An Asymptotic Reduction of a Painleve' VI equation to a Painleve' III (January 2011)

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by  Davide Guzzetti

An Asymptotic Reduction of
a Painlev´e VI equation to a Painlev´e III
Davide Guzzetti
1
1
Abstract: When the independent variable is close to a critical point, it is shown that PVI can be
1
asymptotically reduced to PIII. In this way, it is possible to compute the leading term of the critical
0
2 behaviors of PVI transcendents starting from the behaviors of PIII transcendents.

n
a
J
1 Introduction
5
2
As it is well known, the first five Painlev´e equations PI, PII, PII, PIV, PV can be obtained from the

] sixth PVI, by a step-by-step degeneration process [5].
A
Here we present a different reduction of PVI to PIII. When the independent variable is close to a
C
critical point, we show that it is possible to reduce PVI, with α=(2µ 1)2/2 C, β =γ =0, δ =1/2,
. − ∈
h namely:
t
a
1 1 1 1 1 1 1
m y = + + (y )2 + + y +
ss s s
2 y y 1 y s − s s 1 y s
[ (cid:20) − − (cid:21) (cid:20) − − (cid:21)

1 + y(y−1)(y−s) (2µ 1)2+ s(s−1) , (1)
v 2s2(s 1)2 − (y s)2
5 − (cid:20) − (cid:21)
0 to the following PIII, with α=β =0, γ = δ =1, namely:
−
7
1 1 1
4 yˆ = (yˆ )2 yˆ +yˆ3 . (2)
. θθ y θ − θ θ − yˆ
1
0
The notationy standsfordy/d . We considerhere,fordefiniteness, the caseofthe criticalpoints=0.
1 ∗ ∗
1 Thereductionwillbedoneaccordinglyfors 0(ands θ2). Theconvergenceisintendedforbounded
→ ∼
: arg(s). This is an asymptotic reduction. We show that, remarkably, it allows to reproduce the correct
v
leading term of the critical behavior of PVI transcendents, which are classified in [4]. The equation (1)
i
X is important in the theory of semi-simple Frobenius manifolds of dimension 3 (see [1] [2]).
r
a
1.1 A 3 3 isomonodromy representation
×
Usually, PVI is regarded as the isomonodromy deformation equation for a 2 2 fuchsian system with
×
four singularities [8]. Here we regard(1) as the isomonodromydeformation condition of the 3 3 linear
×
system (L1) below, having a fuchsian singularity at z = 0 and an irregular singularity of rank one at
z = . This system for the general PVI is described in [7]. For the particular PVI we consider here,
∞
related to Frobenius manifolds, (L1) is introduced and studied in [1] [2], where the following 3 3 Lax
×
Pair is given:
dY V ∂Y
(L1): = U + Y, (L2): = zE +V Y, i=1,2,3.
i i
dz z ∂u
(cid:20) (cid:21) i
(cid:2) (cid:3)
11)International School ofAdvancedStudySISSA/ISAS, Trieste,Italy.
2)KoreaInstitute ofAdvancedStudyKIAS,Seoul,SouthKorea.
E-mail: davide [email protected]
1
The 3 3 matrix coefficients are:
×
U =diag(u ,u ,u ).
1 2 3
V =V(u ,u ,u ), VT = V, V has diagonal form = diag(µ,0, µ).
1 2 3
− −
δ δ
ki kj
(E ) =1, (E ) =0, i,j =k. (V ) = − V .
k kk k ij k ij ij
6 u u
i j
−
A fundamental solution of (L1) at z = 0 has representation Y(z) = ∞ φ (u)zp zdiag(µ,0,−µ)zΛ,
p=0 p
where Λij = 0 if µi−µj 6= n > 0, n ∈ Z. The necessary and sufficienthPcondition for tihe dependence of
the system (L1) on (u ,u ,u ) to be isomonodromic is [8]:
1 2 3
∂V ∂φ
0
=[V ,V], =V φ .
i i 0
∂u ∂u
i i
The equations ∂V =0 and u ∂V =0, imply that V =V(s), where s= u3−u1. Thus, if we let:
i ∂ui i i∂ui u2−u1
P P
0 Ω Ω
3 2
−
V(s)= Ω 0 Ω , (3)
3 1
 − 
Ω Ω 0
2 1
−
 
the equation for V becomes:
dΩ1 = 1 Ω Ω
ds s 2 3
 ddΩs2 = 1−1s Ω1Ω3 (4)
dΩ3 = 1 Ω Ω
ds s(s−1) 1 2
Ω2+Ω2+Ω2 = µ2
1 2 3 −
This system is equivalent to (1), and the following holds [see [3] for details]:
2 2
sA(s) Ω Ω +µΩ Ω Ω +µΩ
1 2 3 1 2 3
y(s)= − , A(s):= = (5)
1 s(1+A(s)) µ2+Ω2 Ω2+Ω2
− (cid:20) 2 (cid:21) (cid:20) 1 3 (cid:21)
This last equation allows to compute the critical behavior of y(s) if we know that of the Ω (s)’s.
j
2 Asymptotic Reduction of PVI to PIII
As s 0 we do the following approximation to the system (4):
→
dΩ(a) 1 dΩ(a) dΩ(a) 1
1 = Ω(a)Ω(a), 2 = Ω(a)Ω(a), 3 = Ω(a)Ω(a).
ds s 2 3 ds 1 3 ds −s 1 2
with Ω (s) Ω(a)(s) for s 0. The superscript (a) stands for ”asymptotic”. The reduced system has
j ∼ j →
a first integral:
(Ω(a))2+(Ω(a))2 =R2 C
1 3 ∈
This implies that we can introduce the new dependent variable φ(s) as follows:
Ω(a) =Rsinφ, Ω(a) =Rcosφ, R=0
3 1 6
The system becomes:
dφ 1 dΩ(a) R2
= Ω(a), 2 = sin(2φ).
ds −s 2 ds 2
Thus:
d2φ 1dφ R2
+ + sinφ=0
ds2 s ds 2s
2
With another change of variables:
x2
u:=2φ, s= , (namely 2φ(s)=u(2R√s))
4R2
we obtain the following particular form of the Painlev´e III equation:
1
u + u +sin(u)=0. (6)
xx x
x
A last change of variables is necessary:
iu
x=2iθ, yˆ=exp .
2
(cid:18) (cid:19)
This gives the PIII equation in standard form (2):
1 1 1
yˆ = (yˆ )2 yˆ +yˆ3
θθ θ θ
yˆ − θ − yˆ
To summarize, the change of variables is:
R iR dyˆ
Ω(a) = yˆ−1+yˆ , Ω(a) = yˆ−1 yˆ , Ω(a) =is
1 2 3 2 − 2 ds
(cid:16) (cid:17) (cid:16) (cid:17)
R√s
yˆ(s):=yˆ(θ(s))=yˆ .
i
(cid:18) (cid:19)
3 From asymptotic behaviors of PIII to behaviors of PVI
Let s 0, argx < π. In [4] we classified the critical behaviors of the PVI transcendents into a few
→ | |
classes, in the case when there is a one to one correspondence between branches of PVI-transcendents
and points in the space of the associated monodromy data. The critical behaviors are decided by the
value of a complex “exponent” σ such that 0 σ 1. Let ν be a real number. Let also a=0 and C
≤ℜ ≤ 6
be twocomplexnumbers,which,togetherwithσ, playthe roleofconstantsofintegration. Accordingto
[4],theequation(1)hassolutionswithbranchesadmittingthethefollowingcriticalbehaviorsfors 0:
→
1) Small-power-type behaviors (Jimbo [6]) – 4 real parameters:
y(s)=ax1−σ(1+O(sσ +s1−σ)), 0< σ <1.
ℜ
2) Sine-type oscillatory behaviors – 3 real parameters:
y(s)=s sin2(νlns+C)+O(s) , σ =2iν.
3) Inverse sine-type oscillatory b(cid:2)ehaviors – 3 real para(cid:3)meters:
1
y(s)= , σ =1+2iν.
1 4ν2+(2µ−1)2 sin2(νlns+C)+O(s)
− 4ν2
4) Log-type behaviors and Taylor expansions– 2 real parameters:
4 1 1
y(s)= 1+O , σ =1.
−(2µ 1)2(lns+C)2 ln2x
− (cid:20) (cid:18) (cid:19)(cid:21)
y(x)=as+O(s2), σ =0.
The last is a convergent Taylor series, degeneration of a log-behavior which occurs when β = 1 1=0
δ −
in the general PVI. The higher order terms in 1), 2), 3) can be written as convergent expansions in s
and sσ, as it is explained in [4].
3
Wearegoingtoshowthatwecanobtaintheleadingtermoftheabovebehaviors,whenwesubstitute
into the Ω(a)’s the asymptotic expansions of the solutions of (2), according to the formulas:
j
2 2
sA(s) Ω(a)Ω(a)+µΩ(a) Ω(a)Ω(a)+µΩ(a)
y(s)= − , A(s) 1 2 3 = 1 2 3 , s 0. (7)
1 s(1+A(s)) ∼" µ2+(Ω(a))2 # "(Ω(a))2+(Ω(a))2# →
− 2 1 3
and:
R iR dyˆ
Ω(a) = yˆ−1+yˆ , Ω(a) = yˆ−1 yˆ , Ω(a) =is .
1 2 3 2 − 2 ds
(cid:16) (cid:17) (cid:16) (cid:17)
At this point, we need the asymptotic behaviors of the solutions of the PIII equation (2). We can
find them in [9].
i) The first expansion we consider is (already in variable s):
yˆ(s)=Bsσ2 1+ 41 BR 2 (1s1−σσ)2 − 41((1R+Bσ))22s1+σ++∞j+1cjks21(j−σ(j+2−k),
(cid:18) (cid:19) − j=3k=2
XX
 
B, σ C, 1< σ <1, c C
jk
∈ − ℜ ∈
B and σ are integration constants. The c ’s are certain rational functions of B and σ. For symmetry
jk
reasons, we can restrict to the case
0 σ <1.
≤ℜ
It follows that:
Ω1(a)(s)=bs−σ2 1− (1 b2σ)2s1−σ+ (1+a2σ)2s1+σ+... +asσ2 1+ (1 b2σ)2s1−σ− (1+a2σ)2s1+σ+... ,
(cid:18) − (cid:19) (cid:18) − (cid:19)
Ω3(a)(s)=i bs−σ2 1− (1 b2σ)2s1−σ+ (1+a2σ)2s1+σ+... −asσ2 1+ (1 b2σ)2s1−σ− (1+a2σ)2s1+σ +... ,
(cid:20) (cid:18) − (cid:19) (cid:18) − (cid:19)(cid:21)
iσ b2 a2
Ω(a)(s)= +i s1−σ i s1+σ+...,
2 2 1 σ − 1+σ
−
where
R RB σ2
b:= , a:= , µ2 = R2+o(1)
2B 2 − 4 −
Thedotsarehigherordercorrections,andtheorderingdependsonthespecificvalueofσ. Ifwesubstitute
into (7) and we keep the dominant term, we obtain:
4b2
y(s) s1−σ, for 0< σ <1 and for σ =0,
∼ (2µ σ)2 ℜ
−
y(s) s sin2(νlns+C(R,B))+O(s) , for σ =2iν, ν R 0 .
∼ ∈ \{ }
h i
Thus, wehaveobtainedthe small-power-type behaviors 1),the Taylorseriesdegenerationofthe log-type
behaviors 4) and the sine-type oscillatory behavior 2).
ii) Next, we consider the case corresponding to σ = 1. Let ω = lnθ +γ, where γ is the Euler’s
4
constant. In [9] we find the solution:
θ5
yˆ(θ)= θω (8ω3 8ω2+4ω 1)+O(θ9ω5)=
− − 128 − −
θ
= θ ln +γ +O(θ5ln3θ), θ >0, θ 0
− 4 →
(cid:18) (cid:19)
4
In the variable s:
R R
yˆ(s)=i s12 (lns+C) 1+O(s2ln2s) , C :=2ln +2γ
2 4i
h i
Now we compute:
1dyˆ i i
Ω(a) =is = + +O(s2lns) 1+O(s2ln2s)
2 yˆds 2 lns+C
(cid:20) (cid:21)(cid:16) (cid:17)
R i R2
Ω(1a)(s)= 2(yˆ−1+yˆ)=(cid:20)−s12(lns+C) +i 4 s12(lns+C)(cid:21)(1+O(s2ln2s))
i
1
= +O(s2 lns)
−s21(lns+C)
R 1 R2
Ω(3a)(s)=i2(y−1−yˆ)=(cid:20)s21(lns+C) + 4 s12(lns+C)(cid:21)(1+O(s2ln2s))
1
1
= +O(s2 lns)
s21(lns+C)
Observe that µ2 1 = R2+o(1). Now, if we substitute into (7) and we keep the dominant term, we
− 4 −
obtain:
4 1
y(s)
∼−(2µ 1)2(lns+C)2
−
Thus, we have obtained a log-type behavior 4).
iii) The last case to be studied is:
σ =1+2iν, ν R, ν =0
∈ 6
In [9] we find the solution:
1 θ
y(θ)= θ sin 2νln +2ϕ(ν) +O(θ3), θ 0, ϕ(ν)=argΓ(iν).
−2ν 4 →
(cid:20) (cid:18) (cid:19) (cid:21)
In variable s:
R R
1 3
y(s)=i s2 sin(νlns+D)+O(s2), s 0, D =2νln i +2ϕ(ν)
2ν → − 4
(cid:18) (cid:19)
Therefore:
1dyˆ i 1
(a)
Ω =is = sin(νlns+D)+νcos(νlns+D)+O(s) =
2 yˆds sin(νlns+D)+O(s) 2
(cid:20) (cid:21)
i cos(νlns+D)+O(s)
= +iν ,
2 sin(lns+D)+O(s)
R 1 iν R2
Ω(1a) = 2(yˆ−1+yˆ)=−s21 sin(νlns+D)+O(s) +i4νs12(sin(νlns+D)+O(s)),
R 1 ν R2
Ω(3a) =i2(yˆ−1−yˆ)= s21 sin(νlns+D)+O(s) + 4νs21(sin(νlns+D)+O(s)).
Remark: Thereisasequenceofpoles(accumulatingats=0)correspondingtotherootsofsin(νlns+
D)+O(s) = 0. We stress that it is not possible to write 1 as 1 (1+O(s)),
sin(νlns+D)+O(s) sin(νlns+D)
because when we collect sin(νlns+D) in the denominator we divide O(s) by sin(νlns+D) itself, so
we introduce poles (the roots of sin(νlns+D)=0) in the O(s) terms!
5
The computation of y(s) is slightly more complicated than before. We have:
Ω(3a) =iΩ(1a)+O(s12),
2 2
Ω(a)Ω(a)+iµΩ(a) O(√s) Ω(a)
A(s) 1 2 1 + = 1 +O(√s) =
∼" Ω(a) 2+µ2 R2 # "Ω(a) iµ #
2 2 −
(cid:0) (cid:1) 1+O(√s)
=
s 2µ−1sin(νlns+D) cos(νlns+D)+O(s) 2
2ν −
Notethatinthecomputationw(cid:0)ehaveusedonlytheleadingtermofΩ(a). Th(cid:1)ehigherordertermsO(√s)
1
have been divided by Ω(a) 2+ Ω(a) 2 =R2, while the leading partΩ(a)Ω(a)+iµΩ(a) has been divided
1 3 1 2 1
by Ω(a) 2+µ2,withtheassumptionthat Ω(a) 2+ Ω(a) 2+ Ω(a) 2 = µ2. Ifwesubstitute theabove
2 (cid:0) (cid:1) (cid:0) (cid:1) 1 3 2 ∼−
result into (7) and we keep the dominant term, we obtain:
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
1
y(s) =
∼ 1 2µ−1sin(νlns+D) cos(νlns+D) 2+O(√s)
− 2ν −
(cid:0) 1 (cid:1) i 2µ 1 2iν
= , C :=D ln − − .
1 4ν2+(2µ−1)2 sin2(νlns+C)+O(√s) − 2 2µ 1+2iν
− 4ν2 −
Thus, we have found an inverse sine-type oscillatory behavior 3).
4 On the error in the approximation of PVI with PIII
We mayestimate the errorin the approximationofΩ with Ω(a). Let’s estinate the errorinΩ . We put
j j 3
Ω :=Ω(a)+δΩ
3 3 3
andwesubstituteintothelefthandsideofthethirdequationof(4). Intherighthandsidewesubstitute
(a) (a)
Ω and Ω :
1 2
dΩ(a) d 1
3 + (δΩ )= (1+s+...) Ω(a)Ω(a)
ds ds 3 −s 1 2
Recalling that dΩ3(a) = 1 Ω(a)Ω(a) we get
ds −s 1 2
dds(δΩ3)=−Ω(1a)Ω(2a)+...∼−iσ2(bs−σ2 +asσ2)+...
which implies
δΩ3 ∼=O(s1−σ2 +s1+σ2)
We write Ω := Ω(a)+δΩ and we substitute in the left hand side of the first equation of (4), while in
1 1 1
the right hand side we substitute the Ω(a), Ω(a). The same procedure yields:
2 3
δΩ1 ∼=O(s1−σ2 +s1+σ2)
In fact, the terms of order s1±σ2 are missing in the approximated solutions Ω(1a) and Ω(3a), but they
appear in the true formal expansion of the Ω and Ω , which is computed in subsection 4.1. As for Ω ,
1 2 2
we proceed as above making use of the second equation of (4), which becomes:
dδΩ
2 =sΩ(a)Ω(a) =ib2s1−σ ia2s1+σ+...
ds 1 3 −
Thus
b2 a2
δΩ =i s2−σ i s2+σ+ higher orders
2
2 σ − 2+σ
−
6
4.1 Expansion with respect to a Small Parameter
The error in the asymptotic reduction is more precisely evaluated if we write the true formal expansion
oftheΩ ’s. Inordertodothis,asmallparameterexpansioncanbeused,asitisdonein[3]. Lets:=ǫz
j
where ǫ is the small parameter. The system (4) becomes:
dΩ 1 dΩ ǫ dΩ 1
1 2 3
= Ω Ω , = Ω Ω , = Ω Ω . (8)
2 3 1 3 1 2
dz z dz 1 ǫz dz z(ǫz 1)
− −
The coefficient of the new system are holomorphic for ǫ E := ǫ C ǫ ǫ and for 0< z < 1 ,
∈ { ∈ | | |≤ 0} | | |ǫ0|
in particular for z D := z C R z R , where R and R are independent of ǫ and satisfy
1 2 1 2
∈ { ∈ | ≤ | | ≤ }
0 < R < R < 1. The small parameter expansion is a formal way to compute the expansions of the
1 2 ǫ0
Ω ’s for s 0. To our knowledge, the procedure does not give a rigorous justification of the uniform
j
→
convergenceofthes-expansionsoftheΩ ’s. Forǫ E andz D wecanexpandthefractionsasfollows:
j
∈ ∈
∞ ∞
dΩ 1 dΩ dΩ 1
1 = Ω Ω , 2 =ǫ znǫn Ω Ω , 3 = znǫn Ω Ω (9)
2 3 1 3 1 2
dz z dz dz −z
n=0 n=0
X X
and we look for a solution expanded in powers of ǫ:
∞
Ω (z,ǫ)= Ω(n)(z) ǫn , j =1,2,3. (10)
j j
n=0
X
We find the Ω(n)’s substituting (10) into (9). At order ǫ0 we find
j
Ω(0)′ =0
2
 Ω(10)′ = z1 Ω(20)Ω(30)
Ω(0)′ = 1 Ω(0)Ω(0)
The prime denotes the derivative w.r.t. z. T3hus: −z 2 1
iσ
Ω(0) =
2 2
Then we solve the linear system for Ω(0) and Ω(0) and find
1 3
Ω(10) =˜b z−σ2 +a˜ zσ2 =(˜bǫσ2) s−σ2 +(a˜ǫ−σ2)sσ2
Ω(30) =i˜b z−σ2 −ia˜ zσ2 =i(˜bǫσ2) s−σ2 −i(a˜ǫ−σ2)sσ2
wherea˜and˜bareintegrationconstants. Wewillrequirethatb:=˜bǫσ2,a:=a˜ǫ−σ2 arefinite,whenǫ 0.
→
The higher orders are:
Ω(n)(z)= zdζ n−1 ζk n−1−k Ω(l)(ζ) Ω(n−1−k−l)(ζ)
2 k=0 l=0 1 3
 R Ω1(Pn)′ = z1 Ω(2P0)Ω3(n)+A1(n)(z)
Ω(n)′ = 1 Ω(0)Ω(n)+A(n)(z)
where:  3 −z 2 1 3
A(n)(z)= 1 n Ω(k)(z) Ω(n−k)(z)
1 z k=1 2 3
A (z)= 1 n Ω(l)(z) Ω(n−Pl)(z)+ n zk n−k Ω(l)(z) Ω(n−k−l)(z)
 3 −z l=1 2 1 k=1 l=0 1 2
h i
 P P P
7
The system for Ω(n), Ω(n) is closed and non-homogeneous. By variation of parameters we find the
1 3
particular solution
Ω(1n)(z)= zσσ/2 zdζ ζ1−σ2R1(n)(ζ) − z−σσ/2 zζ1+σ2R1(n)(ζ)
Z Z
z
Ω(n)(z)= Ω(n)(z)′ A(n)(z)
3 iσ/2 1 − 1
(cid:16) (cid:17)
where
1 iσ
R(1)(z)= A(n)(z)+ A(n)(z)+A(n)(z)′
1 z 1 2z 3 1
Thus:
Ωj(s)=s−σ2 ∞k, q=0b(kjq) sk+(1−σ)q +sσ2 ∞k, q=0a(kjq) sk+(1+σ)q j =1,3
(11)
 Ω2 = P∞k, q=0b(k2q) sk+(1−σ)q + ∞k, qP=0a(k2q) sk+(1+σ)q
The coefficients a(kjq) and b(kPjq) contain ǫ. In fact, thPey are functions of a := a˜ǫ−σ2, b := ˜bǫσ2. The re-
normalizationafter restoring s is possible if σ =2n+1, n Z (no lnz terms in Ω ), and if the additive
2
6 ∈
constant in the integrationof Ω(n) is zero. If this is not the case, some coefficients of the expansions for
2
the Ω ’s diverge.
j
We can fix the range of σ according to the condition that the first term Ω(0) in Ω be the leading
2 2
one. The approximationat order 0 for Ω is:
2
iσ
Ω constant
2
≈ 2 ≡
Theapproximationatorder1containspowersz1−σ,z1+σ. Ifweassumethattheapproximationatorder
0 in ǫ is actually the limit of Ω as s=ǫ z 0, than we need
2
→
1< σ <1
− ℜ
The ordering of the expansion (11) is somehow conventional: namely, we could transfer some terms
multiplied by sσ2 in the series multiplied by s−σ2, and conversely. I report the first terms:
Ω1(s)=bs−σ2 1− (1 b2σ)2s1−σ+ 4(1σ2 σ)s+ (1+a2σ)2s1+σ +...
(cid:18) − − (cid:19)
+asσ2 1+ b2 s1−σ+ σ2 s a2 s1+σ+...
(1 σ)2 4(1+σ) − (1+σ)2
(cid:18) − (cid:19)
Ω3(s)=ibs−σ2 1− (1 b2σ)2s1−σ+ σ4((1σ−σ2))s+ (1+a2σ)2s1+σ+...
(cid:18) − − (cid:19)
iasσ2 1+ b2 s1−σ+ σ(σ+2)s a2 s1+σ+...
− (1 σ)2 4(1+σ) − (1+σ)2
(cid:18) − (cid:19)
σ b2 a2
Ω (s)=i +i s1−σ i s1+σ+...
2
2 1 σ − 1+σ
−
Note that the dots do not mean higher order terms. There may be terms bigger than those written
above (which are computed through the expansion in the small parameter up to order ǫ) depending on
the particular value of σ in ( 1,1). Finally, we note that we can always assume:
ℜ −
0 σ <1,
≤ℜ
because that would not affect the expansion of the solutions but for the change of two signs.
8
It is worth observing that if we pretend that the solutions (11) are still valid for σ = 1 and if we
ℜ
extract the terms where s has exponent with negative or vanishing real part, we have:
∞ b2 q
Ω1 =bs−21−iν (−1)q (1 σ)2 s(1−σ)q|σ=1+2iν +...
q=0 (cid:20) − (cid:21)
X
∞ b2 q
Ω3 =ibs−21−iν (−1)q (1 σ)2 s(1−σ)q|σ=1+2iν +...
q=0 (cid:20) − (cid:21)
X
iσ b2 ∞ b2 q
Ω = +i s−2iν ( 1)q s(1−σ)q +...
2 2 1 σ − (1 σ)2 |σ=1+2iν
− q=0 (cid:18) − (cid:19)
X
For σ <1 an s small, we sum the series:
|ℜ |
∞ b 2 q 1
( 1)q s1−σ =
q=0 − "(cid:18)1−σ(cid:19) # 1+ b 2s1−σ
X 1−σ
(cid:16) (cid:17)
Then, we analytically extend the result at σ =1+2iν. In this way
iν
Ω = +...
1 −s21 sin(νlns+C)
ν
Ω = +...
3 s12 sin(νlns+C)
i
Ω = +iνcot(νlns+C)+...
2
2
where C = iln(2ν/b). The result is similar to that obtained for the Ω(a)’s, but the O(s)-terms in the
− j
denominator do not appear.
NOTE: Consider thesystem (4) and expand thefractions as s→0. We find
∞ ∞
dΩ 1 dΩ dΩ 1
1 = Ω Ω , 2 = sn Ω Ω , 3 =− sn Ω Ω (12)
ds s 2 3 ds 1 3 ds s 1 2
Xn=0 Xn=0
We can look for a formal solution:
∞ ∞
Ωj(s)=s−σ2 b(kjq) sk+(1−σ)q+sσ2 a(kjq) sk+(1+σ)q , j =1,3
kX, q=0 kX, q=0
∞ ∞
Ω = b(2) sk+(1−σ)q+ a(2) sk+(1+σ)q
2 kq kq
kX, q=0 kX, q=0
Plugging the series into the equation we find solvable relations between the coefficients and we can determine
them. Forexample, thefirst relations give
Ω = iσ + i[b(010)]2s1−σ+... − i[a(010)]2s1+σ+...
2 2 1−σ 1+σ
(cid:18) (cid:19) (cid:18) (cid:19)
Ω1 =(b(010)s−σ2 +...)+(a(010)sσ2 +...)
Ω3 =(ib(010)s−σ2 +...)+(−ia(010)sσ2 +...)
Allthecoefficientsdeterminedbysuccessiverelationsarefunctionsofσ,b(1),a(1). Thesearethethreeparameters
00 00
on which thesolution of (4) must depend. Wecan identify b(1) with b and a(1) with a.
00 00
9
The case σ = 1 in the small parameter formalism is more complicated. If we perform the small
parameter expansions as before, we find the same Ω(0) than before. But due to the exponent z−1/2 the
j
integration for Ω(1) gives
2
i
Ω(1) = a˜2z2+i˜b2ln(z)
2 −2
In this way,we find for Ω and Ω anexpansionin powerof ǫ with coefficients which are polynomialsin
1 3
ln(z);alsothe powersz−1/2, z1/2,... ,zn/2,n>0appearinthecoefficients. Ω isanexpansioninpower
2
of ǫ with coefficients which are polynomials in ln(z) and z. It is not obvious how to recombine z and ǫ
when logarithms appear. We can put ǫ=1. Anyway, we see that the first correction to the constant i
2
in Ω is ln(s), which is not a correction to the constant when s 0, because it diverges.
2
→
We can try an expansion which already contains logarithms of ǫ:
Ω (z,ǫ)= +∞ +∞Ω(k)(z) ǫ2k2+1 , j =1,3 (13)
j j,n (lnǫ)n
k=−1n=0
X X
+∞+∞ ǫk
Ω = Ω(k)(z) (14)
2 2,n (lnǫ)n
k=0n=0
XX
Thenwesubstitutein(9)andweequatepowersofǫandlnǫ. Therequirementthatwecouldre-compose
thepowersoflnǫandlnz appearingintheexpansionintheformln(zǫ)imposesverystrongrelationson
the integration constants. The result which we obtain, when we solve the equations for the coefficients
Ω(k) equating powers up to 1 and ǫ−1/2, is :
j,n (lnǫ)n
i 1
1
Ω1 = s21(ln(s)+C) +O(cid:18)(lnǫ)n(cid:19)+O(ǫ2),
1 1
1
Ω3 = s21(ln(−s)+C) +O(cid:18)(lnǫ)n(cid:19)+O(ǫ2),
i i 1
1
Ω2 = 2 + lns+C +O (lnǫ)n +O(ǫ2).
(cid:18) (cid:19)
These are the log-type behaviors.
5 Conclusions
The asymptotic reduction of PVI to PIII produces the correct leading term of the critical behavior of
branches of PVI transcendents, classified in [4], starting from the asymptotic behaviors of PIII tran-
scendents, computed in [9]. We have evaluated the error of the asymptotic reduction, showing that it
does not affect the leading term of the PVI transcendents obtained from asymptotic behaviors of PIII
transcendents.
References
[1] B.Dubrovin: Geometry of 2D topological field theories, Lecture Notes in Math, 1620, (1996), 120-
348.
[2] B.Dubrovin: Painlev´e trascendents in two-dimensional topological field theory, in “The Painlev´e
Property, One Century later” edited by R.Conte, Springer (1999).
[3] D.Guzzetti: Inverse Problem and Monodromy Data for three-dimensional Frobenius Manifolds.
Math.Phys.Analysis and Geometry 4, (2001), 245-291.
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