Electronic Journal of Differential Equations, Vol. 2004(2004), No. 48, pp. 1–24.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
DIRICHLET PROBLEM FOR DEGENERATE ELLIPTIC
COMPLEX MONGE-AMPÈRE EQUATION
SAOUSSEN KALLEL-JALLOULI
Abstract. We consider the Dirichlet problem
∂2u det
= g(z, u) in Ω ,
∂zi ∂zj
u∂Ω = ϕ ,
where Ω is a bounded open set of Cn with regular boundary, g and ϕ are
sufficiently smooth functions, and g is non-negative. We prove that, under
additional hypotheses on g and ϕ, if | det ϕij − g|C s∗ is sufficiently small the
problem has a plurisubharmonic solution.
1. Introduction
Let Ω be a bounded domain in R2n with smooth boundary and let zi = xi +
ixi+n (1 ≤ i ≤ n). We shall also denote by Ω the set of z = (z1 , z2 , . . . , zn ) satisfying (Re z, Im z) ∈ Ω. We study the problem of finding a sufficiently smooth
plurisubharmonic solution to the degenerate problem
det
∂2φ = g(z, φ) in Ω ,
∂zi ∂zj
φ = ϕ .
(1.1)
∂Ω
In [8, 9], the author studies local solutions, while, here we consider global solutions.
This problem has received considerable attention both in the non-degenerate case
(g > 0) and in the degenerate case (g ≥ 0). In particular, Caffarelli, Kohn, Nirenberg and Spruck [4] established some existence results in strongly pseudoconvex
domains based on the construction of a subsolution. The recent work of Guan [6],
extends some of these results to arbitrary smooth bounded domains. Guan proved
for the nondegenerate case that a sufficient condition for the classical solvability is
the existence of a subsolution. Here we are concerned with degenerate problems in
an arbitrary smooth bounded domain, which need not be Pseudoconvex.
Counterexamples due to Bedford and Fornaes [2] show that the Dirichlet problem, in general, does not have a regular solution. This implies that we should place
some restrictions on g and ϕ.
2000 Mathematics Subject Classification. 35J70, 32W20, 32W05.
Key words and phrases. Degenerate elliptic, omplex Monge-Ampère,
Plurisubharmonic function.
c
2004
Texas State University - San Marcos.
Submitted May 15, 2003. Published April 6, 2004.
1
2
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Let us assume that ϕ is a real function defined in Ω, Σ is a finite set of points in
Ω, and g(z, φ) = K(z)f (Re z, Im z, φ). We further assume the following hypotheses.
(A1) K ≥ 0 in Ω, and K −1 (0) = Σ
∂f
(A2) f (x,
u) > 0 in Ω × R, and ∂u ≥ −% in Ω× R, with 0 ≤ % << 1
(A3) ϕ Ω\Σ is strictly plurisubharmonic, (ϕij )Σ is of rank (n − 1), and the eigenvalues of (ϕij ) on Σ are distinct.
Our main results are the following theorems:
Theorem 1.1. Let s∗ ≥ 7 + 2n be an integer, α ∈]0, 1[, and Γ > 1. If ϕ ∈
C s∗ +2,α (Ω) satisfies the condition (A3), then one can find a constant ε0 > 0 (depending on s∗ , α, Γ, Ω and ϕ) such that for any g = Kf ∈ C s∗ satisfying (A1),
(A2),
| det ϕij − g(ϕ)|C s∗ ≤ ε0
(1.2)
∂g
|C s∗ ≤ Γ, then problem (1.1) has a plurisubharmonic (real valued) solution
and | ∂u
s∗ −3−n
φ∈C
(Ω), which is unique when ρ = 0.
Let lα (x) denote α-th row the matrix of cofactors of (ϕij ), and
Dk K(x)(lα (x), lβ (x))(k) = Dk K(x) lα (x), lβ (x); . . . ; lα (x), lβ (x) .
Theorem 1.2. Under the assumptions in Theorem 1.1, suppose that ϕ ∈ C ∞ (Ω)
and for any point x0 ∈ Σ one can find an integer k such that Dj K(x0 ) = 0 for all
j ≤ k−1 and there exists α 6= β ∈ {1, . . . , n} such that Dk K(x0 )(lα (x0 ), lβ (x0 ))(k) 6=
0. Then there exists an integer s∗ > 0 and a constant ε0 > 0 such that for any
function g ∈ C ∞ satisfying (A2), (A3) and (1.2), the plurisubharmonic solution φ
to the problem (1.1) is in C ∞ (Ω).
In Theorem 1.1, the assumption concerning Σ leads to a-priori estimates and
the assumption on g and ϕ ensures the convergence of an iteration scheme of
Nash-Moser type. It is to be noted that we do not require demonstrating that
a subsolution exists as in [4] and [6].
Under some additional conditions on g, we can prove the smoothness of the
solution, using the works of Xu [12] and Xu and Zuily [13].
This paper is organized as follows. In Section 2 we state some preliminary
results. In Section 3, we state fundamental global a-priori estimates for degenerate
linearized operators that are crucial to establish an iteration scheme of Nash-Moser
type. We then prove Theorem 1.1 in Section 4. We prove Theorem 1.2 in Section
5. Finally, we prove the a-priori estimates stated in Section 3.
2. Preliminary results
We shall use the norms
| · |k = k · kC k (Ω) ,
k · kk = k · kH k (Ω) ,
| · |k,τ = k · kC k,τ (Ω)
where k ∈ N and τ ∈]0, α[.
In this work, we need some technical lemmas which play important roles in the
proof of convergence of our iteration scheme.
Lemma 2.1. Let s∗ be an integer, s∗ ≥ 7 + 2n. We can find a constant β ≥ 2
such that for any 0 ≤ i, j, k ≤ s∗ + 2, n∗ = n + τ and u ∈ C s∗ +2,α (Ω) we have: The
Sobolev inequality
|u|i,τ ≤ βkuki+n∗
(2.1)
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
3
The Gagliardo-Nirenberg inequality
k−j
j−i
kukj ≤ βkukik−i kukkk−i ,
i<j<k
(2.2)
The inequality
kuks∗ ≤ β|u|s∗
(2.3)
For any λ ≥ 1, there exists a family of smoothing linear operators Sλ : ∪i≥0 H i (Ω) →
∩j≥0 H j (Ω), satisfying
kSλ uki ≤ βkukj ,
kSλ uki ≤ βλ
i−j
kSλ u − uki ≤ βλ
if i ≤ j
kukj ,
i−j
kukj ,
if i ≥ j
if i ≤ j
(2.4)
(2.5)
(2.6)
Lemma 2.2 ([1, 7]). (1) For t > 0; if u, v ∈ L∞ ∩ H t , then uv ∈ L∞ ∩ H t and
kuvkt ≤ K1 (|u|0 kvkt + kukt |v|0 ),
(2.7)
where, K1 is a constant ≥ 1 independent of u and v.
(2) Let H : Rm → C be a function C ∞ of its arguments.
For s > 0, if ω ∈ (L∞ ∩ H s )m and |ω|0 ≤ M , then
kH(ω)ks ≤ K2 (s, H, M )(kωks + 1),
(2.8)
where K2 ≥ 1 and is a constant independent of ω.
If ω ∈ (C i,µ )m , µ ∈]0, 1[ and i ∈ N, then H(ω) ∈ C i,µ .
If we suppose that |ω|0 ≤ M , then we can find a constant K3 = K3 (i, µ, H, M ) ≥
1 such that
|H(ω)|i,µ ≤ K3 (|ω|i,µ + 1).
(2.9)
We shall also need the following technical lemma.
Lemma 2.3 ([8, Lemma]). Let F (uzi zj ) = det(uzi zj ). For 1 ≤ i, j, a, b ≤ n, we
have
∂2F
∂F
∂F
∂F
∂F
F
=
−
.
(2.10)
∂uza zb ∂uzi zj
∂uza zb ∂uzi zj
∂uzi zb ∂uza zj
3. A priori estimates for the linearized operator
Defining φ = ϕ + εw, (1.1) becomes
det(φzi zj ) = det(ϕzi zj + εwzi zj ) = g.
Let
G(w) =
1
[det Φ − g].
ε
(3.1)
(3.2)
Then the linearization of G at w is
LG (w) =
n
X
φij ∂zi ∂zj + b,
(3.3)
i,j=1
e = (φij ) is the matrix of cofactors of Φ = (φz z (z, ε, w)) and b = ∂g .
where Φ
i j
∂u
Now we construct linear elliptic operators, maybe degenerate, related to linearized operators. For any smooth real valued function w, the matrix (φij ) is
Hermitian and we can find a unitary matrix T (z, ε) satisfying
T (z, ε)(φzi zj )t T (z, ε) = diag(λ1 , . . . , λn ).
(3.4)
4
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Without loss of generality we may assume that Σ is reduced to one point, the origin.
By means of change of variables we may assume, using (A3), that
ϕzi zj (0) = σi δij
i, j = 1, . . . , n,
where σi > 0 for i = 1, . . . , n − 1, σn = 0 and σi 6= σj for i 6= j. Let 0 < τ ≤
(3.5)
α
4.
Lemma 3.1. There exist constants ε1 > 0, δ1 > 0 and M > 0 depending only on
ϕ, n, Ω such that when
V0 = {(z, ε, w)/|z| ≤ δ1 , 0 ≤ ε ≤ ε1 , w ∈ C 3,τ (Ω), |w|3,τ ≤ 1},
we have: (i) The eigenvalues λi , i = 1, . . . , n of Φ are distinct on V0 and of class
C 1 in V̊0 . Moreover, λi > 0 in V0 , for i = 1, . . . , n − 1.
(ii) For (z, ε, w) ∈V0 ,
n
X
|σi − λi (z, ε, w)| + |Φnn (z, ε, w) −
i=1
n−1
Y
σi | ≤ M (ε + |z|).
(3.6)
i=1
(iii) For (z, ε, w) ∈ V0 and i = 1, . . . , n − 1,
λi ≥
inf
1≤i≤n−1
σi − M δ1 − (M + 1)ε1 > 0 and Φnn ≥
n−1
Y
σi − M δ1 − M ε1 > 0. (3.7)
i=1
Proof. Let us consider the function H(z, ε, w, λ) = det(ϕzi zj + εwzi zj − λδij ). Then
H ∈ C 1 and by (3.5), we have
∂H
(0, 0, 0, σi ) 6= 0, ∀i ∈ {1, . . . , n}.
∂λ
By the implicit function theorem, one can find two constants ε1 > 0 and δ1 > 0
such that (i) holds. Moreover by (3.5) we have
H(0, 0, 0, σi ) = 0 and
n−1
Y
∂F
(ϕij )(0) = Φnn (0, 0, w) =
σi > 0,
∂unn
i=1
which gives (ii) and (iii).
Lemma 3.2. There exists a positive constant ε2 such that for any 0 < ε < ε2 , any
real valued function w ∈ C 3,τ (Ω) satisfying |w|3,τ ≤ 1 and θ = maxz∈Ω |G(w)|, the
operator
L = −LG (w) − θ4
(3.8)
Pn
∂2
∂2
is elliptic, maybe degenerate. (Here 4 = i=1 ( ∂x2 + ∂y2 ))
i
i
Proof. Let
A = θ|ξ|2 +
n
X
φij ξi ξj ≥ 0,
∀(z, ξ) ∈ Ω × Cn .
(3.9)
i,j=1
If z ∈ Ω\{0}, as ϕ is strictly plurisubharmonic, then A > 0 for all ξ ∈ Cn \{0}.
e Then we have
If z = 0, for ξ ∈ Cn , we let ξ =t T (τ, ε)ξ.
e Φ
e
e = θ|ξ|2 +t ξT
e t T ξ.
A = θ|ξ|2 +t ξ Φξ
e = det Φ Id, by (3.4),
Since ΦΦ
e t T = diag(λi )T Φ
e tT ,
det Φ Id = T Φt T T Φ
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
e t T = det Φ diag(
TΦ
5
n
Y
1
1
1
)=
λi diag( ) = (εG + g) diag( ).
λi
λ
λ
i
i
i=1
Thus,
e 2 + det Φ
A = θ|ξ|
n
X
|ξei |2
i=1
e2 +
= θ|ξ|
n−1
X
det Φ
i=1
= (θ +
n−1
Y
λi
n−1
Y
|ξei |2
+
λi |ξen |2
λi
i=1
λi )|ξen |2 +
i=1
n−1
X
i=1
εG + g + θλi e 2
|ξi | .
λi
By (3.7), for i = 1, . . . , n − 1, ε ≤ ε1 and |w|3,τ ≤ 1, we have
εG + θλi ≥ θ(σi − M δ1 − (M + 1)ε1 ) ≥ 0.
Therefore, A ≥ 0, which proves the lemma.
Now we study a boundary-value problem for the degenrate elliptic operator
L = −LG (w) − θ4 =
n
X
bij ∂zi ∂zj + b,
i,j=1
where
bij = −
∂F
(ϕ + εwij ) − θδij = −Φij − θδij
∂uij ij
∂f
. For k, s ∈ N we let
and b = K ∂u
A(k) = max(1, max |bij |k , |b|k )
1≤i,j≤n
(3.10)
Λs = {(i, j) : 0 ≤ i, j ≤ s, i + j ≤ s, and i + 2 ≤ max(s, 2)}
Now from Lemma 3.2 we have the following statement.
Theorem 3.3. Suppose that θ ≤ 1 and A(2) ≤ M0 , for some constant M0 > 0. One
can find ε3 > 0 such that for any ε ∈]0, ε3 ], any real valued function w ∈ C s∗ +2,τ (Ω)
satisfying the inequality |w|3,τ ≤ 1 and any real valued function h ∈ H s∗ , the
problem
Lu = h in Ω
(3.11)
u = 0
∂Ω
has a unique solution u ∈ H s∗ . Moreover for 0 ≤ s ≤ s∗ ,
kuk0 ≤ C0 khk0
kuks ≤ Cs {khks +
kuk1 ≤ C1 (khk1 + kuk0 )
X
(1 + |ϕ + εw|i+4,τ )kukj },
(3.12)
(3.13)
s≥2
(3.14)
j≤s−1, (i,j)∈Λs
for some constant Cs = Cs (ϕ, s, Ω, M0 , ε3 ) independent of w and ε.
For ν ∈]0, 1[, we denote Lν = L − ν4. To solve the Dirichlet problem (3.11), we
first establish the following proposition.
6
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Propositon 3.4. Let θ ≤ 1 and, for some constant M0 > 0, A(2) ≤ M0 . Then
there exists ε3 > 0 such that for any ε ∈]0, ε3 ], any real valued function w ∈
C s∗ +2,τ (Ω) satisfying the inequality |w|3,τ ≤ 1 and any real valued function h ∈
H s∗ (Ω), the regularized problem
Lν u = h in Ω,
u = 0,
(3.15)
∂Ω
has a unique (real valued) solution u ∈ H s∗ +1 (Ω).
Proof. Since LG (w) is a second order operator with real coefficients, from Lemma
3.2, Lν is uniformly elliptic with coefficients in C s∗ ,τ (Ω). Thus by [3, Theorems
6.14 and 8.13] we see that (3.15) has a real valued solution.
If (3.12)–(3.14) hold for the regularized problem (3.15) with an uniform constant
Cs independent of ν ∈]0, 1], then by letting ν tend to zero we get a solution u ∈
H s∗ (Ω) to the original problem which of course satisfies (3.12)–(3.14).
Using Theorem 3.3, we prove Theorem 1.1 by constructing a sequence of approximating solutions and a priori estimates for linearized operators. The hypothesis
(1.2) will play an important role in the proof of the convergence of our iteration
scheme of Nash-Moser type.
4. Proof of Theorem 1.1
Part 1: An iteration scheme of Nash-Moser type. In this section, we use
the Nash-Moser procedure [7, 10] and the results of Section 3 to prove Theorem
1.1. We construct a sequence which converges to a solution to our problem. We
define
M0 = 1 + maxK3 (2, τ, H, (1 + |ϕ|2 ))(1 + |ϕ|4,τ ),
(4.1)
H∈F
where F
∂g
∂F
, ∂u
/1
={ ∂u
ij
≤ i, j ≤ n} and K3 is the constant introduced in (2.9). (i.e:
|H(u)|j,µ ≤ K3 (j, µ, H, M )|u|j,µ ). We also define
D = max
max Cs , 1 .
0≤s≤s∗
(4.2)
Here Cs is the constant (depending only on s, ϕ, Ω, M0 ) given by Theorem 3.3. We
let
1
µ = max(β, 3Ds2∗ (1 + |ϕ|s∗ +2,τ ), n, 2 τ ) and µ
e = β 2 µs∗ ,
(4.3)
a1 = 9K0 µ5 ,
a2 = 5a1 µs∗ +1 ,
a3 = 7K0 µ5 ,
(4.4)
were K0 is the constant given by Proposition 6.1. Also, we fix εe satisfying
εe ≤ min[1, (εi )1≤i≤4 , (3D2 a2 + 6e
µD2 )−2 ],
(4.5)
were εi are given in Lemma 3.2, Theorem 3.3, the proof of Theorem 3.3 and the
proof of (3.13).
As a consequence of these inequalities, we have 6e
εµs∗ ≤ 1/4. Let g ∈ C s∗ satisfy
| det ϕij − g(ϕ)|s∗ ≤ εe2 ,
with ε0 in Theorem 1.1 equal to εe2 . Let Sn = Sµn the family of operators given by
Lemma 2.1, with µn = µn (µ is given by (4.3)).
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
7
Using Theorem 3.3, we construct wn , n = 0, 1, . . . , by induction on n as follows.
We let u0 , w0 = 0, and assume w0 , w1 , . . . , wn have been chosen and define wn+1
by
wn+1 = wn + un+1 ,
(4.6)
where un+1 is the solution to the Dirichlet problem
LG (w
en )un+1 + θn 4un+1 = gn ,
un+1 = 0,
in Ω
(4.7)
∂Ω
given by Theorem 3.3. Here
w
en = Sn wn ,
(4.8)
θn = |G(w
en )|0 ,
(4.9)
g0 = −S0 G(0), gn = Sn−1 Rn−1 − Sn Rn + Sn−1 G(0) − Sn G(0),
R0 = 0,
Rn =
n
X
rj ,
(4.10)
(4.11)
j=1
r0 = 0,
rj = [LG (wj−1 ) − LG (w
ej−1 )]uj + Qj − θj−1 4uj ,
Qj = G(wj ) − G(wj−1 ) − LG (wj−1 )uj ,
1 ≤ j ≤ n,
1 ≤ j ≤ n.
(4.12)
(4.13)
To ensure that the wn ’s are well defined, we prove the following proposition.
Propositon 4.1. Let s ∈ N. If s∗ ≥ 7 + 2n and 4 + 2n + 2τ ≤ σ < s∗ − 2, we have
√
kuj ks ≤ εe[max(µ, µj−1 )]s−σ , j ∈ N∗ , 0 ≤ s ≤ s∗ ,
(4.14)
( √
2 εe,
for s ≤ σ − τ
j ∈ N∗ ,
(4.15)
kwj ks ≤ √ s−σ
εeµj , for σ − τ ≤ s ≤ s∗
|w
ej |4,τ ≤ 1, j ∈ N∗ ,
√
kwj − w
ej ks ≤ 2β εeµs−σ
, 0 ≤ s ≤ s∗ , j ∈ N∗ ,
j
s−σ
krj ks ≤ εea1 [max(µ, µj−1 )]
kgj ks ≤
εea2 µs−σ
,
j
√
,
(4.16)
(4.17)
∗
0 ≤ s ≤ s∗ − 2, j ∈ N ,
0 ≤ s ≤ s∗ , j ∈ N,
θj ≤ a3 εeµ−2
j ≤ 1,
Aj (2) ≤ M0 ,
j ∈ N,
(4.18)
(4.19)
(4.20)
j ∈ N.
(4.21)
Here, Aj (k) is defined by using the definition of A(k) in (3.10), where the coefficients correspond to w
ej .
Let us first show how that Proposition 4.1 implies Theorem 1.1. The proof of
this proposition will be given later in Appendix 1.
Part 2: Proof of Theorem 1.1. We prove the convergence of the sequence (wn )
using Proposition 4.1. Set σ = s∗ − 2 − τ and s = σ − τ . By (4.6) and (4.14), for
any i, k ∈ N∗ , i > k,
kwi − wk ks ≤
i
X
j=k+1
i
i
√ X
√ X
µ−τ
=
β
ε
e
(µ−τ )j−1 .
kuj ks ≤ β εe
j−1
j=k+1
j=k+1
Since µ ≥ 2 and τ > 0, then kwi −wk ks → 0 as i, k → ∞. Hence, there is a function
w ∈ H s∗ −2−2τ (Ω) satisfying wn → w in H s∗ −2−2τ (Ω).
8
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Since H s∗ −2−2τ (Ω) ⊂ C s∗ −2−n−3τ (Ω), it follows that w ∈ C s∗ −3−n (Ω). On the
other hand, combining (4.7), (4.12) and (4.13), we obtain
rj = G(wj ) − G(wj−1 ) − gj−1
Taking the sum between j = 1 and j = n, using (4.10) and (4.11), we get
G(wn ) = (I − Sn−1 )Rn−1 + (I − Sn−1 )G(0) + rn .
(4.22)
For n ≥ 2, using (2.2) and (4.18), we have
∗ −2−2τ −σ
krn ks∗ −2−2τ ≤ a1 βe
εµsn−1
= a1 βe
εµ−τ
n−1 .
Combining (2.3) with (2.6) and (1.2), we get
−2−2τ
k(I − Sn−1 )G(0)ks∗ −2−2τ ≤ βµn−1
kG(0)ks∗ ≤ β 2 µ−2−2τ
e.
n−1 ε
Combining (2.6), (4.11) and (4.18), we can write
−2τ
k(I − Sn−1 )Rn−1 ks∗ −2−2τ ≤ βµ−2τ
n−1 kRn−1 ks∗ −2 ≤ βµn−1
n−1
X
krj ks∗ −2
j=1
≤ βµ−2τ
ea1 Big(µs∗ −2−σ +
n−1 ε
n−1
X
∗ −2−σ
µsj−1
j=2
≤
s∗ −2−σ
εeβa1 µ−2τ
n−1 µn−1
≤ βa1 εeµ−τ
n−1 .
These inequalities imply G(wn ) → 0 in H s∗ −2−2τ (Ω) as n → ∞.
s∗ −2−2τ
(Ω) ⊂ C 2 (Ω) and wn|∂Ω = 0, we conclude that G(w) = 0 and
Since H
w ∂Ω = 0. That is u = ϕ + εw is a solution to the original Monge-Ampère equation
which is by Lemma 3.1 plurisubharmonic since g is nonnegative. If we suppose that
ρ = 0, in (A2), then the uniqueness of the solution follows immediately from [4].
5. Proof of Theorem 1.2
We shall use the result of Xu and Zuily [12, 13] that we recall briefly. Let us
consider a non linear partial differential equation
F (x, y, u, ∇u, D2 u) = 0 ,
where F is C ∞ . To any solution u we can associate the vector fields Xj =
P ∂F
k ∂ujk ∂k . Then
ρ
Theorem 5.1 ([12]). Suppose u ∈ Cloc
(Ω) with ρ > M ax(4, r+2) for some constant
r ≥ 0 and that the brackets of the Xj , up to the order r, span the tangent space at
each point of Ω, then u belongs to C ∞ (Ω).
To prove this theorem, it is sufficient to prove that the solution of Theorem 1.1
satisfies Theorem 5.1 at any point in Σ. Suppose Σ = {0}. For i = 1 . . . n;
Xi = φii
Xi+n = φii
∂
+
∂xi
n
X
φij + φij ∂
+
2
∂xj
j6=i, j=1
n
X
∂
+
∂xi+n
j6=i, j=1
n
X
j6=i, j=1
φij + φij ∂
−
2
∂xj+n
iφij − iφij ∂
,
2
∂xj+n
n
X
j6=i, j=1
iφij − iφij ∂
.
2
∂xj
(5.1)
(5.2)
For computing the Lie algebra generated by the Xi , we need the following result.
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
9
Lemma 5.2. For any integer 1 ≤ m ≤ k,
(adXn )m−1 [Xn − iX2n , Xi − iXi+n ]
=
2n
X
X
(Ciβp )∂xβ g + εdpij ∂xl
(5.3)
l=1 |β|≤m, i6=j
+ [An (ϕij )]m−1 Ai (ϕij ) (∂xmn g + i∂xm−1
∂x2n g)(∂xi + i∂xi +n ) ,
n
where Ciβp and dpij are C s∗ −m,τ (Ω) (depending on w and ϕ bounded for ε small
enough) satisfying for |β| = m, Ciβp (0) = 0, p = 1, . . . , n if n ≥ 3 and Ciβ1 (0) = 0
∂F
∂2F
if n = 2. An = ∂u
and Ai = ∂unn
∂u .
nn
ii
Proof. We use induction on the size of the brackets. First we calculate Din =
[Xn + iX2n , Xi + iXi+n ], for i ≤ n − 1.
n
n
n
n
i
hX
X
X
X
Din =
Φnj ∂xj + i
Φnj ∂xj+n ,
Φil ∂xl + i
Φil ∂xl+n
=
j=1
n
n
XX
j=1
l=1
l=1
{Φnj ∂xj (Φil ) − Φij ∂xj (Φnl )}∂xl
l=1 j=1
|
{z
}
(1)
n X
n
X
+i
{Φnj ∂xj+n (Φil ) − Φij ∂xj+n (Φnl )}∂xl
l=1 j=1
|
−
n X
n
X
{z
}
(2)
{Φnj ∂xj+n (Φil ) − Φij ∂xj+n (Φnl )}∂xl+n
l=1 j=1
|
{z
}
(2)
n X
n
X
+i
{Φnj ∂xj (Φil ) − Φij ∂xj (Φnl )}∂xl +n ,
l=1 j=1
{z
|
where
(1) =
n X
n
X
j=1 p,q=1
}
(1)
{
∂F
∂2F
∂2F
∂F
}∂xj upq .
−
∂unj ∂uil ∂upq
∂uij ∂unl ∂upq
Using (2.10), we get
F.(1) =
n X
n
X
∂F ∂F ∂F
∂F ∂F
)∂xj upq
(
−
∂unj ∂uil ∂upq
∂uiq ∂upl
j=1 p,q=1
−
=
n X
n
X
∂F ∂F ∂F
∂F ∂F
(
−
)∂xj upq
∂u
∂u
∂u
∂u
pq
nq ∂upl
ij
nl
j=1 p,q=1
n X
n
X
∂F
∂F ∂F
∂F ∂F
)
∂xj upq (
−
∂u
∂u
∂u
∂u
pq
nj
ij ∂unl
il
j=1 p,q=1
|
{z
}
(5)
10
SAOUSSEN KALLEL-JALLOULI
+
EJDE-2004/48
n
X
∂F ∂F ∂F
∂F ∂F
(
−
)∂xj upq .
∂u
∂u
∂u
∂u
nq
ij
nj ∂uiq
pl
j,p,q=1
|
{z
}
(6)
Using (2.10), we have
(5) = ∂xj (F )F
∂2F
.
∂unj ∂uil
Similarly, we prove that
F.(2) =
n
X
∂xj+n (F )F
j=1
+
∂2F
∂unj ∂uil
n
X
∂F ∂F ∂F
∂F ∂F
(
−
)∂xj+n upq .
∂unj ∂uiq
∂upl ∂uij ∂unq
j,p,q=1
|
{z
}
(7)
We can easily see that (6) + i(7) = 0, so,
(1) + i(2) =
n
X
(∂xj (F ) + i∂xj+n (F ))
j=1
and
Din =
n X
n
X
(∂xj (f ) + i∂xj+n (f ))
l=1 j=1
Since F is the determinant function, then,
l = 1, . . . , n. Therefore
Din =
2
∂ F
∂uij ∂upq
X
∂2F
∂unj ∂uil
∂2F
[∂x + i∂xl +n ].
∂unj ∂uil l
∂F
∂uij
is independent of uil and ulj for
vanishes unless i 6= p, j 6= q. So,
(∂xj (f ) + i∂xj+n (f ))
(l,j)6=(i,n), l,j≤n
∂2F
[∂x + i∂xl +n ].
∂unj ∂uil l
We have ϕij (0) = (1 − δin )σi δij ; Therefore, if n ≥ 3 and (l, s) 6= (i, n),
∂2F
(ϕ )(0) = 0.
∂uns ∂uil ij
If n = 2 and l = 1, then s = 1 and we also have
∂2F
(ϕ )(0) = 0.
∂u21 ∂u11 ij
So, (5.3) is proved for m = 1. By a recursion on m, we deduce this lemma.
On the other hand, we have by (3.5)
Φij (ϕij )(0) = 0,
for (i, j) 6= (n, n),
An (ϕij )(0) =
n−1
Y
σi > 0,
i=1
Ai (ϕij )(0) =
n−1
Y
j6=i, i=1
σi > 0.
(5.4)
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
11
Or by the hypothesis, ∂xβ g(0) = 0 for all |β| < k, and by (5.4), we can suppose
that ∂xkn g(0) 6= 0 (∂xk2n g(0) 6= 0 leads to the same result, just consider (adX2n )m−1
instead of (adXn )m−1 ).
So, by taking the real and the imaginary parts of (5.3) at the origin, we obtain
(adXn )k−1 ([Xn , Xi ] − [X2n , Xi+n ])
=
2n X
X
εd0pij (0)∂xl + [An (ϕij )(0)]k−1 Ai (ϕij )(0)[∂xkn g∂xi − ∂xk−1
∂x2n g∂xi +n ]
n
l=1 j6=i
and
(adXn )k−1 ([X2n , Xi ] + [Xn , Xi+n ])
=
2n X
X
∂x2n g∂xi + ∂xkn g∂xi +n ].
εd00pij (0)∂xl − [An (ϕij )(0)]k−1 Ai (ϕij )(0)[∂xk−1
n
l=1 j6=i
Suppose now that |w|k+2 ≤ 1. We will get at the origin for ε ≤ εe small enough
the determinant of the vectors
(adXn )k−1 ([Xn , Xi ] − [X2n , Xi+n ]),
(adXn )k−1 ([X2n , Xi ] + [Xn , Xi+n ])i=1,...,n−1 ,
(5.5)
Xn , X2n is different from zero.
Now, choose s∗ so big that s∗ ≥ max(7 + 2n, 6 + k + n) by means of Theorem
1.1 there exists ε0 < εe2 such that for any g satisfying (1.2) there exists a unique
1
solution u = ϕ+ε02 w ∈ C k+3 (Ω) to the problem (1.1). Moreover; by (2.1), |w|k+2 ≤
βkwkk+2+n+τ . Since σ = s∗ − 2 − τ , s∗ ≥ 6 + k + n and τ ≤ α4 < 14 , then
k + 2 + n + τ ≤ s∗ − 4 + τ = σ − 2 + 2τ ≤ σ − τ.
We have then, using (4.3), (4.5) and (4.15),
√
|w|k+2 ≤ 2β εe ≤ 1.
So, by (5.5), we can conclude that for εe sufficiently small, the vector fields at the
origin; [(adXn )k−1 ([Xδn , Xi ])]δ=1,2;i=1,...,2n−1 , Xn and X2n span all the tangent
space. Theorem 1.2 follows then from Theorem 5.1.
6. Appendix 1
To prove proposition 4.1, we need the following result.
Propositon 6.1. There exists a constant K0 ≥ 1 such that for any function wi ∈
C s∗ +2,τ (Ω), |wi |2 ≤ 1, i = 1, 2, 3 and for any ε ≤ 1 we have
|G(w1 ) − G(w2 )|0 ≤ K0 |w1 − w2 |2 kϕk2+n∗ + kw1 k2+n∗ + kw2 k2+n∗ + 1 . (6.1)
Also for t ∈ [0, 1], s ∈ [0, s∗ ],
d
[LG (w1 + tw2 )w3 ]ks
dt
≤ εK0 [(kϕk2+s + εkw1 k2+s + εkw2 k2+s + 1)|w2 |2 |w3 |2
k
+ (kϕk2+n∗ + εkw1 k2+n∗ + εkw2 k2+n∗ + 1)(|w2 |2 kw3 k2+s + |w3 |2 kw2 k2+s )].
(6.2)
12
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Proof. Just write
G(w1 ) − G(w2 )
1
1
2
= [det(ϕij + εwij
) − det(ϕij + εwij
) + g(w1 ) − g(w2 )]
ε
Z 1 X
n
∂F
2
1
2
1
2
=
(ϕij + εwij
+ tε(wij
− wij
))(wij
− wij
)dt
∂u
0 i,j=1
ij
Z 1
∂g
+
(ϕ + εw2 + tε(w1 − w2 ))(w1 − w2 )
∂u
0
Z 1
∂g
+
(ϕ + εw2 + tε(w1 − w2 ))(wi1 − wi2 ),
0 ∂pi
and
d
[LG (w1 + tw2 )w3 ]
dt
n
d X ∂F
∂g
1
2
3
= [
(ϕij + εwij
+ tεwij
)wij
+
(ϕ + εw1 + tεw2 )w3 + . . . ]
dt i,j=1 ∂uij
∂u
=ε
n
X
∂2F
1
3
2
2
wij
(ϕ + εwij
+ tεwij
)wpq
+ ....
∂uij ∂upq ij
i,j,p,q=1
Combining (2.1), (2.7), (2.8) and (2.9) with the inequalities
2
1
2
|ϕij + εwij
+ tε(wij
− wij
)|0 ≤ |ϕ|2 + 2|w2 |2 + |w1 |2 ≤ 3 + |ϕ|2
and
1
2
|ϕij + εwij
+ tεwij
|0 ≤ |ϕ|2 + ε|w1 |2 + tε|w2 |2 ≤ 2 + |ϕ|2 ,
we deduce (6.1) and (6.2).
Proof of the proposition 4.1. The proposition is proved by induction. We have u0 =
0. Let begin by proving (4.19)0 to (4.21)0 . (i.e. (4.19) to (4.21) corresponding to
j = 0).
(a) (4.19)0 : Using (3.2) and (4.10), we have
1
g0 = −S0 G(0) and G(0) = (det(ϕij ) − g(ϕ)).
εe
But ϕ ∈ C s∗ +2,α (Ω), g ∈ C s∗ and Sn are smoothing operators, so g0 ∈ H s∗ (Ω).
(2.3), (2.4), (3.2) and (1.2) show that
β
β2
k det(ϕij ) − g(ϕ)ks∗ ≤
| det(ϕij ) − g(ϕ)|s∗ ≤ β 2 εe.
εe
εe
Using (4.4) and β ≤ µ, we get kg0 ks ≤ µ2 εe ≤ a2 εe
(b) (4.20)0 : (3.2), (4.4), (4.5) and (1.2) give
√
1
θ0 = |G(0)|0 ≤ | det(ϕij ) − g(ϕ)|s∗ ≤ εe ≤ εea3 ≤ 1.
εe
(c) (4.21)0 : We have
kg0 ks ≤ βkG(0)ks ≤
A0 (2) = max(1, |
∂g
∂F
(ϕlq )|2 + θ0 ).
(ϕ)|2 , max|
i,j
∂u
∂ϕij
Then, by (2.9), (4.1) and (4.20)0 , A0 (2) ≤ M0 .
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
13
Assume that u0 , u1 , . . . , un−1 ∈ H s∗ (Ω) satisfy (3.12)–(3.14) and (4.14)–(4.21)
for j ≤ n − 1. We shall construct un ∈ H s∗ (Ω) satisfying (3.12)–(3.14) and prove
that (4.14)–(4.21) are satisfied for j = n.
Combining (4.16)n−1 –(4.21)n−1 , we have |w
en−1 |4,κ ≤ 1, θn−1 ≤ 1, An−1 (2) ≤
M0 and gn−1 ∈ H s∗ (Ω). We can then apply Theorem 3.3 to get a solution un ∈
H s∗ (Ω) to the problem (4.7)n satisfying (3.12)–(3.14). Then:
(a) (4.14)n : For n = 1, using (1.2), (2.3), (3.2), (3.12), and (4.2), we have
ku1 k0 ≤ Dkg0 k0 ≤ DβkG(0)k0 ≤ D
(4.3), (4.5), and s∗ ≥ σ give
ku1 k0 ≤
√
β2
| det(ϕij ) − g(ϕ)|s∗ ≤ Dβ 2 εe.
εe
εeµ−σ .
(6.3)
By (3.13), we have ku1 k1 ≤ D(kg0 k1 + ku1 k0 ). Therefore, using (1.2), (2.3), (6.3),
and s∗ ≥ σ, we get
√
√
ku1 k1 ≤ D(β 2 εe + εeµ−σ ) ≤ εeµ1−σ .
Suppose that for 0 ≤ l ≤ s and s ≥ 2 we have
√
ku1 kl ≤ εeµl−σ .
(6.4)
Using (3.14), we have, for s ≥ 2,
ku1 ks ≤ D kg0 ks +
X
(1 + |ϕ|i+4,τ )ku1 kl .
l≤s−1, (i,l)∈Λs
(1.2), (2.3), (2.4), (4.3), (4.10), and s∗ ≥ σ imply
kg0 ks ≤ βkG(0)ks ≤ β 2 |G(0)|s ≤ β 2 εe ≤ µ
eεeµs−σ ,
which by (6.3) and (6.4) gives
ku1 ks ≤ D µ
eεeµs−σ +
X
√
(1 + |ϕ|i+4,τ ) εeµl−σ
l≤s−1, (i,l)∈Λs
√
+ s2∗ (1 + |ϕ|i+4,τ )µ−1 εeµs−σ ,
√
which by (4.3) and (4.5) shows that ku1 ks ≤ εeµs−σ .
For n ≥ 2, (3.12), (4.2), (4.5), and (4.19)n−1 imply
√ −σ
kun k0 ≤ Dkgn−1 k0 ≤ De
εa2 µ−σ
εeµn−1 .
n−1 ≤
s−σ
≤D µ
eεeµ
In the same way; (3.13), (4.2), (4.5), (4.19)n−1 and (6.5) give
√
kun k1 ≤ εeµ1−σ
n−1 .
√
Suppose that, for 0 ≤ l < s and s ≥ 2, kun kl ≤ εeµl−σ
n−1 . By (3.14), we have
X
kun ks ≤ D(kgn−1 ks +
(1 + |ϕ + εew
en−1 |i+4,τ )kun kl ).
l≤s−1, (i,l)∈Λs
But, (2.1), (2.5), (4.15)n−1 , and 4 + n∗ ≤ σ − τ imply that, for 0 ≤ i ≤ s − 2,
√
|w
en−1 |i+4,τ ≤ βkw
en−1 k4+n∗ +i ≤ β 2 µin−1 kw
en−1 k4+n∗ ≤ 2β 2 εeµin−1 .
Therefore, using (4.19)n−1 , we get
X
√
√
kun ks ≤ D εea2 µs−σ
(1 + |ϕ|s∗ +2,τ + 2β 2 εeµin−1 ) εeµl−σ
n−1 +
n−1
√ s−1−σ 2 2
2
≤ D εea2 µs−σ
eµs−σ
eµn−1
,
n−1 + 2β s∗ ε
n−1 + (1 + |ϕ|s∗ +2,τ )s∗ ε
(6.5)
14
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
√
which combined with (4.4) and (4.5) gives kun ks ≤ εeµs−σ
n−1 .
Pn
(b) (4.15)n : (4.6) shows that wn = j=1 uj . By (4.14)j , 1 ≤ j ≤ n, we have
kwn ks ≤
n
X
kuj ks ≤
√
εeµs−σ +
j=1
n √
X
εeµs−σ
j−1 ≤
√
εeµs−σ +
j=2
1/τ
For s ≤ σ − τ , since µ ≥ 2
kwn ks ≤
≥ 2, we have
n−1
X√
n−1
X√
εeµs−σ
.
j
j=1
µs−σ
j
εeµs−σ
≤
j
j=0
≤
µ−τ
j
≤
1
2j
and
X 1
√ n−1
√
εe
≤ 2 εe.
j
2
j=0
For s ≥ σ − τ , we have
kwn ks ≤
√
εeµs−σ +
√ µn(s−σ) − µs−σ
εe
.
µs−σ − 1
√
Since µ ≥ 21/τ , it follows that µs−σ ≥ µτ ≥ 2. Therefore, kwn ks ≤ εeµs−σ
n .
(c) (4.16)n : Combining (2.1), (2.4), (4.5), (4.15)n and 4 + n∗ ≤ σ − τ , we obtain
√
|w
en |4,τ ≤ βkw
en k4+n∗ ≤ β 2 kwn k4+n∗ ≤ 2β 2 εe ≤ 1.
(d) (4.17)n ): In the case s ≤ σ − τ , using (2.6) and (4.15)n , we obtain
√
√
]−1
]+1−σ
kwn − w
en ks ≤ βµns−[σ+τ ]−1 kwn k[σ+τ ]+1 ≤ βµs−[σ+τ
εeµ[σ+τ
≤ β εeµs−σ
n
n
n .
In the case s > σ − τ , (2.6) (4.15)n ) and β ≥ 1 give
√
kwn − w
en ks ≤ βkwn ks ≤ β εeµs−σ
n .
(e) (4.18)n : By (4.12), we have
rn = [LG (wn−1 ) − LG (w
en−1 )]un − θn−1 4un + Qn
|
{z
} | {z } |{z}
(1)
(2)
(3)
When n = 1, (1) = 0. In the case n ≥ 2, since
Z 1
d
(1) =
[LG (w
en−1 + t(wn−1 − w
en−1 ))un ]dt,
dt
0
by (2.1) and (4.17)n−1 , we get
√
∗ −σ
|wn−1 − w
en−1 |2 ≤ βkwn−1 − w
en−1 k2+n∗ ≤ 2β 2 εeµ3+n
.
n−1
√
But 2β 2 εe ≤ 1 and 3 + n∗ ≤ 4 + 2n∗ ≤ σ, so, |wn−1 − w
en−1 |2 ≤ 1. In the same
way, (2.1), (4.5) and (4.14)n give
√
∗ −σ
|un |2 ≤ βkun k2+n∗ ≤ β εeµ3+n
≤ 1.
n−1
By (4.16)n−1 , we also have |w
en−1 |2 ≤ 1. Hence, we can apply Proposition 6.1 to
get
k(1)ks ≤e
εK0 {[kϕks+2 + kw
en−1 ks+2 + kwn−1 ks+2 + 1]|wn−1 − w
en−1 |2 |un |2
+ (kϕk2+n∗ + kw
en−1 k2+n∗ + kwn−1 k2+n∗ + 1)
× (|wn−1 − w
en−1 |2 kun ks+2 + kwn−1 − w
en−1 ks+2 |un |2 )}.
Using (2.3) and (4.3), we get for 0 ≤ s ≤ s∗ ,
kϕks+2 ≤ β|ϕ|s∗ +2 ≤ βµ ≤ µ2 .
By (2.2), it suffices to prove (4.18)n for s = 0 and s = s∗ − 2.
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
Case s = 0: combining (2.1), (4.14)n , (4.15)n−1 and (4.17)n−1 , we have
√
√
∗ −2σ
k(1)k0 ≤ εeK0 {(µ2 + 2β εe + 2 εe + 1)2β 3 εeµ4+2n
n−1
√
√
∗ −2σ
+ (µ2 + 2β εe + 2 εe + 1)4β 3 εeµ4+n
},
n−1
which using (4.5) and σ ≥ 4 + 2n∗ ≥ 4 + n∗ gives k(1)k0 ≤ εeK0 µ−σ
n−1 .
Case s = s∗ − 2: (4.5) and s∗ ≥ σ + τ , as in the previous case, imply
∗ −2−σ
k(1)ks∗ −2 ≤ εeK0 µsn−1
.
By (2.2), we obtain for 0 ≤ s ≤ s∗ − 2,
k(1)ks ≤ βe
εK0 µs−σ
n−1 .
Next,
k(2)ks ≤ θn−1 kun ks+2 .
If n = 1 combining (4.5), (4.9) and (4.14)n , we obtain
√
k(2)ks ≤ |G(0)|0 ku1 ks+2 ≤ εe εeµs+2−σ ≤ εeµs−σ .
In the case n ≥ 2: (4.14)n and (4.20)n−1 imply
s+2−σ
k(2)ks ≤ a3 εeµ−2
= a3 εeµs−σ
n−1 µn−1
n−1 .
Finally, since by (4.13),
(3) = Qn = G(wn−1 + un ) − G(wn−1 ) − LG (wn−1 )un
Z 1 Z t
d
=
(
[LG (wn−1 + hun )un ]dh)dt .
0
0 dh
Then, using (2.1), (4.5) and (4.15)n−1 , we obtain
√
|wn−1 |2 ≤ βkwn−1 k2+n∗ ≤ 2β εe ≤ 1.
Since we proved that |un |2 ≤ 1, we can apply proposition 6.1 to have
k(3)ks ≤ εeK0 [(kϕks+2 + kun ks+2 + kwn−1 ks+2 + 1)|un |22
+ 2|un |2 kun ks+2 (kϕk2+n∗ + kun k2+n∗ + kwn−1 k2+n∗ + 1)].
Combining (2.1), (4.14)n and (4.15)n−1 , we get
For s = 0:
k(3)k0
√
√
≤ εeK0 {(µ2 + εe[max(µ, µn−1 )]2−σ + 2 εe + 1)β 2 εe[max(µ, µn−1 )]4+2n∗ −2σ
√
√
+ 8(µ2 + εeβ[max(µ, µn−1 )]2+n∗ −σ + 2 εe + 1)e
εβ[max(µ, µn−1 )]4+n∗ −2σ },
which combined with (4.5) and σ ≥ 4 + 2n∗ gives
k(3)k0 ≤ εeK0 [max(µ, µn−1 )]−σ .
For s = s∗ − 2; since σ ≥ 4 + 2n∗ , we also get
k(3)ks∗ −2 ≤ εeK0 [max(µ, µn−1 )]s∗ −2−σ .
Then (2.2) shows that, for 0 ≤ s ≤ s∗ − 2,
k(3)ks ≤ βe
εK0 [max(µ, µn−1 )]s−σ ,
and we conclude that
krn ks ≤ (2βK0 + a3 )e
ε[max(µ, µn−1 )]s−σ
15
16
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
≤ 9K0 µ5 εe[max(µ, µn−1 )]s−σ
= a1 εe[max(µ, µn−1 )]s−σ .
(f) (4.19)n ): By (4.10) and (4.11),
gn = Sn−1 Rn−1 − Sn Rn + (Sn−1 − Sn )G(0)
= (Sn−1 Rn−1 − Sn Rn−1 ) − Sn rn + (Sn−1 − Sn )G(0).
|
{z
} | {z } |
{z
}
(5)
(4)
(6)
Case s = 0: (2.6), (4.11) and (4.18)j , j ≤ n − 1, imply
k(4)k0 ≤ k(I − Sn−1 )Rn−1 k0 + k(I − Sn )Rn−1 k0
2−s∗
∗
≤ βkRn−1 ks∗ −2 µ2−s
kRn−1 ks∗ −2
n−1 + βµn
∗
∗
≤ (βa1 εeµ2−s
eµ2−s
)(µs∗ −2−σ +
n
n−1 + βa1 ε
n−1
X
∗ −2−σ
µsj−1
).
j=2
Since s∗ − 2 > σ and β ≤ µ, then
2−s∗
∗
∗ −2−σ
k(4)k0 ≤ βa1 εe(µ2−s
)µsn−1
≤ 2a1 µ2 εeµ−σ
n .
n−1 + µn
On the other hand, combining (2.4), (4.18)n , σ < s∗ − 2 and β ≤ µ, we obtain
k(5)k0 ≤ βkrn k0 ≤ βa1 εe[max(µ, µn−1 )]−σ ≤ a1 µ2 εeµ−σ
n .
We also have by (1.2), (2.3), (2.6). and σ < s∗ − 2,
k(6)k0 ≤ k(I − Sn−1 )G(0)k0 + k(I − Sn )G(0)k0
−σ
≤ βµ−σ
n−1 kG(0)kσ + βµn kG(0)kσ
2 −σ
≤ β 2 µ−σ
n−1 |G(0)|s∗ + β µn |G(0)|s∗
σ
s∗
≤ β 2 εeµ−σ
eµ−σ
n (µ + 1) ≤ 2µ ε
n .
We finally get
kgn k0 ≤ (2 + 3a1 )µs∗ εeµ−σ
n .
Case s = s∗ : (2.5), (4.11), (4.18)j , 1 ≤ j ≤ n, and σ < s∗ − 2 show that
k(4) + (5)ks∗
≤ kSn−1 Rn−1 ks∗ + kSn Rn ks∗
≤ βµ2n−1 kRn−1 ks∗ −2 + βµ2n kRn ks∗ −2
≤ βµ2n−1 a1 εe(µs∗ −2−σ +
n−1
X
∗ −2−σ
µsj−1
) + βµ2n a1 εe(µs∗ −2−σ +
j=2
≤
≤
∗ −2−σ
βa1 εe(µ2n−1 µsn−1
+ µ2n µsn∗ −2−σ )
2βa1 εeµsn∗ −σ ≤ 2µa1 εeµsn∗ −σ .
Next, by (1.2), (2.5), (2.3), and β ≤ µ, we have
k(6)ks∗ ≤ kSn G(0)ks∗ + kSn−1 G(0)ks∗
∗ −σ
≤ βµsn∗ −σ kG(0)kσ + βµsn−1
kG(0)kσ
≤ 2β 2 εeµns∗ −σ ≤ 2µ2 εeµsn∗ −σ .
Therefore,
kgn ks∗ ≤ 2µ(a1 + µ)e
εµsn∗ −σ .
n
X
j=2
∗ −2−σ
µsj−1
)
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
17
We can finally conclude using (4.4) and µ ≤ a1 , that
kgn ks∗ ≤ 4a1 µ2 εeµsn∗ −σ ≤ a2 εeµsn∗ −σ .
(g) (4.20)n : By (4.9), we have
θn = |G(w
en )|0 ≤ |G(wn ) − G(w
en )|0 + |G(wn )|0
Using (4.22):
G(wn ) = (I − Sn−1 )Rn−1 + (I − Sn−1 )G(0) + rn .
Then
θn ≤ |G(wn ) − G(w
en )|0 + |(I − Sn−1 )Rn−1 |0 + |(I − Sn−1 )G(0)|0 + |rn |0 .
{z
} |
{z
} |
{z
} |{z}
|
(7)
(8)
(9)
(10)
Since we proved that |wn |2 ≤ 1 and |w
en |2 ≤ 1, we can apply Proposition 6.1 to get
(7) ≤ βK0 kwn − w
en k2+n∗ (kϕk2+n∗ + kwn k2+n∗ + kw
en k2+n∗ + 1).
Equations (2.4), (4.15)n , (4.17)n , and 3 + n∗ ≤ 4 + 2n∗ − τ ≤ σ − τ imply
√
√
√
∗ −σ
(7) ≤ 2β 2 K0 εeµ2+n
(µ2 + 2 εe + 2β εe + 1).
n
Since εe ≤
1
(6β 2 )2 ,
β ≤ µ and 4 + n∗ − σ ≤ 4 + 2n∗ − σ ≤ 0 then
√
(7) ≤ 4µ5 K0 εeµ−2
n .
√
In √
the case n = 1, (8) = 0. For n ≥ 2, since β ≤ µ, n∗ − σ ≤ −2 and µ4 a1 εe ≤
a2 εe ≤ 1, combining (2.1), (2.6), (4.11), and (4.18)j , j ≤ n − 1, we obtain
(8) ≤ βk(I − Sn−1 )Rn−1 kn∗
∗ −s∗ +2
≤ β 2 µnn−1
a1 εe µs∗ −2−σ +
≤β
2
∗ −σ
a1 εeµsn−1
≤
√
n−1
X
∗ −2−σ
µsj−1
j=2
εeµ−2
n .
Equations (1.2), (2.1), (2.3), (2.6), (4.5), and β ≤ µ imply
∗ −s∗
(9) ≤ βk(I − Sn−1 )G(0)kn∗ ≤ β 2 µnn−1
kG(0)ks∗
√
−2
3 −2
3 2
−2
≤ β µn−1 εe ≤ β µ εeµn ≤ εeµn .
Finally, by (2.1) and (4.18)n ,
(10) ≤ βkrn kn∗ ≤ βa1 εe[max(µ, µn−1 )]n∗ −σ
≤ µa1 εe[max(µ, µn−1 )]−2 ≤
√
εeµ−2
n .
Thus, we conclude that
√
√
θn ≤ 7K0 µ5 εeµ−2
eµ−2
n = a3 ε
n ≤ 1.
(h) (4.21): We have
∂g
∂F
An (2) ≤ max 1, | (ϕ + εew
(ϕkl + εe(w
en )kl )|2 + θn .
en )|2 , max |
1≤i,j≤n ∂uij
∂u
Using (2.9), (4.1), (4.16)n and (4.20)n , we get An (2) ≤ M0 .
18
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
7. Appendix 2
In the rest of this paper, we prove estimates (3.12)–(3.14) for Lν . We shall need
the following result.
Propositon 7.1. The operator
P =
n
X
∂F
(uzi zj )∂zi ∂zj ,
∂u
zi zj
i,j=1
where u ∈ C 3 (Ω), is formally self-adjoint.
Proof. Let Σ = {z ∈ Ω/F (uzi zj )(z) = 0}. Since
n
X
n
n
X
X
∂F
∂F
P =
∂zi
∂zj −
∂zi
(uij ) ∂zj ,
∂u
∂u
zi zj
zi zj
i=1
j=1
i,j=1
it is sufficient to prove that for j = 1, . . . , n and z ∈ Ω,
Aj (z) =
=
n
X
∂zi
i=1
n
X
∂F
(uzi zj )(z)
∂uzi zj
∂2F
(uzi zj )uzi zp zq (z) = 0.
∂uzi zj ∂uzp zq
i,p,q=1
Using the relation (2.10), we get Aj (z) = 0 for any z ∈
/ Σ. The continuity of the
determinant function allow as to have the conclusion when z ∈ Σ.
7.1. Estimates in the elliptic Zone of L. Let Q =
2n
P
bij Dxi Dxj + b be a
i,j=1
ji
degenerate elliptic operator with real coefficients b, bij = b ∈ C s∗ ,τ (Ω). Assume
that there is a continuous function λ(x) ≥ 0 defined in Ω such that
2n
X
bij ξi ξj ≥ λ(x)|ξ|2 .
i,j=1
Let S be a subset of Ω satisfying {x ∈ Ω : λ(x) = 0} ⊂ S.
Lemma 7.2. Assume that Q is uniformly elliptic in Ω; that is λ(x) ≥ λ0 , λ0 is
a positive constant Then for any integer 1 ≤ s ≤ s∗ there exists a constant Cs0
depending only on s, λ0 and A(0) such that for any real function u ∈ C s∗ ,τ (Ω) ∩
H01 (Ω),
kuk1 ≤ C10 (kQuk0 + A(2)kuk0 ),
X
kuks ≤ Cs0 (kQuks−1 +
A(i + 2)kukj ), s ≥ 2.
(7.1)
(7.2)
i≤s−2, i+j≤s−1
It is not difficult to prove (7.1). In fact, we need only to apply well-known standard techniques to the linear elliptic operator Q and to calculate several constants
precisely. By induction with respect to s and patient calculation, (7.2) follows from
(7.1).
For δ > 0, we define the set Sδ by
Sδ = {x ∈ Ω, d(x, S) < δ}.
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
19
Lemma 7.3. Assume that S is a compact C ∞ submanifold of Ω and Ω\S is connected. Then there exists a function µ ∈ L∞ (Ω) and a constant C > 0 such that
µ = 0 on S, mδ = inf Ω\Sδ µ > 0 for any sufficiently small δ and
Z
1
2
(7.3)
µu2 dx ≤ C kQuk0 kuk0 + sup[bij
ij − 2b]kuk0 ,
2
Ω
for u ∈ C s∗ ,τ (Ω) ∩ H01 (Ω).
Proof. Standard techniques of elliptic operators give
Z
1
2
λ|Du|2 dx ≤ C kQuk0 kuk0 + sup[bij
ij − 2b]kuk0 .
2
R
R
Hence, it suffices to show that µu2 dx ≤ λ|Du|2 dx. First, let us fix a point
p ∈ Ω\S arbitrarily.
By virtue of the fundamental theorem of ordinary differential equations, we can
construct a family of curves c(t, x) ∈ C ∞ ([0, Tp ]×Up ) such that c(0, x) = x, c(t, x) ∈
/
S for 0 < t < Tp when x ∈ Ω\S, c(Tp , x) ∈
/ Ω, |ċ(t, x)| ≡ 1, supx∈Up τx < ∞, and
c(t, .) is a local C ∞ diffeomorphism defined in Up for any fixed t.
Here, Tp is a positive constant, Up is a sufficiently small open neighborhood of
p, and, τx = inf{t ≥ 0 : c(t, x) ∈
/ Ω } We define a function µp (x) by
µp (x) = inf{λ(c(t, x)) : 0 ≤ t ≤ τx }.
For u ∈ C 1 (Ω) satisfying u∂Ω = 0, since
Z τx
u(x) = u(c(0, x)) − u(c(τx , x)) = −
Du(c(t, x)).ċ(t, x)dt,
0
we have
2
Z
|u(x)| ≤ C
τx
|Du(c(t, x))|2 dt.
0
Multiplying this inequality by µp and using its definition, we obtain
Z τx
µp (x)|u(x)|2 ≤ C
λ(c(t, x))|Du(c(t, x))|2 dt,
0
which implies
Z
Up
µp |u|2 ≤ C
Z
λ|Du|2 dt.
Ω
Secondly, we note that the above argument ensures the existence of a finite number
of points p1 , . . . , pN such that Ω\S ⊂ ∪N
i=1 Upi and
Z
Z
2
µpi |u| ≤ C
λ|Du|2 dt.
Upi
Ω
Therefore, we have only to define µ by
(
min{µpi (x) : x ∈ Upi , 1 ≤ i ≤ n}, if x ∈ Ω\S,
µ(x) =
0,
if x ∈ S.
20
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Lemma 7.4. For u ∈ C01 (Ω),
X
k[∂k , Q]uk20 ≤ C(A(2)kQuk1 kuk1 + A(2)2 kuk21 ),
(7.4)
k
X
k[∂k , Q]uk2s ≤ C(A(2)kQuks+1 kuks+1 +
k
X
A(i + 2)2 kuk2j ) s ≥ 1. (7.5)
(i,j)∈Λs+1
Proof. [11, Lemma 1.7.1] shows that
2
ij
(bij
k uij ) ≤ CA(2)b uli ulj ,
which implies
X
k[∂k , Q]uk20 ≤ C
k
XZ
2
2
{(bij
k uij ) + (bk u) }
k
≤ CA(2)
XZ
bij uli ulj + CA(1)2 kuk21 .
k
Integrating by parts
Z
1
− 2b)ul , ul i,
bij uli ulj = −h(Qu)l , ul i + h[∂l , Q]u, ul i + h(bij
2 ij
which implies
Z
X
k[∂k , Q]uk0 kuk1 + A(2)kuk21 .
bij uli ulj ≤ C kQuk1 kuk1 +
k
From these inequalities, and using the inequality αβ ≤ εα2 + 1ε β 2 it follows that
X
k[∂k , Q]uk20 ≤ C(A(2)kQuks+1 kuks+1 + A(2)2 kuk21 ).
k
For s ≥ 1, (6.5) is proved by recursion on s using (6.4).
∞
Lemma 7.5. Let χ ∈ C satisfy supp ∇χ ⊂ Ω. For any integer 0 ≤ s ≤ s∗ , there
exists a constant Cs > 0 such that for all u ∈ C s∗ ,τ (Ω),
X
k[χ, Q]uk2s ≤ Cs A(2)kQuks kuks +
A(i + 2)2 kuk2j .
(7.6)
(i,j)∈Λs
Proof. Let us consider a cut-off function χ
e ∈ C0∞ (Ω) satisfying 0 ≤ χ
e ≤ 1 and
e = ebij Dx Dx + eb by Q
e=χ
χ
e = 1 on ∪i supp ∂i χ, and define an operator Q
eQ. Since
i
j
e
e
e
[χ, Q]u = [χ, Q]u and kQuks ≤ CkQuks , it will suffice to prove (7.6) for Q.
For s = 0: The corollary to Lemma 1.7.1 in [11] shows that
X
ebij uj 2 ≤ 2A(0)ebij ui uj .
i,j
which gives
e 20 ≤ CA(0)
k[χ, Q]uk
Z
ebij ui uj + CA(0)2 kuk2 ,
0
Integrating by parts we have
Z
ebij ui uj = −hQu,
e 0 kuk0 + CA(2)kuk20 ,
e ui + 1 h(ebij − 2eb)u, ui ≤ kQuk
2 ij
which implies (7.6)0 .
Note that (7.6)s≥1 follows from (7.6)0 by induction with respect to s
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
21
7.2. Estimates near the degenerate points of L. For t ≥ t0 ≥ 1, we define
1
Vt (0) = {x ∈ Ω, |xn | < } ∩ B(0, δ1 ).
t
Propositon 7.6. For any integer 0 ≤ s ≤ s∗ and any function u ∈ C0s∗ ,τ (Vt (0)),
there exists a constant Cs00 = Cs00 (n, Ω, ϕ, δ1 ) > 0 such that
00
kuk0 ≤ C0 t−1 kLν uk0 ,
X
kuks ≤ Cs00 t−1 (kLν uks +
A(i + 2)kukj ),
(7.7)
s ≥ 1,
(7.8)
(i,j)∈Λs
where δ1 is as in Lemma 3.1.
Proof. Let v = (T − etxn )−1 u, and T > 5e a constant. A direct computation gives
Qu = (T − etxn )Qv − tetxn {2bnj vj + tbnn v},
Z
(T − etxn )−1 Qu.v = −I + II − III − IV,
where
Z
I=
III = t
2
Z
bij vi vj ,
etxn bnn (T − etxn )−1 v 2 ,
1
2
Z
2
{bij
ij − 2b}v ,
Z
IV = 2t etxn (T − etxn )−1 vbnj vj .
II =
Using the Cauchy-Schwartz inequality, we get
Z
Z
|IV | ≤ bij vi vj + 4t2 e2txn (T − etxn )−2 bnn v 2 .
Since
etxn (T − etxn )−1 − 4e2txn (T − etxn )−2 = etxn (T − etxn )−2 (T − 5etxn ),
it follows that
Z
Z
2
2txn
txn −4
txn nn 2
t
e
(T − e ) (T − 5e )b u ≤ − (T − etxn )−2 Qu.u − II.
Also
e−1 ≤ etxn ≤ e,
(T − e−1 )−1 ≤ (T − etxn )−1 ≤ (T − e)−1 ;
therefore,
1
C0 t2 inf (bnn )kuk20 ≤ C kQuk0 kuk0 + sup [bij
− 2b]kuk20 .
2 Vt (0) ij
Vt (0)
(7.9)
To prove (7.7), we apply (7.8). So, for u ∈ C0s∗ ,τ (Vt (0)), we can write
C
2
t tC0 inf (bnn ) −
sup |bij
ij − 2b| kuk0 ≤ CkQuk0 kuk0 ,
2 Vt (0)
Vt (0)
with Q = Lν and bnn = (Φnn + 4(θ + ν)). If |w|3,τ ≤ 1, |x| ≤ δ0 and ε ≤ ε1 , we
have
n−1
Y
nn
Φ ≥
σi − M δ1 − M ε1 = α > 0.
i=1
4(C+1)A(2)
, 1),
max(
αC0
(7.7) is proved. To prove (7.8), we use (7.7)
Taking t ≥ t0 =
and recursion on s. We now estimate kχuks .
22
SAOUSSEN KALLEL-JALLOULI
EJDE-2004/48
Propositon 7.7. For any cut-off function χ ∈ C0∞ (Vt (0)), u ∈ C s∗ ,τ (Ω) ∩ H01 (Ω)
and 1 ≤ s ≤ s∗ ,
X
(|ϕ + εw|i+4,τ + 1)kukj ). (7.10)
kχuks ≤ 2Cs00 (kLν uks + k[χ, Lν ]uks +
j<s, (i,j)∈Λs
Proof. Let us consider a cut-off function χ ∈ C0∞ (Vt (0)). For u ∈ C s∗ ,τ ∩ H01 (Ω),
since supp χ ⊂ Vt (0), we have by (7.9) for any 1 ≤ s ≤ s∗ ,
X
kχuks ≤ Cs00 t−1 kχLν uks + k[χ, Lν ]uks +
A(i + 2)kukj
j<s, (i,j)∈Λs
+
Cs00 t−1 A(2)kχuks .
We have A(2) ≤ M0 . We fix t ≥ t0 such that for 1 ≤ s ≤ s∗ , Cs00 t−1 A(2) ≤ 21 . On
the other hand,
∂g
∂F
A(i + 2) = max 1, | (ϕ + εw)|i+2 , max |
(ϕkl + εwkl )|i+2 + θ .
1≤p,q≤n ∂upq
∂u
But, for k ∈ {0, 1, 2}, |∂ k ϕ + ε∂ k w|0 ≤ |ϕ|2 + 1, then by (2.9), since θ ≤ 1, we get,
for 0 ≤ i ≤ s∗ − 2,
A(i + 2) ≤ C(ϕ) |ϕ + εw|i+4,τ + 1 .
(7.11)
and we deduce (7.10).
7.3. Proof of the estimates (3.12)–(3.14) for Lν . . Since kuks ≤ k(1 − χ)uks +
kχuks , it will suffice to estimate k(1 − χ)uks and kχuks .
Proof of (3.12). Since χ = 1 in a neighborhood of zero in V , then, there exists
δ > 0 such that Supp(1 − χ) ⊂ Ω\B(0, δ).
e
e
Let us consider the cut-off functions: χ
e, χ
e ∈ C0∞ (Ω\S), 0 ≤ χ
e, χ
e ≤ 1 and such
e
that χ
e = 1 on supp ∂i χ and χ
e = 1 on supp χ
e. Let µ be the function given by
Lemma 7.3 (mδ depends only on ϕ, Ω, n).
By (7.3), there exists C0 = C0 (ϕ, Ω, n) > 0 such that
Z
Z
1
2
2
µu2 dx ≤ C0 (kuk0 kLν uk0 + Bkuk20 ),
u dx ≤
k(1 − χ)uk0 =
mδ
Ω\B(0,δ)
P ij
where B = 12 sup[bij
ij − 2b]. By proposition 7.1,
ij bij = 0, and the hypothesis
(A2) imply that −2b ≤ %. So, B ≤ % and we have
k(1 − χ)uk20 ≤ C1 (kuk0 kLν uk0 + %kuk20 ).
e
Since Supp χ
e ⊂ Ω\{0}, we also have by the same way,
e
kχ
euk20 ≤ C1 (kuk0 kLν uk0 + %kuk20 ).
On the other hand, by (7.8),
kχuk20 ≤ C2 kLν χuk20 ≤ C2 (kLν uk20 + k[χ, Lν ]uk20 ),
e
e
but χ
eLν χ
eu = χ
eLν u and [χ, Lν ]u = [χ, χ
eLν ]χ
eu. Since A(2) ≤ M0 and ν ≤ 1, using
Lemma 7.5, we get
e
e
e
e
k[χ, Lν ]uk20 = k[χ, χ
eLν ]χ
euk20 ≤ C ke
χLν χ
euk0 kχ
euk0 + (M0 + 1)2 kχ
euk20
e
e
≤ C 0 kLν uk0 kχ
euk0 + kχ
euk20 .
EJDE-2004/48
COMPLEX MONGE-AMPÈRE EQUATION
23
Combining these inequalities with the fact that % << 1, and using the inequality
αβ ≤ εα2 + 1ε β 2 , we get (3.12)
Proof of (3.13). We have supp(1 − χ) ⊂ Ω\B(0, δ). Or ϕ is strictly plurisubharmonic on E = supp(1 − χ), then for ε ≤ ε4 small enough, L is uniformly elliptic on
E. Using (7.1) and the estimation A(2) ≤ M0 , we have
k(1 − χ)uk1 ≤ C10 (kLν uk0 + (M0 + 1)kuk0 + k[χ, Lν ]uk0 ).
Applying Lemma 7.5, we get
k[χ, Lν ]uk0 ≤ C0 (kLν uk0 + (M0 + 1)kuk0 ),
therefore,
k(1 − χ)uk1 ≤ C1 (M0 )(kLν uk0 + kuk0 ).
On the other hand, since A(2) ≤ M0 , we get using (7.10),
kχuk1 ≤ C1 (M0 )(kLν uk1 + k[χ, Lν ]uk1 + kuk0 ) .
e
e
But χ
eLν χ
eu = χ
eLν u and [χ, Lν ]u = [χ, χ
eLν ]χ
eu, so, since A(2) ≤ M0 , Lemma 7.5
gives
e
e
k[χ, Lν ]uk1 ≤ C1 (ke
χLν χ
euk1 + (1 + M0 )kχ
euk1 )
e
≤ C1 (kLν uk1 + (1 + M0 )kχ
euk1 ).
e
Since Lν is uniformly elliptic on supp χ
e and A(2) ≤ M0 , then we have by (7.1),
e
e
kχ
euk1 ≤ C10 (kLν uk0 + (M0 + 1)kuk0 + k[χ
e, Lν ]uk0 ),
which using (7.6) gives
e
kχ
euk1 ≤ C1 (M0 )(kLν uk1 + kuk0 ).
Combining these inequalities, we get (3.13).
The proof of (3.14) is identical to that of (3.13) using the inequalities (7.1), (7.2),
(7.6), and (7.10).
Acknowledgments. The author wishes to thank the anonymous referee for his or
her helpful comments.
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SAOUSSEN KALLEL-JALLOULI
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[13] C. J. Xu, C. Zuily: Smoothness up to the boundary for solutions of non linear and non
elliptic Dirichlet problem, Trans. of the Amer. Math. Soc. Vol. 308, (1), 243-257, 1988.
Saoussen Kallel-Jallouli
Faculté des Sciences, Campus Universitaire, 1060 Tunis, Tunisie
E-mail address: Saoussen.Kallel@fst.rnu.tn