GENERAL SIXTH PAINLEVÉ EQUATION

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GENERAL SIXTH PAINLEVÉ EQUATION

HUIZENG QIN AND YOUMIN LU

Received 9 May 2006; Revised 3 September 2006; Accepted 1 October 2006

We study the general sixth Painlevé equation, develop, and justify the existence of several groups of asymptotics of its real solutions. Our methods also justify the differentiability of the asymptotics. Particular attention is paid to the solutions between 0 and 1. We find the asymptotics of all real solutions between 0 and 1 of the sixth Painlevé equation as x→+∞.

1. Introduction

The mathematical and physical significance of the six Painlevé transcendents has been well established. In the last 20 to 30 years, many mathematicians have spent dramatic effort on studying the properties of these transcendents. Although it is the most com- plicated one among the six Painlevé equations, there have been many results about the sixth Painlevé transcendent. In fact, the asymptotics problem of the sixth Painlevé transcendent has been studied in many papers such as [1,2,4–7,9–12], and the connection problem is also studied in the papers [1,4–7,10,11]. In this paper, we study the general sixth Painlevé equation

d²y dx² ⁼

1 2

dy dx

21 y+ 1

y−1+ 1 y−x

− 1

x+ 1 x−1+ 1

y−x dy

dx + y(y−1)(y−x)

x²(x−1)²

α+βx

y² +γ(x−1)

(y−1)² +δx(x−1) (y−x)²

,

(PVI)

whereα,β,γ, andδare parameters. Heuristically, if yis a “small” solution of(PVI), the following equation truncated from(PVI)would be its “major” part asx→+∞:

d²y dx² ⁼

1 2

dy dx

21 y+ 1

y−1

−1 x

dy dx⁻

β(y−1) x²y ⁻

γy

x²(y−1). (1.1)

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 69562, Pages1–10

DOI 10.1155/IJMMS/2006/69562

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Lettingt=lnxand applying elementary techniques to (1.1), one may solve it to get three diﬀerent solutions:

y=A+Bsin(alnx+b), (1.2)

whereA=1/2 + (β+γ)/a²andB²=(1/2 + (β+γ)/a²)²−2β/a²; y=b

2x^a+B²

2bx⁻^a+A, (1.3)

whereA=1/2−(β+γ)/a²andB²=2β/a²+ (1/2−(β+γ)/a²)²; and y=β+γ

2 (lnx+b)²+ β

β+γ. (1.4)

It is reasonable to expect that some solutions of(PVI) take (1.2), (1.3), or (1.4) as their asymptotics. Indeed, many authors [4–7,9–12] have obtained the corresponding asymptotics asx→0 that can be used to obtain (1.2), (1.3), and (1.4) by applying the well- known symmetry transformations. In this paper, we will prove the following theorems.

The diﬀerences of our results are pointed out following each theorem.

Theorem 1.1. Letβ >0 andγ <0. Ifx0>1, 0< y0<1, andyis a solution of(PVI)with y(x0)=y0, then 0< y <1 for allx > x0and it satisfies, asx→+∞,

y=A+Bcos(alnx+b) +Ox⁻¹,

y= −aBx⁻¹sin(alnx+b) +Ox⁻², (1.5) whereA=1/2 + (β+γ)/a²andB²=(1/2 + (β+γ)/a²)²−2β/a².

It is clear that the parameters need to satisfy the condition (1/2 + (β+γ)/a²)²≥2β/a². The complex form of the asymptotics asx→0 corresponding to this asymptotics has been obtained in many papers [7,10], but our result provides the conditions on the coeﬃcients of the equation for the real solutions to exist, together with the bound 0< y <1. Our proof of this theorem is also elementary and simple.

Theorem 1.2. Equation(PVI)has a group of solutions with the following asymptotics:

y=b

2x^a+A+Ox⁽³^/²⁾^a⁻¹^/²), y=ab

2 x^a⁻¹+Ox⁽³^/²⁾^a⁻³^/²), asx−→+∞,

(1.6)

where 0< a <1/4 or 3/4< a <1, andA=1/2−(β+γ)/a².

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We have noticed that it makes more sense for this result to be true when 0< a <1.

But, it seems to be impossible to prove it using our method. This asymptotics is actually a well-known result [4–6,9–12], but our result here also estimates the second term of the leading behavior of the solution as well as the diﬀerentiability of the asymptotics.

Theorem 1.3. Ifγ+β=0, then(PVI)has a group of solutions with the following asymp- totics:

y∼β+γ

2 (lnx+b)²+ β β+γ, y∼(β+γ)x⁻¹(lnx+b), asx−→+∞.

(1.7)

Various forms of this result occur in the literature. For example, in [7] Guzzetti has the following result forx→0:

y(x)∼

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ x

θ_x²−θ²0

4

lnx+4r+ 2θ0

θ0²−θ²_x 2

+ θ²0

θ0²−θ_x²

, θ0= ±θ_x,

x(r±θ0lnx), θ0= ±θx.

(1.8)

The coeﬃcientsθ0andθ_xcome from the isomonodromy deformation theory andris a free complex parameter. Applying the symmetry transformations y(x)=xz(t) andx= t⁻¹to this result, our result inTheorem 1.3can be obtained.

It is well known that the transformation y=x/z transforms (PVI) to itself with (α,β,γ,δ) changed to (−β,−α, 1/2−δ, 1/2−γ). Hence, based on the previous theorems, one may easily obtain the following one.

Theorem 1.4. Equation(PVI)has solutions with the following asymptotics:

y∼ x

A+Bsin(alnx+b), asx−→+∞, provided thatα <0,δ >1

2, (1.9) whereA=1/2 + (1−2α−2δ)/2a²andB²=(1/2 + (1−2α−2δ)/2a²)²+ 2α/a², and

y∼ 2x

(1/2−α−δ)(lnx+b)², asx−→+∞, provided that 1

2⁼α+δ. (1.10) 2. Proof of Theorems1.2and1.3

In this section, we use the classical successive approximation method to proveTheorem 1.2. The proof ofTheorem 1.3is similar. In fact, Shimomura [12] studied a more general nonlinear ordinary diﬀerential equation, applied the successive approximation method to it, and obtained the result as an application. We first denote the functions (1.2), (1.3),

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and (1.4) asy0and substitutet=lnxandy=y0+y1into(PVI). The new equation is d²y1

dt² ⁻ 1

y0

+ 1

y0−1 dy0

dt dy1

dt + β

y²0

− γ y0−1²

y1+1

2 dy0

dt 2

1 y0²

+ 1

y0−1²

y1

=Gy0,y1

y²1+Hy0,y1

y1dy1

dt +Ky0,y1

dy1

dt 2

+e⁻^tIy0,y1

dy1

dt +e⁻^tFy0,y1,t, (2.1) where

Gy0,y1

=

1 2y²0

y0+y1

+ 1

2y0−1²y0+y1−1 dy0

dt 2

+ β

y0²

y0+y1

− γ

y0−1²y0+y1−1, Hy0,y1

= − 1

y0

y0+y1

+ 1

y0−1y0+y1−1) dy0

dt , Ky0,y1

=1 2

1

y0+y1+ 1 y0+y1−1

+1

2

e⁻^t y0+y1

e⁻^t−1², Iy0,y1

= 1 y0+y1

e⁻^t−1² dy0

dt ⁻ 1

y0+y1

e⁻^t−11−e⁻^t, Fy0,y1,t=1

2 1

y0+y1

e⁻^t−1² dy0

dt 2

− 1

y0+y1

e⁻^t−11−e⁻^t dy0

dt +αy0+y1

y0+y1−1y0+y1

e⁻^t−1

1−e⁻^t² +βy0+y1−1y0+y1−2+e⁻^t y0+y1

1−e⁻^t² +γy0+y1

1−e⁻^t ^±

δy0+y1

y0+y1−1 1−e⁻^ty0+y1

e⁻^t−1.

(2.2) As a routine, we introduce a new functionzby using the standard transformation

y1= y0

y0−1z. (2.3)

Now, (2.1) is changed to d²z

dt² +

β3/2−y0

y0²

−γy0+ 1/2 y0−1²

z

=My0,y1

z²+Ny0,y1

zdz

dt +Py0,y1

dz dt

2

+e⁻^tQy0,y1

z+e⁻^tIy0,y1

dz

dt+ e⁻^t y0

y0−1Fy0,y1,t,

(2.4)

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where My0,y1

= y0

y0−1Gy0,y1

+ 2y0−1 2y0

y0−1Hy0,y1

+

2y0−1²

4y^3/20 (y0−1)³^/²Ky0,y1

,

Ny0,y1

= y0

y0−1Hy0,y1

+ 2y0−1

y0

y0−1Ky0,y1

,

Py0,y1

= y0

y0−1Ky0,y1 , Qy0,y1

= 2y0−1 2y0

y0−1 dy0

dt Iy0,y1

.

(2.5) To proveTheorem 1.2, we assume thaty0takes the function in (1.3). Then, there exist constantsLandt0such that, for|a|<1,t > t0, and|y1| |y0|, the following estimates are true:

My0,y1< L, Ny0,y1< L, Py0,y1< L,

Fy0,y1,t y0

y0−1 < e^atL, Qy0,y1< e^atL, Iy0,y1

|< e^atL.

(2.6)

We convert (2.4) into the following integral equation:

z1

z2

= _∞

t

t−τ 1

Rτ,z1(τ),z2(τ)dτ, (2.7)

wherez1=z,z2=z, and

R(τ,z,z)=

γy0+1/2 y0−1² ⁻

β3/2−y0

y²0

z+My0,y1

z²+Ny0,y1

zdz

dt+Py0,y1

dz dt

2

+e⁻^tQy0,y1

z+e⁻^tIy0,y1

dz

dt+ e⁻^t y0

y0−1Fy0,y1,t.

(2.8) In order to apply the successive approximation method to (2.7), we rewrite it as

Z(t)=Lt,Z(t), (2.9)

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whereZ(t)=(^z_z¹2) andL(t,τ,z) is the right-hand side of (2.7). Now, we can define the sequence

Z−1(t)=0,

Zn(t)=Lt,Zn−1(t), n=0, 1, 2,.... (2.10) We first take care of the case when|a|<1/4. Lett0be large enough such that, when t≥t0,

y0

y0−1e⁻^(1/2⁻^a/2)t<1

2, 5L

(1−a)²e⁻^(1/2⁻^a/2)t<1 6, 4e⁻^at<1

6, 8Le⁻^at<1

6, 4L

3(a−1)²e⁽^a⁻¹⁾^t<1 6, 2

(1−a)²e⁻^at<1

6, 8L

9(a−1)²e⁽^a⁻¹⁾^t≤1 6.

(2.11)

Then,

Z1(t)≤

_∞

t

τ−t 1

e⁻^τ y0

y0−1Fy0, 0,τdτ

≤ L

(1−a)²e⁻⁽¹⁻^a⁾^t≤1

2e⁽^a/²⁻¹^/²⁾^t, fort≥t0, Z1(t)−Z0(t)≤1

2e⁽^a/²⁻¹^/²⁾^t, fort≥t0.

(2.12)

Assume that

Zn(t)≤1

2e⁽^a/²⁻¹^/²⁾^t, Zn(t)−Zn−1(t)≤

1 2

_n−1

e^(a/2⁻^1/2)t, fort≥t0.

(2.13)

Then, fort≥t0, Z_n+1(t)≤

_∞

t

τ−t 1

5Le⁽^a⁻¹⁾^τ+1

2e⁻⁽⁽^a⁺¹⁾^/²⁾^τ+ 2Le⁽³^/²⁾⁽^a⁻¹⁾^τ

dτ

≤1

2e⁽^a/²⁻¹^/²⁾^t.

(2.14)

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Using the mean value theorem, we can also get Zn+1(t)−Zn(t)≤

_∞

t

τ−t 1

e⁻âτ+ 2Le⁻âτ+ 3Le⁽â⁻¹⁾^τZn(τ)−Zn−1(τ)dτ

≤ 1

2

n−1_∞

t

τ−t 1

e⁻⁽⁽â⁺¹⁾^/²⁾^τ+ 2Le⁻⁽⁽â⁺¹⁾^/²⁾^τ+ 3Le⁽³^/²⁾⁽â⁻¹⁾^τdτ

≤ 1

2 _n

e^(a/2⁻^1/2)t.

(2.15) Therefore, the sequence{Zn(t)}converges uniformly toZ(t) and

Z(t)≤1

2e^(a/2⁻^1/2)t ∀t≥t0. (2.16)

Hence, we have proved that(PVI)has a solution satisfying y=b

2e^at+A+Oe⁽⁽³^/²⁾^a⁻¹^/²⁾^t=b

2x^a+A+Ox⁽³^/²⁾^a⁻¹^/², y=ab

2 x^a⁻¹+Ox⁽³^/²⁾^a⁻³^/²), asx−→ ∞, 0< a <1 4.

(2.17)

Applying the transformationy=x/zto(PVI)and using the result we have obtained, we can get the asymptotics for 3/4< a <1 and finish the proof ofTheorem 1.2.

3. Proof ofTheorem 1.1

We can easily prove Theorems1.2and1.3using the successive approximation method since the corresponding homogeneous equation is easy to solve. When y0takes the ex- pression in (1.2), the corresponding homogeneous equation to (2.4) becomes one of the famous Hill equations [3] whose solutions are very hard to analyze. Thus, we have dif- ficulties to apply the successive approximation method to this case. Fortunately, we can manage to manipulate(PVI)a little bit and apply a method used by Hastings and McLeod [8] to it.

We first prove the first part of the theorem. Suppose that y(x1)=0 for somex1> x0. Sincey(x) is analytic nearx1, we have the expansion

y(x)=cx−x1

_n

+Ox−x1

_n+1

, (3.1)

wherec=0 andn >0. Substituting (3.1) into(PVI), we get the equation cn(n−1)x−x1

_n₋2

+Ox−x1

_n₋1

=c

2n²x−x1

_n₋2

+Ox−x1

_n₋1

+ β

cx1−1²

x−x1−n+Ox−x1−n+1 .

(3.2)

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Thus, we haven=1 andc/2 +β/c(x1−1)²=0. This is impossible whenβ >0 and therefore,y(x)>0 for allx > x0. Similarly, we can prove thaty(x)<1 for allx > x0whenγ <0.

This result enables us to assume thatyis a solution between 0 and 1 in this section. We first apply the transformation

t=lnx (3.3)

to(PVI)and obtain d²y

dt² ⁼ 1 2

1 y+ 1

y−1 dy

dt 2

+β(1−y)

y + γy

1−y +e⁻^t

1 2ye⁻^t−1

dy dt

2

− y−1 1−e⁻^tye⁻^t−1

dy

dt +αy(y−1)ye⁻^t−1 1−e⁻^t² +β(y−1)y−2 +e⁻^t

y1−e⁻^t² + γy

1−e⁻^t+ δ y(y−1) 1−e⁻^tye⁻^t−1

.

(3.4) Since 0< y <1, we can rewrite (3.4) into

d dt

y(1−y)⁻¹^/²dy dt

·2y⁻¹^/²(1−y)⁻¹^/²dy dt

= 2 y(1−y)

β(1−y)

y + γy

1−y dy

dt + 2e⁻^t

y(1−y)Q(y,t)dy dt.

(3.5)

Integrating both sides of (3.5), we get 1

y(1−y) dy

dt 2

+2β y ⁻

2γ

1−y⁼C+ 1

y(1−y)Oe⁻^t. (3.6) Sinceβ >0,γ <0, and 2β/y−2γ/(1−y) dominates (1/y(1−y))O(e⁻^t) whentis large, C=a²>0, 1/y, 1/(1−y), anddy/dt are all bounded astgoes to infinity. Multiplying both sides of (3.6) byy(1−y) and lettingy=z+rwhereris a constant to be determined later, we get

dz dt

2

+a²z²+2ra²−2β−2γ−a²z+ 2β−

2β+ 2γ+a²r+a²r²=Oe⁻^t. (3.7) We selectr=1/2 + (β+γ)/a²=A, then

dz dt

2

+a²z²=a²B²+Oe⁻^t. (3.8) To solve (3.8), we let

z(t)=ρ(t) cosφ(t),

z(t)=aρ(t) sinφ(t). (3.9)

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We are using two functionsρ(t) andφ(t) inz(t) andz(t) to describe the relationship of the functionz(t) and its derivative. Because of this relationship, one ofρ(t) andφ(t) should be depending on another as the product rule of derivatives prescribes. Substituting (3.9) into (3.8), one first gets

ρ²(t)=B²+Oe⁻^t. (3.10)

Following (3.8), we may also get dφ

dt ^{= −}a+azz+a²z²

ρ²(t) ^{= −}a+Oe⁻^t. (3.11) Integrating (3.11) and combining the result with (3.9) and (3.10), one gets

z(t)=

B+Oe⁻^tcosat+b+Oe⁻^t,

z(t)=aB+Oe⁻^tsinat+b+Oe⁻^t, (3.12) and finishes the proof ofTheorem 1.1.

Acknowledgment

The authors are deeply grateful to the referee for his/her valuable comments and sugges- tions.

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Huizeng Qin: Department of Mathematics and Information Science, Shandong University of Technology, ZiBo, Shandong 255049, China

E-mail address:qinhz [email protected]

Youmin Lu: Department of Mathematics and Computer Science, Bloomsburg University, Bloomsburg, PA 17815, USA

E-mail address:[email protected]