ON THE ASYMPTOTICS OF THE REAL SOLUTIONS TO THE GENERAL SIXTH PAINLEVÉ EQUATION

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 complicated 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


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 complicated 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][5][6][7][9][10][11][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 dy dx 2 where α, β, γ, and δ are parameters.Heuristically, if y is a "small" solution of (PVI), the following equation truncated from (PVI) would be its "major" part as x → +∞: . (1.1) Letting t = lnx and applying elementary techniques to (1.1), one may solve it to get three different solutions: where where 2 ; and 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][5][6][7][9][10][11][12] have obtained the corresponding asymptotics as x → 0 that can be used to obtain (1.2), (1.3), and (1.4) by applying the wellknown 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.Let β > 0 and γ < 0. If x 0 > 1, 0 < y 0 < 1, and y is a solution of (PVI) with y(x 0 ) = y 0 , then 0 < y < 1 for all x > x 0 and it satisfies, as x → +∞, where It is clear that the parameters need to satisfy the condition (1/2 + (β + γ)/a 2 ) 2 ≥ 2β/a 2 .The complex form of the asymptotics as x → 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: where 0 < a < 1/4 or 3/4 < a < 1, and H. Qin and Y. Lu 3 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][5][6][9][10][11][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.If γ + β = 0, then (PVI) has a group of solutions with the following asymptotics: Various forms of this result occur in the literature.For example, in [7] Guzzetti has the following result for x → 0: ( The coefficients θ 0 and θ x come from the isomonodromy deformation theory and r is a free complex parameter.Applying the symmetry transformations y(x) = xz(t) and x = t −1 to this result, our result in Theorem 1.3 can be obtained.
Theorem 1.4.Equation (PVI) has solutions with the following asymptotics: where , and

Proof of Theorem 1.1
We can easily prove Theorems 1.2 and 1.3 using the successive approximation method since the corresponding homogeneous equation is easy to solve.When y 0 takes the expression 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 difficulties 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(x 1 ) = 0 for some x 1 > x 0 .Since y(x) is analytic near x 1 , we have the expansion where c = 0 and n > 0. Substituting (3.1) into (PVI), we get the equation Thus, we have n = 1 and c/2 + β/c(x 1 − 1) 2 = 0.This is impossible when β > 0 and therefore, y(x) > 0 for all x > x 0 .Similarly, we can prove that y(x) < 1 for all x > x 0 when γ < 0. This result enables us to assume that y is a solution between 0 and 1 in this section.We first apply the transformation to (PVI) and obtain Integrating both sides of (3.5), we get and finishes the proof of Theorem 1.1.