A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
T HOMAS K ARLSSON
Hölder continuity and thin obstacle problems for vector valued functions
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 13, n
o2 (1986), p. 281-297
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http://www.numdam.org/
Continuity
for Vector Valued Functions.
THOMAS KARLSSON
1. - Introduction.
The
object
of this paper is tostudy
theregularity
of solutions u, defined in a set Qc R",
of a thin obstacleproblem
for vector valued functions. Weare concerned with the Holder
continuity
for solutions of a nonlineardia,go-
nal
system
of variationalinequalities.
The convex set of admissible varia- tions isgiven by
an obstacle which is active on a smooth n - 1-dimensional manifold .E c.Q.Our main result is that if the obstacle
function ?p
has distributional derivatives of order one which are of class EP for some p > n and is Holder continuous withexponent
p, 0p 1,
then solutions are Holder continuous with the sameexponent It
on the manifold E. In order to achieve this wefirst prove the
following,
a result of interest in itself. If the setE,
wherethe constraint is
introduced,
is ahyperplane
then solutions are Holdercontinuous with
exponent
fl in directionsparallel
to thehyperplane.
Inthe
special
case p = 1 weget Lipschitz continuity
in those directions and this is one of the results in the author’s doctoral thesis[7].
In
[8]
the author studied a, in some sense, moregeneral problem.
It follows from the results there that bounded weak solutions of ourproblem
are Holder continuous with some
exponent,
at least if the bound is suf-ficiently
small.There are few
existing
papersdealing
with thin obstacleproblems
forvector valued functions.
However,
we can compare our situation with that in[11].
There Kinderlehrer was concerned withcontinuity properties
for the solution of a
boundary
obstacleproblem
with a variationalinequality pertaining
to anon-diagonal system
with smoothcoefficients,
aproblem
Pervenuto alla Redazione il 28 Dicembre 1984.
which arises in the
theory
of linearelasticity.
Let us also mention[13]
where Necas
proved W2,2-regularity
for solutions of a class of non-linearboundary
obstacleproblems.
The scalar valued case has been more
extensively
studied and this wasinitiated
by Lewy [12].
He considered theproblem
ofminimizing
the Di-richlet
integral
in two dimensions and showed that if the thin obstacle function has a continuous derivative then thesolution,
y restricted to the set where the constraint isintroduced,
has a continuous derivative. After that this kind of interior thin obstacleproblems
as well asboundary
ob-stacle
problems
have been studiedby
several authors. Here we mention Kinderlehrer[9]
and[10],
Beirao daVeiga
and Conti[1], Giaquinta
andModica
[5],
Frehse[3],
Frehse and Mosco[4]
and Caffarelli[2].
Acknowledgements.
I am muchobliged
to Dr.Bengt
Winzell for his carefulreading
and for his manysuggestions
forimprovements
of themanuscript.
I am also verygrateful
to Mrs.Birgitta Arnsby
for her efficienttyping
of themanuscript.
2. -
Notations.
Let Q be a bounded open set in
Rn, n > 3,
ofpoints x
=(xl,
.., xn_l,xn)
= (x’ xn),
andput Br(xo)
=(x c- Rn: Ix-x,,Ir}.
Denoteby w1’P(Q), p:>l,
the Sobolev space of
functions n
such thatand
by wl’°(Q)
the closure ofOr;(.Q)
in the wl,p-norm. In the notation for afunction space we add the
symbol
RN to denote thecorresponding
spaceof RN-valued functions. For instance
OO,P(S, RN)
stands for the space of RN-valued functions withcomponents
inOO,P(S),
the space oflocally
Holdercontinuous functions with
exponent
,u. We also use the notations u =(u1,
... , y...,
2GN-1, uN)
=(Pu, uN), Da, === ô/ôxa,
and V u =(..., Daui, ...),
where 1 lXnand
liN.
Forpositive h
and for unit vector e =(e’, 0), e’ERn-l,
we letu_(z)
=u(x::f:he)
andWhen ,u
= I we writeL1hu
instead ofdh u. Moreover,
we use a summationconvention such that
means that
where
us, v2
andf are
thecomponents of u, v
andf, respectively. Finally,r
different constants
appearing
in the text willmostly
be denotedby
thesame letter c.
3. - Results.
We will look at solutions 2c =
(u1,
...,uN)
of the variationalinequality
for all v E K =
{V
eWI,2 (Q, RN): (v -1f’) (x)
EF(x)
for x E Eand V _ 99
EWI,2(,Q, RN)}, where q
is aprescribed
RN-valuedfunction, {F(x)lEn
is a set of halfspaces in RN and E c Q is an n - 1-dimensional manifold without
boundary
in Q. The obstacle
function V
is of class w1,p (’BC°"(Q, RN)
for some p > nand some
It, 0 u 1. Furthermore,
the coefficientsa’# Ioc, #n, give
an
elliptic operator,
that is there exists apositive
constant A such thatIn the three first theorems we consider the situation where the sets
.(x),.
x E
Q,
all are the same,namely
the upper half space. This means that the constraint(u - 1J’)(x)
EF(x)
for x E .E takes the form :uN(x):> ’lpN(X)
for x E E.Moreover,
theright
hand side is of the formFor g, which grows
linearly
inVu, only measurability
is assumed. On the other hand for go, which growsquadratically
inVu,
we need someregularity-
condition. More
precisely,
gotogether with ga
andatX,fJ, I I cx, Pn,
are con-tinuous functions
satisfying
thefollowing
conditions. For bounded u e K.and with uj ==
u(xj)
andVuj - Vu(xj), j - 1, 2,
we suppose thatfor some
u’, p p 1,
and for almost every x,, x2 E Q. Here6,, it
= ul - u2 ,Vb,u
=Vui- Vu,
andbju - PUI-PU2, Vðiu
=VPUI- VPu2
forI i N
-1. For the function g =
g(x,
u,Vu)
we assumemeasurability
in Q foru E K and the existence of
positive
functions a ELOO(D)
and b ELP(D)
suchthat
for x
E Q, U
E Kand p
e RnN.Note
that, except
for the restriction thatg§, I I N - I,
are inde-pendent
ofBfuN,
condition(4)
allows the functiongo == (go, ..., g:-l, g:)
togrow
quadratically
in Vi.The first theorem treats the case when E = Q’=
{x
=(x’, xn)
e Q:xn =
0),
and states that in directionsparallel
to E solutions are Holder continuous with theexponent
/-l, 0,u c 1, given by
the obstaclefunction
From
[8]
we know that bounded weaksolutions,
y withsufficiently
smallLoo-norm,
are Holder continuous with someexponent
so we will consideronly
Holder continuous solutions of(1).
THEOREM 1.
If uEcO,Y(Q,RN), Oyp,l, is a
solutionof (1)
with_.E = Q’ then
for
everyQocc Q,
where the
supremum is
taken over all(x’, xn)
and(y’, xn)
EDo, x’ =1= y’.
REMARK. Let us for a moment assume that 0
Iz
1. Then it follows from Theorem 1 that u is Holder continuous withexponent
in all direc- tions if we for thesystem
of differentialequations pertaining
to the varia-tional
inequality
have relevant Schauder estimates.The
proof
of Theorem 1 alsogives
thefollowing
theoremdealing
withthe
corresponding
thick obstacleproblem.
DefineKn == {v E W1,2(Q, RN):
(v
-1p)(x)
e F for x E Qand v - cp
EW2(Q, RN)}.
THEOREM 2.
I f u
eOO,Y(Q, RN), 0 Y #1 is a
solutionof (1)
withKn
instead
of
K then u is Hölder continuous withexponent u in
Q.Next we consider n - 1-dimensional manifolds E with the
following
smoothness
property.
PROPERTY E. For every
XoE E
there is aneighbourhood
U and abijec-
tion 6 of U onto
BR(O), R> 0,
such that theimage
of U n E isB£(0)
=== BR(O) n {y ERn: Yn == O}. Moreover,
0 and its invers (j-l are of class(Jl,p,
andc( U)
denotes apositive
constant for which the Jacobian matrix(dO/dx)(0153)
satisfiesTHEOREM 3. Let the set .E in the
definition of
K haveProperty
.E and letu E
CO’Il(Q, RN),
0 Y p, be a solutionof (1).
Then u EcO’Il(E, RN).
Finally,
we are concerned with theregularity
of solutions of(1)
whenthe sets
F(x),
x EQ,
are definedby
with
v(x)
E(Jl’Jl(Q, RN)
andIv(x) =
1 for x E Q. Also in this case we may havequadratic growth
in theright
handside,
but here the restriction on thisgrowth depends
on the directionsv(x),
x E Q. There is a similar restric-tion on the coefficients
a(XfJ, 1 ex, fJ n,
and theirdependence
on u. Forsimplicity,
wedispense
with the details here and assume thata(XfJ,
1 c a,pn,
areindependent
of u and that go = 0. Thatis,
we look at solu-tions u of
for all v E K =
{V
EWI,2 (S?9 RN): (v
-V) (x)
EI’(x)
for x E E and v - cp EWOI,2(,Q,
RN)},
wherea(XfJ E (J°,P(Q, RN)
and the functions g(X, 1 tX n, and g are as before.THEOREM 4. Let E
satisfy Property
E.I f
u EcO’’’(Q, RN),
0 y/-l,
isa solution
o f (6)
then u EcO’P(E, RN).
4. -
Auxiliary
lemmata.When
mentioning Cauchy’s inequality
andYoung’s inequality
we meanthe
inequalities
and
respectively .
Here 8 >
01 sy t > 0
andI lp + I lp’- 1,
p > 1. We writeYoung’s inequality
in the form
which is more convenient for our purposes.
We will utilize the Green function G for the
operator E = - D,, (b’x" DO),
where
b’O
are Holder continuous in a ballB,
and its mollification G-e definedby
We list some useful
properties
of G and G(1 cf. Widman[14]
and[15].
LEMMA 1. For z E B and e > 0 small
enough G(1(., z)
Ewo2(B)
andsatisfies
and
for
Moreover,
sincebtX{J
are Hölder continuous G will beregular
and we have theinequalities
and
Finally, if B2rc
BandBr
is the concentric ballof
radius r thenNext we
generalize
a lemma from Hildebrandt and Widman[6],
p.203,
which will be needed in the
proof
of Theorem 1.LEMMA 2. Let XoE Q acnd suppose that u E
OO,Y(Q, RN),
0 yp,
is ac so- lutiono f
theinequality ( 1 ) .
Thenfor
0 x2y’, y’ -
min(y,
1 -n/p),
for
all r, 0rl
dist(xo, ôQ).
Here the constant Cdepends
onparameters of
theproblem,
on x and on the OO,Y -normo f
u.PROOF. Fix xo E o and let r be such that
B2r(xo)
C Q.Moreover,
let,q c-
Co (B2r(xo))
be a function which satisfiesq (x)
= 1 forx E Br(xo), 0 ,q I and I Vq c/r.
IfB2r(xo)
(1E =1= ø
choose an x, inB2r(xo)
f1 .E andput
where 8 > 0 and
W,
0(! r/2,
is the mollification of the Green func- tion C for theoperator
L = -DfX(atXfJDp). Here,
the Hölder continuous coefficientsatXfJ === atXfJ(0153, u)
are extended to Hölder continuous functions de- fined in an open ball B DQ
such that theellipticity property
stillholds,
at least for a
slightly
smallerellipticity
constant Â. It isreadily
seen that vis a test function if c is
sufficiently
small.Now,
use this test function in(1),
cancel 8 and rearrange terms to obtainStandard calculation now
gives
For convenience of the reader we sketch the
proof
of(8). Expand
theleft hand side in
(7),
use theellipticity condition,
thecontinuity of u,
knownproperties
of Ge andCauchy’s inequality
to obtainTreating
the firstintegral
on theright
hand side in(7)
in a similar mannerwe find
From
(4)
it follows that there are constants C1 and e,depending
on theLoo-norm of u such that
Igo(x,
u,VU) I e, IVU 12 +
C2 for almost every x E Q.Using this,
the fact that the functions g,,,1 ex n,
arebounded,
condi-tion
(5)
and as before suitableproperties of u, V
and G" we see that the secondintegral
on theright
hand side in(7)
is bounded from aboveby
Summing
up andusing
Lemma 1 weget
and
together
with Fatou’s lemma thisgives
the estimate(8).
Next we consider the case when
B2r(xo) r1 E =
0. Here we can use thesame test
function,
where x1 now is anarbitrary point
inB2r(xo),
for instance xo . The sameprocedure
as aboveyields
theinequality (8)
also in this case.It is not hard to see that the statement in the lemma follows from
(8).
Infact,
and we are done.
5. - Proofs of the results.
PROOF OF THEOREM 1. We will use a rather
special
test function and in order to construct this we need some notations. Introduce the setsQT == {x E Q:
dist(0153, aQ)
>T}
and I ={x E Q’ n QT/2: uN(x)
=1pN(X)},
and let0 h
T/4.
Choose 1 > 0 such thatL1’ljJN (x) + l:> ð
> 0 for some (5independent
of z EQT/2’ hand e
=(e’, 0 ) .
Here e’ is a unit vector in RN-i. Put
where
Fix
XoE QT
and let 0rðT(xo)/4,
where6,(x,,)
= max(dist (xo, ô,Q) -
T,0).
As a test function we use
where e > 0
and’
_b(xo) q.
Here q, e, G2 are as in theproof
of Lemma 2.We will comment on this choice in an
appendix
at the end of the paper.This v is an admissible test vector since v E
’U)1,2(Q, RN)
and v - rp Ew’,’(Q, RY),
and since it is
possible
to show that(vN- v N) (X) > 0
for X C S2’ if 8 is suf-ficiently
small. Infact,
if x c- I c S2’ we haveand it follows that
A" uN + 1
> 0 in some n - 1-dimensionalneighbourhood .N(I )
cQ’ of I.Consequently,
This
implies
that for .r EN(l),
since
w’v(x)
0 andWN (x)
= 0 if xbelongs
toN(l).
For
x c- Q’BQ,12, v = u
sotrivially vN(0153):> ’lpN(0153). Finally,
for x E [J’r)
Qï/2""’N(I)
there is apositive
constant c such that(UN_1pN)(X):>C
andif E is chosen small
enough. By
use of this v in the variationalinequality (1)
we can show thatwith a constant c which is
independent
of e and h. Theproof
of(9)
ispost- poned
to somewhat later.From
(9)
and the definition of it,, we obtain thatAccording
to the definition ofuN, luv 1 c yields
thateither
from which the
following
estimate from below is deduced:However,
due to the fact that the constant ci isindependent of h, e
and x,we are allowed to reverse the direction of e in
(11),
whenceThis
together
with(10) gives
the statement in Theorem(1).
It remainsto prove
(9).
To do so weexploit
thetechnique
usedby
Hildebrandt-Wid-man in
[6],
pp. 205-207.First we assume
that g .
== 0.Now,
insert theproposed
test function in(1)
move theoperator J"h
to the other factor and delete c to arrive atBy
use of theidentity
we
split
theintegral
on the left hand side in(12)
into twoparts,
move the second one to theright
hand side andexploit
the conditions(2)-(4)
onaa#
and on the
right
hand side to obtainwhere Qz
==(s
e QT:L1uN(x) +
lO}
==(z e .QT: u § # 0 ).
By
use ofCauchy’s
andYoung’s inequalities
and the facts that inQz, lL1ul «l + l)(luL11 + 1)
and)V4j/ u
==)Vuj the right
hand side in(13)
canbe estimated
by
When
expanding
the left and side in(13)
theintegral
will show up. Since
and
we find
Now,
wesplit
theintegral
on the left hand side in(13)
into threeparts
thefirst of which is estimated from below
by
use of theellipticity condition,
the second of which is treated as indicated above and the third of which is treated in the
following
way.Utilizing Cauchy’s inequality
we obtainthat
Summarizing
andusing luj I lzl," u
andJUA I I + JUA 12
weget
Since r
ôT(xo)/4
we have6,,(x,,,) 26,(x)
for x EB2r(xo)
and thisimplies
thatfor all x E
B2r(xo).
Combine this and knowninequalities
for Gfl andVG,..
with
(14) whereupon
Furthermore,
if h is so small thatindependently
of e s-upp 77+
hec B2r(xo)
then
and, taking
Lemma 2 intoaccount,
Combine these
inequalities
and theinequality
with
(15)
it iseasily
seen thatNow,
let e -> 0 and inferIf we choose
xo E S2z
such thatand let f = min
(6,(x.)14, (2C.)-11")
we findwith a constant C
independent
of hand e. Thisimplies
thatwhich is the desired
inequality (9)
and thiscompletes
theproof
for thecase
where g =
0. If this is not the case the termwill show up. If we as before
exchange
the differenceoperator A_,
for thegradient operator V
and use condition(5)
it is not hard to see that thisintegral
is bounded from aboveby
Now
looking
at(14)
we see that it ispossible
to have also these terms in theright
hand side and stillget
the desiredinequality.
Thiscompletes
the
proof
of Theorem 1.By inspection
it isreadily
seen that Theorem 2 follows from thisproof
if we let e in the definition of
J"u
be anarbitrary
unit vector inRn,
instead of
being
of the form(e’, 0).
PROOF OF THEOREM 3. Fix
x,, c- B.
Let F be aneighbourhood
and 0be a transformation such that
Property
E holds. Observe that our solu-tion u of
(1)
is also a solution of theinequality
and
After an
elementary
calculation we see that ft = UO(J-l is a solution ofand
for all
vEKBR=== {VEW1,2(BR(O),RN): VN(y»ijJN(y)
fory E BR(o)
and v-ùE w,2(B.R(O), RN)},
where v === voO-1and 1jJ == 1pOO-l.
It is not hard to seethat the new coefficients
àfX’fJ’
and the newright
hand sidef satisfy
condi-tions like those in
(2)-(5).
Noteespecially
that the functionsgive
anelliptic operator
with anellipticity
constantdepending
on 2 andC( U),
the constantappearing
inProperty
E.Since 1p
is Holder continuous withexponent
u Theorem 1implies
that ft is Holder continuous with thesame
exponent
onBR
andconsequently u
is Holder continuous on U r) E.Now the collection of sets
U n E, x,c- B,
constitute an opencovering
of Eand since every
compact
subset of E is coveredby
a finite subcollection this proves the theorem.PROOF OF THEOREM 4. We will show that it is
possible
to introduce functions uand 1p
such that the constraint(u - 1p)(0153)
EF(x)
for x E B takesthe form
fiN(X) > fo(X)
for x E B. Infact,
let fi =Qu where Q
is anorthogonal
matrix with entries of class
C1,p
such thatfiN(x)
=y(x)u(x),
and where u is our solution of(6).
Then it is not hard to show that u is a solution of a similarproblem
where all the relevant conditions still hold and where the constraint isgiven by iN(X) > 1jJ(x)
=y(x) 1p(x)
for x E E. From Theorem 3we conclude that U E
OO,P(E, RN),
andconsequently
u EOO’P(E, RN),
which isthe statement in Theorem 4.
6. -
Appendix.
In
[6],
Hildebrandt and Widmanconsidered
a similarproblem
but with athick
obstacle,
that is when the constraint is of the form(u - 1p)(0153) E F
for all x E D.
There ip
was assumed to be three timescontinuously
differen-tiable and then a
simple argument
shows that one can supposethat ip
=0,
at least if the conditions on the
right
hand side are less restrictive thanthose in
[6].
In this case v - it+ EJ-h(JhUGQ’2)
is a test function if 8 issufficiently small,
since it can be written as a convex combination of ele- ments in K. If the obstaclefunction V)
isonly
Holder continuous theargument
above is not available and a natural choice in this case would have seemed to be v = u+ 8,Jl" (Jll(lt ’lp)GQ’2),
whichcertainly
is an ad-missible test function if 8 > 0 is small
enough. Unfortunately,
terms ofthe form
DOJ"yy
will show up and we have no uniform bound for these kind of terms.By introducing
the function u4 andusing
as a test vector we eliminate this
difficulty.
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