IJMMS 2003:13, 845–851
PII. S0161171203206220
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS
TO THE FOURTH GENERAL PAINLEVÉ EQUATION
YOUMIN LU
Received 5 June 2002
There have been many results on the asymptotics of the Painlevé transcendents in
recent years, but the asymptotics of the fourth Painlevé transcendent has not been
studied much. In this note, we study the general fourth Painlevé equation and develop an asymptotic representation of a group of its solutions as the independent
variable approaches infinity along a straight line.
2000 Mathematics Subject Classification: 34E99.
1. Introduction. Asymptotic behaviour of the second and the third Painlevé
transcendents has been much studied [1, 3, 4, 7]. But there are very few results
about the fourth Painlevé equation
1 dη 2 3 3
β
d2 η
=
+ η + 4ξη2 + ξ 2 − α η + ,
dξ 2
2η dξ
2
η
(1.1)
where α and β are parameters. In [2], Clarkson and McLeod studied the special
case of (1.1) for the parameter β = 0 and obtained the complete asymptotic
representation of its solution with η(∞) = 0. In [6], we studied the general
case of (1.1) and proved the following theorem.
Theorem 1.1. Under the assumption that iα ≥ 0 and β > 0, (1.1) has a
solution η(ξ) with the following asymptotics:
(i) as ξ → eπ i/4 ∞,


−2 ir
β
i
2
 cos φ,
η(ξ) =
r − +O ξ
±
ξ
ξ
2


(1.2)
−2 β
 sin φ,
η (ξ) = ±2 r 2 − + O ξ
2
where φ = iξ 2 − (1/2)(3r + iα) log(−iξ 2 ) + θ0 + O(ξ −2 ), and r > 0 and
θ0 are real parameters satisfying r 2 − β/2 > 0;
(ii) as ξ → eπ i/4 0,
η(ξ) = eπ i/4 a + be−π i/4 ξ + O ξ 2 ,
where a and b are real parameters with a > 0.
(1.3)
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YOUMIN LU
Furthermore, any solution of (1.1) of the form η(ξ) = eπ i/4 W (ξ) with
W (ξ) ≥ 0 has the above asymptotics.
A natural question following this result is how this solution behaves when ξ
goes to infinity in the opposite direction of the line. As we know, through the
transformations
η(ξ) = e−π i/4 y,
ξ = e−π i/4 x,
(1.4)
3
β
y+ ,
− y 3 − 4xy 2 − 2 x 2 + α
2
y
(1.5)
(1.1) is related to the equation
1
d2 y
=
dx 2
2y
dy
dx
2
Applying the transformation
where α = iα.
y(x) = −xw 2 (t),
t = −x 2 ,
(1.6)
we can obtain the equation
w + t −1 w =
α
1 −2
3 5 1 3 1
β
t w−
w + w − w + t −1 w + 2 3 .
16
16
2
4
4
8t w
(1.7)
It is easy to prove that w, w , and 1/t 2 w 2 are all bounded as t → +∞ and
≤ 0 and β ≥ 0. Inspecting (1.7) carefully, we can conclude that there
when α
are possibly four cases for the behaviour of w as t → +∞ :
(1) w(t) → 0 as t → +∞;
√
(2) w(t) → 2 as t → +∞;
(3) w(t) → 2/3 as t → +∞;
(4) w does not have a limit as t → +∞.
Case (1) has been studied and eliminated in [5]. In this note, we pay our attention to case (2) and develop the asymptotic representation of η corresponding
to this case as ξ → e5π i/4 (+∞).
2. Development of the asymptotic representation. From now on, we as√
≤ 0, β ≥ 0, and w(t) → 2 as t → +∞. We first prove the following
sume that α
theorem.
√
Theorem 2.1. The function W is bounded as t → +∞ if w = 2 + t −1/2 W .
√
Proof. Applying the transformation w = 2 + t −1/2 W to (1.7), we obtain
W =
√
√
3 −2
9 2 −1/2 2 13 −1 3
2 −3/2
t
t W −W −
t
t W
−
W −
16
16
4
4
√
√
2 −1/2 α
15 2 −3/2 4
3 −2 5 α
βt −3/2
t
t W +
t
−
W −
+ t −1 W +
.
16
16
4
4
8w 3
(2.1)
ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS . . .
847
Multiplying both sides of (2.1) by 2W and integrating it, we have
√
W 2 + W 2 =
√ 2 −3/2
3 2 t −5/2
3 −2 2 3 t −3 2
t
t W −
W+
t
W dt −
t W dt
8
16 t0
16
8 t0
√
√ t
3 2 −1/2 3 3 2
13 −1 4 13 t −2 4
t
t W −
−
W −
t −3/2 W 3 dt −
t W dt
2
4
8
8 t0
t0
√
√ t
t
3 2 −3/2 5 9 2
1 −2 6 1
t
t W −
−
W −
t −5/2 W 5 dt −
t −3 W 6 dt
8
16 t0
16
8 t0
√
√ 2 −1/2
2 t −3/2
t −2 2
α
α
α
α
t
+
W+
t
W dt + t −1 W 2 +
t W dt
2
4
4
4 t0
t0
√ t
β
β 2
dt
−
−
+ C1 .
2 3
8tw 2
8
t0 t w
(2.2)
Clearly, W 2 (t) = o(t), and therefore C1 = o(t0 ). Because t −1/2 W = o(1) as
t
t → +∞, t0 t −5/2 W dt is bounded and the following inequalities are true for
large t0 and t ≥ t0 :
√ 9 2 t −5/2 5
1 t −2 4
t W dt +
t
W dt ≥
t W dt,
16 t0
2 t0
t0
√
√
3 2 −1/2 3 3 2 −3/2 5 1 2
W2+
W +
W ≥ W ,
t
t
2
8
2
√ t
√ t
3 2
α
2
t −5/2 |W | − W dt −
t −3/2 |W | − W dt
32 t0
8
t0
√ 2 t −3/2 α
t
≤−
|W | − W .
4
t0
13
8
t
−2
4
(2.3)
Hence, there exists a constant C2 and a large t0 such that
1
3 −2 2 3
t W +
W 2 + W 2 +
2
16
8
t
3
1
t −3 W 2 dt + t −1 W 4 +
8
2
t0
t
t0
t −2 W 4 dt
t −2 2
1 −2 6 1 t −3 6
α
α
t W +
t W dt − t −1 W 2 −
t W dt
16
8 t0
4
4 t0
√ √ 3
2 t −3/2 3 2 t −3/2 α
|W | + W dt +
|W | + W dt
+
t
t
32 t0
8
t0
√ √ 3
2 t −3/2 α
3 2 t −3/2 t
t
≤
|W | − W dt −
|W | − W dt + C2 .
32 t0
4
t0
+
(2.4)
t
From (2.1), we can claim that limt→+∞ W (t) ≠ −∞. If t0 t −3/2 W 3 dt ≥ 0 and
t −3/2
W dt ≥ 0, then every term, especially W 2 , in inequality (2.4) is bounded.
t0 t
If W is bounded below, the same argument can be applied to obtain that W 2 is
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YOUMIN LU
bounded for large t. If W is not bounded below, from inequality (2.4), we have
√ 3
2 1
3 2 t −3/2 1
t
|W | − W dt
|W | − W ≤ W 2 ≤
8
2
32 t0
(2.5)
√ 2 t −3/2 α
−
t
|W | − W dt + C2 .
8
t0
Since (|W | − W )3 > −(2α/3)(|W
| − W ) when (|W | − W ) is larger than a certain
constant, there exists a constant C3 > 0 and C3 = o(t0 ) such that
√ 3
2 3 2 t −3/2 t
(2.6)
|W | − W dt + C3 .
|W | − W ≤
2
t0
Now, let I be the right-hand side of (2.6). Then,
√
√
3 3 2 −3/2 3/2
3 2 −3/2 t
t
I .
I =
|W | − W ≤
2
2
(2.7)
√
1/2
1/2
Integrating both sides of (2.7), we obtain 1/I 1/2 ≥ 1/C3 − 3 2/2t0 +
√
1/2
> 0 for large t0 . Thus, I is bounded, and we reach a contradiction.
3 2/2t
Hence, W is bounded below and the theorem is proved.
Theorem 2.2. As ξ → ∞ along arg ξ = 5π /4,
η(ξ) = −2ξ ± 2eπ i/4 2d + O ξ −1 cos φ,
η (ξ) = −2 ∓ 4e3π i/4 2dξ sin φ + O ξ −1 ,
(2.8)
where φ = −iξ 2 + (3c 2 + iα) log(−iξ 2 ) + φ0 + O(ξ −1 ).
Proof. We need to rewrite (2.2) to be as follows:
√
√ 2 −3/2
3 2 t −5/2
3 −2 2 3 t −3 2
W 2 + W 2 =
W+
t
W dt −
t W dt
t
t W −
8
16
16
8
√ t
√
13 −1 4 13 t −2 4
3 2 −1/2 3 3 2
W −
t −3/2 W 3 dt −
t W dt
−
t
t W −
2
4
8
8
√
√ t
t
3 2 −3/2 5 9 2
1 −2 6 1
−
t
t W −
W −
t −5/2 W 5 dt −
t −3 W 6 dt
8
16
16
8
√ √
2 t −3/2
t −2 2
2 −1/2
α
α
α
α
t
W+
t
W dt + t −1 W 2 +
t W dt
+
2
4
4
4
√ t
dt
β 2
β
−
+ C.
−
8tw 2
8
t2w 3
(2.9)
Then, C = limt→∞ (W 2 (t)+W 2 (t)). If C = 0, then W → 0 as t → +∞. In this case,
√
we can solve (2.1) to obtain that W (t) ∼ (α 2/4)t −1/2 , and W = O(t −3/2 ).
√
√
2/4)t −1 and w = −(1/2)t −3/2 W + t −1/2 W = O(t −2 ). If
Thus, w ∼ 2 + (α
C ≠ 0, we let
W (t) = ρ(t) cos φ,
W (t) = ρ(t) sin φ.
(2.10)
ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS . . .
849
Then,
ρ 2 (t) = C + O t −1/2 .
(2.11)
Now, we let C = c 2 and find the asymptotics of φ
dφ
dt
W W + W 2
W 2 + W 2
√
√
−1
2/4)t−1/2 W+(α/4)t
W 2+O t−3/2
−(9 2/4)t−1/2 W 3 −(13/4)t−1 W 4+(α
√
√
= −1+
2/2)t−1/2 W+O(t−1 )
c 2 −(3 2/2)t−1/2 W 3+(α
√
√
2 −1/2
13
α
9 2 −1/2 3 α
t
W +
t
W − 2 t −1 W 4 + 2 t −1 W 2
= −1 −
4c 2
4c 2
4c
4c
2
27
3α
α
− 4 t −1 W 6 + 4 t −1 W 4 − 4 t −1 W 2 + O t −3/2 .
4c
c
4c
(2.12)
= −1 +
This implies that W behaves like cos t roughly. Thus, we need to pay attention
to the terms with even power of W which will give significant contribution to
the expression of φ when we use integration by parts later. Using (2.12), we
can find the asymptotic representation of dt in terms of dφ as follows:
dt = 1 −
√
√
−1 2 −1/2
9 2 −1/2 3 α
(−dφ).
t
W
+
t
W
+
O
t
4c 2
4c 2
(2.13)
Plugging (2.13) back to (2.12), we have
√
√
√
2 −1/2
9 2 −1/2 3 α
9 2 −1/2 3
dφ = −dt + −
t
W
+
t
W
1
−
t
W
4c 2
4c 2
4c 2
√
−1 2 −1/2
α
(−dφ)
+
t
W
+
O
t
4c 2
α
27
3α
13
+ − 2 t −1 W 4 + 2 t −1 W 2 − 4 t −1 W 6 + 4 t −1 W 4
4c
4c
4c
c
2
α
− 4 t −1 W 2 + O t −3/2 dt
4c
√
√
2 −1/2
9 2 −1/2 3 α
t
W
+
t
W
(−dφ)
= −dt + −
4c 2
4c 2
−3/2 −1 4 α
2 −1 2
81 −1 6 9α
(−dφ)
+
t
W
−
t
W
+
t
W
+
O
t
8c 4
4c 4
8c 4
13
α
27
3α
+ − 2 t −1 W 4 + 2 t −1 W 2 − 4 t −1 W 6 + 4 t −1 W 4
4c
4c
4c
c
2
α
− 4 t −1 W 2 + O t −3/2 dt.
4c
(2.14)
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YOUMIN LU
By (2.9) and (2.10),
√
√
1/2
2 −1/2
3 2 −1/2 3 α
W = C−
W +
W + O t −1
cos φ
t
t
2
2
√
√
2 −1/2
3 2 −1/2 3 α
W +
W + O t −1 cos φ.
= c−
t
t
4c
4c
(2.15)
Plugging (2.15) back to (2.14), we obtain
√
−1/2
9c 2 −1/2
α
dφ = −dt + −
t
t
cos3 φ +
cos φ (−dφ)
4
4c
81 −1 3
27α −1
3α
+
t W cos3 φ −
t W cos3 φ − 3 t −1 W 3 cos φ
8c
8c
8c
2 −1
α
+ 3 t W cos φ (−dφ)
8c
−3/2 −1 4 α
2 −1 2
81 −1 6 9α
(−dφ)
+
t
W
−
t
W
+
t
W
+
O
t
8c 4
4c 4
8c 4
α
27
3α
13
+ − 2 t −1 W 4 + 2 t −1 W 2 − 4 t −1 W 6 + 4 t −1 W 4
4c
4c
4c
c
2
α
− 4 t −1 W 2 + O t −3/2 dt.
4c
√
−1/2
9c 2 −1/2
α
= −dt + −
t
t
cos3 φ +
cos φ (−dφ)
4
4c
27c 2 −1
13c 2 −1
6
−1 cos4 φ + O t −3/2 dφ
t cos φ +
t cos4 φ + 3αt
+ −
2
4
t −1 dt
= −dt + 3c 2 − α
√
−1/2
α
27c 2 −1
5
9c 2 −1/2
t
t
t
(−dφ)
cos3 φ +
cos φ +
cos6 φ −
+ −
4
4c
2
16
2
13c
3
1
α
t −1 cos4 φ −
+
+ 3α
+ t −1 cos2 φ −
+ O t −3/2 dφ.
4
8
4
2
(2.16)
log t + φ0 + O(t −1/2 ).
Therefore, φ = −t + (3c 2 − α)
Using transformations (1.4), (1.6), (2.10), and the one in Theorem 2.1, we
get the asymptotic representation in the theorem and finish the proof of the
theorem.
References
[1]
[2]
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asymptotics to the second Painlevé transcendent, Arch. Rational Mech. Anal.
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ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS . . .
[3]
[4]
[5]
[6]
[7]
851
A. R. Its and V. Yu. Novokshenov, The Isomonodromic Deformation Method in
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personal communication.
Youmin Lu: Department of Mathematics, Computer Science and Statistics, College
of Science and Technology, Bloomsburg University of Pennsylvania, Bloomsburg, PA
17815, USA
E-mail address: ylu@bloomu.edu