The Dirichlet Problem for Degenerate Elliptic Darboux Equation

The Dirichlet Problem for Degenerate EllipticDarboux Equation

Saoussen Kallel-Jallouli*

Faculte des Sciences, Campus Universitaire, Tunis, Tunisie

ABSTRACT

We study the Dirichlet problem

det@2u

@xi@xjþ aij

� �¼ K xð Þf x; u;Huð Þ in O � Rn

u j @O ¼ j

8><>:where K vanishes on a compact set of O. We prove under some hypothesis on

K; f , j and if jdetðjij þ aijÞ � KfðjÞjCs� is sufficiently small, that we can find aunique sufficiently smooth solution u to the Dirichlet problem such that

ðuij þ aijÞ � 0:

Key Words: Darboux equations; Degenerate elliptic.

Mathematics Subject Classification: 35J70; 35Mxx.

*Correspondence: Saoussen Kallel-Jallouli, Faculte des Sciences, Campus Universitaire, Tunis1060, Tunisie; Fax: (216)-71-885-350; E-mail: [email protected].

COMMUNICATIONS IN PARTIAL DIFFERENTIAL EQUATIONS

Vol. 29, Nos. 7 & 8, pp. 1097–1125, 2004

1097

DOI: 10.1081/PDE-200033756 0360-5302 (Print); 1532-4133 (Online)

Copyright # 2004 by Marcel Dekker, Inc. www.dekker.com

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I. INTRODUCTION

Let O be a bounded domain in Rn with smooth boundary. We shall consider theDarboux equation

det@2u

@xi@xjþ aij

� �¼ K xð Þf x; u;Huð Þ in O � Rn

f j @O ¼ j:

8><>:where aij ¼ aji are sufficiently smooth real functions satisfying

ðiÞ @iakp ¼ @ka

ip; 8i; k;pðiiÞ ðaijÞ � 0 in O:

ðI:2Þ

We assume that j is a real function defined on O and

gðx;fÞ ¼ KðxÞfðx;f;HfÞ ¼ KðxÞ½f1ðx;fÞ þ f2ðHfÞ�:Let S be a compact C1 submanifold of O such that OnS is connected. We need

the following assumptions:

ðiÞ K � 0 in O

ðiiÞ K�1ð0Þ ¼ S

(ðI:3Þ

ðiÞ fðx; u;pÞ > 0 in O�R�Rn

ðiiÞ f2 is concave:

ðiiiÞ 2K@f1@u

�Xk

@K

@xk

@f2@pk

in O�R�Rn

8>>><>>>: ðI:4Þ

ðiÞ ðjij þ aijÞ jOnS is strictly positive

ðiiÞ ðjij þ aijÞ jS is of rankðn� 1Þ

ðiiiÞ The eigenvalues of ðjij þ aijÞ on S are distincts:

8>>><>>>: ðI:5Þ

Our main result is the following

Theorem A. For any integer s� � 7þ n; a 2 �0; 1½, G > 1, aij 2 Cs�;aðOÞ satisfying(I.2) and j 2 Cs�þ2;aðOÞ satisfying the condition (I.5), there exists a constante0 > 0 such that for any g ¼ Kf ; Cs� of its arguments satisfying (I.3) (I.4), if

jdetðjij þ aijÞ � gðjÞjCs� � e0 ðI:6Þ

and

jHðu;pÞgjCs�þa � G:

1098 Kallel-Jallouli

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Then the problem (I.1) admits a unique solution f 2 C½s��3�n2�ðOÞ satisfying

ðfij þ aijÞ � 0 in O: ðI:7Þ

In the case when aij ¼ 0, f � 1, related results were found by Amano (1988) forthe dimension n ¼ 2, then generalized, in the special case when S is an isolated pointin O, by Atallah (2000) for higher dimension.

Equations of this type appear in many geometrical (see Pogorelov, 1978) andphysical problems (see Oliker and Waltman).

Their strong nonlinearity makes their study rather difficult, and this difficulty isgreater when the equation imposes the vanishing of the Monge–Ampere determi-nant, since then the equation is degenerate.

Among the constructive existence proofs in analysis, the Implicit FunctionTheorem is well known; it furnishes the construction of the solutions of manynonlinear problems in the theory of ordinary and partial differential equations (seeAlinhac and Gerard, 1991).

In our case this method fails, however, because of a phenomenon which may bedescribed as ‘‘loss of derivatives’’. This loss will be circumvented by an appropriatesmoothing process and the use of a new type of Implicit Function Theorem as a basicmethod in the proof of our Theorem.

In the first step of the proof, we shall introduce the basic norms and state somesimple calculus lemmas which play important roles in the proof of convergence ofour iteration scheme.

Section 3 is devoted to the study of a priori estimates, for modified linearizedoperators, in suitable functional spaces.

In fact, although we have a strong estimate in an elliptic region (Lemma III.6), itis quite hard to show certain a priori estimates (III.15–III.17) in a neighborhood ofzero points of K since the linearized operators are degenerating. The compactness offK ¼ 0g, Lemma III.7 and the hypothesis on g and j will play an important role inthe proof of Theorem III.4.

Finally, in Sec. 4, one must use an iteration algorithm of Nash–Moser type(see Moser,1961) (a new type of the inverse function theorem) and takes into accountthe fact that one does not solve the linear problem exactly. We construct a sequence ofgood approximating solutions that converges, du to the key estimates (III.15–III.17)established previously and the hypothesis on g and j, to a solution to our problem.

II. SOME TECHNICAL LEMMAS

We use three kinds of norms

j jk ¼ k kCkðOÞ; k kk ¼ k kHkðOÞ and j jk;t ¼ k k

Ck;tðOÞ

where k 2 N and t 2 �0; a½.Let us denote n� ¼ n

2 þ t, and recall some technical lemmas.

Dirichlet Problem for Degenerate Elliptic Darboux Equation 1099

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Lemma II.1 (see Alinhac and Gerard, 1991). Let s� be an integer �7þ n. We canfind a constant b � 2 such that for any 0 � i; j; k � s� þ 2; we have

juji;t � bkukiþn� ðII:1Þ

kukj � bkukk�j

k�i

i kukj�i

k�i

k ; i < j < k ðII:2Þ

kuks� � bjujs� ðII:3Þ

For any l � 1, we can find a family of smoothing operators Sl: HiðOÞ ! HjðOÞ

satisfying

kSluki � bkukj; if i � j ðII:4Þ

kSluki � bli�jkukj; if i � j ðII:5Þ

kSlu� uki � bli�jkukj; if i � j ðII:6Þ

((II.1) and (II.2) are Sobolev and Gagliardo–Nirenberg interpolation inequal-ities, respectively.)

Lemma II.2 (see Alinhac and Gerard, 1991; Hormander, 1984).

(1) If u; v 2 L1 \ Ht (t > 0) then uv 21 \Ht and

kuvkt � K1ðjuj0kvkt þ kuktjvj0Þ ðII:7Þ

K1 is a constant � 1 independent of u, v.(2) Let H : Rm ! C be a function C1 of its arguments.

– If o 2 ðL1 \ HsÞm (s > 0) and joj0 � M, then

kHðoÞks � K2ðs;H;MÞðkoks þ 1Þ ðII:8Þ

K2 is a constant �1 independent of o.– If o 2 ðCi;mÞm, m 2 �0; 1½, i 2 N then HðoÞ 2 Ci;m and if we suppose that

joj0 � M, then we can find a constant K3 ¼ K3ðm; i;H;MÞ � 1 such that

jHðoÞji;m � K3ðjoji;m þ 1Þ: ðII:9Þ

Following Zuily (1988, Lemma 1.6), we get

Lemma II.3. We denote by F the determinant function.

FðuijÞ ¼ detðuij þ aijÞ


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For 1 � i; j; a; b � n, we have

F@2F

@uij@upq¼ @F

@upq

@F

@uij� @F

@uip

@F

@uqjðII:10Þ

Proposition II.4. Under the hypothesis (I.2) the operator P ¼Pni;j¼1

@F@uij

uij� �

@i@jwhere u 2 C3ðOÞ is a formally self-adjoint operator.

Proof. Let us set G ¼ fx 2 O=FðuijÞðxÞ ¼ 0g: Since

P ¼Xni¼1

@i

Xnj¼1

@F

@uijðuijÞ@j

!�Xni;j¼1

@i

�@F

@uij

�ðuijÞ@j

it is sufficient to prove that for j ¼ 1; . . . ; n and x 2 O

AjðxÞ ¼Xni¼1

@i

�@F

@uij

�ðuijÞðxÞ

¼Xn

i;p;q¼1

@2F

@uij@upqðuijÞ½uipqðxÞ þ @ia

pqðxÞ� ¼ 0

Using the identity (II.10), and the hypothesis (I.2) we get AjðxÞ ¼ 0 for anyx 62 G. The continuity of the determinant function allow as to conclude either whenx 2 G.

III. A PRIORI ESTIMATES FOR LINEAR OPERATORS

Following Atallah (2000), by means of the change of unknown f ¼ jþ ew wecan reduce (I.1) to the following equation

detðfijÞ ¼ detðjij þ aij þ ewijÞ ¼ g ðIII:1Þ

We shall consider

GðwÞ ¼ 1

e½detF� g� ðIII:2Þ

The linearized operator of G at w is given by

LGðwÞ ¼Xni;j¼1

fij@xi@xj þXni¼1

bi@xi þ b ðIII:3Þ

where eFF ¼ ðfijÞ is the matrix of cofactors of F ¼ ðfijðx; e;wÞÞ, bi ¼ � @g@pi

andb ¼ � @g

@u.


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Now, for any smooth real valued function w, the matrix ðfijÞ is symmetric andwe can find a unitary matrix Tðx; eÞ satisfying

Tðx; eÞðfijÞtTðx; eÞ ¼ diagðl1; . . . ; lnÞ ðIII:4Þ

Let us consider a point x0 2 S that can be supposed to be the origin. By means ofchange of variables, we may assume using (I.5) that

jijð0Þ þ aijð0Þ ¼ sidji ; i; j ¼ 1; . . . ; n ðIII:5Þ

where dji is the Kronecker symbol. The constants si are chosen such that

si > 0 for i ¼ 1; . . . ; n� 1; sn ¼ 0 and si 6¼ sj for i 6¼ j ðIII:6ÞLet 0 < t � a

4 : We have

Lemma III.1. We can find three constants e1ðx0Þ > 0, d0ðx0Þ > 0 and M > 0depending only on j; aij; g, n, O such that when

Vx0 ¼ fðx; e;wÞ=jx� x0j � d0; 0 � e � e1; w 2 C3;tðOÞ; jwj3;t � 1g;

we have

(i) The eigenvalues li, i ¼ 1; . . . ; n of F are distinct on Vx0 and of class C1 in�VVx0 . Moreover, li > 0 in Vx0 , for i ¼ 1; . . . ; n� 1:

(ii) For ðx; e;wÞ 2 Vx0

Xni¼1

jsi � liðx; e;wÞj þ��Fnnðx; e;wÞ �

Yn�1

i¼1

si

�� Mðeþ jxjÞ ðIII:7Þ

(iii) For ðx; e;wÞ 2 Vx0 and i ¼ 1; . . . ; n� 1

li � inf1�i�n�1

si �Md0 � ðM þ 1Þe1 > 0 and

Fnn �Yn�1

i¼1

si �Md0 �Me1 > 0 ðIII:8Þ

Proof. Let us consider the function

Hðx; e;w; lÞ ¼ detðjij þ aij þ ewij � ldji Þ

H is C1 of its arguments, and by (III.5) (III.6) we have for i 2 f1; . . . ; ng

Hð0; 0; 0; siÞ ¼ 0 and@H

@lð0; 0; 0; siÞ 6¼ 0


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By means of the implicit mapping theorem, there exists three constants e1 > 0,d0 > 0 and M > 0 such that we have (i). Moreover by (III.5), we have

@F

@unnðjijÞð0Þ ¼ Fnnð0; 0;wÞ ¼

Yn�1

i¼1

si > 0

which gives (ii) and (iii).Moreover, we have

Lemma III.2. One can find a positive constant e2 such that for 0 < e < e2,w 2 C3;tðOÞ a real valued function satisfying jwj3;t � 1 and y ¼ max

x2O jGðwÞj,the operator

L ¼ �LGðwÞ � y4 ðIII:9Þis a degenerate elliptic operator

�here ¼ 4 ¼Pn

i¼1@2

@xi2

�:

Proof. We have to prove

A ¼ yjxj2 þXni;j¼1

fijxixj � 0; 8ðz; xÞ 2 O�Rn ðIII:10Þ

(�) If x 2 OnS, by (I.5.i), we can deduce that A > 0 for all x 2 Rnnf0g.(�) If x 2 S, we can suppose x ¼ x0 ¼ 0:

For x 2 Rn, let us set x ¼ tTðt; eÞexx. We have

A ¼ yjxj2 þt xeFFx ¼ yjxj2 þt exxT eFFtTexxbut FeFF ¼ detF Id so by (III.4),

detF Id ¼ TFtTT eFFtT ¼ diagðliÞT eFFtT

T eFFtT ¼ detF diag�

1

li

�¼Yni¼1

li diag�

1

li

�¼ ðeGþ gÞdiag

�1

li

�Then,

A ¼ yjexxj2 þ detF Xni¼1

exx2ili

¼ yjexxj2 þXn�1

i¼1

detFjexxij2li

þYn�1

i¼1

liexx2n¼�yþ

Yn�1

i¼1

li

�exx2n þXn�1

i¼1

eGþ g þ ylili

exx2i


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by (III.8), for i ¼ 1; . . . ; n� 1, e � e1ðx0Þ, jwj3;t � 1 eGþ yli � yðsi �Md0�ðM þ 1Þe1Þ � 0 and A � 0.

Since S is compact, then S � SLi¼0 Bðxi; diÞ, where diðxiÞ is constructed in the

same way us d0ðx0Þ given by Lemma III.1. It suffices then to takee2 ¼ min0�i�L e1ðxiÞ.

We shall also need the result

Proposition III.3. There exists a constant K0 � 1 such that for any functionwi 2 Cs�þ2;tðOÞ, jwij2 � 1, i ¼ 1; 2; 3 and any e � 1 we have

jGðw1Þ �Gðw2Þj0 � K0jw1 �w2j2�kjk2þn� þ kw1k2þn� þ kw2k2þn� þ 1

�ðIII:11Þ

and for t 2 ½0; 1�, s 2 ½0; s�� d

dt½LGðw1 þ tw2Þw3�

��s� eK0½ðkjk2þs þ ekw1k2þs þ ekw2k2þs þ 1Þjw2j2jw3j2þ ðkjk2þn� þ ekw1k2þn� þ ekw2k2þn� þ 1Þ� ðjw2j2kw3k2þs þ jw3j2kw2k2þsÞ� ðIII:12Þ

Proof. Just write

Gðw1Þ �Gðw2Þ

¼ 1

e½detðjij þ ew1

ij þ aijÞ � detðjij þ ew2ij þ aijÞ þ gðw1Þ � gðw2Þ�

¼Z 1

0

Xni;j¼1

@F

@uijðjij þ ew2

ij þ teðw1ij � w2

ijÞÞðw1ij � w2

ijÞdt

þ þZ 1

0

@g

@uðjþ ew2 þ teðw1 � w2ÞÞðw1 � w2Þ d

dt½LGðw1 þ tw2Þw3�

¼ d

dt

"Xni;j¼1

@F

@uijðjij þ ew1

ij þ tew2ijÞw3

ij þ@g

@uðjþ ew1 þ tew2Þw3 þ

#

¼ eXn

i;j;p;q¼1

@2F

@uij@upqðjij þ ew1

ij þ tew2ijÞw2

pqw3ij þ

then combining (II.1) (II.7) (II.8) and (II.9) with the inequalities

jjij þ ew2ij þ teðw1

ij � w2ijÞj0 � jjj2 þ 2jw2j2 þ jw1j2 � 3þ jjj2

and

jjij þ ew1ij þ tew2

ijj0 � jjj2 þ ejw1j2 þ tejw2j2 � 2þ jjj2;

we can deduce (III.11) (III.12).


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Now, we will study a boundary value problem for the degenerate ellipticoperator

L ¼ �LGðwÞ � y4 ¼Xni;j¼1

bijDxiDxj þXni¼1

bi@xi þ b

bij ¼ @F

@uij

ðjij þ ewijÞ þ ydji ¼ Fij þ ydji

bi ¼ K@f

@pi

b ¼ K@f

@u

For the sake of simplicity, we put

AðkÞ ¼ max�1; max

1�i;j�njbijjk; max

1�i�njbijk; jbjk

�Ls ¼ fði; jÞ; 0 � i; j � s; iþ j � s and iþ 2 � maxðs; 2Þg:

8<: ðIII:13Þ

for k; s 2 NThe main result of this section is the following

Theorem III.4. Suppose that y � 1 and for some constant M0 > 0, we haveAð2Þ � M0: One can find e3 > 0 such that for any e 2 �0; e3� any w 2 Cs�þ2;tðOÞ a realvalued function satisfying the inequality jwj3;t � 1 and any real valued functionh 2 Hs� , the problem

Lu ¼ h in O

u j @O ¼ 0:

ðIII:14Þ

possess a unique solution u 2 Hs� : Moreover for 0 � s � s�, the inequalities

kuk0 � C0khk0 ðIII:15Þ

kuk1 � C1ðkhk1 þ kuk0Þ ðIII:16Þ

kuks � Cs

(khks þ

Xði;jÞ2Ls

j�s�1

ð1þ jjþ ewjiþ4;tÞkukj); s � 2 ðIII:17Þ

for some constant Cs ¼ Csðj; aij; g; s;O;M0; e3Þ, independent of w and e holds.

In order to solve the Dirichlet problem (III.14), we need the following. Forn 2 �0; 1½; we denote Ln ¼ L� n4.


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Proposition III.5. Suppose that y � 1 and for some constant M0 > 0, we haveAð2Þ � M0: One can find e3 > 0 such that for any e 2 �0; e3� any real valued functionw 2 Cs�þ2;tðOÞ satisfying the inequality jwj3;t � 1 and any real valued functionh 2 Hs� ðOÞ the regularized problem

Lnu ¼ h in O

u j @O ¼ 0:

ðIII:140Þ

admits a unique solution u 2 Hs�þ1ðOÞ.

Proof. First of all, we recall that LGðwÞ is a real second order operator with realcoefficients. From the Lemma III:2, Ln is uniformly elliptic with coefficients inCs�;tðOÞ then we conclude using Gilbarg and Trudinger (1983) (Theorems 6.14 and8.13).

Now, suppose (III.15) to (III.17) are valid for the regularized problem (III.140)with an uniform constantCs for n 2 �0; 1�, then by letting n go to zero we shall get a solu-tion u 2 Hs� ðOÞ to the original problem which of course will satisfy (III.15) to (III.17).

In the rest of this paragraph, we shall prove the estimates (III.15) to (III.17) forthe modified linearized operators Ln .

Unless otherwise specified, Q denotes a degenerate elliptic operator of the form

Q ¼Xni;j¼1

bijDiDj þXni¼1

bi@i þ b

with Cs�;t real coefficients b, bi, bij ¼ bji defined in O.Assume that there is a continuous function lðxÞ � 0 defined in O such that

Xni;j¼1

bijxixj � lðxÞ

III.1. Estimates in the Elliptic Zone of L

Lemma III.6. Assume that Q is uniformly elliptic in O, that is

Xni;j¼1

bjiðxÞxixj � l0jxj2; l0 is a constant >0

Then for any integer 1 � s � s�, there exists a constant C 0s depending only on

s; l0 and Að0Þ such that for any real function u 2 Cs�;tðOÞ \ H10ðOÞ

kuk1 � C 01ðkQuk0 þ Að2Þkuk0Þ ðIII:18Þ

kuks � C 0s

kQuks�1 þ

Xiþj�s�1i�s�2

Aðiþ 2Þkukj!; s � 2 ðIII:19Þ


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It is not difficult to prove (III.18). In fact, we need only to apply well-knownstandard techniques to the linear elliptic operator Q and to calculate several con-stants precisely. By induction with respect to s and patient calculation, (III.19)follows from (III.18).

Let S be a subset of O satisfying

fx 2 O; lðxÞ ¼ 0g � S

and for d > 0; we define a set Sd by

Sd ¼ fx 2 O;dðx; SÞ < dg

Lemma III.7. Assume that S is a compact C1 submanifold of O and OnS is con-nected. Then there exists a function m 2 L1ðOÞ and a constant C > 0 such thatm ¼ 0 on S, md ¼ infOnSd m > 0 for any sufficiently small d andZ

Omu2 dx � C

kQuk0kuk0 þ

1

2sup½bijij þ bii � 2b�kuk20

ðIII:20Þ

for u 2 Cs�;tðOÞ \ H10ðOÞ.

Proof. Stand techniques of elliptic operators giveZljDuj2dx � C

kQuk0kuk0 þ

1

2sup½bijij þ bii � 2b�kuk20

Hence, using the inequalityZmu2 dx �

ZljDuj2 dx

established by (see Amano, 1988), we get (III.20)

Lemma III.8. We have for u 2 C10ðOÞX

k

k½@k;Q�uk20 � CðAð2ÞkQuk1kuk1 þ Að2Þ2kuk21Þ ðIII:21ÞXk

k½@k;Q�uk2s � C

�Að2ÞkQuksþ11kuksþ1 þ

Xði;jÞ2Lsþ1

Aðiþ 2Þ2kuk2j�; s � 1

ðIII:22Þ

Proof. ð�Þ Lemma 1.7.1 of Oleınik and Radkevic shows that

ðbijk uijÞ2 � CAð2Þbijuliulj


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which implies

Xk

k½@k;Q�uk20 � CXk

Z nðbijk uijÞ2 þ ðbkuÞ2

o� CAð2Þ

Xk

Zbijuliulj þ CAð1Þ2kuk20

By integrating by parts, we haveZbijuliulj ¼ �hðQuÞl; uli þ h½@l;Q�u; uli þ 1

2hðbijij � 2bÞul; uli

which meansZbijuliulj � C

�kQuk1kuk1 þ

Xk

k½@k;Q�uk0kuk1 þ Að2Þkuk21�

From these inequalities, it follows that

Xk

k½@k;Q�uk0 � CðAð2ÞkQuk1kuk1 þ Að2Þ2kuk21Þ

(�) For s � 1, (III.22) is proved by recursion on s using (III.21).

Lemma III.9. Let w 2 C1ðOÞ satisfying supp Hw � O. For any integer 0 � s � s�,there exists a constant Cs > 0 such that 8u 2 Cs�;tðOÞ

k½w;Q�uk2s � Cs

�Að2ÞkQukskuks þ

Xði;jÞ2Ls

Aðiþ 2Þ2kuk2j�

ðIII:23Þ

Proof. Let us consider a cut-off function eww 2 C10 ðOÞ satisfying 0 � eww � 1 and eww ¼ 1

onS

i supp @iw, and define an operator eQQ ¼ ~bbijDxiDxj þ ~bb by eQQ ¼ ewwQ. Since

½w; eQQ�u ¼ ½w;Q�u and keQQuks � CkQuks, it will suffice to prove ðIII:23Þ for eQQ.

(�) For s ¼ 0, Corollary of Lemma 1.7.1 of Oleınik and Radkevic showsthat

�Xj

~bbijuj

�2

� 2Að0Þ~bbijuiuj

which gives

k½w; eQQ�uk20 � CAð0ÞZ

~bbijuiuj þ CAð0Þ2kuk20


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by integrating by parts, we haveZ~bbijuiuj ¼ heQQu; ui þ 1

2

D�~bbijij � 2~bb

�u; uE

� keQQuk0kuk0 þ CAð2Þkuk20

which implies (III.23)0.

(�) For s � 1, we prove (III.23)s using (III.23)0 and an induction withrespect to s.

III.2. Estimates Near the Degenerate Points of L

For t � 1, we define

Ut ¼(x 2 O; jxnj < 1

t

)

Lemma III.10. We can find two constants C0 > 0 and C � 0 independent of Q andt such that; 8u 2 C2ðUtÞ, u j @Ut

¼ 0

C0t2 inf

Ut

ðbnnÞkuk20 þ t infUt

ðbnÞkuk20 � C

(kQuk0kuk0 þ

1

2supUt

½bijij þ bii � 2b�kuk20)

ðIII:24ÞProof. For a real-valued function u 2 C2ðUtÞ satisfying u j @Ut

¼ 0, we put

v ¼ ðT � etxnÞ�1u; T = constant >5e

Direct computation gives

Qu ¼ ðT � etxnÞQv� tetxnf2bnjvj þ bnvþ tbnnvgZðT � etxnÞ�1

Qu:v ¼ �Iþ II� III� IV� V

with

I ¼Z

bijvivj

II ¼ 1

2

Zfbijij � bii þ 2bgv2

III ¼ t2ZetxnbnnðT � etxnÞ�1

v2

IV ¼ t

ZetxnbnðT � etxnÞ�1

v2


ORDER REPRINTS

V ¼ 2t

ZetxnðT � etxnÞ�1

vbnjvj

Using the Cauchy–Schwartz inequality, we get

jVj �Zbijvivj þ 4t2

Ze2txnðT � etxnÞ�2

bnnv2

Since

etxnðT � etxnÞ�1 � 4e2txnðT � etxnÞ�2 ¼ etxnðT � etxnÞ�2ðT � 5etxnÞThen

t2Ze2txnðT � etxnÞ�4ðT � 5etxnÞbnnu2 � �

ZðT � etxnÞ�2

Qu u� II� IV

but, e�1 � etxn � e, ðT � e�1Þ�1 � ðT � etxnÞ�1 � ðT � eÞ�1. So, we obtain (III.24).Let x be any point of S and t � tx � 1. The main result of this section is

Proposition III.11. We can find a neighborhood VtðxÞ of x, such that: For anyinteger s 2 ½0; s�� and any function u 2 C

s�;t0 ðVtðxÞÞ, there exists a constant

C00s ¼ C00

s ðn;O;j; dxÞ > 0 such that

kuk0 � C000t

�1kLnuk0 ðIII:25Þ

kuks � C00s t

�1

�kLnuks þ

Xði;jÞ2Ls

Aðiþ 2Þkukj�; s � 1 ðIII:26Þ

(where dx is given in the same way us d0ðx0Þ of Lemma III.1).

Proof. We can suppose x ¼ x0 ¼ 0. Let us set Vtðx0Þ ¼ Ut \ Bðx0; d0Þ.ð�Þ To get (III.25), we apply Lemma III.10, so we can write for

u 2 Cs�;t0 ðVtðx0ÞÞ,

t tC0 infUðtÞ

ðbnnÞ � C

2supUðtÞ

jbijij þ bii þ 2bj( )

kuk20 � CkLnuk0kuk0

but

bnn ¼ ðFnn þ 4ðyþ nÞÞ

If jwj3;t � 1, jxj � d0 and e � e1ðx0Þ, we have

Fnn �Yn�1

i¼1

si �Md0 �Me1 ¼ a > 0


ORDER REPRINTS

We have only to take

t � t0 ¼ 1þ 2CAð2ÞaC0

and (III.25) is proved.

ð�Þ To prove (III.26), we use (III.25) and a recursion on s.

III.3. Proof of the Estimates (III.15) to (III.17) for Ln

Proposition III.12. For any point x 2 S, we can find a neighborhood V ðxÞ such that:for any cut-off function w 2 C1

0 ðV ðxÞÞ, u 2 Cs�;tðOÞ \H10ðOÞ and 1 � s � s�

kwuks � 2C00s

�kLnuks þ k½w;Ln �uks þ

Xði;jÞ2Ls

j<s

ðjjþ ewjiþ4;t þ 1Þkukj�

ðIII:27Þ

Proof. It suffices to consider the point x0. We shall take the neighborhood Vtðx0Þconstructed in the proof of Proposition III.11.

Let us consider a cut-off function w 2 C10 ðVtðx0ÞÞ,

For u 2 Cs�;t \H10 ðOÞ. Since supp w � Vtðx0Þ, we have by (III.26) for any

1 � s � s�

kwuks � C00s t

�1

�kwLnuks þ k½w;Ln �uks þ

Xði;jÞ2Ls

j<s

Aðiþ 2Þkukj�

þ C00s t

�1Að2Þkwukswe have Að2Þ � M0. We fix t � t0 such that for 1 � s � s�

C00s t

�1Að2Þ � 1

2

In the other hand,

Aðiþ 2Þ

¼ max1�p;q�n

�1;

�� @g@u ðjþ ewÞ��iþ2

;

�� @g@pp

ðjþ ewÞ��iþ2

;

�� @F@upqðjkl þ ewklÞ

��iþ2

þ y�;

but for k 2 f0; 1; 2g, j@kjþ e@kwj0 � jjj2 þ 1, then by (II.9), since y � 1, we get for0 � i � s� � 2;

Aðiþ 2Þ � CðjÞðjjþ ewjiþ4;t þ 1Þ ðIII:28Þ

and we deduce (III.27).


ORDER REPRINTS

Proof of Theorem III.4. Since S is compact, then

S �[

0�i�N

VðxiÞ ¼ V

(where VðxiÞ is defined in Proposition III.12). Later we shall denotee1 ¼ inf0�i�N e1ðxiÞ, where e1ðxiÞ is constructed in the same way as in Lemma III.1.

Using partition of unity, there exists a family of regular functions wi such that,for each i 2 f1; . . . ;Ng

supp wi � VðxiÞ; 0 � wi � 1 andXi

wi ¼ w � 1 in a neighborhood of S:

Since kuks � kð1� wÞuks þP

i kwiuks; it will suffice to estimate kð1� wÞuks andkwiuks, respectively.

Proof of III.15. Since w ¼ 1 in a neighborhood of S in V , then, Suppð1� wÞ � OnS.Let us consider the following cut-off functions: eww, eewweww 2 C1

0 ðOnSÞ, 0 � eww, eewweww;� 1 andsuch that; eww ¼ 1 on

Si;j supp @iwj and

eewweww ¼ 1 on supp eww. We consider now the functionm given by Lemma III.7 (md depends only on j;O; n)

By III.20, there exists C0 ¼ C0ðj;O; nÞ > 0 such that

kð1� wÞuk20 ¼ZOnS

u2dx � 1

md

Zmu2dx � C0ðkuk0kLuk0 þ Bkuk20Þ

where B ¼ 12 sup½bijij þ bii � 2b�:

By Proposition II.4,P

ij bijij ¼ 0, and the hypothesis (I.4) imply that

Xi

bii � 2b ¼Xi

@K

@xi

@f2@pk

þ K

Xij

@2f2

@pi@pj

ðjij þ ewijÞ � 2@f1@u

� eXij

�� @2f2

@pi@pj

��so B � en2kf2kC2 ¼ r. Hence, we have

kð1� wÞuk20 � C0ðkuk0kLuk0 þ rkuk20ÞSince Suppeewweww � OnS, we also have by the same way,

keewwewwuk20 � C0ðkuk0kLuk0 þ rkuk20ÞOn the other hand by (III.25) and fixing t � max0�i�N txi , we have

kwiuk20 � C2kLnwiuk20 � C2ðkLnuk20 þ k½wi;Ln �uk20Þ

but ewwLneewwewwu ¼ ewwLnu and ½wi;Ln �u ¼ ½wi;ewwLn �eewwewwu, so since Að2Þ � M0 and n � 1 we get


ORDER REPRINTS

using Lemma III.8,

k½wi;Ln �uk20 ¼ k½wi;ewwLn �eewwewwuk20 � C½kewwLnw2uk20 þ ðM0 þ 1Þ2keewwewwuk20�� C0ðkLnuk20 þ keewwewwuk20Þ

Combining these inequalities we get (III.15), for e small enough.

Proof of ðIII.16Þ. Since Suppð1� wÞ � OnS. then by (I.5.i), for e � e4 smallenough, L is uniformly elliptic on E: Using (III.18) and the estimation Að2Þ � M0,we have

kð1� wÞuk1 � C01ðkLnuk0 þ ðM0 þ 1Þkuk0 þ k½w;Ln �uk0Þ

Applying Lemma III.8, we get

k½w;Ln �uk0 � C0ðkLnuk0 þ ðM0 þ 1Þkuk0Þ

so,

kð1� wÞuk1 � C1ðM0ÞðkLnuk0 þ kuk0Þ

On the other hand, since Að2Þ � M0, we get using (III.27)

kwiuk1 � C1ðM0ÞðkLnuk1 þ k½wi;Ln �uk1 þ kuk0ÞBut ewwLn

eewwewwu ¼ ewwLnu and ½wi;Ln �u ¼ ½wi;ewwLn �eewwewwu, so since Að2Þ � M0, Lemma III.8gives

k½wi;Ln �uk1 � C1ðkewwLneewwewwuk1 þ ð1þM0Þkeewwewwuk1Þ

� C1ðkLnuk1 þ ð1þM0Þkeewwewwuk1ÞSince L is uniformly elliptic on Suppeewweww and Að2Þ � M0, then we have by (III.18)

keewwewwuk1 � C01ðkLnuk0 þ ðM0 þ 1Þkuk0 þ k½eewweww;Ln �uk0Þ

that gives using (III.23)

keewwewwuk1 � C1ðM0ÞðkLnuk1 þ kuk0Þ

The proof of (III.17) is identical to that of (III.16) using the inequalities (III.18)(III.19) (III.23) and (III.27).

IV. PROOF OF THEOREM A

Now we shall construct a sequence which converges to a solution of ourproblem.


ORDER REPRINTS

We begin by fixing the constants. We choose first

M0 ¼ 1þmaxH2F

K3ð2; t;H; ð1þ jjj2ÞÞð1þ jjj4;tÞ ðIV:1Þ

where

F ¼

@F

@uij;@g

@u;@g

@pi

�1 � i; j � n

and K3 is the constant introduced in (II.9) (i.e., jHðuÞjj;m � K3ðj; m;H;MÞjujj;m).Then

D ¼ max�max0�s�s�

Cs; 1�

ðIV:2Þ

Cs being the constant (depending only on s;j; g; aij;O;M0) given by theTheorem III.4

m ¼ maxðb; 3Ds2�ð1þ jjjs�þ2;tÞ; 21tÞ and ~mm ¼ b2ms� ðIV:3Þ

a1 ¼ 9K0m5

a2 ¼ 5a1ms�þ1

a3 ¼ 7K0m5:

8><>: ðIV:4Þ

where K0 is the constant determined by Proposition III.3.Finally, we fix e0 satisfying

ffiffiffiffie0

p � minh

inf1�i�4

ð1; eiÞ; ð3D2a2 þ 6~mmD2Þ�2i

ðIV:5Þ

where ei are given in the proof of Theorem III.4, Lemma III.2, Theorem III.4 and theproof of (III.16).

After the constants being fixed, let g be Cs� of its arguments and satisfy

jdetðjij þ aijÞ � gðjÞjs� � e0

Let denote by Sn ¼ Smn the family of operators given by Lemma II.1, whenmn ¼ mn (m is given by IV.3).

We will construct wn, n ¼ 0; 1; . . . ; by induction on n as follows; starting with u0,w0 ¼ 0, suppose w0, w1; . . . ;wn have been chosen and define wnþ1 as follows

wnþ1 ¼ wn þ unþ1 ðIV:6Þ

where unþ1 is the solution of the Dirichlet problem

LGð~wwnÞunþ1 þ yn4unþ1 ¼ gn; in Ounþ1 j @O ¼ 0 ðIV:7Þ


ORDER REPRINTS

satisfying

~wwn ¼ Snwn ðIV:8Þyn ¼ jGð~wwnÞj0 ðIV:9Þg0 ¼ �S0Gð0Þ; gn ¼ Sn�1Rn�1 � SnRn þ Sn�1Gð0Þ � SnGð0Þ ðIV:10Þ

R0 ¼ 0; Rn ¼Xnj¼1

rj ðIV:11Þ

r0 ¼ 0; rj ¼ ½LGðwj�1Þ � LGð~wwj�1Þ�uj þQj � yj�14uj; 1 � j � n ðIV:12ÞQj ¼ GðwjÞ �Gðwj�1Þ � LGðwj�1Þuj; 1 � j � n ðIV:13ÞThe wk are well defined. We give first the following result.Let us denote ~ee ¼ ffiffiffiffi

e0p

:

Proposition IV.1. Under assumptions

s� � 7þ n

4þ nþ 2t � s < s� � 2

then we have for any s 2 N,

kujks �ffiffi~ee

p½maxðm; mj�1Þ�s�s; j 2 N�; 0 � s � s� ðIV:14Þ

kwjks �2ffiffi~ee

p; for s � s� tffiffi

~eep

ms�sj ; for s� t � s � s�; j 2 N�

(ðIV:15Þ

j~wwjj4;t � 1; j 2 N� ðIV:16Þ

kwj � ~wwjks � 2bffiffi~ee

pms�sj ; 0 � s � s�; j 2 N� ðIV:17Þ

krjks � ~eea1½maxðm; mj�1Þ�s�s; 0 � s � s� � 2; j 2 N� ðIV:18Þkgjks � ~eea2ms�s

j ; 0 � s � s�; j 2 N ðIV:19Þ

yj � a3ffiffi~ee

pm�2j � 1; j 2 N ðIV:20Þ

Ajð2Þ � M0; j 2 N ðIV:21Þ

Proof. We have u0 ¼ 0: Let begin by proving (IV.19)0 to (IV.21)0.

(a) (IV.19)0: Using (III.2) and (IV.10), we have

g0 ¼ �S0Gð0Þ and Gð0Þ ¼ 1

e½detðjij þ aijÞ � gðjÞ�


ORDER REPRINTS

But j 2 Cs�þ2;aðOÞ, g is Cs� of its arguments and Sn are smoothing operators, sog0 2 Hs� ðOÞ. (II.3) (II.4) (III.2) and (I.6) show that

kg0ks � bkGð0Þks �b~eek detðjij þ aijÞ � gðjÞks�

� b2

~eej detðjij þ aijÞ � gðjÞjs� � b2~ee

using (IV.4) and b � m, we get

kg0ks � m2~ee � a2~ee

(b) (IV.20)0: (III.2) (IV.4) (IV.5) and (I.6) give

y0 ¼ jGð0Þj0 �1

~eej detðjij þ aijÞ � gðjÞjs� � ~ee �

ffiffi~ee

pa2 � 1

(c) (IV.21)0: We have

A0ð2Þ ¼ maxi;j

1;

�� @g@pi

ðjÞ��2

;

�� @g@u ðjÞ��2

;

�� @F@jij

ðjlqÞ��2

þ y0

!

Then, by (II.9) (IV.1) and (IV.20)0, A0ð2Þ � M0.

Assume that we have constructed u0, u1; . . . ; un�1 2 Hs� ðOÞ satisfying (III.15) to(III.17) and we have proved (IV.14) to (IV.21) for j � n� 1. We shall constructun 2 Hs� ðOÞ satisfying (III.15) to (III.17) and prove that (IV.14) to (IV.21) are satis-fied for j ¼ n. Combining (IV.16)n�1 to (IV.21)n�1, we have j~wwn�1j4;k � 1, yn�1 � 1,An�1ð2Þ � M0 and gn�1 2 Hs� ðOÞ. We can then apply Theorem III.4 to get a solutionun 2 Hs� ðOÞ to the problem (IV.7)n satisfying (III.15) to (III.17). Then:

(a) (IV.14)n: ð�Þ For n ¼ 1, using (I.6) (II.3) (III.2) (III.15) and (IV.2), weget

ku1k0 � Dkg0k0 � DbkGð0Þk0 � Db2

~eejdetðjij þ aijÞ � gðjÞjs� � Db2~ee

(IV.3) (IV.5) and s� � s give

ku1k0 �ffiffi~ee

pm�s ðIV:22Þ

We have by (III.16)

ku1k1 � Dðkg0k1 þ ku1k0Þ

so, using (I.6) (II.3) (IV.22) and s� � s, we get

ku1k1 � Dðb2~eeþffiffi~ee

pm�sÞ �

ffiffi~ee

pm1�s


ORDER REPRINTS

Suppose that we have for 0 � l � s; s � 2 :

ku1kl �ffiffi~ee

pml�s ðIV:23Þ

Using (III.17), we have for s � 2,

ku1ks � D

kg0ks þ

Xði;lÞ2Ls

l�s�1

ð1þ jjjiþ4;tÞku1kl!

ðIV:24Þ

(I.6) (II.3) (II.4) (IV.3) (IV.10) and s� � s imply

kg0ks � bkGð0Þks � b2jGð0Þjs � b2~ee � ~mm~eems�s

that gives by (IV.23) (IV.24)

ku1ks � D

~mm~eems�s þ

Xði;lÞ2Ls

l�s�1

ð1þ jjjiþ4;tÞffiffi~ee

pml�s

!

� D

�~mm~eems�s þ s2�ð1þ jjjiþ4;tÞm�1

ffiffi~ee

pms�s

�which shows by (IV.3) (IV.5) that

ku1ks �ffiffi~ee

pms�s

ð�Þ For n � 2, (III.15) (IV.2) (IV.5) (IV.19)n�1 imply that

kunk0 � Dkgn�1k0 � D~eea2m�sn�1 �

ffiffi~ee

pm�sn�1

ðIV:25Þ

By the same way, (III.16) (IV.2) (IV.5) ((IV.19)n�1) and (IV.25) give

kunk1 �ffiffi~ee

pm1�sn�1

Suppose that for 0 � l < s, s � 2

kunkl �ffiffi~ee

pml�sn�1

By (III.17), we have

kunks � D

kgn�1ks þ

Xði;lÞ2Ls

l�s�1

ð1þ jjþ ~ee~wwn�1jiþ4;tÞkunkl!

But, (II.1) (II.5) (IV.15)n�1 and 4þ n� � s� t imply that for 0 � i � s� 2

j~wwn�1jiþ4;t � bk~wwn�1k4þn�þi � b2min�1k~wwn�1k4þn� � 2b2ffiffi~ee

pmin�1


ORDER REPRINTS

so, using ðIV:19Þn�1

kunks � D

�~eea2ms�s

n�1 þX�

1þ jjjs�þ2;t þ 2b2ffiffi~ee

pmin�1

� ffiffi~ee

pml�sn�1

�� D

�~eea2ms�s

n�1 þ 2b2s2�~eems�sn�1 þ

�1þ jjjs�þ2;t

�s2�

ffiffi~ee

pms�1�sn�1

�combined with (IV.4) (IV.5), that gives

kunks �ffiffi~ee

pms�sn�1

(b) (IV.15)n: (IV.6) shows that wn ¼Pn

j¼1 uj. By (IV.14)j, 1 � j � n we have

kwnks �Xnj¼1

kujks �ffiffi~ee

pms�s þ

Xnj¼2

ffiffi~ee

pms�sj�1 �

ffiffi~ee

pms�s þ

Xn�1

j¼2

ffiffi~ee

pms�sj

For s � s� t, since m � 21t � 2 then ms�s

j � m�tj � 1

2j and

kwnks �Xn�1

j¼0

ffiffi~ee

pms�sj �

ffiffi~ee

p Xn�1

j¼0

1

2j� 2

ffiffi~ee

p

For s � s� t, we have

kwnks �ffiffi~ee

pms�s þ

ffiffi~ee

p mnðs�sÞ � ms�s

ms�s � 1

Since m � 21t then ms�s � mt � 2 so,

kwnks �ffiffi~ee

pms�sn

(c) (IV.16)n: Combining (II.1) (II.4) (IV.5) (IV.15)n and 4þ n� � s� t, weobtain

j~wwnj4;t � bk~wwnk4þn� � b2kwnk4þn� � 2b2ffiffi~ee

p� 1

(d) (IV.17)n:

In the case s � s� t, we obtain using (II.6) and (IV.15)n

kwn � ~wwnks � bms�½sþt��1n kwnk½sþt�þ1

� bms�½sþt��1n

ffiffi~ee

pm½sþt�þ1�sn � b

ffiffi~ee

pms�sn


ORDER REPRINTS

In the case s > s� t, (II.6) (IV.15n) and b � 1 give

kwn � ~wwnks � bkwnks � bffiffi~ee

pms�sn

(e) (IV.18)n: We have by (IV.12)

rn ¼ ½LGðwn�1Þ � LGð~wwn�1Þ�un|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}ð1Þ

� yn�14un|fflfflfflfflffl{zfflfflfflfflffl}ð2Þ

þ Qn|{z}ð3Þ

When n ¼ 1, ð1Þ ¼ 0. In the case n � 2, since

ð1Þ ¼Z 1

0

d

dt½LGð~wwn�1 þ tðwn�1 � ~wwn�1ÞÞun�dt

We have by (II.1) and (IV.17)n�1

jwn�1 � ~wwn�1j2 � bkwn�1 � ~wwn�1k2þn� � 2b2ffiffi~ee

pm3þn��sn�1

But 2b2ffiffi~ee

p � 1 and 3þ n� � 4þ 2n� � s, so jwn�1 � ~wwn�1j2 � 1. By the sameway, (II.1) (IV.5) and (IV.14)n give

junj2 � bkunk2þn� � bffiffi~ee

pm3þn��sn�1 � 1

We also have by (IV.16)n�1, j~wwn�1j2 � 1:Hence, we can apply Proposition III.3 to get

kð1Þks � ~eeK0f½kjksþ2 þ k~wwn�1ksþ2 þ kwn�1ksþ2 þ 1�jwn�1 � ~wwn�1j2junj2þ ðkjk2þn� þ k~wwn�1k2þn� þ kwn�1k2þn� þ 1Þ� ðjwn�1 � ~wwn�1j2kunksþ2 þ kwn�1 � ~wwn�1ksþ2junj2Þg

Using (II.3) (IV.3), we get for 0 � s � s�

kjksþ2 � bjjjs�þ2 � bm � m2

By (II.2), it suffices to prove (IV.18)n for s ¼ 0 and s ¼ s� � 2:

s ¼ 0: Combining (II.1) (IV.14)n(IV.15)n�1 and (IV.17)n�1, we have

kð1Þk0 � ~eeK0

h�m2 þ 2b

ffiffi~ee

pþ 2

ffiffi~ee

pþ 1�2b3~eem4þ2n��2s

n�1

þ�m2 þ 2bffiffi~ee

pþ 2

ffiffi~ee

pþ 1�4b3~eem4þn��2s

n�1

ithat gives using (IV.5) and s � 4þ 2n� � 4þ n�

kð1Þk0 � ~eeK0m�sn�1:


ORDER REPRINTS

s ¼ s� � 2: (IV.5) and s� � sþ t imply as in the previous case

kð1Þks��2 � ~eeK0ms��2�sn�1

By (II.2), we obtain for 0 � s � s� � 2

kð1Þks � b~eeK0ms�sn�1

Next,

kð2Þks � yn�1kunksþ2

In the case n ¼ 1: combining (IV.5) (IV.9) and (IV.14)n; we obtain

kð2Þks � jGð0Þj0ku1ksþ2 � ~eeffiffi~ee

pmsþ2�s � ~eems�s

In the case n � 2: (IV.14)n and (IV.20)n�1 imply

kð2Þks � a3~eem�2n�1m

sþ2�sn�1 ¼ a3~eems�s

n�1

Finally, since by (IV.13)

ð3Þ ¼ Qn ¼ Gðwn�1 þ unÞ �Gðwn�1Þ � LGðwn�1Þun

¼Z 1

0

Z t

0

d

dh½LGðwn�1 þ hunÞun�dh

!dt

Then using (II.1) (IV.5) and (IV.15)n�1, we obtain

jwn�1j2 � bkwn�1k2þn� � 2bffiffi~ee

p� 1

Since we proved that junj2 � 1, we can apply Proposition III.3 to have

kð3Þks � ~eeK0½ðkjksþ2 þ kunksþ2 þ kwn�1ksþ2 þ 1Þjunj22þ 2junj2kunksþ2ðkjk2þn� þ kunk2þn� þ kwn�1k2þn� þ 1Þ�

By combining (II.1) (IV.14)n(IV.15)n�1, we get For s ¼ 0:

kð3Þk0 � ~eeK0

��m2 þ

ffiffi~ee

p½maxðm;mn�1Þ�2�s þ 2

ffiffi~ee

pþ 1�b2~ee½maxðm;mn�1Þ�4þ2n��2s

þ 8�m2 þ

ffiffi~ee

pb½maxðm;mn�1Þ�2þn��s þ 2

ffiffi~ee

pþ 1�~eeb½maxðm;mn�1Þ�4þn��2s�

which gives combined with (IV.5) and s � 4þ 2n�

kð3Þk0 � ~eeK0½maxðm; mn�1Þ��s:


ORDER REPRINTS

For s ¼ s� � 2: we get also since s � 4þ 2n�

kð3Þks��2 � ~eeK0½maxðm; mn�1Þ�s��2�s

Then (II.2) shows that for 0 � s � s� � 2

kð3Þks � b~eeK0½maxðm; mn�1Þ�s�s

and we conclude that

krnks � ð2bK0 þ a3Þ~ee½maxðm; mn�1Þ�s�s

� 9K0m5~ee½maxðm; mn�1Þ�s�s ¼ a1~ee½maxðm; mn�1Þ�s�s

(f) (IV.19n): By (IV.10) and (IV.11)

gn ¼ Sn�1Rn�1 � SnRn þ ðSn�1 � SnÞGð0Þ¼ ðSn�1Rn�1 � SnRn�1Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð4Þ

� Snrn|{z}ð5Þ

þ ðSn�1 � SnÞGð0Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}ð6Þ

The case s ¼ 0: (II.6) (IV.11) and (IV.18)j, j � n� 1 imply

kð4Þk0 � kðI � Sn�1ÞRn�1k0 þ kðI � SnÞRn�1k0� bkRn�1ks��2m

2�s�n�1 þ bm2�s�

n kRn�1ks��2

� ðba1~eem2�s�n�1 þ ba1~eem2�s�

n Þ ms��2�s þ

Xn�1

j¼2

ms��2�sj�1

!

Since s� � 2 > s and b � m, then

kð4Þk0 � ba1~eeðm2�s�n�1 þ m2�s�

n Þms��2�sn�1

� 2a1m2~eem�sn

In the other hand, combining (II.4) (IV.18)n, s < s� � 2 and b � m we obtain

kð5Þk0 � bkrnk0 � ba1~ee½maxðm; mn�1Þ��s � a1m2~eem�sn

We also have by (I.6) (II.3) (II.6) and s < s� � 2

kð6Þk0 � kðI � Sn�1ÞGð0Þk0 þ kðI � SnÞGð0Þk0� bm�s

n�1kGð0Þks þ bm�sn kGð0Þks

� b2m�sn�1jGð0Þjs� þ b2m�s

n jGð0Þjs�� b2~eem�s

n ðms þ 1Þ � 2ms�~eem�sn


ORDER REPRINTS

We finally get

kgnk0 � ð2þ 3a1Þms�~eem�sn

The case s ¼ s�: (II.5) (IV.11) (IV.18)j, 1 � j � n and s < s� � 2 show that

kð4Þ þ ð5Þks� � kSn�1Rn�1ks� þ kSnRnks�� bm2n�1kRn�1ks��2 þ bm2nkRnks��2

� bm2n�1a1~ee

ms��2�s þ

Xn�1

j¼2

ms��2�sj�1

!

þ bm2na1~ee

ms��2�s þ

Xnj¼2

ms��2�sj�1

!� ba1~eeðm2n�1m

s��2�sn�1 þ m2nm

s��2�sn Þ

� 2ba1~eems��sn � 2ma1~eems��s

n

Next, we have by (I.5) (II.5) (II.3) and b � m

kð6Þks� � kSnGð0Þks� þ kSn�1Gð0Þks�� bms��s

n kGð0Þks þ bms��sn�1 kGð0Þks

� 2b2~eems��sn � 2m2~eems��s

n

so,

kfnks� � 2mða1 þ mÞ~eems��sn

We can finally conclude using (IV.4) and m � a1

kfnks� � 4a1m2~eems��sn � a2~eems��s

n

(g) (IV.20)n: We have by (IV.9)

yn ¼ jGð~wwnÞj0 � jGðwnÞ �Gð~wwnÞj0 þ jGðwnÞj0

Combining (IV.7) to (IV.13), we have

GðwnÞ ¼ ðI � Sn�1ÞRn�1 þ ðI � Sn�1ÞGð0Þ þ rn

Then

yn � jGðwnÞ �Gð~wwnÞj0|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}ð7Þ

þ jðI � Sn�1ÞRn�1j0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}ð8Þ

þ jðI � Sn�1ÞGð0Þj0|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}ð9Þ

þ jrnj0|{z}ð10Þ


ORDER REPRINTS

Since we proved that jwnj2 � 1 and j~wwnj2 � 1, we can apply Proposition III.3to get

ð7Þ � bK0kwn � ~wwnk2þn� ðkjk2þn� þ kwnk2þn� þ k~wwnk2þn� þ 1Þ

(II.4) (IV.15)n(IV.17)n and 3þ n� � 4þ 2n� � t � s� t give

ð7Þ � 2b2K0

ffiffi~ee

pm2þn��sn ðm2 þ 2

ffiffi~ee

pþ 2b

ffiffi~ee

pþ 1Þ

Since ~ee � 1ð6b2Þ2, b � m and 4þ n� � s � 4þ 2n� � s � 0 then

ð7Þ � 4m5K0

ffiffi~ee

pm�2n

In the case n ¼ 1, ð8Þ ¼ 0. For n � 2, since b � m, n� � s � �2 andm4a1

ffiffi~ee

p � a2ffiffi~ee

p � 1, combining (II.1) (II.6) (IV.11) and ðIV:18Þj, j � n� 1 we obtain

ð8Þ_� bkðI � Sn�1ÞRn�1kn�� b2mn��s�þ2

n�1 a1~ee

ms��2�s þ

Xn�1

j¼2

ms��2�sj�1

!� b2a1~eem

s��sn�1 �

ffiffi~ee

pm�2n

(I.6) (II.1) (II.3) (II.6) (IV.5) and b � m imply

ð9Þ � bkðI � Sn�1ÞGð0Þkn� � b2mn��s�n�1 kGð0Þks�

� b3m�2n�1~ee � b3m2~eem�2

n �ffiffi~ee

pm�2n

Finally, we get by (II.1) and (IV.18)n

ð10Þ � bkrnkn� � ba1~ee½maxðm; mn�1Þ�n��s

� ma1~ee½maxðm; mn�1Þ��2 �ffiffi~ee

pm�2n

Thus, we conclude that

yn � 7K0m5ffiffi~ee

pm�2n ¼ a3

ffiffi~ee

pm�2n � 1

(h) (IV.21): We have

Anð2Þ � max1�i;j�n

�1;

�� @g@pi

ðj þ ~ee~wwnÞ��2

;

�� @g@u ðj þ ~ee~wwnÞ��2

;

�� @F@uij

ðjkl þ ~eeð~wwnÞklÞ��2

þ yn

�Using (II.9) (IV.1) (IV.16)n and (IV.20)n, we get Anð2Þ � M0:


ORDER REPRINTS

Proof of Theorem A. We shall prove using Proposition IV.1 the convergence of thesequence ðwnÞ.

Set s ¼ s� � 2� t and s ¼ s� t. By (IV.6) (IV.14), for any i; k 2 N� i > k

kwi � wkks �Xij¼kþ1

kujks � bffiffi~ee

p Xij¼kþ1

m�tj�1 ¼ b

ffiffi~ee

p Xij¼kþ1

ðm�tÞj�1

Since m � 2, t > 0 then kwi � wkks ! 0 as i; j ! 1. Hence, there is a functionw 2 Hs��2�2tðOÞ satisfying wn ! w in Hs��2�2tðOÞ. Since Hs��2�2tðOÞ �Cs��2�n�3tðOÞ then w 2 Cs��3�nðOÞ.

On the other hand,

GðwnÞ ¼ ðI � Sn�1ÞRn�1 þ ðI � Sn�1ÞGð0Þ þ rn

For n � 2, we have using (II.2) and (IV.18)

krnks��2�2t � a1b~eems��2�2t�sn�1 ¼ a1b~eem�t

n�1

Combining (II.3) with (II.6) and (I.6), we get

kðI � Sn�1ÞGð0Þks��2�2t � bm�2�2tn�1 kGð0Þks� � b2m�2�2t

n�1 ~ee

Combining (II.6) (IV.11) and (IV.18), we can write

kðI � Sn�1ÞRn�1ks��2�2t � bm�2tn�1kRn�1ks��2

� bm�2tn�1

Xn�1

j¼1

krjks��2

� bm�2tn�1~eea1

ms��2�s þ

Xn�1

j¼2

ms��2�sj�1

!� ~eeba1m�2t

n�1ms��2�sn�1

� ba1~eem�tn�1

These inequalities imply GðwnÞ ! 0 in Hs��2�2tðOÞ as n ! 1. SinceHs��2�2tðOÞ � C2ðOÞ and wn j @O ¼ 0, we conclude that GðwÞ ¼ 0 and w j @O ¼ 0. Thatis u ¼ jþ ew is a solution to the original Monge–Ampere equation which satisfies byLemma III.1, (I.7) since g is nonnegative. The uniqueness of the solution followsimmediately from Caffarelli et al. (1984).

REFERENCES

Alinhac, S., Gerard, P. (1991). Operateurs pseudo-differentiels et theoreme de Nash-Moser, Inter Editions et Editions du CNRS.

Amano, K. (1988). The Dirichlet problem for degenerate elliptic 2-dimensionalMonge-Ampere equation. Bull. Austral. Math. Soc. 37:389–410.


ORDER REPRINTS

Atallah, A. (2000). Probleme de Dirichlet pour une equation de Monge-Amperereelle elliptique degeneree en dimension n. Trans. A.M.S. 352(6):2701–2721.

Caffarelli, L., Nirenberg, L., Spruck, J. (1984). The Dirichlet problem for nonlinearsecond order elliptic equations I, Monge–Ampere equation. Comm. Pure Appl.Math. 37:369–402.

Gilbarg, D., Trudinger, N. S. (1983). Elliptic Partial Differential Equations ofsecond order. 2nd ed. Berlin, Heidelberg, New York: Springer Verlag.

Hormander, L. (1984). On the Nash–Moser implicit function theorem. AcademiaScientiarum. Serie A I, Math., 10. Fennicae, pp. 67–97.

Moser, J. (1961). A new technique for the construction of solutions of non linearpartial differential equations. Proc. Natl. Acad. Sc. USA 47:1824–1831.

Oleınik, O. A., Radkevic, E. V. Second Order Equations with Nonnegative Charac-teristic Form. New York-London: A.M.S., Plenum Press. Translated fromRussian.

Oliker, V., Waltman, P. Radially symmetric solutions of a Monge-Ampere equationarising in a reflector mapping problem. In: Dold et, A., Eckmann, B., eds.Differential Eqs. and Mathematical Physics. Lecture Notes in Mathematics1285. Springer Verlag.

Pogorelov, A. (1978). The Minkowski Multidimentional Problem. New York: Wiley.Zuily, C. (1988). Sur la Regularite Des Solutions Non Strictement Convexes de

l’equation de Monge–Ampere Reelle, Estratto Dagli Annali Della SculaNormale Superiore di Pisa. Vol. XV, Fasc. IV. Scienze Fisiche e MatematicheSerie IV.

Received April 2003Accepted April 2004


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The Dirichlet Problem for Degenerate Elliptic Darboux Equation

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