ON THE A-LAPLACIAN
NOUREDDINE A¨ISSAOUI Received 25 January 2003
We prove, for Orlicz spacesLA(RN) such thatAsatisfies the∆2 condition, the nonresolvability of theA-Laplacian equation∆Au+h=0 onRN, whereh=0, ifRNisA-parabolic. For a large class of Orlicz spaces including Lebesgue spaces Lp(p >1), we also prove that the same equation, with any bounded measurable functionhwith compact support, has a solution with gradient inLA(RN) ifRN isA-hyperbolic.
1. Introduction
An important application of the nonlinear potential theory is the resolution of some equations involving the p-Laplacian operator. In [6], Gol’dshtein and Troyanov proved that the p-Laplace equation∆pu+h=0 onRN,N≤p, has no solution ifhhas a nonzero average. This result remains true for the same equation on anyp-parabolic manifold. The proof is essentially based on a ca- pacity argument. Later, Troyanov proved in [9] that the equation∆pu+h=0, on ap-hyperbolic manifoldM, has a solution withp-integrable gradient for any bounded measurable functionh:M→Rwith compact support.
Since the strongly nonlinear potential theory is sufficiently developed, we propose in this paper the generalization of these two equations onRNto the set- ting of Orlicz spaces. For this goal, we introduce, for a givenᏺ-functionA, the notion ofA-parabolicity andA-hyperbolicity which reduces top-parabolicity and p-hyperbolicity when A(t)=p−1|t|p. We also consider the so-called A- Laplacian ∆A, which is the p-Laplacian ∆p, when the Orlicz space LA is the Lebesgue spaceLp. If theᏺ-functionAsatisfies the∆2condition andRN isA- parabolic, then the equation∆Au+h=0 has no weak solution for any function hhaving a nonzero average.
Copyright©2003 Hindawi Publishing Corporation Abstract and Applied Analysis 2003:13 (2003) 743–755 2000 Mathematics Subject Classification: 46E35, 31B15 URL:http://dx.doi.org/10.1155/S1085337503303069
744 On the
For reflexive Orlicz spacesLA, withAsatisfying the conditions(A)>0, where s(A) :=inf
logA◦f dλ
log|f|A −1, f ∈LA,|f|A>1
, (1.1)
if the functionhis inL∞and has a compact support, then the equation∆Au+ h=0 has a weak solution when RN isA-hyperbolic. We give large classes of Orlicz spacesLA, including Lebesgue spacesLp(p >1), which satisfiess(A)>0.
This paper is organized as follows. InSection 2, we list the prerequisites from the Orlicz spaces and we introduce the notion ofA-hyperbolicity.Section 3is reserved to the resolution of the equation∆Au+h=0 whenhhas a nonzero average or bounded with compact support.
2. Preliminaries
2.1. Orlicz spaces. We recall some definitions and results about Orlicz spaces.
For more details, one can consult [1,7,8].
LetA:R→R+be anᏺ-function, that is,Ais continuous, convex, withA(t)>
0 fort >0, limt→0A(t)/t=0, limt→+∞A(t)/t=+∞, andAis even.
Equivalently,Aadmits the representation:A(t)=|t|
0 a(x)dx, wherea:R+→ R+ is nondecreasing, right continuous, witha(0)=0, a(t)>0 for t >0, and limt→+∞a(t)=+∞.
Theᏺ-functionA∗conjugatetoAis defined byA∗(t)=|t|
0 a∗(x)dx, where a∗is given bya∗(s)=sup{t:a(t)≤s}.
LetAbe anᏺ-function, letλbe the Lebesgue measure onRN, and letΩbe an open set inRN. We denote byᏸA(Ω) the set, called anOrlicz class, of measurable functions f, onΩ, such that
ρ(f , A,Ω)=
ΩAf(x)dλ(x)<∞. (2.1) LetAandA∗be two conjugateᏺ-functions and let f be a measurable func- tion defined almost everywhere inΩ. TheOrlicz norm of f,fA,Ω, orfA, if there is no confusion, is defined by
fA=sup
Ω
f(x)g(x)dλ(x) :g∈ᏸA∗(Ω), ρg, A∗,Ω≤1 . (2.2)
The setLA(Ω) of measurable functions f such thatfA<∞ is called an Orlicz space. WhenΩ=RN, we setLAin place ofLA(RN).
Iff ∈LA(Ω), then fA=inf
k−1
1 +
ΩAk|f|(x)dλ(x)
:k >0 . (2.3)
TheLuxemburg norm|f|A,Ωor|f|A, if there is no confusion, is defined inLA(Ω) by
|f|A=inf
r >0 :
ΩA f(x)
r
dλ(x)≤1
. (2.4)
Orlicz and Luxemburg norms are equivalent. More precisely, if f ∈LA(Ω), then
|f|A≤ fA≤2|f|A. (2.5) It is well known that we can suppose thataanda∗are continuous and strictly increasing. Hence theᏺ-functionsAandA∗are strictly convex anda∗=a−1.
LetAbe anᏺ-function. We say thatAverifies the∆2conditionif there exists a constantC >0 such thatA(2t)≤CA(t) for allt≥0.
Recall thatAverifies the∆2conditionif and only ifᏸA=LA. Moreover,LAis reflexive if and only ifAandA∗satisfy the∆2condition.
H¨older inequality in Orlicz spaces is expressed in the following way:
|f·g|dλ≤ |f|A· gA∗, f ∈LA, g∈LA∗. (2.6) We recall the following results. LetAbe anᏺ-function andaits derivative.
Then the following occurs.
(1) Theᏺ-functionAverifies the∆2condition if and only if one of the fol- lowing holds:
(i) for allr >1, there existsk=k(r) (for allt≥0,A(rt)≤kA(t)) ; (ii) there existsα >1 (for allt≥0,ta(t)≤αA(t)) ;
(iii) there existsβ >1 (for allt≥0,ta∗(t)≥βA∗(t)) ; (iv) there existsd >0 (for allt≥0, (A∗(t)/t)≥d(a∗(t)/t)).
Moreover,αin (ii) andβin (iii) can be chosen such thatα−1+β−1=1.
We note thatα(A) is the smallestαsuch that (ii) holds.
(2) IfAverifies the∆2condition, then
A(t)≤A(1)tα, ∀t≥1, A(t)≥A(1)tα, ∀t≤1,
A∗(t)≥A∗(1)tβ, ∀t≥1, A∗(t)≤A∗(1)tβ, ∀t≤1. (2.7) We setα∗=α(A∗).
Recall also that ifAverifies the∆2condition, then
A f
|f|A
(x)dλ(x)=1. (2.8)
746 On the 2.2.A-hyperbolicity
Definition 2.1. Let A be anᏺ-function andK a compact set in RN. The A- capacity ofKis defined by
ΓA(K)=inf|∇u|A:u∈C∞0RN, u=1 in a neighborhood ofK. (2.9) The spaceRN is said to beA-parabolic ifΓA(K)=0 for all compact subsets K⊂RNandA-hyperbolic otherwise.
Remark 2.2. In the definition ofΓA, a simple truncation argument shows that we may restrict ourselves to functionsu∈C∞0(RN) such that 0≤u≤1.
Form < N, the Riesz kernel is defined onRNbyRm(x)= |x|m−N. ForX⊂RN, we defineRm,A(X) by
Rm,A(X)=inf|f|A:f ∈LA, f ≥0, Rm∗f ≥1 onX. (2.10) The following lemma is proved in [3, Lemma 3.6].
Lemma2.3. LetLAbe a reflexive Orlicz space. Then there is a positive constantC such that
C−1R1,A(K)≤ΓA(K)≤CR1,A(K), (2.11) for all compactK,Cindependent ofK.
We recall the following result proved in [4, Theorem 3.1].
Lemma2.4. LetAbe anᏺ-function such that|Rm|A∗,{|x|>1}= ∞. Then for all X,Rm,A(X)=0.
We will need the following lemma in the sequel.
Lemma2.5. LetAbe anyᏺ-function such thatA∗verifies the∆2condition and let mbe a positive integer such thatm < N andα∗≤N/(N−m). ThenRm,A(X)=0 for allX.
Proof. FromLemma 2.4, it suffices to prove that|Rm|A∗,{|x|>1}= ∞. SinceA∗ verifies the∆2condition, we must establish that
{|x|>1}A∗|x|m−N
dλ(x)= ∞. (2.12)
By a change of variable, there is a positive constantCsuch that
{|x|>1}A∗|x|m−N
dλ(x)=C ∞
1 A∗tm−N·tN−1dt. (2.13)
From the inequalityA∗(tm−N)≥A∗(1)·tα∗(m−N), we get
{|x|>1}A∗|x|m−N
dλ(x)≥CA∗(1)· ∞
1 tα∗(m−N)+N−1dt. (2.14) Now, the inequality
α∗(m−N) +N−1≥ N
N−m(m−N) +N−1= −1 (2.15)
gives the desired result.
3. On theA-Laplacian
The Orlicz-Sobolev spaceW1LA(RN) is defined as the space of functionsusuch thatuand its derivatives, in a distributional sense, of order less or equal to one are inLA. The spaceW1LA(RN) is a Banach space when equipped with the norm
|u|1,A=
|γ|≤1
DγuA. (3.1)
Recall thatW1LA(RN) is reflexive if and only ifAandA∗satisfy the∆2con- dition.
TheA-Dirichlet spaceL1A(RN) is the space of functionsu∈WA,loc1 (RN) (i.e., uis locally inW1LA(RN)) admitting a weak gradient such that|∇u|A<∞.
LetAbe anyᏺ-function and letabe its derivative. Forx∈RN, we define MA(x)=a(|x|)
|x| ·x ifx=0, MA(0)=0. (3.2) TheA-Laplacian of a functionf onRNis defined by∆Af =divMA(∇f).
A functionu∈WA,loc1 (RN) is said to be a weak solution to the equation
∆Au+h=0 (3.3)
if, for allϕ∈C10(RN), we have
MA(∇u),∇ϕdλ=
hϕ dλ. (3.4)
LetD⊂RNbe a nonempty bounded domain. The Banach spaceᏱA(D) is the space of functionsu∈WA,loc1 (RN) such that
|u|DA:= |u|A,D+|∇u|A<∞. (3.5) We denote byᏱ0A(D) the closure ofC10(RN) inᏱA(D).
748 On the
3.1. A nonresolvability result
Theorem3.1. LetAbe anᏺ-function satisfying the∆2condition. Suppose that RNisA-parabolic and leth∈L1(RN)be such thath dλ=0. Then the equation
∆Au+h=0 (3.6)
has no weak solution onL1A(RN).
Proof. We may suppose thath dλ >0. Hence there is a bounded set D⊂RN such thatλ(D)>0,s:=infDh >0, andDh dλ >|
h−dλ|. Let 0< c <1 be such that 0≤ −
h−dλ < cDh dλ.
By the definition ofΓ1,A(D), forε >0, we can find a functionv∈C0∞(RN) such that 0≤v≤1,v=1, onDand
|∇v|A≤ΓA(D) +ε. (3.7)
On the other hand, we have−cDvh dλ <vh−dλ≤0. Hence (1−c)
Dvh dλ <
Dvh dλ+
vh−dλ
<
Dvh dλ+
vh−dλ+
cDvh+dλ
≤
vh dλ.
(3.8)
Buts·λ(D)≤
Dvh dλ. Thus
(1−c)·s·λ(D)≤
vh dλ. (3.9)
Now suppose that u∈L1A(RN) is a weak solution of (3.6) and let ξ:=
−(a(|∇u|)/|∇u|)· ∇u. Then div(ξ)= −∆Au=h, and sinceAsatisfies the∆2
condition,|ξ| ∈LA∗(RN).
An integration by part and H¨older inequality in Orlicz spaces applied to in- equality (3.9) imply that
(1−c)·s·λ(D)≤
v·div(ξ)dλ
= −
∇v, ξdλ≤ ξA∗|∇v|A.
(3.10)
From (3.7), and sinceεis arbitrary, we get 0< λ(D)≤ ξA∗
(1−c)·s·ΓA(D). (3.11)
This is impossible, and the theorem is proved.
Corollary3.2. LetLAbe a reflexive Orlicz space such thatα∗≤N/(N−1). Let h∈L1(RN)be such thath dλ=0. Then (3.6) has no weak solution onL1A(RN).
Proof. By Lemmas2.5and2.3,RN is thenA-parabolic. We applyTheorem 3.1
to get the result.
Remark 3.3. WhenA(t)=p−1|t|p,LA=Lpis the usual Lebesgue space andα∗= p/(p−1). Hence the conditionα∗≤N/(N−1) is exactly the conditionN≤p.
Thus our result recovers the one in [6, Th´eor`eme 1].
3.2. A resolvability result. In this section, we resolve the equation∆Au+h=0 under some assumptions on theᏺ-functionAand on the functionh.
We begin by recalling the following Poincar´e inequality for Orlicz-Sobolev functions, which is a combination of [5, Theorem 3.3] and [5, Proposition 3.9].
Lemma3.4. LetAbe anᏺ-function such thatAandA∗satisfy the∆2condition.
Let Ebe any measurable set inRN such that0< λ(E)<∞. Then there exists a positive constantCsuch that
u−uEA,E≤C|∇u|A,E, (3.12) for allu∈WA,loc1 (RN), whereuE=(1/λ(E))Eu dλis the mean value ofuonE.
An application of H¨older inequality in Orlicz spaces gives
E
u−uEdλ≤χEA∗u−uEA,E, (3.13)
whereχEis the characteristic function ofE.
Recall that
χEA∗=λ(E)·A−1 1
λ(E)
,
|1|A,E=χEA= 1 A−11/λ(E).
(3.14)
Hence we obtain the following proposition.
Proposition3.5. LetAbe anᏺ-function such thatAandA∗satisfy the∆2con- dition. LetEbe any measurable set inRNsuch that0< λ(E)<∞. Then there exists a positive constantCsuch that
E
u−uEdλ≤C|∇u|A,E, (3.15)
for allu∈WA,loc1 (RN).
We will need the following proposition in what follows.
750 On the
Proposition3.6. LetAbe anᏺ-function such thatAandA∗satisfy the∆2con- dition. Suppose thatRNisA-hyperbolic. LetEbe any nonempty bounded domain inRN. Then there exists a positive constantCsuch that, for allu∈Ᏹ0A(E),
E|u|dλ≤C|∇u|A. (3.16) Proof. Suppose that such constant does not exist. Then for allε >0, we can find a functionu∈Ᏹ0A(E) such that
E|u|dλ=λ(E), |∇u|A≤ε. (3.17) We may assume thatu≥0.Proposition 3.5implies that
E|u|dλ≤Cε. (3.18)
We now choose a ballBEand a functionϕ∈C10 such that 0≤ϕ≤2−1, supp(ϕ)⊂E, and ϕ=2−1 on B. Define the function v∈Ᏹ0A(E) by v= 2 max(u, ϕ). Thenv≥1 onB. Now, define the sets
S=
x∈E:ϕ(x)≥u(x), S=
x∈E:u(x)−1≥2−1. (3.19) We haveS⊂Sand, by (3.18), 2−1λ(S)≤Cε. Thus
λ(S)≤2Cε. (3.20)
On the other hand, we have almost everywhere
∇v=
2∇u oncS,
2∇ϕ onS. (3.21)
This implies that
|∇v| ≤2|∇u|+ 2χS|∇ϕ| a.e. (3.22) Sincev≥1 onBandεis arbitrary, we deduce thatΓA(B)=0. This contradicts the fact thatRNisA-hyperbolic. The proof is complete.
Lemma3.7. LetAbe anᏺ-function. If RNisA-parabolic, then1∈Ᏹ0A(D)for any nonempty bounded domainD.
Proof. SinceRN isA-parabolic,ΓA(D)=0. Hence for all ε >0, there exists a functionu∈C01such thatu=1 onDand|∇u|A≤ε. Thus
|1−u|A= |1−u|A,D+|∇u|A= |∇u|A≤ε. (3.23)
This means that 1∈Ᏹ0A(D).
Theorem3.8. LetAbe anᏺ-function such thatAandA∗satisfy the∆2condi- tion. LetDbe nonempty bounded domain inRN. Then the following assertions are equivalent
(i)RNisA-hyperbolic;
(ii)there exists a constantCsuch that, for allu∈Ᏹ0A(D),
|u|A,D≤C|∇u|A; (3.24)
(iii) 1∈/ Ᏹ0A(D).
Proof. It is easy to verify that (ii) implies (iii). The implication (iii)⇒(i) isLemma 3.7. It remains to prove that (i) implies (ii).
Writeu=(u−uD) +uD.Proposition 3.6andLemma 3.4give
|u|A,D≤u−uD
A,D+uD
A,D
≤C|∇u|A,D+uD· |1|A,D
≤C|∇u|A,D+ 1
A−11/λ(D)·λ(D)−1
D|u|dλ
≤C|∇u|A,D+ 1
A−11/λ(D)·λ(D)−1C|∇u|A
≤C|∇u|A.
(3.25)
The proof is complete.
Recall that for all f ∈LAsuch that|f|A>1, we haveA◦f dλ >|f|A. We set
s(A)=inf
logA◦f dλ
log|f|A −1, f ∈LA,|f|A>1
. (3.26)
Hences(A)≥0.
Now we are ready to solve theA-Laplace equation.
Theorem 3.9. Let LA be a reflexive Orlicz space such that s(A)>0. Let h∈ L∞(RN)have compact support. Then the equation∆Au+h=0has a weak solution u∈L1A(RN)if RNisA-hyperbolic.
Proof. LetDbe a bounded domain such that supp(h)⊂D. Define the functional Ᏺ:Ᏹ0A(D)→Rby
Ᏺ(u)=
A|∇u| dλ−
hu dλ. (3.27)
752 On the Hence
Ᏺ(u)≥
A|∇u| dλ−
hu dλ
≥
A|∇u|
dλ− h∞· uL1(D).
(3.28)
SinceRNisA-hyperbolic, byProposition 3.6, we get Ᏺ(u)≥
A|∇u|
dλ−Ch∞· |∇u|A. (3.29) Hence there is a constantC1such that
Ᏺ(u)≥
A|∇u|
dλ−C1· ∇u|A. (3.30) By (2.3) and (2.5), there is a constantC2such that, for allk >0,
Ᏺ(u)≥
A|∇u| dλ−C2
k
Ak|∇u| dλ−C2
k . (3.31)
Now, lett >0 and consider the continuous function ψt defined onR+ by ψt(k)=(C2/k)A(kt)−A(t). Since
xa(x)≥A(x), ∀x≥0, limt→0
A(t)
t =0, lim
t→+∞
A(t)
t =+∞, (3.32)
the functionψtincreases from−A(t) to +∞. Hence there is ak0such thatψt(k0)
=0. Thus
Ᏺ(u)≥ −C2
k0. (3.33)
We conclude that the functionalᏲis bounded below on the spaceᏱ0A(D).
NowᏱA(D) is a reflexive Banach space andᏱ0A(D) is a closed convex subspace ofᏱA(D). We first prove thatᏲis lower semicontinuous. Lett∈R, and consider the setᏲt= {u∈Ᏹ0A(D) :Ᏺ(u)≤t}. Let (ui)i⊂Ᏹ0A(D) be such thatᏲ(ui)≤t, for alli, and (ui)iconverges touinᏱ0A(D). By the compactness of the imbedding Ᏹ0A(D)⊂L1(D), we may assume that (ui)iconverges strongly inL1(D). Hence
Dhuidλ−→
Dhu dλ. (3.34)
Theorem 3.8implies thatu→ |∇u|Ais an equivalent norm onᏱ0A(D).
Hence |∇u− ∇ui|A→0. Since A verifies the ∆2 condition, A(|∇u−
∇ui|)dλ→0. Hence there is a subsequence of the sequence (A(|∇u− ∇ui|))i, still denoted by (A(|∇u− ∇ui|))i, which convergesλ-almost everywhere to 0.
Thus (|∇ui|)i convergesλ-almost everywhere to|∇u|. By the continuity ofA, Fatou’s lemma, and (3.34), we get
Ᏺ(u)=
ilim→∞A∇uidλ−lim
i→∞
huidλ
≤lim inf
i→∞
A∇uidλ−lim
i→∞
huidλ≤t.
(3.35)
HenceᏲis lower semicontinuous.
Now, s(A)>0 implies that A(|∇u|)dλ≥ |∇u|s(A)+1A for |∇u|A>1.
Hence
Ᏺ(u)≥ |∇u|s(A)+1A −C1· |∇u|A for|∇u|A>1. (3.36) This proves thatᏲis coercive.
ThusᏲ attains its minimum onᏱ0A(D); that is, there is u∗∈Ᏹ0A(D) such thatᏲ(u∗)=min{Ᏺ(u) :u∈Ᏹ0A(D)}. By the usual arguments from variational calculus, we deduce thatu∗is a weak solution to the equation∆Au+h=0. The
proof is complete.
Remark 3.10. We have in fact solved the equation in the spaceᏱ0A(D)⊂L1A(RN).
Remark 3.11. WhenA(t)=p−1|t|p, p >1, andLA=Lp is the usual Lebesgue space, we haves(A)=p−1>0. Thus we recover the result in [9, Theorem 2]
when the manifoldMisRN.
Recall the following result in [2, Lemma 3].
Lemma3.12. LetAbe anᏺ-function satisfying the∆2condition. Ifα < N, then R1,A(B(x, r))>0, whereB(x, r)is the open ball of radiusr >0, with center atx.
Corollary3.13. LetLAbe a reflexive Orlicz space such thats(A)>0andα < N.
Suppose thath∈L∞(RN)has compact support. Then the equation ∆Au+h=0 has a weak solutionu∈L1A(RN).
Proof. By Lemmas3.12and 2.3, we deduce that RN isA-hyperbolic, and we
applyTheorem 3.9to get the result.
3.3. Some examples. In addition to theLpLebesgue case corresponding toA(t)
=p−1|t|p,p >1, we consider the followingᏺ-functions:
(1)
A1(t)=
tp for 0≤ |t| ≤1,
tq for 1<|t|, 1< p≤q <∞, (3.37) (2)A2(t)= |t|plog(1 +|t|),p >1,
(3)A3(t)= |t|plog(1 +|t|p),p >1,
754 On the
(4)A4(t)= |t|plogp(1 +|t|),p >1,
(5)Ap,q,r(t)= |t|plogq(1 +|t|r),p >1,q >0, andr >0.
All theseᏺ-functions and their conjugates satisfy the∆2condition. We show thats(Ai)>0,i=1,2,3,4, ands(Ap,q,r)>0.
First remark thatA2=Ap,1,1 andA3=Ap,1,p. Thus it suffices to show that s(Ap,q,r)>0 and for allp >1,q >0,r >0.
(1) Let f ∈LA1be such that|f|A1>1. Then, by (2.8), 1=
A1
f
|f|A1
(x)dλ(x)
≤ 1
|f|Ap1
{|f|≤|f|A1}|f|pdλ
+ 1
|f|qA1
{|f|>|f|A1}|f|qdλ
≤ 1
|f|Ap1
{|f|≤|f|A1}|f|pdλ+
{|f|>|f|A1}|f|qdλ
≤ 1
|f|Ap1
{|f|≤1}|f|pdλ+
{1<|f|≤|f|A1}|f|pdλ +
{|f|>|f|A1}|f|qdλ
≤ 1
|f|Ap1
{|f|≤1}|f|pdλ+
{1<|f|≤|f|A1}|f|qdλ +
{|f|>|f|A1}|f|qdλ
≤ 1
|f|Ap1
A1(f)(x)dλ(x).
(3.38)
Hence|f|Ap1≤
A1(f)(x)dλ(x). This implies thats(A1)>0.
(2) Let p >1, q >0, andr >0 and setA=Ap,q,r. Let f ∈LA be such that
|f|A>1. Then by (2.8), 1=
A
f
|f|A
(x)dλ(x)
≤ 1
|f|Ap
|f|plogq
1 + |f|r
|f|rA
dλ
≤ 1
|f|Ap
|f|plogq1 +|f|r dλ
≤ 1
|f|Ap
A(f)(x)dλ(x).
(3.39)
Thus|f|Ap≤
A(f)(x)dλ(x) and hences(A)>0.
Remark 3.14. AlthoughTheorem 3.9gives a solution for large classes of Orlicz spacesLA, includingLpLebesgue spaces,p >1, it would be sharp if we can drop the conditions(A)>0. This question is open.
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Noureddine A¨ıssaoui: D´epartement de Math´ematiques ´Ecole Normale Sup´erieure, BP 5206, Ben Souda, F`es, Morocco
E-mail address:[email protected]
