IJMMS 31:1 (2002) 43–49
PII. S0161171202108222
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
APPLICATION OF UNIFORM ASYMPTOTICS TO THE FIFTH
PAINLEVÉ TRANSCENDANT
YOUMIN LU and ZHOUDE SHAO
Received 30 August 2001
We apply the uniform asymptotics method to the fifth Painlevé transcendants, find its
asymptotics of the form y = −1 + t −1/2 A(t) as t → ∞ along the positive t-axis, and obtain
the corresponding monodromy data.
2000 Mathematics Subject Classification: 34E05.
1. Introduction. We study the general fifth Painlevé equation
d2 y
1
1
dy 2 1 dy
=
−
+
dt 2
2y y − 1
dt
t dt
2
(y − 1)
γy δy(y + 1)
β
+
+
+
,
αy +
t2
y
t
y −1
(1.1)
where α, β, γ, and δ are parameters, and its solution of the form
y(t) = −1 + 4t −1/2 A(t),
y (t) = −2t −3/2 A(t) + 4t −1/2 A (t) = 4t −1/2 A (t) + O t −3/2 ,
(1.2)
with A(t) = O(1) as t → ∞.
The fifth Painlevé equation (1.1) can be obtained as the compatibility condition of
the following linear systems of equations (see [2, 3]):


t 2v + θ0
w
u v + θ0
uy 2w − θ1
+
−
−
+
2
2z
z−1
z
2(z − 1) 


 Y (z),
(1.3)
Yz (z) = 

 v
2w + θ1
t 2v + θ0
w
−
− −
+
uz 2uy(z − 1)
2
2z
z−1

1
2


Yt (z) = 
 1
1
θ1
v−
w+
uz
y
2

u
θ1
v + θ0 − y w −

z
2

 Y (z),

1
−
2
(1.4)
where
w =v+
1
θ0 + θ∞ ,
2
1
dy
= ty − 2v(y − 1)2 − (y − 1) θ0 − θ1 + θ∞ y − 3θ0 + θ1 + θ∞ ,
dt
2
1
du
1
1
t
= u − 2v − θ0 + y v + θ0 − θ1 + θ∞ +
v + θ0 + θ1 + θ∞
,
dt
2
y
2
(1.5)
t
(1.6)
44
Y. LU AND Z. SHAO
with
α=
1 θ0 − θ1 + θ∞ 2
,
2
2
β=−
1 θ 0 − θ1 − θ∞ 2
,
2
2
1
δ=− .
2
γ = 1 − θ0 − θ1 ,
(1.7)
The canonical solutions of system (1.3) are defined in [2] by
−
π
3π
+ kπ ≤ arg λ < − + kπ ,
2
2
(1.8)
Yk (λ) ∼ Ŷ∞ (λ)e(tλ/2−log λ)σ3 ,
where σ3 =
1
0
0 −1
, and the Stokes multiplier G1 is defined in [2] by
Y2 (λ) = Y1 (λ)G1 ,
where G1 =
1 0
s 1
(1.9)
and its entry s is independent of t and y.
Y1
B
, φ = B −1/2 Y1 ,
2. Reduction of the problem. Generally, if (d/dz) YY1 = A
C −A
Y
2
and ψ = C −1/2 Y2 , then
2
d2 φ
3 −2 2 1 −1 2
−1
=
A
+
BC
+
A
−
B
B
A
+
B
−
B
φ,
B
B
dz2
4
2
3 −2 2 1 −1 d2 ψ
2
−1
C
C
=
A
+
BC
−
A
+
C
C
A
+
C
−
C
ψ.
dz2
4
2
(2.1)
We first apply the transformation
1
Ŷ =
i
i
u−(1/2)σ3 Y
1
(2.2)
to system (1.3) to get
1 L
dŶ
=
dz
2 M
N
Ŷ ,
−L
(2.3)
where
i 2v + θ0
i 1/y + y w + i 1/y − y θ1 /2
L=
−
,
z
z−1
i 2v + θ0 − θ0
y − 1/y − 2i w − y + 1/y (θ1 /2)
+
+ it,
M=
z
z−1
y − 1/y + 2i w − y + 1/y (θ1 /2)
i 2v + θ0 + θ0
+
− it.
N =−
z
z−1
(2.4)
APPLICATION OF UNIFORM ASYMPTOTICS . . .
45
Applying (2.1) to (2.3), we get



 t
d φ
2v + θ0
w
=
+
−

dz2
2
2z
z
−1


2
2
v
u v + θ0
uy 2w − θ1
2w + θ1
+ −
+
−
z
2(z − 1)
uz 2uy(z − 1)
1
θ1
1
i
+y w +i
−y
i 2v + θ0
y
y
2
−
+
2z2
2(z − 1)2

1 θ1
1
y − + 2i w − y +
 2iv + (i + 1)θ
y
y 2

0

−

z2
(z − 1)2
−

L


2
N

1 θ1
1
y − + 2i w − y +
 2iv + (i + 1)θ
y
y 2

0

−

z2
(z − 1)2
+
3
4
−
−
2




N2
2iv + (i + 1)θ0
1
1 θ1
+
y
−
+
2i
w
−
y
+
z3
y
y 2
N









φ.
(2.5)
Now, using (1.6), the following asymptotics can be obtained:
1
1
1
1
1
v(t) = − t − t 1/2 A (t) + A2 (t) − 2A(t)A (t) − θ0 − θ∞ + O t −1/2 ,
8
2
2
2
4
−1/2 t/2
1+O t
.
u(t) = Ce
(2.6)
Substituting (2.6) into (2.5), we get the following second-order equation:
"
#
d2 φ
(2z − 1)2
(2z − 1)θ∞ A2 + 4A2 i 2z2 − 2z + 1
2
−1
−t
−
+
= −t −
−
dz2
16z(z − 1)
4z(z − 1)
4z(z − 1)
8z2 (z − 1)2
+
$
%
i(2z − 1) (it/4)(2z − 1) + t 1/2 Az2
+ O t −3/2 φ
8z2 (z − 1)2 − it 1/2 A − t 1/2 Az − it(z − 1/2)2
(2.7)
= −t 2 F (z, t)φ.
Equation (2.7) has two turning points
zj =
&
1
± t −1/2 A2 + 4A2 + i 1 + o(1) ,
2
j = 1, 2
(2.8)
46
Y. LU AND Z. SHAO
which merge to 1/2 as t → ∞, and Stokes directions
'
Re
z(z − 1) = 0.
(2.9)
Now, we define a constant α by
1
π iα2 =
2
(α
−α
2
1/2
τ − α2
dτ =
( z2
z1
F 1/2 (z, t)dz
(2.10)
and a new variable ζ by
(ζ
α
2
1/2
dτ =
τ − α2
(z
z1
F 1/2 (s, t)ds.
(2.11)
Using [1, Theorem 1], we have the following theorem.
Theorem 2.1. Given any solution φ of (2.7), there exist constants c1 and c2 such
that, uniformly for z on the Stokes curve, as t → ∞,
ζ 2 − α2
F (z, t)
−1/4
φ(z, t) =
*
)
&
&
c1 +o(1) Dν eπ i/4 2tζ + c2 +o(1) D−ν−1 e−π i/4 2tζ ,
(2.12)
where ν = −1/2 + (1/2)itα2 and Dν (z), D−ν−1 (z) are solutions of the parabolic cylinder equation.
3. Monodromy data and asymptotics
Theorem 3.1. For large t and z,
1 2 A2 + 4A2 + i
ζ −
log ζ + o t −1
2
2t
=
iz iθ∞
i
2it 1/2
1 − i π θ∞
−
log(4z) +
log
+
−
2
2t
2t
2iA + A
4
4t
−1 −1 π iα2
+o t
−
+O z
.
4
(3.1)
Proof. Carrying out the integration on the left-hand side of (2.11), we have
α2 α2
1 2 α2
ζ −
log(2ζ) −
+
log α + O α4 ζ −2 =
2
2
4
2
(z
z1
F 1/2 (s, t)ds.
(3.2)
Because we are going to calculate the higher-order part of the right-hand side, we
will simply ignore the lower-order part in F (z, t), and split the right-hand side into
two integrals
(z
z1
F 1/2 (s, t)ds =
( z∗
z1
+
(z z∗
F 1/2 (x, t)dx = I1 + I2 ,
(3.3)
APPLICATION OF UNIFORM ASYMPTOTICS . . .
47
where z∗ = 1/2 + T t −1/2 and T is a large parameter to be specified later. Using the
substitution
x−
1
= st −1/2 ,
2
(3.4)
I1 can be evaluated as follows:
I1 =
=
(
'
1 T
s 2 − A2 + 4A2 + i + o(1) ds
√
2
2
t
A +4A
T 2 A2 + 4A2 + i A2 + 4A2 + i
−
−
log(2T )
2t
4t
2t
+
(3.5)
A2 + 4A2 + i
log A2 + 4A2 + i + o t −1 .
4t
Using the formula
(
2ax + b
+
ax 2 + bx + c
x2 −
1
4
dx
'

a + 2bx − 2 b2 − 4ac x

 arctanh ,
'
'
2b2 − 2b b2 − 4ac − 4ac − a2
4x 2 − 1
,
= 2a
'
2b2 − 2b b2 − 4ac − 4ac − a2

(3.6)
'
a + 2bx + 2 b2 − 4ac x
arctanh ,
'
'
2b2 + 2b b2 − 4ac − 4ac − a2
4x 2 − 1
,
,
+ 2a
'
2b2 + 2b b2 − 4ac − 4ac − a2
we find the asymptotic expression of I2

(z 

(2s − 1)2
(2s − 1)θ∞ A2 + 4A2 i 2s 2 − 2s + 1
−1 
I2 =
−
−
−t
−
+
16s(s − 1)
4s(s − 1)
4s(s − 1)
8s 2 (s − 1)2
z∗ 
1/2
i(2s − 1) (1/2)it 1/2 (s − 1/2) + As 2 
− 2
ds
8z (s − 1)2 iA + As + it 1/2 (s − 1/2)2 
=
(z
z∗


i(2s−1)  16s(s−1)  (2s−1)θ∞ A2 +4A2 i 2s 2 −2s+1
&
−
+
−
1+
4s(s−1)
4s(s−1)
8s 2 (s−1)2
4 s(s−1)  t(2s−1)2
1/2

i(2s−1) (it/4)(2s−1)+t 1/2 As 2

+ 2
ds
2
1/2
1/2
8s (s−1) − it/4−t A −t As−its(s−1) 
48
Y. LU AND Z. SHAO
=
(z
z∗
#
i A2 + 4A2 + i
iθ∞
i(2s − 1)
&
&
− &
−
4 s(s − 1) 2t s(s − 1) 2t(2s − 1) s(s − 1)
$
2it 1/2 (s − 1/2) + A
&
dz
−
4t (s − 1/2)2 − 1/4 iA + (1/2)A + A(s − 1/2) + it 1/2 (s − 1/2)2
(z
+ O t −2
z∗
=
|ds|
1
+
O
|2s − 1|3
z
iπ A2 + 4A2 + i
iz i iθ∞
1 T 2 π θ∞
− −
log(4z) −
+ −
−
2
4
2t
4t
4 2t
4t
−
A2 + 4A2 + i
t
2it 1/2
i
+ O z−1
log 2 +
log
4t
T
2t
2iA + A + O t −1/4
+ O T −2 t −1 + O T 4 t −2 .
(3.7)
√
Using definition (2.10) and setting T = − A2 + 4A2 in I1 , we have the following
expression for α:
α2 = −
A2 + 4A2 + i
+ o t −1 .
t
(3.8)
Substituting (3.8) into (3.2), setting T < t 1/4 , and combining it with (3.5) and (3.7),
the theorem is proved.
Knowing [4] that
Dν (z) ∼


ν −(1/4)z2

,
z e


2


zν e−(1/4)z −
if | arg z| <
√
2π iπ ν −ν−1 (1/4)z2
e z
e
,
Γ (−ν)
if arg z =
3
π,
4
3
π,
4
(3.9)
√
and arg(eπ i/4 2tζ) ∼ π /4 when z → −∞, we can choose, as z → −∞,
(11)
Ŷ2
(z) ∼ z−θ∞ /2 etz/2
&
2
2
2
∼ 2θ∞ t 1/4 2iA + Ae−(π i/4)(1/2+(3it/2)α ) e(1/4)t+i[t/4−(1/4)(A +4A ) log(2t)+θ∞ /4]
&
× ζ 1/2 Dν eπ i/4 2tζ .
(3.10)
√
Because arg(eπ i/4 2tζ) ∼ 3π /4 when z → ∞, we have the following asymptotics
(11)
for Ŷ2 (z) as z → ∞:
(11)
Ŷ2
(z) ∼ z−θ∞ /2 etz/2
− zθ∞ /2 e−tz/2
√ 4θ∞ π A2 + 2iA2 (π i/4)(2tiα2 −3)+(1/2)t+i[t/2−(1/2)(A+4A2 ) log(2t)]
e
.
Γ 1/2 − (ti/2)α2
(3.11)
APPLICATION OF UNIFORM ASYMPTOTICS . . .
49
By (1.9) and (2.2), the Stokes multiplier can be defined by Ŷ2 = Ŷ1 G1 . Therefore, the
monodromy data is
√ 4θ∞ π A + 2iA (π i/4)(2tiα2 −5)+i[t/2−(1/2)(A+4A2 ) log(2t)]
e
.
s= Γ 1/2 − (ti/2)α2
(3.12)
Taking the square of the absolute value of both sides of this equation, we find that
A2 + 4A2 ∼ d2 where d is a constant. Solving (3.12) for A + 2iA , we have
'
2
A + 2iA = ± A2 + 4A2 + O t −1/2 ei[t/2−(1/2)(A+4A ) log(2t)+θ] .
(3.13)
Taking the real part and the imaginary part of (3.13), we obtain the following theorem.
Theorem 3.2. Equation (1.1) has a solution with the following asymptotics:
t 1 2
− d log(2t) + θ ,
y(t) ∼ −1 + 4t −1/2 d + O t −1/2 cos
2 2
t 1 2
y (t) ∼ 2t −1/2 d + O t −1/2 sin
− d log(2t) + θ , as t → ∞.
2 2
(3.14)
References
[1]
[2]
[3]
[4]
A. P. Bassom, P. A. Clarkson, C. K. Law, and J. B. McLeod, Application of uniform asymptotics
to the second Painlevé transcendent, Arch. Rational Mech. Anal. 143 (1998), no. 3,
241–271.
A. S. Fokas, U. Muğan, and M. J. Ablowitz, A method of linearization for Painlevé equations:
Painlevé IV, V, Phys. D 30 (1988), no. 3, 247–283.
A. R. Its and V. Y. Novokshenov, The Isomonodromic Deformation Method in the Theory of
Painlevé Equations, Lect. Notes in Math., vol. 1191, Springer-Verlag, Berlin, 1986.
E. T. Whittaker and G. M. Watson, Modern Analysis, 4th ed., Cambridge University Press,
Cambridge, 1927.
Youmin Lu: Department of Mathematics and Computer Science, Bloomsburg University, Bloomsburg, PA 17815, USA
Zhoude Shao: Department of Mathematics, Millersville University, Millersville, PA
17551-0302, USA