16) L (cos θ e + (1 − cos θ )( n · e ) n ) (cid:17)(cid:19) n d n = 13 cos θL e + (1 − cos θ ) (cid:18)(cid:90) S n · L n ( n ⊗ n ) d n (cid:19) e , where the term involving [ n ] × vanishes since its integrand is odd with respect to n (cid:55)→ − n .To compute the second term of this expression we will make use of the followinglemma proved at the end of this section: Lemma 4.17.
For a given matrix L ∈ M , we have (cid:90) S n · L n ( n ⊗ n ) d n = 115 ( L + L T ) + 115 tr( L )Id . Using this lemma we have that y ( θ ) = cos θ L e + (1 − cos θ ) (cid:0) ( L + L T ) + tr( L )Id (cid:1) e = (1 + 4 cos θ ) L e + (1 − cos θ ) (cid:0) L T e + tr( L ) e (cid:1) . Therefore we obtain X = 4 ρd − πZ (cid:90) π (cid:101) m ( θ ) sin ( θ/ y ( θ )] × dθ = 4 ρd − πZ (cid:90) π (cid:101) m ( θ ) sin ( θ/ (cid:0) (1 + 4 cos θ )[ L e ] × + (1 − cos θ )[ L T e + tr( L ) e ] × (cid:1) dθ = ρ (cid:0) C [ L e ] × + C [ L T e + tr( L ) e ] × (cid:1) for C := 4 d − πZ (cid:90) π (cid:101) m ( θ ) sin ( θ/ θ ) dθ,C := 4 d − πZ (cid:90) π (cid:101) m ( θ ) sin ( θ/ − cos θ ) dθ. X = X + X + X + X = C ρ Λ T ∂ t Λ + C [ e × Λ T ∇ x ρ ] × + ρC [ L e ] × + ρC [ L T e + tr( L ) e ] × . In particular Λ X = 0 and from the fact that Λ[ w ] × = [Λ w ] × Λ for any w ∈ R we get X = C ρ∂ t Λ + C [(Λ e ) × ∇ x ρ ] × Λ + C ρ [Λ L e ] × Λ + C ρ [Λ L T e + tr( L )Λ e ] × Λ . (47)Since we have taken L = Λ T D x (Λ)Λ , we get that tr( L ) = tr (cid:0) D x (Λ) (cid:1) = δ x (Λ) and, thanksto (41): [Λ L T e ] × = [ D x (Λ) T Λ e ] × = [( D x (Λ) − [ r x (Λ)] × )Λ e ] × Furthermore, we have [Λ L e ] × Λ = [ D x (Λ)Λ e ] × Λ = (cid:0) (Λ e ) · ∇ x (cid:1) Λ thanks to the definitionof D x given in (40). Finally, inserting these expressions into (47) and dividing by C , weget the equation ρ (cid:16) ∂ t Λ + c (cid:0) (Λ e ) · ∇ x (cid:1) Λ (cid:17) + c [(Λ e ) × ∇ x ρ ] × Λ + c ρ [ − r x (Λ) × (Λ e ) + δ x (Λ) Λ e ] × Λ = 0 , for c = C + C C = (cid:104) θ (cid:105) (cid:101) m ( θ ) sin ( θ/ ,c = C C = d (cid:104) ν ( + cos θ ) − (cid:105) (cid:101) m ( θ ) sin ( θ/ ,c = C C = (cid:104) − cos θ (cid:105) (cid:101) m ( θ ) sin ( θ/ , which ends the proof. Proof of Lemma 4.17.
Denote by I ( L ) the integral that we want to compute I ( L ) := (cid:90) S n · L n ( n ⊗ n ) d n , then, written in components, we have I ( L ) ij = (cid:90) S ( n · L n ) ( e i · n ) ( e j · n ) d n = (cid:26) ( L ij + L ji ) (cid:82) S ( e i · n ) ( e j · n ) d n if i (cid:54) = j (cid:80) k L kk (cid:82) S ( e k · n ) ( e i · n ) d n if i = j = (cid:26) ( L ij + L ji ) if i (cid:54) = j (cid:80) k L kk + L ii if i = j = 115 ( L ij + L ji ) + (cid:26) if i (cid:54) = j (cid:80) k L kk if i = j , i (cid:54) = j, (cid:90) S ( e i · n ) ( e j · n ) d n = 14 π (cid:90) [0 ,π ] × [0 , π ] sin φ cos ψ cos φ dφdψ = 115 ; for k = i, (cid:90) S ( e k · n ) d n = 14 π (cid:90) [0 ,π ] × [0 , π ] cos φ sin φ dφdψ = 15 . Finally, we consider the orthonormal basis given by { Λ e =: Ω , Λ e =: u , Λ e =: v } , where { e , e , e } is the canonical basis of R . We can have an expression of the opera-tors δ x and r x in terms of these unit vectors { Ω , u , v } , which allows to rewrite the evolutionequation of Λ as three evolution equations for these vectors. Proposition 4.18.
We have δ x (Λ) = [(Ω · ∇ x ) u ] · v + [( u · ∇ x ) v ] · Ω + [( v · ∇ x )Ω] · u , (48) r x (Λ) = ( ∇ x · Ω)Ω + ( ∇ x · u ) u + ( ∇ x · v ) v . (49) Consequently, we have the following evolution equations for Ω , u , and v , corresponding to theevolution equation of Λ given in (43) : ρD t Ω + P Ω ⊥ (cid:16) c ∇ x ρ + c ρ (cid:0) ( ∇ x · u ) u + ( ∇ x · v ) v (cid:1)(cid:17) = 0 , (50) ρD t u − ( c u · ∇ x ρ + c ρ ∇ x · u ) Ω + c ρ δ x (Ω , u , v ) v = 0 , (51) ρD t v − ( c v · ∇ x ρ + c ρ ∇ x · v ) Ω − c ρ δ x (Ω , u , v ) u = 0 , (52) where D t := ∂ t + c (Ω · ∇ x ) , and where δ x (Ω , u , v ) is the expression of δ x (Λ) given by (48) .Proof. We first prove (48). We have δ x (Λ) = tr( D x (Λ)) = tr(Λ T D x (Λ)Λ) = (cid:88) k Λ T D x (Λ)Λ e k · e k = (cid:88) k (cid:0) D x (Λ)Λ e k (cid:1) · Λ e k = (cid:88) k [ D x (Λ)Λ e k ] × · [Λ e k ] × = (cid:88) k [ D x (Λ)Λ e k ] × Λ · [Λ e k ] × Λ= (cid:88) k ((Λ e k · ∇ x )Λ) · [Λ e k ] × Λ , thanks to the definition of D x given in (40). Now we use the fact that for two matrices A , B ,we have A · B = tr( A T B ) = (cid:80) i A e i · B e i (half the sum of the scalar products of the33orresponding columns of the matrices A and B ), to get δ x (Λ) = 12 (cid:88) k (cid:88) i (cid:2) (Λ e k · ∇ x ) (Λ e i ) (cid:3) · (cid:2) (Λ e k ) × (Λ e i ) (cid:3) = 12 (cid:16) (Ω · ∇ x ) u · v − ( u · ∇ x )Ω · v − (Ω · ∇ x ) v · u +( v · ∇ x )Ω · u + ( u · ∇ x ) v · Ω − ( v · ∇ x ) u · Ω (cid:17) = [(Ω · ∇ x ) u ] · v + [( u · ∇ x ) v ] · Ω + [( v · ∇ x )Ω] · u . For this last equality we used the fact that · ∇ x )( u · v ) = (Ω · ∇ x ) u · v + (Ω · ∇ x ) v · u since u ⊥ v and analogously for the other components.We proceed next to proving the expression of r x (Λ) given by (49). We first provethat r x (Λ) · Ω = ∇ x · Ω . We have (recall that [ r x (Λ)] × = D x (Λ) − D x (Λ) T and that for all w in R , w · ∇ x Λ = [ D x (Λ) w ] × Λ ): r x (Λ) · Ω = r x (Λ) · ( u × v ) = v · ([ r x (Λ)] × u ) = v · (cid:0) D x (Λ) − D x (Λ) T (cid:1) u = v · D x (Λ) u − u · D x (Λ) v = (Ω × u ) · D x (Λ) u + (Ω × v ) · D x (Λ) u = [ D x (Λ) u ] × Ω · u + [ D x (Λ) v ] × Ω · v = [ D x (Λ) u ] × Λ e · u + [ D x (Λ) v ] × Λ e · v = (cid:0) ( u · ∇ x )Λ e (cid:1) · u + (cid:0) ( v · ∇ x )Λ e (cid:1) · v = (cid:0) ( u · ∇ x )Ω (cid:1) · u + (cid:0) ( v · ∇ x )Ω (cid:1) · v . Since (Ω · ∇ x )Ω is orthogonal to Ω , we therefore get r x (Λ) · Ω = (cid:0) (Ω · ∇ x )Ω (cid:1) · Ω + (cid:0) ( u · ∇ x )Ω (cid:1) · u + (cid:0) ( v · ∇ x )Ω (cid:1) · v = (cid:88) i,k,j Λ ik ∂ i Ω j Λ jk = (cid:88) i,j ∂ i Ω j (cid:88) k Λ ik Λ Tkj = (cid:88) i ∂ i Ω i = ∇ x · Ω , since ΛΛ T = Id (the first line is actually the expression of the divergence of Ω in thebasis { Ω , u , v } ). For the other two components of r x (Λ) , we perform exactly the samecomputations with a circular permutation of the roles of Ω , u , v to get r x (Λ) · u = ∇ x · u and r x (Λ) · v = ∇ x · v . Therefore we obtain (49).Finally we rewrite the equation for Λ as the evolution of the basis { Ω , u , v } . To obtainthe evolution of Λ e k for k = 1 , , , we multiply the Eq. (43) by e k and compute to obtain: ρD t Ω + P Ω ⊥ ( c ∇ x ρ + c ρ r x (Λ)) = 0 ,ρD t u − u · ( c ∇ x ρ + c ρ r x (Λ)) Ω + c ρδ x (Λ) v = 0 ,ρD t v − v · ( c ∇ x ρ + c ρ r x (Λ)) Ω − c ρδ x (Λ) u = 0 , D t = ∂ t + c (Ω · ∇ x ) . To perform the computations we have used, for w = ∇ x ρ or w = r that [ w × Ω] × Ω = − P Ω ⊥ ( w ) and ( w × Ω) × u = ( u · w )Ω since Ω ⊥ u (analogously for v ). From here, using (49) we obtain straightforwardlyEqs. (50), (51), and (52) for Ω , u and v respectively. In the present work we have presented a new flocking model through body attitude co-ordination. We have proposed an Individual Based Model where agents are describedby their position and a rotation matrix (corresponding to the body attitude). From theIndividual Based Model we have derived the macroscopic equations via the mean-fieldequations. We observe that the macroscopic equation gives rise to a new class of models,the Self-Organized Hydrodynamics for body attitude coordination (SOHB). This modeldoes not reduce to the more classical Self-Organized Hydrodynamics (SOH), which is thecontinuum version of the Vicsek model. The dynamics of the SOHB system are more com-plex than those of the SOH ones of the Vicsek model. In a future work, we will carry outsimulations of the Individual Based Model and the SOHB model and study the patternsthat arise to compare them with the ones of the Vicsek and SOH model.Also, there exist yet many open questions on the modelling side. For instance, onecould consider that agents have a limited angle of vision, thus the so-called influencekernel K (see Section 3.1) is not isotropic any more, see [29] for the case of the Vicsekand SOH models. One could also consider a different interaction range for the influencekernel K that may give rise to a diffusive term in the macroscopic equations, see [19].Moreover, in the case of the SOH model, when the coordination frequency and noiseintensity (quantities ν and D in the Individual Based Model (16)-(17)) are functions ofthe flux of the agents, then phase transitions occur at the macroscopic level [19], (see also[4, 6, 20, 44]). An analogous feature is expected to happen in the present case. Finally, onecould think of elaborating on the model by adding repulsive effects at short range andattraction effects at large range.On the analytical side, this model opens also many questions like making Prop. 3.4 rig-orous, which means dealing with Stochastic Differential Equations with non-Lipschitz co-efficients. In the context of the Vicsek model, the global well-posedness has been provenfor the homogeneous mean-field Vicsek equation and also its convergence to the vonMises equilibria in [28], see also [31]; an analogous result for our model will be desirable.The convergence of the Vicsek model to the model which was formally done in [24] hasbeen recently achieved rigorously in [39]. Again, one could also think of generalizingthese results to our case. 35 cknowledgements. P.D. acknowledges support from the Royal Society and the Wolf-son foundation through a Royal Society Wolfson Research Merit Award; the NationalScience Foundation under NSF Grant RNMS11-07444 (KI-Net); the British “Engineeringand Physical Research Council” under grant ref: EP/M006883/1. P.D. is on leave fromCNRS, Institut de Math´ematiques de Toulouse, France.A.F acknowledges support for the ANR projet “KIBORD”, ref: ANR-13-BS01-0004funded by the French Ministry of Research.S.M.A. was supported by the British “Engineering and Physical Research Council” un-der grant ref: EP/M006883/1. S.M.A. gratefully acknowledges the hospitality of CERE-MADE, Universit´e Paris Dauphine, where part of this research was conducted.
A Special Orthogonal Group SO (3) Throughout the text, we used repeatedly the following properties:
Proposition A.1 (Space decomposition in symmetric and antisymmetric matrices) . Denoteby S the set of symmetric matrices in M and by A the set of antisymmetric ones. Then S ⊕ A = M and A ⊥ S . Proof.
For A ∈ M we have A = ( A + A T ) + ( A − A T ) , the first term being symmetricand the second antisymmetric. The orthogonality comes from the properties of the trace,namely tr( A T ) = tr( B ) , and tr( AB ) = tr( BA ) for B ∈ M . Indeed if P ∈ A and S ∈ S then tr( P T S ) = tr( SP T ) = tr( P S T ) = − tr( P T S ) . Hence P · S = tr( P T S ) = 0 . Proposition A.2 (Tangent space to SO (3) ) . For A ∈ SO (3) , denote by T A the tangent spaceto SO (3) at A . Then M ∈ T A if and only if there exists P ∈ A s.t M = AP, or equivalently the same statement with M = P A . Consequently, we have that M ∈ T ⊥ A if and only if there exists S ∈ S s.t. M = AS, or equivalently the same statement with M = SA .Proof. We have that M ∈ T A if and only if there exists a curve Λ( t ) from the neighborhoodof in R to SO (3) such that Λ(0) = A and Λ (cid:48) (0) = M . We then have Id = Λ( t )Λ T ( t ) = ( A + tM + o ( t ))( A T + tM T + o ( t )) = Id + t ( A T M + M T A ) + o ( t ) . So if M ∈ T A , we must have ( A T M + M T A ) = 0 , that is to say that P = A T M ∈ A .Conversely if M = AP with P ∈ A , the solution of the linear differential equa-tion Λ (cid:48) ( t ) = Λ( t ) P with Λ(0) = A is given by Λ( t ) = Ae tP so it is a curve in SO (3) .Indeed we have Λ( t ) T Λ = ( e tP ) T e tP = e tP T e tP = e − tP e tP = Id . Since Λ (cid:48) (0) = AP = M ,36e get that M ∈ T A . The equivalent condition comes from the fact that if M = AP ,with P ∈ A , then M = AP A T A = (cid:101) P A with (cid:101) P ∈ A . Finally the last part is obtained thanksto Prop. A.1) and the fact that the dot product is left (and right) invariant with respectto SO (3) : if B, C ∈ M and A ∈ SO (3) , then AB · AC = tr( B T A T AC ) = B · C . Proposition A.3 (Projection operator on the tangent space) . Let A ∈ SO (3) and M ∈ M (set of square matrices). Let P T A be the orthogonal projection on T A (tangent space at A ), then P T A ( M ) = 12 (cid:0) M − AM T A (cid:1) . (53) Notice that then P T ⊥ A ( M ) = 12 (cid:0) M + AM T A (cid:1) . Proof.
It suffices to verify that the expression given for P T A ( M ) satisfies P T A ( M ) ∈ T A and M − P T A ( M ) ∈ T ⊥ A , that is to say A T P T A M ∈ A and A T ( M − P T A ( M )) ∈ S thanksto Prop. A.2. We have indeed A T ( M − AM T A ) = ( A T M − M T A ) which is clearlyantisymmetric, and A T ( M + AM T A ) = ( A T M + M T A ) which is symmetric.To compute the gradient in SO (3) of a function ψ : SO (3) → R we will consider A ( ε ) adifferentiable curve in SO (3) such that A (0) = A, ddε A ( ε ) (cid:12)(cid:12)(cid:12)(cid:12) ε =0 = δ A ∈ T A then ∇ A ψ ( A ) is the element of T A such that for any δ A ∈ T A , we have lim ε → ψ ( A ( ε )) − ψ ( A ) ε = ∇ A ψ ( A ) · δ A . In particular, one can check that ∇ A ( A · M ) = P T A ( M ) , M ∈ M . (54)We now show that the differential equation corresponding to following this gradient hastrajectories supported on geodesics. Proposition A.4. If B ∈ SO (3) and A ∈ SO (3) , the trajectory of the solution of the differentialequation dAdt = ν ( A · B ) P T A B = ν ( A · B ) ∇ A ( A · B ) with A (0) = A (and with ν smooth andpositive) is supported on a geodesic from A to B .Proof. Indeed, write B T A = exp( θ [ n ] × ) thanks to Rodrigues’ formula (10) with [ n ] × anantisymmetric matrix of unit norm and θ ∈ [0 , π ] . If we set A ( t ) = B exp( θ ( t )[ n ] × ) where θ satisfies the equation θ (cid:48) = − ν ( + cos θ ) sin θ with θ (0) = θ , we get dAdt = B exp( θ ( t )[ n ] × ) θ (cid:48) ( t )[ n ] × = − ν ( + cos θ ( t )) B exp( θ ( t )[ n ] × ) sin θ ( t )[ n ] × . sin θ [ n ] × = (exp( θ [ n ] × ) − exp( θ [ n ] × ) T ) = ( B T A − A T B ) , and A · B = Id · AB T = tr(exp( θ [ n ] × )) = + cos θ thanks to (7). Therefore we obtain dAdt = − ν ( A · B ) A ( B T A − A T B ) = ν ( A · B ) P T A B, thanks to (53) and we have A (0) = A . Since θ ∈ [0 , θ ] (cid:55)→ exp( θ [ n ] × ) is a geodesic be-tween Id and B T A , then θ (cid:55)→ B exp( θ [ n ] × ) is a geodesic between B and A , and the solu-tion A ( t ) is supported on this geodesic. It is also easy to see that, except in the case θ = π or θ = 0 , for which the solution is constant, the function t (cid:55)→ θ ( t ) (solution of the one-dimensional differential equation θ (cid:48) = − ν ( + cos θ ) sin θ ) is positive, decreasing, andconverge exponentially fast to , with an asymptotic exponential rate ν ( ) . Therefore, astime goes to infinity, the trajectory covers the whole geodesic from A to B (excluded).We now turn to the proofs of the expressions of the gradient, the volume form and thedivergence in SO (3) in the so-called Euler axis-angle coordinates, that were presented insection 4.2. Proof of Prop. 4.3: expression of the gradient in SO (3) . Consider a curve in SO (3) given by A ( t ) = exp( θ ( t )[ n ] × ( t )) = Id + sin( θ ( t ))[ n ] × ( t ) + (1 − cos( θ ( t )))[ n ] × ( t ) (following (9)-(8)) with A (0) = A , θ (0) = θ and [ n ] × ( t ) = [ n ( t )] × , n (0) = n . Define: δ A = A (cid:48) (0) ∈ T A ,δ θ = θ (cid:48) (0) ∈ R ,δ n = n (cid:48) (0) ,δ [ n ] × = [ n ] ×(cid:48) (0) = [ δ n ] × . With these notations, for a function f = f ( A ( θ, n )) it holds: ∇ A f · δ A = ∂f∂θ δ θ + ∇ n f · δ n . (55)On the other hand, it holds true that δ A = A [ n ] × δ θ + sin θδ [ n ] × + (1 − cos θ ) (cid:0) [ n ] × δ [ n ] × + δ [ n ] × [ n ] × (cid:1) = A [ n ] × δ θ + AA T (cid:16) sin θδ [ n ] × + (1 − cos θ ) (cid:0) [ n ] × δ [ n ] × + δ [ n ] × [ n ] × (cid:1) (cid:17) = A [ n ] × δ θ + A (cid:0) Id − sin θ [ n ] × + (1 − cos θ )[ n ] × (cid:1)(cid:16) sin θδ [ n ] × + (1 − cos θ ) (cid:0) [ n ] × δ [ n ] × + δ [ n ] × [ n ] × (cid:1) (cid:17) = A [ n ] × δ θ + A (cid:16) sin θδ [ n ] × + (1 − cos θ ) (cid:0) δ [ n ] × [ n ] × − [ n ] × δ [ n ] × (cid:1) (cid:17) = A [ n ] × δ θ + 2 sin( θ/ A (cid:0) cos( θ/
2) [ δ n ] × + sin( θ/
2) [ n × δ n ] × (cid:1) , = A [ n ] × δ θ + L [ n ] × ( δ [ n ] × ) , (56)38here the last line defines L [ n ] × . In the first line, the term in δ θ is obtained by differentiat-ing the exponential form (10) of A ( t ) assuming that [ n ] × ( t ) is constant. The term in δ [ n ] × isobtained by differentiating Rodrigues’ formula (9). To do the computation we have usedRodrigues’ formula (9) to express A T and the facts that [ n ] × = − [ n ] × ; [ n ] × δ [ n ] × [ n ] × = 0 ;and δ [ n ] × [ n ] × − [ n ] × δ [ n ] × = [ δ n × n ] × .In particular notice that { [ n ] × , [ δ n ] × , [ n × δ n ] × } is an orthogonal basis of A (antisym-metric matrices) from which we obtain a basis of T A (by Prop. A.2). So, we just need tocompute the components of ∇ A f in span { A [ n ] × } and span { ( A [ n ] × ) ⊥ } .We will show that the component in span { A [ n ] × } is given by P A [ n ] × ( ∇ A f ) = ∂f∂θ A [ n ] × (57)and the one on span { ( A [ n ] × ) ⊥ } is P ( A [ n ] × ) ⊥ ( ∇ A f ) = 12 sin( θ/ A (cid:0) cos( θ/
2) [ ∇ n f ] × + sin( θ/
2) [ n × ∇ n f ] × (cid:1) . (58)The sum of the two previous expressions gives (23) ( ∇ A f = P A [ n ] × ( ∇ A f )+ P ( A [ n ] × ) ⊥ ( ∇ A f ) ).The component (57) is computed considering the case where δ n = 0 in (56)-(55), so that ∇ A f · δ A = ∇ A f · A [ n ] × δ θ = ∂f∂θ δ θ . Expression (57) is obtained by noticing that ( A [ n ] × ) · ( A [ n ] × ) = [ n ] × · [ n ] × = n · n = 1 (using (13)).To obtain the component (58), consider the case δ θ = 0 in (56) and (55) so that ∇ A f · δ A = ∇ A f · L [ n ] × ( δ [ n ] × ) = ∇ n f · δ n , (59)where L [ n ] × is given in (56).We have that P ( A [ n ] × ) ⊥ ( ∇ A f ) = A [ u ] × for some u ⊥ n . The goal is to compute u as a function of v := ∇ n f . By (59) we have that A [ u ] × · L [ n ] × ( δ [ n ] × ) = ∇ n f · δ n . This implies that θ/
2) [ u ] × · (cid:0) cos( θ/
2) [ δ n ] × + sin( θ/
2) [ n × δ n ] × (cid:1) = v · δ n for all δ n ⊥ n , so (see (13)) we get θ/
2) (cos( θ/ u + sin( θ/ u × n ) · δ n = v · δ n . Since this is true for all δ n orthogonal to n , we get v = 2 sin( θ/
2) (cos( θ/ u + sin( θ/ u × n ) . n × v in terms of u and n × u . After some computationswe finally obtain that u = 12 sin( θ/
2) (cos( θ/ v + sin( θ/ n × v ) . Proof of the volume form, Lemma 4.4.
We denote by g the metric of the Riemannian mani-fold SO (3) associated to the inner product A · B = 12 tr( A T B ) , A, B ∈ SO (3) . The volume form is proportional to (cid:112) det( g ) [30]. We compute the volume form usingspherical coordinates, i.e., we consider the coordinates ( θ, φ, ψ ) ∈ [0 , π ] × [0 , π ] × [0 , π ] .Given the Euler axis-angle coordinates ( θ, n ) we have that n = sin φ cos ψ sin φ sin ψ cos φ . For the spherical coordinate system, we consider the vector field (cid:16) ∂∂θ , ∂∂φ , ∂∂ψ (cid:17) . Denoting Y = ∂A∂θ , Y = ∂A∂φ , Y = ∂A∂ψ , A ∈ SO (3) , we get that ( Y i ) i =1 , , ∈ T A ( SO (3)) forms a basis of vectors fields at A .The metric g is defined as g ij = g ( Y i , Y j ) = tr( Y Ti Y j ) , i, j = 1 , , . We compute nexteach term. Firstly, we know that for a given δ A ∈ T A , there exists δ θ , δ ψ , δ φ ∈ R such that δ A = ∂A∂θ δ θ + ∂A∂φ δ φ + ∂A∂ψ δ ψ and also for a given δ n ∈ T n ( S ) (the tangent plane to the sphere at n ), there exists δ (cid:48) ψ , δ (cid:48) ψ such that δ n = ∂ n ∂φ δ (cid:48) φ + ∂ n ∂ψ δ (cid:48) ψ . Now, following the computation given in (56) we have that, for δ θ = 1 , δ φ = 0 , δ ψ = 0 ∂A∂θ = δ A = A [ n ] × . Now, if δ θ = 0 , δ φ = 1 , δ ψ = 0 then, using that δ n = ∂ n ∂φ we have that ∂A∂φ = δ A = 2 sin( θ/ A (cid:20) R n ,θ/ (cid:18) ∂ n ∂φ (cid:19)(cid:21) × , R n ,θ/ ( v ) = cos( θ/ v + sin( θ/ n × v ) , which corresponds to the rotation of the vector v around n by an angle θ/ (anticlockwise)as long as v · n = 0 . Analogously one can also deduce that ∂A∂ψ = 2 sin( θ/ (cid:20) R n,θ/ (cid:18) ∂ n ∂ψ (cid:19)(cid:21) × . From here, using that (cid:107) ∂ n ∂φ (cid:107) = 1 and (cid:107) ∂ n ∂ψ (cid:107) = sin φ , we conclude that g = ( θ/
2) 00 0 4 sin ( θ/
2) sin φ . Notice that to compute g (cid:16) ∂A∂θ , ∂A∂φ (cid:17) we use that R n ,θ/ (cid:16) ∂ n ∂φ (cid:17) ⊥ n .Finally we have that (cid:112) det ( g ) = 4 sin ( θ/
2) sin φ and therefore (cid:90) SO (3) f ( A ) dA = (cid:90) [0 ,π ] × [0 ,π ] × [0 , π ] (cid:101) f ( θ, φ, ψ )4 sin ( θ/
2) sin φ dθdφdψ = 4 (cid:90) θ ∈ [0 ,π ] (cid:18)(cid:90) [0 ,π ] × [0 . π ] ˜ f ( θ, φ, ψ ) sin φdφdψ (cid:19) sin ( θ/ dθ. The term sin φdφdψ is the volume element in the sphere S so we have that (cid:90) S ˆ f ( θ, n ) d n = (cid:90) [0 ,π ] × [0 , π ] ˜ f ( θ, φ, ψ ) sin φdφdψ. Therefore, the volume element corresponding to the Euler axis-angle coordinates is pro-portional to sin ( θ/ dθd n . Since the volume element is defined up to a constant, wechoose the constant c such that (cid:90) π c sin ( θ/ dθ = 1 , i.e., c = 2 /π . In conclusion, the volume element in the Euler axis-angle coordinates corre-sponds to π sin ( θ/ dθd n . roof of divergence formula, Prop. 4.5. We compute the divergence by duality of the gradi-ent, Prop. 4.3. Let f = f ( A ) be a function and consider (cid:90) SO (3) ∇ A · B ( A ) f ( A ) dA = − (cid:90) SO (3) B ( A ) · ∇ A f ( A ) dA = − (cid:90) (0 ,π ) × S W ( θ ) b ( θ, n ) ∂ θ f ( θ, n ) dθd n − (cid:90) (0 ,π ) × S W ( θ )2 sin( θ/ v ( θ, n ) · (cid:0) cos( θ/ ∇ n f ( n , θ ) + sin( θ/ n × ∇ n f ( n , θ ) (cid:1) dθd n = (cid:90) (0 ,π ) × S f ( θ, n )sin ( θ/ ∂ θ (cid:0) sin ( θ/ b ( θ, n ) (cid:1) W ( θ ) dθd n + (cid:90) (0 ,π ) × S f ( θ, n )2 sin( θ/ ∇ n · (cid:0) v ( θ, n ) cos( θ/
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