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Removable singularities of holomorphic solutions of linear partial di¤erential equations

By Katsuju Igari

(Received Feb. 20, 2002) (Revised Jul. 10, 2002)

Abstract. In a complex domain V H Cn, let P be a linear holomorphic partial dif- ferential operator and K be its characteristic hypersurface. When the localization of P at K is a Fuchsian operator having a non-negative integral characteristic index, it is proved, under some conditions, that every holomorphic solution to Pu¼ 0 in V nK has a holo- morphic extension in V . Besides, it is applied to the propagation of singularities for equations with non-involutive double characteristics.

1. Introduction.

We employ the following notation in this paper. z¼ ðz1; . . . ; znÞ A Cn, z0 ¼ ðz2; . . . ; znÞ, z00 ¼ ðz2; . . . ; zn 1Þ, Dj ¼ q=qzj, D¼ ðD1; . . . ; DnÞ, Da ¼ D1a1    Dnan and jaj ¼ a1þ    þ an for multi-index a ¼ ða1; . . . ; anÞ, D00 and a00 are same as z00, N ¼ f0; 1; 2; . . .g.

Let n b 2, m b r b 1 and 0 a s a 1. For simplicity, we call Pðz; DÞ a locally Fuchsian operator of class ðm; r; sÞ and write P A Lm; r; s if it is written in the form

Pðz; DÞ ¼ Xr

s¼0

asðzÞðz1D1Þr s

!

Dnm rþ X

0asam; a A AðsÞ

z1lðaÞaaðzÞDa ð1:1Þ

where AðsÞ ¼ fa; jaj ¼ m s; anam r; a 0ðr s; 0; . . . ; 0; m rÞg, coe‰cients are all holomorphic in V ¼ fz; jzij < ri; Eig ðri > 0Þ, a0ð0Þ 0 0 and lðaÞ are non-negative inte- gers satisfying

slðaÞ þ ð1 sÞa1br s for jaj ¼ m s: ð1:2Þ We note that the hyperplane zn ¼ h is characteristic for P for every h. Set

Qjðz; D1; D00Þ ¼ X

a¼ða1; a00; m jÞ

z1lðaÞaaðzÞDða1; a00; 0Þ: ð1:3Þ

By Iðz; z1D1ÞDnm r denote the first term of the right hand side of (1.1), then one can write P¼ MDnm rþPm

j¼rþ1QjDnm j, where

2000 Mathematics Subject Classification. Primary 35A20; Secondary 35A10, 35A21.

Key Words and Phrases. Removable singularities, singular solutions, propagation of singularities, Cauchy-Kovalevskaya type theorem.

This research was partially supported by Grant-in-Aid for Scientific Research (No. 12440029), Japan Society for the Promotion of Science.

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Mðz; D1; D00Þ ¼ I ðz; z1D1Þ þ Qrðz; D1; D00Þ: ð1:4Þ Mðz1; z00; h; D1; D00Þ is called the localization of P at zn ¼ h and, by the condition (1.2), it is easy to see that Mðz1; z00; h; D1; D00Þ is a Fuchsian partial di¤erential operator on the characteristic hyperplane zn ¼ h with respect to z1 of weight 0 for every jhj < rn. (This is why we call P a locally Fuchsian operator.) As we saw in [6], this class is closely related with operators having non-involutive multiple characteristics.

Denote K ¼ fz; zn ¼ 0g and define the indicial polynomial by

Iðz; lÞ ¼Xr

s¼0

asðzÞlr s ð1:5Þ

then the following theorem holds.

Theorem 1.1 ([4]). Let Pðz; DÞ be a locally Fuchsian operator of class ðm; r; sÞ. If

Ið0; kÞ 0 0 Ek A N ð1:6Þ

then every holomorphic solution of Pu¼ 0 on the universal covering of V nK has a unique holomorphic extension in V.

Our purpose in this paper is to investigate the case where the indicial polynomial Ið0; lÞ has a non-negative integral root. We assume

(A.1) there exists n A N such that Iðz; nÞ ¼ 0 on z1 ¼ 0 but I ð0; kÞ 0 0 for all k A Nnfng,

(A.2) s¼ 1=p with some p A f1; 2; . . . ; r þ 1g,

(A.3) slðaÞ þ ð1 sÞa1 > r s when jaj ¼ m s and r s b a1. Let p denote the restriction of functions onto z1 ¼ 0, namely

pfðz0Þ ¼ f ð0; z0Þ then the following proposition holds.

Proposition 1.2. Let M be defined by (1.4) and suppose (A.1). Then 1) there exists a unique operator Lðz0; D1; D00Þ ¼Pn

j¼0Ljðz0; D00ÞD1j with Ln ¼ 1 such that pLM ¼ 0,

2) there exists a unique operator Rðz1; z0; D00Þ ¼P

jbnRjðz0; D00Þz1j=j! with Rn ¼ 1 such that MRp¼ 0,

3) if (A.3) is also supposed, the operator pLQrþ1Rp is free from D00, namely a function of z0.

Set

qðz0Þ ¼ pLQrþ1Rp: ð1:7Þ

Then the main result of this paper is the following theorem.

Theorem 1.3. Let Pðz; DÞ be a locally Fuchsian operator of class ðm; r; sÞ and sup- pose (A.1), (A.2) and (A.3). Then, if

qð0Þ 0 0 ð1:8Þ

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every holomorphic solution uðzÞ to Pu ¼ 0 on the universal covering of V nK has a unique holomorphic extension in V.

This theorem will be applied to the propagation of singularities for equations with non-involutive double characteristics in §6. The definition (1.7) looks implicit, however, an algorithm to obtain qðz0Þ will be given in §2. The following example explains the role of the condition (1.8).

Example 1.4. Let b; c A C and consider

P¼ D12þ z1D1Dn bDnþ c:

K ¼ fz; zn ¼ 0g is characteristic. Since P¼ ðz1D1 bÞDnþ D12þ c, it is easy to see that P is a locally Fuchsian operator of class ð2; 1; 1=2Þ and I ðlÞ ¼ l b. Therefore Theorem 1.1 applies when b B N.

When b A N, the conditions (A.1), (A.2) and (A.3) hold. Since M ¼ z1D1 b, one can easily see that L¼ D1b, R¼ z1b=b! and q¼ c. Therefore Theorem 1.3 applies if c 00.

When c¼ 0, the equation Pu ¼ 0 has two solutions

u1 ¼ znðbþ1Þ=2ðzn z12=2Þb=2; u2 ¼ z1znðbþ2Þ=2ðzn z12=2Þðb 1Þ=2:

If b A N, one of u1; u2 is singular only on zn ¼ 0, namely the conclusion of Theorem 1.3 is not true when b A N and c¼ 0.

Remarks. 1) Let a< b and denote Vab ¼ fz A V ; a < arg zn < bg. Under the same assumptions as in Theorem 1.3, one can prove that every holomorphic solution to Pu¼ 0 in Vab has a holomorphic extension to a neighborhood of z¼ 0.

2) Let P be a linear holomorphic partial di¤erential operator in V for which K is characteristic. The existence of a singular solution, namely a solution to Pu¼ 0 holo- morphic in VnK but singular at K, has been studied by several authors. (See [8], [10]

and their references. Cf. [2], [7], [12] as well.) Especially, there exist singular solutions when the localization of the principal part of P at K is a non-degenerate partial di¤er- ential operator on K of order b 1. (See S. Ouchi [8].) In our case, the localization Mðz1; z00; 0; D1; D00Þ of whole P at K degenerates at z1 ¼ 0 to be a Fuchsian operator, and the Theorems 1.1 and 1.3 imply the non-existence of singular solutions.

3) In C2, let P¼ D21 þ z1D1D2 bðzÞD2þ cðzÞ. In [5], the corresponding result was obtained when bð0; z2Þ ¼ 0; 1; 2; . . . : When bð0Þ ¼ 0; 1; 2; . . . but D2bð0Þ 0 0, the existence of a singular solution was also verified, namely Theorem 1.3 does not hold in general by assuming only Ið0; nÞ ¼ 0 instead of (A.1).

4) In C3, let P¼ z1D1D3þ z12D2D3þ D2þ c. Then P is a locally Fuchsian op- erator whose characteristic index is 0 but does not satisfy the condition (A.3). Since u¼ ð1=z3Þ exp½ cðz2 z12=2ފ is a solution to the equation Pu ¼ 0, we see Theorem 1.3 is not true in general without assuming (A.3).

The theorem will be proved as follows. Take 0<jhj < rn and a branch of uðzÞ near zn ¼ h arbitrarily. We will prove that if h is su‰ciently small, there exists a unique holomorphic solution in some neighborhood W of z¼ ð0; . . . ; 0; hÞ containing the origin z¼ 0 to the characteristic Cauchy problem

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Pv¼ 0; Dnjv¼ Dnju on zn ¼ h; j ¼ 0; 1; . . . ; m r 1: ð1:9Þ (Theorem 3.1.) Since uðzÞ is a solution of Pu ¼ 0, the uniqueness means vðzÞ gives a holomorphic extension of u up to the origin. Then it is well known that uðzÞ has a unique holomorphic extension in V . (Cf. [6, Proposition 3.3].)

In §2, the operator M will be studied. The characteristic Cauchy problem (1.9) will be reduced to a pair of integro-di¤erential equations in §3. We will introduce two Banach spaces B and B0 of holomorphic functions in §4 and prove Theorem 3.1 by means of contraction principle in §5.

2. Operator M.

2.1. Proof of Proposition 1.2.

First, in this subsection, we prove Proposition 1.2. When an ¼ m rand a0000, it follows from (A.3) that

sðl a1Þ > r s a1 ¼ ja00j > 0 ð2:1Þ and therefore l> a1. (It means M is a Fuchsian operator.)

Since z1lD1a1 ¼ z1l a1z1a1D1a1 and z1jD1j ¼Pj

k¼1cj; kðz1D1Þk with certain constants cj; k

ðcj; j ¼ 1Þ, one can write

Mðz; D1; D00Þ ¼X

r

s¼0

Msðz; D00Þðz1D1Þr s

with Msðz; D00Þ ¼ P

ja00jasz1ll~ðs; a00Þbs; a00ðzÞDa00, where bs; a00 are all holomorphic in Vr, Msðz; 0Þ ¼ asðzÞ and ja00j < s ~ll for a0000 because of (2.1).

Denote

Ms; iðz0; z00Þ ¼ qiMs

qz1i ð0; z0; z00Þ:

If i< ~ll, then qiðz1ll~bza00Þ=qz1i ¼ 0 on z1 ¼ 0 for any function b. Hence si b s ~ll>ja00j for a0000. It means

orderD00Ms; i < si for i> 0:

Using D1jðz1D1Þr s ¼ ðz1D1þ jÞr sD1j, we have

pLMf ¼X

n

i¼0

X

r

s¼0

X

i

j¼0

Li

i j

 

Ms; i jjr spD1jf:

Set

mijðz0; D00Þ ¼X

r

s¼0

i j

 

Ms; i jðz0; D00Þ jr s ð2:2Þ

then we have pLM ¼ 0 if and only if

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Xn

i¼j

Liðz0;D00Þmijðz0;D00Þ ¼ 0; j ¼ 0; . . . ; n: ð2:3Þ

Note that mj; j ¼ I ð0; z0; jÞ. Since Ið0; z0; nÞ ¼ 0 and I ð0; 0; jÞ 0 0 for j 0 n by the assumption (A.1), we see Lj are uniquely determined by these relations with Ln ¼ 1.

Besides, because orderD00mij < sði jÞ when i > j, one can easily see that

orderD00Lj < sðn jÞ for j < n: ð2:4Þ Thus the first part 1) of Proposition 1.2 has been proved.

Since

pD1iMRpf ¼X

i

j¼n

X

r

s¼0

i j

 

Ms; i jjr sRjpf ¼X

i

j¼n

mijRjpf we have MRp¼ 0 if and only if

X

i

j¼n

mijðz0;D00ÞRjðz0;D00Þ ¼ 0 for i¼ n; n þ 1; n þ 2; . . . : ð2:5Þ Therefore Ri ði > nÞ are uniquely determined with Rn ¼ 1 and

orderD00Ri < sði nÞ for i > n: ð2:6Þ Thus the second part 2) has been proved, too.

The operator Qrþ1 is the sum of zl1ðaÞaaDða1; a00; with a¼ ða1; a00;m r 1Þ, a1þ ja00j ¼ r þ 1 s and 0 a s a rþ 1. When a0000, we have r s a1 ¼ ja00j 1 b 0. It then follows from the assumption (A.3) that sðl a1Þ > r s a1b0, and hence l > a1. Because i a n a j, the power of z1 in the term

LiD1iz1laaDða1; a00;z1j j!Rj

is at least lþ j i a1b l a1 >0, and therefore we have pLz1laaDða1; a00;Rp ¼ 0:

When a00¼ 0, we have r s a1 ¼ 1. From the assumption (1.2) it follows that sðl a1Þ b r s a1 ¼ 1, and therefore sða1 lÞ a 1. If n i >0 or a1þ j > n, because of (2.3) and (2.4), we have

orderD00 Li i j

 

D1i jðz1laaÞRjþa1

 

< sða1þ j iÞ:

Since the terms with i j < l all vanish on z1 ¼ 0, one may suppose i j b l. Then sða1þ j iÞ a sða1 lÞ a 1:

It means the order of D00 is zero. If n i¼ 0 and a1þ j ¼ n, the term is evidently free from D00. Thus the last part 3) has been proved.

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2.2. Function q(zM).

Next, we explain how to calculate qðz0Þ defined by (1.7). Since Msðz; 0Þ ¼ asðzÞ, we have

mijðz0; 0Þ ¼ X

r

s¼0

i j

  qi jas

qz1i j ð0; z0Þ jr s ¼ i j

  qi jI

qz1i j ð0; z0; jÞ:

Next, set

liðz0Þ ¼ Liðz0; 0Þ; riðz0Þ ¼ Riðz0; 0Þ:

Then li ðn p a i a nÞ and rj ðn a j a n þ pÞ are given by the relations X

n

i¼j

liðz0Þmijðz0; 0Þ ¼ 0; j ¼ n 1; . . . ; n p

X

i

j¼n

mijðz0; 0Þrjðz0Þ ¼ 0; i¼ n þ 1; . . . ; n þ p

with ln ¼ rn ¼ 1.

When a¼ ðk; 0; m r 1Þ, as seen above, l k b 1=s ¼ p. Therefore one can write

Qrþ1ðz; D1; 0Þ ¼Xrþ1

k¼0

ckðzÞD1k; ck ¼ Oðzðk pÞ1 þÞ as z1 ! 0

where ðk pÞþ¼ maxfk p; 0g. Hence qðz0Þ ¼ pX

i; j; k

Liðz0; 0ÞD1ickðzÞD1kRjðz0; 0Þz1j=j!

¼X

i; j; k

Liðz0; 0Þ i j

 

ck; i jðz0ÞRjþkðz0; 0Þ

where ck; iðz0Þ ¼ ðqick=qz1iÞð0; z0Þ and the sum is taken for

0 a k a rþ 1; i a n; jþ k b n; i j bðk pÞþ:

The last inequality is equal to that jþ k i aminfk; pg. Especially we have jþ k a nþ p and i b n p and therefore qðz0Þ is given by

qðz0Þ ¼X

i; j; k

liðz0Þ i j

 

ck; i jðz0Þrjþkðz0Þ: ð2:7Þ

2.3. Equation Mhv F g.

We next consider the problem

MhvðzÞ ¼ gðzÞ; pD1nv¼ jðz0Þ ð2:8Þ

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near z¼ ð0;. . . ; 0; hÞ, where M

hðz1;z00;D1;D00Þ ¼ Mðz1;z00;h; D1;D00Þ. So are Lh;Rh;qh and mijh.

Proposition 2.1. Let fðzÞ and jðz0Þ be holomorphic functions near ð0;. . . ; 0; hÞ.

Then the problem (2.8) has a unique holomorphic solution vðzÞ if and only if

pLhg¼ 0: ð2:9Þ

If we denote the solution to the equation Mhv¼ g satisfying pD1nv¼ 0 by ðMhÞ 1g, then the solution to (2.8) is given by

v¼ ðMhÞ 1gþ Rhj: ð2:10Þ

The proof will be given in §5. For its preparation, we here explain the role of the condition (2.9). Since pLhMh ¼ 0, we have pLhg¼ pLhMhv¼ 0. Hence the condi- tion (2.9) is necessary.

Denote pD1ig¼ gi, then we have Xi

j¼0

i j

  Xr

s¼0

Ms; i jh jr svj ¼Xi

j¼0

mijhvj ¼ gi; i ¼ 0; 1; 2;. . . :

Note that miih ¼ I ð0; z00;h; iÞ, miih00 for i 0 n and mnnh ¼ 0. Then we see that, by setting vn ¼ j, vj ð j 0 nÞ are uniquely determined by these relations if and only if

X

i

j¼0

mijhvj ¼ gi; i¼ 0; 1; 2;. . . ; n 1

X

n 1

j¼0

mnjhvj ¼ gn .

8

>>

>>

><

>>

>>

>:

The first n relations give one to one correspondence between ðv0;. . . ; vn 1Þ and ðg0;. . . ; gn 1Þ. Under these correspondence, the last relation is equivalent to pLhg¼ 0.

In fact,

gn ¼ Xn 1

i¼0

Lihgi ¼ Xn 1

i¼0

Xi

j¼0

Lihmijhvj

¼ X

n 1

j¼0

X

n 1

i¼j

Lihmijh

!

vj ¼X

n 1

j¼0

Lnhmnjhvj ¼X

n 1

j¼0

mnjhvj

where we have used pLhg¼Pn

i¼0Lihgi and Pn

i¼jLihmij ¼ 0.

Lastly consider the case where g ¼ 0 and pD1nv¼ j. In this case, vj ¼ 0 for j < n, vn ¼ j and vj ð j > nÞ are determined uniquely by

X

i

j¼n

mijhvj ¼ 0:

Because Pi

j¼nmijhRjh ¼ 0 by the definition of Rjh, we see vj ¼ Rjhj and hence v¼ Rhj.

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3. A characteristic Cauchy problem.

Let h A C be a parameter and consider the Cauchy problem with characteristic initial hyperplane zn ¼ h.

Pu¼ f

u¼ Ofðznm rg as zn ! h



ð3:1Þ Recall that

Pu¼ MhDnm ruþ ðM MhÞDnm ruþ X

m

j¼rþ1

QjDnm ju

where M ¼ I þ Qr. Since pLhMh ¼ 0, pLhðM MhÞ ¼ 0 on zn ¼ h and the third term vanishes on zn ¼ h by the initial condition, it is necessary for f to satisfy the compati- bility condition

pLhf ¼ 0 on zn ¼ h: ð3:2Þ

Theorem 3.1. Let P be a locally Fuchsian operator of class ðm; r; sÞ and assume (A.1), (A.2) and (A.3). Then there exists a constant d >0 such that, for any jhj < d and any holomorphic function fðzÞ in Vr satisfying the compatibility condition (3.2), there exists a unique holomorphic solution uðzÞ to the Cauchy problem (3.1) in Wh; d ¼ fz; jzij <

d ði < nÞ; jzn hj < dg.

Here we explain how to apply the above theorem to the Cauchy problem (1.9). Set

u ¼ m rX1

j¼0

Dnjuðz1;z00;hÞðznj j!

and v¼ u w. Then (1.9) is equivalent to

Pw¼ Pu; w¼ Ofðznm rg:

Now that pLhMh ¼ 0 and pLhðM MhÞ ¼ 0 on zn ¼ h, we see

pLhPu ¼ Xm

j¼rþ1

pLhQjDnm ju ¼ Xm

j¼rþ1

pLhQjDnm ju¼ pLhPu¼ 0

on zn ¼ h. Thus the compatibility condition (3.2) is fulfilled and therefore the Theorem 3.1 is applicable.

Theorem 3.1 will be proved by using the contraction principle. For that purpose we rewrite the Cauchy problem (3.1). Denote

~ u

u¼ Dnm ru Dn1fðzÞ ¼

ðzn

h

fðz1;z00;znÞ dzn

Q¼ I þ Qr Mhþ Qrþ1Dn1 þ    þ QmD mþr: 8

>>

>>

<

>>

>>

:

ð3:3Þ

Then the Cauchy problem (3.1) is written

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Mhuu~þ Q~uu¼ f : ð3:10Þ Set

c¼ pD1nuu~; v¼ ~uu Rhc: ð3:4Þ Then, because MhRhc¼ 0, we have

Mhvþ Qv þ QRhc¼ f : Operating pLh and noting pLhMh ¼ 0, we have

pLhQvþ pLhQRhc¼ pLhf:

It is easy to see that pLhIRhc¼ I ð0; z00; zn; nÞc ¼ 0. Since each term of Qr has the form z1laDða1; a00; 0Þ with a1þ ja00j ¼ r s and a0000, it follows from the condition (A.3) that l a1bsðl a1Þ > r s a1 ¼ ja00j > 0 and therefore we see pLhQrRhc¼ 0. Be- cause Rh and Dn1 are commutative, we have

pLhQrþ1h Dn1Rhc¼ qhDn1c where Qrþ1h ¼ Qrþ1ðz1; z00; h; D1; D00Þ. Thus, if we denote

j¼ Dn1c; w¼ Mhv Q~

Q¼ ðQrþ1 Qrþ1h ÞDn1þ Qrþ2Dn2þ    þ QmD mþr (

ð3:5Þ then the Cauchy problem (3.1) is reduced to

wþ QðMhÞ 1wþ QDnRhj¼ f

jþ ~qqhpLhQðMhÞ 1wþ ~qqhpLhQQD~ nRhj¼ ~qqhpLhf (

ð3:6Þ where ~qqh ¼ 1=qh.

4. Banach spaces B and B0.

To consider the equations (3.6), Banach spaces are introduced and some of their fundamental properties are proved in this section.

4.1. A lemma.

The following lemma will play an important role.

Lemma 4.1. Let r b 1, p b 1 and n b 0 be integers. Then there exists a corre- spondence i! i from N to N such that the following 1)–6) hold with some i0 A N.

1Þ i1ai2 if i1 < i2

2Þ ðn pÞ ¼ n 3Þ nþ p ¼ ðn þ 1Þ 4Þ ði þ pÞ p a i

5Þ ði kp 1Þþ kp a i if 1 a k a rþ 1 or i a n 6Þ ibi Ei and i ¼ i for i b i0

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Proof. Define i by

 n ¼ n;

 ðn þ pj þ lÞ ¼ nþ pj þ p if ðr þ 1Þl > j nþ pj þ l if ðr þ 1Þl a j



for j ¼ 0; 1; 2; . . . and l ¼ 1; 2; . . . ; p;

 ðn 1 pjþ lÞ ¼ n pjþ p

for j ¼ 1; 2; . . . and l ¼ 1; 2; . . . ; p:

Then it is easy to show 1), 2), 3) and the first half of 6).

Set A¼ ði þ pÞ i. When i > n, write i¼ n þ pj þ l ð j b 0; 1 a l a pÞ. Since ðr þ 1Þl a j implies ðr þ 1Þl a j þ 1, we have

A¼ ðn þ jp þ p þ lÞ ðn þ jp þ lÞapþ maxf0; l pg ¼ p:

When iþ p < n, write i þ p ¼ n 1 jpþ l ð j b 1; 1 a l a pÞ. Then A¼ ðn 1 jpþ lÞ ðn 1 jp pþ lÞ ¼ p:

When i a n a iþ p, note i þ p a n þ p and i b n p. Then A¼ ði þ pÞ naðn þ pÞ n ¼ p:

Thus 4) has been proved.

Denote B¼ i ði kp 1Þ. When i kp 1 > n, write i kp 1¼ n þ jp þ l ð j b 0; 1 a l a pÞ. Then

B¼ ðn þ ð j þ kÞp þ l þ 1Þ ðn þ jp þ lÞ: If l¼ p, we have

B b nþ ð j þ k þ 1Þp þ minf1; pg ðn þ jp þ pÞ b kp:

If l < p, we have

B b nþ ð j þ kÞ p ðn þ jpÞ þ minf1; 0; p lg b kp:

Here we have used thatðr þ 1Þl a j when ðr þ 1Þðl þ 1Þ a j þ k. When i kp 1¼ n, we have

B¼ ðn þ kp þ 1Þ nb nþ kp þ 1 n>kp:

When i kp 1 < n < i, write i kp 1¼ n 1 jpþ l ð j b 1; 1 a l a pÞ. Since i¼ n þ ðk jÞp þ l > n, we have k b j and therefore ðr þ 1Þl b r þ 1 b k > k j b0.

Hence

B¼ ðn þ ðk jÞp þ lÞ ðn 1 jpþ lÞ

¼ n þ ðk jÞ p þ p ðn jpþ pÞ ¼ kp:

When i¼ n,

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B¼ n ðn kp 1Þ ¼ n ðn kpÞ ¼ kp for all k b 0.

When i < n, write i¼ n 1 jpþ l ð j b 1; 1 a l a pÞ. Then B b i ði kpÞ ¼ i ðn 1 jp kpþ lÞ

¼ n jpþ p ðn jp kpþ pÞ ¼ kp for all k b 0. Thus we get 5).

For i > n, one can write i ¼ n þ pj þ l with j b 0 and 1 a l a p. Set i0 ¼ n þ p2ðr þ 1Þ þ 1. Then, if i b i0, we have j b pðr þ 1Þ b lðr þ 1Þ and therefore the second half of 6) holds. We have thus finished the proof the lemma. r

4.2. Definition of B and BM

Letting i! i be a correspondence satisfying this lemma, we introduce two Banach spaces B and B0. Let 0< r < 1, R1b1, Rcb1, Rnb1, and set

gðbÞ ¼ b1!jbY00jþbn

j¼1

ðb1þ jpÞ ð4:1Þ

GðbÞ ¼ gðbÞrb1 b1R1b1Rjbc 00jRnbn ð4:2Þ for multi-index b¼ ðb1; . . . ; bnÞ. Denote b! ¼ b1!   bn!, ^hh¼ ð0; . . . ; 0; hÞ and ðz hh^Þb ¼ z1b1    znbn11ðznbn. We say a power series

fðzÞ ¼X

b

fbðz hhÞ^ b=b! ð fb A CÞ

belongs to B and write f A B if sup

b

j fbj

GðbÞ< y: ð4:3Þ

Denote the left hand side by k f k, then it defines a norm, with which B is a Banach space.

In the same way, denoting z0 ¼ ðz2; . . . ; znÞ, we say a power series jðz0Þ ¼X

b0

jb0ðz0 hh^0Þb0=b0! ðjb0 A CÞ

belongs to B0 and write j A B0 if sup

b0

jjb0j

Gðn; b0Þ< y: ð4:4Þ

Denote the left hand side by kjk0, then it also defines a norm, with which B0 is a Banach space.

If fðzÞ is holomorphic in fz; jzij a ri0 ð1 a i a n 1Þ; jzn hj a rn0g, then it belongs to B if R1b1=r10, Rcb1=ðprc0Þ and Rnb1=ðprn0Þ, where rc ¼ minfr2; . . . ; rn 1g. In fact

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jDbfðcÞj a C 1 r10

 b1

   1 rn0

 bn

b1!   bn!

aCb1! Y

jb00jþbn

k¼1

ðb1þ kpÞrb1 b1 1 r10

 b1

1 prc0

 jb00j

1 prn0

 bn

:

In the same way, one can see that if jðz0Þ is holomorphic in fz; jzij a ri0 ð2 a i a n 1Þ;

jzn hj a rn0g, then it belongs to B0 if Rcb1=ðprc0Þ and Rnb1=ðprn0Þ.

Conversely, if f A B, then it is holomorphic in

R1jz1j þ ðn 2Þ pRcjz00j þ pRnjzn hj < 1 ð4:5Þ where jz00j ¼ max2aian 1jzij. In fact, jP

b fbðz cÞb=b!j is estimated

aC1k f kX

b

b1! b!

Y

jb00jþbn

k¼1

ðb1þ kpÞrb1 b1ðR1jz1b1ðRcjz00jb00jðRnjzn hjÞbn

aC2k f k X

b1; bc; bn

b1!Qðb1þ kpÞ

b1!bc!bn! ðR1jz1b1ððn 2ÞRcjz00bcðRnjzn hjÞbn

aC3k f kX

y

m¼0

ðR1jz1j þ ðn 2ÞpRcjz00j þ pRnjzn hjÞm:

If jðz0Þ A B0, in the same way, one can verify that it is holomorphic in ðn 2ÞpRcjz00j þ pRnjzn hj < 1:

4.3. Propositions.

To apply the contraction principle to (3.6), we will need to estimate kQðMhÞ 1wk, kQDnRhjk, kpLhfk0 and k ~QQDnRhjk0. We make here some preparations, employing the notation ð f Þb ¼ Dbfð^hhÞ and ðkÞl ¼ kðk 1Þ    ðk lþ 1Þ.

Denote

I0ðz1D1Þ ¼X

r

s¼0

asð0Þðz1D1Þr s;

I0ðlÞ ¼Xr

s¼0

asð0ÞðlÞr s: 8

>>

>>

<

>>

>>

:

ð4:6Þ

Then, by the assumption (A.1), there exists a positive constant c0 such that

jI0ðkÞj b c0ðk þ 1Þr Ek A Nnfng: ð4:7Þ Therefore, for any holomorphic function gðzÞ with pD1ng¼ 0, the problem

I0ðz1D1ÞvðzÞ ¼ gðzÞ; pD1nv¼ 0 ð4:8Þ has a unique holomorphic solution vðzÞ, which we denote by v ¼ I01g.

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Proposition 4.2. Let ðl; aÞ with anam r satisfy the conditions (1.2), (A.2) and (A.3), or let l¼ a1 ¼ r s and an ¼ m r. Assume (A.1) and rR1 >1, then there exists a positive constant C independent of r; R1;Rc;Rn such that

kz1lDaDnmþrI0 1wk a C Rc R1p

 ja00j

R1p Rn

 m r an

kwk ð4:9Þ

for any w A B.

Proof. Since D1kðz1D1Þ ¼ ðz1D1þ kÞD1k, we have ðz1lDaDnmþrI01b ¼ ðb1Þl

I0ðb1 lþ a1Þwb

1 lþa1;b00þa00;bnþan mþr: From the definition of I0 1, one may suppose b1 lþ a10n.

Set

A¼ ðb1Þl

jI0ðb1 lþ a1Þjgðb1 lþ a1;b00þ a00;bnþ an mþ rÞ and note ja00j þ an mþ r ¼ r s a1. Then we have

A¼ ðb1Þl

jI0ðb1 lþ a1Þjðb1 lþ a1Þ! Y

jb00jþbnþr s a1

k¼1

fðb1 lþ a1Þþ kpg:

It follows from the conditions (1.2), (A.2) and (A.3) that ðb1 l a1Þþ ðr s a1Þp a b1:

In fact, when r s a1b0, we have l a1 > pðr s a1Þ by (A.2) and (A.3), and therefore l a1bpðr s a1Þ þ 1 (because both sides are integers). When r s a1<

0, we have l a1bpðr s a1Þ by (1.2) and (A.2), namely a1 l apða1 rþ sÞ. In both cases, the inequality follows from 4) and 5) of Lemma 4.1.

When r s a1 >0, it holds that ðb1 lþ a1Þa1 jI0ðb1 lþ a1Þj

Y

r s a1

k¼1

fðb1 lþ a1Þþ kpg a C1

with a positive constant C1 independent of b1. Therefore

A a C1ðb1Þl a1ðb1 lþ a1Þ! Y

jb00jþbn

k¼1

fðb1 lþ a1Þþ ðr s a1Þ p þ kpg

aC2b1! Y

jb00jþbn

k¼1

fðb1þ kpÞg ¼ C2gðbÞ

where we used b1 ¼ b1 for large b1 (Lemma 4.1).

When r s a1a0, it holds that

fðb1 lþ a1Þga1 jI0ðb1 lþ a1ÞjQ0

k¼r s a1þ1fðb1 lþ a1Þþ kpg aC3

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with a positive constant C3 independent of b1. Therefore

A a C3ðb1Þlfðb1 lþ a1Þ a1g! Y

jb00jþbn

k¼1

fðb1 lþ a1Þþ ðr s a1Þp þ kpg

aC4b1! Y

jb00jþbn

k¼1

fðb1þ kpÞg ¼ C4gðbÞ where we used b1 ¼ b1 for large b1 again.

On the other hand, if 0< r < 1 and rR1 > 1, then rð b1 lþa1Þ ð b1 lþa1ÞRð b1 1 lþa1ÞRjbc 00þa00jRbnnþan mþr

¼ rb1 b1Rjbc 00jRnbnrl a1ðrR1Þð b1 lþa1Þ b1Rjac00jRann mþr

arb1 b1Rjbc 00jRnbn Rc

R1p

 ja00j

R1p Rn

 m r an

where we used l a1bpðr s a1Þ ¼ pðja00j þ an mþ rÞ. Thus the proof has been

finished. r

Proposition 4.3. Suppose rR1 > 1. Then there exists a positive constant C inde- pendent of r; R1; Rc and Rn such that

kz1g1ðz00Þg00ðzngnwk a Crg1 1 pRc

 jg00j

1 pRn

 gn

kwk ð4:10Þ

for any g¼ ðg1; g00; gnÞ and any w A B.

Proof.

jðz1g1ðz00Þg00ðzngnbj ¼ ðb1Þg

1ðb00Þg00ðbnÞg

njwb1 g1; b00 g00; bn gnj aðb1Þg

1ðb00Þg00ðbnÞg

nGðb1 g1; b00 g00; bn gnÞkwk:

Since b1bb1, we have ðb1Þg1ðb1 g1Þ!

b1! a ðb1 g1Þ!

ðb1 g1Þ! aðb1 g1Þ!

ðb1 g1Þ! a sup

k

k!

k! ¼ max

kai0

k! k!

and therefore

ðb1Þg1ðb00Þg00ðbnÞg

ngðb1 g1; b00 g00; bn gnÞ

aCb1!ðb00Þg00ðbnÞg

n

Y

jb00 g00jþbn gn

k¼1

ðb1þ kpÞ

aCp jg00j gnb1! Y

jb00jþbn

k¼1

ðb1þ kpÞ where C ¼ maxkai0k!=k!.

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On the other hand, rR1 > 1 implies

rð b1 g1Þ ð b1 g1ÞRð b1 1 g1ÞRjbc00 g00jRbnn gn

¼ rg1ðrR1Þð b1 g1Þ b1Rcjg00jRngnrb1 b1R1b1Rjbc 00jRnbn arg1Rcjg00jRngnrb1 b1Rb

 1

1 Rjbc00jRnbn:

Thus we get (4.10). r

Proposition 4.4. Suppose aðzÞ satisfies

jDgað^hhÞj a A0g!R0g ð4:11Þ for all g with positive constants A0 and R0. Then, if

rR0 <1

2; R0 pRc <1

2; R0 pRn <1

2; rR1 > 1 ð4:12Þ there exists a constant C independent of r; R1; Rc; Rn such that

kawk a CA0kwk; w A B: ð4:13Þ

Proof. In the same way as in the proof of Proposition 4.3, we have jðawÞbj ¼ X

g

b g

 

agwb g

a X

g

A0ðb1Þg

1ðb00Þg00ðbnÞg

nRg01þjg00jþgnGðb1 g1; b00 g00; bn gnÞkwk

aA0C1X

g

ðrR0Þg1 R0 pRc

 jg00j

R0 pRn

 gn

GðbÞkwk:

Therefore, if (4.12) is fulfilled, the inequality (4.13) holds with a positive constant C

independent of r; R1; Rc and Rn. r

Now, one can write Lh appearing in Proposition 2.1 as Lh ¼ X

a1þja00jan

bah

1; a00ðz00ÞD1a1D00a00

where bn; 0h ¼ 1, ja00j < sðn a1Þ if a1 < n and bah1; a00ðz00Þ are holomorphic functions.

Taking d su‰ciently small, one may suppose jDg00bah

1; a00ð0Þj a A0g00!Rjg000j ð4:14Þ for all g00 with positive constants A0; R0 independent of jhj < d and ða1; a00Þ.

Proposition 4.5. Take d su‰ciently small and suppose (4.12), (4.14), ðRc=R1pÞ < 1 and jhj < d. Then there is a positive constant C independent of h; r; R1; Rc and Rn such that

kpLhfk0aCk f k: ð4:15Þ

References

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