Poisson brackets for the dynamically coupled system of a free boundary and a neutrally buoyant rigid body in a body-fixed frame
PPoisson brackets for the dynamically coupled system ofa free boundary and a neutrally buoyant rigid body in abody-fixed frame.
Banavara N. Shashikanth ∗ Abstract
The fully coupled dynamic interaction problem of the free surface of anincompressible fluid and a rigid body beneath it, in an inviscid, irrotationalframework and in the absence of surface tension, is considered. Evolutionequations of the global momenta of the body+fluid system are derived. Itis then shown that, under fairly general assumptions, these evolution equa-tions combined with the evolution equation of the free-surface, referred to abody-fixed frame, is a Hamiltonian system. The Poisson brackets of the sys-tem are the sum of the canonical Zakharov bracket and the non-canonicalLie-Poisson bracket. Variations are performed consistent with the mixedDirichlet-Neumann problem governing the system. ∗ Mechanical and Aerospace Engineering Department, MSC 3450, PO Box 30001, New Mex-ico State University, Las Cruces, NM 88003, USA. E-mail:[email protected] a r X i v : . [ phy s i c s . f l u - dyn ] J a n ontents Introduction.
Water wave dynamics may be described as both a classical and modern topic ofresearch in fluid dynamics. Lamb’s book [1] (Chapter VIII on ‘Tidal Waves’) con-tains references to several important papers of the classical oeuvre. The relationof the Korteweg-de Vries (KdV) equation to integrability and soliton theories hasbeen a relatively more recent development [2] but has spawned a very large andactive field of research [3]. Johnson’s book [4] provides a nice introduction toboth classical and modern aspects of the subject of water waves.Theoretical investigations of rigid bodies in water also has a rich history, inparticular, the topic of bodies floating on the water surface; some landmark pub-lications being [5, 6, 7, 8, 9]. Most approaches to these problems–with applica-tions to ship and marine vehicle motions— are typically in a linearized frameworkor/and with the body executing prescribed motions, or with the objective of deriv-ing expressions for the hydrodynamic loads on the body [10]. The fully couplednonlinear problem has not been as well investigated theoretically. Most investiga-tions of the nonlinear problem are numerical; see, for example, [11] and referencestherein.Papers on Lagrangian and Hamiltonian formulations of the coupled problemare even fewer. Miloh presented Lagrangian formulations in cases where the bodyis executing oscillations or is set in impulsive motion, on or below the free surface[12, 13]. Miloh also derived expressions for the hydrodynamical reaction forceson the body. To the best of the author’s knowledge, the Hamiltonian formulationof the fully coupled system was first presented by van Daalen, van Groesen andZandbergen [14, 15]. Working in a spatially fixed frame and without any explicitreference to Poisson brackets, they showed that the combined system is a canoni-cal Hamiltonian system, with the Hamiltonian being the sum of the fluid and bodykinetic+potential energies.In this paper, the Poisson brackets of the combined system in a body-fixedframe are presented for the case when the body is completely beneath the free sur-face and surface tension is ignored. It is shown that the brackets are the sum of theZakharov bracket [16], written in a body-fixed frame, and the non-canonical Lie-Poisson bracket [17]. The paper is organized as follows. In Section 2, the setupof the physical problem and some assumptions made are described. The fluid do-main has a flat bottom stationary boundary that extends to infinity in all horizontaldirections. In Section 3, the evolution equations for the combined momenta of thesystem are presented. First these are derived in a spatially-fixed frame, following atraditional momentum balance analysis, without any assumptions on the buoyancy3f the rigid body. As one would expect, the spatial momenta are not conserved.Conservation is obtained by moving the bottom boundary to infinity and assumingneutral buoyancy. The momentum equations are then transformed to a body-fixedframe. The details of the momentum balance analysis are relegated to Appen-dices A and B. Section 4 is the main section of the paper in which the variablesin the body fixed frame are presented, and it is shown how the variations can beperformed consistent with the mixed Neumann-Dirchlet boundary-value problem.The equations of the system are derived and shown to be Hamiltonian relative tothe brackets above. Section 5 has some future directions for research.Apart from standard assumptions such as the far-field decay rates of the ve-locity potential function, and existence and uniqueness of solutions of the mixedDirichlet-Neumann problem, certain other assumptions are made in the paper. Thefour main ones are those given by equations (6), (7) and (9), and the invertibilityof the mass matrix given by (49).
A schematic sketch of the system being considered is shown in Figures 1 and 2.Introduce the notations Σ f for the free boundary of the incompressible fluid, Σ b ≡ ∂B for the fluid-body boundary and S for the stationary flat bottom (spanning thehorizontal directions). Denote the half-space bounded by S by R . The non-compact fluid domain D ⊂ R therefore has a boundary which is the disjointunion of three pieces ∂D = Σ b ∪ Σ f ∪ S . The uniform density fields of the rigidbody and the fluid are denoted by ρ b and ρ f , respectively. Convention for unit normals.
Before proceeding, the convention for the unitnormal field on the different boundary components is established. n f points away from D , n b points into D (and away from the body B ) and n s also points into D .The function Φ satisfies the following mixed Dirichlet-Neumann problem ateach time instant t : ∇ Φ = 0 in D, Φ | Σ f prescribed , ∇ Φ · n b | Σ b = U · n b , ∇ Φ · n s | S = 0Φ → x, y → ±∞ (1)where U is the rigid body velocity field in a spatially-fixed frame xyz . Both Φ | Σ f and U are prescribed at initial time but fixed at all later times by the evolutionequations. 4 C R S b f x y z B Figure 1: Schematic perspective of a rigid body B beneath the free surface Σ f ofwater. The bottom flat surface S is shown by the dashed rectangle. Both Σ f and S extend to infinity in the x and y (horizontal) directions. In the text, the origin ofthe spatial frame xyz is located at the center of the disc C R ⊂ S . Far-field assumptions.
Far from the body, it will be assumed that the fluid sur-face is undisturbed and has a constant elevation η . The velocity field also goes tozero. Since the fluid flow field is vorticity-free, the decay rate of Φ is assumed tobe [18] Φ ∼ / | r | (2)Applying B.E., ∂ Φ ∂t + pρ f + ∇ Φ · ∇ Φ2 + gz = f ( t ) to a point on the free surface at far infinity, obtain f ( t ) = p atm ρ f + gη = constant D R S f η W x z g R D R Figure 2: A vertical slice of the setup in Figure 1. The body-fixed frame is shown.Using the above, B.E. then gives that the pressure at any point as p = p atm + ρ f g ( η − z ) − ρ f (cid:18) ∂ Φ ∂t + ∇ Φ · ∇ Φ2 (cid:19) Using (2) and cylindrical coordinates ( R, θ, z ) , one can write the far-field pressuredistribution as p ( R, θ, z, t ) = p atm + ρ f g ( η − z ) + A ( R, θ, z, t ) , A ( R, θ, z, t ) = O (1 /R ) (3)At the flat bottom S ( z = 0) , one obtains p = p atm + ρ f gη − ρ f (cid:18) ∂ Φ ∂t + ∇ Φ · ∇ Φ2 (cid:19) | S (4)6nd at the free surface Σ f ( z = η ) , one obtains p = p atm + ρ f g ( η − η ) − ρ f (cid:18) ∂ Φ ∂t + ∇ Φ · ∇ Φ2 (cid:19) | Σ f (5)Note that surface tension effects are absent for the undisturbed surface.Moreover, the waves at the free surface are assumed to satisfy the followingtwo conditions: ρ f g (cid:90) S ( η − η ) ν s = 0 (6) ρ f g (cid:90) S r × ( η − η ) ν s = 0 , (7)where r is a position vector (defined later). The bottom flat surface S has volume(area) form ν s and is taken as the datum for the potential energies, and η is the free-surface elevation with respect this datum. The first equation may be obviouslyinterpreted as a conservation of mass condition satisfied by the waves on the freesurface. The second equation may be viewed as the angular analog of (6). Itcould perhaps be interpreted as a zero global moment, about the gravity axis, dueto the waves. Total Energy.
Ignoring surface tension and surface energy, the total energy isthe kinetic plus potential energy of the fluid+body system,
K.E. + P.E. = ρ f (cid:18)(cid:90) D (cid:104)(cid:104)∇ Φ , ∇ Φ (cid:105)(cid:105) µ + g (cid:18)(cid:90) S η ν s − (cid:90) S η ν s (cid:19)(cid:19) + ρ b (cid:104)(cid:104) U, U (cid:105)(cid:105) R + ( ρ b − ρ f ) V B gz c , = ρ f (cid:32)(cid:90) Σ f Φ ∇ Φ · n f ν − (cid:90) Σ b Φ ∇ Φ · n b ν + g (cid:18)(cid:90) S η ν s − (cid:90) S η ν s (cid:19)(cid:19) + 12 (cid:104)(cid:104) ( V, Ω) , ρ b M b · ( V, Ω) (cid:105)(cid:105) + ( ρ b − ρ f ) V B gz c , (8)where V B is the volume of the rigid body, U ≡ ( V, Ω) is the vector of rigid bodyvelocities, M b is the body mass tensor and z c is the elevation of the body centroidwith respect to the datum S . 7 emark. Note that the potential energy of the fluid is relative to the undisturbedstate. This is done to subtract the infinite potential energy of the fluid in the undis-turbed state which is due to the unbounded domain in the horizontal directions,and irrespective of the location of the datum. However, subtracting the rest po-tential energy still does not guarantee that the fluid potential energy is finite. Anadditional assumption is needed about the rate at which η decays in the horizontaldirections: (cid:90) S η ν s − (cid:90) S η ν s < ∞ (9) Fluid Momentum.
Denote by P T ≡ ( L T , A T ) the momenta of the total system,i.e. body+fluid system, and by P T f ≡ ( L T f , A
T f ) the contribution to these fromthe fluid. Using well-known vector identities (see, for example, [19] ) allow thelatter to be written as follows. Considering the linear momentum first, L T f := lim R →∞ ρ f (cid:90) D R ∇ Φ µ, = ρ f (cid:90) ∂D R Φ n ν, [n outward] , = ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + lim R →∞ ρ f (cid:18)(cid:90) Σ R Φ n f ν − (cid:90) C R Φ n s ν + (cid:90) W Φ e R ν (cid:19) , where, as in Figures 1 and 2, D R ⊂ D is a a vertical cylindrical domain of (vary-ing) height η and radius R , and bounding surfaces Σ b , Σ R ⊂ Σ f , C R ⊂ S andlateral surface W (with outward normal in the radial direction e R ). C R is a cir-cular disc of radius R and r is position vector measured from the origin of thespatially-fixed frame which is taken, wlog, to lie at the center of C R . To avoidnotational clutter, the same symbol ν is used to denote the volume form on anybounding surface.For the first integral on the right, use is made of the vector identity (see SSKM8r Saffman) to write it in a different way from the other terms, (cid:90) A r × ( n × ∇ Φ) ν = − (cid:90) A Φ n ν, (10) Remark.
Note that as R → ∞ , W recedes uniformly from the body whereas Σ R and C R do not.From the Far-Field Assumptions Φ → constant as R → ∞ , so that the lastintegral on the right vanishes in this limit, and the expression reduces to L T f = ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + lim R →∞ ρ f (cid:18)(cid:90) Σ R Φ n f ν − (cid:90) C R Φ n s ν (cid:19) , (11)Similarly, considering angular momentum of the flow about the origin of thespatially-fixed frame, A T f := lim R →∞ ρ f (cid:90) D R r × ∇ Φ µ, = lim R →∞ ρ f (cid:90) ∂D R r ( n × ∇ Φ) ν, [n outward] , = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + lim R →∞ ρ f (cid:18)(cid:90) Σ R r × Φ n f ν − (cid:90) C R r × Φ n s ν + (cid:90) W r × Φ e R ν (cid:19) , where for the last three integrals use has been made of the vector identity (cid:90) ∂A r ( n × ∇ Φ) ν = (cid:90) ∂A r × Φ n ν, (12)where ∂A denotes the smooth boundary of a domain A ⊂ R . As per the As-sumption above, r = Re R on W , so that the integral over W vanishes, leaving A T f = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + lim R →∞ ρ f (cid:18)(cid:90) Σ R r × Φ n f ν − (cid:90) C R r × Φ n s ν (cid:19) , (13)9 otal Momentum. The total body+fluid momenta in a spatially-fixed frame is P T ≡ ( L T , A T ) = ( L T f + L b , A T f + A b ) , (14)where L b = M b V, A b = r c × M b V + I Ω (15)are the rigid body momenta, with r c being the position vector of the center ofmass in a spatially-fixed frame and I its moment of inertia tensor in a principal-axes frame.Carrying out a traditional momentum analysis, details of which are presentedin the Appendix A, one then obtains the following evolution equations in a spatially-fixed frame, d L dt = − ρ f (cid:90) S ∇ Φ · ∇ Φ2 n s ν + ( ρ f − ρ b ) g V B k, (16) d A dt = (cid:90) S r × (cid:18) − ρ f ∇ Φ · ∇ Φ2 (cid:19) ν + r c × ( ρ f − ρ b ) g V B k, (17)where L = ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + L b + ρ f (cid:90) Σ f Φ n f ν, A = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + A b + ρ f (cid:90) Σ f r × Φ n f ν The contributions to the momentum change come from the presence of the fixedsurface S and the lack of neutral buoyancy of the rigid body. The contributionof S to the momentum change is represented by the integrals in (16) and (17).From the Far-field assumptions, it is easily seen that these integrals go to zero as S → z = −∞ . To obtain global momentum conservation, one therefore needs tomake the following assumptions: Assumptions. (a) The surface S is at z = −∞ and (b) the rigid body is neutrallybuoyant. 10 pecial case. Under the above assumptions, d L dt = 0 (18) d A dt = 0 (19)Henceforth, the paper will only deal with this special case. Body-fixed frame.
Equations (18) and (19) are now transformed to a body-fixedframe, with origin at the center of mass of the body and axes aligned with theprincipal axes, using r = R ( t ) · l + r c ( t ) (20)and the general rule for transforming any vector a ∈ R located at r in thespatially-fixed frame a ( r ) = R ( t )¯ a ( l ) , (21)where ¯ a is the vector located at l in the body-fixed frame. Using this (20) can alsobe written as r = R ( t ) ( l + ¯ r c ) Real-valued functions transform as Φ( r, t ) = Φ (cid:48) ( l, t ) , (22)etc. It follows that ∇ Φ( r, t ) = R ( t ) ∇ b Φ (cid:48) ( l, t ) , (23)etc . Note that under this orthogonal transformation volume forms are preserved.The details of the transformation are presented in Appendix B. The equationstake the form d L dt + ¯Ω × L = 0 , (24) d A dt + ¯Ω × A + ¯ V × L = 0 (25)where L = ρ f (cid:90) Σ b l × (¯ n b × ∇ b Φ (cid:48) ) ν + ¯ L b + ρ f (cid:90) Σ f Φ (cid:48) ¯ n f ¯ ν, (26) A = − ρ f (cid:90) Σ b l (¯ n b × ∇ b Φ (cid:48) ) ν + ¯ I Ω + ρ f (cid:90) Σ f l × Φ (cid:48) ¯ n f ¯ ν, (27)11 Variations and Hamiltonian structure in the body-fixed frame.
The total energy function (8) is now written in terms of the variables in the body-fixed frame, keeping in mind the special case( (18) and (19)) and the associatedassumptions.Consider the kinetic energy terms first. For h ≡ ( R, r c ) , let Ψ h : R → R bethe map defined by (20). ψ h ( l ) = r. Using relations (21), (22) and (23), (cid:90) Σ f Φ( r ) ∇ Φ( r ) · n f ( r ) ν = (cid:90) Ψ h ( ¯Σ f ) Φ( r ) ∇ Φ( r ) · n f ( r ) ν, = (cid:90) ¯Σ b Φ( ψ h ( l )) ∇ Φ( ψ h ( l )) · n b ( ψ h ( l )) ¯ ν, [change of variables Theorem] , = (cid:90) ¯Σ b Φ (cid:48) ( l ) R ( t ) ∇ b Φ (cid:48) ( l ) · R ( t )¯ n b ( l ) ¯ ν, [using (21) and (23)] , = (cid:90) ¯Σ b Φ (cid:48) ( l ) ∇ b Φ (cid:48) ( l ) · ¯ n b ( l ) ¯ ν, The other fluid kinetic energy term in (8) transforms in a similar way.Next, the potential energy term in (8) has to be written in a body-fixed frame.For this first write the potential energy term in its original form, ρ f g (cid:90) S (cid:90) η z dzν s ≡ (cid:90) D ∪ B f µ, where f : D ∪ B → R and µ = dzν s . Think of D ∪ B as transformed domainfrom the domain in the body-fixed frame, which is denoted by ¯ D ∪ ¯ B , under themap ψ h . Using the change of variables theorem again, (cid:90) D ∪ B f ( r ) µ = (cid:90) ψ h ( ¯ D ∪ ¯ B ) f ( r ) µ, = (cid:90) ¯ D ∪ ¯ B f ( ψ h ( l )) ¯ µ.
12t is not hard to see that f ◦ ψ h denotes the perpendicular distance from the trans-formed surface ¯ S in the body-fixed frame, and so the potential energy term in thebody-fixed frame can be written as ρ f g (cid:90) ¯ S ¯ η ¯ ν s , where ¯ η is the value of f ◦ ψ h for a point on the free surface. The rest potentialenergy transforms in a similar way.The total energy, for the neutrally buoyant case (with ρ f = ρ b = ρ ), referredto the body fixed frame, is therefore K.E. + P.E. = ρ (cid:32)(cid:90) ¯Σ f Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν (cid:33) + 12 ρg (cid:18)(cid:90) ¯ S ¯ η ¯ ν s − (cid:90) ¯ S ¯ η ¯ ν s (cid:19) + 12 (cid:10)(cid:10) ( ¯ V , ¯Ω) , ρM b · ( ¯ V , ¯Ω) (cid:11)(cid:11) , (28)Now write this using the variables ( L , A ) . To do this, first rewrite (26) and (27)using (10) and (12) as ( L , A ) = ρM b · (cid:0) ¯ V , ¯Ω (cid:1) + ρ ¯ P f (29)where ¯ P f ≡ (cid:0) ¯ L f , ¯ A f (cid:1) := − (cid:90) ¯Σ b Φ (cid:48) ¯ n b ¯ ν (cid:124) (cid:123)(cid:122) (cid:125) ¯ L f + (cid:90) ¯Σ f Φ (cid:48) ¯ n f ¯ ν (cid:124) (cid:123)(cid:122) (cid:125) ¯ L f , − (cid:90) ¯Σ b l × Φ (cid:48) ¯ n b ¯ ν (cid:124) (cid:123)(cid:122) (cid:125) ¯ A f + (cid:90) ¯Σ f l × Φ (cid:48) ¯ n f ¯ ν (cid:124) (cid:123)(cid:122) (cid:125) ¯ A f , (30)Inverting, obtain (cid:0) ¯ V , ¯Ω (cid:1) = ( M b ) − · (cid:20) ρ ( L , A ) − ¯ P f (cid:21) , (31)13 .1 The variables and the variations in the body-fixed frame. Consider now the variables (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1) , where φ (cid:48) f := Φ (cid:48)| ¯Σ f and ¯Σ f , rather than ¯ η , is the variable that will be used to characterize the freesurface. As in [20] and [21], view ¯Σ f as the image of a smooth embedding of areference configuration of the free surface Σ which could, without loss of gen-erality, be taken as the undisturbed surface. Note that, as in [21], the variationsin ¯Σ f are those that are normal to the fluid surface, and will be denoted either bythe vector (cid:0) δ ¯Σ f (cid:1) n or its magnitude (cid:0) δ ¯Σ f (cid:1) n := (cid:0) δ ¯Σ f (cid:1) n · ¯ n f . These variations arerelated to δ ¯ η by (cid:0) δ ¯Σ f (cid:1) n = δ ¯ η (cid:0) ¯ k · ¯ n f (cid:1) , ⇒ (cid:0) δ ¯Σ f (cid:1) n ¯ ν = δ ¯ η ¯ ν s , (32)where ¯ k is the unit vector k in the body-fixed frame.In the body-fixed frame, the mixed Dirichlet-Neumann problem of (1) takesthe form ∇ b Φ (cid:48) = 0 in ¯ D, Φ (cid:48) | ¯Σ f prescribed , ∇ b Φ (cid:48) · ¯ n b | ¯Σ b = ¯ U · ¯ n b , Φ (cid:48) → x l , y l → ±∞ , z l → −∞ (33)where ( x l , y l , z l ) = ψ − h ( x, y, z ) .Examining relations (29) and (30) it is seen that, due to the mixed Dirichlet-Neumann problem, ¯ P f is not independent of the rigid body’s velocities ( ¯ V , ¯Ω) .Otherwise, variations in (cid:0) ¯Σ f , φ (cid:48) f (cid:1) could be performed keeping ( L , A ) constantand vice-versa, by making appropriate variations in (cid:0) ¯ V , ¯Ω (cid:1) . Indeed such is thecase in the problem of a rigid body dynamically interacting with singular vortices, op. cit. The variations therefore need to be performed more carefully and this warrantsa discussion.Consider the following linear maps. First, L K : R → C ∞ (cid:0) R , R (cid:1) Kirchhoff problem in R : ∇ b Φ (cid:48) = 0 in R \ B, ∇ b Φ (cid:48) · ¯ n b | ¯Σ b = ¯ U · ¯ n b , Φ (cid:48) → l → ∞ (34)As is well-known in the Kirchhoff problem [22, 23], Φ (cid:48) ( l, t ) = (Ψ (cid:48) ( l ) , ζ (cid:48) ( l )) · ( ¯ V ( t ) , ¯Ω( t )) , (35)where Ψ (cid:48) ( l ) ≡ (cid:0) Ψ (cid:48) x ( l ) , Ψ (cid:48) y ( l ) , Ψ (cid:48) z ( l ) (cid:1) , ζ (cid:48) ( l ) ≡ (cid:0) ζ (cid:48) x ( l ) , ζ (cid:48) y ( l ) , ζ (cid:48) z ( l ) (cid:1) ) (36)are 3-vectors each of whose components satisfy the following Neumann problems ∇ b Ψ (cid:48) x = 0 in ¯ D, ∇ b Ψ (cid:48) x · ¯ n b | ¯Σ b = i · ¯ n b , Ψ (cid:48) x | ∞ = constant (37) ∇ b ζ (cid:48) x = 0 in ¯ D, ∇ b ζ (cid:48) x · ¯ n b | ¯Σ b = ( i × l ) · ¯ n b , ζ (cid:48) x | ∞ = constant (38)and similarly in the y l - and z l -directions (of the body-fixed frame). Next, consider the linear map L S : C ∞ (cid:0) ¯Σ f , R (cid:1) → C ∞ (cid:0) ¯ D, R (cid:1) , associated with the following mixed Dirichlet-Neumann problem for a stationarybody: ∇ b Φ (cid:48) = 0 in ¯ D, Φ (cid:48) | ¯Σ f prescribed , ∇ b Φ (cid:48) · ¯ n b | ¯Σ b = 0 , Φ (cid:48) → x l , y l → ±∞ , z l → −∞ (39)Each of these linear maps further gives rise to other linear maps by restricting tothe boundaries of ¯ D : L K,b : R → C ∞ (cid:0) ¯Σ b , R (cid:1) , (40) L K,f : R → C ∞ (cid:0) ¯Σ f , R (cid:1) , (41) L S,b : C ∞ (cid:0) ¯Σ f , R (cid:1) → C ∞ (cid:0) ¯Σ b , R (cid:1) (42)Finally, consider the linear maps I B : C ∞ (cid:0) ¯Σ b , R (cid:1) → R , I F : C ∞ (cid:0) ¯Σ f , R (cid:1) → R , (43) To avoid notation clutter, Ψ (cid:48) x etc. is used instead of Ψ (cid:48) x l etc. (cid:16)(cid:82) ¯Σ b Φ (cid:48) ¯ n b ¯ ν, (cid:82) ¯Σ b l × Φ (cid:48) ¯ n b ¯ ν (cid:17) and (cid:16)(cid:82) ¯Σ f Φ (cid:48) ¯ n f ¯ ν, (cid:82) ¯Σ f l × Φ (cid:48) ¯ n f ¯ ν (cid:17) ,respectively. Restricting to the ‘linear’ and ‘angular’ components, respectively,each of these maps can also be identified with a pair of maps: I B ≡ ( I B , I B ) and I F ≡ ( I F , I F ) .With these maps in place, the arbitrary and independent variation of each vari-able in the set (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1) will now be discussed.1. An arbitrary variation δ o ( L , A ) , with δφ (cid:48) f = (cid:0) δ ¯Σ f (cid:1) n =0. Here δ o denotesthat only one in the pair ( L , A ) is varied while the other is kept fixed. Toachieve this requires an appropriate variation δ o ( ¯ V , ¯Ω) (for otherwise, Φ (cid:48) inthe domain ¯ D remains unchanged and hence also ¯ P f , making it impossible,by (29), to achieve the variation δ o ( L , A ) ). But this induces a variation δ Φ (cid:48) in ¯ D , including at the boundaries, given by δ Φ (cid:48)| ¯Σ b = δ Φ (cid:48) K + δ Φ (cid:48) S ,δ Φ (cid:48)| ¯Σ f = δφ (cid:48) f = 0 where δ Φ (cid:48) K = L K,b (cid:0) δ o ( ¯ V , ¯Ω) (cid:1) ,δ Φ (cid:48) S = L S,b ◦ (cid:0) − L K,f (cid:0) δ o ( ¯ V , ¯Ω) (cid:1)(cid:1) , the minus sign ensuring that the constraint δφ (cid:48) f = 0 is respected. Generallytherefore, one has an induced variation δ ¯ P f , which is given by δ ¯ P f = δ o (cid:0) ¯ L f , ¯ A f (cid:1) = I B ◦ ( L K,b − L S,b ◦ L K,f ) (cid:0) δ o ( ¯ V , ¯Ω) (cid:1) , the map I B acting through one of its components I B or I B . The arbi-trary variation δ o ( L , A ) is therefore possible only if the variation δ o ( ¯ V , ¯Ω) is chosen such that the equation δ o ( L , A ) = ρM b · δ o (cid:0) ¯ V , ¯Ω (cid:1) + ρδ o (cid:0) ¯ L f , ¯ A f (cid:1) = ρ ( M b + I B ◦ ( L K,b − L S,b ◦ L K,f )) · δ o (cid:0) ¯ V , ¯Ω (cid:1) (44)is satisfied. To show that it is possible to choose such a δ o ( ¯ V , ¯Ω) , it isnecessary and sufficient that the linear map M b + I B ◦ ( L K,b − L S,b ◦ L K,f ) is invertible. 16. Next, an arbitrary variation δφ (cid:48) f , with δ ( L , A ) = (cid:0) δ ¯Σ f (cid:1) n =0. The variationsinduced are similar to case 1, but there is an extra term due to the imposedvariation δφ (cid:48) f . The boundary variations are therefore given by δ Φ (cid:48)| ¯Σ b = δ Φ (cid:48) K + δ Φ (cid:48) S ,δ Φ (cid:48)| ¯Σ f = δφ (cid:48) f (45)where δ Φ (cid:48) K = L K,b (cid:0) δ ( ¯ V , ¯Ω) (cid:1) ,δ Φ (cid:48) S = L S,b ◦ (cid:0) δφ (cid:48) f − L K,f (cid:0) δ ( ¯ V , ¯Ω) (cid:1)(cid:1) , The induced variation δ ¯ P f is now given by δ ¯ P f = δ (cid:0) ¯ L f , ¯ A f (cid:1) , = δ (cid:0) ¯ L f , ¯ A f (cid:1) + δ (cid:0) ¯ L f , ¯ A f (cid:1) = I B ◦ ( L K,b − L S,b ◦ L K,f ) (cid:0) δ ( ¯ V , ¯Ω) (cid:1) + I B ◦ L S,b (cid:0) δφ (cid:48) f (cid:1) + I F (cid:0) δφ (cid:48) f (cid:1) In such a case again, a choice of the variation δ ( ¯ V , ¯Ω) is required such thatthe equation ρM b · δ (cid:0) ¯ V , ¯Ω (cid:1) + ρδ (cid:0) ¯ L f , ¯ A f (cid:1) ⇒ I B ◦ L S,b (cid:0) δφ (cid:48) f (cid:1) + I F (cid:0) δφ (cid:48) f (cid:1) = ( M b + I B ◦ ( L S,b ◦ L K,f − L K,b )) (cid:0) δ ( ¯ V , ¯Ω) (cid:1) (46)is satisfied. As in case 1, to show such a δ ( ¯ V , ¯Ω) exists for any choice of δφ (cid:48) f requires the map M b + I B ◦ ( L S,b ◦ L K,f − L K,b ) to be invertible.3. Finally, an arbitrary variation (cid:0) δ ¯Σ f (cid:1) n , with δ ( L , A ) = δφ (cid:48) f =0. The meaningof δφ (cid:48) f =0 is explained in [21]. φ (cid:48) f is viewed as a function of the referenceconfiguration. There is thus an induced change δ Φ (cid:48) | ¯Σ f given by δ Φ (cid:48) | ¯Σ f = −∇ b Φ (cid:48) | ¯Σ f · (cid:0) δ ¯Σ f (cid:1) n (47)There is a perturbed fluid domain ˜ D in this case and, generally speaking, O ( (cid:15) ) -sized subdomains of ˜ D could lie outside ¯ D . Considerations of thesesubdomains is, however, not necessary to compute the variational derivativewith respect to (cid:0) δ ¯Σ f (cid:1) n .This case is, therefore, treated just like case 2 on the unperturbed domain,with equation (45) replaced by (47). The equation determining the choice of δ ( ¯ V , ¯Ω) is given by (46) with δφ (cid:48) f replaced by the δ Φ (cid:48) | ¯Σ f of equation (47).17he map I B ◦ ( L K,b − L S,b ◦ L K,f ) which appears in all three cases above, is now examined.First, it should be obvious from the definitions of the maps (35), (36), (40)and (43) that I B ◦ L K,b : R → R is nothing but the symmetric added massmatrix: M a := (cid:82) ¯Σ b Ψ (cid:48) x n x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y n x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z n x ¯ ν (cid:82) ¯Σ b ζ (cid:48) x n x ¯ ν (cid:82) ¯Σ b ζ (cid:48) y n x ¯ ν (cid:82) ¯Σ b ζ (cid:48) z n x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) x n y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y n y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z n y ¯ ν (cid:82) ¯Σ b ζ (cid:48) x n y ¯ ν (cid:82) ¯Σ b ζ (cid:48) y n y ¯ ν (cid:82) ¯Σ b ζ (cid:48) z n y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) x n z ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y n z ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z n z ¯ ν (cid:82) ¯Σ b ζ (cid:48) x n z ¯ ν (cid:82) ¯Σ b ζ (cid:48) y n z ¯ ν (cid:82) ¯Σ b ζ (cid:48) z n z ¯ ν (cid:82) ¯Σ b Ψ (cid:48) x ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b ζ (cid:48) x ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b ζ (cid:48) y ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b ζ (cid:48) z ( l × ¯ n b ) x ¯ ν (cid:82) ¯Σ b Ψ (cid:48) x ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b ζ (cid:48) x ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b ζ (cid:48) y ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b ζ (cid:48) z ( l × ¯ n b ) y ¯ ν (cid:82) ¯Σ b Ψ (cid:48) x ( l × ¯ n b ) z ¯ ν (cid:82) ¯Σ b Ψ (cid:48) y ( l × ¯ n b ) z ¯ ν (cid:82) ¯Σ b Ψ (cid:48) z ( l × ¯ n b ) z ¯ ν (cid:82) ¯Σ b ζ (cid:48) x ( l × ¯ n b ) z ¯ ν (cid:82) ¯Σ b ζ (cid:48) y ( l × ¯ n b ) z ¯ ν (cid:82) ¯Σ b ζ (cid:48) z ( l × ¯ n b ) z ¯ ν Recall, that the symmetry is shown using the boundary conditions in (37) and (38)and invoking the following well-known reciprocity result for any two harmonicfunctions f and g in R satisfying the Kirchhoff problem: (cid:90) ¯Σ b ( f ∇ b g · ¯ n b − g ∇ b f · ¯ n b ) ¯ ν = (cid:90) R (cid:0) f ∇ b g − g ∇ b f (cid:1) µ, (48) ⇒ (cid:90) ¯Σ b ( f ∇ b g · ¯ n b − g ∇ b f · ¯ n b ) ¯ ν = 0 Next, consider the map −I B ◦ ( L S,b ◦ L K,f ) : R → R . Referring to (35), (36), (39), (40)and (43), this map is given by a coupling matrix, denoted by M c , whose elementsare the elements of M a replaced in the following manner: (cid:90) ¯Σ b Ψ (cid:48) x n x ¯ ν → − (cid:90) ¯Σ b L S,b (cid:16) Ψ (cid:48) x | ¯Σ f (cid:17) n x ¯ ν, (cid:90) ¯Σ b Ψ (cid:48) x ( l × ¯ n b ) x ¯ ν → − (cid:90) ¯Σ b L S,b (cid:16) Ψ (cid:48) x | ¯Σ f (cid:17) ( l × ¯ n b ) x ¯ ν, etc. Now from (39), (cid:90) ¯Σ b ∇ b (cid:16) L S (cid:16) Ψ (cid:48) x | ¯Σ f (cid:17)(cid:17) · ¯ n b ¯ ν = 0 , etc. Use this boundary condition in the identity (53), with f := Ψ (cid:48) x + L S (cid:16) Ψ (cid:48) x | ¯Σ f (cid:17) , g := Ψ (cid:48) y + L S (cid:16) Ψ (cid:48) y | ¯Σ f (cid:17) Ψ (cid:48) x and Ψ (cid:48) y already satisfy the reciprocity result, one obtains: (cid:90) ¯Σ b (cid:16) L S,b (cid:16) Ψ (cid:48) x | ¯Σ f (cid:17) ∇ b Ψ (cid:48) y · ¯ n b − L S,b (cid:16) Ψ (cid:48) y | ¯Σ f (cid:17) ∇ b Ψ (cid:48) x · ¯ n b (cid:17) ¯ ν = 0 , etc . Using (37) and (38) again, this shows that M c is also a symmetric matrix.Therefore, the arbitrary and independent variations discussed previously arepossible if and only if the × symmetric matrix M := M b + M a + M c , (49)is invertible. Note that case 1 requires only the invertibility of the upper left orlower right × blocks of M . The invertibility of M is not examined in thispaper, and it is assumed to be invertible. Consider the space of (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1) , denoted by Z b × se (3) ∗ . On this space define the Hamiltonian function as the total energy function (28)written in terms of these variables: H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1) = ρ (cid:32)(cid:90) ¯Σ f Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν (cid:33) + 12 ρg (cid:90) ¯ S ¯ η ¯ ν s + 12 (cid:28)(cid:28) ( M b ) − · (cid:20) ρ ( L , A ) − ¯ P f (cid:21) , ρM b · ( M b ) − · (cid:20) ρ ( L , A ) − ¯ P f (cid:21)(cid:29)(cid:29) , = ρ (cid:32)(cid:90) ¯Σ f Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν (cid:33) + 12 ρg (cid:90) ¯ S ¯ η ¯ ν s + 12 (cid:28)(cid:28) ( M b ) − · (cid:20) ρ ( L , A ) − ¯ P f (cid:21) , ρ (cid:20) ρ ( L , A ) − ¯ P f (cid:21)(cid:29)(cid:29) , (50)Now consider the following Poisson brackets on Z b × se (3) ∗ , { F, G } := { F | Z b , G | Z b } Zakharov + { F | se (3) ∗ , G | se (3) ∗ } Lie − Poisson (51)19here { F | Z b , G | Z b } Zakharov is the Zakharov bracket but written in the variables (cid:0) ¯Σ f , φ (cid:48) f (cid:1) , and given by { ˆ F , ˆ G } Zakharov := 1 ρ (cid:90) ¯Σ f (cid:32) δ ˆ f (cid:0) δ ¯Σ f (cid:1) n δ ˆ gδφ (cid:48) f − δ ˆ g (cid:0) δ ¯Σ f (cid:1) n δ ˆ fδφ (cid:48) f (cid:33) ¯ ν, for functions ˆ F , ˆ G : Z b → R of the form ˆ F = (cid:90) ¯Σ f ˆ f ¯ ν, etc. And { F | se (3) ∗ , G | se (3) ∗ } Lie − Poisson is the negative Lie-Poisson bracket on se (3) ∗ ≡ R ∗ × R ∗ , given by { ˜ F , ˜ G } Lie − Poisson := − (cid:42) µ, (cid:34) ∂ ˜ F∂µ , ∂ ˜ G∂µ (cid:35)(cid:43) , = (cid:42) ∂ ˜ F∂µ , ad ∗ ∂ ˜ G/∂µ µ (cid:43) , (52)for ˜ F , ˜ G : se (3) ∗ → R and µ ∈ se (3) ∗ ≡ R [17]. Functional Derivatives.
Compute now the various functional derivatives of H corresponding to the variations 1, 2 and 3, described previously.Starting with case 1, the variation in the Hamiltonian is computed as (cid:28)(cid:28) δHδ ( L , A ) , δ ( L , A ) (cid:29)(cid:29) = 1 (cid:15) lim (cid:15) → (cid:2) H (cid:0) ¯Σ f , φ (cid:48) f , ( L , A ) + (cid:15)δ ( L , A ) (cid:1) − H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1)(cid:3) = 1 (cid:15) lim (cid:15) → (cid:34) (cid:15)ρ (cid:32)(cid:90) ¯Σ f Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b − (cid:90) ¯Σ b Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n b ¯ ν b (cid:33) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:20) ρ δ ( L , A ) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:21) , ρ (cid:20) ρ ( L , A ) − ¯ P f (cid:21)(cid:29)(cid:29) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:20) ρ ( L , A ) − (cid:0) ¯ L f , ¯ A f (cid:1)(cid:21) , ρ (cid:20) ρ δ ( L , A ) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:21)(cid:29)(cid:29)(cid:21) , δ Φ (cid:48) is the induced variation in this case, as discussed previously. In thefirst integral on the right, it should be noted that though δ Φ (cid:48) | ¯Σ f =0 (since δφ (cid:48) f = δ ¯Σ f =0), ∇ b δ Φ (cid:48) could be non-zero. Now use the well-known identity for two har-monic functions f, g in a domain with boundaries (cid:90) ∂D f ∇ g · n ν − (cid:90) ∂D g ∇ f · n ν = 0 Apply this to the functions Φ (cid:48) and δ Φ (cid:48) and with ∂D ≡ ¯Σ f ∪ ¯Σ b ∪ S . The normalderivatives of both the functions vanish at S , leading to the relation (cid:90) ¯Σ f Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n b ¯ ν b = (cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b , (53) = − (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b And so (cid:15) lim (cid:15) → (cid:2) H (cid:0) ¯Σ f , φ (cid:48) f , L , A + (cid:15)δ ( L , A ) (cid:1) − H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1)(cid:3) = 1 (cid:15) lim (cid:15) → (cid:20) − (cid:15)ρ (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b + (cid:15)ρ (cid:28)(cid:28)(cid:0) ¯ V , ¯Ω (cid:1) , ρ δ ( L , A ) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:29)(cid:29)(cid:21) , = (cid:20) − ρ (cid:90) ¯Σ b δ Φ (cid:48) (cid:0) ¯ V + ¯Ω × l (cid:1) · ¯ n b ¯ ν b + (cid:10)(cid:10)(cid:0) ¯ V , ¯Ω (cid:1) , δ ( L , A ) − ρδ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:11)(cid:11)(cid:21) , = (cid:2) ρ (cid:10)(cid:10)(cid:0) ¯ V , ¯Ω (cid:1) , δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:11)(cid:11) + (cid:10)(cid:10)(cid:0) ¯ V , ¯Ω (cid:1) , δ ( L , A ) − ρδ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:11)(cid:11)(cid:3) , [from (30)] from which is obtained ∂H∂ ( A , L ) = (cid:0) ¯Ω , ¯ V (cid:1) (cid:90) ¯Σ f δhδφ (cid:48) f δφ (cid:48) f ν ≡ (cid:90) ¯Σ f δhδφ (cid:48) f δ Φ (cid:48) ν = 1 (cid:15) lim (cid:15) → (cid:2) H (cid:0) ¯Σ f , φ (cid:48) f + (cid:15)δφ (cid:48) f , L , A (cid:1) − H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1)(cid:3) = 1 (cid:15) lim (cid:15) → (cid:34) (cid:15)ρ (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν + (cid:90) ¯Σ f Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b − (cid:90) ¯Σ b Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n b ¯ ν b (cid:33) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3) , ρ (cid:20) ρ ( L , A ) − ¯ P f (cid:21)(cid:29)(cid:29) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:20) ρ ( L , A ) − (cid:0) ¯ L f , ¯ A f (cid:1)(cid:21) , ρ (cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3)(cid:29)(cid:29)(cid:21) , = 1 (cid:15) lim (cid:15) → (cid:34) (cid:15)ρ (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b (cid:33) + (cid:15) (cid:10)(cid:10)(cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3) , ρ (cid:0) ¯ V , ¯Ω (cid:1)(cid:11)(cid:11)(cid:35) , [using (53)]= ρ (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν + (cid:10)(cid:10)(cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3) , (cid:0) ¯ V , ¯Ω (cid:1)(cid:11)(cid:11)(cid:33) , = ρ (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν + (cid:42)(cid:42)(cid:34) − (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) ¯ n f ¯ ν, (cid:90) ¯Σ f l × δ Φ (cid:48) ¯ n f ¯ ν (cid:33)(cid:35) , (cid:0) ¯ V , ¯Ω (cid:1)(cid:43)(cid:43)(cid:33) , = ρ (cid:32)(cid:90) ¯Σ f δ Φ (cid:48) (cid:0) ∇ b Φ (cid:48) − ¯ V − ¯Ω × l (cid:1) · ¯ n f ¯ ν (cid:33) , which implies that δhδφ (cid:48) f = ρ (cid:0) ∇ b Φ (cid:48) − ¯ V − ¯Ω × l (cid:1) · ¯ n f ρg (cid:90) ¯ S δ ¯ η ¯ ν s = 0 , (54) (cid:90) ¯Σ f δh (cid:0) δ ¯Σ f (cid:1) n (cid:0) δ ¯Σ f (cid:1) n ¯ ν = 1 (cid:15) lim (cid:15) → (cid:2) H (cid:0) ¯Σ f + (cid:15)δ ¯Σ f , φ (cid:48) f , L , A (cid:1) − H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1)(cid:3) = 1 (cid:15) lim (cid:15) → (cid:34) ρ (cid:32) δ (cid:90) ¯Σ f Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν − (cid:15) (cid:18)(cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν + (cid:90) ¯Σ b Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n b ¯ ν (cid:19)(cid:33) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3) , ρ (cid:20) ρ ( L , A ) − ¯ P f (cid:21)(cid:29)(cid:29) + (cid:15) (cid:28)(cid:28) ( M b ) − · (cid:20) ρ ( L , A ) − (cid:0) ¯ L f , ¯ A f (cid:1)(cid:21) , ρ (cid:2) − δ (cid:0) ¯ L f , ¯ A f (cid:1)(cid:3)(cid:29)(cid:29)(cid:21) + ρg (cid:90) ¯ S ¯ η δ ¯ η ¯ ν s , where the δ (cid:82) ¯Σ f Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f ¯ ν term is calculated as in the problem without therigid body [16]; see also [21]. 23mporting this term, obtain (cid:90) ¯Σ f δh (cid:0) δ ¯Σ f (cid:1) n (cid:0) δ ¯Σ f (cid:1) n ν = 1 (cid:15) lim (cid:15) → (cid:2) H (cid:0) ¯Σ f + (cid:15)δ ¯Σ f , φ (cid:48) f , L , A (cid:1) − H (cid:0) ¯Σ f , φ (cid:48) f , L , A (cid:1)(cid:3) = 1 (cid:15) lim (cid:15) → (cid:34) ρ (cid:32) ± (cid:90) V Σ ∇ b Φ (cid:48) · ∇ b Φ (cid:48) ¯ µ + (cid:15) (cid:90) ¯Σ f ( δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f + Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n f ) ¯ ν − (cid:15) (cid:18)(cid:90) ¯Σ b δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n b ¯ ν b + (cid:90) ¯Σ b Φ (cid:48) ∇ b δ Φ (cid:48) · ¯ n b ¯ ν b (cid:19)(cid:19) + (cid:15)ρ (cid:10)(cid:10)(cid:0) ¯ V , ¯Ω (cid:1) , [ − δ ( L f , A f )] (cid:11)(cid:11) + (cid:15)ρg (cid:90) ¯ S ¯ η (cid:0) δ ¯Σ f (cid:1) n ¯ ν (cid:21) , [invoking (32) for the potential energy term]= 1 (cid:15) lim (cid:15) → (cid:34) ± ρ (cid:90) V Σ ∇ b Φ (cid:48) · ∇ b Φ (cid:48) ¯ µ + (cid:15)ρ (cid:90) ¯Σ f δ Φ (cid:48) ∇ b Φ (cid:48) · ¯ n f + (cid:15)ρ (cid:10)(cid:10)(cid:0) ¯ V , ¯Ω (cid:1) , [ − δ ( L f , A f )] (cid:11)(cid:11) + (cid:15)ρg (cid:90) ¯ S ¯ η (cid:0) δ ¯Σ f (cid:1) n ¯ ν (cid:21) , [using (53) and proceeding in the same way as in case 2]= 1 (cid:15) lim (cid:15) → (cid:34) ± ρ (cid:90) V Σ ∇ b Φ (cid:48) · ∇ b Φ (cid:48) ¯ µ + (cid:15)ρ (cid:90) ¯Σ f δ Φ (cid:48) (cid:0) ∇ b Φ (cid:48) − (cid:0) ¯ V + ¯Ω × l (cid:1)(cid:1) · ¯ n f + (cid:15)ρg (cid:90) ¯ S ¯ η (cid:0) δ ¯Σ f (cid:1) n ¯ ν (cid:21) , (cid:15) lim (cid:15) → (cid:34) ρ (cid:15) ( δ ¯Σ f ) n (cid:90) ¯Σ f ∇ b Φ (cid:48) · ∇ b Φ (cid:48) ¯ ν + (cid:15)ρ (cid:90) ¯Σ f δ Φ (cid:48) (cid:0) ∇ b Φ (cid:48) − (cid:0) ¯ V + ¯Ω × l (cid:1)(cid:1) · ¯ n f + (cid:15)ρg (cid:90) ¯ S ¯ η (cid:0) δ ¯Σ f (cid:1) n ¯ ν (cid:21) , [as in the problem without the rigid body]= 1 (cid:15) lim (cid:15) → (cid:34) ρ (cid:15) ( δ ¯Σ f ) n (cid:90) ¯Σ f ∇ b Φ (cid:48) · ∇ b Φ (cid:48) ¯ ν − (cid:15)ρ (cid:90) ¯Σ f ∇ b Φ (cid:48) · ¯ n f (cid:0) δ ¯Σ f (cid:1) n (cid:0) ∇ b Φ (cid:48) − (cid:0) ¯ V + ¯Ω × l (cid:1)(cid:1) · ¯ n f + (cid:15)ρg (cid:90) ¯ S (¯ η − ¯ η ) (cid:0) δ ¯Σ f (cid:1) n ¯ ν (cid:21) , [using (47) , (54) and (32)] And so δhδ ¯Σ f = ρ (cid:18) ∇ b Φ (cid:48) · ∇ b Φ (cid:48) − ( ∇ b Φ (cid:48) · ¯ n f ) + ( ∇ b Φ (cid:48) · ¯ n f ) (cid:0) ¯ V + ¯Ω × l (cid:1) · ¯ n f + g (¯ η − ¯ η ) (cid:19) , Collecting all the functional derivatives, δhδφ (cid:48) f = ρ (cid:0) ∇ b Φ (cid:48) − ¯ V − ¯Ω × l (cid:1) · ¯ n f ,δh (cid:0) δ ¯Σ f (cid:1) n = ρ (cid:18) ∇ b Φ (cid:48) · ∇ b Φ (cid:48) − ( ∇ b Φ (cid:48) · ¯ n f ) + ( ∇ b Φ (cid:48) · ¯ n f ) (cid:0) ¯ V + ¯Ω × l (cid:1) · ¯ n f + g (¯ η − ¯ η ) (cid:19) ,∂H∂ ( A , L ) = (cid:0) ¯Ω , ¯ V (cid:1) The Hamiltonian equations of the motion of the coupled system, with respect25o the Poisson brackets (51), are: ∂ ¯Σ f ∂t = 1 ρ δhδφ (cid:48) f = (cid:0) ∇ b Φ (cid:48) − ¯ V − ¯Ω × l (cid:1) · ¯ n f , (55) ∂φ (cid:48) f ∂t = − ρ δh (cid:0) δ ¯Σ f (cid:1) n = − (cid:18) ∇ b Φ (cid:48) · ∇ b Φ (cid:48) − ( ∇ b Φ (cid:48) · ¯ n f ) + ( ∇ b Φ (cid:48) · ¯ n f ) (cid:0) ¯ V + ¯Ω × l (cid:1) · ¯ n f + g (¯ η − ¯ η ) (cid:19) , (56) d ( A , L ) dt = ad ∗ ∂H/∂ ( A , L ) ( A , L ) = (cid:0) A × ¯Ω + L × ¯ V , L × ¯Ω (cid:1) (57)It is easily checked that equation (57) is the same as equations (24) and (25), ob-tained from the global momentum analysis. Equation (56) is Bernoulli’s equationat the free surface (5) in the absence of surface tension ( p = p atm ), after using thefollowing relation [23, 16, 21] ∂φ (cid:48) f ∂t = ∂∂t (cid:16) Φ (cid:48)| ¯Σ f (cid:17) + ( ∇ b Φ (cid:48) · ¯ n f ) | ¯Σ f ∂ ¯Σ f ∂t The problem presented in this paper is in a general framework. It would be ofparticular interest to seek some special configurations, for example, moving equi-librium configurations involving a rigid body and traveling wave(s), and examinetheir associated stability. The Hamiltonian formalism would allow a nonlinearstability analysis to be performed, analogous to that done for the F¨oppl equilib-rium in the problem of a 2D rigid cylinder and point vortices [24]. Examiningthe dynamically coupled interaction of a soliton approaching a neutrally buoyantrigid body would be another interesting direction.From a Hamiltonian and geometric mechanics perspective, it would also be ofinterest to derive the Poisson brackets of this paper from well-formulated theoriesof symmetry and reduction of Hamiltonian systems [17], along the lines of [20,25].Vortices can be generated by free surfaces, and the problem of the dynamicallycoupled interaction of a free surface and vortices has also been examined from a26amiltonian perspective [20, 26]. The same is true for the problem of a neutrallybuoyant rigid body and vortices [24, 27, 28]. It would be a natural extensiontherefore to examine the dynamics interaction problem of a rigid body close toa free surface and in the presence of vortices. Indeed, in the viscous Navier-Stokes setting, this problem for stationary rigid bodies has quite a few interestingfeatures; see, for example, [29] and references therein.Apart from linearization approaches, free surface dynamics has also been stud-ied in various asymptotic limits. The shallow water approximation in particularhas proved to be very popular. It would be interesting to see how the presenceof a dynamically interacting rigid body could be accommodated in such approxi-mations. Presumably, including parameters based on the body size, could lead tosome new asymptotic limits. 27
Appendix A: Global momentum evolution equa-tions in a spatially-fixed frame.
Details of the derivation of equations (16) and (17), valid in a spatially-fixed frame(whose origin is taken at the center of the disc C R ), are presented in this appendix. Linear Momentum.
Applying Newton’s second law for the evolution of L T first, at any given time instant t , dL T dt = lim R →∞ (cid:18) − (cid:90) Σ R p atm n f ν + (cid:90) C R pn s ν − ρ f gk (cid:90) C R η ν − (cid:90) W pe R ν (cid:19) + ρ f g V B k − ρ b g V B k, where k is unit vector opposite to the gravity direction and coincides with n s . Notethat the third integral on the right in original form is (cid:82) η (cid:82) C R dz ν .28eferring to (11) and (14), the equation becomes ddt (cid:18) ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + L b (cid:19) = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R Φ n f ν − (cid:90) Σ R p atm n f ν + ρ f ddt (cid:90) C R Φ n s ν + (cid:90) C R pn s ν − ρ f gk (cid:90) C R η ν − (cid:90) W pe R ν (cid:19) + ρ f g V B k − ρ b g V B k, = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R Φ n f ν − (cid:90) Σ R p atm n f ν + (cid:90) C R (cid:18) ρ f ∂ Φ ∂t + p (cid:19) n s ν − ρ f gk (cid:90) C R η ν − (cid:90) W pe R ν (cid:19) + g V B k ( ρ f − ρ b ) , [at any R fixed in time]= lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R Φ n f ν − (cid:90) Σ R p atm n f ν + (cid:90) C R (cid:18) p atm + ρ f gη − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν − ρ f gk (cid:90) C R η ν − (cid:90) W pe R ν (cid:19) + g V B k ( ρ f − ρ b ) , [using (4)]= lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R Φ n f ν − (cid:90) Σ R p atm n f ν + (cid:90) C R (cid:18) p atm − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν + ρ f gk (cid:90) C R ( η − η ) ν − (cid:90) W pe R ν (cid:19) + g V B k ( ρ f − ρ b ) , Using (6) and re-arranging terms, one gets ddt (cid:32) ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + L b + ρ f (cid:90) Σ f Φ n f ν (cid:33) = lim R →∞ (cid:18) − (cid:90) Σ R p atm n f ν + (cid:90) C R (cid:18) p atm − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν − (cid:90) W pe R ν (cid:19) + g V B k ( ρ f − ρ b ) , lim R →∞ (cid:90) W pe R ν = lim R →∞ (cid:90) W ( p atm + ρ f g ( η − z ) + A ( R, θ, z, t )) e R ν. The integral of the A term is O (1 /R ) . To resolve all the terms containing p atm ,apply Stokes theorem in D R to obtain the result lim R →∞ (cid:90) D R ∇ (1) µ = 0 = lim R →∞ (cid:18)(cid:90) Σ R n f ν − (cid:90) C R n s ν − (cid:90) Σ b n b ν b + (cid:90) W e R ν (cid:19) , where the third integral on the right is zero due to the closedness of the body.Incorporating this result into the linear momentum equation, and since (cid:82) W ρ f g ( η − z ) e R ν = 0 due to the cylindrical geometry, obtain in the limit R → ∞ , ddt (cid:32) ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + L b + ρ f (cid:90) Σ f Φ n f ν (cid:33) = (cid:90) S (cid:18) − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν + g V B k ( ρ f − ρ b ) , Defining L = ρ f (cid:90) Σ b r × ( n b × ∇ Φ) ν + L b + ρ f (cid:90) Σ f Φ n f ν (58)the linear momentum equation becomes d L dt = − ρ f (cid:90) S ∇ Φ · ∇ Φ2 n s ν s + ( ρ f − ρ b ) g V B k. Angular Momentum.
Similarly, apply Newton’s second law for the evolutionof A T , dA T dt = lim R →∞ (cid:18) − (cid:90) Σ R r × p atm n f ν + (cid:90) C R r × pn s ν − ρ f g (cid:90) η (cid:90) C R r × kdz ν + (cid:90) W r × pe R ν (cid:19) + r c × ρ f g V B k − r c × ρ b g V B k, r c is the position vector of the centroid of the rigid body.Referring to (13) and (14), obtain ddt (cid:18) − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + A b (cid:19) = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R r × Φ n f ν − (cid:90) Σ R r × p atm n f ν + ρ f ddt (cid:90) C R r × Φ n s ν + (cid:90) C R r × pn s ν s − ρ f g (cid:90) η (cid:90) C R r × kdz ν + (cid:90) W r × pe R ν (cid:19) + r c × ( ρ f − ρ b ) g V B k, = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R r × Φ n f ν − (cid:90) Σ R r × p atm n f ν + (cid:90) C R r × (cid:18) ρ f ∂ Φ ∂t + p (cid:19) n s ν − ρ f g (cid:90) η (cid:90) C R r × kdz ν + (cid:90) W r × pe R ν (cid:19) + r c × ( ρ f − ρ b ) g V B k, [at any R fixed in time] , = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R r × Φ n f ν − (cid:90) Σ R r × p atm n f ν + (cid:90) C R r × (cid:18) p atm + ρ f gη − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν − ρ f g (cid:90) η (cid:90) C R r × kdz ν + (cid:90) W r × pe R ν (cid:19) + r c × ( ρ f − ρ b ) g V B k, [using (4)]= lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R r × Φ n f ν − (cid:90) Σ R r × p atm n f ν + (cid:90) C R r × (cid:18) p atm + ρ f g ( η − η ) − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν s + ρ f g (cid:90) C R r × ηn s ν − ρ f g (cid:90) η (cid:90) C R r × kdz ν + (cid:90) W r × pe R ν (cid:19) + r c × ( ρ f − ρ b ) g V B k, = lim R →∞ (cid:18) − ρ f ddt (cid:90) Σ R r × Φ n f ν − (cid:90) Σ R r × p atm n f ν + (cid:90) C R r × (cid:18) p atm + ρ f g ( η − η ) − ρ f ∇ Φ · ∇ Φ2 (cid:19) n s ν + (cid:90) W r × pe R ν (cid:19) + r c × ( ρ f − ρ b ) g V B k, [since r × k is independent of z, (cid:90) η (cid:90) C R r × kdz ν = (cid:90) C R r × ηn s ν ] lim R →∞ (cid:90) W r × pe R ν = lim R →∞ (cid:18)(cid:90) W r × ( p atm + ρ f g ( η − z ) + A ( R, θ, z, t )) e R ν (cid:19) , The integral of the A term is again O (1 /R ) , since r × e R filters off the R -coordinate.Now use another result obtained from Stokes’ theorem applied in D R (cid:90) D R ∇ × r µ = 0 = (cid:90) Σ R n f × r ν − (cid:90) C R n s × r ν − (cid:90) Σ b n b × r ν + (cid:90) W e R × r ν R The third integral can be shown to be equal to zero by applying the same integraltheorem again with the body as the domain. Incorporating this result into theangular momentum equation, and noting that due to the cylindrical geometry of W , (cid:90) W r × ρ f g ( η − z ) e R ν = (cid:90) W ρ f gz ( η − z ) e t ν = 0 ,ddt (cid:32) − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + A b + ρ f (cid:90) Σ f r × Φ n f ν (cid:33) = lim R →∞ (cid:18)(cid:90) C R r × (cid:18) ρ f g ( η − η ) n s − ρ f ∇ Φ · ∇ Φ2 (cid:19) ν s (cid:19) + r c × ( ρ f − ρ b ) g V B k Invoking (7) and defining A = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + A b + ρ f (cid:90) Σ f r × Φ n f ν obtain in the limit R → ∞ , d A dt = (cid:90) S r × (cid:18) − ρ f ∇ Φ · ∇ Φ2 (cid:19) ν s + r c × ( ρ f − ρ b ) g V B k Appendix B: Global momentum evolution equa-tions in a body-fixed frame.
Linear momentum.
Using again the relations (20)– (23) and the change of vari-ables theorem, the linear momentum L transforms as L = ρ f R ( t ) (cid:90) ¯Σ b l × (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + R ( t ) (cid:18) ¯ b ( t ) × (cid:90) ¯Σ b (¯ n b × ∇ b Φ (cid:48) b ) ¯ ν (cid:19) + R ( t ) ¯ L b + ρ f R ( t ) (cid:90) ¯Σ f Φ (cid:48) ¯ n f ¯ ν. Next, show that (cid:90) ¯Σ b (¯ n b × ∇ b Φ (cid:48) ) ¯ ν = 0 by using the following argument. Let ¯Φ (cid:48) b satisfy the following Dirichlet problem ∇ ¯Φ (cid:48) b = 0 in B, ¯Φ (cid:48) b = Φ (cid:48) b in Σ b . Then (cid:90) ¯Σ b (¯ n b × ∇ b Φ (cid:48) b ) ¯ ν = (cid:90) ¯Σ b (cid:16) ¯ n b × ∇ b ¯Φ b (cid:48) (cid:17) ¯ ν = (cid:90) B ∇ b × ∇ b ¯Φ b (cid:48) ¯ µ = 0 , (59)and therefore L = R ( t ) (cid:32) ρ f (cid:90) ¯Σ b l × (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + ¯ L b + ρ f (cid:90) ¯Σ f Φ (cid:48) ¯ n f ¯ ν (cid:33) , =: R ( t ) L where ¯ L b = ρ b m b ¯ V . The linear momentum equation in L becomes d L dt + ¯Ω × L = 0 Angular momentum.
Similarly, transform the angular momentum A , A = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + A b + ρ f (cid:90) Σ f r × Φ n f ν, = − ρ f (cid:90) Σ b r ( n b × ∇ Φ) ν + b × M b V + I Ω + ρ f (cid:90) Σ f r × Φ n f ν, [using; 15] A = − ρ f R ( t ) (cid:90) ¯Σ b (cid:0) l + 2 l · ¯ r c + ¯ r c (cid:1) (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + R ( t ) (cid:0) ¯ r c × M b ¯ V + ¯ I Ω (cid:1) + ρ f R ( t ) (cid:90) ¯Σ f ( l + ¯ r c ) × Φ (cid:48) ¯ n f ¯ ν, = − ρ f R ( t ) (cid:90) ¯Σ b (cid:0) l + 2 l · ¯ r c (cid:1) (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + R ( t ) (cid:0) ¯ r c × M b ¯ V + ¯ I Ω (cid:1) + ρ f R ( t ) (cid:90) ¯Σ f ( l + ¯ r c ) × Φ (cid:48) ¯ n f ¯ ν, [using result (59)]= R ( t ) (cid:32) − ρ f (cid:90) ¯Σ b l (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + ¯ I Ω + ρ f (cid:90) ¯Σ f l × Φ (cid:48) ¯ n f ¯ ν (cid:33) − ρ f R ( t ) (cid:90) ¯Σ b l · ¯ r c (¯ n b × ∇ b Φ (cid:48) ) ¯ ν + R ( t ) (cid:0) ¯ r c × M b ¯ V (cid:1) + ρ f R ( t ) (cid:90) ¯Σ f ¯ r c × Φ (cid:48) ¯ n f ¯ ν. At this point, a Proposition proved in [28] is invoked (adapted to the notation ofthis paper):
Proposition 7.1