# ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS YOUMIN LU ```IJMMS 2003:13, 845–851
PII. S0161171203206220
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ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS
TO THE FOURTH GENERAL PAINLEV&Eacute; EQUATION
YOUMIN LU
There have been many results on the asymptotics of the Painlev&eacute; transcendents in
recent years, but the asymptotics of the fourth Painlev&eacute; transcendent has not been
studied much. In this note, we study the general fourth Painlev&eacute; 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 Painlev&eacute;
transcendents has been much studied [1, 3, 4, 7]. But there are very few results
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 , 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 , we studied the general
case of (1.1) and proved the following theorem.
Theorem 1.1. Under the assumption that iα ≥ 0 and β &gt; 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
&plusmn;
ξ
ξ
2


(1.2)
−2 β
 sin φ,
η (ξ) = &plusmn;2 r 2 − + O ξ
2
where φ = iξ 2 − (1/2)(3r + iα) log(−iξ 2 ) + θ0 + O(ξ −2 ), and r &gt; 0 and
θ0 are real parameters satisfying r 2 − β/2 &gt; 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 &gt; 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 . 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 &gt; −(2α/3)(|W
| − W ) when (|W | − W ) is larger than a certain
constant, there exists a constant C3 &gt; 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
&gt; 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 &plusmn; 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.
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ON THE ASYMPTOTIC REPRESENTATION OF THE SOLUTIONS . . .





851
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Youmin Lu: Department of Mathematics, Computer Science and Statistics, College
of Science and Technology, Bloomsburg University of Pennsylvania, Bloomsburg, PA
17815, USA