The Poisson equations in the nonholonomic Suslov problem: Integrability, meromorphic and hypergeometric solutions

0.43 MB·English

by  Yuri Fedorov

The Poisson equations in the nonholonomic Suslov problem:
9 Integrability, meromorphic and hypergeometric solutions
0
0
2
Yuri N. Fedorov

n
Department de Matem´atica Aplicada I,
a
J Universitat Politecnica de Catalunya,

1 Barcelona, E-08028 Spain
3
e-mail: [email protected]

] and
h
Andrzej J. Maciejewski
p
- Institute of Astronomy, University of Zielona G´ora,
h
PL-65-246, Zielona G´ora, Poland
t
a
e-mail: [email protected]
m
and
[
Maria Przybylska

1
Torun´ Centre for Astronomy, Nicolaus Copernicus University
v
9 PL-87-100, Torun´, Poland
7
e-mail: [email protected]
0
0
. January 31, 2009
2
0
9
0
AMS Subject Classification 70F25; 37J60;34M35; 70E40
:
v
i
X Abstract
r
We consider the problem of integrability of the Poissonequations describing spatial mo-
a
tion of a rigid body in the classical nonholonomic Suslov problem. We obtain necessary
conditions for their solutions to be meromorphic and show that under some further restric-
tions these conditions arealsosufficient. The latter leadtoa family ofexplicit meromorphic
solutions, which correspond to rather special motions of the body in space. We also give
explicit extra polynomial integrals in this case.
In the more general case (but under one restriction), the Poisson equations are trans-
formed into a generalized third order hypergeometric equation. A study of its monodromy
group allows us also to calculate the “scattering” angle: the angle between the axes of limit
permanent rotations of the body in space.
1
1 Introduction
In some cases of the rigid body dynamics, in particular, in the problem of motion of a solid
about a fixedpoint, the Euler equations for the angularvelocityvectorω R3 separateand
∈
can be integrated. Then, given a generic solution ω(t), to determine the motion of the solid
in space it is necessary to solve the reconstruction problem, that is, to find 3 independent
solutions of the linear Poisson equations
γ˙ =γ ω(t), (1.1)
×
γ R3 being a unit vector fixed in space.
∈
ThemostknownexampleofsolvablePoissonequationsgivestheEulertopproblem,when
generic ω(t) are elliptic (i.e., doubly periodic) functions and one particular solution γ(t) is
also elliptic, whereas the other two are quasiperiodic (see, e.g., [7, 19]).
Similar,butformallymorecomplicatedsolutionsγ(t)appearinthecaseoftheZhukovsky–
Volterra gyrostat(see [18, 22]).
A nontrivial integrable generalization of the Euler–Poisson equations was found in [3],
where the Euler equations have the standard form and (1.1) are replaced by the equations
γ˙ =κγ ω(t),
×
κ being an arbitraryodd integer number. It was shown that, like in the Euler top problem,
the latter equations possess an extra algebraicintegral, howevera complete solution for γ is
still unknown.
In present paper, following Suslov [16], we consider the motion of the rigid body about
a fixed point in presence of constraint ω,a = 0, a being a fixed vector in the body frame.
Let I : R3 R3 be the symmetric inehrtia tiensor of the body. The Euler equations for the
7→
angular velocity vector ω separate and take the following simple form
d
(Iω)=Iω ω+λa, (1.2)
dt ×
where denotes the vector product in R3 and λ is the Lagrange multiplier. Differentiating
the con×straint,we findλ= Iω ω,I−1a / a,I−1a .Therefore,(1.2)canbe representedas
−h × i h i
d 1
(Iω)= I−1a ((Iω ω) a),
dt a,I−1a × × ×
h i
which, in view of ω,a =0, is equivalent to
h i
d
(Iω)= Iω,a ω I−1a. (1.3)
dt h i ×
In the sequel, without loss of generality,we assume that a=(0,0,1)T, which, in view of the
constraint, implies ω 0. This simplifies the Poissonequations to the form
3
≡
γ˙ = ω (t)γ , γ˙ =ω (t)γ , γ˙ =ω (t)γ ω (t)γ . (1.4)
1 2 3 2 1 3 3 2 1 1 2
− −
2
We alsoassume thatthe tensorI is disbalanced,i.e., is not diagonalin the chosenframe. (If
a is an eigenvector of I, then all the solutions of (1.3) are equilibria.)
It is knownthat under these assumptions the system(1.3) restrictedto the planeω =0
3
has a line of equilibria points I ω +I ω = 0 and that the trajectories ω(t) are elliptic
13 1 23 2
arcs that form the heteroclinic connection between the asymptotically unstable and stable
equilibria (see Fig. 1 below).
Whereas the reduced system (1.3) is elementary integrable in terms of hyperbolic func-
tions, it is believed that the corresponding Poissonequations (1.4) are not, although we did
not find a proof of that in the literature. A study of complex solutions of these equations
was done in [11]. A qualitative analysis of the behavior of γ(t) in the classical Suslov prob-
lem, as well as in its multi-dimensional generalizationwas made in [21], whereas some other
interesting generalizations of the problem were studied in [8].
Contents of the paper. InSection2wepresentgenericsolutionsoftheEulerequations
for the Suslov problem and formulate the problem of integrability of the system (1.3), (1.4)
by the Euler–Jacobi theorem.
In Section 3 necessary and sufficient conditions of meromorphicity of solutions are ob-
tained, they both require that one of the components I ,I of the inertia tensor must be
13 23
zero.
Section4discussesgeneralpropertiesofsolutions ofthe Poissonequationsinthe specific
case I = 0, which are compared with generic solutions of another famous nonholonomic
13
system, the Chaplygin sleigh. It is also shown that the meromorphicity conditions on I are
compartible with the restrictions on the inertia tensor of a physical rigid body.
In Section 5 we show that in the case I =0 the Poisson equations can be transformed
13
to a generalizedthird orderhypergeometricequation. Using its monodromygroup,we solve
the classicalproblem of calculating the angle between the axes of limit permanent rotations
of the body in space.
When the sufficient conditions of meromophicityare satisfied, we observethat the corre-
sponding hypergeometric series solutions reduce to products of polynomials and exponents,
which are explicitly calculated.
Next, we apply the differential Galois analysis to the Poisson equations and prove that
when the parameters of the problem satisfy the necessary conditions of meromophisity, but
not the sufficient ones, these equations and, therefore, the whole Suslov system, are not
solvable in the class of Liouvillian functions.
InSection6wepresentallthemeromorphicsolutionsoftheproblemandthecorrespond-
ing extra polynomial integrals in the explicit form.
In Conclusion some relevant open problems are briefly discussed.
2 Generic solutions of the Euler and the Poisson equa-
tions
In the general case the components I ,I of the inertia tensor I are not zero, but, by an
13 12
appropriate choice of the frame that preserves the constraint one can always make I = 0.
12
ForWealsoimposethenormalisationdetI=1. ConditionthatIispositivelydefinedmeans
3
that all main minors are greater than zero and we obtain I >0 and I I >0 that gives
11 11 22
I >0.
22
Then the Euler-Poissonequations (1.3), (1.4) have the form
ω˙ =I (I ω +I ω )ω , ω˙ = I (I ω +I ω )ω , (2.1)
1 22 13 1 23 2 2 2 11 13 1 23 2 1
−
and
γ˙ = ω (t)γ ,
1 2 3
−
γ˙ =ω (t)γ , (2.2)
2 1 3
γ˙ =ω (t)γ ω (t)γ .
3 2 1 1 2
−
They have two first integrals, the energy and the trivial geometric one:
F =I ω2+I ω2, F = γ,γ =γ2+γ2+γ2. (2.3)
1 11 1 22 2 2 h i 1 2 3
Inthispaper the mainproblemweconsideris: For which values of I the system (2.1),
i,j
(2.2) is integrable? Here the integrability will be understood in the context of the classical
Euler–Jacobitheorem,whichreliesuponthe existenceofaninvariantmeasureandsufficient
number of independent integrals.
Although the system (2.1), (2.2) does not have an invariant measure in the strict sense
(the density of the volume form tends to 0 as t ), we can consider its restriction on
→ ±∞
the energy level F = E > 0, which consists of two open components. On each of them we
1
choose time t and γ as coordinates and then the restricted equations read
t˙=1,
γ˙ = ω (t)γ ,
1 2 3
− (2.4)
γ˙ =ω (t)γ ,
2 1 3
γ˙ =ω (t)γ ω (t)γ .
3 2 1 1 2
−
They have the trivial integral F , as well as the invariant volume form
2
µ=dt dγ dγ dγ ,
1 2 3
∧ ∧ ∧
Thus, for the integrability in the Jacobi sense only one additional first integral F(t,γ) is
required.
The existence of such an integral is closely related to the existence of single-valued solu-
tionsofthePoissonequations. Namely,letP(t)=(P ,P ,P )asingle-valuedvector-function
1 2 3
satisfying P˙ =P ω. Then the system possesses the single-valued integral
×
F = P(t),γ , (2.5)
h i
Indeed,
F˙ = P˙(t),γ + P(t),γ˙ = P˙(t),γ + P(t),γ ω = P˙(t)+ω P(t),γ =0.
h i h i h i h × i h × i
If,moreover,the componentsofP(t)aresingle-valuedfunctionsofthesolutionsω (t),ω (t),
1 2
then the system (2.1), (2.2) admits an additional first integral
F (ω ,ω ,γ ,γ ,γ )= P(ω),γ
3 1 2 1 2 3
h i
4
functionally independent with F and F .
1 2
We start with generic solution of the dynamic equations (2.1), which have the form
a(eAt e−At)+c b(eAt e−At)+c I
1 2 13
ω (t)= − , ω (t)= − , b= a (2.6)
1 eAt+e−At 2 eAt+e−At − I
23
where
AI AI I AI I
23 13 22 23 11
a= , c = 2 , c = 2 ,
I2 I +I2 I 1 ± I2 I +I2 I I 2 ± I2 I +I2 I I
13 22 23 11 13 22 23 11r 11 13 22 23 11r 22
Abeing anarbitrarypositive constantrelatedto the energyintegral. Note thatfort
→±∞
these expressions give points on the equilibria line I ω +I ω =0, as required.
13 1 23 2
Remark 2.1 For eachA fixed, the Poissonequations (2.2) and the above functions ω(t) are
invariant with respect to time rescaling t t/A, which reduces the solution of (2.1) to the
→
form
a(et e−t)+c b(et e−t)+c I
1 2 13
ω (t)= − , ω (t)= − , b= a , (2.7)
1 et+e−t 2 et+e−t − I
23
I 2I I 2I I
23 13 22 23 11
a= , c = ± , c = ± . (2.8)
I2 I +I2 I 1 I2 I +I2 I I 2 I2 I +I2 I I
13 22 23 11 13 22 23 11r 11 13 22 23 11r 22
This implies that, without loss of generality, one can study solutions of (2.2) with the
coefficients ω ,ω having the form (2.7), (2.8). This will be assumed in the sequel.
1 2
Note that choosing ω(t) in the form (2.7) implies that the energy integral is fixed to be
1
F = . (2.9)
1 I2 I +I2 I
13 22 23 11
3 Painlev´e property
In this section we investigate the following problem.
Problem 3.1 For which values of the parameters I all solutions of the system (2.2) are
ij
meromorphic in C.
The answer is contained in the following theorem.
Theorem 3.2. All solutions of (2.2) are meromorphic solutions iff either
I I
I =0, and 11− 22 =p2, (3.1)
13
I I (I I I 1)
11 22 11 22 33
−
I I
or I =0, and 11− 22 = p2, (3.2)
23
I I (I I I 1) −
11 22 11 22 33
−
where p is a nonzero integer.
5
Figure 1: Generic trajectory ω(t) on the plane (ω ,ω ). Black and white dots indicate stable and unstable
1 2
equilibria.
Below we also show that the second case (3.2) can be obtain from (3.1) by the linear
transformation
ω ω , ω ω , γ γ , γ γ , γ γ ,
1 2 2 1 1 2 2 1 3 3
→− →− → → → (3.3)
I I , I I , I I , I I , I I .
11 22 22 11 13 23 23 13 33 33
→ → → → →
Note that the conditions (3.1) expressedin terms of the parameters (a,b,c ,c ) take the
1 2
following form
c =b=0 and c2 4a2 =4p2, p Z∗. (3.4)
1 2− ∈
The proof of Theorem 3.2 is given in two next subsections.
3.1 Necessary conditions—Kovalevskaya analysis
Let us show that if all solutions of the system (2.2) are single-valued,then one of the condi-
tions (3.1), (3.2) is satisfied.
First observe that the right hand sides of the system (2.1), (2.2) are homogeneous poly-
nomials of degree 2 in the components of x = (ω ,ω ,γ ,γ ,γ ). Then we can apply the
1 2 1 2 3
Kovalevskaya-Lyapunov analysis (see [10] for details) to find that the system admits two
solutions of the form
1
x(t)= d, where d=(ω¯ ,ω¯ ,γ¯ ,γ¯ ,γ¯ ). (3.5)
1 2 1 2 3
t
The first of them is given by
1 1
ω¯ = , ω¯ = ,
1 2
I11I23+iI13√I11I22 −I13I22 iI23√I11I22 (3.6)
−
γ¯ =γ¯ =γ¯ =0,
1 2 3
6
and the second one is its complex conjugate. According to the Lyapunov theorem, if all
solutions of (2.2) are single-valued, then
1. all eigenvalues of the Kovalevskaya matrix
∂v
K(d)= (d)+Id,
∂x
are integers, and
2. the Kovalevskayamatrix is semi-simple.
For the solution d given by (3.6) the characteristic polynomial of K(d) is
(λ 2)(λ 1)(λ+1)W(λ)
det(K(d) λId)= − − , (3.7)
− I I (I I I 1)2
11 22 11 22 33
−
where
W(λ)=( I I iI I )[I +I ( 1 I (I I i I I )2(λ 1)2)].
11 23 13 22 22 11 22 13 22 11 23
− − − − −
p p p p
The roots of W(λ) have the form
(I I )(√I I +iI √I )2(√I I iI √I )2
11 22 11 23 13 22 11 23 13 22
λ =1 Λ, Λ= − − .
1,2
± √I I (I I I 1)2
p 11 22 11 22 33
−
If λ and λ are integer, then, in particular, they are real, which implies that
1 2
I I (I I )=0. (3.8)
13 23 11 22
−
Moreover,if λ ,λ Z, then Λ2 =p2, where p Z⋆. This gives
1 2
∈ ∈
(I I )(I I2 I I2 )
11− 22 11 23− 22 13 =p2. (3.9)
I I (I I I 1)2
11 22 11 22 33
−
The condition (3.8) gives three possibilities. Either I = 0, and this case gives (3.1), or
13
I = 0, which gives (3.2), or, finally, I = I . In the last case, the Kovalevskaya matrix
23 11 22
hasthemultiple eigenvalueλ=1andisnotsemi-simple. Asaresult,weprovedthe‘if’ part
of Theorem 3.2.
Remark 3.3 In terms parameters (a,b,c ,c ) conditions (3.8) and (3.9) have the form
1 2
ac +bc =0 and c2+c2 4(a2+b2)=4p2, p Z. (3.10)
1 2 1 2− ∈
3.2 Sufficient conditions—analysis of the monodromy group
The general solution (2.7) of the Euler equations is single-valued. Thus, all solutions of the
Euler-Poisson equations (2.1), (2.2) are single-valued if and only if all the solutions of the
linear Poisson equations (2.2) with ω (t) and ω (t) as in (2.7) are single-valued.
1 2
7
Notethattheonlysingularpointsof (2.2)aresimplepoleslocatedatt = πi mod πi.
0 ±2
Then a necessary condition for the single-valuedness is that the monodromy matrices at all
the singular points are identities.
To show this, rewrite the system (2.2) in the form
0 0 ω (t)
2
−
γ˙ =A(t)γ, A(t)= 0 0 ω (t) (3.11)
1
 
ω (t) ω (t) 0
2 1
−
 
and observe that
0 0 b+ 1ic
− 2 2
1
A(t)= A +O(t t ), A = 0 0 a+ 1ic . (3.12)
t t 0 − 0 0  2 1
0
− b 1ic a+ 1ic 0
 − 2 2 − 2 1 
 
The monodromy matrix of the canonical loop around t is M := exp2πiA . A direct
0 t0 0
calculation shows that the eigenvalues of A are
0
1
ρ =0 and ρ = c2+c2 4(a2+b2)+4i(ac +bc ). (3.13)
1 2,3 ±2 1 2− 1 2
q
Hence, if the conditions (3.10) are satisfied, then A has eigenvalues ρ = 0, ρ = p, and
0 1 2
ρ = p with p Z⋆. In this case M = Id provided that there is no local logarithmic
3 − ∈ t0
solutions in a neighborhood of t . In order to check this, we use the following theorem (see,
0
e.g., [1] for the details).
Theorem 3.4. Assume that in the linear system
∞
Y˙ =B(t)Y, B(t)=t−1 B ti, (3.14)
i
i=0
X
the matrix coefficient B does not have a pair of eigenvalues such that their difference is a
0
non-zero integer. Then there exists a matrix T given by a convergent power series
∞
T = T ti, T =Id,
i 0
i=0
X
such that the linear map Y =TZ transforms (3.14) into the following form
1
′
Z = B Z. (3.15)
0
t
The form of the fundamental matrix Z of (3.15) is well known. Namely, let P be the
B0
similarity transformation reducing B into its Jordan form J = P−1B P, where P is an
0 0
invertible matrixwithconstantcoefficients,andletZ denote the new fundamentalmatrix.
J
Then Z = PZ P−1, and Z has a block-diagonal structure such that the Jordan block
B0 J J
8
(of dimension ν) corresponding to an eigenvalue λ has the form
1 lnt lnν−1t
··· (ν−1)!
tλ0 1 ... ... .
... ... ... lnt 
 
0 ... 0 1 
 
 
Now assume that conditions (3.1) or (3.2) are satisfied. By the existence of the linear
transformation (3.3), without loss of the generality we can assume that conditions (3.1) are
satisfied.
Expressed in terms of parameters (a,b,c ,c ), the latter take the form (3.4), or, setting
1 2
c = 2ca, c2 =I /I , the following form
2 11 22
−
(c2 1)a2 =p2. (3.16)
−
We also denote d2 =c2 1.
−
Now we check the presence of logarithmic terms in the solutions.
Theorem 3.5. The Poisson equations (2.2) with ω given by (2.7) with c = b = 0, c =
i 1 2
2ca = 2√d2+1p/d, p Z, do not have logarithmic terms in local solutions around
− − ∈
singular points t .
0
Proof. For c =b=0, c = 2ca= 2√d2+1p/d, the matrix A reads
1 2 0
− −
ip√d2+1
0 0
− d
 p 
A = 0 0 . (3.17)
0 d
 
ip√d2+1 p 
 0 
 
 d −d 
 
Using the similarity transformation S, one obtains its diagonal form: A˜ =S−1A S with
0 0
i√d2+1 i i√d2+1
p 0 0
− d −√d2+1 d
A˜ = 0 0 0 , S = 1 1 .
0
  1
0 0 p  d −d 
 −   1 0 1 
 
Thus we see that all the eigenvalues of A are integer and their differences are also nonzero
0
integers. Under this transformation the matrix A of the linear Poisson equations takes the
form
ipsech(t)(d2+1+isinh(t)) ip(sech(t)+itanh(t))
0
d2 2d
A˜=(d2+1)p(−isech(t)+tanh(t)) 0 (d2+1)p(−isech(t)+tanh(t)),
 d3 d3 
 ip(sech(t)+itanh(t)) p[ i(d2+1)sech(t)+tanh(t)] 
 0 − 
 2d d2 
 
 
9
Transformation γ Γ defined by γ =UΓ with
→
2p
iπ
t 0 0
− 2
U =(cid:18) (cid:19) iπ p ,
0 t 0
 − 2 
 (cid:18) (cid:19) 
 0 0 1
 
 
gives the equivalent linear system Γ˙ =B(t)Γ with
B(t)=U−1AU U−1U˙.
−
This matrix has residual part diagonal
p 0 0
−
B = 0 p 0 ,
0
 − 
0 0 p
−
 
which satisfies the assumptions of Theorem 3.4. Since all the Jordan blocks of B are one-
0
dimensional, the logarithmic terms do not appear in the solutions of the Poisson equations.
4 Special case of the inertia tensor. General properties
of solutions
Assume now that, apart from I = 0, the first meromorphisity condition in (3.1) is also
12
satisfied, that is, I =0, which imposes an essential restriction on I.
13
Then the reparametrized solutions (2.7), (2.8) are reduced to
a(et e−t) 2ac
ω (t)= − , ω (t)= , (4.1)
1 et+e−t 2 −et+e−t
I I I2 I
a= 22 33− 23, c= 11.
I I
23 r 22
It is seen that they have equilibria along the line ω =0.
2
Usingtheparameterpintroducedin(3.1),andsettingd=p/a,wecanrewritethe whole
system (2.1), (2.2) in the form
d
ω˙ = ω2,
1 ±p(d2+1) 2
d
ω˙ = ω ω ,
2 1 2
−p (4.2)
γ˙ = ω γ ,
1 2 3
−
γ˙ =ω γ ,
2 1 3
γ˙ =ω γ ω γ ,
3 2 1 1 2
−
10