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N.M. Tri

ON THE GEVREY ANALYTICITY OF SOLUTIONS OF

SEMILINEAR

PERTURBATIONS OF POWERS OF THE MIZOHATA

OPERATOR

Abstract. We consider the Gevrey hypoellipticity of nonlinear equations with mul-tiple characteristics, consisting of powers of the Mizohata operator and a Gevrey analytic nonlinear perturbation. The main theorem states that under some con-ditions on the perturbation every Hs solution (with ss0) of the equation is a Gevrey function. To prove the result we do not use the technique of cut-off functions, instead we have to control the behavior of the solutions through a fun-damental solution. We combine some ideas of Grushin and Friedman.

1. Introduction

Since a Hilbert’s conjecture the analyticity of solutions of partial differential equations has at-tracted considerable interest of specialists. The conjecture stated that every solution of an an-alytic nonlinear elliptic equation is anan-alytic (see [1]). This conjecture caused considerable re-search by several authors. In 1904 Bernstein solved the problem for second order nonlinear elliptic equations in two variables [2]. For equations with an arbitrary number of variables the problem was established by Hopf [3], Giraud [4]. For a general elliptic system it was proved by Petrowskii [5], Morrey [6], and Friedman [7]. The last author also obtained results on the Gevrey analyticity of solutions. At present the results on the analyticity and Gevrey analyticity of solutions of linear partial differential equations have gone beyond the frame of the elliptic theory. A linear partial differential equation is called Gevrey hypoelliptic if every its solution is Gevrey analytic (in this context we can assume a solution in a very large space such as the space of distributions), provided the data is Gevrey analytic. Recently the class of linear Gevrey hypoelliptic equations with multiple characteristics has drawn much interest of mathematicians (see [8], and therein references). In this paper we consider the Gevrey analyticity of solutions in the Sobolev spaces of the nonlinear equation with multiple characteristics:

M2khux1,x2,u, . . . ,

l1+l2ul1

x1∂

l2

x2

!

l1+l2≤(h−1)

=0,

where u = u(x1,x2), M2k = x1 +i x2k1 x2, the Mizohata operator inR2. The Gevrey hy-poellipticity for the equation was obtained in [9] in the linear case. We will prove a theorem for a so-called Grushin type perturbationϕ(see the notations in §2 below). The analyticity of

Dedicated to my mother on her sixtieth birthday.

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C∞solutions was considered in [10]. The paper is organized as follows. In §2 we introduce notations used in the paper and state some auxiliary lemmas. In §3 we state the main theorem and prove a theorem on C∞regularity of the solutions. In §4 we establish the Gevrey analyticity of the solutions.

2. Some notations and lemmas

First let us define the function

F2kh(x1,x2,y1,y2)= 1 2π(h−1)!

(x1−y1)h−1 x12k+1−y2k1+1

2k+1 +i(x2−y2)

.

For j=1, . . . ,h−1 we have (see [11])

M2kj F2kh = 1 2π(hj−1)!

(x1y1)hj−1

x12k+1−y12k+1

2k+1 +i(x2−y2)

and M2kh F2kh =δ(xy) ,

where x=(x1,x2), y=(y1,y2)andδ(·)is the Dirac function. We will denote bya bounded domain inR2with piece-wise smooth boundary. We now state a lemma on the representation formula. Its proof can be found in [10].

LEMMA4.1. Assume that uCh()¯ then we have

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u(x)= Z



(−1)hF2kh(x,y)M2khu(y)d y1d y2

+ Z

∂

 

hX−1

j=0

(−1)jM2kj u M2khj−1F2kh(x,y)

(n1+i y12kn2)ds.

For reason of convenience we shall use the following Heaviside function

θ (z)=

1 if z≥0,

0 if z<0.

We then have the following formuladtdmm(zn)=n. . . (nm+1)θ (nm+1)znm.

We will use∂1α,∂2β,γ∂α,β,∂1yα ,∂

β

2yandγ∂ y

α,βinstead of ∂

α

x1α,

∂β ∂x2β, x

γ

1 ∂ α+β

x1α∂x2β,

∂α ∂y1α,

∂β ∂y2β and

yγ1 ∂α+β

yα

1∂y

β

2

, respectively. Throughout the paper we use the following notation

n!=

n! if n≥1,

1 if n≤0, and C

n= Cn if n≥1, 1 if n≤0.

Furthermore all constants Ciwhich appear in the paper are taken such that they are greater than 1. Put h(2k+1)=r0. For any integer r≥0 letŴrdenote the set of pair of multi-indices(α, β) such thatŴrr1∪Ŵr2where

(3)

For a pair(α, β)we denote by(α, β)∗ the minimum of r such that (α, β) ∈ Ŵr. For any nonnegative integer t we define the following sieves

4t = {(α, β, γ ):α+β≤t,2kt≥γ≥α+(2k+1)β−t},

40t = {(α, β, γ ):(α, β, γ )∈4t, γ =α+(2k+1)β−t}.

Later on we will use the following properties of4t:

4tcontains a finite number of elements (less than 2k(t+1)3elements). If(α, β, γ )∈4t,β≥1,γ ≥1, then(α, β−1, γ−1)∈4t−1.

For every(α, β, γ )∈ 4t,(α0, β0, γ0)∈4t0 we can expressγ∂α,β(γ0∂α0,β0)as a linear

combination ofγ′∂α′, where(α′, β′, γ′)∈4t+t0.

For every (α, β, γ ) ∈ 4t, (α1, β1) ∈ Ŵr and a nonnegative integer m we can rewrite

γ−m∂α+α1−m,β+β1 asγ−m∂α2,β2(∂1α3∂

β3

2 )where(α2, β2, γ−m)∈4tand(α3, β3)∈Ŵrm (see [12]).

For any nonnegative integer r let us define the norm

|u, |r = max

(α1,β1)∈Ŵr

∂α1

1 ∂

β1

2 u, 

+ max (α11)∈Ŵr

α1≥1,β1≥1

max x∈ ¯

1h

∂α1

1 ∂

β1

2 u(x),

where|w, | =P(α,β,γ )4

h−1maxx∈ ¯|γ∂α,βw(x)|.

For any nonnegative integer l letHlloc()denote the space of all u such that for any compact K inwe haveP(α,β,γ )4lkγ∂α,βukL2(K)<∞. We note the following properties ofHlloc():

Hlocl ()⊂Hlloc()where Hlocl ()stands for the standard Sobolev spaces.

H4kloc+2()⊂Hloc2 ()⊂C().

If u∈Hlloc()and(α, β, γ )∈4t, tl thenγ∂α,βu∈Hlloct().

The following lemma is due to Grushin (see [12])

LEMMA4.2. Assume that uD′()and M2khu∈Hlloc()then u∈Hlloc+h().

Next we define a space generalizing the space of analytic functions (see for example [7]). Let Ln and Ln¯ be two sequences of positive numbers, satisfying the monotonicity condition

i n

LiLniALn(i = 1,2. . .; n = 1,2. . .), where A is a positive constant. A function F(x, v), defined for x = (x1,x2)and forv = (v1, . . . , vµ)in aµ-dimensional open set E ,

is said to belong to the class C{Lna;| ¯Lna;E}(a is an integer) if and only if F(x, v)is infinitely differentiable and to every pair of compact subsets0⊂and E0⊂E there corre-spond constants A1and A2such that for x∈0andv∈E0

j+kF(x, v) ∂xj1

1∂x j2

2 ∂v k1

1 . . . ∂v kµ

µ

A1A

j+k

2 LjaL¯ka

j1+ j2= j, X

ki =k; j,k=0,1,2. . .

.

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LEMMA4.3. There exists a constant C1such that if g(z)is a positive monotone decreasing function, defined in the interval 0z1 and satisfying

g(z)≤ 1

812kg z 1−

6k n

!! + C

znr0−1 (nr0+2,C>0) ,

then g(z) <CC1/znr0−1.

3. The main theorem

We will consider the following problem

(2) M2khu+ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1=0 in  .

We now state our main

THEOREM4.1. Let s4k2+6k+h+1. Assume that u is aHsloc()solution of the equation (2) andϕ ∈C{Lna2;|Lna2;E}for every a ∈ [0,r0]. Then uC{Lnr02;}. In particular, ifϕis a s-Gevrey function then so is u.

The proof of this theorem consists of Theorem 4.2 and Theorem 4.3.

THEOREM4.2. Let s4k2+6k+h+1. Assume that u is aHsloc()solution of the equation (2) andϕ∈C. Then u is a C∞()function.

Proof. Note that M2kh is elliptic onR2, except the line(x1,x2)= (0,x2). Therefore from the elliptic theory we have already uC∞(∩R2\(0,x2)).

LEMMA4.4. Let s4k2+6k+h+1. Assume that uHsloc()andϕ ∈ Cthen

ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1 ∈H

sh+1 loc ().

Proof. It is sufficient to prove thatγ1∂α1,β1ϕ(x1,x2,u, . . . ,γ∂α,βu)∈ L

2

loc()for every(α1,

β1, γ1)∈4sh+1. Let us denote(u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1by(w1, w2, . . . , wµ)withµ≤

2kh3. Since s4k2+6k+h+1 it follows thatw1, . . . , wµ ∈ C(). It is easy to verify that∂1α1∂β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu)is a linear combination with positive coefficients of terms of the form

(3) ∂

kϕ

xk1

1 ∂x k2

2 ∂w k3

1 . . . ∂w kµ+2 µ

µ

Y

j=1 Y

(α1,j,β1,j)

1α1,j2β1,jwj

ζ (α1,j,β1,j)

,

where k=k1+k2+ · · · +kµ+2≤α1+β1;ζ (α1,j, β1,j)may be a multivalued function of

α1,j, β1,j;α1,j, β1,jmay be a multivalued function of j , and

X

j

α1,j·ζ (α1,j, β1,j) ≤ α1,

X

j

(5)

Hence xγ11 ∂α1

1 ∂

β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu)is a linear combination with positive coefficients of

terms of the form

kϕ

xk1

1 ∂x k2

2 ∂w k3

1 . . . ∂w kµ+2 µ

xγ1

1

µ

Y

j=1 Y

(α1,j,β1,j)

1α1,j2β1,jwj

ζ (α1,j,β1,j)

.

Therefore the theorem is proved if we can show this general terms are in L2loc(). If all

ζ (α1,j, β1,j)vanish then it is immediate that∂kϕ/∂x1k1∂x k2

2 ∂w k3

1 . . . ∂w kµ+2

µ ∈ C, sinceϕ ∈

C∞,w1, . . . , wµ ∈C(). Then we can assume that there exists at least one ofζ (α1,j, β1,j) that differs from 0. Choose j0such that there existsα1,j0, β1,j0withζ (α1,j0, β1,j0)≥1 and

α1,j0+(2k+1)β1,j0 = j=max1,...,µ

ζ (α1,j1,j)≥1

α1,j+(2k+1)β1,j.

Consider the following possibilities

I)ζ (α1,j0, β1,j0)≥2. We then haveα1,j+β1,jl−(h−1)−(4k+2). Indeed, if j 6= j0

andα1,j+β1,j >l−(h−1)−(4k+2)thenα1,j0+β1,j0 ≥2k. Therefore

l−(h−1)−(4k+2) < α1,j+β1,j≤α1,j+(2k+1)β1,j

≤α1,j0+(2k+1)β1,j0 ≤(2k+1)(α1,j01,j0)

2k(2k+1) .

Thus l< (2k+2)(2k+1)+(h−1), a contradiction.

If j= j0andα1,j0+β1,j0 >l−(h−1)−(4k+2)then we have

l−(h−1)≥α1+β1≥2(α1,j0+β1,j0) >2(l−(h−1)−(4k+2)) .

Therefore l< (h+1)+4(2k+1), a contradiction. Next define

γ (α1,j, β1,j)=max{0, α1,j+(2k+1)β1,j+(h−1)+(4k+2)−l}.

We claim thatγ (α1,j, β1,j)≤2k(l−(h−1)−(4k+2)). Indeed, if j6= j0andγ (α1,j, β1,j) >

2k(l−(h−1)−(4k+2))then

(2k+1)(l−(h−1)) ≥ α1+(2k+1)β1

≥ (α1,j+2α1,j0)+(2k+1)(β1,j0+2β1,j0)

> 3(2k+1)(l−(h−1)−(4k+2)) .

Thus l< (h−1)+3(2k+1), a contradiction.

If j= j0andγ (α1,j0, β1,j0) >2k(l−(h−1)−(4k+2))then it follows that

(2k+1)(l−(h−1))≥α1+(2k+1)β1≥2(α1,j0+(2k+1)β1,j0)

>2(2k+1)(l−(h−1)−(4k+2)) .

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From all above arguments we deduce that(α1,j, β1,j, γ (α1,j, β1,j))∈4l−(h−1)−(4k+2). Next

we claim thatPγ (α1,j, β1,j)ζ (α1,j, β1,j)≤γ1. Indeed, ifPγ (α1,j, β1,j)ζ (α1,j, β1,j) >

γ1then we deduce that

α1+(2k+1)β1−2(l−(h−1)−(4k+2))≥ X

γ (α1,j, β1,j)ζ (α1,j, β1,j) > γ1

≥α1+(2k+1)β1−(l−(h−1)) .

Therefore l< (h−1)+4(2k+1), a contradiction. Now we have

xγ1

1

µ

Y

j=1 Y

(α1,j,β1,j)

1α1,j2β1,jwj

ζ (α1,j,β1,j)

=x1γ1

µ

Y

j=1 Y

(α1,j,β1,j)

x1γ (α1,j,β1,j)∂1α1,j2β1,jwj

ζ (α1,j,β1,j)

C()

since x1γ (α1,j,β1,j)∂1α1,j2β1,jwj ∈H4kloc+2()⊂C().

II)ζ (α1,j0, β1,j0)=1 andζ (α1,j, β1,j)=0 for j6= j0. We have

x1γ1

µ

Y

j=1 Y

(α1,j,β1,j)

1α1,j2β1,jwj

ζ (α1,j,β1,j)

=xγ111α1,j02β1,j0wj0 ∈L

2 loc() .

III)ζ (α1,j0, β1,j0)=1 and there exists j16=j0such thatζ (α1,j1, β1,j1)6=0. Define

¯

γ (α1,j0, β1,j0)=max{0, α1,j0+(2k+1)β1,j0+(h−1)−l}.

As in part I) we can prove(α1,j, β1,j, γ (α1,j, β1,j)) ∈ 4l−(h−1)−(4k+2) for j 6= j0 and

1,j0, β1,j0,γ (α¯ 1,j0, β1,j0))∈ 4l(h1). Therefore xγ (α1 1,j,β1,j)∂1α1,j2β1,jwj ∈ H4kloc+2()

C()for j6= j0and x ¯

γ (α1,j0,β1,j0)

1 ∂

α1,j0 1 ∂

β1,j0 2 wj0∈L

2

loc(). We also have P

j6=j0γ (α1,j, β1,j)ζ (α1,j, β1,j)+ ¯γ (α1,j0, β1,j0)≤γ1as in part I). Now the desired result follows from the decomposition of the general terms.

(continuing the proof of Theorem 4.2) uHlocl (), l4k2+6k+h+1H⇒u∈Hlloc(),

l4k2+6k+h+1H⇒ϕ(x1,x2,u, . . . ,γ∂α,βu)∈Hlloch+1()(by Lemma 4.4). Therefore

by Lemma 4.2 we deduce that u ∈Hlloc+1(). Repeat the argument again and again we finally arrive at u∈Hlloc+t()for every positive t , i.e.

u∈ ∩lHlloc()=C∞() .

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4. Gevrey analyticity of the solutions

PROPOSITION4.1. Assume that

ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1∈C{Lna−2;|Lna−2;E}

for every a∈[0,r0]. Then there exist constantsHe0,He1,C2,C3such that for every H0≥ He0, H1He1, H1C2H2r0+3

0 if

|u, |qH0H( qr0−2)

1 Lqr0−2, 0≤qN+1, r0+2≤N

then

max x∈ ¯

1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)

C3H0H1Nr0−1LNr0−1

for every1, β1)∈ŴN+1.

Proof. ∂α1

1 ∂

β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu)as in Lemma 4.4 is a linear combination with positive

coefficients of terms of the form

kϕ

xk1

1 ∂x k2

2 ∂w k3

1 . . . ∂w kµ+2 µ

µ

Y

j=1 Y

(α1,j,β1,j)

1α1,j2β1,jwj

ζ (α1,j,β1,j)

.

Substitutingwjby one of the termsγ∂α,βu with(α, β, γ )∈4h−1we obtain

1α1,j2β1,j(γ∂α,βu)= α1,j

X

m=0

m

α1,j

γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1,jm,β+β1,ju.

We can decompose(γ−m)∂α+α1,jm,β+β1,ju into(γ−m)∂α2,β2

1α3∂β3

2 u

with(α2, β2, γ − m)∈4h−1and(α3, β3)∈Ŵ(α1,j,β1,j)∗−m. Put S = N+1−α1−β1. Define R =r0−S.

It is easy to see that R is a nonnegative integer and less than r0. Sinceα1,j ≤ α1we deduce that(α1,j, β1,j)∈Ŵ(α1,j+β1,j+S). Using the monotonicity condition on Lnand the inductive

assumption we have

m

α1,j

γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1,jm,β+β1,ju

m

α1,j

γ· · ·(γ−m+1)θ (γ−m+1)H0H

α1,j+β1,jmR−2

1 Lα1,j+β1,jmR−2

C4H0H

α1,j+β1,jmR−2

1 Lα1,j+β1,jR−2.

Therefore we deduce that

1α1,j2β1,j(γ∂α,βu)

α1,j

X

m=0

m

α1,j

γ· · ·(γ−m+1)θ (γ−m+1)m)∂α+α1,jm,β+β1,ju

C5H0H

α1,j+β1,jR−2

(8)

and the general terms can be estimated by

µ

Y

j=1 C5H0H

α1,j+β1,jR−2

1 Lα1,j+β1,jR−2.

Sinceϕ∈C{Lna;|Lna;E}, there exist constants C6,C7such that

kϕ

xq1

1 ∂x q2

2 ∂w r1

1 . . . ∂w rµ

µ

C6C

qR

7 LqR−2C7rLrR−2 (q=q1+q2,r =r1+ · · · +rµ) .

Now we take X(ξ, v)=X1(v)·X2(ξ )where

X1(v)=C6 p X

i=0

C8iLiR−2vi

i ! , X2(ξ )= p X

i=0

C7iRLiR−2ξi

i ! .

and

v(ξ )=C5H0 p X

i=1

H1iR−2LiR−2ξi

i ! .

By comparing terms of the form (3) with the corresponding terms inddξppX(ξ, v)it follows that

∂α1

1 ∂

β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu)

x=x0

d p

dξpX(ξ, v)

ξ=0

.

Next we introduce the following notation: v(ξ ) ≪ h(ξ )if and only ifv(j)(0) ≤ h(j)(0)for 1≤ jp. It is not difficult to check that there exists a constant C9(independent of p) such that

v2(ξ )≪(C5H0)2C9 p X

i=2

H1iR−3LiR−3ξi

(i−1)! .

And by induction we have

vj(ξ )≪(C5H0)jC9j−1 p X

i=j

H1ijR−1LijR−1ξ i

(ij+1)! .

Next, it is easy to verify that X1(0) = 0, d Xd1(ξ )ξ

ξ=0 ≤ C5C6C8H0, and djX2(ξ )

dξj

ξ=0 = C7jRLjR−2.

We now compute djX1(ξ )

dξj

ξ=0 when 2 ≤ jp. If 2 ≤ j2R + 4 we can al-ways choose a constant C10 such that d

jX

1(ξ )

dξj

ξ=0 ≤ C10H0H jR−2

(9)

If j2R+5, we have

(4)

djX1(ξ ) dξj

ξ=0 ≤C6

RX+2

i=1

Ci8(C5H0)iC9i−1H1jiR−1LjiR−1j ! i !(ji+1)!

+ jXR−2

i=R+3

C8i(C5H0)iCi9−1H1jiR−1LiR−2LjiR−1j ! i !(ji+1)!

+ j X

i=jR−1

C8i(C5H0)iC9i−1LiR−2j ! i !(ji+1)!

 .

The first sum in (4) is estimated by

(5) C10H0H1jR−2LjR−2

provided H1≥(C5C8C9H0)2r0+3as for 2≤ j2R+4.

By using the monotonicity condition on Lnthe second sum in (4) is estimated by (6)

C6 jXR−2

i=R+3

Ci8(C5H0)iC9i−1H1jiR−1LiR−2LjiR−1j ! i !(ji+1)!

C11H0H jR−2

1 j !Lj2R−3

(j2R−3)!

jXR−2

i=R+3

1 i· · ·(iR−1)

1

(ji+1)· · ·(jiR)

C12H0H1jR−2LjR−2, provided H1≥C5C8C9H0. For the third sum we see that

(7)

C6 j X

i=jR−1

C8i(C5H0)iC9i−1LiR−2j ! i !(ji+1)!

C5C6C8H0H1jR−2j ! j X

i=jR−1

LiR−2 i !(ji+1)!

C13H0H1jR−2LjR−2,

if H1≥(C5C8C9H0)2.

By (4), (5), (6), (7) and taking H1≥(C5C7C8C9H0)2r0+3=C2H02r0+3we obtain

dpX(ξ, v)

dξp

ξ=0 ≤C14

p X

j=0

j p

H0H1jR−2LjR−2C7pjRLpjR−2

C15H0H1pR−2LpR2.

Hence

max x∈ ¯

1α1∂β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu)

C15H0H1pR−2LpR−2

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REMARK4.2. The constantsHe0,He1,C2,C3depend onincreasingly in the sense that with the sameHe0,He1,C2,C3Proposition 4.1 remains valid if we substituteby any′⊂.

COROLLARY4.1. Under the same hypotheses of Proposition 4.1 with qN+1 replaced by qN , then

max x∈ ¯

1α1∂β1

2 ϕ(x1,x2,u, . . . ,γ∂α,βu) ≤C3

|u, |N+1+H0H1Nr0−1LNr0−1

for every1, β1)∈ŴN+1.

Proof. Indeed, as in the proof of Proposition 4.1 all typical terms, except ∂w∂ϕ

j0

α1

1 ∂

β1

2 wj0 can

be estimated by| · |N. Replacingwj0by one ofγ∂α,βu, we have

∂ϕ ∂wj0

∂α1

1 ∂

β1

2 wj0= ∂ϕ ∂wj0

γ∂α+α1,β+β1u

+ ∂ϕ

∂wj0 α1

X

m=1

m

α1

γ· · ·(γ−m+1)θ (γ−m+1)m)∂α+α1−m,β+β1u.

The first summand is estimated by C3|u, |N+1. The second sum is majorized as in Proposition 4.1.

THEOREM 4.3. Let u be a Csolution of the equation (2) and ϕ ∈ C{Lna−2;|Lna−2;E} for every a ∈ [0,r0]. Then uC{Lnr0−2;}. In

particu-lar, ifϕis a s-Gevrey function then so is u.

Proof. It suffices to consider the case(0,0)∈. Let us define a distanceρ((y1,y2), (x1,x2))=

max

|x12k+1−y12k+1|

2k+1 ,|x2−y2|

. For two sets S1,S2the distance between them is defined as

ρ(S1,S2)=infxS1,yS2ρ(x,y). Let V

T be the cube with edges of size (in theρmetric) 2T ,

which are parallel to the coordinate axes and centered at(0,0). Denote by VδTthe subcube which is homothetic with VT and such that the distance between its boundary and the boundary of VT isδ. We shall prove by induction that if T is small enough then there exist constants H0,H1 with H1≥C2H02r0+3such that

(8)

u,VδT

nH0 for 0≤n≤max{r0+2,6 k+1}

and

(9)

u,VδT

nH0

H1

δ

nr0−2

Lnr0−2 for n≥max{r0+2,6

k+1}

andδsufficiently small. Hence the Gevrey analyticity of u follows. (8) follows easily from the Csmoothness assumption on u. Assume that (9) holds for n = N . We shall prove it for n= N+1. Putδ′=δ(1−1/N). Fix xVδT and then defineσ =ρ(x, ∂VT)andσ˜ =σ/N.

Let Vσ˜(x)denote the cube with center at(x)and edges of length 2σ˜ which are parallel to the co-ordinate axes, and Sσ˜(x)the boundary of Vσ˜(x). Let E1,E3(E2,E4) be edges of Sσ˜(x)which are parallel to O x1(O x2) respectively. We have to estimate maxxVT

δ γ∂α,β

1α1∂β1

(11)

for all(α, β, γ )∈4h−1,(α1, β1)∈ŴN+1, and maxxVT

δ ∂1h

∂α1

1 ∂

β1

2 u(x)for all(α1, β1)∈

ŴN+1,α1≥1,β1≥1. But when(α1, β1)∈ŴNwe have already the desired estimate. Hence it suffices to obtain the estimate only for(α1, β1)∈ŴN+1\ŴN.

LEMMA4.5. Assume that(α, β, γ )∈4h−1and(α1, β1)∈ŴN+1. Then ifα1≥1,β1≥1 there exists a constant C16such that

max xVδT γ∂α,β

∂α1

1 ∂

β1

2 u(x)

C16

T2k1+1

u,VδT

N+1+H0

H1

δ

Nr0−1

LNr0−1

T2k1+1 + 1

H1

.

Proof. First we observe that M2khu = ∂x1+i x

2k 1 ∂x2

h

u = P

(α,˜β,˜γ )˜ ∈40hC

1 ˜

αβ˜γ˜ γ˜∂α,˜β˜u,

where C1 ˜

αβ˜γ˜ are some complex constants. Differentiating the equation (2)α1times in x1andβ1 times in x2then applying the representation formula (1) for=Vσ˜(x)we have

1α1∂2β1u(x)= Z

Vσ˜(x)

(−1)hF2kh(y,x) (A(y)+B(y))d y1d y2

+ Z

Sσ˜(x)  

hX−1

j=0

(−1)jM2kj ∂α1

1y

β1

2yu(y)

M2khj−1F2kh

(n1+i y2k1 n2)ds,

where

(10) A(y)= − X

(α,˜β,˜γ )˜∈40h ˜ β≥1,γ˜≥1

min{X2kh,α1}

m=1 C1

˜

αβ˜γ˜m

m

α1

(γ˜−m)∂α1+ ˜α−m,β1+ ˜βu,

and

B(y)= −∂α1

1y

β1

2yϕ(y1,y2,u, . . . ,γ∂α,β(y)u) .

Therefore differentiatingγ∂α,βgives

(11)

γ∂α,β

∂α1

1 ∂

β1

2 u(x)

= Z

Vσ˜(x)

(−1)hγ∂α,βF2kh(y,x) (A(y)+B(y))d y1d y2

+ Z

Sσ˜(x)  

hX−1

j=0

(−1)jM2kj

∂α1

1y

β1

2yu(y)

γ∂α,βM2khj−1F2kh

(n1+i y12kn2)ds.

It is easy to verify that

(12)

γ∂α,βF2kh

C17

x12k+1−y12k+1 2k+1

+ |x2−y2|

.

(12)

I)α1≤r0(=(2k+1)h). The typical terms(γ˜−m)∂αy

1+ ˜α−m,β1+ ˜β

u(y)in (10) can be rewritten

as(γ−m)∂αy2,β2

∂α3

1y

β3

2yu(y)

with(α2, β2, γ−m)∈4h−1and(α3, β3)∈ŴN+1−m. Hence we have

|A(y)||V ˜

σ(x)≤C18

min{X2kh,α1}

m=1 H0

H1

δ′

Nr0−m−1

LNr0−m−1

C19H0

H1

δ′

Nr0−2

LNr0−1.

II) Ifα1≥r0+1 then we can decompose(γ˜−m)∂αy

1+ ˜α−m,β1+ ˜β

u(y)into

˜

γ−mαy˜+ ˜β,0

1yα1− ˜β−m2yβ1+ ˜βu(y)

,

where(α1− ˜β−m, β1+ ˜β)∈ ŴN+1−m andα1− ˜β−m ≥1,β1+ ˜β ≥ 1. Therefore by inductive assumptions we see that

|A(y)||V ˜

σ(x)≤C20

min{X2kh,α1}

m=1

m

α1

H0

H1

δ′

Nr0−m−1

LNr0−m−1

C21H0

H1

δ′

Nr0−2

LNr0−1.

Hence in both cases we have

(13) |A(y)||V ˜

σ(x)≤C22H0

H1

δ′

Nr0−2

LNr0−1.

On the other hand, from Proposition 4.1 and the inductive assumption we have

(14) |B(y)| ≤C3 |u,VδT′|N+1+H0

H1

δ′

Nr0−1

LNr0−1

!

.

Combining (12), (13), (14) we obtain

Z

Vσ˜(x)

γ∂α,βF2kh(y,x) (A(y)+B(y))d y1d y2

C23 |u,VδT′|N+1+H0

H1

δ′

Nr0−1

LNr01 !

× Z

Vσ˜(x)

1

x2k1+1−y12k+1 2k+1

+ |x2−y2| d y1d y2

C24T2k1+1 |u,VT

δ′|N+1+H0

H1

δ

Nr0−1

LNr0−1

!

.

To estimate the second integral in (11) we first note that

γ∂α,βM2khj−1F2kh = minX{j,α}

λ=0

αX−λ

˜

λ≥2kα−+λ1 C3

αβλ˜λ

(x1−y1)j−λx(2k+1) ˜

λ+λ−α+γ

1

x2k1+1−y12k+1

2k+1 +i(x2−y2)

(13)

Let us consider two cases:

I) 0≤ jh−2. We then split into sub cases:

A)|x1| ≤2σ˜

1

2k+1. We have the following estimate

can be expressed as a linear combination of γ′∂y

α′

2k+1 is not an integer. Here [·] denotes the integer part of the argument. Therefore in both cases the integral can be estimated in the following way

where{·}denotes the fractional part of the argument.

B)|x1| ≥2σ˜

can be decomposed into

(14)

that

A) We first estimate the integrals along the edges E2and E4. Integrating by parts gives

(15)

LEMMA4.6. Assume that(α, β, γ )∈4h−1. Then there exists a constant C38such that

following formula holds

(15)

Therefore the first integral in (15) is estimated by

C39T

We now estimate the second integral in (15).

(16)

B) Consider now integrals along the edgeseE1andEe3, whereeE1,Ee3denote the edges of S6kσ˜,

which are parallel toO x1. 1) 0≤ jh−2. We have

It follows that

(16)

By Proposition 4.1 the first integral in the right hand side of (16) can be estimated by

(17)

To estimate the second integral in the right hand side of (16) we now integrate by parts: above the volume integral can be estimated by

C52T

For the boundary integral we again split into 2 cases

A) j=h−2. As in Lemma 4.5 we obtain the following estimate

By Proposition 4.1 the first integral in the right hand side of (17) can be estimated by

(18)

To estimate the second integral in the right hand side of (17) we now integrate by parts

Proof. We have the following representation formula forγ∂α,β

Therefore, as in Lemma 4.5, the first integral in (18) is estimated by

C59T

(19)

I) 0≤ jh−2. As in Lemma 4.5 I) A) 2) we obtain

A) We first estimate the integrals along E2and E4. Note that

B) The integrals along E1 and E3 now can be estimated in the same manner as in Lemma 4.5.

LEMMA4.8. Assume that1, β1) ∈ ŴN+1\ŴN,α1≥ 1,β1 ≥1. Then there exists a constant C64such that

(20)

Proof. This lemma can be proved as in [10]. We omit the details.

(Continuing the proof of Theorem 4.3). Put|u,VδT|N+1 =g(δ). Combining Lemmas 4.5-4.8 gives

g(δ)≤C65 T

1 2k+1g

δ(1−6k/N)

+H0

H1

δ

Nr0−1

LNr0−1

T2k1+1 + 1

H1 !

.

Choosing T

1/812kC65 2k+1

then by Lemma 4.3 we deduce that

g(δ)≤C66H0

H1

δ

Nr0−1

LNr0−1

T2k1+1 + 1

H1

.

If T is chosen to be small enough such that T ≤ (1/2C66)2k+1and choosing H1 ≥2C66(in addition to H1≥C2H02r0+3) we arrive at

g(δ)≤H0

H1

δ

Nr0−1

LNr0−1.

That means

u,VδT

N+1≤H0

H1

δ

Nr0−1

LNr0−1.

The proof of Theorem 4.3 is therefore completed.

Acknowledgment. The author would like to express his gratitude to the Consiglio Nazionale delle Richerche for support. He also would like to thank Professor L. Rodino who helped to prove Theorem 4.2.

References

[1] MIRANDAC., Partial Differential Equations of Elliptic Type, Springer-Verlag 1970, 370.

[2] BERNSTEINS., Sur la Nature Analytique des Solutions des ´Equations aux D´erive´es Par-tielles du Second Ordre, Math. Annal. 59 (1904), 20–76.

[3] HOPFE., Ub¨er den Funktionalen Insbesondere den Analytischen Charakter der L¨osungen Elliptischer Differentialgleichungen Zweiter Ordnung, Math. Zeit. 34 (1931), 194–233.

[4] GIRAUDG., Sur la Probl`eme de Dirichlet G´en´eralis´ee; ´Equations non Lin´eaires `a m Vari-ables, Ann. ´Ec. N. Sup. 43 (1926), 1–128.

[5] PETROWSKIII., Sur l’analyticit´e des Solutions des Syst`emes d’ ´Equations Diff´erentielles, Mat. Sbornik (N. S.) 5(47) (1939), 3–70.

[6] MORREYC.B., On the Analyticity of the Solutions of Analytic Nonlinear Systems of Partial Differential Equations I, II, Amer. J. Math. 80 (1958), 198–218, 219–237.

[7] FRIEDMANA., On the Regularity of the Solutions of Non-Linear Elliptic and Parabolic Systems of Partial Differential Equations, J. Math. Mech. 7 (1958), 43–59.

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[9] CATTABRIGAL., RODINO L., ZANGHIRATI L., Analytic-Gevrey Hypoellipticity for a class of Pseudo-Differential operators with multiple characteristics, Comm. P. D. E. 15 (1990), 81–96.

[10] TRIN.M., Semilinear Perturbation of Powers of the Mizohata Operators, to appear in Comm. P. D. E.

[11] CALANCHIM., RODINO L., TRIN.M., Solutions of Logarithmic Type for Elliptic and Hypoelliptic Equations, Ann. Univ. Ferrara XLI (1997), 111–127.

[12] GRUSHINV.V., On a Class Elliptic Pseudo differential Operators Degenerate on a Sub-manifold, Math. USSR Sbornik 13 (1971), 155–183.

[13] KOMATSUH., Ultradistribution, I Structure Theorems and a Characterization, J. Fac. Sci. Univ. Tokyo Sect. I A Math. 20 (1973), 25–105.

[14] KOMATSUH., Ultradistribution, II The Kernel Theorem and Ultradistributions with Sup-port in a Submanifold, J. Fac. Sci. Univ. Tokyo Sect. I A Math. 24 (1977), 607–628.

AMS Subject Classification: 35H10, 35B65, 35C15.

Nguyen Minh TRI Institute of Mathematics P.O. Box 631, Boho 10000 Hanoi, Vietnam

and Dipartimento di Matematica Universit`a di Torino

Via Carlo Alberto 10 10123 Torino, Italia

(22)

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