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0Russian Journal of Nonlinear Dynamics, 2022, vol. 18, no. 4, pp. 609-627. Full-texts are available at http://nd.ics.org.ru DOI: 10.20537/nd221210
NONLINEAR PHYSICS AND MECHANICS
MSC 2010: 70F25, 70E40
Gyrostatic Suslov Problem
A. J. Maciejewski, M. Przybylska
To Anatoly Pavlovich Markeev on his 80th birthday
In this paper, we investigate the gyrostat under influence of an external potential force with the Suslov nonholonomic constraint: the projection of the total angular velocity onto a vector fixed in the body vanishes. We investigate cases of free gyrostat, the heavy gyrostat in the constant gravity field, and we discuss certain properties for general potential forces. In all these cases, the system has two first integrals: the energy and the geometric first integral. For its integrability, either two additional first integrals or one additional first integral and an invariant n-form are necessary. For the free gyrostat we identify three cases integrable in the Jacobi sense. In the case of heavy gyrostat three cases with one additional first integral are identified. Among them, one case is integrable and the non-integrability of the remaining cases is proved by means of the differential Galois methods. Moreover, for a distinguished case of the heavy gyrostat a co-dimension one invariant subspace is identified. It was shown that the system restricted to this subspace is super-integrable, and solvable in elliptic functions. For the gyrostat in general potential force field conditions of the existence of an invariant n-form defined by a special form of the Jacobi last multiplier are derived. The class of potentials satisfying them is identified, and then the system restricted to the corresponding invariant subspace of co-dimension one appears to be integrable in the Jacobi sense.
Keywords: gyrostat, Suslov constraint, integrability, nonholonomic systems
Received November 10, 2022 Accepted December 09, 2022
The work has been partially founded by The National Science Center of Poland under Grant No. 2020/39/D/ST1/01632.
Andrzej J. Maciejewski a.maciejewski@ia.uz.zgora.pl
Janusz Gil Institute of Astronomy, University of Zielona Gora, Licealna 9, 65-417, Zielona Gora, Poland
Maria Przybylska m.przybylska@if.uz.zgora.pl
Institute of Physics, University of Zielona Gora, Licealna 9, 65-417, Zielona Gora, Poland
I11 0 I13
I= 0 I22 3 I23
I13 3 I23 I33
1. Introduction
The classical heavy top problem admits several generalizations. For example, instead of the constant gravity field, one can consider just an arbitrary potential force field. Another generalization is the so-called heavy gyrostat. It is a rigid body with a fixed point under the action of gravity and gyroscopic forces. The gyroscopic forces appear when the body has moving parts as rotors, or it has cavities filled with an ideal incompressible fluid, see [7, 8, 20].
The Euler-Poisson equations of a gyrostat have the form
d
-(/i]) = (/0 + A)xi] + rxl/'(r), r = rxfl, (1.1)
where Q and IQ are the angular velocity, and the angular momentum of the body; A is the constant gyrostatic moment, r is the unit vertical vector; V(r) is the potential and V'(r) denotes its gradient. In the classical case V(r) = (B, r), where B is the unit vector from the fixed point of the body towards its mass centre, and (■, ■) denotes the scalar product.
The Suslov nonholonomic constraint has the form (Q, A) = 0, where A is the unit vector that is fixed in the body. Without loss of the generality, we can assume that the body fixed frame is chosen in such a way that A = [0, 0, 1]T. Additionally, with this choice of vector A, we can assume that the tensor of inertia is
(1.2)
Then the Suslov constraint is just Q3 = 0, so the equations of motion for the gyrostatic Suslov problem are
dV dV
hi^i = -(/i3«i + + A3)^2 + — r2 - — r3,
dV dV (i 3)
I22a2 = (/^ + /23Q2 + A3)f21 + — r3 - —T1}
, r3 = —Q2r3, r2 = ^^3, r3 = —
see [5, 10, 11]. The classical nonholonomic Suslov problem corresponds to A3 = 0 and its numerous generalizations were investigated by many authors, see for example [3-5] and references therein. In particular, in [5] the possibility of writing the classical Suslov system in generalized Hamiltonian form with a degenerate Poisson structure for various potentials V(r) was discussed. System (1.3) has two first integrals: the energy and the geometric first integral
Fl = -2 (Jnfi? + I22Qi) + V(T), F2 = <r, r). (1.4)
The inertia tensor is, by assumption, positively defined, thus the following conditions have to be simultaneously fulfilled
Jn > 0, I22 > 0, det I = /11/22/33 — 111123 — I22Zi23 > 0. (1.5)
Notice, that the right-hand sides of system (1.3) does not depend on I33. Thus, if I11 > 0 and I22 > 0, then for arbitrary I13 and I23 we can always find I33 such that the inertia tensor is positively defined. Thus, later, we assume only that I11 > 0 and I22 > 0.
We assume that A3 = 0, otherwise we obtain the classical Suslov problem.
In the above formulation, independent variables are X = (Q1, Q2, ri; r2, r3). The divergence of the vector field W(X) defined by the system (1.3) is
di = (1.6) I22 I11
In this paper, we present our results concerning the integrability of system (1.3). As it is not a Hamiltonian system, a definition of integrability is needed. We will use two notions of integrability.
We say that system of differential equations
x = v(x), x e Mra, (1.7)
is super-integrable if it has (n — 1) functionally independent first integrals.
Let us recall that a differential n-form
p(x) = J (x) dx1 a-- - a dxn, (1.8)
where J(x) is a smooth function, is invariant with respect to the system (1.7) if J(x) satisfies
n pi
Y^^[J(x)vl(x)}= 0. (1.9)
i=1 i
Traditionally, the function J(x) is called the Jacobi last multiplier.
We say that the system (1.7) is integrable in the Jacobi sense if it possesses a Jacobi last multiplier and admits (n — 2) functionally independent first integrals.
The plan of this paper is following. In Section 2 we consider the case when V(r) = 0. In this case, the system has the Jacobi last multiplier. We distinguish three cases integrable in the Jacobi sense.
If the body is located in a constant gravity, that is when V(r) = (B, r), system (1.3) does not admit Jacobi last multiplier in general. Nevertheless, we found three cases with one additional first integral. The question whether these cases are integrable in the Jacobi sense is answered in Section 3.
Moreover, for a special choice of vector B, in this case system admits an invariant subspace and its restriction to it appears to be super-integrable and solvable in elliptic functions that is described in Section 4.
The general case of the gyrostat with Suslov constraint in a general potential force field is considered in Section 5. In particular, conditions for an invariant n-measure with specific the Jacobi last multiplier are formulated and one potential satisfying them is identified. Then exists an invariant subspace of the phase space, and the system restricted to it becomes integrable.
The article also contains two appendices. Appendix A contains the non-integrability proof of the classical non-diagonal Suslov problem with linear potential V = (B, r) which is used in Section 3. For the convenience of readers, Appendix B is added containing basic facts about Riemann P equation and its differential Galois group applied in non-integrability proofs.
2. Free gyrostat
If V(r) = 0, then system (1.3) splits because the Euler equations for Q1 and Q2 do not depend on r
' h^ 1 = —(I13+ I23l2 + A3)^2, Inil 2 = (I13^1 + I23 Q2 + A3)^1, (2.1)
r 1 = — l2Гз, r 2 = llГз, r 3 = ^2r1 — 0^2.
Now, the first integrals of this system simplify to
F1 = InQ1 + I22Q2, F2 = (r, r) =r1 + r2 +r2. (2.2)
Moreover, the system has the last Jacobi multiplier
J{X) 1
A3 + I23^2 + A3
For I13 = I23 = 0 the standard volume form is an invariant measure, so J(X) = 1.
In our paper [6], see also [5], about the classical nonholonomic Suslov problem with A3 = 0 we found integrable cases with additional first integral linear in r1, r2, r3 with coefficients that are polynomials in Q1, Q2 with constant coefficients. Applying a direct method for searching polynomial first integrals of the gyrostatic Suslov problem, we found only three cases:
1) if I13 = 0 and I11I22 = I|2 + I|3, then the additional first integral is
F3 = I2 + 4) + (4^2 " I23A3) r2 + I22(A3 + ^^ (2-3)
2) if I23 = 0 and I11I22 = I21 + I23, then the additional first integral is
F3 = {Ih^1 - I13A3) r + (I?1 + 4) ^2 + In(A3 + I^)^; (2-4)
3) if I13 = I23 = 0 and I22 = I11, then the additional first integral is
F3 = In^r + I11 ^2 +A3r3- (2.5)
In all these cases the system is integrable in the Jacobi sense. We conjecture that if A3 = 0, then for any admissible values of parameters, the system does not admit a first integral of the form
F3(X) = P1 (Q1, + ^2(^1, ^2 + ^3(^1, ^2^3, (2.6)
with polynomials Pi(H1, Q2) of degree higher than one.
The first two equations in (2.1) have non-isolated equilibria. In fact, equations
(I13 Q1 + I23^2 +A3)^2 = 0, (I13 Q1 + I23^2 + A3R =0,
have an isolated solution Q1 = Q2 = 0, and a straight line of non-isolated solutions given by I13H1 + I23ü2 + A3 = 0. Let us notice that for A3 = 0 this line of equilibria does not pass through origin. Thus, the phase portrait of the Euler equations for Q1 and Q2 looks as it is shown in Fig. 1.
Fig. 1. Generic trajectories ^ on the plane with marked the line of equilibria
All these equilibria are not stable. In the classical Suslov system, the line of the equilibria passes through the origin. Thus, on each energy level F1 = e with e > 0 there are two equilibria and two heteroclinic orbits joining them. When A3 = 0, then the intersection of the line I13^ + + I23+ A3 = 0 with the energy level F1 = e with e > 0, if not empty, has two or one point.
If A3 = 0, then for a fixed solution of Euler equations for Q1 and Q2 the monodromy group of the Poisson equation can be effectively determined. Using it one calculates the "scattering" angle: the angle between the axes of limit permanent rotations of the body in space, for details, see [16]. It will be interesting to perform a similar analysis when A3 = 0, however, the problem is much more difficult than in the case with A3 = 0.
3. Gyrostat in a constant gravity field
In this section we assume that the Suslov gyrostat is located in the constant gravity field, so V(r) = (B, r). The equations of motion in explicit form reads
InQ 1 = -(I13^1 + I23Q2 + A3)^2 + B3r2 — B2r3> ^22^2 = (I13+ I23^2 + A3)^1 + B1r3 - B3r1, r 2 = ^1r3, r 3 = ^2r1 — ^1r2"
(3.1)
r 1 = —^2r3,
For this case, we searched polynomial first integral using the direct method. We found the following three cases:
1) if I13 = 0, I22I11 = I222 + 1223, and B = B2[0, I23, — I22]T, then the additional first integral
is
F3 = (I22 + 4) + (I222^2 — I23A3) r2 + I22(I23^2 + A3^; (3.2)
2) if I23 = 0, I22I11 = If1 + If3, and B = B1[I13, 0, — I11]T, then the additional first integral is
F3 = (I21Q1 — I13A3) r1 + (I!21Q2 + Ii23^^ r2 + In(I13^1 + A3)r; (3.3)
3) if B = [0, 0, B3]T, I13 = I23 = 0, and I22 = In, then the additional first integral is
F3 = Iii^r + Ill ^ +A3^. (3.4)
The last case is the generalization of the Kozlov case of the classical Suslov problem [13, 14], however it seems that has no Hamiltonian formulation. Moreover, in this last case when I13 = = I23 = 0 the system has a constant last Jacobi multiplier. Thus, for its integrability, just one additional first integral is sufficient. Hence, this case is integrable in the Jacobi sense. We prove the following theorem.
Theorem 1. Assume that I13 = I23 = 0 and B1 = B2 = 0, B3 = 0. Then the system (3.1) admits additional rational first integral F3 functionally independent with F1 and F2 if and only if I11 = I22.
Proof. We assign weights (1, 1, 2, 2, 2) to variables (^1, ^2, Then the domi-
nant part of the system with respect to these weights correspond to A3 = 0, i.e., to the classical Suslov problem. If the system admits an additional first integral F3 then the dominant part admits an additional first integral F3°. However, F3° can be functionally dependent with F° and F2°. But the Ziglin Lemma [21] guarantees that we can find such F3 such that F3° and F° and F2° are functionally independent. We end reasoning recalling the non-integrability result of the standard Suslov problem in the case I13 = I23 =0 and B1 = B2 = 0, B3 = 0 except, the case I11 = I22 proved in [22] and [23] and in more general case in [15]. □
It is interesting to notice that at the same time we proved that the investigated case is not integrable in the Jacobi sense. More precisely, we have the following.
Corollary 1. Assume that I13 = I23 = 0 and B1 = B2 = 0, B3 = 0. Then the system (3.1) is integrable in the Jacobi sense if and only if I11 = I22.
Proof. As in the proof of the above theorem, we deduce that if the system (3.1) is integrable in the Jacobi sense, then it is also integrable in the Jacobi sense when A3 = 0. However, it is not true. In [15] we investigated the system with A3 = 0, and showed among other things, that the identity component of the differential Galois group of variational equations along the chosen particular solution is not commutative. This fact implies the lack of an additional first integral. Moreover, by Theorem 3.1 in [17] it implies also that the system is not integrable in the Jacobi sense. □
Now it is natural to ask if in the first two above cases the system is integrable. This problem has two versions:
1) find the parameters of the system for which it is integrable in the Jacobi sense, or
2) find the parameters of the system for which it admits four functionally independent first integral i. e. it is super-integrable.
Notice that in the first version, we ask whether the system admits the Jacobi last multiplier while in the second ask whether the system has an additional first integral. Theorem 1 and Corollary 1 show that using appropriate tools, we can answer both questions simultaneously. In fact, we show that it is possible.
Theorem 2. Assume that I13 = 0, I23 = 0, I22I11 = I22 + I23, and B = B[0, I23, —122]T. Then the system (3.1) does not admit an additional first integral F4 functionally independent with F1, F2 and F3.
Proof. The reasoning is the same as in the proof of Theorem 1. If for the assumed values of parameters the system admits an additional first integral F4 such that four functions F1, F2, F3 and F4 are functionally independent, then the same is true for A3 = 0. However, by our Theorem A.1 in Appendix A, it is impossible. This contradiction finishes the proof. □
Using the same argument as in the proof of Corollary 1 we deduce the following.
Corollary 2. Assume that I13 = 0, I23 = 0, I22111 = I|2 + I|3, and B = B[0, I23, -I22]T. Then the system (3.1) is not integrable in the Jacobi sense.
For the second case we have similar facts as for the first case.
Theorem 3. Assume that I23 = 0, I13 = 0, I22I11 = If1 + I23, and B = B[I13, 0, —111 ]T. Then the Suslov system (3.1) does not admit an additional first integral F4 functionally independent with F1, F2 and F3.
Corollary 3. Assume that I23 = 0, I13 = 0, I22111 = If1 + If3, and B = B[I13, 0, —111 ]T. Then the system (3.1) is not integrable in the Jacobi sense.
4. Reduced system
In this section, we still assume that V(r) = (B, r).
Integrable, or partially integrable systems are rare. This is why we look for such values of parameters for which the system admits invariant sets of possibly high dimensions. For example, we can look for a function such that its zero level is invariant with respect to the flow of the considered system. If this function is a polynomial, then it is called a Darboux polynomial. If its degree in not too big, then we can find it by a direct method.
The formal definition is the following. Assume that the vector field v defining system (1.7) is polynomial. Then a polynomial F(x) is a Darboux polynomial of this system if LvF(x) = = P(x)F(x) for a certain polynomial P(x) which is called the cofactor of F(x). Here Lv denotes the Lie derivative.
A search of Darboux polynomials F up to the fourth degree gives only one case with B = 0, namely, for an arbitrary inertia tensor I, and gyrostatic moment A = (A1, A2, A3), and B = = B(I22I13, I11I23, 0), there exists a linear Darboux polynomial
F(X) = i^n, + I23a2 + A3, P(X) = ^n, - ^a2, (4.i)
I22 I11
where P(X) is the cofactor of F(X). In this case the system has the Jacobi last multiplier
J(X) = -J— = 1
F(X) I13n1 + I23a2+ A3
Let us assume that the level F(X) =0 is not empty. Thus, either I13 = 0, or I23 = 0. Without loss of generality we can assume that. I23 / 0, and we introduce parameters a = A = j2-, and /3 = I23B. Then, equality F(X) = 0 gives Q2 = —aQ1 — A, and the restriction of the system (1.3) with potential V(r) = (B, r) to the level F(X) = 0 reads
ñ 1 = —/r3, r 1 = № + A)r,
r 2 = ^3, r 3 = —(añ1 + A)r — 0^2. .
We will call it the reduced system. It preserves that standard volume form in R4 with coordinates (Q-, rl; r2, r3). Moreover, it is super-integrable i.e. it possesses the maximal number of independent first integrals.
Lemma 1. System (4.2) is super-integrable and its solutions are given by elliptic functions and their integrals.
Proof. From (4.2) we find that
dr-, 1 . ^ dr2 1 .
Thus, we can express components of r as functions of Q-
a£l\ + 2AQ1 + C1 O2 + C2 _ y/Ptfh) ( .
l 2/] >2 2/i ' 3 2/3 ' [ ' '
where
P4(Q-) = 4p2 - (C2 + Ql)2 - [C- + Q-(aQ- + 2A)]2. (4.5)
Then, the integration of the first equation in (4.2) reduces to the inversion of the following quadrature
Q1(i)
«i ^P4 (Ql) 2
As the degree of polynomial P4 (Q,) is four, the inversion can be performed explicitly. This inversion is described in [19, Section 20.6], and it expresses Q,(i) in terms of the Weierstrass elliptic p function and its derivative. This shows that Q,(i), r,(i) and r2(i) are elliptic functions. The last equation in (4.2) shows that r3(i) is an integral from such a function. System (4.2) has the energy and geometric first integrals
F- = I--Ql + l22(aQ- + A)2 + 2P(/n r2 + a/22 F2 = T +r2 + r2. (4.7)
Moreover, from (4.4) we find that also functions
G- = 2^r- + aQ- + 2AQ-, G2 = 2pr2 + Q2, (4.8)
are first integrals of system (4.2), and as three functionally independent first integrals either F-, F2, and G-, or F-, F2 and G2 can be chosen. Taking a combination of these integrals we find that
F3 = AQ- + P(r- - ar2) (4.9)
is also a first integral. □
In order to present the phase portrait of the reduced system it is convenient to consider a constant level of the first integral F3 = 7P. The phase curves on this level uniquely project to the unit sphere S2 = {r e R3 | <r, r) = 1}. In fact, from F3 = 7P we find that Q- = ¿(ar2 -— T1 + 7), where 5 = j, so after elimination of Q1 from system (4.2) we obtain
r- = ^<^2 - r- + 7) + A],
r2 = <^3(ar2 - r- + 7), (4.10)
r3 = + ar-)(ar2 - r- + 7) - AT-.
We look for equilibria of this system on the sphere S2. If (r1, r2, r3) are coordinates of an equilibrium, then necessarily r3 = 0, because by assumption A = 0. Thus, the equilibria are given by points of intersections of two quadrics in the plane R2
—¿(r + ar )№ — r + y) — AT1 = 0, r2 + r2 = 1. (4.11)
Hence, there exist at most four real equilibria of this system. Moreover, as the vector field defined by the right-hand sides of (4.10) is smooth there exists at least one equilibrium, see § 36 of [1].
y
Fig. 2. Selected phase portraits for reduced system in plane (x, y) in cases with four equilibria
(a) a = S = 1, y = 0 and A =1
_ 99^(9401+99^601)
\U) u. ln, V ±, y__in 100
and A = -10 799-99^601
1000 J2 (9401+99 V860T)
Fig. 3. Selected phase portraits for reduced system in plane (x, y) in cases with three equilibria
Assume that there are four non-degenerated equilibria. As the sum of indices of these equilibria must be two, one equilibrium must be a saddle. The system is integrable on the sphere S2, thus none of the equilibria is a node. In effect, if there exist four non-degenerated equilibria, then three of them are centres and one is a saddle, see Figs. 2a, 2b, 2c and 2d. There are also possibilities that system will have three singular points: two with index one and one with index zero, see Figs. 3a and 3b with two centers and one point around that the vector field does not rotate and its index is zero. In the next two Figs. 4a and 4b our system has two centers of index one each.
In order to present the phase portraits of the system in Figs. 2-4 we use coordinates (x, y) defined as
V. y = (4.12)
1 " 1 " Ts
which determine the stereographic projection of a point r £ S2. The inverse transformation is given by
2x 2y x2 + y2 — 1
Tl = 1 +.r2+y2' r2 = 1 +.r2 + y2' = 1 + x'2+y'2' (4'13)
5. Gyrostat in a general potential force field
In this section we consider a more general Suslov system in potential field V = V(r). Then the equations of motion have the form (1.3). Only two first integrals F1 and F2 given by (1.4) are known.
(a) a = -f, S = 7 = -f and A = —I
(b) a = -1, Ô = 1, 7 = 0 and A = 2
Fig. 4. Selected phase portraits for reduced system in plane (x, y) in cases with two equilibria
If I13 = I23 = 0, then the system admits constant Jacobi last multiplier J(X) = 1. In this section, we assume that If3 +1|3 = 0, and we ask if there exist cases when the system possesses a Jacobi last multiplier. In general, there is no a general method for solving this problem. Thus, guided by the constant gravity field case, we postulate that the Jacobi last multiplier has the form
W = r n ■ A , (5.1)
Il3 Q + +A
23 2
where G(r) is a smooth function. Then
i=1
P(Qi, Q2)
/11 ^22(I13Q1 + /23Q2 + A3)2
(5-2)
where P(Q1, Q2) is the polynomial of degree two with respect to Q1, Q2. This polynomial has to vanish. Hence, all its coefficients which are functions of r have to vanish. This condition gives the following system of partial differential equations for function G(r) and potential V(r):
I11I22113 G - ^g) = 0, /11/22A3 G - ^G) = 0,
/11/22/23 (r^G - ^G = 0, /11/22A3 (r^G - ^G) = 0,
I11/22
(/13r1 - /23^)^G + r3 ( /23dr2G - /^ G
(5.3)
G
(/11 /23r1 - W^)^, V + T3 /22/13ÔP V - /11/23dr V
0.
The general solution of the first four equations in (5.3) is G(r) = G (r2 + r2 +r2). But we restrict our results to the level rf + T + r3 = 1, so we set G(r) = 1 and then also the fifth equation in (5.3) is also satisfied.
If I23 = 0, then the last equation in (5.3) has the general solution of the form V(r) = U (In№ + W^i, I2i4 (r2 + r3) - Ii3l22ri(/22li3ri + 2In, (5.4)
where U is an arbitrary smooth function of two arguments. In particular, if Ii3 = 0 this potential simplifies to V(r) = U (r2, r2 + r3). If I23 = 0, then the last equation in (5.3) simplifies to
I22I13
^ V " ^ V
0,
which has the solution V(r) = U (r^ r2 + r2). Thus, in general, the last eqution in (5.3) has always a solution for a potential, an invariant form exists and for this potential only one additional first integral is sufficient for the integrability.
Let us notice that on the level rf + r2 + r3 = 1 formula (5.4) simplifies to
V(r) = U (/n/23r2 + /22-(/11/23r2 + /22W)2).
Thus, U is a function of one argument U = U(Y), where Y = /11 /23r2 + /22/13r1.
Notice also that, fact that. J(X) = j^x), with F(X) = /13Í21 + /23ü2 + A3 is the last. Jacobi multiplier of the system implies that F(X) satisfies
LWF(X)= P(X)F(X), P(X)=div W(X). (5.5)
That is F(X) is a Darboux polynomial of the system.
As in the previous section, we restrict the system to the zero level of F(X), which is by assumption non-empty. As (/13H1 + /23ñ2 +A3) = 0, assuming that /23 = 0 we can eliminate ñ2.
It. is also convenient, to introduce new variable T = -X- and parameters
J23
a = P = 7 = ^, A = 8 = ah- (5.6)
^11 J23 J23 J23
Then the reduced system can be written in the following form
ñ 1 = -r3U'(r),
r 1 =r3(7^1 + A), ■ 1 (5.7)
r = -r3(Q1 + á(7Q1 + A)), a
r3 = -arñ1 + r1((S - Y)ñ1 - A),
where U(r) = /23U(Y). Obviously, the above system has two first integrals which correspond to the energy and geometric integrals (1.4). In explicit form, they read
G1 = ^(1 + + A5ÍÍ! + aU(T), G2 = + (aT - ÓT^2 + T'¡. (5.8)
Moreover, just from the form of the right-hand sides of the reduced system (5.7) it is visible that it has a constant Jacobi multiplier. Thus, the reduced system is integrable in the Jacobi sense. So, we can conclude this section with the following.
Lemma 2. System (1.3) with potential V(r) = U(Y), where Y = /11/23r2 + /22/13r1, has Darboux polynomial F(X) = /13ñ1 + /23ñ2 + A3, and restricted to its zero level F(X) = 0 is integrable in the Jacobi sense.
Acknowledgments
The authors thank an anonymous reviewer for the indication of article [5] containing references to important papers [10, 11].
Conflict of interest
The authors declare that they have no conflict of interest.
Appendix A. Non-integrability of non-diagonal Suslov problem
Here we consider the classical Suslov problem with linear potential V = (B, r). The equations of motion have the form
InQ! = -(/13+ /23^2)^2 + B3r2 - B2r3> ^22^2 = (I13+ I23^2)^1 + Bir3 - B3T1, f1 = -^2^3, r
^1^3,
- ^1r2 •
(A.1)
They coincide with equations (3.1) for A3 = 0. We prove the following theorem.
Theorem A.1. The non-diagonal Suslov system (A.1) with I13 = 0, I23 = 0, and I|2 + + I223 = I22111 and B = B[0, I23, -I22]T does not admit four functionally independent first integrals.
Our proof is based on the following theorem.
Theorem A.2. If a complex analytic system x = v(x), x e Cn admits (n - 1) functionally independent complex meromorphic first integrals, then the variational equations along a particular solution have the differential Galois group with Abelian identity component.
This theorem is a particular version of the theorem given in [2].
At the beginning we assume only that I13 = 0 and B1 = 0. With these assumptions, system (A.1) has the invariant subspace
N = {(Q1, Q2, r1, r2, e C5 I Q2 = r = 0} , and its restriction to N is
InQ 1
B3r2 - B2Г3,
r0
^1r3, r 3 = - ^1r2.
(A.2)
The variational equations along a solution (Q1, r2, r3) of the above equations have the form
0
UJ1
U 2
7 1 =
7 2
j 3.
0 0 0
T3
-r0
n B3 B2
0 j— - —
23 O 3
J22 1 122 -T
111 0
0 0 0 0 0 -Q1
h. 0
0
Q1 0
U1
U2
1
2
3
(A.3)
Equations for variables w2 and 71 form a closed subsystem called the normal variational equations
I B
uj2 = - 7i = —r3w2. (A.4)
II
22 22
We rewrite them as the second-order scalar equation
<»2 " + + ^23 ~ BshlT \ ^ = 0 (A 5)
J22 V 111*22 111*22 /
For further analysis we have to parametrize properly chosen particular solution. To this end we set r2 = cos p and r3 = sin p. Then, from (A.2) we get Q1 + p = 0, and
B2 B3
(p—— Smt£> + — COS Lp = 0. (A.6)
A *u
So, it is natural to introduce angle a defined by
B2 B3 2 1 / 2 2\ BB
cosa = -rT' sin« = -rT' k =Tr(B2+B3j = yr-In effect, we obtain a pendulum equation
■ip + fc sin ■ = 0, ■ = p — a. (A.7)
Now the normal variational equation (A.5) transforms into
¿¿>2 + a(t)ui2 + b(t)w2 = 0, (A.8)
with coefficients
a(t) =/3ip, b(t) = — hp sin ip + fc7sinasin(0 + a), /3 = y^-, 7 = y^-.
*22 *22
Let us notice that the pendulum-type particular solution is completely defined by the choice of the value e of its energy
1 • 2
E = -tp — k cos ip.
To rationalize the normal variational equation we use the change of the independent vari-
able z = tan fyp-) which gives also
2
After this transformation we obtain
¿2 = J (l + z2) (e + k + (e - k)z2), z = z(e + (e- k)z2).
d
üü!i+p(z)üü'2 + q(z)uj2 = 0, ' = —, (A.9)
z + az _ e (1 + z2) {z + ß)~ k (zs + (-1 + z2) ß) P{-Z> ~ z2 (1 + z2)(e + k + (e-k)z2)
b —Azß — 2 (—1 + z2) 7 sin a2 + 2*7 sin(2a) q(-Z' = ¿2 = (1 + z2) (e + fc + (e - k)z2) '
This equation has in general five singularities including infinity but on energy level e = k we have coalescence of singularities and coefficients simplify to
2 + 2/ —2z/ — (z2 - 1)7 sin a2 + ZY sin(2a)
p{z) = tt^' q{z) =--'
All three singularities are regular and on this energy level equation (A.9) becomes the hyperge-ometric equation with differences of exponents
p = isJa(1 -e"2i»)7 + (l -2if3)2 = i\Jl - A(32 + 87sin2a - Ai{(3 - 7sin(2a)), a = i \J A(1 — e2ia) 7 + (1 + 2if3)'2 = i ^ 1 - 4/i2 + 87 sin2 a + 4i(/3 - 7 sin(2a)),
One can transform our normal variational equation to the canonical Riemann P equation (B.1) by the transformation of the independent variable z —> y = + iz).
The explicit forms of the real and imaginary parts of exponents differences are following
where
Rep = Red = \JA + 87sin2(a) - A/32 + 1, Im p = — Im a = — 87 sin2(o;) + Aj32 — 1,
A = \JI67 sin2(a) (-4f32 + 47 + 1) - 32/^7 sin(2a) + (4f32 + l)2,
(A.12)
and 5 = sign(-4(^ — Ysin(2a))). The technical lemma below gives sufficient conditions for non-commutativity of the identity component of the differential Galois group of Eq. (A.9) with coefficients (A. 10).
Lemma A.1. If
1) /3 — Y sin(2a) = 0 and Re p + Re a ± t is not odd integer, or
2) 3 — y sin(2a) = 0 and Re p + Re a ± t is odd integer, and t is not integer, or
3) 3 — y sin(2a) = 0 and 1 — 4/2 + 8y sin2 a < 0, and t is not odd integer, or
4) 3 — y sin(2a) = 0 and 1 — 4/2 + 8y sin2 a < 0, and t is odd integer, but the infinity is a logarithmic singularity,
then the identity component of the differential Galois group of equation (A.9) with coefficients (A. 10) is not Abelian.
Proof. The proof is just a direct application of three lemmas given in Appendix B. The assumption / — y sin(2a) =0 in the two first items of this lemma, guarantees that the imaginary parts of differences of exponents p and a do not vanish. Thus, if the identity components of the differential Galois group is Abelian, the equation must be reducible. The assumption that Re p + Rea ± t is not odd integer in the first item means that this equation is not reducible. The condition that Re p + Re a ± t is an odd integer guarantees the reducibility but the fact that t is not an integer gives reducible non-Abelian case. The assumption in the remaining two
items f — y sin(2a) = 0 implies that the numbers under square roots in p and a siven in (A.11) are reals. Condition 1 — 4f2 + 8y sin2 a < 0 implies that p and a are purely imaginary numbers, thus they are not rational. Hence, if the identity components of the differential Galois group is Abelian, then the equation has to be reducible. But the assumption that t is not an odd integer in the third item excludes reducibility. In the fourth item, our Riemann P equation is reducible but the presence of logarithmic terms in local solutions at infinity excludes that the identity component of the differential group will be Abelian. □
Note that if the conditions specified in the above lemma are fulfilled, then the system (A.1) is not super-integrable under assumptions I13 = 0 and B1 = 0.
Now let us consider the special subcase with additional assumptions specified in theorem A.1. Namely, we assume that B2 = BI23 and B3 = —BI22, and I|2+I|3 = I22111. The last assumption in terms of introduced parameters reads 1 + f2 = y. Moreover, we have also
k a k d2 7 2
coso; = ——p, sina = -, B = k 7. BB
Thus, in this case differences of exponents simplify to
3 3
p = --+if3, <r = ---ip, t = 2.
Let us notice that p + a + t = —1, thus this Riemann P equation is reducible. Since, by assumption two differences of exponents p and a are not rational and the
third t = 2 corresponding to exponents at infinity is integer. Direct calculations show, see Eqs. (B.11)-(B.12) in [18], that the logarithmic term in local solutions at infinity is proportional to g2 = —| (l + 4/32) / 0. Hence, by Lemma B.2 the identity component of the differential Galois group of the normal variational equation (A.9) is not Abelian. Thus, by theorem A.2 system (A.1) in this case is not super-integrable and this finishes the proof of Theorem A.1.
Appendix B. Riemann P equation
The Riemann P equation is the most general linear second order differential equation with three regular singularities which can be placed at 2 = 0, 2 = 1 and 2 = to. Its canonical form is following
l~ Pl~ P'2 + 1 h£l + + T1T2 -P1P2 V = Q
dz2 \ z z — 1 j dz \ z2 (z — l)2 z(z — 1)
(B.l)
where (pl, p2), (a1, a2) and (r1, t2) are the exponents at the respective singular points.
The properties of this equation which are needed for our considerations will be expressed in terms of the differences of exponents
P = Pi — P2, ° = — T = T1 — T2-Using Tschirnhaus transformation of the dependent variable
v(z) = w(z) exp
z
1
f p(C)d(
where
p(z) = ^ ^ +
1 - Pi - P2 , 1 - u1 - u2
Z - 1
we transform Eq. (B.1) to its reduced form
,, 1/1 - p2 1 - u2 p2 + u2 - t2 - 1
w + 7 ~if" + 7-7^2 + -77- W> = 0. (B.2)
4 V z2 (z -1)2 z(z - 1) J
The advantage of the application of the reduced form of a linear second order differential equation is the fact that its differential Galois group is an algebraic subgroup of SL(2, C).
The exponents of the hypergeometric equation (B.2) at singularities 0, 1 and to are
Pl,2 = ^(l±P), <71,2 = ^(1 ±<7), Tli2 = i(-1±T), (B.3)
respectively, so the differences of exponents are the same as for the Riemann P equation (B.1).
Here by G we denote the differential Galois group of the Riemann P equation and by G° its identity component.
The Kimura Theorem [12] gives the necessary and sufficient conditions which guarantee that the Riemann P equation (B.1) has the differential Galois group with solvable identity component. However, we need a stronger result. Namely, we have to distinguish the cases when this group is solvable but non-commutative.
By Kimura Theorem [12], if all numbers p, u, and t are not rational, then G° is not solvable. If two between numbers p, u, and t are not rational and G° is solvable, then Eq. (B.1) is reducible. If G° is solvable and not commutative, then Eq. (B.1) is reducible.
The lemma below gives the necessary and sufficient conditions for Eq. (B.1) to be reducible, for more details see [9].
Lemma B.1. The equation (B.1) is reducible if and only if at least one among numbers p + u + t, -p + u + t, p - u + t, p + u - t, (B.4)
is an odd integer.
Let us notice that if our equation is reducible, then all numbers p, u, and t are rational, or one is rational and the remaining two are not rational. Tools to distinguish between these cases provide the following two lemmas.
Lemma B.2. Assume that the Eq. (B.1) is reducible. If the identity component of its differential Galois group is Abelian, then either all numbers p, u and t are rational, or two of them are not rational and one is an integer and in this last case the corresponding singularity is not logarithmic.
Lemma B.3. Suppose the Eq. (B.1) is not reducible and G° is solvable. Then, either all numbers p, u and t are rational, or one of them is not rational and the remaining two are half integers.
For details and proofs see Appendix B in [17].
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