Electronic Journal of Differential Equations, Vol. 2018 (2018), No. 45, pp. 1–23. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
NONLINEAR ANISOTROPIC ELLIPTIC EQUATIONS WITH VARIABLE EXPONENTS AND DEGENERATE COERCIVITY HOCINE AYADI, FARES MOKHTARI Communicated by Jes´ us Ildefonso D´ıaz
Abstract. In this article, we prove the existence and the regularity of distributional solutions for a class of nonlinear anisotropic elliptic equations with pi (x) growth conditions, degenerate coercivity and Lm(·) data, with m(·) being small, in appropriate Lebesgue-Sobolev spaces with variable exponents. The obtained results extend some existing ones [8, 10].
1. Introduction We consider the problem −
N X
Di ai (x, u)|Di u|pi (x)−2 Di u + |u|s(x)−1 u = f
in Ω,
i=1
u=0
(1.1)
on ∂Ω,
where Ω is a bounded open domain in RN (N ≥ 2) with Lipschitz boundary ∂Ω, and the right-hand side f in L1 (Ω) (or Lm(·) (Ω)). We assume that the variable exponents s : Ω → (0, +∞), pi : Ω → (1, +∞), i = 1, . . . , N are continuous functions such that 1 < p(x) ≤ N
where
N 1 1 X 1 = , p(x) N i=1 pi (x)
∀x ∈ Ω.
(1.2)
Here, we suppose that ai : Ω × R → R, i = 1, . . . , N are Carath´eodory functions such that for a.e. x ∈ Ω, for every t ∈ R, we have α ≤ ai (x, t) ≤ β, (1 + |t|)γi (x)
i = 1, . . . , N,
(1.3)
where α, β are strictly positive real numbers and γi ∈ C(Ω), γi (x) ≥ 0 for all x ∈ Ω and i = 1, . . . , N . 2010 Mathematics Subject Classification. 35J70, 35J60, 35B65. Key words and phrases. Anisotropic elliptic equations; variable exponents; degenerate coercivity; distributional solutions; irregular data. c
2018 Texas State University. Submitted May 18, 2017. Published February 12, 2018. 1
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The main difficulty in dealing with problem (1.1) is the fact that, because of assumption (1.3), the differential operator A(u) = −
N X
Di ai (x, u)|Di u|pi (x)−2 Di u
i=1
˚ 1,~p(·) (Ω) and its dual but it fails to be coercive if u is is well defined between W large (see [17]). This shows that the classical methods for elliptic operators can’t be applied. To overcome this problem, we will proceed by approximation by means of truncatures in a(x, t) to get a coercive differential operator. We cite some papers that have dealt with the equation (1.1) or similar problems, see [1, 3, 5, 6, 7, 8, 9, 10, 14, 20] and the references therein. In case of a constant exponent pi (x) = 2, s(x) = q and γi (x) = γ (resp. pi (x) = p) similar results can be found in [8, 10]. The problem was also considered in [14] when pi (x) = p(x) and γi (x) = γ(x) ≥ 0, where the authors supposed that ai (x, u) = a(x, u) ≤ β(|u|) with β : [0, +∞) → (0, +∞) is a continuous function. The lack of growth condition on a(x, u) prompted them to consider only the renormalized and entropy solutions. The corresponding results in the isotropic case and without lower order term are developed in [1, 5, 6, 7, 20]. More general results are obtained in [1] in the constant case pi (x) = p, γi (x) = θ(p − 1), − −1 the θ ∈ [0, 1]. In the case pi (x) = p(x) and γi (x) = θ(p(x) − 1) where 0 ≤ θ ≤ pp+ −1 main results are collected in the paper [20]. Recently, the mathematical researchers paid attention to the anisotropic nonlinear problems with variable exponents. For instance, Problem (1.1) was investigated in [3, 4, 15, 16] under uniform ellipticity condition i.e γi (x) = 0. In this article we assume that the condition (1.3) holds and f ∈ Lm(·) (Ω) where pi is assumed to be merely a continuous function, and we treat the regularity of distributional solution u depending simultaneously on s(·) and m(·). 2. Preliminaries In this section we recall some facts on anisotropic spaces with variable exponents and we give some of their properties. For further details on the Lebesgue-Sobolev spaces with variable exponents, we refer to [2, 11, 12] and references therein. In this article we set C+ (Ω) = {p ∈ C(Ω) : p(x) > 1, for any x in Ω}. For any p ∈ C+ (Ω), we denote p+ = max p(x)
and p− = min p(x).
x∈Ω
x∈Ω
We define the Lebesgue space with variable exponent Lp(·) (Ω) as the set of all measurable functions u : Ω → R for which the convex modular Z ρp(·) (u) = |u|p(x) dx, Ω
is finite. The expression u kukp(·) := kukLp(·) (Ω) = inf λ > 0 : ρp(·) ≤1 , λ
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defines a norm on Lp(·) (Ω), called the Luxemburg norm. The space (Lp(·) (Ω), kukp(·) ) is a separable Banach space. Moreover, if 1 < p− ≤ p+ < +∞, then Lp(·) (Ω) is uni0 formly convex, hence reflexive and its dual space is isomorphic to Lp (·) (Ω) where 0 1 1 p(·) (Ω) and v ∈ Lp (·) (Ω), the H¨older type inequality p(x) + p0 (x) = 1. For all u ∈ L Z 1 1 uv dx ≤ − + 0− kukp(·) kvkp0 (·) ≤ 2kukp(·) kvkp0 (·) , p p Ω holds. We define also the Banach space W 1,p(·) (Ω) = u ∈ Lp(·) (Ω) : |∇u| ∈ Lp(·) (Ω) , which is equipped with the norm kuk1,p(·) = kukW 1,p(·) (Ω) = kukp(·) + k∇ukp(·) . 1,p(·)
The space (W 1,p(·) (Ω), kuk1,p(·) ) is a Banach space. Next, we define W0 Sobolev space with zero boundary values by 1,p(·) W0 (Ω) = u ∈ W 1,p(·) (Ω) : u = 0 on ∂Ω , 1,p(·)
endowed with the norm k · k1,p(·) . The space W0 −
provided that 1 < p ≤ p Poincar´e inequality holds
+
< +∞. For u ∈
(Ω) the
(Ω) is separable and reflexive
1,p(·) W0 (Ω)
with p ∈ C+ (Ω), the
kukp(·) ≤ Ck∇ukp(·) , (2.1) for some C > 0 which depends on Ω and p(·). Therefore, k∇ukp(·) and kuk1,p(·) are equivalent norms. An important role in manipulating the generalized Lebesgue and Sobolev spaces is played by the modular ρp(·) (u) of the space Lp(·) (Ω). We have the following results. Proposition 2.1 ([2, 11]). If un , u ∈ Lp(·) (Ω) and p+ < +∞, then the following properties hold: • kukp(·) < 1 (resp. = 1, > 1) ⇐⇒ ρp(·) (u) < 1 (resp. = 1, > 1), 1 1 1 1 + • min ρp(·) (u) p+ , ρp(·) (u) p− ≤ kukp(·) ≤ max ρp(·) (u) p , ρp(·) (u) p− , − + − + • min kukpp(·) , kukpp(·) ≤ ρp(·) (u) ≤ max kukpp(·) , kukpp(·) , • kukp(·) ≤ ρp(·) (u) + 1, • kun − ukp(·) → 0 ⇐⇒ ρp(·) (un − u) → 0. Remark 2.2. Note that the inequality Z Z |f |p(x) dx ≤ C |Df |p(x) dx, Ω
Ω
in general does not hold (see [13]). But by Proposition 2.1 and (2.1) we have Z + − |f |p(x) dx ≤ C max{kDf kpp(·) , kDf kpp(·) }. (2.2) Ω
Now, we present the anisotropic Sobolev space with variable exponent which is used for the study of problem (1.1). First of all, let pi (·) : Ω → [1, +∞), i = 1, . . . , N be continuous functions, we set p~(·) = (p1 (·), . . . , pN (·)) and p+ (x) = max1≤i≤N pi (x), for all x ∈ Ω. The anisotropic variable exponent Sobolev space W 1,~p(·) (Ω) is defined as W 1,~p(·) (Ω) = u ∈ Lp+ (·) (Ω) : Di u ∈ Lpi (·) (Ω), i = 1, . . . , N ,
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which is Banach space with respect to the norm kukW 1,~p(·) (Ω) = kukp+ (·) +
N X
kDi ukpi (·) .
i=1 1,~ p(·)
We denote by W0
(Ω) the closure of C0∞ (Ω) in W 1,~p(·) (Ω), and we define ˚ 1,~p(·) (Ω) = W 1,~p(·) (Ω) ∩ W 1,1 (Ω). W 0
If Ω is a bounded open set with Lipschitz boundary ∂Ω, then ˚ 1,~p(·) (Ω) = u ∈ W 1,~p(·) (Ω) : u|∂Ω = 0 . W It is well-known that in the constant exponent case, that is, when p~(·) = p~ ∈ ˚ 1,~p (Ω). However in the variable exponent case, in general [1, +∞)N , W01,~p (Ω) = W 1,~ p(·) 1,~ p (·) ˚ W0 (Ω) ⊂ W (Ω) and the smooth functions are in general not dense in ˚ 1,~p(·) (Ω), but if for each i = 1, . . . , N, pi is log-H¨older continuous, that is, there W exists a positive constant L such that |pi (x) − pi (y)| ≤
L , − ln |x − y|
∀x, y ∈ Ω, |x − y| < 1.
˚ 1,~p(·) (Ω), thus W 1,~p(·) (Ω) = W ˚ 1,~p(·) (Ω). We set for all Then C0∞ (Ω) is dense in W 0 x∈Ω N p(x) = PN , 1 i=1 pi (x)
and we define ( ?
p (x) =
N p(x) N −p(x) ,
for p(x) < N,
+∞,
for p(x) ≥ N.
We have the following embedding results. Lemma 2.3 ([12]). Let Ω ⊂ RN be a bounded domain, and p~(·) ∈ (C+ (Ω))N . If q ∈ C+ (Ω) and for all x ∈ Ω, q(x) < max(p+ (x), p? (x)). Then the embedding ˚ 1,~p(·) (Ω) ,→ Lq(·) (Ω), W is compact. Lemma 2.4 ([12]). Let Ω ⊂ RN be a bounded domain, and p~(·) ∈ (C+ (Ω))N . Suppose that ∀x ∈ Ω, p+ (x) < p? (x). (2.3) Then the following Poincar´e-type inequality holds kukLp+ (·) (Ω) ≤ C
N X
kDi ukLpi (·) (Ω) ,
˚ 1,~p(·) (Ω), ∀u ∈ W
(2.4)
i=1
where C is a positive constant independent of u. Thus ˚ 1,~p(·) (Ω). equivalent norm on W
PN
i=1
kDi ukLpi (·) (Ω) is an
The following embedding results for the anisotropic constant exponent Sobolev space are well-known [18, 19].
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Lemma 2.5. Let αi ≥ 1, i = 1, . . . , N , we pose α ~ = (α1 , . . . , αN ). Suppose u ∈ W01,~α (Ω), and set ( N α α? = NN−α if α < N, 1 1 X 1 = , r= α N i=1 αi any number in [1, +∞) if α ≥ N. Then, there exists a constant C depending on N, p1 , . . . , pN if α < N and also on r and |Ω| if α ≥ N , such that kukLr (Ω) ≤ C
N Y
1/N
kDi ukLαi (Ω) .
(2.5)
i=1
Lemma 2.6. Let Q be a cube of RN with faces parallel to the coordinate planes and αi ≥ 1, i = 1, . . . , N . Suppose u ∈ W 1,~α (Q), and set r = α?
if α < N,
r ∈ [1, +∞)
if α ≥ N.
Then, there exists a constant C depending on N, α1 , . . . , αN if α < N and also on r and |Q| if r ≥ N , such that kukLr (Q) ≤ C
N Y
kukLαi (Q) + kDi ukLαi (Q)
1/N
.
(2.6)
i=1
We will use through the paper, the truncation function Tk at height k (k > 0), that is Tk (t) = max{−k, min{k, t}}. Proposition 2.7. If u : Ω → R is a measurable function such that Tk (u) ∈ ˚ 1,~p(·) (Ω) for all k > 0, then there exists a unique measurable function v : Ω → RN W such that ∇Tk (u) = vχ{|u|≤k} a.e. in Ω, (2.7) where χA denotes the characteristic function of a measurable set A. Moreover, if u ∈ W01,1 (Ω) then v coincides with the standard distributional gradient of u. ˚ 1,~p(·) (Ω) for any k > 0, does not necessarily A function u such that Tk (u) ∈ W 1,1 belong to W0 (Ω). However, according to the above proposition, it is possible to define its weak gradient, still denoted by ∇u, as the unique function v which satisfies (2.7). Definition 2.8. For 0 < r < +∞, the set of all measurable functions v : Ω → R such that the functional [u]r = supk>0 k meas{x ∈ Ω : |u(x)| > k}1/r is finite is called a Marcinkiewicz space and is denoted by M r (Ω). If |Ω| < ∞ and 0 < < r−1, we can show that Lr (Ω) ⊂ M r (Ω) ⊂ Lr− (Ω). 3. Statement of results Definition 3.1. We say that u is a distributional solution for problem (1.1) if u ∈ W01,1 (Ω), |u|s(·) ∈ L1 (Ω) and Z Z N Z X ai (x, u)|Di u|pi (x)−2 Di uDi ϕ dx + |u|s(x)−1 uϕ dx = f ϕ dx, (3.1) i=1
Ω
for every ϕ ∈
Ω
C0∞ (Ω).
Ω
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Our main results are the following Theorem 3.2. Let f ∈ L1 (Ω), pi : Ω → (1, +∞), s : Ω → (0, +∞) and γi : Ω → [0, +∞) be continuous functions such that for all x ∈ Ω, s(x) ≥ pi (x), p(x)(N − 1 − N (p(x) − 1 −
+ γ+ ) + γ+ )
i = 1, . . . , N,
(3.2)
< pi (x) < k(x),
(3.3)
where k(x) =
+ p(x)(N −1−γ+ ) + (1+γ+ )(N −p(x))
,
+∞,
if p(x) < N if p(x) = N
and
+ γ+ = max max γi (x). 1≤i≤N x∈Ω
Let ai be Carath´eodory functions for i = 1, · · · , N satisfying (1.3). Then, prob˚ 1,~q(·) (Ω) where qi (·) are lem (1.1) has at least one distributional solution u ∈ W continuous functions on Ω satisfying 1 ≤ qi (x)
max ; (1 + γi (x))(pi (x) − 1) , i = 1, . . . , N. (3.5) pi (x) − 1 Let ai be Carath´eodory functions satisfying (1.3). Then, problem (1.1) has at least ˚ 1,~q(·) (Ω) ∩ Ls(·) (Ω) where qi (·) are continuous one distributional solution u ∈ W functions on Ω satisfying pi (x)s(x) 1 < qi (x) < (3.6) , ∀x ∈ Ω, i = 1, . . . , N. s(x) + 1 + γi (x) Theorem 3.4. Let m : Ω → (1, +∞), pi : Ω → (1, +∞), γi : Ω → [0, +∞) and s : Ω → (0, +∞) be continuous functions such that (2.3) holds and for all x ∈ Ω 1 < m(x) < h(x), where
( h(x) =
∇m ∈ L∞ (Ω),
N p(x) N p(x)+p(x)−N , p+ (x) p+ (x)−1 ,
(3.7)
if p(x) < N if p(x) = N,
and s(x) ≥
1 + γ+ (x) , m(x) − 1
∇s ∈ L∞ (Ω),
∇γ+ ∈ L∞ (Ω),
γ+ (x) = max γi (x). 1≤i≤N
(3.8) Let f ∈ Lm(·) (Ω) and let ai be Carath´eodory functions satisfying (1.3). Then, ˚ 1,~p(·) (Ω) ∩ Ls(·)m(·) (Ω). problem (1.1) has at least one distributional solution u ∈ W Theorem 3.5. Let f ∈ Lm(·) (Ω) with m as in (3.7), pi : Ω → (1, +∞), γi : Ω → [0, +∞), and s : Ω → (0, +∞) be continuous functions. Assume (2.3), and for all x ∈ Ω, 1 + γ (x) 1 + γ+ (x) i > s(x) > max ; (1 + γi (x))(pi (x) − 1) , (3.9) m(x) − 1 pi (x)m(x) − 1
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∇s ∈ L∞ (Ω), and i = 1, . . . , N , where γ+ (·) = max1≤i≤N γi (·). Let ai be Carath´eodory functions satisfying (1.3). Then, problem (1.1) has at least one distributional solu˚ 1,~q(·) (Ω) where qi (·) are continuous tion u such that |u|m(x)s(x) ∈ L1 (Ω) and u ∈ W functions on Ω satisfying 1 < qi (x) =
pi (x)m(x)s(x) , s(x) + 1 + γi (x)
∀x ∈ Ω, i = 1, . . . , N.
(3.10)
Remark 3.6. In Theorem 3.2, it is clear that the conditions (1.2) and (3.3) imply that (2.3) holds since we have k(x) ≤ p? (x), ∀x ∈ Ω. Remark 3.7. Observe that the conditions (3.7), (3.8), and (2.3) guarantee that s(x) > (1 + γ+ (x))(pi (x) − 1),
∀x ∈ Ω, i = 1, . . . , N.
Remark 3.8. In Theorem 3.5, the conditions (3.7) and (2.3) imply that the assumption (3.9) is not empty since we have 1 > pi (x) − 1, m(x) − 1
∀x ∈ Ω, i = 1, . . . , N.
(3.11)
Remark 3.9. Let f ∈ L1 (Ω). Assume that for all x ∈ Ω, p(x) < N and s(x) > + ) (1+γi (x))N (p(x)−1−γ+ + )(N −p(x)) (1+γ+
for all i = 1, . . . , N . Then, assumption (3.3) implies (3.5) and
+ )pi (x) N (p(x) − 1 − γ+ pi (x)s(x) , > + s(x) + 1 + γi (x) p(x)(N − 1 − γ+ )
∀x ∈ Ω, i = 1, . . . , N,
so Theorem 3.3 improves Theorem 3.2 (and [3, Theorem 3.1]). 4. Approximate equation We will use the following approximating problem −
N X
Di ai (x, Tn (un ))|Di un |pi (x)−2 Di un + |un |s(x)−1 un = Tn (f )
i=1
un = 0
in Ω,
(4.1)
on ∂Ω.
We are going to prove the existence of solution un to problem (4.1). Lemma 4.1. Let f ∈ L1 (Ω) and let s : Ω → (0, +∞), pi : Ω → (1, +∞), i = 1, . . . , N be continuous functions. Assume that (2.3) holds. Then, there exists at ˚ 1,~p(·) (Ω) to problem (4.1) in the sense that least one solution un ∈ W Z N Z X ai (x, Tn (un ))|Di un |pi (x)−2 Di un Di ϕ dx + |un |s(x)−1 un ϕ dx Ω Ω i=1 (4.2) Z =
Tn (f )ϕ dx, Ω
˚ 1,~p(·) (Ω) ∩ L∞ (Ω). Moreover for every ϕ ∈ W Z Z s(x) |un | dx ≤ |f | dx. Ω
Ω
(4.3)
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Proof. Consider the problem −
N X
Di ai (x, Tn (unk ))|Di unk |pi (x)−2 Di unk + Tk |unk |s(x)−1 unk
i=1
(4.4)
= Tn (f )
in Ω, unk = 0
on ∂Ω.
˚ 1,~p(·) (Ω) to problem It has been proved in [12] that there exists a solution unk ∈ W (4.4), which satisfies N Z X ai (x, Tn (unk ))|Di unk |pi (x)−2 Di unk Di ϕ dx Ω
i=1
Z
(4.5)
Tk |unk |s(x)−1 unk ϕ dx
+ Ω
Z =
Tn (f )ϕ dx,
˚ 1,~p(·) (Ω). ∀ϕ ∈ W
Ω
Choosing ϕ = unk in (4.5), by (1.3) and using that Tk (|unk |s(x)−1 )unk ≥ 0, we have Z N Z X α pi (x) |Di unk | dx ≤ |unk | dx. + Ω n(1 + n)γ+ i=1 Ω Using Young’s inequality for all ε > 0, we obtain Z N Z X − |Di unk |pi (x) dx ≤ ε |unk |p− dx + C1 i=1
Ω
Ω
Z ≤ εC2
−
|Di unk |p− dx + C1
Ω
≤ εC2
N Z X i=1
|Di unk |pi (x) dx + C3 ,
Ω
where C1 , C2 , and C3 are positive constants not depending on k. Now, we choose ε = 1/(2C2 ), then N Z X |Di unk |pi (x) dx ≤ C(n). i=1
Ω
˚ 1,~p(·) (Ω). So, there exists a It follows that the sequence {unk }k is bounded in W ˚ 1,~p(·) (Ω) and a subsequence (still denoted by un ) such that function un ∈ W k unk * un
˚ 1,~p(·) (Ω) and a.e in Ω. weakly in W
(4.6)
Choosing ϕ = unk − un in (4.5) as a test function, we can easily prove that, for all i = 1, . . . , N , Z h i ai (x, Tn (unk )) |Di unk |pi (x)−2 Di unk − |Di un |pi (x)−2 Di un Di (unk − un ) dx → 0 Ω
as k → +∞. By (1.3), we obtain Z Ei (k) = |Di unk |pi (x)−2 Di unk − |Di un |pi (x)−2 Di un Di (unk − un ) dx → 0 Ω
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as k → +∞. We recall the following well-known inequalities, that hold for any two real vectors ξ, η and a real p > 1: (|ξ|
p−2
p−2
ξ − |η|
η)(ξ − η) ≥
( 22−p |ξ − η|p , (p −
if p ≥ 2,
|ξ−η|2 1) (|ξ|+|η|) 2−p ,
if 1 < p < 2.
(4.7)
Therefore, Z + 22−pi |Di (unk − un )|pi (x) dx {x∈Ω,pi (x)≥2} Z ≤ |Di unk |pi (x)−2 Di unk − |Di un |pi (x)−2 Di un Di (unk − un ) dx {x∈Ω,pi (x)≥2} Z ≤ |Di unk |pi (x)−2 Di unk − |Di un |pi (x)−2 Di un Di (unk − un ) dx Ω
= Ei (k). (4.8) On the set Ωi = {x ∈ Ω, 1 < pi (x) < 2}, we employ (4.7) as follows Z
|Di unk − Di un |pi (x) dx
Ωi
|Di unk − Di un |pi (x)
Z ≤ Ωi
(|Di unk | + |Di un |)
|Di unk − Di un |pi (x)
≤ 2k
(|Di unk | + |Di un |)
pi (x)(2−pi (x)) 2
× k (|Di unk | + |Di un |)
(|Di unk | + |Di un |)
pi (x)(2−pi (x)) 2
k
pi (x)(2−pi (x)) 2
pi (x)(2−pi (x)) 2
dx
2
L pi (·) (Ωi )
k
2
L 2−pi (·) (Ω) −
≤ 2 max
n Z Ωi
p2i |Di unk − Di un |pi (x) dx , i (x)) pi (x)(2−p 2 |Di unk | + |Di un | +
|Di unk − Di un |pi (x)
Z
pi (x)(2−pi (x))
p2i o dx
2 (|Di unk | + |Di un |) + 2−p Z n pi (x) 2 i × max |Di unk | + |Di un | dx ,
Ωi
Ω
Z
pi (x) |Di unk | + |Di un | dx
− 2−p i 2
o
Ω + − − o + n p p p p i i − − 2i − 2i 2 , (p 2 ≤ 2 max (p− − 1) E (k) − 1) E (k) i i i i +
× max
n Z
2−p pi (x) 2 i |Di unk | + |Di un | dx ,
Ω
Z Ω
pi (x) |Di unk | + |Di un | dx
− 2−p i 2
o .
(4.9)
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˚ 1,~p(·) (Ω) and un ∈ W ˚ 1,~p(·) (Ω), after letting k → +∞ in Since unk is bounded in W (4.8) and (4.9), we find Z lim |Di unk − Di un |pi (x) dx = 0, k→+∞
Ω
which implies, for all i = 1, . . . , N , Di unk → Di un
strongly in Lpi (·) (Ω) and a.e. in Ω.
(4.10)
We are going to prove (4.2) by passing to the limit in (4.5). By (4.10) we have |Di unk |pi (x)−2 Di unk * |Di un |pi (x)−2 Di un
0
weakly in Lpi (·) (Ω), p0i (·) =
pi (·) . pi (·) − 1
From (4.6) and Lebesgue’s dominated convergence theorem, we obtain ai (x, Tn (unk ))Di ϕ → ai (x, Tn (un ))Di ϕ strongly in Lpi (·) (Ω), 1 ≤ i ≤ N. Let ρj (t) be an increasing, uniformly bounded Lipschitz function [5] (or W 1,∞ (Ω) function), such that ρj (σ) → χ{|σ|>t} sign(σ), as j → +∞. Taking ρj (unk ) as a test function in (4.5), we obtain Z N Z X ρ0j (unk )ai (x, Tn (unk ))|Di unk |pi (x) dx + Tk (|unk |s(x)−1 unk )ρj (unk ) dx Ω
i=1
Ω
Z =
Tn (f )ρj (unk ) dx. Ω
As j → +∞, we obtain Z |Tk (|unk |
s(x)−1
Z unk )| dx ≤
{|unk |>t}
|f | dx.
(4.11)
{|unk |>t}
Let E ⊂ Ω be any measurable set, using (4.11), we have Z |Tk (|unk |s(x)−1 unk )| dx E Z Z s(x)−1 = |Tk (|unk | unk )| dx + |Tk (|unk |s(x)−1 unk )| dx E∩{|unk |≤t} +
−
E∩{|unk |>t}
≤ (ts + ts ) meas(E) +
Z |f | dx. {|unk |>t}
Then we deduce that the sequence Tk (|unk |s(x)−1 unk ) is equi-integrable in L1 (Ω), and since Tk (|unk |s(x)−1 unk ) → |un |s(x)−1 un a.e. in Ω, Vitali’s theorem implies that Tk (|unk |s(x)−1 unk ) → |un |s(x)−1 un
in L1 (Ω).
Therefore, we can obtain (4.2) by passing to the limit in (4.5). n) in (4.2) as a test function, we have To show (4.3), we choose ϕ = Tk (u k Z Z Z Tk (un ) Tk (un ) |un |s(x)−1 un dx ≤ Tn (f ) dx ≤ |f | dx. k k Ω Ω Ω Fatou’s lemma implies that estimate (4.3) holds as k → 0.
In the rest of this paper, we will denote by Ci (or C) the positive constants depending only on the data of the problem, but not on n.
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5. Uniform estimates In this section, we assume that un is a solution of (4.1). Lemma 5.1. Let pi : Ω → (1, ∞), s : Ω → (0, ∞) and γi : Ω → [0, ∞) be continuous functions. Then, there exists a constant C > 0 such that N Z X |Di un |pi (x) dx ≤ C, ∀λ > 1, (5.1) (1 + |un |)γi (x)+λ i=1 Ω Z + k |Di Tk (un )|pi (x) dx ≤ (1 + k)γ+ kf kL1 (Ω) , i = 1, . . . , N. (5.2) α Ω Proof. We introduce the function ψ : R → R by Z t dx 1 = [(1 + |t|)1−λ − 1] sign(t), λ > 1. ψλ (t) = λ 1−λ 0 (1 + |x|) Note that ψλ is a continuous function satisfies ψλ (0) = 0 and |ψλ0 (·)| ≤ 1. We take ψλ (un ) as a test function in (4.2) and we use the assumption (1.3), we obtain Z N Z X |Di un |pi (x) dx ≤ C |f | dx, 1 (1 + |un |)γi (x)+λ Ω i=1 Ω In particular, there exists C2 > 0 such that − Z |Di un |pi dx ≤ C2 , γ + +λ Ω (1 + |un |) +
∀i = 1, . . . , N.
(5.3)
We take ϕ = Tk (un ) in (4.2), we find Z |Di Tk (un )|pi (x) k dx ≤ kf kL1 (Ω) . γi (x) α (1 + |u |) n Ω Hence Z
|Di Tk (un )|pi (x) dx =
Ω
Z Ω
|Di Tk (un )|pi (x) (1 + |Tk (un )|)γi (x) dx (1 + |Tk (un )|)γi (x)
+ k ≤ (1 + k)γ+ kf kL1 (Ω) . α
Which yields (5.2).
Lemma 5.2. Assume that s(·), pi (·) and γi (·) are restricted as in Theorem 3.2. Then, there exists a constant C > 0 such that for all continuous functions qi (·), i = 1, . . . , N on Ω as in (3.4), we have kDi un kLqi (·) (Ω) ≤ C,
(5.4)
kun kLq? (·) (Ω) ≤ C.
(5.5)
Proof. Firstly, for pi is defined in (3.3), we have 1
1 such that 1 − θ +? + q − γ+ > λ > 1, θ so, θ + (5.7) (γ+ + λ) < q +? . 1−θ Using H¨ older’s inequality and (5.3), we obtain + Z Z + + |Di un |qi |Di un |qi dx = (1 + |un |)(γ+ +λ)θ dx + (γ+ +λ)θ Ω (1 + |un |) Ω − Z θ Z 1−θ + θ |Di un |pi (γ+ +λ) 1−θ ≤ dx (1 + |u |) n γ + +λ Ω (1 + |un |) + Ω Z 1−θ + θ , ≤C (1 + |un |)(γ+ +λ) 1−θ Ω
so that N Y
1/N
kDi un k
i=1
+ Lqi (Ω)
1
≤ C q+
Z
+
θ
(1 + |un |)(γ+ +λ) 1−θ
1−θ + q
.
Ω
Therefore, by (5.7) and Young’s inequality, we can write N 1/N Z 1−θ Y +? q+ kDi un k q+ ≤ C1 (ε) + ε |un |q . L
i=1
i
(Ω)
(5.8)
Ω
In view of (2.5), with r = q +? , we obtain kun kLq+? (Ω) ≤ C0
N Y
kDi un k
i=1
1/N q
L
+ i (Ω)
≤ C2 (ε) + εC0 kun kLq+? (Ω)
N (1−θ) + N −q
.
(5.9) We choose ε = 1/(2C0 ), then 1 kun kLq+? (Ω) ≤ C3 + kun kηLq+? (Ω) , 2
η = (1 − θ)
N . N − q+
(5.10)
The assumption (1.2) implies that η ∈ (0, 1]. Hence, the estimate (5.10) implies (5.5), and by (5.8) we deduce that (5.4) holds. This completes the proof of the case (a).
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Case (b): In the second, we suppose that (3.4) holds and qi+ ≥
+ − N (p− − 1 − γ+ )pi . + − p (N − 1 − γ+ )
By the continuity of pi (·) and qi (·) on Ω, there exists a constant δ > 0 such that max
qi (t)
0 such that δ > |Ωj | = meas(Ωj ) > ν,
Ωj = Qj ∩ Ω
for all j = 1, . . . , k.
+ We denote by qi,j the local maximum of qi (·) on Ωj (respectively p− i,j the local minimum of pi (·) on Ωj ), such that + qi,j
0 such that N Z X Ω
i=1
|Di un |pi (x) dx + (1 + |un |)γi (x)+1−[(m(x)−1)s(x)]
Z
|un |m(x)s(x) dx ≤ C.
(5.17)
Ω
Proof. Taking ψ(x, un ) = (1 + |un |)(m(x)−1)s(x) − 1 sign(un ) in (4.1) as a test function, by (1.3) and the fact that for a.e. x ∈ Ω and for all i = 1, . . . , N Di ψ(x, un ) = (m(x) − 1)(1 + |un |)(m(x)−1)s(x) sign(un )Di s(x) ln(1 + |un |) +
(m(x) − 1)s(x)Di un (1 + |un |)1−(m(x)−1)s(x)
+ Di m(x)(1 + |un |)(m(x)−1)s(x) sign(un )s(x) ln(1 + |un |), we obtain αs− (m− − 1)
N Z X i=1
Z
Ω
(1 + |un
|Di un |pi (x) γ |) i (x)+1−((m(x)−1)s(x))
dx
|un |s(x) (1 + |un |)(m(x)−1)s(x) − 1 dx
+ ZΩ ≤
|f | (1 + |un |)(m(x)−1)s(x) − 1 dx
Ω
+ C1
N Z X i=1
(1 + |un |)(m(x)−1)s(x) ln(1 + |un |)|Di un |pi (x)−1 dx.
Ω
+
Using that |un |s(x) ≥ min{1; 21−s }(1 + |un |)s(x) − 1, Proposition 2.1, and Young inequality, we have N Z X i=1
Ω
|Di un |pi (x) (1 + |un |)γi (x)+1−((m(x)−1)s(x))
dx +
1 2
Z
(1 + |un |)m(x)s(x) dx
Ω
+ m− ≤ C2 + C3 max kf km Lm(·) (Ω) , kf kLm(·) (Ω) N Z X + C4 (1 + |un |)(m(x)−1)s(x) ln(1 + |un |)|Di un |pi (x)−1 dx. i=1
Ω
(5.18)
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We can estimate the last term in (5.18) by applying Young’s inequality Z (1 + |un |)(m(x)−1)s(x) ln(1 + |un |)|Di un |pi (x)−1 dx Ω Z (m(x)−1)s(x)+(pi (x)−1)(γi (x)+1) pi (x) = (1 + |un |) ln(1 + |un |) Ω
|Di un |pi (x)−1
×
γi (x)+1−(m(x)−1)s(x) p0 (x) i
dx
(5.19)
(1 + |un |) Z p (x) ≤ C5 (1 + |un |)(m(x)−1)s(x)+(pi (x)−1)(γi (x)+1) (ln(1 + |un |)) i dx Ω
1 + 4C4
Z Ω
|Di un |pi (x) dx. (1 + |un |)γi (x)+1−(m(x)−1)s(x)
We combine (5.18) and (5.19), we obtain Z N Z 3X |Di un |pi (x) 1 (1 + |un |)m(x)s(x) dx dx + 4 i=1 Ω (1 + |un |)γi (x)+1−(m(x)−1)s(x) 2 Ω N Z X ≤ C6 + C7 (1 + |un |)(m(x)−1)s(x)+(pi (x)−1)(γi (x)+1) i=1
(5.20)
Ω
× ln(1 + |un |)pi (x) dx = I. Since s(x) > (pi (x) − 1)(γi (x) + 1), we have (pi (x) − 1)(γi (x) + 1) − s(x) ≤ ((pi (x) − 1)(γi (x) + 1) − s(x))+ = bi
1. Choosing s(x)(p (x) − q (x)) i i λ = min min − γi (x) > 1 . 1≤i≤N x∈Ω qi (x)
Then assumption (3.6) implies
Thanks to the choice of λ and (3.6), we have qi (x)(γi (x) + λ) ≤ s(x), pi (x) − qi (x)
∀x ∈ Ω, ∀i = 1, . . . , N.
(5.24)
Combining (5.24), (5.23), and (4.3) results (5.22).
Lemma 5.5. Let m, s, pi , and γi be restricted as in Theorem 3.5. Then, there exists a constant C > 0 such that kDi un kLqi (·) (Ω) ≤ C,
(5.25)
for all continuous functions qi on Ω satisfying (3.10). Proof. Note that s(x) < inequality, we have Z |Di un |qi (x) dx
1+γ+ (x) m(x)−1
and (3.10) imply qi (x) < pi (x). Then by Young’s
Ω
|Di un |qi (x)
Z =
(1 + Z q (x) Ω
≤
q (x)
(γi (x)+1−[(m(x)−1)s(x)]) pi (x)
|un |)(γi (x)+1−[(m(x)−1)s(x)]) pqii(x) (x)
(1 + |un |)
i
dx
|Di un |pi (x) dx γi (x)+1−[(m(x)−1)s(x)] Ω pi (x) (1 + |un |) Z (γi (x)+1−[(m(x)−1)s(x)])qi (x) qi (x) pi (x)−qi (x) (1 + |un |) dx, + 1− pi (x) Ω i
and by (3.10), we obtain Z |Di un |qi (x) dx Ω
Z ≤ C1 Ω
|Di un |pi (x) dx + C2 (1 + |un |)γi (x)+1−[(m(x)−1)s(x)]
Z
(5.26) (1 + |un |)m(x)s(x) dx.
Ω
Therefore, (5.26) and (5.17) imply the desired result.
Lemma 5.6. Let m, s, pi , and γi be restricted as in Theorem 3.4. Then, there exists a constant C > 0 such that Z N Z X |Di un |pi (x) dx + |un |s(x)+1+γ+ (x) dx ≤ C. (5.27) i=1
Ω
Ω
Proof. Taking ψ(x, un ) = (1 + |un |)1+γ+ (x) − 1 sign(un ) in (4.1) as a test function, by (1.3) and the fact that for a.e. x ∈ Ω, for all i = 1, . . . , N , Di ψ(x, un ) = (1 + γ+ (x))(1 + |un |)γ+ (x) Di un + Di γ+ (x)(1 + |un |)1+γ+ (x) ln(1 + |un |) sign(un ),
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− we set γ+ = max1≤i≤N minx∈Ω γi (x), we obtain − α(1 + γ+ )
N Z X i=1
Z ≤
|Di un |pi (x) dx +
Z
Ω
|un |s(x) (1 + |un |)1+γ+ (x) − 1 dx
Ω
|f | (1 + |un |)1+γ+ (x) − 1 dx
Ω
+ C1
N Z X i=1
(1 + |un |)1+γ+ (x) ln(1 + |un |)|Di un |pi (x)−1 dx.
Ω +
Using that |un |s(x) ≥ min{1, 21−s }(1 + |un |)s(x) − 1, Proposition 2.1 and Young’s inequality and since m0 (·)(1 + γ+ (x)) ≤ s(x) + 1 + γ+ (x), we have Z N Z X 1 (1 + |un |)s(x)+1+γ+ (x) dx |Di un |pi (x) + 2 Ω Ω i=1 + m− ≤ C2 + C3 max kf km , kf k (5.28) m(·) m(·) L (Ω) L (Ω) + C4
N Z X i=1
(1 + |un |)1+γ+ (x) ln(1 + |un |)|Di un |pi (x)−1 dx.
Ω
We can estimate the last term in (5.28) by applying Young’s inequality, Z (1 + |un |)1+γ+ (x) ln(1 + |un |)|Di un |pi (x)−1 dx Ω Z Z (5.29) 1 p (x) |Di un |pi (x) dx. ≤ C5 (1 + |un |)pi (x)(1+γ+ (x)) (ln(1 + |un |)) i + 2C4 Ω Ω We combine (5.28) and (5.29) to obtain Z N Z X |Di un |pi (x) dx + (1 + |un |)s(x)+1+γ+ (x) dx i=1
Ω
≤ C6 + C7
Ω N Z X i=1
(5.30) pi (x)(γi (x)+1)
(1 + |un |)
pi (x)
ln(1 + |un |)
dx = J.
Ω
Thanks to Remark 3.7 we have s(x) > (pi (x) − 1)(γ+ (x) + 1), so (pi (x) − 1)(γ+ (x) + 1) − s(x) ≤ ((pi (x) − 1)(γ+ (x) + 1) − s(x))+ = di
0. (i) un is such that Tk (un ) ∈ W (ii) un converges almost everywhere in Ω to some measurable function u which ˚ 1,~p(·) (Ω) for all k > is finite almost everywhere, and such that Tk (u) ∈ W 0 (note that (i) and (ii) imply that Tk (un ) weakly converges to Tk (u) in ˚ 1,~p(·) (Ω)). W (iii) un is bounded in M r1 (Ω) for some r1 > 0 and u ∈ M r1 (Ω). (iv) There exists θi > 0 , i = 1, . . . , N such that |Di un |θi is bounded in Lr2 (Ω), for some r2 > 1 and |Di u|θi ∈ Lr2 (Ω). Then, up to a subsequence, Di un converges to Di u almost everywhere in Ω for all i = 1, . . . , N . ˚ 1,~p(·) (Ω) to Proof. It has been proved in [12] that there exists a solution un ∈ W problem (5.32). We follow the technique in [1, 20] with some modifications, since our method depends on the anisotropic variable exponent. Define the vector-valued function a ˆ(x, s, ξ) : Ω × R × RN → RN , where a ˆ(x, s, ξ) = {ˆ ai (x, s, ξ)}i=1,...,N with a ˆi (x, s, ξ) = ai (x, s)|ξi |pi (x)−2 ξi . Let θ be a real number between 0 and 1, which will be chosen later, and Z θ I(n) = a ˆ(x, Tn (un ), ∇un ) − a ˆ(x, Tn (un ), ∇u) ∇(un − u) dx. Ω
Note that I(n) is well defined and I(n) ≥ 0. We fixe k > 0 and split the integral I(n) on the sets {|u| ≥ k} and {|u| < k}, obtaining Z θ I1 (n, k) = a ˆ(x, Tn (un ), ∇un ) − a ˆ(x, Tn (un ), ∇u) ∇(un − u) dx, {|u|≥k} Z θ I2 (n, k) = a ˆ(x, Tn (un ), ∇un ) − a ˆ(x, Tn (un ), ∇u) ∇(un − u) dx. {|u| 1, we have | {|u| ≥ k} | ≤ | {|u| > k − 1} | ≤
C . (k − 1)r1
Using the above inequality, we obtain lim lim sup I1 (n, k) = 0.
(5.33)
k→+∞ n→+∞
As ∇u = ∇Tk (u) on the set {|u| < k}, we obtain Z θ I2 (n, k) = a ˆ(x, Tn (un ), ∇un )−ˆ a(x, Tn (un ), ∇Tk (u)) ∇(un −Tk (u)) dx. {|u| k + 1 and split the integral I2 (n, k) on the sets {|un − Tk (u)| ≥ h} and {|un − Tk (u)| < h}, obtaining I3 (n, k, h) Z =
θ a ˆ(x, Tn (un ), ∇un ) − a ˆ(x, Tn (un ), ∇Tk (u)) ∇(un − Tk (u)) dx,
{|un −Tk (u)|≥h}
and Z
n a ˆ(x, Tn (un ), ∇un )
I4 (n, k, h) = {|un −Tk (u)|k}
(5.37)
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Let E ⊂ Ω be any measurable set, using (5.37). We have Z Z Z s(x) s(x) |un | dx = |un | dx + |un |s(x) dx E E∩{|un |≤k} E∩{|un |>k} Z s+ s− ≤ (k + k ) meas(E) + |f | dx. E∩{|un |>k} s(x)−1
Then we deduce that (|un | un ) is equi-integrable in L1 (Ω), and since un → u a.e. in Ω, then Vitali’s theorem implies |un |s(x)−1 un → |u|s(x)−1 u in L1 (Ω).
(5.38)
6. Proof of main results In this section, using the uniform estimates of Section 4, we prove Theorem 3.2, 3.3, 3.4 and 3.5. 6.1. Proof of theorems 3.2, 3.3. By Lemma 5.2 the sequence (un ) is bounded ˚ 1,~q(·) (Ω) where qi (·) is defined as (3.4). Without loss of generality, we can in W therefore assume that ˚ 1,~q(·) (Ω), un * u weakly in W un → u strongly in Lq0 (Ω),
q0 = min min qi (x),
(6.1)
1≤i≤N x∈Ω
un → u a.e. in
Ω.
It follows from (4.3) and Fatou’s lemma that Z Z s(x) |un |s(x) dx ≤ C, |u| dx ≤ lim inf Ω s(x)
thus |u| such that
n→+∞
Ω s−
1
∈ L (Ω), furthermore, u ∈ M kun kM r1 (Ω) ≤ C
(Ω). Then, there exists r1 = s− > 0
and u ∈ M r1 (Ω).
(6.2)
Let fn = Tn (f )−Tn (|un |s(x)−1 un ) ∈ L∞ (Ω), where un is a solution of (5.32). Then, from (5.2), (5.4), (6.2), (6.1), and lemma 5.7 we can deduce that Di un → Di u a.e. in Ω, for all i = 1, . . . , N. So, by (5.4), we have qi (·)
|Di un |pi (x)−2 Di un * |Di u|pi (x)−2 Di u weakly in L pi (·)−1 (Ω), ∀i = 1, . . . , N, (6.3) qi (·) where qi is defined as in (3.4). The choice of pi (·)−1 > 1 is possible since we have (3.3). From (1.3) and (6.1), we obtain ai (x, Tn (un )) → ai (x, u)
weak∗ in L∞ (Ω).
(6.4)
C0∞ (Ω),
For any given ϕ ∈ using ϕ as a test function in (4.1), we have Z Z Z N X pi (x)−2 s(x)−1 ai (x, Tn (un ))|Di un | Di un Di ϕ dx+ |un | un ϕ dx = Tn (f )ϕ dx, i=1
Ω
Ω
Ω
(6.5)
22
H. AYADI, F. MOKHTARI
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Letting n → +∞ in (6.5), by (6.3), (6.4), and (5.38), we obtain Z Z N Z X pi (x)−2 s(x)−1 ai (x, u)|Di u| Di uDi ϕ dx + |u| uϕ dx = f ϕ dx. i=1
Ω
Ω
Ω
For the proof of Theorem 3.3, we only replace (6.3) with qi (·)
|Di un |pi (x)−2 Di un * |Di u|pi (x)−2 Di u weakly in L pi (·)−1 (Ω), ∀i = 1, . . . , N. where qi is defined as in (3.6). The choice of (3.5).
qi (·) pi (·)−1
> 1 is possible since we have
6.2. Proof of theorem 3.4, 3.5. Because the proof of Theorem 3.5 is similar to that of Theorem 3.2, here we only give the proof of Theorem 3.4. According to ˚ 1,~p(·) (Ω). This implies that we can Lemma 5.6, the sequence (un ) is bounded in W extract a subsequence (denote again by (un )), such that ˚ 1,~p(·) (Ω), un * u weakly in W un → u strongly in Lp0 (Ω),
p0 = min min pi (x), 1≤i≤N x∈Ω
un → u a.e. in Ω. Arguing as the proof of Theorem 3.2, by using (5.27), we conclude that 0
|Di un |pi (x)−2 Di un * |Di u|pi (x)−2 Di u weakly in Lpi (·) (Ω), ∀i = 1, . . . , N. The proof of Theorem 3.4 is complete. Remark 6.1. All the results in this work also hold if our problem is exchanged by a more general one, − div(a(x, u, ∇u)) + g(x, u) = f u=0
in Ω,
on ∂Ω,
where a(x, t, ξ) = {ai (x, t, ξ)}i=1,...,N : Ω × R × RN → RN is a Carath´eodory vectorvalued function such that for a.e. x ∈ Ω and for every (t, ξ) ∈ R × RN , the following assumptions hold: a(x, t, ξ)ξ ≥ α
N X i=1
|ξi |pi (x) , (1 + |t|)γi (x)
N 1− p 1(x) X i , |ai (x, t, ξ)| ≤ β 1 + |ξj |pj (x)
α > 0,
i = 1, . . . , N, β > 0,
j=1
(a(x, s, ξ) − a(x, s, ξ 0 ))(ξ − ξ 0 ) > 0,
∀ξ 6= ξ 0 .
Assume that g : Ω × R → R is a Carath´eodory function satisfying sup |g(x, t)| = hk (x) ∈ L1 (Ω),
∀k > 0,
|t|≤k
g(x, t) sign(t) ≥ |t|s(x) . Acknowledgements. The authors would like to thank the referees for their comments and suggestions. They are also grateful to Mr. Lakhdar Benaissa, the head of the department of Mathematics and Informatics at Benyoucef Benkhedda University, who provided all the means to the research team.
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[email protected] Fares Mokhtari ´partement de Mathe ´matiques et Informatique, Universite ´ Benyoucef Benkhedda, AlDe ger 1, 2 Rue Didouche Mourad, Algeria. Laboratoire des EDPNL, ENS-Kouba, Alger, Algeria E-mail address: fares
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