A NNALES DE L ’I. H. P., SECTION C
A RRIGO C ELLINA
Minimizing a functional depending on ∇u and on u
Annales de l’I. H. P., section C, tome 14, n
o3 (1997), p. 339-352
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Minimizing
afunctional depending
on~u and
on uArrigo
CELLINAS.I.S.S.A./I.S.A.S., via Beirut 2/4, 34013 Trieste (I)
Ann. Inst. Henri Poincaré,
Vol. 14, n° 3, 1997, p. 339-352 Analyse non linéaire
ABSTRACT. - We prove existence of solutions for a class of minimum
problems
of the Calculus ofVariations,
where theintegrand depends
bothon and on ui.
Key words: Minimization, gradient.
RESUME. - On donne un theoreme d’existence de solutions pour un
probleme
du calcul des variations ou la fonctionnelledepend
de u et deINTRODUCTION
The purpose of the
present
paper is to contribute to thetheory
ofexistence of solutions to minimum
problems
of the Calculus of Variations when there is noassumptions
ofconvexity
withrespect
to the variablegradient.
Whenconvexity
is notassumed,
to prove existence of a solutionone cannot
rely
onpassing
to the limitalong minimizing
sequences, but in most cases one has toactually provide
a constructionyielding
the solution.Several
examples
of thisapproach
exist:[1], [3], [4], [5], [8], [13].
Theconstructions
appearing
in these papers are, so to say,local,
in the sense thatthe
problem
is solvedlocally
and then the construction is extended to the fullregion by
means ofcovering arguments.
These constructions are usedA.M.S. Classification: 49 A 21, 49 B 22.
Annales de l’Institut Henri Poincaré - Analyse non linéaire - 0294-1449 Vol. 14/97/03/$ 7.00/@ Gauthier-Villars
to solve non-convex
problems
both in the scalar and in the vector cases of the Calculus of Variations.However,
when the functional to be minimizeddepends,
besides on thegradient
of the function u, on the function uitself,
a local constructions in
general
cannotyield
the solution. The purpose of this paper is toprovide
one non-local construction for a class ofproblems depending
both on ~u and on u.In a paper
by
B.Kawohl,
J. Stara and G. Wittum[ 11 ],
a non convexfunctional, arising
in aproblem
ofshape optimization,
isinvestigated, partially by
numerical methods. The functional to be minimized is of theform _
under the condition u = 0 at and H is a two dimensional square. The function h is non convex, the infimum of two
parabolas (the investigation
ofthis functional has
actually
alonger history,
see[ 10]
and[ 14]).
Numericalevidence
suggests
that(for
theparameters
considered in the numericalevaluations)
solutions to the minimumproblem
do not exist. Noticethat,
in case 0 was adisk,
solutions to the aboveproblem
would exist and beunique, essentially
without anyassumptions
onh, except
some lowersemicontinuity
andgrowth
condition[2], [15], [16],
conditions satisfiedby
the function h in
[ 11 ] .
In the case of adisk,
infact,
one can show theexistence of a
radially symmetric
solution to the convexifiedproblem
andmodify
this solution to obtain a(radially symmetric)
solution to theoriginal
non convex
problem.
The
problem
we consider is the minimizationproblem
statedabove,
where h
belongs
to a class of(not necessarily convex) functions,
and H is anybounded,
open, convex subset ofR2
with apiecewise
smoothboundary (so
as to include a square as aspecial case).
Weprovide
a result that states that when the width of the set H is small(depending
on aproperty
of the functionh),
a solution to theproblem
exists.Moreover,
this estimate on the width cannot beimproved,
in a sense to be madeprecise,
as it is shownby
an
example.
Anothercondition, connecting
moreaccurately
theproperties
of h and of
H,
is alsopresented. Again,
this condition cannot beimproved.
Our
results,
assuch, give
no informations for themap h
consideredin
[11]. However,
were the lowerparabola, appearing
in the definition ofh,
bereplaced by
a half line(a degenerate parabola;
we obtain the functional consideredby
Kohn andStrang
in[12]),
our result wouldguarantee
existence of solutions on squares whose side is oflength
notlarger
than twice theangular
coefficient of the convexification of h. It is conceivablethat,
under similar conditions on thelength
of the side of the square, a solution to the trueproblem
should exist.Annales de l’Institut Henri Poincaré - Analyse non linéaire
MINIMIZING A FUNCTIONAL DEPENDING ON VM AND ON 2G 341 NOTATIONS AND BASIC
ASSUMPTIONS
we mean the euclidean norm of x. When B is a
matrix, B ~
denotes the determinant of B. The
complement
of a set A is denotedby C(A)
and the interior of Aby int(A). By II(x)
we mean the set of nearestprojections
on ~SZ from x inH,
i.e.{y
E =d(x , Throughout
this paper we make thefollowing assumptions
on H.The set H is convex, open and bounded. Moreover there exist: M
points Oi , i
=1, ... , M,
in such that H admits aunique (inward)
normal nat M functions each defined on a closed interval
Ii containing
theorigin
in itsinterior,
such that:a)
withrespect
to thepair
of coordinate axis withorigin
in0
i anddirections defined
by
thetangent
and the(inward)
normal atOi,
thepoints (03BE, 03C6i(03BE)) belong
to ~03A9 forevery (
inIi; for 03BE
inint (Ii), points ((, 03BE) belong
to 0for (
>~.; (~) sufficiently
small.b)
the functions~,
i are of classC2
on an open setcontaining Ii;
as aconsequence of
a)
and of theregularity
of we have that~.L (0)
= 0and
~i (0)
= 0.c) Every point
of 0fi isrepresented
in the formabove;
moreover therepresentation
isunique except
for thefinitely
manypoints
ofimages
of the
endpoints
of the intervalsI; .
We recall that the radius of curvature R at a
point (~, ~.i (~) )
of theboundary
ofSZ, ~
inimt (I; ),
is R = when~i’ (~)
= 0 andwhen
~) ~
0.The width
Ws2
of 03A9 is defined to beWo
= e03A9}.
Notice that the word width has been used in convex
anaysis
with a somewhatdifferent
meaning.
We will consider a class of
maps h satisfying
thefollowing assumption:
Assumption
AThe map :
[0, (0)
-~[0, oo]
is anon-negative
lower semicontinuous extended valued function with minimum value 0.Moreover, sup~r
> 0 :==
~~
is finite.Definitions of p and A
Set p
=sup{r
> 0 : h(r) =0},
and call A the set ofsupporting
linearfunctions at p: A =
~~x, : > ~,(s - p),
forevery s 2: 0~.
We have that0 E A. Set A in
[0, oo]
to be : A = sup A.Vol. 14. n° 3-1997.
Whenever A oo, A is in A. Whenever h is
smooth,
A = 0. In the casethe
map h
is convex, we have thatMAIN RESULTS
It is our purpose to consider the
following problem
Problem
P):
Minimizeon u E
LEMMA l. - Let y E E Q. Then ~03A9 is
differentiable
at y.Proof. -
The open disk centered at xhaving radius p contains
no
points
of Since x is interior toH,
there is a ball about x contained in H. Each half lineissuing
frompoints
in this ball in the directionof y -
xmeets the
boundary
of H inexactly
onepoint. By moving
theorigin
of thecoordinates
to y
and the axis oriented as the normalto x - y
and as x - y, there exists an open interval Icontaining
0 such that theboundary
of H isrepresented by ( = ~(~) where §
is a convex function defined on I. Theconvex
function §
has both a left derivative and aright
derivative~+
at0. Since there are no
points
of in the disk centered at(0, p)
and radiusp we must have
~
> 0 and~+
0while, being §
a convexfunction,
’ ~+,
so that~’ (0)
exists andequals
0.DEFINITION. - Let x E ~03A9 be a
point
ofdifferentiability
of SetI(x)
to beThe set inside
parenthesis always
contains at least A = 0. About theproperties
of.~,
we have:LEMMA 2. -
a) .~(~) W~;
b) For y
in 03A9 and x in03A0(y), d(y,
c)
Let ~03A9 beof
classC2
in aneighborhood of
x.Then; l(x) ~ R(x);
moreover;
d)
whenR(x),
there exists z Ec~SZ, z ~ ~
x, such thatd(x
+~n(x), z) _ .~(x).
Annales de I’Institut Henri Poincaré - Analyse non linéaire
343
MINIMIZING A FUNCTIONAL DEPENDING ON AND ON U
Proof. - a)
Obvious when P = 0.For A1’(:r); ~ = C(S2)),
so that
P(x)
= +WS2;
b)
Forevery A
is theunique point
in dS2 nearest tox + hence the supremum of such A is >
d(y, C(52));
c)
There isnothing
to prove when = oo. AssumeR(x)
oc and=
R(x)
+ r~, 7l >0,
and choose ~ =R(x)
+2 .
The open disk centered at x +An(x)
of radius A does not containpoints
ofC(S2).
Move theorigin
of the coordinates to x and the axis in the directions of the
tangent
and normal in x, so that ~03A9 isrepresented,
in aneighborhood
N of(0, 0), as ( = ~(~),
with~(0)
= 0 and~’(0)
= 0.Hence,
for~~~ sufficiently small,
we havec~(~)
=2~"(~)~z with~~~
In aneighborhood
of0,
~)
>R+~r/4°
whileand,
for~~~ small,
1-(R+r~)c~"(~) -~-o(~)) ~ 0,
so thatd((~, ~{~)), (0, R~- r~))
R-~- r~,
a contradiction to the definition off(x)
for x = 0.d)
Let = andagain
consider theorigin
of the coordinatesto be in x. From the definition
of f’)
there exist: En1 0,
zn E0,
such thatd(zn, 0
+(.~
+ .~ + For nlarge,
all the disks centered at 0 +(.~
+en)n(0)
with radius f + En are contained in the disk with center 0 +(.~ -~ 2 )n{0)
and radius f +~.
In aneighborhood
of0,
theequation
of this circle isIn a
neighborhood
of(0,0)
thepoints
of ~03A9 arerepresented by
Hence there exists a
neighborhood
N of(0,0)
such thatpoints
zn =(çn, ~.,~)
cannot be in
N,
since otherwise we would haveThen,
asubsequence
of the zn converges to apoint z
E ~03A9 nC(N),
and we have
Vol. 14, n° 3-1997.
LEMMA 3. - For each
i, the function ~.i (~) )
is continuous onIi.
Proof - (i) Continuity
onint (Ii).
a)
It cannothappen
that there exists(xn)n
with xn in ~S~ and x*such that --~ .~ _
.~(x~) -
r~. If this is the case, infact,
threesituations can
happen,
in view of Lemma 2:1)
on asubsequence,
_
R(xn); 2)
there arepoints
z~ in xn butd(xn, zr,, )
--~ 0and
d(zn, xn
+ and3)
thepoints
zn are such thatd(xn, zn)
are bounded away from zero. In case1),
sinceR(x*),
then
R(x~) ~(x~)
and this contradicts Lemma 2c).
Consider case
2).
In aneighborhood
N of x~,R(x)
>R(x~) - 2
when
R ( x ~ )
isfinite,
orlarger
than the diameter of 0 whenR ( x ~ )
= oo .Since the open disk with center xn + and radius has
empty
intersection with c~~ and has the twopoints
zn and x.n on itsboundary,
at somepoint
on theboundary
intermediate between zn and xn, the radius is notlarger
than When both zn and xn are inN,
wehave
.~(x*) - 2 .R(x~) - 2
a contradiction.Case
3)
cannothappen:
the sequence(zn)n
would converge to z* inx*
having
distance R from x *-+-.~n (~ * ) .
Thepoint
z* would havedistance less than
f(x*)
from x*b)
It cannothappen
either that there exists(xn)n
with xn in ~03A9 andx* such that -~ .~ _
.~(x*) ~
r~. Let theorigin
be in x* and theaxis oriented as the
tangent
and normal. The distancedx’
from x’ _(~’, (’)
in ~03A9 to 0 +
n(0)l
is(03BE’)2
+(1 - (’)2.
When1(0)
=R(0), locally
~03A9is
represented by
so that
for ç’
small. For nlarge, d(xn
+x’)
.~ so that forlarge
~zone would have
d(xn
+x’) .~(~n). Finally,
in the case thereexists such that
.~(0)
=d(0
+n(0).~(0), x’), again
one would haved~~
r~ +.~(0) = 1
and the same conclusion would follow.(ii) Continuity
at theboundary points
ofIi .
When ~
is aboundary point
ofIi,
either the normal at ~SZ exists at(~, (~) )
or it does not. In the first case £ is defined and the considerations above
apply
as one-sided considerations. In the second case it is easy to see thatAnnales de l’Institut Henri Poincaré - Analyse non linéaire
345
MINIMIZING A FUNCTIONAL DEPENDING ON AND ON U
for ~
close to theboundary
ofIi, .~(~)
tends to zero. In this case,by defining £
to be zero at theseboundary points,
one achieves theproof
ofthe
continuity
onIi-
DEFINITION AND PROPERTIES OF THE MAPS gi AND With
respect
to thesystem
of coordinates centered atOi,
consider thetransformation gi
thatassociates to the
pair (~, l) : ~
inint(Ii)
and 0 l.~( (~, ~i (~) ) ) ,
thevector of
components ~2 ) given by
Here
-03C6’i(03BE) 1+03C6’i2
and are thecomponents
of the normal n at(03BE, 03C6i(03BE)).
We will set
f(ç)
to bel((03BE, 03C6i(03BE)).
Set6’,
to be theimage
of the map gi on its domain andS°
to be theimage of ~ (~, l ) : ~
E l.~(~) ~ . On S0i
the map gi isinvertible,
since(03BE, 03C6i(03BE))
is theunique point
innearest to
Z).
Call~2
the map fromS°
to R definedby
the firstcomponent
ofg-l,
sothat, for ~
inint(Ii),
For the derivatives of
fi, setting (gi )
and(g2 )
to be the twocomponents
of gi, we have thesystem
so that
Hence the norm of the
gradient of f
2 isVol. 14, n ° 3-1997.
computed
atgi 1 (~l ., ~2 ) . Computing ]
we obtainso
that,
sincel(03BE) ~ R(03BE), |~gi|
is~
0 onint(Ii)
x{0 ~
ll(03BE)}.
Sinceour
assumptions imply
that and1
+( ~i ) 2
areuniformly
bounded onIi,
the map gi islipschitzean. Moreover, by setting Sf
to be theimage
of
int(Ii) {0 ~
ll(03BE) - ~},
one has that the mapf
2 isLipschitzean
on We will need both
properties
in what follows.About the sets
Sf
andS,i
we have thefollowing
result.LEMMA 4. - We have: = 0. In
particular,
= 0.
Proof -
Fix y E 0 and let x be inII(y).
At x,by
Lemma1,
the normaln(x) exists,
and y can be written as y = x +n(x)d(y, CH)
andby b)
ofLemma
2, C ( SZ ) ) f(x).
Hence y may fail to be in~S~i
when eitherthe
points
x in arerepresented
as(03BE, 03C6i(03BE)) with 03BE
in~Ii
or when thereis some x in
II(y) represented by ~
inint(Li)
butC(SZ)) .~(~~) -
c.Points of the first
type
are contained in the union offinitely
manysegments (on
the normal linesthrough x),
a set of measure zero. About the otherpoints,
noticethat,
in the space~2
xR,
theset ~ (~, l ) : .~(~) - ~
l~i (~) ~
has measure c Its
image by
thelipschitzean
map g2 is of measurethat can be made
arbitrarily
small in Hby decreasing
c. The union of theseimages
contains allpoints
of the secondtype.
The
following
is our existence Theorem. A moreprecise
condition isexpressed
in Theorem 2.THEOREM 1. - Let 0 be an open,
bounded,
convex subsetof 5~2
withpiecewise
smoothboundary, having
widthWq.
Let hsatis, fy Assumption A)
and let p and A bedefined
as above. WhenW~ A,
thefunction
_ -p solution to the minimization
problem
P.Proof. - a)
The map ;; -~d(x,
is differentiable a.e. and itsgradient
is
(a.e.) -n(II(x)) (see [9],
p.354) .
Inparticular,
issingle
valuedfor a.e. x in Q. The map u
is -pd(x, C(SZ))
so that a.e.,_ -pn(g),
y the
unique point
inII(x).
In the case p >0,
jj =-n(y),
while for03C1 =
0 we set to be-n(y) by
definition.Annales de l’Institut Henri Poincaré - Analyse non linéaire
347
MINIMIZING A FUNCTIONAL DEPENDING ON AND ON u
Let c~ be a function in and such that: 0
a ( x )
A for a.e. x in S2when A oo, and 0 c~ when A = oo .
Then,
for any vector v, when p > 0and,
for p = 0Hence,
for every p and for every function r~ inWo ~ 1 ( SZ ) ,
we haveWe are
planning
to show that there exists a function 03B1 in with theproperties
statedabove,
and such that for everyfunction ~
inCo (0),
Finding
this function c~,then,
amounts toproving
that the function u is asolution to the minimization
problem
P. Infact, by approximating
a functionr~ in
by
standardmollifiers,
one sees that the aboveequation
mustactually
be true for everyfunction ~
in(SZ),
so that u solves P.b)
Thefunction |~gi (03BE, s) | is uniformly
bounded onIi
x{0 ~
ll(03BE)}.
Consider the function defined
by
so that
Gi
> 0 and.~(~) )
= 0. Since .~ is a continuous function ofç
EIi, G
i is a continuous function of its variables(~, l).
Set the function~
to beVol. 14, n° 3-1997.
so that:
(i) l)
is continuousfor ~
in and 0 G LP(~), ~3;(~, ~(~))
= 0and
,(3i(~; 0) ((~~.
These last assertions followby actually computing
themap
~3~(~, l) using
theexpression
found for~:
one obtainsFrom the above
expression
one can see that the derivative withrespect
to l ofl)
exists and a smallcomputation
shows that it isnegative: ,~L
achieves its maximum at l = 0. We have: =
0; l)
G~(~) -l
Iand
f3; ~~; «) _ f(ç)
when~/~"(~)
= 0 and!y~~; O) _ ~~~) - ~~~~))2~2~~~)
when
%"(~) ~
0. In either case,l3,(~,.0) G P(~).
(ii)
Forevery ~ and l,
we haveG~(~, l)
= 0.c) Having defined /3,(~),
define cx; onS’° by setting
The map cxL is continuous on the open
(relative
toQ)
setO? By
ourprevious claim,
the setSZ~
UOf
has measure zero. The map c~, defined to be a.L onO.°
and 0elsewhere,
ismeasurable,
nonnegative
anduniformly
bounded.d) Let ~
be any function in and let uscompute
Since the
integrand
is inby
our claim onOf
we have alsoOn
f
i is aLipschitzean
map with values inR; by
the coarea formula([7],
p.117)
we havewhere H is the one-dimensional Hausdorff measure.
Annales de l’Institut Henri Poincaré - Analyse non linéaire
349
MINIMIZING A FUNCTIONAL DEPENDING ON AND ON v,
The set
f-1i (03BE)
is thesegment
describedby
for 0
P(~).
On it the Hausdorff measure coincides with theLebesgue
measure. We have:
By integrating by parts,
since!V~, ~ -- --1Gi(~; l)
and-~yr~((~. c~y(~))-~
ln(~)) _ (n,
we haveSince
~|~03A9
=0,
Then
where n =
n(03BE),|~gi| = |~gi (03BE, l)| and
the functions and cxiappearing
inside the
integral
arecomputed
+l n(~) ) :
0 l.~(~) - ~~ .
At thesepoints
the function c~iequals l ) .
Vol. 14, n° 3-1997.
By point i)
ofb) above, hence, by
Lemma2, a), and, by
theassumption
of theTheorem,
A.Moreover,
sinceby point ii)
of
b)
we have+ 0,
we obtainHence
Each
integrand
is a continuous functionuniformly converging
to 0 so thatThe map c~ satisfies all the
requirements
ofa).
Thiscompletes
theproof.
EXAMPLE 1. - Let h be the indicator function of any closed and bounded set. Then A = oo and u is a solution on any bounded convex set
H,
witha
piecewise
smoothboundary.
Next
example
shows that the conditionexpressed
in terms of cannotbe
improved.
EXAMPLE 2. - Consider the function h defined
by h(r)
= r, for 0 r 1and
h ( r )
= oc for r > 1. Then p = 0 and A = 1. Henceproblem
Pconsists in
minimizing
a convex coercitive functional on and admitsa solution v.
By
ourprevious theorem, u -
0 is a solution whenever 1. Let us showthat, given
anypositive
c, there are convex sets 0 with = 1 + E such that a solution must have itsgradient
different fromzero on a set of
positive
measure.Consider the
rectangle
with sides oflength 2(1 ~- ~ )
and2 ( 1--~ ~ ) +A
and a second concentric
rectangle
with sides 2 and 2 + A. For SZ =W~
= 1 + c,independent
of A. The value of the functionalcomputed along
the map Uo - 0 is zero. Consider where u,l isnegative
withgradient
in norm =1,
andorthogonal
to thesides,
on thestrip
difference of therectangle
and therectangle Ro,A.
andgradient
0 onRo,A. Computing
the functionalalong
ui we have the value(8 +
+(4~2 ) - f(4
+ +(4
+~)~2 ~ (4/3)~3l
=4~ -11~2 - (4/3)~3
and,
for Alarge,
this value isnegative.
Hence Uo is not a solution.Annales de l’Institut Henri Poincaré - Analyse non linéaire
351
MINIMIZING A FUNCTIONAL DEPENDING ON ~u AND ON 2L
Remark. - The condition
appearing
in thepreceding
Theorem isexpressed
in terms of a
quantity easily computed.
For thevalidity
of theresult, however,
thefollowing
condition isactually sufficient,
as one can see fromproperty (ii)
ofb)
in the Proof of Theorem 1:THEOREM 2. - Under the same conditions on hand SZ assume
that, for
a. e.x in
.~(x)
A whenR(x)
= oo, and.~(x) - (.~(x))2/21~(x)
A whenR(x)
oo. Thenproblem
P admits the solutionu(x) _ -
pd(x,
Next
Example
shows that this second condition cannot beimproved.
EXAMPLE 3. - Consider the case where H is a disk of radius R.
Then,
for every x in ~03A9 we have =
R,
andl(x) - (l(x))2/2R(x)
=R/2.
So the condition becomes:
.R/2
A. Let h be as inExample
2 andset H to be
B2+2~,
a disk of radius 2 + 2~. SinceR/2
= 1 + c and A =1,
the above condition is violated.Again
for 0 the value of the functional is 0. Consider a concentric diskB2
of radius 2 and let ~cl be such that thegradient
is in norm 1 on the annulus from r = 2 + 6- tor = 2 and 0 otherwise.
Computing
the value of the functional one obtains-~7r8[3c
+3c2
+c3]
+47r[2c
+~3~,
anegative
number.Hence Uo
is nota solution.
It is obvious that the
examples
above refer to our function u notbeing
asolution. That other solutions
might
exist is not aproblem easily
solvable.REFERENCES
[1] A. CELLINA, On minima of a functional of the gradient: sufficient conditions, Nonlinear
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(Manuscript received May 15, 1995;
Revised October 6, 1995. )
Annales de l’Institut Henri Poincaré - Analyse non linéaire