Application of the Kovacic algorithm for the investigation of motion of a heavy rigid body with a fixed point in the Hess case
aa r X i v : . [ n li n . S I] N ov Application of the Kovacic algorithm for theinvestigation of motion of a heavy rigid body with afixed point in the Hess case.
Boris S. Bardin ∗ , Alexander S. Kuleshov ∗∗ ∗ Moscow Aviation Institute (national research university),Moscow 125080, Russia
E-mail: [email protected] ∗∗ M. V. Lomonosov Moscow State University,Moscow 119234.
E-mail: [email protected]
Аннотация
In 1890 German mathematician and physicist W. Hess [4] found new special case ofintegrability of Euler – Poisson equations of motion of a heavy rigid body with a fixed point.In 1892 P. A. Nekrasov [17, 18] proved that the solution of the problem of motion of a heavyrigid body with a fixed point under Hess conditions reduces to integrating the second orderlinear differential equation. In this paper the corresponding linear differential equation isderived and its coefficients are presented in the rational form. Using the Kovacic algorithm [8],we proved that the liouvillian solutions of the corresponding second order linear differentialequation exists only in the case, when the moving rigid body is the Lagrange top, or in thecase when the constant of the area integral is zero.
Let us consider the problem of motion of a heavy rigid body with the fixed point O . To describemotion of the body we introduce two orthogonal coordinate systems: the fixed system Oxyz andthe moving system Ox x x . The Oz axis of the fixed system is directed along the upward vertical.The Ox x x system is rigidly connected with the moving body and its axes are directed alongthe principal axes of inertia at O . We denote the unit vectors of Ox x x system by e , e , e . Let γ = γ e + γ e + γ e be the unit vector of the Oz axis, where γ , γ , γ are projections of thisvector onto the Ox x x axes. We denote by P the gravity force (directed vertically downwardand applied at the center of gravity of the body G ), then we have P = − M g γ , where M is themass of the body and g is the gravity acceleration. We will define the position of the center ofmass of the body by the radius – vector r = −→ OG = x e + x e + x e , where x , x , x are projections of this vector onto Ox x x axes.We will apply the principle of angular momentum, taking with respect to the point O , to derivethe equations of motion of the body. The corresponding vector equation with respect to the fixedcoordinate system Oxyz takes the following form: d K dt = M , (1)1here M is the total moment of the external forces (the force P in the considered case) about thefixed point O , and K is the angular momentum of the body with respect to the fixed point O .This vector can be represented as follows (see. [21, p. 331]): K = J O ω , (2)where ω is the angular velocity vector of the body and J O is the tensor of inertia of the body at O . It is well known that the rate of change of the vector K – the vector d K /dt , calculated in thefixed coordinate system Oxyz , equals the rate of growth of this vector δ K /δt (calculated in themoving coordinate system Ox x x ) plus the rate of transport, i.e. (see [21, p. 347]): d K dt = δ K δt + [ ω × K ] . (3)Taking into account (3), we can rewrite equation (1) as follows: δ K δt + [ ω × K ] = [ r × P ] . (4)In the moving coordinate system Ox x x we will write the vector ω , the vector K and thetensor J O as follows: ω = ω e + ω e + ω e , J O = A A
00 0 A , K = J O ω = A ω e + A ω e + A ω e , where A , A , A – are principal moments of inertia of the body at O .If we take the projections of vector equation (4) onto the Ox x x axes we obtain the threescalar equations: A ˙ ω + ( A − A ) ω ω = M g ( x γ − x γ ) ,A ˙ ω + ( A − A ) ω ω = M g ( x γ − x γ ) ,A ˙ ω + ( A − A ) ω ω = M g ( x γ − x γ ) . (5)These equations are called the Euler’s equations of motion of a rigid body with a fixed point.These equations actually involve six parameters A , A , A , x , x , x , characterizing the massdistribution of the body relative to the principal axes of inertia at O , and six unknowns ω , ω , ω , γ , γ , γ to be solved as functions of time t . To be able to solve for them we need three moreindependent equations relating the ω , ω , ω , γ , γ , γ . We have, since γ is a space – fixed vector(i.e. it is fixed in the Oxyz coordinate system), then δ γ δt + [ ω × γ ] = 0 . Ox x x axes ˙ γ = ω γ − ω γ , ˙ γ = ω γ − ω γ , ˙ γ = ω γ − ω γ . (6)These equations are called the Poisson equations. Finally, we obtain the close system of sixnonlinear differential equations (5), (6) with respect to the six unknown functions of time t : ω , ω , ω , γ , γ , γ . These equations are called the Euler – Poisson equations.It is well known that to solve the Euler – Poisson equations we need to find four independentautonomous first integrals of the system (5), (6) [2, 14]. For any values of parameters A , A , A , x , x , x of the body and for any initial conditions we have three first integrals of the Euler –Poisson equations.1. The energy integral T + V = h , where T is the kinetic energy of the body, defined by theformula T = 12 (cid:0) A ω + A ω + A ω (cid:1) , and V is the potential energy of the body which has the form: V = M g ( x γ + x γ + x γ ) . Thus with respect to the axes Ox x x the energy integral takes the form: (cid:0) A ω + A ω + A ω (cid:1) + M g ( x γ + x γ + x γ ) = h.
2. The area integral ( K · γ ) = C, which can be written with respect to the axes Ox x x as follows: A ω γ + A ω γ + A ω γ = C.
3. The geometrical integral γ + γ + γ = 1 . Thus for integrability of the Euler – Poisson equations we need to find only one additionalautonomous first integral. Under appropriate conditions on the parameters A , A , A , x , x , x this integral can exist only in the following three cases [2, 14].1. The Euler case ( x = x = x = 0) . The additional integral has the form: A ω + A ω + A ω = k .
2. The Lagrange case ( A = A , x = x = 0) . In this case the additional integral has the form: ω = ω = const .
3. The Kovalevskaya case ( A = A = 2 A , x = x = 0) . If we denote M gx A = n, then the additional first integral can be written as follows: (cid:0) ω − ω − nγ (cid:1) + (2 ω ω − nγ ) = j. There are no other general cases of integrability of Euler – Poisson equations. However, thereare several cases, when for the special initial conditions we can find the additional special integral.One of these cases is the Hess case which has been firstly described in 1890 by W. Hess [4]. Letus consider this case in more details.
We suppose that the parameters of the body A , A , A , x , x , x satisfy the conditions x = 0 , A ( A − A ) x = A ( A − A ) x , A ≥ A ≥ A . (7)Let us prove that under conditions (7) the Euler – Poisson equations (5), (6) have an additionalspecial integral (the Hess integral) of the form: A ω x + A ω x = 0 . (8)To prove the existence of the integral (8) we write two first equations of the system (5) takinginto account conditions (7): A ˙ ω + ( A − A ) ω ω = − M gx γ ,A ˙ ω + ( A − A ) ω ω = M gx γ . (9)Multiplying the first of equations (9) by x and the second by x and taking their sum weobtain: ddt ( A ω x + A ω x ) = ( A − A ) ω ω x + ( A − A ) ω ω x . (10)The right hand part of equation (10) can be transformed using (7) to the form: ( A − A ) ω ω x + ( A − A ) ω ω x = ω ( A − A ) x A x ( A ω x + A ω x ) . (11)Taking into account equation (11) we can conclude from the equation (10), that if at initialinstant the equation (8) is valid then it holds during the whole time of motion of the rigidbody. Therefore under conditions (7) the Euler – Poisson equations (5), (6) possess the specialintegral (8).For the first time this case of special integrability of Euler – Poisson equations was discoveredby W. Hess [4]. Hess found this case of integrability when he searched for singular solutions ofhis own form of the Euler – Poisson equation [4]. In 1892 the Hess case was rediscovered byG. G. Appelroth [1] when he analysed the branching of the general solution of Euler – Poisson4quations on the complex plane of time using the ideas of S. V. Kovalevskaya [12,13]. The detailedanalytical investigation of the Hess case has been made by P. A. Nekrasov [17,18]. In his papers [17,18] Nekrasov presented both the Hess conditions and the Hess integral and reduced the solutionof the problem to the integration of a second order linear differential equation. Nekrasov provedthat in the Hess case the solution branches out on the complex plane of time. He investigatedthe analytical properties of the obtained second order linear differential equation and pointedout the basic properties of trajectories on the Poisson sphere. He proved also that on the zerolevel of the area integral the problem is integrable in elliptic functions. The Hess integral as wellas the reduction to the second order linear differential equation was independently rediscoveredin 1895 by Roger Liouville [15]. The geometrical analysis and the modelling of the Hess top(in particular, on the zero level of the area integral) were given by N E. Zhukovsky [5]. In thepaper [16] B. K. Mlodzeevskii and P. A. Nekrasov proved that under special restrictions solutionsof the problem can have asymptotic behaviour.The complete analysis of the phase trajectories in a Hess case was made by A. M. Kovalev [9].In the paper [11] which substantially uses the results obtained in [9], the authors presented detailedclassification of possible hodographs of the angular momentum vector in dependence of constantsof the energy integral h and the area integral C . In the paper [10] the kinematic descriptionof motion of a rigid body in the Hess case at zero level of the area integral was given by thehodographs method [6, 7]. In the paper [3] at zero level of the area integral the temporal andspatial evolution of the angular velocity vector and the angular moment vector was investigated.The motion of the body with a fixed point in the Hess case was represented by the rolling motionof the ellipsoid of inertia on a fixed plane.Since the solution of the problem of motion of a heavy rigid body with a fixed point inthe Hess case is reduced to solving the second order linear differential equation it is possible toset up the problem of existence of liouvillian solutions of the corresponding linear differentialequation. For this purpose it is possible to apply the so-called Kovacic algorithm [8], which allowsto find liouvillian solutions of a second order linear differential equation in explicit form. If a lineardifferential equation has no liouvillian solutions, the Kovacic algorithm also allows one to ascertainthis fact. The necessary condition for the application of the Kovacic algorithm to a second orderlinear differential equation is that the coefficients of this equation should be rational functions ofindependent variable.Below we derive the second order linear differential equation in the Hess case and reduceits coefficients to the form of rational functions. Further we study the problem of existence ofliouvillian solutions for the obtained second order linear differential equation using the Kovacicalgorithm.As the first step to derive the corresponding second order linear differential equation in theHess case let us write the Euler – Poisson equations in the special coordinate system, proposedby P. V. Kharlamov [6, 7]. Instead of the principal axes of inertia at O (the Ox x x coordinate system) let us choose thearbitrary trihedral coordinate system Oξ ξ ξ which is rigidly connected with the moving body.We denote the unit vectors of the Oξ ξ ξ system by e I , e II , e III . In this coordinate system wecan write the angular momentum K as follows: K = K e I + K e II + K e III . (12)5n the other hand, the angular momentum is defined by (2) K = J O ω , where ω = Ω e I + Ω e II + Ω e III , (13) J O = L − L − L − L L − L − L − L L . (14)Here the components L ij of inertia tensor J O satisfy the condition L ij = L ji . In this tensorcomponents L , L , L are the moments of inertia of the body with respect to Oξ , Oξ , Oξ respectively and L ij , ( i = j ) are products of inertia. Using (12), (13), (14) we can conclude thatthe angular momentum K has the following components onto Oξ ξ ξ axes: K = L Ω − L Ω − L Ω ,K = − L Ω + L Ω − L Ω ,K = − L Ω − L Ω + L Ω . (15)We can write the kinetic energy of the moving body using the angular momentum K by theformula: T = 12 ( K · ω ) . Taking into account (13), (15) the expression for the kinetic energy can be written in the scalarform: T = 12 (cid:0) L Ω + L Ω + L Ω (cid:1) − L Ω Ω − L Ω Ω − L Ω Ω . (16)The inertia tensor (14) is a non-singular matrix. Therefore equations (15) can be representedas the linear nonhomogeneous equations with respect to Ω , Ω , Ω . To solve equations (15) withrespect to Ω , Ω , Ω , we find: Ω = l K + l K + l K , Ω = l K + l K + l K , Ω = l K + l K + l K . (17)Here we denote: l = L L − L ∆ , l = L L − L ∆ , l = L L − L ∆ ; l = l = L L + L L ∆ , l = l = L L + L L ∆ , l = l = L L + L L ∆ ;∆ = det ( L ij ) = L L L − L L − L L − L L − L L L . T , we present it as the function of K , K , K : T = 12 (cid:0) l K + l K + l K (cid:1) + l K K + l K K + l K K . (18)Now we simplify the obtained expression (18) for the kinetic energy. For this purpose we rotatethe coordinate system Oξ ξ ξ about the Oξ axis counter clockwise through an angle α . We denotethe obtained coordinate system by Oξ ∗ ξ ∗ ξ ∗ . In this case the new components of the vector K ,which we denote by K ∗ , K ∗ , K ∗ , can be expressed through the components K , K , K by thefollowing formulae: K = K ∗ , K = K ∗ cos α − K ∗ sin α, K = K ∗ sin α + K ∗ cos α. The expression for the kinetic energy T as a function of components K ∗ , K ∗ , K ∗ of the vector K takes the form: T = 12 (cid:0) l ∗ ( K ∗ ) + l ∗ ( K ∗ ) + l ∗ ( K ∗ ) (cid:1) + l ∗ K ∗ K ∗ + l ∗ K ∗ K ∗ + l ∗ K ∗ K ∗ . (19)In this formula the coefficients l ∗ ij are expressed through the coefficients l ij and the angle α bythe formulae: l ∗ = l , l ∗ = l cos α + l sin α + l sin 2 α,l ∗ = l sin α + l cos α − l sin 2 α,l ∗ = l sin α + l cos α, l ∗ = l cos α − l sin α,l ∗ = l cos 2 α −
12 ( l − l ) sin 2 α. We assume now that the angle α is defined by the formula: α = 12 arctg 2 l l − l . In this case the coefficient l ∗ in (19) becomes zero and the expression for the kinetic energytakes the form: T = 12 (cid:0) l ∗ ( K ∗ ) + l ∗ ( K ∗ ) + l ∗ ( K ∗ ) (cid:1) + ( l ∗ K ∗ + l ∗ K ∗ ) K ∗ . (20)Assuming that the center of mass G of the body does not coincide with the fixed point O we choose the coordinate system Oη η η as follows. Let the Oη axis passes through the centerof mass G of the body (i.e. it is directed along the vector r ) and let the two other axes Oη and Oη be directed in such a way that the expression for the kinetic energy T has a form (20),i.e. it does not contain the product K ∗ K ∗ . This coordinate system was firstly introduced byP. V. Kharlamov [6, 7]. He called the Oη η η system as the special coordinate system. Obviously,this coordinate system is uniquely determined, and it will be rigidly connected with a rigid body.Using the special coordinate system we will write a , a , a , b , b instead of l ∗ , l ∗ , l ∗ , l ∗ , l ∗ andwe will write L , L , L , ω I , ω II , ω III instead of K ∗ , K ∗ , K ∗ , Ω , Ω , Ω respectively. Finally we7ave the following expression for the kinetic energy T and components of the angular velocity ω of the body onto Oη η η axes: T = 12 (cid:0) aL + a L + a L (cid:1) + ( b L + b L ) L . (21) ω I = aL + b L + b L , ω II = a L + b L , ω III = a L + b L . (22)Using (5), (6), (21), (22) we can write the Euler – Poisson equations in the specialP. V. Kharlamov coordinate system as follows (see [6, 7]): ˙ L = ( a − a ) L L + ( b L − b L ) L , ˙ L = ( a − a ) L L + ( b L + b L ) L − b L + Γ ν , ˙ L = − ( a − a ) L L − ( b L + b L ) L + b L − Γ ν , ˙ ν = ( a L + b L ) ν − ( a L + b L ) ν , ˙ ν = ( aL + b L + b L ) ν − ( a L + b L ) ν , ˙ ν = − ( aL + b L + b L ) ν + ( a L + b L ) ν . (23)Here ν , ν , ν are projections of the vector γ onto the Oη η η axes and Γ =
M gρ , where ρ = p x + x + x .We will use the special coordinate system Oη η η for the description of motion of a heavyrigid body with a fixed point in the Hess case. In the problem of motion of a heavy rigid body with a fixed point in the Hess case thetransformation from the principal axes of inertia Ox x x (the unit vectors e , e , e ) to thespecial P. V. Kharlamov coordinate system Oη η η (the unit vectors e I , e II , e III ) is defined bythe formulae: e I = e cos α + e sin α, e II = − e sin α + e cos α, e III = e , where cos α and sin α equal cos α = x p x + x , sin α = x p x + x . (24)Let us prove now that the unit vectors e I , e II , e III are indeed the basis vectors of the specialP. V. Kharlamov coordinate system. We note< first of all, that x = 0 according to (7), andtherefore the unit vector e I of the axis Oη is indeed collinear to the vector −→ OG = r . Now let usmake sure that the kinetic energy of the body in this case is represented in the form (21). For thispurpose we write the angular momentum as follows: K = L e I + L e II + L e III = ( L cos α − L sin α ) e + ( L sin α + L cos α ) e + L e . (25)8n the principal axes of inertia the angular momentum has the form K = K e + K e + K e = A ω e + A ω e + A ω e , (26)and the kinetic energy of the body can be written as follows: T = 12 (cid:18) K A + K A + K A (cid:19) . (27)It follows from (25), (26) that the components L , L , L of the angular momentum K withrespect to the Oη η η coordinate system are connected with the components K , K , K of thisvector with respect to the principal axes of inertia by the formulae: K = L cos α − L sin α, K = L sin α + L cos α, K = L . (28)If we substitute expressions (28) to the formula (27) for the kinetic energy of the body and takeinto account the explicit expressions (24) for sin α and cos α , we obtain the following expressionfor the kinetic energy of the body as a function of L , L , L : T = 12 (cid:18) x A + x A (cid:19) L ( x + x ) + 12 (cid:18) x A + x A (cid:19) L ( x + x ) + 12 L A + (cid:18) A − A (cid:19) x x ( x + x ) L L . (29)Note that expression (29) for the kinetic energy of the rigid body does not contain the products L L and L L . This means, that the kinetic energy of the body in the considered case has theform (21), i.e. the unit vectors e I , e II , e III are indeed the basis vectors of the special coordinatesystem by P. V. Kharlamov [6, 7]. The coefficients of the kinetic energy (29) are such, that b = 0 , l ∗ = 0 . Moreover, according to (7) we have: A x + A x A A ( x + x ) = 1 A , and therefore a = a . Finally expression (29) for the kinetic energy of the body with a fixed point in the Hess casecan be written as follows: T = 12 aL + 12 c (cid:0) L + L (cid:1) + bL L , (30)where we denote a = A x + A x A A ( x + x ) , b = ( A − A ) x x A A ( x + x ) , c = 1 A . ˙ L = − bL L , ˙ L = ( a − c ) L L + bL L + ν Γ , ˙ L = − ( a − c ) L L + bL − bL − ν Γ , ˙ ν = cL ν − ( cL + bL ) ν , ˙ ν = − cL ν + ( aL + bL ) ν , ˙ ν = ( bL + cL ) ν − ( aL + bL ) ν . (31)To find the additional first integral, existing in the Hess case, we consider the first equation ofthe system (31) ˙ L = − bL L . In this equation, the right – hand side is equal to the variable L itself, multiplied by thecoefficient − bL bounded in absolute value. This means that if at the initial instant of time thequantity L = 0 , then we have L ≡ . (32)The invariant manifold (32) (or, in other notations (8)) together with (7) defines the Hess case.Under conditions (7), (32) equations (31) are noticeably simplified and take the form ˙ L = bL L + ν Γ , ˙ L = − bL − ν Γ , ˙ ν = cL ν − cL ν , ˙ ν = bL ν − cL ν , ˙ ν = cL ν − bL ν . (33)Equations (33) possess the following first integrals: c (cid:0) L + L (cid:1) + Γ ν = E ; L ν + L ν = k ; ν + ν + ν = 1 . (34)Note that condition b = 0 corresponds to the Lagrange integrable case in the problem of motionof a heavy rigid body with a fixed point. Now let us write the equations (33) and the first integrals (34) in dimensionless form. For thispurpose we introduce the dimensionless components of angular momentum L = r Γ c y, L = r Γ c z, and the dimensionless time τ : t = τ √ Γ c .
10e introduce also the dimensionless parameter d = bc . and the dimensionless constants of the first integrals h = E Γ , k = k r c Γ . Now we can write equations (33) in dimensionless form: dydτ = d yz + ν , dzdτ = − d y − ν ,dν dτ = zν − yν , dν dτ = d yν − zν , dν dτ = yν − d yν . (35)System (35) possesses the following first integrals: y + z ν = h, yν + zν = k , ν + ν + ν = 1 . (36)From the system (35) using (36) we will obtain the second order linear differential equation.Before we obtain this equation, let us determine the range of parameters d , h , k . It is easy tosee that the parameter k ranges in the infinite interval ( −∞ , + ∞ ) . Since the expression y + z is nonnegative, then we have for the parameter h the following inequality h − ν ≥ , or h ≥ ν . The minimal value of the first component ν of the vector ν equals − . Therefore the parameter h ranges in the interval h ∈ [ − , + ∞ ) . The parameter d can be represented as follows: d = bc = ( A − A ) x x ( A x + A x ) . First of all we note that d < according to (7). We transform the expression for d to theform: d = ( A − A ) x x ( A x + A x ) = ( A − A ) x x (cid:16) A x x + A (cid:17) . (37)From the conditions (7) of existence of the Hess integral, we find x x = A ( A − A ) A ( A − A ) , то есть x x = p A ( A − A ) p A ( A − A ) . d as follows: d = − s ( A − A ) ( A − A ) A A , or d = ( A − A ) ( A − A ) A A . Since the moments of inertia satisfy the triangle inequality A + A > A , i.e. A − A A < ,A + A > A , i.e. A − A A < , then taking into account these inequalities we have d = ( A − A ) A · ( A − A ) A < . Since d < then we finally obtain that the parameter d ranges in the interval d ∈ ( − , . We obtain now the second order linear differential equation from the system (35) using (36).Multiplying the first equation of the system (35) by y and the second by z and adding them, weget: ddτ (cid:18) y + z (cid:19) = yν − zν . (38)Using the following identity (cid:0) y + z (cid:1) (cid:0) ν + ν (cid:1) = ( yν + zν ) + ( yν − zν ) , we find from the first integrals (36): ν = h − y + z . Therefore we have ν + ν = 1 − (cid:18) h − y + z (cid:19) = 1 − (cid:18) y + z − h (cid:19) ,yν + zν = k . Finally we obtain ( yν − zν ) = (cid:0) y + z (cid:1) − (cid:18) y + z − h (cid:19) ! − k . We will take that yν − zν = − vuut ( y + z ) − (cid:18) y + z − h (cid:19) ! − k (39)12we can choose the arbitrary sign before the square root in (39)). Taking into account (39) we canrewrite (38) in the form: ddτ (cid:18) y + z (cid:19) = − vuut ( y + z ) " − (cid:18) y + z − h (cid:19) − k . Now we multiply the first equation of the system (35) by z and the second – by y and subtractthe first equation from the second equation. As a result we obtain: y dzdτ − z dydτ = − d y − yν − d yz − zν , or, taking into account (36) y dzdτ − z dydτ = − d y (cid:0) y + z (cid:1) − k . We pass now from the variables y and z to the polar coordinates x and ϕ by putting: y = x cos ϕ, z = x sin ϕ. Then for the variables x and ϕ we have the following system of two differential equations: x dxdτ = − vuut x " − (cid:18) x − h (cid:19) − k ,x dϕdτ = − d x cos ϕ − k . (40)From this system we obtain the single first order differential equation for the function ϕ = ϕ ( x ) : dϕdx = d x cos ϕ + k x vuut x " − (cid:18) x − h (cid:19) − k . (41)Note that when we pass from the system (40) to the equation (41) we exclude the case x = const that is, y + z = const or ν = const from consideration. Meanwhile for a heavy rigid body witha fixed point in the Hess case there are steady motions for which ν = ν = const (see, forexample [19]).The substitution w = tan ϕ reduces (41) to the Riccati equation: dwdx + ( d x − k ) w − d x − k x vuut x " − (cid:18) x − h (cid:19) − k = 0 .
13r similarly dwdx = − d x − k x vuut x " − (cid:18) x − h (cid:19) − k w + d x + k x vuut x " − (cid:18) x − h (cid:19) − k . It is well known from the general theory of ordinary differential equations (see, for example [20]),that if the Riccati equation has the form: dwdx = f ( x ) w + f ( x ) w + f ( x ) , then the substitution of the form u ( x ) = exp (cid:18) − Z f ( x ) w ( x ) dx (cid:19) reduces it to the second order linear differential equation f d udx − (cid:18) df dx + f f (cid:19) dudx + f f u = 0 (42)or, if we divide this equation by f : d udx − (cid:18) f df dx + f (cid:19) dudx + f f u = 0 . (43)In our case f = − d x − k x vuut x " − (cid:18) x − h (cid:19) − k , f = 0 , f = d x + k x vuut x " − (cid:18) x − h (cid:19) − k . Note that the transition from the equation (42) to the equation (43) is possible only when f = 0 . Taking into account the fact that x = const , the condition f = 0 is equivalent to thesimultaneous fulfillment of the conditions d = 0 , k = 0 . (44)Under the conditions (44), the equation (41) gives ϕ = ϕ = const . It can be shown (seeAppendix) that, under the conditions (44) a heavy rigid body with a fixed point in the Hess casewill perform pendulum nutational oscillations. We will assume further that f = 0 . Thus, theproblem of motion of a heavy rigid body with a fixed point in the Hess case is reduced to solvingthe following second order linear differential equation with the rational coefficients: d udx + a ( x ) dudx + b ( x ) u = 0 , (45) a ( x ) = d x − k x − d ( h − x + 12 k hx − k d x − k ( h − x − k x ( x − hx + 4 ( h − x + 4 k ) ( d x − k ) ,b ( x ) = ( d x + k ) ( d x − k ) x ( x − hx + 4 ( h − x + 4 k ) . Now we can study the problem of existence of liouvillian solutions for the second order lineardifferential equation (45). To solve this problem we can use the Kovacic algorithm [8]. Below wegive a brief description of this algorithm. 14
Description of the Kovacic algorithm
Let us consider the differential field C ( x ) of rational functions of one (in general case complex)variable x . We accept the standard notations Z and Q for the sets of integer and rational numbersrespectively. Our goal is to find a solution of the differential equation d zdx + a ( x ) dzdx + b ( x ) z = 0 , (46)where a ( x ) , b ( x ) ∈ C ( x ) . In the paper [8] an algorithm has been proposed that allows one tofind explicitly the so-called liouvillian solutions of differential equation (46), i.e. solutions, thatcan be expressed in terms of liouvillian functions. The main advantage of the Kovacic algorithmis precisely that it allows one not only to establish the existence or nonexistence of a solutionof differential equation (46) expressed in terms of liouvillian functions, but also to present thissolution in an explicit form when it exists. In turn, liouvillian functions are elements of a liouvillianfield, which is defined in the following way. Определение 1
Let F be a differential field of functions of one (in general case complex) variable x that contains C ( x ) ; namely F is a field of characteristic zero with a differentiation operator () ′ with the following two properties: ( a + b ) ′ = a ′ + b ′ and ( ab ) ′ = a ′ b + ab ′ for any a and b in F . Thefield F is liouvillian if there exists a sequence (tower) of differential fields C ( x ) = F ⊆ F ⊆ . . . ⊆ F n = F, obtained by adjoining one element such that for any i = 1 , , . . . , n we have: F i = F i − ( α ) , with α ′ α ∈ F i − (i.e. F i is generated by an exponential of an indefinite integral over F i − ); or F i = F i − ( α ) , with α ′ ∈ F i − (i.e. F i is generated by an indefinite integral over F i − ); or F i is finite algebraic over F i − (i.e. F i = F i − ( α ) and α satisfies a polynomial equation of the form a + a α + · · · + a n α n = 0 , where a j ∈ F i − , j = 0 , , , . . . , n and are not all zero). Thus, liouvillian functions are built up sequentially from rational functions by using algebraicoperations and the operation of indefinite integration and by taking the exponential of a givenexpression. A solution of equation (46) that is expressed in terms of liouvillian functions mostclosely correspond to the notion of a "close-form solution" or a "solution in quadratures". Toreduce differential equation (46) to a simpler form, we use the following formula y ( x ) = z ( x ) exp (cid:18) Z a ( x ) dx (cid:19) . (47)Then equation (46) takes the form: y ′′ = R ( x ) y, R ( x ) = 12 a ′ + 14 a − b, R ( x ) ∈ C ( x ) . (48)15ereinafter, it is assumed that that the second order linear differential equation with whichthe Kovacic algorithm deals is written in the form (48). The following theorem which has beenproved by J. Kovacic [8], determines the structure of a solution of this differential equation. Теорема 1
For the differential equation (48) only the following four cases are true.1. The differential equation (48) has a solution of the form η = exp (cid:18)Z ω ( x ) dx (cid:19) e ω ( x ) ∈ C ( x ) (liouvillian solution of type ).2. The differential equation (48) has a solution of the form η = exp (cid:18)Z ω ( x ) dx (cid:19) , where ω ( x ) is an algebraic function of degree over C ( x ) and case does not hold (liouvilliansolution of type ).3. All solutions of differential equation (48) are algebraic over C ( x ) and cases and do nothold. In this situation a solution of the differential equation (48) has the form η = exp (cid:18)Z ω ( x ) dx (cid:19) where ω ( x ) is an algebraic function of degree , or over C ( x ) (liouvillian solution oftype ).4. Differential equation (48) has no liouvillian solutions. In order for one of the first three cases listed in Theorem 1 to take place the function R ( x ) inthe right hand side of equation (48) must satisfy certain conditions. These conditions are necessarybut not sufficient. For example, if the conditions corresponding to Case 1 of Theorem 1 are violated,then we must turn to the verification of the conditions corresponding to Cases 2 and 3. If theseconditions are fulfilled, then we must search for solutions of equation (48) exactly in the form,indicated for the corresponding case. However, the existence of such a solution is not guaranteed.In order to explain the sense of the necessary conditions mentioned, we recall some facts fromcomplex analysis.Recall that any analytic function f of a complex variable z can be expanded in a Laurent seriesin a neighborhood of any point a as follows: f ( z ) = a + a ( z − a ) + a ( z − a ) + · · · + a − z − a + a − ( z − a ) + · · · . The part of this series a + a ( z − a ) + a ( z − a ) + · · · containing nonnegative powers of z − a is called the analytic part of the Laurent series. Whereasthe other part, namely a − z − a + a − ( z − a ) + · · ·
16s called the principal part of the expansion. By definition, a point a is called a pole of f ( z ) oforder n if the principal part of the Laurent expansion contains a finite number of terms and thelast term has the form a − n ( z − a ) n . If f ( z ) is a rational function of z , then a point a is a pole of f ( z ) of order n if it is a root ofthe denominator of f ( z ) of multiplicity n .Let z = ∞ be a zero of a function f ( z ) of order n (i.e., n is the order of the pole at z = 0 of f ( z ) ). Then we say that n is the order of f ( z ) at z = ∞ . If f ( z ) is a rational function, then itsorder at z = ∞ is the difference between the degrees of the denominator and the numerator.The following theorem, which has been proved in [8], specifies conditions, that are necessaryfor one of the first three cases listed in Theorem 1 can hold. Теорема 2
For the differential equation (48) the following conditions are necessary for one ofthe first cases listed in Theorem 1 to hold, i.e. for equation (48) to have a liouvillian solution ofthe type specified in description of the corresponding case.1. Each pole of the function R ( x ) must have even order or else have order . The order of R ( x ) at x = ∞ must be even or else be greater than 2.2. The function R ( x ) must have at least one pole that either has odd order greater than orelse has order .3. The order of a pole of R ( x ) cannot exceed and the order of R ( x ) at x = ∞ must be atleast . If the partial fraction expansion of R ( x ) has the form R ( x ) = X i α i ( x − c i ) + X j β j x − d j , then for each i √ α i ∈ Q , X j β j = 0 and if γ = X i α i + X j β j d j , then p γ ∈ Q . To find a liouvillian solution of type 1 of the differential equation (48), the Kovacic algorithmis stated in the following way (see [8] for details). We assume that the necessary conditions forthe existence of a solution in case 1 are satisfied and denote the set of finite poles of the function R ( x ) by Γ . Step 1.
For each c ∈ Γ S {∞} we define a rational function h √ R i c and two complex numbers α + c and α − c as described below. 17 c ) If c ∈ Γ is a pole of order 1, then h √ R i c = 0 , α + c = α − c = 1 . ( c ) If c ∈ Γ is a pole of order 2, then h √ R i c = 0 . Let b be the coefficient of x − c ) in the partial fraction expansion of R ( x ) . Then α ± c = 12 ± √ b. ( c ) If c ∈ Γ is a pole of order ν ≥ (the order must be even due to the necessaryconditions, stated in Theorem 2), then h √ R i c is the sum of terms involving x − c ) i for ≤ i ≤ ν in the Laurent expansion of √ R at c . There are two possibilities for h √ R i c that differ by sign; we can choose one of them. Thus, h √ R i c = a ( x − c ) ν + · · · + d ( x − c ) . Let b be the coefficient of x − c ) ν +1 in R − h √ R i c . Then α ± c = 12 (cid:18) ± ba + ν (cid:19) . ( ∞ ) If the order of R ( x ) at x = ∞ is greater than 2, then h √ R i ∞ = 0 , α + ∞ = 1 , α −∞ = 0 . ( ∞ ) If the order of R ( x ) at x = ∞ is 2, then h √ R i ∞ = 0 . Let b be the coefficient of x in the Laurent series expansion of R ( x ) at x = ∞ . Then α ±∞ = 12 ± √ b. ( ∞ ) If the order R ( x ) at x = ∞ is − ν ≤ (it is even due to the necessary conditions statedin Theorem 2), then the function h √ R i ∞ is the sum of terms involving x i , ≤ i ≤ ν
18f the Laurent expansion of √ R at x = ∞ (one of the two possibilities can be chosen).Thus, h √ R i ∞ = ax ν + · · · + d. Let b be the coefficient of x ν − in R − (cid:16)h √ R i ∞ (cid:17) . Then we have: α ±∞ = 12 (cid:18) ± ba − ν (cid:19) . Step 2.
For each family s = ( s ( c )) c ∈ Γ S {∞} , where s ( c ) are either + or − let d = α s ( ∞ ) ∞ − X c ∈ Γ α s ( c ) c . (49)If d is a non – negative integer, then we introduce the function θ = X c ∈ Γ s ( c ) h √ R i c + α s ( c ) c x − c ! + s ( ∞ ) h √ R i ∞ (50)If d is not a non – negative integer, then the family s should be discarded. If all tuples s have been rejected, then Case 1 cannot hold. Step 3.
For each family s from Step 2, we search for a monic polynomial P of degree d (theconstant d is defined by the formula (49)), satisfying the differential equation P ′′ + 2 θP ′ + (cid:0) θ ′ + θ − R (cid:1) P = 0 . (51)If such a polynomial exists, then η = P exp (cid:18)Z θ ( x ) dx (cid:19) is the solution of the differential equation (48). If for each tuple s found on Step 2, we cannotfind such a polynomial P , then Case 1 cannot hold for the differential equation (48).Now we state the Kovacic algorithm to search for a solution of type 2 of differentialequation (48). We denote the set of finite poles of the function R ( x ) by Γ . Step 1.
For each c ∈ Γ S {∞} we define the set E c as follows. ( c ) If c ∈ Γ is a pole of order 1, then E c = { } . ( c ) If c ∈ Γ is a pole of order 2 and if b is the coefficient of x − c ) in the partial fractionexpansion of R ( x ) , then E c = n(cid:16) k √ b (cid:17) \ Z o , k = 0 , ± . c ) If c ∈ Γ is a pole of order ν > , then E c = { ν } . ( ∞ ) If R ( x ) has order > at x = ∞ , then E ∞ = { , , } . ( ∞ ) If R ( x ) has order 2 at x = ∞ and b is the coefficient of x in the Laurent seriesexpansion of R at x = ∞ , then E ∞ = n(cid:16) k √ b (cid:17) \ Z o , k = 0 , ± . ( ∞ ) If R ( x ) has order ν < at x = ∞ , then E ∞ = { ν } . Step 2.
Let us consider the families s = ( e ∞ , e c ) , c ∈ Γ , where e c ∈ E c , e ∞ ∈ E ∞ and at leastone of these numbers is odd. Let d = 12 e ∞ − X c ∈ Γ e c ! . (52)If d is a non – negative integer, the family should be retained, otherwise it should bediscarded. Step 3.
For each family retained from Step 2, we form the rational function θ = 12 X c ∈ Γ e c x − c (53)and search for a monic polynomial P of degree d (the constant d is defined by theformula (52)) such, that P ′′′ + 3 θP ′′ + (cid:0) θ + 3 θ ′ − R (cid:1) P ′ + (cid:0) θ ′′ + 3 θθ ′ + θ − Rθ − R ′ (cid:1) P = 0 . (54)If success is achieved, we set ϕ = θ + P ′ P and let ω be a solution of the quadratic equation (algebraic equation of degree 2) of theform: ω − ϕω + 12 ϕ ′ + 12 ϕ − R = 0 . Then η = exp (cid:18)Z ω ( x ) dx (cid:19) − is a solution of the differential equation (48). If success is not achieved, then Case 2 cannothold for the differential equation (48).Similarly the Kovacic algorithm is stated to search for a liouvillian solutions of type 3 of thedifferential equation (48). Let us apply now this algorithm to search liouvillian solutions of thesecond order linear differential equation (45). 20 Application of the Kovacic algorithm to the differential equation (45) .General case.
So, the differential equation being investigated has the form (45). In this equation we make asubstitution according to (47) and reduce it to the form (48): d ydx = R ( x ) y. (55)Here the function R ( x ) takes the form: R ( x ) = U ( x ) V ( x ) , (56) U ( x ) = − (1 + 4 d ) d x + 8 (2 d − d hx + 4 (2 d + 5) d k x − d −
7) ( h − d x −− d + 19) d k hx + 8 ((1 + 17 d − d ) k −
12 ( h − d h ) x + 576 ( h − k d x ++16 (2 ( h − d + 29 h − d k x + 8 (cid:16) h − d − (7 + 40 d ) k h (cid:17) x − k hx ++32 (( d − k −
21 ( h − h ) d k x + 16 (14 ( h − d + 8 h − k x + 288 k d x ++96 (cid:16) h − − k h (cid:17) d k x + 4 ((32 d + 35) k −
24 ( h − h ) k x + 48 ( h − k .V ( x ) = 4 (cid:0) d x − k (cid:1) (cid:0) x − hx + 4 (cid:0) h − (cid:1) x + 4 k (cid:1) . Thus, it is easy to see, that the function R ( x ) has nine finite poles of the second order. Let usdenote the roots of the polynomial x − hx + 4 (cid:0) h − (cid:1) x + 4 k = 0 (57)by x , x , x , x , x , x . Note that this polynomial contains only the terms of even degree, thereforeits roots satisfy the conditions: x = − x , x = − x , x = − x . Let us denote the roots of the polynomial d x − k = 0 (58)by x , x , x . Now let us consider the partial fraction expansion of the function R ( x ) . It has theform: R ( x ) = − X i =1 x − x i ) + X i =1 γ i ( x i ) x − x i + 34 X i =7 x − x i ) . The coefficients γ i ( x i ) , i = 1 , , . . . have a very complicated form and we do not write themexplicitly here. It is possible to note the following properties of the partial fraction expansion ofthe function R ( x ) . 21. The coefficients b , . . . , b of b i ( x − x i ) , i = 1 , . . . , are all equal b i = − , i = 1 , . . . , .
2. The coefficients b , b , b of b i ( x − x i ) , i = 7 , , are all equal b i = 34 , i = 7 , , .
3. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = − (1 + 4 d )4 x + O (cid:18) x (cid:19) Thus, we have b ∞ = − − d , and therefore b ∞ = − d This means that the numbers α ±∞ calculating during the application of the Kovacic algorithmfor searching the liouvillian solutions of type 1, are complex numbers if d = 0 . All the remainingnumbers α ± c are rational. They are presented in the following Table. Therefore, the number d ,calculated by formula (49) in the process of searching for liouvillian solutions of type 1, is acomplex number for d = 0 . This fact indicates the absence of liouvillian solutions of type 1 for d = 0 . x x x x x x x x x α + c
34 34 34 34 34 34 32 32 32 α − c
14 14 14 14 14 14 − − − Table. Numbers α ± c for searching liouvillian solutions of type 1.Moreover, the coefficient b ∞ coincides with the number γ calculating during the checking of thenecessary conditions of existence of liouvillian solutions of type 3 for the differential equation (48).According to this necessary conditions for existence of liouvillian solutions of type 3 the number p γ = p b ∞ should be rational. However, when d = 0 this number is pure imaginary. Thus, we can statethat for d = 0 the second order linear differential equation (55) (or (45)) do not have liouvilliansolutions of type 3. Thus, the following Theorem is valid.22 еорема 3 If all roots of polynomials (57) and (58) are distinct and d = 0 , then the problemof motion of a heavy rigid body with a fixed point in the Hess case has not liouvillian solutions oftype 1 and type 3. According to Theorem 3, equation (55) can have liouvillian solutions of type 1 only when d = 0 , i.e. when the moving rigid body with a fixed point is the Lagrange top. To search liouvilliansolutions of type 1 for the differential equation (55) in the Lagrange integrable case we put d = 0 in this equation. When d = 0 we can write equation (55) as follows: d ydx = R ( x ) y = U ( x ) V ( x ) y, (59) U ( x ) = 2 x − hx + 4 (8 h − x + (35 k − h ( h − x − k hx + 12 k ( h − ,V ( x ) = (cid:0) x − hx + 4 (cid:0) h − (cid:1) x + 4 k (cid:1) . We assume that in the equation (59) there should be k = 0 . Otherwise, the conditions (44) aresatisfied and it is impossible to obtain from the equation (41) the second-order linear differentialequation (45) (or (55)), from which for d = 0 we obtain (59).It is easy to see that the poles of the function R ( x ) are the roots of the polynomial (57). Weassume that the polynomial (57) has no multiple roots (the possibility of multiple roots for thepolynomial (57) is considered below, in Section 12). Then the function R ( x ) has six finite poles ofthe second order. We denote these poles by x , x , x , x , x , x . Now let us consider the partialfraction expansion of the function R ( x ) . It has the form: R ( x ) = − X i =1 x − x i ) + X i =1 γ i ( x i ) x − x i ,γ i ( x i ) = (cid:16)
48 ( h − − k h ( h −
1) ( h − − (121 h − k (cid:17) x i k (cid:0) k + 8 k h ( h − −
16 ( h − (cid:1) ++ (cid:16) k ( h − − h ( h − + 3 k h (61 k − h ) (cid:17) x i k (cid:0) k + 8 k h ( h − −
16 ( h − (cid:1) ++ (cid:16)
48 ( h − − k h ( h − − k − h ) k (cid:17) x i k (cid:0) k + 8 k h ( h − −
16 ( h − (cid:1) . It is possible to note the following properties of the partial fraction expansion of the function R ( x ) .1. The coefficients b , . . . , b of b i ( x − x i ) , i = 1 , . . . , are all equal b i = − , i = 1 , . . . , .
23. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = 2 x + O (cid:18) x (cid:19) . Our goal is to find liouvillian solutions of type 1, type 2 and type 3 of the differentialequation (59). To find liouvillian solutions of (59) we will use the Kovacic algorithm. First wesearch liouvillian solutions of type 1. According to the Kovacic algorithm, we calculate the numbers α ± c . For all the roots x = x i , i = 1 , . . . , these numbers are equal α + x i = 34 , α − x i = 14 , i = 1 , . . . . (60)The numbers α ±∞ are equal α + ∞ = 2 , α −∞ = − . (61)Note that the polynomial (57) contains the terms of even degree, therefore its roots can berepresented as follows: x = y , x = − x = − y , x = y , x = − x = − y , x = iy , x = − x = − iy . Since the numbers α ± x i are determined by (60) and the numbers α ±∞ are determined by (61)therefore the number d calculating according to (49) can be only zero. This takes place for thefollowing sets of signs + and − : s = ( s ( ∞ ) , s ( x ) , s ( x ) , s ( x ) , s ( x ) , s ( x ) , s ( x )) s = (+ , + , − , − , − , − , − ) , s = (+ , − , + , − , − , − , − ) , s = (+ , − , − , + , − , − , − ) ,s = (+ , − , − , − , + , − , − ) , s = (+ , − , − , − , − , + , − ) , s = (+ , − , − , − , − , − , +) . We must check all these sets. Let us check now the set s . For this set of signs let us find thefunction θ according to the formula (50). This function has the form: θ = 34 ( x − y ) + 14 ( x + y ) + 14 ( x − y ) + 14 ( x + y ) + 14 ( x − iy ) + 14 ( x + iy ) . The polynomial P of degree d = 0 has P ≡ . The substitution of the polynomial P and thefunction θ with the differential equation (51) reduces it to the form: dθdx + θ − R ( x ) = 0 . (62)The conditions on the parameters k and h for which the left hand side of (62) becomes zeroare the conditions of existence of liouvillian solutions of type 1 for the differential equation (59).Before we substitute the function θ to the equation (62), let us simplify it. We can rewrite thisfunction in the form: θ = 12 ( x − y ) + 14 (cid:20) x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy (cid:21) . Expression in square brackets can be represented as follows: x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy = 2 x (3 x − hx + 4 ( h − x − hx + 4 ( h − x + 4 k . θ has the form: θ = 12 ( x − y ) + x (3 x − hx + 4 ( h − x − hx + 4 ( h − x + 4 k ) . If we substitute this function θ to (62) we obtain in the left hand side of (62) the rationalexpression. The numerator of this expression is the fifth degree polynomial of the form: P ( x ) = 4 y x + · · · The necessary condition for this polynomial to vanish is the condition y = 0 , that is, thepolynomial (57) must have a zero root. This condition is equivalent to the condition k = 0 .However, we previously suggested that k = 0 . Therefore, y = 0 and we can state that theexpression in the left side of the equation (62) does not vanish. Thus, for the set s there are noliouvillian solutions of type 1 in the problem of the motion of a heavy rigid body with a fixedpoint in the Hess case.Similarly, the fact of the absence of liouvillian solutions of type 1 for the sets s , . . . , s isestablished. So, we can state that the following theorem is true. Теорема 4
If all roots of polynomial (57) are distinct and d = 0 , k = 0 , then the problem ofmotion of a heavy rigid body with a fixed point in the Hess case has not liouvillian solutions oftype . Now we continue the study of the existence of liouvillian solutions in the problem of the motionof the Hess top under the additional condition d = 0 , that is, in the problem of the motion of theLagrange top when the corresponding second-order linear differential equation has the form (59).Let us study here the problem of the existence of liouvillian solutions of type 2 for the differentialequation (59).According to the Kovacic algorithm for searching the liouvillian solutions of type 2 we shouldfind the sets E x i , i = 1 , . . . , which correspond to finite poles of R ( x ) and the set E ∞ whichcorresponds to the pole of R ( x ) at x = ∞ . For the finite poles x = x i , i = 1 , . . . , , which are theroots of the polynomial (57), the sets E x i are all equal E x i = { , , } , i = 1 , . . . , . The set E ∞ has the following form: E ∞ = {− , , } . Now we should calculate the constant d using (52). Note that the minimal value of the sum X x i ∈ Γ e x i equals 6. Therefore, the maximal value of d , calculated according to (52), equals d = 1 . The family s = ( e ∞ , e , e , e , e , e , e ) corresponding to d = 1 is s = (8 , , , , , , .
25e also have several families, for which we have d = 0 . The following families correspond to d = 0 : s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , ,s = (8 , , , , , , , s = (8 , , , , , , , s = (8 , , , , , , . We must check all these families. We start with the family s . According to the algorithm letus find the function θ by the formula (53). Since the coefficients at all finite poles are the same,we can write the function θ in explicit form. For the family s this function has the form: θ = 3 x − hx + 4 ( h − xx − hx + 4 ( h − x + 4 k . The polynomial P of degree d = 1 ( P ≡ x + b ) should identically satisfy differentialequation (54). After substitution of the polynomial P and the functions θ and R ( x ) in (54),we obtain in the left hand side of it the following expression: Bx (6 h − x ) x − hx + 4 ( h − x + 4 k . This expression becomes identically zero when B = 0 . Therefore, the polynomial P ≡ x existsfor all values h and k . Thus, we can state the following theorem. Теорема 5
In the Lagrange integrable case of motion d = 0 under the Hess conditions (7) andalso under condition that all the roots of polynomial (57) are distinct and k = 0 , all solutions ofa linear differential equation (55) are liouvillian solutions of type . Indeed, it is easy to find the general solution of equation (45) for d = 0 . In this case thedifferential equation (45) takes the form d udx + 4 ( x − hx + 2 ( h − x + k ) x ( x − hx + 4 ( h − x + 4 k ) dudx − k ux ( x − hx + 4 ( h − x + 4 k ) = 0 , and its general solution can be written as follows: u ( x ) = C exp Z k dxx p x − hx + 4 ( h − x + 4 k ! ++C exp − Z k dxx p x − hx + 4 ( h − x + 4 k ! .
26f we check the other families s , . . . , s we obtain the same results on the existing of liouvilliansolutions of type 2 for the differential equation (59).Now we come back to the investigation of the general case d = 0 and consider the problem ofexistence of liouvillian solutions of type 2 for the differential equation (55). So we are going to study the problem of existence of liouvillian solutions of type 2 for the differentialequation (45) (or (55)). According to the Kovacic algorithm we firstly define the sets E c and E ∞ for every pole of the function R ( x ) . For the finite poles x = x i , i = 1 , . . . , , which are roots ofthe polynomial (57), these sets E x i have the form: E x i = { , , } , i = 1 , . . . , . For the finite poles x = x i , i = 7 , , , which are roots of the polynomial (58), these sets E x i have the form: E x i = {− , , } , i = 7 , , . The set E ∞ contains only one element and this set equals E ∞ = { } . Now we should calculate the constant d using the formula (52). Note that the minimal valuesof the sum of the elements of sets corresponding to finite poles is zero. Therefore the maximalvalue of d , calculated according to (52), equals d = 1 . The value d = 1 corresponds to the set s = ( e ∞ , e , e , e , e , e , e , e , e , e ) , in which the elements e ∞ and e i , i = 1 , , . . . are equal s = (2 , , , , , , , − , − , − . We also have several families for which the constant d , calculated by the formula (52), equalsto zero. The following families correspond to d = 0 : s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − , = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − , s = (2 , , , , , , , − , − , − ,s = (2 , , , , , , , − , − , − . We must check all these families. We start with the family s . According to the algorithm,let us find the function θ by the formula (53). Since the coefficients at all finite poles x = x i , i = 1 , , . . . , are the same, and the coefficients at all poles x = x i , i = 7 , , are the same, thenwe can write the function θ in explicit form. For the set s this function has the form θ = 3 x − hx + 4 ( h − xx − hx + 4 ( h − x + 4 k − d x d x − k . The polynomial P of degree d = 1 P = x + B should identically satisfy the differential equation (54). After substitution of P = x + B and thefunctions θ and R ( x ) to the equation (54), we obtain in the left hand side of (54) the rationalexpression. The numerator of this expression has a form of the ninth degree polynomial: P = − Bd (cid:0) d (cid:1) x + · · · Let us put B = 0 . Then the numerator of the rational expression in the left hand side of (54)takes the form: P = − xk d (cid:0) d x − k (cid:1) + · · · This polynomial becomes zero when k = 0 or when d = 0 . Therefore we can state thefollowing Theorem based on the verification of the family s . Теорема 6
For the existence of liouvillian solutions of type 2 in the problem of motion of a heavyrigid body with a fixed point in the Hess case one of the two conditions must be satisfied: d = 0 or k = 0 . In other words, liouvillian solutions of type can exist either in the case, when the movingrigid body is the Lagrange top, or in the Hess case, if the constant of the area integral is zero. The fact, that the problem of motion of a heavy rigid body with a fixed point in the Hess casefor k = 0 is integrable in elliptic functions (which are liouvillian functions) was firstly discoveredby P. A. Nekrasov [17, 18].To prove Theorem 6 we need to check the families s , . . . , s , and to consider various criticalcases, which we will discuss further, in Sections 11 – 15. We start with families s , . . . , s . For thefamily s the function θ , calculated by the formula (53), has the form: θ = 12 (cid:18) x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy (cid:19) − d x d x − k . We slightly simplify this expression for the function θ . We can rewrite it in the form: θ = 12 (cid:20) x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy (cid:21) + 1 x − y − d x d x − k . x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy = 2 x (3 x − hx + 4 ( h − x − hx + 4 ( h − x + 4 k . Finally we have the following expression for the function θ : θ = 3 x − hx + 4 ( h − xx − hx + 4 ( h − x + 4 k + 1 x − y − d x d x − k . The polynomial P of degree d = 0 has the form P ≡ . This polynomial should identicallysatisfy the differential equation (54). After substitution of P = 1 and the functions θ and R ( x ) inthis equation, we obtain in the left hand side of (54) the rational expression. The numerator ofthis expression has a form of the ninth degree polynomial of x , the leading coefficient of which isequal to P = y d (cid:0) d (cid:1) x + · · · If we put y = 0 (this condition is equivalent to the condition k = 0 ), then this polynomial P becomes zero. Thus, checking of the family s gives the same conditions as the checking of thefamily s . These conditions have been formulated in Theorem 6. Similarly we can check the otherfamilies s , . . . , s .Let us check now the families s , . . . , s . For the family s the function θ , calculated by theformula (53), has the form: θ = 1 x − y + 1 x + y + 12 (cid:18) x − y + 1 x + y + 1 x − iy + 1 x + iy (cid:19) − d x d x − k . We rewrite this function as follows: θ = 12 ( x − y ) + 12 ( x + y ) + 12 (cid:18) x − y + 1 x + y + 1 x − y + 1 x + y + 1 x − iy + 1 x + iy (cid:19) − d x d x − k . The latter expression can be transformed to the form: θ = 3 x − hx + 4 ( h − xx − hx + 4 ( h − x + 4 k − d x d x − k + xx − y . The polynomial P of degree d = 0 has the form P ≡ . This polynomial should identicallysatisfy the differential equation (54). After substituting P = 1 and the functions θ and R ( x ) inthis equation, we obtain in the left hand side of (54) the rational expression. The numerator ofthis expression has a form of 13th degree polynomial P = 4 y d (cid:0) d (cid:1) x + · · · If we put y = 0 and k = 0 then this polynomial becomes zero. Thus, checking of the family s gives the same conditions as checking of the family s . These conditions have been formulated inTheorem 6. Similarly we can check the other families s , . . . , s . Checking of these families givesus the same conditions as checking of the family s .Finally we can state that the conditions of existence of liouvillian solutions of type 2 for thedifferential equation (55) are formulated in Theorem 6. To confirm the obtained results let usconsider the equation (55) in the case when k = 0 .29 k = 0 . In the case k = 0 , the function R ( x ) , defined by (56), has the form: R ( x ) = U ( x ) V ( x ) ,U ( x ) = − (cid:0) d (cid:1) x + 8 h (cid:0) d − (cid:1) x + 8 (cid:0) − d (cid:1) (cid:0) h − (cid:1) x − h (cid:0) h − (cid:1) x + 48 (cid:0) h − (cid:1) ,V ( x ) = 4 x (cid:0) x − h − (cid:1) (cid:0) x − h + 2 (cid:1) . Thus, the function R ( x ) has five finite poles of the second order. One of these finite poles is x = 0 and the other four x , x , x , x are the roots of two polynomials: x − h − (we denote its roots by x , x ) and x − h + 2 = 0 (we denote its roots by x , x ). Partial fraction expansion of the function R ( x ) has the form: R ( x ) = 34 x − X i =1
316 ( x − x i ) − X i =1 (4 d ( h + 1) − h + 1) x i
32 ( h + 1) ( x − x i ) + X i =3 (4 d ( h − − h − x i
32 ( h −
1) ( x − x i ) . It is possible to note the following properties on the partial fraction expansion of the function R ( x ) .1. The coefficient b of b x equals b = 34 .
2. The coefficients b , . . . , b of b i ( x − x i ) , i = 1 , , , are all equal b i = − , i = 1 , , , .
3. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = − (1 + 4 d )4 x + O (cid:18) x (cid:19) . To find liouvillian solutions of type 2 of the differential equation (55) in the case k = 0 we willuse the Kovacic algorithm. According to the algorithm, let us define the sets E x i , correspondingto finite poles of the function R ( x ) and the set E ∞ corresponding to the pole of R ( x ) at x = ∞ .For the pole x = 0 the corresponding set has the form E = {− , , } . x i , i = 1 , , , the sets E x i have the form: E x i = { , , } , i = 1 , , , . The set E ∞ contains only one element and this set equals E ∞ = { } . It is easy to see, that the constant d , calculated by the formula (52) is a non-negative integeronly for the family s = { , − , , , , } , for which we have d = 0 . Using this family, let us find the function θ by the formula (53). Thisfunction has a form θ = − x + 2 ( x − h ) x ( x − h −
2) ( x − h + 2) . The polynomial P of degree d = 0 has the form P ≡ . This polynomial should identicallysatisfy the differential equation (54). After substitution of P = 1 and the functions θ and R ( x ) to the differential equation (54) we find that this equation is satisfied identically for any values ofparameters d and h . Thus, we can state the following theorem. Теорема 7
In the case of motion of a heavy rigid body with a fixed point in the Hess case ( d = 0) ,equation (55) has liouvillian solutions only in the case k = 0 . Indeed, when k = 0 the differential equation (45) takes the form: d udx + x − h − x ( x − h −
2) ( x − h + 2) dudx + d x ( x − h −
2) ( x − h + 2) u = 0 . (63)The general solution of the equation (63) has the form: u ( x ) = C exp (cid:18) id Z xdx √ x − hx + 4 h − (cid:19) ++C exp (cid:18) − id Z xdx √ x − hx + 4 h − (cid:19) . It is easy to see, that this function is a liouvillian function of type 2. The general solution ofthe equation (63) can be represented also as follows: u ( x ) = K sin (cid:18) d (cid:16) x − h + √ x − hx + 4 h − (cid:17)(cid:19) ++K cos (cid:18) d (cid:16) x − h + √ x − hx + 4 h − (cid:17)(cid:19) , where K and K are arbitrary constants.Thus we can conclude that conditions of existence of liouvillian solutions of type 2 in theproblem of motion of a heavy rigid body with a fixed point in a Hess case are formulated in31heorem 6. When each of these conditions is fulfilled, liouvillian solutions of type 2 exist regardlessof what values the other parameters of the problem take.Let us summarize the results. In the problem of motion of a heavy rigid body with a fixedpoint in the Hess case there are no liouvillian solutions of type 1 and type 3. Liouvillian solutionsof type 2 exist when d = 0 or when k = 0 .These results have been obtained in general case, when we suppose that all the nine finite polesof the function R ( x ) defined by (56), are distinct. Now we will consider the special cases, whenthese poles can coincide with each other, i.e. the function R ( x ) can have multiply roots.
11 Description of the special cases.
The function R ( x ) defined by (56) has a denominator which is the square of the product oftwo polynomials: the sixth degree polynomial (57) and the cubic polynomial (58). Let us studynow the polynomial (57). As we already noted, this polynomial includes only even degrees of anindependent variable, as a result of which by changing x = z we can reduce it to a third-degreepolynomial with respect to z : P z = z − hz + 4 (cid:0) h − (cid:1) z + 4 k = 0 . (64)Let us study the type of roots of the polynomial (64). If we put z = y + 4 h then the polynomial (64) takes the form y − (cid:18) h (cid:19) y + 4 (cid:18) k + 4 h − h (cid:19) = 0 . (65)Thus the third-degree polynomial (65) has the form y + 3 py + 2 q = 0 , where we denote p = − (cid:18) h (cid:19) , q = 2 (cid:18) k + 4 h − h (cid:19) . The character of the roots of the polynomial (65) is determined by the sign of the expression D = q + p . If D > then the polynomial (65) has one real root and two complex-conjugate roots. If D < then the polynomial (65) has three distinct real roots. If D = 0 then the polynomial (65) has amultiple root and all of its roots are real. In the explicit form expression D can be written asfollows: D = 4 (cid:18) k + 827 (cid:0) h − (cid:1) k h − (cid:0) h − (cid:1) (cid:19) . (66)Thus, if (66) is zero, the polynomial (57) has multiple roots.32e consider now the cubic polynomial (58). The coefficient d is such that d ∈ ( − , . Letus denote k = c d , where c is a new parameter. Then we can write the polynomial (58) as follows: d (cid:0) x − c (cid:1) = 0 . (67)The roots of (67) have the following form: x = c, x = −
12 + i √ ! c, x = − − i √ ! c. (68)It is easy to see, that all these roots x , x , x are distinct for c = 0 . Therefore, since wedo not consider the case k = 0 here, the polynomial (67) cannot have multiple roots. Howeverthe situation is possible when one of the roots (68) of the polynomial (67) is the root of thepolynomial (57). Below we consider in details all possible special cases described in this Section.
12 The special case of multiple roots of the polynomial (57) . Let us consider the first special case when the polynomial (57) has a multiple root. First of all weconsider the case, when d = 0 , i.e. we consider the case of motion of the Lagrange top. For d = 0 the cubic polynomial (58) becomes a constant. Therefore in this case polynomials (57) and (58)have not the common roots. Further in this Section we assume, that k = 0 .The polynomial (57) has multiple roots if expression D defined by (66), is zero. This conditioncan be represented in the form of a biquadratic equation with respect to k : k + 827 (cid:0) h − (cid:1) k h − (cid:0) h − (cid:1) = 0 . (69)From the equation (69) we find the value of k : k = 427 (cid:16)(cid:0) h + 3 (cid:1) √ h + 3 − h (cid:0) h − (cid:1)(cid:17) . In this case the cubic polynomial (64) has three real roots: z = 4 h − √ h + 3 ,z = z = 4 h √ h + 3 . It is easy to see that the roots z and z are positive. This means, that the polynomial (57)has one real positive root of multiplicity 2 and one negative root of multiplicity 2. In addition, thepolynomial (57) has two pure imaginary roots, which are square roots of the negative expression z . Finally in the considered special case the polynomial (57) can be represented as follows: (cid:0) x − z (cid:1) (cid:0) x − z (cid:1) . z and z are connected by the relation z = z − z . This relation allows us to essentially simplify the coefficients a ( x ) and b ( x ) of the differentialequation (45). Indeed, we choose z as a parameter, included in the coefficients a ( x ) and b ( x ) .The parameters h and k are expressed in terms of z as follows: h = 38 z − z , k = (cid:18) z − z (cid:19) √− z . (70)Substitution of expressions (70) together with the condition d = 0 to the coefficients a ( x ) and b ( x ) of the differential equation (45) gives the following expressions for these coefficients: a ( x ) = 16 z x + 2 (16 − z ) x + z ( z − x − z ) (4 z x + 16 − z ) x ,b ( x ) = z ( z − ( z + 4) x − z ) (4 z x + 16 − z ) x . By the formula (48) let us find the function R ( x ) . It is the rational function which denominatorhas the form: V ( x ) = (cid:0) x − z (cid:1) (cid:0) z x − z + 16 (cid:1) , Thus it is easy to see, that the function R ( x ) has four finite poles of the second order. Itspartial fraction expansion has a very complicated form and we do not write it explicitly here.However it is possible to note the following properties of the partial fraction expansion of thefunction R ( x ) .1. The coefficients b i of b i ( x − x i ) , where x i , i = 1 , are roots of the polynomial x − z = 0 , are all equal b i = − , i = 1 , .
2. The coefficients b i of b i ( x − x i ) , where x i , i = 3 , are roots of the polynomial z x − z + 16 = 0 , (71)are all equal b i = − z + 82 (3 z + 16) , i = 3 , .
3. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = 2 x + O (cid:18) x (cid:19) . h and k was proved above). According to the Kovacic algorithm for searchingliouvillian solutions of type 1, let us calculate the constants α ± c . For the finite poles x = x i , i = 1 , these constants are all equal α + x i = 34 , α − x i = 14 , i = 1 , . The constants α ±∞ are equal α + ∞ = 2 , α −∞ = − . It is easy to see, that the coefficients b i of b i ( x − x i ) , where x = x i , i = 3 , are the roots ofthe polynomial (71), depend on the parameter z . Let us estimate these coefficients. First of all,we can state, that b i < , i = 3 , . Let us prove now that the following inequality is valid for thesecoefficients: b i < − , i = 3 , . Indeed, the inequality − z + 82 (3 z + 16) < − can be rewritten in the form: z + 83 z + 16 − > . The latter inequality is equivalent to the inequality − z z + 16) > , which is valid for all z such that k = 0 .Moreover, analysis of the expressions b i , i = 3 , shows, that these expressions take a minimumfor z = 0 . This minimum equals b i = − . Thus, for the coefficients b i , i = 3 , we have the following estimate: − < b i < − . Then the constant α + x i = 12 + 12 p b i = 12 − z p z + 16 , i = 3 , satisfies the estimate < α + x i < , and the constant α − x i = 12 − p b i = 12 + z p z + 16 , i = 3 , < α − x i < , Therefore, in the considered case the constant d , calculated by the formula (49), can be onlyzero. The value d = 0 corresponds to the following sets of signs plus (+) and minus (-): s =( s ( ∞ ) , s ( x ) , s ( x ) , s ( x ) , s ( x )) s = (+ , + , − , + , − ) , s = (+ , − , + , + , − ) ,s = (+ , + , − , − , +) , s = (+ , − , + , − , +) . We need to check all these sets. Let us start with the set s . For the set s we calculate thefunction θ by the formula (50). For the set s this function has the form: θ = 34 ( x − x ) + 14 ( x − x ) + α + x ( x − x ) + α − x ( x − x ) . The polynomial P of degree d = 0 has the form P ≡ . The substitution of this polynomial tothe differential equation (51) reduces it to the equation (62). The substitution of the function θ to the differential equation (62) gives in the left hand side of this equation the rational expressionwhich vanishes only for z = 0 . But we assume that z = 0 . Therefore we can state that therational expression in the left hand side of (62) is not zero. Thus, in the considered critical casefor the set s there are no liouvillian solutions of type 1 in the problem of motion of a heavy rigidbody with a fixed point in the Hess case. The fact of nonexistence of liouvillian solutions of type 1for the sets s , s , s in the considered critical case is established similarly. Finally, we can state,that there are no liouvillian solutions of type 1 in the problem of motion of a heavy rigid bodywith a fixed point in the Hess case.Now let us consider the case d = 0 . Let us choose d and z as a parameters, included in thecoefficients a ( x ) and b ( x ) of the differential equation (45). Substitution of the expressions (70) tothe coefficients a ( x ) and b ( x ) of the differential equation (45) gives the following expressions forthese coefficients: a ( x ) = U ( x ) V ( x ) ,U ( x ) = 32 d z x + ( z − (cid:2) d z x + 16 z √− z x − d z x − z − √− z x ++ z ( z − √− z (cid:3) ,V ( x ) = x (cid:0) x − z (cid:1) (cid:0) z x − z + 16 (cid:1) (cid:0) d z x + (cid:0) z − (cid:1) √− z (cid:1) ,b ( x ) = (8 d z x + 16 √− z − z √− z ) (8 d z x − √− z + z √− z )4 x ( x − z ) (4 z x + 16 − z ) . By the formula (48) let us find the function R ( x ) . It is the rational function which denominatorhas the form: V ( x ) = (cid:0) x − z (cid:1) (cid:0) z x − z + 16 (cid:1) (cid:0) d z x − √− z + z √− z (cid:1) . Thus, it is easy to see that the function R ( x ) has seven finite poles of the second order. Itspartial fraction expansion has a very complicated form and we do not write it explicitly here.However, it is possible to note the following properties of the partial fraction expansion of thefunction R ( x ) . 36. The coefficients b i of b i ( x − x i ) , where x i , i = 1 , , are roots of the cubic polynomial d z x − √− z + z √− z = 0 , are all equal b i = 34 , i = 1 , , .
2. The coefficients b i of b i ( x − x i ) , where x i , i = 4 , are roots of the polynomial x − z = 0 , are all equal b i = − , i = 4 , .
3. The coefficients b i of b i ( x − x i ) , where x i , i = 6 , are roots of the polynomial (71) are allequal b i = 14 (cid:18) z ( d − −
16 ( d + 1)16 + 3 z (cid:19) , i = 6 , .
4. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = − (1 + 4 d )4 x + O (cid:18) x (cid:19) . The Laurent expansion of R ( x ) at x = ∞ allows us to state that in this special case thedifferential equation (45) can have liouvillian solutions of type 2 only. Moreover, it is easy to seethat the coefficients b i of x = x i , i = 6 , depend on the parameters d and z . Let us estimatethese coefficients. According to the restrictions on the parameter d we can state that b i < , i = 6 , . Let us prove now that the following inequality is valid for these coefficients: b i < − , i = 6 , . We rewrite the latter inequality in the form:
16 ( d + 1) + (2 − d ) z
16 + 3 z > . (72)The denominator of expression in the left hand side of (72) is positive. Therefore we canrewrite (72) as follows: d + 16 + 2 z − d z >
12 + 94 z . This inequality is equivalent to the inequality (cid:18) d + 14 (cid:19) > z (cid:18) d + 14 (cid:19) . > z . The latter inequality is valid for all z such that k = 0 .To find liouvillian solutions of the differential equation (45) in the considered special case wewill use the Kovacic algorithm. Let us define the sets E x i , i = 1 , . . . , corresponding to finite polesand the set E ∞ corresponding to the pole at x = ∞ . For the finite poles x = x i , i = 1 , , thecorresponding sets have the form: E x i = {− , , } , i = 1 , , . For the finite poles x i , i = 4 , the corresponding sets E x i have the form: E x i = { , , } , i = 4 , . Taking into account estimations made for the coefficients b i , i = 6 , , we can state that thesets E x i , corresponding to these poles, contain only one element and have the form E x i = { } , i = 6 , . The set E ∞ corresponding to the pole of R ( x ) at x = ∞ also contains only one element andhas the form E ∞ = { } . Now we should determine the families s = ( e ∞ , e x , e x , e x , e x , e x , e x , e x ) for which theconstant d calculating by the formula (52), is a non-negative integer. It is easy to see, that d is anon-negative integer for three families s only. For the family s = (2 , − , − , − , , , , we have d = 1 and for the families s = (2 , − , − , − , , , , , s = (2 , − , − , − , , , , , we have d = 0 . Let us check the family s . Using the elements of this family we calculate thefunction θ by the formula (53). For the family s this function has the form: θ = − d z x d z x − √− z + z √− z + xx − z + 8 z x z x − z + 16 . The polynomial P of degree d = 1 has the form P = x + B , where B indefinite coefficient.This polynomial should identically satisfy differential equation (54). After substitution of thepolynomial P and the functions θ and R ( x ) with this equation we obtain in the left hand sideof (54) the rational expression. The numerator of this expression is the 12th degree polynomial P = − d z (cid:0) d (cid:1) Bx + · · · For the leading coefficient of this polynomial to vanish we put B = 0 . Then the numerator ofthe rational expression in the left hand side of (54) takes the form of the 10th degree polynomial P = 12288 z d √− z ( z −
4) ( z + 4) x + · · · z < , then z − < . We do not consider the cases z = 0 (i.e. k = 0 ) and d = 0 ,because these cases have been already investigated. The in case z = − we have z = 0 , i.e. k = 0 . Therefore the polynomial P is not identically zero. This means that for the for thefamily s the differential equation (45) does not have liouvillian solutions of type 2. Similarly wecan prove that equation (45) does not have liouvillian solutions for the families s and s . Finallywe can state the following Theorem. Теорема 8
In the special case of multiple roots of the polynomial (57) the second-order lineardifferential equation (55) (or (45) ) does not have liouvillian solutions.
Thus, in the first special case we have considered, when the polynomial (57) has multiple roots,equation (55) does not have liouvillian solutions. We turn now to the second special case whenthe polynomials (57) and (58) have a common root.
13 The special case of the common root of the polynomials (57) and (58) . Let us consider now the case, when the root x = c the real root of the polynomial (58) from theroots (68) is also the root of the polynomial (57). The parameter k is expressed in terms of x = c and d as follows: k = c d . The condition that x = c is a root of the polynomial (57) has the form: (cid:0) d (cid:1) c − hc + 4 h − . (73)We express d from the condition (73). As a result we obtain: d = 4 − ( c − h ) c . (74)We will use (74) to simplify the coefficients a ( x ) and b ( x ) of the linear differential equation (45).We will consider c and h as parameters included in these coefficients. Taking into account (74),we can write a ( x ) and b ( x ) as follows: a ( x ) = x ( x − c )( x − c ) ( x + c ) ( x + ( c − h ) x + c − c h + 4 h − −− h − x + (2 h − c + 2) (2 h − c − c x ( x − c ) ( x + xc + c ) ( x + ( c − h ) x + c − c h + 4 h −
4) ++ c x (12 chx + 4 h − − c + 4 c h )( x − c ) ( x + c ) ( x + xc + c ) ( x + ( c − h ) x + c − c h + 4 h − ,b ( x ) = − ( x − xc + c ) ( x + xc + c ) (2 h − c + 2) (2 h − c − c x ( x + ( c − h ) x + c − c h + 4 h − . Now we find the function R ( x ) by the formula (48). It is the rational function whichdenominator has the form: V ( x ) = ( x − c ) (cid:0) x + xc + c (cid:1) ( x + c ) (cid:0) x + (cid:0) c − h (cid:1) x + c − c h + 4 h − (cid:1) c . R ( x ) has eight finite poles of the second order. It has the poles at x = c and x = − c , at x = x i , i = 1 , . . . , , where x i , i = 1 , . . . , are roots of the fourth degree polynomial x + (cid:0) c − h (cid:1) x + c − c h + 4 h − , (75)and at x = x i , i = 5 , , where x i , i = 5 , are roots of the polynomial x + xc + c = 0 . (76)The partial fraction expansion of R ( x ) has a very complicated form and we do not write itexplicitly. However it is possible to note the following properties of the partial fraction expansionof the function R ( x ) .1. The coefficient b c of b c ( x − c ) equals b c = 516 .
2. The coefficient b − c of b − c ( x + c ) equals b − c = − .
3. The coefficients b i of b i ( x − x i ) , where x i , i = 1 , , , are roots of the polynomial (75) areall equal b i = − , i = 1 , , , .
4. The coefficients b i of b i ( x − x i ) , where x i , i = 5 , are roots of the polynomial (76) are allequal b i = 34 , i = 5 , .
5. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = − (1 + 4 d )4 x + O (cid:18) x (cid:19) . The Laurent expansion of R ( x ) at x = ∞ allows us to state that the differential equation (55)in the considered special case can have liouvillian solutions of type 2 only. To find these liouvilliansolutions we will use the Kovacic algorithm. To find this liouvillian solutions we will use theKovacic algorithm. Let us define the sets E x i , i = 1 , . . . , corresponding to finite poles and theset E ∞ corresponding to the pole of R ( x ) at x = ∞ . For the finite pole x = c the correspondingset has the form: E c = {− , , } . x = − c the corresponding set has the form: E − c = { , , } . For the finite poles x = x i , i = 1 , , , the corresponding sets E x i have the form: E x i = { , , } , i = 1 , , , . At last, for the finite poles x = x i , i = 5 , the corresponding sets E x i have the form: E x i = {− , , } , i = 5 , . The set E ∞ corresponding to the pole of R ( x ) at x = ∞ contains only one element and hasthe form: E ∞ = { } . Now we should determine the families s = ( e ∞ , e c , e − c , e x , e x , e x , e x , e x , e x ) , for which the constant d calculated by the formula (52), is a non-negative integer. It is easy to see,that the minimal value of the sum of the elements of sets, corresponding to finite poles, is zero.Therefore, for the family s = (2 , − , , , , , , − , − we have d = 1 , and for the families s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − , s = (2 , − , , , , , , − , − ,s = (2 , − , , , , , , − , − we have d = 0 . We must check all these families. We start with the family s . According to thealgorithm, let us find the function θ by the formula (53). For the family s this function has theform: θ = 2 x + ( c − h ) xx + ( c − h ) x + c − c h + 4 h − − x + cx + xc + c + 12 ( x + c ) −
12 ( x − c ) . P of degree d = 1 equals P ≡ x + B , where B is indefinite coefficient.This polynomial should identically satisfy the differential equation (54). After substitution of thepolynomial P = x + b and the functions θ and R ( x ) to this equation, we obtain in the left handside of (54) the rational expression. Its numerator has the form of the 8th degree polynomial P = 4 (cid:0) h − c h − (cid:1) Bx + · · · If we put B = 0 then the polynomial P takes the form: P = 3 c (cid:0) h − c + 2 (cid:1) (cid:0) h − c − (cid:1) x + · · · The coefficient at x of the polynomial P vanishes only if c = 0 or d = 0 . None of thesepossibilities is considered by us; therefore, the polynomial P does not identically vanish. We canverify similarly that the sixth degree polynomial obtained from P does not identically vanish, ifwe assume that h − c h − . (77)Finally this means that equation (55) has not liouvillian solutions of type 2 for the family s in the considered special case.Now let us consider the families s , . . . , s . For the family s the function θ calculated by theformula (53), has the form θ = 2 x + ( c − h ) xx + ( c − h ) x + c − c h + 4 h − − x + cx + xc + c + 32 ( x + c ) −
12 ( x − c ) . The polynomial P of degree d = 0 has the form P ≡ . This polynomial should identicallysatisfy differential equation (54). After substitution of P = 1 and the functions θ and R ( x ) to thedifferential equation (54) we obtain in the left hand side of this equation the rational expression.Its numerator has the form of 8th degree polynomial P = 4 (cid:0) h − c h − (cid:1) x + · · · If we assume that condition (77) is valid, then the substitution of this condition with P reduces it to the polynomial of the 6th degree, which is not zero for any c and d such that c = 0 and d = 0 . Thus, the differential equation (55) has not liouvillian solutions of type 2 for thefamily s . Similarly we can check the families s , . . . , s . For all these families the differentialequation (55) has not liouvillian solutions of type 2. Thus we can state that in the special casewhen the polynomials (57) and (58) have a common root, which is not the a multiple root of thepolynomial (57), the differential equation (55) has not liouvillian solutions.
14 The special case of the multiple common root of the polynomials (57) and (58) . Now let us consider the case, when the root x = c – the real root of the polynomial (58) fromthe roots (68) id the multiple root of the polynomial (57). In this case the cubic polynomial (64)obtained from the polynomial (57), has the form P z = z − hz + 4 (cid:0) h − (cid:1) z + 4 d c = (cid:0) z − c (cid:1) ( z + z ) . z in two different representations of thepolynomial P z we obtain the following system of equations for the parameters of the problem: c − z − h = 0 , c z − c + 4 h − , d c − c z = 0 . (78)Solving the system (78) we find the following expressions for the parameters h , z e d in termsof the parameter c : d = √ c + 42 c − ,z = 4 d c = 2 √ c + 4 − c ,h = c (cid:18) − d (cid:19) = c − √ c + 42 c ! = c − √ c + 42 . (79)We will use (79) to simplify the coefficients a ( x ) and b ( x ) of the differential equation (45).Taking into account (79) we can write the coefficients a ( x ) and b ( x ) of the second order lineardifferential equation (45) as follows: a ( x ) = x + 2 cx + 4 c x + 2 c x + 2 c (2 x + c ) (cid:0) √ c + 4 − c (cid:1) x ( x + c ) ( x + xc + c ) (cid:0) x + 2 √ c + 4 − c (cid:1) ,b ( x ) = ( x + xc + c ) ( x − xc + c ) (cid:0) √ c + 4 − c (cid:1) c x ( x − c ) ( x + c ) (cid:0) x + 2 √ c + 4 − c (cid:1) . Now we find the function R ( x ) for this special case by the formula (48). It is the rationalfunction of the rather complicated form which denominator can be written as follows: V ( x ) = 4 c ( x + c ) (cid:0) x + xc + c (cid:1) (cid:16) x + 2 √ c + 4 − c (cid:17) ( x − c ) . Thus, the function R ( x ) has six finite poles. At x = c function R ( x ) has the first order pole.At x = − c it has the second order pole. The function R ( x ) has also the second order poles at x = x i , i = 1 , , , , where x and x are roots of the polynomial x + xc + c = 0 , (80)and x and x are roots of the polynomial x + 2 √ c + 4 − c = 0 . (81)The partial fraction expansion of R ( x ) has a very complicated form and we do not write it hereexplicitly. However it is possible to note the following properties of the partial fraction expansionof the function R ( x ) . 43. The coefficients b i of b i ( x − x i ) , where x i , i = 1 , are roots of the polynomial (80) equal b i = 34 , i = 1 , .
2. The coefficients b i of b i ( x − x i ) , where x i , i = 3 , are roots of the polynomial (81) equal b i = − , i = 3 , .
3. The coefficient b − c of b − c ( x + c ) equals b − c = − .
4. The Laurent expansion of R ( x ) at x = ∞ has the form: R ( x ) = − (1 + 4 d )4 x + O (cid:18) x (cid:19) , where parameter d are expressed in terms of c by (79).The Laurent expansion of R ( x ) at x = ∞ allows us to state that the differential equation (55)in the considered special case can have liouvillian solutions of type 2 only. To find these liouvilliansolutions, we will use the Kovacic algorithm. According to the algorithm, let us find the sets E x i , i = 1 , , . . . , , corresponding to the finite poles and the set E ∞ corresponding to the pole of R ( x ) at x = ∞ . For the first order finite pole x = c the corresponding set has the form: E c = { } . For the finite pole of the second order x = − c the corresponding set has the form: E − c = { } . For the finite poles of the second order x = x i , i = 1 , the corresponding sets E x i have theform: E x i = {− , , } , i = 1 , . At last, for the finite poles of the second order x = x i , i = 3 , the corresponding sets E x i havethe form E x i = { , , } , i = 3 , . The set E ∞ corresponding to the pole of R ( x ) at x = ∞ contains only one element and hasthe form: E ∞ = { } . Now we should determine the families s = ( e ∞ , e c , e − c , e x , e x , e x , e x , e x , e x ) , for which the constant d calculated by the formula (52), is a non-negative integer. It is easy tosee, that the minimal value of the sum of elements of sets, corresponding to finite poles equals4. This means that there are no families s , for which the constant d , calculated by (52), is anon-negative integer. Thus, in the considered special case the differential equation (55) has noliouvillian solutions. 44 (57) and (58) . Now let us consider the case when one of the roots x or x from the roots (68) of the cubicpolynomial (58) is the root of the polynomial (57). If we substitute any of these roots to (57) andtake into account that k = c d , we can find, that parameters of the problem should satisfy thesystem of two conditions: h + hc − , (cid:0) d (cid:1) c + 2 hc − (cid:0) h − (cid:1) = 0 . (82)Let us consider the first of these conditions (82). This is quadratic equation with respect to h .Its solutions have the following form: h = 12 √ c + 4 − c , h = − √ c + 4 − c . It is easy to see that for h the following inequality holds: − √ c + 4 − c ≤ − , Taking into account the range of the parameter h we obtain that h can take only a value h = h . Thus, we have h = √ c + 42 − c . Substituting this expression for h to the second equation of the system (82) and solving it withrespect to d , we obtain d = − √ c + 4 − c c . (83)The right hand side of the equation (83) is negative for all values of c = 0 . This means that thecondition (83) cannot be valid for all values of parameters c and d . Thus, the polynomials (57)and (58) cannot have the common non-real root. This means, that the considered special casedoes not hold.
16 Conclusions
In this paper we considered the problem of motion of a heavy rigid body with a fixed pointin the Hess case. The integration of this problem is reduced to solving the second order lineardifferential equation. We reduce this equation to the equation with rational coefficients (45).Using the Kovacic algorithm we studied the problem of existence of luouvillian solutions of thedifferential equation (45). We proved that this equation has liouvillian solutions of type 2 only andonly if the moving rigid body is the Lagrange top ( d = 0) or if the constant of the area integralis zero ( k = 0) . Thus the problem of existence of liouvillian solutions for the second order lineardifferential equation (45), the integration of which solves the problem of motion of a heavy rigidbody with a fixed point in the Hess case, is completely solved.45 (44) holds. Let us find out the motion of the Hess top in the case, when the conditions (44) holds in additionto the conditions (7). Condition d = 0 is equivalent to the condition ( A − A ) x x = 0 . (84)Further we will assume that A = A . The condition A = A together with (7) correspondsto the case of kinetic symmetry. Then (84) is equivalent to x = 0 or x = 0 . We will considerthe case x = 0 (the case x = 0 is studied similarly). The condition x = 0 together with (7),corresponds to the Lagrange integrable case x = 0 , x = 0 , A = A , x = a = 0 . (85)The Hess integral (8) takes the form ω = 0 , (86)and the area integral can be rewritten as follows: ω γ + ω γ = 0 , (87)where we taking into account that the constant of this integral is zero.The Euler equations (5) under conditions (85) takes the form: A ˙ ω = − M gaγ , A ˙ ω = M gaγ . If we introduce the standard Euler angles by the formulae γ = sin θ cos ϕ, γ = cos θ, γ = sin θ sin ϕ,ω = ˙ ψ sin θ cos ϕ − ˙ θ sin ϕ, ω = ˙ ψ cos θ + ˙ ϕ, ω = ˙ ψ sin θ sin ϕ + ˙ θ cos ϕ we can rewrite equation (86) in the form ˙ ψ cos θ + ˙ ϕ = 0 , (88)and the equation (87) takes the form: ˙ ψ sin θ = 0 . (89)We will consider the case sin θ = 0 (if θ = 0 , then all the components ω , ω , and ω of angularvelocity are zero). Then, from the condition (89) we have ˙ ψ = 0 . Condition (88) gives ˙ ϕ = 0 . This means that for the considered motions of the Lagrange top we have ψ = ψ = const , ϕ = ϕ = const . ω and ω of angular velocity of the body takes the form: ω = − ˙ θ sin ϕ , ω = ˙ θ cos ϕ . The Euler equations (5) are reduced to the differential equation ¨ θ − M gaA sin θ = 0 , which describes the pendulum nutational oscillations of the Lagrange top in the vicinity of θ = π .Thus the motions of a heavy rigid body with a fixed point in the Hess case for which d = 0 and k = 0 correspond to the nutational oscillations of the body. Acknowledgements.
This work was supported financially by the Russian Foundation for Basic Researches (grants no.19-01-00140, 20-01-00637).
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