strong maximum principles for supersolutions of quasilinear elliptic equations

Nonlinear Analysis 37 (1999) 431–448

Strong maximum principles for supersolutionsof quasilinear elliptic equations

Marcelo MontenegroDepartamento de Matem�atica, Caixa Postal 6065, Universidade Estadual de Campinas,

13081-970 Campinas, SP, Brasil

Received 4 September 1996; accepted 13 October 1997

Keywords: Strong maximum principles; Nonunique continuation principles; Orlicz–Sobolev spaces

1. Introduction

The quasilinear elliptic operators second order we are going to treat are of the form

Lu=div(’(|∇u|)∇u) (1)

and the function u satis�es

− Lu+ d(x)f(u)≥ 0; (2)

weakly, i.e., in D′(), where d∈L∞loc() and d≥ 0 a.e. in . In other words∫{’(|∇u|)∇u · ∇v+ d(x)f(u)v} dx≥ 0;

for every v∈D() :=C∞c () and v≥ 0.

Let the functions aj(�)=’(|�|)�j; j=1; : : : ; N , be de�ned for all �∈RN , with’(t)¿0 (3)

for t¿0. We assume the following structural conditions related to the regularity resultsof [13]:

aj(0)= 0; (4)

aj ∈C1(RN\{0}) ∩ C 0(RN ); (5)

0362-546X/99/$ – see front matter ? 1999 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(98)00057 -1

432 M. Montenegro / Nonlinear Analysis 37 (1999) 431–448

N∑i; j=1

@aj@�i(�)�i�j ≥�1’(|�|)|�|2; (6)

N∑i; j=1

∣∣∣∣@aj@�i (�)∣∣∣∣ ≤�2’(|�|) (7)

for some positive constants �1 and �2, all �∈RN\{0} and all �∈RN .Let g(t)=’(|t|)t for t ∈R. Note that Eqs. (3)–(7) imply the unidimensional estimate�1’(|t|)≤ g′(t)≤�2’(|t|) (8)

for t=0. Also g(0)= 0; g′ ∈L1(R) and g∈C1(R\{0}) ∩ C 0(R). So that g strictlyincreases for t≥ 0. Clearly, g is odd and bijective in R. The primitive G(t)= ∫ t

0 g(s) dsis strictly convex for t≥ 0. Thus, G is in fact an N -function.The natural environment to deal with the operator L is the Orlicz–Sobolev space

W 1; G()= {u∈LG(): ∇u∈ (LG())N};where

LG()={u : → R measurable: ∃�¿0 such that

∫G( |u(x)|

�

)dx¡∞

}

is the so-called Orlicz space and the domain ⊂RN ; N ≥ 1, is arbitrary. They areBanach spaces under their respective norms

‖u‖W 1; G() =N∑i=0

‖Diu‖LG()

and

‖u‖LG() = inf{�¿0:

∫G( |u(x)|

�

)dx≤ 1

}:

Analogously to the classical Sobolev spaces we de�ne W 1; G0 ()=C∞

c ()‖·‖W1; G () .

As we shall see, assumptions (3)–(7) imply the �2-condition for G, that is, thereexist constants k¿0 and T ≥ 0 such that G(2t)≤ kG(t) for all t≥T . Also, it is easyto see that the same condition holds for the conjugate N -function G(t)=

∫ t0 g

−1(s) ds.Therefore, the above spaces are re exive. The classical references about the spacesconsidered here are the books [7, 11].The reader can easily see that the p-Laplacian is included in the present work.

Namely, ’(t)= |t|p−2, with 1¡p¡∞, in this case. However, the Orli1cz–Sobolevsetting allow us to consider a much larger class of operators than in [5, 19]. Forinstance, the operators below are included:

Lu=div(’(|∇u|)∇u);

M. Montenegro / Nonlinear Analysis 37 (1999) 431–448 433

where

’(t)=1tddt(t� + t� log(t� + 1)); �¿1 and �≥ 0

and

Mu=div(|∇u|q−2(1 + |∇u|q)(p=q)−1∇u);where p; q∈ (1;∞). This second operator is studied in [18], it is a variant of theone treated in [15]. Note that L behaves asymptotically like an �-Laplacian, when|∇u| is near 0, because ’(t)t−(�−2) tends to � as t → 0+. But, if |∇u| is su�-ciently large, the function ’ does not behave asymptotically like a power, because’(t)t−m tends to 0 as t→+∞, if m¿�− 2. And, on the other hand, if m≤ �− 2, thelimit is +∞. In the second example, if p= q, then M behaves asymptotically like ap-Laplacian if |∇u| is near ∞ and like a q-Laplacian if |∇u| is near 0. If p= q, thenM is the p-Laplacian. The respective N -functions associated to the above operatorsare G(t)= t�+ t� log(t�+1) and G(t)=

∫ t0 s

q−1(1+ sq)p=q−1 ds for t≥ 0. In the aboveexamples the functions ’; g and G are asymptotically power functions near t=0. Thereader is invited to search an example avoiding such behaviour in order to create anoperator L satisfying Eqs. (3)–(7), which is not an m-Laplacian, 1¡m¡∞, when |∇u|is near 0.In some steps we use the monotone operator theory as in [2, 14], so we impose a

monotonicity assumption on f. We assume that there exists an �∈ (0;∞) such thatf : [0; �]→ R is a continuous and nondecreasing function with f(0)= 0: (9)

The above local de�nition is sometimes replaced by appropriate extensions of f to[0;∞) on R. We also suppose that either

f(s0)= 0 (10)

for some s0 ∈ (0; �] orf(s)¿0 (11)

for s∈ (0; �]. In the latter case we assume that∫0+[�−1(F(s))]−1 ds=∞; (12)

where F(t)=∫ t0 f(s) ds; �(t)=�(t)−G(t) and �(t)= g(t)t for t≥ 0. Note that � is

strictly increasing for t≥ 0, by Eq. (8). Condition (12) is necessary and su�cient fora strong maximum principle. We prove a nonunique continuation property as in [19]if we assume the following convergence integral condition:∫ �

0[�−1(F(s))]−1 ds¡∞; (13)

for some �¿0, instead of Eq. (12).


Let ⊂RN ; N ≥ 1, be an arbitrary domain unless otherwise stated. Our main resultsare the following:

Theorem 1. Suppose that Eqs. (3)–(7) and Eqs. (9)–(12) are ful�lled and letu∈C1() satisfying −Lu+d(x)f(u)≥ 0 in D′(0; �) where 0; �= {x∈: 0¡u(x)¡�}.If u≥ 0 in ; then either u≡ 0 in or for every compact set X ⊂ there exist aconstant c= c(X )¿0 such that u≥ c in X . In particular; if u vanishes on a subsetof positive Lebesgue measure of ; then u≡ 0 in .

Theorem 2. Suppose that Eqs. (3)–(7) and Eqs. (9)–(12) are ful�lled with [0;∞)instead of [0; �] in Eq. (9). Let u∈W 1;G

loc (); u≥ 0 a.e. in ; u ≡ 0 in satisfy-ing f(u)∈LGloc() and −Lu + d(x)f(u)≥ 0 in D′(). Given a compact set X ⊂;there exist a constant c= c(X )¿0 such that u≥ c a.e. in X . In particular; if u van-ishes on a subset of positive Lebesgue measure of ; then u=0 a.e. in . Clearly;if u∈W 1; G

loc () ∩ L∞(); assumption (9) can be made with �= ‖u‖L∞(); so thatf(u)∈L∞().

Theorem 3. Let x0 ∈ @ be a point satisfying the interior sphere condition and let Bbe one such sphere and � be the inward unit normal vector at x0. If u satis�es thehypotheses of Theorem 1; u∈C1(∪{x0}) and u(x0)= 0; then there exists a constant�¿0 such that

@u(x0)@�

≥ �: (14)

If u satis�es the hypotheses of Theorem 2; then there exists a constant �¿0 such that

ess lim infx→x0x∈ B

u(x)(x − x0)� ≥ �: (15)

Remark. If u∈L∞(); u+≡ 0 and u− ≡ 0 in Theorems 1 and 2; then either u≡ inf uor u¿ inf u in . Therefore, rigorously speaking the subject of this paper is strongminimum principles, not strong maximum principles. But it has been noted in theliterature that theorems stated as above are called strong maximum principles.

Remark. In Theorem 1 and in the second part of Theorem 2; the function f couldhave any behaviour at ∞.

Now, we construct a solution vanishing on a set of positive Lebesgue measure.If condition (12) is not satis�ed we have the following boundary point version of anonunique continuation property instead of a strong maximum principle.

Theorem 4. Suppose that Eqs. (3)–(7); (9) and (13) are veri�ed. Then for everyx0 ∈ @(RN\)⊂ @ and all R¿0 there exists a function u∈C1() such that u≥ 0in ; u≡ 0 in and Lu ∈ L∞() satisfying −Lu+ f(u)= 0 a.e. in and u=0 in\BR(x0).


For points inside the domain we have the following counterpart:

Theorem 5. Suppose that Eqs. (3)–(7); (9) and (13) are veri�ed that N ≥ 2. LetAR=BR(0)\{0}⊂RN . If there exists w∈C1(AR); w¿0 in AR; w radially symmet-ric and w(x)→+∞ as |x|→ 0; satisfying −Lw + f(w)= 0 in D′(AR); then for allx0 ∈ there exists u∈C1(); u≥ 0 in and u ≡ 0 in such that Lu∈L∞() (orLu∈L∞loc() with [0;+∞) in (8)); −Lu+ f(u)= 0 a.e. in and u=0 in \BR(x0).

Remark. If there exists a constant k¿0 such that d(x)≥ k a.e. in ; then we mayreplace the function f in Theorems 4 and 5 by kf. So that we can construct afunction u satisfying −Lu + kf(u)= 0 a.e. in and u=0 in \BR(x0). Since u≥ 0in ; we have d(x)f(u)≥ kf(u) a.e. in ; thus −Lu+ d(x)f(u)≥ 0 a.e. in .

Theorem 1 has been proved in [19] for the Laplacian and stated without proof forthe p-Laplacian without the term d(x). The remark concerning the de�nition of f onlyon the interval [0; �] is made only for the Laplacian case. But in our Theorem 1 andin the second part of Theorem 2 the function f could have any behaviour at ∞. OurTheorem 2 is also proved in [19] only for the Laplacian case, but in a more generalsituation, namely, u∈L1loc() there. Essentially, the same results of Theorem 1 andEq. (14) in Theorem 3 have been obtained in [5] for an operator more general thanthe p-Laplacian, but less general than our operator L. More precisely, the operatorstudied in [5] is supposed to satisfy a local condition, that is, for every compact setK ⊂RN there exists a constant c= c(K)¿0 such that

N∑j=1

(aj(�)− aj(�′))(�j − �′j)≥ c|�− �′|p

for every �; �′ ∈K and some 1¡p¡∞. This gives rise to a locally strongly monotoneoperator from a Sobolev space into its topological dual, that is, for every u1; u2 ∈W 1; p(B), where B⊂Rn is a ball, we have the following norm estimate:∫

B[’(|∇u1|)∇u1 − ’(|∇u2|)∇u2] · ∇(u1 − u2)≥ c‖∇u1 −∇u2‖pLp(B):

Generally our operator L is not locally strongly monotone. For example, the p-Laplacianis strongly monotone if p≥ 2, but it is not locally strongly monotone if 1¡p¡2, asit can be seen in the proof of Theorem 6, stated below. Furthermore, generally, whenwe are in the Orlicz–Sobolev setting, our operator L is not locally strongly monotone,as it can be seen in the proof of Lemma 8 and in the proof of Theorem 2. But in thepresent work we can prove that L is coercive, this is a key point in some proofs here.A result analogue to our Theorem 2 is not mentioned in [5]. It is easy to see

that a reasoning similar to the proofs of our Theorems 4 and 5 is applicable under theconditions introduced in [5]. Theorems 3–5 are proved in [19], when L is the Laplacianand it is remarked that they are valid for the p-Laplacian. Therefore, we extend theresults of [5–19] to a much larger class of operators and functions u belonging to moregeneral spaces.


The Orlicz–Sobolev approach naturally adds intrinsic di�culties for proving our the-orems. First, we develop some general results of Orlicz–Sobolev spaces. Second, westudy especi�cally the function G and the operator L with the aid of the conditions(3)–(7). Once our preliminary results are accomplished, we use the theory of mono-tone operators and a similar construction to the one employed in the Classical Hopf’sMaximum Principle to prove Theorems 1–3. In synthesis, if u vanishes at some pointin , we construct a regular subsolution v, i.e., −Lv+ d(x)f(v)≤ 0 in D′(), whichvanishes at some point x∈ and has a nonvanishing gradient at x. We argue witha weak comparison principle in order to get a contradiction. The construction of thefunction in Theorems 4 and 5 is done by inverting Eq. (13). The problems we aretreating here appear in various mathematical models, see [4].The above theorems work on another situation. We replace the conditions (6) and

(7) above by the regularity assumptions considered in [6, 12, 17], namely,

N∑i; j=1

@aj(�)@�i

�i�j ≥ 1(� + |�|)p−2|�|2 (16)

and

N∑i; j=1

∣∣∣∣@aj(�)@�i

∣∣∣∣ ≤ 2(� + |�|)p−2 (17)

for some �∈ [0; 1], some positive constants 1 and 2, all �∈RN\{0} and all �∈RN .Thus Eqs. (16) and (17) give rise to an approach in the classical Lp() spaces and

Sobolev spaces W 1; p() and W 1; p0 (), with 1¡p¡∞. So we can reformulate the

above results.

Theorem 6. The previous theorems are true if we replace the conditions (6) and (7)by (16) and (17) and the space W 1; G

loc () by W1; ploc ().

Clearly, the p-Laplacian is included in conditions (16) and (17). But Eqs. (6) and(7) are more general than Eqs. (16) and (17), as it can be seen in the examples givenabove, namely L and M for p= q. Although, if q=2 or p= q, then M satis�esEqs. (16) and (17). The operator L under conditions (16) and (17) is more general thanthose studied in [5, 19]. Note that Eq. (16) does not imply local strong monotonicityif 1¡p¡2, see also the proof of Theorem 6.

2. Some general results

In this section we develop some preliminary tools of a general character beforeproving the main theorems. In the following lemmas ⊂RN ; N ≥ 1, is a boundeddomain, M is an arbitrary N -function and M its conjugate N -function, see the gene-ral de�nitions and properties in [7, 11]. Analogously to the last section we de�ne


LM (); LM (); W 1;M () and W 1;M0 (). The set

LM ()={u : →R measurable :

∫M (|u(x)|) dx¡∞

}

is not too relevant for our purposes, it is called Orlicz class. We have the inclusionLM ()⊂LM (). Moreover, in the case that is bounded, the equality LM ()=LM () holds as sets, i� M satis�es the �2-condition. But this is the case of ourfunction G. For this reason we do not make any distinction between Orlicz spaceand Orlicz class. In the case that has in�nite Lebesgue measure we require the�2-condition in the whole interval [0;∞). The domain may be unbounded in thetheorems stated above. However, our arguments in their proofs are only local. So wecan apply the following lemmas, where the assumption of boundedness is made.The following result will be useful for proving the coercivity of some operators, we

refer to [9] for a proof.

Lemma 1. The function M satis�es the �2-condition i�

1‖u‖LM ()

∫M (|u(x)|) dx→∞ when ‖u‖LM ()→∞: (18)

We say that a sequence un ∈LM () converges to u∈LM () in M -mean, i�∫M (|un(x)−u(x)|) dx→ 0. If M satis�es the �2-condition, then M -mean convergenceand norm convergence are equivalent in LM (). Adequate test functions play an essen-tial role in our arguments. In particular, we need u+ ∈W 1;M () provided u∈W 1;M (),so we formulate some analogue results in [16]. The proof follows step by step by usingthe previous equivalence of convergence types, if M satis�es �2-condition. A moreelaborated proof in [10] shows that the �2-condition is unnecessary.

Lemma 2. If u∈W 1;M () and � is lipschitz continuous; then �(u)∈W 1;M (). Ifmoreover; u∈W 1;M

0 () and �(0)=0; then �(u)∈W 1;M0 ().

The following result is well known in the Lp() setting, we reformulate it forLM (). It is possible to prove it by the above-mentioned equivalence between normand M -mean convergence and following the Lp() approach just as in [1].

Lemma 3. Suppose that M satis�es the �2-condition. If a sequence un ∈LM () con-verges to u∈LM () in M -mean; then it has a subsequence unk and there existsh∈LM () such that unk → u a.e. in in |unk | ≤ h a.e. in .

Analogously to the classical Sobolev spaces we have stronger properties and imbed-dings in dimension 1.

Lemma 4. Let I ⊂R be an open interval. For every function u∈W 1;M (I) there existsa function u∈C0( �I) such that u= u a.e. in I and u(x) − u(y)= ∫ x

y u′(t) dt for all


x; y∈ �I ; that is; the function u is absolutely continuous. And if I is bounded; theimbedding W 1;M (I) ,→ C0( �I) is continuous and compact.

Proof. The assertion concerning the continuous representative is true; becauseW 1;M (I)⊂W 1;1(I). The H�older inequality implies ‖u‖L1(I)≤‖u‖LM (I)‖1I‖LM (I), so‖u‖L∞(I)≤ c1‖u‖W 1; 1(I)≤ c2‖u‖W 1; M (I), where c1 and c2 are positive constants depend-ing only on |I | and M is the conjugate N -function of M . So the inclusion W 1;M (I) ,→C0( �I) is continuous. Let U be the unit ball of W 1;M (I). Given u∈U we have

|u(x)− u(y)|=∣∣∣∣∫ x

yu′(t) dt

∣∣∣∣≤‖u′‖LM (I)‖1[x; y]‖LM (I)

= ‖u′‖LM (I)|x − y|M−1(

1|x − y|

)≤ |x − y|M−1

(1

|x − y|)

=M−1(1=(x − y))M (M−1(1=(x − y))

for all x; y∈ I . Remembering that and N -function satis�es M (t)=t→+∞ as t→+∞ weconclude that U is uniformly equicontinuous, and by Ascoli’s Theorem U is relativelycompact in C0( �I).

Remark. The above lemmas are valid if the N -function M is substituted by G underthe hypotheses (3)–(7). As we shall see; in the next section; the N -functions G and�G satisfy the �2-condition.

3. Proof of theorems

At �rst, we are going to establish some especi�c properties concerning the function Gand the operator L. Secondly, we proceed with the proofs of our main results.Note that Eq. (8) implies

0¡�1g(t)≤ tg′(t)≤�2g(t) (19)

for t¿0; which by integration gives

�1G(t)≤ tg(t)− G(t)≤�2G(t) (20)

for t≥ 0. Then

1¡1 +�12≤ lim inf

t→+∞tg(t)G(t)

≤ lim supt→+∞

tg(t)G(t)

≤ 1 + �2;

which implies the �2-condition for both G and �G. By virtue of the relation G(t)≤ tg(t)for t≥ 0 we get

(1 + �2)−1tg(t)≤G(t)≤ tg(t) (21)


for t≥ 0. Applying �G(t)=∫ t0 g

−1(s) ds in Eq. (21) we obtain a useful inequality,namely

�G((1 + �2)−1g(t))≤ �G(G(t)t−1)≤G(t) (22)

for t¿0.

Remark. There are equivalent conditions to Eqs. (12) and (13). By Eq. (20) and thede�nition of � we obtain(

�11 + �1

)�(t)≤ �(t)≤�(t) (23)

for t≥ 0. So that∫0+[�−1(F(s))]−1 ds=∞ (24)

and Eq. (12) are equivalent. Moreover; Eq. (20) says that

�1G(t)≤ �(t)≤�2G(t) (25)

for t≥ 0. Thus∫0+[G−1(F(s))]−1 ds=∞ (26)

and Eq. (12) are equivalent. Note that Eq. (8) implies that ’∈L1(0;∞); and sointegrating twice Eq. (8) we get

�1H (t)≤G(t)≤�2H (t) (27)

for t≥ 0 where H (t)= ∫ t0

∫ �0 ’(s) ds d�: So that∫

0+[H−1(F(s))]−1 ds=∞ (28)

is equivalent to Eq. (12).Clearly Eq. (13) is equivalent to∫ �

0[�−1(F(s))]−1 ds¡∞; (29)

∫ �

0[G−1(F(s))]−1 ds¡∞ (30)

and ∫ �

0[H−1(F(s))]−1 ds¡∞ (31)

for some �¿0; by Eqs. (23), (25) and (27), respectively.

In the next lemmas ⊂RN ; N ≥ 1; is a bounded domain.


The operator L is strictly monotone in the following sense:

Lemma 5. Suppose that Eqs. (3)–(7) are satis�ed. Let S :W 1; G0 ()→W 1; G

0 ()∗ bede�ned by

〈Su; v〉=∫’(|∇u|)∇u · ∇v dx (32)

for all v∈W 1; G0 (). Then S is well de�ned and

〈Su− Sv; u− v〉≥ 0 (33)

for all u; v∈W 1; G0 (). The equality holds i� u= v a.e. in . Also, S satis�es Eq. (33)

if we replace W 1; G0 () above by W 1; G().

Proof. By Eq. (22), ’(|∇u|)∇u∈L �G(). So that S is well de�ned by H�older’s in-equality. We assert that L is strictly monotone. Indeed, without loss of generality, take�; �′ ∈RN such that |�| ≤ |�′|. Then

14 |�− �′| ≤ |t�+ (1− t)�′| ≤ |�|+ |�′|

for t ∈ [0; 14 ]. A simple calculation using Eqs. (3)–(7) and similar reasoning to [17]furnish

N∑j=1

(aj(�)− aj(�′))(�j − �′j)≥�14|�− �′|2(|�|+ |�′|)−1g

(14|�− �′|

): (34)

Hence, for any w1; w2 ∈W 1; G0 (), we have

〈Sw1 − Sw2; w1 − w2〉 =∫[’(|∇w1|)∇w1 − ’(|∇w2|)∇w2] · (∇w1 −∇w2) dx

≥ �14

∫|∇w1 −∇w2|2(|∇w1|+ |∇w2|)−1

× g(14|∇w1 −∇w2|

)dx≥ 0

and 〈Sw1 − Sw2; w1 − w2〉=0 i� w1 =w2 a.e. in . The above inequality also holdsfor w1; w2 ∈W 1; G().

Usually a Nyemitskii mapping between Orlicz spaces is not continuous or even wellde�ned on the whole space. The peculiar form of g and the previous lemmas allow usto state the following result similarly to the ones in [3].

Lemma 6. Suppose that Eqs. (3)–(6) are satis�ed. The Nyemitskii operator u→g(u) from LG() to L �G() is continuous and bounded.

Proof. Let un be a sequence in LG() such that un→ u in LG(). By Lemma 3, wemay assume that g(un)→ g(u) a.e. in and g(un)≤ g(h) a.e. in for some h∈LG().


By Eq. (22) we conclude that g(h)∈L �G(), then g(un)∈L �G() and g(un)→ g(u) inL �G() by dominated convergence theorem. The boundedness is easy to see.

We need the following weak comparison principle:

Lemma 7. Suppose that Eqs. (3)–(7) and (9) are ful�lled and f is extended non-decreasing and bounded to R. If u1; u2 ∈W 1; G() satisfy

− Lu1 + d(x)f(u1)≤−Lu2 + d(x)f(u2) in D′() (35)

and u1≤ u2 on 9; then u1≤ u2 a.e. in .

Proof. If |{u1¿u2}|¿0, then by Eqs. (34) and (35) we get

0 ≤ �14

∫{u1¿u2}

|∇u1 −∇u2|2(|∇u1|+ |∇u2|)−1g(14|∇u1 −∇u2|

)dx

≤∫{u1¿u2}

[’(|∇u1|)∇u1 − ’(|∇u2|)∇u2] · (∇u1 −∇u2) dx

≤ −∫{u1¿u2}

d(x)(f(u1)− f(u2))(u1 − u2) dx≤ 0:

Hence ∇(u1−u2)+ =0 a.e. in {u1¿u2}, implying (u1−u2)+ =0 a.e. in {u1¿u2}.

The following two-point boundary value problem is fundamental in order to constructa subsolution, i.e., a function v satisfying −Lv+ d(x)f(v)≤ 0 in D′().

Lemma 8. Suppose that Eqs. (3)–(7) are valid in dimension 1. Assume also Eq. (9)and that f is extended continuously and nondecreasing to R. For any positive realnumbers k1; k2; T; a there exists a unique solution u= u(r; k1; k2; T; a)∈C2(0; T ]∩C1[0; T ] of the problem

−(e−k1rg(u′(r)))′ + k2e−k1rf(u(r))= 0 in (0; T );

u(0)= 0; u(T )= a:(36)

Moreover u; u′ ≥ 0 in [0; T ]; u′′ ≥ 0 in (0; T ]. And if Eqs. (10)–(12) are ful�lled, thenu′(0)¿0 and 0¡u(r)¡a for 0¡r¡T .

Proof. We shall use the theory of variational inequalities. For that matter we introducethe set K =W 1; G

0 (0; T )+ h, where h(r)= (a=T )r, and the operator A :K→W 1; G(0; T )∗

de�ned by

〈Au; v〉=∫ T

0e−k1r{g(u′(r))v′(r) + k2f(u(r))v(r)} dr

for all v∈W 1; G(0; T ). The operator A is well de�ned in view of Lemma 5 and theimbedding W 1; G(0; T ) ,→C0[0; T ] of Lemma 4. It is easy to see that K = ∅ and convex.


The previous imbedding is compact, so K is weakly closed. The weak topology weare considering is �(W 1; G(0; T ); W 1; G(0; T )∗). So that K is closed.The operator A is strictly monotone. Indeed, the monotonicity of g(t)=’(|t|)t; f and

e−k1r imply 〈Au − Av; u − v〉≥ 0 for all u; v∈K and it vanishes i� u= v=0, becauseu and v are continuous in [0; T ] and vanish in 0, see Lemmas 4 and 5.The operator A is hemicontinuous. Indeed, let un be a sequence in W 1; G(0; T ) such

that un→ u in W 1; G(0; T ), so un→ u uniformly in C0[0; T ], see Lemma 4. We have〈Au−Aun; v〉→ 0 for all v∈K , because the Nyemitskii mapping u→ g(u) is continuousfrom LG(0; T ) to L �G(0; T ) by Lemma 6. The second term in the integral goes to 0 byuniform convergence.The operator A is bounded. Indeed, if ‖u‖W 1; G(0; T )≤ c, then there exists a constant

M¿0 such that |〈Au; v〉|≤M for all v∈K such that ‖v‖W 1; G(0; T )≤ 1, by Eq. (22) orLemma 6, H�older’s inequality and Lemma 4.It remains to prove the coercivity of A. Since g(t)t≥G(t) and g(t)t≥ �G(g(t)) for

t≥ 0, by Young’s inequality g(t)s≤ �G(g(t)) + G(s) for all t; s≥ 0, we have2g(t)(t − s)≥G(t)− G(2s) (37)

for all t; s∈R, because of the oddness of g in R. Choosing �u∈W 1; G(0; T ) we concludethat

〈Au− Av; u− v〉 ≥ e−k1T∫ T

0g(v′)(u′ − v′)− g(v′)u′ dr

≥ e−k1T∫ T

0

12{G(|u′|)− G(2|v′|)} dr − ‖g(v′)‖L �G(0; T )‖u′‖LG(0; T ):

Now, we divide the above expression by ‖u − v‖W 1; G(0; T ) and take into accountthe equivalence to ‖u′ − v′‖LG(0; T ), by the unidimensional Poincar�e’s inequality. FromLemma 1 we obtain

1‖u− v‖W 1; G(0; T )

〈Au− Av; u− v〉→+∞ as ‖u‖W 1; G(0; T )→+∞; u∈K:

Therefore, we conclude that there exists a unique solution u∈K such that 〈Au; v−u〉≥ 0 for all v∈K . Taking v= u±w, with w∈W 1; G

0 (0; T ), we conclude that u is indeeda solution of problem (36). Similar to Lemma 7 we conclude that u≥ 0 in [0; T ]. Theregularity of u follows from a bootstrap reasoning. Integrating the equation in Eq. (36)and applying the inverse of g(t)=’(|t|)t we obtain

u′(r)= g−1{ek1r

[k2

∫ r

0e−k1tf(u(t)) dt + g(u′(0))

]}(38)

for 0≤ r≤T . Taking into account Eq. (5) we conclude that u∈C1[0; T ]. And,a fortiori, u∈C2(0; T ] from Eq. (38). If g(u′(0))¡0, then u′(0)¡0, implying thatu¡0 on a neighbourhood (0; �), which is a contradiction. So that Eq. (38) implyu′ ≥ 0 in [0; T ]. Since (’′(u′)u′ + ’(u′))u′′=(’(u′)u′)′ ≥ 0 in [0; T ] by Eq. (8), wehave u′′ ≥ 0 in (0; T ].


It remains to prove that u′(0)¿0. At �rst, we are going to assume Eqs. (11) and (12).Suppose on the contrary that u′(0)= 0. Let T0 ∈ [0; T ] be the largest value in which uvanishes. Clearly, 0≤T0¡T; u(r)¿0 in (T0; T ]; u′(T0)= 0 and u is a bijection from[T0; T ] to [0; a]. Multiplying Eq. (36) by u′ we rewrite it in the following form:

(�(u′(r)))′ − k1�(u′(r))= k2(F(u(r)))′ + (G(u′(r)))′ (39)

for T0≤ r≤T . Integrating Eq. (36) we obtain

e−k1rg(u′(r))= k2∫ r

T0e−k1tf(u(t)) dt

for T0≤ r≤T . Using the monotonicity of e−k1r and f we get

e−k1rg(u′(r))≤ k2f(u(r))∫ r

T0e−k1t dt=

k2k1f(u(r))(e−k1T0 − e−k1r)

for T0≤ r≤T . Multiplying the above expression by u′ we have the following estimate:

�(u′(r))≤ ck2k1(F(u(r)))′ (40)

for T0≤ r≤T and some constant c= c(T )¿0. Joining Eqs. (39) and (40) we obtain(�(u′(r)))′ − (G(u′(r)))′ ≤ c(F(u(r)))′

for T0≤ r≤T and some constant c= c(k1; k2; T )¿0. Integrating we get�(u′(r))≤ cF(u(r))

for T0≤ r≤T and c= c(k1; k2; T )¿0.Consequently,∫ T

T0[�−1(F(u(r)))]−1u′(r) dr=

∫ a

0[�−1(F(s))]−1 ds¡∞;

a contradiction.Suppose now that f satis�es Eq. (10), then f(s)= 0 for 0≤ s≤ s1 where s0≤ s1≤ �

and s0 is the largest number such that f vanishes. By the previous calculations Eq. (36)has a unique solution u∈C2(0; T ]∩C1[0; T ] such that u; u′ ≥ 0 in [0; T ] and u′′ ≥ 0 in(0; T ]. Then

e−k1rg(u′(r))= g(u′(0)) (41)

for 0≤ r≤T ′ and some T ′ ≤T . There exists a constant M¿0 such that 0≤ u(r)≤Mfor 0≤ r≤T . If M ≤ s1, then f(u)= 0 in [0; T ] and we make T =T ′, so there exists�∗ ∈ [0; T ] such that u′(�∗)¿0, then u′(0)¿0 by Eq. (41). If M¿s1, then f(u)= 0 in[0; T ′], where u(T ′)= s1 and T ′¡T . Since u′(T ′)¿0, by Eq. (41) we have u′(0)¿0.The proof is complete.

The main work for proving the theorems is complete. Now ⊂RN ; N ≥ 1, is anarbitrary domain. Up till now we have proved our preliminary results in bounded


domains. We have taken advantage of the �2-condition and other properties satis�edby G and L. It is worth to mention that the domain , may be unbounded in thestatements of the theorems. But our further arguments in their proofs are only local.We are going to work on balls, and the spaces will be modelled in those balls. SinceG and �G satisfy the �2-condition, the above lemmas will apply successfully.

Proof of Theorem 1. If 0; �= ∅, then u≥ �¿0 in , so the theorem is valid. If0; �= ∅, we suppose by contradiction that there exists �x∈ such that u( �x)= 0. Thenit is possible to �nd two points x1; x2 ∈ and a ball B=BR(x2)⊂⊂ such thatx1 ∈ 9B; u(x1)= 0 and 0¡u(x)¡� in B. We observe that ∇u(x1)= 0. Consider theannulus Y = {x∈: R=2¡|x−x2|¡R} with R¿0 su�ciently small in order to 0¡a=inf{u(x)∈R: |x − x2|=R=2}¡�. Let r= |x − x2| for x∈ �Y . Using Lemma 8 and theradial symmetric expression of L, if we choose

k1≥ 2(N − 1)R

≥ N − 1r

and k2≥‖d‖L∞(Y )≥d(x) for x∈ �Y , then the function v(x)= v(r)= u(R−r; k1; k2; R=2; a)is a C2 subsolution in Y , i.e., −Lv+ d(x)f(v)≤ 0 a.e. in Y . Indeed,

Lv(x) = (g(v′(r)))′ +(N − 1r

)g(v′(r))≥ (g(v′(r)))′ + k1g(v′(r))

= k2f(v(r))≥d(x)f(v(x))for x∈Y . Since v≤ u on @Y , applying Lemma 7 we obtain v≤ u in �Y . Thus

lim inft→0+

u(x1 + t(x2 − x1))− u(x1)t

≥ limt→0+

v(R− tR)− v(R)t

=−Rv′(0)= u′(0)¿0;then ∇u(x1)= 0, a contradiction.

Proof of Theorem 2. By Fubini’s Theorem we can select a ball B⊂⊂ such thatu ≡ 0 on 9B. Taking b¿0, we restrict the function f to the interval [0; b] and we seth=min(u; b)∈W 1; G(B)∩L∞(B). Clearly, if u∈L∞() we may take b= �= ‖u‖L∞()

in Eq. (9). Consider the problem

−L(w) + d(x)f(w)= 0 in B;

w= h on 9B; (42)

which is equivalent to the translated problem by w= z + h

−L(z + h) + f(z + h)= 0 in B;

z=0 on 9B: (43)


We de�ne the operator A :W 1; G0 (B)→W 1; G

0 (B)∗ by

〈Az; v〉=∫B{’(|∇z +∇h|)(∇z +∇h) · ∇v+ d(x)f(z + h)v} dx;

for all v∈W 1; G0 (B), where f is extended nondecreasing and bounded to R.

We repeat an analogue procedure of Lemma 8. Again we appeal to the monotoneoperator theory. Note that the Niemytskii mapping u→f(u) from LG(B) to L �G(B)is continuous and bounded. By virtue of Lemma 5, A is well de�ned and bounded.Again Lemma 5 and the monotonicity of f imply that A is strictly monotone. TheNiemytskii mapping de�ned by f and Lemma 6 assert that A is hemicontinuous. Itremains to prove the coercivity. Let w= z + h and �w= �z + h for z; �z ∈W 1; G

0 (B). ByCauchy–Schuartz inequality, Eq. (37) and the monotonicity of f we get

〈Az; z − �z〉=∫B’(|∇w|)∇w:(∇w −∇ �w) + d(x)f(w)(w − �w) dx

≥∫B[g(|∇w|)|∇w| − g(|∇w|)|∇ �w|]− d(x)f(w) �w dx

≥ 12

∫B[G(|∇w|)− G(2|∇ �w|)]− d(x)f(w) �w dx:

Since ‖∇w‖LG(B) increases as ‖z‖W 1; G0 (B) increases, we get

1‖z‖W 1; G

0 (B)〈Az; z − �z〉→+∞ as ‖z‖W 1; G

0 (B)→+∞

by Lemma 1. The results in [2,14] show that Eq. (43) has a unique solution in W 1; G0 (B).

Then there exists a unique w belonging to the closed convex set K =W 1; G0 (B)+h such

that ∫B{’(|∇w|)∇w · ∇v+ d(x)f(w)v} dx=0

for all v∈W 1; G0 (B), that is, w is the weak solution of Eq. (42). Note that f(u)∈L1(B)

and −Lu+ d(x)f(u)≥ 0 in D′(B). Since w − h and (h− u)+≡ 0 belong to W 1; G0 (B),

we have w≤ u on @B. The weak comparison principle of Lemma 7 furnishes 0≤w≤ ba.e. in B, then w∈W 1; G(B) ∩ L∞(B). The regularity result of [13] imply w∈C1; �loc (B)for some 0¡�¡1. Theorem 1 imply w¿0 in B. Again, using Lemma 7 we get w≤ ua.e. in B. Thus, there exists a constant c= c(B)¿0 such that u≥ c a.e. in B.It is worth to remark that here we did not assume previously that f is extended to

R as in the past lemmas, because w is the weak solution of Eq. (42) with f de�nedonly on [0; b].Every compact set X ⊂ can be covered by a �nite union BX =B ∪ B1 ∪ · · · ∪ Bm

of overlapping balls. Since u≡ 0 on the boundary of every component ball Bi for


1≤ i≤m, we repeat the above ideas solving Eq. (42) on each ball Bi. Thus, thereexists a constant c= c(X )¿0 such that u≥ c a.e in BX .

Proof of Theorem 3. It follows from Lemmas 7, 8 and a similar reasoning as in [19].Let Y be the annulus, as in Theorem 1, corresponding to the ball of the hypotheses.As before −Lv+ d(x)f(v)≤ 0≤−Lu+ d(x)f(u) in D′(Y ) and v≤ u on @Y . Since Ytouches @ at x0 and v− u∈W 1; G

loc (), we move Y along � replacing the center x2 ofY by x2 + �� for su�ciently small � and keep R �xed. The new annulus Y� is suchthat Y�⊂⊂. And as above u(x)≥ v(x− ��) = u(R− |x− x2− ��|; k1; k2; R=2; a) a.e. inY and k1; k2 and a do not depend on �. Now, we make �→ 0+ to get Eq. (15) with�= u′(0)¿0 just as in the end of the proof of Theorem 1. The item (14) follows asa consequence.

Proof of Theorem 4. First we handle the case N =1. The function M (�)=∫ �0 [�

−1(F(s))]−1 ds de�ned in [0; �] is invertible, then v(r)=M−1(r) is a C1 solution of

−(g(v′(r)))′ + f(v(r)) = 0;v(0)= v′(0)= 0;

(44)

de�ned in [0; I ], where I =∫ �0 [�

−1(F(s))]−1 ds. Moreover v; v′¿0 in (0; I ]. Take�∈ (0; I) and de�ne the function w�(t)= v(−t + �) for t ∈ [� − I; �] and w�(t)= 0for t≥ �. Thus w� satis�es the equation in Eq. (44) in [0;∞), is a C1 function andC2 except possibly in t= �. Moreover, w� is strictly increasing for 0¡t¡�. Givenx0 ∈ @ for all 0¡�¡I such that �≤R the function u(x)=w�(|x − x0|) is a solutionof the equation in Eq. (44) in R\{x0} which vanishes i� |x − x0| ≥ �.Now, we deal with the case N ≥ 2. If x0 ∈ @(RN\)⊂ @ and R¿0, there exists

x1 ∈ such that 0¡�=dist(x1;)¡R=2. The problem is solved if we construct afunction u≥ 0 satisfying −Lu + f(u)= 0 in RN\B�(x1) and such that u(x)= 0 i�|x− x1| ≥ b for some b∈ (�; R=2). We consider the two-point boundary value problem

−(rN−1g(z′(r)))′ + rN−1f(z(r))= 0 in (�; �);

z(�)=w�(�); z(�)= 0;(45)

where �=min(�+ I=2; R=2). Let A :K→W 1; G(�; �)∗ be de�ned by

〈Az; v〉=∫ �

�rN−1{’(|z′(r)|)z′(r)v′(r) + f(z(r))v(r)} dr

for all v∈W 1; G(�; �) in K =W 1; G0 (�; �) + h, where

h(r)=w�(�)� − � (r − �)

and K ⊂W 1; G(�; �). By the same reasoning of Lemma 8, this problem has aunique solution z ∈C2[�; �) ∩ C1[�; �] and z≥ 0 in [�; �]. Let �z(r)=w�(r). Since−(rN−1g( �z′(r)))′+rN−1f( �z(r))≥ 0 for �¡r¡� and �z(�)= z(�), �z(�)= z(�) it follows


from Lemma 7 that z(r)≤ �z(r) for �¡r¡�. Since �z′(�)= 0 we have z′(�−)= 0 and zcan be extended by 0 for r≥ � to a solution in (�;∞). Now, we set u(x)= z(|x− x1|)for x∈ and b=sup{r ∈ [�;+∞): z(r)¿0}(≤ �) to get the desired result.

Proof of Theorem 5. Writing r= |x| and integrating the equation −(rN−1g(w′(r)))′+rN−1f(w(r))= 0 we get rN−12 g(w′(r2))≥ rN−11 g(w′(r1)) for 0¡r1≤ r2. Assume thatw′(r)≥ 0 in some interval (0; r′], then w(r)≤w(r′) for all r ∈ (0; r′]. This contra-dicts the assumption w(r)→+∞ for r→ 0. Hence there exists an R′ ∈ (0; r′] such thatw(R′)¡0. We may assume that R=R′, then g(w′(r))¡0 in (0; R]. Thus, w′(r)¡0 in(0; R]. Let z be the solution of Eq. (45) since z(�)= 0¡w(�) and −z′(�)= 0¡−w′(�)we obtain z(r)≤w(r) as long as both functions are de�ned if r≤ �. Note that z ismonotone. Extending z to (0;∞), the function u(x)= z(|x − x0|) is exactly we arelooking for.

Proof of Theorem 6. The general steps are given in the above proofs. We mentiononly the di�erent properties, which can be derived from Eqs. (16) and (17). Assumingthat the above lemmas are stated with the conditions Eqs. (16) and (17) instead ofEqs. (6) and (7), we can prove them in the same manner, once we establish thefollowing relations. Analogous to the inequality (34) we can prove that

N∑j=1

(aj(�)− aj(�′)) (�j − �′j)≥ c{

|�− �′|2(� + |�|+ |�′|)p−2 for 1¡p≤ 2;|�− �′|p for 2¡p≤∞;

(46)

for some �∈ [0; 1] and every �; �′ ∈RN , see [17]. This implies the following normestimate for functions u1; u2 ∈ W 1;p(), namely:

∫[’(|∇u1|)∇u1 − ’(|∇u2|)∇u2] · ∇(u1 − u2)

≥ c{ ‖∇u1 −∇u2‖2Lp()(� + ‖∇u1‖Lp() + ‖∇u2‖Lp())p−2 for 1¡p≤ 2;

‖∇u1 −∇u2‖pLp() for 2≤p¡∞;(47)

for some constant c= c(p; 1)¿0, where is a bounded domain. The proof of Eq. (47)follows from an idea similar to the one employed in [8]. Thus Eq. (47) replaces ourLemma 1. Clearly, Eq. (46) allow us to prove a weak comparison principle just asin Lemma 7, for functions belonging to W 1;p(). The results of Lemmas 2–6 canbe readily established. We use estimate (47) in order to prove the coercivity propertyrequired in the proof of Lemma 8 and in Theorem 2. Note that Eq. (47) impliesthat the p-Laplacian is strongly monotone only if p≥ 2. The remaining results followeasily.


Acknowledgements

The author was supported by a CNPq=Brazil. He is grateful to Professors DjairoGuedes de Figueiredo and Jean-Pierre Gossez for many fruitful conversations and toan anonymous referee for his valuable comments.

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