Introduction to Shape optimization

Noureddine Igbida
1
1
Institut de recherche XLIM, UMR-CNRS 6172, Faculte des Sciences et Techniques, Universite de Limoges 123, Avenue Albert Thomas
87060 Limoges, France. Email : noureddine.igbida@unilim.fr
Topologies for Families of Sets
A shape optimization problem
min
K
J()
where K is a set of subsets of IR
N
. A rst mathematical approach for this problem consists of :
To verify that m = inf
K
J() is nite
to show that this inf is reached ; that is m is a minimum.
This we say that we have an existence theorem.
A systematic method consist in
taking a minimizing sequence
n
, in a sense to be precise (Topology on families of
domains)
to show that J is lower semi-continuous on K, with respect to the previous topology
then
J(

) liminf
n
J(
n
) and J(

) = inf
K
J().
This shows naturally the necessity of a topology on set of subsets of IR
N
. Moreover, the
convergence of
n
needs a compactness results for this topology. And, nally the fact that
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J(

) m, corresponds to the lower semi-continuity of J with respect to this topology. So,

n
J() liminf
n
J(
n
)
Each sequence Jbounded is sequentially compact ; i.e.
sup
n
|J(
n
)| < (
n
k
)
kIN
, K ;
n
k
.
Then, if J is lower bounded, then there exists

K, such that
J(

) = min
K
J().
Proof : Just rewrite the proceeding arguments.
On see that the two requirements of the proposition are conicting, since compactness is
subject to less ne topologies (that contains less open domains), as to the continuity or lower
semi-continuity of J is subject to more ne topology. The art of the analyst will be to navigate
between these two requirements of the problem.
Here we give the main lines concerning three topologies on families of subsets of IR
N
. The rst
one (convergence of characteristic functions) concerns all measurable sets of IR
N
. The others
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concerns families of opens sets of IR
N
. However, respect to some situations one can be brought
to work with others topologies more adapted.
Contents
1. Convergence with respect to characteristic functions . . . . . . . . . . . . . . . 4
2. Convergence of open sets in the sense of Hausdor . . . . . . . . . . . . . . . . 7
3. Convergence in the sense of compact sets . . . . . . . . . . . . . . . . . . . . . 16
4. Other notions of convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1. Convergence with respect to characteristic functions
A rst idea consists in using the bijection between the measurable sets (with respect to Lebesgue)
of IR
N
and the characteristic function
E
(which is equal to 1 on E and 0 outside). It is clear
that
E
L

(IR
N
), and
E
L
p
(IR
N
), for any 1 p , whenever E is bounded (or
more generally |E| < ). Then, the idea is to use the standard topologies on characteristic
functions.
Denition 0.1 Let (
n
)
nIN
and be measurable sets of IR
N
. We say that
n
converges
to in the sense of characteristics if and only if

in L
p
loc
(IR
N
), for any p [1, ). (0.1)
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For the convergence in the sense of characteristic function, the compactness property is
immediate.
Theorem 0.1 If (E
n
)
nIN
is a sequence of measurable sets of IR
N
, then there exists
L
(IR
N
) and a subsequence
E
n
such that

E
n
in L

(IR
N
) weak

.
That is
lim
n

IR
N
E
n
=

IR
N for any L
1
(IR
N
).
Moreover, we have
0 1 a.e. in IR
N
.
if there exists a measurable set such that =
E
, a.e. in IR
N
, then

E
n

E
in L
p
loc
(IR
N
), for any 1 p < .
Lemma 0.1 (Banach-Alaogo-Bourbakis Theorem) The unit closed ball of the dual
space X

of a Banach space X is compact for the weak

topology (X

, X). If moreover,
X is separable then it is sequentially compact.
Lemma 0.2 Let E
n
and E be measurable sets of IR
N
. If

E
n

E
in L

(IR
N
) weak

,
then

E
n

E
in L
p
loc
(IR
N
), for any 1 p < . and a.e. in IR
N
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Proof : It is not dicult to see that, for any 1 p < and B
R
= B(0, R),

B
R
|
E
n

E
|
p
= |B
R
E \ E
n
| + |B
R
E
n
\ E|.
Thanks to the assumption, we know that
lim
n

IR
N(
E
n

E
) = 0, for any L
1
(IR
N
).
This implies that
lim
n
|B
R
E
n
\ E| = lim
n

IR
N(
E
n

E
)
B
R
E
= 0.
Moreover, since

IR
N(
E
n

E
) = |B
R
E
n
\ E| |B
R
E \ E
n
|,
then
lim
n
|B
R
E
n
\ E| = lim
n

B
R
(
E
n

E
) + lim
n
|B
R
E \ E
n
|
= 0.
Thus,
lim
n

B
R
|
E
n

E
|
p
= 0,
and the result of the lemma follows.
Proof of Theorem 0.1 : The proof is a direct consequence of the preceding lemmas.
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Remark 0.1 Obviously,
E
is dened only a.e. . So, this representation of the
domains does not make any dierence between an open set and the same open set
private of a negligible compact set. Yet, we know that the solution of the PDE in
the modied set may be dierent. So, this approach may have serious limitations for
certain shape optimization problems.
The least favorable aspect remains in the fact that in general =
E
. But, its value
leaves in the interval [0, 1]. If the convergence holds to be true in L
P
(IR
N
), for some
1 p , then the limit is a characteristic function. Indeed, this implies the
convergence a.e., so that the limit takes only 0 or 1 as values.
2. Convergence of open sets in the sense of Hausdor
In this section we restrict ourself to open domains and we assume that they are conned in a
bounded domain ; that is there exists a compact set xed B that contains all the open domain
we are considering. Let us denote by K
B
the set of all the nonempty compact sets included in
B. We denote by d(., .) the standard Euclidean distance in IR
N
.
The Hausdor distance, or Hausdor metric, measures how far two subsets (of a metric space)
are from each other. It turns the set of non-empty compact subsets (of a metric space) into a
metric space in its own right. It is named after Felix Hausdor.
Informally, two sets are close in the Hausdor distance if every point of either set is close to
some point of the other set. The Hausdor distance is the longest distance you can be forced
to travel by an adversary who chooses a point in one of the two sets, from where you then must
travel to the other set.
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Let K
1
, K
2
K
B
, we set
for any x B, d(x, K
1
) = inf
yK
1
d(x, y)
(K
1
, K
2
) = sup
xK
1
d(x, K
2
)
d
H
(K
1
, K
2
) = max((K
1
, K
2
), (K
2
, K
1
)).
Proposition 0.2 d
H
is a distance in in K
B
.
Proof : Exercise. See that, since K is compact, for any x B, there exist k K such that
d(x, K) = d(x, k), and
d(x, K) d(x, y) + d(y, K) for any y B.
Denition 0.2 d
H
is called the Hausdor distance.
Proposition 0.3 If K
1
and K
2
are two compact sets, then
d
H
(K
1
, K
2
) = d
K
1
d
K
2

(IR
N
)
= d
K
1
d
K
2

(K
1
K
2
)
.
In particular,
K
n
K if and only if d
K
n
d
K

(IR
N
)
0.
Proof : Let us denote by
(K
1
, K
2
) = d(., K
1
) d(., k
2
)
L

()
.
For any x K
2
, we have
d(x, K
1
) = |d(x, K
1
) d(x, K
2
)| (k
1
, K
2
),
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so that
(K
2
, K
1
) (K
1
, K
2
).
In the same we can prove that
(K
1
, K
2
) (K
1
, K
2
)
and deduce that
d
H
(K
1
, K
2
) (K
1
, K
2
).
Now, for i = 1, 2, let us consider k
i
K
i
, such that
d(x, K
i
) = d(x, k
i
).
We have
d(x, K
1
) d(x, k
2
) + d(k
2
, K
1
)
d(x, K
2
) + +d(k
2
, K
1
).
So,
d(x, K
1
) d(x, K
2
) d(k
2
, K
1
)
(K
2
, K
1
)
d
H
(K
1
, K
2
).
In the same we can prove that
d(x, K
2
) d(x, K
1
) d
H
(K
1
, K
2
),
and we deduce that
d(x, K
2
) d(x, k
1
)
L

(K
1
K
2
)
d
H
(K
1
, K
2
),
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This ends up the proof of the proposition.
Now, let
B
be the set of open domains included in B, given by
d
H
(
1
,
2
) = d
H
(B \
1
, B \
2
), for any
1
,
2

B
.
Notice that, if
B
. In
B
, we consider the distance d
H
dened by
d
H
(
1
,
2
) = d
H
(B \
1
, B \
2
), for any
1
,
2

B
.
Notice that, if
B
, then B \ K
B
. Moreover, it is interesting to see that d
H
is
independent of B as well see in the following proposition.
Proposition 0.4 For any
1
and
2
open sets of IR
N
,
d
H
(
1
,
2
) = max((
2
\
1
,
2
), (
1
\
2
,
1
).
Proof : For any x B, we have
d(x, B \
2
) =

0 if x B \
2
d(x,
2
) if x
2
.
So, if
2
\
1
= , then
(B \
1
, B \
2
) = sup
xB\
1
d(x, B \
2
) = sup
x
2
\
1
d(x,
2
) = (
2
\
1
,
2
).
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Denition 0.3 Let (
n
)
nIN
and be open domains in B. We say that
n
in the
sense of Hausdor, if
d
H
(
n
, ) 0, as n .
We denote this by

n
.
Remark 0.2 Even if the denitions is given only for compacts sets, well use the notion
of Hausdor distance also open sets. However, well denote it by d
H
for open sets and d
H
for compact sets.
Remark 0.3 If
1
= , then
d
H
(,
2
) = max {r > 0 ; x IR
N
, B(x, r)
2
}.
Exercise 0.1 For compact sets (K
n
) and K, prove the following Hausdor convergence :
1. If (K
n
) is a sequence of non increasing compact sets of IR
N
, then
K
n

nIN
K
n
.
2. If (K
n
) is a sequence of non decreasing compact sets of IR
N
contained in a compact
set, then
K
n

nIN
K
n
.
3. If K
n
K, then
K =
n
(
pn
K
p
)
= {x : inB ; x
n
p
K
n
p
, x
n
p
x}
= {x : inB ; x
n
p
K
n
p
, x
n
p
x}.
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4. If k
n
k

n
, K
n
K and K

n
K

, then K K

.
5. If K
1
, K
2
are two compacts, then
d
H
(K
1
, K
2
) = inf { > 0 ; K
2
K

1
and K
1
K

2
},
where
K

= {x IR
N
; d(x, K) }.
6. We have
d
H
(K
n
, K) 0 if and only if d(., K
n
) d(., K) in L

(B).
7. If (K
n
) is a sequence of connected bounded compact sets. If K
n
K, in the sense of
Hausdor, then K is connexe. In general, if has p connected component independent
of n, then it is the same for K.
Exercise 0.2 Let (
n
) and be bounded open sets. Prove the following Hausdor con-
vergence :
1. If (
n
) is a sequence of non increasing bounded open sets of IR
N
, then

n
int(
nIN
K
n
).
2. If (
n
) is a sequence of non decreasing compact sets of IR
N
contained in a compact
set, then

n

nIN

n
.
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3. The inclusion is stable in the sense of Hausdor :

1
n

1

2
n

2

1
n

2
n
n

1

2
.
4. If (
n
) is a sequence of open bounded sets converging in the sense of Hausdor to ,
then for any x , there exists a sequence x
n
, such that x
n
x.
5. The nite intersection is stable with respect to Hausdro convergence.
6. For the inclusion, we have

1
n

1

2
n

2

1
n

2
n

1

2
.
7. Let (
n
) is a sequence of open bounded sets converging in the sense of Hausdor to
. If
K is compact and K ,
then, there exists N IN, such that
K
n
for any n N.
In particular, if is a function compactly supported in , is compactly supprted in

n
, for large n.
8. The convexity is preserved by the Hausdor convergence of open sets.
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9. The connectedness is preserved by the Hausdor convergence of open sets.
Remark 0.4 1. The volume is not preserved by Hausdor convergence. But, for open
bounded set, the function volume is l.s.c.
2. The perimeter is not preserved by Hausdor convergence. It neither l.s.c neither u.s.c.
3. the diameter is continuous for Hausdor convergence.
Theorem 0.2 If K
n
is a sequence of compact sets contained in a xed bounded compact
set B, then there exists a compact set Kcontained in B and a subsequence that we denote
again by n such that
K
n
K in Hausdor sense.
Corollary 0.1 If
n
is a sequence of open sets contained in a xed bounded compact set
B, then there exists a, open set contained in B and a subsequence that we denote again
by n such that
K
n
K in Hausdor sense.
Proof of Theorem 0.2 : Thanks to Proposition 0.3, it is enough to prove that d(., K
n
)
d(., K) uniformly in B. Let us divide the proof in three steps.
Step 1 : There exists a Lipschitz function f such that d(., K
n
) f uniformly in B. Since, for
any n, d(., K
n
) diam(B), we deduce that d(., K
n
) is bounded. Moreover, it is clear that
|d(x, K
n
) d(y, K
n
)| d(x, y),
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which implies that the function d(., K
n
) is equicontinuous. Then the rst step of the proof is a
simple consequence of Arzel`a-Ascoli Theorem
1
.
Step 2 : Taking K = {x B ; f(x) = 0}, we have f(x) d(x, K) for any x B. Since f is
continuous, for any y K, we have
f(x) = |f(x) f(y)|
d(x, y).
This implies that
f(x) d(x, K) for any x B.
Step 3 : d(x, K) f(x) for any x B. For a xed x B, let y
n
K
n
be such that
d(x, K
n
) = d(x, y
n
).
SInce B is compact and K
n
B, then there exists y B and a subsequence that we denote
again by n such that y
n
y. Moreover, we see that
f(y) = lim
n
d(y, K
n
)
lim
n
d(y, y
n
) + d(y
n
, K
n
)
lim
n
d(y, y
n
)
0,
which implies that y K. At last, since d(x, K
n
) = d(x, y
n
) d(x, y) we deduce that
f(x) = d(x, y) d(x, K). And the proof is nished.
1
Let be a bounded subset of IR
N
and a sequence of functions f
k
. If f
k
is equibounded and uniformly equicontinuous then there exists a uniformly
convergent subsequence f
km
.
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3. Convergence in the sense of compact sets
This the less natural topology, but it could be useful in some situation. In particular, when
dealing with the continuous dependence of the solution of a PDE with respect to the domains.
Denition 0.4 Let (
n
) and be bounded open sets of IR
N
. We say that
n
converges
to in the sense of compact sets, denoted by
n
, if
(i) K comapct , K
n
for large n
(ii) L compact
c
, L
c
n
for large n.
Remark 0.5 One of the major inconvenience of this topology is the non uniqueness of
the limit. Indee, if
n
in the sense of compact set, then
n
converges to any open
set satisfying and = . This is not a separable topology. However, we can get
uniqueness if we work modulo an adequate equivalence relation :

1

2

1
=
2
.
For compactness property with respect to this topology the situation is less favorable. Let us
study an example in two dimension of space :
For a given function f : [a, b] IR, let us denote by

f
= {(x, y) IR IR ; a < x < b and 0 < y < f(x)}.
Proposition 0.5 We have
If
f
n

f
, in the sense of compact sets, then f
n
f uniformly in ]a, b[.
If f
n
f uniformly in [a, b] then
f
n

f
in the sense of compact sets.
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Remark 0.6 We do not have equivalence in the the proposition. Indeed, for the rst part
take a = 0, b = 1, f 1 and f
n
(x) = 1 + x
n
.And for the second part of the proposition,
take a = 0, b = 1, f 2 and f
n
(x) = 2 x
n
.
The proposition shows that the compactness result like for Hausdor measure is far to be
true for the topology generated by convergence in the sense of compact sets. Indeed, even if we
assume that f
n
is uniformly bounded, we know that we need some equicontinuity in order to get
a uniform convergence through a subsequence (even on any compact sets). The equicontinuity
can be obtained by estimating the gradient of f
n
(this could be the case if we assume that f
n
are uniformly continuous Lipschitz.
Proof of Proposition 0.5 : For a given > 0, let us consider
K

= {(x, y) IR
N
; a + x b and y f(x) }.
Assuming that
f
n

f
in compact senses, we deduce that
K

f
n
for large n,
so that
for any x [a + , b ], f(x) f
n
(x) (since (x, f(x) )
f
n
).
In the same way, by taking the exterior compact
L

= {(x, y) IR
N
; a x b and f(x) + y 1/},
and using the convergence in compact sens of
f
n

f
we deduce that
L

f
n
c
for large n,
I
n
t
r
o
d
u
c
t
i
o
n
t
o
S
h
a
p
e
o
p
t
i
m
i
z
a
t
i
o
n
N
.
I
g
b
i
d
a
1
8
and
for any x [a, b], f
n
(x) f(x) + (since (x, f(x) + )
c
f
n
).
This ends up the proof of the rst part. For the second part, we see that for any compact set
K of
f
, there exists > 0 such that K K

. Moreover, using the uniform convergence of f

f
n
for large n. Indeed, if a + x b and y f(x)
then a < x < b and 0 < y < f
n
(x) for n large enough. In the same way, for any compact set
L of
c
, the compact
L [a, b) [0, ) L

for a certain (since f > 0).

f
n
c
for large n.
At last, since
f
n
[a, b] [0, ), then it is clear that
L \ L [a, b) [0, )
f
n
c
for large n.
And the proof is nished.
4. Other notions of convergence
There are many other notion of convergence for domains that well not abort here :
convergence of the boundaries
convergence in the sense of Kuratowski
I
n
t
r
o
d
u
c
t
i
o
n
t
o
S
h
a
p
e
o
p
t
i
m
i
z
a
t
i
o
n
N
.
I
g
b
i
d
a
1
9
convergence in the topological sense
convergence
convergence in the sense of C
k
dieomorphism of IR
N
.
Remark 0.7 In general, none of the convergence implies the two others. Only some
relations exists between the three convergence that we list bellow.
Exercise 0.3 Let
n
and be bounded open sets contained in a bounded set B. If
n

in the sense of Hausdor, prove that
1. | \
n
| 0
2.

liminf
n

n
a.e.
3. if, moreover,

n
in L

weak

, then