A NNALES DE L ’I. H. P., SECTION C
K AISING T SO
Remarks on critical exponents for hessian operators
Annales de l’I. H. P., section C, tome 7, n
o2 (1990), p. 113-122
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Remarks
oncritical exponents for Hessian operators
Department of Mathematics, The Chinese University of Honk Kong,
Shatin, Hong Kong
Ann. Inst. Henri Poincaré,
Vol. 7, n° 2, 1990, p. 113-122. Analyse non linéaire
ABSTRACT. - The critical exponent for the k-th Hessian operator
(k = 2,
...,n)
is determined and thesolvability
of the associated Dirichletproblem
with sub-criticalnonhomogeneous
term is discussed.Key words : Hessian operators, Critical exponent, Associated Dirichlet problem.
RESUME. - On determine
l’exposant critique
pour lesoperateurs
deHess,
et l’on etudie leprobleme
de Dirichletassocie,
avec un terme nonhomogene sous-critique.
INTRODUCTION
Let Q be a bounded domain in 2. It is well-known that the Sobolev exponent p
n + 2 plays
a critical roleconcerning
thesolvability
ofn-2 p y g y
the Dirichlet
problem
in
Q,
u = 0 on aQ.Namely,
thisproblem
admits nopositive
solutions in astar-shaped
Qwhen
/? ~ n + 2
and it has apositive
solution in any Q whenn + 2
> p > 0n-2 p y n-2
Annales de l’Institut Henri Poincaré - Analyse non linéaire - 0294-1449 Vol. 7/90/02/113/10/$3.00/0 Gauthier-Villars
Kaising TSO
and p is not
equal
to one. In aprevious
work[6]
we have studied acorresponding problem
for theMonge-Ampere
operator on a convex Q:We found that in contrast to the semilinear case this
problem
admits non-zero convex solutions
provided
p ispositive
and is notequal
to n. Sincethe
Laplace
operator and theMonge-Ampere
operator arerespectively
the first and the last Hessian operators, we are led to the
question
ofdetermining
the "criticalexponents"
for theremaining
Hessian operators.Let
Sk (O2 u), k =1,
..., n, be the k-th Hessian operator, i. e., it is the k-thelementary symmetric
function of the Hessian matrix of u. ConsiderIn this note we shall establish the
following
result:(i) (1)
has nonegative
solution in C 1(S2)
n C4(Q)
whenp >_
y(k); (ii)
Itadmits a
negative
solution which isradially symmetric
and is in C2(SZ)
when0 p
y (k),
p is notequal
to k.The non-existence result
actually
holds in moregeneral
situation. SeeProposition
1 in the next section. Part(ii)
of this theorem is contained inPropositions
2(0
pk)
and 3(k
py (k))
in Section 2. Whenp = k
one should
study
aneigenvalue problem. However,
we shall not consider it here.NOTATION. -
Subscripts
like those in ui, uy,F z’
..., stand forpartial
differentiations.
Also,
summation convention isalways
in effect.SECTION 1. NON-EXISTENCE Consider the
following
Dirichletproblem
115
CRITICAL EXPONENTS FOR HESSIAN OPERATORS
Many
authors have studied the existence ofnegative
solutions for thisproblem
in the case k =1.Negative
solutions aresought
becausethey
areimportant
in manyapplications. However,
when k is greater than one,negative
solutions will besought
notonly
as ananalogue
of the semilinearcase but
mainly
becausethey
areprecisely
the "admissible ones". When k is greater than one a Hessian operator is nonlinear and so its typedepends
on the Hessian of u.
According
toCaffarelli-Nirenberg-Spruck [1]
aC2-
function u is called admissible
(with
respect toSk, k >__ 2)
if theeigenvalues
of lie in the connected component
rk
of the set~~, _ (~,1,
...,~,n):
The k-thelementary symmetric
function of À ispositive}
which contains the cone
~~, : ~,i
> 0 for alli~.
It can be shown that for anadmissible u is
elliptic, i. e., aSk (V2
u) ~i ~~
>0,
for all~, I ~ I > 0, Orij
and We claim that any solution of
(2)
whichbelongs
toC
(n) n
C2(Q)
is admissible if andonly
if it isnegative. For,
at a non-negative
maximum of u,Sk (V2 u)
isnon-positive.
But sincef
ispositive
this is
impossible. Conversely,
if u isnegative,
it attains a minimum in Q.At this minimum the
eigenvalues
of the Hessian of u lie inrk.
Henceby
the
continuity
ofV2
u and thepositivity of f
u is also admissible.The
equation
in(2)
has a variational structure. To describe it weneed to express
Sk
in terms of the Newtonian tensor. Recall that fork = o,
...,n -1,
the k-th Newtonian tensor isgiven by
Here is the
generalised
Kronecker delta: It isequal
to 1(resp. -1 )
if
ii,
...,ik,
i are distinct and is an even(resp. odd) permutation;
Otherwise it is
equal
to zero. A fundamental property of the Newtonian tensor is that it is ofdivergence free,
i. e.,Sk
is related toTk _ 1
viaand
(5)
follows from Euler’sidentity.
Forproofs
of(3)
and(4)
seeReilly [5].
Vol. 7, n° 2-1990.
Using (3)-(5)
onereadily
verifies that theequation
in(2)
is the Euler-Lagrange equation
for the functionalwhere F
(x, z)
=f (x, t)
dt.Now,
consider ageneral integrand f7 (x,
z, Let u be a C4-solution of the
Euler-Lagrange equation
for F. Then for any constant a thefollowing identity
holds:This
identity,
which can be verifieddirectly,
was first obtained in Pucci- Serrin[3]. Applying
thisidentity
to F=-zSk(rij)
+ F(x, z),
wehave, by
A;+ 1
(3)-(5),
Next we formulate the non-existence result. Recall that a bounded domain Q is called
star-shaped
if for somexo, ~
x - xo,v ~ >_
0 on aSZ.(Here
v is the unit outer normaland ( . , . )
is the Euclidean innerproduct.)
PROPOSITION l. - Let Q be a
star-shaped
C2-domain with xobeing
the
origin. Suppose
thatf belongs
to C(SZ
X( -
~,0])
~ C1(Q
X( -
~,0)), positive
in Q X(- 0)
andequal
to zero on Q X~0}.
There are nonegative
solutions to
(2)
whichbelong
to C1(SZ)
(~ C4(Q) f
CRITICAL EXPONENTS FOR HESSIAN OPERATORS 117 in Q X
( -
~,0).
The same conclusion holds underif (
x,v )
> 0 on lQ.Proof -
Letu(x) - E) . By
a strong maximumprinciple (see
Lemmabelow) 03A9~
hasC4-boundary
for small s > 0. Choosea = so that
k (a + 2) + a - n = 0
in(6)
and thenintegrate
both sidek + I
of
(6)
over03A9~. By divergence
theorem we haveA
repeated application
of(3)-(5) gives
and
Consequently,
If u is a
negative
solution of(2), (V2 u)i J
vi v~ > 0by ellipticity.
Hencethe left hand side of the above
identity
isnon-negative. However,
this contradicts with(7).
On the otherhand, if ( x, v ~ >_ ~
for somepositive
8 on
an, by (8)
and the lemma below the left hand side of the aboveVol. 7, n° 2-1990.
identity
ispositive.
But then this contradicts with(7)’. Proposition
1 isthus
proved.
Proof -
Let x E aS2 and let B be a ball which is inscribed inside Q and touches aS2 at x. For anon-negative radially symmetric function g
withg
(I x I)
~f (x,
u(x))
let w be the solution forw can be solved
explicitly
in terms of g; see(10)
in the next section.By
the
maximum principle
w ~ M in B. Hence~u ~v(x)~ ~w ~v(x)
> 0. NSECTION 3. EXISTENCE
Throughout
this section Q denotes a ball of radius R centered at theorigin and f (x, z)
in(2)
isequal
tog (r, z), r= x I where
We shall look for
radially symmetric
solutions for(2).
Denote such a solutionby y (r). y
satisfiesHere and
Ck -1
are combinatorial constants. It is easy to see thatwhenever y
solves(10), u (x) = y (~ x ~)
is a solution for(3).
For a
negative radially symmetric
u, the functionalJk (u)
isequal
towhere
G (r, z) = t) dt,
a =03C4 k(k+1) Ck - i
and T is the volume of the unitsphere
in Rn. It is more convenient to letJk
act on the whole spacerather than on
negative
functions. To this end we extend g to be an even119
CRITICAL EXPONENTS FOR HESSIAN OPERATORS
function in z
(maintaining
the samenotation)
and consider the functionalLEMMA 1. - Let
(i)
For0 03B4 y
(k),
there exists a constant C = C(b, k, R, n)
such thatfor
all y EE; (ii)
For n >_ k >n ,
there exists a constant C = C(k, R, n)
such2 that
Proof. - Applying
Holderinequality
to y(r)
=lr y’
R(s)
ds we haveRaising
both side to their(8+l)-th
power,multiplying by
and thenintegrading
from 0 to R we obtain(i). (ii)
can beproved
in a similarway..
Remark. - In
fact,
asharper
result holds: There exists a constantC = C (k, n)
such thatfor
all y
in E. For aproof
see Lin[2].
Vol. 7, n° 2-1990.
Let
Wk
be the Banach space obtainedby completing
E under the normR 1/(k+ 1)
( R0 I y’ )1/(k
+ , .Using
Lemma lone caneasily
show that thefollowing
lemma holds:LEMMA 2. -
(i) Wk
iscontinuously
embedded in([r, R]) : y (R) = o. ~ _ for
r > o;(ii) Wk
iscompactly
embed-ded in
(rn -1 dr) for
py (k)
when k-- ~
and inC ([o, R])
when2
We also have
LEMMA 3. -
Suppose
gsatisfies
for
some p,0 p y (k). Then jk belongs to C1 (Wk; R). Moreover,
y ~
R0
G(r, y)
isweakly
continuous onWk.
The
proof
of this lemma isparallel
to theproof
for thecorresponding
result in the semilinear case; See Rabinowitz
[4], Appendix B,
for details.Any
criticalpoint of jk
will be called ageneralised
solution of(10).
Inother
words, y
is ageneralised
solution of(10)
if andonly
iffor all cp in
Wk.
Theregularity
of such solutions isgiven
inLEMMA 4. - Under
(9)
and( 11 ),
anygeneralised
solutionof (10)
is inC2 ([0, R]),
and solves(10)
in classical sense. Moreover, it isnegative
in[0, R)
unless it vanishesidentically.
Proof. -
We shall first show that anygeneralised
solution is bounded.When k is greater
than -
this follows from Lemma 1. So we assume k is less than orequal to -.
.
Let y
be ageneralised
solution.For s >__
1 and N ~ 1 we define a.
function C in C1
[1, oo ) by
forz E [ 1, N]
andtaking
itto be linear for z ~ N. Select cp =
y++11 |03A6’ (t) IH 1 dt where y +
= max (y, 0)
121
CRITICAL EXPONENTS FOR HESSIAN OPERATORS
as a test function.
Substituting
cp into(12)
we haveafter
using cp __ I d~’ (y + + 1 ) (k + 1 y + .
It follows from Lemma1,
Holderinequality
and(11)
that for a fixed~3, max (p, k) y (k),
where
q = p if p > k
andq = k if p _ k
and C is a constantdepending
onIlyllwk’ Letting N-+oo,
we see thatimplies
thestronger inclusion
y + E LS « + l (r’~ -1 dr). Also,
we havewhere K = 1 T ~’
> 1.
Setting
s = Km, m >_1,
an iterationyields
~ -~- n -
Similarly
one can show that the above estimate holdsfor y - - -
min(y, 0)
in
place of y + . Thus y
is bounded.Take
R
r
~(t) dt where ~
is smooth as a test function in(12).
Afteran
integration by
parts, we haveHenceforth
Since g
isnon-negative, y
isincreasing (hence
isnegative)
unless it vanishesidentically.
From this the desired conclusion of Lemma 4 can be drawn..Now we can state the existence results.
PROPOSITION 2. - Under
(9)
anduniformly on [0, R], (2)
admits anegative radially symmetric
solution in CVol. 7, n° 2-1990.
This solution is obtained
by minimizing j~
overW~. (1 3)
ensures thatj~
is bounded from below and is
negative
somewhere. It follows from the directmethod,
which works in view of Lemma3,
that a non-zero absolute minimum exists.PROPOSITION 3. - In addition to
(9)
and(11),
suppose gsatisfies lim g(r,z)
= 0
unformly
in[0, R], and,further,
there exist0 e (0, 1)
and az - o z
positive
constant M such thatThen
(2)
admits anegative radially symmetric
solution whichbelongs
toC2 (n).
(14) implies that jk
fulfills the Palais-Smale condition.Proposition
3follows from the mountain pass lemma. For details see the
corresponding proof
for the semilinear case in[4], Chapter
2.ACKNOWLEDGEMENT
The author would like to thank Professor
Wei-Ming
Ni fordrawing
his attention to a paper of Pucci and Serrin
[3].
REFERENCES
[1] L. CAFFARELLI, L. NIRENBERG and J. SPRUCK, The Dirichlet Problem for Nonlinear Second Order Elliptic Equations, III: Functions of the Eigenvalues of the Hessian,
Acta Math., Vol. 155, 1985, pp. 261-301.
[2] C.-S. LIN, Interpolation Inequalities with Weights, Comm. P.D.E., Vol. 11, 1986,
pp. 1515-1539.
[3] P. PUCCI and J. SERRIN, A General Variational Identity, Indiana Univ. Math. J., Vol. 35, 1986, pp. 681-703.
[4] P. RABINOWITZ, Minimax Methods in Critical Point Theory with Applications to Differ-
ential Equations, No. 65, Expository Lectures from the C.B.M.S. Regional Conference, A.M.S. 1986.
[5] R. C. REILLY, On the Hessian of a Function and the Curvatures of its Graph, Michigan
Math. J., Vol. 20, 1973-1974, pp. 373-383.
[6] K. Tso, Real Monge-Ampère Functionals, Preprint, 1987.
(Manuscript received December 6, 1988.)