Elements of Nonlinear Analysis (Birkhäuser Advanced Texts Basler Lehrbücher)

in- Au-Wdx=

fVdx.

Noting that

(-Au) V we get (n div(VVu) dx + Jo Vu V p dx = jn f W dx.

Now, due to the divergence theorem, we see that the first integral in the above equality vanishes. So, we have

jVu.Vwdx=jfpdx

dp E D(tl).

(3.2)

REMARK 3.1. Integrating once more would show that u satisfies the first equation of (3.1) in the distributional sense. So, instead of looking for a solution of (3.1), one can look for a u vanishing on r in a sense yet to be made precise and satisfying (3.2). One advantage is dearly that the integral in the left-hand side of (3.2) makes sense under weaker assumptions

on u than u E Z(SFor Cl). instance, it clearly makes sense for u E HI (0). So, in order to match the 0 boundary condition, it is reasonable to look for u E Ho (n) so that (3.2) holds. Then clearly, D(ST) being dense in HH((), we will have Vv E H1(fT).

3. Elliptic Linear Problems

40

We call it a weak formulation of (3.1) when u is such that u E Ho (f2),

in Vu-VV dX =

jfvdx

dv E H(fl).

We just have seen above that, in the case where u is a strong solution to (3.1), then u is also a weak solution. So, if we are lucky enough and if (3.3) has a unique solution it will be the solution to (3.1) - or a suitable ersatz if our datum f is not smooth enough for the pointwise equality (3.1) to be true. This is indeed the case, and the framework of Sobolev spaces allows us to solve (3.1) without difficulty. Indeed we have

THEOREM 3.1. Let 1 be an open set of R" that we assume to be bounded in

one direction. Let f E H-1(1) - the dual space of Ho(f2). Then, there exists a unique solution u to

Iu E Ho (ft)

fro, Vu.Vvdx=(f,v) VvEHa(ft) where () denotes the duality bracket between H-1(l) and Ho (f2). PROOF. We have seen before (see Theorem 2.11) that

In

Vu Vv dx

is a scalar product on Ho (fl). The result is then just an application of the Riesz representation theorem (see [171).

1. One can show very easily that u is also the unique minimizer on Ha (ft) of the functional

REMARKS 3.2.

J(v)

2

jjVvI2_(f,v)

(see also the next section).

2. In the case where for instance Cl is bounded, one has L2(f2) C H-1(0) and (3.4) solves (3.1) for any L2-function and in particular for a function in C°(f2). The above technique, applied here to a simple example, allows us to treat more

general cases that we would like to put now into the framework of an abstract setting.

3.2. The Lax-Milgram theorem and its applications

41

3.2. The Lax-Milgrram theorem and its applications The Lax-Milgram Theorem allows us to solve problems of the type (3.4) in the case where the left-hand side of the second equation is not a scalar product but a more general bilinear form. More precisely we have:

THEOREM 3.2 (Lax-Milgram). Let H be a real Hilbert space. Let a(u,v) be a bilinear form on H such that: 1. a is continuous; that is to say there exists a constant C such that

Ia(u,v)I < Clujjvi V u,v E H,

(3.5)

2. a is coercive, which means that for some constant a > 0

a(u, u) > ajul2 Vu E H.

(3.6)

denotes either the absolute value or the norm in H that derives from the scalar product ( )). Then for any f E H' the dual of H there exists a unique solution u to

a(u,v)=(f,v) VvEH.

(3.7)

Moreover, in the case where a is symmetric - i.e. satisfies a(u, v) = a(v, u) Vu, v E H,

(3.8)

then u is the unique minimizer of the functional

J(v) = Z a(v, v) - (f, v)

(3.9)

on the space H.

PROOF. First remark that v

a(u,v)

is a linear form on H that - by (3.5) - is continuous. So, by the Riesz representation theorem, there exists a unique Au E H so that

a(u, v) _ (Au, v) Vv E H. Clearly, the mapping u H Au is linear. Next, we claim that AH is a dense subspace

of H. Indeed, if v is in (AH)l, the orthogonal of AH, then 0 = (Av, v) = a(v, v) > alvl2

and thus v = 0. Let f E H' - from the Riesz representation theorem again - we know that there exists a unique f E H so that (1 , v) _ (f, v) V v E H.

Since AH is dense in H, there exists a sequence u E H such that

Au,, -+ 1 in H.

(3.10)

3. Elliptic Linear Problems

42

So, for some constant C, we have in particular (note that a converging sequence is bounded): I Aun I < C.

It follows that

(Au.,un) <Using (3.6) we deduce easily that alun12 -< CIu,I.

Thus, un is a bounded sequence in H and we can extract a subsequence - that we will still label by n - such that un -- u weakly in H. From (3.10) we have for every v E H

lim(Aun,v) = (f,v). n

a linear form continuous

But (Aun, v) = a(un, v) - a(u, v) since on H. Thus u is the solution to

a(u,v)=(f,v) VvEH.

(3.11)

This completes the proof of existence. To show that u is unique it is enough to notice that if u' denotes another solution to (3.11), then a(u, v) = a(u', v) `dv E H. Taking v = u - u' and using (3.6) we deduce

a1u-u'12=a(u-u',u-u')=0. Hence u = u' and the uniqueness of a solution to (3.11) is shown. Let us assume now that a is symmetric and let u be the unique solution to (3.11). We have for every v E H

J(u+v) = 2a(u+v, u+v) - (fu+v) a(u,u)+a(u,v)+ 2a(v,v) - (f,u) - (f,v) = J(u) + 2a(v, v). Replacing v with v - u gives us

J(v)=J(u)+2a(v-u,v-u) VVEH. Thus, due to the fact that a(v - u, v - u) > 0 when v # 0 (see (3.6)), u is the unique minimizer of J and the proof of the theorem is complete.

0

We turn now to some applications. Let 0 be an open subset of R". Let us denote by aij(x), i, j = 1,...,n, a(x), functions in L°°(fl) such that there exists a positive constant a such that

a'j(x)&tj > aItI2 V E Rn, a.e. x E fl, a(x) > 0

a.e. x E Q.

(3.12) (3.13)

3.2. The Lax-Milgram theorem and its applications

43

In (3.12) we have used the Einstein convention, that is to say we have dropped the sign E to write n

for i,)=1

In what follows we will do so each time repeated indices - i.e. here i, j - occur. In (3.12), I I denotes the euclidean norm in IRn. (3.12) is called a uniform ellipticity

condition. It is clear that it means that the matrix aij (x) has, for almost every x, eigenvalues greater than or equal to a > 0, hence the word elliptic since n 1,2=1

would be the equation of an ellipsoid in Rn. Then we can show: THEOREM 3.3. Let Sl be an open set in W with boundary I'. Assume that (3.12) and (3.13) hold and

(a)

1Z is bounded in one direction

or (b) a(x) > 6 > 0

a. e. x E 12,

(3.14)

(3.15)

then, for every f E H-1(12) there exists a unique solution u to the Dirichlet problem u E Ho (S2),

I ai, (x) a 8 + a(x)uv dx = (f , v) Vu E Ho (fl).

(3.16)

PRooF. Of course, in the integral occuring in (3.16) we used the Einstein convention. We set for u, v E Ho (n)

a(u, v) =

& 8xj 8xi Jo ai j (x) ou

+ auv dx.

(3.17)

Clearly, a is a bilinear form on Ho (Sl). In order to apply the Lax-Milgram theorem we just need to check the assumptions (3.5) and (3.6). First, for (3.5) if we denote by A a positive constant so that

Ia,j(x)I, Ia(x)I < A a.e. X E 12,

and taking into account the fact that for every i = 1, ... , n

iix-i

3. Elliptic Linear Problems

44

we obtain la(u, v)I <_ in Ia;s(x)I 18xj I I 8xi I + lal Jul IvI dx

< n2A

IVvI + Jul Ivl} dx

< n2Allulll,2llvlll,2

which proves (3.5). To verify (3.6) we notice that by (3.12)

jai(z) g-z & dx>a fa(Dul2dx for every u E Ho (12). So, if 12 is bounded in one direction, we have

a(u, u) = in aij (x)

8x- 8xi

+ a(x)u2 > a In IVul2 dx

which by (2.76) gives us (3.6). In the case of (3.15) we easily deduce

a(u, u) > min(a, 0)

J in

I Vul2 + u2 dx

0

which completes the proof. REMARKS 3.3.

1. It is clear that 8u E L2(0). aij

-

J

Moreover, in the distributional sense we have (take v E D(O) in (3.16)):

++

(ai?i) ax au=f

and thus u is a weak solution to the Dirichlet problem

' Ti Iu=O

/a,J

'

Ox

I + au = f

/

in 12,

on F.

2. In the case where a is symmetric - that is to say when

ajj(x)=ajj(x) Vi,j=1,...,n,a.e.xEft, then u is also the unique minimizer on Ho (St) for

J(v) = 1 1 aij (x) .

j

i

+av2 dx - (f, v).

See also Remark 3.2, 2, for the choice of f .

3.2. The Lax-Milgram theorem and its applications

45

3. In the case where a is only assumed to be nonnegative, then the assumption (a) is necessary. To see it consider S1= (1, +oo) and for f E L2(S1) let us look for u a solution to

l -u" = f

in S1,

u E HJ (il). Integrating the first equation of (3.17) we have clearly X

-u' = J f (a) ds + Cst. Thus, taking for instance f so that AX) =

1

X

forx > 1

we deduce that

-u' = Log x + Cst forx > 1. But clearly this above function is not in L2((1,+oo)) and u' cannot be in L2(S1) so that u cannot be in Hl (fl). 4. In the case (3.14) one can even allow a to take negative values. Indeed,

assuming

a(x)>-,0 a.e.xEIl we have for a(u, v) given by (3.17)

a(u, u) > a

I Vu12 dx

n

u2 dx.

Knowing that

fu2dx

< c J IDu12 dx

n

for some constant c we get

a(u, u) > (a - fle) J I VuI2 dx

in

which implies coerciveness in case a - f3c > 0.

Instead of considering the bilinear form a given by (3.17) on Ho (S1) we can consider it on H'(0). Then this leads for instance to: THEOREM 3.4. Let Cl be an open set in Rn with boundary r. Assume that (3.12) holds and that a(x) > A > 0 a.e. x E it.

Then for every f E L2(1l) there exists a unique solution u to: 1 u E H1(11)

1 faj(x)L.+a(x)uvdx=jfvdx

(3.20)

3. Elliptic Linear Problems

46

PROOF. Let a be the bilinear form defined by (3.17). Clearly the continuity of a on Hl (11) follows as in the proof of Theorem 3..3. Next, we have also

a(u,u) > aJ IVuI2dx+0 f u2dx a

n

mina, i3) IIuIIi,2

for every u E Hl (0). Finally

v' -

r fvdx

in

is a linear form on H3 (12) that is continuous by fafvdx1 <-1f121"12 <- If1211V111,2. I

3o, applying the Lax-Milgram theorem with H = H1(fl), the proof is complete. REMARK 3.4. Taking v E D(f) in (3.20) we have as before

-

a

ail (x)

5i (

au ) +au

bij-

f

(3.21)

in the distributional sense.

Of course u f H01(0) and even so u satisfies, in the distributional sense, the same equation as the solution to (3.16), there is no reason for u to be equal to this solution. Indeed, as we are about to see, the solution to (3.20) satisfies some boundary conditions that are not explicit in (3.20) but hidden in the formulation. To see this let us assume that u and our data are smooth - i.e. the aij's are smooth. Let us also choose v smooth in (3.20). Noting that

8u 8v

8

f

8u

a+i (x} ax1 v a; (x) & f exi = Oxi

x 1

4xi (0ij (x)

;) v

we obtain

I

8

8uv

(aij(.T)-11

8uf)+au vdx=Jafvdz.

+

Using then (3.21) it follows that

J E

(affi(x)

v) dx = 0

for every v smooth. Using the divergence theorem we obtain

f ajj(x) 'vjvdo,(x) = 0 ax-

(3.22)

3.2. The Lax-Milgram theorem and its applications

47

for every v smooth. v = (v1, ... , v,) denotes the outward unit normal to r. It is clear that if our data are chosen smooth enough in such a way that a,; -v, E L2(r), then (3.22) implies that

a,;(x)

2 v; = 0 da a.e. on F

(3 . 23)

.

J

Conversely if it E H' (S2 ) is a smooth solution to

8xt

(a,;

in

0,

on

r

(3.24)

t..Ou it is clear

8 + au = f

v,

=o

,

going back to the various equalities above - that (3.20) holds. So, the

solution to (3.20) is the weak solution to (3.24). a,;v, is called the conormal derivative of u. Problems of the type (3.24) are called Neumann problems. In the particular case where (a,;) = Id,

a = 1

where Id denotes the Identity matrix in R' , we have solved in a weak sense the Neumann problem:

-Du + u = f

in fl,

Ou

on r,

5;

(3.25)

=0

where Ou denotes the normal derivative of u. It is possible to consider also non-homogeneous boundary conditions. Let us, for instance, consider g E Hl (12) and suppose that for f E H-1(11) one wishes to solve in a weak sense the problem:

\ / la;;(x) )+au=f u=g

in 12,

(3.26)

on r.

Assuming we are under the hypotheses of Theorem 3.3, then u-g = in is a solution to ax;

w=0

8 (a,, (x) -IV ) + au = f + 1 a;; (x) 22ax; ax; \ ;/

1

in it,

(3.27)

on r.

Clearly the right-hand side of the first equation in (3.27) belongs to H-1(12). So, by Theorem 3.3. there exists a unique weak solution to (3.27) and u = in + g is

3. Elliptic Linear Problems

48

the unique weak solution to (3.26) that satisfies:

Ju-9EHo'(fl)

rr,

Jnai;(x}

(3.28)

Oxi OX.

+auvdx = (f, v) Vv E Ho (0).

3.3. Exercises 1. Show that (3.12) implies that every eigenvalue of (aii) is greater than or equal to a. 2. Let u be a smooth solution to

-Au = f

=o

in f2, *

()

on r.

Assuming that f E L' (A), show that one has necessarily

fn f

0.

Show that for any constant c, u - c is also a solution to M. 3. Let ft be a Lipschitz domain in R". For u r= LI(f2) we set fc

Inl

f u(x)

dx.

101 is the measure of n. fi is called the average of is on n. Show that for some constant C it holds that Iu - u12 < CIIVuII2 Vu E HI (11).

Prove that for f E L2(f2) with f = 0, there exists a unique u E Vo = (v E Hl (S2) I v = 0 } that is a solution to

Vu Vv dx = (f, v) dv E Vo. Give an interpretation of the problem solved by u. 4. Let 0 be a Lipschitz domain in R". Let V be the subspace of H1(f2) defined by (2.36) with the measure of ro 54 0. Show that under the assumptions (3.12), (3.13) for f E L2(f2) there exists a unique solution u E V to

uEV,

J aii(x)±-

r

+a(x)uvdx =1n fvda Vv E V.

Give an interpretation of the problem solved by is.

Chapter 4

Elliptic Variational Inequalities 4.1. A generalization of the Lax-Milgram theorem Let us first recall: DEFINITION 4.1. A subset K of a real vector space is called convex if

b'a E (0,1),

b'u, v E K,

au + (1 - a)v E K.

(4.1)

Let us then prove the following result that extends the Lax-Milgram theorem as well as the theorem of the projection on a closed convex set in a Hilbert space.

THEOREM 4.1. Let K # 0 be a closed convex set of a real Hilbert space H. Let a = a(u, v) be a bilinear form on H such that (3.5) and (3.6) hold. Then, for every f E H', the dual of H, there exists a unique solution u to

JuEK, 1a(u,v - u) > (f,v - u) V v E K.

(4.2)

Moreover, in the case where a is symmetric, u is the unique minimizer on K of the functional

J(v) = 2a(v,v) - (f,v). PROOF. We consider as and aA respectively the symmetric and antisymmetric part of a given by a(u, v) + a(v, u) a(u, v) - a(v, u) as ( u, v) = aA(u,v) 2 , = 2 By (3.5), (3.6) we have

Ias(u,v)I 5

2{Ia(u,v)I + Ia(v,u)I} < Cluj IvI,

as (u, u) = a(u, u) > aIuI2, and as is a scalar product on H which is equivalent to the initial one. By the Riesz representation theorem there exists a unique f E H such that

as(f,v)=(f,v) VvEH. Then, by the theorem of the projection on a closed convex set in a Hilbert space (see for instance (651), there exists a unique u E K minimizing the distance (associated

to as(-, )) between 1 and K.

4. Elliptic Variational Inequalities

50

FIGURE 4.1

Of course it is well known - see also Figure 4.1 - that u is the unique solution to

Ju E K,

as(u,v-u)>as(f,v-u)=(f,v-u) VvEK.

(4.3)

Let us then introduce for t E R at (u, v) = as (u, v) + taA (u, V)

and T the set T={ t E R I V f E H' the problem Pt has a unique solution } where Pt is the variational inequality

Ju E K,

at(u,v -u) > (f,v-u) Vv E K.

(Pt)

Due to (4.3) T is not empty. It contains indeed t = 0. Let us assume now that T contains a point to. Then, we claim that T contains the segment [to - f , to + ] where a and C are the constants in (3.5) and (3.6). Indeed, if to E T, then for any f E H' and w E H there exists u = F(w), unique such that

ruEK, ato (u, v - u) > (f, v - u) + (to - t)aA(w, v - u) Vv E K. This is indeed due to the fact that

(4.4)

v ,-- (f, v) + (to - t)aA (w, v)

is an element of H'. Then we claim that, provided to-t is small enough, w i-' F(w) is a contraction on H. Indeed, let W1, w2 E H and let u1, u2 be the corresponding solutions to (4.4). One has u; E K and moreover

ato(ul,v-u1)? (f,v-ul)+(to-t)aA(wl,v-u1) VvEK, at. (u2, v - u2) ? (f, v - u2) + (to - t)aA (w2, v - u2) V v E K.

4.1. A generalization of the Lax-Milgram theorem

51

Taking v = u2 in the first inequality and v = ul in the second one and adding leads to

at, (u1 - u2, u2 - ul) 2! (to - t)aA(wl - W2, U2 - u1) or equivalently

at<, (ul - u2,ul - u2) < (to - t)aA(wl - w2,ul -U2)-

(4.5)

It is clear - due to the fact that aA(v, v) = 0 -- that

ato(v,v) = as(v,v) = a(v,v) d v E H. Hence, using (3.5) and (3.6) one deduces easily from (4.5) alai - U212 < Ito - tIClwl - wzl Iu1 - u21.

Thus liz1 - U21 = IF(wi) - F(w2)I 5 Ito - tI

Iwl -"I <-

1 Iw1 - w21

principle it follows that for t E [to - zC, to + ]. From the Banach contraction a when t belongs to the above interval then F has a fixed point. Now, dearly, if u = F(w) = to is a fixed point for F we see from (4.4) that u satisfies

uEK,

at(u,v-u)> (f,v-u) YvEK, i.e. u is a solution to Pt. Of course this problem has only one solution. For instance if u' is another solution, then one has u, u' E K and

at(u,v-u) > (f,v-u) dv E K,

at(u',v-u')>(f,v-u') 'dvEK. Taking v = u' in the first inequality, v = u in the second, leads after adding to

alu-u'12
J(v) = J(u+v - u) = 2a(u+v - u,u+v - u) - (f,u +v - u). Using the bilinearity of a we obtain easily

J(v) = J(u) + a(u, v - u) + 2a(v-u,v-u)-(f,v-u) = J(u)+ 2a(v-u,v-u). Since a(v - u, v - u) vanishes if and only if v _> u, it is clear that u is the only minimizer of J on K. This completes the proof of Theorem 4.1.

0

4. Elliptic Variational Inequalities

52

REMARKS 4.1.

1. As we mentioned before a problem of the type (4.2) is called a Variational Inequality. We refer to [57], [48], [67], (20], [63] for a more complete study of these problems. 2. In the case where K is a vector space, for every v in K, u ± v is a suitable test function in (4.2). So, we obtain

a(u, v) ? (f, v), a(u, -v) > (f, -v) for every v E K. In other words u is a solution to

uEK, a(u,v)_(f,v) VVEK, which is the Lax-Milgram theorem.

It is sometimes convenient - in particular when dealing with nonlinear opera, tors - to use an equivalent form to (4.2). We have

THEOREM 4.2 (Minty). The problem (4.2) is equivalent to the search for u such that

fu E K

I a(v,v-u)>(f,v-u) bvEK. PROOF. (a) (4.2)

(4.6). Indeed, let u be a solution to (4.2). Then,

a(v,v-u)=a(v-u,v-u)+a(u,v-u)>a(u,v-u)>(f,v-u) VvEK. (4.2). Let u be a solution to (4.6). Taking into (4.6) v = u+t(v-u) (b) (4.6) that belongs to K for any t E (0, 1) leads to

a(u+t(v-u),t(v-u))>t(f,v-u) VvEK. Dividing by t and letting t go to 0 leads to

a(u,v-u) > (f,v-u) Vv E K. This completes the proof.

0

4.2. Some applications 4.2.1. The obstacle problems. Consider an elastic membrane (see Figure 4.2) that is fixed to a rigid support and pushed from below by an obstacle. ct is the reference configuration of the unspanned membrane. Then, the equilibrium position u of the membrane is going to be a minimizer of the elastic energy (see Chapter 1) (4.7) J(v) = 2 f I VvI2 dz. n Now, if cp is a function defined on fl and describing the obstacle, not all the

4.2. Some applications

53

FIGURE 4.2

deformations u will be admissible but only the ones such that

v>cp in f2. (Vv = (vy, , v,.2) but the problem can be stated in dimension n as well.) So, if one sets

K={v: f1-+Rjv=Oonr,v>cpon11} - r denoting the boundary of fl - the equilibrium position of an elastic membrane put under tension by an obstacle W is given by u E K that minimizes the Dirichlet integral (4.7) over K, that is to say u such that

JuEK,

(4.8)

J(u) < J(v) Vv E K.

Perhaps it is time to rephrase our problem in terms of Sobolev spaces. So, let Cl be an open subset of R" with boundary r. For a function p E H'(0) recall that we define - if max denotes the maximum of two numbers or two functions V+ = = (-w)+.

max(O, gyp),

'P

Then we have DEFINITION 4.2. Let cp E HI (&I); we will say that

w
(4.10)

if and only if ;p+ E Ho (Cl).

Then, for ,p E HI (12), P < 0 on r we can introduce K = { v E Ho (f2) I v(x) > V(x) a.e. x E fl }. Then we have:

(4.11)

4. Elliptic Variation[ Inequalities

54

THEOREM 4.3 (Obstacle problem). Assume that f1 is an open subset of Rn bounded in one direction. Let SP E H' (f2) such that V:5 0 on T. Then, there exists a unique solution u to U E K,

fn VuV(v-u)dx>0 `dvEK.

(4.12)

Moreover, u is the unique minimizer of the Dirichlet integral (4.7) over K.

PROOF. It is enough to verify the assumptions of Theorem 4.1 with H = Ho (5l), f = 0

a(u, v) = jVuVvdx and K given by (4.11). We have seen previously that (3.5) and (3.6) hold true. So, we need only to check that K is closed and not empty. It is indeed obvious to check that K is convex. The fact that K is not empty follows from our assumption on W. We have indeed i*+ E K. Next, to see that K is closed, consider a sequence cp,, -" Spo,

in Ha (f2).

(4.13)

We would like to show that ip E K. Rom (4.13) we deduce that cpn."",V,,.

inL2())

and that there exists a subsequence cpn,, such that ipnk (x) --+ V,,,, (x)

ae. in Q,

i.e. we have

Vx E ft \ U (4.14) where U is a set of Lebesgue measure 0. Since the sequence conk is in K, we have cpnk (x)

also conk (x) >_
Vx E Cl \ Uk

(4.15)

where Uk is a set of Lebesgue measure 0. Combining (4.14) and (4.15) we have clearly

coa(x)>cp(x) YxEf1\UkUkUU, that is to say almost everywhere. So, cp E K and K is closed. Applying Theorem 4.1 the proof is complete.

0

REMARK 4.2. The set

A = Ix E f2 I u(x) = ip(x) } (4.16) is called the coincidence set. This as well as u is an unknown of the problem. The obstacle problem is one of the most popular free boundary problems. The boundary of A, a priori unknown, is a so-called free boundary. A huge amount of literature has been devoted lately to this kind of problems (see [63], [20], [48], [44], [70]).

4.2. Some applications

55

Regarding the solution u to (4.12) we note the following properties:

THEOREM 4.4. Assume that we are under the assumptions of Theorem 4.8. Then if u denotes the solution to (4.12), then

-Au > 0 in fZ i.e.

Jo

VuVtdx>0 dl; ED(1l),£>0.

(4.17)

If u - W is continuous in f1 so that the set

)+=IXr=01u-W(x)>0} is open, then

-Au = 0 in D'(f2+).

(4.18)

PROOF. Let l: E D(fl), C > 0. Then plugging

v=u+t;EK into (4.12) will lead exactly to (4.17). To prove (4.18) consider t E D(fl+). Then, for a small enough one has

n±eC-cp>0. So, plugging u ± e£ into (4.12) will lead to

L which gives (4.18).

REMARKS 4.3.

1. One can see that for cp continuous, u is continuous (see 1201).

2. One can write - at this point only formally - that u is the solution to

u-gyp>0,

-'&u>0, (u-cp)Du=0.

Let us quote now a general result that will include Theorem 4.3 as a particular case.

THEOREM 4.5. Let A, B be two measurable subsets of fl and gyp, 0 two measurable functions on A, B respectively. Set K = K(W, ib) = { v E Hp (St) I v(x) > w(x) a.e. X E A v(x) < O(x) a.e. x E B }.

(4.19)

Moreover, let aid, a be LOD(f2) functions satisfying (3.12) and (3.13). Let us assume that (3.14) or (3.15) holds. Then, if K 54 0, for every f E H-I(f2) there is a unique

solution u to

uEK,

J

8u 8(v - u) +au(v-u)dx>(f,v-u) VvEK. a; 8xi 8x;

{420)

4. Elliptic Variational Inequalities

56

PROOF. It is enough to show that K is closed and convex in H01(fl) and that the bilinear form (3.17) satisfies the assumptions of Theorem 4.1. The latter has already been done in Theorem 3.3. To show that K is a closed convex set of Ho (f)) proceeds from the same arguments as in Theorem 4.3. 0 REMARK 4.4. In the case where n = 2,

(aij)=Id, a=0, f =0, u can be seen as the elastic deformation of an elastic membrane forced by two punches - see Figure 4.3. Of course, in the case where B = 0, A = ft, this is the

F1auR.E 4.3

one obstacle problem.

The interpretation given in Figure 4.3 makes it easy to understand the following comparison principle:

THEOREM 4.6. Let us assume that we are under the assumptions of Theorem 4.5. Let uk, k = 1, 2 be the solution to

uk E K(,pk,,k), aii(x)Suk 8v -

L

j

uk +

ox{

auk(v - uk) dx ! (fk,4J - uk) Vv E K(Vk,iPk). (4.21)

Then - assuming the convex sets K(Wk, tbk) 56 0 - if

Vl(x)>rp2(x) a.e.xEA 01(x) > 2(x) a. e. x E B,

f, ? f2 in the X-1(ft)-sense

(4.22)

we have

ul(x) > u2(x) a.e. x E ft. ft >- f2 in the H-1(ft)-sense means (fl - f2iv) > 0 Vv E H0(f)), v(x) > 0 a.e.

xEft.

4.2. Some applications

57

PROOF. One has to show that

(ul - u2)- =

0.

For that, note that we have by (3.17):

a(ul,v-u1) > (f1,v-u1) a(u2iv-u2) > (f2,v-u2)

Vv E K(co1,+G1),

(4.23)

Vv E K(,p2ivP2).

(4.24)

Then we claim that

u1 + (u1 - u2)- E K(W1, '1),

u2 - (u1 - u2)- E K(W2,02).

(4.25)

Indeed, by Theorem 2.8 these two functions are in Ho (0). Moreover, on the subset A we have 141 + (u1 - u2)- > u1 > (O1,

u2 - (u1 - u2)- = min(ul,u2) ? w2

and on B u1 + (u1 - u2) = max(ul,u2) < 01,

u2 - (ul - u2)- < u2 S

2

almost everywhere in A and B. This proves (4.25). Plugging the first function of (4.25) into (4.23) and the second one into (4.24) leads to

a(ul, (ul - u2)-) a(u2, -(ul - U2)-)

(fl, (ul - U2)-), (f2, -(ul - U2)-)-

Adding the two inequalities we get a((uI -

<*

u2),-(ul

- u2) ) : (f2 - fl, (u1 - u2)-) < 0

a((u1 - u2)-,(ul - u2)-) < 0

see (4.22) and Theorem 2.8.

Thus (u1 - u2)- = 0 by (3.6) and the proof is complete.

0

REMARK 4.5. In the case where A, B = 0, the convex set K reduces to Ho (f1). In this case the theorem says that the less is the density of forces working against the spanning of the membrane, the less this membrane moves up.

In the case B = 0, A = ft, one has the case of the one obstacle problem pushing from below. Again in this case our result has a very simple physical explanation.

4.2.2. Other variational inequalities. We consider a(u, v) given by (3.17) under the assumptions of Theorem 4.5. Then clearly for this bilinear form one has a variational inequality each time one takes a convex set K closed in H01(11). Suitable

K are (4.26) K = {v E Ho(1l) I IVv(x)I < 1 a.e. X E n }. The corresponding problem is called the elastic plastic torsion problem. One can also introduce non-local constraints, for instance

K = 1v E H01(0) Ila i (v(x)) dx < a1

(4.27)

4. Elliptic Variational Inequalities

58 or

K = IV E Ho (1l) I ji(Vv(x))dx

(4.28)

t

where j denotes a convex function in R in (4.27) and a convex function in R' in (4.28). We refer to [151, [271, [63] for more on this topic.

4.3. Exercises 1. Prove that the map

f -+u where u is the solution to (4.2) is Lipschitz continuous. 2. Prove that when a fails to be symmetric, u can fail to be a minimizer of J (see Theorem 4.1). 3. Let cp E Ho (12) and u be the solution to (4.12). Show that the map is continuous on Ho (fl).

4. Assume n = 1, 52 = (-1,1), 'p(x) = -x2 + a Compute the solution to

EK={vEH0(SZ)I v(x)> ,(x)a.e.xES2}, ulju'1

-u')dx>OdvEK. 5. Prove that (4.19) defines a closed convex set of Ha (52). 6. Prove that (4.27), (4.28) define closed convex sets of Ho (52). 7. Let 521 C 522 be two bounded open subsets of R1. Let cp be a function of H1(112) such that

'P=O onf22\521. Let u;, i = 1, 2 be the solutions to

f

u;EK;={vEHa(52;)Iv(x)>'(x)a.e.xE1Z }, jvui

V(v-uj)dx>0VVEProve

1 that u2 > ul where ul denotes the extension of the function ul in 522 \ 521.

Chapter 5

Nonlinear Elliptic Problems Ib solve a nonlinear elliptic problem the technique is almost unique: one has to rely ca a fixed point argument. Th do so one can always first solve the problem at hand to a finite dimensional space - this is where the computer stops its investigations

- and in practice this is sufficient. Then, one has to pass to the limit. For this purpose few techniques are available. We will consider in the first sections two of then: compactness and monotonicity (see [56]). In Section 5.3 we will consider a ataaiant of variational inequalities and an application to the solution of monotone problems. Then in a last section we will introduce monotone multivalued problems.

L1. A compactness method Let us consider the simple problem of finding a solution u to

r

f

8x;

n,

(5.1)

U E Ho(S2).

lR is here some bounded open set of R" and f some element of H-n(f), the dual of Hp (0).

For a we make the following assumptions - a is a Caratheodory function that is to say a.e. X E ft, Vu E R,

is .- a(x, u) is continuous from R into R, x i- a(x, u) is measurable.

(5.2) (5.3)

Moreover, we make the assumption that there exist two constants in and M such that

0<m
(5.4)

Than we can show:

THEOREM 5.1. Under the above assumptions there is a solution to (5.1). PROOF. Again we get inspiration from numerical analysis or from an approxhnatlon of the problem. So, let

Vh, hER

5. Nonlinear Elliptic Problems

60

be a family of finite dimensional subspaces of Ho (11) such that

Vv E HH(fl), 3Vh E Vh such that (in Ha (1)), rim vh = V.

(5.5)

Clearly, one can find plenty of such families (see for instance Chapter 8). Then the

first step is to find an approximation of u. For that we claim that there exists a solution Uh to

VvE Vh,

(5.1h)

I uh E Vh.

Indeed, consider w E Vh. Then, due to the Lax-Milgram theorem applied to Vh with

a(u, v) =

J in

a(x, w)Vu Vv dx,

there exists a unique solution u = T(w) to

1n a(x, w)Vu Ov dx = (f v) VV E Vh,

(5.6)

Iu E Vh. On Vh - space of finite dimension - all the norms are equivalent, so we will consider the restriction of the one considered on Ho (0), i.e. we will consider I1av1l

^

L12dx.

Taking v = u in (5.6) and using (5.4) we easily obtain

mlIVull2 < f where

a(x,

w)JVuI2 dx _ {f, u)

11f I.11Vu[12

(5.7)

denotes the dual norm of Ha (f). Then clearly IIVuII2 < 11/11+/m

= Il m

ll..

(5.8)

So provided we choose w in B(0, II the ball of Vh of center 0 and radius 117- 11., u = T(w) will also be in this ball. By the Brouwer fixed point theorem the mapping T will have a fixed point in Vh, provided we can show its continuity (for Brouwer's fixed point theorem - see [46]). For this purpose, consider a sequence wn E Vh such that

w -, w in Vh - i.e. in Ho (St).

(5.9)

Let u,, be the solution to (5.6) corresponding to w,,. Of course the bound (5.8) holds for u,, and - since we are in a finite dimensional space - one can extract a subsequence nk from n such that unie - u E Vh.

(5.10)

5.1. A compactness method

61

Up to a subsequence one can also assume that wnk

w a.e. in ft.

Thus, we consider the equality

Vv dx = (f, v)

in

`dVEVh.

(5.11)

We see that

Vunk -' Vu,

a(x,wnk)Vv -' a(x,w)Vv

both convergences taking place - for each component - in L2(ft). Thus passing to the limit in (5.11) we get

a(x, w)Vu Vv dx = (f, v)

V V E Vh,

i.e. u is equal to T(w). Since T(w) is the only possible limit for the subsequences of u,,, the sequence u = T(w) T is continuous. This establishes the existence of an approximate solution to (5.1), i.e. the existence of a solution to (5.1h). Then one would like to show that, at the limit, uh will provide us with a solution to (5.1). For that let v E Ha(ft). Let vh E Vh be a "sequence" such that lim vh = V.

It is clear that the estimate (5.8) holds for uh - so that uh is bounded in Ho (ft). Using the compactness of the embedding of Ho (ft) into L2(ft) (see Section 2.3) one

can - up to the extraction of a subsequence - assume that there exists u E Ha(ft) such that

it,, - u in Ho (ft),

uh

u in L2(ft),

uh - u a.e. in it.

But then it is easy to show that

a(x, uh)Vvh - a(x, u)Vv in L2(fl), thus passing to the limit in a(x, uh)Vuh - Vvh dx = (f, vh) we obtain

ja(x, u) Vu Vvdx = (f,v) dv E Ha(ft), i.e. we have established the existence of a solution to (5.1). REMARK 5.1. The intermediate result (5.16) is itself very important. It established the existence of a discrete solution for the weak formulation of (5.1).

5. Nonlinear Elliptic Problems

62

REMARK 5.2. With the same assumptions on a and the same technique it is clear that one can solve non-local problems of the type

f

(a(x,l(u))` lu E Ho' (Q)

in it,

(5.12)

//

I being for instance a continuous linear form on Ha (it) - say for instance 1(u) =

(5.13)

(see [29J, (31J and Chapter 13). REMARK 5.3. In order to pass to the limit above we have used the fact that

Vuh - Vu,

a(x,uh)Vvh - a(x,u)Vv

in L2(it). This is a particular case of the weak-strong convergence limit - i.e. LEMMA 5.1. Let H be a Hilbert space and xn, yn two sequences such that xn -a X,

Yn --'I Y;

then we have lira (xn,yn) = (x,y)

n-.+oo

PROOF. We can write

I(xn,yn) - (x,y)I = I(xn -x,y)+(xn,vn-y)I <_ On - x,y)I + IxnlIyn --YIPassing to the limit, the result is due to the fact that xn is bounded in H (see

0

1651).

5.2. A >trmonotonicity method Let us consider first the nonlinear generalization of solving the system n

E aj

= bi,

i = 1, ... , n.

j=1

So, let us introduce continuous functions C H Ai(k) from Rn into R. Moreover assume that for some constants C and a,

IA.(S) - A1(t')I

CJC - C'J

Vi, VC,C' E

R°,

(5.14)

n

E(A1(e) - A.(`'))(ti - Si) >_ aIC - e'I2 VC'tl E R".

(5.15)

i=1

Then we can show THEOREM 5.2.

Under the above assumptions, for every b E R" the system

A(e)=b i i = 1,...,n admits a unique solution.

(5.16)

5.2. A monotonicity method

63

PROOF. The system is clearly equivalent to finding a 4 E Rn such that

Yi=1,...,n, i.e. to finding a fixed point for a certain mapping. So, consider the iteration defined by

° arbitrary

= ? - eAf(tP) + ebf Vi = 1, ... , n, p > 0. It is clear that by subtraction one has tp+1

fir1 - ' = ° - er1 - e(A=(EP) -

`d i = 1, ... , n.

Taking the euclidean norm of the vectors whose entries are given in the above equality gives - after using ((5.14), (5.15)), n Ie

1 - CPI2 < ICP - CP-1 I2 - 2e

is

(Ai(SP) - A1(SP-'))(Cr -

eP-1)

G (1 -2ea+e2C')le-e-1I2 where C' is some constant. Taking a small enough will cause

0=1-2Ea+e2C'
b; = implies - see (5.15)

a1C,-C2I2=0. Note that above we just rewrote the proof of the Banach contraction principle for 0 the mapping )--+ (Ci - eAi(C) + eb;). So, for f E H-1(S2) we would like now to attack the problem of finding u such

that

j__(As(Vu))=f in 11,

(5.17)

u E Ho (11).

Clearly, replacing if necessary Af by A; - A,(0) one can always assume - and we will do so from now on - that A;(0) = 0.

(5.18)

Then we are looking for a weak solution u to (5.17) - i.e. for u such that

f

Ai (Vu)

&V

8xi

u E Ha (SZ).

dx = (f, v) Vv E Ho (S2), (5.19)

5. Nonlinear Elliptic Problems

64

Clearly, under our assumptions - see (5.14), (5.18) - the integral in the first equation of (5.19) makes sense. Then we have THEOREM 5.3. Under the assumptions of Theorem 5.2 there exists a unique solution to (5.19).

PROOF. Let us perhaps consider first the question of uniqueness, which is easy. Let u1, u2 be two weak solutions. By substraction we get

{Ai(Vu1) - Ai(Vu2)}bj dx = 0 Vv E H l). So, taking v = ul - u2 and recalling (5.15) we get

a f IV(ul-u2)12dx< f (A(Vui) - A(VU2))( oxi

-

adx=0,

hence Ui = u2.

Let us then pass to the resolution of the approximated problem. We claim that there exists a unique solution uh to n

Ai(Vuh)

8v OZ.

dx = (f, v) Vv E Vh,

(5.19h)

uh E Vp.

The uniqueness could be proved as above. (Vh is a subspace of Ho (0) satisfying (5.5)). For the existence one considers a basis W1, ... , WN of Vh. Then the problem reduces to finding N

fiwi

uh = i=1 such that

I A,( N 4jVWi1.dx= (f, wi) d j = 1, ... , N.

\ta,

n

Let us set

N

t

Bj(C) = .[

(V(Et Siwi)) ,a1 /////

dx,

Then one is reduced to solving the system

B1(t)=bj dj=1,...,N.

bj = (f,wi)

(5.20)

5.2. A monotonicity method

65

Of course B3 (0) = 0. Moreover, the Ai being Lipschitz continuous, the same holds

for the B. Finally

B())(-

j[A(v

)=

X

ivli

)) - AiCV(r w;)]

(t(feiwi) N C'Wi

r n

l

icl

8x=

/F

'Wi) 1 2

>,61

t,12

\\\._i

since on Vh all the norms are equivalent. Existence and uniqueness of uh follows then from Theorem 5.2. Then we pass to the existence of a solution to (5.19). Taking v = uh in (5.19h) we get (see (5.15), (5.18)) aIIvuhII2 f Ai(Vuh)` `h dx = < (f,uh) <- IIfII.IIVuhII2. 8xi n Hence uh is bounded in Ho (Sl) and up to a subsequence we can assume that there exists u E H H (S2) such that

uh - u

in Ho (St).

(5.21)

in Ho (l),

(5.22)

If Vh E Vh is a sequence such that vh

v

we have from (5.19h) (5.23)

Ai(Vuh) 8 h dx = (f, vh).

However, we are not able to pass to the limit in the left-hand side of this inequality! Replacing vh by Vh - Uh - (5.23) reads Ai(Vuh)8(yh -uh)

dx = (f,vh - uh)

8xi Thanks to the monotonicity of Ai we get also ft

Uh)

n { AL (V uh) - A+ (V vh) } 49()h _ Uh) dx + L Ai (vvh) O(V h and thus Ai(Ovh)

dx = (f, vh - uh )

8(yh - uh)

dx > (f,vh - uh) 8xi Assuming - which is suitable by (5.22) - that Vvh Vv, a.e., we can pass to the limit and get

J in

I Ai (Vv) tt

8v - u 8xi

dx > (f, v - u) V v E Ho (Sl).

5. Nonlinear Elliptic Problems

66

Then taking v =

+ tw, w e Ha (R) we get uu

Jn

dx>(f,w) VwEHa(f2).

A;(Vu+tVw)

Letting t -' 0 we obtain Out > (f, w) V W E Ho (1l). jAi(Vu)dx

Changing w into -w shows finally that u is a solution to (5.19) which is what we wanted to achieve. 0

5.3. A generalisation of variational inequalities Let H be a real Hilbert space. We denote by the associated norm.

the scalar product in H and by

Let a(u, v) be a bilinear form on H that will be assumed continuous and coercive, i.e. such that (3.5), (3.6) hold. Let K # 0 be a closed convex set of H. Let from K into [0, +oo] such that

be a proper convex function

is lower semicontinuous for the weak topology. By "proper" we mean: there exists vo E K such that J[va] 36 oo.

(5.24) (5.25)

For f E H', the strong dual of H, we consider the problem of finding u such that

Ja(u,v-u)+{J[v]-J(u]}>(f,v-u) VvEK, (5.26)

uEK. Under the above assumption we have THEOREM 5.4. There exists a unique solution u to (5.26).

PROOF. We proceed as in the proof of Theorem 4.1. First let us assume a symmetric, i.e. with the notation of Theorem 4.1

a = as. Then set

I(v) = 2 a(v, v) - (f, v) + J[v]. We claim that for every f E H' there exists a unique u minimizing I on K - i.e. such that

I(u)
Let us consider u a minimizing sequence of I on K - i.e. a sequence such

that u,EKand

1(u,,) -' Inf I(u). UEK

(5.27)

5.3. A generalization of variational inequalities

67

Due to (3.6). (5.25) we have 2

I(vo) + 1

Iu6I2 - If

for n large enough (I I. denotes the strong dual norm of H). Thus

for some constant C independent of n. One can now extract a subsequence from u, that we still label u,a such that u1, - u in H. Since K is closed and convex, u E K. Moreover using the lower semicontinuity of J and a(., ) we deduce that

iim inf l(un) > 1 a(u, u) - (f, u) + J[u] = I (u), ?I-+00

i.e. u is a minimizer of I on K (see (5.27)). We have also for every t E (0,1)

1(u) < I (u + t(v - u)) Vv E K. Thus we can write 2

2a(u, u) - (f. u) + J[u] < 2a(u, u) + ta(u, v - u) + 2 a(v - u, v - u)

(5.28)

- (f, u + t(v - u)) + J[u + t(v - u)]. By the convexity of .I this leads to 2

t(f, v - u) + J(u]


Hence

(f, v - u) < a(u, v - u) + 2a(v - u, v - u) + (J[v] - J[u]),

d v E K.

Letting t - 0, it is clear that u is a solution to (5.26). Conversely if u is a solution to (5.26) we have for every v E K I a(u, u) - (f. U) + J[u] < 2a(u, u) + a(u, v - u) - (f, v) + J[v]

< 2a(u,u)+a(u,v-u) + a(v - u, v - u) - (f, v) + J[v]. 2 But this is (5.28) written for t = 1 and thus u is also a solution to I(u) < I(v) Vv E K.

(5.29)

Thus the problems (5.26) and (5.29) are equivalent. If now ul, u2 are two solutions to (5.26) we have

a(ul.zv-ul)+{J[v]-J[ul]}>(f,v-ui) VvEK, a(1u2.n-u2)+(J[v] -J[u2])> (f,v-u2) VvE K.

5. Nonlinear Elliptic Problems

68

Taking v = u2 in the first inequality, v = ul in the second and adding leads to

a(ul - u2,ul - u2) < 0 which by (3.6) leads to ul = u2 and uniqueness. (Note that when u is a solution

to (5.26), taking v = vo leads to J[u] < +oo). Then we drop the symmetry assumption and proceed as in Theorem 4.1. That is to say we introduce the set

T={ t E R I V f E H' the problem Pt has a unique solution} where Pt is the "variational inequality" (see Theorem 4.1 for the notation regarding at)

uEK,

jae(u,v-u)+{J[v]-J[uJ}>(f,v-u) VVEK. As we have just seen, T contains 0. Next for to E T, f E H', w c- H there exists a unique u = F(w) E K that is a solution to

at,,(u, v - u) + {J[v] - J[u] } > (f, v - u) + (to - t)aA(w, v - u) V V E K. One shows exactly as in Theorem 4.1 that (4.5) holds and that F has a fixed point for every t E [to - , , to +]. This implies that T = R and for t = 1, PI is the problem one wanted to solve. REMARK 5.4. As we saw in the above proof, in the case where a is symmetric, u is the unique minimizer on K of

I(u) = 2 a(u, u) + J[u] - (f, u). REMARK 5.5. In the case where K = H and J is differentiable,

a(u,v-u)+{J[v]-J[u]}>(f,v-u) dvEH. Taking v = u + tw we get for any t > 0,

a(u, tw) + {J[u + tw] - J[u] } > (f,tw) Vw E H. Dividing by t and letting t -, 0 we obtain

a(u, w) + (J'(u), w) > (f, w) dw E H, hence after changing w into -w to get the reverse inequality, it turns out that

a(u, w) + (J'(u), w) _ (f, w) dw E H.

(5.30)

We consider now the bilinear form a defined by (3.17) and we assume that we are under the hypothesis of Theorem 3.3. Let $(x, u) be a Caratheodory function defined on i2 x R verifying x F- Q(x, u)

is measurable on 11, V u E R,

(5.31}

5.3. A generalization of variational inequalities

69

3 C > 0 such that 113(x, u) - ,3(x, v) I < Cju - vi Vu, V E R, a.e. x E Sl, ,3(x,0) = 0,

and u - 13(x, u) is monotone in the sense that P(x, u) -13(x, v))(u - v) > 0 Vu, v E R, a.e. x E Sl. Then we have

(5.32) (5.33) (5.34)

THEOREM 5.5. Under the assumptions of Theorem 3.3, i.e. if (3.12)-(3.15) holds and if (5.31)-(5.34) also hold, there exists a unique weak solution to the problem

(Of - a;j 8 ) (x)

o-x ,

+ a(x)u + 13 (x, u) = f

in n ,

(5.35)

u E Ho (Sl).

PROOF. N ote first that due to (5.32), (5.33) we have 1/3(x, u)I = 1/3(x, u) - (3(x, 0) 1 < CIul,

(5.36)

and clearly in (5.35) we have

$(x,u) E L2(fl) So, with the notation (3.17) the first equation of (5.35) means:

a(u, v) + j (3(x, u) v dx = (f, v) Vv E Ho (S)).

in

(5.37)

Suppose first that ul, u2 are two solutions to (5.37). By subtraction we obtain

a(ul - u2, v) +

j

n

(3(x, ul) - Q(x, u2)}v dx = 0 b v E Hp (Sl).

Taking v = ul - u2 and using (5.34) we get a(ul - U2, U1 - U2) < 0 and thus ut = u2 by the coerciveness of a - see the proof of Theorem 3.3. Next

let us set

j (x, u) = j 13(x, u s) ds.

(5.38)

0

By (5.33), (5.34) we have

j (x, u) > 0 a.e. x E (l, d u E R. Moreover we claim that j is a Caratheodory function - indeed

x i- j(x,u) is measurable -- since writing the integral as a Riemann sum Jim

j(x, u) =

" E 13

n-4-00 k=1

(x,ku\.1 n

n

(5.39)

5. Nonlinear Elliptic Problems

70

and

u) is the limit of a sequence of measurable functions. Moreover

u i- j (x, u) is clearly continuous a.e. x E ft and j is a Caratheodory function. In addition one has, due to the monotonicity of d and (5.36),

0 < j(x,u) < If(x,u)[ lug < C[u['. Thus, for any u E L'(0) we can define

J[u] = 1 j(x,u(x))dx > 0.

(5.40)

in

Due to the monotonicity of f, j(x, ) is convex for almost every x and so is J. Moreover, when u,a - u in Ho (fl) we can extract a subsequence of un - still labelled un - such that lim J[un] = lim inf J[un], t n -' u a.e. in ft. n-+oo By Fatou's Lemma we deduce that

lim+nfJ[un] =lim / j(x,un(x))dx n

= lliim inf

_f

> lim

fn

j(x,un(x))dx

j(x, u(x)) dx = J[u].

This shows the weak lower semioontinuity of J on Ho (ft). Since a is bilinear, continuous, coercive - see the proof of Theorem 3.3 - and J(O) = 0, we can apply Theorem 5.4 to get the existence of a solution u to

Ja(u, v - u) + {J[v] - J[u]} > (f, v - u) by E Hp (ft), u E Ho (ft).

Taking v = u + tw with w E Ho (i)) we obtain

r ru(s)+tw(x)

a(u,tw)+ f J

ft u(m)

Q(x,s)dsdx> (f,tw) /wE HH(ft)

and thus due to the mean value theorem and for 0(x) E (0,1) a(u, tw) + J fl(x, u(x) + O(x)tw(x))tw(x) dx > (f, tw) `d w E Ha (ft).

a

Dividing by t and letting t - 0 we get a(u, w) + f Q(x, u(x))w(x) dx > (f, w) Yin E Hp (ft). Changing in into -w we obtain the equality and thus (5.37). This completes the proof of the theorem.

5.4. Some multivalued problems

71

5.4. Some multivalued problems We would like to address next a multivalued case. This theory was introduced by H. Brezis in [14]. We will consider here some new extensions of it that one can find in [26].

Let S2 be a bounded domain of R", n > 1 with boundary 1:'. Let us consider A an elliptic operator given by

Au = E

8xi

(aii(x)'_).

(5.41)

We use the Einstein summation convention of repeated indices and assume as usual aij E L°O(SZ)

di, j = 1,...,n,

ajt12 dt E It", a.e. x E SZ,

(5.42) (5.43)

where I I denotes the usual euclidean norm in R".

For f E LP(Q) we would like to consider problems of the form:

I-Au + f3(x, u)

u=0

f

in SZ,

on r,

(5.44)

where 6(x, ) is some monotone graph. The formulation (5.44) is understood in a weak sense (cf. [17], [48]) which means that one looks for a couple (u, g) E Ho (f)) x L2(fZ)

(5.45)

-Au + g = f in f?

(5.46)

such that in a weak sense and g(x) E f3(x, u(x))

If

a.e. x E Cl.

(5.47)

denotes the duality bracket betwH1(fl) and H(SZ) and if we set

a(u, v) = (-Au, v) =

jau .9xjOxi dx V u, v E H(I),

the meaning of (5.46) is simply (-Au, v) + (g, v) = (f, v) Vv E Hp (SZ)

(5.48)

(5.49)

denotes the usual scalar product in L2(SZ). The reader is referred to [13], [16] for the theory of maximal monotone operators. For simplicity we will suppose in what follows that where

0 E Q(x, 0)

a.e. X E SZ.

(5.50)

In this case and when ,0 is independent of x, a theory of existence for (5.44) has been developed in [15], [14]. It is based on an L2(SZ)-estimate for /3(u). More

5. Nonlinear Elliptic Problems

72

precisely, multiplying (5.44) by /3(u) = /3(x, u) - when for instance $ is a smooth function independent of x - and integrating on fl will lead to (5.51)

h(u)la 5 if12

thanks to the inequality

(-Au, J3(u)) = in a,'

8xs

f3'(u) dx > 0.

(5.52)

In the case where S depends on x, an extra term appears in the integral above and the problem has to be handled differently. This is what we would like to consider now.

So, we assume that for a.e. x, /3(x, ) is a maximal monotone graph in R x R, (see [16]), with OE,O(x,O)

a.e.xE12.

(5.53)

For a maximal monotone graph /3 we denote by D(/3) = { t E R I p(t)

0}

(5.54)

its domain. For every t E D(/3), /3(t) is a closed interval and ,lo(t)

(5.55)

(reap. (3°(t))

the element of /3(t) of smallest (reap. largest) absolute value when t 96 0, 0(0) = [Oo(O), X8°(0)] (see [13], [16] for all the notions about maximal monotone operators).

Let #I, f3 be two maximal monotone graphs in R x R with domains D(r8;), i = 1, 2 and such that 0 E A- (0), i = 1, 2. We adopt the following definition introduced in [26). We will say that - for b > 0 (5.56)

1011 < I$21 + b

if and only if

D(#j) = D($2),

d (t, zl) E pl

3(t, z2) E /32

with [zi l 5 [z2I + 5.

(5.57)

If (3 is a maximal monotone graph on R x R, then the Yosida approximation of is the function defined for every A > 0 by

PA(t) = (t - JA(t))/A

(5.58)

JA(t) = V+ as)-10).

(5.59)

where Ja is given by

Recall that due to the monotony of 0, JA is a contraction defined on R and 0a is a Lipschitz continuous monotone function - with Lipschitz constant i (see [16]). Then we can show LEMMA 5.2. Let P1, #2 be two maximal monotone graphs in R x R satisfying (5.56), 0 E Jli(0), i = 1,2. It holds that (5.60) Ith.X(t)I < jI2,(t)I +5 bt E R.

5.4. Some multivalued problems

73

PROOF. Set

Ja(t) _ (I + AQ;)-1(t)

i=1,2.

(5.61)

Due to (5.58) it is enough to show first

Ja(t) > J\ (t) _,\b Vt > 0.

(5.62)

But due to the definition (5.61) it holds that

y; = J1(t) a t E (I + AQ;)(y;).

(5.63)

Suppose that 0 < y1 < y2 - M. Then, due to (5.63) one has for some z1, z2

yi+Azl=t=y2+Az2i z;EQ,(yt).

(5.64)

Due to (5.56) one can find z3 with b + z3 ? zl,

z3 E 02(yl).

From (5.64) we get

A(z3-z2)?7/2-l/1 Multiplying by 112 - y1 > Ab > 0,

A(Z3-z2)-(y2-1/1)>0. This contradicts the monotonicity of 02 and thus it holds that

112-Ab<1h which is (5.62).

When t < 0 one has to show Ja(t) < j.%2 (t) + Ab.

Adopting the notation (5.63), suppose y2 + A6 < y1 < 0. From (5.63) we derive (5.64) for some z1, z2. Then due to (5.56) we can find z3 < 0 such that z3 < z1 + 6,

z3 E 02(yl)

Then from (5.64) we get

1\42 - z3) ? y1 - y2 - Ab. Multiplying by yl - y2 > A6 we get a contradiction to the monotonicity of 02. This completes the proof of the Lemma. 0 Then we can show:

THEOREM 5.6. Assume that for a.e. x E fl, Q(x, ), Q are maximal monotone graphs in R x IR satisfying (5.53) with domain D = D(f3) = D(Q(x, )) a.e. x, and such that the functions

x " Qo(x, t), O

O

x i-- Q°(x, t)

(5.65)

are measurable Vt ED where D denotes the interior of D. Moreover assume that for some constants 0 < a < b and the maximal monotone graph Q, with 0 E,6(0),

5. Nonlinear Elliptic Problems

74

it holds that Ia,6I <- IR(x, -)I < IbfI

a.e. X E ft.

(5.66)

Then, for any f E L2(f2) there exists a unique solution u to

Au +,6(x, u) 3 f in ft, (5.67)

lu E Ha (fi),

in the sense of (5.45)-(5.47). Moreover, we have the estimates I/3(x,u)12 <- QIf12 IAuI2 <_

(i+ a) V12-

(5.68)

(5.69)

12 denotes the usual L2(fl)-norm, S(x, u) the unique function g in L2(it) such that (5.46) holds. I

PROOF. 1. Uniqueness Consider two couples (u, g), (u', g') E Ho (0) x L2(fl) such that (5.46) and (5.47) hold. By subtraction we obtain

-A(u-u')+g-g'=O inf2.

(5.70)

Multiplying this equation by u - u' and integrating on ft we get

(-A(u - u'),u - u')= by (5.47). It follows from (5.43) that u = u'. Then, by (5.46), g = g'. This completes the proof of uniqueness. Note that since g is unique there is no ambiguity to denote

it by f3(x, u). We will do that in what follows. We turn then to the question of existence, which will be done in several steps. 2. Approximation of the problem

Set fora.e.xEfl,A>0 JA(x,t} _ (I+A(3(x, ))-1(t) Vt E R,

(5.71)

(3a(x,t)_(t-JA(x,t))/A VtER.

(5.72)

As mentioned above #A (x, t) is the Yosida approximation of P(x, t). We claim that

(x, t) '-- 8,,(x, t)

is a Caratheodory function,

(5.73)

i.e. we have for every A > 0 a.e. X E Cl,

V t E R,

t i-' la(x, t) is continuous,

(5.74)

x

(5.75)

flu (x, t) is measurable in Q.

To establish (5.74) we remark that for every t, t' E R, t > t' Sa (x, t) -

= {(t -e) - (J,,(x, t) - JA(x, t'))}/A < t - t'/A

5.4. Some multivahied problems

75

-this due to the monotonicity of 8(x, ) and JA (x, -). It follows easily that (see

*o [18])

-,ea(x,t')I < It ael vt,t' E R

(5.76)

izid (5.74) follows. To establish (5.75) it is enough to show that for every t the n x 1-4 A(x,t)

is measurable - for instance that for every c # 0 the set

[Ja>c]={xEfl I Ja(x,t)>c},

(5.77)

is measurable. Recall that

z=J.\(z,t)gtEz+XO(x,z).

(5.78)

Thus one has - suppose first c > 0:

z>CO t>c+as°(x,c).

(5.79)

(x = JA (x, t) implies that z E D(,6(x, -)) = D and so c ED by (5.53)). Thus one derives easily that

[Ja>c]={zEf2I10e(x,c)
(5.80)

and by (5.65) the measurability of this set follows. Suppose now c < 0. Then it is enough to show that the set

[JA
[Ja
z
=f

in S2,

u, E Ho(0)-

(5.81)

P rst, remark that due to (5.53) one has f3A (x, 0) = 0 a.e. x E Sl

(5.82)

and from (5.76) it follows by Theorem 5.5 that there exists a unique solution ua to (5.78).

5. Nonlinear Elliptic Problems

76

S. Estimates First it is easy to show that one has - for any maximal monotone graph # .1i > A2 =13a, (t) 5 fla, (t) d t > 0.

(5.83)

Moreover, for any positive constant c - it holds that

(cl)a = cfi.

(5.84)

Thus, due to Lemma 5.2 it holds that, since a < b for a.e. x E 12, afiba(t) < afloa(t) < fa(x,t) < bfba(t) Vt > 0.

(5.85)

lb simplify our notation, let us set

7=#M

(5.86)

and multiply the first equation of (5.81) by 7(u.) where denotes the positive part of a function. Clearly, 'y(ua) E Ho (i2) since 7 is Lipschitz continuous and 7(0) = 0. Thus, after integration on 12 we get (-Aua, 7(ua )) + (QA(x, ua), 7(ua )) = (f, 7(ua )) Since

= jau

7(u,+) dx > 0 ,

(5.87)

it follows that (5.88) (!3a (x, ut ), 7(ua )) < (f,7(ua )). Using (5.85) it turns out easily - if If 1z,+ = f[..' '01 If 12 dx (with [u,\ > 01 the set

of xin 11such that ua(x)>0)-that aI7(ua )I2

5

IfI2,+I7(ua )I2,

i.e.

17(ua )12 < If1"2 +.

(5.89)

From (5.85) and (5.89) we get

IQa(x,ut)12
(5.90)

;IfI2,+.

As in (5.85) we establish easily that for a.e. x E 12 bf3aa(t) < bfiba(t) <,6\(x,t) < afjaa(t)

dt < 0.

(5.91)

As above let us set 7 = Qoa

and multiply the first equation of (5.81) by y(-ua) where denotes the negative part of a function. 7{-u-) is a nonpositive function belonging to H01(11). Thus integrating on 12 it comes (-Aua, 7(-ua )) + (,Oa (x, ua), 7(-u; ))

7(-ua )).

Since

(-Au.\, 7(-ua )) = r

Jn a,

8u- 8u-

ax,

7'(-ua) dx > 0

(5.92)

5.4. Some multivalued problems

77

we obtain

V% (X, -ua ), 7(-ua )) < (f,')'(-ua )). From (5.91) we derive easily, with I f I2,- = f(uA

Hence, as in (5.90) IQa(x,-ua)12 <- aIfI2,-

Next it is dear - due to the definition of the L2(SZ)-norm - that IQa(x, ua)12 = Iaa(x, ua - ua )12 = faa(x, ua )12 +

(} b

2 2

+

1#,\(X,

-uA )12

()2IfIL 2

(5.93)

(5)2

<-

If12

We now multiply (5.81) by ua. Integrating over fl and using the fact that Qa(x,ua(x)) u,,(x) > 0 a.e. in Il we get

(-Aua, ua) < (f, ua) from which we derive easily that for some constant C, IIua111.2 <- CIA

where II of A.

' 111,2

(5.94)

denotes a norm on HO '(P) and C is a positive constant independent

4. Passage to the limit Considering (5.93) and (5.94) we can extract from A a "subsequence" - still denoted

A - such that in HO '(Q),

in L2(Sl), a.e. in fZ,

(5.95)

in L2(Sl).

From the first equation of (5.81) it follows that

-Au+g=f infZ as one can see taking the limit in the distributional or weak sense. Next, it is well known that #,\ (X, t) E O(Ja(x, t)) a.e. X E 0, d t,

(5.96)

(see (5.71) and (5.72)). Moreover we deduce from (5.72) that, when A -* 0, IJJ(x,ua) - ua12 <- AIAa(x,ua)12 - 0.

(5.97)

5. Nonlinear Elliptic Problems

78

This combined with (5.95) implies

JA=JA(x,ua)-u inL2(fl)

(5.98)

when A -+ 0. Next, from the monotonicity of ,0 we deduce that

- ,0°(x, t)) (J- t) dx > 0

(5.99)

j(t9A(xuA) - flo(x, t)) (J- t) dx > 0

(5.100)

°

for every measurable subset w E St, and t ED (Bee (5.96), (5.98)). From the mean surability of the function x -+ /6°(x,t), x r- 00(x,t), and from the bound (5.66), it is clear that these functions are in L°O(f) C L2((l). Thus letting A -' 0 in (5.99) and (5.100) we get

bt ED . Since this holds for any measurable subset of w we get a.e. x E 12

tItED. Multiplying the first equation by v E [0,1], the second by 1 - v and adding, it comes a.e. in 11 and for every v E [0,11

dtED, i.e. it follows that °

VtED,

VaE86(x,t), this of course a.e. x E 1l. Next we claim that

u(x) E D a.e. x E n. Indeed, if not, then one of the sets u(x) 0 D U (-oo, 01, u(x) 0 D U [0, +oo) has a positive measure. Assume for instance that this is the first one. Then from (5.89) by Fhtou's lemma we have

I lmi

I f1bA(ut (x)) 12 dx <

V12

(5.101) .

We have also

AA(ut(x)) E Q(JbA(ut(x))).

(5.102)

Of course (see (5.97))

IJbA(ua)-ut12-5 bAIJ9bA(ua)12-'0

when A - 0. Since ua - u+ a.e. on f2 up to a subsequence, we can assume that JbA(UA -' u+

5.4. Some multivalued problems

79

a.e. in Q. But then by (5.102) and when u(x) V D U (-oo, 0] we have lira inf f3ba (ua (x)) _ +oo

for x such that u(x) 0 D U (-oo, 0],

i.e. on a set of positive measure. This contradicts (5.101). The argument would be the same if u(x) V D U [0, +oo) is a set of positive measure in which case one would consider instead of (5.89) the inequality IQaA(-ua )Iz <

I

az

0

E f3(x, ), t,a ED such Thus u(x) E D a.c. x. Then considering a sequence (o,,, that t, t E OD (OD denotes the boundary of D), o -- 7(x, t) where

7(x, t) =

Sup f3(x, t)

for t < 0 D c [t, +oo),

Inf f3(x, t)

for t > 0 D C (-oo, t),

t

we derive from

(9(x) - a.) - (u(x) - t,,) > 0 the inequality

(9(x) - 7(x, t)) - (u(x) - t) > 0. If t < 0 and u(x) = t this implies

(g(x) - o) (u(x) - t) > 0

`d a E f3(x, t).

If t < 0 and u(x) > t then this implies 9(x) > 7(x, t) > o V a E f3(x, t),

i.e. it holds also that

(g(x) - o) (u(x) - t) > 0 V o E f3(x, t). One would argue the same way to obtain this if t > 0. Thus we have

VoEl(x,t), VtED. It follows from the maximality of 13(x, ) that u(x) E D(f3(x, )) and g(x) E /3(x, u(x)) a.e. x E 0.

This gives (5.47) and completes the proof of existence to (5.67). To get (5.68) and (5.69) one uses (5.93) and (5.95) and the lower semicontinuity of the L2(Q)-norm for the weak topology to get

I0(x,u)I2 < QIf[z-

(5.103)

Considering then (5.67) we have

Au = #(x, u) - f and the result follows from (5.103). This completes the proof of the theorem. 0

5. Nonlinear Elliptic Problems

80

We assume now that the coefficients a,2 of A and the domain ft are smooth enough in such a way that the LP-regularity theory applies for the Dirichlet problem. More precisely if u is the weak solution to

-Au = f

in 0,

(5.104)

u E Ha (S2)

and f E LP(i2), p > 2, then we assume that u belongs to W2.P(() and that we have the estimate (5.105)

IIUII2,p <- CIfIP

where C is a positive constant depending only on A and Q. In (5.105)

II

II2,p

denotes a W2,P(f1)-norm, I Ip the usual LP(fl)-norm. We refer the reader to [46] for precise conditions for (5.105) to hold. Then we can complete Theorem 5.6 by the following:

THEOREM 5.7. Under the assumptions of Theorem 5.6, let u be the solution to (5.67). If f E LP(fl), p > 2 then it holds that /3(x,

u)Ip

<

IAuIp <

(5.106)

a IfIP+

(1+

)iii.

(5.107)

Moreover, if A, U are such that the solution of (5.104) satisfies (5.105), then u E W2"P(f2) and it holds that II1I12,p <- CIfIP

(5.108)

for some constant C.

PROOF. In (5.86) instead of multiplying by 7(u,') we multiply by

7(ua) I7(ua )1,2 by its truncation. Then arguing replacing if necessary the function 7( )I7( exactly as above (see also [14j for this technique) we obtain instead of (5.89) and (5.90)

I7(u, )IP <

Ifa,

with an obvious notation for If [p.+ = of Y(-uf )

1,6(x,

ua )Ip < a If IP.+

If I P dx. In (5.92) using now instead

7(-u )I7(-u,, )Ip-2 we obtain easily

Ioa(x,-ua)IP < blflp,, Then (5.106) follows as in (5.93), after passing to the limit in A. The rest of the 0 proof of the Theorem is trivial.

5.4. Some multivalued problems

81

EXAMPLES 5.1. 1. Let /3 be a maximal monotone graph on R x R, v/, a function

in L°°(fl) such that for some constants a, b it holds that

0
(5.109)

Then for f E LZ(Sl) there exists a unique solution to the problem

I -Au + t,(x)Q(u) 3 f,

uEHo()).

(5.110)

Indeed it is clear that all the assumptions of the theorem are satisfied for /3(x, t) _ &(x)l(t). This follows from

(3°(x, t) = i(x)/3°(t),

/3°(x, t) = O(x)/3°(t)

(5.111)

and the estimates for a.e. x E fI Iaa(t)I 5 It'(x)A(t)I < Iba(t)I Vt E it It should be noticed in the case where /3(t) = t that (5.110) reduces to

(5.112)

-Au + i,b(x)u = f in Sl, u E HO (0).

Then in order to make sense of i(x)u one needs V' bounded - moreover if one wants to get an estimate for I/3(u)12 = Iu12 - multiplying the equation by u it is clear that a hound for 0 from below is needed. This justifies our assumption (5.66). 2. Suppose that 6(x, t) is a Caratheodory function monotone in t. Moreover,

suppose that for some monotone function 0 and two constants a, b > 0 it holds that a1/3(u)l < 10(x,u)l < bIf3(u)J.

Then if /3(x, 0) = 8(0) = 0 a.e. X E Sl, f E L2(1) the problem

-Au + /3(x, u) = f in fl, lu E Ho (11),

admits a unique solution, with in addition /3(x, u) E L2(Sl). 3. Let 3 be a monotone continuous function on R such that for some positive constants a, b

all - t'l < J/3(t) - /3(t')I < blt - t'9 bit, t' E R.

(5.113)

For 0 E LP(I), f E LP(fl) and under the assumptions (5.104) and (5.105) the problem

l-Au + Q(u - +,(x)) = f in fl,

uEHo(0),

(5.114)

5. Nonlinear Elliptic Problems

82

admits a unique solution which belongs also to W2'P(fl). Indeed by (5.113) it is easy to see that 10(-+G)1 <- IQ(0)1 + blPI

so that /3(-tai) E LP(fl) since fl is bounded. Then the first equation of (5.114) can be written

-Au + (3(u -,,(x)) - #(-O(x)) = f - 0(-+'(x))

in Cl.

(5.115)

Then clearly the operator

7(x, t) = Q(t - t'(x)) - Q(-O(x)) satisfies

altI :5 1'r(x,t)I <- b1t1

Since f - (3(-tli) E LP(fl) the result follows from Theorem 5.7. It is clear that (5.114) could be the approximation of an obstacle problem (with @ in place of 13). The solution is in W2-P(f)), even so vp is only in LP(fl). But of course the estimate of the W2' (fl)-norm of u depends on e and "blows up" when e -+ 0. 4. /3(x, t) = /3(t) a.e. X E fl. Then in this case a = b = 1 and (5.68) and (5.106) give back the estimates of [14]. 5. Let there be, for i = 1,. .. , p, p maximal monotone graphs /Ji on R x R with

domain D(/) such that D(/3;)=D di=1,...,p,

0E/3i(0)

di=1,...,p.

Let aj, i = 1, ... , p, be p functions measurable in fl and such that for some positive constants a, b

0
P

Q(x,t) = 1 ai(x)fi(t) i=1

satisfies

I aEA3-W <- IQ(x,t)I !5 I bEA-(t)I i=1

i=1

and the result applies. For some extensions and complements we refer the interested reader to (26].

5.5. Exercises 1. Extend Theorem 5.1 to the case of the problem Caij (x, u) &u

+ ai(x, u) =fin fl,

L9xj

u E Ha (C),

under suitable assumptions on the functions ai?, a..

5.5. Exercises

83

2. Let i be a bounded open subset of R". For f E H'1(f) show that there exists a unique solution u to: u E Ha (12),

fn IVuldx

>(f,v-u) VvEHa(12). 3. Let a(x. u) be a Caratheodory function such that (5.4) holds. For V E H1(12), < 0 on I', show that for f E H-1(12) there exists a solution to the variational inequality

uE K = { v E Ho (1) I v(x) > w(x) a.e. in fl },,

ja(x u) Vu V(v - u)dx> (f,v -u) /vEK. 4. Let J,,, [ia he defined by (5.58), (5.59). Show that JA is a contraction on R

- i.e. that IJA(t) - JA(t')I < It - t'j Vt, t' E R and that ;3a is a Lipschitz continuous, monotone function with Lipschitz constant equal to 5. Establish (5.83), (5.84). 6. Show that the inclusion (5.96) holds true.

Chapter 6

A Regularity Theory for Nonlocal Variational Inequalities 6.1. Some general results the scalar product in H and by Let H be a real Hilbert space. We denote by the norm associated. Let a(u, v) be a bilinear form on H that will be assumed to be continuous and coercive, i.e. such that (3.5), (3.6) hold. Let J[] be a proper convex function on H into [0,+00] such that is lower semicontinuous for the weak topology, there exists vo E H such that J[vo] = 0.

(6.1) (6.2)

Let p > 0, e _> 0 be two constants - for f E H' the strong dual of H - we consider the problem of finding uE such that

fa(ue,v-ue)+e{J[v]-J[ue]}?(f,v-ue) VvEH, U. E H,

and for

K,,={vEHIJ[v]
fa(u, v - u) > (f , v - u) d v E K,,, uEKM. First we have

THEOREM 6.1. There exists a unique solution of to (6.3) and a unique solution u to (6.5). PROOF. The existence of ur is an immediate consequence of Theorem 5.4. It is easy to see that K. is convex and non-empty since vo E Kµ. Moreover, due to (6.1), KM is weakly closed - and so also closed for the topology of H since convex. Due to (3.5) and (3.6) the existence of a unique solution to (6.5) follows from Theorem 4.1.

0

6. A Regularity Theory for Nonlocal Variational Inequalities

86

In what follows we would like to show that for p > 0 there exists always an e = e,, such that: U = U.

(6.6)

For that let us first prove THEOREM 6.2. Let ue be the solution to (6.3). Then the mapping

CHue (6.7)

(0, +oo) -' H is continuous when H is equipped with its strong topology.

PROOF. Taking v = vo in (6.3) we get

a(u., u`) + eJ[u.] 5 (fled) + a(u., vo) - (f, vb).

(6.8)

Recall that is the duality bracket between H' and H. Since E, J > 0, it follows from (3.5), (3.6) that alue I' <- If I.I I + CIvol Iusl + Ifl.Ivol Recall that I I, denotes the strong dual norm in H'. This leads easily to

lull <- C'.

(6.9)

where C' is a constant independent of e. Going back to (6.8) we get also for some constant Co

0<J[ue]<

o

(6.10)

So, let us consider E, c, a -* e'. From (6.3) we have

a(ut,v-us)+E{J[v]-J[ua]}>(f,v-u6) VvEH,

(6.11)

a (ue , v - ue) + e' { J [v] - J [ue] } > (f , v - us) Vu E H. Taking v = u,' in (6.11), v = u, in (6.12) and adding, we obtain

(6.12)

a(uf - u', ue - u.) + (e - e'){J[ue] - J[ug]} > 0. Hence

a[uz - ueI' < a(u., - ue, u. - ue) < (e - E'){J[uel - J[u,]}.

(6.13)

Clearly when e' > 0, then due to (6.10), J[u£], J[ur'] remain bounded when e --+ c. Then passing to the limit in (6.13) leads to

uE - ti

in H

which is what we wanted. For c' = 0, by the solution to (6.3) we mean the solution uo to

a(uo,v - uo) > (f,v - uo)

d v E H.

One notices that by (6.9) u,, is bounded independently of e, e -' 0. Thus up to a subsequence one can assume that when e - 0, for some uo E H we have: u, 60 in H.

6.1. Some general results

87

From (6.3) we deduce

VVEH. Taking the lim inf of both sides it turns out that a(uo, uo) <_ lim anf a(ue, ue) 5 a(uo, v)

- (f, v - uo),

i.e. uo = uo. Since ue has only one possible limit, then ue - uo when £ -+ 0. To show strong convergence it is enough to notice that from above one gets

limsupa(ue7ue)
Taking v = uo it holds that

limsupa(u,,ue)
Since

lim inf a(uei ue) > a(uo, uo)

we deduce that a(ue, ue) - a(uo, u0)

when £ - ., 0 and the strong convergence follows. This completes the proof of the theorem. 0 REMARK 6.1. For E = 0, (6.3) is just

a(uo, v - uo) ? (f, v - uo) Vu E H. Taking v = u° ± v in this inequality - it becomes clear that uo is the solution to J a(uo, v) = (f, v)

V v E If,

uoEH.

(6.14)

THEOREM 6.3. The mapping [0, +oo) -+ (0, +oo),

£'-' J[ue] is nonincreasing, continuous. Moreover it holds that

0:5 J[ue] <°,

(6.15)

(6.16)

where Co is a constant independent of E.

PROOF. Suppose £ > e'. From (6.13) we deduce

(E - E'){J[ue'] - J[ue]} > 0, i.e. J[ue,] > J[ue] and e i -+ JIue] is nonincreasing. Taking v = ue, in (6.3), EJ[ue] < (f, ue - uet) + £J[ue,] + a(ue, ue# - ue) (611) <- IfI.[ue - ue,I +EJ[ue'] +CIuel lue, - uel.

6. A Regularity Theory for Nonlocal Variational Inequalities

88

From (6.9) we know that Jue I is uniformly bounded. Thus letting E -' ein (6.17) and using Theorem 6.2 we get

limsupEJ[uE] < e'J[u'].

e-e'

(6.18)

Using the lower semicontinuity of J for the weak topology we would get

1 emu -J[u] ? e'J[u,,].

(6.19)

Thus combining (6.18) and (6.19) we get

lim CJ[U] = e'J[ue"];

C-e when e> 0 this implies also

el J[ue] = J[ue,] and the continuity of E - J[ue] follows. When e = 0, from the monotonicity of E - J[ua] we have J[u`] < J[U0]Hence

lim sup J[ua] <_ J[uo]

e-0

This inequality joined to limes J[ue] ? J[uo] due to the lower semicontinuity of J, proves the continuity in 0. (6.16) is nothing O but (6.10). This completes the proof of the theorem. Then we can now show:

THEOREM 6.4. For p > 0 let is be the solution to (6.5). Then there exists e = e,, such that is = ueN .

(6.20)

PROOF. Let uo be the solution to (6.3) for e = 0. We have seen that uo is a solution to (6.14). Suppose that J[uo] < A.

Then uo E K. and by (6.14) uo satisfies

a(uo,v-uo)=(f,v-uo) VvEKK so that uo is the solution of (6.5) in this case and e$A = 0. So, we assume now that

J[uo] > µ > 0.

(6.21)

By Theorem 6.3 the mapping e - J[ue] is continuous on the interval [0, +oo) and non-increasing. Moreover by (6.16) one has lien J[ue] = 0. e+00

(6.22)

6.1. Some general results

89

Also by Theorem 6.2 and the continuity of e'- J[ue]

uoJ[u`] =

J[u0] > A.

So, clearly fore small enough one would have

J[uE]>p>0.

(6.23)

Then combining (6.22) and (6.23) and applying the intermediate value theorem we can find e = ep such that J[ue] = A.

(6.24)

For this e, (6.3) reads

a(ue,v-ue) > (f,v-ue)+e{J[ue] -J[v]} = (f,v - uE) + E{p - J[v]1

(6.25)

(f, v - uf) dv E KK,. Since by (6.24), u, E K,,, u£ is the solution to (6.5). This completes the proof of 0 the theorem.

REMARK 6.2. Assume J to be differentiable - replacing in (6.3) v by uE + t(v - uE) it turns out that

ta(u£. v - uf) + e{J[ue + t(v - ue)] - J[uf]} > t(f, v - ue). Assuming t > 0. dividing by t and letting t -' 0 we obtain

a(u£,v-u,)+E(J'(u,),v-u,)>(f,v-u,,) VvEH. Changing v into uE ± v we get

a(uf,v) +e(J'(uf),v) _ (f,v) dv E H.

(6.26)

In the case where a is symmetric, the solution to (6.5) is also the minimizer of 1

v'-- 2a(v,v) - (f,v) (cf. Theorem 4.1) and e appears in (6.26) as a Lagrange multiplier.

To end this section we turn also to the behavior of u, when e -+ +oo. THEOREM 6.5. Let u... be the solution to the variational inequality

`dvEK0, a(u,o,v-u.)>(f,v-u,,.) u.EKo={VEHIJ(v)=0};

(6.27)

uE - u,,. in H

(6.28)

then it holds that

when e - +oc.

90

6. A Regularity Theory for Nonlocal Variational Inequalities

PROOF. From (6.9) we know that ue is uniformly bounded independently of e and up to a subsequence we can assume that there exists a u E H such that ue

in H.

u00

(6.29)

From (6.10) and the lower semicontinuity of J we have

0 =e+00 lim J[uf] >

0.

So, u E Ko. Moreover, from (6.3) we have for any v E Ko

a(tz, v - ue) >- (f, v - ue) + eJ[ur] ? (f, v - ue).

(6.30)

This implies a(ut,u!)

a(ue,v) - (f, v - uf)

+oo and using the lower semicontinuity of a(u, u) it

Taking the lim inf when a

turns out that a(uao,v) - (f,v - u00)

'-v

a(ue,ue) ! a(uoo,uoo)

}

In other words u. satisfies

f a(u,,, v - uoo) > (f, v -

u E Ko. Thus this is the unique solution to (6.27). By uniqueness of its limit this is the whole sequence ue that satisfies (6.29). Next we show that the convergence is strong. Indeed from (6.30) we have

a(u, uQ) < a(ua, v) - (f, v - ui), Vv E Ko.

Thus taking v = u

u - ue).

a(ue, ue) < a(uef

Taking the lim sup when e - +oo and using (6.29) we get

lim sup a(ua, ua) < a(u., u.). e-+0o

Joined with

liminfa(ua,uf) > we obtain

e+00

a(u,,, u.).

Then it is easy to show that

lim a(ue - u,,, u, -

e-'+00

0

and the strong convergence follows by (3.6). This completes the proof of the the0 orem.

6.2. Applications to second order variational inequalities

91

REMARK 6.3. The results developed here extend of course when a(u, v) is replaced by (Au, v), A being a suitable nonlinear operator from H into H' - eventually from a Banach space into its dual. For the sake of simplicity we restricted ourselves to the linear case.

6.2. Applications to second order variational inequalities In this section f2 is a bounded open subset of R", n > 1, with boundary F. Let us denote by a, aij functions in fl such that

a,aij E L°°(fl) i,j = 1,...,n, a,jt;l j > aItI2 a.e. x E fl, Vt E R", a > 0 a.e. x E fl,

(6.31)

(6.32) (6.33)

a is a positive constant. In (6.32) we make the summation convention of repeated indices, I I denotes the euclideanrn(orm in Rn. For u, vl E H' (Q) we will set

a(u,v) =

+auv } dx.

(6.34)

In this paragraph we will consider problems in Ho (fl). We will denote by A the operator

A =

a

axj

aij

a l - a ax; J

(6.35)

such that

a(u, v) _ (-Au, v) Vu, v E Ho (fl)

(6.36)

denotes the duality bracket between H-1(fl) and HH(S)). We will assume A, Sl smooth enough in such a way that if u denotes the weak solution to the Dirichlet problem where

-Au = f

in fI, (6.37)

u E Ho (Sl),

then for any f E L"(11), p > 2, one has u E IIUII2,p <-CIflp,-

and the estimate (6.38)

In the above inequalities I - Ip, II . II2,p denote the norms in LP(fl), W -P(11) respectively, C constants depending only on A, fl. (One will find in [46] conditions for this to hold). Let ,3 be a maximal monotone graph of R x R satisfying (see [13]) 0 E Q(0).

(6.39)

92

6. A Regularity Theory for Nonlocal Variational Inequalities

Let j be the positive convex function such that 8j = 0

j(0) = 0,

(6.40)

and define J[u] on Ho (f2) by

+ (u) dx

J[u] = For

l

e(u) E L1(n),

(6.41)

>0 set

K,={vEHo(f2) I J[v]
(6.42)

Then one can show:

THEOREM 6.6. Let f E H-1(0). Then the variational inequality

(-Au,v-u)>(f,v-u) VvEK,,, u E Kµ,

(6.43)

has a unique solution. Moreover, assuming that (6.37) and (6.38) hold, if f E LP (11), p > 2, then u E W 'P(l) and one has an estimate (6.44)

I1UI12,v <- C]fI,.

PROOF. Let us first prove that J is lower semicontinuous for the weak topology of Ho (f)). For that consider a sequence un E Ho (l) such that

u-u

in Ha (S2).

Extracting if necessary a subsequence we can assume that limo J[un] = lim inf J[un]. n

(6.45) (6.46)

Then from this subsequence still denoted u we can extract a subsequence such that

u,,--+ u

in L2 (S2),

(6.47)

un - u

a.e. in St.

(6.48)

Applying Fatou's lemma we get - recall that j is continuous -

J[u] =

< liminf J[u]. n-}oo n- 4-oo Jn j(u) dx = jliminf3(un)

(6.49)

J being lower semicontinuous - one can apply Theorem 6.1 with a(u, v) defined by (6.34) and H = Ha (S2). This shows existence and uniqueness for a solution to (6.3). K. is not empty, since by (6.39), (6.40) J[0] = 0 and 0 E K. This shows existence and uniqueness for a solution to (6.43). Next for c > 0, f E L2(1l) we consider the solution ut to (-Au` + e$(ur) -3 u6 E Ho (f).

f,

(6.50)

6.2. Applications to second order variational inequalities

93

The meaning of the first equation was explained in Chapter 5. Multiplying the and integrating on SZ we get

first equation of (6.50) by v - uE, vr/

(-Au,, v-ue)+EJ f3(uf)(v-%.)dx > (f,v-ue)

(6.51)

denotes the usual scalar product in L2(0). Since /3 = Oj we derive

where

- J nj(u)dx >- Jst 6(u)(v - u)dx and (6.51) leads to

(-Ant. v - u£) + e{J[v] -

(f, v - ue) Vv E HH(Sl), (6.52)

u£ E Ho (S2).

Thus u, is also a solution to (6.52). Applying then Theorem 6.4 for some e = Eµ we have u = uf, but due to the regularity theory developed in [14] and Theorem 5.7 for f E LP(Q) we have u = uE E W2,p(Sl) and an estimate of the type (3.15). This completes the proof. 0

REMARK 6.4. One can apply Theorem 6.6 with for instance

j(u) = IuI° q > 1, i.e. for convex sets

K,, = { v E HH (n) 111

f IvlQ dx < µ}.

Jn

It is possible to consider also convex sets of the type

K,,=IV EHo(1)Ij)(xv)dx

S

where for a.e. x. j (x, u) is a convex function such that

j(x, 0) = 0,

8j = /3(x, ) a.e. x E ft

(6.53)

/3(x, ) being for a.e. x a maximal monotone graph. Indeed the W2'p(II)-regularity theory of [14] can be extended in some cases (see theorem 5.7 and [27], [49]). For example, assuming that t/i E L°°(fl)

O
K,, = {v E Ho(Sl) I belongs to

jb(x)i(v)dx < µ}

provided j is a convex function such that j(0) = 0,

Oj = 0

6. A Regularity Theory for Nonlocal Variational inequalities

94

where,3 is a maximal monotone graph. The proof is carried out as in Theorem 3.1 using Theorem 5.7.

The above technique of regularity can be carried out for various other problems. The reader is referred to [27], [49].

6.3. Exercises 1. Extend the result of Section 6.1 to the case of a nonlinear opterator A (see Remark 6.3). 2. Show the results claimed in Remark 6.4. 3. Let bi3, i, j = 1, ... , n be smooth functions such that bij E L°°(i2),

bi}Cij>0 VEER". Set

r

J[u]= J bi; n

8,

Ou

VU E HO' (n),

1

Kµ= {vEHo(fl)I J[v]<µ). For A defined by (6.35) show that for f E L"(12), p > 2 the variational inequality

f µEK,, (-Au, v-u)

f v-udr VvE Kµ

has a unique solution in N""(S ).

Chapter 7

I.iqueness and Nonuniqueness Issues T.I. Uniqueness result for local nonlinear problems i this section we address the question of uniqueness for nonlinear problems. In Ct we will consider the question of monotonicity as well since the two questions ate related. Assume that ft is a smooth open set of R" with boundary r. We assume that 'P is divided in two parts ro, rl and - assuming that the measure area of ro is jxwitive - we set I

V={vEH1(fl)Iv=0onro}.

(7.1)

In for f E V' we consider u the weak solution to the problem

{ _(A(zuVu))

=f

in fl, (7.2)

u E V.

In other words we assume that u is a solution to

r

1 A; (x, u, Vu)

8v

dx = (f, v)

Vv E V, (7.3)

U E V.

We have seen in the previous chapters examples of such problems. Note that (see Chapter 3) - if we assume our data smooth then u is a weak solution to the probAern

-

(Ai(x, u, Vu)) = f in n,

u=0

on ro,

A{(x,u, Vu)vf = 0

on r1,

ovhere v = (vl, ... , v") denotes the outward unit normal to r. We could include the ease where 1ra1 = 0, but we would like to confine ourselves to a simple situation. We refer the interested reader to (SO] for variants or generalizations.

7. Uniqueness and Nonuniqueness Issues

96

Let us assume that the vector field A satisfies the following assumptions: there

exists a constant a > 0 suds that n

C `` -ti') > aISt - Ai (x, u, t')) (Si E(Ai (x, u, S) CC

- t,12

i=1

b'uER, dl:,C ER", a.e.xEQ.

(7.5)

Moreover there exists a non-decreasing function w (w is the modulus of continuity

of A in the variable u), a function C(x) E L2(f3), C(x) > 0, a constant C such that I At(x, u, C) - Ai(x, v, )I <- (C(x) + CICUw(lu - vl )

Vi, Vf ER",bu,vER, a.e.xEi2 with

L

ds

w(s)

= + 00.

(7.6)

(7 . 7)

(We can always take in (7.6) w larger so that w > 0 on (0, +oo) and W remain bounded on each subinterval (e, +oo) ). In these conditions we can show:

THEOREM 7.1. Assume that (7.5)-(7.7) hold. Let fi E V', i = 1, 2 and let ui, i = 1, 2 be weak solutions of the problem (7.2) corresponding to fi. Then if

h < f2

(7.8)

u1 < u2.

(7.9)

in the sense of V' we have

In particular the problem (7.2) has at most one solution. REMARK 7.1. The inequality f1 < f2 in the sense of V' means only that

(fliv)<(f2iv) VvEV, v>0.

(7.10)

In the case where Ai is Lipschitz continuous in u - with eventually a Lipschitz constant depending on x and ty - then one can take w(t) = t (7.11) so that (7.7) holds in this case. PROOF OF THEOREM 7.1. By subtraction of the equations satisfied by ui we have

f{Ai(xuiVui) - Ai (x,u2i Vu2)} Let us set

fx F`(x) =

-

(7.12)

for x > e (7.13)

w2(s)

0

dx = 0 Vv E V.

else.

7.1. Uniqueness result for local nonlinear problems

97

clearly makes sense if w > 0 on (0, +oo). Moreover - see below (7.7) - if Obe assumes 1 /w2 bounded on each interval [e, +oo) then dearly Fs is a Lipschitz xuous function such that 1

Fax= w2(x) 0

for x > E, (7.14)

else.

Then it is easy to see that the function v = Fe(ul - u2)

belongs to V (see Chapter 2). Going back to (7.12) we obtain

j{Ai(xui,Vui)_Ai(x,u2,Vu2)}_Fe(ui -u2)dx=0 t

which can be written as

in

jAj(x,uj,Vuj) -Ai(x,ul,Vu2)}58-xiFc(u1 -u2)dx

= j{Ai(x, u2, Vu2) - Ai (x, u1, Vu2)}= F, E(ul - u2) dx. Using (7.5), (7.6), (7.14) we derive

j{Ai(x1 u, Vui) - Ai(x,ul,Vu2)}BxiFQ(ul -u2)dx

- k,-u,>a} >a

1 10

- u2)

{AE (x, u1, Vu1) - Ai(x,ul, Vu2)}8xi (ul - U2)

IV(u1 - u2)I2 dx

J{W (ul - u2)

and

f{Ai(xu2

Vu2) - A.(x,u1, Vu2)}

{ui-u2Je}

_I

w(Iu1 - U21)411j(ul1-

F.(u1 - u2)dx U21

V(u1 - u2)I(C(x) + CIVu2I) dx

IV(ul - u2)1(C(x) + CIVu2(x)I) dx

- J{ul-U2>_4) w(ul - u2)

7. Uniqueness and Nonuniqueness Issues

98

where {ul - u2 > E} denotes the set where ul - u2 is greater than E. Combining these two inequalities and applying the Cauchy-Schwarz inequality will lead to

a

IV(ul - u2)I2 J{ u1-u2>e} (J2(ul - U2) r

dx

I.

1/2

jV(u1-u2)I2J1/2

/'

<

L[{ui-u2>e} (C(x)+CIVU2(x)I)2dx]

J{u1-u2>e} w2(u1 - u2)

Hence it follows that

2f Recall that

r

IV((u1 - u2)I2 dx < (C(x) + CIVU2I)2 dti. JS1 ul-uz>e} w 2(UI - u2) IVu2I E L2(fl). Then, let us set

Ge(x) _

J 0

s

(7.15)

for x > E,

W(s) else.

Then, (7.15) reads now

Jn

IVGe(u1 -u2)I2dx < C'

where C' is a constant independent of e. Using the Poincard Inequality that is valid for functions in V we get G,2 (ul - u2) dx < C".

Due to (7.7), letting c -' 0 we will get a contradiction if u1-u2 is positive on a set of positive measure. We have thus established (7.9) and the proof of the theorem is complete. 0 APPLICATION. Suppose that we consider the question of uniqueness of a weak

solution u to

{_(aii(xu+u)) = f 8xi

in f2,

(7.16)

u E V.

Then we have here Ai (x, u, t) = aii (x, u)fi + (3 (u). In this case the condition (7.5) reads

ai (x, u)t,EE > alt l' Vu E R, d£ E P", a.e. x E fl,

(7.17)

(7.18)

i.e. we have an ellipticity condition on the operator. Now clearly (7.6), (7.7) holds, provided for instance the functions

u"Qi(u),

u,-- aij(x,u)

7.2. Nonuniquenees issues

99

ore Lipschitz continuous - i.e.

la(u) - Q+(v)I, Iaci(x, u) - ait(x, v)I < Clu - vl far all u, v E R and a.e. X E 11. For instance the problem

J-' (inf2 uEV,

11nss a unique solution provided u

a(x,u) is Lipschitz continuous.

;17.2. Nonuniqueness issues In the preceding section we developed a very powerful technique to prove uniquenws. Unfortunately, dropping our structure assumptions, it is not always possible ,to prove uniqueness. It even may fail in very simple situations as we will see below. Prom a natural point of view this is reassuring: a system reacting in a complex manner can have several stable states! More striking, we will see later that, even If a problem has a unique solution, its discrete version might have several. Let us start by a simple non-uniqueness result in one dimension.

Y.2.1. A simple non-uniqueness result. We follow here some ideas of [6]. So, let Us consider on [0, 1] two functions u and v that we will assume to be smooth. Moreover we assume that

v'(x)>0

bxE[0,1]

u(x) > v(x)

Vx E (0, z) u(J) = v(}) > 0,

U(O) = v(0) = 0,

u'(0) = v'(0),

u1(2.) = v'(z).

We suppose also to have extended u and v by symmetry on [0,1] - i.e. we have set, for every x E [0, 1],

ux +

=U(1 - x),

v(2 -x). (+) =v('

The picture of v and u are represented in Figure 7.1. Then in the (x, z)-plane we define a function a(x, z) in the following way:

a(x, z) =

1

for x E 10, 1], z < v(x),

v'(x)/u'(x)

for x E [0, 1], z > u(x),

t + (1 - a fix)

for x E 10, 11, z = tu(x) + (1 - t)v(x).

It is clear that for x 36 0,1,1 the function z' - a(x, z)

100

7. Uniqueness and Nonuniquenese Issues

FtouaE 7.1 is Lipschitz continuous - it is indeed constant or linear. Now, it is also a continuous function in (x, z) so that it is a Caratheodory function. Now, clearly, u, v are two functions in Ho (Sl) - recall that we assumed them smooth - and one has, due to the definition of a,

a(x, u)u' = a(x, v)v'. Also, a(x,v)v' E L'(0, 1) so that

f = -(a(x, v)v')' E H-'(0,1) and u, v are both solutions to the nonlinear Dirichlet problem

(a(x, u)u')' = f in (0,1), u E Ho(0,1). Working on the behaviour of u and v around 0, 2,1 one can even arrange things so that for every x u }-+ a(x, u)

is Holder continuous with exponent ry - arbitrary between 0,1. We refer the reader to [6] for details.

7.2.2. Nonlocal problems. We have seen in Section 7.1 that, provided the dependence in u is sufficiently smooth, local problems 8` I 8x \a(x, u) 8xi 1 = f in St u E Ha (ft),

8

have a unique solution. This local dependence in u - i.e. a(x, u) reads a(x, u(x)) and depends only on the value of u at the point x - is here local. In the case where we allow dependence of neighbouring points, uniqueness can be immediately lost. In fact the uniqueness can be lost even if a is smooth.

7.2. Nonuniqueness issues

101

Let us for instance consider a weak solution u of the following problem:

-a\\n `J udx'itu+u=f

l

-

infl, (7.19)

= g on 1'.

We refer the reader to Chapter 1 for some physical motivation regarding this problem. It turns out that its resolution - as in general for such nonlocal problems - reduces to the resolution of a nonlinear problem in R (see [34J). We suppose here that g is some smooth function such that

jg(x)do(x) 96 0. Then integrating the first equation on 0 we obtain

-a

(J

u dx)

Du dx + J u dx = J f dx.

J

Then, by the Green formula,

f Du dx = u

ou

Jr

da(x) = - r g dor(x).

r

Thus one obtains a

(J u dx} r g da(x) + J u dx =

f

f dx,

i.e.

i=1 udx n

is a solution to

a(µ)

f

r

g da(x) + µ=

jfdx.

In fact we have

THEOREM 7.2. The problem (7.19) has as many solutions as the equation in

pEll a(µ)

J

g da(x) +,u =

jfdx.

PROOF. Indeed we already saw that if (7.19) has a solution then

l'=J udx n

is a solution to (7.21).

(7.21)

7. Uniqueness and Nonuniqueness Issues

102

Conversely, let u be a solution to (7.21). Then there exists a unique weak solution u to

f-a(p)tiu+u = f inn,

-5;;=g onl. Integrating the first equation above on S2 and using the Green formula (7.20) it

turns out that

r

r

a(u) J

9 da'(x) +

Jin

r

r

r

u dx = J f dx = a(u) f g do(x) + An

lt follows that It

u=J udx in

and u is a solution to (7.19). This concludes the proof of the theorem.

Note that in the above result we did not use the continuity of a. We refer the reader to Chapter 13 for some other nonlocal problems of this type. Let us return to the question of uniqueness. (7.21) reads also a(u) =

(j9&T(x))'{_P+jfdx},

and a solution p of this equation is determined by the intersection of the graph of a with the straight line

u~

(j&(x)){-+fisr}.

(7.22)

Clearly one can choose a - even very smooth - such that this equation has as many solutions as one wishes. Letting a equal this straight line on some interval will even produce a continuum of solutions. If one supposes now

a continuous,

0< m < a(C) < M V C E 1R for some positive constants m, M, then the problem always has a solution (indeed assuming for instance fr g do > 0 for u large positive, the graph of a will be above the straight line (7.22) and below for p negative large enough). The problem could have no solution at all for a discontinuous a or an a which is unbounded, as one can see easily considering again the graphs of the two curves.

7.3. Exercises 1. In theorem 7.1 the assumptions are adapted to L2(fl). Extend this to the L1(Sl)-case assuming for instance (7.5) replaced by n

Ai(x,u,e))(C - C) ? alt - {'Jp. 2. Extend Theorem 7.1 in the case of variational inequalities (see [301).

7.3. Exercises

103

3. Let a(x, u) be a Caratheodory function satisfying (5.4) and such that for some constant A it holds that: (a(x, u) - a(x, v) < Abu - vi Vu, v E R. For V E Hl(fl), W:5 0 on F, f E H-1(f)) show that the problem

fu K = {v E

(x) a.e. XE fl},

I v(x)

//j a(x, u)Du V(v - u) dx > (f v - u) V v E K has a unique solution (see Exercise 3 in Chapter 5). 4. Study the uniqueness of the solution of the problem

(a(x, u)u')' + u = f in (0,1), lu(0) = u(1) = 0. 5. Study the problem

/

-a / J dx}

in ft,

n

On

( fcn

where a, b are two ' real functions.

)

I

Chapter 8

Finite Element Methods for Elliptic Problems 8.1. An abstract setting We consider the framework introduced Chapter 3 for the Lax-Milgram Theorem. In other words let H denote a real Hilbert space and a(u, v) a bilinear form on

H satisfying (3.5), (3.6). For f E H', the dual space of H, we would like to approximate the solution u of the problem

u E H,

ja(u,v)=(f,v) VvEH. For that - in numerical analysis - one replaces the problem (8.1), which is posed in an infinite dimensional space, by a problem on a finite dimensional space. Thus let Vh be a finite dimensional subspace of H. Vh, being closed is a Hilbert space and, by applying the Lax-Milgram Theorem to Vh, we conclude the existence of a unique solution uh to:

JuhEVh ja(uh,v) = (f,v) Vv E Vh. h - the mesh size in the application - is a parameter that is supposed to tend to 0. The question is then - under what condition do we have Bin uh = u?

(8.3)

h-0

If this happens it is clear that uh is a "sequence" of elements of Vh converging toward u. Now, any u E H is the solution of a problem of type (8.1) - since clearly a(u,v) = a(u,v)

V v E H

and v '-+ a(u, v) is a continuous linear form on H. So, in order to have (8.3) for any solution u of a problem of type (8.1), we need to assume that the spaces Vh are such that Vv E H, 3 vh E Vh such that lim vh = v.

h-0

Under this assumption we can show:

(8.4)

8. Finite Element Methods for Elliptic Problems

106

THEOREM 8.1. Let u (reap. Uh) be the solution to (8.1) (reap. (8.2)). Assuming that (8.4) holds, we have hlimm Uh = U,

the convergence taking place in H. PROOF. Combining (8.1), (8.2) we have

a(u - uh, v) = 0

Vv E Vh.

Hence

a(u - uh,v - uh) = 0

VV E Vh.

This can be written

a(u-uh,v-u+u-uh) =0 hence

Vv E Vh,

a(u-uh,u-uh)=a(u-uh,u-v)

VvEVh.

Using (3.5), (3.6) it follows that

alu - Uhl'
VV E Vh

or

lu - uhl < flu-vl

VVE Vh.

Taking v = Phu to be the orthogonal projection of u on Vh, it is clear that (8.4) implies

Flu-Phul = 0

and the result follows. REMARK 8.1. We have shown that

Iu - uhl<

Q VEV4

lu - vl

which is known as the C6a lemma.

8.2. Some simple finite elements 8.2.1. The one dimensional can. Let f2 = (A, B) be a bounded interval. Let us then consider

xo=A<x1 <x2

<xn=B

(8.6)

a subdivision of (A, B). In this context the number h = Max (xi}1 - xi) i=O,...,n-1

(8.7)

is the mesh size of the subdivision. Then set Vh = { v : (A, B) - R continuous, v(A) = v(B) = 0, v is adine

on (xi,xi+1) Vi = 0,...,n - 11.

(8.8)

8.2. Some simple finite elements

107

3t is easy to verify that Vh is a subspace of Ho (0) - see for instance Proposition 8.1 below. Then for a a bounded measurable function such that

0
a(x) < C a.e. x E Il,

(8.9)

we set

a(u,v) =

J

a(x)u' v' dx d u, v E H' (St).

(8.10)

We can then state THEOREM 8.2. Under the above as8umptions, for f E H-1(f), there exists a unique u (reap. uh) solution to

u E H'(tt), (ReSp.

a(u,v) _ (f,v) dv E H'(f1) a(uh, v) _ (f, v) dv E Vh).

Uh E Vh,

(8.11) (8.12)

PROOF. Due to (8.9) this is just a trivial application of the Lax-Milgram

0

theorem. REMARK 8.2. Of course u is the weak solution to 11,

(a(x)u,)' = f in

(8.13)

lu(A)=u(B)=0. We will see later on that the approximation uh of u converges toward u when the mesh size h goes to 0. 0.333. The two dimensional case. Consider SZ a polygonal domain of RI (see Figure 8.1). A family of triangulations (Th)h is a family of subdivisions of fl into triangles. For such a subdivision we assume that N 1) = U Ti,

Ti E Th Vi=1,...,N,

(8.14)

i=1

i.e. the triangles cover n - note that we assumed the triangles to be dosed. For

ayTEiwe get

hT = diam(T)

(8.15)

where diem denotes the diameter of a set - hT is simply the length of the largest side of T. h=TMaxhT

(8.16)

Is the mesh size of the triangulation. Besides (8.14) we will assume that for every pair of triangles TI, T2 we have T1 n T2 = 8,

one point or one entire side,

i.e. we will not allow the situation of Figure 8.2.

(8.17)

8. Finite Element Methods for Elliptic Problems

108

0 FIGURE 8.1

FIGURE 8.2

Then we set

Vh={v:S2--+RIviscontinuous, v=0on1',vIKEPIVKETi,}

(8.18)

where VIK denotes the restriction of v to K, P1 the set of polynomials of degree smaller or equal to 1 - i.e. v is a0'ine on each triangle of the triangulation. We will denote by ni

i = 1,..., M

(8.19)

the inside nodes of the triangulation. Then we have PROPOSITION 8.1. Vh is a finite dimensional subspace of Ho (S2) and dim Vh = M.

(8.20)

PROOF. First let us show that v E Vh is uniquely determined by the knowledge

of v(ni) Vi = 1, ... , M. Indeed consider a triangle T with, for instance, vertices

8.2. Some simple finite elements

109

denoted

n1 = (n11,n12), On T, v is given by

n3 = (n31,n32)

n2 = (n21,n22),

v(x1 i x2) = ax1 + bx2 + c.

If we assume that v takes the values v(ni) at the vertices ni, i = 1, 2, 3, we must have

Ian11 + bn12 + c = v(nl ), an2l + bn22 + c = V(n2), an31 + bn32 + c = v(n3).

Clearly, a, b, c are uniquely determined provided that the determinant n11

n12

1

n21

n22

1

n31

n32

1

e

n11

n12

1

n21 - nil

n22 - n12

0

n31 - nil

n32 - n12

0

00,

that is to say provided n21

fl 22

ri31- -nll n11 n32-n12 - n121

960,

(8.21)

i.e. provided the vectors

n3 - nl are linearly independent. If one does not allow flat triangles, this is of course the n2 - n1,

case. This shows that v - vanishing on the boundary of f) - is uniquely determined by its values v(n,) at the nodes of the triangulation. In particular there is a unique Ai E Vh such that

)(n)=o,,

(8.22)

Then these functions are a basis of Vh. Indeed they are linearly independent since

M E v;Ai(nj) = vj = 0

`dj = 1,..., M.

i=1

Moreover, for any v E Vh the function M

.L v(n`)Ai i=1 is a function of Vh which coincides with v at the nodes ni. It follows from what we have seen above that it holds everywhere: M

v(ni)Ai,

v= +=1

8. Finite Element Methods for Elliptic Problems

110

i.e. any function of Vh can be expressed in a unique way as a linear combination of A,'s. This shows that the Ai's are a basis of Vh and that (8.20) holds. Next, let us show that (8.23)

Vh C Ho (52).

For this purpose let us note first that inside each triangle T E rh a function v E Vh is C'. We denote by Sv

i = 1,2

5i

T

its derivative in the usual sense. Then we set

v1

vi = TErk

Ox, T

XT

where XT is the characteristic function of T. Clearly since vi is constant on each triangle of T we have v, v1, v2 E 9 (f2).

(8.24)

We will also have (8.23) provided one can show that - in the distributional sense

- it holds that: i

=vi i=1,2.

(8.25)

To show this equality, let (p E D((l). We have for i = 1, 2: Ox`

TErk

TErk

-1 v. Oxi dx T

-JT { Oxi(v'W)-8xi.V Idx JT 0xs (v '

TErk

v)

+

L JTi

dZ.

TErk

Using the Greenr formula it turns out

-Jv

&e =TErk 1: - JaTT v

&,i

&7(x) + in vi




(8.26)

where v = (vi, P2) denotes the outward unit normal to OT the boundary of T. When one integrates on a side S of T included in r,

Js

vi du(x)=0

since


f v9

1 T1

v

da(x)

+ fT. v

v do(x)

0,

8.2. Some simple finite elements

111

since v is continuous and v; on Tl is equal to -v, on T2. From these considerations it follows from (8.26) that

-fv

a(pdx =

f

v; cp dx

`d V E D(fl).

n ex, a This is precisely (8.25) and this completes the proof of Proposition 8.1.

0

For a a nonnegative constant let us set

a(u, v) = j Vu Vv + auv dx.

(8.27)

Then we have:

THEOREM 8.3. Under the above assumptions, for f E H-'(11), there exists a unique solution u - (resp. uh) - to

u E Ho(ft), (resp.

uh E Vh,

a(u,v) = (f,v) dv E HH(fl), a(uh,v) = (f,v) 'v E Vh).

(8.28)

(8.29)

0

PROOF. This is just a trivial application of the Lax-Milgram theorem. REMARK 8.3. Of course u is the weak solution to the Dirichlet problem

-Du + au = f in fl, lu = 0 on I'.

(8.30)

We will see that uh is an approximation of u. To approximate the Neumann problem one can introduce

Vh={v:fl-*RIvis continuous, VIKEPI VKErh}.

(8.31)

Then with the same proof as above we can show: PROPOSITION 8.2. Vh is a finite dimensional subspace of H'(f)) and dim Vh = Ml

(8.32)

where Ml denotes the number of nodes of rh (in fl and on t). Moreover,

THEOREM 8.4. Under the assumptions of Theorem 8.3 and for a > 0, f E L2(S1), there exists a unique solution u (neap. uh) to

u E H' (fl), (resp.

uh E V h ,

a(u, v) = ji v dx dv E Hl), a(uh, v) =

JQ

f v d x V v E Vh).

(8.33) (8.34)

8. Finite Element Methods for Elliptic Problems

112

REMARK 8.4. u is the weak solution to the Neumann problem

-Au + au = f in it, 8-n=0

(8.35)

on r.

We will see that uh is an approximation of u.

tonal cue.

8.2.3. The n-di

be n+1 vectors of Rn such

DEFINITION 8.1 (ti-simplex). Let that

P2 - P1,P3 -PI'- - -'P"+I -PI

are linearly independent. One calls the closed convex hull of pi an n-simplex of vertices P1, ... , Pn+l. EXAMPLES. A 1-simplex is simply a segment, a 2-simplex is a triangle, a 3simplex is a tetrahedron.

Let n be a bounded domain of Rn whose boundary is a union of pieces of hyperplanes.

DEFINITION 8.2 (T iangulation). A triangulation r is a collection of simplices

K such that

S2 = U K,

(8.36)

KEr

K, r1 K2 = 8 or a p-simplex of vertices in K1 fl K2

(8.37)

(see Figure 8.2).

For KErwe set hK = diam K

(8.38)

h = Sup hx,

(8.39)

and

KEr

the mesh size of the triangulation. Then we set

Vh={v:fl -'RIv is continuous, v=0on r,

VIKEP1'KEr=rh}

(8 . 40)

where P, denotes the set of polynomials of degree 1 - i.e. the set of affine functions. Denote by

M the vertices of the simplices of rh that are inside 12. Then we have PRoPOSITiON 8.3. Vh is a finite dimensional subspace of H0' (it) and dim Vh = M.

(8.41)

8.3. Interpolation error

113

PROOF. It is enough to follow the proof of Proposition 8.1 with the obvious 0 changes due to the higher dimension.

For a;j constants satisfying (3.12) and a > 0, set

a(u, v) = I a, u

8V + auv dx.

(8.42)

Then we have

THEOREM 8.5. Under the above assumptions, for f E H-1(0), there exists a unique solution u (resp. uh) to

a(u, v) _ (f, v) Vv E Ho (S2), a(uh, V) _ (f, v) Vv E Vh).

u E Ho (1), (,esp.

uh E Vh,

(8.43) (8.44)

0

PROOF. Apply the Lax-Milgram theorem. REMARK 8.5. u is the weak solution to the Dirichlet problem

a (a`j ou) + au = f inn,

(8.45)

u=0 our and, as we will see, uh is an approximation of u. 8.3. Interpolation error

Let K be a n-simplex. Let pk be the diameter of the largest euclidean ball that one can fit into K. Let v be a function of class C2 on K. Denote by the vertices of K and by Ai the affine functions such that Aj(pi) = bti

Vi=1,...,n+1.

(8.46)

If n+1

1_

1: v(pk)Ak

(8.47)

k=1

denotes the interpolant of v on the pi's, then we have that THEOREM 8.6. It holds for every x E K, i = 1, ... , n that 8v Bxi

(x) -

8v (n + 1)n2 hK Sup I D"v(x),. (x) < 8xi 2 pK IQI=2 I

xEK

(8.48)

8. Finite Element Methods for Elliptic Problems

114

PROOF. Assume that

Vi=1,...,n+1.

pc = (psi,...,P{n)

By the Taylor expansion we have for k = 1, ... , n + 1, n

8x! (x) (Pkl - XI)

V (PA;) = V (X) + !=1

(8.49) 82v

+

2

5x7

8x4

i.q=1

=)(pk! - x!)(pkq - xq)

is a point belonging to K. Setting

where

Rk(x) =

1

82V

2

8x18x4

1,q=1

(tz)(Pkl - xl)(pkq - xq),

we have clearly

IRk(x)I <

2

E

x1I Ipkq - x.I

Iq 1 8xiOxq

(8.50)

< n2hK Sup I D°v(x)I I°I=2

2

zEK

and this for every k = 1, ... , n + 1. From (8.47) it follows that n+l

(x) = E V(Pk) 8xk(x). k=1

Thus, multiplying the equation (8.49) by V(z) and summing up, we obtain: af)

n+1 (x) = E V(Pk)

`(x)

k=1

= v(x)

E k=1

814 (x) +

8xi

E

E' (x)(pk! -

k=1 1=1 OX,

n+1

+ E Rk(x) 8k (x). k=1

xl)8-(x) 8x{

(8.51)

115

8.3. Interpolation error

Ow can check easily that the Ak's are solutions of the system n+1 1=1,...,n,

E Ak(x)Pkt = x! k=1

n+1

E Ak(x) = 1. k=1

taking the derivative in i of these equations, we get n+ 1 8A

1: axk (x)Pkt = Sit

1 = 1,...,n,

f

k=1

n+18a +1

E 8xk (x) = Q. k=1

wing back to (8.51) we obtain: n

n+l

n+1 L9A

(x) _ !=1 n

_

(x) ' E (Plkl - xl) 8x{ {x) + E Rk(x) x (x) k=1 k=1 /

n+1 (8.52)

(x)aa + E Rk(x)exi(x) k=1

l=1

n+1

(Z) +, Rk(x) Oxs (Z). k=1

Next, consider the orthogonal projection qk of pk on the hyperplane of equation Ak(x) = 0.

(8.53)

This gives us

1 = Ak(Pk) - Ak(gk) =

qk

(8.54)

(recall that Ak is a polynomial of degree 1). Now, both vectors VA*,

Pk - qk

are orthogonal to the hyperplane of equation (8.53) and are thus colinear. Then (.54) becomes

1=VAk'Pk-qk=IVAaIIPA: -gkI?QKIVAk1 Thus we obtain for every i = I,... , n 8Ak

c IVAk 1 < 1 .

-

8K

8. Finite Element Methods for Elliptic Problems

116

Going back to (8.52), (8.50) we get SfJ i

n+1

< F IRk(x)II8ak (x)l < 8xi

8v oxi

(x) - -(x)I

(n + 1) 2

n2 hK sup ID°v(x)I ek

xEK

which gives (8.48). DEFINITION 8.3. Let (rh) be a family of triangulation of Sl - see definition 8.2.

We say that the family is regular - or quasi-uniform - iff there exists a constant v such that hK

< v V K E Th

(8.55)

QK

(i.e. the flatness of our simplices is bounded from below).

Then one can show

THEOREM 8.7. Let (Th) be a regular family of triangulations of Cl. Let V E C2(i) and vh the interpolant of v on Th - i.e. M

vh = Fv(ni)a;.

(8.56)

i=1

Then for some constant C = C(n, Cl, v, v) it holds that

Ilvv - Vvhll2 < Ch.

(8.57)

PROOF. By (8.48) for each simplex K E rh it holds that

8 (x) - lei W1 < chK < Ch V x E K. Thus,

r

dz. Dv - av(x)I2 dx < ncah2 JK JK Making the summation in K E rh we get I

IIvv - Ovll2 < nc2h2In, where IS21 denotes the measure of Cl and the result follows.

0

8.4. Convergence results Let (Th) be a regular family of triangulation of Cl - note that in dimension 1 any family of triangulation is regular. Then we have THEOREM 8.8. Let u (neap. uh) be a solution to (8.11), (8.12) or (8.43), (8.44)

then it holds that lli-0m uh = u h

in Ho (i2).

(8.58)

8.4. Convergence results

117

PROOF. Let v E Ho (1). Denote by Ph the orthogonal projection from Ho (i2) onto Vh - recall that Ho (11) is a Hilbert space. Then we claim vh = Phv

satisfies (8.4). Indeed lets > 0 be given. Due to the definition of HH(fl) there exists w E D(f2) such that Ilv - w111,2

3.

Then we have IIv - Phvll 1,2 = IIv - w + w - PhW + PhW - PhvIl1,2 liv - w111,2 + 11w - PhwII1,2 + IIPhw - PhvIl1,2-

Let us denote by 1uh the interpolant of w on rh. Due to the definition of Ph and from the fact that it is a contraction, we derive from above IIv - PhvII1,2 <- IIv - w1I1,2 + 11w - whll l,2 + IIw - v1I1,2 Zr

<

+ llw - whlll,2.

By theorem 8.7 we have

IIw-whiI1,2
<3

for h small enough, i.e. IIv-PhvII1.2<<-e. Thus (8.4) follows and also (8.58) - by Theorem 8.1.

0

It is possible to estimate 1Iu - uhI11,2

in terms of the mesh size. For a complete analysis of this problem, which involves the regularity of the solution u, we refer the reader to specialized books like [36], [41], [62]. We will restrict ourselves here to the one dimensional case. So, as in

Paragraph 8.2.1 let us consider an interval it = (A, B) - bounded. Let us then introduce a subdivision

xo=A<x1 <...<xn-1 <xn=B

(8.59)

of (A, B). Set

h=

Max

==0....,n-1

(xi+1 - xi)

(8.60)

and

Vh = { v : (A, B) -r R, continuous,

v' = cst = v, on (xi,x;+1)Vi = 0,...,n- 1 }. Let us first show:

(8.61)

8. Finite Element Methods for Elliptic Problems

118

PROPOSITION 8.4. Let U E H2 (a, fl). Then we have

r0 f

u(0) -- a

Eu'(-1dz<

$

(p - a)2 2

J

IQ

J

Iu"(x)I2dx.

(8.62)

PROOF. We know that H2 (a,,6) C C1Ia, flJ so that we have for some C E (a, Q)

UP) - u(a)

#-a

Then we can write for every x E (a, 0)

U1(X) - u'(0 = J x u- (a) da. By the Cauchy-Schwas inequality it turns out that

(JX ds l 1J2Ix

Iu'(x) - u'(O I <-

l/

- I1/2.

Squaring everything we obtain

[u'(x) - u'(f)J2 < Ix - fl 1 lu'(s)12 da. a

Integrating in x leads to 0

J

C

[u (x) -

x) dx + fc (x -

dx < L

f `xJ J

Iu"(s)I2 ds

which leads after further integration to

- u'(0J2 dx < a j[u'(x)

(e 2a)2 +

l 2zll! being achieved for

The maximum of the function the result follows.

a or O

For a continuous function u on (A, B) we define by u the interpolant of u -i.e. u is the unique function of Vh such that

u(xi) = u(xi)

Vi = 0,..., n.

(8.63)

Then we can show:

THEOREM 8.9. Let U E H2(f) - it holds that

lu - 02 < where 1 12 denotes the usual Le(ft)-norm.

h

-72=Iu"I2

(8.64)

8.4. Convergence results

119

PROOF. From (8.62) we deduce for every i =JJ 0,..., n - 1 x,}1

J zI

h2

(u'(x) - u'(x))2 dx <

x{+I

Jxi

2

u'(x)2 dx.

0

Summing in i and taking the square root leads to (8.64). If one tries to estimate the L2 or Loo-norm one gets THEOREM 8.10. Let u E H1(0), then

1U-u12 <

h flu

ti12 (8.65)

lu - ul,o <- hl'/21u' V - u'12. loo denotes the usual Lo°-norm on (A, B).

PROOF. For x E (xi, xi+1) it holds that u(x) - 71(x) =

J

x(u - 4)'(s) ds.

Hence x [u' - u ]2(s) ds](x 1/2 - xs1/2.

Iu(x)

- u(x)I <

Vi

(8.66)

Squaring we obtain x,+'

(u' - u')2(8) da x - xi.

(u(x) - tf(x))2 < Hence, integrating on (xi, xi+1)

j

J=

(u(x) - u(x))2 dx < (X,+1 -

X,)2

(u'

J=

2

h2 rx:+ 2 f

I

(u'

- u')2 ds

u')2 ds.

x;

This gives the first inequality of (8.65). Next, going back to (8.66) we obtain lu(x) - u(x)12 < (x - xi)

j(u x -

u)'($)2 ds

z'

rz.+i

-
(u-u)'(s)2ds dxE(xi,xi+1)

x{

This leads to the second inequality of (8.65).

Let us then consider u and uh the solutions to (8.11), (8.12). We can state

8. Finite Element Methods for Elliptic Problems

120

THEOREM 8.11. Let us assume f E L2(fl), a E C'(0). Then it holds that

Iu' - 02:5 ChIf12, Iu - 612:5 Ch21f 12,

(8.67)

Iu - u1.:5 Ch312If 12 where C = C(Sl, a).

PROOF. From (8.13) it holds that

-a(x)u" - a (x)u' = f so that u E H2(1l) and Iu"12 <- 1 {Ia'IoIu'I2+ If12}.

Taking v = u in (8.11) we derive that aIu'I2 <- If 121uI2 5 CIf I21u 12

by the Poincare Inequality. Thus

Iu'I2
0

We can now estimate the error between u and uh in terms of the mesh size. We have

THEOREM 8.12. Assume that f E L2(f), a E C' (N). Then it holds that

Iu'-uh12 5ChIf12, Iu - uh12 < Ch2I f I2,

(8.68)

Iu - Uhl. < Ch3121f 12 where C = C(1Z, a).

PROOF. Recalling (8.5), (8.10) one derives easily

1u'-v'12 VvE Vh.

Iu'-uh12 <-

a

Taking v = u in the above inequality we get

C Iu'-uh12 <- aIu'-u'12. Then the first inequality of (8.68) follows from the first inequality in (8.67). Next let us prove that:

Iu' - uh12 5 Iaal Iu - u12.

(8.69)

8.4. Convergence results

121

From (8.11), (8.12) we derive (see also (8.10)):

a(u - uh, v) = 0 Vv E Vh. Taking v = u - uh it results that a(u - uh, u - uh) = 0, hence

a(u-u+u-uh,u-uh)=0.

This we obtain

aIu'-u%I2
x,+i

a(u - u)'(u - uh)' dx.

(8.70)

ainoe

(ft - uh)" = 0 On (xi, xi+1) it holds that

a(u - u)'(u - uh)' = (a(u - u) (6 - uh)')' - a'(u - u)(u - uh)'. Thns +1

a(u - u)'(u - uh)' dx = rx`+l (a(u - u)(u - uh)')' - a'(u - u)(u - uh)' dx

dnm (u - u)(x3) = 0 Vj.

Fhm (8.70) it follows that

alu -uhl25- f a(u-u)(u'-uh)dx n

< la'I.lu - tI2Iu' - 1thl2 'l`hls proves (8.69). By the Poincare inequality we have Iv12 <5

(B_A) 102 = I fll 7Iu I2 dv E Ho(st)

(8.71)

where Istl is the length of 11. Also as in (8.66) we derive IVI0 <_ In11/2Iv'I2 VV E Ho(st).

Combining (8.69)-(8.72) we obtain

Iu - uhl2 < Clfi - tI2, Iu - uhIoo 5 Clu - UI2.

(8.72)

8. Finite Element Methods for Elliptic Problems

122

Then it follows that

lu-uhl2 = Iu - u+fi - uhlz< (1+C)lu-ulz, lu - uhIoo
8.5. Appraaimation of nonlinear problems In this section we come back to problem (5.1) and its approximation (5.1b). In Chapter 7 we saw that the problem (5.1) has a unique solution, for instance when

u i- a(x, u) is for a.e. x E Sl Lipschitz continuous, i.e. when for some constant C it holds that

la(x, u) - a(x, v)I < Clu - vi a.e. x E 11, u, v E R.

(8.73)

We would like to show that in this condition the problem (5.1h) can however have several solutions. The notation is the one of section 8.2.2, i.e. we consider fl a polygonal domain of Bs with boundary r and (rh) a family of triangulation of Cl. We make on m an

assumption slightly stronger than regularity, namely if OF i = 1,2,3 denote the angles of the triangle K, we assume that for some 6 > 0 it holds that

0<6<eK < 2 -b di=1,2,3, VKErh,

(8.74)

i.e. each of the angles of the triangles are smaller than z (see also [38] for such a condition). Let Vh be the space defined by (8.18). Then one can show (see [7]): function THEOREM 8.13. Assume that f E H-1(12) and a is a such that (5.4), (8.73) holds. Then for h small enough the problem (5.1h) has a unique solution. (The existence of a solution follows from the proof of Theorem 5.1). Before giving the proof of the theorem let us establish the following lemmas:

LEMMA 8.1. Let K be a triangle with vertices K1, K2, K3 satisfying (8.74). Let )q be the of ne functions such that Ai(K5) = bit di, j = 1, 2,3.

(8.75)

For i, j = 1, 2,3 there exist positive constants c1, c2 such that

VA; 0.1;

hK

< 0 d i j,

(8.76)

96

hK < IVAl

hK

Vi.

(8.77)

8.5. Approximation of nonlinear problems

123

PROOF. Without loss of generality let us assume

i=2, j=3,

K1=0.

Let R be the rotation around 0 of angle z. Denote also by product on R2. We have A2(x) _ (x, R(K3)) (K2, R(K3))

A3 (X) _

the canonical scalar

(x, R(K2)) (K3, R(K2))

and thus

VA2 VA3 = (R(K2), R(K3))/(K2, R(K3))(K3, R(K2)). We have also clearly - R being an isometry - 81 the positive measure of the angle between K2, K3 (R(K2), R(K3)) = (K2, K3) = cos01IK2I IK31 > 0, (K2, R(K3)) = -sin 011 K21 IK31 < 0, (K3, R(K2)) = sin91IK21 IK3I > 0. (I

I denotes the length - recall that K1 = 0). It follows that VA2 VA3 =

cos B1

sin 021 IK2I IK3I

and (8.76) follows from (8.74). To obtain (8.77) it is enough to notice that

Ioa2I =

-

IR(K3)I

IK31

I(K2,R(K3))I

sinejIK211K31

1

sine1IK21

0 LEMMA 8.2. Let u be the solution to (5.1) anduh be a solution to (5.ih). Then it holds that hl

0 Huh - x111,2 = 0

PROOF. Since the solution u to (5.1) is unique, it follows from the proof of Theorem 5.1 that uh - u in Ho (SZ),

uh - u in L2(ft).

(Note that (5.5) holds as shown along the lines of the proof of Theorem 8.8). Then from (5.4) it follows that

mfn IVuh

- Vu12dx < /r a(x,uh)1IVUh - Vu12dx =ja(x,uh)IVuaIZdx - 2

fa

+ Ja(x,uh)IVuI2dx.

8. Finite Element Methods for Elliptic Problems

124

We have, when h - 0,

a(x,uh)IVuhl2dx = (f,uh) -i (f,u) = ja(x,u)IVuI2dx. Thus if one shows that Jo

a(x,uh)VuhVudx - Ja(x,u)IVul2dx, a(x,uh)IVuI2dx --' Ja(x,u)IVuI2dx,

the result will follow. But - up to a subsequence - we have

uh -' u a.e. in fl so that

a(x, uh)Vu - a(x, u)Vu in L2(Q). Since Vuh -. Vu in L2(1l) the only possible limit for and

ja(x,uh)IVuI2dx

is

a(x, u) I VuI2 dx.

0

This completes the proof of the lemma. We turn now to the proof of Theorem 8.13.

PROOF OF THEOREM 8.13. Let u; = u;,h, i = 1, 2 be two solutions to (5.1,,). We have

a(x,ul)Vul Vvdx =

Jn

a(x,u2)Vu2 Vvdx Vv E Vh

which leads to

Ja(x,ui)V(ui - u2) Vvdx = J(a(xiia) -a(x,ul))Vu2 Vvdx

Vv E Vh.

Setting w = of - u2 and using (8.73) it turns out that

L a(x, ul)Vw Vvdx < C inf Iw) IVu2I IVvl dx Vv E Vh.

(8.78)

Consider then the function v E Vh defined by v(KK)

1

if w(K,) > 0,

0

else.

The Kj's are the nodes of the triangulation. Clearly we have

Vv=O on K,

(8.79)

8.5. Approximation of nonlinear problems

125

if K denotes a triangle of vertices K1, Kj, Kk, unless we have

w(K;) < 0,

w(K1) > 0,

w(Kk) < 0

(8.80)

or

w(Ki) > 0, w(K,) > 0, w(Kk) < 0. (The values i, j, k are distinct and for simplicity taken in t 1, 2,31).

(8.81)

From (8.78) we deduce:

r E / a(x,ul)Vw Vvdx
(8.82)

where the summation is taken on the triangles K satisfying (8.80) or (8.81). Now, if K is a triangle such that (8.80) or (8.81) holds, to vanishes at some point in K and it holds for some y E K that

w(x) = w(x) - w(y) = Vw (x - y) V x E K, hence

Iw(x)I <- IpwlhK

`dx E K.

Recalling (8.82) we get

JK a(x,ul)Vw Vvdx
hK J IVwI IDu2I IVvIdx.

(8.83)

K

Let us set aK = w(Ki) so that on each K E Ti, it holds that to = aKA1 + 4A2 + a3 )13.

(8.84)

Let us consider a triangle K such that (8.80) holds - i.e. assume for instance that a3K<0. aK>0, :5 0' Then, on K we have (see (8.79)) v = A 1 and it holds that >

c

K

(8.85)

(Iai`I+Ia2I+Ia3KI)

where c = min(1, c1) is independent of K. If now K is a triangle such that (8.81) holds - i.e. if for instance

aK > 0,

a2K >0,

a3 < 0,

then

v=Al+A2=1-A3 and

Vw Vv = -aKpa1 VA3 - a2Kpa2 V'\3 - a31pasI2

hK(IaKI+I4I+I43KI) where c is as above.

(8.86)

8. Finite Element Methods for Elliptic Problems

126

By (8.84), (8.77) we have IVwl <_ -EL(JaKI

K

+ Ias l + Ia'I),

IVvl

kK

hence from (5.4), (8.83), (8.84), (8.85) we deduce

tWE hK

f

IaKI+141+laf Idx
hK

K

fr

(IaKI+lafI+lafI)IoU2Idx.

Recall that (8.74) holds so that

IKI?fihK for some constant $. It then follows for some constant M that

(IaKI+IaKI+Ias l) <ME K

IaKI+IaKI+Ia3Kf

hK

K

L121

S ME(Iai`I+Ias I+Ias I)11Vu2112,K K

where I- I2,K denotes the usual L2(K}norm. From Lemma 8.2 we deduce that IIVu21I2 K < 1IVu2 - VUII2,K + IIDuII2,K

< 1Iou2 - VuI12 + I IVUII2,K

= e(h)

with limp-,O,-(h) = 0. It follows that

(IaKI + Ia'I + Iaf I) < Me(h) E(IaKI + 141 +14 I). K

K

Thus, for h small enough, the set of K such that (8.80) or (8.81) holds is empty, i.e. one has

w=u1 -u2<0.

Exchanging the role of ul and u2 the result follows. However one can show:

THEOREM 8.14. Under the assumptions of the preceding theorem and when h is not assumed to be small, then uniqueness might fail.

PROOF. Let T be a fixed triangle of R2 of measure 1- i.e. such that ITI = 1.

(8.87)

Let us denote by +p a function of class C°° on T such that

P =1 on 8T, ,V

= 0 on OT, 1

Co(y) dy =

21

8-88)

8.5. Approximation of nonlinear problems

127

8T denotes here the boundary of T. It is clear that it is possible to find such a function. For K E rh let us denote by YK the affine bijection

YK:K -+ T that associates to a point of barycentric coordinates r the point yK (x) having in T the same barycentric coordinates in T. One has clearly - see (8.87)

IhI=JKdx=JTIDyK1(y)Idy=IDy)(y)I.

(8.89)

(I DyKI (y) I denotes the determinant of the Jacobian matrix DyK' ). Consider then a function u E Vh such that

u>0 on the inside nodes of the triangulation rh. Then, let us define on 11 x R a function

a(x,z) by for z < u(x), for z > 2u(x). for z = pu(r) + (1 -,u)2u(x)

1

a(x z) =

p(YK(x)) P + (1 - µ)1,2(YK(x))

(8.90)

p E (0, 11.

for every x E K. One has clearly

Ja(xtu(z))dx = I KI, Ja(x2u(x))dx = J'(YK(x))dx =

j(Y)IDv.'(Y)IdY

=IAIJTv(y)dy= I2I Thus for every v E Vh

f

Vu Vv

a(x, u)Vu Vv dx = KEr,,

E

r

Jh

a(x, u) dx

J a(.r,2u)dx

KErr,

=

J ft .I

a(x. 2u)O2u Vv dx.

Clearly for u fixed in Vh as above the mapping

f : v'-.

Jn

A

a(x, u)Vu Vi'dx

(8.91)

8. Finite Element Methods for Elliptic Problems

128

is an element of H-'(Q). It follows from (8.91) that both It and 2u are solutions to

uEV,,

ja(xn)VuVvdx_(I,v)

VvEVh

(8.92)

which shows non-uniqueness. To complete the proof of the theorem let us show that a is continuous on S2 x R and Lipschitz continuous in z and also satisfies (5.4). For the last condition, due to (8.90) and the last inequality of (8.88) we have

1
and a is continuous. To see that a is Lipschitz continuous we notice that z - a(x, z) is piecewise affine on K and we have (x)) - 1 as (x, z) = 0

or

(8.93)

da (x, z) =

So, if K is a triangle that does not touch the boundary of S2, then as

< u(x) < C

az(x'z) on K. If K is a triangle with a side on the boundary make a corner of the boundary of Il

but such that K does not

we have for some in > 0

u(x) > in dist(x, I') = mjx - ryI

(8.94)

for some ry E r. Moreover

IV(yK(x)) -

lI = IP(yK(x)) - P(YK(-Y))l

< CIYK(x) - YK(-Y)I

(8.95)

< C'Ix - ryI

for some constant C'. Combining (8.93) (8.95) it follows that in this case as (x, z)
REMARK 8.6. The above construction works in any dimension for the P1 method. It is also easy to see that (5.1 h) can have as many solutions as one wishes. For instance to have 3u also a solution it is enough to introduce a second function iP satisfying (8.88) with the third line replaced by 1

0(y) dy =

8.6. Exercises

129

Then, defining a(x. ) on K by a(x, u(x)) = 1, a(x, 2u(x)) = V(yK (x) ), a(x, 3u(x)) = V'(yK (.r)) and interpolating it linearly (see (8.90)) will force 3u to satisfy also (8.92).

8.6. Exercises 1. Provide the details of the proofs of Proposition 8.2 and Proposition 8.3. 2. A rectangle is a set n

Q = fl(a;,b;J a; < b,. a,. b, E R. A triangulation of S2 is a decomposition of f in rectangles such that

S2= U Q QE rh

with the property that for any Q1. Q2 E rh Q1 fl QZ = 0 or a lower dimensional face of Q,. We set (

1=SPIP(x)_ E aoai',.....r"

l

0
Show that dim Q1 = 2". Set

Vh={v:S2-.RIviscontinuous, vIQEQ1 VQErh, v=Oonl'}. Develop the corresponding finite element method when h is given by

h = Max diam(Q). QE rh

3. Let u, uh be the solutions of (8.33), (8.34) respectively. Show that lim uh = u.

h-0

(Follow the steps of Theorem 8.8). 4. Work out the estimates of Theorem 8.12 in the case of a one dimensional Neumann problem.

Chapter 9

Minimizers 9.1. Introduction Many problems in physics can be formulated as minimization problems. i.e. one minimizes some quantity - for instance some energy on a suitable class of functions and one searches for a point achieving the infiumin. As seen in Chapter 1 this is the case in elasticity theory (see [9], [12], [37], [47], [43]) and problems in this field. Let us recall the definition of a minimizer. DEFINITION 9.1. Let C be a set and (p a function from C into R. in E C is a minimizer of on C if Ins p(r) = y7(na).

Note that the existence of a minimizer requires p to be bounded from below.

and an arbitrary set, various situations can occur. The For a function archetype of the situation occurs already for C = R, i.e. for a function from IR into R. Indeed we can have: 1. p has a unique minimizer on R. This is the case for instance for the function wp(x) = x2.

2. V has infinitely many minimizers. This is the case for y7(r) = (r - 1)+ ()+ denotes the positive part of the function.

3. cp has no minimizer. This is the case for instance for y;(r) = c-. The infimuni of this function on Qt is 0, but there is no point where it is achieved.

In the calculus of variations where one minimizes on classes of functions, the same situations arise however they are sometimes not so easy to see. But there is no reason for the situation occurring in dimension I not to be replicated in higher dimensions. We shall try to convince the reader with some simple examples. Let Q be the interval (0, 1). Let

C={vE4",-(0,1)1v(0)=0.v(l)=1/2} where W1-°°(0, 1) denotes the set of Lipschitz continuous functions on !1 (recall that we refer for instance to [17], [46], [48] for notions on Sobolev spaces). Then we can show

9. Minimizers

132

THEOREM 9.1. The problems

I

Inf f 1112 (x) dx,

(9.1)

Inf

1(1 - v2(x))2 dx,

(9.2)

Arctan2 v'(x) dx

(9.3)

vEC

t

vEC

Inf

vEC

J t

have respectively a unique minimizer, infinitely many minimizers, no minimizer.

PROOF. Let us start with (9.1). Then we claim that

u(x) =

I

2

is the only minimizer. To see this fact. it is enough to remark that for any v E C t

10 v2dx- fo

t

t

u2dx(zi-u') (v'+u')dx 0

_

(v' - u') (v' - u' + 2u') dx 0

=

j(' -u')2dx>_0. J(v' -u')2dx+v'-u'dx= 0

Now this last integral vanishes if and only if (v - u)' = 0 ' v - u = cst, but due to our boundary conditions this means if and only if v = u. For (9.2), it is

FIGURE 9.1

clear that the infimum 0 is achieved for any function u E C such that u' = ±1. Infinitely many such functions exist, as depicted in Figure 9.1. Each of them uses

9.2. The direct method

133

one path - made of broken lines of slope ±1 - and located in the rectangle shown in Figures 9.1 and 9.2. For (9.3), if we show that the infimum is 0, then we are done. Indeed, if u is a minimizer, we must have u' = 0 identically, hence u = cst which is impossible due to the boundary conditions to be satisfied. Now to show that the infimum is 0 one considers the "sequence" of functions defined in Figure 9.2. Clearly since Arctan2(v) is bounded from above by nz/4, z

0 < r Arctanz U, (x) dx <_ e Jst

with E -' 0.

0

uf: 1/2

1-e

0

c

1

FIGURE 9.2

REMARK 9.1. One could of course minimize on the space of piecewise affine functions to get the same result. Clearly the only useful properties for the integrands 1/' in (9.2), (9.3) are respectively 0 > 0,

except

V'(±1) = 0,

' > 0,

except

V'(0) = 0,

/, bounded.

In this section we shall now consider the case where the minimum is achieved.

9.2. The direct method Suppose that one considers a function : X II2 where X is some metric space. If C is a subset of X, then all the problems we are going to deal with have the form

Find u E C such that o(u) = Inf cp(v). C

(9.4)

In other words, does such a u exist and, if not, what can we say about the problem?

9. Minimizers

134

The direct method of the calculus of variations is the most natural to think of in order to solve (9.4). That is to say, one considers a minimizing sequence in C -

i.e. a sequence u such that

u E C,

Inf cp(v)

(9.5)

1, EC

and one tries to show that u converges toward the solution u to (9.4). The two difficulties are to show that

u converges toward u E C u is a minimizer of the problem. A simple frequent situation in the calculus of variations where this can be achieved is the following:

THEOREM 9.2. Let X be a reflexive Banach space. Let C be a weakly closed subset of X and p : C --s R be a function such that cp has a bounded minimizing sequence in C, cp is lower semicontinuous on C;

(9.6)

(9.7)

then there exists a such that

p(u) = Inf p(v). vEC

(9.8)

PROOF. Let us denote by u a minimizing sequence - see (9.5) which is bounded. We can extract a subsequence, which we still label by u.,, such that it,, converges toward u weakly. C being weakly closed and cp weakly lower semicontinuous we see that u E C,

Inf p(v)

lim inf cp(u,,) > cp(u)

and the result follows. REMARK 9.2. The assumption (9.6) holds for instance in the cases

C is bounded, I1r1lim V(X) = +00, rEC

this last condition is often called coerciveness. One can also replace it by lim 11-11

+-

rEC

yo(x) > Inf W(x).

r

The assumption (9.7) holds when p is convex and continuous (see 1401). Finally, C convex and closed implies that C is weakly closed. Let us now consider a few examples.

9.3. Applications

135

9.3. Applications We first introduce the Sobolev Space (see also Chapter 2) W1.2(1; Ry")

v : S2 -

I

E L2(Sl),

v,,

(9.9)

in, Vi = l.....n}.

Vi

Recall that in the above definition the derivatives are taken in the distributional sense. 0 is a bounded open set of 1R" with boundary r = &i. We shall assume W1,2 (!Q; Rt) to be equipped with the norm 8vt

1/2

IIt'IIi.z= (E I1'=I2+EIB.

I(9.10)

,,i

J 2

,

Iz denotes the usual Lz(Q)-norm. It is easy to see that, when equipped (ft R') is a Hilbert space and thus reflexive. Let C be a with this norm, where I

closed convex subset of W1.2 (Q; R"') Let r!, he a function such that i

:f

(9.11)

where Alm,,, denotes the space of matrices with in rows and n columns. We assume

that V' is a Caratheodory function, i.e. that x '--+ V'(x, v, Al) is measurable on Q. t1 (v. Al) E 1R' x

(v, M) '--+ r'(x, v, Al) is continuous a.e. x E f1.

(9.12) (9.13)

In fact we assume a little bit more than (9.13) in the sense that we suppose also that (v. M)

rtp(x, v, M) is convex a.e. X E 12.

(9.14)

Moreover we shall impose

a.e. x E 1 ,

Irt'(x, v, Al)I <_ cl (IvIZ + IAIIZ) + 91(x)

V (v, M) E

x Al,,.,,

(9.15)

where cl is some positive constant 91 E L1(Q)

and I I denotes the usual euclidean norm "i

IvI2=va [ ,=

IA1Is

(9.16)

i.e.

V'11=(v1.... .vm)ER"

(9.17)

V Al = (Al,3) E

(9.18)

1

= E AI i

AI,,,,,,.

,.)

Then under the above assumptions - that is to say if C' is a closed convex subset of W1.2 (Q; R') and if (9.11)-(9.16) holds - we can state

136

9. Minimizers

THEOREM 9.3. If C is bounded and

tp(x, v, M) > -92 (x) a. e. x E 0, V (v, M) E Rm x M.,,,

(9.19)

or

tp(x, v, M) > c2(Iv12 + IMI2) - 92(x) a.e. x E 11, V(v, M) E Rm x Mm,,

(9.20)

where c2 > 0 is some constant and 92 E Ll(S2),

(9.21)

Id r O(x, v(x), Vv(x)) dx

(9.22)

then there exists a minimizer to v

where Vv = (

n ) is the Jacobian matrix of v. Moreover, if (v, M)

- tp(x, v, M ) (9. 23 ) is strictly convex for a.e. x E fl, then this minimizer is unique. (Recall that the function defined by (9.23) is strictly convex if r'(x, a(vl, Ml) + (1 - a)(v2, M2)) < ar/i(x, vi, MO) + (1 - a)d'(x, v2, M2)

for every a E (0, 1), a. e. x E 0, and every (vi, Ml), (v2i M2)). PROOF. First one should notice that, since tp is a Caratheodory function, x i- tp(x,v(x),Vv(x)) is measurable for any v E C. Next due to (9.15)

r 10(X, V(X), VV(X)l ciIIvIIi.2 - fn92(x)dx

P(v) =

and thus lim1,.,,.-.,, cp(v) = +oo which requires to every minimizing vEC

sequence to be bounded. It remains to show that v -* ,p(v) = ff tp(x, v, Vv(x)) dx is weakly lower semicontinuous. Since clearly by (9.13) W(v) is convex, it is enough to show that V(v) is lower semioontinuous for the strong topology. So, let v --' v in C. Due to (9.19), (9.20) O(x, vn, Vv,) + 92 > 0.

(9.24)

9.3. Applications

137

Moreover, one can extract from vn a subsequence - still denoted vn - such

that vn, Ovn - v, Vv a,e. in fl,

lim inf V(vn). lim cp(Vn) = n-+oo n-.+oo

(9.25)

Applying Fatou's lemma, lim inf {?'(x, vn (x), Vvn(x)) + 92}

J12 n-++oo

r


{u'(r,v(x),Vv(x)) +g2(x)}dx < liminftp(vn) +J g2(x) dx - n-+oo

in

which means that w(v) < lim inf tp(vn),

n-+oo the result we were looking for. To show uniqueness, we notice that if u1, u2 are two distinct minimizers, then by the strict convexity of the integrand, a1+u2 CU1+u2 1) x dx < 1jt,(x,uiVui)dx s1

2

2

2

+

21n 0(x, u2, Vu2) dx

vnf f n41(x,v(x),Vv(x))dx which is a contradiction. This completes the proof of the theorem.

0 It is in fact possible to drop the convexity assumption in v. Indeed assume that (9.11) (9.13) hold, but instead of (9.14) let us assume only that a.e. X E 11, VV ERn M -' ,b(x, v, M) is convex. (9.26) Moreover, we assume that there exists a function to such that I?G(x, v, A!) - O (X, v', M)I <- w(Iv - v'I)(I MI2 + Iv12 + Iv I2 + f (x))

(9.27)

for a.e. x E Q, V v, v' E R, V M E M.,n with ili mw(t) = 0,

f E L'(0).

(9.28)

Then we can prove:

THEOREM 9.4. Assume that (9.11)-(9.13), (9.15), (9.26)-(9.28) hold. Then if

C is bounded or

lim

IIn11,2--+m

W(v) = +oo,

(9.29)

vEC

t'.'(x, v, M) > c3IMI2 - C4IVI2 - g3

(9.30)

9. Minimizers

138

for ae. x E St, `d (v, M) E R- x Mn,,n, g3 E Ll ()). Then there exists a minimizer to QInf

Jn

b(x, v(x), Vv(x)) dx.

(Recall that we have set cp(v) = fo 1P(x, v(x),Vv(x)) dx).

PROOF. C, being closed and convex, is weakly closed and by (9.29) we need to show only that W is weakly lower semicontinuous. So, let us consider a sequence

vn such that

vn _ v in W1,2(fl; Rm).

(9.31)

Due to the compactness of the embedding of W1"2(fl; Rm) into L2(11; Rm) one can

extract from vn a subsequence - still labelled vn - such that

vn - v v,, - v

in L2 (11; Rm)

(9.32)

a.e. in Cl.

(9.33)

Due to the Egorov theorem, for every e > 0 there exists a compact set Cl in fZ such that vn -+ v uniformly in C. It) \ fl.I < E, Denote by X. the characteristic function Cli. Note that by (9.30) (9.34)

,b(x, v, M) + ra Iv12 + g3 >_ 0

and thus we have Jo

y,(x,vn,Vvn)+c41vnI2+g3dx (9.35)

fro, XeV(x,v,,Vvn)+XelvnI2+Xeg3dx. Due to (9.32), (9.33), we conclude that UM n-+oo

n lun

fa

Glvn12+g3dx=J nc41vI2+g3dx Xec4Ivn I2 + Xeg3 dx = in

+ X.93 dx.

Next

JXt*(XVnVVn) = J X{i(x, v, Vv) 2

n

+ JXe,1'(zvVVn)dx. Clearly, since the last integral above is convex in Vvn, we can show that p fo Xtp(x, v, p) is lower semicontinuous for the strong topology of (L2(12))mxn as

we did in Theorem 9.3 using Fatou's lemma Just copy that proof using (9.34).

9.3. Applications

139

Then due to its convexity the integral is also weakly lower semicontinuous. So, one has lim inf Xe'i/'(x, v, Vvn) dx >- f XeO(x, v, Vv) dz. n -++oo

J

n

S2

Next by (9.27) f XF(,G(x,vn,Vvn)-tt(x,v,Vv)}dr S2

jX'(x1vnVvn)-(x,v,Vv )Idx

Jn w(Ivn - vI)(IVvnI2 + IVnI2 + IVI2 + f) < Cw(b) for n large enough and any 6 > 0. (Recall that vn is bounded in W1,2 (f); Rm) since it converges weakly). Then passing to the liminf in (9.35) we get lim inf fu iG(x, vn, Vvn) dx +

n--}x u

-C(6)+

Jn c4Ivl2 +.93 dx 9)dx.

Jn

Up to an extraction of a subsequence one can of course always assume that the above lim inf is the lim inf of the whole sequence. 6 being arbitrary, we derive from (9.28)

liminf J tt'(x,vn,Vvn)dx> / Xet/b(x,v,Vv)dx- / (1-Xe)(c4IV12+g3). n-.+oc n

s2

n

Letting a --+ 0, the result follows by the Lebesgue theorem.

0

EXAMPLE. For i, j = 1, ... , m, h, k = 1, ... , n let us consider Carathbodory functions

Assume that these functions define a positive, bounded quadratic form on the space of m x n matrices in the sense that it holds for some positive constants, A, A and, with the usual summation convention for a.e. x E Q, Vv E RI '\ (9.36) Aijhk(x, v)M:hMjk S AIMI2 V M E Mm,n. 11,112 <

Moreover, assume that

v '- Aijhk(x,v) is uniformly continuous in such a way that there exists a function w satisfying (9.28) such that IAijhk(x, v)MihMjk - Aijhk(x,V')MihMjkI < m(Iv - v'I)IMI2, for a.e. x E S2, V

v' E R', V M E Mm,n

(9.37)

Next for i, j =1,... , n consider Carathbodory functions

ail (x,v):IlxRm-FL Assume that these functions define a positive, bounded quadratic form on R"' in the sense that we have, for some positive constants b and A and a.e. x E it, b'vER""

bI(I2 < aj(x,v)ttj s Altl2 V4 E Rm.

(9.38)

Moreover, assume that

v' - aij (x, v) are uniformly continuous in such a way that, for some w satisfying (9.28), for a.e.

xESl,Vv,v'ER"',dC ER" Iaj(x,v)e;Ej-aj(x,v')Cgjl<_w(Iv-v'I)ItI2.

(9.39)

For f E La (sl; R"') - i.e. f = (fl.. . . , f,,) with fi E L2(fl) - set &(x, v, M) = Ajhk(x, v)MihMik + aij(x, v)vivj - (f, v) where

(9.40)

denotes the euclidean scalar product in R"' . Then we can state

THEOREM 9.5. Let C be a closed convex set of W1.2(c2,Rm). Under the above assumption there exists a minimizes to vInf

Ja P(x,v(x),Vv(x))dx.

(9.41)

Moreover if C C W01, 2(0,Rm) _ {v E W1.2(fl;Rm)Iv = 0 on 09), we can assume

b = 0 in (9.38). PROOF. It is enough to check all the assumptions of Theorem 9.4. First note

that clearly (9.11)-(9.13) hold. Next, due to (9.36), (9.26) holds. By CauchySchwarz inequality, (9.42) 1fI2+ 2IvI2. Combining with (9.36), (9.38) it is clear that (9.15), (9.30) hold. To get (9.27) we

(f,v) <- Ifllvl <-

remark that V,(x, v, M) - tp(x, v', M) = Aijhtt(x,V)MtMik - Aijkk(x, v')MihMjk (aij (x, v) - aij (x, v') )vivj + aij (x, v')(vivj - vivj) + (f - v - v') Taking = v in (9.39) and using (9.37) one easily derives

I+'(x,v,M)- fi(x,v',M)I <w(Iv-v'I){IMI2+IvI2)+IfIIv-v'l + ail (x, v'){(vi - vi)vj + vi'(vj - vJ)} Since by (9.38) the ai j are uniformly bounded it follows easily that

I j'(x, v, M) - (x, v', M)I <'(Iv - v'I){IMI2 + Iv12} + If llv _V11 + CIv - v'lly l + CIv - v'llvl

9.3. Applications

141

where C is some constant. Using the Young inequality, for instance

,z

z

it follows that

I1/'(x,v,M) - (x, v',M)I

<.(Iv-v'I){IMI2+Iv12}+IV -v'1{121 + < {w(iv-v'1)+ 211,-v

122

+C2+ 212 +2}

I}{IMI2+11X12+11'112+g(x)}

where g E L1(S1). This is clearly (9.27). So, it remains only to show (9.29) when C C W01'2(SZ;IR°'). But from (9.40), (9.36), (9.38) it follows, using Young's inequality, that

v(x,v,M) > )IMI2+tIv12

- If111'I

>AIM12+61x12-EI1712-

fI 4E z

\IMI2 + (6 - E)Ivl2 - 14E This implies

j

l4E ff12

'(x, v(x), Vv(x)) dx > A f Ivv12 dx + (6 - e) f Iv12 dx -

it, > f Ivvl2 dx n

4E

I f I2 dx

dx.

2

(This inequality holds for c small enough and by the Poincare inequality). This completes the proof of the theorem since 1/2

(in

Ivvlz dx)

is a norm on W0'2(S2;Rm)

REMARK 9.3. Using the same structure assumptions as above one can transport the L2(S2) theory into LP(S2) - i.e. find minimizers for problems of the type Inf C

f

V)(x, v(x), Vv(x)) dx

it,

where C is a closed convex set of IV'"P(S1:lRt) or W"P(Q,RI). The details are left to the reader.

9. Minimizers

142

9.4. The Euler Equation Let C be a convex set of W1.2(S2; R"') and consider as above the problem Inf

vEC

J in

V)(x, v(x), Vv(x)) dx.

(9.43)

It is well known that at a point where a real function achieves a local minimum, provided it is differentiable, its derivative vanishes. The phenomenon occurs of course also for functionals of the type of (9.43). More generally we can state THEOREM 9.6. Let u be a minimizer to (9.43). Let us assume that the function

v E R11 M E M",.n is differentiable in v and M for a.e. x E Q. Then, provided the integral below makes (v, M) - V)(x, v, M),

sense,

2'(x,u,Vu)(v;(x) - ui(x)) + 8 (x,u,Vu)a(11' - u`)(x)dx by E C. 0 < j1av, ax, aM,, (9.44)

Moreover if for some ua E W1.2(S2 R"') C = uu + W01'2(Q; W')

(9.45)

then in the distributional sense (9.46) u(x), Vu(x)) = 0 in Q, axe a v,for every i = 1, ... , in. We assumed of course the usual summation convention of

(90 (x, u(x)' Vu(x)) -

(x,

repeated indices.

PROOF. Since C is assumed to be convex, for every A E [0, 11,

u+A(v-u)EC VvEC. Thus the function A-

r

Jn

O(x, u + A(v - u), Vu + AV(v - u)) dx = J(A)

achieves its minimum at A = 0 and, provided it is differentiable,

daI

J(A)> 0

n

which leads precisely to (9.44) if differentiation under the integral sign is possible. If (9.45) holds, then for every w E W01,2 (1l,QRm),

u±wEC and plugging these functions into (9.44) leads to Jst

LIP (x' u, Du)-u;, +

dM (x, u, Vu)

' dx = 0 dw E OXJ

This is (9.46). This completes the proof of the theorem.

Wo.2(f; R-)

9.5. Exercises

143

9.5. Exercises 1. Prove that g'1.2(1; R'°) is a Hilbert space. 2. One can for p > 2 introduce W'.P(Q: Rm) =

{t' : Q - Rm

ate' E LP(Q)

Vi = 1,...,m, Vj = 1.....n}. Prove that W' P(Q; Rm) is a Banach space for the norm

t

Ell"

'I

P1 P

Y

and extend Theorems 9.3, 9.4 to the case of C being a closed convex set of this space (see Remark 9.3). 3. Let a(x, u) be a Caratheodory function satisfying

a
= 21 a(x, v)IVvI2 dx - J f . vdx.

Show that there exists u E Ho (S2) such that

J[u] < J[v] Vv E Ho (S2). What is the Euler equation of the problem? 4. Under the assumptions of Theorem 9.5 find the Euler equation of the problem (9.41). Consider the particular case where m = 1 and rephrase Theorem 9.5 in this setting.

Chapter 10

Minimizing Sequences 10.1. Some model problems In this section we consider again problems of the type Inf /(x, v(x), Vv(x)) dx vECJ t

but we drop the convexity assumption. This will force our problems to have no minimizer. Then it is very natural to turn to the study of the minimizing sequences to see in particular if they present common features that could describe properties of the underlying physical problem. A tool for constructing minimizing sequences is the notion of Young measure that we will briefly explain. Let Q be a bounded subset of R2. Then consider the following model problems: Inf

vEW'o(n) I

nf

J vr + (v=2 - 1)2 dx,

(10.1)

Jv1+(v2_1)2+v2dx.

(10.2)

=vEWI.x(O)

t

r inf (n) vE W0

Inf

v=1 + (v=2 - xi)2 dx,

(10.3)

J

vEW'--(f2)Jn

v=, + (v=2 - x1)2 + v2 dx.

(10.4)

In the framework of the preceding chapter we are in the case m = 1, n = 2. This is the so-called scalar case in the calculus of variations. The points in 1R2 are denoted (X1,x2), Vv = (v.-,, v.,,), and dx = dx1dx2 is the usual Lebesgue measure. Setting 0(f1, W = 1 +

1)2

(10.5)

it is easy to see that V) is not convex and that the direct method of the calculus of variations does not apply here. In fact we can state

THEOREM 10.1. Each infima listed above is equal to 0. None of the above problems admits a minimizer.

146

10. Minimizing Sequences

PROOF. Let us show that the infimum (10.1) is equal to 0. It is well known we refer for instance to [40] - that vEwo'OD(f2)

Jt t'(Vv)dx = I)I,"(0,0)

(10.6)

where 0** denotes the convex envelope of t/,, i.e. the largest convex function less than or equal tot', Jill the Lebesgue measure of Q. Now clearly, see (10.5),

11(0,±1) =0 and also, since 0 is a convex function less than or equal to t/i

0 < 0-- (0, ±1) :5 0(0, ±1) = 0. Next using the convexity of tG" we deduce that 0

<'P"(0,0)='P** 1(0,1)+

2(0,-1))

< 1 P**(o,1)+ 2P"(o,-1) =0

and thus the infimum (10.6) is equal to 0. It will be clear later that the other infima are also equal to 0. Next suppose that u is a minimizer of one of these problems. For instance for the problems (10.1) and (10.3) we would have

f a4 +(uz2 -

1)2dz=0,

'2-1+(u2-2-x2)2dx=0

and in particular

usl = 0 a.e. in fl. This will force u to be independent of xl; but since u vanishes at the boundary of Cl, this will imply that u = 0 which is impossible since for u = 0 the above integrals do not vanish. In the case of problems (10.2), (10.4) if u is a minimizer, then the term in u2 in the integral will again force u to be equal to 0 which is impossible. This completes the proof of the theorem.

With no hope of finding a minimizer we turn to the study of the minimizing sequences. As we just saw, the only possible minimizer for our problems could be 0. So, this function has to play a certain role in our situation. Indeed we have THEOREM 10.2. Let v, be a bounded minimizing sequence of the problems (10.1) -(10.4) - i. e. such that for some constant independent of e,

Ivil0, IIVV'l1. < C (]

(10.7)

Ioo denotes the usual L°°(fl)-noun, I I the euclidean norm). Then

of -. 0 uniformly in fl,

Vvt - 0 in (L°°(fl))2-weak * .

(10.8)

(10.9)

10.2. Young measures

147

PROOF. It is well known that W','10(11), the space of Lipschitz continuous functions in Q, is compactly embedded in C(1l), the space of continuous functions in S1. Thus, we can extract from vs a "subsequence" still labelled v.. such that Ve - V uniformly in S1,

Vvf - Vv in (L°°(11))2-weak for some v E W'°°°(0) (see Remark 10.1 below for the definition of the weak convergence). Since of is a minimizing sequence, it follows from the preceding theorem that (v[)_, --+ 0 in L2(0), in the case of problems (10.1), (10.3) and

v-'0 in L2(1l) in the case of problems (10.2), (10.4). By the uniqueness of the limit in 1Y(i1), for instance, we deduce that v=0 v=1 =0, respectively which in both cases leads to v = 0. Thus, any subsequence of vt has to converge toward 0. This shows that vE -+ 0 uniformly in 0,

Vv,

0

O

in (L°°(f1))2-weak * .

10.2. Young measures DEFINITION 10.1. (see [11], [591). Let UE : fl-'R v be a sequence of functions in L°° (1; RP). Suppose that this sequence is bounded in the sense that iIUeIioo < C

where C is a constant independent of e,

[ the euclidean norm in RP, [ I the

usual LOD(S1)-norm. Then:

there exists a subsequence of OF - still labelled U, there exists a probability measure on Rp parameterized by x E SZ such that for any function F continuous on RP

F(Uf)

f F(A) dvx(A) in L°O(1l)-weak * .

(10.10)

o

vx is called a Young measure associated with or generated by the sequence U, REMARK 10.1. Recall that

f- f

in L°°(il)-weak *

r

li mJ ffgdx = n

jfodx

Vg E L'(l). (10.11)

Note that (10.10) implies that the mapping x'--. jP F(A) dv.(A)

is measurable. For more on Young measures we refer the reader to [59].

10. Minimizing Sequences

148

EXAMPLE. We have seen in the previous theorem that every minimizing sequence of the problems we are dealing with satisfies

vt - 0 uniformly in n. Since v6 is bounded it defines a Young measure on R. Due to the definition and to (10.8), for this Young measure vx and any continuous F,

F(vv) -. F(O) uniformly in Cl, F(ve)

in L°°(i2)-weak * .

Thus,

F(0) =

JR F(A) dvx (A)

V F E C(R).

We denoted by C(R) the space of continuous functions on R. This above equality implies simply that 60 where 6o denotes the Dirac measure at 0.

REMARK 10.2. The formula (10.10) can be interpreted as follows: at the limit - i.e. when a -- 0 - the function F(U1) takes at x the value F(A) with probability vx(A). A less heuristic interpretation can be found in (59). From this point of view the above equality vx = 6p conveys no more information than (10.8). So, we turn to the gradient of vs . Regarding the gradient of the minimizing sequences we can show: THEOREM 10.3. Let va be a bounded minimizing sequence of problems (10.1)-

(10.4), i.e. satisfying (10.7). Then Vvs defines a unique Young measure on R2 given in the can of problems (10.1), (10.2) by.

vx=26W-+26W+ ae.xEf, W+=(0,1), W_=(0,-1)

(10.12)

where bw denotes the Dinnc mass or measure at W. In the case of problems (10.3), (10.4) vx =

1

1

26"'-(:,) + 26W+(xl)

a.e. x E i2,

(10.13)

W+(xi) = (O,-Ti), W(XI) = (0,-xi). PROOF. Consider first the case of problems (10.1), (10.2). Recall the definition of 0 given in (10.5). Since ve is a minimizing sequence in both cases when e -- 0,

it,

1-

(Vvv)dx-.0.

Due to (10.10), (10.11) it follows also, up to a subsequence, that

Ji (Vv) dz -+ JJ(Ai)2)dvzi)dx.

10.2. Young measures

149

Thus we have A1,A2)dv1(A) =

r a+

- 1)2dvl(a) = 0

R'

JR2

for almost every x E Q. But then this implies that the support of v1 is contained in the set of zeros of >, - i.e. in { R'-, W+ } where W-, W+ are given by (10.12). Since v1 is a probability measure, we have for a E [0, 11. ci measurable, (1 - a(.r))bµ,+

v1 =

a.e..r E Q.

Next from (10.9),

(uF)1, -. 0 in L"(c)-weak * .

(10.14)

Using again (10.10),

I A2dvs(A) = 0 a.e..r E St R2

which means also

-a(x) + 1 - a(s) = 0 q a(x) =

2

a.e. x E Q.

This completes the proof of (10.12). To prove (10.13) consider again a bounded minimizing sequence of problems (10.3), (10.4) such that Vve defines a measure v1 on R2. Since v, is a minimizing sequence, we have for (10.3), (10.4) 2 2 lim / vex +(vE12 -xi)2dx=0. $

C-0 in

Expanding the second term we get

llm J {veIj+vE1,-2xjv1j+4}dx=0. e-.O " Using (10.10) we derive that If2

JR

2

Aj+(,12-11)2dv1=0 a.e. x E S2.

Thus we derive as above v1 = a(x)bw_(1,) + (1 -

a.e. x E S2 and with W_(x1). W+(xj) defined in (10.13). Next using (10.14) it follows a.e. x E S2

f A2 dv.(A) = 0 a (1 - 2a(.r))xI = 0.

'

If xl 36 0, i.e. a.e. x E Q. we get

a(x) = and the proof is complete.

2

10. Minimizing Sequences

150

REMARK 10.3. The interpretation of (10.12) is the following. At the limit, i.e.

when E -' 0, around each point x Vv, takes only the values W_, W+ and each of them with the same probability 2. Regarding (10.13), we conclude that, at the limit, around each point .r, Vv, takes only the values W_ (x1), W+(xi) each of them with the same probability 2. Together with (10.8) this information will help us to construct the minimizing sequences of these problems.

10.3. Construction of the minimizing sequences For e > 0 let us set TI (X, :C2) = E

x2 12e - X2

1

on (0, E), on (,-, 2E),

and assume uE extended to ]R2 by periodicity of period 2e in the x2 direction. uE is depicted in Figure 10.1.

X2

FIGURE 10.1

Clearly

0 < uE < e,

V UE = W_ or W+

a.e. in Q.

(10.15)

If we denote by vE the restriction to S2 of uE, then

0
where JQJ denotes the measure of 11 and thus vE is a minimizing sequence of (10.2).

This shows at the same time that the infimum is 0.

10.3. Construction of the minimizing sequences

151

Let d(x) be the distance function to the boundary of 1 - i.e. set

d(x) = d(x, 09) = l f Ix - VI

(10.16)

VEOO

where

is the euclidean norm in R2. Set also

yr(x) = ue(x) Ad(x) dx E Sl where A stands for the minimum of two numbers. Clearly vE E Wo'°°(Q). Moreover, due to (10.15), vE(x) = ue(x) V x such that d(x) > e.

Thus if we set (d < el = (X

S

I d(x) < 6),

then

J

L(Vve)dx = f

W(Vv,)dx.

(10.17)

JJ [d<e)

1

Now jVdf, I Vu, l < 1

a.e. in 0

and thus (see (10.5))

0
0<

jtP(VvE)dx < 2 fn xld<e] (x) dx

where X[d<e] denotes the characteristic function of the set Id :5 eJ. Since Xid<.l - 0

a.e. in R when e - 0, it follows from the Lebesgue theorem that

lim eo fn

i(Vve) dx = 0

i.e. vE is a minimizing sequence to (10.1). Note that vE is also a minimizing sequence to (10.2).

We turn now to the construction of a minimizing sequence to (10.3), (10.4). We recall (10.13) that was treated in more detail in Remark 10.3. We are going to cover fZ with squares of size b that will be chosen later (see Figure 10.2). Let Qt be a square of side 6 included in Cl. Let us denote by (xi,x2) its center. Clearly choosing properly the grid covering 11 (see Figure 10.2) one can always assume that xi # 0. Then for a small enough, i.e. such that

e «b,

(10.18)

define uE by ue(x1, x2)

x' Ix2 12.-jxfl-jX'jjX2

on (0, e), on (e,2e),

10. Minimizing Sequences

152

Flour 10.2 and assume us to be extended by periodicity with period 2E in the x2 direction. Denote also by d(x, 0Q1) the distance function to the boundary 8Q, of Q, - i.e.

dist(x, 8Q,) = of Ix - v1. Y Then let us define v, on SZ by

_ (us A dist(x, 8Q,) in Qi, if Q, C SZ, v`

if Q, ¢ SZ.

0

(10.19)

Clearly, due to our definition, vv E W0'OO(SZ).. Let us first estimate, for Q, C fl, vcx1 +

(vx,

- (x1i )2)2 dx.

Qi

Due to the definition of u,,,

0- Ix1 le

it follows that

v` = us and Vve = (0, fxl ). [d < el xi 11 denotes the set

{ x E Qi I di(x, 8Qi) <_ EIxI 11, then

f V.2' + (v =2 - (xi )2)2 dx = qi

f (d<4Ixi0)

(vim - (xi )2)2 dx

10.3. Construction of the minimizing sequences

153

Now

Vve=Vue=(0,±x1)

IVvI=IVd(x, Q)I <1,

or

thus we derive f V'2X3 + (VFy - (xi)2)2 dx < {1 + (1 + (x1)2)2}1[d < __Ixi III

where I I denotes the usual Lebesgue measure in R2. Since Q; has sides of length

6 it follows that (V. - (x1)2)2 dx < 4{1 + (1 + (x1)2)2}Ixile6 < Ce6

(10.20)

, ,

where C is some constant depending on ft only. Let Q;, i = I,-, N be the squares included in SZ and let fZ' = SZ \ U', Q;. Then (recall the definition (10.19) of vs) jo 7 exi + (VI-12 - X2)2 dx

I

=

N

xl dx + i=1

JQ+ Veal

(vex2 - xi)2 dx.

Clearly if ft has a finite perimeter, then

fn -

4dxSC6

where C is some constant depending on ft. Next 2 22 dx 2 vexi + (Vexz - x1)

i=1

Q;

N

j i=1

2 Q.

2

Vexl + (V21.2

i 22 2 2 2)2 - xi- Vea - (Vex2 - (x1) ) dx

N 2

+ s=1

N

r

i=1

JQ.

<

s 2 2

2

VeXj + (VE22 - (x1) ) dx Q.

(v 2 - x1)2 - (vEx - (:e )2)2 dx + NCES

(by (10.20)). But (see the estimate (10.20)) (ve

-x1)2 - (vrx2 - (x1)2)2 < Ixl

-

(xi)211ZVex2 - x1 - (x)21

Ixi -xiIIzi +xUI2+xl +(x1)21 < C6

where C is a constant depending on 0 only. Collecting the above estimates we obtain easily:

in vex + (v2

- xi )2 dx < Co + NCb3 + NCeb

10. Minimizing Sequences

154

for some constant C depending on S1 only. But

N62 < Jill where 10I denotes the measure of St. Thus

- x2)2 dx < c(a + a).

+ (V,2

in

Choosing 6 = e112, which is such that (10.18) holds for a 4K 1, it follows that

JRv62j +(V.2. -xi)2dx
0
10.4. A more elaborate issue We address here a vectorial case i.e., our class of functions is vector valued so that our integrand 0 is defined on matrices. We shall consider here the case of 2 x 2 matrices. Such a matrix is given by

_

(XI 212M

X21

X22

In the x11, x22-plane consider the matrices given by (see Figure 10.3) Wl

= (0 2)

W2

=

(0 01)

W3

=

(-01 2)

W4

= (_o2 1)

It will be also useful to introduce the "intermediate points"

wl=(a i) wz=(o

01)

w4=(ol 1).

One should notice - if one identifies Ws with Wi - that

W,' = 3 W+,+ 3 W{ Vi=l,...,4.

(10.21)

We consider then a function t/i defined on the space M2,2 of 2 x 2 matrices such

that

tp(Wi)=o di=1,...,4, ty(W)>0 VW iAW1, i=1,...,4. Moreover we suppose that 0 is continuous. Then we set WO"- (Q) = Wo'0o(Sl; R2) = { v = (vl, v2) I v; E Wo'0°(S2), i = 1, 2}

(10.22)

10.4. A more elaborate issue

155

A X22

W

1

4'

W4

----------

0

W2

W2

3 6

FIGURE 10.3

and we consider the problem Inf

JsI

O(Vv(x)) dx

(10.23)

where Vv denotes the Jacobian matrix of v given by

v=

8z

and dx the Lebesgue measure on R2. Let us first recall DEFINITION 10.2. Two matrices A, B E Mm,n are said to be rank one compatible if and only

Rk(A-B)<1. Then, it is easy to see that there exist two vectors a, b such that

A-B=a®b=a-J where in the above formula we write the vectors in columns. bT denotes the transposed vector of b. Clearly among the Wi's there is none which is compatible with another.

Then we can state

10. Minimizing Sequences

156

THEOREM 10.4. For the problem introduced above it holds

J 0(Vv(x)) dx = 0,

Inf

(10.24)

tiEwo.00(n} n

moreover the infimum is not achieved.

PROOF. The result is basically due to L. Tartar who introduced this four well problem for another purpose (see [68), [71]). Note that the reader can skip the proof since the fact that the infimum is 0 will be a consequence of the minimizing sequences constructed below. Let us denote by Qt(i(A) the quasiconvexification of

. It is well known that (see [40)) (10.25) -} r'(A + Vv(x)) dx n where f n denotes the average over Il. Moreover Qt, is a rank-1 convex function defined on 2 x 2 matrices (see [40)), in the sense that if A, B are two matrices such

Qtp(A) =

Inf

that rk(A - B) < 1,

(10.26)

then

(10.27) Q,O(tA + (1 - t)B) < tQc&(A) + (1- t)QP(B) Vt E (0, 1). Considering these preliminaries it is clear that (10.24) reduces to showing that (10.28)

Q>G(0) = 0.

We remark immediately that Wi and W{+1 are rank-1 compatible, i.e.

rk(WW-W' 1)=1 11i=1,...,4 with of course the convention Wa = W11. Then from (10.27) we derive (see (10.21))

Qb(W,) <- 3Q&(Wi1+1) + 3Q(Wi) Vi=l,...,4. Noting that by (10.25)

0
Qt(W,) <_ IQO(Wi+1) Vi = 1,...,4, hence

0<

(Wi)

3Qei (WD 5

3Qtb(W)

34QtG(Wi

It follows that

Qt(W!)=0 Then, using rank-1 convexity again, we deduce that Q0 vanishes on the sides of the square with vertices Wi, Wa, Ws, W4 then inside this square (see Figure 10.3). Thus (10.28) and (10.24) follow.

10.4. A more elaborate issue

157

Assume now that this infimum is achieved. Then, there exists u E such that t''(Vu(x)) dx = 0. f1

It follows that

fu(r) = Lt', _ (0 *)

a.e..r E Sl.

In particular 112.,, = 0

u1,r2 = 0,

and - as in the preceding section due to our boundary condition this is possible only in the case where it = 0: but then of course v,(0) = IS2It1'(0) j4 0

in

(see (10.22)).

O

Then we have an analogue to Theorem 10.2 and 10.3. More precisely: THEOREM 10.5. Let uE be a rninzmizing sequence to (10.23) such that there exists a constant C such that aEIX

I

I V 1, 1

/ 1 < C VE:

10. 2 9)

then uE

uniformly in Q.

0

(10.30)

(10.31) in (L'(Sl))'z weak *. Moreover, the sequence of gradient Du, generates a unique Young measure quell by

Vue

0

.1

1

vs = E6w.

(10.32)

1=1

where h1v, denotes the Dirac mass at It-,

(In (10.29) IovEI denotes the euclidean norm of the matrix Cu, and (10.30) means that each component of it, converges toward 0 uniformly.)

PROOF. From (10.29) it is clear that one can extract from - a sequence Tt such that there exists u E IV '"(Q) such that uEk

Vu,,

u

uniformly in Q,

(10.33)

Vu

in (L' (Sl))'f-weak *,

(10.31)

and there exists a parameterized probability measure yr on 1112 2 the space of 2 x 2 matrices such that F(VuEk)

for any continuous F : 1112.2 -. R.

F(A)dv1(A)

in L"(Sl:)-weak *

(10.35)

10. Minimizing Stxluences

158

Applying (10.35) with F = w, and since it, is a minimizing sequence, we obtain

f

0

ff

1.1/,(Vi,,)dx

0

Thus vr is supported by the four wells

q)(.1)dv7(A)dr=0.

Idly 1

recall (10.22)

i.e.

4

V, _ E(ti(s)btt',

(10.36)

1=1

where a i = 1, .... 4 are four functions such that oa(r) = 1

(10.37)

,=1

(recall that v, is a probability measure). Next, applying (10.35) with F(A) the function equal to the different entries of the matrix A we obtain

Vu", - f

Advr(A)

in (L'(52))4-weak * .

Thus combining with (10.34) we obtain

Vu(x) =

f

tr,(x)R',

Adv,.(A) _

A122

(10.38)

,=1

a.e..r E Q. In particular one gets C

8112

dx2

thr1

Thus, since it = 0 on the boundary of S2, it = 0. Hence, since 0 is the limit of any subsequence extracted from it,. of -+ 0

uniformly in S2,

Vu, - 0

in (Lx(S2))4-weak *

.

Going back to (10.38) we have then 0=

P.) 141".

Hence, see the definition of W,:

a1(:r.) +202(x) - o3(x) - 204(x) = 0, 2a1(x) - a2(x) - 203(x) +o4(x) = 0.

(10.39) (10.40)

Even with (10.37) we still are short all equation in o,. In order to find it one uses the continuity of the determinant function in L° (S2)-weak *. In other words (see (10.35))

det(Vu,) - 0 in L'(S2)-weak *

10.4. A more elaborate issue

159

(see [401) and by (10.35) 4

0=

ai (x) det Wi

det A dvv (A) JAM2,2

i=1

from which we derive

2ai(x) - 2a2(x) + 2Q3(x) - 2Q4(x) = 0.

(10.41)

Multiplying (10.39) by 2 and subtracting (10.40) leads to

a2(x) = a4(x). Then subtracting (10.40) from (10.41) we obtain Q3 (x) = a2(x) = Q4(x)-

This and (10.39), (10.37) leads then to 1

al(x) = a2(x) = a3(x) = a4(Z) = 4

0 REMARK 10.4. A consequence of Theorem 10.5 is that - at the limit - any bounded minimizing sequence will have a gradient that takes only the values Ws and this with an equal probability of 4. In the previous section we constructed explicit minimizing sequences to problems (10.1)-(10.4). We repeat this construction for (10.24). However, the problem is here more involved. This is in particular due to the incompatibility of the wells. Let us recall the following observation due to J. Hadamard: PROPOSITION 10.1. Let V be a domain in Rri having two nonempty open con-

nected components V+ and V- (see Figure 10.4).

FIGURE 10.4

10. Minimizing Sequences

160

Let u be a continuous mapping from V into RP such that the derivative of u in the distributional sense in V is given by

Vu = M- in V-.

Vu = M+ in V+,

(10.42)

Nu = ( 9ji ), i = 1, ... , p, 7 = 1, ... , n). Then,

rk(M+ - M-) < 1.

(10.43)

If M+ - M-

0, then V+, V- are separated by a hyperplane with unit normal n and, for some a E RP,

M+ - M- =a(& n.

(10.44)

(When we write the vectors in columns, a ® n is the matrix anT). PROOF. Integrating in V+ and V- respectively, u(x) = M-x + C- in Vu(x) = M+x + C+ in V+, where C+ and C- are two constant vectors of RP. On the boundary between V+ and V-, since u is continuous, one has

(M+ - M- )x + C+ - C- = 0.

(10.45)

Thus, if M+ - M- 96 0, (10.45) defines an affine space that is necessarily a hyperplane. This implies that the rows in the system (10.45) are linearly independent i.e. one has

rk(M+ - M-) = 1 and then

M+-M-=a®n 0

for some a E RP.

In our search for minimizing sequences to (10.24), Theorem 10.5 and Proposition 10.1 together convey some puzzling information: on one hand a minimizing sequence should have its gradient using - at the

limit - only the four wells W each of them with the same probability (see (10.32)), on the other hand it is not possible for a minimizing sequence uE to be such

that Vuc = Wi, Due = W; (i 96 j) in two contiguous regions of Il, due to the incompatibility of the wells Wi (see (10.42)-(10.44)). This causes - as we are about to see - twisted oscillations between the different gradients. Let us fix e > 0. Then, consider the function vl = Vie given by

vl(XI,x2) =

xl 12c-xl

on (0,E) x R, on (e, 2E) x R,

(10.46)

10.4. A more elaborate issue

161

and suppose this function extended to all of R2 by periodicity in the direction xI with period 2e. Set also, (see Figure 10.5), v = (vi, v2) with v2(x1, x2) = VI (x2i x1).

We will denote this function v by v°. Its distribution of gradients is given by

zl

FIGURE 10.5

Figure 10.6 that repeats itself on all of llt2. What we would like to do now is to x2 A

2e

W

W

3'

2'

£

W

W4'

1'

0

FIGURE 10.6

modify v° on each of the squares of the lattice of size £ in order to have a new function v1 with more of its gradients on the wells. We proceed first on the square Q = (0. £)Z where Vv° = Wi. First, remark that 2

1

Wi=3W1+3W2.

162

Moreover, if v°

10. Minimizing Sequences

v2) then 14 (X) = v4(xl,x2) = 22.

The function v2 goes up with a slope 1. Let us replace it by a function that makes oscillations going up repeatedly with a slope 2 and then down with a slope -1 (see Figure 10.7). More precisely, let X be the periodic function of period 1 defined on (0, 1) by 1

X(B)

when 0 E (0, s )

0 when 0 E (2, 1).

W2,v2

i 0

elk

11

C

22

FIGURE 10.7

Then, for k E N' to be determined later on, set on Q = (0, e)2 Ez U'2(x)=U4(x2)=kJ0 (3X(0)-1)d0. It is clear that almost everywhere Vio2(x) = (0, 2) or (0, -1)

(10.47)

and W2 is the function depicted in Figure 10.7. If now 81Q denotes the part of the boundary of Q located on the set {xl = 0} U {x1 = el we set w2(x) = w2 A (v2

81Q))

in Q

where diet(., 81Q) is the distance function to 81Q and A the minimum between functions. We finally set v1(2) = (vi(x),w2(x))

in Q.

162

Moreover, if v°

10. Minimizing Sequences

v2) then 14 (X) = v4(xl,x2) = 22.

The function v2 goes up with a slope 1. Let us replace it by a function that makes oscillations going up repeatedly with a slope 2 and then down with a slope -1 (see Figure 10.7). More precisely, let X be the periodic function of period 1 defined on (0, 1) by 1

X(B)

when 0 E (0, s )

0 when 0 E (2, 1).

W2,v2

i 0

elk

11

C

22

FIGURE 10.7

Then, for k E N' to be determined later on, set on Q = (0, e)2 Ez U'2(x)=U4(x2)=kJ0 (3X(0)-1)d0. It is clear that almost everywhere Vio2(x) = (0, 2) or (0, -1)

(10.47)

and W2 is the function depicted in Figure 10.7. If now 81Q denotes the part of the boundary of Q located on the set {xl = 0} U {x1 = el we set w2(x) = w2 A (v2

81Q))

in Q

where diet(., 81Q) is the distance function to 81Q and A the minimum between functions. We finally set v1(2) = (vi(x),w2(x))

in Q.

10. Minimizing Sequences

164 X2

e

a FIGURE 10.10

function obtained. It is clear that on each of the squares of the lattice of size c where we have VvO = W; , we can proceed also to p laminations, again calling the function obtained that way vP (p will be fixed later on). Finally let us denote also by vP the restriction of vP to 12. Clearly vP does not vanish on the boundary 1' of SZ - so set

uP = tt' Let us denote by A, the set

r) V VP A

r).

(10.48)

It is clear that the gradient of uP = uP is uniformly bounded independently of E. So, recall (10.22), since tP is continuous, one hasr for some constant C

r tk(VuP) dx < C J

a dx.

in

(10.49)

Let us now estimate this last integral. First, due to the truncation (10.48) A. contains a neighborhood of the boundary of 11. More precisely

0
-e
(10.50)

Set

BE= {xEfl dist(x,F)<2E},

Q

Wi,i=1,...,4

(10.51)

10.4. A more elaborate issue

165

It follows from (10.48), (10.50) that for dist(x, r) > le, up = vP so that

A.cBrUC. end the inequality (10.49) implies

r f rb(Vu')dx
(10.52)

n n where XB. denotes the characteristic function of Ba and I I the usual Lebesgue

measure. To estimate the measure of CC we remark first that during the ntblamination we have, on the border of each square of side Mf- --T , a set of bad gradients (i.e. a set of points x where VvP # W;) of measure e _ 4e2 2 e

(10.53)

.

3k(3k)2(n-1) 23k (3k nn-1 (3k nn-1 Now, if M denotes the measure of a big square containing f2 whose sides are on

lines of the lattice xl = zle, x2 = z2 for some z;, it is dear that the number N, of squares affected by the first lamination is such that

N,e2 < M a N1 <

M.

(10.54)

Regarding the second lamination, the number N2 of squares affected is

Arguing by induction it is clear that the number N,a of squares where the ntblamination occurs is (10.55) Nn = Nn-,k(k - 2):5 N,k2(n-r). Combining (10.53)-(10.55) we see that the set of bad gradients located on the boundary of the squares used during the p laminations has a measure bounded by P

2 (n-1)<2k 3kEn=1 31

(10.56)

Now it is clear that, except for those gradients not on the wells and located on the boundary of the squares where lamination occurs,

vvP=w; 0w; an a set of measure bounded by (10.57)

3P 1111-

Combining (10.52), (10.56), (10.57) we obtain for some new constant C: t


Due to the Lebesgue convergence theorem it is clear that

lim f Xa.dx=0. n

J

10. Minimizing Sequences

166

Moreover, selecting k = k(E), p = p(E) converging toward +oo when e - 0, we obtain a minimizing sequence to our problem.

10.5. Numerical analysis of oscillations In this section we show how minimizing sequences can be obtained numerically. This is of course an important issue. We address only a model problem, referring the reader to [58], [18], [21], [24] 1231, [281, [391, [54], [551, [55] for more.

Consider for instance the problem (see [SS])

Inf

JlVxJ+IIvz2H1kzidx2

(10.58)

t

where fl is a bounded open subset of R2 and W01"(12) is as before the set of Lipschitz continuous functions vanishing on the boundary of ft. Our density of energy is of the type

J in

*(Vv) dxl dx2

where Vv = (vx, , vx,) and to is the function defined by

'V) =O(C1,6) = I6I + II6I - 11-

(10-59)

This function is not smooth (see Figure 10.1), but this does not affect the generality of our results.

FIGURE 10.11

First let us show that this problem is like the ones encountered in Section 10.1. In fact one could avoid what follows by introducing a general framework for (10.58) and the problems of Section 10.1 (see [21]). But we repeat the argument of Section 10.1 for didactic purposes. First we state THEOREM 10.6. The infimum of (10.58) is equal to 0. PROOF. Let £ be positive fixed. Then consider the function - periodic of period 2e in the x2 direction - defined by ue(XI, x2)

x2

2F - x2

on R x 10, E], on R x [E, 2e].

(10.60)

10.5. Numerical analysis of oscillations

167

Clearly one has Vu, = (0, ±1) so that tk(Vu.) = 0 a.e. in f2. However this function does not belong to 4t "(fl). So, modify it slightly by setting

t)

u, = u, A where

(10.61)

denotes the distance of a point to the boundary r of Q. Then IVucI < 1 a.e. in 0, (10.62)

and also since

0 e.

(10.63) (10.64)

Thus, if Al denote a constant such that

OW = I6I+II6I-11 <M for every , such that .sl

1, then v;(Dur) dx1 dx2 < M

rn Xld
(10.65)

where [d < EJ = { .r I dist (x, t) < E } denotes the set of points in 11 whose distance

from the boundary is smaller than a and Xl.l the characteristic function of sets. Clearly, XId<EJ converges toward 0 a.e. in f2 and thus, by the Lebesgue Theorem, passing to the limit in (10.65) one gets lim

C-ofa

O(Vi) dx1 dx2 < 0.

(10.66)

Since, l' > 0, it is clear that the infimum (10.58) is non-negative, so combined with (10.66) we see that it is 0. REMARK 10.5. The sequence ue is a minimizing sequence of the problem.

Now, let us show that the infimum (10.58) is never achieved. Indeed one can

state THEOREM 10.7. The infimum (10.58) is never achieved.

PROOF. Assume that there is a u such that

JIux1I +IIussI-lldx1dx2=0; then ux, = 0 = u = 0. But for u = 0 the integrand above reads (if IS1I denotes the measure of ft), n

dx1dz2=A

0. 11

REMARK 10.6. If the problem has a minimizer, then it is 0. So, as in previous sections, 0 has to play some kind of role here. This is clarified in the next theorem.

10. Minimizing Sequences

168

THEOREM 10.8. Let u£ be a minimizing sequence of (10.58) such that IIVII, I L < C

IuE100,

(10.67)

YE,

then uE converges uniformly toward 0.

PROOF. Let uE be a sequence satisfying (10.67). Then, by compactness of the set of Lipschitz continuous functions in the set of continuous functions, we can extract a sequence from e which we will still label E, such that

u£ - u

uniformly.

(10.68)

From the fact that u, is a minimizing sequence, f(Vue)dxd11 > J1UE.rlidxtdx2

0.

(10.69)

1

2

But due to the Poincare Inequality

fluids < C

Jn

u,., dx,

uE - 0 in L' (a). Thus from (10.68) we deduce that u = 0 (note that uniform convergence implies convergence in L'(SZ)). Since every subsequence of uE has to converge toward 0, the whole sequence uE converges toward 0. 0 Now the question is What happens to Vu,'? Clearly from (10.67) one can again extract a subsequence such that

Vu,,

I

in L' (Sj)2 -weak

Now, due to Theorem 10.8,

VuE - 0 in D'(11)2 so that I = 0 and

Vu, - 0 in L' (p)2 -weak

(10.70)

Of course, as we have seen in Theorem 10.7 uE -+ 0 uniformly, Vu, - 0 in L0c(S2)2-weak *

is not enough to insure that 0 is a minimizer for our problem. In other words it is not the case that

JIoi/'(0)dx.

lies

n +x fn r'(Vu£)dx

(10.71)

But regarding the behaviour of the minimizing sequences we can state THEOREM 10.9. Let u£ be a minimizing sequence of (10.58) satisfying (10.67).

Then Vuf generates a unique Young measure yr and 1

v,r = 1hvs" +

1

where 6w* denotes the Dirac mass of the point Wf = (0, ±1).

(10.72)

10.5. Numerical analysis of oscillations

169

PROOF. Due to (10.67) and Definition 10.1, there exists a probability measure on R1, vy(A) such that - up to an extracted sequence - for any continuous function F : R2 --+ R we have in L' (0)-weak *

F(Vu,) -

j. F(A) dvx(A).

(10.73)

lying (10.73) with F = +/ leads to (note that ua is minimizing) fit2O(A)dvx(A)=0

a.e.xER.

%nce 0 vanishes only on W+ and W_, it is clear that vx is supported by W+, W_, i.e.

vx = a(x)bw+ + (1 - a(x))6w_

(10.74)

for a.e. x E S2. Recall that vy is a probability measure. Next we use the fact that

&Ua - 0 in L'(0)-weak

*.

We derive a

O=f

Azdvs(A) = a(x) - (1 - a(x)}. 2

It follows of course that

a(x)=2 a.e.xE0, and the result follows from (10.74).

REMARK 10.7. In the case where u is a minimizer of some problem of the calculus of variations, one can write in tP(Vu(x)) dx

J L io(A) d6v,.(x)(A) dx

= ewe bvu(x) denotes the Dirac measure at the point Vu(x). In the case of no minimizer but uniqueness of Young measure this measure can be considered as e, minimizer in some class of measures (the ones generated by gradients). Also vecall that an important meaning of the uniqueness result (10.72) is that every

$iiAimizing sequence - at the limit - will have a gradient that takes only the '4alus W+, W_, each of them with the same probability J. Indeed recall that rr..(A) is the probability that at the limit the sequence Vu, takes the value A. For results in this direction we refer also to [25], (59]. We shall also explain this concept from a numerical point of view. We now pass to the approximation of problem (10.58) (see (39]). So, for this purpose we will assume that Cl is a polygonal domain of R2. As shown in [22] the estimates that are obtained for this kind of problems are somehow independent of the class of finite element considered, and we will restrict ourselves to Pl finite

10. Minimizing Sequences

170

elements. So, let us consider rh a family of triangulations of Q. If K denotes a triangle of rh, recall that the mesh size h is given by

h= Maxdiam(K) KErh

(10.75)

where diam(K) denotes the diameter of K - i.e. the size of the longest side of K. We will assume rh regular in the sense that when h 0 the triangles are not allowed to flatten - i.e. there exists a v such that for any angle 0K of K E rh

v-eK

(10.76)

independently of It and K. Then we introduce

0 ={v:0

RIvis continuous, VI KEP,dKErh}.

(10.77)

(vIK denotes the restriction of v to K, Pl the space of polynomial of degree 1 in R' i.e. P1 = { a + bxl + cz2 I a, b, c E R }. Vh is the set. of continuous functions affine on each triangle of the triangulation.) Moreover we define

VJ`={vEV"Iv=0on Xl}.

(10.78)

(Note that in Chapter 8, Vh was denoted by Vh and Vol` by Vh. For clear didactic reasons we change our notation here). Then the discrete analogue of (10.58) is

lnf vEVnh

J In,

1vr, I + I Iv,., I - I d:r.

(10.79)

As we are about to see - and in contrast with (10.58) this problem admits a minimizer. First recall (see Figure 10.12) that the dimension of Vhis exactly

FIGURE 10.12

10.5. Numerical analysis of oscillations

171

equal to the number M of vertices of the triangulation inside Cl. Indeed if n i = 1, ... , M denote these vertices, and if Ai denotes the function of Yh such that Ai(nj) = bid

di, j

(10.80)

bi, = 0 or 1 if i 34 j or i = j, then clearly the Ai's constitute a basis of Vh and for any v E V we can write (10.81)

(see Proposition 8.1). Therefore, for v E Vh

if,

Ivx, l + I Iv=z l - 1I dx = J(v1, .... vM1

(10.82)

where J is a continuous function on RM. Then we can state THEOREM 10.10. The problem

Inf f Iv,I+Iiv.2l-1Idx

vEVo 0

(10.83)

admits a minimizer. PROOF. Due to (10.82) we have to minimize a continuous function J on RM. By the Poincare inequality we have for some constant c

Ivx,Idx>cJ Ivldx

in

(10.84)

for any v E Wo'1(St). Now,

v'-'

f

lvi dx = livIl

(10.85)

is a norm on Va (this space is of finite dimension and all the norms defined on it are equivalent). Thus from (10.82) we get J(v1i...,VM1 > Cllvll

- +oo

when IIvII - +oc. Thus, to minimize J on RM it is enough to minimize it on a compact set of RM where the minimum is achieved since J is continuous. This completes the proof of the theorem.

Next we can state THEOREM 10.11. There exists a constant C > 0, independent of h 4Z 1, such that

Inf f lu=1I+IIvZZI-1Idx
vEVo 0

(10.86)

10. Minimizing Sequences

172

PROOF. Let a E (0, 1) be a number that we will fix later on. Define

x2) =

if x2 E (0, h"1.

1X2

(10.87)

2h" - x2 if x2 E (h", 2h"1,

and suppose uh to be extended by periodicity in the .r.2 direction with period 2h". Note that when h K 1, h « h' see also Figure 10.12. If we recall that

W_ = (0, -1),

W+ = (0,1).

(10.88)

then clearly

r/'(Vuh) ='r/'(W±) = 0 a.e..r E R2.

(10.89)

If uh denotes the restriction of uh to S2, then Uh

l/0,

since Uh does not vanish on the boundary of 1) and since also Uh is not necessarily affine on each triangle of 1h. Indeed such a triangle can intersect a line r2 = kh" in a neighbourhood of which Uh is not affine see Figure 10.12. So, one sets a,, = u,, A dist(x, Oft)

where A denotes the infimum between numbers and

u,, = the interpolant of it,,.

(10.90)

The interpolant of idh is the function of Vh that coincides with u,, at the nodes of the triangulation - i.e.

uh =

(10.91)

11h(70A,.

Clearly ul, E V0'. Moreover, from (10.87) it follows that

0
for dist(:c, OS1) > h" + It.

(10.92)

Since Th is assumed to be regular and since IVuh1, IVd(x,ast)I

1,

it follows that for sortie constant C independent of It (see [211)

Ipu,,I < C.

(10.93)

Thus -j(Vu)dx for some other constant - see (10.89) CI[uh

uhIwhere

uhJ = {x E

f1h ut

is the Lebesgue measure. Clearly see (10.90), (10.92) the above set is contained in a neighbourhood of size h"+h of the boundary or in a neighbourhood

10.5. Numerical analysis of oscillations

173

of size h around each of the lines x2 = kh°. Thus it follows that for some constant

fn

I/'(Vith) dx < C(h° + h) + N.2h.D

(10.94)

where D denotes the diameter of ft, N the number of lines x2 = kh° cutting 12. It is clear that

(N - 1)h° < D, i.e. if we choose h° < D, then

Nh° < 2D. Since for It << 1. h° >> h, (10.94) becomes for some constant C

Jo

IO(Vuh) dx < C{h° + hl-°}

and the result follows by choosing a = s - i.e. Info

JQ

z/'(Vv) dx < in 'P(Vuh) dx < Chl/2vEVo

REMARK 10.8. Clearly uh is a minimizing sequence to the problem (10.58). Producing such minimizing sequences is the main interest of estimates like (10.86). Compare also this minimizing sequence to the one exhibited in Theorem 10.6. 1

ih'=h/2 0

1

FIGURE 10.13

REMARK 10.9. Note first that for some particular triangulations - related to (10.86) could be improved. Indeed, choose for instance the triangulation of Figure 10.13 and set the wells

u(xl, x2)

rx2

on (0,h')

= 12h' - x2 on (h', 2h').

10. Minimizing Sequences

174

ci is here the unit square. Extending u periodically with period 211', and setting fil, = interpolant of u A dist(, 0Q) one clearly has Inf

Vh

vE a

Jt

(V v) dx <

f

dx dy

Ch << Ch 112

n

for h small. However, this estimate depends on the triangulation, as we are about to see.

We show now that the estimate (10.86) is sharp in the sense that, for some family of triangulations an estimate from below in h''2 holds in such a way that the exponent z in (10.86) cannot be improved - except as we just saw for some very peculiar triangulation.

xo

U

t

FIGURE 10.14

For simplicity, assume now that Q = (0, 1)2 and consider the family of triangulations depicted in Figure 10.14. Then we can state THEOREM 10.12. (see [33)). Under the above assumptions there exists a constant c > 0 independent of h << such that

cht/2< Inf J IvX,I+IIvx,I-iidx,. vE Vn'

n

(10.95)

The proof will be constructed through several lenunas. For x0 = kh/2, k E N we are going to consider the strip (xo, x() + 2) x (0, 1). Let u E VOh and 0 < b < 3. Recall that Vu is constant on each of the triangles of r,,. Then for a given u let us adopt the following definition:

10.5. Numerical analysis of oscillations

175

DEFINITION 10.3. Let T E Trh with one side at least of length h. We will say

that T is of

lux, < 6, Iux, - 11 < 6, lux,l<6, lux,+1I<6,

type + if

type- if type 0

else.

In other words T is of type + or - if Vu is close to W+ or W_ respectively. Then on the strip (xo, xo + 2) x (0,1) we have triangles of various types. Let

No = No(xo) be the number of triangles of type 0. We say that we have a change of phase along the line x0 when two triangles change from type + to type - (see Figure 10.15). Then we can prove,

LEMMA 10.1. At each change of phase along the line x = xo, the triangle having only one vertex on xI = xo is of type 0 (see Figure 10.15).

FIGURE 10.15

PROOF. Let us denote by (u,,+,, u.+2), (u.-,, use) the gradients of u in the triangle of type + and of type -, respectively. If A = (xo, yo), then (without loss of generality we can suppose that we are in the case of Figure 10.15) h hl +h +h

+ux,2,

u xo+ 2,1/0+2 J u (xo +

h

h

h

h

2, yo - 2 = u(xo, yo) + U.-' 2 - u=, 2 .

So, in the triangle whose only vertex on x = xo is A, ( u + u- ).

u,2 = 2 ( u x i - uy,) +

2

10. Minimizing Sequences

176

Since u +, u;1 E (-6,b], u z E [1-5,1+6], u;2 E (-1-b,-1+6] one deduces that luy, I < 26 so that

Iu,,±11>1-26>6,

since 0 < b < 1. The considered triangle is thus of type 0. Then we can show

LEMMA 10.2. Let u be a piecewise C' function on (a,b) such that u' has a constant sign and lu'I > C. Then

f Ju(z)l dz > C\(b 4a)2}. PROOF. If u does not vanish on [a, b], then a

Iu(z)I dz > min (1 b lu(z) - u(a)J dz,

`a

r

Ju(z) - u(b) I dz ).

/

a

So, considering possibly u - u(a) or u - u(b), instead of u we can assume that u vanishes at some point {. Since u' has a constant sign,

lu(z)1=Iu()+Icz u'(z')dz'I= J Ju'(z)Idz>CIA-zl. t,=1

Integration in z on (a, b) leads to the result. LEMMA 10.3. Assume that (No + 1)h < Z, then Iu(xo,x2)Idx2>

a

1-a

(10.96)

16(No+1)

PROOF. Consider a maximal chain of triangles of type + or -. In other words

consider (a b;) such that there is a change of phase or a "boundary point" at (xo, aj ), (xo, b;) and all sides of triangles of the strip (xo, xo + ) x (0, i) located on x = x0 between a; and b, belong to triangles of the same type + or -. (A "boundary point" could be a vertex of one triangle of the boundary with all its sides strictly smaller than h). Since u,,, has a constant sign and luy, I > 1 - b, we deduce from Lemma 10.2 that

f

bi

lu(xo, x2)1 dx2 > (I - S) (b'

4a)2

(10.97)

Denote by N1 the number of maximal chains as above, i.e. i = 1, ... , N1. Then from (10.97) we deduce

(1 - b)Ni

N

4

2

(b, _ a)2

and by a convexity argument Ni

f 1 Iu(xo,x2)ldx2 >o

J

)N'

Iu(xo,x2)Idx2 > (1

4

N

()2(10.98)

10.5. Numerical analysis of oscillations

177

Due to Lemma 10.1 the sides of triangles on x1 = xo which do not belong to a chain belong to a triangle of type 0 or a small triangle of the boundary. Therefore N,

Ebi - ai> 1 -(No+1)h> i=1

1

2

and from (10.98) we deduce that Iu(x0 x2)Idx2 >

A

(1-6) 16N1

The result follows from the fact that, by Lemma 10.1, N1 < No + 1. LEMMA 10.4. Let U E Vi`, then

j(Vu)dx.

Iu(xo,x2)Idx2 <-

(10.99)

fo1

PROOF. We see that o

u(x0.x2) = u(0, x2) +

170

x2) d =

ux1

10--u"

x2) dt.

Thus 111(xo,x2)I <

J0

1

Iux,(S,x2)Id

<-

J0

1

0

and the result follows by integration in x2.

LEMMA 10.5 (Estimate of the number of changes of phase). Let u E Vo such that TG(Du) dx < Ch 1/2

(10.100)

where C is a constant. If h < 64C2 then one has

No+1>

1-6 h-1/ 2 1

(10 101) .

6C

(Recall that X) is the number of triangles of type 0 in the strip (xo, xo + 2) x (0,1)).

PROOF. If (10.101) does not hold, then

(No + 1)h <

(1 - b)h1/2 16C

h1/2

< 16C

<

1

2'

Thus combining (10.96), (10.99) and (10.100) we obtain

1b 16(No + 1)

which leads to (10.101).

1

< A Iu(xo, x2)I dx2 < Ch' /2

10. Minimizing Sequences

178

END OF THE PROOF OF THEOREM 10.12. Knowing that (10.86) holds, con-

sider a it E Vh such that

Z

O(Vu) dx < Ch''2.

(10.102)

On a triangle of type 0 for such a u,

lur,

Iur, I? h or

i.e.:'(Vu) > b. Thus if S,z, = (.rp.xp + 2 x (0, 1), then fz/,(Vu) dx >

Assuming that h is small more precisely

2.

(l_6)h

that

1/2

32C.

>I

we deduce from (10.101)

1 -6 No ? 32C't '/2 which leads to

fro

/,(Du) dx >

Since the number of stripes Sr is

(1 - b)hh2/2 64C

h, we deduce

V,(Vu)dx >

that

(1 - 6)bh'l2 32C

si

which completes the proof.

REMARK 10.10. The exponent. 1/2 appearing here is related to the linear growth of 0. Other growth conditions would produce other exponents. Our argument extends also to general polygonal domains and more general functioiials. For all this we refer to 1321.

Let us denote by I

I

i , the norm on R2 given by I(a,b)II = Ial + Ibl.

Moreover, define by x the following projection: 7r (a, b)

whenb > 0,

- W+ W_ when b < 0.

(10.103)

It is easy to see that I(a,b) - 7r(a,b)It = *(a, b)

V(a,b) E R 2

(10.104)

10.5. Numerical analysis of oscillations

179

Let us now denote by Q an arbitrary square with sides parallel to the axes - see

Figure 10.16-and for 0<6<1,vEWa'°°(n)define Q+={xEQI IVv(x)-W+I1 <6}, Q- = {x E Q I IVv(x) - W_I1 <

},

(10.105)

Q°={xEQI x0Q-UQ+}. One should notice that Q+, Q-, Q0 are disjoint subsets of Q and that

Q+uQ-uQ°=Q. So - if I

I

is the Lebesgue measure IQ+1

+

IQ

IQI

= 1

I

IQ°I

IQI

(10.106)

1QI

Next we can state PROPOSITION 10.2. Let V E Wa'OO(c1) and Q° be defined as above. Then

WI <

1 /' P(Vv(x)) dx.

IQI - 61QI Jn PROOF. Note that on Q°,

(10.107)

6 < IVv(x) - W+I1, IVv(x) - W-11Thus by (10.104)

6 < O(Vv(x)). Then

>

in

LP(Vv(x)) dx > 61Q°I

and the resu lt follows by dividing both sides of the inequality by IQI. Next note the following PROPOSITION 10.3. Let V E Wo'°°(11). Then

/ Iv(x)Ido(x) <

f Iv=I(x)Idx.

(10.108)

(8Q denotes the boundary of Q, do(x) the measure area on 8Q). PROOF. This is an easy trace theorem. Consider, with the notation of Figure 10.16, x = (x1, x2) E 8Q, y = (y1, x2) E t. Since v vanishes on the boundary of Il, v(x1, x2) = -

vx, (8, x2) ds.

Thus we get:

Iv(x1,x2)I < j

Iv:,(8,xa)I do.

10. Minimizing Sequences

180

a

0

X1

FIGURE 10.16

Integrating in X2 E (a, b) we obtain fb

Ja

1b

Iv(r1,x2)Idx2 <_ n

d.sdx2

1

0

and the result follows by integrating on the other sides of 8Q. Then we can show PROPOSITION 10.4. Let Q be as above. Then for v E

there exists a

constant C = C(b) independent of v and Q such that

IQ+I_1 < IQI

IQI

2

o(Vv(x)) dx,

(10.109)

C f il'(Vv(x)) dx. IQI

(10.110)

IQI Jo

2

IQ-I1

C

<

PROOF. We know that fQ

Vv - ir(Vv) =

J,IQ

1, v&(x) - IQ+IW+ - IQ IW_ fQ n(Vv(x))dx.

(v denotes the unit outward normal to OQ). This can also be written

IQ+IW+ + IQ IW_ _ J v vdd(x) - f Vv - 7r(Vv) aq

(10.111)

7r(Vv(x)) d.r. QO

10.5. Numerical analysis of oscillations

181

Now the vector V = (V1, V2) on the right-hand side of the above inequality is such

that IV21 < IVII <2 18Q Iv1do(x)+ f IVv-7r(Vv)I1+IQ°I

-

a

< 2 fn Iv ,, (x)1 dx + j ?P(Vv(x)) dx + f

V'(Vv(x)) dx.

(We used Propositions 10.2, 10.3). Thus IV21 < 2ln Iv x, (x)I dx + (1

+ s) Ib(Vv(x)) dx (10 . 112)

1

< (+_) bf O(Vv(x)) dx.

n Going back to (10.111) we see that

IQ+1

1Q-1

V2

IQI

IQI

1Q1

Combining with (10.106) we obtain IQ+1

1=IV2

IQ°1

21 1QI

1Q1

2

IQI

IQ-1-1=-l 2

IQI

V2

IQ°I

2l IQI + IQI I. 0

The result follows from (10.107), (10.112).

REMARK 10.11. is the probability for Vv to have its gradient on Q in a 6-neighbourhood of W+, W_ respectively. It follows that if in (10.109), (10.110) one considers a minimizing sequence, then the probabilities go to!' This is another interpretation of Theorem 10.9.

One can then show:

THEORLX1 10.13. Let vh be a minimizer to (10.83). Set - for Q a square as above:

Qr+, ={xEQI IVvh-W+I
Qh=Q\Qh

II

then there exists a constant C = C(b) independent of Q and h such that IQI

2I' I IQI

-

1

'Qh-<_ Ch,/z

2I'

IQI

IQI

PROOF. This is a trivial consequence of the above results.

0

10. Minimizing Sequences

182

10.6. Exercises 1. Let i2 = (0,1)2. With the notation of Section 10.1, we consider the problem

f

In vEWQ LM) fl

t,.2,

+ (vx2 - 1)2(vv3 - 4)2 dx.

Show that the above infimum is equal to 0. Show that it is not achieved. Determine all the Young measures generated by the minimizing sequences. 2. Let W1i W2, W3 be three vectors in R2 such that 0=a1W1+a2W2+a3W3 for some a, E [0,11, E3i=1 a, = 1. Let 0 be a function such that

rp(W{) =0 V Show that the problem Inf

vEWV (fl)

J((Vv)dz

has infinitely many minimizers. 3. Let W1 i W2,- - -, W E R". Suppose that 0 E Co(WW) the convex hull of these vectors and that

Wig60 Vi. Let

R"

R+ be a function such that

¢(WW)=0 Vi=1,...,n,

+p(W)>0 VW#Wi.

Show that for Sl a bounded open set of R":

f

Inf

v(x)) dx = 0.

vEW.--(a) l Show that this lnfimum is not achieved. Assuming 0 polygonal and considering a triangulation of fl, show that for some constant C (with an obvious definition for 0 )

Inf J V(Vv(x)) dx < Ch'12 vE Vol'

l

(see [211 and Chapter 8).

4. The notation is the one of Section 10.4. Let A be the 2 x 2 matrix

A_(all 0

0 a22

with a11ia22 E [-1,1). Set

WA"' (fl) = { v = (v1, v2) I vi E W1'°°(0), i =1, 2, v(x)r = Ax on r). If ap is a function satisfying (10.22) show that Inf

vEW(0) J

(Vv(x)) dx = 0.

10.6. Exercises

183

Show that this infimum is not achieved and construct the corresponding minimizing sequences. 5. Let Q be the unit cube of 1EY2 given by Q = [0,112. Show that I

vdx=J v(xi,0)dx1+fol v(xl,1)dxi+ faQ

I

0 1

+

v(1,x2)d.xz 0

(see Proposition 10.3).

f l v(O,x2)dx2

Chapter 11

Linear Parabolic Equations 11.1. Introduction The problems we shall solve now are of the so-called heat equation type (see chap. 1). That is to say, 1 being a domain of R". we would like, for instance, to find a function u(x,t) such that

ui - Du = f

in S1 x (O, T),

u(.r, t) = 0 on t x (0, T), tu(.r,0) = u0(X) on Q.

(11.1)

f (x. t). u0(x) are two given data. A strong solution to (11.1) could be a function it such that all the above equalities hold in a usual sense. Clearly as we did already for the Dirichlet problem the second equation of (11.1) can also be interpreted as (11.2)

E Ha(S1).

Moreover - assuming that all the integrations that we are performing make sense - if for v E Ho (Q) we multiply the first equation of (11.1) by v and integrate over 0 we get:

jutvdx+Jvu.Vvd.r=ff.i'd.r

Vi E H(S1).

rl

Introducing

(u. v) = / u.vdx, fn

Vu.Vt'dr.

((u, i')) st

the above equality reads t (u, r) + ((u, v))

v)

V v E H( (f1).

(11.3)

So, we can weaken our initial problem and look for u such that (11.3) holds for instance in E)'(0, T) see Chapter 2. It is true that (11.3) makes perfect sense for smooth functions. However, if one wants to attack the problem from the point of view of functional analysis, one sees that some new spaces have to be introduced. This is what we shall do now.

11.2. Functional analysis for parabolic problems Let us denote by X a Banach space. with norm 11 jjx.

11. Linear Parabolic Equations

186

11.2.1. LP spaces. For the notions below a good reference is the appendix of [16] and the bibliography there. DEFINITION 11.1. For a, b E R we denote by (11.4) L"(a,b;X), 1
11f(t)[IXdt <+oo,

{11.5)

and by LO (a, b; X) the space of functions essentially bounded on (a, b) - i.e. such

that there exists M such that II f (t) I [x < M a.e. t E (a, b).

(11.6)

Then it is easy to show: THEOREM 11.1. When equipped with the norms i/p f 1
`a

Ill IIL."(a,b;X) =Inf{M ER I IIf(t)IIx <- Ma.e. t E (a,b)}, the spaces LP(a, b; X), 1 < p < +oo are Banach spaces. REMARK 11.1. For every f E LP(a, b; X) we can define

fbf(t)dtEX (see [161)-

11.2.2. Vector valued distributions. Let -oo < a < b < +oo. DEFINITION 11.2. A vector valued distribution on (a, b) with values in X is a linear mapping

V(a,b)-IX, ,P -' (T, cP),

continuous on D(a, b) in the sense that f

lim (T, cPf) _ (T" P) +00

in X

for any sequence Wi such that caj - cp in D(a, b) (see Chapter 2). EXAMPLES 11.1.

1. Let u E LL (a, b; X); then +P - r u(s)co(s) ds a

is a distribution on (a, b) with values in X. We say that this is the distribution defined by u. Two distinct functions define the same distribution if they agree almost everywhere.

11.2. Functional analysis for parabolic problems

187

2. ForxEX, cp- W(0)x is a distribution with values in X for any interval (a, b) containing 0 - this is a Dirac mass at 0. NOTATION. We will denote by D'(a, b; X) the space of distributions on (a, b) with values in X.

DEFINITION 11.3. Let T E D'(a, b; X). Then for every integer k we define a distribution T(k) by the formula

/

k

(T(kl,(p) _ (-1)k (T, dtk ).

(11.7)

T(k) is called the k-th derivative of T on (a, b) and will also be denoted dkT

T() =

dtk

EXAMPLES. Let u E L'(a,b;X). Then u(') will also be denoted u' or ut if t is the variable in (a, b). If X = L' (S2) for some domain £ of R", then u E L' (a, b; L' (f 1)) can be identified with a function of two variables by the formula

u(t)(x) = u(x, t). We can easily show that, for smooth functions,

ut(t)(x) = 8t (x,t), thus we shall use both notations. REMARK 11.2. If X, Y are two Banach spaces such that

X y Y (continuous embedding),

(11.8)

then clearly

D'(a,b;X)

D'(a,b;Y)

and

LY(a, b; X)

L'(a, b; Y) d l < p < +oo.

DEFINITION 11.4. Let X, Y be as in (11.8). Let it E Ll (a, b; X ). We say that u' E L2(a, b; Y) if there exists v E L2(a, b; Y) such that

- Ia

b

u(t)cp'(t) dt = f b

dt dE D(a, b)

(11.9)

(this equality takes place in Y). In what follows we will have to consider the situation of two Hilbert spaces V,

H such that

V- H'-'V',

(11.10)

V dense in H,

(11.11)

11. Linear Parabolic Equations

188

where V' is the dual of V. We will denote by

(( )), (

),

I

II respectively the scalar product and the norm in V, I respectively the scalar product and the norm in H. II

As explained in the introduction of this chapter a suitable choice could be:

HH(f) C L2(Q) C H-1(0). Then we can define

H1(a,b;V,V') _ {u E L2(a,b;V)Iut E L2(a,b; V')}.

(11.12)

To define a Hilbertian norm in V' by the Riesz representation theorem, we identify V' and V through the formula

(f,v)v-,v = V, V)) and we set

IfIV,=II/11=IIIV Of course this norm is equivalent to the norm of strong dual. Now we can state THEOREM 11.2. HI (a, b; V, V') is a Hilbert space for the norm IIUIIi = IIUIIL2(a,b;V) + IIUtIIL2(a.b;V1)

(11.13)

PROOF. Let un be a Cauchy sequence in H'(a, b; V, V'). Then un is a Cauchy sequence in L2(a, b; V) and, (un)t is a Cauchy sequence in L2(a, b, V'). Thus, there exist u E L2 (a, b; V), u' E L2 (a, b; V') such that

(un)t - u' in L2(a, b; V').

un - u in L2 (a, b; V)

Of course un - u in D'(a, b, V) and in D'(a, b, V') and thus (u )t

Ut in D' (a, b, V'). This implies that u' = ut and the proof of the theorem is completed since it is easy to show that u, - u in H1 (a, b; V, V'). 0

Let us denote by D(Ia, b]; V) the space of restrictions to [a, b] of functions of D(R; V). (D(R; V) is the space of functions indefinitely differentiable with compact support and values in V). Therefore we can proceed to

THEOREM 11.3. D(Ia,b];V) is dense in H'(a,b,V,V').

PROOF. We suppose first that a = -oo, b = +oo - i.e. we want to show that D(R; V) is dense in H1(-coo, +oo; V, V'). Let u E H'(-oo, +oa; V, V'). We can approximate u by functions in H'(-oo, +oo; V, V') with compact support. Indeed, let p be a smooth function such that

_ LOW

1

0

x E (-1,1), x (-2,2),

11.2. Functional analysis for parabolic problems

189

then u,, (t) = o(.L)u(t) is such that u,, -+ u in H1(-oo, +oo; V, V'). So, without loss of generality we can assume u with compact support. Then, if B is a smooth positive function with compact support such that

fe(t)dt=1,

e > 0,

it is not difficult to show that the usual convolution technique allows us to conclude the proof, i.e. if for e > 0 one sets

Qe(t) = ep(e), then

Be * u(t) = J p (t - s)u(s) ds E D(U ; V) R

and converges toward u in H'(-oo, +oo; V, V'). Let us now turn to H' (a, +oo; V, V'), with a > -oo. Then, if Oh denotes a smooth function such that

h (t)

1

fort>a - Z,

0

for t < a - h,

h is here a positive number, then it is easy to see that tLh(t) = bh(t)u(t + h)

is a function of Hl (-oc, +oo; V, V') that, when h -' 0, converges toward u in H1 (a, +oo; V, V'). Thus D([a, +oo); V) is dense in H' (a, +oo; V; V'). We would argue the same way for H 1(-oo, b; V, V') with b < +oc. Then, if 01, 02 are two smooth functions such that 01 = 0 on (-oo, a],

82 = 0 on [b, +oo),

01 + 02 = 1 on [a, b],

then

u=Btu+02u=u1+u2 with u, E H'(-oc, b; V, V'), U2 E H'(a, +oo; V, V') and approximating u, leads to the conclusion. Now we will need the following lemma:

LEMMA 11.1. There exists a continuous linear operator of extension that we denote by P, from H1(a, b; V, V') into H1(-oo, +oo; V, V') such that

Pu = u a.e. on (a, b) Vv E H1 (a, b; V, V'). PROOF. We can always assume b finite. If a = -oo then one sets

Pu(t) =

t < b, u(2b - t) t > b. U(t)

11. Linear Parabolic Equations

190

It is easy to show that this solves the problem. If a is finite, then setting as in Theorem 11.3

u=01u+0271 we have clearly - recall (11.13): 1101U1I11 5 C11u111, 1I02i%11II' <- C117111i,

and we conclude using the previous step. Then we can show THEOREM 11.4. Let u E 11'(a, b; V, V'). Then u can be identified with a continuous function on [a, b] with values in H. Moreover

C([a,b];H)

H' (a. b; V, V')

(11.14)

where C([a, b]; H) denotes the space of continuous function on [a. b] with values in H equipped with the topology of the uniform convergence on [a, b].

PROOF. Let u E H' (a. b; V, V') and Pu E H' (-oo, +x; V, V') the extension introduced in Lenitna 11.1. Let p E D(R, V) such that

p,, -' Pu in H'(-oc,+oo;V,V'). Then we have: /

Cdn

)_

(dc'n) <

(the duality in the bracket is the V'-V duality; note that for two vectors of V, by identifying it with the linear forth V' we have (u, v) = V H v '--; (u, v) that it defines on V in V'). Integrating we obtain 2

f

Using the Young inequality

ab<

a2 2

+

b2

2

it follows that Iwn(t)12 <

1l,Pn11

i.e.

(11.15) dt E R. Thus if we apply this inequality to V,, - p,n, we see that yp is a Cauchy sequence in C((-oo, +oo), H) so converges uniformly in this space. This implies that Pu E C(-oo, +oo; H) - thus it E C([a, b], H) and passing to the limit in (11.15) we get IV.,(t)12 --

II'PnIIH'(-o,+oo;V.V')

1711C(la.b).H) 5 IIPi11H-(-oo.+oo;V,V') S

which gives (11.14).

CII7111H1(a.b.t,V')

11.3. The resolution of parabolic problems

191

A fundamental consequence of Theorem 11.4 is that it allows us to define for u E H1 (a, b; V, V') the trace u(a), u(b) - i.e. these values are the values in H of the function u E C([a, b]; H). From that argument we can deduce the Green formula: THEOREM 11.5. Let

fa

uu, v E

H1 (a, b; V, V'), a, b finite, then

b(u'(t), v(t)) dt + fa b (v'(t), u(t)) dt = (u(b), v(b)} - (u(n),v(a)).

(11.16)

PROOF. For u, v Er D([a, b], V) we have

f(u'(t)v(t)) dt +

Ja

(v'(t), u(t))dt = J

(u(t), v(t)) = u(b)v(b) - u(a)v(a).

a

0

Then the result follows by density.

We can also state THEOREM 11.6. If u E Hl (a, b; V, V'), then for all v E V

v) _ (ut(-), v) in D'(a, b). dt PROOF. Let p E D(a,b). From (11.16) we derive

J b(u'(t), W(t)v) +

u(t)) dt = 0 VVEV.

a

This can also be written b

b

b

f (u'(t), v)V(t) dt = a

Ja

(v, u(t))V'(t) dt = -

Ja

(u(t),v)v'(t) dt,

hence the result.

11.3. The resolution of parabolic problems Let us analyze the setting of (11.1)-(11.3). We shall use the three spaces: Ho(Q) C L2(f2) C H-1(f2).

Indeed, one could choose f in H-1(1). Moreover we also consider a bilinear form which by multiplying it by a function of t, we could let depend on t. So, let us consider the following abstract situation already seen in the previous section. Let V, H be two real Hilbert spaces with the following properties: V

V, H are separable,

(11.18)

H '.-' V', V dense in H.

(11.19)

(Recall that ti means that the canonical embedding is continuous.) For the norm and scalar product we will mimic the Hfl-L2 situation, in other words: 11 will denote the scalar product and the norm in V, { , ), will denote the scalar product and the norm in H.

11. Linear Parabolic Equations

192

Then, fort E (0. T) we denote by a(t; u, v) a bilinear form on V with the following properties: a(t; u, v) is measurable,

V u, v E V, t

(11.20)

3111 a constant such that

Vu,v E V, a.e. t E (0,7'),

(11.21)

a(t;u,u) +A1u12 > crjIujI2 Vu E V, a.e. t E (0,T).

(11.22)

Ia(t;u,v)) < 1111jull JIvMM

3 two positive constants or, ,1 such that

The assumption (11.21) is a hypothesis of continuity, the assumption (11.22) a hypothesis of coerciveness. Then for no E II,

f E L2(0,T;V')

(11.23)

we would like to find a function it solution to the problem

u, E L2(O.T:V'),

it E L2(0,T; V),

(11.24) (11.25)

it (0) = nO,

(f, v) inV'(0,T) VvE V.

(11.26)

Note that (11.25) makes sense by (11.14). Also, since (see (11.17)) d (it

the equation (11.26) can be written ut

f

in L2(0,T;V').

11.27)

Then we can prove:

THEOREM 11.7. Under the above assumptions (11.18) (11.22) there exists a unique solution to (11.24) (11.26).

PROOF. 1. We can suppose A = 0 in (11.22). Indeed, assume that. we know that. a solution exists in this case. Then, we can find a solution it, to dt

(w, v) + a(.; w, v) + A(w, u) _ (e-,\t f, v)

V v E V.

(11.28)

w(0) = uO

(a(t; u, v) +.\(u, v) satisfies (11.22) with A = 0 and also (11.21)). Then multiplying both sides of the first equation of (11.28) by eat we get in P'(0, T) dt (eAtu', v) + a(.;

ca'u,, v)

v)

V11 E V

and thus it = elktw is the solution we are looking for. Note that for brevity we dropped the conditions u E L2(0,T;V), III E L2(0,T;V') that are easy to check here.

11.3. The resolution of parabolic problems

19:3

2. We construct an appmrimation of u. For that let w1,. .. , u',,.... be a basis in V, i.e. the w, are linearly independent and the span of these vectors is dense in V and also in H since V is dense in 11. Note that the separability of V insures the existence of such a basis. Then, since u,) E 11, there exists a sequence of numbers such that in H.

(11.29)

Then we claim that there are functions rl (t)..... r (t) defined on (0. T) that are solutions of the linear differential system r,(O) = r,

Vi = 1,....n.

(11.30)

un(t) _ T,(t)w7.

(11.31)

i=1

u2)+a(t;un(t).w3) = (f(t),u,,) a.e. t E (0. T) Vj = 1.....n.

(11.32)

Note that the equation (11.32) holds also in D'(O,T). To see the existence and uniqueness of .r(t) = (x,(I)) note that, the matrix ((w,. uwj)),., being invertible, the equations (11.32) read

r'(t) = A(t)r(t) + F(t)

(11.33)

where A(t) is a matrix with coefficients in L"(0.T). F a vector with entries in L2(0.T). Then, x is clearly the only solution to the integral equation .r(t) = r(0) +

JU

f

{A(s)x(s) + F(s)} ds

(when the data are smoother the solution is the classical solution). 3. We get "a priori" estimates for u,,. Multiplying each equation of (11.32) by ra(t) and summing in j we obtain(it', (t), u,, (t)) + a(t; u,. (1), u,, (t)) = (f(t), it,, (t)) or else

1

a(t;

d

un(t)) _ (f(t). un(I)) s 11f(1)11.Ilu"(0I1

(11.34)

11. denotes the strong dual north in W. Noting that we have supposed that (11.22) holds for A = 0, we get: where 11

.

2dtlu"(f)12 +a1lun(t)112 <_

We can then apply the Young inequality

ab < 2 a2 + -b2 V a, b to obtain:

2 dt

Iun(t)12 + allu"(1)112 <

2n IIf(t)11:.

11. Linear Parabolic Equations

194

Hence, finally

IIf(t)II. 2 at lu-(t)I2 + 2 Integrating between 0 and t we get for every t E [0, 71: llu.(t)II2 <_

2

IIw.(e)II2de < IUn(0)12

+

jIf(8)II ds.

J Now, we have seen (see (11.29)) that

u,(0)-4u0 in H, so that for some constant C, Iu,.(0)I < Clual.

Thus, the above inequality reads, for some other constant C, t


For some constant C, we have obtained C.

IunI L°.(O,T;H), I

(11.36)

4. Passage to the limit. Before passing to the limit, we make a remark. It follows from (11.21) that

v+-+a(t;u,v) is a linear continuous form on V for a.e. t, and every u. So, by the R.iesz Theorem, there exists an element that we will denote A(t)u in V' such that

a(t; u, v) = (A(t)u, v) b u, v, a.e. t.

(11.37)

It is not difficult to see that for the is for which such definition makes sense

u '- A(t)u is a linear operator from V into W. Moreover, due to (11.21), for a.e. t

IIA(t)ull < MIIufl Vu E V. It follows then from (11.36) that IA(t)unI L2(o,T;Vl) 5 C

(11.38) (11.39)

for some constant C. Considering (11.36), (11.39), we see that we can extract a which for simplicity we label u,. - such that

sequence from

u,. - u in L2(0,T;V),

(11.40)

u,, - u

(11.41)

in L°D(0,T; H) weak *,

A(t)u,,-=u in L2(0,T;V'), for suitable u, u,

(11.42)

But it is not difficult then to show that

u = u,

u = A(t)u.

(11.43)

11.3. The resolution of parabolic problems

195

Indeed, if v E L2(0, T; V), then v E L2(0, T; H) due to the inequality (see (11.19) ) Vv E V.

Iv} < CI}vjI

(11.44)

Thus, for any v E L2 (0, T; V), by (11.41),

1T (un, v) dt - f (ii, T v) dt. 0

0

Now, for v fixed

jT

is a linear form on L2(O, T; V) - this is also due to (11.44) - and from (11.40) we deduce ) T (tl, v) dt,

T (un, v) dt

J0

0

thus

T

L for every v E L2(O, T; V) and by density of V in H we conclude that it = u. Next, for v E L2 (0, T; V) we have

I

rT

T

J(u,v)dt,

(A(t)un,v)dt'

0

but since (A(t)un, v) = a(t; un, v) and since for v fixed

t " a(t; , v) belongs to L2 (0, T; V'), we have also

J

(A(t)un, v) dt 0

f

T

(A(t)u, v) dt.

0

So

rT

J

fT (A(t)u, v) dt = J (u, v) dt

for all v E L2 (0, T; V) and (11.43) follows. So, we have now a sequence u suc

that

u -U

in

L2 (0, T; V),

u

in

L°D(0, T; H) weak *,

un

A(t)un - A(t)u in

L2(0,T; V'),

and we will be done, provided we show that u is a solution to (11.24)-(11.26).

11. Linear Parabolic Equations

196

5. u is a solution to (11.24)-(11.26). Let V E V. Then, dearly, we can find a sequence

n

vn = E a..' w, such that vn - v in V - and so also in H. Then for every (p E VQO, TD p(t)vn --, (p(t)v in L2(O,T;V) and so also in L2(0, T; H). Now, clearly, multiplying each equation of (11.32) by 4(t)a7, summing up and integrating on (0, T) we derive =1T(f

I 0T -(un(t), (P (t)vn) + (A(t)un, rp(t)vn) dt 0

(t), w(t)vn) dt.

0

(11.45)

Passing to the limit we get immediately t 0

-(u(t), sa'(t)v) + a(t; u, V(t)v) dt = f (f (t), W(t)v) dt. e

This can also be written 0

T -(u(t), v)(p + a(t; u, v)(P dt

=1T (f (t), v)(p dt, 0

(11.46)

i.e. (11.26) holds. It remains to show that Ut E L2 (0, T; V'),

u(0) = uo.

Fbr the first point we remark that (11.46) can also be written

d (u(t), v) = (f (t) - A(t)u, v)

in D'(0, T)

for every v E V. This also means that

-J T (u(t), v)('(t) dt = J T(f (t) - A(t)u, v)So(t) dt 0

d v E V, V (P E D(0, T)

0

and, considering integrals with values in V', T(u(t), T(f )sv' dt = f (t) - A(t)u, )(p dt V V E D(0, T).

- f0

0

Thus:

ut = f (t) - A(t)u E L2(0, T; V').

(11.47)

Next, to establish that u(O) = uo we consider a smooth function (p such that (p(0) 96 0, (p(T) = 0. From (11.47) and from the fact that (p(t)v E L2(0,T;V) `d v r= V we derive

(ut, v)(p(t) = (f (t) - A(t)u, v)(p(t) a.e. t E (0, T). Integrating in t between 0,T we conclude that (see (11.16), (11.17)) T(u,v)t(pdt T(f(t)

f I0

=

f JO

- A(t)u,v)(pdt.

11.3. The resolution of parabolic problems

197

Hence after integration

- J0T (u, v)cp' dt - (u(0),

f T (f (t) - A(t)u, v)cp dt. 0

Using (11.45) with our 'p and integrating by parts, we get

- J0

T

(u,, (0), vn)cp(O) = f T (f (t) - A(t)u,,, vn)pdt.

(un,

0

Passing to the limit in n will give

- J0T (u, v)cp' dt - (ufl, v)cp(O) = J0T (f (t) - A(t )u, v)ip dt. Thus

(uo - u(0), v) = 0 VV E V and by density of V in H we conclude that u(O) = uo. 6. Uniqueness. Assuming that u1, u2 are two solutions to (11.24)-(11.26), we have by subtraction d (u1 - u2, v) + dt

ul - u2, v) = 0 in 1Y(0, T), Vv E V

or else

d (ul - u2) + A(t)(ul - U2) = 0

in L2(O,T; V').

Multiplying both sides by ul - U2 and integrating on (0, t) gives - u2)(t)12 +

hence the result since a > 0.

j

a(s;ul - u2iu1 - u2)ds = 0

0

Regarding the convergence of u,, we can be a little bit more precise and show u strongly in L2(0,T;V). Indeed, from (11.35) written for t = T, it is

that u

clear that u (T) remains bounded in H. So, there exists U. E H such that

u.

in H.

To show that u,,,,, = u(T) we argue as to prove that u(O) = vo, choosing ip smooth and such that V(0) = 0, p(T) 0. Then we consider the quantity T

U. (t) =

u(T)I2 +

fa(t; u(t)

u. (t) - u(t)) dt (11.48)

2Iu.(T)I2+ foT

11. Linear Parabolic Equations

198

where T

R. (T) = 21u(T)12 - (u. (T), u(T)) + fo a(t; u(t), u(t)) dt

+ f T a(t; -u(t), un (t) - u(t)) dt + (T a(t; u(t) -

dt.

0

0

Knowing that un(T) - u(T) and due to the weak convergence of un, it is clear that n

lim Rn (T) _ - 21u(T)12 - ja(t;u(t),u(t))dt.

(11.49)

Now from (11.34) integrated on (0, T) we derive: 1

f a(t; u, (t), un(t)) dt = 2lun(0)12 + f (f (t), u. (t)) dt.

Hence, passing to the limit

.1 +oo 2lun(T)I2+jTa(t;u..(t),u,.(t))dt= 2IuoI2+j(f(t),u(t))dt. But we also have from (11.26) written with v = u(t) and after integration

21u(T) I' + f a(t; u(t), u(t)) dt = 21uo12 + 0

Io

(f (t), u(t)) dt.

(11.50)

Combining (11.48), (11.49), (11.50) we deduce that lira U,,, (T) = 0. n-.+oo

Then the result follows from the fact that

fT

a

IIu,.(t) - u(t)II2 < Un(T).

11.4. Applications 11.4.1. Existence and uniqueness results. Let f be a bounded open set of R" with boundary r. Let

H=L 2(n)

(11.51)

and V be a closed subspace of Hl (i2) such that

Hp(0) C V C H'(ft).

(11.52)

Let a' J, a j, i, j = 1, ... , n, ao be functions of LOO (f) x (0, T)) such that

aij(x,t)Ej ? aIC12 a.e. (x,t) E f2 x (0,T), VC E R.

(11.53)

11.4. Applications

199

Then we can set for u, v E V

a(t; u, v) = J { a,;

+ a;

v + aouv I dx.

(11.54)

The hypothesis (11.21) is easy to check. Moreover, see (11.53), (

a t;u,u)

8u 8 u 8u +a,ax`u+aflu 1dx fn ail 8xj axi a1

>a

fa

r

IVuI2dx-IjaLI fo IVulIuldx-IaolooJ u2dx fn

denotes the L°°(Q) norm and I I the euclidean norm, a the vector where I with entries a,. Using Young's inequality we get a(t; u, u)

a fn I VuI2 dx - I lal I(,

jI

VuI2' dx

2

- I lal I- e J u2 d x - laoloo f u2 dx, n n i.e.

a(t; u, u) +

/

+

lI) f u2 x >_ (a\ - eIlaI I1 f n

2

/J

n

Setting

A= II2e°°+laol00+aEll21' this equation can also be written, for a.e. t E (0, T),

a(t;u,u)+A

n

u2dx>

(C,-EllaL)f {JVu12+u2}dx 2

n

> 2 for e small enough.

which is exactly (11.22) with a replaced by a - £ a Taking

uo E L2(ft),

f E L2(0,T; V'),

(11.55)

Theorem 11.7 leads directly to THEOREM 11.8. Under the above assumptions there exists a unique solution u to

u E L2 (0, T; V ),

u t E L2 (0, T; V'),

u(0) = u0,

(11.56)

inV'(0,T),

V v E V.

11. Linear Parabolic Equations

200

Let us interpret the problem (11.56). More precisely, let us assume that all the functions at stake are smooth in such a way that all the following computations make sense. The last equation of (11.56) reads

(ut, v) + a(t; u, v) = (f, v) b t E (0, T),

Vv E V,

v smooth.

Taking for instance v E V(fl)rw(e obtain

1utv dx + J j ai l

11

l

+ ai

v + aouv } dx JJJ

(11.57)

f vdx Vt E (O,T).

n Integrating by parts in the second integral we get In S ue

8x- )

8x,

+ a;

Z + aou } v dx = J f v dx b t E (O, T),

i.e. it follows that Su

8

8u

(aij

8xi

+ a:5t + aou =fin 0 x (0, T).

(11.58)

Going back to (11.57) and taking a smooth v E V - but not necessarily vanishing on I' - we get after the use of the Green formula for every t E (0, T): Infut

8 8u - A(atJ

+ aou v dx +

//ri

rj a{1-v;vda(x)=0 j

bvEV

+

a1J

8u

v;v da(x)

f u dx.

Using (11.58) we derive that (11.59)

v smooth. (Recall that v = (P,,. . . , v,,) denotes the outward unit normal to 1,.) Of course to this we have to add the initial condition u(x,0) = uo(x).

(11.60)

We can apply this to different cases: V = Ha (f2). In this case (11.56) is a weak formulation to the parabolic problem Ou

-88-x,

(a,

a; 1

u(x, t) = 0 on r x (0, T), u(x, 0) = uo(x)

in iZ.

, + aou =fin St x (0,1'), (11.61)

11.4. Application

201

When a,, = b, j. d i, j = 1, .... n, ai = 0, V i = 0_-n, we get the weak formulation of the classical heat equation

J inc2x(0,T),

Au

u(x, t) = 0 on r x (0, T),

u(x.0) = no(r)

(11.62)

in Q.

V = H'(S2).

Then - as seen in Chapter 3 - in this case (11.59) can he interpreted as a,

Ou"

V, = 0

on r

and we have solved in a weak sense &u

f

Of

On

-

aa,(a,,ar,) +a;a +aeu=finS2x (0,T), ai,(x)au

(11.63)

v,=0 onrx(0,T),

u(x,0) = uo(x). This is a parabolic problem of Neumann type. In the particular case where a,, = b,, V i, j = 1, ... , n, a, = 0 d i = 0, .... n we have solved in a weak sense On

at au an

- Au = f __

in S2 x (0. T).

on r x (0, T),

U

u(r, 0) = uo(x)

(11.64)

in S2.

Note that in the parabolic version of the Neumann problem ao = 0 is allowed.

V = (vEHr(S2)I v=0onr1)} where rn is some part of the boundary r. We denote by

rN = r \ rD the complementary part of r (recall Theorem 2.6). Then, in this case, it is easy to see that (11.56) is a weak version of the problem On _

a (a,., au 1

Su

ar,J+a,fix,+aeu=JiuS2x(U,T).

at ax, it = 0 on rv x (0, T). all a;, (x) ax, v, = 0 on rN x (0. T). 11(x. U) = 110(1)

in SZ.

(11.65)

11. Linear Parabolic Equations

202

11.4.2. Weak maximum principlee. We proceed under the assumptions of Theorem 11.8. More precisely we consider a(t; u, v) the bilinear form given by (11.54). Let uo,uo E L2(S2), fl,f2 E L2(0,T;V'). (11.66)

Then we can state THEOREM 11.9. Under the above assumptions let uj, i = 1, 2 be the solution to

(ui)t E L2(0,T;V'),

Ui.E E L2(0,T;V),

u,(0) = uo,

(11.67)

in1Y(O,T), bvEV. Assume that uo < ua,

fl < f2;

(11.68)

then

u1 < u2.

(11.69)

PROOF. Let us first make more precise the meanings of (11.68), (11.69). The first inequality of (11.68) is simply an inequality a.e. in fl. (11.69) is meant a.e. in S2 x (0, T). The second inequality of (11.68) means that for any v E L2 (0, T; V),

v>0a.e.inS1x(0,T),

(fl, v) < U2,0-

(11.70)

These natural definitions being adopted, from (11.67) by subtraction,

(ul - u2i v) + a(t; ul - u2, v) = (fl - f2, v) in 7Y(O,T) for every v E V. This means also that in L2(0,T; V')

jj (ul - u2) + A(t)(u1 - u2) = fi - f2.

(11.71)

Clearly (see Theorem 2.8)

(u1 -u2)+ E L2(O,T;V). Thus we obtain from (11.71)

((u1 - U2)t, (u1 - u2)+) + a(t; (ul - u2), (u1 - U2)+) = (f1 - f2, (ul - U2)+) a.e. t E (0, T). We next claim that (see the Lemma below) ((ul - U2)t, (U1 - U2)+) = 2

I(u1 - u2)+I2 in V(0, T)

(11.72)

where 1- 12 denotes the usual L2(f2)-norm and thus u2)+I2 + a(t;(ul - u2)+,(Ul - U2)+) = (fl - f2, (u1 - u2)+) 2 dtI(u1 -

11.4. Applications

203

in V'(0, T) but also a.e. It follows that

d I(ul - U2)'12 <- 0 and thus

I(ui - uz)+Iz(t) I(ul - uz)+I2(0) = 0 a.e. t E (0,T). This implies (11.69) and completes the proof of the theorem.

D

The equation (11.72) is a consequence of the following lemma: LFr4N1A 11.2. Let U E H' (0, T; V, V') where V is given by V = { v E H' (fZ) I

v = 0 on I'D }. Then u+ E L2(0,T; V) and

(ut,u+) =

d Iu+i2

2

in D'(O,T)

(11.73)

where 1- 12 is the usual L2(ft)-norm.

PROOF. Let us first assume that u E D([0, T]; V). Then consider the functions

when x < 0,

0 He(x) _

when 0 < x < e,

Fi(x) =

(11.74)

when < x,

1

(11.75)

f_:(8)

G(x) = jFe(8)dS.

(11.76)

Clearly for every x (x+)2

Ge(x) -

Fe(x) -4 x+,

(11.77)

2

when a -* 0. Moreover

0
0
for ut and v smooth we have (ut, v) _ (ut, v)

and thus for every t

(ut,F.(u)) = f utFF(u)dx =

d

fG(u)dx.

From (11.77), (11.78) it follows that for every t when e -' 0 Ge(u)

2)2 (u

a.e. x E fl,

IGe(u)I <

2

a.e. x E Q.

(11.78)

11. Linear Parabolic Equations

204

Hence

G`(u)dz -+

2

(u+)2dx.

(11.80)

Since, again by (11.78),

fG(u)dzs

E L'(0,T)

the convergence (11.80) takes place also in L' (0, T) and thus in D' (0, T). Moreover

it is easy to see that when c - 0 Ff(u) -1 u+ in L2(O,T;V),

(11.81)

so that

(ut,Fs(u)) - (ut,u+) in L'(O,T). Passing to the limit in E in (11.79) we get in D'(O, T)

(ut,u+) =

2

jju+js Vu E V([0,T],V).

This is also for u E D([0, T], V) r 1 (ut,u+)cpdt = -

T

[u+[z9`dt

2 J0

0

b+cp E

D(O,T).

Next consider u E H' (0, T; V, V'). By Theorem 11.3 there exists a sequence u E D([0, T]; V) such that u,, - u in H' (0, T; V, V').

It is easy to show that

u:-+ u+ Thus passing to the limit in

J

T ((u0)t,un)wdt

= -2

the result follows - i.e. we get (11.73).

(11.82)

in L2 (0, T; V ).

Ir

T

dcp E D'(O,T)

0

REMARK 11.3. We developed in this chapter only one aspect of the linear theory for parabolic problems (see also [501). A second fruitful approach relies on the notion of semigroup. We refer the reader to [13], [16], [17], [4], [19] for details on this topic. Note also that recently the semigroup theory developed by H. Amann (see [5], [2], [3]) for nonlinear problems provides a very efficient tool for obtaining existence and uniqueness for such problems. In the next section we will however confine ourselves to the variational theory, referring the interested reader to the references quoted above.

205

11.5. Exercises

11.5. Exercises 1. Show Theorem 11.1.

2. Show that for a smooth function the derivative ut in D'(a, b; C'(0)) and the usual partial derivative ei coincide. 3. Establish the assertions whose proofs are missing in Theorem 11.3. 4. Show that when V is dense in H, then L2(O,T; V) is dense in L2(0,T; H). 5. Establish (11.81), (11.82). 6. For uo E H, f EL 2 (0, T; V') let u be the solution to (11.24)-(11.26). Show that the mapping u (uo, f) is continuous from H x L2(0,T;V') into L2(0,T;V). 7. Let u be the weak solution to Iut - Au + Au = f in f? x (0, T), u(x, t) = 0 on F x (0, T), lu(x,0) = uo(x) in Q.

Assuming A > 0, uo E LI (Q), f E L°°(11 x (0, T)) show that

u E L°°(! x (0,T)) and that

lulu < Max(Iuol,o, If ) I1

with an obvious notation for I EOD. 8. On f = (0,1) we consider the problem

(P)

ut -a 2u:r = 0 in S2 x (0, +oo), u(0, t) = u(1, t) = 0 for t E (0,+00), u(x,0) = uo(x) in Q.

a) Show in a formal way that +00

u(x, t) _

cn sin

nir

x e-tom }s«'t

n=1

with

c=

2

f uo(Q)sin( t

o)da.

o

b) Justify this for a suitable class of functions u0.

Chapter 12

Nonlinear Parabolic Problems 12.1. Local problems Let Sl be a Lipschitz bounded open subset of 1R" with boundary I'. Denote by I'D some measurable subset of r (for the measure dt(x)) and by rN the complement

of I'D in I'- that is to say

I'N =it\I'D.

(12.1)

V={vE H'(c) I -y(v) = 0 da a.e.onl'D}.

(12.2)

Set

(See Chapter 2 for the definition of y.)

Let a(x, t; u), b(x, t; u) he Caratheodory functions - i.e. such that u a(x, t; u), (resp. b(x, t; u)) is continuous a.e. (x, t) E Sl x R+, (x, t) -+ a(x, t; u), (resp. b(x. t; u)) is measurable d u E R.

(12.3) (12.4)

(a, b are defined on c x R+ x 1R, R+ = (0, +oo)). Moreover we assume that for some constants m, M: a.e. (x, t) E i x 1R+, (12.5) 0 < m < a(x, t; u) M Vu E IR,

Ib(x,t;u)I < M

a.e. (x,t) E c x II2+,

`du E R.

(12.6)

-

Then we would like to investigate the existence of a weak solution to the problem 8u a I a(x, t; u) I + b(x, t; u)u = fin Q x (0, T), at 8x,

u = 0 on I'D x (0, T), art = 0 on I'N x (0, T),

(12.7)

u(x, 0) = uo(x) in i, where uo E L2 (n), f E L2(0,T; V'). We have: THEOREM 12.1. Under the above assumptions there exists a solution u to

u E L2(0,T;V), ut E L2(0,T;V'), u(0) = uo,

d(u,v)+a(u,v) _ (f, v) in D'(0, T), Vv E V.

(12.8)

12. Nonlinear Parabolic Problems

208

(,) denotes the scalar product in L2(0) and

j{a(xt;u)JJJ

}

a(u, v) =

+ b(x, t; u)u v } dx

(12.9)

(clearly (12.8) is a weak formulation of (12.7)).

PRooF. We argue using the Schauder fixed point theorem. For that purpose, consider to E L2(0,T; L2(1)) and u = T(w) the solution to u E L2(0,T;V), ut E L2(0,T;V'),

U(O) = tb,

(12.10)

d (u, v)+a.(u,v)

(f,v) in D'(0, T), Vv E V,

where

av,(u,v) = j{o(xt;w)-f

s

+b(x,t;w)u-v}dx.

(12.11)

We know from Theorem 11.8 that such a u = T(w) exists. If we can show that the mapping w H T(w)

from L2(0,T; L2(C1)) into itself has a fixed point we will be done. First let us remark that, if we consider (11.47), and with an obvious notation

for A = A, ut + Au = f in L2(0,T; V') and thus for every v E L2(0 , T ; V) (ut, v) + am (u, v) = (f, v)

a.e. t E (0, T).

that 0 < m < b(x, t; u):5 M a.e. (x, t) E 0 x R+, Vu E R.

Let us assume

(12.12)

firs t

(12.13)

Then taking v = u in (12.12) we obtain by (11.17) 2

(u, u) +a,v(u,u) _ (f,u) a.e. t E (0,T).

Using (12.13), (12.5) we obtain 2i (u, u) + mjIuIIv < jjf II. IIuIIv

a.e. t E (0, T)

(12.14)

where we have set

IIuIR' =j{1Vu12 +u2}dx,

(12.15)

Of 11. =Sup (f,v).

(12.16)

Ilvpv<1

12.1. Local problems

209

Using the Young inequality for the right-hand side of (12.14) ctb <

m 2

b2 +

i.e.

u2. 2111

we obtain 1d 1n 111i112. (u, u) + 1nlIulli. < IIIII?. + 1 2 dt 2 2n1

dt

(U.. 1i) + m111i11i' 1n

IIf11

a.e. t E (0. T).

(12.17)

Integrating on (0. T) it follows that Iu(T)Iz + 111 JT Ilulli dt < Iuol2 +

In particular

due to (12.15)

111

Jr lull: dt.

(12.18)

o

we obtain

II I'll L=(O.T:V) Ilulit 1(O.T.L1(s1)) <_ C

(12.19)

C2

(12.20)

with (ucI2 + =1 M

11f11%=((J.7'.%'1)11n2

Setting

C)

B = { u E L2(0,T; L2(S2))

(12.21)

it is clear from (12.12) that u'

U = T(u' )

is a mapping from B into itself. Moreover, going back to the third equation of (12.10)

see also (12.12) - we have IlutIIL to.r.t"> < 2%! IIu11L2(o.T:V) + 11111L2(O.T.V') (12.22)

< RIC + Ill IIL2(II.T.') = C'. Thus, a belongs to a bounded subset of H1 (0. T: V. V') which is relatively compact in L2(0,T; L2(S2)) (see (411). Thus, in order to conclude by applying the Schauder fixed point theorem in B, we just need to show that T is continuous on B. So. let

us consider a sequence u' E B such that u' - it,

Let us denote by u =

in B.

the solution to (12.10) corresponding to it, = 11',,. Due to (12.19) and (12.22) we can extract from n a subsequence. still labeled by

12. Nonlinear Parabolic Problems

210

n, such that

ww

a.e. in f2 x (0, T),

u -- u00

in L2(0,T; L2(1)),

8tt

O+QO

un,

in

L2(0,T;L2(1l)),

(12.23)

in L2(0,T; V'),

u00t

where u,,, E L2 (0, T; V). From the third equation of (12.10) we deduce T

f

v)c'(t) dt +

+

1

T I cp(t)

a(x, t; w,.)

dx dt i

fT

(12.24)

J

t

=

f(f, v)c (t) dt VV E V, V V E 1)(0, T). 0

By the Lebesgue convergence theorem we have W(t)b(x, t; w.)v

w(t)a(x, t; wn)

8x,

P(t)b(x, t; w)v,

- p(t)a(x, t; w)

iii-I

,

in L2(0,T; L2 (0)) = L2(a x (0,T)). Thus passing to the limit in (12.24) leads to (uQ,, v) + aw(u,o, v) _ (f, v)

in 1Y(0, T).

(12.25)

Of course

ua, E L2(O,T;V),

U ,,.,t E L2(0,T;V').

(12.26)

Now, we can write for a.e. t E (0,T), Vv E V

(U. (t), v) - (uo, v) = f (u,,,, v) dt. t

(Recall that by (12.19), (12.22), (12.23), un, u E C([0, T], L2(f2)) - see also (11.17) - and if we pass to the limit we get

(u.(t),v) - (uo,v)

t 0

(u.(t),v) - (uco(0),v)

a.e. t E (0, T), by E V. Note that by the second line of (12.23) one can assume without loss of generality that t4, (t) -' ua,(t)

in L2 (f2), a.e. t E (0, T).

12.1. Local problems

211

Thus u,o(0) = uo.

(12.27)

Combining (12.25)-(12.27) one sees that u,,. = T(w) - since u, has only T(w) as possible limit,

u., = T(w) in L2(0, T; L2(51)) and the continuity of T is established. This proves Theorem 12.1 in the case of (12.13). In the general case we notice that if we set

u = e-ktu, then u is a solution to (12.7) if u satisfies.

(a(x, t; ektu)ekt "4 1 + b(x, t; a u)ektu = f, axi 8u = 0 on rN x (0, T), u = 0 on rD x (O, T); 6

ax i

(ekeu)

u(0) = uo. The first above equation can also be written afi

+ ku - 8,

,t , ektu)

) + b(x, t; ektu)u = f e kl

Thus, choosing k large enough due to (12.6), we can always assume that (12.5), 0 (12.13) hold. This completes the proof of the theorem. Consider now a Lipschitz continuous function b - i.e. such that Jb(u) - b(v) I < Blu - vi Vu,v E R

(12.28)

for some positive constant B. Moreover assume that b(0) = 0,

b'(0)

exists.

(12.29)

Then we can state

THEOREM 12.2. Under the assumptions of Theorem 12.1 and if (12.28), (12.29) hold, there exists a weak solution to

8 au (ax, t; u) a ) /+ b(u) = f in Cl x (0, T), 8t 8x; = 0 on rN x (0, T), u = 0 on ro x (O, T), On

u(x, 0) = uo(x) in Cl, where uo E L2 (Q), f E L2(O,T;V').

(12.30)

12. Nonlinear Parabolic Problems

212

PROOF. It is enough to notice that one can write

u

b(u) = b(u ' u; )

then setting

b(u) for u # 0,

u b'(0)

b(x,t;u) =

for u = 0

by (12.28), (12.29), b fulfills the assumptions of Theorem 12.1 and the result fol-

0

lows.

We now turn to the question of uniqueness. For this purpose we consider the problem (12.30) even though our analysis applies to more general situations. For simplicity, in addition to (12.28), (12.29) we assume that

u ' b(u)

is monotone.

(12.31)

Regarding a, we apply the assumptions of Theorem 12.1. In addition we will suppose that

Ia(x, t; u) - a(x, t; u)I < w(]u - uI)

j

(12.32)

for every u, u E R where w is a continuous positive function satisfying dB

= +00.

(12.33)

w2 (8)

(w can be the modulus of continuity of a.) Then under the above assumptions we can show: THEOREM 12.3. If the above assumptions hold, then there exists a unique solution to (12.30).

PROOF. Let u, u be two solutions to (12.30). We know that, in a weak sense in L2(O,T;V'), Su i

8 (a(x, t; u) 8x;

{

\

1 + b(u)

//

8t

8x;

(a(x, t; u)

+ b(u)

in S1 x (0, T). This can also be written

(u - u) -

/ i

a(x, t; u) tom= (u - u)) + b(u) - b(u) (12.34)

I (a(x, t; u) - a(x, t; u)]

1.

Without loss of generality we can assume that

0 < w(a) da > 0,

t+00

J

da < +oo. w2(e)

(12.35)

12.1. Local problems

213

Then, for e > 0. we set

if t<e,

0

(12.36)

1 11 da T w2 (s) ift>E,

Hr(f) where

If = 1

ds

+O°

a

(12.37)

w2 (s)

For d> 0 and e small enough we have e < £and ds w2(s)

-+1

(12.38)

as a -+ 0 (see (12.35)). Moreover

He(y) = 0 for

(12.39)

< 0. We consider then the function H4(u - u)

that belongs to V. Using this function with (12.34), we get

((u - u),.IIE(u-u))

+ja(.;

u) IV(u-u)12H(u-f(b(u)

- b(u))HE(u - u) dx

+

=

f (a(.; u) -

u))VU - V(u - u)HH(u - u) dx.

Due to the monotonicity of b the third term above is nonnegative. Using now (12.5), (12.32) we get:

((u - ii)j. He(u - u)) + m fn IV(u - u)12HH dx s!

< Jw(u_u)Vu.V(u_u)H(u_u)dx. Note that by the definition of He we are integrating on the set

S=[u-u>E]={xEilI(u-'u)(x)>E}. Thus it follows that

((u - )t,11e(u - u)) +

m

IV(u - u)12

If Js w (u - u)

dx <

1

- I, Js jVUl

Using now the Young inequality

ab<

2a2+-b2

ba,b>O

IV(u - u)I dx. w(u - u)

12. Nonlinear Parabolic Problems

214

in the last integral it follows easily that (( u

- u )r, H,(u - u)) +

m / 1o(u - u)l2 dx <

21E

is

w2(u - u)

/ 1

2niIE is

dx

from which follows

((u - u) r, Hf(u - u)) <

IDulz dr.

1

2nnI£

(12.40)

su

Setting

f HE(s)ds, 0

this inequality can also be written (see (11.79) for a justification) dt

Vu12 d.r.

1J E

u

s

Integrating on (0, t) we get for every t r

f K,(u-il)d.r.< 2

Now due to (12.38), (12.39), when E Ke(y)

I 2 mIE .o

f

IVu12dx.

(12.41)

1

0

v0,

if

+00

where V denotes the maximum of numbers. Letting a -+ 0 in (12.41) leads to 61

(x,t)dx < 0

for a.e. t. Thus we obtain

u-u<0. Exchanging the roles of u and u leads to the result.

0

REMARK 12.1. If E L2(St) and if u, a denote the solutions to (12.30) corresponding to uo and uo respectively the above proof shows in fact that

fo [u - fil' (x., t) dx <

(uo - ft()]'

In particular ufi < fio would imply

u < u, i.e. the solution to (12.30) is monotone in uo.

(12.42)

12.2. Nonlocal problems

215

12.2. Nonlocal problems We denote as above by n a Lipschitz open subset of R" with boundary F. We have seen in Chapter 1 that in various diffusion problems the speed of diffusion is given by the Fourier law

V = -aVu, where a is some constant attached to the medium where the process is taking place. In the case of diffusion of bacteria for instance we can assume that a depends also on the population itself, i.e. on quantities like ju(x)dx

or

t

r

(12.43)

Q' beingsome subdomain of U where some influential group is leaving. More generally one can weight these factors and consider a dependence of a in

1(u) = f g(x)u(x)dx

(12.44)

where g E L2(ft) - i.e. one can consider I to be a linear form on L2(fl). So, for FD a subset of r of dc-measure positive let us set as before

V = {vEH'(U)I r(v)=0onFD}.

(12.45)

Then for 1 a linear form on L2(U) we consider the problem

8

St

Su)

= f in R x (0, T), (a(1 (u)) 8xi 8xi / t) E V for t E (0, T),

a u(.,t)=0onF\I'D,

(12.46)

u(. 0) = uo, where f(x,t), uo(x) are some data. Of course we have in mind to solve (12.46) under its weak form. For a we will assume that a:R (0, +oo) is continuous and that for some constant m, M,

0
(12.47) (12.48)

Then we can show: THEOREM 12.4. Let f E L2(0, T; V') where V' denotes the strong dual of v and let uo E L2 (f2). Then, under the above assumptions, there exists a solution u to u E L2(0, T; V) f1 C([0, 71; L2(fl)), ut E L2(0, T; V'),

u(0) = uo, dt (u, v) + a(1(u))

(12.49)

r Vu Vv dx = (f, v) in D'(0, T), Vv E V.

in

12. Nonlinear Parabolic Problems

216

denotes the usual scalar product in L2(fl), V' and V),

the duality bracket between

PROOF. We are going to rely again on the Schauder fixed point theorem. For this let w E L2(O,T;L2(fl)) = L2(1l x (0,T)). Then - see the representation formula (12.44) - the mapping is measurable and due to (12.47) so is

t -- a(l(w)). By Theorem 11.8 there is a unique u = Tw solution to u E L2(O,T;V) nC([O,71;L2(il)), ut E L2(0, T; V'), u(O) = uo,

r jj(u,v)+a(I(w)) / Vu.Vvdx=(f,v)inD'(O,T), VvEV. in

(12.50)

(Note that a(l(w)) E L°0 (0,T) by (12.48)). Thus we would like to show that the mapping has a fixed point - this will be dearly a solution to our problem. The third equation of (12.50) reads in L2(0,T;V') dt

- a(l(w))Du = f.

(12.51)

Thus, it follows that

(ut, u) - (a(l(w))Du, u) = (f, u) a.e. t E (0, T) which leads to

r 2dtluls+a(I(w))j IVu12dx={f, u)

a.e. t E (0, T).

Choosing as norm in V, I I v such that

lulu = f IVuI2dx n

and denoting by If I. the strong dual norm of f by (12.48) it follows that

2dtIuI +mluly <- IfI.Iuly Using now the Young inequality

ab< Za2+2L62 Va,b>O we obtain

1 d lu[z + m

2dt

2

July <_2m 1 If i.

12.2. Nonlocal problems

217

Hence, integrating between (0, t) we get for every t c

2IuI22 + 2

f luly dt <_ 2 Iuolz + 2m J'If12 dt, o

(12.52)

in particular IuIL2(o,T;V) <_ C

(12.53)

for some constant C = C(m, uOi f) independent of to. Going back to (12.51) and noting that

1(-ou,v)I =

j

Vu- Vvdxl < Iulvlvly VV E V

we get by (12.48) for almost every t E (0, T) Iu:1. <- MIuJv + If I. and thus

(12.54)

I ut I L2(O,T;V') < C'

where C' = C'(m, M, uo, f) is independent of to. Going back to (11.13) we see that u E H 1 (0, T; V, V') and for some constant C = C(m, M, u0i f ),

Ilulli 5 C independently of to. Clearly for some other constant independent of to,

(12.55) (12.56)

IIkIIL2(O,T;L3(n)) S C.

Thus by (12.56) the mapping T goes from the ball B(0, C) of L2(0,T; L2(11)) into itself. Moreover, see (41] or (561, the balls of H1(0, T; V, V') are relatively compact in L2(0,T; L2(1l)) and by (12.55) T(B(0, C)) is relatively compact in B(0, C). In order to be able to apply the Schauder fixed point theorem, we now just need to prove that T is continuous from B(0, C) into itself. For that let w be a sequence in B(0, C) such that

w,, -, to in L2(0,T;L2(1l)).

(12.57)

Set u,, = T(w,,). From (12.57) we derive that

1(w,,) - 1(w) in L2(0,T).

(12.58)

IlunhIl < C

(12.59)

Moreover, by (12.55)

where C is a constant independent of n. By the last equation of (12.50), for any p E D(0, T)), v E V

-J o

f

f r (t) dx dt + J 1

_ fT (f, v),p(t) dt.

o

n

Vv p(t) dx dt (12.60)

12. Nonlinear Parabolic Problems

218

From (12.58), (12.59) we can find u,, E H1(0, T; V, V') and a subsequence from n that we will label also n such that a.e. t E (0,T),

1(w)

u - u,,

in H1 (0, T; V, V'),

u

in L2(0,T; L2(1)).

u00

Since a is supposed to be continuous, then also

a(l(w))

t E (0. T)

and thus

a(l(w))Vv

in L2(fl x (0, T)). Passing to the limit in (12.60) we see that u,,. satisfies

u. E L2(0,T;V),

dt(u.,v) +a(l(w))J

(u.)t E L2(0,T;V'),

(12.61)

Vvda = (f,v) in D'(0,T), by E V.

(12.62)

v

Arguing exactly as we did to establish (12.27) we obtain easily that

u,o(0) =

u1.

Thus by the uniqueness of the solution to (12.50), u00 = U.

Since any subsequence of u,, has the same limit,

u = T(w)

L2(0,T; L2(1 )).

This completes the proof of the continuity of T and the proof of the theorem. Regarding uniqueness we can show:

THEOREM 12.5. Let us assume that a satisfies for some constant A > 0

ja(l;) - a({')j5 Ajl; - £'j dl;,l;' E IR.

(12.63)

Then, under the assumption of Theorem 12.4. the problem (12.49) has a unique solution.

PROOF. Let u, u be two solutions to (12.49). In L2(0.T;V'),

q

ut - a(l(u))Du = fit - a(l(u)Du (u - u)i - a(l(u))0(u - u) = (a(l(u)) - a(l(u))Du.

Multiplying both sides byru - u E L2(0,T;V) it holds that a.e.ft E (0,T)

((u-u)t,u-u)+a(l(u))J JV(u-u)I2dx = {a(l(u))-a(l(u)) n

Jn

Vu.V(u-u)dx.

12.2. Nonlocal problems

219

Using (12.48), (12.63) we obtain for some constant C:

2dtlu - u12 + inJ IV(u - u)I2d.r < A11(u) - 1(u)I f IVi IIV(u - u)Idr fl

S2

< CIu - u12lulVIu - ult'(Recall that Iukk = IIVuII2). Thus setting C' = C'luly we get


ub< 71 u2+2 -b2 t1a.bEIR2 we get 1

dIu-u12+m1u-uiV
2 dt

Thus we obtain d

z

dtla-ul2
l'u-u12=9(t)lu-i12

(12.64)

where

2mluly E L'(O,T).

9

We derive from (12.64) that d

it(expS- f 9(s)ds}1u-zil2) -U. 11

o

JJ

Thus the function

exp{- fo `g(s)ds}lu-ul2 JJJ

is non-increasing. Since it vanishes at 0. it vanishes for every t > 0 and we obtain

u=u. This completes the proof of the theorem.

0

REMARK 12.2. In the nonlocal case that we consider the comparison principle (12.42) does not hold anymore - see (311.

12. Nonlinear Parabolic Problems

12.3. Exercises 1. Show that L2(0,T;L2(f2)) = L2 (!Q x (0,T)). 2. Extend Theorem 12.1 in the case of the operator

-;

(ai(x,t;u)gfix ] +a(x, t; u) jj- +ao(x,t;u)u

under suitable conditions on the functions a,j, at i,

n and ao.

3. Under the assumptions of Theorems 12.2, 12.3 let uk, k = 1, 2 be the solutions to auk - 88-x, Ca(x, t; uk)v

/

+ b(uk) = A in f2 x (0, T),

uk=0onrox(0,T), -k=0onrNx(0,T), uk(x,0) = uk(x) in f2. Show that < u°iimplies that

u2°,

h <- f2

ul < U2

(see Theorem 11.9 for the meaning of these inequalities). 4. Extend theorem 12.4 in the case where

rD=0. 5. Let a, b be two positive continuous functions satisfying for some constant CO

la(8)I, Ib(8)I <- Co(1 + 181).

We consider the problem

Ut - Au = -a [ J u(x, t) dx) in fix (0, T), n Ou u = 0 on 1'D x (0,T), = b(

n

u(x, .) dx I on rN x (O, T),

u(x, 0) = u° in Cl.

Give the weak formulation of this problem. Under reasonable assumptions on a, b, show that it admits a solution (see [35]).

Chapter 13

Asymptotic Analysis The goal of this section is to analyze the behavior of the solution of parabolic equations when t goes to infinity. Let us start with a simple example.

13.1. The case of one stationary point Let u be the weak solution (see Theorem 11.8 and (11.62)) to

Iut=Du+f u(x,t)=0

in0xR+, onrxR+,

(13.1)

tu(x,0) = uo(x). We assume here that f = f (x) E H-1(11), no E L2(t2). It is clear that the problem (so-called steady st ate, i.e. obtained for ut = 0)

Du = f in g, on r, lu=0

(13.2)

has a unique solution u = u,,. Then we can show: THEOREM 13.1. If u, u are the weak solutions of (13.1), (13.2) then t

lim 1U(-,t)-U,,.12=0, +00

more precisely. them exists a positive constant v such that e-"tluo - U.12lu(', t) - U 12
intlxR+. Multiplying this equation by u - u and integrating on tt, we find that 2

djIu-u,oI2+J

n From the Poincare Inequality follows that dtJu - uoo12 + vIu - uooI2 - 0

(13.3)

(13.4)

13. Asymptotic Analysis

222

where v is the first eigenvalue of the Laplace operator with Dirichlet boundary conditions. This also means that, after multiplying by c2it, (ezvtlu _U"'122) dt

< 0,

so the function t , e2vtJze - u.. ,12 is lion-increasing and we have e2vt lu - 1 1 ",1< 1110

- ux 12

which is exactly (13.4).

O

13.2. The case of several stationary points Even in a linear situation the stationary problem associated to the parabolic equation at hand can have infinitely many solutions. For instance, let its consider a the solution to (see (11.64) and Theorem 11.8):

ut=Azt+f e921

=0

in 12xR-' onrx)FY+,

art u(x, 0) = uo

(13.5)

in Q.

Then it is clear that the steady state problem

1-Au = f

in S2,

T =0

on r

On

has infinitely many solutions

(13.6)

i.e. if we fix one of them. say u,,_ such that

fit

u,,, (x) dx = 0,

(13.7)

then, for every constant C, ux + C is another solution (see also Exercise 3 of Chapter 3). We assume here that f. uo E L2(Sl). This enables its to state THEOREM 13.2. There exists a constant v so that

Iu-u'x-C'012<_e-"17h)-ux-Cb12

(13.8)

where Co is the constant given by Cu = JUO(X)dI..

(13.9)

u

PROOF. From (13.5), (13.6) we deduce

(u-u.)t -0(u-ux,) =0. So, integrating this equation on S2 we notice that d dt

(u-u"')dx=0

"

(13.10)

13.3. A nonlinear case

223

so that by (13.7) we see that the integral of u on f2 is preserved in time and we have

fuds = J uodx=Co1. a

(13.11)

We can now write (13.10) as

(u-u00 -Co)t -i(u-u". -Co) =0 and, multiplying by u - ucO - Co and integrating on ft, we obtain

I'c-u,o-CoI +IV(u-u". -Co)I2 dx=0.

2 dt Now, due to the fact that

fn

in u - u,,O - Codx = 0, we have a Poincare Inequality (see for instance [46] and Exercise 3, Chapter 3).

IV(u-u". -Co)I2dx 1 Llu - u". - C0I2 Then, we conclude as in the proof of Theorem 13.1.

0

REMARK 13.1. In the above case the steady state is selected by the initial condition.

13.3. A nonlinear case We suppose that we are under the assumptions of Section 12.2, in particular I is a linear foriii on ! 2(f ), and we consider the problem (12.46) - i.e. the problem where u satisfies in a weak sense (see (12.49)) 8u a(l(u))Au = f in Cl x (0, T), St (13.12) t) E V in (0, T),

-

U(., 0) = uo.

t) when t -' +oo. As in Then we would like to find out the behaviour of the previous sections we start by studying the steady state equation associated with (13.12).

13.3.1. Steady states. For simplicity we suppose that in (13.12) f is independent of t - i.e. f E V' the dual of V. Then we consider the problem of finding a weak solution it to

a(l(u))Du =fin 0, u E V.

(13.13)

As we saw previously. solving a nonlinear problem relies often on a fixed point argument in infinite dimensional space. However, in the case of this nonlocal problem

13. Asymptotic Analysis

224

- as more generally for a large class of nonlocal problems - it is enough to develop a fixed point argument in a finite dimensional space, in fact, in this particular case in R. To see this, let us first introduce 'p, the weak solution to

A


(13.14)

The existence and uniqueness of a solution, i.e. the existence and uniqueness of a

p such that

VEV,

jVco.Vvdz=(f,v) bvEV,

(13.15)

is a direct consequence of the Lax-Milgram theorem, due to the fact that V is a Hilbert space for the norm IvIv =

U

\ 1/2

IVvI2d )

(13.16)

(see Theorem 2.15 - note that we suppose rD of positive measure). Under these conditions we can state THEOREM 13.3. Let a be a mapping from R into (0, +oo). The problem (13.13)

has as many solutions as the problem in R a(p)p =1(cp).

(13.17)

PROOF. Let us first consider a solution u to (13.13). Since a(1(u)) is a constant the first equation of (13.13) can also be written

-A(a(l(u))u) = f in ft. Thus by (13.14) we have a(t(u))u = cp. Applying 1 to both sides of the equation, we obtain due to the linearity of 1

a(l(u))I(u) = 1(v), i.e. 1(u) E R is solution to (13.17). Conversely, let µ be a solution to (13.17). Then, there exists a unique weak solution to

a(p)Au = fin n, 1uEV.

(13.18)

Of course due to the uniqueness of the solution to (13.14) it also follows that a(A)u = cp.

Then, taking I of both sides we obtain

a(u)l(u) =1(c) = a(ju)p

225

13.3. A nonlinear case

and thus since a > 0 I(u) = A. Going back to (13.18) we see that u is a solution to (13.13). This completes the

0

proof of the 1 heoreni.

Of course many situations can then occur. Let us list a few of them. First, if I(W) = 0, the only solution to (13.17) is p = 0 and thus the only solution to (13.13) is

u=

(

a 0)

(13.19)

.

If a is a continuous function such that

0<m
(13.20)

then the range of the mapping

p - a(p)p is the whole real line and by the intermediate value theorem one is always insured of the existence of a solution to (13.13). To fix the ideas, let us now assume that (13.21)

I(pp) > 0.

Then any p such that a(p) =

I(MP) 11

will produce a solution to (13.13). In other words, any point at the intersection of will provide a solution to (13.13). Thus the graph of a and the hyperbola p " for instance in the case of Figure 13.1,

y = a(p)

µ F[GUn.E 13.1

the problem (13.13) has three solutions. The problem can also have a complete continuum of solutions as shown in Figure 13.2.

13. Asymptotic Analysis

226

y y = a(µ)

A

FIGURE 13.2

In order to study the asymptotic behaviour of the solution of (13.12), we need to introduce the concept of a dynamical system. 13.3.2. An introduction to dynamical systems. Let us denote by (X, d) a complete metric space equipped with the distance d. DEFINITION 13.1. A dynamicall system on X is a family of mappings {S(t)}t>o

from X into itself such that (i) S(t) : X - X is continuous, (ii) S(O)x = x Vx E X, (iii) S(t + s) = S(t) o S(s) Vs, t > 0, (iv) dx E X, t H S(t)x is continuous from (0, oc) into X. This notion emerged from various examples of time dependent problems. We shall consider one in the next section. The concept has been designed to study asymptotic behaviour. In particular, for any x E X, the curve t .--* S(t)x

is called the trajectory issued from x. Next, with the asymptotic behaviour in mind, we introduce the notion of an w-limit set. More precisely we have DEFINITION 13.2. Let x E X. The set

w(x)yEX 13t,, -*+oosuch that

n

= n,>oUt>,{S(t)x} is called the w-limit set of x. Then we can show if {S(t)}t>o is a dynamical system on X:

(13.22)

13.3. A nonlinear case

227

THEOREM 13.4. For any .r E X, t > 0,

w(S(t )r) = W(r), S(t)w(r) C w(r).

(13.23) (13.24)

Moreover, if Ut>(){S(t)r} is relatively compact in X. then S(t),U(.r) = W(r) 7-1 0.

(13.25)

w(r) is compact, connected in X,

(13.26)

d(S(t)r,W(r)) -. 0 when t - +x,

(13.27)

the distance between two sets A. B being defined as

d(A,B) =

aE

lnfElyd(a.b).

(13.28)

PROOF. Let us first prove (13.23). By (13.22) we obtain w(S(tcl)r) = n,>c)Ur>,{S(r)(S(t())r)} =

nQUp>g{S(r + tt,)(r)}

=

w(.r)

Vt()>0.

This proves (13.23)

Next, if y E .o(r). there exists t - +3c with

y. It follows by (i) of

Definition 13.1 that

-y S(t)y

S(t +

when it - +oo. This shows that S(t)y E v(y) and thus (13.24) holds. Assume now that

Ut»{S(t)r} is relatively compact in X. Then we can find a sequence t -. +x such that

S(t )r

.y

(13.29)

in X.

Clearly, this y E .,(x) and w(r) -A 0. y for some sequence t Next consider y E W(r). By definition. +x. Due to (13.29), for some subsequence still labeled by n.

S(t - t )x

z E w(x).

Then

S(t ),'. = lim S(t)(S(t - t)r) = lim S(t ).r = Y. n

n

i.e. y = S(t)z E S(t)w(r). This joined to (13.24) proves (13.25). For every s since the function t S(t)r is continuous

Ut>,{S(t)r}

is connected, relatively compact in X

(see (13.29)). The compactness of v(x) and (13.26) follow then from (13.22).

13. Asymptotic Analysis

228

Finally, if (13.27) fails, then for some e > 0 there exists a sequence t -a +oo with e

`du.

(13.30)

But due to (13.29) one can extract a subsequence of t such that

S(t,,)r-,yEw(x). This contradicts (13.30) and completes the proof of the theorem.

Let (X, d) be a complete metric space and {S(t)})>O be a dynatnical system on X. DEFINITION 13.3. A continuous function (D : X , iR is called a Lyapunov function for {S(t)},>)) if 4i(x)

`dx E X, `dt > 0.

(13.31)

In particular for any x E X the function t H 4b(S(t)x)

is non-increasing.

The following result is known as the invariance principle of Lasalle: THEOREM 13.5. Let {S(1)})>() be a dynamical system on X, x E X such that (13.29) holds. Let (D be a Lyapunov function for this dynamical system, then: 4i(S(t)x) = C,

(i)

them, exists C such that

(ii)

4)(y) = C'

(iii)

41(S(t)y) _ ID(y) = C `d ,y E w(x) V t > 0.

Bin

(13.32) (13.33)

V Y E w(x),

(13.34)

PROOF. The function I .. 4)(S(t)x) is non-increasing and, by (13.29), bounded. Hence C exists and (13.32) is proved. If y E w(x) then for some sequence t,,, Y.

Hence

4i(y) = lim t(S(t )x) = C. n

This proves (13.33). (13.34) is then an immediate consequence of (13.24). This completes the proof of the theorem.

13.3. A nonlinear case

229

13.3.3. Asymptotic analysis. In this section we suppose that we are under the assumptions of Theorem 12.5. In particular we suppose that a is Lipschitz continuous so that (13.12) admits a unique weak solution. Let us start with the following stability result: LEMMA 13.1. Let U0" E L2(l) be a sequence such that when n - +oo

uo - uo in L2(it).

(13.35)

Let u", u be the solution to (12.49) corresponding to the initial data up, uo respectively. Then

u"(t) - u(t) dt > 0

in L2(Sl).

(13.36)

(We denote by u"(t), U(t) the functions u"(,t), U(., t) respectively). PROOF. Since uo - uo in L2(S2), uo is bounded in L2(1l) independently of n and from (12.52)-(12.54) we deduce that, for some constant C independent of n, Iu"IL2(o,T;V) <_ C,

(13.37)

Iu"ILoo(o,T;L2(O)) C C,

(13.38)

IU: I L-(o,T;V') <- C.

(13.39)

Thus one can extract a subsequence from n - which we still label n - such that, when n --+ +oc:

u" - u°O in L2(0, T; V), U" -+ U°O

in L2(0, T; L2(fl)), in LO0(O, T; L2(St)) -weak *,

u" - u°° ut u- in L (0, T; V').

(13.40)

(We used the compactness of the embedding from HI (0,T; V, V') into L2(0, T; L2(c?))). By definition un satisfies for every v E V

- f T f nu"vcp'(t)dxdt+ Jj a(l(u"))(Vu- Vv),(t)dxo =

f

(13.41)

VV E D(O,T).

0

Clearly from (13.40)

l(u") -, l(u0O) in L2(0,T) and up to a subsequence one can assume that this convergence holds for a.e. t E (0, T). By the Lebesgue theorem it follows that a(l(u"))WVv - a(l(uO°))coVv

(13.42)

in L2(Sl x (0,T)) = L2(0, T; L2(fl)) and passing to the limit in (13.41) we get that uO0 satisfies for every v E V

d (u°O, v) + a(l(u0O)) f VuOO Vvdx = (f,v)

in D'(0,T).

13. Asymptotic Analysis

230

Moreover, for every v E V

(u" (t), v) - (uo, v) =

j

(u , v) di

a.e. I.

up to a subsequence, one can assume that

u"(t) -, u°O(t) a.e. t in L2(1Z), passing to the limit in the above inequality leads to

(U' (t), v) - (u0, v) =

j

v) dt = (u°° (t), v) - (u°°(0), v).

t

0

Thus u°O(0) = u0 and by uniqueness of the solution to (12.49) it follows that u°O = U.

By uniqueness of the limit we obtain then that the whole sequence u" satisfies (13.40) with u°O = u. Thus we have

u" - u

in LOD(0,T; L2(fl)) -weak

and in particular for every v E V (u" (t), v)

(u(t), v) in L°° (0, T) -weak * .

Now, for every tl, t2 E 10, TJ, t2 > tl it holds that tz

(u"(t2),v)-(u"(tl),v)= J(u,,v)dt tt

<

Jt2 ui.iviv

<- 02 - tl)' IVIV IU7IL2(O,T;V')


(U' (t), v) - (u(t), v)) in C([0,71) for every v E V. V being dense in L2 and u"(t) being bounded, it follows easily that (u"(t), V) -+ (u(t), v) V V E L 2(11), V t > 0 which gives (13.36) and completes the proof of the lemma.

Let us now suppose that I is nonnegative in the sense that

f l(v) > 0 Vv > 0, v E L2(St),

t00.

(13.43)

13.3. A nonlinear case

231

For f we suppose that f is independent of t and that

(f,v)>0 Vv>O, vEV,

I f EV',

(13.44)

f # 0, in short we suppose f > 0 and f $ 0. Going back to (13.14) it results from the maximum principle that

2>0 inff

and thus l(ip) > 0

(13.45)

(see the representation formula (12.44)).

Then we consider µl, 02 two intersection points of the graph of a with the graph of the hyperbola y = 1(cp)/µ and we suppose that we are in one of the cases described by Figures 13.3 or 13.4.

FIGURE 13.3

In other words we assume that a(/A2) <_ a(µ) <_ a(pi) b'µ E [111,'2], 1(O)

< a(µ)

ViA E (µ1,µ2)

in case of Figure 13.3,

(13.46)

- a(µ) d µ E {µl, µ2] i n case of Fi gure 13 .4 . µ To µl, µZ correspond two stationary points (see Theorem 13.3) given by u

l

=

112

a(µt)

<

=

(13 . 47) a(µ2)

13. Asymptotic Analysis

232

y= y = a(p)

P]

112

F1cURE 13.4

Then we would like to analyze the asymptotic behaviour of u. the solution to (13.12). We will restrict ourselves to the case where

ut < uo < U2

a.e. in Q.

(13.48)

First, let us show

THEOREM 13.6. Let a(t) _> in > 0 be a continuous junction. Let a' be a solution to

'AE L2(0,T;V)nC([O,T],L2(S2)), wi E L2(0,T;V'). a,(0) < 0

w(0)

0,

(13.49)

/

VvEV, u>0inV'(0,T).

f(w,v)+a(t) J

st

Then

w(.e, t) < 0

t1 t > 0, a.e..r E Q.

(13.50)

PROOF. Let U' be a smooth subdomain of 11 and v he the weak solution to

ui-Au=0 in51'x(O,T), v(0) = w(0), v(-, t) E 11O(U'),

(13.51)

I E (0,T).

(See (11.61)). We assume U' large enough so that. Ia'(0)Idx > 0. It is well known (see [171) that for any e > 0

v E C'((e,T) x U').

13.3. A nonlinear case

233

Moreover,

v(x, t) < 0 V (x, t) E 12' x (0, T].

(13.52)

To see this last point, it is enough to notice that by the weak maximum principle v(x, t) < 0 in 12' x [0,T]. (13.53) Moreover, for E small enough [v(', e)[2 > 0.

This is due to the fact that w(0) # 0 and v E C([0, T], L2(12)). The usual maximum principle applied to the domain 12' x [e,T] implies then (13.52) - see [60]. Next by the weak maximum principle - see Theorem 11.9 - we have to < 0. Moreover. setting vI

Jto a(s) ds)

we have in a weak sense vt = ve a(t) = a(t)Ov = a(t)AV,

with ii(0) = v(0) = w(0).

(The first above equation holds in 12' x (0, T') with f ' a(s) ds = T but T, T' can be chosen arbitrarily large). Then the weak maximum principle leads to

w
Then we turn to the following stability result that holds true when one is under the assumptions described by Figures 13.3, 13.4. Thus we have:

THEOREM 13.7. Under the assumptions above - in particular if (13.43), (13.44), (13.46). (13.48) hold, then, if u denotes the weak solution to (13.12),

U1 0.

(13.54)

PROOF. Let us assume first that

u1 < uo < u2 a.e. in 12.

(13.55)

Then, due to (13.43).

Al = 1(u0 < l(uo) < l(u2) =/U2Let us then denote by t. (t. > 0 by (13.56)) the supremum t` = Sup{ t > 0 [

s)) E [µl g2] `d s E [0, t] I.

(13.56)

(13.57)

We claim that t' = +,Do. Indeed, if not, then l(u(', t' )) =101 or µ2 (recall that u E C([0. +oo); L2(12))). Suppose for instance that

l(u(', t`)) = I`2

13.58)

13. Asymptotic.Analysis

234

(the proof in the case of equality with p, would be the same). Then, for almost every t it holds in L2(0, t': V') that u, - a(1(u))Du = f = -a(A2)Au12 (u - 112)1 - a(l(u))O(u - u2) = (a(112) - a(1(u)))(-Du2)

(u - u2)1 - a(l(u))A(u - u2) = (a(lit) - a(l(u)) a({1z)

Since 1(u) E [ zl,1121, by (13.44), (13.46) we deduce that.

(u - u2), - a(l(u))O(u - u2) < 0 a.e. t E (0, V). Setting w = u - u2 it follows that

ui - a(l(u))Ow < 0,

w(0) = uo - u2 < 0.

(13.59)

By Theorem 13.6 we deduce that

u'(t') < 0 which contradicts (13.58). Thus t' = +oc and by the above argument ul < u(t) < 112

in S2, VVt.

(13.60)

If now uo satisfies 42

= uI < uo <

112 =

a(pl) then for n - +oo it holds, that

ul < 0 =

(u1

-1-

n

a(I12)'

V uo A 1u2

-

n

< 112

(V denotes the maximum, A the minimum of two functions). Thus if u" is the solution to (13.12) corresponding to u", then

Un(t)EX={vEL2(f2)Iu,
(13.61)

for any t. X convex, closed in L2(!1) is also weakly closed and by (13.36), u(t) E X V t > 0. This completes the proof of the theorem. We return now to our original motivation for introducing the theory of dynam-

ical systems, to use this theory in our study of the asymptotic behaviour of the solution of (13.12). In what follows we assume that we are under the assumptions described in Figures 13.3 or 13.4. Moreover, X denotes the set given by (13.61). It is clear that X is a closed, bounded, convex open set of L2(!1). One can then equip it with the weak topology. This topology is metrizable (see for instance [65] ) for some distanced for which X is a complete and compact space. For uo E X, we define then

S(t)uo = tc(t)

where u(t) =

t,) denotes the solution to (13.12).

(13.62)

13.3. A nonlinear case

235

We can now state THEOREM 13.8. {S(t)}t>o defined by (13.62) is a dynamical system on X. PROOF. First the fact that S(t) maps X into itself follows from Theorem 13.7. The point (i) of the definition 13.1 follows from Lemma 13.1. The points (ii), (iii) are easy to check. Finally the last point follows from the fact that u E c([0, 71, L2(Sl)).

This completes the proof of the theorem.

Let us now find out a Lyapunov function for {S(t)}t>o. For this purpose we introduce 4) the solution to

O0 = I in fl, 4'E V.

(13.63)

Since I E L2(S2) C V' it is clear that (13.63) admits a unique solution. Moreover due to (13.43)

4)>0 in Q.

(13.64)

If now we take v = 4) in the last requation of (12.49) it follows that dt

u) + a(l(u)) J Vu vfi dx = (f, ) _ (-OAP, 4') n

Integrating by parts, we conclude that a.e. t E (0, +oo): d (4), u) = I(wp) - a(l(u))I(u).

(13.65)

If we choose uo E X it is clear that by Theorem 13.7

d

u) < 0

(13.66)

in the case of Figure 13.3 and, in the case of Figure 13.4,

d (4p, u) = 0.

(13.67)

Thus u' (4>, u) is a Lyapunov function on X. Let us now deduce from all of this the asymptotic behaviour of the solution u to (13.12). Let us assume first that we are in the case of Figure 13.3. Then one has:

THEOREM 13.9. Under the assumptions described in Figure 13.3 in particular if (13.46) holds, then for uo r= X, uo 0 u2

u(t) = S(t)uo - ul in L2(f2) when t --y +oc.

(13.68)

13. Asymptotic Analysis

236

PROOF. Let us denote by w(uo) the w-limit set of uo. By (13.65) and Theorem 13.7 we have (see (13.63) for the definition of (D)

d (,t, u) =1(p) - a(l(u))1(u) < 0.

(13.69)

Thus the scalar product with fi is a Lyapunov function for our problem. Clearly by (13.54) we have (13.70)

(4), u(t)) ? (4), ui ).

Moreover, since by (13.69) the function (4i, u) is nonincreasing, there exists C such

that urn (4?, u(t)) = C = (4), w)

(13.71)

for any w E w(uo) - (see Theorem 13.5). Going back to (13.69) for any w E w(uo) it holds by (13.33), (13.34) that

dt (4i, S(t)w) = 0 = l(ye) - a(l(S(t)w))l(S(t)w), i.e. due to the continuity of the map t '--+ l(S(t)w) 1(S(t)w) = p

Vt

or l(S(t)w) =142

Vt.

This implies that

l(w) = p,

µ2 Vw E w(uo).

or

Thus setting

W, = { w E w(uo) I l(w) =,u; wi are two disjoint closed sets of X such that w(uo) = w1 Uw2.

It follows that (see (13.26)) w(uo) = wi

or w2,

i.e. if u denotes the solution to (12.49). then

l(u(t)) -+ z1

or l(u(t)) -+ p2.

Let us assume that we have proved the following lemma: LEMMA 13.2. Let u be the weak solution to (13.12). If

l(u(t)) - pi (i = 1, 2)

(13.72)

when t -* +oc then u(t) -' ui

in L2(1l) strong.

(i = 1,2)

(13.73)

237

13.3. A nonlinear case

Thus if l(u(t)) -- tb2i then u(t) -+ u2. Since (4',u(t)) is non-increasing and '1 > 0, this implies that (4b, u(t)) _ (4', u2) Vt. Thus by (13.64), u(t) = u2 Vt which contradicts UO 0 u2. Thus

1(u(t)) - ki

0

and by Lemma 13.2 the result follows.

PROOF OF LEMMA 13.2. From (13.12) and the definition of u; we have a.e. t

in V'

(u - u;)t - a(l(u))Au = -a(µ:)Du; (u - u;)t - a(!(u))A(u - u;) = -(a(µ;) - a(1(u)))Du;. Taking the scalar product with u - u; and setting e(t) = la(pi) - a(l(u))I we get by (13.20)

2dtl"-u:l2+mf IV(u-u;)12dx<e(t)f IVu;IIV(u-u;)Idx.

in

in

(13.74)

By the Young inequality we get for any b > 0

f IVu,IIV(u-u;)IdxS 2 f IV(u-u.)12dx+2 j

IVu;I2dx.

(13.75)

Since e is continuous and bounded we can choose 6 such that 6IEI... = m

where IeI, = Supt E(t). Then the conjunction of (13.74), (13.75) leads to 2dt1u-u;I2+

2

IV(u-u;)I2dx
for some constant C independent of t. Thus we obtain for some constant c dt Iu - u;I22 + clu - u;12 2 <- Ce(t) (see Theorem 2.15). Setting

y(t) = Iu - u; Iz the above inequality reads y'(t) + cy(t) < CC(t) (e`ty(t))' < Ce(t)e`t.

(13.76)

13. Asymptotic Analysis

238

Integrating between to, t it follows that ecty(t) - ectoy(to) < C r e(s)e" ds

y(t) 5

rt +CJ e(s)ec(8-t) ds.

y(to)ec(t0-t)

to

co > 0 being given we choose to large enough such that

e(8) < C 2 ds > to. Then for t > to it holds that y(t) <

y(to)ec(1-t) +

<

1/(to)ec(tO-t)

<-

y(to)ec(to-t)

c

L It.t eH°-t) dg

+c

ec(C t) It to

22 +

Choosing now t large enough we get clearly

y(t) < eo

0

which completes the proof of the lemma.

Let us now assume that we are under the assumptions of Figure 13.4. Thus for any p E [JAI, µ2] (P

a(p) is a stationary point. Let us consider uo E X. Then by Theorem 13.7

u(t)EX

(13.77)

tl t E (0, +oo)

and by (13.65)

a (t, u) = 0 a (Cu(t)) = ((, uo) d t E (0, +oo).

(13.78)

From (13.78) it is clear that a natural candidate for the limit of u(t) is given by (13.77) with a(p) such that (4,w) a(y) = (t, uo), i.e. a natural limit of u(t) is

U. =

uo)

This is what we would like to establish now.

cp.

(13.79)

239

13.3. A nonlinear case

First we remark that, since w(uo) is compact in X equipped with the weak topology, there exists wo E w(u0) such that

a(l(wo)) = min a(l(w)).

(13.80)

wEw(uo)

(Recall that a and l are continuous). Let us set U*:_

W

(13.81)

a(l(wo)) Then we have:

LEMMA 13.3. For any w E w(uo) it holds that w < u*

in Q.

(13.82)

PROOF. Let w E w(uo), v(t) = S(t)w. Recall that by (13.81) -a(l(wo))Du' _

-AW = f, then in L2(0,T;V')

(v - u')= - a(1(v))0(v - u') = -(a(l(wo)) - a(1(v))Du'. Multiplying by (v - u* )+ it follows that 2 dt

I(v - u`)+I2 + 0(v))

J in

IV(v - u`)+ I2 dr

= a(l(wo)) - a(l(v)) (--Ou , (v - u')+) < 0.

This is indeed due to the fact that v E w(uo), (13.80), -Au- > 0. Thus for some constant c,

d I(v - u`)+I2 + cI(v - u')+I2 < 0 te°tI(v-u')+I2} <0 i.e.

I(v - u')+I2 < e-''Iv(0) - u'12, e-1Iw - U* 12

I(S(t)w - u')+ I2 <

< e-d Sup Iw - u` I2 wEw(o)

< Ke-`t. Due to (13.25) any z E w(uo) can be written in the form

z = S(t)w for t arbitrary. It follows that (z - u')+ = 0 Vz E w(uo) - which is (13.82). We can now show

O

13. Asymptotic Analysis

240

TiEoR,Ebt 13.10. Under the assumptions described in Figure 13.4, in particular if

t

a{ )

=

1(rp)

dµ E

(13 . 83)

1,Wi , 1

then for uo E X it holds that

u(t) - u

(13.84)

(where u is given by (13.79)) in L2(Sl). PROOF. We claim first that

u'
(13.85)

where u is given by (13.79). Indeed, if not - since both functions are multiples of

>0 - it holds that

u' > u,,. Let w E w(uo). Set w(t) = S(t)w. By (13.78), (13.34) we have then

(+,w) = (',uo) _ (),u00) <

(13.86)

Moreover v = w - u' satisfies

(w - u')t = Wt = a(l(w))Ow + f

= a(l(w))A(w - u') + (a(I(w)) - a(I(u')))Duweakly. But since we are in the case of Figure 13.4 by (13.82),

a(l(w)) > a(l(u')). Thus v = w - u' satisfies vt - a(l(w))Av < 0,

v(0) = w - u' < 0,

v(0) 0 0.

(v(0) 96 0 by (13.86)). Using Theorem 13.6 it follows that

w(t)-u' <0 Vt this for any w E w(uo). Thus I(w) < 1(u5) 11w E w(uo).

This contradicts a(1(w)o))

=

l(wo),

and proves (13.85). Thus for any w E w(uo) it holds that

w
u Vw E w(uo) - i.e.

(0, w(uo)

Thus the solution u(t) to (13.12) satisfies u(t) -- u.,, in L2(fl) and thus a(l(u(t)) -. a(1(ua,)). The result follows then from Lemma 13.2. 0

13.4. Blow-up

241

13.4. Blow-up Parabolic equations, like ordinary differential equations, may experience solutions

that do not exist globally. We say then that they "blow up" in finite time. The diagnosis of such an occurrence proceeds always from the same principle: one exhibits a scalar quantity that cannot exist globally - due for instance to the fact that this quantity satisfies an ordinary differential equation whose solution blows up in finite time. This is the reason why we consider first the case of an ordinary differential equation.

13.4.1. The case of an ordinary differential equation. Let us consider the differential equation ut = f(u)

dt > 0, (13.87)

U(0) = uo

where t

u(t) is a function defined on [0, T) with values in R. Then we have:

THEOREM 13.11. Assume that f is a continuous, positive function on [uo, + oc) such that P+00 ds

t' = J

(13.88)

+00.

s

Then there exists a unique solution to (13.87) with maximal existence time to -i.e. which blows up at to given by (13.88).

PROOF. From the positivity of f we deduce that solving (13.87) is equivalent to finding u(t) such that 1,

f(u)

U(0) _ U0,

i.e. after integration in t such that

.

(t) ds JU.U

f (8)

t-t

/'+«,

J,.

ds

f(8)

Since

fu' a f (s) is increasing this equation is uniquely solvable for any t < t` and Bra u(t) _ +00.

tr t' is called the blow-up time of the solution.

EXAMPLE 1 /3.1. Consider uo > 0, f (u) = up with p > 1. Then

-p+ 1 "

s-p ds =

1

s-n+1I= +U0

1

(p-

1)uo-ds

(13.89)

13. Asymptotic Analysis

242

In the particular case where p = 2, it is well known that the solution to the problem

Jut = u2 t > 0, lu(0) = ua > 0.

blows up. In fact the blow-up time is t' = uo-t as we have just seen above. As a corollary we can show: COROLLARY 13.1. Let f be a continuous, positive function from Iuti, +oo) such

that

t' =

J+m ds

< +ou.

fg )

u

(13.90)

Let u be a function satisfying

fut > f (u) a.e. t > 0, u(0) = U0,

(13.91)

then u is defined at most until the time t'. PROOF. From (13.91) it follows that Ut) > u

f

1

a.e. t > 0,

hence after integration in t

lu") ds `

u

which completes the proof.

13.4.2. Some blow-up techniques in PDE's. The goal of this section is to provide some examples where the solution of an evolution problem fails to exist for every time t. Also we will insist on different methods. Let us begin with a simple example. 13.4.2.1. A simple example. Let us consider u a weak solution to

ut - Au = f (u) in n x (0, T), C9u

on r x (0, T),

t77a=0 u(., { 0) = un,

(13.92)

(see (12.30)). If f > 0 then the source term is positive i.e. the system is provided with heat. On the other hand the boundary condition du

=0

prevents any heat from escaping. So, in such a situation one expects the solution to blow up if the heat brought into the system is sufficient. More precisely one has

13.4. Blow-up

243

THEOREM 13.12. Assuming uo E L2(1) set

uo =

ll f uo(x) dx.

Let f be a positive, convex function on [uo, +oo) such that

= +oo ds to

f(3)

< +oo,

then (13.92) cannot have a smooth solution after the time to. PROOF. If u is a weak solution, then for any v E H' (f2) dt (u, v) +

Jn

Vu Vv dx =

Jn

f (u)v dx

a.e. t

(note that the smoothness required here is just f (u) E L2(11)). Taking v = 1

fudx=jf(u)dx. Dividing by lftl and setting u

Jnudx lffil

we get by Jensen's inequality for a.e. t

f

dtu

( Jnf(u)dx'f`InlJnudx)=f(a). I1

0

Since u(0) = uo the result follows by Corollary 13.1. REMARK 13.2. A suitable choice would be

p > 1, f(u) = uP, Then, by the Example 13.1, to < +oo.

uo > 0.

13.4.2.2. The Kaplan technique. Let us consider u a smooth solution to

Iut - Du = up t) E Ho (ft)

in fZ x (0, T), p > 1, t E (0, T),

(13.93)

By the maximum principle

u>0. Moreover, let us introduce V the first eigenvalue of the Dirichlet problem - i.e. ep is the solution to

App = \IV in ft, (13.94)

E Ho(fl)

13. Asymptotic Analysis

244

It is well known that co > 0 and without loss of generality we can assume that

Vdx=1.

(13.95)

We have:

THEOREM 13.13. Under the above assumptions and if

fn

(13.96)

uocp dx

is large enough, then (13.93) cannot have a smooth solution for all time.

PROOF. We multiply the first equation of (13.93) by V and integrate on Q. We get

in utp dx -

J in

Au

dx = J u'cp dx. a

Integrating the second integral by parts and using the Jensen inequality in the

third integral (since by (13.95) cp dx is a probability measure) it follows that

Hence by (13.94)

v= ju.sodx satisfies

Vt>vA - Aiv. Clearly, for v large enough the function v *- vA - alv satisfies the assumption of Corollary 13.1. Thus, for (13.96) large enough, the solution to (13.93) cannot exist 0 for all time. This completes the proof of the theorem.

13.4.2.3. The energy method. We consider again a solution to (13.93). Then we have

THEOREM 13.14. Assume that

E(uo)= 1j jVuoj2dx-p+1

jurldX<0,

(13.97)

then the problem (13.93) cannot have a smooth solution for all time. PROOF. By smooth we mean a solution such that all the computations below make sense. First we note that since

u(x,t)=0 onrx(O,T) we have

ut(x, t) = 0 on t x (0, T).

13.4. Blow-up

245

Thus, multiplying the first equation of (13.93) by tit and integrating on Q. we obtain Jsz

uidx=j

J u''uidx. st

:

Integrating by parts in the second integral we obtain

\

Judx=-

uP+'dx)

n

st

t

1

I- 1 I IVuI2dx+ P + I J updr 1

2

st

s1

)t*

Here we call energy the quantity

- '

E = E(u) = 1 J IVu12dx

20

P+ I

J up+1 dx. si

Then from the above relation it is clear that dE

(13.98)

u2, dx < 0.

dt

s1

Thus the energy is non-increasing along the trajectories. Consider then

v=

1 u2 21 S2I

dx.

st

One has vt ISI 1

=

o

IS21

u (Au + up) dx

J up+l dxl

I IVU12dx+

IS 21

1

2E -

(by (13.93))

2

uP+' dx +

up+I dx

P+ 1 in Assuming (13.97). by (13.98) E remains negative and it holds that IQ-I

U1 >

_L (I If21\

Recalling that

P21

-

1

2

P+I Ju

UP+s dx =

\

I-

2

P+1) illIJs:

(112)

> 1 since p > 1. by Jensen's inequality we get

V,>l

P+Il`A""2`

i.e.

11>I1

2

p+1

?''

)

dx.

13. Asymptotic Analysis

246

Since by (13.97) we clearly get

uo dx > 0,

v(0) = Ink fa then by applying Corollary 13.1 the result follows This completes the proof of the theorem.

REMARK 13.3. It is easy to find up such that (13.97) holds. Indeed choosing

u'>0,

u100

it is clear that UO = Au1

satisfies (13.97) for A > 0 large enough.

13.4.2.4. The concavity method. Let A denote a positive number. Consider M = M(t) a positive function satisfying

(M-a)" < 0,

M'(0) > 0,

(13.99)

M_a

is concave with an initial positive derivative. Then clearly i.e. the function such a function has to come back to 0 - i.e. M has to blow up. Indeed using the concavity of M-' we have

M-A(t) < M-.,(0) + (M--`)'(0)t, i.e.

M-'`(t)< M-1(0) -AM-a-'(0)M'(0)t which imposes

t < t' = M-'`(0)/AM-a-1(0)M'(0) = M(0)/AM'(0),

(13.100)

and t' provides an upper bound for the blow-up time of M. Let us now compute a sufficient condition for M-a to be concave. We have

(M-1 _ -w-A-10

-

(M--\)"

_

1)M-a-2Mrs

-AM-A-s{MM"

- AM-a-1M"

(13.101)

- (A+ 1)M'2}.

Thus a sufficient condition for concavity reads

MM"-(A+1)M12>0. Let us go back to the problem (13.93) and consider for some constant 6 > 0

M(t) = M =

j

ju2(r,8)dxda +;8. t

(13.102)

13.4. Blow-up

247

Note that Al is positive when uo 0 0 and increases with time. Moreover

All = r u2 dx = n

r

re

r

Jo\r Jnu2 dx)t ds + / nuo dx rt

2J

r r uutdxds+1 u,2odx n

o

(13.103)

n

= 2 J / uut dx ds + Co o n ft

with an obvious definition for Co. We derive then

M" = 2 / uut dx n

r (rn uut dx)

=2

o

e

ds + 2 J uut dxl n o

(13.104)

re / r / uut dx ds + C1.

2J

I

n

o

t

Note that (for uo E Ho (St)):

Cl = 2 J

nuut

dxl 0

= 2 (u(Du + un) dxl n

0

(13.105)

=211 uo''dx - J IVuol2dx}. Let us write Al" differently. Going back to (13.104) it holds for any C that e ft r ( ft l r M" = C ut2 dx ds } + Cl. dx ds + { 2 J ( uut dx It ds - C /

J J nut

ll

o

/

n

o

do

Jn

1J

Next remark that 2

(M')2 = (21 r uutdxds+Co l 2

=4

t

+ 4Co (fot

o rt

< (4 + E)

f Juutdxds-i-Co

\

J 0

2

C,C02 Jnuut dx da) +

for some constant Q. (we used the Young inequality). Using Cauchy-Schwarz inequality it follows that roe

rot

r

M'2 < (4+e) / f u= dads. / J u2dxds+CCC n

n

rt

(4+E)M f Juidads+CeCo. o

n

13. Asymptotic Analysis

248

Hence

MM" - 4 +e

M'2

re

>

M{2

Then remark that

rt / r

2J

uut\

Je dx

J / o\n I

2f

d3 - C

J

r

(1

n

uut dz) ds - C

rc

Jo

ui da da + C1o

fn

rtr utut dx ds J Jn o

t(fU(tU+U1'))dXd8 -C fJ ut(Au+u")dxd3 o

fitf

o

t

n

r =21t(-IIVulI2 ),da+2(p+1)10tI u Put dx ds IIVu1

ds.

Choosing C = 2(p + 1) we get

rt

e

2f fn(uut)tdxds-C

fo

f

utdxds=(p-1)

f

t

21

IIVuII2 ds

= (p- 1){I,Vu,I2

- IIVuoil2}.

Since M > Q we obtain r

+2IuoIP+;-2IIVuOj12 - 2Cr((4 + 1)IUI21 +c)Q 1

>0 for some data uo for which blow-up occurs. There are of course many other techniques regarding blow-up. One could for instance consult [10], (45], [511, [52], [611, 1661.

13.5. Exercises 1. Justify the computations made in the proofs of Theorem 13.1 and Theorem 13.2 using the weak formulation of these problems. 2. Consider l : V R, homogeneous of degree a - i.e. 1(.1v) = ). I(v) Vv E V, bA > 0.

249

13.5. Exercises

Show that the problem

a(l(u))Au = f in (1, uEV

has as many solutions as the equation

a"(! A = 1M in R, p being the solution to (13.14). are m functions as above - i.e. from V into R Assume that It, ... and homogeneous of degree a;, i = 1, ... , m. Let a be a function from RI into (0, +oa). Show that solving the problem /

a(Ij (U),12(u), ... , I,n(U))AU = J

m Q,

uEV reduces to solving a nonlinear system in R. 3. Prove the equality (13.22). 4. Show that the family {S(t)}t>o defined by (13.62) satisfies the points (ii), (iii) of the definition 13.1. 5. if one replaces the case of Figure 13.3 by a case where

a(µ1) < a(µ) < a(,U2) Vu E [Al, A21,

a(µ) <

bµ E (Al, µ2),

show that for ul < ILO !5 U2,

tLO 34 UI

the solution u to (13.12) converges toward u2 in L2(S1). 6. Consider the so-called Thermistor problem - i.e. (u, gyp) is a solution to:

Ut = V (rc(u)Vu) + au = 0 on r x (0,T),

in (1 x (0, T),

(x, 0) = uo(x) in S2,

(T)

V (a(u)VV) = 0 in 0 x (0, T), cp = po on r x (0, T). (Here u is a temperature, V an electrical potential (see [8] for details)). One assumes that S2 is a smooth bounded open set of R' and 0 < tc(s), or (s) < +oo V s > 0, o differentiable, d (s) > 0 V s > 0,

00 ds

f

() <+oo, uo>0.

os

13. Asymptotic Analysis

250

a) For p EHI(1)set `p(x) d-r(x)

Irl

where IrI denotes the measure of r, dy the measure area on r. Show that there exists a constant C such that

Jr

I(p -

0I2 d7(x) < C In Ivy012 dx V p E H' (Q).

(*)

b) Denote by C the smallest constant such that (*) holds. Show that if /'+'° p 1 ds

rr

dx <

s)

C

J

o

J

r

l

Col dry (r) dt

where cpo = - L fr `po(x) d-y(x), then the problem (T) cannot have a smooth

solution for all time. (Hint: consider Y(t) = f f }O° `t" dx). i(xj) WM

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Index n.eimplex, L12

0-limit set, 22fi

Heat equation, 185

Asymptotic analysis, 22; 229.

Interpolation, 113 Interpolation error, 113 Invariance principle, 228

mow-up, 242, 248 Mow-up time, 241 Brouwer fixed point, ¢Q

Lax-Milgram, 41 Linear parabolic equations, 185

calculus, 7

Local nonlinear problems, 95 Local problems, 207

Corathbodory function, 5s

Coercive, 41 Coincidence set, 54

compactness, 59 Caanormal derivative, 47 Convex, 49

Deformation, 1 5 Density, 3 Density of population, 9 D1Husion, 9

Direct method of calculus of variations, 134

Diithlet, 12, 39 Dirichlet problem, 9 Dlvrocgence, 12

Divergence Theorem, 211

Dynamical system, 22fi

mastic membrane, 52 Elasticity, 1 Elliptic linear problems, 39 Ellipticity condition, 43

Local, 1110

Maximal monotone graph, 72 Membrane, 5 Mesh size, 105, 107, L12 Minimizers, 131 Minty, 52 Monotonicity, 62 Multivalued problem, 71

Neumann problem, 47 Nonlinear parabolic problems, 207 Nonlocal, 1QIZ

Nonlocal problems, 215 Nonlocal variational inequalities, 85 Numerical analysis of oscillations, 166

Obstacle problem, 52. 54

Parabolic equation, 13 Poincare Inequality, 32 Quasi-uniform, 110

lay, 2 245

Reference configuration, 1 5 Regular, 116

Finite element, 1115

Sobolev spaces, 21

Pfee boundary Problem, 54

Stationary point, 221

256

Index

Steady state, 221-223 Stored energy, 3 Strong solution, 39

Variational inequality, 49, 52 Variations, 7 Vector valued distributions, 1S6

Trajectory, 226 Triangulation, 1R2

Weak formulation, 40 Weak maximum principle, 202, 233 Well posed, 14

Uniqueness, 95 Unit normal, 20

Yosida approximation, Y2 Young inequality, 193

diffusion, serve to motivate the theoretical parts that are then applied to various aspects of elliptic and parabolic problems. In particular, existence, uniqueness, regularity and approximation of solutions for quasilinear and monotone problems are studied, as well as some new aspects of the calculus of variations including Young measures or approximation of minimizing sequences.

783764 364069

1 *1 SON-S, NMI-111-S,