Variational Inequalities and Flow in Porous Media (Applied Mathematical Sciences)

<-(Au - Av0),u - v0>/IlvuIlp - + as

IlvulIp

This verifies the coerciveness assumption (1.9) for Ilv(u-v0)I1p - IlVullp

when

-A

(note that

Ilvullp -> +°°), and the result follows from

20

EXAMPLES AND APPLICATIONS

2.

Theorem 1.10, since

K0

and K are closed and convex.

As a second example, one can consider for

1 < a < 2

V E Rn.

ai(C) _ (1 + IE12)a-2/2 Ci

(2.38)

W-1'a

We define an operation from

into

W1'a(O)

(a + a1 = 1)

(O)

by

setting Au =

(ai(Vu)).

2x. i

Then given

u1 E K

with

f E W-I'a (O), there exists a unique solution

- ul)dx >- (PQ ai(ul)8x(v i ,

Vv E K

J

a.e. in

_ {v E W1'a(O)lv(x) > (x)

K

convex set

K0

O}

of

u1

(2.39) 10

provided that the closed

is not empty (see [76] for a proof).

Let us note that when

a = 1, A

is the prescribed mean curvature

operator, and it is known that even for the Dirichlet problem, the existWhen

ence of a solution may fail. 0 < 0

O

is smooth and convex, ¢ E W1'm(O),

r, the interested reader will find a proof of existence and

on

uniqueness of the solution of (2.39) in [76].

2.6.

Fourth Order Variational Inequalities Recently (see [38],

[44)) variational inequalities with fourth order

operators have been investigated.

For instance, if

prove that there exists a unique solution u E K,

u)dx >

1

f

with, for example, K

a

f E H-1(O)

one can

of (see [38])

Vv E K

equal to

K = {u E H01 (o) D W2'2(0)

where

u

I

a.e. in

IAu(x)I < a

O},

is some positive constant.

More generally, if one can consider, for

Q

is a positive quadratic form of

a > 0,

K = {u E HI (O) 0 W2'2(c2)IQ(uxx

<

1

A relevant physical example is given in K = {u E H0(O) n W2'2(O)I(uxx

-

a}.

]R2

(see [79]) by

uyy) 2 + 4uXy < 1}.

n2

variables,

Comments for Chapter 2

21

Comments

The reader interested in more results on the Sobolev spaces is referred to Adams [1].

See also Necas [95].

We have restricted our examples to elliptic problems and elliptic variational inequalities in divergence form:

For parabolic variational

inequalities see Mignot-Puel [89] or Lions [84], Bensoussan-Lions [19]; for problems which are not on divergence form see Bensoussan-Lions [19] or P. L. Lions [87].

The relationship between variational inequalities

and convex programming are investigated in Mancino-Stampacchia [88].

For

nonlinear variational inequalities, see, for instance, M. F. Bidaut-Veron [21] and the bibliography of this paper.

In this chapter we have restricted our examples to the simplest and, maybe, most famous ones.

Many other applications related to mechanics or

optimal control theory can be found in [19], [57], [76] and in the two volumes of [13].

Note also that throughout these notes we will not discuss the numerical analysis of the problems investigated.

Much of this material is cov-

ered by the books:

R. Glowinski, J. L. Lions, R. Tremolieres:

Numerical Analysis of

Variational Inequalities, North Holland, 1981. P. G. Ciarlet:

North Holland, 1978.

The Finite Element Method for Elliptic Problems,

Chapter 3

The Obstacle Problems: A Regularity Theory

Throughout this chapter, 0

will be a bounded domain of IRn

with

Lm(52)

boundary

r

and

aij

functions in

which satisfy the ellipticity

assumption

a..(x)Ei j > vM2

Vx E 0, V E Mn.

We will denote by A

the operator

A

3.1.

2x.

(3.2)

(aij(x)ax.).

Monotonicity Results For

and

(3.1)

ui

measurable functions, let us denote by

f,f' E

ul

the solutions of

ul E K,, <-Aul,v - u1> >

Vv E K

(3.3)

ui E K, <-Aui,v - ui> >

Vv E KC

(3.4)

where

K _ {v E H,(52) Iv(x) > $(x)

a.e. in

c}

and

K0, _ {v E Hp(52) Iv(x) > 01(x)

a.e. in

0}

are not empty.

Then we have Proposition 3.1:

'(x)

a.e., then

Assume that

f > f'

ul(x) > ui(x)

a.e.

22

(in the

H-1

sense) and

fi(x) >

3.1.

Monotonicity Results

Proof:

(See Brezis [28]).

23

We have

(u1 - ui)

E H1(0)

(see Theorem 2.4)

and it is easy to check that ul +

(u1

E K.,

- ui)

ui - (u1 - ui)

E K01.

Using (3.3) and (3.4), this leads to <-Aul,(ul - ui) > > <-Au',-(ul - ui) > > .

Now by adding the above inequalities, we obtain <-A(ul - ui),(ul - ui) > > > 0. That is to say, (see (3.1), (2.16)) II2 < -<-A(ul - ui),(ul - Up -> < 0

vIIV(ul - ul) and so

= 0, which gives the result.

(u1 - ui)

With a similar proof, one can show: Proposition 3.2.

empty and if

U1 E KV ul'

E KV

Assume that

ul,ui

0,0' E L (Q).

If

K

and

are non-

denote respectively the solutions of

> 0

<-Au ,v - u> > 0

Vv E K0

(3.5)

Vv E K,

(3.6)

then we have I0+I.

0 < ul -<

(3.7)

lul - uilm < Im - O'lm.

(3.8)

Proof:

For the first inequality, let

v = ul > u1 > 0

and

k =

and take in (3.5)

v = uI - (u1 - k)+ = min(ul,k) >

which leads to 0 < <-Aul,ui - uI> = -<-Au l,uI> - ul = 0 -- u1 > 0 0 < <-Aul,-(u1,-k)+> = - -* (u1 - k) + = 0 *-* u1 < k.

The other inequality follows in the same way, and the proof is left to the reader.

24

3.

Remark 3.3.

THE OBSTACLE PROBLEMS:

A REGULARITY THEORY

One can, of course, give similar results for the two obstacles Reasonable assumptions are found by considering the physical

problems.

problem as stated in Remark 2.10.

Indeed, Proposition 3.1 (with

f =

f' = 0) means only that the more the membrane is pushed, the more it goes up.

The Penalty Method

3.2.

Let

be two functions satisfying

81,82

1

is Lipschitz, monotone, 81(t) = 0

Vt > 0

(3.9)

82

is Lipschitz, monotone, 62(t) = 0

Vt < 0.

(3.10)

Then for $, E L2 (S2) , f E H-1(S2) e > 0, there exists a unique solution

al

Aue +

and by Theorem (2.7), for all ue

of:

E a2

(3.11)

u e E Hp (Q)(Indeed, from the Lipschitz character of for some constants

and since

8i

8i(0) = 0, we have,

C1,C2,

IF(x,y)I = 181(y-C + 82(y4)I < Clly - $(x)I + C21y - i,(x)I

< Clyl + C'(Im(x)I + IP(x)I).) Now we want to show that for a suitable choice of small enough, u£

provides us with an approximation of

81,82 u1

or

and

e

u2, the

solutions of u1 E K.,

<-Aul,v - u1> >

Vv E Kf

(3.12)

u2 E

<-Au2,v - u2> >

Vv E 4.

(3.13)

As we will see later, this approximation is very convenient in regularity theory, since it allows us to deal with an equality instead of an inequality.

So we have Theorem 3.4. ue + u1

in

(ii)

(i)

K

# 0, 61(t) < 0

If K # 0, 81(t) < 0

Choose

in v

HO(S2), u2

in

Vt < 0, 82 = 0, then when

c + 0,

being the solution of (3.12).

H0(0), ul

e + 0, ue + u2 Proof:

If

K

Vt < 0, 02(t) > 0

Vt > 0, then when

being the solution of (3.13).

in case (i) and

v

in

K*

in case (ii).

For

The Penalty Method

3.2.

25

v, we have by (3.9), (3.10)

such a

=0

01(v-$)

pi, we get

and by the monotonicity assumptions on the

Jn(01(v-b) + 82(v-*)).(ue-v) = 0.

Jn(B1(ue-t) +

But by (3.11), this can also be written as <-Auc,uc - v> <

(3.14)

and after an easy computation, using (3.2) and Theorem 2.1, we get "IIuEII1,2 < CIIucIi1,2 + C' where

and

C

are constants which don't depend on

C'

for some constant

e.

So, we have

C

(3.15)

H ue'Il,2 < C,

by Theorem 2.5, we can construct a sequence

uE - u

in

H01 (Q)

weakly

uen - u

in

L2(52)

strongly.

en -+ 0

such that

n

By Lebesgue's Theorem and from the Lipschitz character of the deduce from

uE - u

in

gi, we

that

L2(0)

n (and so in

in

B1(u-0) + 02(u-i)

+ 02(ue

slue

n

n

But by (3.11), (3.15), we also have

+ 02(uE -iy) = en(AuE +f)

1(uE n

n

-

0

in -9'(S2)

n

and by the uniqueness of the limit in g'(S2), we get 51(u-$) + 82(u-VP) = 0,

which gives

u E K.

is true for all

v

in case (i) and in

K

u E K*

in case (ii).

in case (i) and all

v

in

we get

<-Au en'

v - u

> en> -


Cn

>

Vv E K (Resp. v E

K*

Since (3.14) in case (ii),

26

THE OBSTACLE PROBLEMS:

3.

By the monotonicity of <-Av,v - u

Letting

-A

> >
En -

go to

en

en

By Minty's Lemma and since

(Resp. v E K*).

Vv E K

(Resp. v E 0).

u E KV we obtain that

in case (ii)).

in H(0).

uc -- ui

Vv E K

>

leads to

0

<-Av,v - u> >

(Resp. u = u2

A REGULARITY THEORY

u = uI

Now by uniqueness of

u1

and

in case (i)

u2, we have

To get strong convergence, it is enough to note

that (see (3.1), (3.14)) vl1u-uE112

2

<-A(u-uE),u - uE> = <-Au,u - uE> +

-e

< <-Au,u - u > -
with

c.

> -

0

E

This concludes the proof. By choosing

Remark 3.5.

convergence of uE

0, = 0, a2(t) > 0

to the solution

u1 E Ku', <-Aul,v

ul> >
-

ul

Vv E K

ul>

-

Vt > 0, we would obtain the

of:

under the assumption that K = {v E H0'(2)lv(x) < i(x) Remark 3.6.

a.e.} , 0.

It is easy to see that the above proof also holds for a non-

linear operator

A

from

H1(H)

into

H-1(0)

which satisfies the assump-

tions of Minty's Lemma and the coercivity property > vllu - vuI1,2

Vu,v E H1(92).

In the next part, we will use

Remark 3.7.

uE

to denote the solution of

a slightly more complicated problem, that is to say 0

-Au. +

(uE

+

e 2

(uE-4)

+

E HO(B).

With the same proof as above, one can easily see that Theorem 3.4 holds without any change for such a

uE.

3.3. W2'p(S2)-Regularity (2 < p < +W).

27

W2'P(S2)-Regularity (2 < p < +.o),

3.3.

In this part, we shall assume that aij

are sufficiently smooth.

any

2 < p < +.o

ax.

1

uE is in

when

r

the boundary of

SZ

and the

More precisely, we shall assume that for

f E LP(53), the solution of the Dirichlet problem f

(aij axj)

(3.17)

H0 (S3)

and that we have the estimate

W2'P(0)

I1u1I2,p < CPIfIp

(see [4), [19]) where

(3.18)

is a constant which doesn't depend on

Cp

Now let ,J be two functions in H1(f) $(x) < V(x)

a.e.

in

u

or

f.

such that (3.19)

S2

and such that < 0 <

on

(3.20)

P.

(3.20) is taken in the trace sense (see [82]) or simply in the usual sense

if $, are smooth enough.

Moreover, assume that

A$

and

Ap

are

measures such that (-AO)', (-A*)

Then if

81,52

are two smooth functions (for instance

(3.9) and (3.10) and if

e

uE, for

C1) satisfying

e > 0, is the solution of (see 3.11)

8

S

-AuE +

(3.21)

E LP(52).

(U C_#) +

E (uE-4) = f (3.22)

uE E H0(0), then we have: Theorem 3.8.

the functions

Under the above assumptions (3.18)-(3,20) and for 1

(uE-*)

and

02

E (u,-*) are in

LP (52)

,sl(uE-e)Ip < If - (-A#)+Ip

s

(3.23)

(-A*)_ip.

(3.24)

I e (uE-*)Ip < If +

Proof:

f E LP(S)),

and

(See [29]).

For

F(t) = Itlp-2t, let us consider

28

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

v = F(E (uC - O))

a(uE - )

As we will see, it is no restriction to assume that So by Theorem 2.3 and the assumptions on

bounded. (3.9)), v

is in

and

$

is

((3.20),

dl

H1(S2), and from (3.22), we deduce that $E(ue-

<-A(uE-f),v> + <-A$,v> + J

fvdx. J

S2

(3.25)

Q

Sl

Now, since the function

Y = F o

is monotone, we get

<-A(uf),v> = <-A(ue-O),Y(ue-$)>

I Y'(U

(3.26)

4+)8z(uE-b)dx > 0.

*)aij 3x1(uE J

Moreover, 91

is negative, so is

v, and

dx.

f

(3.27)

J

Finally, we note that

v

0

only when

ue < $ < *

and hence when

92 E

(ue-V) = 0.

From this it follows that

(!e (u£-#) +

02(u e-V')) v dx (3.28)

Is£(uE-O)Ip.

= Jo

Combining (3.25)-(3.28), we get I8E

J (f -

By applying Holder's Inequality, we get with

q = p/p-l

(-A$)+IplvIq

ICE (uE-t)Ip t If -

But recalling that

Iviq = ISl(uE

v =

ISl(uC_$)Ip 2.

l(uewe have

Ip-1

-

01

and (3.23) follows.

Now if a (ue-p)

the truncated function if

ItI

is not bounded, it is clear that

F (t) = F(t) if n > n, leads to a v = Fn(S1(ue-$))

ItI < n, F (t) = Sign n

which is in

H0((2).

The result

3.3. W2'p ((2) -Regularity (2 < p < +)

29

t in then follows after an easy passage to the lim!2(,,

Finally, to ob-

n.

v = F(E i))

tain (3.24), it is enough to consider

(or

s

v = Fn(e (uE - )) which concludes the proof.

Assuming that

Remark 3.9. as

p .+ +-

EL

((2)

f E L%1),#,* E W2'w(f2)

in (3.23), (3.24) leads to the fact that

and taking the limit £1(uE-+), s?(uE-1P)

with

l- (uE a) l , _ I f -

I

< I f + (-AP)

I

IW.

(3.29)

For the obstacle problems, we claim now that under the assumption (3.18), we have:

Theorem 3.10:

solution

uI

be in

Let

measure such that

with

H1(()

(-A$)+ E LP((2)

and if

on

f E LP(S2)

If A is a

r.

(p > 2), then the

of

u1 E KV <-Au1,v - ul> > is in

0 < 0

Vv E K0

and there exists a constant

W2'P(SZ)

Cp

(3.30)

independent of ul,f,$

such that

IIulII2,P

Theorem 3.11. (3.21).

(3.31)

< Cp(IfIp + I(-A$)+Ip). Let

Then for

O,ip

be in

f E LP((2)

H1(0)

and satisfy (3.19), (3.20) and

the solution

u2 E KV, <-Au2,v - U2> > is in

of the problem

Vv E K*

and there exists a constant

W2'P(c)

u2

CP

(3.32)

independent of

u2,f,$,ip

such that (3.33)

Cp(Iflp + I(-Ab)+Ip + I(-AMY) Ip)

IIu2II2,P <<

Proof:

First, for the proof of Theorem 3.10, consider

of (3.22) with

S2 - 0

and

81(t) < 0

Vt < 0.

uE

the solution

Of course, for this

uE

(3.23) is satisfied and from (3.22), we get

IAuEIp < Iflp + If - (-Ao)+lp. By (3.18), we get

IIUEII2,P < Cp(Iflp + I(-Ao)+lp)

(3.34)

30

where

doesn't depend on

Cp

So we can extract a subsequence

c.

which converges weakly to some

u

in

ue

when

a - 0. Now by 2,p(Q) which is in W and

W2'P(92)

is necessarily equal to

Theorem 3.4, u

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

uI

(3.30) results from the lower semicontinuity of the

norm in a Banach

To prove Theorem 3.11, it is enough to consider the solution

space.

of (3.22) with

Vt < 0, 02(t) > 0

81(t) < 0,

uE

From (3.18), (3.22),

Vt > 0.

(3.23), and (3.24), we get Iluell2,p

Since

Remark 3.12.

fied for

a -

when

ue - u2

When the

0

a..

il

(by Theorem 3.4), the result follows as above.

are smooth, the above assumptions are satis-

f E LP(S2),0,* E W2'P(c2).

Remark 3.13.

(3.35)

Cp(iflp + I(-A$)+IP + 1(-A*) Ip)

<

If

(See [28], p. 8 for further comments.)

p > n, by the Sobolev embedding theorem, we get that

u 1

and

u2

are in

with

Cl'a(S2)

a = 1 - P.

Note that in general ui f C2(0)

(see below).

For optimal results of regularity see [29].

Remark 3.14.

Estimates (3.34), (3.35) could be used in place of (3.15) to

go to the limit in (3.22).

Some Complementary Results

3.4.

In the unidimensional case, the solution of (3.30) or (3.32) is always continuous since

H1(52) a C(SC).

In higher dimension, under the assumptions

on the previous part, we can prove:

Theorem 3.15.

Assume that

such that the solution tion

u1

u

and

¢ E C(S2)

If f E H(S2) is

of (3.17) is continuous on

S2, then the solu-

of

ul E KV <-Aul,v - u1> > is continuous in Proof:

K4 # 0.

(3.36)

SZ.

(See [84].)

smooth functions

Vv E K

0n

Assume first that such that

0n < $

f = 0

and

and choose a sequence of $n i $

in

L%2).

be the sequence defined by un E Ken, <-Aun,v - un> > 0 By Theorem 3.10, we have

Iun So

- ull <

un - u1

IOn

Vv E Kin.

un E C(S1)

and by the Proposition 3.2

-

uniformly and

ul

is continuous.

Let

un

3.4.

Now to relate the case ing device due to Brezis. u E HQ(Q),

u

in

be the solution of (3.17) i.e.,

Q.

ul - u

(u1-u) E K# - u =

satisfies

f = 0

we have that

is continuous in ? so is

u

V(v-u) E

(u1-u)> > 0

<-A(u1-u),(v-u) -

and by the result in the case and since

to the general one, we use the follow-

f = 0 Let

-Au = f

We see from (3.36) that

in

31

Some Complementary Results

is continuous

ul - u

u1.

usualy regularity results, 3.15 holds, in particular for

Remark 3.16.

By

f E W-1'p(c)

with

being smooth enough.

p > n, Q

Under the assumptions of Theorem 3.15, we can divide

in two sets

0

0 = {x E OIul(x) > t(x)}

(3.37)

I = {x E OIul(x) = (x)}.

(3.38)

By Theorem 3.15, 0

is open and

I

is closed.

I

is called "the

coincidence set" and its boundary is the free boundary of the obstacle problem.

The interested reader will find a study of this free boundary

and also an extended definition of (3.37), and (3.38) in (76 Note that in

.

0, we have

-AuI > f

(3.39)

in the sense of distributions and so in the sense of measures. for

C E .9(0), C > 0, choose <-Aul,C> >

v = ul + C

Indeed,

in (3.36), we get

C > 0

C E -9(Q),

and so (3.39) holds.

Moreover, we have -AuI = f Indeed, for

in

C E,9(0), C > 0

test function for (3.36).

<-Aul,-EC> > <-Aul,C>

(3.40)

0,

<

and

c > 0

small enough

v = ul - cC

is a

So, we get VC E-9(0),

C > 0

VC E-9(0)1

C > 0

and this inequality with (3.39) leads to (3.40) and so for smooth data, u1

is the solution of the nonlinear problem

32

THE OBSTACLE PROBLEMS:

3.

-Aul > f,

ul E H01 (Q),

A REGULARITY THEORY

ul > 4,

(3.41)

0,

that is to say:

u1 E

Max(f + Aul, 4 - u1) = 0.

HI (S2)

(3.42)

(See [33] to compare with the Bellman-Dirichlet problem.)

We now use this characterization of that is to say, to prove that

to complete our Remark 3.13,

4(x) = -x2 +

and

3

even if

C2(S2)

Consider the one dimensional case with

the data are smooth.

A = d2/dx2, f = 0

u1

is not generally in

u1

D = (-1,1),

(see the figure below).

First note that the solution will be symmetric and we have to find it From (3.40), we have

for instance on

(0,1).

ul = ax + b, on

u1 > 4.

Since

ui = 0, and hence

u1(1) = 0, this gives

ul=a(1 -x) u1 > 4

on the component of point

x0

1.

The fact that

ul

touches

4

at a

can be expressed by

u1(x0) = 4(x0), (u1 - 4

near

ui(x0) = 4'(x0).

achieves its minimum at

x0) which leads to consider

ul

defined

by

ul(x) =

x + 1

if

x E [-1,-

4(x)

if

x E [- 2,+ 2].

-x + 1 if

(3.43)

x E [2,1]

Now it is easy to check that

ul

is a solution of

+1

u1 E K4,

ui(v' - ui)dx > 0

Vv E K4.

(3.44)

3.5. W'-U) -Regularity

33

Indeed, from (3.43), we deduce that

ul =

1

if

x E (-1,- Z)

-2x

if

x E (- ?,Z)

-1

if

x E (-,1)

(3.45)

2

and so for all smooth

v

+l

we get

in

f-1/2 (v'-ul)dx

f

+ 1f

-2x(v-u l)'dx +

J1/2

-1/2

-1

1

fl

-(v'-ui)dx

11/2

1/2

(v-u1)(-

+

f_1/22(v-ul)dx - (v-u1)(?)

(v-ul) (- 2) + (v-u1) (z)

1/2

2 (v-$) dx > 0, satisfies (3.44).

and thus

u1

not

However, from (3.45), we deduce that

C

.

Now, it is clear from (3.45) that

u1

is

ul E L (S1), and as we will

see now this property is always true for sufficiently smooth data and has an extension to more than one dimension.

3.5.

W2'-

(id)-Regularity

As we mentioned in the preceding section, we now give a proof that for smooth data the second derivatives of solutions of obstacles problems are bounded up to the boundary.

To be precise, we will assume here that a.. E

For ,i E W

r

2'm (S2)

first consider

u

is of class

C4.

(3.46)

satisfying (3.19), (3.20) and for

e > 0

we will

the solution of 07

l Au. + e(uE-f) + E(uE-$) + E

E(uE-V') = 0

(3.47) uE E H'(52) . With

01102

two Lipschitz functions of class

C2

satisfying

B1(t) = 0

Vt > 0;

si(t) > 0,

81 11(t) < 0

Vt

(3.48)

02 (t) = 0

Vt < 0;

0 u(t) > 0,

Bu(t) > 0

Vt

(3.49)

34

3.

THE OBSTACLE PROBLEMS:

A REGULARITY THEORY

The advantage of (3.47) over (3.11) (see Remark 3.7) is that if we put B.

i = 1,2

ai(t) = e (t) + Et,

(3.50)

we have ai(t) > e > 0

Vt.

(3.51)

This will be useful in our estimates.

(One could avoid this "et" by using

a weak maximum principle.)

Now (3.47) can be rewritten as

Au. = aI (uE - $) + a2 (u6 (3.52)

uE E HO (l).

I

Our first goal will be to obtain bounds for the second derivatives of ue which are independent of

Then, an argument similar to the proofs

E E (0,1].

W2'°°-regularity for ob-

of Theorems 3.10, and 3.11 will provide us with stacle problems.

The estimates of bound

will be performed in three steps, first we will

uE

Au., then the second derivatives of

and finally inside 3.5.1.

Q.

on the boundary

of

F

S2

So let us consider first:

A Uniform Bound for

We first note that

uE

Au

E

Indeed, this follows immediately

AuE E L (l).

from (3.47) and the fact that

uE E L (Q)

and satisfies

(3.53)

Iu6I., < To prove this last inequality we put (uE - k)+ E HO(Q)

Since

k =

and by (3.52) we get

<-Au E,(ue-k)+> = -JS2[aI(uC-C + a2(uE4)](u6-k)+dx < 0,

since when

uE > k

we have

uE > , uE > 4

and so

a1(u6-#) + a2(uE-*) > 0.

Now we deduce from the above inequality (see Proposition 3.2) that The inequality Aue E L (52).

uE > -k

uE

u

< k. e

We thus have 3.53 and

Moreover, we have:

Proposition 3.17. solution

follows in a similar way.

If

,ip E W2'w(l)

of (3.47) is such that

satisfy (3.19) and (3.20), then the

Aue E L%)

and satisfies the

estimate:

IAuEI., < Max[I

(-A*)

(3.54)

W2'p(1)-Regularity

3.5.

35

f = e(ue-$) + e(uwe obtain,

Applying the Remark 3.9 with

Proof:

since

and

a1 < 0

s2 > 0

Se

101(u6

(-A0)+I., I f

I., < Max[ I f -

+ (-A,) I.,]

< Max[1(-A0)+1.,,1(-A1P)

Ifl_.

Now the result follows from (3.47) since by (3.53) we have

Ifl <

It results from (3.54) that under the assumptions of Proposi-

Remark 3.18.

tion 3.17, for

c E (0,1]: (3.55)

IAuel., < C(I 10112,., + 11V'112,.,) where

depends on the

C

aid's but not on

uE E W2'P(S2)

(3.18) we have

E.

Moreover, by (3.55) (see

and

I IuEI 12,p < CPIAuEIP < Cp(I 10112,., + 11*112,,,) for all

p > 2.

for

C1(52)

only on

Using the continuity of the embedding of W

p > n and the

52

I Iu111,.

we deduce that there is some constant

C

2,p(Q)

in

depending

aid's such that

(3.56)

C(110112,,, + I 1*112,.,).

(See also the Section 3.6 for an improvement of this.)

Taking the limit when

Remark 3.19.

e -

0, (3.54) provides us with

IAull. < I (-A0)+Im, IAu21., < Max[I (-A0)+I,., I (-AVM) where

uI

and

Remark 3.20.

u2

are the solutions of (3.30), (3.32) with

In the case where

f = 0.

n = 1, it is clear that (3.55) already

provides us with the estimate

I Iu112,., < C(I 10112,., + I IP112,0 ) and so in the sequel we shall assume that 3.5.2.

n > 2.

Boundary Estimate

First let us note that since deduce from (3.46), (3.47) that

uE E W2'p(52)

(see Remark 3.18) we

ue E W4'p(1) (p > n).

This will allow

us in the sequel to apply the maximum principle on the second derivatives of

uE

in the sense of Bony [25].

and use an approximation argument.)

(One could choose at first smooth data

36

THE OBSTACLE PROBLEMS:

3.

Now

u

e

is at least in

A REGULARITY THEORY

so the following proposition makes

C2(S2)

sense:

Under the assumptions of the Proposition 3.17, if uE

Proposition 3.21.

denotes the solution of (3.47) there exists a constant depend on

or on $, such that for all

c E (0,1]

C

which doesn't

i,j = 1,...,n

we

have 2 a

u

i

E

ax ax j

Yx E r.

C(I*I2,m + IIkHI2,m)

(x)

(3.57)

We will use here ideas from H. Brezis - D.,Kinderlehrer [34] and

Proof:

R. Jensen [69]. Step 1:

We will first show that there is no loss of

x0 E r.

Let

is flat at

aij's satisfy

generality in assuming that

r

some additional properties.

(The arguments below are classical in bound-

and that the

x0

ary value problems and to follow the idea of the proof instead of technical tools the reader is invited to go directly to (3.67) and (3.68).) Since

is smooth, there exists a diffeomorphism

r

from a neighborhood centered at

of

V

onto

0 = (61162,...,6n)

an open ball of IRn

B

and

6(x0) = 0

such that

0

in in

x0

0(SZ n v) = B+ = {x = (X1,.... Xn) E BIXn > 0}

0(r n v) = a0B+ = {X = (X1,.... Xn) E BIXn = 0}.

For each function on

defined on

v

v n 52, we define a function

(and conversely) by setting

B+ U a0B+

v' (X) = v(6-1(X)) - v(x) = v'(6(x)). So we have (setting n

ax. (x) _

i

v'

k=1

k

axiaxj(x) =

in place of

(3.59)

(6 (x)) . axk (x)

i

a2u,

Be

axit

n

aek

x(x)

axj(x)

axk(6(x))'

k,t=l

uE)

a6

,

aX

n

a2u

u

(3.58)

(3.60)

i

aXk(6(x)).

+ kIl

axiax.(x). 7

2

Since 8X'

and

aXaX

are given by a similar formula it results are bounded on

from (3.56) that the derivatives 2X

i

B+

by the quantity

'a

3.5.

W

(s)-Regularity

37

Moreover, from this bound and (3.59) (3,60) we

C(110112,- + 1I'P112,a).

easily see that to have (3.57) in a neighborhood of

in

x0

it is

r fl v

2

a u ax.ax.

enough for

to satisfy an inequality of type (3.57) with 30B+.

in place of formation

on a neighborhood of we have to study

0

u'

0

Thus, via this trans-

in

satisfying (see (3.52), (3.59), (3.60))

with the summation convention: 2

aij BXiaxj + bk aXk =

a1('-$') + a2(u'-,p')

on

B+ (3.61)

I& = 0 on

a0B+

the coefficients (X) a!

being given by (see (3.60))

aij

ao. ae.

n

.

13

k,k= l (akR axk ax9)(e

Replacing

(a!. + a!.)/2

by

a!.

(3.62)

(x))

it is clear that there is no loss of

generality in assuming that in (3.61) we have (3.63)

Vi,j = 1,...,n.

aij =

(Note also that from (3.62) we can deduce that (3.1) holds for some

in place of

v'

T = (T1,...ITn)

v.)

with

aij

Now let us perform a second transformation

defined by (see [61], [69]). if

x = (X1....,xn) _ (x,xn),

Ti(x) = X. + xn$i(x)

n

i

Tn(x) = xn where

is given by

ai

fi(x) =

(3.64)

ani(x'0)/ann(X,0).

An easy computation shows that the Jacobian of T a neighborhood of maps

into

B+

0, T

B+.

at

is

0

is an isomorphism which preserves

1

and so in and

B0B+

Because of this second transformation, and with ob-

vious notations for functions the study (in some ball

AB+

v"

(see (3.58)), we are now concerned with

(a < 1) still denoted by

B+) of

u"

the

solution of

a'..

a2ulf

i3 axiaxj

+ b" LL LL = a (uif- fit) + a (urt-prr) k aXk

1

2

on

B+

(3.65) lull

= 0

a0B+

on

38

THE OBSTACLE PROBLEMS:

3.

A REGULARITY THEORY

aj = ani, but now (see (3.62), (3.64)) for all

with (see (3.62), (3.63)) i = 1,...,n-1 n

ain = ani

aT

BT.

annbi = 0

axX = ani +

ak1C' aXk

k,t=1

on

BOB+

Rewriting the first equality in (3.65) as a

aXi

(a

+ (b -

au'l )

ij aXj

aXi

k

aXk

+a

(u"-®"")

2

1

be the solution of

w

and letting

Balik) au It = a

a' .

ij aX) = (bk

ax.

B+

aXk)3Xk on

(3.66)

w=0

aB+,

on

u* = u" - w, * = V - w, j* = 1" - w

we see by setting

l(u* - *) + a2(u* -*) on

aX(sij aX.) = 1

u* = 0

that we have

B

30B+

on

Now from (3.56), (3.59), (3.60) it follows that the right-hand side of the first equation in (3.66) is bounded in quantity we have

W

C(110I12,- + II'II2 .,), and so on w E W3'p I_

I IwI

W2' if p > n

12,,,_ C(11#112,- +

1,p

(B+), for all

or on some

B+

p, by the AB+ (A < 1)

and an estimation of type

I I*112,,,)

Thus to prove (3.57) we have only to prove it on a neighborhood of

in

in place of

and with

a0B+

u,b,i.

0

in

Reverting to our notation

u, we have proved that in place of (3.52) we can assume that

u = uE

satisfies

as

(aij 2-U-

= a1(u-$) + a2(u-*)

on

B+

az)

(3.67) 1

with

u = 0

on

a.. = a.. 13

31

30B+

and

ain = ani = 0 $,i

being in

on

W2,w(B+)

(3.56) now being true on

30 B+

Vi = 1,...,n-1,

(3.68)

satisfying always (3.19), (3.20) and (3.55), B+.

W2,w

3.5.

(S2) -Regularity

39

The fact that (3.55) is preserved by the different trans-

Remark 3.22.

formation results from (3.56), (3.59), (3.60), (3.62). the fact

u E W4'p

Note also that

is preserved since by (3.46), (3.64), ToB

is of class

C4.

We deal now with

Step 2:

u = 0

First since

on

satisfying (3.67), and (3.68) holding.

u

we have:

a0B+

2

on

= 0

aa.u

aOB+

Vi,j = 1,...,n-1.

J

Using the first equation in (3.67) and (3.68) we get (see (3.19), (3.48), (3.49), (3.50)) 2

as

a

Yx E a0B+.

ann(x) Z(x) + axn WD u (x) = n ax n

(3.69)

n

(i.e., ann > v > 0) and from (3.56)

A

And thus by the ellipticity of E E (0,1]

we get for 2

()

<

Yx E DOB+

112,,,)

11

(3.70)

ax2

n

where

doesn't depend on

C

vatives of type

Thus it remains only to bound the deri-

e.

a2u/axkaxn, k = 1,...,n-1

in a neighborhood of

in

0

a0B+.

We thus want to estimate

Step 3:

32u/axkaxn.

Before entering into

the technical argument let us briefly outline the proof.

Assume that

is a smooth solution of:

u = u

E

Au = a1(u - m) + a2 (u -

J)

on 1R+ (3.71)

with Rn

equal to the half space

(3.56) being satisfied in + with on

x

n

xn = 0

on

I

A = A, al,a2

as above, and

< 0 <

Let us set

= 0. 2

Bu =

{(xl,x2,...,xn)Ixn > 0}, (3.55) and

a

u

axkaxn

+ au axk

To get 2

ax axn(x) < C(11#112,- + 11*112,,,) k it is clearly sufficient (by (3.55) and (3.56)) to bound

(3.72)

40

A+u = Bu + Au

or

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

A_u = Bu - Au

by

C(11m112,,, + l1*112 m).

x+

(Resp. x) a point of

gives us two chances!)

(±

where

A+u

Thus denoting by

(Resp. A-u) achieves its maxi-

mum it is enough to prove that we have

or

A+u(x+)

A_u(x) < C(1 *12,- + 114'112,m)

(3.73)

Two cases are possible Case 1:

Then of course in this case we have

x+ E {xn = 0}. 3

(A+u)(x+) < 0

ax

n

But on

\

2

Au (X+) < 0. ( a u 2 + ax ax + ax axkaxn k n n

(3.74)

//

we have (since

xn = 0

< 0 < :y, and by (3.48), (3.49))

u = 0,

2

Au=a,=-E(t+ii). ax-

n

Thus taking the derivative in the

xk

direction

3

a u 2(x+

= -E as ( + *)(X+).

axkaxn

k

Using (3.71) we also easily obtain

(2u -

Au (X+) = E --ax

n

- P) (x+) .

n

By (3.74), (3.56) we have now 2

a (x) ax kaxn +

with

C

I C(I I*I 12,.0 +

independent of

leads to (3.73).

I

IiPl 12,w) and by (3.55), (3.56) this clearly

c E (0,1]

The case where

x

follows in the same way,

E {xn = 01

so we can assume: Case 2:

x+

and

x

E R+.

Then applying

A+

and

A

to the first

equation of (3.71) we get:

A(A+u) = ai(u-$)(A+u-A+$) + ai(u-P)(A+u-A+P) (3.75)

+ al(u-0)Q+(u-C +

az(u-V')Q+(u-V')

3.5.

W

2

'(Sl)-Regularity

41

A(A_u) = (3.76)

al (u-$) Q_ (u-f) + aZ (u-Vi) Q (u-*) where we have set Q+(v) = av

.

axk

Ivv!2 av ± 2xn

(clearly

is positive, Q

Q

+

is negative).

By the maximum principle we have in this case A(A+u)(x+) < 0 If

and

A(A-u)(x) < 0.

(3.77)

x+ E [u > #] = {x E1RI u(x) > (x)} we have (see (3.48), (3.49))

a'11i (u - 0)(x+) = 0

(x+)

aZ (u - V)

and

Thus from (3.75), (3.77) we deduce (since

> 0.

Q+ > 0)

ai (u - ) (A+u - A+$) + aZ (u - S) (A+u - A+W) (x+) < 0

and by (3.51) we get A+u(x+) < A+t(x+)

or

which clearly leads to (3.73).

and

al (u - $)(x< 0

and

since in this case

Q

A-u(x) <

or

and

But this cannot occur.

x

E [u < ] we have

aZ (u - *) (x) = 0 < 0, we get now as above

which leads again to (3.73).

x+ E [u < $]

If

x

A-O(x )

So the only remaining case is when

E [u > ].

(3.78)

We have indeed from the definition of x+,x

Bu + Au(x+) = A+u(x+) > A+u(x) = Bu + Au(x ) Bu - Au(x-) = A u(x-) > A u(x+) = Bu - Au(x+). Adding these inequalities, we get [al(u-0) + a2(u-*)](x+) = Au(x+) > Au(x ) = [a

1(u-#)

+ a2(u-V')](x )

42

THE OBSTACLE PROBLEMS:

3.

A REGULARITY THEORY

But if (3.78) holds the above left hand side is strictly negative (see (3.48), (3.49)) and the right side is strictly positive.

This is a con-

The inequality in the

tradiction and hence (3.72) holds in all cases.

other side follows in the same way; note for instance that tion of a problem of the same type as (3.71) with

is a solu-

-u

in place of

Let us now return to (3.67).

With an eye to deriving the analogs of (3.75), (3.76), set 2

Bu =

where

k

ax ax + c ax , k n k

is a positive constant satisfying

c

as c

nn

Applying

(3.79)

axnn > n

al

Au

to

B

we easily obtain aa..

B(Au) = A(Bu) +

axi

B(aij)axj

xk

+

aaij

+ axn

Now, let

ax n

axj

(3.80)

a2u axkaxj)

be a smooth function such that

a

0 < a < 1

2

on

a = 1

B+

in a neighborhood of

0

(3.81) 30B+

a = 0

on

BB+ - B0B+

= 0

on

n

ax

(One of the roles of this function is to avoid the problems created by the fact that

u

is not prescribed on a part of the boundary of

Setting for

(with the summation convention in

i = 1,...,n aaiJ .

au

Si = B(aij)ax.

+ axk

and multiplying (3.80) by

B+.)

2

j)

2

aai] .

a u axnaxj + axn

a u axkaxj

a, we get after an easy computation:

a -Si) + as (CFSj - 2aij Bx3 Bu)

oB(Au) = A(ciBu) +

ax

x

(3.82)

af. = A(aBu) + f0 + ax.

i

with obvious definition for the

f.'s. 1

3.5.

W2''A-Regularity

43

Now let us temporarily assume that we have proved the following lemma: Lemma 3. in

If the

denote the functions of (3.82), there exists a

fi

W2'p(B+) (p > n)

w

which is a unique solution of the problem

af.

-Aw = f0 + x1

on

B+

(3.83)

i

w = 0

on

BB+-BOB+,

= 0

a,D-

on

90B+

n

Moreover, there exists a constant b,lp

which doesn't depend on

C

e E (0,1],

such that:

i lwl lW_ C(I 1+112,., + I III 12,1,).

(3.84)

Now from (3.82) we deduce A(aBu - w) = a8(Au).

Using (3.67) to compute

we get

B(Au)

A(aBu-w) = a1(u-d)(oBu-aB+) + a2(u-1P)(aBu-a8P) (3.85)

+ al(u-0)a as (u-$)BX (u-0) +

On the other hand, applying

axk

n

k

A

to

Au

ax (u-V+)

n

given by (3.67) leads to

A(Au) = ai(u-f)(Au-A+) + a2 (u-*)(Au-A*) (3.86)

+ a1(u-#) aij

Multiplying (3.85) by

Xi(u-$)aa

(u4) + a2(u-*)a ij

v, adding and subtracting

A(Au)

give us now the

following equations to which we will apply the maximum principle: A(V(aBu-w) + Au) = al'(u-O)(A+u-A+*) + a21(u-')(A+u-A+*)

+ al(u-$)Q+(u-0) + a2"(u-4V)Q+(u-'P) (3.87)

A(v(OBu-w) - Au) = ai(u-+)(A_u-A_+) + a21(u-4)(A-u-A-*)

+a

(u-0) + a2 (u-1P)Q _

In these equalities we have set by definition:

A+v = vaBv ± Av Q+v = va L Vaxk

By '

+ a.. av av i7 azi axj

44

THE OBSTACLE PROBLEMS:

3.

A REGULARITY THEORY

We shall note that by (3.1), (3.81) we have: av axk

8z. >_ vI0v12 > vv

aij

X.

i

av axn

7

is positive and

and thus, Q+

Let us now denote by (Resp. v(oBu - w)

Q-

is negative.

(Resp. x) a point where

x+

achieves its maximum in

- Au)

B+.

v(QBu - w) + Au We want to prove

that 2

-

aXkax (x) k n

c(i1*1

By the definition of

Vx E B.

2,m +

x+,x

(3.89)

it suffices to (see (3.55), (3.56), (3.84))

,

prove that (v(UBu-w) + Au)(x+)

(v(1Bu-w) - Au)(x-)

or

(3.90)

< c(I I01I2,m + I I1I I2,,,) Different cases are possible: x+ E BB+ - 90B+.

Case 0:

since in this case

We have then, by definition of

v(x+) = 0

and

x+

and

w(x+) = 0:

(v (QBu - w) + Au) (x+) = Au (x+) , Clearly the case

and by (3.55) we get (3.90).

x

E aB+ - a0B+

follows

in the same way. +

Case 1:

(see Case 1 of the heuristic proof).

E a B

x

0

then (recall that

We have

v(a'Bu-w) + Au E W2'p(B+), (p > n))

axn(v (QBu - w) + Au) (x+) < 0. Using (3.67), (3.81), (3.83), the above inequality can be rewritten as

[va.

But on

ax

n

Bu + ai(u-O)aX (u-0) + az(u-*)aX (u-V)](x+) < 0. we have

9OB+

n

n

u = 0

and thus, ai(u $)(x+) =

Combined with the definition of 3

a u2 Lvcr

axk axn

e

this leads to

B

2

+ vac ax k

+ E 8x (2u - $ -

ax

n

0.

n

To estimate the third derivative of

u

the derivative of (3.69) in the

direction and we get

xk

in the above inequality we take

3.5.

W2,cQM -Regularity

3 a

45

2

aann

u

a

u

_

ann axkaxn E axkaxn

2

3a

22u

nn axk

a ann 8u 8xk8xn axn -

_

axn

ann, which is greater or equal to

Dividing by

8

e aXk($ + )

v, and replacing

a3u 2 axkaxn

by its value in our inequality above leads after using (3.56), (3.70) to as

vc7(c

where

C

"Q ..

2

- al

axnn) axk axn (x+) n nn

doesn't depend on u

k

n

(x+)

< C0101 2,W +

e E (0,1].

- C(I l,I

I

But now from (3.79) we deduce that

I'PI 12,00).

Clearly the same argument holds if

and (3.90) follows easily.

x

E a0B+

and thus we can assume that we are in the: Case 2.

x+

and

are in

x

Let us assume first that

B+.

x+ E [u > #].

Then we have (see (3.48),

(3.49))

and

a1" (u - $)(x+) = 0

(3.91)

az(u - i)(x+) > 0.

Now the maximum principle in

W2'p(n) (p > n)

(see [25]) leads to

if A(v(aBu - w) + Au)(x) < 0.

x-'x But from (3.91), (3.87) and the fact that

ess l im inf [ai (u-$) (A+u-A+m) + az

Q+

is positive we get

(A+u-A+V') ] (x) < 0.

From (3.51) one easily deduces that this implies

A+u(x+) < IA+,IW

or

IA+V'I..

Using now (3.84) we get (3.90).

The case

x

E [u < V']

follows in

the same way, see Case 2 in the heuristic proof, and the case where x+ E [u < 4]

and

x

E [u > ]

is impossible.

To see this, just add the

two inequalities v(ciBu-w)(x+) + Au(x+) > v(aBu-w)(x-) + Au(x-)

V(aBu-w)(x) - Au(x) > V(aBu-w)(x+) + Au(x+) to get a contradiction.

(As after 3.78) thus (3.90) holds in all cases.

46

THE OBSTACLE PROBLEMS:

3.

The lower bound can be obtained by noting that

-u

satisfies a prob-

-m, 4.

lem of the form (3.67) with now m, changed to

By (3.89) the

and a compactness argument

x0

estimate (3.57) holds in a neighborhood of

A REGULARITY THEORY

Thus it remains only to prove Lemma 3.

shows that (3.57) is true globally. Proof of Lemma 3.

Step 4:

can be written as

f0 + afi/axi

First we remark that agi go + ax.

where the bounded in

gi's are of the same type as the

(i.e., in

fi

W2'p(B+)

and

C(Iloll2,- + 11*112,,,) -- see (3.82),

by the quantity

Lp(B+)

(3.55) but also with gn(x) = 0

on

(3.92)

a0B+.

in (3.82)) in

fn

But (see the form of

can be written like

afn/axn

Clearly, it is enough to check that this.

afn/axn

only the terms

2

2

and

(a axiax.

5

# n

j

n

To see that they can be suitably rewritten,

may not be of the above type. note that we have: a2u

a

ax

92U

a

n

axn

n

aann

1

(a 2) = ax a(ax

n

+ a

nn

(a

au ax + e4 + CC) n

n

as ax

as an (nn n

nn

2

au a _ a u axi axn(a xj) axn(B axiaxj) a

3u

+ et + EVp)

n

au as a axn (axi ax

as (3.69) allows us to conclude.

We then set for

denoting the sign of

x = (x,xn) E B, sign(xn)

(i.e., ±1),

in(x) = ani(x) =

sign(xn)ain(x,Ixn1)

9ij(x) = aji(x) = aij(x,Ixn1)

Vi = 1,...,n-1

for all pair

(i,j) # (k,n)

(k = 1,...,n-1)

gi(x) = gi(x,lxnl)

Vi = 0,...,n-1,

gn(x) = sign(xn)'gn(x,Ixnl)

xn

W

3.5.

2,w

(I)-Regularity

47

Now it is easy to check that the

iij(x)

satisfy the ellipticity

condition (3.1) and thus, there exists an unique solution

w E H1(B)

of

the problem

x.(aij 1,_

= 0

on

axj) = g0 +

B

P.

The gi's are in

W1'p(B+)

g0 + agi/axi E LP(B)

thus

on

z1 gi

and the

gi's are in

w(x,xn) = w(x,-xn)

tions are solutions of the above problem. on

(see (3.92)),

and (see [4], [65]) w E W2'p(B) a C1(B).

over, it is easy to check that

a (x) = 0

W1'p(B)

More-

since these two func-

Hence

a0B+

n

and the restriction of w

to

results simply from the theorem Remark 3.22.

satisfies (3.83).

B+

In the case where

The estimate (3.84)

of the appendix combined with (3.55).

Al

on

= 0

$ =

r

this proposition is

simpler to prove using an unpublished result of H. Brezis.

(See [51].)

We can now conclude the estimates of the second derivatives of u

e

by:

3.5.3.

Global Estimates.

Proposition 3.23.

If

$,V'

W2'-(S2)

satisfy (3.19), (3.20), and under

the assumptions of Proposition 3.17, then the solution in

W2'"(l)

e E (0,1],$,*

1 lu6112,Proof:

and there exists a constant such that

C(1 ICI 12,E + I ICI

(3.93)

First, let us estimate the derivative

in place of

Bu =

of (3.47) is

uE

which doesn't depend on

C

ague/axkaxt.

Set (with

ue)

a2u axkaxR

We have

aa.. B(Au) = A(Bu) + axi(B(aij)axj

By introducing the solution

2

as J

2

axk]

axjaxR + axR

+

w

of

axjaxk

u

48

3.

Aw

_

au (B(aij)ax

a

axi

THE OBSTACLE PROBLEMS:

2

aai.

a

u

a2u

aai

+ axk ax.ax

A REGULARITY THEORY

+ axk ax.axk)

w=0 on r the above inequality becomes A(Bu - w) = B(Au)

with, moreover, (see the Theorem AI of the Appendix and (3.55))

11w11-

C(11#112,,, + I I*I I2,o)

Now computing

(3.94)

by replacing Au by its value (see (3.52))

B(Au)

we get

A(Bu - w) = ai(u-0)(Bu-B0) + +

a2(u-Vi)(Bu-BW)

(u-$)ax (u-*) +

z(u

ax

R

k

Multiplying the two sides of this equality by

ax

k

t

v, then adding and

subtracting (3.86) we get the following two equations: A(v(Bu-w) + Au) = al(u-CQ+(u-$) + a2(u-lP)Q+(u-'P)

A(v(Bu-w) - Au) = al(u-O)(A_u-A_t) + a2(u-*)(A_u-A_Vp)

+ al(u-#)Q_(u-$) + aZ

_

where we have set A+v = vBv ± Av av

Q+v = v axk (Note that

of the sign

aij

av

av

av

axR ± aij axi axj

av Lv_

av

axi axj ± v axk axR

v(Iovl

av

2

av

± axk axt) > 0

Q+

is

±.)

The proof is now as in the preceding proposition. we denote by

and

x+

(Resp. x) a point in

v(Bu - w) - Au) achieves its maximum.

where

S2

If

x+

then by (3.55), (3.57) and the definition of

That is to say,

v(Bu - w) + Au

(or x) belongs to x+

(Resp. r,

we have

v(Bu-w) (x) + Au(x) < v(Bu-w) (x+) + Au(x+) < C(l I+I 12,,, + I hPI ! 2,,,) , VxESZ.

3,5.

2,w

W,

(Std-Regularity

49

Applying (3.55) and (3.94) we get

1B-(-) I < C( I 1$ 12,,, + I ['P112,.)' Thus we can assume

x+,x

(3.95)

E Q. but applying verbatim (with

B = 52,

a = 1) the arguments of Case 2 of the previous proposition leads again to (3.95) which completes the proof (for a variant of this technique, see

[52]). As a consequence for the obstacle problems we now have: Under the assumptions (3.46) if

(See [34] and [69]).

Theorem 3.24.

0 E W2'p(S2), 0 < 0

on

P, f E Wl'p(a), p > n

then the solution

u1

of

the problem <-Au1,v - u1> >

u

1

EK

0

is in W2(St) u1,0,f

Vv E K0

= {v E Hl0 (S2) v(x) > 0(x)

a. e. in

and there exists a constant

C

S2)

which doesn't depend on

such that (3.96)

11u11 12,- < C(IIfIIl,p + 11+112,,,).

Under the assumptions (3.46) if

Theorem 3.25.

(See [51].)

0 < 0 < *

r, f E W1'p(S2), p > n, then the solution

on

{<-Au2,v - u2> >

u is in

2

a

W2'(S2)

u2,0,'P,f

0,i, E W2'w(S2),

of the problem

Vv E KV

(x) > v(x) > 0(x) a.e.

E K* _ {v E H1(52) 0

u2

and there exists a constant

C

in

S2)

which doesn't depend on

such that

I1.2112,,, < C(I Ifl Il,p + 11+112,., + I1*112,m) Proof:

w

Let

ue

be the solution of (3.16).

By introducing the solution

of

-Aw = f,

w E H1(S2)

we easily see that of

(3.97)

0,*.

ue - w

satisfies (3.47) with

Thus from the Theorem 3.23 we get that

0 - w, 'P - w

in place

uc - w E W2'-'(1)

and

by (3.93) I Iue - wl12,W < C(IIO-wl

1'-w112,.)

12,-- + I

By regularity results for the Dirichlet problem (see [4]) and the Sobolev

50

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

embedding theorem (see [1])

11-1 12,-

C'1lfl Il,p

C1 1-113,p

and the above inequality leads now to

(3.98)

I IuEI12,., < C(l Ifl1l,P + 1101 12,- + I I*I Under the assumptions (3.48), (3.49) taking

s(t) < o l

Vt < o,

8 2(t)

= 0,

v+ = 110112,,

then we have that (see Proposition 3.4 and Remark 3.7) uE + u1

as

e + 0

and a standard argument gives us Theorem 3.24, (3.96) resulting from Taking

(3.98).

Vt < 0,

a1(t) < 0

Vt > 0

02(t) > 0

leads to Theorem 3.25.

W1'°(S2)-Regularity

3.6.

First let us take a look at the case of an operator with constant In this case the translation method works and we are even

coefficients.

able to get

here that

0 < *.

Let $, be two functions in Assume that

we have:

H0I(S2) n C0'a(37)

is a constant and

u

Indeed, assuming

0 < a < 1.

is a bounded Lipschitz domain of IRn

n

Theorem 3.26. with

regularity for all

CO'a(S2)

u

(0 < a < 1)

is the solution of

V..V(v-u)dx >

u E f 12

J

S1

and we have

u E CO'a(S2)

lula, < Max(IOla,I*1,,) where IvIa

=

Iv(x) - v(y) I

sup

yIa

x#y

Ix -

(x,y)E52 Proof:

(See H. Brezis - N. Sibony [36] and P. Hartman - G. Stampacchia

[68].)

For

outside

S.

v E H1(1) for

let us denote by

v E C0'a(52)

1"v(x + h) -v(x)I < Ivlalhla Now fix

h EIRn

v

the extension of

v

by

0

we have clearly

and consider for

Vx,h EIRn. C = Max(lO1a,1v,la)

(3.100)

the functions

3.6. W1'p(S2) -Regularity

51

v(x) = Max(u(x),u(x+h) - CIhIa), By (3.100) we have in

w(x) = Min (u(x),ii (x-h) + CIhIa)

IRn

$(x) < u(x) < v(x) < Max(i(x),i(x+h) - CIhIa) = (x) $(x) < Min($(x),+(x-h) + CIhIa) < w(x) < u(x) < (x) and so the restrictions to and

of

52

v(x) = u(x) + (u(x+h) - u(x) - CIhIa)+

w(x) = u(x) - (u(x-h) - u(x) + CIhIa)

are in

0.

Applying (3.99)

we get IRn

u(x) - CIhIa)+dx > fe u(u(x+h) - u(x)

fe

-

CIhIa)+dx

u(x) + CIhIa) dx > fin -1(u(x-h) - u(x) + CIhIa) dx.

Now changing

x

into

x + h

in the integrals of the

second inequality

gives us

e

Vu(x+h)-V-(u(x) - u(x+h) + CIhIa) dx

fIR

> -f

n

}i(u(x) - u(x+h) + CIhIa) dx.

]R

Thus adding with the first one, using that

fn V(u(x) -

(-f)

= f+, leads to

u(x) - CIhIa)+dx > 0

IR

tia

CIhIa)+I2

fn IV(u(x+h) - u(x) -

< 0

(Vh EIRn)

IR

So for almost all

x

we have

u(x + h) - u(x) < CIhIa

Vh EIRn.

That is to say, after possibly changing

lu(x) - u(y)I

Remark 3.27.

x

and

on a set of measure

y

CIx - yla,

gives the result, i.e., Vx,y E N.

The above method works also, of course, for inequalities of

type u E K*,

0

Vx,y EIRn.

u(x) - u(y) < Clx - yla,

Now a permutation of

u

<-Au,v - u> >

Vv E 0

52

THE OBSTACLE PROBLEMS:

3.

is an operator with constant coefficients and

A

where

A REGULARITY THEORY

(Indeed, if

able space.

problem to the case

is in a suit-

f

is smooth enough one can easily reduce the

S2

f = 0.)

to give a well-known application of this re-

f = u

We now choose

Let us denote by

sult to the elastic-plastic torsion problem.

the

6

Euclidean distance to the boundary, i.e., 6(x) = Infjx - yj yEP

with

for the usual Euclidean norm in IRn.

I

I

Vx E 52

Then under the previous assumptions upon Theorem 3.28.

If

we have:

12

is a positive constant the problems

.i

u E K = {v E H(12)Iv(x) < 6(x)

in

a. e.

S2}

(3.101)

Vv E K

v(v-u)dx J52

1S2

and

u E K' = {v E

H1(c2) I I Ov(x)

in

a. e.

<1

52}

(3.102)

f p(v-u)dx

I

J

Vv E K'

have the same solution.

By uniqueness of the solution of both problems it is

See [36].

Proof:

enough to prove that the solution First take

u

of (3.101) satisfies also (3.102).

in (3.101) we get that

v = u+

u > 0

and so

K

can be re-

placed in (3.101) by

K+ = {v E HI(S2)JO < v(x) < 6(x) But now

0,6

u

12}.

are Lipschitz with Lipschitz constant less than

Theorem 3.26 tells us that that

a.e. in

satisfies (3.102).

if the above operator

A = A

u

is in

K'

and the inclusion

1, so

K' c K

shows

One can again remark that the theorem holds is replaced by an elliptic one with con-

stant coefficients. Remark 3.29. [77],

One can show also that the following equalities hold (see

[42])

{x E 1Iu(x) < 6(x)) _ {x E Qlu(x)

ESI

I

< 11

= 6(x)} _ {xE521IVu(x)I = 11.

3.6.

A

W

(S2)-Regularity

53

For applications of Theorem 3.28 see [77], [42].

Note also that some

extensions can be given to the non simply connected case (i.e., in the case of (2.33)) see [42], [43], [53].

Let us now consider the case where We shall assume in this case that of ]Rn

has variable coefficients.

A S2

(that is to say at least of class

is a smooth bounded open set CI) and that

satisfies the

f2

following exterior sphere condition (see [55]): There exists

In the case

Remark 3.30.

x0 E r

such that for all

R > 0

exists a ball of radius

R

which meets

n = 1

there

only at

x0.

clearly (3.103) holds.

For

can see easily that this holds for

3

37

of class

(3.103)

n > 2

one

C2.

The coefficients of A will be chosen such that

aij E WI'°(S2)

(3.104)

and as previously, for

$,Ip E W1'-(S2)

we shall first consider

c > 0

Au. = C (uE

-)+

uE

-(uE - )

satisfying (3.19), (3.20) and for

the solution of the problem

C

(3.105)

uE E H1(0) where

S1

and

S2

are two smooth functions satisfying (3.9), (3.10).

In what follows we shall assume that the

aij's are in

CI(52).

Since

only through the norms

the different estimates depend on the aij

III

it is clear that the results extend under the assumption (3.104). 1

Using now the same arguments as in 3.5.1 we have clearly (3.53) and thus AuE E Lp(S2).

But the usual regularity results for elliptic problems in-

sure us that

uE E W2'p(S2) - CI(S2)

if

p > n

and so the following pro-

position makes sense; 'm

For

Proposition 3.31.

there exists a constant

$,t4 E W C

(S2)

IVuE(x)I < C Max(IIVO+II,,,IIVP Let

x0 E P

and the norms

x

in

3,

(3.106)

dx E r.

and choose our coordinates such that

with the center of the ball of radius Set, for

c,v,R

such that:

Ilaijlll,b

Proof:

and under the above assumptions

which depends only on

R which meets

52

0

coincides

only at

x0.

54

d+(x)

k(RP

=

k,p

d

Ixlp),

-

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

(x) = -d+(x)

being positive constants which will be chosen later on, Ixj

the usual Euclidean norm of

denoting

x.

First we have

6-(x 0 ) = uE(x0) = 6+(x0)

(3.107)

< uE (x) < 6+(X),

6-(X)

Vx E r.

Moreover, an easy computation gives for

x

in

n

adr(x) = kpxllxI-(P+2)

(3.108)

ax.1 a 2 d+

kp6..Ixl-(P+2)

ax.ax.(x) = i

(with

if

Sij = 1

Since

i = j,0

(3.109)

kp(P+2)xixjlxl-(P+4)

otherwise).

is bounded, there exists a real number

S2

included in the ball of center k,p

-

7

and radius

0

D

such that

f2

is

So assume first that

D.

are chosen such that: Iod+(x)l = kplxl-(P+1) > kpD-(P+l) = Max(IIoo+II,0,1lV

IIm)

(3.110)

VxE52. Thus, if (x,0)

and if

x E S2

meets

y

denotes the first point where the segment

we have (note that

r

6+

is radial) 4+(x)-O+(r)=0+(x)

d+(x) > d+(x)

-

6+(y) > Max(IIO4+II.,,IIOV+ IiIx-yj >

Hence if (3.110) holds we have in < b+ < d+

and

6

a

<

(3.111)

Now using (3.108), (3.109) we get -Ad+(x)

n

=-

2+

+

I

(aij (x)axaz.) + aX. aij WL. (x)) i i i,j=1 j

n

=

kpjxI-(p+2) .l,I [(P+2)aij(x)xixjlxl-2 - dijaij(x) - aX aij(x)xj]

Hence from (3.1) we deduce -Ad+(x) > kplxl-(P+2)Iv(P+2)

- M]

(3.112)

55

W1'"(Sd)-Regularity

3.6.

where

M

is some constant which bounds from above on

the quantity

SZ

n E

i,j=1

[6ijaij(x)!!a(x) + _x.]I. 1

Assuming M

is greater than

2v, we can clearly choose

k,p

such that

(see (3.110))

v(p+2) - M = 0,

kpD-(P+1)

= Max(IIO++II.0,1 vg

From (3.111), (3.112) we then deduce in

+ s +

I,.).

(3.113)

(see (3.9), (3.10))

0

s -A6+

-A6

S2

I

- (6 -b)

SE

(6

+

!2

-A6+ > 0 = -Aue + !-,(u e-

e

s2

-A6- < 0 = -Aue +

Hence using (2.15), (2.16) and test functions

(ue

(ue-1P)

(ue - 6 )

one deduces from these inequalities that

< u(x) < 6+ (x)

6 -( X )

Vx E S2

and thus IDuE(x0)1 < lV6+(x0)I = kpR (P+1). By (3.113) this finishes the proof.

Let us now deduce global estimates for Proposition 3.32. if

u

E

:

Under the assumptions of the previous proposition and

is the solution of (3.105) there exists a constant

ue

pends only on

52,v,R

and the norms

Ilaijlll,.o

C

which de-

(in particular not on

E)

such that: (3.114)

Iv.110, I Io*I I.,)

I IVuel 1p

First let us prove that we have

Proof:

loo+I

where

C

(3.115)

I.,, I IV i I

I IVu6112 <

depends only on R,v and the

11-ij1I i)

Indeed

v=0+is a function in

$
H01 (Q)

(see 3.20) such that

l,p'

56

(Since b+

THE OBSTACLE PROBLEMS:

3.

b < *+

- p- <

t>0

and

and

<

>

b

and so

,

b+

b =

Thus from (3.105) we get - since

- - = iy)

- b

S1 = 0

<

on

t < 0:

on

$2=0

b+

b+

we have

A REGULARITY THEORY

v> _ -J52[sE(uE-b)

<-Au E,uE

0

(uE4) - El(v-b) -

+

and so au

au f aij(x)

xi i x i

SZ

au dx < J sij(x) axe

ax

i

dx. 1

Applying now (3.1) and the Cauchy-Schwarz inequality leads to (3.116)

'JIlVUE11 2 < CI IVujj I2I IovI I2 which gives (3.114) by our choice of

v.

Now let us take the derivative of the first equation of (3.105) in the

We get (by writing

direction.

xk

u

in place of

uE)

-A(axk) + S1(u-b) (ate (3.117) a

(aaij au

axi axk

Let us define

axj

by

T

(3.118)

T =

where

C

sets (u

is the constant which occurs in (3.106) and

is in

W2,p (S2))

521,522

the open

:

a, = {x E 52I ax > T} k

S22 = {x E 52I aX < -T}. k Moreover, in au

521,522

vl = axk - T,

respectively, let us set: au

v2 = - xk - T.

We thus obtain two positive functions in fy (see (3.117) and note that

Si > 0)

HO(521)

and

H1(c2)

which satis-

3.6.

Wl'W (0)-Regularity

57

as

-Avl = -Au - T)

-A(8x

k A(

-Av2 =

k

ax(a i

)

au aa axkl

auk - T) = -A(- axk)

ax.(

But now we can apply the Corollary A2

in

S2

k

(3.119)

aX)

of the Appendix.

in

522.

More precisely,

if n = 2, using (iii) of the Corollary A2 from (3.115), (3.119) it re-

vi E L(c)

sults that

for all

(i = 1,2) and that

q

Iviiq,s2i < C Max(IIV II,,,IIo*II.,) denotes the

(I

Iq,S2

depend on

Lq-norm in

By dividing

E.)

provides us with au E Lq(1)

52

and

SZi

into

S21,S22

some constant which doesn't

C

and

this

{x E SZI Iau I < T}

and an estimate

k

auax

k

for all

-< C Max(I Iaol I, , I IV*1 L,),

q

q > 2.

choosing

q > n

But now since this estimate holds clearly for all

Thus with the same decomposition of au axk

k,

we can apply the point (i) of Corollary A2 to (3.119). 0

as above we now get

< C Max(IIo0IIm,1IVOI.)

which is (3.114). In the case

n = 1, by (3.115) we are in case (i) of Corollary A2 and

one can conclude as above (of course, in this case one can provide a shorter proof).

vi E L

If now

n > 2, (3.115) and (ii) of Corollary A2 show that

and an estimate

(SZ.)

IviI2*,s2 < C Max(IIooII,,,IIopjL).

i

But the decomposition of

S1

into

5211522

and

{x E lI I aX I < T}

k-

combined

with the above inequality provide us now with: Du axk

Thus if viously.

2* < C Max(IIv$II.0.IIvl)II.) 2* > n If

we can apply (i) of Corollary A2 to conclude as pre-

2* < n, and since the above inequality holds for all

get by applying the point (ii) of the Corollary A2 with

I Uk p

* < C.Max(I IooIIm,I IVpII.,)

p < 2*:

k, we

58

p < 2*.

for all n

THE OBSTACLE PROBLEMS:

3.

for some

p*

A REGULARITY THEORY

Repeating the procedure, it is clear that we can exceed and the result is now an application of (i) of Corollary

A2 which concludes the proof. Because of (3.106) and the estimate (see (3.115))

Remark 3.33.

IVuII2 < C Max(IIoo+II2'IIo I12) one might expect that an estimate of type

JIVueI Itl < C Max(I Ioo+I I.o, I IVI-I I, )

(3.120)

holds (this would lead to a similar estimate for the solution of the homogeneous obstacle problems).

But this is not the case in general (see Re-

mark 3.36).

Note that with the same proof as in (3.53) we have

u£ I .o < Max (I c+ I

.,,

I 'P

I.,) .

As a consequence of the above estimates, we are now able to prove for the obstacle problems:

Let

Theorem 3.34.

<0

o E W1'°(52),

on

r

and

f E Lp(52)

with

If (3.17), (3.18) hold then under the above assumptions the solution

p > n. uI

of

u1 E K.

<-Aul,v - u > >

Vv E K

I

1'

is in

W

(0)

and there exists a constant

C

which doesn't depend on

such that

f,$, ul

I IVull I. < C(Iflp + I Idol Im). Let

Theorem 3.35.

u2

of

u2 E

0,

is in

satisfy (3.19), (3.20), f E LP(Q)

with

If (3.17), (318) hold then under the above assumptions the solu-

p > n.

tion

(3.121)

W

'

<-Au2,v - u2> >

(52)

Vv E KIP

and there exists a constant

C

which doesn't depend on

f, ",u2 such that

IVu2I I® < C(I flp + I ldol l + ! IVPI IW). Proof:

For

81.82

(3.122)

satisfying (3.9), (3.10) and 4, satisfying (3.19),

(3.20) let us consider the solution

u6

of:

WJ,W(Q)-Regularity

3.6.

-AuE +

59

e (°E - O) +

uE E H (S2)

SE (ue - V+)

=f

.

Introducing the solution

of

w

w E H1

0(52)

uE - w

we see that

satisfies (3.105) i.e.,:

A(uE - w) =ES ((uE - w) - ($

w) ) +

s((uE -w) -

w) )

E

uE - w E H1(S2) and by Proposition 3.32 we get

< C Max(IIV(e - w)II.,,IIV(* -

I lv(uE - w)II

But by (3.18) and the Sobolev embedding theorems we have

1IOwII. < Ciiui'2,p `- CIfip

and thus we get easily

IlvuCIL < C(Iflp + iIveli + where when

doesn't depend on

C

(3.123)

Taking now

E.

0, , = IOIm, pl < 0

s2

we get (3.121) from (3.123) by applying Theorem 3.4 and by

t < 0

taking the limit of (3.105) when 3.35 follow by choosing

c

0.

when

01 < 0

Similarly (3.122) and the Theorem

t < 0, S2 > 0

when

t > 0

and by

passing to the limit in (3.123). Remark 3.36.

where

ul

It is clear from (3.121) that the mapping

+ + u1($) = ul

is the solution of

ul E KO, <-Aul,v-u1> > 0

_ (v E H01 (S2)Iv(x) > 4(x)

Vv E K

a.e. on Wl'm

is continuous at

On + 0

in

0

W1'W (52)

on

(3.124) S2}

W1'm

(0)

- i.e., if n E

then ul (+n) * 0

in H (ft) n

is not generally continuous in W(S2)

when

[30] we easily see that in ]Rn (n > 1) and for

(S2), On < 0 W1'm

n > 1.

(S2) .

But

on

r,

+ + ul (#)

Indeed following

60

THE OBSTACLE PROBLEMS:

3.

SZ= {xE]Rnl

A REGULARITY THEORY

lxi
(E>0)

lxl

the solution of (3.124) is given by (use similar arguments that those right after (3.45))

if r<1-/

r

(E

u, ($)

)2(1

(1 Now clearly 1'0 in

W

if r > 1

- 1)

But when

11vu1(0)1jp = 1.

goes to

a

and the solution of (3.124) for

(S])

vl_E.

-

m0

0, 0 + $0(r) _ -r

is given by

ul($0) = 0.

Thus we cannot have

u1($) + ul(#0). A similar argument shows the impossibility of (3.120) in general.

Indeed

D

being as above and

A

equal to

A

let us consider

f = 0

solution of the two obstacle problem where

u2

the

and the obstacles are

given for instance by

(x) = c(g - It - Zl)+.

Ve(x) = r = ixi

(Then (3.120) implies that Iv0+l

I lvu211- < C Maxi and

u2

show that

3.7.

would go to

0

W

in

C'E This is impossible since one can

(S2).

on a set of positive measure around

touches

u2

l., i Iv I

0.)

W1'p(Sl)-Regularity (2 < p < +-).

The key arguments in this section will combine the results of 3.6 So let us begin with a few words about

with interpolation techniques.

Our statements will be as simple as possible, and we will

interpolation.

just explain what is useful for our purpose.

Details and extensions can

be found in [20], [39] or [108]. Let

A0,A1

be two Banach spaces with ncrms

and

lA l

res-

lA l

0

1

pectively and such that AI c A0.

We will confine ourself to the t E]R+, a E A0 let us define K(t,a) =

inf a.EAi

a0+aI=a

K

K-interpolation method.

by

+ tlallA

(laDIA 0

1

So for

W1'p(U)-Regularity (2
3.7.

0 < 8 < 1,

Then for

61

1 < q < +m

(other values of

6,q

are suitable

but we will not use them here) set AOIt-8K(t,a) E Lq(IIt+, dt)},

(A0'A1)8,q = {a E

denotes the space of equivalence classes of functions of

(Lq(JR+, dt)

the qth power integrable with respect to the measure a E (AO,A1)8,q

For

dt/t.)

we will set 1/q

rf

IaIO,q -

LJjR+

8K(t,a)]q dtJ

[t

Then we first can prove that

polation" spaces between

and

(AO,A1)6,q

I

I8,q

gives a family of "inter-

A1, that is to say we have:

Under the above assumptions

Theorem 3.37.

for the norm

A0

(3.125)

(AO,A1)0,q

is a Banach space

and we have the algebraical and topological embeddings:

c (AO,A1)O,q c A0.

A 1

Proof:

See [20], [39] or [108].

Now it is well known that linear bounded operators mapping a pair of Banach spaces to another pair also map continuously their interpolation spaces.

As we will see, this result extends to the case of Lipschitz

functions.

(For more general results, see [104].) A1,AO,B1,B0

So let

Banach spaces with norms denoted by

I

IA.,

such that

B.

1

Al c A0,

B

Under the above assumptions, let

Theorem 3.38. B0 C

T(A1) c B1.

such that and

C'

B

T

be a map from

VaO,a6 E A0

< CIa0 - a6 IA

0

IT(al)IB

< C1Ia1IA

0

Vat E A1.

1

Then for all

to

Let us assume that there exist two constants

such that

IT(a0) - T(ao)IB

A0

a E (AO,A1)8,q , T(a) E (BO,B1)8,q

IT(a)IB,q < Max(C,C')lal8,q.

and we have

62

THE OBSTACLE PROBLEMS:

3.

A REGULARITY THEORY

Let us write

Proof:

VaI E Al,

T(a) _ (T(a) - TCa1)) + T(al) By definition of

K

(between

B0

Va E A0.

BI) we get

and

K(t,T(a)) < IT(a) - T(al)1B0 + tIT(al)IB 1

+ tC'IalIA

< CIa - allA 0

1

< Max(C,C')[Ia - aIIA

+ tIaIIA ].

0

Since this inequality holds for all

1

aI E AI

taking the

Inf on

a1

we

obtain:

K(t,T(a)) < Max(C,C')K(t,a)

and the result follows from our definition of the interpolation spaces and from (3.125). on

Ie,q

Note that we have used the same notation for the norm

(AO,AI)e,q

and

(BO,B1)e,4

but there should be no confusion.

In the framework of the one obstacle problem we now want to apply the above results with W01,2

A0 = B0 =

(Q),

Al = B1 = W0'm(0)

where we have set (see (2.4)) II

Wl'W

(0) = W"02 (Q) fl W1'tl(0)

with the norm

III,., As a consequence of the interpolation results of [56] we have if

S2

is a bounded, Lipschitz open set of 19": Theorem 3.39. (W01,2

(0),

Proof.

For all

p

such that

w0''° (c))1-

wo,P(c).

2,p = P

See [56].

From now on, we will assume that of 1Rn (3.2)

2 < p < +.o

(for instance of class

0

is a smooth, bounded open set

C2) and that the operator A defined by

satisfies (3.1) and has its coefficients in

Cl(S2).

Thus we have (see [23]).

Theorem 3.40.

Let

f E W-1'p(S2), 2 < p < +.o. Then for all

E W1'P(Sl),

$ < 0 on r, the solution uI of u1 E KV

<-Aul,v - u1> >

Vv E K

(3.126)

3.7. Wl'PCS2)-Regularity (2
is in

W0'P(S2)

63

and we have for some constant

which doesn't depend on

C

O,f,ul

I IujI I1,P

C(I IfI I_l,P +

Let us assume first that

by

W-1'P(SZ).

denoting the strong dual norm on

IIfHI-l,p Proof:

(3.127)

11i,p)

I1

f = 0.

Then (see (3.7)) one can replace

so that in this case one can assume

0+

E W0' 2 (0)

0 E WD'P(S3).

Thus for

let us consider

T: O - T(O), where

T(O)

For

is the solution of (3.126) with

0,0' E H1(S2)

we have

T(0)+(0' -0)>0',

T(O) > 0

<-A(T(O)), v - T(O)> > 0 <-A(T(O')), v - T($')> > 0 v

equal to

and

T($') > O'

thus

T(0') + (0-0')>0,

and the above functions are all in

with

f = 0.

T($') +

HI(S2).

Thus using the inequalities

Vv E KO Vv E KOI

in the first and

v = T(0) + (0'

-0)

in the second one leads to <-A(T(O)),(T(O') - T(O)) + (0 - O')> > 0 <-A(T(O')),(T(O) - T(O')) + (0' - O)> > 0.

By adding these we get

<-A(T(O) - T(O')),T(O) - T($')> < <-A(T(O) - T($')),($ - O')> and (3.1) with the Cauchy-Schwarz inequality leads to

IIT(O)

Now if

(3.128)

- T(O')I11,2 ' CIIO - O'1I1,2. 0 E WO'p(c)

by Theorem 3.34 we have

T(O) E W0J'W(0)

and

I IT(O) 111, < C' I I OI 11,,, Thus these two last inequalities combined with Theorems 3.38 and 3.39 give us that for

0 E WQ'P(S2), T(O) E

IIT(O)Ill,p -r C11#1 11'p.

and an estimate

64

This proves the theorem when

solution w

A REGULARITY THEORY

THE OBSTACLE PROBLEMS:

3.

For the general case introduce the

f - 0.

of

Aw = f w E H1(St). We can check (see the proof of Theorem 3.15) that of (3.126) with where

uI - w

is the solution

in place of . But from our study of the case

0 - w

f = 0, and since by regularity result for the Dirichlet problem

w E W0I'P(S2), it results that

u1 - w E WI'P(c)

with the estimate

- C11# - -111,p,

I lul - w11l,p

and (3.127) follows from the fact that we have

IIw111,p < CI If11_l,p Remark 3.41.

In the case of two obstacles, due to difficulty of obtaining

the equivalent of (3.128) and due to the fact that in this case it is not possible to assume that

= 0

_

r, the above technique doesn't

on

apply.

To conclude this section, let us take a look at the continuity in of the mapping

Wl'p(S2)

(3.126). 0 -

u1(0)

For

where

0 - u ($)

W

fixed in

f

is the solution of

W1'p(SZ)-strong (2 < p < +°)

W1'p(0)-strong into

from

u1(0) = ul

the problem of the continuity of

}'P(s2)

is

However, some interesting results

actually open (see Remark 3.44 below).

In particular, one can

have been obtained by L. Boccardo - F. Murat [24]. prove:

Theorem 3.42.

Let

stacles such that

f

be fixed in

on E

H01 (Q).

and

H-1(S2)

If

¢n

0

in

be a sequence of ob-

n W1'p(c)

weak (2 < p <

n - +.o, then

when

in

ul (en) -* uI (e)

HO (SZ)

strong.

To prove Theorem 3.42 we will use the following result (see [93],

[31]

for a proof).

Theorem 3.43.

Let

S2

a sequence suchthat fn

be a bounded open set in 1Rn. fn > 0

f in H_ (Q)- weak,

If

fn E H-I(S2)

(in the measure or distributional sense), ( ) then n in W

f

f

strong for all

q < 2. Proof of Theorem 3.42:

Let

u1 = ul(,n)

be the solution of

is

W1'P(H)-Regularity (2
3.7.

65

a.e. in

{v E H1(S2)lv(x) > en(x)

ui E K,

n}

n 1 <-Au i ,

(3.129)

ui> >

v

Vv E K-0 K.

n

in (3.129) we easily obtain

v = On

Taking

C I

where

C

doesn't depend on

IlOnIl1,2

exists a subsequence of

ul -I u n

From

(We note that by our assumptions

n.

is bounded independently of

in

H1

(u1,On)

n.)

in

L2(S2)

and pointwise a.e.

On

in

L2(S2)

and pointwise a.e.

un(x) > - pn(x)

(u1,On)

such that

R

ul + u m

Thus, (see Theorem 2.5), there

still denoted by

a.e. we easily deduce that

u E K,.

Moreover, re-

1

writing (3.129) as <-Aui - f,v - ui> > 0

W E K4, On

and taking

-Au1

v = ui +

where

C > 0,

E -9(5))

we obtain that

- f > 0 in 0.

Applying Theorem 3.43 we obtain (using for instance the fact that a limit

is unique in ' (S2) )

-Aui - f -+ -Au - f Now taking for

in W'() (p1 + p = 1).

w E K,, v = w + n -

<-Aui,w> > <-Aui,u1> +
-

in (3.129) we have

4,

ui>

(3.130)

-

<-AuI - f, 0n

(3.131)

Taking the limit - inf of both sides we obtain using the lower semicontinuity

of u - <-Au,u>

and the fact that

(3.130) and from our assumption on <-Au,w> > <-Au,u> + This proves that

u = ul

<-Aui -

4,> - 0

(this is due to

n)

Vw E K,.

and the weak convergence of

ui

The strong convergence is obtained by noting first that:

toward

u1.

66

THE OBSTACLE PROBLEMS:

3.

vllun-U11112 1

A REGULARITY THEORY

< <-A(un-u),u n - u > = <-Au n un-u > - <-Au

,2

1

Then applying (3.131) with

1

1

w = uI

1

1'

1

1

1

un-u >. 1

1

we deduce:

vllun-ul1I1,2 < + <-Aui-f,,n - > - <-Aui,ui-u1> -+ 0 with

n

and the result follows.

Remark 3.44.

As a consequence of Theorem 3.42 and under the same assump-

tions we obtain also (see [24])

ul($n) -r u1(0) for all

q

in Wl,q(p)

satisfying

strong

2 < q < p.

Indeed, as a consequence of Holder's inequality, we have

u(0))IIq < IIV(ul(@n) -

< CIIO(u1n) - ul(4))II2 with

r =

2(p-q) q(p-2)

and

s =

p(q-2) q(p-2)

uI(O))IIP (by (3.127))

Appendix

LP-Estimates for the Solution of the Dirichlet Problem

In this section ® will be a bounded open set of IRn

m L (®)

be functions in

vj with

V > 0.

f0,...,fn

a..

will

1J

Vx E® , VE E,n

2

(A. 1)

A will be the operator defined by (3.2). Under the above assumptions, if

(Stampacchia [100]).

Theorem AI.

and

satisfying the usual coerciveness assumption

Lp(®)

are functions in

with

p > 2

and

u

the solution of

the Dirichlet problem n

2f

-Au = f0 +

.

in

2x1

i=1

1

(A.2)

Then

Le(®) (i)

If

u E

p > n, we have

and

u

satisfies an estimate

of type n jul.. < C

Ifil

E

i=0 (ii)

If

2 < p < n, then we have

by P =

and

-

P

n

u

u E L

(®)

where

P*

is defined

satisfies an estimate of type

n

Ifilp

lulp* I 0 i= 0 (iii)

If

p = n = 2, then we have

67

u E Lq(®)

for all

q

and

APPENDIX

68

n

lulq < C lIo 1fil2' The constants of the

f i

;

First, let us assume that

Proof:

u

and

LP(Q).

LP-norms in

denotes

lp

i

in (i), (ii), and (iii) are independent of

C

p = 2.

Then from (A.1), (A.2), and

(2.7), we deduce that

2

n

t

r vllvul12 < J

0

a..u u ij x) xi

f u - f.u 0 1 xi

e

< C l IfiJ I1Du(A.3) 2112' i=0

Hence, it follows that n iiVu112 < v

(A.4)

lfil2

I

'

i=0

Thus, if

p = 2 = n, (iii) follows immediately from Sobolev's inequality

(see for instance [82]).

Now if

n = 1, &= (x0,x1), we have

lu(x)i < fx lu'(t)ldt < lu'l2ioil/2

-

x0

and this combined with (A.4) gives

n

nn

lulp < C

E i= 0

n

ifil2 < C'

L

i =0

which proves (i) in this case.

ifilp

Thus we now assume

n > 2

and that we are

not in the case (iii). k > 0, E

Let us consider, for

E= (u-k)+- (u+k)

defined by

.

As the sum of two functions of H1('), it is clear (see Theorem 2.4) that

E

equal to

is a function in H(&) which ±(lul

- k)

on

Jul > k.

vanishes on

Let us denote by

Jul < k A(k)

and is

the set

A(k) _ {x E Ol lu(x) l > k}. An application of the same arguments as in (A.3) with Holder's inequality leads to

viivEii2 <

!e

ai)xJxi

<-Au.E> =

1e

f0E (A.5)

n < C fiEx.

ifilp'liVEilp 1I0

Appendix

where

69

depends only on &.

C

But now since

p > 2, that is to say, p' <.2,

we have by Holder's inequality

IIOEIipi p

Ip,-2/P')p'/2IA(k)ll-(P'12)

Ip' < (JI

'A(k) IO

=

1

A(k) 1

where

Ip, <

denotes the measure of A(k).

IA(k)I

1

IIOEII2IA(k)I7_

2

ab Ilv

From (A.5), we then deduce

that

n

IMII

C

2

p

i=0

p

-1

Ifil

E

But using now the Sobolev's inequality (2.8), we deduce that -1 IfiIP'IA(k)IP.

< C

IEI

(A.6)

E

i=0

P

pi* - - n = 1

with

(Let us remark that

C

depending only on & and Indeed, if we have

is well defined.

p'*

n > 2, we would have

since

and

- P - n

Now, let us note that for

p' = p = 2 = n

h > k, we have

p.

p' = n,

which is the case (iii).) and so for such a

A(h) c A(k)

h, we have k)P'*j1P

(h-k)IA(h)l"p '* < Lj

(Jul

-

A(h) (Jul

< If

-

2

n

k)P ICI

A (k)

* P

C I If i=0

I

1 P

IA(k)IP,

1

This can be written as

n /(h-k))p*

IA(h)I < (C

n

i0

Ifil

IA(k)IP

'*( 2r

-1) (A.7)

P

Let us now use the following lemma which will be proved later on. Let : [0,+.0) -+]R+

Lemma.

$(h) < (h'k)a'$(k)o where

c,a,S

(i)

If

be a nonincreasing function satisfying

h>k

are positive constants.

0 > 1, we have (d) = 0

(A.8) Then

for

d = c. (0)S-1/a

2S/S-l

70

APPENDIX

/2 1/1-S

a < 1, we have

if

(ii)

(h)

h

u

cl

l

with

u = a/1 - s

for

\

h > 0.

More precisely, if we put (t) = IA(t)l, then we have by (A.7) an inequalNow if

ity of type (A.8).

is greater than

(1 - P)/(1 - P - n)

(d) = 0

that

p > n, the exponent

0-1

E

IfilP 'I ®Ipi*

and by (i) of the lemma we deduce

1

a.e.) with

Jul < d

(i.e.

n

1) =

0 =

0 25-1

d =

C

i=0

Now if

Hence (i) results.

p < n, we have

and by the

$ = p'*(P, - 1) < 1

part (ii) of the lemma and since

p'*

1-P

=

-

1

=

1

1

-l) p,-*-P,+1 P, - n - P, + 1 pn

2 (__

1*

2

1

1

1

2

1

1

= P,

we deduce that

n C 11Olfilp P* I[Jul > h] I

(A.9)

h

i = 0,...,n

Consider now for

the mapping

Ti: f -+ ui = Ti(f)

where

ui

is the solution of

-Au. = of 1 axi

ui E H1(®) (with

of

Then by (A.9) for all

f).

-gx

2 < q < n, Ti

is a linear map

0

from

Lq(B)

Lq (®)-weak.

into

for instance, 2 < p < q < n,

Choose,

into L2*(®) and Lq(&) into Lq*(&)-weak.

then Ti maps L2(Q)

Therefore by the Marcinkiewicz Theorem (see [20], p. 6) T.

maps

LP(®)

*

into

LP (B)

continuously and we have by (A.9)

ITi(f)Ip* < Clflp n Since now

u =

T(f.), (ii) follows.

I

1

i=0

What remains now is to prove

1

the lemma.

Proof of the Lemma:

By (A.8), we have for

hn = d - n (n E AI) 2

(hn+l) < (h

c - h

n+1

n

)0'0(hn)S =

ca 2(n+l)a,(hn)g d

Appendix

71

na

$(hn) < ,(0)21-8

Now we prove by induction that

.

If

n = 0, it is clear.

n, from the inequality above we deduce

If it is true at order

nab 21-8

ca (

(hn+l) < d

But by definition of

this leads to

d

nsa -0 t(0)2(n+1)a.21-8.28-1

$(hn+l) <

(n+l)a

= (0)2 1 8

Thus, we have na 21-8

0 < $(d) < $(hn) < 0(0)

when

n * + if

For (ii) put (h) _ (i)u$(h)

and (i) results. (A.8), since

-0 0

(h) =

we have for

with

8 > 1

p =

18

> 0.

By

h > k > 0

(h)up(h) < (A.10) hu

(h) <

(h-k)ak

Moreover, since

$

is bounded by

0

when

h + 0.

So let

be such that 0 < p(t) < 1 Let

h

be in

Vt E (0,T).

(0,T), from (A.10), we deduce that for all

n

h)P V+(2nh) <

n (2 h-2

h)

(2

n_l

h)

us

-- *(2'h) < 2"[p(?-1h)]0. And by iteration this leads to

'Y(2n

Since

h) < 2u C1+8+...+8n 1)(V(h))8 n

h E (0,T), we have u

1

21-8

(2h) <

Vn E IN

Vh E (0,T)

21-5

and so

'

is bounded on

(0,+m)

by

which gives the result.

T

72

APPENDIX

Remark 1.

Let us note that the theorem can be slightly improved by taking

f0 E Lq(c), 9 = 1 + n

and by replacing

by

If01q

in all in-

If01p

equalities (indeed it suffices in (A.5) to apply Holder's inequality and the Sobolev embedding theorem). As a corollary we have: Corollary A2.

Under the assumptions of the Theorem Al and if

u

satis-

fies n

af.

-Au < fo +

ax1

i=1

u > 0

in

in & (in a weak sense)

i

(A.10)

'

u E H1(0) then (i), (ii), and (iii) hold without any change. Proof:

Let us call

v

the solution of (A.2).

Then from (A.10), we have

-Au < -Av.

By considering

(A. 11)

(u - v)+, we have from (A.11)

vI iV(u-v)+I Iz < <-A(u-v),(u-v)+> < 0 and so the result follows from the Theorem A

since we have 1

0
Note that the above results are true without assumptions of

regularity on B and

except that they have to be bounded. a1

Comments for Chapter 3

73

COMMENTS

Before commenting on the regularity theory itself, let us note that an axiomatisation of the penalty method can be found in Lions [83]. Regularity in

W2'F(0) (2 < p < +') has been initiated for obstacle

problems by Lewy-Stampacchia [80] (c.f. also, [76] for a description of this method).

Other proofs and results are available in Brezis-Stampacchia

[37] and Stampacchia [101] and also via the Lewy-Stampacchia inequality (see [81]) in Mosco - Troianiello [92]. For

W

2 'b

(0)-regularity the pioneering papers are those of Brezis-

Kinderlehrer [34] and Gehrardt [66].

Caffarelli-Kinderlehrer [46] give

other techniques using potential theory. Let us note that

W2'w(0)-regularity is a useful tool to study the

free boundary of the problem. Kinderlehrer et. al.

We refer the reader to Caffarelli [40] and

[74], [75], and [76] for this kind of result.

In the case of variable coefficients

WI'-(0)-regularity theory follows

Chipot [50], see also [54] for other variants. for

Other techniques and ideas

C0'a(0)-regularity can be found in [22], [46] and [64],

[96] when

a = 1. An

Lp(0)-regularity theory is done in Adams [2].

for interesting results via capacity theory.

See also Adams [3]

Finally we note that we have

not included here regularity theory for thin obstacle problems. reader may refer mainly to [41], [60], [70], and [73].

The

Chapter 4

The Dam Problem

Among the applications of Variational Inequalities with obstacles, the most famous one is probably that of the Dam Problem.

The modern ap-

proach started with the pioneering work of C. Baiocchi (see [9], [10]). Many developments have followed, many of them due to the Pavia school, [16], [17] and [13], [15] where a complete bibliography

(see [11], [12], can be found).

Unfortunately this approach is only possible in the case For instance, the case of a rectangu-

of porous media with vertical walls.

lar dam (as in the figure (4.1(B)) is suitable in dimension 2 (see [76], [102] for the case of porous media in three dimensions).

To overcome

this restriction several attempts have been made including formulations via Quasivariational Inequalities (see [11], [12], "maximal" solution (see [5]) proach).

[15]) and the use of a

(see also [109] for a very interesting ap-

More recently H. Brezis, D. Kinderlehrer, and G. Stampacchia in

[35] and H. W. Alt in [7] introduced a new formulation which can treat the case of general domains.

We shall adopt the setting of [35] and we will

show how it can be reduced to a Variational Inequality with obstacle in the rectangular case.

Let us first begin by presenting the physical prob-

lem (see [32], [49]).

4.1.

Statement of the Problem 4.1.1.

Notation

We will restrict ourselves to the case of two dimensions; 52 will be

a bounded domain with a locally Lipschitz continuous boundary

S.

represents the section of a porous medium.

We subdivide

S

into the following subsets (see Fig. (4.1)):

74

0

4.1.

75

Statement of the Problem

(A)

(B)

Figure 4.1

S1

the impervious part of the dam

S2

the part of the boundary

S3

the part of the dam under water.

b2

of

S2

in contact with the open air

For convenience in some theorems we shall assume that subset of

and that

ab2

S3

SI

is open in the complement of

is a closed S1

in

S.

Moreover, in the case of several reservoirs we shall also denote by the different connected components of S3. For inS3 1'S3 2'" .'S3 n stance, in the above figure S3 = S3,1 U S322. Now, instead of giving directly the proof of our theorem in full generalities we shall often restrict ourselves to the following simpler case:

The domain say

I

S2

will be assumed to be vertically convex, that is to

will satisfy:

d(x,y),(x,y') E 52

the segment

{x} x [y,y'] c Q.

(4.2)

Moreover we will assume that: S1

is a closed connected subset of S 52

S2 U S3 S

which surrounds (4.3)

from below surrounds

52

from above

(4.4)

has only a finite number of vertical walls.

(4.5)

Note that both examples given in the Figure (4.1) satisfy these assumptions.

More precisely, if we denote by the

x

axis, for

'rtx

the usual projection of IR2

x E nx(0), we can define

on

76

THE DAM PROBLEM

4.

S -(x) = Inf{yl (x,y) E 92},

S(x) = Sup{yl (x,y) E SZ}.

(4.6)

Then, (4.2) is clearly equivalent to 92 = {(x,Y)lx E wx(f2),

(4.7)

S _(x) < y < S+(x)}.

Now (4.3), (4.4) mean respectively that is a connected piece of

S1

of

(x,y)

such that

S

contains all points

S, U S,

which contains all points

S

(4.8)

y < S (x) (x,y) of

S

such that

(4.9)

Y > S+(x) Moreover, (4.5) implies clearly that is continuous on

S

7x(92)

except perhaps on a finite (4.10)

subset 9 is continuous on

S+

7x(92)

except perhaps on a finite (4.11)

subset S°+. (Note that vertical walls can occur on 9' the end of the dam; of course at these points

U.97+

S-, S+

and eventually at have a right and

left limit -- when it makes sense.)

These assumptions are by no means necessary to study the mathematical problem in question.

However, they should help the reader in a first

After each statement under the assumptions (4.2)-(4.5) we will

reading.

indicate how to

treat the general case.

The basic problem now is to find the pressure fluid in

92

4.1.2.

and that portion of 2

FI

r3 = S3

A

is divided in four parts (see the Figure

the impervious part, then

P3

r2

the free boundary of

the part covered by the fluid and finally

the dam in contact with the air. only

of the A.

Strong Formulation

The boundary of First

p = p(x,y)

which is wet -- i.e., the wet set

r4

(4.1(A)).

A,

the wet part of

Among these pieces of the boundary,

is completely known.

Experimentally, the velocity v of the fluid, which we assume for instance to be water with a specific weight equal to

1, is given in

A

by Darcy's law

v = -k0 (p + y) -11

with

k

being the coefficient of permeability and

V

the operator (-L, ax

ay)

77

Statement of the Problem

4.1.

If the medium is assumed to be homogeneous, then

is a constant

k

which we shall choose strictly positive and in this case the incompressibility of the fluid leads to div(v) = 0

in

A.

Hence Ap = 0

in

(4.12)

A.

Assume now that the atmospheric pressure is chosen to be lect the effects of capillarity and of evaporation. by

and neg-

0

Then if we denote

hi (i = 1,...,n) the level of the water in the reservoir with bottom

the pressure on S2 U S3 S3,i ous function 0, where on

0

has to be given by the Lipschitz continu-

S2 (4.13)

0(x,Y) = i -y p

Thus

on

Vi = 1,...,n.

S3,i

must fulfill the following Dirichlet boundary conditions

p = 0

F2 U F4,

on

on

p =

(4.14)

t3.

Moreover the fact that no water can flow through

and

r1

F2

leads

to conditions of Neumann type viz: -kv(p + y).

= 0

on

U F2,

r 1

where v = (vx,vy)

denotes the unit outward normal to

A.

This also

gives:

8v (p + y) =

0

on

F1 U F2.

(4.15)

Finally, we can express the fluid flow through 0

k

(p + y) < 0

is strictly positive, by

on

(4.16)

r4.

The problem is now to find and (4.16) hold.

by

F4,

on

and thus, since

I'4

(p,A)

such that (4.12), (4.14), (4.15),

78

THE DAM PROBLEM

4.

4.1.3.

Weak Formulation

We introduce here the ideas of H. Brezis, D. Kinderlehrer, and G. Stampacchia [35] (compare with [7]).

Assume that we have found a pair

which is a solution of (4.12), (4.14), (4.15) and (4.16) with

(p,A)

and the boundary of we have for

E E CI(S2)

Vp.VE + E

f

A

A

Y

=

p

By applying the Green's formula,

smooth enough.

(see Section 2.1):

1AoP

j8A 8v L (p + Y) E

E

the derivative in the

(We have denoted by

y

Here and

direction.

Ey

in the following we omit the measures in the integrals).

From (4.12),

(4.15), we obtain 2

ap-VE + E A Y = JP3ur4 If we choose now be enough but

r

4

E = 0

on

E > 0

and

r3

on

S2

(E > 0

on

would

P4

is unknown!) we obtain from (4.16)

Op-DE + Ey < 0

J

(p + Y)'E.

B

VE E C1(St),

E = 0

on

> 0

r3,

on

S2.

A Hence

being equal to

p

VE E C1(S2),

12 Vp-VE + X(A)Ey < 0

where here and in the sequel of the set

A.

A

outside

0

X(A)

we can rewrite this inequality as E = 0

on

S3,

E > 0

on

S2

denotes the characteristic function

But now, by the maximum principle, A

can be characterized

as the set,

[p > 0] _ {(x,Y) ES2J P(x,Y) > 0}. is positive on

p

Indeed

< 0

Y -

on

r1

(see

rl,

(4.17)

the strong maximum principal gives us a pair

on

Since by our assumptions

(4.15)).

V

and av = -vy

P2 U r3 U r4

(p,A)

or equivalently

p > 0

(p,X(A))

inside

A.

Thus to look for

leads to considering the weaker

problem:

Find a pair (p,X) E HI (0) x L%2)

(i) p > 0 a.e. in S2, p = (P)

(ii) 0 < X < 1 a.e. in iii)

f JS2

on

52, X = 1

such that S2 U S3

a.e. on [p > 0]

VE E H1(S2), E = 0

on

S3, E > 0

on

S2.

Statement of the Problem

4.1.

79

First let us prove that this problem has a solution. 4.1.4. If

Existence of a Solution denotes the following maximal monotone graph:

H0(p)

if p> 0

1

H0(p) _

[0,1]

if

p = 0

0

if

p < 0,

then, clearly, the line (ii) of (P) simply expresses that

X E HO(P) Introducing 1

HE(p) =

defined for

He(p)

if

e > 0

as:

p>e

EP

if 0
0

if p < 0,

it is natural to look for

as the limit of

p

where

pE

pE

is the solu-

tion of pE E H1(H),

PE =

on

4>

S2 U S3

(Pe)

j

V EH1(2),

VPE'V + HE(pE)'cy = 0

First, let us study the penalization

= 0

on

S2 U S3.

of the problem (P) and

(PE)

prove:

Theorem 4.1.

([35])

There exists a solution

Let us consider the mapping

Proof:

uE = FE(v)

pE

which for

Fe

of (P E).

v E

L2(SZ)

associates

the solution of the problem:

uE = 0

on

S2 U S3 (4.18)

DuE VE + HE(v).

, =

0

VE E H1(0), E = 0

on

S2 U S3.

J

(Existence and uniqueness of

ue

results from Corollary 1.13 and

Poincare's inequality which is valid for functions which vanish only on S2 U S3.)

If we still denote by on

S2 U S3

since

HE(v)

a function in

a

and if we substitute is bounded by

1

= ue -

Hl (0) 4>

which agrees with

4>

in (4.18) we easily obtain

80

THE DAM PROBLEM

4.

j IV(u6 - 0) 12 < C($) where

(4.19)

is a constant which depends on

C(0)

u

Thus

only.

0

is bounded

e

in

Hl(f2)

and from the compactness of the identity mapping from

to

L2(S2)

(see Theorem 2.5) it follows that

Hl(S2)

is completely continuous.

Fe

Moreover, according to Poincare inequality, we have $12

< K

faI ut -

So, for (in

large enough, FE

R

into itself.

L2(52))

of PE

IV (uE - 0)j2 <

maps the ball of center

and radius

0

R

Applying Corollary 1.2, we get the existence

as a fixed point of

FE.

Moreover, we have:

The solution

Theorem 4.2.

where in

of

pE

(P e)

are two Lipschitz continuous functions on corresponding solutions

p1'p2

of

is unique, p6 > 0

almost every-

is nondecreasing (namely, if

0 - pE

52, and the mapping

(PE)

S2 U S3, if satisfy

O1'02

> O1, then the

0

P£ > pi -almost every-

where in 0). (As in the previous theorem the Lipschitz character of

Proof:

needed to insure the existence of an extension of general quires If 0l

and

can be considered.

0

to be defined by (4.13)). are two solutions of (P

02' respectively, we set

(1 -

+

to

Q.

0

is only

Thus a more

However the physically relevant case re-

0

pE,pE

0

corresponding to boundary values

q = qE = pE - pE.

if

x>0

if

x<0

Setting for

d > 0:

f6 (x) = 0

we easily see that since which vanishes on

H1(11)

of = fa(q).Vq =

q < 0

on

S2 U S3.

S2 U S3,

and thus by (4.20)

(4.20)

5]).Vq/q2.

j Vq-VE = f (HE(p£) - HE(Pe))Ey, Q

is a function in

Moreover, we have (see Theorem 2.3)

From the equality satisfied by

52

= fd(q)

pE,pE

we then deduce:

Statement of the Problem

4.1.

Oq

q22

= I [q>6] 6'(H E(P2) E

6'

I [q>6]

81

- H

E

(P1))E

.

q2

Using the Lipschitz continuity of the

HE

this implies:

IgyI

I

q2 - E

[q>5]

q

[q>6]

Thus by the Cauchy-Schwarz inequality, we obtain

qL

(q-6)+)12

f

=

f ID Log(l +

E2

q2

[q>6]

where

denotes the measure of

1521

< .LL

6

11.

Applying now the Poincare inequality, we get:

with

-

101

6.

But when

Log(l + ) )2 < K

In

independent of

K

E2

d

goes to zero we deduce from the

above inequality

q = pe - P2 < 0

a.e.

on

(4.21)

S2.

Applying this inequality for

0 = Ol = 02

and

pe,pe

leads to the uniqueness of the solution of (p E).

of

(pe)

of

0 -+ PC

then follows from (4.21), and since

the solution of (PE) when

0 = 0).

two solutions

Monotonicity

0 > 0, pE > 0

CO

being

This completes the proof.

Using this approximation we are able to prove: Theorem 4.3.

Under the above assumptions, there exists a pair

(p,X)

which is a solution of (P). From (4.19), pE

Proof:

independently of

H1(S2)

e

n

being some c.

ue, we deduce that

pE

such that:

-+ 0

pE - p

in

Hl (52)

Pe

in

L2 (52)

n

P

n

HE (pE ) -a X n

in

L

n

2

R

is bounded in L7(S2) and hence also in L2 (Sl)). Moreover, if in (PE ) we choose E E 19(52) we obtain:

(HE (pe)

n

is bounded in

Thus (see Theorem 2.5) we can find a sequence

82

THE DAM PROBLEM

4.

ApE

n

_ -He (pE ) (pE ) y E L2(0) n

n

n

and by an extended Green's formula (see [15]) we easily have: 8p

Op S2

(P en

En

n

/

+H

V

=J

) y

En

E < 0 d E H l(0) ,

8v

S2

E > 0

=0

on

on

S , 3

S2.

aPE

pE

(Note that

Letting

> 0

and

PE

= 0

on

imply

S2

n < 0

on

S2.)

8v

n n En -+ 0, (iii) of (P) follows.

On the other hand, since the convex sets a.e. in

K = {v E H1(S2)1v > 0

0, v = $

S2 U S3}

on

K' = {v E L2 (Q) J0 < v < 1 a. e. in 0} are weakly closed in

and

H1(S2)

L2(S2), respectively, we conclude that

This gives (i) and the first part of (ii) in (P).

(p,X) E K x K'.

Finally, on the set

we have almost everywhere (after ex-

[p > 0]

traction of a subsequence) HE (PC )

n

2

HE (PC ) - 1 in L ([p > 0]). n

1.

Thus by Lebesgue's Theorem

n

From the fact that HE (PC )

n

n

and by uniqueness of the limit we deduce that on

[p > 0].

Remark 4.4.

X = 1

-+ X

in

L

2

almost everywhere

The theorem follows.

Except in our formulation (see 4.17) we have not used the

assumptions (4.2)-(4.5).

Thus the problems (P ) and (P) make sense in a

general Lipschitz domain

Q.

to this case. X

that

In particular the Theorems 4.1 - 4.3 extend

Note that in the sequel an important point will be to prove

is the characteristic function of the set

[p > 0].

We now want to point out some properties of a solution pair

(p,X)

of (P).

Some Properties of (p,X) Solution of (P).

4.2.

Under the assumptions (4.2)-(4.5) we have: Theorem 4.5. (i)

Let

(p,X)

Lp + Xy = 0

in

(ii)

pp > 0

in

(iii)

Xy < 0

and thus

Proof:

(52)

n

be a solution of (P) then: 9

1

X

is a nonincreasing function of

For (i) it is enough to choose

E E-9(0)

in (P).

y

in

Q.

For (ii) we

4.2.

Some Properties of (p,X) Solution of (P)

use a technique of [5].

Since

E_9(0), C > 0

Choose

S2 U S3, C

on

> 0

83

are both test functions for (P).

vanishes on

and for

S2 U S3

set

e > 0

and

C

and

-

Thus substituting these two functions

(iii) we get:

in (P)

jnX-[min(P,z;)]y = 0.

Since

a.e. on

X = 1

[p > 0]

and

[min(-,g)]y = 0

a.e. on

[p = 0]

(see Remark 2.5) the above equality becomes: 0,

and thus: (since

0

min(P,t;) = 0

on

S).

S2

Writing this equality as Vp.V 1

IDpI2 = 0

+ J

[p>EO]

e

[p<Et]

we get fSZ

X([P>c 1)VP.V < 0.

Now using the fact tain, by letting

c

X([p>eC]) - X([p > 0])

go to

0

c + 0, we ob-

a.e. when

in the above inequality and applying the

Lebesgue Theorem

fX(P > 0).VpV = J Vp.V < 0

VC

C > 0.

From (i), (iii) is then obvious (see [98] and

This is precisely (ii).

note that by our assumption on the shape of II

(S2),

S2, every vertical strip in

is connected).

Remark 4.6. way.

In the case of a general domain (i), (ii) follow in the same

Then, we can choose all at once a decomposition of

,Sdp,

52

in

521,522,

vertically convex open subsets of 0

(see (4.22)) each S2

(see (4.2)).

become, Xy < 0

In this situation the part (iii) of the previous theorem and

X

is a nonincreasing function of

y

in each

52j

4.

84

THE DAM PROBLEM

(4.22)

In the sequel, in the general case, we shall tacitly assume that such a decomposition of

52

into open sets

(4.10), (4.11), setting for

S(x) = Inf{yI(x,y) E c

1.

is possible.

Similar to (4.6),

x E nx(S2i )

},

S+(x)

= Sup{yl(x,y) E Q]},

we shall assume that S and S are continuous except on a finite set Sod

and Sod

respectively.

The next statement does not depend on the shape of Theorem 4.7.

For all

(p,X)

and on

S2 U S3, thus the set

Proof:

By localization near

solution of (P), p

is continuous on all

n

is open.

[p > 0]

S2 U S3

We have:

52.

this follows easily from (i) of

Theorem 4.5 and from the usual regularity results (see for instance [65], p. 196). Remark 4.8. s > 1

for all belongs to WI's(c) (see [10], and the recent results

Note that by Theorem 4.5 (i), p Usually

(see [97]).

u * W1'"(1)

of [8]). Note also that (see Theorem 4.5 (i) and (P)(ii)) we have the open set

Thus

[p > 0].

p

is analytic in

[p > 0].

Ap = 0

This will be

used later.

Let us now point out the properties of the set Theorem 4.9.

Let

(p,X)

then there exists an cc: = {(x,y) E QI

lies in

[p > 0].

be a solution of (P).

e > 0

If

such that the cylinder

Ix-xoI < e, y < yo + E)

[p > 0].

First:

(x0,y0) E [p > 0]

in

4.2.

Some Properties of (p,X) Solution of (P)

If

Proof;

85

and since this set is open, for

(x0,y0) E [p > 0]

small

E

enough the square QE = {(x,y) E )1

is also included in

tonicity of CE

Thus we have

[p > 0].

X, X =

ly-y0I < s}

1x-x01 < e,

CE

on

1

is connected by vertical segments). Op = 0

obtain

CE, and if

on

maximum principle

p

on

y = 1

QE

and, by mono-

(we have used here the fact that by (4.2),

p

But from Theorem 4.5 (i) we now

takes the value

is identically

on

0

by the

CE

CE, which contradicts

on

0

Q. c

[p>0]. As a consequence we have: Corollary 4.10.

Let

p(x0,y) = 0

then Proof:

be a solution of (P).

(p,X)

for all

(x0,y) E S)

Otherwise, we would have

with

If

p(x0,y0) = 0,

y > y0.

p(x0,y) > 0

and, from

y > y0

for some

Theorem 4.9, a contradiction. In the general case (see Remark 4.6) Theorems 3.9 and Cor-

Remark 4.11.

ollary 4.10 are replaced by the following: [p > 0], then for

a

QE

ment coming down, starting in and

p(x0,y0) = 0

y > y0

and included in

p(x0,y) = 0

for all

(x0,y0) E

on each straight seg-

p > 0

S.

(x0,y0) E 0

If

(x0,y) E SZ

such that

The proof is the same as above.

Physically the meaning of Theorem 4.9 is clear:

Remark 4.12.

point

then

{x0} x [y0,y] c 52.

and

choose a point

small enough, we have

is wet, then by

(x0,y0)

when a

gravity acting, all is wet below.

How-

ever note that this is no longer the case when the permeability is not constant (see [17]).

Concerning the wet set, under the assumptions (4.2)-(4.5) we can prove: Theorem 4.13.

Let

p(x,y) > 0

V(x,y) E n,

in other words below Proof:

If

be a solution of (P), then

(p,X)

S3

x E Trx(S3)

the dam is wet.

(x,y) E S3, we have

p(x,y) = hi - y > 0.

is strictly positive in a neighborhood of from Theorem 4.9.

(x,y)

(In the general case we have only

up to the first point of

SI U S2

By Theorem 4.7, p

and the result follows p > 0

vertical lines.)

About the set

[p = 0]

below

S3

that we encounter by coming down along

we have in complete generality:

86

4.

Theorem 4.14.

Let

ball of center

be a solution of (P).

(p,X)

(x0,y0)

r

and radius

THE DAM PROBLEM

Denote by

included in

52.

B

an open

r

If p = 0

in

Br, then

in Br.

p=X=0

By Corollary 4.10 and Remark 4.11 we have

Proof:

on the part

of

C

Choose a strip For

of points connected to

S2

included in

E = R x (h,++)

a E 9(R), a > 0,

Br

p = 0

Br U (x0-r,x0+r) x (y0,+n).

is a test function for

=

Br, i.e.,

above

by vertical segments.

P.

Thus

we get

f Vp.V

+ X.

=

Y

C

Since

a > 0

X = 0

a.e. on

0.

I

11 C(lZ

is arbitrary in _9(t)

and

and thus on

E fl C

At this stage, let us note that if are able to define the free boundary the assumptions (4.2)-(4.5) set for Sup{yjp(x,y) > 0}

X > 0, this implies clearly

C.

0

(p,X)

is a solution of (P) we

of our problem.

Indeed under

x E nx(S2)

if this set is not empty

0(x) _

(4.23)

otherwise.

IS -(x)

By Theorem 4.9, Corollary 4.10 this definition makes sense and we have: For all solutions

Theorem 4.15.

semi-continuous (l.s.c.) on Thus

(p,X)

nx(0)

of (P), the function

except perhaps on .5.'

is lower

(see (4.10)).

is measurable and we have:

0

(4.24)

[p > 0] = { (x, Y) E D y < @(x) } = [y < '(x) ] . Proof:

The l.s.c. of

not in .Y S (x0)

for

is clear (by (4.10)) on the points which are

O(x) = S (x) .

e > 0, let

y0

p(x0,y0) > 0

0

(S2)

and

O(x0) >

O(x0) > y0 > O(x0) - c

with

and Corollary 4.10 we have of course

p(x,y) > 0

and thus

This means that for

a.

If now x0 E 'r

be such that

By definition of

(x0,y0) E Q.

and radius

0

and where

on a ball

Ba

of center

(x0,y0)

x E (x0-a,x0+a), '(x) > y0 > 41(x0)-e.

Hence the result follows since (4.24) is easy to check. Remark 4.16.

One of the advantages of assumptions (4.2)-(4.5) is that in

the decomposition of one 0

52j, namely

Sa

52

into

itself.

2j

(see Remark 4.6) we actually have only

So if one wants to extend the definition of

to the general case it is easily done by defining one

4) }

in each

0

87

Some Properties of (p,X) Solution of (P)

4.2.

The formula is the same as in (4.23) with now

ferent

in place of

Std

0, S (x)

Under the assumptions of the Remark 4.6 the dif-

being replaced by

are l.s.c. on their domain of definition except perhaps on

(D )

(J =

'p) -

Let us now state a result which will play a fundamental role in the sequel:

Theorem 4.17. C

Let

be a solution of (P), h

(p,X)

be a connected component of the set is an interval.

wrx(C) = (x0,x1)

Zh = SZ

fl

Clearly

Set

we have

+X2< fZ pY +X<0.

XpY

J

h

(4.25)

h

(Here as in the following, we denote by a bar the closure in meaning of the set

]R2.

The

is obvious, i.e., [y > h] = {(x,y) E Sty > h}).

[y > h]

Let us assume that we have proved the following inequality for all

Proof:

{ E H1(Zh) fl C(Zh), C

positive, vanishing on

y

Zh

1

(x0 + e,x1 - E), e > 0

on

IZ py + X = h

(

h) +

JZh

(4.26)

(x0,x1)

aE a function in -9((x0,x1)), between

and choose equal to

[y = h]:

C(x,O(x))dx

< 1

X([p > 0])C

f

( = JZ

and

0

1, and which is

We have:

being small.

h)y X.[ac(Y-h)]y + lZ

h

h +

Zh fl S3 = 0, clearly

Since

Let

a real number. [y > h].

fl

(x0,x1) x (h,+-).

If Zh fl S3 = 0 Z

[p > 0]

X(Zh)ac.(y - h)

E)(y-h)]y.

is a test function for (P)

and the first integral in the above formula is negative.

Applying (4.26)

to the second leads to j

p y

X < J

Zh Y

Letting

(1-a (x0,x1)

a - 0

that (4.26) holds.

E

Zh

(X-X([p > 0]))(I-a E ).

the result follows by Lebesgue's Theorem, provided (The first inequality is clear since

X2 < X

and

88

THE DAM PROBLEM

4.

Xpy = py.)

To prove (4.26) let us first remark that for

C(Zh), with

C > 0, and

= 0

on

f1

[y = h], the function

(E > 0)

= X(Zh)min(e,C)

is a test function for (P).

(Note that

p(xi,y) = 0

This follows from the definition of

i = 0,1.

C E HI(Zh)

C

V(xi,y) E 1, y > h,

and from Theorem 4.9.)

Thus from (P)(iii) we get:

Ej

Since

f

IVPI2 + j

Zhn [P<EC]

Zhft [P>eC]

[min(E,C)]y = 0

f Zh

4-0

h

almost everywhere on

this leads to

[p = 0]

Y([P>eC])OP'VC + X([P > 0])[min(E,C)]y < 0

X([P > 0])[C - min(e,C)] y

X([P>EC])OP'OC + )(([P > 0])Cy 1Z

0.

1Z

Z

h

h (4.27)

JZhX([P>EC])OP'OC + )(([p > O])Cy < JZhX([P > 0])(C - e)Y

By Fubini's Theorem the last integral is equal to (C - P)Y(x,y)dy dx, (h,O(x))

(xO,xl)

where for simplicity we have still denoted by S (x).

(S1

can indeed intersect

continuity in (x0,x1)

all

of

y

such that

(C - E)+

y = h.)

(C

Now letting

6

the maximum of

d

h

and

But now using the absolute

(see [98], p. 57) for almost all

(D(x) > h, and for

f

h

x

in

small enough we have

E)Y(x,y)dy < (C - e)+(x,Vx)-6) < C(x,(P (x) d). go to zero, by continuity of

C

we get for almost

x E (x0,x1)

(C - E)y(x,Y)dY '< C(x,O(x)) (h,'(x))

and clearly (4.26) follows.

(Note that in (4.26) the right-hand side makes

sense by Theorem 4.15.) Remark 4.18.

In the general case of Remark 4.6 the result is as follows.

As above let

C

be a connected component of

[p > 0]

11

[y > h].

4.2.

Some Properties of (p,X) Solution of (P)

If

denotes the decomposition of

(EL )

S2 = {(x,y) ESL x E 7Tx(C

fl S2)),

89

in

S2

52j, set:

Y>h}

Zh= (Note that each

S2

Roughly speaking

is a connected open set which is vertically convex.

0

is obtained as follows:

Zh

the vertical segment from counters

through

S, then

Zh

pick

is the union of such segments

then (4.25) holds.

C, draw

running

(x,y)

Now the result of Theorem 4.16 is the same:

C.)

in

(x,y)

up to the first points where it en-

(x,y)

If

Zh fl S3 = 0,

First, with the same proof as above, (4.26) is

replaced by: P

J OP'V + X([P > 0])C < y Zh

J1

f,x(52'.) 7r J

C E H1(Zh) fl C(Zh), C > 0, C = 0

for all

boundary defined in each

on

[y = h].

as in the Remark 4.15.

12j

tained by splitting (4.27) on the different function

on each

aE

nx(Q')

is the free

(0)

The above sum is ob-

Introducing then a

the result follows as above.

Once more

the simplification under the assumptions (4.2)-(4.5) lies in the fact that Zh

is, in this case, comprised of only one

Q'.

Finally, to conclude this section let us prove the following result valid for any

0:

Theorem 4.19.

Let

and radius

(x0,y0)

(p,X)

r

be a solution of (P), B

r

included in

0.

an open ball of center

Then the following cannot occur:

P(x0,Y) = 0

V(xO,Y) E Br

p(x,y) > 0

V(x,y) E Br,

(i)

(ii)

Proof:

x # x0

P(x,Y) = 0

(x, y) E Br n [x < x0]

(Resp. Br n [x > x0])

P(x,Y) > 0

(x,y) E Br fl [x > x0]

(Resp. B. n [x < x0])

+l

Let

E 9(Br).

Since

E

and - are test functions for (P)

we have: (4.28)

0. J

B

r

Under the assumption (i) we have

X = 1

a.e. on

Br

and thus

90

4.

JB

Ey = 0.

))IB

r

r

Under the assumption (ii) by Theorem 4.14 we have [x < x0]

f / BrX

X = 0

a.e. on

Br fl

and thus

= 0.

v

Ey ='Brfl [x>x0] Ey

B (Brfl [x>x0])

denotes the component in

(v

THE DAM PROBLEM

Y

of the outward unit normal

y

v

to

Y 8(Br ll

[x > x0])). 0

J

Thus in both cases by (4.28)

C E 9(Br) - Ap = 0

in

Br.

But (i) and (ii) are now in contradiction with the maximum principle since p

achieves its minimum inside

4.3.

B

S3-Connected Solutions

An important point, that we have not yet discussed is of course: does (P) have a unique solution?

The answer is no.

The simplest example

is the following:

Assume that we are in the case of Figure (4.29) with noting the regions indicated in (4.29). (hl - y,l)

on

C

and

C'

de-

Then clearly

C

4.30)

(P,X)

I(0'0)

outside

is a solution of (P), but so too is

C

S3-Connected Solutions

4.3.

(P,X)

91

(hl - y,l)

on

C

(k - y,l)

on

C'

(0,0)

elsewhere.

Of course, the choice of

C'

is not unique.

One of our goals will now be to show that the Figure (4.29) provides us in fact with the only pathology which leads to nonuniqueness.

More

precisely (4.30) seems to be the only physically relevant solution of (P). It will be called, for obvious reasons, the

S3-connected solution of (P).

Then we will show that all other solutions are obtained by adding to the S3-connected solution pairs defined by

on sets of type

(k - y,l)

C'.

As we will see such a splitting of the solutions of (P) is quite general, no matter the shape of the dam considered. Let us first begin with a precise definition.

We shall say that a solution

Definition 4.20.

connected if for all connected components

(Recall that

positive.

denotes the closure of

C

Assume that

Remark 4.21. P

we have

[p > 0]

0.

Cns3

a point

of

C

of (P) is S3-

(p,)()

of Thus

which touches

S

3,j' C

S3,j.

c n S3,j

$

C

in

1R2

for some

Then C contains

j.

p is strictly

Around this point (see Theorem 4.7)

contains all the connected component of This shows also that if

[p > 0]

then

c n S3 = 0

wet without supply of water from the different reservoirs.

is

C

This, of

course, seems to be incorrect from a physical point of view.

Although the formulation (P) does not rule out completely the possibility of connected components of

[p > 0]

which do not touch

S3,

(see (4.29)) the shape of such a component as well as the pressure inside is easily described.

Indeed, as claimed in the beginning of this section,

we have: Theorem 4.22.

ponent of

Let

[p > 0]

(p,X)

be a solution of (P) and

such that

C n S3 = 0.

C

a connected com-

Then there is some

h

such

that: C

is a connected component of

Proof:

Recall that

y < h.

Set

C

[y < h]

[y < h],

p = h - y

is the set of points

k = Inf{yl(x,y) E C}

which is a connected component of

in

(x,y)

and consider the set [p > 0] n [y > k]

here holds for a general dam, see Remark 4.18).

on

Zk

C.

0

such that

associated to

(the proof given

By Theorem 4.17 we have

92

THE DAM PROBLEM

4.

< 0.

X2

I

1Zk On the other hand (see Remark 4.18) which are in

f0p12

Zk

(Recall that

This leads to:

Zk fl S3 = 0.)

J

is a test function for (P).

±X(Zk)p

and

52

on all boundary points of

p = 0

+

0.

Zk

Adding this to the above inequality we get p2 + (X + py) 2 < 0. 1

Zk

Since

X =

on

1

on

p = (h - y)

this leads to

[p > 0]

for some

C

the bottom of such a

plies that

in the remainder of

(p,X) = 0

Definition 4.23.

Op = (0,-1)

on

is surrounded by

C

C

and that

SI

Zk.

Let us call a "pool" a pair of functions

on a connected component of

and thus

Note that this im-

The result follows.

h.

(p,X)

defined

by

[y < h]

(P,X) = (h - y,l).

Then we have:

Any solution

Theorem 4.24.

(p,X)

of (P) is the sum of an

S3-connected

solution and "pools". Let

Proof:

(p,X)

be a solution of (P).

Denote by

the connected

C i

component of

such that

[p > 0]

(P',X') = (P,X)

C. fl

fl S3 # ID.

f 0p'.V 52

for all

and set

(i E I)

(X(C1)P,X(Ci))

-

Then clearly all connected components C'

S3 = 0

C'

of

are such that

[p' > 0]

Moreover, by Theorem 4.22 we have

X -

+ Y

152

E E HI(S2), E > 0

i

C.

on

S2, E = 0

on

< 0,

+

-

f

Y

Y

S3.

y Thus

(p',X')

is a

S3-connected solution of (P) which concludes the proof.

This theorem allows us to restrict our attention to solutions.

S3-connected

S3-Connected Solutions

4.3.

93

First let us assume that the numbering of the S3 i (i = 1,...,n)

reservoirs with bottom

is chosen so that

hn < hn-1 < ... < hl.

(4.31)

Then the first natural result is to show that for an

S3-connected

solution the level of the wet set cannot exceed the level

hl, and in

fact, we have more precisely: Theorem 4.25.

Let

be an

(p,X)

S3-connected solution of (P) then:

(i) P=X=0 in

[Y>h1] 0 < p < (h1 - y)+ in 0.

(ii)

Set

Proof: (P)

E = (p -

(hl-y)+)+

Clearly

C

and by

S2 U S3

vanishes on

(iii) we get:

(hl-y)+)y < 0

1 Op-0(P - (hl-Y)+))+ +

f OP-0(P - (hl-Y)+)++ X (P - (hl-Y)+)+ + J 1

[Y
But on

[y < h1]

IVp12

+

Y -

[y>h11

y

0.

we have almost everywhere

X-(P - (hl-y)+)Y = V - (hl-y)+.V(P - (hl-y)+)+. (This results from the fact that is equal to 0

when

p

(hl-y)+)y

-(hl-y)+ = 0, and since

(p - (hl-y)+)y = -(hl-y)y(p - (h1-y)+)y

where

p > 0

and to

vanishes.)

Thus the above inequality becomes: (h1-y)+)+12

1V(p -

f

+

IVP12 + X.Py < 0.

f

[Y
(4.32)

[Y>h1]

In the second integral it is enough to integrate on the different con-

nected components of

[p > 0]

fl

[y > hl]

or still on the different

Zh 1

generated by these components (see Remark 4.18).

But on these sets we

have by Theorem 4.17

<0. Zh 1

Summing in

i

and adding to (4.26) leads to:

1o(P - (hl-Y)+)+12 + I

I

[y
f

pX + (Py + X)2 < 0.

i

i JZh 1

94

THE DAM PROBLEM

4.

Thus we deduce from this inequality that nected component

of

C1

[p > 0]

being of course strictly greater than

on such a component, k

by analytic continuation (see Remark 4.8) we must have the connected components of

[p > 0]

containing

S3-connected, such a component must touch k > hI.

Thus

p = 0

on each con-

Op = (0,-1)

p = k - y

This implies that

[y > h1].

fl

But

on all

p = k - y Since

C1.

hl.

is

(p,X)

which is impossible since

S3

[y > hl], and (i) results from Theorem 4.14.

on

(ii) results from the fact that (4.32) can now be written Y)+)+I2

0.

f jv(p - (hl -

Remark 4.26.

The necessity of assuming that

(p,X)

is

S3-connected is

clear from the Figure (4.29).

Before describing more completely the properties of S3-connected solutions let us take a look at the case of Baiocchi's dam, that is to say the rectangular one (see Figure (4.1)(B)).

In this case the main tool is to integrate along vertical lines (hence the necessity of vertical walls).

More precisely, let

(p,X)

be a solu-

tion of (P) in the case of the Figure (4.1)(B) set

u(x,Y) = j I p(x,y)dy.

(4.33)

y (This transformation is called the Baiocchi Transform), then we have: Theorem 4.27.

(i)

(ii)

u

defined by (4.33) satisfies:

[u > 0] = [P > 0] -Au +X= 0 in Q.

Proof:

For (i), u(x,y) > 0

(x,y')

with

p(x,y) > 0.

if and only if

p

is positive for some point

But by Theorem 4.9 it is the case if and only if

y' > y.

For (ii) now, we first deduce from 4.33 that

p = -u

Thus Y'

by Theorem 4.5(i) (-Au + X)y = 0 and

-Au + X

in

SZ

depends only on

x.

But on a level greater than

(i) in the previous theorem) we have Remark 4.28.

p = u = X = 0

hl

(see

and thus (ii) follows.

At this stage (ii) allows us to answer an important ques-

tion (this technique applies to more general domains than rectangular ones but not in all generality).

Indeed from regularity theory of elliptic

problems we deduce from (i) that

2

u E wloc(O)

for all

s, and thus (see

4.3.

S3-Connected Solutions

Remark 2.5) Au = X = 0 (i) proves that

95

almost everywhere on

[u = 0].

This combined with

is a characteristic function, i.e., that we have

X

X=X([P> 0]).

(4.34)

To obtain a variational inequality, it is enough to note that we have now

u > 0,

-Au > -1,

Moreover, 4.33 allows us to compute the boundary

(compare with 3.41).

value of

We have:

u.

u=0

(-Au + 1).U = 0

on

S2 lnl - YJ

rnl

-

U

(hl - y)dy =

J

y

on

S3 3,1

on

S3,2.

(4.35)

2

ln2 - YJ

In n2

U

2

2

(h2 - Y)dY =

2

y

value of u

Now to get the i.e., that

S1

p > 0

around

on

S1, assume that the dam is wet around

(this can be shown easily).

S1

Then we

have

Au = around

1

S1.

Moreover, by (4.33) and (4.15)

uyy = -py = 1 and from

Au = 1

on

S1

it follows that

that is to say, u

is linear on

uxx S1.

has to be equal

to

0

on

Sl

This leads with the notation of

(4.1) (B) to h2

h2 on

u = (a - x) 2a + x. 2a

Thus, introducing

(4.36)

as the set

K

K = {u E H1(f2)ju > 0, u

S1.

satisfy (4.35), (4.36)}

u

has to be a solution of <-du,v - u> > <-l,v - u>

Vv E K

uE K. This problem was investigated for the first time by Baiocchi in [9], [10].

The uniqueness of

in this case.

u

of course leads to the uniqueness of

Note that the knowledge of u

gives us

p

as

-uy

(p,X)

(see

96

4.

THE DAM PROBLEM

For extensions of this method to higher dimensions

[10], [15], [76]).

the reader is referred to [76] and [102].

To prove (4.34) in all cases, as well as to establish some monotonicity results about the free boundary of our problem, the next two theorems will be very convenient tools. Theorem 4.29.

k = 1,...,n

Let

(p,)()

and all

h

First we have in all generality:

be a

S3-connected solution of (P).

satisfying

y)+}

ph = { (x,Y) E 01 p(x,y) > Ch has at most

satisfies

connected components.

k

we denote by Ch

S3

i =

More precisely, if for

,

i = 1,...,k

whose closure in

the connected component of Ph

Ch,i

For all

hk+l < h < hk, the set

one has:

i

Ph = Ch,I U Ch,2 U ... U Ch,k.

(4.37)

(Of course in the above formula some of the Chi also that when to be read

k

can be the same.

Note

not being defined the assumption has simply

n, hn+I

h < hn

IR2

In the case

h > h1

it has already been proved that

See Theorem 4.25.)

Ph = (d.

k

Set

Proof:

Ch = a -

U

Ch i.

i=1

'

Then

k (h-y)')+

= (I -

= X(Ch)-(P - (h-y)+)+

L

i=1

is a test function for (P) (see [49] for a complete justification.

that in the above sum the same Chi Thus we get:

once).

(h-y)+)+ + X-(P - (h-y)+)Y < 0 f Ch'

ab

j

Vp.V(p - (h-y)+)+ + X-(p - (h-y)+)+ + Y

IOpI2

+

Crif[y>h]

But, (see the proof of Theorem 4.25) on X-(p - (h-y)+)Y = V and thus the above inequality becomes:

Note

are assumed to be taken into account

[y < h]

(h-y)+)+,

+

0.

Y

one has

4.3.

S3-Connected Solutions

97

IVp12

(h-y)+)+I2

IV(p 1

+

f

+

(4.38)

0.

C,ir1 [y>h]

Chn [y
In the second integral it is clearly enough to integrate on the different connected components in some Zh

C7

of

[p > 0]

fl

which are not included

[y > h]

Of course, this is still the same as integrating over the

Ch i.

generated by these connected components, (see Remark 4.18) and since

such a

satisfies h fl S3 = 0

Zh

we have by Theorem 4.17

X2 < 0. JZ] h

Combining this with (4.32) we get

of

By analytic continuation containing

[p > 0]

eluded in some

Z

i

Thus on the connected components

Chi

If

C..

P2 +

f

+

Chfl[y
k > h.

(p y

C

(h-y)+)+I2

IV(p -

f

C3

X)2 < 0.

Y

x

h

we must have

p = k - y, where

p = k - y

on all the connected component

(p,X)

S -connected and

is

this is impossible.

3

So all the

C.

not in-

C.

are empty and

(4.38) becomes (h-y)+)+I2

< 0.

IV(P -

f C

h

Thus

p < (h - y)+

and (4.37) follows.

Ch

on

Let us now point out an interesting feature of these first that (4.2)-(4.5) holds. Theorem 4.30. (xl,h)

of

Ph.

and

Let

(p,X)

Assume

Ch i.

Then we have:

be an

S3-connected solution of (P).

Let

be two points in the same connected component

(x2,h)

Then if the segment

[xl,x2] x (h}

does not intersect

S2

Ch

i

one has

for (x,y) E 2: p(x,y) > 0 Proof:

Vx E [xl,x2],

Let us assume

By definition of

p(x0)y0) = 0

Vy < h.

for some point below

[xl,x2] x (h}.

Ph

(4.39)

98

THE DAM PROBLEM

4.

one has

Thus, by Theorem 4.9, one can assume

p(xi,h) > 0.

Now since Chi

(x1,x2) x (--,h).

is connected and open, it is arcwise

Thus, one can find a path

connected.

(x0,y0) E

joining

r

(x1,h)

to

(x2,h)

(see Figure (4.39)) which is built with straight line segments and moreover has no double point. Thus since

being in

P

p(x0,y) = 0

has to cut the vertical line

[y < h]

below

x = x0

the connected component of

w0

is in particular in

Ch,i (x0,y) E S2, y > y0

for

Let us denote by

(x0,y0).

in the complement of

(x0,y0)

[p > 0].

(see Corollary 4.10), P

r

in

and set:

_ -X (w0) (P - (h - y) ) Clearly

tion of

= 0

[x1,x2] x {h}

and in a neighborhood of

[y = h]

on

Moreover, E = 0

Ph.

on the

since for such a

r, by defini-

S3,j, which eventually intersects

we have necessarily

S3,j

h. > h.

Thus,

is a test function for (P) and we get

Op-V[-(p - (h-y))

]

+ X-[-(p - (h-y)) ]y < 0.

w0

w0 c [y < h]

Since

fw0f[P>Ol Vp.V[-(p Now since on X > 0

w0 fl

- (h-y))

]

+ X'[-(P - Ch-y)) -]

[p > 0]

we have

- (h-y))

I2 < 0.

Thus

+

(O,X) = V - [(h-y)]

X < 0. ( w0f[P=0]

and since

Since

is constant on all connected components of

(h-y))

(p -

w0 n [p > 0].

Let

be the connected component whose boundary contains

C

p > (h - y)

is equal to

w0.

on

r

we have

Indeed otherwise

and at this point we would have both have

y

we have

fw0fl [p>0] IV(p

P.

this inequality becomes

p > h - y

in

w0

C

p > h - y

on

C.

But clearly

could have a boundary point in p = 0

which contradicts

and

p > h - y > 0.

p(x0,y0) = 0

C

w0

Thus we

and concludes

the proof.

Remark 4.31. ture:

The general case can be illustrated by the following pic-

4.3.

99

S3-Connected Solutions

(4.40)

In this case clearly the result is the following (and the proof is the same as above):

Let

(p,y)

be an

S3-connected solution of (P) and

(xi,h); i = 1,2, be two points in the same

ponents of in

s nIR x (-o,h)

S1 U S3

then

first point of

S

[xl,x2] x {h}

is strictly positive below

p

are included

[xl,x2] x {h}

up to the

encountered by coming down along vertical lines.

particular, this is the case when cluded in

If all connected com-

Ch i.

with end point on

In

are both in a ball in-

(xl,h), (x2,h)

Q.

To conclude this section and assuming first that we are in the case (4.2)-(4.5) we have: Theorem 4.32.

Let

(p,y)

by (4.23) is continuous on

The function

be a solution of (P). wx(7)

except perhaps on 3"

c

defined

U Y+. Moreover

_V+ '

has right and left limits on

limit at the endpoints of

U 3`

as well as a right or a left

ax(p).

Clearly by Theorem 4.24 one can assume that

Proof:

(p,x)

is

S3-connected.

Now let us first prove the following: (x,y) E 2, y > O(x), then

If

neighborhood of

Indeed assume that (4.41) fails. 0

of center (i)

(x,y)

and of radius

For all ball

BE,

x < x, p(x,y) > 0 For

(Resp.

c'

vanishes in a (4.41)

Let us denote by E < y - cD(x).

centered at

there exist two points

(ii)

p

(x,y).

(x,Z)

and

a ball included in

and of radius c' < E

(x,y)

and

BE

Two cases could occur:

(x,y)

in

BE,

x < x, p(x,y) > 0.

small enough, less than

a

one has

p(x,y) = 0

V(x,L) E Be

x < x

p(x,y) = 0

V(x,y) E BE

x < x)

such that

100

THE DAM PROBLEM

4.

and for all

e" < e'

centered at

(x,y))

(x,v) E Be

there exists a point

(a ball

such that

X < X. P(X,y) > 0 (Resp.

x < x, P(x,y) > 0).

Let us consider for instance case (i).

We thus can find two se-

quences (see Figure (4.42)).

(xn,Yn) (4.42)

(Xn.Y-E)

and

(x ,yn)

(xn,yn)

in

such that

B.

x

< x,

P(xn,,vn) > 0,

(x

yn) - (x, y)

when

n -+ +m

x

<xn,

P(xn,yn) > 0,

(xn,yn) - (x,y)

when

n -+ +m.

Moreover, we can assume () nondecreasing and us then consider the sequence

(xn,y - E).

(xn)

(If we choose

one can assume without loss of generality that this the straight segment

[x1,x] x {y - e}

P(xn,y - E) > 0

deduce that

for all

nonincreasing.

are in

small enough,

a

sequence as well as From Theorem 4.9 we

0).

(x y - e)

Thus the points

n.

Let

belong to

P which (by Theorem 4.29) has only a finite number of cony-E nected components. Thus there is an infinity of (xn,y - e) (starting

for instance from

ponent of {y - E}.

Py-E.

(xl,y - e))

which belong to the same connected com-

By Theorem 4.30 this leads

to

p > 0

below

But clearly the same applies on the other side of

are in the case (i) of Theorem 4.19 (recall that e])

and thus a contradiction.

x

p(x,y') = 0

(x1,x) x

and we for

y' E

It is not difficult to see that

case (ii) would lead to case (ii) of Theorem 4.19, which proves (4.41) in all cases.

To conclude the proof, let us first remark that by Theorem 4.15 we already know that -V-.

is l.s.c. (lower semicontinuous) except perhaps on

So let us prove that

perhaps on it is clear that

0

is u.s.c. (upper semicontinuous) except

If we are on a point where '

is u.s.c. since

0 < S+

O(x) = S+(x) and

S+

and

x t 5ol+

is continuous at

x.

4.4.

Uniqueness of S3-Connected Solutions

If now

the u.s.c. of

4'(x) < S+ (x)

the proof of Theorem 4.15).

Thus

at

0 0

101

x

results from (4.41) (see

is continuous except perhaps on

To prove the existence of a left (or right) limit on a point of 5

or So+ (as well as the end of

wrx(S2))

assume:

1 = lim inf O(x) < lim sup '(x) = L. x+xO X')-X0

x<XO

x<x0

Then choose a sequence infinity.

For

a

x

such that O (x ) - L

< x0

n

small enough such that

L - e > t

n

when

argument as above that an infinity of terms of the sequence belongs to the same connected component of V(x',y') E 0, y' < L - e $(x') > L - E > t

near

for

near

x'

PL-e'

goes to

we get by the same

Thus

(x

L - E)

p(x',y') > 0

This clearly implies that

x0.

and a contradiction since then

x0

lim inf '(x) > L - e > L. x<x0 Remark 4.33.

In the general case, and with the same proof as above, the

Theorem 4.32 has simply to be replaced by: perhaps on

x(S2

is continuous on each

(see Remark 4.6).

limits exists on

points of

cj

Moreover, right and left

as well as right or left limit at the end

U-91"+

irx (S2.) .

As a consequence we have now in complete generality: Theorem 4.34.

Let

be a solution of (P), then

(p,X)

(4.43)

X = Up > OD. Proof:

This follows directly from the previous theorem and from Theorem

4.14 according to the fact that union of the sets

{(x,o.(x))Ix E 7rx(S2j))

is of measure zero.

Let us now complete Theorem 4.24 by proving that there is only one S3-connected solution of (P).

Uniqueness of S3-Connected Solutions

4.4.

This time let us argue directly in the general case. be two S3-connected solutions of (P).

a decomposition of 0 denote

in

s2j,

(D ,0 the function

0j

j = 1,...,p

Let

(p 1,X1),

Let us assume to be chosen

(see Remark 4.6) and let us

(see Remark 4.16) corresponding to

respectively in this decomposition.

pl'p2

102

THE DAM PROBLEM

4.

Now set p0 = min(P1,P2),

the functions

being defined in

00

= min('

XO = min(Xl,X2), irx(S2.),

First we have:

For all

Lemma 4.35.

C E Hl (2)

C(C), c > 0 we have for i = 1 or

fl

(Pi-PO)-VC +

(4.44)

JDi c(x,oj(x))dx

JSIV

2

J

where

D _ {x First note that the sets D are clearly measurable and the inte-

Proof:

grals on the right side of (4.44) make sense. instance for

=

i = 1.

min(P1

Clearly

Thus for

Let us give the proof for

as above and

C

e > 0

consider

a p0

E = 0

on

S2 U S3

and

±

is a test function for (P).

Applying (P) (iii) corresponding respectively to

pl

and

p2

we get by

subtraction: j 0(P1 - P2)OE + (X1 -

0.

But in this integral it is enough to integrate on the set

[p1 - p0 > 0]

p2 = pO, thus we have

where

f 0(P1-PO).V

1

0

+

IV(pl-p0) I2 +

J

0(pl-p0) Vc <

(

[P1-P0>ec]

[pl-pO<Ec] + J

(X1-XO)

[min(Pi

PoC)ly

This implies that >ec]O(P1 - PO )-VC +

f

jr)

P1 PO

(Xl-XO)[C - min(pl IS2

s

pO,c)]y

= 0.

Uniqueness of S3-Connected Solutions

4.4.

But now splitting

Sl

into the different

hand side we get (note that on

103

in the integral on the right

92. J

[p0 > 0], )(1 = X0 = 1

and use (4.43)) 7,

f (Xl-X0)[

-

min(p1Ep0,C)]y =

52

f

j-1

0

(L

1

-)y

(4.45)

where (x)] _ {(x,y) E 01lD0(x) < y < @(x)

[D0(x) < y <

Now, using the same arguments as in the conclusion of the proof of (4.26), i.e., integrating in

y

>Ec]

f

f

on the right hand side of (4.45) we get

ID

[p1-p0

D.

j=l l

and the lemma follows by letting

go to zero.

e

We are now able to prove:

There is one and only one

Theorem 4.36.

The existence results obviously from Theorem 2.42.

Proof:

is always that of the preceding lemma. S3 k

S3-connected solution of (P).

for some

ponent of

such that

k

[p1 > 0]

Choose

Br

The notation

a ball centered on

is included both in the connected com-

Br

and in those of

[p2 > 0]

touching

S3,k.

(See

Theorems 4.7 and 4.13 and the figure below.)

(4.46)

Let

and

rl

r0

be an open part of

the complement of

r0

BBr in

Br) outside

0

Moreover, let us denote by

a

(the bonddary of

BBr.

a smooth function such that Aa = 0

in

Br,

a = 1

on

r1,

By Green's formula we have for

0 < a < 1 i = 1

or

on

2

r0.

(4.47)

104

THE DAM PROBLEM

4.

f

rlno

(pi-p ). 2a __ 2V 0

r

V(Pi-P )'Va +

(p i-p

f

0

f2fBr

0

c2fBr

v(pi-po) Va.

2nB

If we still denote by

which agrees with the fact that

o

and is equal to

Br

in

Xi = XO =

on

1

r

the function (clearly in

a

H1(D) n C(?))

outside, we get using

1

Br:

nc(pi-p0).

(4.48)

2v = 12

!P 1

Let us now denote by

A0

be a smooth function in IR aE(x) = 0

when

suitable.)

satisfying

d(x,AO) > e

0 < aE < 1, aE = 1

on

A0,

would be

(1 - d(x,AO)/c)+

(for instance

aE

let

0 < a < 1.

is a test function for (P) since it is a positive func-

tion which vanishes on

This leads to:

A0.

aa)a]y < 0

10 VPi.0[(1 - aE)a] +

Since

e > 0

For

[p0 > 0].

By the maximum principle we deduce from (4.47) that

(1 - aa)a

Thus

the set

on

(1 - a,)a = 0

and also

A0

(i = 1,2).

p0 = XO = 0

outside

A0

we

have:

aE)a]y = 0.

f VP0'O[(1 - aE)a] +

By subtracting from the above inequality we thus get: I0v(pi-pO)'va + (Xi-XO)'ay < f

(Pi-PO)'V(aca) + (Xi-XO)(aca)y.

Thus using (4.48) and (4.44) we obtain: P

P 1no Now letting on

(pl-p0)'W <

r

E1 ID

i(aca)(x,D (x))dx

(i = 1,2).

j J

E -* 0, we have by Lebesgue's Theorem and since

aa(x,&'(x)) i 0

D1 (pi-p0). La

8v

rlnQ

0

(i = 1,2).

But (4.47) and the maximum principle imply inequality implies P1 = P2

on

pi = p0 rI n o.

on

rI n P

for

8v > 0

on

i = 1,2

F1.

and

So the above

4.4.

Uniqueness of S3-Connected Solutions

105

Repeating the procedure for balls included in

Br, we get

By analytic continuation we obtain that

Br n Q.

connected component of

pl = p2

p1 = p2

C, the

in

which contains

[p1 > 0] n [p2 > 0]

in

And

Br n 0.

so we easily obtain that

C=C where S3,k C1.

1 =C2 denote the connected component of

Ci

in its boundary.

Now if

p2(x) > 0

xn + x and thus

in C

Indeed, C c C1 C1

from

we deduce that

pI(xn) = p2(xn)

is closed in

C1.

Hence

One should note that the

(see Theorem 2.24).

solution of [5]

(see [54]).

k'

k = 1,...,n.

S3-connected solution

(P) is also the minimal one, i.e., any other solution

pl > p, X1 > X

S3

pl(x) _

and the

C = C1 = C2

result follows since we can do the same proof for all Remark 4.37.

which contains

[pi > 0]

and is an open nonempty subset of

of

(p,X)

(p',X')

satisfies

It coincides also with the minimal

Moreover, when the shape of

0

is such that

no pools can appear (see Theorem 2.24) then clearly the solution of (P) is unique.

This is the case for the rectangular dam and more generally

for a dam with horizontal bottom as studied in [11]

(see Figure (4.49)(A)).

We have drawn on the Figure (4.49)(B) a case of a more complicated

0

where such uniqueness holds too, since no pool can appear!

(4.49)

(A)

(G)

The reader is referred to [49] for more results in this direction.

106

THE DAM PROBLEM

4.

Let us conclude this section by a theorem which is useful in deciding what part of the dam is wet as well as whether uniqueness holds or not. Physically, the result is simply that the pressure of the

S3-connected

solution increases if the levels of the reservoirs increase. Theorem 4.38.

Let

(p1)X1), (P2,X2)

be two

associated respectively to the functions

viously

S1

data of

01,02.)

S3-connected solutions of (P)

1 and

Then if

we have

02 > 01

as in (4.13).

02

S3,S2

is assumed to be prescribed and

p2 > pl, X2 > X1

From (4.43) it is obviously enough to prove that

Proof:

denote by

p

E:

1

, p

e

the solutions of (P e)

En -+ 0

p2 > pl.

If we

(see Section 4.1) associated with

2

6

respectively, by Theorem 4.2 we have 4.3) we can find a sequence

(Ob-

are defined by the

6

p2 > pi.

Now (see Theorem

such thate pin -. p!

L2(O)

in

E

(i = 1,2).

Thus letting

(since the

S3-connected solution is the minimal one).

clearly that

p2 > pl

have

and thus

to

p2 > 0

in both cases

S3

Remark 4.39.

cn

go to zero in

p2n > p1n

we get

p2 > P1 > pl But this implies

since on the connected components of p2 = p2.

[p1 > 0]

we

(Note that such a component is connected

i = 1,2.)

As a consequence of this theorem consider for instance a dam

as in the Figure (4.1)(A).

S3 = S3,2), then clearly the tion is given by

If we assume the above reservoir empty (i.e.,

S3-connected solution of (P) in this situa-

((h2 - y)+, X([y < h2]).

Thus, if we assume now that

both reservoirs are full, then the Theorem 4.38 implies that the

S3-

connected solution of (P) satisfies p > (h2 - Y)+.

Since no pools can appear on the part of set

[y < h2]

S1

which is outside the

this implies also that the solution of (P) in this case is

unique.

4.5.

Some Monotonicity Results for the Free Boundary In this section we shall study only the case of one or two reservoirs. 4.5.1.

The Case of One Reservoir

The notation will be that of (4.50).

4.5.

Some Monotonicity Results for the Free Boundary

107

(4.50)

0

is here a general domain satisfying, for simplicity, (4.2) and

For all

such that

(x,y), (x',y) E 52

we have

y < hl

(4.51)

(x,x') x {y}

fl S2 = 0.

Then we have: Theorem 4.40.

Let

be the

(p,X)

the function defined by (4.23).

S3-connected solution of (P) and

t

Then under the assumption (4.51) we have:

0

is a non-decreasing function on the interval

(--,x1) fl trx([p > 0])

0

is a non-increasing function on the interval

(x2,+-) 0 ax([p > 0]).

Proof:

Let

a point

which is below

p(x0,y0) = 0, we would have in 4.7 one can find a point S > h, p(a,s) > 0. Theorem 4.29).

h < O(x) < hl.

If there exists

(x2,x) x {h}, and such that

S2, p(x0,y) = 0

Vy > y0.

Now by Theorem

near

S3

such that

Clearly this point is in

Ph

which is connected (see

in

(a,s)

So since

(-tl,x0) x {h}

52

is also in

(x,h)

Vy > y0, an arc connecting half line

and

x E (x 2,+m) 0 nx([p > 0])

(x0,y0) E 0

(x,h)

to

(a,B)

Ph

and since

a < x0,

p(x0,y) = 0

must intersect in

at some point which would be thus in

S2

Ph.

the But

using (4.51), Theorem 4.30 and Remark 4.31 this leads to a contradiction of

p(x0,y0) = 0.

(D(x') > h

for

Thus in

x' E (x2,x).

52

(x2,x) x {h}

below Since

h

we have

p > 0

is arbitrary we have

and

4(x') > O(x)

and the result follows as the proof is the same on the other side. Remark 4.41.

In fact, except for some trivial cases where

(h1 - y)+, the free boundary can never be stationary in also true for more than one reservoir.

52.

p

is given by

This fact is

A proof is given in [49].

Note

that this leads to some nontrivial examples where the uniqueness of the solution of (P) fails.

(See also [49].)

108

4.

THE DAM PROBLEM

The Case of Two Reservoirs

4.5.2.

The notation is that of (4.52) and we assume that (4.2), (4.51) hold.

(4.52)

Let

Theorem 4.42.

be the

(p,X)

the function defined by (4.23). (i)

0

S3-connected solution of (P) and

Then under the assumption (4.51) we have:

is a non-decreasing function of the interval (--,xI) q

Yx([P > 0]).

Moreover, if we assume

is connected (otherwise it would

[p > 0]

be the case of one reservoir) then: (ii)

(iii)

If

O(x) > h2

on

(x2,x3)

for all

If there exists an

Here for

(ail$i) E Ch,i.

First let (x0,y0) E 0

have in

al > x0.

h < hi

x < x1

below

and

Assume

is

(xm,x3).

a point near an end

dy > y0.

(x0,+o) x {h)

Clearly by definition of and

(x,h)

(a1,R1) to

h < h2

0

we have

as above with moreover

(ail$1)

at some point.

by considering the arc connecting [x,x1) x {h}

If there exists a point

p(x0,y0) = 0, then we would

such that

(x,h) E Ch 1

we would necessarily have

below

(ai,Si)

0

p(ai,$i) > 0, that is to say

h < 4D(x) < h1.

(x,xl) x {h}

and Remark 4.31 this contradicts

p > 0

and

(Clearly such a point exists by Theorem 4.7.)

Then any arc binding

the half line

O(x) < h2, then

O(xm) < h2

is non-decreasing on

we shall denote by

0, p(x0,y) = 0

(x,h) E Ph.

such that

(x2,xm), 0

and which satisfies

S3 i

is non-increasing

0

such that

x E (x2,x3)

xm E (x2,x3)

non-increasing on

point of

then

(see 4.52).

there exists

Proof:

x E (x2,x3)

must intersect in

0

But by (4.51), Theorem 4.30

p(x0,y0) = 0.

By assuming

(x,h) E Ch

,2

and a contradiction would be obtained (x,h)

to

and consequently

(a2,s2).

O(x') > h,

Thus we have dx' > x, but

h

4.5.

Some Monotonicity Results for the Free Boundary

being arbitrary leads to

and to (i).

(D(x)

with the same proof by considering a point to prove (iii) let us denote by 4P(xm) =

xm

so let us assume

First if

c(x

)

< @(x)

m

(x,h) E Ch,l

the arc connecting

to

(x,h)

(--,xm] x {h}

Vx' > x

(x,h)

to some

with

(a1,Sl)

h < h2. (a2,82)

'(x)

such that

al < xm

If

Vx' > x,

O(x ) < h < m

p(xm,y) = 0

Vy > h

must intersect the

But by (4.51) and Theorem 4.30

p(xm,O(xm)) = 0, thus

(x,h) E Ch,2'

By considering as previously an arc we deduce then that

and the proof is complete since

Remark 4.43.

h

we have

Sl

x E (xm,x3).

xm, (D(x') >

and consider

at some point.

which implies in particular

Let

[xm,x3).

since in

we would have a contradiction with

connected

Now

such that

Let us prove then that, for instance,

O(x) = @(xm), then clearly by definition of

half line

O(x) > h2.

(x2,x3)

@(x) < h2. inf xE(x2,x3)

is nondecreasing on the interval

cD(x).

Clearly (ii) holds

such that

x

a point in

Such a point exists by Theorem 4.15. $

109

h

4P(x1) > h

is arbitrary.

The Figure (4.53) gives an example where (iii) occurs.

(4.53)

Here O(x)

hl = h2 hl

and thus

fi(x) < hl.

But if between the reservoirs

we have a contradiction with the Remark 4.41.

case (iii), see [49].

Thus we are in

COMMENTS

110

Comments

The presentation here follows Brezis [32] and Carrillo-Menendez-Chipot [49].

When the problem reduces to a variational inequality a nice exposi-

tion is given in Kinderlehrer-Stampacchia [76]. [15].

See also Baiocchi-Capelo

Other approaches are given in Alt [5] and Visintin [109].

Recently

Alt and Gilardi [8] studied the manner in which the free boundary meets the fixed boundary of the domain; they give also other proofs of some of our results.

Most of the matter exposed here has an extension to of course, the physical case, the dimension in canals, St

IRP; p = 3

is,

being a good model for flow

2

being the plane section of the porous medium (see [49]).

Finally, we did not study the evolution problem.

Details on this subject

as well as complete references can be found in the proceedings [13]. Our model studies the case where the pressure is prescribed on the pervious boundary.

Some other models are also valuable.

In particular,

the interested reader will find in [13] the derivation of such a model from the macroscopic point of view.

We note in conclusion that the prescription of

p

on the pervious

boundary leads to realistic prediction on location and shape of the wet set (see Theorem 4.38, Section 4.5 and [49]). in hydrodynamics.

This should be of interest

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J. Funct.

Index

Baiocchi transform, 94

Elastic-plastic torsion problem, 17, 52

Bellman-Dirichlet problem, 32 Bilinear form, 8

Elliptic equation existence of solutions, 14, 1S

Chain rule for piecewise smooth functions, 13 Coercive operator, 4

Lp-estimates, 67-72 nonlinear, 15

Exterior sphere condition, 53

Coincidence set, 31

Compact embedding, 13

Fixed point theorem, 1

Continuous on finite dimensional subspaces, 5

Fourth order variational inequality,

Convex function, 3

Free boundary, 31, 86, 106, 107

21

Convex set, 3 Hilbert space, 8 Dam problem, 74-110

Holder continuous functions, 14

existence of solutions, 79, 81 monotonicity of free boundary,

Interpolation spaces, 60-62

106

nonuniqueness of solutions,

properties of solutions, 82-90 strong formulation, 76

uniqueness of S3-connected solutions, 10l weak formulation, 78

Laplacian, 14 Lax-Milgram theorem, 8 Lp-spaces, 10 Minty's Lemma, 6 Monotone operator, 5 strictly, 5

Darcy's law, 76

surjectivity of, 7

Dirichlet problem, (see elliptic equation)

117

INDEX

118

Negative part of a function, 13 Norms, 10 Numerical analysis, 21

Variational inequality, 2, (see also dam problem, elastic-plastic torsion problem, and obstacle problems)

existence of solutions, 3, 4, 6 Obstacle problems, 16, 18, 22-66 boundary estimates, 3S comparison principle, 22

fourth order, 20

parabolic, 21

uniqueness of solutions, 7

continuity of solutions, 30 existence of solutions, 16, 18 free boundary, 31

W1'p regularity, 60-66 W1'° regularity, 50-60 W2'P regularity, 27-30

W2'- regularity, 33-50 Parabolic variational inequalities, 21

Penalization, 79 Penalty method, 24-26, 73 Permeability, 76 Poincarg's inequality, 11 Pool, 92

Positive part of a function, 13 Porous medium, 74 Prescribed mean curvature operator, 20

Projection on a convex set, 3 Quasi-variational inequality, 9 Regularity, (see obstacle problems)

Reservoir, 75

Schauder fixed point theorem, 1 Sobolev embedding theorems, 12, 13

Sobolev spaces, 10-13, 21

Spaces of continuous functions, 14

Wet set, 76, 84, 85

Applied Mathematical Sciences cont. from page ii

36. Bengtsson/Ghil/Kallen: Dynamic Meterology: Data Assimilation Methods. 37. Saperstone: Semidynamical Systems in Infinite Dimensional Spaces. 38. Lichtenberg/Lieberman: Regular and Stochastic Motion. 39. Piccinini/Stampacchia/Vidossich: Ordinary Differential Equations in R". 40. Naylor/Sell: Linear Operator Theory in Engineering and Science. 41. Sparrow: The Lorenz Equations: Bifurcations, Chaos, and Strange Attractors. 42. Guckenheimer/Holmes: Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields.

43. Ockendon/Tayler: Inviscid Fluid Flows.

44. Pazy:

Semigroups of Linear Operators and Applications to Partial Differential Equations.

45. Glashoff/Gustafson: Linear Optimization and Approximation: An Introduction to the Theoretical Analysis and Numerical Treatment of Semi-Infinite Programs. 46. Wilcox: Scattering Theory for Diffraction Gratings. 47. Hale et al.: An Introduction to Infinite Dimensional Dynamical Systems - Geometric Theory.

48. Murray: Asymptotic Analysis. 49. Ladyzhenskaya: The Boundary-Value Problems of Mathematical Physics. 50. Wilcox: Sound Propagation in Stratified Fluids. 51. Golubitsky/Schaeffer: Bifurcation and Groups in Bifurcation Theory, Vol. I. 52. Chipot: Variational Inequalities and Flow in Porous Media.

53. Majda:

Compressible Fluid Flow and Systems of Conservation Laws in Several Space Variables.