Value distribution of the fifth Painlevé transcendents in sectorial domains

Transcription

1 J. Math. Anal. Appl. 33 (7) Value distribution of the fifth Painlevé transcendents in sectorial domains Yoshikatsu Sasaki Graduate School of Mathematics, Kyushu University, 6-1-1, Hakozaki, Higashiku, Fukuoka-city , Japan Received 9 November 5 Available online 7 September 6 Submitted by A.V. Isaev Abstract This article is concerned with a value distribution of the fifth Painlevé transcendents in sectorial domains around a fixed singular point. We show that the cardinality of the 1-points of a fifth Painlevé transcendent in a sector has an asymptotic growth of finite order, thereby giving an improvement of the known estimates. 6 Elsevier Inc. All rights reserved. Keywords: Painlevé transcendents; Value distribution; Sector. Introduction Finding a new special function is one of the main incentives to study differential equations in a complex domain. In fact this led P. Painlevé and B. Gambier to discover six kinds of differential equations, which later became known as Painlevé equations, at the beginning of the last century. For many years since then, a general solution of each Painlevé equation has been believed to be irreducible, that is, it would be never reducible to any classical or known special function and hence it would define a truly new special function. With this belief, such a general solution has been called a Painlevé transcendent. Eventually, in the last two decades, several irreducibility results have been obtained on rigorous mathematical foundations. Since a Painlevé transcendent is thought of as a new special function, it should be quite natural to explore its function-theoretical properties. Because of difficulty, however, this direction of * Fax: address: sasaki@math.kyushu-u.ac.jp. -47X/$ see front matter 6 Elsevier Inc. All rights reserved. doi:1.116/j.jmaa

2 818 Y. Sasaki / J. Math. Anal. Appl. 33 (7) research has caught little attention until quite recently, although its importance has certainly been recognized. Meanwhile, the main stream of research has been toward geometrical or algebraic aspects of the subject, such as spaces of initial conditions, irreducibility, special solutions, as well as their generalizations to several variables. Today there are growing interests in the functiontheoretical investigations of Painlevé transcendents, with already remarkable developments as may be seen in [4] and in the references therein. Among them, we should mention recent results of Shimomura [5,6] concerning value distributions and growth estimates of those transcendents. In this article, focusing our attention on the fifth Painlevé transcendents, we shall substantially improve some of his results, basically adopting his approach but working in sectorial domains around a fixed singular point. Recall that the fifth Painlevé equation P V is a second-order nonlinear differential equation d y dx = ( 1 y + 1 y 1 )( dy dx ) 1 dy (y 1) + x dx x ( αy + β y ) + γy δy(y + 1) + x y 1, (1) depending on complex parameters (α,β,γ,δ) C 4. We remark that if δ = then P V is reduced to the third Painlevé equation P III. We shall make a generic assumption on (α,β,γ,δ) under which P V is neither reduced to P III nor to any equation solvable by classical methods. Namely that we treat the case (α, β) (, ) and (γ, δ) (, ). If(α, β) = (, ) then P V is reduced to P III, and if (γ, δ) = (, ) then P V is solvable by quadrature [1,7]. The aim of this article is to discuss a value distribution of a solution to (1) in a sector S(,r,R) = { x arg x <, r< x <R }, () with a sectorial angle <<π, fixed throughout the rest of this paper. Let us formulate our problem more precisely. Given a solution y = y(x) to (1), a point x is said to be a 1-point of y if y(x) = 1. We are interested in the asymptotic distribution of the 1-points of y in the above sector as r orasr. If we define N(y,,r,R)= { x S(,r,R) y(x) = 1 }, where the number is counted with multiplicities, then our main theorem is stated as follows. Theorem. If (α, β) (, ) and (γ, δ) (, ), then there exists a positive constant C, independent of (α,β,γ,δ), such that for any solution y = y(x) to (1) one has N(y,,r,R)= O ( r C) as r, N(y,,r,R)= O ( R C) as R. A similar result can be obtained for a-points for any nonzero complex number a. To prove the theorem, we have to improve the methods used in [5,6]. The theorem above gives more precise estimation than [5], instead of the view point becoming local at a fixed singularity. It is known that every solution w(z) of the modified fifth Painlevé equation, which is derived from P V by the change of variable x = e z, is meromorphic on C []. In [5], by considering the modified Painlevé equation P V : d w dz = ( 1 w + 1 w 1 )( dw dz ) + (w 1) ( αw + β w derived from P V, Shimomura established the following result: ) + γ e z + δez w(w + 1), w 1

3 Y. Sasaki / J. Math. Anal. Appl. 33 (7) Proposition. [5] For each fixed (α,β,γ,δ) C 4, denote by w(z) an arbitrary solution of P V. Then T(r,w)= O(e Λr ), where Λ = Λ α,β,γ,δ is a positive number independent of w(z). Here T(r,f)= m(r, f ) + N(r,f) is the characteristic function used in Nevanlinna theory (see [3]): m(r, f ) := 1 π N(r,f):= r π log + f ( re ıθ ) dθ, { n(t, f ) n(,f) }dt t log + x := max{log x,}, + n(,f)log r, n(r, f ) denoting the number of poles in the open disk z <r; each counted according to its multiplicity. In [5], a similar result is given for the modified third Painlevé equation. Though P V has two fixed singularities x =,, we concentrate our attention only around x = in case δ in the following part of this article. The argument in case δ = or around x = is similar. For brevity, we denote S R := {x arg x <, R x }, so sector S(,r,R) is written as S(,r,R) = interior of S R \ Sr.Ifσ SR and y(σ) = 1, we may use a local parameter t defined by x = σ + t and express the solution of P V in the form y(σ + t)= 1 + ( δ) 1/ t + j c j t j. Put y(x) = 1 + Y σ (t) and dy σ (t)/dt = ( δ) 1/ {1 + h σ (t)}, and define η := sup { η Y σ (t) b x P is valid for t <η<1/3 }. Note that <η 1/3 because y(σ) = 1 implies Y σ () =. Then we can establish the following lemma: Key lemma. Suppose Y σ (t) b x P for t <η, with b a sufficiently small positive number independent of σ and P>1 a positive number independent of σ. Then h σ (t) < 1/ for t <η. We will give the proof of this lemma in the next section. Now we show Theorem under Key lemma. Proof of Theorem. For t satisfying t <η, which indicates Y σ (t) b x P, Key lemma gives us dy σ (t) dt = ( δ) 1/{ 1 + h σ (t) }, hσ (t) < 1. Integrating the above, we have the following estimation: Y σ (t) ( δ) 1/ t t δ 1/ h σ (t) dt 1 δ 1/ t, which indicates 1 4 δ 1/ t Yσ (t) 7 4 δ 1/ t.

4 8 Y. Sasaki / J. Math. Anal. Appl. 33 (7) Provided that σ >M:= L + (3b/ δ 1/ ) 1/P with a sufficiently large number L, we have η κ(σ) := b σ P / δ 1/, κ(σ) < 1/3. Because for sufficiently large σ, σ > L + (3b/ δ 1/ ) 1/P implies σ > L and 1/3 >b σ P / δ 1/. Suppose η < b σ P / δ 1/, then Y σ (t) 7 4 δ 1/ t 7 4 δ 1/ η 7 8 b σ P for t <η < 1/3, which contradicts the definition of η.sowehaveη b σ P / δ 1/ = κ(σ). Hence, for σ >M and t <κ(σ),wehave 1 4 δ 1/ t Yσ (t) 7 4 δ 1/ t, which implies y(σ) 1 > for< x σ <κ(σ). Therefore, the number of 1-points is estimated as follows: { σ y(σ) = 1, σ S r \ } Area(S SL r \ SL min σ S r \S L π(κ(σ)/) 1. Proof of Key lemma 1.1. Construction of a neighborhood Area(Sr ) π(κ(r)/) = ) r = O πb 3 δ r P ( r P + ). Lemma 1.1. For each (α,β,γ,δ) C 4 satisfying δ, there exists a quartet of positive numbers T, μ 1, Δ and A, each of them independent of y(x), with the properties: for x = a satisfying a >T,if y(a) μ Δ, then (i) y(x) μ Δ on the circle x a =ɛ a ; (ii) y(x) 1 in the disk x a ɛ a.here,ɛ a > satisfies ɛ a A, ɛ 1 a A ( 1 + y (a) ). (3) To prove this lemma, first we show the following: Lemma 1.. Let u(t) beasolutionof ü = g 1 (t, u) u g (t, u) u g (t, u) ( = d ), (4) dt around t =. Suppose g j (t, u) (j =, 1, ) is analytic in D ={(t, u) C t < 1, u < R }, <R < 1. Suppose g (t, u) < 1/, g 1 (t, u) <K, g (t, u) <K in D, where K is some positive number. Put θ := min{4 1 R 1/,(K) 1/,(K) 1 }.If u() θ /8, then u(t) 15θ in the disk t <ρ and u(t) θ /4 on the circle t =3ρ /4, where { 4θ if u() θ, ρ = (4/3)θ u() if u() >θ. Proof. Case u() θ. Define η := sup { η<1: u(t) 15θ, } u(t) 6θ for t <η.

5 Y. Sasaki / J. Math. Anal. Appl. 33 (7) Suppose η < 4θ. Note that η < 4θ R 1/ < 1. Then, for t <η, (t, u(t)) D is valid, satisfying g (t, u) < 1/, g 1 (t, u) <K, g (t, u) <K, which indicates, with v(t) = u(t) t / u()t u(), that t v(t) = ds { ( ) g 1 s,u(s) u(s) ( ) ( )} g s,u(s) u(s) + g s,u(s) t [ ds K { (6θ) + 6θ } + 1 ] 1 ( ) t < 4 ( ) θ, (5) v(t) 1 ( ) t 4 1 ( ) θ. (6) Hence, provided that t <η, u(t) u() + t + v(t) ( ( ) 4/ ) θ 5.86θ, u(t) u() + u() t + 1 t + v(t) ( ( ) ) 4 / θ θ, which contradicts the definition of η. Therefore η 4θ. Then as far as t < 4θ ( η ), (5) (6) are valid. Moreover, u(t) t u() u() t v(t) ( ( ) ) 3 / θ =.475θ 1 4 θ on the circle t =3θ. Case u() =κθ, κ>1. Define η 1 := sup { η<1: u(t) 15θ, u(t) 6κθ for t <η }. Suppose η 1 < 4θ/κ. Note that η 1 < 4θ R 1/ < 1. Then, for t <η 1, (t, u(t)) D is valid, satisfying g (t, u) < 1/, g 1 (t, u) <K, g (t, u) <K, which indicates t v(t) = ds { ( ) g 1 s,u(s) u(s) ( ) ( )} g s,u(s) u(s) + g s,u(s) t [ ds K { (6κθ) + 6κθ } + 1 ] κ ( ) t < 4κ ( ) θ, (7)

6 8 Y. Sasaki / J. Math. Anal. Appl. 33 (7) v(t) κ 1 ( ) t 4 1 ( ) θ. (8) Hence, provided that t <η 1, u(t) u() + t + v(t) ( ( ) 4/ ) κθ 5.86κθ, u(t) u() + u() t + 1 t + v(t) ( ( ) ) 4 / θ θ, which contradicts the definition of η 1. Therefore η 1 4θ/κ. Then as far as t < 4θ/κ( η 1 ), (7) (8) are valid. Moreover, u(t) u() t u() t v(t) ( ( ) ) 1 / θ =.675θ 1 4 θ on the circle t =θ/κ. Next, we shall prove Lemma 1.1 under the condition δ. Proof. Consider the sector S L and x >T, with sufficiently large T.OnP V, putting u(x) = y(x), we have d ( u dx = 1 (u + ) + 1 u + 1 γ(u+ ) + + x Denote ( = dx d ), we obtain )( ) du 1 ( du (u + 1) + dx x dx x α(u + ) + δ(u + )(u + 3). (u + 1) u = G 1 (u)u x 1 u + 6δ ( 1 + H(x,u) ), H(x,u)= uh (u) + h 1(u) x h (u) := u 1 6(u + 1), h 1 (u) := (6δ) 1 γ(u+ ), + h (u) x, h (u) := (6δ) 1 (u + 1) { α(u + ) + β/(u + ) }. β ) (u + ) Note that G 1 (u), h j (u) (j =, 1, ) is bounded for u < 1/. Change the scaling by x = a + (6δ) 1/ t, then ü = G 1 (u) u G (t) u G (x, u),

7 Y. Sasaki / J. Math. Anal. Appl. 33 (7) where G (t) = (6δ) 1/ a + (6δ) 1/ t, G (t) = H ( a + (6δ) 1/ t,u ). Taking T sufficiently large, h 1 (u)/x and h (u)/x are sufficiently small, and taking R sufficiently small, uh (u) is sufficiently small for u R. Then apply Lemma 1.; concretely, for T, take the numbers as μ =, Δ = θ /8, K = max u =R { G 1, G }, and { 3 δ ɛ a = 1/ θ if y (a) 3 δ 1/ θ, θ / y (a) if y (a) > 3 δ 1/ θ, which supplies the desired conclusion. Also note that dy dx (a) = 1 du (6δ) 1/ dt () u() = (6δ) 1/ y (a), where θ = min{4 1 R 1/, 1 K 1/ } 1/3. For the case δ =,γ, we can verify the lemma in a similar way. 1.. Construction of a path We construct a path which will be used in estimating integrals. In what follows, T, μ, Δ and A denote the constants given in Lemma 1.1. Lemma 1.3. Put S L := {x arg x <, L x } with L := min{t,a / sin }, <<π. Let σ S L be an arbitrary point satisfying y(σ) = 1, σ > L. Then there exists a path Ɣ(σ) with the properties: (i) Ɣ(σ) S L starts from a (σ ) S L {x x =L } and ends at σ ; (ii) the length of Ɣ(σ) does not exceed (π + 1) σ ; (iii) y(x) μ >Δalong Ɣ(σ). Remark. Even if the sectorial angle is small, taking sufficiently large L, a circle with radius A crosses at most only one of two lines; arg x = and arg x =. Proof. Suppose that σ > L. Put s = L e 1argσ. Draw the segment I =[s,σ]. If (iii) is satisfied along I S L =[s,σ], then Ɣ(σ) =[s,σ] is a desired path, where a = s. Now suppose that (iii) is not satisfied, namely that there exists a point a 1 [s,σ] satisfying the following conditions: y(a 1 ) μ =Δ; and y(x) μ >Δfor x [s,a 1 ]\{a 1 }. Applying Lemma 1.1, we have a circle x a 1 =ɛ a1 on which y(x) μ Δ. s may be inside of the circle, or outside of it. Note that σ lies outside of the circle. Denote a 1,a+ 1 two points satisfying { x a 1 =ɛ a1 } {x = re 1ϑ r>, ϑ= arg σ }={a 1,a+ 1 } and a 1 σ > a+ 1 σ. Take a semi-circle c 1 such that { x x a1 =ɛ a1, arg ( (x a 1 )/σ ) [( 1 sgn(arg σ) ) π/, ( 1 sgn(arg σ) ) π/ ]},

8 84 Y. Sasaki / J. Math. Anal. Appl. 33 (7) where { 1 for x, sgn x = 1 for x<. a 1,a+ 1 are two end points of c 1 and the segment [a 1,a+ 1 ] is a diameter. If s is outside of the circle or on the circle, put a (1) = s.ifs is inside of the circle, replace the segment [a 1,a+ 1 ] [s,σ] by the arc c 1 S L. Because ɛ a 1 A L, c 1 crosses the arc S L {x x =L }. Put the crossing point a (1). Now we have the path Ɣ 1 = ( [s,σ] { x }) ( x a 1 >ɛ a1 c1 S L ) ( σ) starting from a (1) S L. Then on the part of Ɣ 1 from a (1) to a + 1, the inequality in (iii) is valid, and in particular, y(a + 1 ) μ Δ. If (iii) is satisfied along Ɣ 1, then we get Ɣ(σ) = Ɣ 1. Otherwise, we start from a + 1 and proceed along [a+ 1,σ] Ɣ 1 until we meet a point a satisfying the following: y(a ) μ =Δ; and y(x) μ >Δon [a + 1,a ]\{a }. Then, take the semi-circle c such that { x x a =ɛ a, arg ( (x a )/σ ) [( 1 sgn(arg σ) ) π/, ( 1 sgn(arg σ) ) π/ ]}. Denote a,a+ two points satisfying { x a =ɛ a } {x = re 1ϑ r>, ϑ = arg σ }= {a,a+ } and a σ > a+ σ. a may be on [a+ 1,a ] Ɣ 1 or not. If a does not lie on [a + 1,a ] Ɣ 1, the intersection of two open disks x a 1 <ɛ a1 and x a <ɛ a is not empty. Consider the path Ɣ = ( Ɣ 1 { x }) ( x a >ɛ a c { x } L x a 1 >ɛ a1 S ), which starts from a () S L and ends at σ. At least on the part of Ɣ from a () to a +,the inequality in (iii) is valid. After repeating this procedure finitely many times, we get a path Ɣ(σ) along which (i) and (iii) are fulfilled. Because, if this procedure must be repeated infinitely many times, there exists a sequence {a n } n=1 I satisfying n=1 ɛ an σ s and y(a n ) Δ + μ, hence, by (3), we can make a choice of a subsequence {a nj } j=1 satisfying a n j a I, y(a nj ) y, y (a nj ) as j. Then y(a ) = y, y (a ) =, which is a contradiction. Let c j (j = 1,,...,l)be all the semi-circles with center at a j which are used in construction of Ɣ(σ). Then the radii ɛ aj satisfy l j=1 ɛ aj σ s. So, the length of Ɣ(σ) does not exceed σ s +π l j=1 ɛ aj (π + 1) σ s (π + 1) σ, which implies (ii). Therefore, we obtain Lemma Auxiliary function Definition. For a solution of P V, we define an auxiliary function as follows: Ψ(μ,x)= where μ, 1,. x y (x) y(x)(y(x) 1) (1 μ)xy (x) (y(x) 1)(y(x) μ) αy(x) + β y(x) + γx y(x) 1 + δx y(x) (y(x) 1), (9)

9 Y. Sasaki / J. Math. Anal. Appl. 33 (7) Ψ(μ,x)satisfies a linear ordinary differential equation, i.e., dψ(x) dx P(x):= P(x)Ψ(x) = Q(x), (1) (1 μ)(y(x) 1)(y(x) + μ) x(y(x) μ), Q(x) := (1 μ) (y(x) + μ)y (x) (y(x) μ) 3 + Θ(x,y(x)) x(y(x) μ), Θ ( x,y(x) ) := 4(1 μ) ( y(x) 1 )( αμy(x) β ) γx ( (1 μ)y(x) + μ ) + 4δμx y(x). Lemma 1.4. If y(x) μ on a path Ɣ (x) starting from x and ending at x, then (1) is satisfied and Ψ(μ,x)is { Ψ(x)= Ψ(μ,x )E(x) (1 μ) y(x) (y(x) μ) + y(x } )E(x) (y(x ) μ) + E(x) E(x) 1 (Ξ 1 Ξ ) dx, where [ E(x) := exp (1 μ) Ɣ Ɣ (x) (y(x) 1)(y(x) + μ) (y(x) μ) ] dx, x (y(x) 1)(αμy(x) β) Ξ 1 := 4(1 μ) x(y(x) μ) ((1 μ)y(x) + μ) y(x) γ (y(x) μ) + 4δμx (y(x) μ), 3 y(x)(y(x) 1)(y(x) + μ) Ξ := (1 μ) x(y(x) μ) Estimation of auxiliary function If there is any point such that y(x) = μ,, 1,, retake L a little bit larger so that y(x) μ,, 1, for x =L. Lemma 1.5. For 1-point x = σ satisfying y(σ) = 1 and σ > L, Ψ(μ,x) is estimated as Ψ(μ,x) K x C in U(σ)={x x σ <η(σ)} with η(σ) := sup { η y(x) 1 < μ 1 / for x σ <η<1 }, C > independent of σ, y(x) and the parameters of P V, K independent of σ. Proof. First, we give an estimation of the exponential part: [ ] dx (y(x) 1)(y(x) + μ) E(x) = exp (1 μ) x (y(x) μ) Ɣ(σ)

10 86 Y. Sasaki / J. Math. Anal. Appl. 33 (7) and 1 μ (y(x) 1)(y(x) + μ) (y(x) μ) B on Ɣ(σ) of Lemma 1.3. Hence E(x) ±1 ] dx exp [B K 1 x C 1. x Ɣ(σ) Therefore, for the auxiliary function { Ψ(μ,x)= Ψ(μ,x )E(x) (1 μ) y(x) (y(x) μ) + y(x } )E(x) (y(x ) μ) + E(x) dxe(x) 1 (Ξ 1 Ξ ), Ξ 1 = 4(1 μ) Ɣ (y(x) 1)(αμy(x) β) x(y(x) μ) γ((1 μ)y(x) + μ) (y(x) μ) + 4δμxy(x) (y(x) μ), 3 y(x)(y(x) 1)(y(x) + μ) Ξ = (1 μ) x(y(x) μ) 4, we have an estimation Ψ(μ,x) K x C. Here note that y μ > Δ = θ /4, and then y 1 y + μ μ + μ 1 μ μ 1 y μ < y μ y μ < 1 + Δ ( μ + μ 1 ) + 4Δ μ μ 1. Therefore B only depends on θ. On the other hand, θ depends on K which is independent of the parameters of P V. This indicates that the constant C is independent of the parameters of P V Completion of the proof of Key lemma For each solution of P V satisfying δ, take a 1-point σ S L. Put y(x) 1 =: Y(x). Then, as the results of the above steps show, we have Ψ(μ,x) K x C and 1 μ + Y(x) in U(σ). Lemma 1.6. Putting y(x) 1 = Y(x)=: Y σ (t) with x σ = t, we obtain dy σ (t) =±( δ) 1/( 1 + h ± σ dt (t)), where h ± σ (t) := ( 1 + Y σ (t) ){ ( 1 + Fσ (t) ) 1/ ( δ) 1/ } (1 μ)y σ (t)(1 Y σ (t)) ± 1, 1 + Y σ (t) 1 μ + Y σ (t) F σ (t) := j 1σ (t) σ + t + j σ (t) (σ + t),

11 Y. Sasaki / J. Math. Anal. Appl. 33 (7) j 1σ (t) := γy σ (t) δ(1 + Y σ (t)), j σ (t) := αy σ (t) + βy σ (t) δ δ(1 + Y σ (t)) (1 μ) Y σ (t) δ(1 μ + Y σ (t)) Y σ (t) Ψ σ (μ, t) δ(1 + Y σ (t)), Ψ σ (μ, t) := Ψ(μ,x). Proof. See the definition of the auxiliary function (9) as the quadratic equation of y : Ay By + C = with A = x y(x){y(x) 1}, B = (1 μ)x {y(x) 1}{y(x) μ}, C = αy(x) + β y(x) + γx y(x) 1 + δx y(x) {y(x) 1} Ψ(μ,x). Put y(x) 1 = Y(x)and solve this quadratic equation, we have Y = B A ± (B 4AC) 1/ A with B A B 4AC (A) = δ(1 + Y) γy(1 + Y) + 1 [ x x αy (1 + Y) βy (1 μ)y (1 Y) =, (1 μ + Y) + (1 μ) Y (1 + Y) (1 μ + Y) + Y (1 + Y)Ψ Observing the above, we find that δ 1/ dominates the right-hand side of (11), which indicates Y δ 1/, as far as x is sufficiently large, and Y is sufficiently small. After a little calculation, we have dy(x) =±( δ) 1/( 1 + h ± (x) ) dx with choice of branch in such a way that (1 + F(x)) 1/ tends to 1 as F(x) and h ± (x) := ( 1 + Y(x) ){ ( ) 1/ ( δ) 1/ } B 1 + F(x) ± 1, 1 + Y(x) A F(x):= j 1(x) + j (x) x x, j 1(x) := γy(x) δ(1 + Y(x)), j (x) := αy(x) δ + βy(x) δ(1 + Y(x)) (1 μ) Y(x) δ(1 μ + Y(x)) Y(x) Ψ(μ,x) δ(1 + Y(x)). So, taking a local parameter t = x σ, we obtain the conclusion of Lemma 1.6: dy σ (t) =±( δ) 1/( 1 + h ± σ dt (t)) with putting Y σ (t) := Y(σ + t), h ± σ (t) := h± (σ + t), F σ (t) := F(σ + t), j 1σ (t) := j 1 (σ + t), j σ (t) := j (σ + t), Ψ σ (μ, t) := Ψ(μ,σ + t). ]. (11)

12 88 Y. Sasaki / J. Math. Anal. Appl. 33 (7) Under the results of the above, Lemma 1.7. Suppose Y σ (t) b x P for t <η with sufficiently small positive number b independent of σ, putting P := C /. Then h ± σ (t) < 1/ for t <η. Proof. Note that σ > L, P = C /. Under the assumption, as far as t <η,wehave Y σ (t) b x P bl P, 1 μ + Y σ (t) 1 μ Y σ (t) 1 μ bl P, 1 + Y σ (t) 1 bl P, Ψ σ (μ, t) K x C. Then terms of F σ (t) are estimated as γy σ (t) δ(1 + Y σ (t)) γ bl P δ 1 bl P, αy σ (t) δ α δ b L P, βy σ (t) δ(1 + Y σ (t)) β/δ b L P (1 bl P ), Y σ (t) Ψ σ (μ, t) δ(1 + Y σ (t)) δ 1 b K 1 bl P, (1 μ) Y σ (t) δ(1 μ + Y σ (t)) (1 μ) /δ b L P ( 1 μ bl P ). Hence we can take b sufficiently small positive number such that F σ (t) is sufficiently small, and how to take b is independent of σ because there is not σ on the right-hand side of each inequality. Therefore, noting h ± (x) = ( 1 + Y(x) ){ ( ) 1/ ( δ) 1/ } B 1 + F(x) ± 1, 1 + Y(x) A we can make h ± σ (t) < 1/ by taking a sufficiently small b. Acknowledgments The author thanks Professor S. Shimomura for helpful discussions. He is also grateful to Professor K. Okamoto for his constant encouragement. References [1] R. Garnier, Contribution a l étude des solutions de l équation (V) de Painlevé, J. Math. Pures Appl. 46 (1967) [] A. Hinkkanen, I. Laine, Solutions of a modified fifth Painlevé equation are meromorphic, Rep. Univ. Jyväskylä 83 (1) [3] I. Laine, Nevanlinna Theory and Complex Differential Equations, de Gruyter, Berlin, [4] I. Laine, V.I. Gromak, S. Shimomura, Nevanlinna Theory and Painlevé Differential Equations, de Gruyter, Berlin,. [5] S. Shimomura, Growth of modified Painlevé transcendents of the fifth and the third kind, Forum Math. 16 (4) [6] S. Shimomura, Growth of the first, the second and the fourth Painlevé transcendents, Math. Proc. Cambridge Philos. Soc. 134 (3) [7] S. Shimomura, Value distribution of Painlevé transcendents of the third kind, Complex Var. 4 (1999) 51 6.