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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 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 (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 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) >_ 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 -
,(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,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) (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)(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 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 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) 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 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 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] 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 0 set
K,={vEHo(f2) I J[v] (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 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 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' 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 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 -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 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
)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, - 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
0G(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,) 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 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
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 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' 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, 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 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
,{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) 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) 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 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
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' 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
jurldXl
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)
