Rend. Sem. Mat. Univ. Pol. Torino
Vol. 57, 1 (1999)
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 multiple characteristics, consisting of powers of the Mizohata operator and a Gevrey
analytic nonlinear perturbation. The main theorem states that under some conditions on the perturbation every H s solution (with s ≥ s0 ) 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 fundamental 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 attracted considerable interest of specialists. The conjecture stated that every solution of an analytic nonlinear elliptic equation is analytic (see [1]). This conjecture caused considerable research 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:
!
∂ l1 +l2 u
h
= 0,
M2k u + ϕ x1 , x2 , u, . . . , l l
∂x11 ∂x22 l +l ≤(h−1)
1
2
where u = u(x1 , x2 ), M2k = ∂∂x + i x12k ∂∂x , the Mizohata operator in R2 . The Gevrey hy1
2
poellipticity for the equation was obtained in [9] in the linear case. We will prove a theorem
for a so-called Grushin type perturbation ϕ (see the notations in §2 below). The analyticity of
∗ Dedicated to my mother on her sixtieth birthday.
37
38
N.M. Tri
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
h (x , x , y , y ) =
F2k
1 2 1 2
(x1 − y1 )h−1
1
.
2k+1
2π(h − 1)! x1 −y12k+1
+
i(x
−
y
)
2
2
2k+1
For j = 1, . . . , h − 1 we have (see [11])
j
h =
M2k F2k
(x1 − y1 )h− j −1
1
h F h = δ(x − y) ,
and M2k
2k
2k+1
2π(h − j − 1)! x1 −y12k+1
+ i(x2 − y2 )
2k+1
where x = (x1 , x2 ), y = (y1 , y2 ) and δ(·) is the Dirac function. We will denote by  a bounded
domain in R2 with piece-wise smooth boundary. We now state a lemma on the representation
formula. Its proof can be found in [10].
¯ then we have
L EMMA 4.1. Assume that u ∈ C h ()
Z
h (x, y)M h u(y) d y d y
u(x) = (−1)h F2k
1 2
2k



Z
(1)
h−1
X
j
h− j −1 h

+
(−1) j M2k u M2k
F2k (x, y) (n 1 + i y12k n 2 ) ds .
∂
j =0
For reason of convenience we shall use the following Heaviside function
1 if
z ≥ 0,
θ(z) =
0 if
z < 0.
d m (z n ) = n . . . (n − m + 1)θ(n − m + 1)z n−m .
We then have the following formula dt
m
β
α , ∂ β and ∂ y instead of ∂ α , ∂ β , x γ ∂ α+β , ∂ α , ∂ β and
We will use ∂1α , ∂2 , γ ∂α,β , ∂1y
γ α,β
2y
∂ x 1α ∂ x β 1 ∂ x α ∂ x β ∂ y1α ∂ y β
2
2
1
2
γ α+β
y1 ∂ α β , respectively. Throughout the paper we use the following notation
∂ y1 ∂ y2
n! =
n!
1
if
if
n ≥ 1,
n ≤ 0,
and C n =
Cn
1
if
if
n ≥ 1,
n ≤ 0.
Furthermore all constants Ci which appear in the paper are taken such that they are greater than
1. Put h(2k + 1) = r0 . For any integer r ≥ 0 let 0r denote the set of pair of multi-indices (α, β)
such that 0r = 0r1 ∪ 0r2 where
0r1 = {(α, β) : α ≤ r0 , 2α + β ≤ r },
0r2 = {(α, β) : α ≥ r0 , α + β ≤ r − r0 } .
39
On the Gevrey analyticity of solutions
For a pair (α, β) we denote by (α, β)∗ the minimum of r such that (α, β) ∈ 0r . 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 of 4t :
4t contains a finite number of elements (less than 2k(t + 1)3 elements).
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 express γ ∂α,β ( γ0 ∂α0 ,β0 ) as a linear
combination of γ 0 ∂α 0 ,β 0 , where (α 0 , β 0 , γ 0 ) ∈ 4t +t0 .
For every (α, β, γ ) ∈ 4t , (α1 , β1 ) ∈ 0r and a nonnegative integer m we can rewrite
α3 β3
γ −m ∂α+α1 −m,β+β1 as γ −m ∂α2 ,β2 (∂1 ∂2 ) where (α2 , β2 , γ − m) ∈ 4t and (α3 , β3 ) ∈ 0r−m
(see [12]).
For any nonnegative integer r let us define the norm
α β
α β
|u, |r = max ∂1 1 ∂2 1 u,  + max max ∂1h ∂1 1 ∂2 1 u(x) ,
(α1 ,β1 )∈0r
where |w, | =
(α1 ,β1 )∈0r
α1 ≥1,β1 ≥1
¯
x∈
P
¯ |γ ∂α,β w(x)|.
(α,β,γ )∈4h−1 maxx∈
l
For any nonnegative integer l let Hloc () denote the space of all u such that for any compact K
P
in  we have (α,β,γ )∈4l kγ ∂α,β uk L 2 (K ) < ∞. We note the following properties of Hlloc ():
l () ⊂ Hl () where H l () stands for the standard Sobolev spaces.
Hloc
loc
loc
2 () ⊂ C().
()
⊂
H
H4k+2
loc
loc
If u ∈ Hlloc () and (α, β, γ ) ∈ 4t , t ≤ l then γ ∂α,β u ∈ Hl−t
loc ().
The following lemma is due to Grushin (see [12])
h u ∈ Hl () then u ∈ Hl+h ().
L EMMA 4.2. Assume that u ∈ D 0 () and M2k
loc
loc
Next we define a space generalizing the space of analytic functions (see for example [7]).
Let L n and L̄ n be two sequences of positive numbers, satisfying the monotonicity condition
i
n L i L n−i ≤ AL n (i = 1, 2 . . . ; n = 1, 2 . . . ), where A is a positive constant. A function
F(x, v), defined for x = (x1 , x2 ) and for v = (v1 , . . . , vµ ) in a µ-dimensional open set E,
is said to belong to the class C{L n−a ; | L̄ n−a ; E} (a is an integer) if and only if F(x, v) is
infinitely differentiable and to every pair of compact subsets 0 ⊂  and E 0 ⊂ E there correspond constants A 1 and A 2 such that for x ∈ 0 and v ∈ E 0
j +k F(x, v)
∂
≤ A 1 A j +k L j −a L̄ k−a
j1 j2 k1
2
kµ ∂ x ∂ x ∂v . . . ∂vµ 1
2
1
X
j1 + j2 = j,
ki = k; j, k = 0, 1, 2 . . . .
We use the notation L −i = 1 (i = 1, 2, . . .). If F(x, v) = f (x), we simply write f (x) ∈
C{L n−a ; }. Note that C{n!; }, (C{n!s ; }) is the space of all analytic functions (s-Gevrey
functions), respectively, in . Extensive treatments of non-quasi analytic functions (in particular,
the Gevrey functions) can be found in [13, 14]. Finally we would like to mention the following
lemma of Friedman [7].
40
N.M. Tri
L EMMA 4.3. There exists a constant C1 such that if g(z) is a positive monotone decreasing
function, defined in the interval 0 ≤ z ≤ 1 and satisfying
!!
1
C
6k
g(z) ≤
+ n−r −1 (n ≥ r0 + 2, C > 0) ,
g
z
1
−
k
n
z 0
812
then g(z) < CC1 /z n−r0 −1 .
3. The main theorem
We will consider the following problem
h u + ϕ(x , x , u, . . . , ∂
M2k
γ α,β u)(α,β,γ )∈4h−1 = 0
1 2
(2)
in
.
We now state our main
T HEOREM 4.1. Let s ≥ 4k 2 +6k+h+1. Assume that u is a Hsloc () solution of the equation
(2) and ϕ ∈ C{L n−a−2 ; |L n−a−2 ; E} for every a ∈ [0, r0 ]. Then u ∈ C{L n−r0 −2 ; }. In
particular, if ϕ is a s-Gevrey function then so is u.
The proof of this theorem consists of Theorem 4.2 and Theorem 4.3.
T HEOREM 4.2. Let s ≥ 4k 2 + 6k + h + 1. Assume that u is a Hsloc () solution of the
equation (2) and ϕ ∈ C ∞ . Then u is a C ∞ () function.
h is elliptic on R2 , except the line (x , x ) = (0, x ). Therefore from the
Proof. Note that M2k
1 2
2
elliptic theory we have already u ∈ C ∞ ( ∩ R2 \ (0, x2 )).
L EMMA 4.4. Let s ≥ 4k 2 + 6k + h + 1. Assume that u ∈ Hsloc () and ϕ ∈ C ∞ then
ϕ(x1 , x2 , u, . . . ,γ ∂α,β u)(α,β,γ )∈4h−1 ∈ Hs−h+1
().
loc
Proof. It is sufficient to prove that γ1 ∂α1 ,β1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) ∈ L 2loc () for every (α1 ,
β1 , γ1 ) ∈ 4s−h+1 . Let us denote (u, . . . ,γ ∂α,β u)(α,β,γ )∈4h−1 by (w1 , w2 , . . . , wµ ) with µ ≤
2kh 3 . Since s ≥ 4k 2 + 6k + h + 1 it follows that w1 , . . . , wµ ∈ C(). It is easy to verify
α
β
that ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) is a linear combination with positive coefficients of terms
of the form
(3)
µ
Y
∂k ϕ
Y
k
k
k
k
∂ x11 ∂ x22 ∂w13 . . . ∂wµµ+2 j =1 (α1, j ,β1, j )
ζ(α1, j ,β1, j )
α
β
,
∂1 1, j ∂2 1, j w 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, j may be a multivalued function of j , and
X
α1, j · ζ (α1, j , β1, j )
≤
α1 ,
β1, j · ζ (α1, j , β1, j )
≤
β1 .
j
X
j
41
On the Gevrey analyticity of solutions
γ
α
β
Hence x1 1 ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) is a linear combination with positive coefficients of
terms of the form
∂k ϕ
k
k
k
k
∂ x11 ∂ x22 ∂w13 . . . ∂wµµ+2
µ
γ Y
Y
x1 1
j =1 (α1, j ,β1, j )
ζ(α1, j ,β1, j )
α
β
.
∂1 1, j ∂2 1, j w j
Therefore the theorem is proved if we can show this general terms are in L 2loc (). If all
k
k
k
k
ζ (α1, j , β1, j ) vanish then it is immediate that ∂ k ϕ/∂ x11 ∂ x22 ∂w13 . . . ∂wµµ+2 ∈ C, since ϕ ∈
∞
C , w1 , . . . , wµ ∈ C(). Then we can assume that there exists at least one of ζ (α1, j , β1, j )
that differs from 0. Choose j0 such that there exists α1, j0 , β1, j0 with ζ (α1, j0 , β1, j0 ) ≥ 1 and
α1, j0 + (2k + 1)β1, j0 =
max
j =1,...,µ
ζ(α1, j ,β1, j )≥1
α1, j + (2k + 1)β1, j .
Consider the following possibilities
I) ζ (α1, j0 , β1, j0 ) ≥ 2. We then have α1, j + β1, j ≤ l − (h − 1) − (4k + 2). Indeed, if j 6= j0
and α1, j + β1, j > l − (h − 1) − (4k + 2) then α1, 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, j0 + β1, j0 )
≤ 2k(2k + 1) .
Thus l < (2k + 2)(2k + 1) + (h − 1), a contradiction.
If j = j0 and α1, 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 that γ (α1, j , β1, j ) ≤ 2k(l − (h − 1) − (4k + 2)). Indeed, if j 6= j0 and γ (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 = j0 and γ (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)) .
Thus l < (h − 1) + 4(2k + 1), a contradiction.
42
N.M. Tri
From all aboveP
arguments we deduce that (α1, j , β1, j , γ (α1, j , βP
1, j )) ∈ 4l−(h−1)−(4k+2) . Next
we claim that
γ (α1, j , β1, j )ζ (α1, j , β1, j ) ≤ γ1 . Indeed, if
γ (α1, j , β1, j )ζ (α1, j , β1, j ) >
γ1 then 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
µ
γ Y
Y
x1 1
j =1 (α1, j ,β1, j )
ζ(α1, j ,β1, j )
α
β
∂1 1, j ∂2 1, j w j
µ
γ Y
= x1 1
Y
j =1 (α1, j ,β1, j )
γ (α
,β
) α
ζ(α1, j ,β1, j )
γ (α ,β ) α
β
∈ C()
x1 1, j 1, j ∂1 1, j ∂2 1, j w j
β
since x1 1, j 1, j ∂1 1, j ∂2 1, j w j ∈ H4k+2
loc () ⊂ C().
II) ζ (α1, j0 , β1, j0 ) = 1 and ζ (α1, j , β1, j ) = 0 for j 6= j0 . We have
µ
γ Y
x1 1
Y
j =1 (α1, j ,β1, j )
ζ(α1, j ,β1, j )
α
β
γ α1, j β1, j
= x1 1 ∂1 0 ∂2 0 w j0 ∈ L 2loc() .
∂1 1, j ∂2 1, j w j
III) ζ (α1, j0 , β1, j0 ) = 1 and there exists j1 6= j0 such that ζ (α1, j1 , β1, j1 ) 6= 0. Define
γ̄ (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
γ (α
,β
) α
β
(α1, j0 , β1, j0 , γ̄ (α1, j0 , β1, j0 )) ∈ 4l−(h−1) . Therefore x1 1, j 1, j ∂1 1, j ∂2 1, j w j ∈ H4k+2
loc ()
P
γ̄ (α1, j0 ,β1, j0 ) α1, j0 β1, j0
2
⊂ C() for j 6= j0 and x1
∂1 ∂2 w j0 ∈ L loc (). We also have j 6= j0 γ (α1, j ,
β1, j )ζ (α1, j , β1, j ) + γ̄ (α1, j0 , β1, j0 ) ≤ γ1 as in part I). Now the desired result follows from the
decomposition of the general terms.
l (), l ≥ 4k 2 + 6k + h + 1 H⇒ u ∈ Hl (),
(continuing the proof of Theorem 4.2) u ∈ Hloc
loc
l ≥ 4k 2 + 6k + h + 1 H⇒ ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) ∈ Hl−h+1
() (by Lemma 4.4). Therefore
loc
by Lemma 4.2 we deduce that u ∈ Hl+1
loc (). Repeat the argument again and again we finally
arrive at u ∈ Hl+t
loc () for every positive t, i.e.
u ∈ ∩l Hlloc () = C ∞ () .
R EMARK 4.1. Theorem 4.2 can be extended to the so-called Grushin type operators (nonlinear version).
43
On the Gevrey analyticity of solutions
4. Gevrey analyticity of the solutions
P ROPOSITION 4.1. Assume that
ϕ(x1 , x2 , u, . . . ,γ ∂α,β u)(α,β,γ )∈4h−1 ∈ C{L n−a−2 ; |L n−a−2 ; E}
e0 , H
e1 , C2 , C3 such that for every H0 ≥ H
e0 ,
for every a ∈ [0, r0 ]. Then there exist constants H
2r0 +3
e
H1 ≥ H1 , H1 ≥ C2 H0
if
(q−r0 −2)
|u, |q ≤ H0 H1
then
0 ≤ q ≤ N + 1, r0 + 2 ≤ N
L q−r0 −2 ,
α β
N−r −1
max ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) ≤ C3 H0 H1 0 L N−r0 −1
¯
x∈
for every (α1 , β1 ) ∈ 0 N+1 .
α
β
Proof. ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) as in Lemma 4.4 is a linear combination with positive
coefficients of terms of the form
µ
Y
∂k ϕ
Y
k
k
k
k
∂ x11 ∂ x22 ∂w13 . . . ∂wµµ+2 j =1 (α1, j ,β1, j )
α
ζ(α1, j ,β1, j )
1, j β1, j
∂1 ∂2 w j
.
Substituting w j by one of the terms γ ∂α,β u with (α, β, γ ) ∈ 4h−1 we obtain
α1, j X
m
α1, j β1, j
∂1 ∂2 (γ ∂α,β u) =
γ · · · (γ − m + 1)θ(γ − m + 1) (γ −m) ∂α+α1, j −m,β+β1, j u .
α1, j
m=0
α β
We can decompose (γ −m) ∂α+α1, j −m,β+β1, j u into (γ −m) ∂α2 ,β2 ∂1 3 ∂2 3 u with (α2 , β2 , γ −
m) ∈ 4h−1 and (α3 , β3 ) ∈ 0(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 . Since α1, j ≤ α1 we deduce
that (α1, j , β1, j ) ∈ 0(α1, j +β1, j +S) . Using the monotonicity condition on L n and the inductive
assumption we have
m γ · · · (γ − m + 1)θ(γ − m + 1) (γ −m) ∂α+α1, j −m,β+β1, j u α1, j
m
α +β −m−R−2
≤
γ · · · (γ − m + 1)θ(γ − m + 1)H0 H1 1, j 1, j
L α1, j +β1, j −m−R−2
α1, j
α
≤ C4 H0 H1 1, j
+β1, j −m−R−2
L α1, j +β1, j −R−2 .
Therefore we deduce that
α
1, j β1, j
∂1 ∂2 (γ ∂α,β u)
α1, j X
m
γ · · · (γ − m + 1)θ(γ − m + 1) (γ −m) ∂α+α1, j −m,β+β1, j u ≤
α
1, j
m=0
α
≤ C5 H0 H1 1, j
+β1, j −R−2
L α1, j +β1, j −R−2 .
44
N.M. Tri
and the general terms can be estimated by
µ
Y
α1, j +β1, j −R−2
C5 H0 H1
L α1, j +β1, j −R−2 .
j =1
Since ϕ ∈ C{L n−a ; |L n−a ; E}, there exist constants C6 , C7 such that
∂k ϕ
q−R
L q−R−2 C7r L r−R−2
q1 q2 r
rµ ≤ C 6 C 7
∂ x ∂ x ∂w 1 . . . ∂wµ
1
2
(q = q1 + q2 , r = r1 + · · ·+rµ ) .
1
Now we take X (ξ, v) = X 1 (v) · X 2 (ξ ) where
X 1 (v) = C6
p
X
C i L i−R−2 v i
8
i=0
i!
,
X 2 (ξ ) =
p
X
C i−R L i−R−2 ξ i
7
i!
i=0
.
and
v(ξ ) = C5 H0
p
X
H1i−R−2 L i−R−2 ξ i
i!
i=1
.
d p X (ξ, v) it follows that
By comparing terms of the form (3) with the corresponding terms in dξ
p
α1 β1
∂1 ∂2 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u)
dp
≤
X (ξ, v)
.
x=x 0
dξ p
ξ =0
Next we introduce the following notation: v(ξ ) h(ξ ) if and only if v ( j ) (0) ≤ h ( j ) (0) for
1 ≤ j ≤ p. It is not difficult to check that there exists a constant C9 (independent of p) such that
v 2 (ξ ) (C5 H0 )2 C9
p
X
H1i−R−3 L i−R−3 ξ i
(i − 1)!
i=2
.
And by induction we have
p
j −1 X
v j (ξ ) (C5 H0 ) j C9
i− j −R−1 L i− j −R−1 ξ
H1
i= j
i
(i − j + 1)!
.
j
1 (ξ ) =
Next, it is easy to verify that X 1 (0) = 0, d Xdξ
≤ C5 C6 C8 H0 , and d X 2j(ξ ) dξ
ξ =0
ξ =0
j −R
C7
L j −R−2 .
j
We now compute d X 1j(ξ ) when 2 ≤ j ≤ p. If 2 ≤ j ≤ 2R + 4 we can al
j
j −R−2
ways choose a constant C10 such that d X 1j(ξ ) ≤ C10 H0 H1
L j −R−2 , provided
dξ
dξ
H1 ≥ (C5 C8 C9 H0 )2r0 +3 .
ξ =0
ξ =0
45
On the Gevrey analyticity of solutions
If j ≥ 2R + 5, we have

R+2
X C i (C5 H0 )i C i−1 H j −i−R−1 L j −i−R−1 j !
d j X 1 (ξ ) 8
9
1

≤C
6
i!( j − i + 1)!
dξ j ξ =0
i=1
+
(4)
j −R−2
X
j −i−R−1
C8i (C5 H0 )i C9i−1 H1
i=R+3
+
j
X
i= j −R−1
The first sum in (4) is estimated by
j −R−2
(5)
L i−R−2 L j −i−R−1 j !
i!( j − i + 1)!

C8i (C5 H0 )i C9i−1 L i−R−2 j !
.
i!( j − i + 1)!
C10 H0 H1
L j −R−2
provided H1 ≥ (C5 C8 C9 H0 )2r0 +3 as for 2 ≤ j ≤ 2R + 4.
By using the monotonicity condition on L n the second sum in (4) is estimated by
(6)
j −R−2
X C i (C5 H0 )i C i−1 H j −i−R−1 L i−R−2 L j −i−R−1 j !
8
9
1
C6
i!( j − i + 1)!
i=R+3
j −R−2
≤
C11 H0 H1
j !L j −2R−3
( j − 2R − 3)!
j −R−2
X
i=R+3
j −R−2
≤ C12 H0 H1
L j −R−2 ,
1
1
i · · · (i − R − 1) ( j − i + 1) · · · ( j − i − R)
provided H1 ≥ C5 C8 C9 H0 .
For the third sum we see that
j
X
C6
C8i (C5 H0 )i C9i−1 L i−R−2 j !
i!( j − i + 1)!
i= j −R−1
(7)
j −R−2
≤ C5 C6 C8 H0 H1
j
X
j!
i= j −R−1
L i−R−2
i!( j − i + 1)!
j −R−2
≤ C13 H0 H1
L j −R−2 ,
if H1 ≥ (C5 C8 C9 H0 )2 .
2r +3
By (4), (5), (6), (7) and taking H1 ≥ (C5 C7 C8 C9 H0 )2r0 +3 = C2 H0 0 we obtain
p X
j
d p X (ξ, v) j −R−2
p− j −R
H H
L j −R−2 C7
L p− j −R−2
≤
C
14
dξ p ξ =0
p 0 1
j =0
p−R−2
≤ C15 H0 H1
Hence
L p−R−2 .
α β
p−R−2
L p−R−2
max ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) ≤ C15 H0 H1
¯
x∈
N−r0 −1
≤ C3 H0 H1
L N−r0 −1 .
46
N.M. Tri
e0 , H
e1 , C2 , C3 depend on  increasingly in the sense that
R EMARK 4.2. The constants H
e0 , H
e1 , C2 , C3 Proposition 4.1 remains valid if we substitute  by any 0 ⊂ .
with the same H
C OROLLARY 4.1. Under the same hypotheses of Proposition 4.1 with q ≤ N + 1 replaced
by q ≤ N, then
α β
N−r −1
max ∂1 1 ∂2 1 ϕ(x1 , x2 , u, . . . ,γ ∂α,β u) ≤ C3 |u, | N+1 + H0 H1 0 L N−r0 −1
¯
x∈
for every (α1 , β1 ) ∈ 0 N+1 .
α
β
∂ϕ
Proof. Indeed, as in the proof of Proposition 4.1 all typical terms, except ∂w
∂ 1 ∂2 1 w j0 can
j0 1
be estimated by | · | N . Replacing w j0 by one of γ ∂α,β u, we have
∂ϕ α1 β1
∂ϕ
∂ ∂ w j0 =
γ ∂α+α1 ,β+β1 u
∂w j0 1 2
∂w j0
α1 ∂ϕ X
m
+
γ · · · (γ − m + 1)θ(γ − m + 1) (γ −m) ∂α+α1 −m,β+β1 u .
∂w j0
α1
m=1
The first summand is estimated by C3 |u, | N+1 . The second sum is majorized as in Proposition
4.1.
T HEOREM 4.3. Let u be a C ∞ solution of the equation (2) and ϕ
∈
C{L n−a−2 ; |L n−a−2 ; E} for every a ∈ [0, r0 ]. Then u ∈ C{L n−r0 −2 ; }. In particular, if ϕ is a s-Gevrey function then so is u.
Proof.
suffices to consider thecase (0, 0) ∈ . Let us define a distance ρ((y1 , y2 ), (x1 , x2 )) =
It2k+1
|x 1
−y12k+1 |
max
, |x2 − y2 | . For two sets S1 , S2 the distance between them is defined as
2k+1
ρ(S1 , S2 ) = infx∈S1,y∈S2 ρ(x, y). Let V T be the cube with edges of size (in the ρ metric) 2T ,
which are parallel to the coordinate axes and centered at (0, 0). Denote by VδT the subcube which
is homothetic with V T and such that the distance between its boundary and the boundary of V T
is δ. We shall prove by induction that if T is small enough then there exist constants H0 , H1
2r +3
with H1 ≥ C2 H0 0 such that
(8)
u, VδT ≤ H0 for 0 ≤ n ≤ max{r0 + 2, 6k + 1}
n
and
(9)
H1 n−r0 −2
L n−r0 −2 for n ≥ max{r0 + 2, 6k + 1}
u, VδT ≤ H0
n
δ
and δ sufficiently small. Hence the Gevrey analyticity of u follows. (8) follows easily from the
C ∞ smoothness assumption on u. Assume that (9) holds for n = N. We shall prove it for
n = N + 1. Put δ 0 = δ(1 − 1/N). Fix x ∈ VδT and then define σ = ρ(x, ∂ V T ) and σ̃ = σ/N.
Let Vσ̃ (x) denote the cube with center at (x) and edges of length 2σ̃ which are parallel to the coordinate axes, and Sσ̃ (x) the boundary of Vσ̃ (x). Let E 1 , E 3 (E 2 , E 4 ) be edges of
Sσ̃ (x) which
α β
are parallel to Ox1 (Ox2 ) respectively. We have to estimate max x∈V T γ ∂α,β ∂1 1 ∂2 1 u(x) δ
47
On the Gevrey analyticity of solutions
α β
for all (α, β, γ ) ∈ 4h−1 , (α1 , β1 ) ∈ 0 N+1 , and max x∈V T ∂1h ∂1 1 ∂2 1 u(x) for all (α1 , β1 ) ∈
δ
0 N+1 , α1 ≥ 1, β1 ≥ 1. But when (α1 , β1 ) ∈ 0 N we have already the desired estimate. Hence
it suffices to obtain the estimate only for (α1 , β1 ) ∈ 0 N+1 \ 0 N .
L EMMA 4.5. Assume that (α, β, γ ) ∈ 4h−1 and (α1 , β1 ) ∈ 0 N+1 . Then if α1 ≥ 1, β1 ≥ 1
there exists a constant C16 such that
α β
max γ ∂α,β ∂1 1 ∂2 1 u(x) x∈VδT
1 ≤ C16 T 2k+1 u, VδT0 N+1
+ H0
h u =
Proof. First we observe that M2k
where C 1
1
1
H1 N−r0 −1
.
L N−r0 −1 T 2k+1 +
δ
H1
∂x1 + i x12k ∂x2
h
u =
P
1
(α̃,β̃,γ̃ )∈40h C α̃ β̃ γ̃ γ̃ ∂α̃,β̃ u,
are some complex constants. Differentiating the equation (2) α times in x1 and β1
1
α̃ β̃ γ̃
times in x2 then applying the representation formula (1) for  = Vσ̃ (x) we have
β
α
∂1 1 ∂2 1 u(x) =
Z
Vσ̃ (x)
+
Z
h (y, x) (A(y) + B(y)) d y d y
(−1)h F2k
1 2
Sσ̃ (x)


h−1
X
j =0

h− j −1 h 
j
α β
F2k (n 1 + i y12k n 2 ) ds ,
(−1) j M2k ∂1y1 ∂2y1 u(y) M2k
where
A(y) = −
(10)
X
(α̃,β̃,γ̃ )∈40h
min{2kh,α
X 1}
m=1
m
C1
α̃ β̃ γ̃ m α1
(γ̃ −m) ∂α1 +α̃−m,β1 +β̃ u ,
β̃≥1,γ̃ ≥1
and
α
β
(y)
B(y) = −∂1y1 ∂2y1 ϕ(y1 , y2 , u, . . . ,γ ∂α,β u) .
Therefore differentiating γ ∂α,β gives
(11)
Z
α1 β1
u(x)
=
∂
∂
∂
γ α,β
1 2
+
Z
Sσ̃ (x)


h−1
X
j =0
Vσ̃ (x)
h (y, x) (A(y) + B(y)) d y d y
(−1)h γ ∂α,β F2k
1 2
j
α β
(−1) j M2k ∂1y1 ∂2y1 u(y)
It is easy to verify that
(12)
C17
h ≤ .
2k+1 2k+1 γ ∂α,β F2k
x1 −y1 + |x2 − y2 |
2k+1
Next we estimate A(y). Consider two cases

h− j −1 h 
F2k (n 1 + i y12k n 2 ) ds .
γ ∂α,β M2k
48
N.M. Tri
y
u(y) in (10) can be rewritten
I) α1 ≤ r0 (= (2k + 1)h). The typical terms (γ̃ −m) ∂
α1 +α̃−m,β1 +β̃
y
α3 β3
as (γ −m) ∂α ,β ∂1y ∂2y u(y) with (α2 , β2 , γ − m) ∈ 4h−1 and (α3 , β3 ) ∈ 0 N+1−m . Hence
2 2
we have
min{2kh,α
X 1}
H1 N−r0 −m−1
|A(y)||Vσ̃ (x) ≤ C18
L N−r0 −m−1
H0
δ0
m=1
H1 N−r0 −2
≤ C19 H0
L N−r0 −1 .
δ0
y
u(y) into
α1 +α̃−m,β1 +β̃
II) If α1 ≥ r0 + 1 then we can decompose (γ̃ −m) ∂
y
α̃+β̃,0
γ̃ −m ∂
α −β̃−m β1 +β̃
∂2y u(y)
∂1y1
,
where (α1 − β̃ − m, β1 + β̃) ∈ 0 N+1−m and α1 − β̃ − m ≥ 1, β1 + β̃ ≥ 1. Therefore by
inductive assumptions we see that
|A(y)||Vσ̃ (x) ≤ C20
min{2kh,α
X 1} m=1
≤ C21 H0
m
H1 N−r0 −m−1
H0
L N−r0 −m−1
α1
δ0
H1 N−r0 −2
L N−r0 −1 .
δ0
Hence in both cases we have
|A(y)||Vσ̃ (x) ≤ C22 H0
(13)
H1 N−r0 −2
L N−r0 −1 .
δ0
On the other hand, from Proposition 4.1 and the inductive assumption we have
!
H1 N−r0 −1
T
L N−r0 −1 .
(14)
|B(y)| ≤ C3 |u, Vδ 0 | N+1 + H0
δ0
Combining (12), (13), (14) we obtain
Z
h (y, x) (A(y) + B(y)) d y d y ∂
F
γ α,β 2k
1 2
Vσ̃ (x)
≤
C23
Z
!
H1 N−r0 −1
L N−r0 −1
δ0
1
2k+1 2k+1 d y1 d y2
x
−y
Vσ̃ (x) 1
1
+ |x2 − y2 |
2k+1
!
1
H1 N−r0 −1
T
2k+1
C24 T
|u, Vδ 0 | N+1 + H0
L N−r0 −1 .
δ
×
≤
|u, VδT0 | N+1 + H0
To estimate the second integral in (11) we first note that
h− j −1 h
F2k =
γ ∂α,β M2k
min{
j,α} α−λ
X
X
λ=0
α−λ
λ̃≥ 2k+1
(2k+1)λ̃+λ−α+γ
C3
αβλλ̃ (x1 − y1 ) j −λ x1
x 12k+1 −y12k+1
+ i(x2 − y2 )
2k+1
β+λ̃+1 .
49
On the Gevrey analyticity of solutions
Let us consider two cases:
I) 0 ≤ j ≤ h − 2. We then split into sub cases:
1
A) |x1 | ≤ 2σ̃ 2k+1 . We have the following estimate
h− j −1 h F2k γ ∂α,β M2k
Sσ̃ (x)
C25
≤
σ̃
h− j −1
2k+1 +1
.
j
y
α β
α β
We note that M2k ∂1y1 ∂2y1 can be expressed as a linear combination of γ 0 ∂α 0 ,β 0 ∂1y2 ∂2y2
j −1
with (α 0 , β 0 , γ 0 ) ∈ 4h−1 , (α2 , β2 ) ∈ 0 N+1− h− j −1 if h−
2k+1 is an integer and (α2 , β2 ) ∈
2k+1
0
h− j −1
i
h
j −1 if 2k+1 is not an integer. Here [·] denotes the integer part of the argument.
N− h−
2k+1
Therefore in both cases the integral can be estimated in the following way


Z
h−2
X
h−
j
−1
β
j
α
h
j
2k
1
1


∂
M
F
∂
∂
u(y)
(−1)
M
(n
+
i
y
n
)
ds
γ α,β 2k
1
2k
1 2
1y 2y
2k
Sσ̃ (x) j =0
≤ C26
h−2
X N
h− j −1
2k+1
j =0
C H
≤ 27 0
H1
H0
h
i
h− j −1
h− j −1
H1 N−r0 −θ { 2k+1 } − 2k+1 −1
δ0
h− j −1
δ 2k+1
L
N−r0 −θ
H1 N−1−r0
L N−r0 −1 ,
δ
n
h− j −1
2k+1
o h
i
j −1
− h−
2k+1 −1
where {·} denotes the fractional part of the argument.
1
B) |x1 | ≥ 2σ̃ 2k+1 . We then have on Sσ̃ (x)
β+λ̃−(h− j −1)
2k+1
|x1 |2k(h− j −1) ≥ C28 |y1 − x1 | j −λ |x1 |(2k+1)λ̃−(α−λ)+γ
− x12k+1 y1
because (α, β, γ ) ∈ 4h−1 . For every (α 0 , β 0 , γ 0 ) ∈ 4 j we now write
γ 0 ∂ y0 0 ∂ α1 ∂ β1 γ ∂α,β M h− j −1 F h 2k S (x) 1y 2y
2k
α ,β
σ̃
γ ∂α,β M h− j −1 F h y
α1 β1
2k
2k 0
= γ +2k(h− j −1) ∂α 0 ,β 0 ∂1y ∂2y
2k(h− j −1)
y1
Sσ̃ (x) C29
y
α1 β1 0 +2k(h− j −1) ∂ 0 0 ∂
≤
×
∂
γ
1y 2y .
α ,β
σ̃ 1+(h− j −1)
y
α β
On the other hand γ 0 +2k(h− j −1) ∂α 0 ,β 0 ∂1y1 ∂2y1 can be decomposed into
y
α3 β3
γ 0 +2k(h− j −1) ∂α 0 +α2 ,β 0 +β2 ∂1y ∂2y
where (α 0 + α2 , β 0 + β2 , γ 0 + 2k(h − j − 1)) ∈ 4h−1 and (α3 , β3 ) ∈ 0 N+1−(h− j −1) . It follows
50
N.M. Tri
that


Z
h−2
X
j
h−
j
−1
β
α
j
2k
h
1
1


(−1)
M
(n
+
i
y
n
)
ds
∂
M
F
∂
∂
u(y)
γ α,β 2k
1
1 2
2k
2k
1y 2y
Sσ̃ (x) j =0
T
h−2
X u, Vδ 0 N+1−(h− j −1) Z
2k n ) ds (n
+
i
y
≤ C30
1
2
1
1+(h−
j
−1)
σ̃
Sσ̃ (x)
j =0
≤
C31 H0
H1
H1 N−1−r0
L N−r0 −1 .
δ
II) j = h − 1.
A) We first estimate the integrals along the edges E 2 and E 4 . Integrating by parts gives
Z
α1 β1
h−1
h dy M
∂
F
∂
∂
u(y)
γ α,β 2k
2
2k
1y 2y
E 2 ∪E 4

Z
X
y
α1 β1 −1
h dy ≤ C32 
0
∂
∂
∂
∂
∂
F
γ
α,β 2k
2y
2
γ α 0 ,β 0
1y 2y
(α 0 ,β 0 ,γ 0 )∈4h−1
X
+
(α 0 ,β 0 ,γ 0 )∈4h−1
≤
≤
C33
u, V T0 δ
E 2 ∪E 4

y
β
−1
α
h

γ 0∂ 0 0 ∂ 1∂ 1
γ ∂α,β F2k 1y 2y
α ,β
∂ E 2 ∪∂ E 4 N
(δ − δ 0 )
C34 H0 H1 N−r0 −1
L N−r0 −1 .
H1
δ
B) We now estimate the integrals along the edges E 1 and E 3 . Integrating by parts gives
Z
α1 β1
h−1
h y 2k d y ∂
M
∂
u(y)
∂
F
γ α,β 2k 1
1
2k
1y 2y
E 1 ∪E 3

Z
X
y
α1 −1 β1
h y 2k−1 d y 0
∂
∂
∂
≤ C35 
∂
F
1
γ α 0 ,β 0
1y
2y γ α,β 2k 1
(α 0 ,β 0 ,γ 0 )∈4h−1
X
+
(α 0 ,β 0 ,γ 0 )∈4h−1
+
X
(α 0 ,β 0 ,γ 0 )∈4h−1
≤
≤
C36
u, VδT0 Z
E 1 ∪E 3
∂ α −1 β
h y 2k d y ∂1y1 ∂2y1
∂
F
γ α,β 2k
1
1
∂ y1
E 1 ∪E 3

α1 −1 β1
h 2k

γ 0 ∂α 0 ,β 0 ∂1y ∂2y γ ∂α,β F2k y1 ∂ E 1 ∪∂ E 3 y
γ 0 ∂α 0 ,β 0
N
(δ − δ 0 )
C37 H0 H1 N−r0 −1
L N−r0 −1 .
H1
δ
51
On the Gevrey analyticity of solutions
L EMMA 4.6. Assume that (α, β, γ ) ∈ 4h−1 . Then there exists a constant C38 such that
1 T
max γ ∂α,β ∂2N+1 u(x) ≤ C38 T 2k+1 u, Vδ(1−6
k /N) N+1
x∈VδT
+ H0
H1
δ
N−r0 −1
L N−r0 −1 T
1
+
H1
1
2k+1
!
.
1
Proof. I) |x1 | ≤ 2σ̃ 2k+1 . Instead of Vσ̃ (x) we take the cube V6k σ̃ (x). As in Lemma 4.5 the
following formula holds
Z
N+1
h (y, x)B(y) d y d y
u(x) =
(−1)h γ ∂α,β F2k
γ ∂α,β ∂2
1 2
(15)
+
where
Z
S6k σ̃ (x)


V6k σ̃ (x)
h−1
X
j

h− j −1 h 
F2k (n 1 + i y12k n 2 ) ds ,
N+1
u(y) γ ∂α,β M2k
(−1) j M2k ∂2y
j =0
B(y) = −
∂ N+1
(y)
ϕ(y1 , y2 , u, . . . ,γ ∂α,β u) .
∂ y2N+1
Therefore the first integral in (15) is estimated by
!
1
H1 N−r0 −1
T
+
H
C39 T 2k+1 u, Vδ(1−6
L
0
N−r0 −1 .
k /N) N+1
δ
We now estimate the second integral in (15).
e4 , where E
e2 , E
e4 denote the edges of S k , which
e2 and E
A) First consider the integrals along E
6 σ̃
e
e
are parallel to Ox2 . We note that |y1 | ≥ |x1 | on E 2 , E 4 .
1) 0 ≤ j ≤ h − 2. As in Lemma 4.5, for every (α 0 , β 0 , γ 0 ) ∈ 4 j the following estimate holds
γ 0 ∂ y0 0 ∂ N+1 γ ∂α,β M h− j −1 F h 2k E
2y
2k
α ,β
e2 ∪ E
e4 C40
N+1−(h− j −1) y
0 +2k(h− j −1) ∂ 0 0
∂
≤
e e ,
γ
α ,β +h− j −1 2y
E2 ∪ E4
σ̃ 1+(h− j −1)
y
N+1−(h− j −1)
where γ 0 +2k(h− j −1) ∂α 0 ,β 0 +h− j −1 ∈ 4h−1 and ∂2y
∈ 0 N+1−(h− j −1) . Therefore


Z
h−2
X
h− j −1 h 
2k n ) ds j M j ∂ α1 ∂ β1 u(y)

(n
+
i
y
(−1)
∂
M
F
γ
α,β
1
2
1
2k
e e
2k
2k
1y 2y
E 2 ∪ E 4 j =0
N−1−r0
C H
H1
≤ 41 0
L N−r0 −1 .
H1
δ
2) j = h − 1. Integrating by parts gives:
Z
N+1
h−1
h
∂2y u(y) γ ∂α,β F2k d y2 e e M2k
E2 ∪ E4
T
N u, Vδ(1−6
k /N) ≤
C42
≤
C43 H0
H1
H1
δ
N
δ
N−r0 −1
L N−r0 −1 .
52
N.M. Tri
e1 and E
e3 , where E
e1 , E
e3 denote the edges of S k ,
B) Consider now integrals along the edges E
6 σ̃
which are parallel toOx1 .
1) 0 ≤ j ≤ h − 2. We have
h− j −1 h ≤
F2k γ ∂α,β M2k
e1 ∪ E
e3
E
C44
σ̃
and
h− j −1
2k+1 +1
1
|y1 || Ee ∪ Ee ≤ C45 σ̃ 2k+1
1
3
Furthermore for h − 2 − (2k + 1)s − 2k ≤ j ≤ h − 2 − (2k + 1)s and every (α 0 , β 0 , γ 0 ) ∈ 4 j
we have
y
h−2− j −(2k+1)s
N+1
y12k γ 0 ∂α 0 ,β 0 ∂2y
u(y)
= y1
y
N−s
u(y) ,
× γ 0 + j +2k−(h−2−(2k+1)s) ∂α 0 ,β 0 +s+1 ∂2y
where (α 0 , β 0 + s + 1, γ 0 + j + 2k − h + (2k + 1)s + 2) ∈ 4h−1 and (0, N − s) ∈ 0 N−s . Hence
we get


Z
h−2
X
j
h−
j
−1
N+1
jM
h  y 2k d y 
∂
u(y)
(−1)
∂
M
F
γ α,β 2k
1
e e
2k
1
2y
2k
E
∪
E
1 3 j =0
≤
C46
h
i
T
X u, Vδ(1−6k /N) N−s Z
h−2
2k+1
s=0
≤
C47 H0
H1
H1
δ
(δ − δ 0 )1+
N−r0 −1
h− j −1
2k+1
L N−r0 −1 .
e1 ∪ E
e3
E
|y1 |h−(2k+1)s− j −2 d y1
2) j = h − 1. We note that
N+1
h−1
∂2y
u
y12k M2k
=
=
h−1
N u = (M − ∂ )M h−1 ∂ N u
∂2y
y12k ∂2y M2k
2k
1y
2y
2k
h ∂ N u − ∂ M h−1 ∂ N u = ∂ N ϕ − ∂ M h−1 ∂ N u .
M2k
1y 2k
1y 2k
2y
2y
2y
2y
It follows that
(16)
Z
N+1
h−1
h
2k
∂2y u(y) γ ∂α,β F2k y1 d y1 e e M2k
E1 ∪ E3
Z
Nϕ ∂
h
≤ ∂2y
γ α,β F2k d y1 e1 ∪ E
e
E
Z 3
h−1
h
N
∂2y u(y) γ ∂α,β F2k d y1 .
+
∂1y M2k
e1 ∪ E
e3
E
By Proposition 4.1 the first integral in the right hand side of (16) can be estimated by
C48 u, V T
C H
δ(1−6k /N) N
≤ 49 0
δ − δ0
H1
H1 N−r0 −1
L N−r0 −1 .
δ
53
On the Gevrey analyticity of solutions
To estimate the second integral in the right hand side of (16) we now integrate by parts:
Z
h−1
h dy N u(y)
∂
F
∂
∂
M
γ
α,β
1
1y
e e
2k
2y
2k
E1 ∪ E3

1
Z
N 1+ 2k+1
N2 N
T
y12k d y1 +
≤ C50 u, Vδ(1−6
+ 2 k /N) 1
N
σ
e1 ∪ E
e3
σ
E
σ 1+ 2k+1
C51 H0 H1 N−r0 −1
≤
L N−r0 −1 .
H1
δ

Z

e e d y1 E1 ∪ E3
1
II) |x1 | ≥ 2σ̃ 2k+1 . We now consider the representation formula for ∂2N+1 u(x) in Vσ̃ (x). As
above the volume integral can be estimated by
1
C52 T 2k+1
u, VδT0 N+1
+
!
H1 N−r0 −1
L N−r0 −1 .
δ
For the boundary integral we again split into 2 cases
A) j = h − 2. As in Lemma 4.5 we obtain the following estimate


Z
h−2
X
j
β
h−
j
−1
α
j
2k
h
1
1


(−1)
M
(n
+
i
y
n
)
ds
∂
∂
∂
M
F
u(y)
γ α,β 2k
1
1 2
2k
2k
1y 2y
Sσ̃ (x) j =0
H1 N−r0 −1
C H
L N−r0 −1 .
≤ 53 0
H1
δ
B) j = h − 1.
1) Along E 2 and E 4 we can estimate exactly as in Lemma 4.5.
2) Consider now the boundary integral along E 1 and E 3 . As in this Lemma I) B) 2) we have
(17)
Z
N+1
h−1
h y 2k d y ∂2y
u(y) γ ∂α,β F2k
M2k
1
1
E 1 ∪E 3
Z
Nϕ ∂
h
∂2y
≤ γ α,β F2k d y1 E 1 ∪E 3
Z
h−1
N u(y)
h
∂2y
∂1y M2k
+ γ ∂α,β F2k d y1 .
E 1 ∪E 3
By Proposition 4.1 the first integral in the right hand side of (17) can be estimated by
C54 u, VδT0 δ − δ0
N ≤ C 55 H0
H1
H1 N−r0 −1
L N−r0 −1 .
δ
54
N.M. Tri
To estimate the second integral in the right hand side of (17) we now integrate by parts
Z
h−1
h dy N u(y)
∂
M
∂
F
∂
γ
α,β 2k
1y 2k
1
2y
E 1 ∪E 3
h−1 N
h ∂2y u(y) γ ∂α,β F2k
≤ M2k
∂ E 1 ∪∂ E 3 Z
∂ h−1
N u(y)
h
∂1y M2k
∂2y
+ γ ∂α,β F2k d y1 ∂ y1
E 1 ∪E 3
!
Z
2
N N
≤ C56 u, VδT0 y12k d y1 + 2 N σ
σ
E 1 ∪E 3
C57 H0
H1
≤
N−r0 −1
H1
δ
L N−r0 −1 .
L EMMA 4.7. Assume that (α, β, γ ) ∈ 4h−1 . Then there exists a constant C58 such that
N−r +1
max γ ∂α,β ∂1 0 u(x) ≤
x∈VδT
1 C58 T 2k+1 u, VδT0 +H0
H1
δ
N+1
N−r0 −1
1
1
L N−r0 −1 T 2k+1 +
.
H1
N−r +1
Proof. We have the following representation formula for γ ∂α,β ∂1 0 u(x) in Vσ̃ (x)
(18)
Z
N−r0 +1
∂
∂
u(x)
=
γ α,β
1

Z
h−1
X
Vσ̃ (x)
h (y, x) (A(y) + B(y)) d y d y
(−1)h γ ∂α,β F2k
1 2
j
N−r +1

+
(−1) j M2k ∂1y 0 u(y)
Sσ̃ (x) j =0

h− j −1 h 
F2k (n 1 + i y12k n 2 ) ds ,
γ ∂α,β M2k
where
A(y) = −
X
(α̃,β̃,γ̃ )∈40
h
β̃≥1,γ̃ ≥1
min{2kh,N−r
X 0 +1}
m=1
C1
α̃ β̃ γ̃ m
m
y
∂
u(y) ,
N − r0 + 1 (γ̃ −m) N−r0 −m+1+α̃,β1 +β̃
and
N−r0 +1
B(y) = −∂1y
(y)
ϕ(y1 , y2 , u, . . . ,γ ∂α,β u) .
Therefore, as in Lemma 4.5, the first integral in (18) is estimated by
C59 T
1
2k+1
u, VδT0 N+1
+ H0
We now estimate the second integral in (18).
!
H1 N−r0 −1
L N−r0 −1 .
δ
55
On the Gevrey analyticity of solutions
I) 0 ≤ j ≤ h − 2. As in Lemma 4.5 I) A) 2) we obtain
Z
Sσ̃ (x)
h−2
X
j =0
≤
j
N−r +1
(−1) j M2k ∂1y 0 u(y)
C60 H0
H1
h− j −1 h
F2k
γ ∂α,β M2k
H1 N−r0 −1
L N−r0 −1 .
δ
n 1 + i y12k n 2 ds II) j = h − 1.
A) We first estimate the integrals along E 2 and E 4 . Note that
h−1 N−r0 +1
∂1y
u(y) M2k
h
N−r
N−r
h−1 2k
= M2k
∂1y 0 u(y) − i M2k
y1 ∂2y ∂1y 0 u(y) min{2kh,N−r
}
X 0
m
N−r0 X
y
1
≤ ∂1y ϕ + C
∂
u
(γ̃ −m) α +α̃−m,β +β̃ α̃
β̃
γ̃
m
N − r0
1
1
(α̃,β̃,γ̃ )∈40
m=1
β̃≥1,γ̃ ≥1h
N−r
h−1 2k
y1 ∂1y 0 u(y) + ∂2y M2k
H1 N−r0 −1
H
N−r
h−1 2k
L N−r0 −1 .
y1 ∂1y 0 u(y) + C61 0
≤ ∂2y M2k
H1
δ
Therefore
Z
E 2 ∪E 4
≤
≤
N−r +1
h−1
∂1y 0 u(y)
M2k
h
γ ∂α,β F2k d y2 H1 N−r0 −1
H
C62 0
L N−r0 −1
H
δ
Z 1
N−r
h−1 2k
h dy + M2k
y1 ∂1y 0 u(y) ∂2y γ ∂α,β F2k
2
E ∪E
2 4 h−1 2k N−r0
h + M2k y1 ∂1y u(y)
γ ∂α,β F2k H
C63 0
H1
H1
δ
N−r0 −1
∂ E 2 ∪∂ E 4
L N−r0 −1 .
B) The integrals along E 1 and E 3 now can be estimated in the same manner as in Lemma
4.5.
L EMMA 4.8. Assume that (α1 , β1 ) ∈ 0 N+1 \ 0 N , α1 ≥ 1, β1 ≥ 1. Then there exists a
constant C64 such that
1 h α1 β1
T
max ∂1 ∂1 ∂2 u(x) ≤ C64 T 2k+1 u, Vδ(1−6
k /N) N+1
x∈VδT
+H0
H1
δ
N−r0 −1
1
1
L N−r0 −1 T 2k+1 +
.
H1
56
N.M. Tri
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
!
1
1
1
H1 N−r0 −1
k
.
g(δ) ≤ C65 T 2k+1 g δ(1 − 6 /N) + H0
L N−r0 −1 T 2k+1 +
δ
H1
2k+1
k
then by Lemma 4.3 we deduce that
Choosing T ≤ 1/812 C65
g(δ) ≤ C66 H0
1
H1 N−r0 −1
1
L N−r0 −1 T 2k+1 +
.
δ
H1
If T is chosen to be small enough such that T ≤ (1/2C66 )2k+1 and choosing H1 ≥ 2C66 (in
2r +3
addition to H1 ≥ C2 H0 0 ) we arrive at
g(δ) ≤ H0
That means
H1 N−r0 −1
L N−r0 −1 .
δ
H1 N−r0 −1
L N−r0 −1 .
N+1
δ
The proof of Theorem 4.3 is therefore completed.
u, VδT ≤ H0
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
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On the Gevrey analyticity of solutions
57
[9] C ATTABRIGA L., RODINO L., Z ANGHIRATI L., Analytic-Gevrey Hypoellipticity for a
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AMS Subject Classification: 35H10, 35B65, 35C15.
Nguyen Minh TRI
Institute of Mathematics
P.O. Box 631, Boho 10000
Hanoi, Vietnam
and Dipartimento di Matematica
Università di Torino
Via Carlo Alberto 10
10123 Torino, Italia
Lavoro pervenuto in redazione il 25.1.1998 e, in forma definitiva, il 10.9.1998.
58
N.M. Tri