Rational solutions of the Sasano system of type $D_5^{(1)}$

0.73 MB·English

by  Kazuhide Matsuda

(1)
Rational solutions of the Sasano system of type D
5
By Kazuhide Matsuda
2 Department of Engineering Science, Niihama National College of Technology,
1
7-1 Yagumo-chou, Niihama, Ehime, 792-8580.
0
2

r Abstract
a
M
(1)
In this paper, we classified the rational solutions of the Sasano system of type D , which
5
3 is given by the coupled P system with the affine Weyl group symmetry of type D(1). The
V 5
1
rational solutions are classified to four types by the B¨acklund transformation group.

]
A
C Introduction
.
h
t In this paper, we classified the rational solutions of the Sasano system of type D(1), which
a 5
m is defined by
[
tx′ = 2x2y +tx2 −2xy −{t+(2α +2α +α +α )}x
2 3 5 4
3
v  +(α2 +α5)+2z{(z −1)w+α3},
7 ty′ = −2xy2 +y2 −2txy +{t+(2α +2α +α +α )}y −α t,
9 D(1)(α )  2 3 5 4 1
26 5 j 0≤j≤5tz′ = 2z2w +tz2 −2zw −{t+(α5 +α4)}z +α5 +2yz(z −1),
. tw′ = −2zw2 +w2 −2tzw +{t+(α +α )}w −α t−2y(−w+2zw +α ),
7 5 4 3 3
0 α +α +2α +2α +α +α = 1,
0  0 1 2 3 4 5

1 
v: where ′ = d/dt and αi(0 ≤ i ≤ 5) are all arbitrary complex parameters. D5(1)(αj)0≤j≤5 is
expressed by the coupled P system:
i V
X
dx ∂H dy ∂H dz ∂H dw ∂H
r t = , t = − , t = , t = − ,
a
dt ∂y dt ∂x dt ∂w dt ∂z
where the Hamiltonian H is given by
H = H (x,y,t;α +α ,α ,α +2α +α )+H (z,w,t;α ,α ,α )+2yz{(z −1)w+α }
V 2 5 1 2 3 4 V 5 3 4 3
= x(x−1)y(y +t)−(2α +2α +α +α )xy +(α +α )y +α tx
2 3 5 4 2 5 1
+z(z −1)w(w+t)−(α +α )zw +α w +α tz
5 4 5 3
+2yz{(z −1)w +α },
3
H (q,p,t;γ ,γ ,γ ) = q(q−1)p(p+t)−(γ +γ )qp+γ p+γ tq.
V 1 2 3 1 3 1 2
(1)
D (α ) possess the B¨acklund transformations, s ,π (0 ≤ i ≤ 5, 1 ≤ j ≤ 4),
5 j 0≤j≤5 i j
which are defined as follows:
1
( ) s ( ) s ( ) s ( ) s ( ) s ( ) s ( ) π ( ) π ( ) π ( ) π ( )
0 1 2 3 4 5 1 2 3 4
x x+ α0 x+ α1 x x x x 1−x y+w+t 1−x x
(y+t) y t
y y y y− α2 y y y −y−t −t(z−1) −y y+t
(x−z)
z z z z z+ α3 z z 1−z y+t 1−z z
w t
w w w w+ α2 w w− α4 w− α5 −w −t(x−z) −w w
(x−z) (z−1) z
t t t t t t t t −t −t −t
α −α α α +α α α α α α α α
0 0 0 0 2 0 0 0 1 5 0 1
α α −α α +α α α α α α α α
1 1 1 1 2 1 1 1 0 4 1 0
α α +α α +α −α α +α α α α α α α
2 2 0 2 1 2 2 3 2 2 2 3 2 2
α α α α +α −α α +α α +α α α α α
3 3 3 3 2 3 3 4 3 5 3 2 3 3
α α α α α +α −α α α α α α
4 4 4 4 4 3 4 4 5 1 5 4
α α α α α +α α −α α α α α
5 5 5 5 5 3 5 5 4 0 4 5
The B¨acklund transformation group hs ,π | 0 ≤ i ≤ 5, 1 ≤ j ≤ 4i is isomorphic to the
i j
(1)
affine Weyl group of type D .
5
In this paper, we define the coefficients of the Laurent series of x,y,z,w at t =
∞, 0, c ∈ C∗ by
a , a , a k ∈ Z, (for x), b , b , b k ∈ Z, (for y)
∞,k 0,k c,k ∞,k 0,k c,k
(c∞,k, c0,k, cc,k k ∈ Z, (for z), d∞,k, d0,k, dc,k k ∈ Z, (for w),
respectively. If b + d 6= −1/2, we say that (x,y,z,w) is a solution of type A. If
∞,1 ∞,1
b +d = −1/2, we say that (x,y,z,w) is a solution of type B. Moreover, we denote
∞,1 ∞,1
the coefficients of the Laurent series of H at t = ∞,0,c ∈ C∗ by
h , h , h , (k ∈ Z),
∞,k 0,k c,k
respectively.
Our main theorem is as follows:
Theorem 0.1. Suppose that for D(1)(α ) , there exists a rational solution. By some
5 j 0≤j≤5
B¨acklund transformations, the parameters and solution can then be transformed so that
one of the following occurs:
(a-1) (α ,α ,α ,α ,α ,α ) = (α ,0,0,0,α ,0) and (x,y,z,w) = (0,0,0,0),
0 1 2 3 4 5 0 4
(a-2) (α ,α ,α ,α ,α ,α ) = (0,0,0,0,1,0) and
0 1 2 3 4 5
(x,y,z,w) = (0,0,0,0),(0,−t,0,0),(0,0,0,−t),(0,−t,0,t),
(b-1) −α +α = −α +α = 0 and (x,y,z,w) = (1/2,−t/2,1/2,0),
0 1 4 5
(b-2) (α ,α ,α ,α ,α ,α ) = (1/2,1/2,0,α ,−α ,−α ) and
0 1 2 3 4 5 3 3 3
(x,y,z,w) = (1/2,−t/2+b,1/2,d),
2
where b, d are both arbitrary complex numbers and satisfy b+d = 0.
This paper is organized as follows. In Sections 1, 2 and 3, we treat the meromorphic
solutions at t = ∞,0,c ∈ C∗ of D(1)(α ) .
5 j 0≤j≤5
We first find that for a meromorphic solution of type A at t = ∞, a (= −Res x)
∞,−1 t=∞
is given by the parameters, and for a meromorphic solution of type B at t = ∞, b +
∞,0
d = 0,−α +α . We next see that for a meromorphic solution at t = 0, b +d =
∞,0 4 5 0,0 0,0
0,−α + α . We last observe that for a meromorphic solution at t = c ∈ C∗, a =
4 5 c,−1
Res x ∈ Z, and b +d = Res (y +w) = nc (n ∈ Z). Therefore, it follows that
t=c c,−1 c,−1 t=c
a −a ∈ Z and (b +d )−(b +d ) ∈ Z.
∞,−1 0,−1 ∞,0 ∞,0 0,0 0,0
In Section 4, for a meromorphic solution at t = ∞,0,c ∈ C∗, we compute h ,h ,
∞,0 0,0
and h = Res H. We then see that h ,h are both expressed by the parameters
c,−1 t=c ∞,0 0,0
and h = nc (n ∈ Z ). Therefore, it follows that
c,−1 ≥0
h −h ∈ Z .
∞,0 0,0 ≥0
In Section 5, we investigate the properties of the B¨acklund transformations and show
the existence of the “infinite solution,” which is given by y ≡ w ≡ ∞ or z ≡ ∞. In Section
6, we determine the infinite solution.
(1)
In Sections 7 and 8, we obtain necessary conditions for D (α ) to have rational
5 j 0≤j≤5
solutions of type A and B. For this purpose, we use
a −a ∈ Z, (b +d )−(b +d ) ∈ Z and h −h ∈ Z ,
∞,−1 0,−1 ∞,0 ∞,0 0,0 0,0 ∞,0 0,0 ≥0
and express the necessary conditions by the parameters.
In Section 9, using the B¨acklund transformations, we transform the parameters so
that one of the following occurs:
(1) α = α = α = α = 0, (2) −α +α = −α +α = 0, (3) −α +α = 0,−α +α = 1.
1 2 3 5 0 1 4 5 0 1 4 5
In this paper, we call cases (2) and (3) the standard forms I and II.
In Sections 10, 11, we treat case (1), that is, determine the rational solutions of type
(1)
A of D (α ,0,0,0,α ,0). In Section 12, we obtain the main theorem for type A.
5 0 4
In Sections 13, 14, 15, 16 and 17, we determine the rational solutions of type B for the
standard form I. In Sections 18, 19, 20, 21 and 22, we determine the rational solutions of
type B for the standard form II. In Section 23, we obtain the main theorem for type B.
Acknowledgments
The author wishes to express his sincere thanks to Professor Yousuke Ohyama.
3
1 Meromorphic solutions at t = ∞
(1)
In this section, for D (α ) , we determine the Laurent series of (x,y,z,w) at t = ∞.
5 j 0≤j≤5
For this purpose, we consider the following five cases:
(0) all of (x,y,z,w) are holomorphic at t = ∞,
(1) one of (x,y,z,w) has a pole at t = ∞,
(2) two of (x,y,z,w) have a pole at t = ∞,
(3) three of (x,y,z,w) have a pole at t = ∞,
(4) all of (x,y,z,w) have a pole at t = ∞.
Moreover, we set
x = a tn0 +a tn0−1 +···+a t−1 +··· ,
∞,n0 ∞,n0−1 ∞,−1
y = b tn1 +b tn1−1 +···+b +··· ,
 ∞,n1 ∞,n1−1 ∞,0
z = c∞,n2tn2 +c∞,n2−1tn2−1 +···+c∞,−1t−1 +··· ,

w = d tn3 +d tn3−1 +···+d +··· ,
∞,n2 ∞,n3−1 ∞,0



where n ,n ,n ,n are all integers.
0 1 2 3
1.1 The case where all of (x,y,z,w) are holomorphic at t = ∞
In this subsection, we treat the case where all of (x,y,z,w) are holomorphic at t = ∞.
Proposition 1.1. Suppose that for D(1)(α ) , there exists a solution such that x, y,
5 j 0≤j≤5
z, w are all holomorphic at t = ∞. Then, a = 0,1 and c = 0,1.
∞,0 ∞,0
(1) If (a ,c ) = (0,0),
∞,0 ∞,0
x = (α +α )t−1 +··· ,
2 5
y = α +α (−α −2α −α +α )t−1 +··· ,
 1 1 1 3 4 5
z = α5t−1 +··· ,

w = α +α (−α −α +α )t−1 +··· .
3 3 3 4 5



(2) If (a ,c ) =(0,1),
∞,0 ∞,0
x = (α +2α +α )t−1 +··· ,
2 3 5
y = α +α (−α +2α −α +α )t−1 +··· ,
 1 1 1 3 4 5
z = 1+α4t−1 +··· ,

w = −α −α (α −α +α )t−1 +··· .
3 3 3 4 5




4
(3) If (a ,c ) = (1,0),
∞,0 ∞,0
x = 1+(α +2α +α )t−1 +··· ,
2 3 4
y = −α −α (α −2α −α +α )t−1 +··· ,
 1 1 1 3 4 5
z = α5t−1··· ,

w = α +α (−α −α +α )t−1 +··· .
3 3 3 4 5



(4) If (a ,c ) =(1,1),
∞,0 ∞,0
x = 1+(α +α )t−1 +··· ,
2 4
y = −α −α (α +2α −α +α )t−1 +··· ,
 1 1 1 3 4 5
z = 1+α4t−1 +··· ,

w = −α −α (α −α +α )t−1 +··· .
3 3 3 4 5



Proof. It can be proved by direct calculation.
Proposition 1.2. Suppose that for D(1)(α ) , there exists a solution such that
5 j 0≤j≤5
x,y,z,w are holomorphic at t = ∞. Moreover, assume that a and c are both deter-
∞,0 ∞,0
mined. The solution is then unique.
Proof. By Proposition 1.1, we can set
x = a +a t−1 +···+a t−k +a t−(k+1) +··· ,
∞,0 ∞,−1 ∞,−k ∞,−(k+1)
y = b +b t−1 +···+b t−k +b t−(k+1) +··· ,
 ∞,0 ∞,−1 ∞,−k ∞,−(k+1)
z = c∞,0 +c∞,−1t−1 +···+c∞,−kt−k +c∞,−(k+1)t−(k+1) +··· ,

w = d +d t−1 +···+d t−k +d t−(k+1) +··· ,
∞,0 ∞,−1 ∞,−k ∞,−(k+1)



where a , a , b , b , c , c , d and d have been already determined.
∞,0 ∞,−1 ∞,0 ∞,−1 ∞,0 ∞,−1 ∞,0 ∞,−1
We treat the case where (a ,c ) = (0,0). The other cases can be proved in the same
∞,0 ∞,0
way.
Comparing the coefficients of the term t−k (k ≥ 1) in
tx′ = 2x2y+tx2−2xy−{t+(2α +2α +α +α )}x+(α +α )+2z{(z −1)w +α },
2 3 5 4 2 5 3
we have
a =ka +2 a a b + a a
∞,−(k+1) ∞,−k ∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
−2 a Xb −(2α +2α +Xα +α )a
∞,−l ∞,−m 2 3 5 4 ∞,−k
+2Xc c d −2 c d , (1.1)
∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
X 5 X
where the first and fourth sums extend over the non-negative integers, l,m,n, such that
l+m+n = k, and the second sum extends over the non-negative integers, l,m, such that
l+m = k+1, and the third and fifth sums extend over the non-negative integers, l,m,n,
such that l+m = k.
Comparing the coefficients of the term t−k (k ≥ 1) in
ty′ = −2xy2 +y2 −2txy +{t+(2α +2α +α +α )}y −α t,
2 3 5 4 1
we obtain
b =−kb +2 a b b − b b
∞,−(k+1) ∞,−k ∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
+2 a bX −(2α +2α +αX+α )b , (1.2)
∞,−l ∞,−m 2 3 5 4 ∞,−k
X
where the first sum extends over thenon-negative integers, l,m,n, such that l+m+n = k,
and the second sum extends over the non-negative integers, l,m, such that l + m = k,
and the third sum extends over the non-negative integers, l,m, such that l+m = k +1.
Comparing the coefficients of the term t−k (k ≥ 1) in
tz′ = 2z2w +tz2 −2zw −{t+(α +α )}z +α +2yz(z −1),
5 4 5
we have
c =kc +2 c c d + c c
∞,−(k+1) ∞,−k ∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
−2 c Xd −(α +α )c X
∞,−l ∞,−m 5 4 ∞,−k
+2Xb c c −2 b c , (1.3)
∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
X X
where the first and fourth sums extend over the non-negative integers, l,m,n, such that
l+m+n = k, and the second sum extends over the non-negative integers, l,m, such that
l +m = k +1, and the third and fifth sums extend over the non-negative integers, l,m,
such that l+m = k.
Comparing the coefficients of the term t−k (k ≥ 1) in
tw′ = −2zw2 +w2 −2tzw +{t+(α +α )}w −α t−2y(−w+2zw +α ),
5 4 3 3
we obtain
d =−kd +2 c d d − d d
∞,−(k+1) ∞,−k ∞,−l ∞,−m ∞,−n ∞,−l ∞,−m
+2 c dX −(α +α )d X
∞,−l ∞,−m 5 4 ∞,−k
−2Xb d +4 b c d +2α b , (1.4)
∞,−l ∞,−m ∞,−l ∞,−m ∞,−n 3 ∞,−k
X 6 X
where the first and fifth sums extend over the non-negative integers, l,m,n, such that
l + m + n = k, and the second and fourth sums extend over the non-negative integers,
l,m, such that l+m = k, and the third sum extends over the non-negative integers, l,m,
such that l+m = k +1.
Equations (1.1), (1.2), (1.3), and (1.4) imply that a , b , c , and d are
∞,−k ∞,−k ∞,−k ∞,−k
inductively determined, which proves the proposition.
Corollary 1.3. Suppose that for D(1)(α ) , there exists a solution such that x,y,z,w
5 j 0≤j≤5
are all holomorphic at t = ∞. Then,
y ≡ 0 if α = 0,
1
(w ≡ 0 if α3 = 0.
1.2 The case where one of (x,y,z,w) has a pole at t = ∞
In this subsection, we treat the case where one of (x,y,z,w) has a pole at t = ∞. For
this purpose, we consider the following four cases:
(1) x has a pole at t = ∞ and y,z,w are all holomorphic at t = ∞,
(2) y has a pole at t = ∞ and x,z,w are all holomorphic at t = ∞,
(3) z has a pole at t = ∞ and x,y,w are all holomorphic at t = ∞,
(4) w has a pole at t = ∞ and x,y,z are all holomorphic at t = ∞.
1.2.1 The case where x has a pole at t = ∞
Proposition 1.4. For D(1)(α ) , there exists no solution such that x has a pole at
5 j 0≤j≤5
t = ∞ and y,z,w are all holomorphic at t = ∞.
Proof. It can be easily checked.
1.2.2 The case in which y has a pole at t = ∞
Lemma 1.5. Suppose that for D(1)(α ) , there exists a solution such that y has a
5 j 0≤j≤5
pole at t = ∞ and x,z,w are all holomorphic at t = ∞. y then has a pole of order one at
t = ∞.
Proof. It can be proved by direct calculation.
Proposition 1.6. Suppose that for D(1)(α ) , there exists a solution such that y has
5 j 0≤j≤5
a pole at t = ∞ and x,z,w are all holomorphic at t = ∞. Then, b = −1,−1/2.
∞,1
7
Proof. By comparing the coefficients of the term t in
tx′ = 2x2y +tx2 −2xy −{t+(2α +2α +α +α )}x
2 3 5 4
+(α +α )+2z{(z −1)w+α },
2 5 3
we have
a (a −1)(2b +1) = 0,
∞,0 ∞,0 ∞,1
which implies that a = 0,1 or b = −1/2.
∞,0 ∞,1
By comparing the coefficients of the term t2 in
ty′ = −2xy2 +y2 −2txy +{t+(2α +2α +α +α )}y −α t,
2 3 5 4 1
we get
(b +1)(−2a +1) = 0,
∞,1 ∞,0
which implies that b = −1 or a = 1/2.
∞,1 ∞,0
Therefore, if a = 0,1, it follows that b = −1. Furthermore, if b = −1/2, it
∞,0 ∞,1 ∞,1
follows that a = 1/2.
∞,0
Let us first treat the case where b = −1.
∞,1
Proposition 1.7. Suppose that for D(1)(α ) , there exists a solution such that y has a
5 j 0≤j≤5
pole at t = ∞ and x,z,w are all holomorphic at t = ∞ and b = −1. Then, a = 0,1
∞,1 ∞,0
and c = 0,1.
∞,0
(1) If (a ,c ) = (0,0),
∞,0 ∞,0
x = −(α +α )t−1 +··· ,
2 5
y = −t+α +α (α +2α +α −α )t−1 +··· ,
 0 0 0 3 4 5
z = −α5t−1 +··· ,

w = α +α (α +α −α )t−1 +··· .
3 3 3 4 5



(2) If (a ,c )= (0,1),
∞,0 ∞,0
x = (−α −2α −α )t−1 +··· ,
2 3 5
y = −t+α +α (α −2α +α −α )t−1 +··· ,
 0 0 0 3 4 5
z = 1−α4t−1 +··· ,

w = −α +α (α −α +α )t−1 +··· .
3 3 3 4 5




8
(3) If (a ,c ) = (1,0),
∞,0 ∞,0
x = 1−(α +2α +α )t−1 +··· ,
2 3 4
y = −t−α +α (α −2α −α +α )t−1 +··· ,
 0 0 0 3 4 5
z = −α5t−1 +··· ,

w = α +α (α +α −α )t−1 +··· .
3 3 3 4 5



(4) If (a ,c ) = (1,1),
∞,0 ∞,0
x = 1−(α +α )t−1 +··· ,
2 4
y = −t−α +α (α +2α −α +α )t−1 +··· ,
 0 0 0 3 4 5
z = 1−α4t−1 +··· ,

w = −α +α (α −α +α )t−1 +··· .
3 3 3 4 5



Proof. It can be proved by direct calculation.
Proposition 1.8. Suppose that for D(1)(α ) , there exists a solution such that y has
5 j 0≤j≤5
a pole at t = ∞ and x,z,w are all holomorphic at t = ∞ and b = −1. Moreover,
∞,1
assume that a and c are both determined. The solution is then unique.
∞,0 ∞,0
Proof. It can be proved in the same way as Proposition 1.2.
Corollary 1.9. Suppose that for D(1)(α ) , there exists a solution such that y has a
5 j 0≤j≤5
pole at t = ∞ and x,z,w are all holomorphic at t = ∞ and b = −1. Then,
∞,1
y ≡ −t if α = 0,
0
(w ≡ 0 if α3 = 0.
Let us deal with the case in which b = −1/2. From the proof of Proposition 1.6, it
∞,1
follows that a = 1/2. We first suppose that c = 1/2.
∞,0 ∞,0
Proposition 1.10. Suppose that for D(1)(α ) , there exists a solution such that y has
5 j 0≤j≤5
a pole of order one at t = ∞ and x,z,w are holomorphic at t = ∞ and c = 1/2. Then,
∞,0
a = 1/2, a = −α +α and either of the following occurs:
∞,0 ∞,−1 0 1
(1) 2α +α +α 6= 0 and b +d = −α +α and
3 4 5 ∞,0 ∞,0 4 5
(α +α )(−α +α ) 2α (−α +α )
4 5 4 5 3 4 5
b = , d = ,
∞,0 ∞,0
2α +α +α 2α +α +α
3 4 5 3 4 5
(1−2α −α −α )c = (α +α )(−α +α ),
3 4 5 ∞,−1 0 1 0 1
9
(2) b +d = −α +α and
∞,0 ∞,0 4 5
α = 0, α +α = 0,
3 4 5
or

−α +α = 0, α +α = 0.
4 5 3 4
Proof. By comparing the constant terms in
tx′ = 2x2y +tx2 −2xy −{t+(2α +2α +α +α )}x
2 3 5 4
+(α +α )+2z{(z −1)w+α },
2 5 3
we get
b +d = −α +α . (1.5)
∞,0 ∞,0 4 5
By comparing the coefficients of the term t in
ty′ = −2xy2 +y2 −2txy +{t+(2α +2α +α +α )}y −α t,
2 3 5 4 1
we have a = −α +α .
∞,−1 0 1
By comparing the constant terms in
tw′ = −2zw2 +w2 −2tzw +{t+(α +α )}w −α t−2y(−w+2zw +α ),
5 4 3 3
we obtain
2α b −(α +α )d = 0. (1.6)
3 ∞,0 4 5 ∞,0
From (1.5) and (1.6), it follows that
(2α +α +α )d = 2α (−α +α ). (1.7)
3 4 5 ∞,0 3 4 5
By comparing the coefficients of the term t−1 in
tx′ = 2x2y +tx2 −2xy −{t+(2α +2α +α +α )}x
2 3 5 4
+(α +α )+2z{(z −1)w+α }
 2 5 3
tz′ = 2z2w+tz2 −2zw −{t+(α +α )}z +α +2yz(z −1),
5 4 5

we obtain 
−2(α +α )a +b −4α c +d = 0,
0 1 ∞,−1 ∞,−1 3 ∞,−1 ∞,−1
(b∞,−1 +2(−1+α4 +α5)c∞,−1+d∞,−1 = 0.
We then find that
−2(α +α )(−α +α )+2(1−2α −α −α )c = 0.
0 1 0 1 3 4 5 ∞,−1
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