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INTEGRABILITY OF HAMILTONIAN SYSTEMS AND TRANSSERIES EXPANSIONS

WERNER BALSER AND MASAFUMI YOSHINO

Abstract. This paper studies analytic Liouville-non-integrable andC-Liouville- integrable Hamiltonian systems with two degrees of freedom. We will show that considerably general Hamiltonians than the one studied in [1] have the property.

We also show that a certain monodromy property of an ordinary differential equa- tion obtained as a subsystem of a given Hamiltonian and the transseries expansion of a first integral play an important role in the analysis. In the former half we will show that the analytic Liouville-non-integrability holds for a rather wider class of Hamiltonians than in [1] under a certain monodromy condition. For these analytic non integrable Hamiltonians we cannot construct nonanalytic first integrals con- cretely as in [1]. In the latter half, we show the nonanalytic integrability from the viewpoint of a transseries expansion of a first integral. We will construct a first inte- gral in transseries formally under general situation. Then we discuss convergence or existence of the first integral which has a given formal transseries as an asymptotic expansion.

1. Introduction

A Hamiltonian system inn degrees of freedom is said to beC-Liouville-integrable if there are n smooth first integrals in involution which are functionally independent on an open dense set. If the first integrals are analytic, then we say that it is analytic- Liouville-integrable. In the paper [1], Gorni and Zampieri showed the existence of a Hamiltonian system with two degrees of freedom which is not analytic-Liouville- integrable, while it is C-Liouville-integrable. The geometrical motivation to study such an example comes from the integrability of a geodesic flows and the Taˇimanov’s problem. (cf. [1]). We note that the proof of analytic-nonintegrability relies on the power series expansion of a first integral, and the C- integrability was proved by constructing concretely a smooth first integral. (cf. Remark after Corollary 2.2.) In this paper, we are interested in the analytical structures which yield nonintegrability in the framework of rather general Hamiltonians than those in previous works. In fact, we will show that the monodromy property of a certain subsystem of a given Hamiltonian plays an important role.

Date: July 26, 2009.

2000Mathematics Subject Classification. Primary 35C10; Secondary 40G10, 35Q15 . Key words and phrases. nonintegrability, Hamiltonian systems, transseries, summability.

Institut f¨ur Angewandte Analysis, Universit¨at Ulm, D–89069 Ulm, Germany

Department of Mathematics, Graduate School of Science, Hiroshima University, 1-3-1 Kagamiyama, Higashi-Hiroshima 739-8526 Japan.

1

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For these analytic-nonintegrable Hamiltonians, we cannot construct a nonanalytic first integral concretely, and instead we make use of a transseries in order to con- struct such an integral. An integral in a transseries expansion is constructed via the Lagrange-Charpit system of a certain vector field obtained by restricting a given Hamiltonian vector field to an invariant manifold. The construction of a first integral as a formal transseries is elementary, while the convergence part is complicated due to the degeneracy of a given Hamiltonian. We will study the convergence from two dif- ferent points of view, transformation of transseries and Borel summability argument for transseries.

This paper is organized as follows. In §2 we study the necessity for the analytic- Liouville-nonintegrability under the monodromy condition. In §3 we give the proof of Theorem 2.1. In §4 we give the transseries expansion of the integrals in a formal sense. The convergence in a curved region and transformation of transseries is dis- cussed in §5. In §6 we study the asymptotic property of transseries in terms of Borel summability method.

2. Analytic nonintegrability

Letσ≥1 be an integer and letr(q1, q2, p1, p2) be an analytic function of (q1, q2, p1, p2) R4 in some neighborhood of the origin 0R4 such that

r≡r(q1, q2, p1, p2) =q12σ+a(q12σ)q22+ ˜r(q1, q2, p1, p2)q23, (2.1)

where ˜r(q1, q2, p1, p2) is analytic at the origin and a(t) (t = q21σ) is a polynomial of t such that a(0) > 0. We are interested in the following analytic Hamiltonian in R4 with two degrees of freedom

H =−q2p2q1r+

r2+q2q2r p1, (2.2)

where q1 = ∂q

1 and q2 = ∂q

2. The associated Hamiltonian system is given by

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

˙

q1 =∂H/(∂p1) =r2+q2q2r+ (2r∂p1r+q2q2p1r)p1−q2p2p1q1r,

˙

q2 =∂H/(∂p2) =−q2q1r−q2p2q1p2r+p1(2r∂p2r+q2q2p2r),

˙

p1 =−∂H/(∂q1) =q2p2q2

1r−(2r∂q1r+q2q1q2r)p1,

˙

p2 =−∂H/(∂q2) =p2q1r+q2p2q1q2r−

2r∂q2r+q2r+q2q22r p1. (2.3)

We need a definition in order to state our theorem.

Definition 2.1. We say that a polynomial a(t) satisfies the monodromy condition (M) if the following equation has no polynomial solution U(t)

2σt2U4σU+ (16σ)tU = (t+ 1)a(t).

(2.4)

Then we have

Theorem 2.1. Assume that a(t) satisfy (M). Then the Hamiltonian system (2.3) is not analytic-Liouville-integrable in any neighborhood of the origin. More precisely, for any analytic first integral u=u(q1, q2, p1, p2) of (2.3) in R4, there exists a function φ of one-variable, being analytic at 0R such that u=φ◦ H.

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If a(t) 1, then we can easily see, from the direct computations or by Lemma 2 of [4] that (M) is satisfied. Hence we have

Corollary 2.2. Suppose that σ = 1, a(t) 1 and r˜ 0 in (2.1). Then the Hamiltonian system (2.3) is not analytic-Liouville-integrable in any neighborhood of the origin.

Remark. (a) As to the fundamental properties of (M) we refer [4]. In this paper, we change the terminology for the sake of simplicity. We remark that (M) is a generic condition.

(b) Theorem 2.1 is a generalization of [4, Theorem 1], where the function rin (2.1) was supposed to be independent ofp1 andp2. Corollary 2.2 was proved in [1]. In this case, it is not difficult to see that (2.3) in Corollary 2.2 is C-Liouville-integrable, because it has a smooth first integral

u=

q2exp

1r

if (q1, q2)= (0,0), 0 if (q1, q2) = (0,0).

(2.5)

On the other hand, it is not known whether (2.3) in a general case has a nonanalytic first integral because one cannot construct the first integral of (2.3) concretely since r also depends onp1 andp2. In §4 we will study the integrability from the viewpoint of transseries.

3. Proof of theorem

The proof of Theorem 2.1 is done by the argument in [4]. For the sake of complete- ness we give the proof.

Proof of Theorem 2.1. By the suitable change of the variableq2one may assume that a(0) = 1. Let u=u(q1, q2, p1, p2) be any analytic first integral of (2.3). We note that u is the first integral of the Hamiltonian system (2.3) if and only ifu is a solution of the following first order equation

{H, u} ≡

q2p2q21r−(2r∂q1r+q2q1q2r)p1 ∂u

∂p1 (3.1)

+

p2q1r+q2p2q1q2r−

2r∂q2r+q2r+q2q22r

p1 ∂u

∂p2

+

r2+q2q2r+ (2r∂p1r+q2q2p1r)p1−q2p2p1q1r ∂u

∂q1 + (−q2q1r−q2p2q1p2r+p1(2r∂p2r+q2q2p2r)) ∂u

∂q2 = 0.

We define

v ≡v(q1, p1, p2) :=u(q1,0, p1, p2).

(3.2)

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By setting q2 = 0 in (3.1) and noting that q2r(q1,0)0 and r(q1,0) =q12σ by (2.1), we obtain

2σp2 ∂v

∂p2 4σq12σp1 ∂v

∂p1 +q21σ+1 ∂v

∂q1 = 0.

(3.3)

We expand v into the power series of p2, v = j=0vj(q1, p1)pj2. Then we see that vj(q1, p1) (j = 0,1, . . .) satisfy

2σjvj 4σq21σp1∂vj

∂p1 +q12σ+1∂vj

∂q1 = 0, j = 0,1,2, . . . (3.4)

We want to show that vj = 0 if j = 0, and v = φ(p1q14σ) for some analytic func- tion φ(t) of one variable. Indeed, by substituting the expansion of vj vj(q1, p1) =

ν=0vj,ν(q1)pν1 into (3.4) we obtain

2σjvj,ν4σνq12σvj,ν+q21σ+1∂vj,ν

∂q1 = 0, j = 0,1,2, . . . (3.5)

If we expand vj,ν into the power series of q1, then we can easily see that vj,ν 0 for allν = 0,1, . . . ,ifj = 0. Hence we havevj = 0 ifj = 0. It follows thatv =v0(q1, p1).

Moreover, by (3.4) v satisfies the equation

4σp1 ∂v

∂p1 +q1 ∂v

∂q1 = 0.

If we substitute the expansion of v into the equation, then, by simple computations, we easily see thatv =φ(p1q14σ) for some analytic function φ(t) of one variable. This proves the assertion.

It follows from (2.2) that v =v0 =φ(p1q14σ) =φ(H|q2=0). We define g(q1, q2, p1, p2) := u(q1, q2, p1, p2)−φ(H).

(3.6)

By (3.2) and by recalling thatHis a first integral we see thatg is an analytic solution of (3.1) such that g(q1,0, p1, p2) 0. In order to prove Theorem 2.1 we shall show g(q1, q2, p1, p2)0 in some neighborhood of the origin. First we will show that

g(q1, q2, p1, p2) =φ1(p1q41σ)p2q2+h2(q1, p1, p2)q22+ ˜h3(q1, q2, p1, p2)q32, (3.7)

for some analytic functionφ1of one variable and analytic functionsh2and ˜h3. Because g is analytic we have the expansion

g(q1, q2, p1, p2) =g1(q1, p1, p2)q2+h2(q1, p1, p2)q22+ ˜h3(q1, q2, p1, p2)q23. (3.8)

We substitute (3.8) withu=g into (3.1) and compare the coefficients ofq2. By (2.1) we have

4σq21σp1∂g1

∂p1 + 2σ

p2∂g1

∂p2 −g1

+q21σ+1∂g1

∂q1 = 0.

(3.9)

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By substituting the expansion g1(q1, p1, p2) = m=0g1,m(q1, p1)pm2 into (3.9) and by comparing the coefficients of pm2 we obtain

4σq12σp1∂g1,m

∂p1 + 2σ(m−1)g1,m+q12σ+1∂g1,m

∂q1 = 0.

(3.10)

Ifm= 1, then we obtain a similar equation as forvj in (3.4). Hence we haveg1,m= 0 if m= 1. It follows thatg1 =g1,1p2, andg1,1 satisfies the equation4σp1∂g∂p1,1

1 +q1∂g∂q1,1

1 =

0. By the same argument as in the above, we see that g1 =g1,1p2 = φ1(p1q14σ)p2 for some analytic function φ1 of one variable. This proves the assertion.

Let us now suppose that

g(q1, q2, p1, p2) = φn1(p1q14σ)pn21qn21 (3.11)

+ hn(q1, p1, p2)q2n+ ˜hn+1(q1, q2, p1, p2)q2n+1,

for some n 2, some analytic function φn1 of one variable and analytic functions hn(q1, p1, p2) and ˜hn+1(q1, q2, p1, p2). Then we substitute (3.11) into (3.1) with u=g and we compare the coefficients of q2n. By (2.1) we have

2pn2σ(2σ−1)q16σ2φn14a(q12σ)(q12σ+ 1)(n−1)p1pn22φn1 (3.12)

4σq14σ1p1∂hn

∂p1 + 2σq12σ1

p2∂hn

∂p2 −nhn

+q41σ∂hn

∂q1 = 0.

By substituting the expansion hn(q1, p1, p2) = m=0hn,m(q1, p1)pm2 into (3.12) and by comparing the coefficients of pn22 we obtain

4σq14σ1p1∂hn,n2

∂p1 4a(q12σ)(q12σ+ 1)(n−1)p1φn1

(3.13)

4σq12σ1hn,n2+q14σ∂hn,n2

∂q1 = 0.

We will show that

hn,n2 = 0, φn1 = 0.

(3.14)

If we can prove φn1 = 0, then it follows from (3.13) that v := hn,n2 satisfies a similar equation as (3.4). Hence, by the same argument as for (3.4) we havehn,n2 = 0. In order to show φn1 = 0 we insert the expansions

φn1(p1q41σ) =

k=0

φn1,kpk1q14σk, hn,n2(q1, p1) = k=0

hn,n2,k(q1)pk1 (3.15)

into (3.13) and we compare the coefficients of pk1. Then we obtain, for k 0

4σq14σ1khn,n2,k 4σq21σ1hn,n2,k+q14σ∂hn,n2,k

∂q1 (3.16)

= 4a(q12σ)(q12σ+ 1)(n−1)φn1,k1q14σ(k1),

where we setφn1,1 = 0. If we set q1= 0 and k= 1 in (3.16), then, by a(0) = 1, we obtain 0 = 4(n−1)φn1,0, which implies φn1,0 = 0.

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Suppose thatφn1,k1 = 0 for somek 2. We divide both sides of (3.16) byq21σ1. Then the right-hand side of (3.16) is divisible by q1N, N = 4σ(k 1) + 1 2σ 2σ+ 1. Because the operator 4σkq12σ +q21σ+1(d/dq1) in the left-hand side of the equation increases the power of q1, it follows that hn,n2,k is divisible by q1N. We set hn,n2,k(q1) = q1NW(q1). Then we have q1(d/dq1)hn,n2,k = qN1 (N +q1(d/dq1))W. It follows from (3.16) that W satisfies

(N 4σk)q12σW 4σW + q12σ+1dW dq1 (3.17)

= 4a(q12σ)(n−1)φn1,k1(q12σ+ 1).

We set W = 2jσ=01qj1Wj(q21σ). Because the right-hand side of (3.17) is a function of q12σ, Wj (1≤j <2σ) satisfy

q12σ(N 4σk+j)Wj4σWj +q12σ+1dWj dq1 = 0.

(3.18)

By a similar argument as for (3.4) we have Wj = 0 for 1 j < 2σ. Hence we have W(q1) = W0(q21σ) =: V(t) (t = q12σ). Because q1(d/dq1)V = 2σt(d/dt)V, it follows from (3.17) that

(16σ)tV 4σV + 2σt2dV

dt = 4a(t)(n−1)(t+ 1)φn1,k1.

If we set U := V /(2(n 1)φn1,k1), then U is an analytic solution of (2.4). This contradicts to the assumption of the theorem, because we assume that (M) is not verified. Hence we have φn1,k1 = 0. Because k is arbitrary we have φn1 = 0.

Next we set φn1 = 0 in (3.12) and consider the coefficients of pm2 (m =n). Then we see that hn,m satisfies a similar equation as for (3.4). Hence we have hn,m = 0 if n = m, and hn,n = φn(p1q14σ) for some analytic function φn of one variable. It follows that hn(q1, p1, p2) = hn,n(q1, p1)pn2 =φn(p1q14σ)pn2. Hence we have (3.11) with n replaced by n+ 1. By induction we obtain (3.11) for an arbitrary integern 2.

It follows from (3.11) with n replaced by n + 2 that, for every n 0 we have

qn2g(q1,0, p1, p2) 0, where (q1, p1, p2) is in some neighborhood of the origin which may depend onn. On the other hand qn2g(q1,0, p1, p2) is analytic in some neighbor- hood of the origin independent ofn. By analytic continuation, we haveqn2g(q1,0, p1, p2)

0 in some neighborhood of the origin independent of n. By the partial Taylor ex- pansiong = nqn2g(q1,0, p1, p2)q2n/n!, we have g = 0.

4. Transseries expansion of first integral

In this section, we shall construct a first integral of (2.3) as a transseries. In order to introduce such a series we consider the terms in (3.1) which preserve the order of q2

Lu:=q12σ1

2σ(p2 ∂u

∂p2 −q2∂u

∂q2) +q21σ(q1∂u

∂q1 4σp1∂u

∂p1)

. (4.1)

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We note that we can write (3.1) in the form

Lu+Ru = 0, Ru :={H, u} − Lu.

(4.2)

In order to construct an inverse of L we consider the Lagrange-Charpit system cor- responding to L

dq1

q41σ = dq2

2σq12σ1q2 = dp1

4σq41σ1p1 = dp2 2σq21σ1p2. (4.3)

We integrate (4.3) by taking q1 as an independent variable. By simple computations we can easily see that the solution of (4.3) is given by

q2 =q20exp q12σ

, p2 =p02exp

−q21 σ

, p1 =p01q14σ, (4.4)

where q02, p02 and p01 are certain constants.

We note that the solution of the homogeneous equation Lv = 0 is given by v =φ(p1q14σ, p2exp

q12σ

, q2exp

−q12σ ), (4.5)

with φ(p01, p02, q20) being an arbitrary function of p01, p02 and q02. We then construct a solutionu of (4.2) in the form

u=

j=0

uj(q1, p1q14σ, p2exp q12σ

) exp

−q12σ q2j

, (4.6)

whereu0(q1, p1q14σ, p2exp q12σ

)≡u0(p1q14σ, p2exp q12σ

). We call (4.6) the transseries solution of (4.2). Then we have

Proposition 4.1. Let u0(p01, p02) be a given analytic function of p01 and p02 such that

∂u0/∂p02 = 0. Then (2.3) is formally Liouville-integrable in the sense that (4.6) is a formal integral of (2.3) which is functionally independent of H.

Proof. We note that R in (4.2) has analytic coefficients and R raises the power of q2 at least by one. On the other hand we have

L uj

exp

−q12σ q2j

= (Luj) exp

−q12σ q2j

. (4.7)

Hence, if we substitute (4.6) into (4.2) and compare the coefficients of qj2 of both sides, then we have

Luj = ( linear functions of uk and their derivatives (k < j), j = 1,2, . . . (4.8)

We note that the right-hand side is a known quantity if we determine uj recursively.

We will solveLv =f, where v =v

q1, p1q14σ, p2exp q12σ

.

By making the change of variables (q1, p1, p2)(q1, p01, p02) given by (4.4), the equa- tion Lv =f(q1, p1, p2) is written in the form

q41σ(∂v/∂q1) =g(q1, p01, p02), (4.9)

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where

g ≡g(q1, p01, p02) =f(q1, p01q14σ, p02exp

−q12σ ).

Hence the solution of (4.9) is given by v =

q1

a0

s4σg(s, p01, p02)ds, (4.10)

wherea0 is an arbitrary complex constant. If we go back to the original variables q1, p1 and p2, then we obtain a solution of Lv = f. Therefore we have a solution u of (4.2) given by (4.6).

Finally, we will show that u converges, then u is an integral of (2.3) functionally independent of H. Hence our Hamiltonian system is formally Liouville-integrable.

Indeed, if this is not the case, then we haveu=φ(H) for some smooth function φ of one variable. If we set q2= 0, then we obtain

u0

p1q14σ, p2exp q12σ

= φ(H)|q2=0 =φ(p1q14σ).

This is a contradiction to the assumption that ∂u0/∂p02 = 0. This ends the proof.

5. Convergence of transseries

In this section we consider the Hamiltonian corresponding tor =q12σ+a(q12σ)q22 in (2.2)

H =2σq21σ1q2p2(1 +q22a) +p1

(q12σ+aq22)2+ 2aq22 , (5.1)

where we assume a(0) = 1 for the sake of simplicity. We study the convergence of transseries solutions (4.6), where we set a0 = 0 in (4.10). Note that uj(0, p01, p02) identically vanishes for any j 1.

Clearly, the integraluis the solution of (4.2), whereLandRare given, respectively, by (4.1) and

Ru = ( ˜αp1+ ˜βp2)∂u

∂p1 +

˜

γp2+ ˜δp1 ∂u

∂p2 (5.2)

+ E(q1, q2)∂u

∂q1 2σq21σ1aq23∂u

∂q2, where

E E(q1, q2) := aq22(aq22+ 2q12σ + 2), (5.3)

˜

α := 4σq21σ1q22(a+a+aq12σ +aaq22), (5.4)

β˜ := 2σ(2σ−1)q2q12σ2(1 +aq22) + 4σ2q41σ2aq23, (5.5)

˜

γ := 6σq12σ1aq22, δ˜:=4q2a(q12σ+aq22+ 1).

(5.6)

Letε0 be a small positive constant. Then we define S0:={q1 C;|q1|< ε0} ∩

q1 C; Req12σ <0 , (5.7)

where Req12σ denotes the real part of q12σ. Then we have

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Theorem 5.1. Letu0(p01, p02)be an analytic function ofp01 andp02 such that∂u0/∂p02 = 0 in some neighborhood of the origin. Then there exist a δ >0, an ε0 >0, neighbor- hoods V1, V2 of the origin in C such that u in (4.6) converges in the domain

{(q1, q2, p1, p2);q1∈S0, p1 ∈V1, p2 ∈V2,|exp

−q12σ

q2|< δ}.

Proof. LetLbe given by (4.1). The Lagrange-Charpit system corresponding toL+R is given by

dq1

q41σ+E = dq2

2σq12σ1q2T = dp1

( ˜α−4σq14σ1)p1+ ˜βp2 = dp2

(2σq21σ1+ ˜γ)p2 + ˜δp1, (5.8)

where we set

T ≡T(q1, q2) := 1 +aq22. (5.9)

We integrate (5.8) by taking q1 S0 as an independent variable. We want to show that the solutions are perturbations of the solutions (4.4) in S0. Namely, we will prove

q2 = q02exp(q21 σ) exp(q1Q2(q1, q20)), (5.10)

p1 = q41 σ

p01(1 +P1(q1, q20)) +p02P˜1(q1, q20)

, (5.11)

p2 = exp(−q12σ)

p01P2(q1, q02) +p02(1 + ˜P2(q1, q20))

, (5.12)

for some functions Q2 Q2(q1, q20), Pj Pj(q1, q20) and ˜Pj P˜j(q1, q20) (j = 1,2), which are holomorphic and bounded when q1 S0 and q20 in some neighborhood of q20 = 0. Herep01 and p02 are arbitrary constants.

In order to verify these properties we first consider the following equation dq2

dq1 = 2σq2q12σ1(1 +aq22)

q41σ+E = 2σq2 q12σ+1

1 + aq22−q14σE 1 +q41 σE

. (5.13)

Clearly, q2 = 0 is a solution of (5.13). We assume q2 0. We note that v :=

q20exp(q12σ) satisfies the equation dv/dq1 =2σvq12σ1. We set U :=q1Q2, and we substitute (5.10) into (5.13). Then we have

dU

dq1 =2σ(aq22−q41 σE)

q12σ+1(1 +q14σE) =:f(q1, U), (5.14)

where q2 = q20exp(q12σ)eU. Because Req21σ < 0 in S0, we see that f(q1, U) is holo- morphic when (q1, U) S0 × Ω, and continuous up to its closure, where Ω is a neighborhood of the origin. Moreover, its maximal norm when (q1, U) S0 ×Ω can be made arbitrarily small if we shrink S0 sufficiently small.

We will solve (5.14) inS0. If we replace U with U +c for a constantc we see from (5.10) that q20 is replaced by ecq20. Hence we may assume that U vanishes at q1 = 0.

We will look for the solution U as the solution of the following equation U =

q1

0

f(s, U)ds, (5.15)

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where the integral is taken along the straight line inS0 which connects 0 andq1. We note that we can make|f(q1, U)|arbitrarily small if we takeε0 inS0 sufficiently small and U is bounded. We can easily show that the right-hand side operator of (5.15) is a contraction mapping on a small ball in the set of functions holomorphic in S0 and continuous on its closure, and U(0) = 0. Hence we have a holomorphic solution U in S0 of (5.15). If we setQ2 :=q11 U we obtain the desired solution. The analyticity of U with respect to q20 is easy to verify because of the definition of f.

Next we study the equations for p1 and p2. It follows from (5.8) that dp1

dq1 =αp1 +βp2, dp2

dq1 =γp2 +δp1, (5.16)

where

α= α˜4σq14σ1 q14σ+E , β=

β˜

q14σ+E, γ = 2σq12σ1+ ˜γ q41σ+E , δ =

δ˜ q41σ+E. (5.17)

We will construct the solution of (5.16) in the following form pν =

j=0

p(νj), ν= 1,2, (5.18)

where

dp(0)1

dq1 =αp(0)1 , dp(0)2

dq1 =γp(0)2 , (5.19)

and p(νj) (ν = 1,2;j = 1,2, . . .) are determined by dp(1j)

dq1 =αp(1j)+βp(2j1), dp(2j)

dq1 =γp(2j)+δp(1j1). (5.20)

First we solve (5.19). By the definition of γ in (5.17) we have the expression γ = 2σq12σ1+γ0, whereγ0 is a bounded holomorphic function inS0. By the change of an unknown function similar in the argument for (5.13) the solution p(0)2 of (5.19) has the following expression

p(0)2 =p02exp(−q21 σ)(1 + ˜P2(0)(q1, q20)) (5.21)

for some ˜P2(0) which is bounded and holomorphic in q1 S0 and q20 in some neigh- borhood of the origin, where p02 is an arbitrary constant. Similarly, noting that α = 4σq11+α0 for some bounded holomorphic function α0 in S0 we see that the solutionp(0)1 of (5.19) has the following expression

p(0)1 =p01q41 σ(1 + ˜P1(0)(q1, q20)) (5.22)

for some ˜P1(0) which is bounded and holomorphic in q1 ∈S0 andq20 in some neighbor- hood of the origin, wherep01 is an arbitrary constant.

(11)

Now we assume thatp(νk)’s (k = 0,1, . . . , j−1,ν = 1,2)

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