A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
F AUSTO A CANFORA S TEFANO M ORTOLA
Non semicontinuous quadratic integral functionals with continuous coefficients
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 25, n
o1-2 (1997), p. 1-9
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Non
Semicontinuous Quadratic Integral Functionals with Continuous Coefficients
FAUSTO ACANFORA - STEFANO MORTOLA
Vol. XXV (1997), pp. 1-9
1. - The
counterexample
The main aim of this note is to construct some
quadratic
forms of the typewith the coefficients
uniformly
continuous in an open bounded set Q ofR~,
such that the matrix(ai,j)
ispositive semidefinite,
this means:for every x E
Q,
forevery ),
ER",
in such a way thatF (u )
is not lower semicontinuous with respect of thetopology
of the uniform convergence.The
example
we will show has beenexplicitily suggested by
DeGiorgi, through
someconjectures
that will beproved
in this note.We look for a
quadratic
form of the typewhere a and b are functions defined in an open bounded set of
JR2, uniformly
continuous and we will suppose that a is
strictly positive.
The
example
will be a consequencce of thefollowing
two results:THEOREM. Given the square Q =
(-l, 1)2
in andf
E withf = (a, b)
wherea (x, y)
> 0for
every(x, y)
EQ, for
every u EC 1 (R 2)
thereexists
fn
E withfn = f uniformly
in S2 and there existsUn E with Un = u
uniformly
in S2 such that2
The
meaning
of the theorem is that theglobal
relaxed functional ofF(u)
=JQ(f. Du)2dxdy
isidentically 0,
whereby global
relaxed we mean thebiggest
functional non greater than
F,
lower semicontinuous with respect to bothf
and u.
LEMMA. Let Y be a metric space, let X be a
complete
metric space and let B : X x Y - R+ =[0,
be afunction
continuous with respect to thefirst
variable and such that
for
every(x, y)
E D x Y, with D dense in X, we haveB (xn , yn )
~ 0for
a suitable(xn, yn )
sequence that converge to(x, y).
Then, for
every(x, y)
E D x Y,for
every E > 0 there exists Z E X such thatd(x, z)
8 andB(z, yn )
~0 for
a suitable yn that converges toy.
PROOF OF THE LEMMA. Let us denote
by dx
the metric in X anddy
the metric in Y. We know that for every 8 > 0 there exists Xl ED,
YI E Y such that1
for our
hypothesis
on B.There exists a
point
x2 E D,Y2 E Y
such thatwhere the last
inequality
turns out from thecontinuity
of B with respect to xand thanks to
1
Proceeding
in the same way we can find two sequences(xn, Yn)
E D x Ysuch that:
for h
= 1, 2, ... , n
.. IThis
implies
that xn converges to apoint z
E X such thatdx(x, z)
8, due tox)
+ E and thecompleteness
of X.Moreover yn converges to
y and,
for thecontinuity
of B, we haveB (z,
for every h. This
gives
our thesis. 2hAs a consequence of the two
previous results,
we have the existence ofquadratic
functionals of thetype F(u) = Du)2dx
withf = (a, b) continuous,
such that F is not lower semicontinuous with respect to the uniform convergence in the variable u. Infact,
we takeB(f, u)
= withf
EC(Q, Jae2),
u and the theorem tells us that wemay
apply
the lemmaand, starting
from apair ( f, u)
E D xC(Q)
such that0,
we consider s > 0 so small thatFg (u )
> 0 for every g ER 2) with II f -
E. The lemma allows us to construct ourcounterexample.
PROOF OF THE THEOREM. Let us fix a, b E u E with compact
support
in Q and let us assume that~=(2013l,l)~cR~,
and that the functiona be
strictly positive
inSZ.
Now,
let us consider the sequence of functions uh solutions of thefollowing
system:
and let
CPh(X, y) = -(aux
+buy) a Duh 2,
where a > 0(a
will be chosenlater).
Let us
verify
these three facts:1 ) Øh
- 0uniformly
2)
uh -~ 0uniformly
-The fact
2)
is immediate: it isenough
to observethat u h (x , y ) ~
forthe maximum
principle,
consequence of the method ofcharacteristics,
as we will see later.For the
point 1),
let us observe that thefunction h y 2
has its maximum valueh+y
equal
to and so-
This
implies
that the sequenceOh
tends to 0uniformly.
Concerning
thepoint 3),
let usdevelop
the functionLet us observe that the second term is
always
0by construction,
the third one tends to 0 for thepoint 1),
and so we haveonly
to check that(q5h - Duh) - -( f ~ Du)
inL2(S2).
Let us consider the
following Cauchy problem:
and let us denote
by Tx (yo)
the solution at thepoint
x.The function yo -
Tx (yo)
is a diffeomorfism and let us denoteby Tx
1 itsinverse.
Let us observe
that,
if y =Tx (yo) :
because u h are solutions of
(2).
This means that
uh (x, Tx (yo))
is constant with respect to x and so its value isUh (0, yo) = 1 sin(hyo).
4
From this it turns out that
In
particular
we have that u h ~ 0uniformly
inQ,
aspreviously
stated.Moreover we have:
Let us
study
the behaviour of the sequence of functionswhere denotes the
gradient
of the function(x, y)
- with respect to the two variables x, y.To prove that this sequence tends to 0 in
L2 (S2),
we may break Q in tworegions,
and in the functioncos2(hTx-l(y))
isbigger
thana fixed s > 0 and so it is small if the exponent of h
(i.e.
2 -3a)
ispositive,
this
happens
when a2/3.
For theintegral corresponding
toQ2,,,h,
whereE, let us observe that its measure is small when 8 is
small, uniformly
with respect toh,
becauseTx
is a diffeomorfism and the sequence isequibounded (due
to the factthat IDTx-I(Y)1 [
isbigger
than0),
thisimplies
that the
integral
is small when h islarge.
REMARK. Even when the functional
F(u)
=Du)2
is not lower semi- continuous in the uniformtopology,
it ispossible always
to prove thatthe
setof functions where the functional assume its minimum
value,
this means theset of the solutions of the first order linear
equation:
is closed in
C 1 (Q)
with respect to the uniformtopology.
Let us write the
simple proof
of thisfact,
due to P.Majer.
If a sequence of functions un solutions of
(3)
convergesuniformly
to a functionu E
C
1then,
if we fix apoint
xo in Q and we fix a solutionx (t )
of theCauchy problem
x’ =f (x), x (o)
= xo we have that un(x (t))
is constant in theneighborood
of 0 and thisimplies
that alsou (x (t))
is constant and so u is asolution of
(3):
this means that the space of solutions is closed with respect to the uniformtopology. Actually
thisproof
shows that the space of solutions is closed also withrespect
topointwise
convergence.As F.
Cipriani pointed
out, this kind ofproof
allows to prove that also for the functionalsthe set of minimum
points
is closed when ai, j arecontinuous,
because it ispossible
to consider the matrix(bi,j)
square root of(ai,j),
i.e. such that ai, j =b i,h b h, j
withb - i,h
=bh,i
continuous and(bi,j)
semidefinitepositive.
Inthis way we have:
where bh
are the vector functions whosecomponents
arebi,h.
So we have that u is a minimum
point
for(4)
if andonly
if u is a solutionof
(b h- Du)
= 0 for every h =1, 2,..., n
and so the set of solutions is closed because it is the intersection of n closed sets.2. - Cases of lower
semicontinuity
In this section we examine the main cases in which one has lower semi-
continuity.
Our functionalsF(u) =
are considered foru E
A)
The coercive caseThe most classical case is when
(ai,j)
ispositive definite,
i.e. when there exists c > 0 such thatfor every x E
Q,
forevery ),
E R.In this case if uh E converges to u E
uniformly
and it is suchthat
F (uh )
isbounded,
it turns out that uh is also bounded in and soit
weakly
converges to u inHI.
The functional F is continuous withrespect
to the strong
topology
ofH
1 and so,being
a convexfunctional,
it isweakly
lower semicontinuous.
Actually
in this case thecontinuity
of the coefficients(ai,j)
is notimportant:
even in the case of measurable and bounded coefficients the
semicontinuity
holds.
6
B)
The unidimensional case and theisotropic
case.Let us suppose that
(ai,j)
isisotropic,
i.e. of the forma(x)I
where I denotesthe
identity
matrix anda(x)
is a continuous function inS2,
nonnegative.
In this case the functional is
pointwise
the least upper bound of the functionalswhere
0,
={x
E1 a(x)
>6~}.
The functionals
F,
are lower semicontinuous in becausethey
arecoercive and it is immediate to observe that
they
are lower semicontinuous also inLet us observe that the unidimensional case is
always
iso-tropic
and so thecontinuity
of a isenough
to prove the lowersemicontinuity
of the functional.
If a is not continuous but
only
in the classL°°,
we can have lack of lowersemicontinuity,
as one canverify
even in the unidimensional case: let A bea dense open set in
(0, 1)
with measure k 1 and let a the function with value 0 in A and 1 in[o, 1 ] B
A. In this case,taking
the functionu (x )
= x, we haveF(u)
= = 1 - h > 0 but it is easy toapproximate
u
uniformly
with functions u h such thatu h
= 0 in[0, 1] B
A: in this way we have = 0 even ifF(u)
> 0.(This example
is consideredby
Carboneand Sbordone in
[2]).
C)
Thediagonal
caseA
generalisation
of theisotropic
case is that ofdiagonal
matrices:where ai
EC(Q)
0 for i =1, 2...,
n.To prove the lower
semicontinuity
of F it isenough
to consider the case ofjust
asingle
term, and inparticular:
If uh ~ u in
L’ (0),
for every x =(~1~2...~) ~ ~
let us consider~(~2~3~...~)={.y~R ! I (y~2,.~...~)=~}cR.
For the result of the unidimensional case we have
If we
integrate
withrespect
to(X2,
...,xn )
andusing
Fatou lemma we findAnother way to show the lower
semicontinuity
for thediagonal
case is thefollowing.
Weapproximate FI
with thefaniily
of functionalswhere
Q,
={x E Q I > ~ } .
For every E >0, F£
are lower semicontin-uous because in
S2E
the function a 1 is the limit of anincreasing
sequence ofC
1positive
functions: for the case ofC
1 coefficients the lowersemicontinuity
will be
proved
in thefollowing paragraph.
D)
Theregular
caseWhen the coefficients ai,j are
C
1 functions(or
even localLipshitzian)
wecan prove that the functionals are lower semicontinuous even with
respect
to the L1 (S2) topology.
In factso it is
enough
to proveL 1-lower semicontinuity
for the functionalswhere 1/1
is aC
1 function with compactsupport
in Q and with values in[0, 1].
For these functionals we have:
Let us note that the functionals
are linear and continuous with respect to the
LI_topology
and so also thefunctional G is
L I-lower
semicontinuous.8
3. - Some open
questions
In this last section we formulate some natural
questions
whose solutions sofar we
ignore.
1)
Is itpossible
to find anintegral quadratic
functional with the coefficients ai,j continuous and notidentically
0 such that the lower semicontinuousenvelope
is the functional
identically
0?2)
It is known that the lower semicontinuousenvelope
of anintegral
qua- dratic functional is still a functional of the sametype
and it is also known aformula for the coefficients of the relaxed
functional,
asproved
in[5]
and in[8].
It is not clear
if, starting
from continuouscoefficients,
the relaxed functional has still continuous coefficients.3)
The set of functions u that are criticalpoints
for our functionalF(u), namely
the functions u such that for everyregular function 0
with compact support in Q we haveis closed in C
1 (Q)
withrespect
to the uniformtopology?
4)
Thefollowing problem
has beensuggested by
DeGiorgi:
let(ai,j)
be asemidefinite
positive
matrix withregular
coefficients and let(bi,j)
be anothermatrix with continuous coefficients such that
for every x E
Q,
for every h ER",
for some C > 1. Then thequadratic integral
functionalcorresponding
to the matrix(bi,j)
is lower semicontinuous with respect to the uniform convergence.REFERENCES
[1] G. BUTTAZZO, "Semicontinuity, relaxation and integral representation in the calculus of variations", Longman Scientific & Technical, 1989.
[2] L. CARBONE - C. SBORDONE, Some properties of 0393-limits of integrals functionals, Ann. Mat.
Pura Appl. 122 (1979), 1-60.
[3] G. DAL MASO, "An Introduction to 0393-convergence", Birkhäuser, 1993.
[4] G. DAL MASO, Integral representation on BV(Q) of r-limits of variational integrals, Manuscripta Math. 30 (1980), 387-413.
[5] N. FUSCO - G. MOSCARIELLO, L2-lower semicontinuity of functionals of quadratic type, Ann. Mat. Pura Appl. 129 (1981), 305-326.
[6] C. Y. PAUC, "La Méthode Métrique en Calcul des Variations", Hermann, Paris, 1941.
[7] J. SERRIN, On the definition and properties of certain variational integrals, Trans. Amer.
Math. Soc. 101 (1961), 139-167.
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