# New Neumann System Associated with a 3 × 3 Matrix Spectral Problem

Authors: Fang Li; Liping Lu
Journal: Advances in Mathematical Physics
 (2014)

---

## Abstract

The nonlinearization approach of Lax pair is applied to the case of the Neumann constraint associated with a 3 × 3 matrix spectral problem, from which a new Neumann system is deduced and proved to be completely integrable in the Liouville sense. As an application, solutions of the first nontrivial equation related to the 3 × 3 matrix spectral problem are decomposed into solving two compatible Hamiltonian systems of ordinary differential equations.


---

## Body

## 1. Introduction

Soliton equations are nonlinear partial differential equations described by infinite-dimensional integrable systems and have various beautiful algebraic and geometric properties [1–5]. It has been shown that the nonlinearization of spectral problems (NSPs) approach is a powerful tool to study soliton equations. According to this method which was first introduced by Cao [6], each (
1
+
1
)-dimensional soliton equation is decomposed into two ordinary differential equations: one is spatial and the other is temporal. The resulting decomposition not only inherits many integrable properties from soliton equations such as possessing Lax pairs, but also provides an effective way to derive explicit solutions of soliton equations.During the 1990s, the method of NSPs has attracted great interest in the soliton field and has been applied to a large number of soliton equations associated with2
×
2 matrix spectral problems [7–14]. Furthermore, based on the Cao’s nonlinearization technique, the work by Zhou and Qiao is the first time to develop the nonlinearization approach to find the algebrogeometric solutions for integrable systems in both continuous and discrete cases [15–20]. However, due to the complexity of higher-order matrix spectral problems, there is not much research on NSPs for soliton equations associated with higher-order matrix spectral problems [21–27]. In addition, because of the limitation of the Riemann theory, the algebrogeometric solutions of the soliton equations associated with 3
×
3 matrix spectral problems cannot be obtained with the aid of the nonlinearization technique.In this paper, we will study the following soliton equation with the help of the method of NSPs:(1)
u
t
=
u
x
x
-
u
x
v
+
2
w
x
,
v
t
=
2
u
x
,
w
t
=
-
w
x
x
-
(
v
w
)
x
.
Equation (1) is first proposed in [28] and associated with the 3
×
3 matrix spectral problem
(2)
Ψ
x
=
U
Ψ
,
U
=
(
0
1
0
u
-
λ
v
1
w
0
0
)
.In [28], Geng and Du have obtained some explicit solutions, which include soliton and periodic solutions. If w
=
0, (1) can be reduced to a couple of equations in u and v, which can be presented as
(3)
u
t
=
u
x
x
-
u
x
v
,
v
t
=
2
u
x
.
The corresponding Lax pair for the reduced system is as follows:
(4)
Φ
x
=
U
^
Φ
,
U
^
=
(
0
1
u
-
λ
v
)
;
Φ
t
=
V
^
Φ
,
V
^
=
(
u
-
λ
0
u
x
u
-
λ
)
.The aim of the present paper is to derive the corresponding finite-dimensional Hamiltonian system associated with the3
×
3 matrix spectral problem, which is proved to be completely integrable in the Liouville sense. As an application, solutions of (1) are decomposed into solving two compatible Hamiltonian systems of ordinary differential equations.The paper is organized as follows. In Section2, we will introduce the Neumann constraint between the potentials and eigenfunctions of the 3
×
3 matrix spectral problem (2). Under this constraint, we obtain a new Neumann system and a generating function of integrals of motion. In Section 3, the generating function approach is used to calculate the involutivity of integrals of motion, by which the Neumann system is further proved to be completely integrable in the Liouville sense. In Section 4, we will reveal the relation between the Neumann system and (1). Solutions of (1) are decomposed into solving two compatible Hamiltonian systems of ordinary differential equations.


## 2. A New Neumann System

In this section, we first consider the stationary zero-curvature equation of the spectral problem (2) and its auxiliary problem; that is,
(5)
V
x
=
[
U
,
V
]
,
V
=
(
V
i
j
)
3
×
3
,
where
(6)
V
11
=
(
v
-
∂
)
A
-
2
B
+
(
-
∂
2
+
v
∂
+
u
)
C
-
λ
C
,
V
12
=
A
,
V
13
=
C
,
V
21
=
(
-
∂
2
+
∂
v
+
u
)
A
-
2
∂
B
+
(
-
∂
3
+
∂
v
∂
+
∂
u
+
w
)
C
-
λ
A
-
λ
∂
C
,
V
22
=
2
v
A
-
2
B
+
(
-
∂
2
+
v
∂
+
u
)
C
-
λ
C
,
V
23
=
A
+
∂
C
,
V
31
=
(
∂
3
+
∂
2
v
+
v
∂
2
+
v
∂
v
+
w
)
A
-
2
(
∂
2
+
v
∂
)
B
-
(
∂
w
+
w
v
)
C
,
V
32
=
-
(
∂
2
+
∂
v
)
A
+
2
∂
B
+
w
C
,
V
33
=
(
∂
+
v
)
A
-
2
B
.
Substituting (6) into (5) yields the Lenard equation:
(7)
(
K
-
λ
J
)
G
=
0
,
G
=
(
A
,
B
,
C
)
T
,
where K and J are two skew-symmetric operators defined by(8)
K
=
(
-
2
∂
3
-
2
v
∂
v
+
∂
u
+
u
∂
4
v
∂
K
13
4
∂
v
-
6
∂
-
2
∂
3
+
2
∂
v
∂
+
2
∂
u
K
31
-
2
∂
3
-
2
∂
v
∂
+
2
u
∂
w
∂
2
-
∂
2
w
-
∂
v
w
-
v
w
∂
)
,
J
=
(
2
∂
0
∂
2
-
v
∂
0
0
2
∂
-
∂
2
-
∂
v
2
∂
0
)
,
K
13
=
-
∂
4
+
v
∂
3
+
∂
2
v
∂
-
v
∂
v
∂
+
∂
2
u
-
v
∂
u
+
2
∂
w
+
w
∂
,
K
31
=
∂
4
+
∂
3
v
+
∂
v
∂
2
+
∂
v
∂
v
-
u
∂
2
-
u
∂
v
+
2
w
∂
+
∂
w
,in which we denote ∂ by ∂
x for convenience. Expanding entries A, B, and C as the Laurent expansions in λ
(9)
A
=
∑
j
≥
0
‍
A
j
λ
-
j
,
B
=
∑
j
≥
0
‍
B
j
λ
-
j
,
C
=
∑
j
≥
0
‍
C
j
λ
-
j
,
then (7) leads to
(10)
K
g
j
-
1
=
J
g
j
,
j
≥
0
,
J
g
-
1
=
0
,
where g
j
-
1
=
(
A
j
,
B
j
,
C
j
)
T. We can choose the first two members as
(11)
g
-
1
=
(
0,0
,
1
)
T
,
g
0
=
(
w
,
0
,
u
)
T
.
In order to calculate the functional gradient of the eigenvalue with regard to the potentials, we introduce the spectral problem
(12)
(
Ψ
Φ
)
x
=
(
U
0
0
-
U
T
)
(
Ψ
Φ
)
,
where Ψ
=
(
Ψ
1
,
Ψ
2
,
Ψ
3
)
T, Φ
=
(
Φ
1
,
Φ
2
,
Φ
3
)
T. Let λ
1
,
…
,
λ
N be N distinct nonzero eigenvalues; then the systems associated with (12) can be written in the form
(13)
(
q
j
1
,
q
j
2
,
q
j
3
)
x
=
(
q
j
1
,
q
j
2
,
q
j
3
)
U
(
λ
j
)
T
,
(
p
j
1
,
p
j
2
,
p
j
3
)
x
=
-
(
p
j
1
,
p
j
2
,
p
j
3
)
U
(
λ
j
)
,
where q
j
i
=
Ψ
i
(
λ
j
), p
j
i
=
Φ
i
(
λ
j
), 1
≤
i
≤
3, 1
≤
j
≤
N, are eigenfunctions. A direct calculation gives rise to the functional gradient of the eigenvalue λ
j with regard to the potentials u, v, and w:
(14)
∇
λ
j
=
(
δ
λ
j
δ
u
δ
λ
j
δ
v
δ
λ
j
δ
w
)
=
(
q
j
1
p
j
2
q
j
2
p
j
2
q
j
1
p
j
3
)
.Now we consider the Neumann constraint(15)
∑
j
=
1
N
‍
∇
λ
j
=
g
0
,
which can be written as
(16)
u
=
〈
q
1
,
p
3
〉
,
w
=
〈
q
1
,
p
2
〉
,
f
1
=
〈
q
2
,
p
2
〉
=
0
,
where 〈
·
,
·
〉 is the standard inner-product in R
N, q
i
=
(
q
1
i
,
…
,
q
N
i
)
T, p
i
=
(
p
1
i
,
…
,
p
N
i
)
T. From (13) and the third expression of (16), it is easy to see that
(17)
f
2
=
〈
q
1
,
p
3
〉
〈
q
1
,
p
2
〉
-
〈
Λ
q
1
,
p
2
〉
+
〈
q
3
,
p
2
〉
-
〈
q
2
,
p
1
〉
=
0
,
v
=
〈
q
2
,
p
1
〉
-
1
(
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
-
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
-
〈
q
3
,
p
1
〉
)
,
where Λ
=
diag
⁡
(
λ
1
,
…
,
λ
N
). Substituting (16) and (17) into (13), we obtain the following new Neumann system on a (
6
N
-
2
)-dimensional manifold M:
(18)
q
x
1
=
q
2
,
q
x
2
=
〈
q
1
,
p
3
〉
q
1
-
Λ
q
1
+
〈
q
2
,
p
1
〉
-
1
×
(
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
-
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
-
〈
q
3
,
p
1
〉
)
q
2
+
q
3
,
q
x
3
=
〈
q
1
,
p
2
〉
q
1
,
p
x
1
=
-
〈
q
1
,
p
3
〉
p
2
+
Λ
p
2
-
〈
q
1
,
p
2
〉
p
3
,
p
x
2
=
-
p
1
-
〈
q
2
,
p
1
〉
-
1
(
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
-
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
-
〈
q
3
,
p
1
〉
)
p
2
,
p
x
3
=
-
p
2
with the manifold M being defined by
(19)
M
=
{
(
q
1
,
q
2
,
q
3
)
T
(
p
,
q
)
∈
R
6
N
∣
f
1
=
f
2
=
0
,
q
=
(
q
1
,
q
2
,
q
3
)
T
,
p
=
(
p
1
,
p
2
,
p
3
)
T
}
.
Consider the following function H:
(20)
H
=
〈
q
1
,
p
3
〉
〈
q
1
,
p
2
〉
-
〈
Λ
q
1
,
p
2
〉
+
〈
q
3
,
p
2
〉
+
〈
q
2
,
p
1
〉
.
Through a direct calculation we have
(21)
{
f
1
,
f
2
}
M
=
2
〈
q
2
,
p
1
〉
,
{
H
,
f
1
}
M
=
0
,
{
H
,
f
2
}
M
=
-
2
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
+
2
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
-
2
〈
Λ
q
1
,
p
1
〉
+
2
〈
Λ
q
2
,
p
2
〉
+
2
〈
q
3
,
p
1
〉
,
in which the poisson bracket of two functions is defined as
(22)
{
f
,
g
}
=
∑
j
=
1
N
‍
∑
i
=
1
3
‍
(
∂
f
∂
q
j
i
∂
g
∂
p
j
i
-
∂
g
∂
q
j
i
∂
f
∂
p
j
i
)
=
∑
i
=
1
3
(
〈
∂
f
∂
q
i
,
∂
g
∂
p
i
〉
-
〈
∂
g
∂
q
i
,
∂
f
∂
p
i
〉
)
.
Introduce a modified function H
^:
(23)
H
^
=
H
+
a
f
1
+
b
f
2
,
where a and b are two Lagrangian multipliers:
(24)
a
=
-
{
H
,
f
2
}
{
f
1
,
f
2
}
=
1
〈
q
2
,
p
1
〉
(
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
-
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
-
〈
q
3
,
p
1
〉
)
,
b
=
{
H
,
f
1
}
{
f
1
,
f
2
}
=
0
.
This means that H
^ is tangent to the manifold M. Therefore, (18) can be represented as the standard canonical equation on M:
(25)
(
q
p
)
x
=
(
∂
H
^
∂
p
-
∂
H
^
∂
q
)
.On the other hand, through tedious calculations we obtain(26)
(
K
-
λ
j
J
)
∇
^
λ
j
=
0
,
where
(27)
∇
^
λ
j
=
(
q
j
1
p
j
2
,
1
2
(
q
j
2
p
j
2
-
q
j
1
p
j
1
-
q
j
3
p
j
3
)
,
q
j
1
p
j
3
)
T
.
Then the solution of the Lenard equation K
G
=
λ
J
G with parameter λ can be written as
(28)
G
λ
=
g
-
1
+
∑
j
=
1
N
‍
∇
^
λ
j
λ
-
λ
j
=
(
Q
λ
12
1
2
(
Q
λ
22
-
Q
λ
11
-
Q
λ
33
)
Q
λ
13
+
1
)
,
where Q
λ
i
j
=
∑
k
=
1
N
(
q
k
i
p
k
j
/
(
λ
-
λ
k
)
), and G
λ satisfies (
K
-
λ
J
)
G
λ
=
0 under the Neumann constraint (16).


## 3. The Liouville Integrability

Now we introduce a Lax matrix by(29)
V
λ
=
(
Q
λ
11
-
λ
Q
λ
12
Q
λ
13
+
1
Q
λ
21
Q
λ
22
-
λ
Q
λ
23
Q
λ
31
+
〈
q
1
,
p
1
〉
Q
λ
32
+
〈
q
1
,
p
2
〉
Q
λ
33
)
.
Through a direct calculation we can prove that V
λ and ζ
I
-
V
λ are two solutions of V
x
=
[
U
,
V
], where I is a 3
×
3 unit matrix and ζ is a parameter. Then F
λ
(
2
)
=
det
⁡
V
λ and F
ζ
λ
=
det
⁡
(
ζ
I
-
V
λ
) are independent of x. It is easy to see that
(30)
F
ζ
λ
=
ζ
3
-
F
λ
(
0
)
ζ
2
+
F
λ
(
1
)
ζ
-
F
λ
(
2
)
,
where
(31)
F
λ
(
0
)
=
Q
λ
11
+
Q
λ
22
+
Q
λ
33
-
2
λ
,
F
λ
(
1
)
=
∑
1
≤
i
<
j
≤
3
‍
|
V
λ
i
i
V
λ
i
j
V
λ
j
i
V
λ
j
j
|
.
In order to generate the Hamiltonians, we take the following notations
(32)
F
^
λ
(
0
)
=
F
λ
(
0
)
+
2
λ
=
Q
λ
11
+
Q
λ
22
+
Q
λ
33
,
F
^
λ
(
1
)
=
F
λ
(
1
)
+
λ
F
λ
(
0
)
+
λ
2
=
∑
1
≤
i
<
j
≤
3
‍
|
Q
λ
i
i
Q
λ
i
j
Q
λ
j
i
Q
λ
j
j
|
-
〈
q
1
,
p
1
〉
(
Q
λ
13
+
1
)
-
〈
q
1
,
p
2
〉
Q
λ
23
-
Q
λ
31
-
λ
Q
λ
33
,
F
^
λ
(
2
)
=
F
λ
(
2
)
+
λ
F
λ
(
1
)
+
λ
2
F
λ
(
0
)
+
λ
3
=
|
Q
λ
11
Q
λ
12
Q
λ
13
Q
λ
21
Q
λ
22
Q
λ
23
Q
λ
31
Q
λ
32
Q
λ
33
|
+
〈
q
1
,
p
1
〉
|
Q
λ
12
Q
λ
13
Q
λ
22
Q
λ
23
|
-
〈
q
1
,
p
2
〉
|
Q
λ
11
Q
λ
13
Q
λ
21
Q
λ
23
|
+
λ
|
Q
λ
11
Q
λ
12
Q
λ
21
Q
λ
22
|
+
|
Q
λ
21
Q
λ
22
Q
λ
31
Q
λ
32
|
-
〈
q
1
,
p
1
〉
Q
λ
22
+
〈
q
1
,
p
2
〉
Q
λ
21
.
Substituting the Laurent expansion of Q
λ
i
k into (32) we have
(33)
F
^
λ
(
0
)
=
∑
m
≥
0
‍
F
m
(
0
)
λ
-
m
-
1
,
F
^
λ
(
1
)
=
∑
m
≥
0
‍
F
m
(
1
)
λ
-
m
,
F
^
λ
(
2
)
=
∑
m
≥
0
‍
F
m
(
2
)
λ
-
m
-
1
,
where
(34)
F
m
(
0
)
=
〈
Λ
m
q
1
,
p
1
〉
+
〈
Λ
m
q
2
,
p
2
〉
+
〈
Λ
m
q
3
,
p
3
〉
,
m
≥
0
,
F
0
(
1
)
=
-
〈
q
1
,
p
1
〉
-
〈
q
3
,
p
3
〉
,
F
m
(
1
)
=
∑
1
≤
i
<
j
≤
3
∑
l
+
k
=
m
-
2
l
,
k
≥
0
|
〈
Λ
l
q
i
,
p
i
〉
〈
Λ
k
q
i
,
p
j
〉
〈
Λ
l
q
j
,
p
i
〉
〈
Λ
k
q
j
,
p
j
〉
|
-
〈
q
1
,
p
1
〉
〈
Λ
m
-
1
q
1
,
p
3
〉
-
〈
q
1
,
p
2
〉
〈
Λ
m
-
1
q
2
,
p
3
〉
-
〈
Λ
m
-
1
q
3
,
p
1
〉
-
〈
Λ
m
q
3
,
p
3
〉
,
m
≥
1
,
F
0
(
2
)
=
-
〈
q
1
,
p
1
〉
〈
q
2
,
p
2
〉
+
〈
q
1
,
p
2
〉
〈
q
2
,
p
1
〉
+
〈
q
1
,
p
1
〉
〈
q
2
,
p
2
〉
-
〈
q
1
,
p
2
〉
〈
q
2
,
p
1
〉
=
0
,
F
m
(
2
)
=
∑
l
+
k
+
s
=
m
-
2
l
,
k
,
s
≥
0
|
〈
Λ
l
q
1
,
p
1
〉
〈
Λ
k
q
1
,
p
2
〉
〈
Λ
s
q
1
,
p
3
〉
〈
Λ
l
q
2
,
p
1
〉
〈
Λ
k
q
2
,
p
2
〉
〈
Λ
s
q
2
,
p
3
〉
〈
Λ
l
q
3
,
p
1
〉
〈
Λ
k
q
3
,
p
2
〉
〈
Λ
s
q
3
,
p
3
〉
|
+
〈
q
1
,
p
1
〉
∑
l
+
k
=
m
-
1
l
,
k
≥
0
|
〈
Λ
l
q
1
,
p
2
〉
〈
Λ
k
q
1
,
p
3
〉
〈
Λ
l
q
2
,
p
2
〉
〈
Λ
k
q
2
,
p
3
〉
|
-
〈
q
1
,
p
2
〉
∑
l
+
k
=
m
-
1
l
,
k
≥
0
|
〈
Λ
l
q
1
,
p
1
〉
〈
Λ
k
q
1
,
p
3
〉
〈
Λ
l
q
2
,
p
1
〉
〈
Λ
k
q
2
,
p
3
〉
|
+
∑
l
+
k
=
m
-
1
l
,
k
≥
0
|
〈
Λ
l
q
2
,
p
1
〉
〈
Λ
k
q
2
,
p
2
〉
〈
Λ
l
q
3
,
p
1
〉
〈
Λ
k
q
3
,
p
2
〉
|
+
∑
l
+
k
=
m
l
,
k
≥
0
|
〈
Λ
l
q
1
,
p
1
〉
〈
Λ
k
q
1
,
p
2
〉
〈
Λ
l
q
2
,
p
1
〉
〈
Λ
k
q
2
,
p
2
〉
|
-
〈
q
1
,
p
1
〉
〈
Λ
m
q
2
,
p
2
〉
+
〈
q
1
,
p
2
〉
〈
Λ
m
q
2
,
p
1
〉
,
m
≥
1
.
We can prove the following assertion.Proposition 1.
Suppose that(
p
,
q
)
∈
R
6
N, then
(35)
{
F
ξ
μ
,
F
ζ
λ
}
=
0
,
∀
λ
,
ζ
,
μ
,
ξ
∈
C
;
(36)
{
F
m
(
i
)
,
F
k
(
j
)
}
=
0
,
∀
m
,
k
≥
0
,
0
≤
i
,
j
≤
2
.Proof.
Through tedious calculation we can obtain(37)
{
F
^
λ
(
i
)
,
F
^
μ
(
j
)
}
=
0
,
∀
λ
,
μ
∈
C
,
0
≤
i
,
j
≤
2
;
then we have
(38)
{
F
λ
(
i
)
,
F
μ
(
j
)
}
=
0
,
∀
λ
,
μ
∈
C
,
0
≤
i
,
j
≤
2
.
Substituting (30) and (38) into the left hand side of (35) shows that (35) holds. Moreover, relation (36) follows by comparison of power of λ
N in (33) with (37) taken into account.In order to guarantee that the Hamiltonians are tangent to the constrained manifoldM, we calculate that
(39)
{
F
j
(
i
)
,
f
1
}
=
0
,
i
=
0,1
,
2
,
{
F
j
(
0
)
,
f
2
}
=
0
,
j
≥
0
,
{
F
0
(
1
)
,
f
2
}
=
0
,
{
F
m
(
1
)
,
f
2
}
M
=
2
〈
q
2
,
p
1
〉
〈
Λ
m
-
1
q
1
,
p
3
〉
,
{
F
m
(
2
)
,
f
2
}
M
=
-
2
〈
q
2
,
p
1
〉
∑
l
+
k
=
m
-
1
l
,
k
≥
0
|
〈
Λ
l
q
1
,
p
2
〉
〈
Λ
k
q
1
,
p
3
〉
〈
Λ
l
q
2
,
p
2
〉
〈
Λ
k
q
2
,
p
3
〉
|
+
2
〈
q
2
,
p
1
〉
〈
Λ
m
q
2
,
p
2
〉
,
m
≥
1
.
The Lagrangian multipliers are given by
(40)
μ
m
(
1
)
=
{
F
m
(
1
)
,
f
2
}
{
f
1
,
f
2
}
=
〈
Λ
m
-
1
q
1
,
p
3
〉
,
μ
m
(
2
)
=
{
F
m
(
2
)
,
f
2
}
{
f
1
,
f
2
}
=
〈
Λ
m
q
2
,
p
2
〉
-
∑
l
+
k
=
m
-
1
l
,
k
≥
0
|
〈
Λ
l
q
1
,
p
2
〉
〈
Λ
k
q
1
,
p
3
〉
〈
Λ
l
q
2
,
p
2
〉
〈
Λ
k
q
2
,
p
3
〉
|
,
m
≥
1
.
Thus the modified functions
(41)
F
^
j
(
0
)
=
F
j
(
0
)
,
F
^
0
(
1
)
=
F
0
(
1
)
,
F
^
m
(
i
)
=
F
m
(
i
)
-
μ
m
(
i
)
f
1
,
i
=
1,2
,
j
≥
0
,
m
≥
1
,
are tangent to the manifold M and are in involution in pairs on M; that is
(42)
{
F
^
m
(
i
)
,
F
^
l
(
j
)
}
M
=
0
,
0
≤
i
,
j
≤
2
,
m
,
l
≥
0
.Proposition 2.
The3
N 1-forms d
F
m
(
0
), d
F
k
(
1
), d
F
l
(
2
) (0
≤
m
≤
N
-
1
,
0
≤
k
≤
N
,
1
≤
l
≤
N
-
1) are linearly independent.Proof.
Assume that there exist3
N constants b
s
(
i
), so that
(43)
∑
m
=
0
N
-
1
‍
b
m
(
0
)
∂
F
m
(
0
)
∂
p
j
+
∑
k
=
0
N
‍
b
k
(
1
)
∂
F
k
(
1
)
∂
p
j
+
∑
l
=
1
N
-
1
‍
b
l
(
2
)
∂
F
l
(
2
)
∂
p
j
=
0
,
1
≤
j
≤
3
.
It is easy to see that
(44)
∂
F
m
(
0
)
∂
p
1
=
Λ
m
q
1
,
∂
F
0
(
1
)
∂
p
1
=
-
q
1
,
∂
F
k
(
1
)
∂
p
1
|
(
p
1
,
p
2
,
p
3
)
=
0
=
-
Λ
k
-
1
q
3
,
k
≥
1
,
∂
F
l
(
2
)
∂
p
1
|
(
p
1
,
p
2
,
p
3
)
=
0
=
0
.
Then we have
(45)
∑
m
=
0
N
-
1
‍
b
m
(
0
)
Λ
m
q
1
-
b
0
(
1
)
q
1
-
∑
k
=
1
N
‍
b
k
(
1
)
Λ
k
-
1
q
3
=
0
,
which gives rise to
(46)
b
0
(
0
)
-
b
0
(
1
)
+
∑
m
=
1
N
-
1
‍
b
m
(
0
)
(
λ
j
)
m
=
0
,
∑
k
=
1
N
‍
b
k
(
1
)
(
λ
j
)
k
-
1
=
0
,
1
≤
j
≤
N
,
by substituting q
3
=
0 and q
1
=
0 into (45), respectively. Therefore, we have b
0
(
0
)
=
b
0
(
1
), b
m
(
0
)
=
b
k
(
1
)
=
0, 1
≤
m
≤
N
-
1, 1
≤
k
≤
N, by utilizing that Vandermonde determinant is not zero. Then (43) is transformed to
(47)
b
0
(
0
)
(
∂
F
0
(
0
)
∂
p
j
+
∂
F
0
(
1
)
∂
p
j
)
+
∑
l
=
1
N
-
1
‍
b
l
(
2
)
∂
F
l
(
2
)
∂
p
j
=
0
,
1
≤
j
≤
3
.
According to (34), we obtain
(48)
∂
F
0
(
0
)
∂
p
2
=
q
2
,
∂
F
0
(
1
)
∂
p
2
=
0
,
∂
F
l
(
2
)
∂
p
2
|
q
1
=
0
=
∑
m
+
k
=
l
-
1
m
,
k
≥
0
(
〈
Λ
k
q
2
,
p
1
〉
Λ
m
q
3
-
〈
Λ
k
q
3
,
p
1
〉
Λ
m
q
2
)
.
Let(49)
q
j
3
=
(
p
j
1
)
-
1
∏
k
=
1
,
k
≠
j
N
‍
(
λ
j
-
λ
k
)
-
1
,
r
k
=
∑
j
=
1
N
‍
(
λ
j
)
k
q
j
2
p
j
1
=
〈
Λ
k
q
2
,
p
1
〉
,
and noticing that
(50)
∑
j
=
1
N
‍
(
μ
j
)
k
∏
k
=
1
,
k
≠
j
N
‍
(
μ
j
-
μ
k
)
-
1
=
{
0
,
0
≤
k
≤
N
-
2
,
∑
m
1
+
⋯
+
m
N
=
k
-
N
+
1
m
1
,
…
,
m
N
≥
0
(
μ
1
)
m
1
⋯
(
μ
N
)
m
N
,
k
≥
N
-
1
,
then we have
(51)
〈
Λ
k
q
3
,
p
1
〉
=
∑
j
=
1
N
‍
(
λ
j
)
k
q
j
3
p
j
1
=
{
0
,
0
≤
k
≤
N
-
2
,
1
,
k
=
N
-
1
.
Thus (47) yields(52)
L
b
=
0
,
b
=
(
b
1
(
2
)
,
…
,
b
N
-
1
(
2
)
,
b
0
(
0
)
)
T
,
where(53)
L
=
(
r
0
q
1
3
(
r
0
λ
1
+
r
1
)
q
1
3
⋯
(
∑
k
=
0
N
-
2
‍
r
k
(
λ
1
)
N
-
2
-
k
)
q
1
3
q
1
2
r
0
q
2
3
(
r
0
λ
2
+
r
1
)
q
2
3
⋯
(
∑
k
=
0
N
-
2
‍
r
k
(
λ
2
)
N
-
2
-
k
)
q
2
3
q
2
2
⋮
⋮
⋱
⋮
⋮
r
0
q
N
3
(
r
0
λ
N
+
r
1
)
q
N
3
⋯
(
∑
k
=
0
N
-
2
‍
r
k
(
λ
N
)
N
-
2
-
k
)
q
N
3
q
N
2
)
.
To arrive atb
=
0, one needs only to prove det L
≠
0. Therefore we introduce a matrix T:
(54)
T
=
(
p
1
1
p
2
1
⋯
p
N
1
λ
1
p
1
1
λ
2
p
2
1
⋯
λ
N
p
N
1
⋮
⋮
⋱
⋮
(
λ
1
)
N
-
1
p
1
1
(
λ
2
)
N
-
1
p
2
1
⋯
(
λ
N
)
N
-
1
p
N
1
)
.
It is easy to verify that detT
=
∏
j
=
1
N
‍
p
j
1
∏
i
<
j
‍
(
λ
i
-
λ
j
)
≠
0, and
(55)
T
L
=
(
0
0
⋯
0
r
0
0
0
⋯
r
0
*
⋮
⋮
⋱
⋮
⋮
0
r
0
⋯
*
*
r
0
*
⋯
*
*
)
,
det
⁡
(
T
J
)
=
(
r
0
)
N
≠
0
,
where * represents the entries which may not be zero, and this proves Proposition 2.Resorting to the two propositions above and noticing thatf
1
=
F
0
(
0
)
+
F
0
(
1
), we obtain the following assertion.Proposition 3.
The Neumann system defined by (18) is completely integrable in the Liouville sense on M.


## 4. The Representation of Solutions

In this section, we will give the representation of solutions for (1). To this end, we denote the variable of H
^
1-flow by t, where
(56)
H
^
1
=
H
1
+
a
1
f
1
,
H
1
=
-
F
1
(
1
)
-
F
1
(
0
)
=
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
q
2
,
p
3
〉
〈
q
1
,
p
2
〉
-
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
+
〈
q
3
,
p
1
〉
,
and a
1 is the Lagrangian multiplier
(57)
a
1
=
-
{
H
1
,
f
2
}
{
f
1
,
f
2
}
=
〈
q
1
,
p
3
〉
.
Then the canonical equation of the H
^
1-flow on M is
(58)
q
t
1
=
∂
H
^
1
∂
p
1
=
〈
q
1
,
p
3
〉
q
1
-
Λ
q
1
+
q
3
,
q
t
2
=
∂
H
^
1
∂
p
2
=
〈
q
2
,
p
3
〉
q
1
+
〈
q
1
,
p
3
〉
q
2
-
Λ
q
2
,
q
t
3
=
∂
H
^
1
∂
p
3
=
〈
q
1
,
p
1
〉
q
1
+
〈
q
1
,
p
2
〉
q
2
,
p
t
1
=
-
∂
H
^
1
∂
q
1
=
Λ
p
1
-
〈
q
1
,
p
3
〉
p
1
-
〈
q
2
,
p
3
〉
p
2
-
〈
q
1
,
p
1
〉
p
3
,
p
t
2
=
-
∂
H
^
1
∂
q
2
=
Λ
p
2
-
〈
q
1
,
p
3
〉
p
2
-
〈
q
1
,
p
2
〉
p
3
,
p
t
3
=
-
∂
H
^
1
∂
q
3
=
-
p
1
.
Using (58) and the Neumann constraints (16) and (17), we derive that
(59)
d
d
t
(
u
v
w
)
=
(
〈
q
1
,
p
3
〉
2
-
〈
Λ
q
1
,
p
3
〉
+
〈
q
3
,
p
3
〉
-
〈
q
1
,
p
1
〉
-
2
〈
q
1
,
p
2
〉
+
2
〈
q
2
,
p
3
〉
〈
q
3
,
p
2
〉
-
〈
q
1
,
p
2
〉
〈
q
1
,
p
3
〉
)
.
On the other hand, combining (18), (16), and (17) we have
(60)
J
g
0
=
(
2
∂
w
+
(
∂
2
-
v
∂
)
u
2
∂
u
-
(
∂
2
+
∂
v
)
w
)
=
(
〈
q
1
,
p
3
〉
2
-
〈
Λ
q
1
,
p
3
〉
+
〈
q
3
,
p
3
〉
-
〈
q
1
,
p
1
〉
-
2
〈
q
1
,
p
2
〉
+
2
〈
q
2
,
p
3
〉
〈
q
3
,
p
2
〉
-
〈
q
1
,
p
2
〉
〈
q
1
,
p
3
〉
)
.
Then we arrive at
(61)
d
d
t
(
u
v
w
)
=
J
(
〈
q
1
,
p
2
〉
0
〈
q
1
,
p
3
〉
)
=
J
g
0
,
which is (1). Therefore we obtain the following result.Proposition 4.
Let(
q
i
(
x
,
t
), p
i
(
x
,
t
)
), 1
≤
i
≤
3 be a compatible solution of the Hamiltonian systems (18) and (58); then the functions
(62)
u
=
〈
q
1
,
p
3
〉
,
w
=
〈
q
1
,
p
2
〉
,
v
=
〈
q
2
,
p
1
〉
-
1
(
〈
q
1
,
p
2
〉
〈
q
2
,
p
3
〉
-
〈
q
1
,
p
3
〉
〈
q
1
,
p
1
〉
+
〈
Λ
q
1
,
p
1
〉
-
〈
Λ
q
2
,
p
2
〉
-
〈
q
3
,
p
1
〉
)
solve (1).Proof.
We only need to prove that the Hamiltonian systems (18) and (58) are compatible. In fact, it is not difficult to verify that {
H
^
,
H
^
1
}
=
0. Hence the proof is completed [29].



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*Source: 708603-2014-07-24.xml*