The characteristic system for the Euler - Poisson's equations
In this paper we investigate the nonlinear system naturally connected with the Euler - Poisson's equations. The solutions of this system may be used for description of the singular points to the Euler - Poisson's equations.
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Інститут прикладної математики і механіки НАН України
1999
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| Назва видання: | Нелинейные граничные задачи |
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| Цитувати: | The characteristic system for the Euler - Poisson's equations / A.V. Belyaev // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 135-147. — Бібліогр.: 8 назв. — англ. |
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irk-123456789-1692852020-06-10T01:26:21Z The characteristic system for the Euler - Poisson's equations Belyaev, A.V. In this paper we investigate the nonlinear system naturally connected with the Euler - Poisson's equations. The solutions of this system may be used for description of the singular points to the Euler - Poisson's equations. 1999 Article The characteristic system for the Euler - Poisson's equations / A.V. Belyaev // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 135-147. — Бібліогр.: 8 назв. — англ. 0236-0497 http://dspace.nbuv.gov.ua/handle/123456789/169285 en Нелинейные граничные задачи Інститут прикладної математики і механіки НАН України |
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In this paper we investigate the nonlinear system naturally connected with the Euler - Poisson's equations. The solutions of this system may be used for description of the singular points to the Euler - Poisson's equations. |
| format |
Article |
| author |
Belyaev, A.V. |
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Belyaev, A.V. The characteristic system for the Euler - Poisson's equations Нелинейные граничные задачи |
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Belyaev, A.V. |
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Belyaev, A.V. |
| title |
The characteristic system for the Euler - Poisson's equations |
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The characteristic system for the Euler - Poisson's equations |
| title_full |
The characteristic system for the Euler - Poisson's equations |
| title_fullStr |
The characteristic system for the Euler - Poisson's equations |
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The characteristic system for the Euler - Poisson's equations |
| title_sort |
characteristic system for the euler - poisson's equations |
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Інститут прикладної математики і механіки НАН України |
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1999 |
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http://dspace.nbuv.gov.ua/handle/123456789/169285 |
| citation_txt |
The characteristic system for the Euler - Poisson's equations / A.V. Belyaev // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 135-147. — Бібліогр.: 8 назв. — англ. |
| series |
Нелинейные граничные задачи |
| work_keys_str_mv |
AT belyaevav thecharacteristicsystemfortheeulerpoissonsequations AT belyaevav characteristicsystemfortheeulerpoissonsequations |
| first_indexed |
2025-07-15T04:02:33Z |
| last_indexed |
2025-07-15T04:02:33Z |
| _version_ |
1837684127620399104 |
| fulltext |
THE CHARACTERISTIC SYSTEM FOR
THE EULER - POISSON’S EQUATIONS
c© A.V.Belyaev
Abstract
In this paper we investigate the nonlinear system naturally connected with the
Euler - Poisson’s equations. The solutions of this system may be used for description
of the singular points to the Euler - Poisson’s equations.
The properties of the solutions of the Euler - Poisson’s equations ([1], [2]) depend on
the singular points of these solutions. For example, the classical case of the S.Kovalevskaya
([3]) was found on the way of investigation of the single-valued solutions. On another hand
it is proved ([4],[5]) that the branching of the solutions implies the absence of single-valued
first integrals.
We hope that complete information about the singular points will permit to get new
results in the solid body problems. In this paper we present in detail the first part of
the method to reseach the singular points ([6],[7]). It is solving of the characteristic
system naturally appearing from the Euler - Poisson’s equations. We emphasize that the
conditions of the Euler, Lagrange, Kovalevskaya and Grioli cases appear in this stage of
the method and without the investigation of the differential equations.
Let’s write the Euler - Poisson’s equations in the following form:
{
A
.
p= Ap× p + γ × r
.
γ= γ × p,
(1)
here p = (p1, p2, p3) ∈ C3, γ = (γ1, γ2, γ3) ∈ C3, Ap = (A1p1, A2p2, A3p3), Ai > 0,
r = (r1, r2, r3) ∈ R3.
We use the notation z(t) = (p(t), γ(t)) too.
Define C-scalar production in C3: 〈x, y〉 =
∑3
i=1 xiyi.
Notate ‖z(t)‖ = 〈p, p̄〉1/2 + 〈γ, γ̄〉1/4.
We use the circle replacement of indices σ = (1, 2, 3) for writing the sumes or products
(for example,
∑
σ A1A2 = A1A2+ A2A3+ A3A1,
∏
σ A1 = A1A2A3), ) and expressions
which differ one from another only by the circle replacement of indices (γ̇ = γ× p, can be
writed γ̇1 = p3γ2 − p2γ3, σ).
Introduce the notations Bij = Ai − Aj, Cij = 2Ai − Aj, Dij = Ai + Aj too.
Let t∗ ∈ C be a singular point of the solution z(t) of the system (1) (i.e. t∗ is a
singular point of the coordinate functions of z(t)). Get rid of the branch in t∗, if any, by
the representation z(t) =
∧
z (ln(t − t∗)), where
∧
z (τ) is single-valued function when Re
τ → −∞.
The system (1) is transformed into:
A
.∧
p= eτ (A
∧
p × ∧
p +
∧
γ ×r)
.∧
γ= eγ (
∧
γ × ∧
p),
where the derivative is taken by τ .
In order to make the right part of the equation independent of τ we make replacement
of variable again, setting
∼
p (τ) = eτ
∧
p (τ),
∼
γ (τ) = e2τ ∧
τ (τ) and then we have:
A
.∼
p= A
∼
p × ∼
p +
∼
γ ×r + A
∼
p
.∼
γ=
∼
γ × ∼
p +2
∼
γ .
(2)
In this case the dependence on between the solutions (1) and (2) is expressed by the
corelations
p(t) =
1
t− t∗
∼
p (ln(t− t∗)), γ(t) =
1
(t− t∗)2
∼
γ (ln(t− t∗)) (3)
Assertion 1 The solution p(t), γ(t) of the system (1) does not have the singularity in the
point t∗, if and only if the corresponding solution (2) by (3) have the asymptotic behaviour
∼
p (τ) ∼ ∼
p0 eτ ,
∼
γ (τ) ∼ ∼
γ0 e2τ , when Re τ → −∞.
Proof. If
∥∥∥∼z (τ)
∥∥∥ isn’t separated from zero then we can neglect the quadratic part of
(2) in the suitable moment, and the solution
∼
z (τ) turns out to be exponently decreasing:
∼
p∼ p0e
τ ,
∼
γ∼ γ0e
2τ if Reτ → −∞. According to (3) we have p ∼ p0, γ ∼ γ0, t → t∗,
consequently (p(t), γ(t)) does not have the singularity in the point t∗.
If
∥∥∥∼z (τ)
∥∥∥ is separated from zero when Reτ → −∞ then according (3) ‖z(t)‖ →
∞, t → t∗ 2
The solution
∼
z (τ) which does not have the asymtotic behaviour (
∼
p0 eτ ,
∼
γ0 e2τ ), Re
τ → −∞, are first, the constant solutions and, second, have trajectories entering singular
points.
This fact is fundamental: we can completely investigate the singular points of the
differential equation but at the same time we cannot say that all singular points of the
solution of (1) can be obtained in such a way.
Definition 1 We call the system (the solution of which are singular points of (2))
A
∼
p
0 × ∼
p
0
+
∼
γ
0 ×r + A
∼
p
0
= 0
∼
γ
0 × ∼
p
0
+2
∼
γ
0
= 0.
(4)
characteristic (for Euler - Poisson’s equations).
Now for convenience we write (p, γ) instead of (
∼
p
0
,
∼
γ
0
) in this paragraph.
Assertion 2 If all Ai are different, r1r2r3 6= 0, then characteristic system is equivalent
to two systems {
Ap× p + Ap = 0
γ = 0.
, (5)
〈Ap,Ap〉 = −〈Ap, r〉2 〈p, r〉−2
〈Ap, p〉 = −2 〈Ap, r〉 〈p, r〉−1
〈p, p〉 = −4
γ = −(Ap× p) 〈p, r〉−1 , p× γ 6= −2γ.
(6)
Proof. Necessity. Let’s obtain the following relations from (4):
0 = 〈γ × p, γ〉+ 2 〈γ, γ〉 = 2 〈γ, γ〉 ,
0 = 〈γ × p, p〉+ 2 〈γ, p〉 = 2 〈γ, p〉 ,
0 = 〈Ap× p, γ〉+ 〈γ × r, γ〉+ 〈Ap, γ〉 = 〈p× γ,Ap〉+ 〈Ap, γ〉 = 3 〈Ap, γ〉 ,
then we see that vectors Ap, p, γ are linearly dependent.
If p and Ap are proportional then two from three coordinates of p (or Ap) are equal
to zero. In this case we should have γ × r + Ap = 0 ⇒ 〈Ap, r〉 = 0, but it is impossible
because all coordinates of r are not equal to zero by condition. So, γ = ν1Ap + ν2p;
multiply this equivalence by Ap and p.
{
ν1 〈Ap,Ap〉+ ν2 〈Ap, p〉 = 0
ν1 〈Ap, p〉+ ν2 〈p, p〉 = 0
Now we have ν1 = ν2 = 0 and the system (5) is true; or we have 〈Ap,Ap〉 〈p, p〉 =
〈Ap, p〉2, where 〈p, p〉 = −4, ( 0,±
√
−〈p, p〉 is eigenvalue of the eigenvector γ of the linear
transformation ξ → p× ξ)
On the one hand we have 〈γ, p〉 = 〈γ,Ap〉 = 0 , therefore γ = νAp× p. On the other
hand we have p× γ = 2γ and γ = ν1Ap + ν2p, thus, ν = −ν1
2
and Ap× p = −2Ap− ν2
ν1
p
(ν1 6= 0, because in other case γ = 0) .
Substitute Ap × p in (4) by this presentation expression and multiply by ν1r . We
obtain the relation −ν1 〈Ap, r〉 − 2ν2 〈p, r〉 = 0.
Add to the obtained equation
ν1 〈Ap, p〉+ ν2 〈p, p〉 = 0,
then we get
−2 〈Ap, r〉 = 〈p, r〉 〈Ap, p〉 .
At last multiply the first equation of characteristic system by p having γ as linear
combination of Ap p . It will give
ν1 〈p× Ap, r〉+ 〈Ap, p〉 = 0
or by substitution p× Ap from (4) we have
ν1 〈Ap, r〉+ 〈Ap, p〉 = 0.
Now it is evident that ν1 = 2 〈p, r〉−1 , and γ = −(Ap× p) 〈p, r〉−1 . Note that 〈p, r〉 6=
0 because in other case 〈Ap, r〉 = 0, and then r × γ = 0 and 〈γ, γ〉 6= 0 that is the
contradiction to the condition.
Sufficiency. If p, γ is the solution of the system (5) then p, γ is the solution of (4) too.
Now let p, γ be the solution of the system (6). Then vector ξ0 = 〈p, p〉Ap−〈Ap, p〉 p is
normal to p and Ap which are not proportional (because of γ = −(Ap× p) 〈p, r〉−1 6= 0).
The vectors ξ0, p × ξ0 are proportional to γ and, therefore, ξ0, γ are eigenvectors of the
linear transformation ξ0 → p× ξ0, namely, p× ξ0 = 2ξ0, p× γ = 2γ. Using the expression
for ξ0 in the last equation we have
Ap× p = −2p− 1
2
〈Ap, p〉 p ⇒
〈Ap× p + γ × r + Ap, r〉 = −〈Ap, r〉 − 1
2
〈Ap, p〉 〈p, r〉 = 0.
It is evident that 〈Ap× p + γ × r + Ap, γ〉 = 0.
We want to prove that 〈Ap× p + γ × r + Ap, p〉 = 0. This equality is equivalent to
〈γ × r, p〉+ 〈Ap, p〉 = 0,
2 〈γ, r〉+ 〈Ap, p〉 = 0,
but γ = −(Ap× p) 〈p, r〉−1, therefore,
−2 〈Ap× p, r〉+ 〈Ap, p〉 〈p, r〉 = −2 〈Ap× p, r〉 − 2 〈Ap, r〉 = 0,
So if the vectors r, γ, p are linear independent then the assertion is proved.
Let 〈p× γ, r〉 = 2 〈γ, r〉 = 0; then this condition is equivalent to 〈Ap× p, r〉 = 0 or
〈Ap, r〉 = 0.
In this case 〈Ap,Ap〉 = 〈Ap, p〉 = 0, consequently Ap× γ = 0 ⇒ Ap× p = −2Ap. The
vector r equals to linear combination of p and Ap. Let r = µ1Ap + µ2p,
µ2 6= 0. Then
Ap× p + γ × r + Ap = 0 ⇐ γ × r = Ap ⇐ 2 〈p, r〉−1 Ap× (µ1Ap + µ2p) = Ap ⇐
2µ2Ap× p
µ2 〈p, p〉 = Ap
So the assertion is proved. 2
Assertion 3 The solution of the system (5) is as follows:
) p = 0;
) (if all Ai are different)
p1 =
√
A2A3
B12B31
, σ,
here if (p1, p2, p3) is solution of (5) then other solutions are
(−p1,−p2, p3), (−p1, p2,−p3), (p1,−p2,−p3).
Proof. The solution z = 0 is evident. The other solutions we find by transformation
the system 〈Ap, p〉 = 〈Ap,Ap〉 = 0 to a non-homogeneous one. 2
Assertion 4 The solution of the system (6) (all Ai are different, r1r2r3 6= 0) may be
found if the solution of the equation
∑
σ
r1(A1 − α)
√
(2A2 − α)(2A3 − α)B23 = 0 (7)
or the equation of the 8th power
∑
σ
[r4
1B
2
23(A1 − α)4(2A2 − α)2(2A3 − α)2−
2r2
2r
2
3B12B31(A2 − α)2(A3 − α)2(2A1 − α)
∏
σ
(2A1 − α)] = 0. (8)
is known.
Proof. Let α = 〈Ap, r〉 〈p, r〉−1 . Then the system (6) has the form
〈Ap,Ap〉 = −α2
〈Ap, p〉 = −2α
〈p, p〉 = −4.
This system is linear as to p2
i . Its solution is
p2
1 =
(2A2 − α)(2A3 − α)
B12B31
, σ (9)
Let’s recall what is α. We obtain the equation
∑
σ
A1r1
√
(2A2 − α)(2A3 − α)
B12B31
= α
∑
σ
r1
√
(2A2 − α)(2A3 − α)
B12B31
,
which equals to (7).
To receive the polynomial by α we use the identity
(
∑
σ
a1)
∏
σ
(a1 − a2 − a3) =
∑
σ
(a4
1 − 2a2
2a
2
3).
2
Now we propose another method for solving the characteristic system which gives a num-
ber of impotant relations.
Assertion 5 The solutions z of the characteristic system satisfy the following relations
H = 1
2
〈Ap, p〉+ 〈γ, r〉 = 0
M = 〈Ap, γ〉 = 0
T = 〈γ, γ〉 = 0
D = 〈p, γ〉 = 0
E =
∑
σ[r1B23γ2γ3(C21B31γ
2
3 − C31B12γ
2
2)] = 0
the last relation takes place if is not true
E0 = 〈Aγ, γ〉 = 0.
Proof. The relations T = 0,M = 0,D = 0 were obtained in the proof of the assertion 2.
0 = 〈Ap× p + γ × r + Ap, p〉 = 2H.
Lets prove the last relation. Note that
p =
2Aγ × γ
〈Aγ, γ〉 , if 〈Aγ, γ〉 6= 0. (10)
Indeed, it is clear that p = λAγ × γ because Aγ and γ are not proportional and
〈p, γ〉 = 〈p,Aγ〉 = 0. Moreover, −4 = 〈p, p〉 = λ 〈p,Aγ × γ〉 = −2λ 〈Aγ, γ〉 .
Substitute p in (4)
4γ1
∏
σ(B12γ3)
〈Aγ, γ〉2 + r3γ2 − r2γ3 +
2A1B23γ2γ3
〈Aγ, γ〉 = 0, σ.
Then by multiplying these equations by r1, σ and adding them we obtain the relation
which is equivalent the relation E = 0.
4
∏
σ
(B12γ3)
∑
σ
(γ1r1) + 2
∑
σ
(A1B23r1γ2γ3)
∑
σ
(A1γ
2
1) = 0. (11)
Then we obtain
∑
σ
[2B12B23B31γ2γ3(−γ2
2 − γ2
3) + A1B23γ2γ3(−B12γ
2
2 + B31γ
2
3)]r1 = 0
∑
σ
r1B23γ2γ3[(A1B31 − 2B12B31)γ
2
3 − (A1B12 + 2B12B31)γ
2
2 ] = 0
∑
σ
r1B23γ2γ3(C21B31γ
2
3 − C31B12γ
2
2) = 0.
2
So, we have the following method for solving the characteristic system:
at first to find the vector λγ, (λ ∈ C), as the solution of the system
{ ∑
σ γ2
1 = 0∑
σ r1B23γ2γ3(C21B31γ
2
3 − C31B12γ
2
2) = 0,
(12)
then to find p,
p =
2Aγ × γ
〈Aγ, γ〉
and, finally, to find γ , using any non-homogeneous relation of the characteristic system.
Remark 1 By condition r3 = 0 the characteristic system (4) has symmetry
S3 : (p1, p2, p3, γ1, γ2, γ3) ←→ (−p1,−p2, p3, γ1, γ2,−γ3). (13)
Theorem 1 The characteristic system (4) has following solutions
0) (p, γ) = 0;
1) by condition
∏
σ B12 6= 0
γ1 = 0, p1 =
√
A2A3
B12B31
, σ
here, if (p1, p2, p3) is the solution of (5) then the other solutions are
(−p1,−p2, p3), (−p1, p2,−p3), (p1,−p2,−p3),
moreover,
a) by condition
∏
σ r1 6= 0, there are 8 solutions (taking into account the multiplisity
of the roots) which can be obtained (see (6), (8), (9)) if we know the roots of the polinomial
(8);
b) by condition r3 = 0, r1r2 6= 0, there are 2 solutions lying on the axes of symmetry
S3 (see (13))
p1 = p2 = γ3 = 0, p3 = ±2i, γ1 =
2A3
r1 ± ir2
, γ2 = ± 2A3i
r1 ± ir2
; (14)
b1) on condition r2
1B23 = r2
2B31 (Grioli’s case), there is a pair os the S3− sym-
metric solutions
p1 =
∓2A2i
D12B12
√
A2
1
B23
B31
+ A2
2, p2 =
∓2A1i
D12B12
√
A2
1 + A2
2
B31
B23
, p3 =
2A1A2
D12
√
B23B31
,
γ1 =
2A2
1A2
D2
12r1
, γ2 =
2A1A
2
2
D2
12r2
, γ3 =
±2A1A2i
D2
12 r1
√
A1 + A2
2
B31
B23
;
b2) by condition r2
1B23 6= r2
2B31 there are 3 pairs S3 – symmetric solutions which
can be obtained as follows: we find the roots γ1
γ2
of the polinomial of the 3rd power
r2B31
γ1
γ2
(
A2B31
γ2
1
γ2
2
− C12B23
)
+ r1B23
(
C21B31
γ2
1
γ2
2
− A1B23
)
= 0,
and then we use the relations
γ2
γ3
=
√√√√1− γ2
1
γ2
2
, p =
2Aγ × γ
〈Aγ, γ〉 , γ =
−Ap× p
〈p, r〉 ;
c) by condition r2 = r3 = 0, r1 6= 0 there are 4 solutions lying on the axises of S2, S3
– symmetries (see (13))
p1 = p2 = γ3 = 0, p3 = ±2i, γ1 =
2A3
r1
, γ2 = ±2A3i
r1
, (15)
p1 = p3 = γ2 = 0, p2 = ±2i, γ1 =
2A2
r1
, γ3 = ∓2A2i
r1
(16)
and the 4 S2, S3 -symmetric solutions
p1 =
√
C21B31
√
C31B12
B12B31
, p2 =
√
A1B23
√
C31B12
B12B23
, p3 =
√
C21B31
√
A1B23
B23B31
,
γ1 =
A1
r1
, γ2 =
A1
√
C21B31
r1
√
A1B23
, γ3 =
A1
√
C31B12
r1
√
A1B23
,
here the signs
√
C21B31,
√
C31B12,
√
A1B23 are taken freely but equally in all the formulas;
d) by condition r = 0 (Euler’s case), other solutions are absent;
2) by condition A1 = A2, r2 = 0,
a) by condition r1 6= 0
p1 = p3 = γ2 = 0, p2 = ±2i, γ1 =
±2A1i
r3 ± ir1
, γ3 =
±2A1
r3 ± ir1
;
a1) by condition C31 6= 0
p1 = ∓2A3r3i
C31r1
, p2 =
2A3r3
C31r1
, p3 = ±2i, γ1 =
2A3
r1
, γ2 = ±2A3i
r1
, γ3 = 0; (17)
a2) by condition C31 = 0
a2.1) by condition r3 6= 0, other solutions are absent;
a2.2) by condition r3 = 0 (Kovalevskaya’s case), an one-parameter set of the
points
γ1 =
A1
r1
, γ2 = ±A1i
r1
, γ3 = 0, p3 = ±2i, p2 = ±p1i; (18)
b) by condition r1 = 0, r3 6= 0 (Lagrange’s case), an one-parameter set of the points
γ3 =
2A1
r3
, p3 = 0, p2 =
γ1r3
A1
, p1 = −γ2r3
A1
, γ2
1 + γ2
2 = −4
A2
1
r2
3
;
c) by condition r = 0 (confluent Euler’s case) any solutions are absent. (the above
mentioned cases see in [1], [2]).
Proof. 1. Is respected to assertion 3.
1. The coefficient at α8 in (8) equals
∑
σ
(r4
1B
2
23 − 2r2
2r
2
3B12B31) = (
∑
σ
r1
√
B23)
∏
σ
(r1
√
B23 − r2
√
B31 − r3
√
B12).
This coefficient isn’t equal to zero because all Ai are different and r1r2r3 6= 0. Hence
there exist exactly 8 roots of the equation (8). Suppose that these roots are different.
Every root α0 corresponds to 4 pairs (p,−p) of the representation (9). But only one pair
satisfies the condition α0 〈p, r〉 = 〈Ap, r〉 .
In fact, if pi = 0 for some i , then by (9) pj = 0 for some another j and then Ap, p are
proportional. According to proof of assertion 2 it is impossible if r1r2r3 6= 0. So all pi are
not equal to zero.
The left part of (8) we can represent in the form (preliminary fixing the branches for
the expressions
√
(2A1 − α)(2A2 − α)B12, σ )
P8(α) = (
∑
σ
r1(A1 − α)
√
(2A2 − α)(2A3 − α)B23
∏
σ
[r1(A1 − α)
√
(2A2 − α)(2A3 − α)B23
−r2(A2 − α)
√
(2A3 − α)(2A1 − α)B31 − r3(A3 − α)
√
(2A1 − α)(2A2 − α)B12] (19)
Since pi 6= 0 then P8(α)′α=α0
6= 0 if and only if only one factor of (19) equals to zero,
when α = α0. But we supposed that all roots of (8) are different, hence only one pair
(−p, p) satisfies the condition α0 〈p, r〉 = 〈Ap, r〉 .
The condition p × γ = 2γ determines the only one possible p. So there exists one to
one correspondence between roots of the polynomial P8(α) and roots of the characteristic
systems.
Suppose now that the polynomial P8(α) has the multiple root. Remined the definition
of the multiplicity of the root α0: it is number of the roots α′0 of the polynomial P8(α)+ ε
(ε ≈ 0) near the α0. Thus it is sufficiently to prove the correspondence between roots of
(6) and (8) in generel case but it is already done.
1.b. Lemma. Let
∏
σ B12 6= 0. Then all solutions (p, γ), γ 6= 0 of the characteristic
system have following form:
p =
2Aγ′ × γ′
〈Aγ′, γ′〉 , γ = λγ′,
where γ′ – is a root of the system (12) such that 〈Aγ′, γ′〉 6= 0, λ – is some constant.
Proof of lemma. Let (p, γ) be a solution of the characteristic system. We want to
prove that 〈Aγ, γ〉 6= 0,
Indeed, if 〈γ, γ〉 = 0, then Aγ × γ 6= 0. But in this case p is proportional to γ because
〈γ, γ〉 = 〈Aγ, γ〉 = 0 = 〈γ, p〉 = 〈Aγ, p〉, consequently, 2γ = p × γ = 0. We have
contradiction with the condition γ 6= 0, hence 〈Aγ, γ〉 6= 0.
So, if (p, γ) is a root of the characteristic system, γ 6= 0 then (γ, p) satisfy (12) and
(10) .
Let now that γ′ be a solution of (12) and 〈Aγ′, γ′〉 6= 0. We take
p =
2Aγ′ × γ′
〈Aγ′, γ′〉 .
Second equation of the system (12) can be represented in the form
〈Ap× p, r〉+ 〈Ap, r〉 = 0
or 〈Ap× p + γ′ × r + Ap, r〉 = 0.
We have too 〈Ap× p + γ′ × r + Ap, γ′〉 = 0,
because Ap× p is propotional γ′. The vectors r, γ′ are non-proportional therefor
Ap× p + γ′ × r + Ap = µγ′ × r;
and taking λ = 1− µ, we get
Ap× p + λγ′ × r + Ap = 0.
Second equation of the characteristic system is satisfied by
(p, λγ′):
p× λγ′ =
2Aγ′ × γ′
〈Aγ′, γ′〉 × λγ′ =
λ
〈Aγ′, γ′〉 [〈2Aγ′, γ′〉 γ′ − 2 〈γ′, γ′〉Aγ′] = 2λγ′.
The lemma is proved.
So let
∏
σ B12 6= 0, r1r2 6= 0, r3 = 0. In this case (12) has the form (if we substitute
γ2
3 = −γ2
1 − γ2
2):
{
r1B23γ2γ3(C21B31γ
2
1 − A1B23γ
2
2)− r2B31γ3γ1(C12B23γ
2
2 − A2B31γ
2
1) = 0
γ2
1 + γ2
2 + γ2
3 = 0.
(20)
We see the root γ3 = 0, hence γ = λ(1,±i, 0) and by (10) p1 = p2 = 0, p3 = ±2i.
Then we have 2A3 = −1
2
〈Ap, p〉 = 〈γ, r〉 = λ(r1 ± ir2), and
γ1 =
2A3
r1 ± ir2
, γ2 =
±2A3i
r1 ± ir2
, γ3 = 0
Let now γ3 6= 0.
1b1) 〈Aγ, γ〉 = 0 is realized for some root γ of system (12). Then 〈γ, r〉 = 0 (see (11))
and r2
1B23 = r2
2B31.
In this case γ1r1 +γ2r2 = γ1r2B31 +γ2r1B23. must devide first equation of system (20).
As a result we get
r2A2B
2
31γ
3
1 + r1B23B31C21γ
2
1γ2 − r2B23B31C12γ1γ
2
2 − r1A1B
2
23γ
3
2 =
(B31r2γ1 + B23r1γ2)
2(A2r2B31γ1 − A1r1B23γ2)r
−2
2 B−1
31 .
So, according the lemma, by condition 〈Aγ′, γ′〉 6= 0 the unique solution of (12) (within
proportionality) is
γ′ = (A1B23r1, A2B31r2,±ir1
√
(A2
1B23 + A2
2B31)B23) ,
p1 = ±2B23A2B31r1r2i
√
(A2
1B23 + A2
2B31)B23
B23B31(−A2
1B23r2
1 + A2
2B31r2
2)
= ∓ 2A2i
(A1 + A2)B12
√
A2
1
B23
B31
+ A2
2 ,
p2 = ±2B31A1B23r
2
1i
√
(A2
1B23 + A2
2B31)B23
B23B31(−A2
1B23r2
1 + A2
2B31r2
2)
= ∓ 2A1i
(A1 + A2)B12
√
A2
1 + A2
2
B31
B23
,
p3 =
2A1A2B12B23B31r1r2
B23B31(−A2
1B23r2
1 + A2
2B31r2
2)
=
2A1A2
(A1 + A2)
√
B23B31
.
By assertion 2 the solution of characteristic system satisfies (5) or (6) if Ap and p are
non-proportional. Hence 〈γ, r〉 = −1
2
〈Ap, p〉 = α, where α is root of (7)
r1(A1 − α)
√
(2A2 − α)(2A3 − α)B23 = −r2(A2 − α)
√
(2A3 − α)(2A1 − α)B31 ⇔
α 6=2A3⇐⇒ r2
1(A1 − α)2(2A2 − α)B23 = r2
2(A2 − α)2(2A1 − α)B31 ⇔
⇔ α =
2A1A2
A1 + A2
.
The root α = 2A3 corresponds to the solution (14), consequently α = 2A1A2(A1 +
A2)
−1 corresponds to the solution which we consider now. Lt γ = λγ′.
2A1A2(A1 + A2)
−1 = λ 〈γ′, r〉 = λ(A1B23r
2
1 + A2B31r
2
2) ⇒ λ =
2A1A2
(A1 + A2)2r2
1B23
,
consequently,
γ1 =
2A2
1A2
(A1 + A2)2r1
, γ2 =
2A1A
2
2
(A1 + A2)2r2
, γ3 = ± 2A1A2
(A1 + A2)2r1
√
A2
1 + A2
2
B31
B23
,
1b2. It follows from the lemma.
1c. Two solutions of (12) are evident:
γ3 = 0 ⇒ γ = λ(1,±i, 0) ⇒ p1 = p2 = 0, p3 = ±2i ⇒ λr1 = −1
2
〈Ap, p〉 = 2A3 ⇒
γ1 =
2A3
r1
, γ2 = ±2A3i
r1
, γ3 = 0,
γ2 = 0 ⇒ γ = λ(1, 0,±i) ⇒ p1 = p3 = 0, p2 = ±2i ⇒ λr1 = −1
2
〈Ap, p〉 = 2A2 ⇒
γ1 =
2A2
r1
, γ3 = ±2A2i
r1
, γ2 = 0,
Moreover, C21B31γ
2
3 − C31B12γ
2
2 = 0 (see (12)) ⇒
γ = λ(
√
A1B23,
√
C21B31,
√
C31B12)
; and we see that
〈Aγ, γ〉 = λ2(A2
1B23 + A2B23C21 + A3B12C31) = −2λ2
∏
σ
B12.
p1 =
√
C21B31
√
C31B12
−B12B31
, p2 =
√
A1B23
√
C31B12
−B12B23
, p3 =
√
C21B31
√
A1B23
−B23B31
From equation (7) we get 〈γ, r〉 = α = A1 (roots α = 2A2, α = 2A3 are correspond to
solutions (15), (16) of the characteristic system).
So, λ
√
A1B23r1 = A1, consequently,
γ1 =
A1
r1
, γ2 =
A1
√
C21B31
r1
√
A1B23
, γ3 =
A1
√
C31B12
r1
√
A1B23
1d. It follows from asseertion 3.
2. If A1 = A2 = A3 then we can choose r1 = r2 = 0, and then we have confluent case
of Euler( r3 = 0) or special case of Lagrange (r3 6= 0), (see lower).
So, A1 = A2 6= A3, r1 6= 0, r3 6= 0 and wee can choose r2 = 0.
Since 〈Ap, γ〉 = 〈p, γ〉 = 0, A1 = A2 6= A3, then p3γ3 = 0.
Let at first p3 = 0, γ3 6= 0, then 〈Aγ, γ〉 6= 0 and from (12) we get r1B23γ2C21B31 =
0 ⇒ γ2 = 0.
From the relation p1γ3−p3γ1 +2γ1 = 0 we get p1 = 0. Consequently,(〈p, p〉 = −4)p2 =
±2i.
From the characteristic system we get γ1/γ3 = p2/p2 = ±i.
The solutions (p, λγ′), γ′ = (±i, 0, 1) which we have within λ satisfy the condition
γ × p = 2γ.
The condition Ap×p+γ× r+Ap = 0 ⇔ γ× r+Ap = 0 is true if λ = 2A1(r3± ir1)
−1.
Now we see the case γ3 = 0 ⇒ λ(1,±i, 0). Since p3γ2− p2γ3 + 2γ1 = 0, then p3 = ±2i;
moreover from p2γ1 − p1γ2 + 2γ3 = 0 we obtain p1/p2 = γ1/γ2 = ∓i.
The condition p × γ = 2γ is equivalent (in this case) to the condition γ1 = λ, γ2 =
±λi, γ3 = 0, p1 = µ, p2 = ±µi, p3 = ±2i, and the condition Ap× p + γ × r + Ap = 0− is
equivalent to
−2µB23 ± λir3 + µA1 = 0
±2µiB31 − λr3 ± µA2i = 0 ⇔
∓λr1i± 2A3i = 0
⇔ λ = 2A3/r1, µC31 ± λir3 = 0.
2a1. We get the root (17).
2a21. Because C31 = 0 the other solutions are absent;
2a22. C31 = 0, r3 = 0, hence µ is an arbitrary constant and we get (18).
2b. The characteristic system in the case r1 = r2 = 0, r3 6= 0 has the form
B23p2p3 + r3γ2 + A1p1 = 0
B31p1p3 − r3γ1 + A2p2 = 0
A3p3 = 0
γ × p + 2γ = 0,
or equivalent
p1 = −A−1
1 γ2r3, p2 = −A−1
1 γ1r3, p3 = 0
−A−1
1 r3γ1γ3 = 2γ1 = 0
−A−1
1 r3γ2γ3 = 2γ2 = 0
A−1
1 r3γ
2
1 + A−1
1 r3γ
2
2 + 2γ3 = 0.
Then we obtain γ1 = γ2 = 0 ⇔ (p, γ) = 0, or
{
γ3 = 2A1r
−1
3
γ2
1 + γ2
2 = −4A2
1r
−2
3
2c. The characteristic system in the case r = 0 has the form
B23p2p3 + A1p1 = 0
B31p1p3 + A2p2 = 0
B12p1p2 + A3p3 = 0
γ × p + 2γ = 0.
Since B12 = 0, then p3 = 0, and p1 = p2 = 0, 2γ = −γ × p = 0.
The theorem is proved. 2
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[3] S.V.Kovalevskaya. The scientific works.- M.: Akad. Nayk, 1948.
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solutions in the dynamics of solids// Prikl. Math. i Mech.- 1978.- T. 42, No 3.-
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[5] S.L.Zigleen. The branching of the solutions and the absence of the first integrals
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Dokl. Akad. Nauk Ukr. SSR.- 1989.- T. 5.- P. 3-6.
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[8] A.V.Belyaev. On the motions of an n-dimentional rigid body with symmetry
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P.1038-1041.
A.V.Belyaev
Dept. of Social policy
Donetsk Institute of market
and social policy,
bulv. Pushkina,32
Donetsk 340050,
E-mail: smith01chat.ru
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