A unilateral problem for elliptic equations with quadratic growth and for L 1 -data
Mazen Saad
MathÃematiques AppliquÃees de Bordeaux, UniversitÃe Bordeaux I 351 cours de la LibÃeration, 33405 Talence, France
Received 20 January 1998; accepted 27 January 1999
Keywords:Nonlinear elliptic equations; L1 theory
1. Introduction
In [3,6,7], the existence of solutions for some quasilinear elliptic equations for a quadratic growth or subquadratic growth with respect to gradient is studied. In these works, the problem is the following:
Au+g(: ; u;∇u) =f in ;
u= 0 on @ (boundary of ); (1.1)
where is a bounded domain of RN(N ≥1), A a second order operator, uniformly elliptic with bounded coecients and g: ×R×RN →Ris a Caratheodory function which satises a sign condition (i.e. sg(x; s; )≥0).
In [7], the existence of solutions is obtained for all integrable function f and the function g has a subquadratic growth with respect to the variable , whereas in [3], the authors establish the existence of solutions where f∈H−1() with a quadratic dependence for the nonlinearity. In the case where the operatorA is strongly nonlinear (i.e. satisfying the hypotheses of Leray–Lions [10,11]) Del Vecchio [6] has obtained a result of existence for problems with gradient-dependent lower-order nonlinearity. In this latter work, the sign condition plays a principal role to obtain a priori estimates and existence of solutions.
E-mail address:saad@math.u-bordeaux.fr (M. Saad).
0362-546X/01/$ - see front matter?2001 Elsevier Science Ltd. All rights reserved.
PII: S0362-546X(99)00260-6
If the function g is given as g(x; u;∇u) =a(x)u+ g(u)|∇u|2 where a(x) ≥a0¿0 and g is continuous, Boccardo et al. [4] establish the existence of solutions inH01()∩ L∞(). This result is recovered by the work of Boccardo et al. [5] where the authors proved the existence of solutions for unilateral problems. In this latter work, the datum f is considered to be bounded, g grows like|∇u|p, and the operator A is an operator of Leray–Lions that contains a term written in a nondivergence form in order to ensure that the obtained solutions are bounded. For example if A is the Laplacian operator, the presence of term of 0th order plays an important role to prove that the solutions are bounded. Indeed, a counter-example due to Kadzan and Kramer [9, p. 637] shows that if g(x; u;∇u) =−|∇u|2, f∈L∞() and f(x) ≥1 a.e in (1 being the rst eigenvalue of −), then problem (1.1) has no solution in H01()∩L∞().
In the case whereg=0,A is the p-Laplacian operator andf∈L1(), Boccardo and Gallouet [2] establish the existence of solutions of a variational inequality.
Also, a pointwise comparison between the solution of nonlinear variational inequality and the solution of a suitable variational inequality dened in a ball with spherical data is proved in [13]. This comparison allows to give a priori estimates for the Lp-norm of u and ∇u in terms of the norm of f in suitable Lorentz spaces. This latter result is obtained for problems with gradient-dependent lower-order nonlinearity.
In this paper the existence of nonbounded solutions to some nonlinear elliptic equa- tions for unilateral problems is investigated. No growth assumption is imposed on the function g(x; s; ) with respect to the variables, but the dependence of can be con- sidered to be quadratic or subquadratic. Furthermore, the function g is assumed to be nonnegative. Finally, the solutions are considered to be also nonnegative. The model problem is to consider
g(x; u;∇u) =|u|q|∇u|r; q ¿0; 0≤r≤2:
The existence of solutions is proved via a sequence of approximated and penalized problems, with solutions u. To obtain some a priori estimates a second approximation is used, with solution u. A major tool in this method is therefore played by suitable compactness properties of the sequence u in some Sobolev space and the use of Fatou’s lemma.
2. Main results
Let be a bounded domain of RN(N ≥ 1), with boundary =@. For i; j∈ {1;2; : : : ; N} the coecients aij dened on verify
aij∈L∞()
∃ ¿0;
XN i;j=1
aij(x)ij≥||2; a:e inx∈: (2.1) The operator
A=XN
i;j=1
@
@xj
aij(x) @
@xi
:W01;p()→W−1;p()
is dened by hAu; vi=
XN i;j=1
Z
aij(x)@u
@xi
@v
@xj dx; wherep0= p p−1:
Furthermore, let g: ×R×RN →R+ be a Caratheodory function:
(i) for all (s; )∈R×RN; the functionx→g(x; s; ) is measurable;
(ii) for almost allx∈;the function (s; )→g(x; s; ) is continuous; (2.2) which satises the growth condition
|g(x; s; )| ≤b(x; s)||r; with 0≤r≤2 where the functionb veries :
for 0≤r ¡2; sup
|s|¡Áb(x; s) =hÁ(x)∈L2=(2−r)(); ∀Á≥0 for r= 2; b(x; s) =b(|s|) andb is an increasing and continuous function fromR+ toR+;
(2.3)
and
∀∈RN; g(x;0; ) = 0 a:e:in : (2.4)
Finally, we recall the denition of a truncated function. For all k∈R+, the function Tk(z) is dened by
Tk(z) =
z if |z| ≤k;
k if z ¿ k;
−k if z ¡−k:
Theorem 1. Assume (2:1)–(2:3). Then; for all f∈L1(); there exists a solution u satisfying:
u∈K={u∈W01;p(); u≥0a:e in }; for allp∈
1; N N−1
; (2.5)
Tk(u)g(: ; u;∇u)∈L1(); g(: ; u;∇u)∈L1(); (2.6)
Tk(u)∈H01(); ∀k ¿0; (2.7)
hAu; v−Tk(u)i+ Z
g(: ; u;∇u)(v−Tk(u)) dx≥ Z
f(v−Tk(u)) dx; (2.8) for all v∈D+() ={v∈D(); v≥0}:
Remark 1. If f∈L2() and under the assumptions of Theorem 1 we can take p= 2 and Tk(u) =u. Precisely, there exists a solution u satisfying
u∈K={u∈H01(); u≥0 a:e:in } ug(: ; u;∇u)∈L1(); g(: ; u;∇u)∈L1() hAu; v−ui+
Z
g(x; u;∇u)(v−u) dx≥ Z
f(v−u) dx; ∀v∈K∩L∞():
Remark 2. The proof of this theorem can be adapted to the case where the coecients aij=aij(x; u) are assumed to be Caratheodory functions and verify (2.1).
3. Proof of the theorem
The proof follows in ve steps. In the rst, we consider a method of penalization and approximation to approach the solutions of the problem. Next, a priori estimates are obtained on the solutions and we give some consequences of these estimates. Finally, to pass to the limit, we distinguish the case where the dependence is quadratic or subquadratic with respect to the gradient.
3.1. Penalization and approximation According to [3,6], let us dene
g(x; s; ) = g(x; s; ) 1 +|g(x; s; )|:
Let also (f) be a sequence of regular functions, particularly in L2(), such that f→f inL1() and|f|L1()≤ |f|L1(): (3.1) Now, consider the approximating and penalizing problem:
Au+g(: ; u;∇u)−1
(u)−=f inD0();
u∈H01(); f∈L2(); (3.2)
where (u)−=−min(0; u):
Using the fact thatg(: ; u;∇u) is bounded, it is possible to use a xed-point theorem of Schauder to prove existence of solutions in H01(). Nevertheless, it seems dicult to obtain a priori estimates, due to the fact that if we multiply Eq. (3.2) by u the quantity ug(x; u;∇u) has no sign.
In order to avoid this inconvenience, we approach the sign function by a continuous and increasing Lipschitz function. Set
S(z) =
z−
z if z≥ ¿0;
−z−
z if z≤ − ¡0;
0 if |z| ≤:
(3.3)
Now, we set
g(x; s; ) =S(s)g(x; s; ) (3.4)
and consider the equation Au+g(: ; u;∇u)−1
(u)−=f inD0();
u∈H01(); f∈L2():
(3.5) Then, the function g veries
sg(x; s; )≥0; ∀(s; ); a:e:in (sign condition);
|g(x; s; )| ≤ |g(x; s; )| ≤b(x; s)||r (growth condition);
|g(x; s; )| ≤ 1
(i:eg∈L∞()):
(3.6)
Since g∈L∞(), for every ¿0 and ¿0, there exists u∈H01() such that XN
i;j=1
Z
aij(x)@u
@xi
@v
@xj dx+ Z
g(x; u;∇u)vdx
−1
Z
(u)−vdx= Z
fvdx ∀v∈H01(): (3.7)
The above result is well known [3,7]. In this case, it is possible to use a xed point of Schauder. For clarity, we give a short proof. Let us introduce the following closed convex subset in H01():
C={w∈H01();|∇w|L2()≤M};
whereM=cp=|f|L2()+cp=||1=2,cp is the Poincare constant and is the constant of coercivity which appears in (2.1).
We denote by T the mapping T(·) :v∈C7→u
where u∈H01() is the solution of XN
i;j=1
Z
aij(x)@u
@xi
@
@xjdx−1
Z
u−dx
= Z
g(x; v;∇v)dx− Z
fdx ∀∈H01(): (3.8)
We shall show that this mapping has a xed point in C.
First of all,T is well dened onCand we show that obviouslyT(C)⊂C. Indeed, choosing=u as test function in (3.8), and owing to the coercivity of the coecients
aij we have
|∇u|2L2()+1
|u−|2L2() ≤ |f|L2()|u|L2()+1
||1=2|u|L2()
≤ cp|f|L2()|∇u|L2()+cp1
||1=2|∇u|L2(); then, from the positivity of the second term we deduce that u belongs in C.
To show the continuity of the operator, let us consider the sequence (vn)n∈C such that vn→vinH01(); we consider the associated sequence un=T(vn) and u=T(v).
Then, for all ∈H01(), we have XN
i;j=1
Z
aij(x) @un
@xi − @u
@xi
@
@xjdx−1
Z
(u−n −u−)dx
= Z
(g(x; vn;∇vn)−g(x; v;∇v))dx:
Taking =un−u, we deduce
|∇(un−u)|2L2()−1 Z
(u−n −u−)(un−u) dx
≤cp|g(: ; vn;∇vn)−g(: ; v;∇v))|L2()|∇(un−u)|L2() and from the positivity of the second term, one gets
|∇(un−u)|L2()≤cp|g(: ; vn;∇vn)−g(: ; v;∇v))|L2():
The last term goes to zero as n goes to innity via the Lebesgue’s theorem, and the continuity of the operator is held.
Finally, to show that the mapping T is compact, we consider a bounded sequence (vn)n∈C, then the sequence Fn=g(x; vn;∇vn)−f is bounded in L2() and conse- quently it is relatively compact in H−1(); thus, the associated sequence un=T(vn) is relatively compact in H01() via the continuity of the operator dened by (3.8) from H−1() to H01().
3.2. A priori estimates
To obtain a priori estimates, we test with the functions used by Boccardo and Gallouet [1,2]. For that, let us dene k(z) by
k(z) =
1 if z≥k+ 1;
z−k if k≤z ¡ k+ 1;
0 if −k ¡ z ¡ k;
z+k if −k−1¡ z≤ −k;
−1 if z≤ −k−1:
(3.9)
The function k is Lipschitz and k(0) = 0. Then according to Stampacchia [14], if u∈W01;p() then k(u)∈W01;p() and ∇k(u) =0k(u)∇u= 1Bk∇u with
Bk={x∈; k ≤ |u|¡ k+ 1}: (3.10) Proposition 1. Assume (2:1)–(2:3)and (3:1). The solutions of (3:7) satisfy
u is bounded in W01;p()uniformly in and ; ∀p ¡ N=(N−1) (3.11) g(: ; u;∇u)is bounded in L1()uniformly in and : (3.12) 1
(u)− is bounded in L1() uniformly in and : (3.13) Tk(u)is bounded in H01()uniformly in and ;∀k ¿0: (3.14) 1
k Z
|∇Tk(u)|2dx→0 as k→ ∞; uniformly in and : (3.15) Tk(u)g(: ; u;∇u)is bounded in L1() uniformly in and ; ∀k ¿0: (3.16) Proof. Taking v=k(u) in (3.7), we get
XN i;j=1
Z
aij(x)@u
@xi
@k(u)
@xj dx+ Z
g(x; u;∇u)∇k(u) dx
−1
Z
(u)−k(u) dx= Z
fk(u) dx: (3.17)
Note that the sign of k(u) is the same as that ofu. Using the sign condition on g and the positivity of the third term, it follows that
|∇u|2L2(Bk)≤XN
i;j=1
Z
aij(x)@u
@xi
@u
@xj1Bkdx≤ |f|L1()|k(u)|L∞()≤ |f|L1(): (3.18) Then using Lemma 1 of [1, p. 155], we have
∀p∈
1; N N−1
;∃c(p;|f|L1()) such that Z
|(∇u)|pdx≤c(p;|f|L1()) which gives (3.11).
Moreover, from (3.17), one gets Z
g(x; u;∇u)k(u) dx= Z
Bk
g(x; u;∇u)k(u) dx +
Z
{x∈;|u|¿k+1}|g(x; u;∇u)|dx≤ |f|L1(); (3.19)
which implies Z
{x∈;|u|¿k+1}|g(x; u;∇u)|dx≤ |f|L1():
Taking k= 0 in the above inequality, shows that to establish (3.12) it is sucient to get
Z
{x∈;|u|≤1}|g(x; u;∇u)|dx≤C;
where C is a constant independent of and . To establish this result, we distinguish the case where the dependence with respect to gradient is subquadratic 0 ≤ r ¡2, from the case where the growth condition is quadratic r= 2.
We haveZ
{x∈;|u|≤1}|g(x; u;∇u)|dx≤ Z
{x∈;|u|≤1}b(x; u)|∇u|rdx: (3.20) If 0≤r ¡2, from hypothesis (2.3), one gets
Z
{x∈;|u|≤1}b(x; u)|∇u|rdx≤ Z
{x∈;|u|≤1}h1(x)|∇u|rdx;
and the Holder’s inequality gives Z
{x∈;|u|≤1}h1(x)|∇u|rdx≤ Z
{x∈;|u|≤1}(h1(x))2=(2−r)dx
!(2−r)=2
Z
{x∈;|u|≤1}|∇u|2dx
!r=2
≤ |h1|Lr=(2−r)(B0)|∇u|rL2(B0);
where B0={x∈;|u|¡1}. Taking k= 0 in (3.18) and from (2.3), follows (3.12).
If r= 2, using assumption (2.3), we have b(|u|)≤b(1) if |u|¡1. From (3.18), one hasZ
{x∈;|u|≤1}b(x; u)|∇u|rdx≤ Z
{x∈;|u|≤1}b(1)|∇u|2dx≤b(1) |f|L1() which leads to (3.12).
We show now estimate (3.13); it is an essential result to obtain compactness properties.
Taking v=k(−((u)−=)) as test function in (3.7), we deduce
−1
Z
(u)−k
−(u)−
dx
≤ 1
XN i;j=1
Z
{x∈; k≤((u)−=)¡k+1}aij(x)@(u)−
@xi
@(u)−
@xj dx
+ Z
g(x; u;∇u)k
−(u)−
dx−1
Z
(u)−k
−(u)−
dx
= Z
fk
−(u)−
dx
≤ |f|L1() and therefore
Z
{x∈; k≤((u)−=)¡k+1}
−(u)−
−(u)−
−k
dx +
Z
{x∈;((u)−=)≥k+1}
(u)−
dx≤ |f|L1():
By taking k= 0 in the above inequality, we deduce that Z
{x∈;((u)−=)¡1}
(u)−
2 dx≤ |f|L1()
and Z
{x∈;((u)−=)≥1}
(u)−
dx≤ |f|L1():
From the rst inequality, the sequence (u=); is bounded inL2({x∈;(u)−= ¡1}), thus bounded in L1({x∈;(u)−= ¡1}), and from the second inequality follows the estimate (3.13).
Finally to obtain estimates (3.14) and (3.16), we test with v=Tk(u) in (3.7).
We have XN i;j=1
Z
aij(x)@u
@xi
@Tk(u)
@xj dx+ Z
g(x; u;∇u)Tk(u) dx
−1
Z
(u)−Tk(u) dx= Z
fTk(u) dx: (3.21)
Set
Ak={x∈;|u|¡ k}:
From the positivity of the second and the third terms in (3.21), we have
Z
Ak
|∇u|2dx= Z
|∇Tk(u)|2dx≤XN
i;j=1
Z
aij(x)@u
@xi
@u
@xj1Akdx
≤ |f|L1()|Tk(u)|L∞()≤k|f|L1() and
|Tk(u)|H1
0()≤ k
|f|L1(); which establishes (3.14).
Estimate (3.15) is obtained as in [12, p. 188]. In fact, (3.21) yields for all M∈(0;∞)
k
Z
|∇Tk(u)|2dx≤1 k
Z
|f||Tk(u)|dx
≤1 k
Z
{x∈;|u|≥M}|f||Tk(u)|dx +1
k Z
{x∈;|u|≤M}|f||Tk(u)|dx
≤ Z
{x∈;|u|≥M}|f|dx+M k
Z
|f|dx:
Our claim then follows since on the one hand, f converges to f in L1(), and meas{x∈;|u| ≥ M} goes to 0 as M goes to ∞ uniformly in and , and thus R
{x∈;|u|≥M}|f|dx goes to 0 as M goes to ∞ uniformly in and ; next, on the other hand, the last term in the above inequality goes to 0 as k goes to ∞ uniformly in .
Finally, estimate (3.16) is a direct consequence of (3.21). One deduces Z
g(x; u;∇u)Tk(u) dx≤k|f|L1(); which completes the proof of Proposition 1.
3.3. Compactness results
We give now a result of almost everywere convergence of gradient of the sequence (u);.
Lemma 1 (Gallouet [7,8], Boccardo et al. [3] and Stampacchia [14]). Let(u);be a sequence of H01() such that
XN i;j=1
Z
aij(x)@u
@xi
@
@xjdx= Z
F; ∀∈D(); (3.22)
where (F); is bounded in L1() uniformly in and . Then; the sequence (u); is relatively compact in W01;p(); for all 1≤p ¡ N=(N−1).
Set F=f−g(: ; u;∇u) + 1(u)−, then (3.7) is equivalent to (3.22). Owing to estimates (3.1), (3.12) and (3.13), the sequence (F); is bounded in L1(). Hence, there exists a subsequence of (u); (still denoted by (u);), such that
∇u →almost everywhere in : (3.23)
3.4. The limit as →0, for ÿxed
In this subsection, we consider a xed positive .
The sequence (u) is bounded in W01;p(), for 1≤ p ¡ N=(N −1) (see (3.11)).
Hence, there exists a subsequence of (u), which for the sake of simplicity will still be denoted by (u), such that
u→u weakly inW01;p();strongly inLp() and a:e in : (3.24) Using Lemma 1 and (3.23), we have, for a subsequence of (u) (still denoted (u))
∇u → ∇u a:e in : (3.25)
As→0, the nonpositive part of the subsequence (u) converges to zero inL1() for the strong topology (see (3.13)), then
u≥0 a:e in ; ∀ ¿0: (3.26)
Now, let us give the inequality satised by the sequence (u).
Let v∈D+() ={v∈D(); v≥0}. We test with =Tk(u)−vin (3.7), to obtain XN
i;j=1
Z
aij(x)@u
@xi
@Tk(u)
@xj − @v
@xj
dx+ Z
g(x; u;∇u)(Tk(u)−v) dx
−1
Z
(u)−(Tk(u)−v) dx= Z
f(Tk(u)−v) dx:
We have XN i;j=1
Z
aij(x)@u
@xi
@Tk(u)
@xj dx+ Z
g(x; u;∇u)Tk(u) dx− Z
fTk(u) dx
− XN i;j=1
Z
aij(x)@u
@xi
@v
@xjdx− Z
g(x; u;∇u)vdx+ Z
fvdx≤0:
Let us dene
h=hAu; Tk(u)i+ Z
g(x; u;∇u)Tk(u) dx− Z
fTk(u) dx: (3.27) Then the above inequality is equivalent to
h− hAu; vi − Z
g(x; u;∇u)vdx+ Z
fvdx≤0; ∀v∈D+(): (3.28) Lemma 2. Let the sequence(u)satisfy(3:1); (3:14); (3:16); (3:24)and (3:25).Then
lim inf
→0 h≥h; where
h=hAu; Tk(u)i+ Z
g(x; u;∇u)Tk(u) dx− Z
fTk(u) dx;
and g(: ; u;∇u) =S(u)g(: ; u;∇u):
Proof. The passage to the inferior limit on the rst and the second terms in (3.27) is obtained thanks to Fatou’s lemma. In fact, from the convergence almost everywhere of the subsequence (u) and of their gradients, we have
∇Tk(u) = 1{|(u)|¡k}∇u→1{|(u)|¡k}∇u=∇Tk(u) a:e in ; as →0:
Next, the boundedness of (Tk(u)) in H01(), leads to lim inf
→0 hAu; Tk(u)i= lim inf
→0
XN i;j=1
Z
aij(x)@u
@xi
@Tk(u)
@xj dx
≥XN
i;j=1
Z
aij(x)@u
@xi
@Tk(u)
@xj dx=hAu; Tk(u)i (3.29) and owing to the coercivity of the coecients aij and estimate (3.13), one gets
Tk(u) is bounded in H01(); ∀k ¿0: (3.30)
For all ¿0, the approximation of the sign function, dened by (3.3), is continuous and from the convergence almost everywere of solutions (see (3.24)), we have
S(u)→S(u) a:e in :
Moreover, g is a Caratheodory function and using (3.24) and (3.25), we have g(: ; u;∇u)→g(: ; u;∇u) =S(u)g(: ; u;∇u) a:e in ; (3.31) and then
Tk(u)g(: ; u;∇u)→Tk(u)g(: ; u;∇u) a:e in : (3.32) Using the sign condition on g (see (3.6)) and estimate (3.12), the Fatou’s lemma yields
lim inf
→0
Z
Tk(u)g(x; u;∇u) dx≥ Z
Tk(u)g(x; u;∇u) dx: (3.33) We then conclude that
Tk(u)g(: ; u;∇u) is bounded in L1(): (3.34) From the convergence of f to f in L1() for the strong topology, the third term of h converges to the desired limit
Z
fTk(u) dx→ Z
fTk(u) dx;
and this completes the proof of Lemma 2.
To complete the passage to the limit as goes to zero, we will distinguish the case where the dependence with respect to the gradient is subquadratic from the case where the growth condition is quadratic.
Proposition 2. Let the sequence (u) satisfy (3:1); (3:12); (3:14); (3:16); (3:24);
(3:25); and (3:28).Moreover if 0≤r ¡2; then we have h− hAu; vi −
Z
g(x; u;∇u)vdx+ Z
fvdx≤0; ∀v∈D+(): (3.35) Proof. Since aij∈L∞(); and the sequence @u=@xi converges to @u=@xi in Lp() for the weak topology, then for v∈D(), the second term in (3.28) converges to the desired limit, i.e.
XN i;j=1
Z
aij(x)@u
@xi
@v
@xj dx→XN
i;j=1
Z
aij(x)@u
@xi
@v
@xjdx:
In the case where 0≤r ¡2, as in [7] we shall prove that
g(: ; u;∇u)→g(: ; u;∇u) strongly inL1(): (3.36) As the sequence (u) is uniformly bounded inL1() then
meas{|u|¿ k} →0 uniformly in; ask→ ∞;
and from (3.12), we have Z
{|u|¿k}|g(x; u;∇u)|dx→0; uniformly in; as k→ ∞:
Let E be a measurable set in a compact in , then the Vitali’s theorem allows to obtain (3.36) if we show that for every k ¿0
Z
E∩{|u|≤k}|g(x; u;∇u)|dx→0 uniformly in ; as meas(E)→0:
From the growth condition (see (3.6)), we have Z
E∩{|u|≤k}|g(x; u;∇u)|dx≤ Z
E∩{|u|≤k}b(x; u)|∇u|rdx:
With the Holder’s inequality, one gets Z
E∩{|u|≤k}|g(x; u;∇u)|dx
≤ Z
E(hk(x))2=(2−r)dx
(2−r)=2 Z
{|u|≤k}|∇u|2dx
!r=2
and with (3.18), we deduce Z
E∩{|u|≤k}|g(x; u;∇u)|dx≤ |hk|L2=(2−r)(E) k
|f|L1() r=2
:
Now as hk belongs to L2=(2−r)() then |hk|L2=(2−r)(E) goes to zero as meas(E) goes to zero, which gives (3.36). Finally, taking the inferior limit in (3.27) and from Lemma 2, Proposition 2 is proved.
To summarize, in the case where the dependence with respect to the gradient is subquadratic we have shown the following result:
Proposition 3. If 0≤r ¡2; there exists a sequence (u) satisfying u∈K; ∀k ¿0; Tk(u)g(: ; u;∇u)∈L1(); g(: ; u;∇u)∈L1()
for allv∈D+();
hAu; Tk(u)−vi+ Z
g(x; u;∇u)(Tk(u)−v) dx≤ Z
f(Tk(u)−v) dx:
(3.37)
Now, we are going to show that the result remains valid in the case where r= 2.
This is a consequence of the following lemma:
Lemma 3. Let (w) be the solution of the following problem:
h+XN
i;j=1
Z
aij(x)@w
@xi
@
@xj dx+ Z
G(x; w;∇w)dx≤ Z
fdx: (3.38) for all ∈D(); ≤0.
We assume the following on the sequences (w); (h); (G) and (f):
• there exists1≤p ¡ N=(N−1)such that the sequence(w) is bounded inW01;p();
∇w→ ∇w a.e in ,and k= sup
1 k
Z
|∇Tk(w)|2dx→0 ask→ ∞;
• the sequence (h) satisÿes lim inf
→0 h≥h;
• the sequence (G) satisÿes Z
|G(x; w;∇w)| ≤c;
G(x; w;∇w)→G(x; w;∇w)a:e in;
sG(x; s; )≥0; ∀(s; ); a:e in (sign condition);
|G(x; s; )| ≤b(|s|)||2 (growth condition);
(3.39)
• and
f→f in L1(): (3.40)
We then have h+hAw; i+
Z
G(x; w;∇w)dx≤ Z
fdx ∀∈D(); ≤0: (3.41)
Proof. The proof of this lemma is based on the particular choice of test functions and the use of Fatou’s lemma. Here, the test functions are inspired from the functions introducted by Boccardo et al. [3, p. 525].
For every k ¿0, let us dene vk =exp
−B(Tk(w+))
Hw
k ;
where ∈D(), ≤0, and where is the constant of coercivity which appears in (2.1). B is the integrand of b dened by
B(t) = Z t
0 b(s) ds; ∀t ≥0
andH is a function fromR toR+, of class C1(), such that 0≤H ≤1; H(z) = 1 if
|z| ≤ 12 andH(z) = 0 if |z| ≥1.
In the rst step, let k be xed. Then the nonpositive sequence (vk) is bounded in H01()∩L∞(), because
|vk(x)| ≤ |(x)|
and
∇vk=∇exp
−B(Tk(w+))
Hw
k
−b(Tk(w+))
∇Tk(w+) exp
−B(Tk(w+))
Hw
k
+1
k∇Tk(w) exp
−B(Tk(w+))
H0w
k :
Moreover
vk →exp
−B(Tk(w+))
Hw
k
a:e in ; as→0:
Using vk as test function in (3.38), we obtain h+XN
i;j=1
Z
aij(x)@w
@xi
@
@xj exp
−B(Tk(w+))
Hw
k dx+
Z
ldx +1
k Z
XN i;j=1
aij(x)@w
@xi
@Tk(w)
@xj exp
−B(Tk(w+))
H0w
k dx
≤ Z
f exp
−B(Tk(w+))
Hw
k
dx; (3.42)
where l=
G(x; w;∇w)−b(Tk(w+))
XN i;j=1
aij(x)@w
@xi
@Tk(w+)
@xj
exp
−B(Tk(w+))
Hw
k :
Now we let go to zero. By virtue of the Lebesgue’s theorem, for allq ¿0 we have the convergence for the strong topology in Lq() of
aij@
@xj exp
−B(Tk(w+))
Hw
k
→aij@
@xj exp
−B(Tk(w+))
Hw
k
inLq():
On the other hand, we have @w=@xi converges to @w=@xi in Lp() for the weak topology, hence
XN i;j=1
Z
aij(x)@w
@xi
@
@xj exp
−B(Tk(w+))
Hw
k dx
→XN
i;j=1
Z
aij(x)@w
@xi
@
@xj exp
−B(Tk(w+))
Hw
k dx:
In the same way, using (3.40), one gets Z
f exp
−B(Tk(w+))
Hw
k
dx→ Z
fexp
−B(Tk(w+))
Hw
k dx:
We now give an upper bound of the fourth term in (3.42).
Since (1=√
k)Tk(w) is uniformly bounded in H01() with respect to, then 1
k Z
XN i;j=1
aij(x)@w
@xi
@Tk(w)
@xj exp
−B(Tk(w+))
H0w
k dx
≤ 1
kN2|aij|L∞()|∇Tk(w)|2L2()|H0|L∞()||L∞()
≤kC||L∞();
where k goes to zero as k tends to innity.
The passage to the inferior limit on the third term is obtained by Fatou’s Lemma.
From the convergence almost everywere of w, ∇w andG(: ; w;∇w) tow,∇w and G(: ; w;∇w), respectively, we have
l→l a:e in ;
where lis given by l=
G(x; w;∇w)−b(Tk(w+))
XN i;j=1
aij(x)@w
@xi
@Tk(w+)
@xj
exp
−B(Tk(w+))
Hw
k :
Moreover the functions l are bounded in L1(). This is due to the fact that G(: ; w;∇w) is bounded in L1() (see (3.39)) and Tk(w) is bounded in H01() and
0≤b(Tk(w+)) exp
−B(Tk(w+))
Hw
k
≤b(k);
because H(w=k) vanishes for |w| ≥k andb is increasing.
Furthermore, if l≥0 a.e in , the Fatou’s lemma permits to conrm that lim inf
→0
Z
ldx≥ Z
ldx: (3.43)
In fact, if w≤0 a.e. in then l=G(x; w;∇w)Hw
k
a:e in {x∈; w≤0}
and the sign condition (3.39) and the nonpositivity of , gives l≥0.
On the other hand, if w≥0, then l=G(x; w;∇w) exp
−B(Tk(w+))
Hw
k
−b(Tk(w+))
XN i;j=1
aij(x)@w
@xi
@Tk(w+)
@xj exp
−B(Tk(w+))
Hw
k :
Using again the sign condition and the growth condition (3.39), one gets G(x; w;∇w)Hw
k
≤b(|w|)|∇w|2Hw k
≤b(Tk(w+))|∇Tk(w)|2Hw k
and from the coercivity of the coecients aij (see (2.1)), we obtain
−b(Tk(w+))
XN i;j=1
aij(x)@w
@xi
@Tk(w+)
@xj
≤ −b(Tk(w+))|∇Tk(w)|2 a:e in {x∈; w≥0};
which ensures the positivity of l and establishes (3.43).
