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Finitely smooth solutions of nonlinear singular partial differential equa- tions

Masafumi Yoshino

Graduate School of Sciences, Hiroshima University, Kagamiyama 1-3-1, Higashi-Hiroshima 739-8526, Japan Received 15 November 2003, revised 30 November 2003, accepted 2 December 2003

Published online 3 December 2003

Key words Fuchsian equations, totally characteristic equations, singular partial differential equations, normal form, vector fields, Monge Amp`ere equation.

MSC (2000) Primary 35G20; Secondary, 37C10, 37J40

This work is partially supported by Grant-in-Aid for Scientific Research (No. 16654028), Ministry of Education, Science and Culture, Japan. This work was partly done when the author stayed in the university of Cagliari.

The author expresses appreciations to Prof. T. Gramchev and Department of Mathematics, Univ. of Cagliari for the invitation and the supports.

We solve a Fuchsian system of singular nonlinear partial differential equations with resonances. These equa- tions have no smooth solutions in general. We show the solvability in a class of finitely smooth functions.

Typical examples are a homology equation for a vector field and a degenerate Monge-Amp`ere equation.

Copyright line will be provided by the publisher

1 Introduction

This paper is concerned with the solvability of a Fuchsian system of a singular nonlinear partial differential equations in a bounded domainΩRn or inRn. These equations naturally appear when we solve a class of Monge-Amp `ere equations or when we linearize a singular vector field by a coordinate change. (See§2). As we can see from the simple exampleLu:= (tdtd 1)u=t, these equations do not have a smooth solution in general.

Indeed, ifu(t) =c0+c1t+v(t),v=O(t2)is a solution, then the relationsL(c0+c1t) =−c0andLv=O(t2) imply thatuis not smooth att = 0. In fact, if we allow a singular solution, then we see thatu=ct+tlogt, (c, constant) gives a solution. Here we take the branch of the logarithm such thatlog 1 = 0. If we restricttto the real line, thenugives a H¨older continuous function on the real line. Similar property holds forLu= x1, whereL=x1∂x

1 +mx2∂x

2 1,(m >1). The equationLu=x1has no smooth solution at the origin, while u=cx1+x1logx1, (c, constant) is a singular solution. It gives a H¨older continuous function on the real line for an appropriate choice of the branch oflogx1. We also note that this phenomenon is closely related with a Grobman-Hartman theorem. (cf. Remark 2.9). These examples are known as a so-called totally characteristic type partial differential equation.(cf. [3]). As to formal solutions of nonlinear first order totally characteristic type equations we refer [3], and as to singular solutions of nonlinear singular partial differential equations we refer [19]. We also remark a related work [11] concerning symbolic calculus on manifolds with edges.

The object of this paper is to solve this type of equations in a class of finitely smooth functions. For this purpose we employ a rapidly convergent iteration method in a class of non smooth functions, because the Fuchsian equations have a loss of regularity. We stress that the usual rapidly convergent iteration scheme is not useful in order to solve this type of equations, because one requires high regularity in the iterative scheme, while our solution does not have such smoothness in general. We introduce a partial smoothing operator which preserves the vanishing order of approximate solutions on every coordinate axis. This smoothing operator is useful in the iterative scheme because the Fuchsian partial differential operators which we study in this paper lose derivatives of the transversal direction of every coordinate axis, although they preserve the vanishing order. Concerning the loss of regularity of nonlinear equations (of multiple characteristics) we refer [5], [7] and [18].

e-mail:yoshino@math.sci.hiroshima-u.ac.jp, Phone: +81 0824 24 7350, Fax: +81 0824 24 0710

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This paper is organized as follows. In§2 we state the main theorem and we give several consequences and applications. In§3 we prepare lemmas which are necessary for the proof of the main theorem. The proof of the main theorem is given in§4 by using a rapidly convergent iteration method.

2 Statement of results

Letx = (x1, . . . , xn) Rn be the variable inRn (n 2). For a multiintegerα = (α1, . . . , αn) Zn+, Z+={0,1,2, . . .}we set|α|=α1+· · ·+αn. We define

j=∂/∂xj, δj =xjj(j= 1, . . . , n), δα=δα11· · ·δαnn. Letm≥1,m≥s≥0,N 1be integers, and let

pj(δ) =

|α|≤m

aαjδα, (aαjR, j= 1, . . . , N)

be Fuchsian partial differential operators. Let aj(x, z), z= (zα)|α|≤s j= 1, . . . , N,

be real-valuedCfunctions of(x, z)Rn×Ω, whereΩRkN,(k= #{α∈Zn+;|α| ≤s})is a neighborhood of the origin.

We study the solvability of the system of equations foru= (u1, . . . , uN)

Gj(u) :=pj(δ)uj+aj(x, δαu;|α| ≤s) = 0, j= 1, . . . , N. (2.1) Letσbe a nonnegative number, andΓbe a domain ofRn. We defineHσ ≡Hσ,Γ as the set of holomorphic (vector) funtionsv(ζ) = (v1(ζ), . . . , vN(ζ))ofζ=η+iξ∈Γ +iRnsuch that

vσ,Γ:= sup

η∈Γ

Rnζσ|v(ζ)|dξ <∞, (2.2) whereζ= 1 +n

j=1j|, and|v(ζ)|= (N

j=1|vj(ζ)|2)1/2. The spaceHσ,Γis a Banach space with the norm (2.2). The fundamental properties ofHσ,Γis given in Proposition 3.1 which follows.

Letf(x)be an integrableN- vector function onRn+, R+ := {t R;t 0} and letfˆ(ζ)be the Mellin transform off

fˆ(ζ)≡M(f)(ζ) =

Rn+f(x)xζ−edx, e= (1, . . . ,1), ζ=η+iξ, η∈Γ, ξ∈Rn, (2.3) wherexζ−e = xζ111· · ·xζnn1, ζ = (ζ1, . . . , ζn). It is easy to see thatfˆ(ζ)is analytic if the integral (2.3) absolutely converges. The inverse Mellin transform is given by

f(x) =M1( ˆf)(x) = (2π)−n

Rn

fˆ(η+)x−η−iξdξ, (2.4) wherexj >0 (j = 1, . . . , n)andηis so taken that the integral converges. We note that these formulas follow from the corresponding ones of the Fourier transform by the change of variableseθj →xj.

We defineHσ,Γas the inverse Mellin transform ofHσ,Γ. We note that the Mellin transform gives the one to one correspondence between the spacesHσ,ΓandHσ,Γ. Foru∈ Hσ,Γwe define the normuσ,Γofuby

uσ,Γ:= M(u)σ,Γ.

For an integerk≥1we denote by(Hσ,Γ)kthe product ofkcopies ofHσ,Γ. The norm in(Hσ,Γ)k is defined as the sum of the norm of each component. For simplicity, we denote the norm in(Hσ,Γ)k by · σ,Γif there is no fear of confusion.

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Letpj(ζ) =

|α|≤maαj(−ζ)αbe the indicial polynomial associated withpj(δ), whereζ= (ζ1, . . . , ζn)is the covariable ofxin the sense of the Mellin transform. We assume

(A.1) There exists a constantc >0such that

|pj(η+)| ≥c(|η|+|ξ|)s, ∀η∈Γ,∀ξ∈Rn, j= 1, . . . , N.

We seta(x, z) = (a1(x, z), . . . , aN(x, z)). Then we assume thata(x, z)(C(Rn×Ω))N and (A.2) ∀α∈Zn+,∀β∈ZkN+ ,∃Cαβ >0such that

|(∂/∂z)βδxαa(x, z)| ≤Cαβ, (x, z)Rn×.

Then our main theorem in this paper is the following

Theorem 2.1 Letσ≥mbe an integer. Suppose that(A.1)hold for some bounded domainΓRncontaining the origin. Assume (A.2). Then there exist an integerν =ν(σ)0and anε=ε(σ)>0depending onσsuch that, if the following conditions are satisfied

a(·,0)ν,Γ < ε, za(·,0)ν,Γ< ε,

then Eq. (2.1) has a solutionu∈(Hσ,Γ)N.

Next we study the local solvability. We say thatu∈(Hν,Γ)N at the origin if there exists aψ∈C(Rn)with compact support and being identically equal to one in some neighborhood of the origin such thatψu∈(Hν,Γ)N. For open setsΓ1 Rn andΓ2 Rnthe relationΓ1 ⊂⊂ Γ2 meansΓ1 Γ2, whereΓ1is the closure ofΓ1. Then we have

Theorem 2.2 Letσ≥mbe an integer. Suppose that(A.1)holds for some bounded domainΓcontaining the origin. Then there exists an integerν 0such that, if

a(x,0)(Hν,Γ)N and za(x,0)(Hν,Γ)kN at the origin,

then there exists a solutionu∈(Hσ,Γ)N of (2.1) in some neighborhood of the origin for everyΓ⊂⊂Γ.

Remark 2.3 a) Theorem 2.1 and Theorem 2.2 yield the solvability of (2.1) in some neighborhood of the origin in a class of finitely smooth functions. Indeed, we can solve (2.1) in the sectorsjxj 0;j= 1, . . . , n}, (εj =±1), after the change of variablesxj →εjxj,(j = 1, . . . , n), becauseδjis invariant under the change of variables. By the assumption0 Γand the definition ofHν,Γ, the solutionutogether with the derivatives δαu,|α| ≤svanishes (to a finite order) on the coordinate planesxj = 0 (j = 1, . . . , n). (See Proposition 3.1.) Hence, by patching the solutions in these sectors we obtain a finitely smooth solution in some neighborhood of the origin.

b) (Bifurcation from a resonance) The uniqueness of solutions in Theorem 2.1 and Corollary 2.2 does not always hold if there is a resonance. Indeed, we consider the equation

p(δ)u+λa(x, u) = 0, a(x, u) =O(|u|2),

whereuis a scalar unknown function,λis a real parameter, and wherep(δ)is an Fuchisian partial differential operator similar topj(δ)in (2.1). We note thatu≡0is a trivial solution of the equation. We assume (A.1) for some domainΓ 0. Then we shall show that the above equation has a non trivial family of solutionsu=uλ, uλ=λu0+vλfor sufficiently smallλ, whereu0satisfiesp(δ)u0= 0.

First we note that there existsu0such thatp(δ)u0 = 0if there is a resonance. (See also Example 2.8 which follows.) If we setv=vλ, thenvsatisfies

p(δ)v+λa(x, λu0+v) = 0.

The conditions in Theorem 2.1 read:

λa(·, λu0)ν,Γ < ε and λ∇ua(·, λu0)ν,Γ< ε.

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These conditions are satisfied for sufficiently small λif a(x, λu0) ∈ Hν,Γ andua(x, λu0) ∈ Hν,Γ for all λclose to 0. For example, if the local solvability is concerned, these conditions are verified if a(x, λu0)and

ua(x, λu0)vanish to some order for all sufficiently smallλ. (We also refer Lemma 4.2 which follows.) It follows from Theorem 2.1 or Corollary 2.2 that there exists a solutionvfor sufficiently smallλ. Moreover, by the constructions of an approximate sequencewkin (4.12), we havev= limkwk and

w1=S0ρ0, L0ρ0=g0=−λa(x, λu0), . . .

It follows from the assumption onathat the vanishing order ofg0at the origin is greater thanu0. Therefore, we see that the vanishing order ofw1at the origin is greater than that ofu0, becauseL10 andS0 preserve the vanishing order. Inductively, we can easily see that the vanishing order at the origin of the solutionv = limwk is greater than that ofu0. It follows thatu= λu0+v = 0. Therefore, we have a family of solutions of our equation.

Remark 2.4 The smallness conditions in Theorem 2.1 for the nonlinear parta(x,0)andza(x,0) of the equation (2.1) are fulfilled if the following conditions are satisfied

a(x,0) = 0, zaj(x,0) = 0, j= 1, . . . , n. (2.5) On the other hand, the condition (A.2) in Theorem 2.1 is fulfilled ifa(x, z)is independent ofxora(x, z)has a compact support with respect tox.

Example 2.5 We give the example which satisfies (A.1). Let

p2(ζ) :=ζ12 n j=2

cjζj2, cj >0.

Letp1(ζ)be a linear function ofζwith real coefficients. We setp(ζ) =p2(ζ) +p1(ζ). We assume that p1(ξ) +η· ∇p2(ξ)= 0 for∀η∈Γ, and∀ξ∈Rn such that p2(ξ)0,|ξ|= 1.

We want to show that there exists real numberKsuch thatp(ζ) +Ksatisfies (A.1) withs= 1. We have p(η+) +K=K−p2(ξ) +p(η) +i(p1(ξ) +η· ∇p2(ξ)).

Becauseη moves in a bounded set it follows that ifK > 0is sufficiently large, the zero set of the polynomial ofξ,p(η+) +K is contained in the setp2(ξ)0,|ξ| ≥ 1, wherepis the real part ofp. On the other hand, by assumption and the homogeneity, the imaginary partp(η+)does not vanish on the setp2(ξ)0,

|ξ| ≥1. It follows thatp(η+) +K= 0for allη Γandξ.

In order to show (A.1) withs= 1it is sufficient to considerξsuch that|ξ| ≥N >0for largeN. Ifξis in a conical neighborhood ofξ0such thatp2(ξ0)= 0, we have (A.1) withs= 2. If otherwise, the assumption implies thatp1(ξ) +η· ∇p2(ξ)= 0. Hence we have

|p(η+)| ≥ |p(η+)| ≥c|ξ| ≥c(|ξ|+|η|) for somec >0andc>0. This proves (A.1) withs= 1.

Example 2.6 We writex1=x,x2=y, and we consider the Monge-Amp `ere operator M(u) :=uxxuyy−u2xy+kxyuxy+cu, 4< k <12, c∈C.

Letu0=x2y2and setf0=M(u0) = (4k−12 +c)x2y2. We want to solve the equation M(u0+v) =f0(x, y) +g(x, y), inR2,

whereg(x, y)is a given function. If we define

Q= 2x2x2+ 2y2y2+ (k−8)xy∂xy+c, M˜(u) =M(u)−kxyuxy−cu,

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then the equation can be written in the form

Qv+ ˜M(v) =g.

In order to write the equation in the form (2.1) we introduce a new unknown functionwbyv(x, y) =x2y2w(x, y).

By simple computations we have x2y2M˜(x2y2w)

= (x2wxx+ 4xwx+ 2w)(y2wyy+ 4ywy+ 2w)(xywxy+ 2xwx+ 2ywy+ 4w)2. x2y2Q(x2y2w)

= 2(δ2x+ 4δx)w+ 2(δy2+ 4δy)w+ (k−8)(δxδy+ 2δx+ 2δy)w+ (4k−24 +c)w,

whereδx =x∂/∂xandδy =y∂/∂y. This proves that our equation can be written in the form (2.1). We note that the condition (A.2) is fulfilled. (cf. Remark 2.4).

The indicial polynomial is given by

p(ζ) := 2(ζ124ζ1) + 2(ζ224ζ2) + (k−8)(ζ1ζ22ζ12ζ2) +c+ 4k−24.

We will show (A.1) withs= 2for some bounded domainΓRn\ {p(η) = 0}containing the origin ifc=iK, K >0is sufficiently large . We note thatp(ξ)is elliptic by the condition4< k <12. It follows that there exist ξ0>0andα >0independent ofKandηsuch thatp(η+)≥α|ξ|2if|ξ|> ξ0andη∈Γ. If|ξ| ≤ξ0, then p(η+)does not vanish ifKis sufficiently large. Therefore we have (A.1) withs= 2.

Next we apply our argument to the normal form theory of a singular hyperbolic vector fieldχ=n

j=1Xj(x)j,

j =∂/∂xjonRn. We say thatχis singular ifXj(0) = 0 (j = 1, . . . , n). We setX = (X1, . . . , Xn). For the sake of simplicity, we assume

X(x) =xΛ +R(x), R(x) = (R1(x), . . . , Rn(x)), (2.6) for a real-valuedCfunctionRj(x)such thatRj(0) = 0,∇Rj(0) = 0, and a diagonal matrixΛ =

diag(λ1,· · ·, λn),λj R. We want to find a change of variablesy →x =y+v(y)which linearizesχ. It follows thatvsatisfies the so-called homology equation

X(y+v(y))(1 +∇v)1=yΛ,

or equivalently,

Lv=R(y+v(y)), Lv:=

n j=1

λjδjv−vΛ. (2.7)

We definep(ζ) =n

j=1ζjλjI−Λ, whereIis an identity matrix. Then we have

Theorem 2.7 Suppose that (A.1) is satisfied fors = 0and some bounded domainΓcontaining the origin.

Assume (2.6). Letσ≥1be an integer. Then there existsν≥0such that, if the following conditions are satisfied R∈(Hν,Γ)n, ∇Rj(Hν,Γ)n2 at the origin (j= 1, . . . , n),

then Eq. (2.7) has a solutionv∈(Hσ,Γ)nfor everyΓ⊂⊂Γ.

Example 2.8 We give examples which satisfy (A.1). Suppose that λ1· · ·λn = 0. By definition thek-th component ofp(ζ) (ζ = η+)is given byn

j=1ηjλj−λk. Hence the set ofη such thatp(ζ) = 0 consists ofnhyperplanes,

jηjλj+λk= 0not passing through the origin. Therefore we have (A.1) withs= 0 for some open setΓcontaining the origin. The followings are typical cases which satisfy (A.1).

(i)Poincar´e case; i.e.,λj>0 (j= 1, . . . , n).

(ii)Nonresonant Siegel case; namely, someλjare positive and others are negative, andp(ζ) = 0 (ζ∈Zn+,|ζ| ≥ 2)has no solution.

(iii)Infinite resonances case; that is,p(ζ) = 0 (ζ∈Zn+)has an infinitely many solutions.

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The third case contains a volume preserving vector fields, namelyn

j=1λj = 0. In the case(i)the set ∈ −Rn+;p(η+) = 0 for someξ∈Rn}

is a compact set not containing the origin. Hence we can takeΓin (A.1) as a bounded domain inRn+\ {p(ζ) = 0}. In the case(ii)the intersection of the hyperplanes

jηjλj +λk = 0andRn+ is noncompact. Hence the setΓin (A.1) may be a smaller set. In the case(iii)there is an additional restriction toΓdue to an infinite resonances apart from the ones caused by a Siegel condition. We note that the larger the setΓis, the more regular the solution is.

Remark 2.9 By Remark 2.3 and Theorem 2.7 we can construct a finitely smooth coordinate change which linearizesχeven in the case of resonances. It is natural to ask wether there exists aCcoordinate change which linearizesχ. The answer to this question is not affirmative. Indeed, if the vector field has a resonance,Lhas a (infinite) kernel. It follows that if (2.7) has aC solutionv, then the Taylor expansion ofvat the origin gives a formal power series solution of (2.7). Hence the Taylor expansion ofR satisfies a compatibility condition.

Because we do not assume any compatibility condition a priori, the solution is not smooth in general. We stress that the regularity of the solution is related with the property of a resonance as we note in the preceeding example. If we assume the weaker conditionλ1· · ·λn = 0, the solution is continuous. We remark that this fact was essentially noted as a Grobman-Hartman theorem for a vector field, which asserts the existence of a continuous solution of a homology equation (cf. [1], p.127 and p191).

Theorem 2.7 can be extended to a commuting system of hyperbolic singular vector fields onRn, χ=µ;µ= 1, . . . , d}, [χµ, χν] = 0 for allνandµ.

We write χµ = n

j=1Xjµ(x)j and setXµ = (X1µ, . . . , Xnµ). For the sake of simplicity we assume that Xµ(x) =xΛµ+Rµ(x)for some real-valuedCvector functionRµsuch that

Rµ(0) = 0, ∇Rµ(0) = 0, and diagonal matrices

Λµ=diag(λµ1,· · ·, λµn), λµj R, µ= 1, . . . , d.

We are interested in the simultaneous linearization ofχby the change of variablesy x= y+v(y). It follows thatvsatisfies an overdetermined system of equations

Lµv=Rµ(x+v),

whereLµ is similarly given by (2.7). LetC be a positive cone generated by the vectors(λ1j, . . . , λdj) Rd, (j= 1, . . . , n), namely

C:={ n j=1

tj(λ1j, . . . , λdj)Rd;tj 0,(j= 1, . . . , n), t21+· · ·+t2n= 0}.

We say thatχ satisfies a simultaneous Poincar´e condition if the cone C does not contain the origin. In case d= 1, this condition is equivalent to that the quantityt1λ11+· · ·+tnλ1n does not vanish fortj 0such that t21+· · ·+t2n= 0. The last condition is equivalent to say thatλ11>0, . . . , λ1n >0. This is a well-known Poincar´e condition for a single vector field. We have

Theorem 2.10 Letσ 1. Suppose that the simultaneous Poincar´e condition is satisfied. Then there exists ν 0such that, if

Rµ(Hν,Γ)n and ∇Rµ(Hν,Γ)n2 at the origin forµ= 1, . . . , d,

thenχis simultaneously linearized in some neighborhood of the origin by the change of the variablesy →x= y+v(y), withv (Hσ,Γ)n,∀Γ⊂⊂Γ.

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3 Some lemmas

In this section we will prepare lemmas which are necessary in the calculus of a class of pseudo-differential operators of totally characteristic type in a Mellin’s sense. We cite [11] concerning symbolic calculus of operators on manifolds with edges.

LetΓbe an open set inRn. First we study fundamental properties ofHs,Γ(s∈R+)defined in§1.

Proposition 3.1 (1)Lets 0be an integer and letuˆ Hs,Γ. Then the inverse Mellin transformu(x) = M1u)(x)ofuˆis a bounded continuous function on Rn+such that for everyα,|α| ≤sandη∈Γ, the function xηδαu(x)is continuous and satisfies

xηδαu(x) 0 as xj 0, j= 1, . . . , n, (3.1)

xηδαu(x) 0 as xj +∞, j= 1, . . . , n. (3.2) Moreover, for everyΓ⊂⊂Γthere existsc >0independent ofuˆsuch that

sup

x∈Rn+,|α|≤s,η∈Γ

|xηδαu(x)| ≤cuˆs,Γ, ∀uˆ∈Hs,Γ. (3.3)

(2)Lets≥0be an integer and letu(x)be any bounded continuous function on Rn+satisfying (3.1) and (3.2) for everyη Γ. Then the Mellin transformu(ζ) =ˆ M(u)(ζ)ofuexists andu(ζis holomorphic inΓ +iRn. Moreover, for everyΓ⊂⊂Γ⊂⊂Γthere existC >0such that

ζs|u(ζ| ≤C sup

x∈Rn+,|α|≤s,η∈Γ|xηδαu(x)|, ∀ζ,ζ∈Γ (3.4) whereζ= 1 +n

j=1j|.

(3) Hs,Γis a Banach space with the norm (2.2).

Proof. We will prove (1). The inverse Mellin transform ofuˆexists becauseuˆ∈Hs,Γ. Moreover we have xηδαu(x) = (2πi)−n

Rn

(−ζ)αu(ζ)xˆ η−ζdξ, η Γ,ζ∈Γ. (3.5) We takeηandζin (3.5) such thatηj− ζj >0ifxj <1,ηj− ζj <0ifxj 1. We easily see that (3.1) and (3.2) hold. The estimate (3.3) follows from (3.5) because|xη−ζ|is bounded by some constant.

We prove (2). The conditions (3.1) and (3.2) withα= 0imply that the Mellin transformM(u)(ζ)exists and it is holomorphic inΓ +iRn. In order to show (3.4), we first note that the right-hand side of (3.4) is finite by (3.1) and (3.2). It follows from (3.1) and (3.2) that, for|α| ≤s

ζαu(ζ) =ˆ

u(x)ζαxζ−edx=

u(x)(x·x)αxζ−edx=

Rn+δαu(x)xζ−edx. (3.6) Letτj(j= 1, . . . , n)be such thatτj = 1orτj=1and defineτ = (τ1, . . . , τn). We defineSτby

Sτ ={x= (x1, . . . , xn)Rn+; 0≤xτjj 1}.

By (3.6), there existsC >0independent ofζsuch that, ifζ∈Γ ζs|u(ζ| ≤C sup

|α|≤s

Rn+xζ−eδαu(x)dx

≤C sup

|α|≤s

τ

Sτ

xζ−eδαu(x)dx

. (3.7)

By assumption, for eachSτwe take anη=η(τ) = (η1, . . . , ηn)and a smallε1>0such thatζj−ηj> ε1>0 ifτj = 1, andζj−ηj ≤ −ε1ifτj =1. For a givenΓ ⊂⊂Γ ⊂⊂Γwe can chooseε1so small that η∈Γ.

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Therefore, there existsC>0independent ofζsuch that

Sτ

xζ−eδαu(x)dx =

Sτ

xζ−e−ηxηδαu(x)dx

sup

x∈Rn+|xηδαu(x)|

Sτ

xζ−e−ηdx

≤C sup

x∈Rn+|xηδαu(x)|. (3.8)

Hence there existsC >0such that ζs|u(ζ| ≤C sup

x∈Rn+,η∈Γ,|α|≤s|xηδαu(x)|. This proves (3.4).

We will prove (3). In order to show thatHs,Γ is complete, suppose that wˆn−wˆm s,Γ 0 (m, n → ∞).

It follows from (3.3) and (3.4) that {wˆn(ζ)} converges compactly uniformly inζ Γ to a functionw(ζ) holomorphic inζ Γ +iRn. Letη Γbe arbitrarily taken and fixed. By assumption, for everyε > 0there existsN 1such that

ζs|wˆn(ζ)−wˆm(ζ)|dξ < ε, ∀n, m≥N.

It follows that, for any compact setK⊂Rnwe have

K

ζs|wˆn(ζ)−wˆm(ζ)|dξ < ε, ∀n, m≥N.

We letm → ∞. Then we have

Kζs|wˆn(ζ)−w(ζ|dξ εfor alln N. Letting K Rn we obtain

Rnζs|wˆn(ζ)−w(ζ|dξ εfor alln N. By taking the supremum with respect toη Γ, we see that wˆn−wˆ∈Hs,Γand{wˆn}converges towˆinHs,Γ.2

Now we define a smoothing operator inHs,Γ. Letφ ∈C(Rn),0 φ 1 be a smooth function with a compact support such thatφ≡1in some neighborhood of the originx= 0and

Rnφ(σ) = 1. LetN 1, 1be integers and letτbe an odd integer,2τ≥. We setψN(ζ) := exp(N2τn

j=1ζj2τ)and define χN(ζ) :=

Rn

φ(σ)

ψN(ζ)

e−σζ/N ν=1

−σζ N

ν 1 ν!

+ (1−ψN(ζ))e−σζ/N

dσ. (3.9)

The functionχN(ζ)is an entire function ofζinCnsuch thatχN(ζ) =χNζ). We define a smoothing operator SN by

SNv:=M1(χN+1(ζv(ζ)), v∈ Hs,Γ (3.10)

wherev(ζ)ˆ is the Mellin transform ofvandM1denotes the inverse Mellin transform. Then we have Proposition 3.2 LetΓbe a bounded domain. ThenSN has the following properties.

(1) For every0≤s≤rsuch thatr−sis an integer, there existsCr>0such that SNvr,Γ≤Cr(N+ 1)r−svs,Γ, v∈ Hs,Γ.

(2) For every0≤s≤rsuch thatr−s≤is an integer, there existsCr>0such that (I−SN)vs,Γ≤Cr(N+ 1)s−rvr,Γ.

(3) SN maps a real-valued function to a real-valued function.

(9)

Proof. Proof of (1). In view of the definition of the normSNvr,Γwe consider

ζrN+1(ζv(ζ)|dξ, ζ=η+iξ, η∈Γ. (3.11)

Writingζr=ζr−sζsand recalling thatr−sis a nonnegative integer we have thatζr−s= (1+

j|)r−s is a polynomial ofj|. Hence we will estimateαχN+1(ζ)|(|α| ≤r−s). In view of (3.9) we consider

φ(σ)(1−ψN+1(ζ))ζαe−σζ/(N+1)

=

φ(σ)(1−ψN+1(ζ))((N + 1)σ)αe−σζ/(N+1)

= (N+ 1)|α|

σαφ(σ)(1−ψN+1(ζ))e−σζ/(N+1)dσ. (3.12)

In order to estimate the right-hand side, we note

ψN+1(ζ) = exp

⎝ 1 (N+ 1)2τ

n j=1

(ηj2+ 2jξj−ξ2j)τ

.

Becauseτis an odd integer,ψN+1(η+)tends to zero forN = 1,2, . . . whenξtends to infinity for a bounded η. Similarly,e−σζ/(N+1)is bounded forN = 1,2, . . . whenξ → ∞andηis bounded. Hence the term (3.12) can be estimated byCr(N+ 1)|α|≤Cr(N+ 1)r−sfor some constantCr>0.

We consider the term

I:=

φ(σ)ψN+1(ζ)

e−σζ/(N+1) ν=1

σζ N+ 1

ν 1 ν!

ζαdσ.

By settingt= (t1, . . . , tn) =ζ/(N+ 1)we have

I= (N+ 1)|α|

φ(σ)ψN+1(t(N+ 1))(e−tσ ν=1

(−σt)ν(ν!)1)tαdσ.

BecauseψN+1(t(N+ 1))is exponentially decreasing to zero whent → ∞forN = 1,2, . . ., the integrand is uniformly bounded forξ Rn andN = 0,1,2, . . .. Therefore we see thatαχN+1(ζ)|(|α| ≤ r−s)is bounded byCr(N+ 1)r−sfor some constantCr >0which is uniform inη∈Γ,N = 0,1,2, . . . andξ∈Rn,

|ξ| → ∞. It follows thatr−sχN+1(ζ)|is bounded byCr(N+ 1)r−sfor some constantCr >0 which is uniform inη Γ,ξ∈RnandN = 0,1,2, . . . By (3.11) we obtain (1).

Proof of (2). By (3.10) we have

(I−SN)vs,Γ= (I−χN+1vs,Γ.

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For the sake of simplicity we sets−r=a≤0. Recalling that

φ(σ)= 1andr−s≤we have

χN+1(ζ)1 =

φ(σ)ψN+1(ζ)

e−σζ/(N+1) ν=1

σζ N+ 1

ν 1 ν!

(3.13)

+

φ(σ)(1−ψN+1(ζ))e−σζ/(N+1)dσ−

φ(σ)

=

φ(σ)ψN+1(ζ)

e−σζ/(N+1)

ν=0

σζ N+ 1

ν 1 ν!

+

φ(σ)(1−ψN+1(ζ))e−σζ/(N+1)dσ−

φ(σ)+

φ(σ)ψN+1(ζ)

=

φ(σ)ψN+1(ζ)

e−σζ/(N+1) ν=0

σζ N+ 1

ν 1 ν!

+

φ(σ)(1−ψN+1(ζ))(e−σζ/(N+1)1)dσ≡I1+I2. By the definition of the norm we consider

(χN+1(ζ)1)ζa=ζaI1+ζaI2.

As to the termζaI1, we setζ=t(N+ 1). Then the integrand is equal to φ(σ)N t+taψN+1(tN +t)(e−tσ

ν=0

(−σt)ν(ν!)1).

We note that

N t+t= 1 + (N+ 1)

j

|tj|. If

j|tj| ≥ε >0for someε, we haveN t+t ≥(N+ 1)ε. Hence it follows thatN t+ta(N+ 1)aεa. BecauseψN+1(tN +t)is an exponentially decreasing function of(tj)2τ whent → ∞the integrand is bounded byC(N+ 1)afor someC >0independent oft.

Next we consider the case

j|tj|< ε. BecauseN t+t ≥(N+ 1)

j|tj|we have N t+ta (N+ 1)a(

j

|tj|)a.

Hence we have

(|t1|+· · ·+|tn|)a(e−tσ ν=0

(−σt)ν(ν!)1) = (|t1|+· · ·+|tn|)a

ν=+1

(−σt)ν(ν!)1. (3.14)

Noting that

|σt| ≤(|t1|+· · ·+|tn|)(1|+· · ·+n|)

and−a , the right-hand side of (3.14) is bounded by some constant independent oft. HenceζaI1 is estimated byC(N+ 1)afor someC >0independent ofζ.

We will estimateζaI2. By settingζ=t(N+ 1)we consider the term J (1exp(

t2τj ))(e−tσ1).

参照

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