Nonl. Analysis and Differential Equations, Vol. 2, 2014, no. 2, 69 - 81
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/nade.2014.31225
Harnack Inequality and Continuity of Solutions
for Quasilinear Elliptic Equations in Sobolev
Spaces with Variable Exponent
Azeddine Baalal and Abdelbaset Qabil
Department of Mathematics - Laboratory MACS
Faculty of Sciences A¨in Chock, University of HASSAN II
B.P. 5366, Casablanca - Morocco
c 2014 Azeddine Baalal and Abdelbaset Qabil. This is an open access article
Copyright distributed under the Creative Commons Attribution License, which permits unrestricted
use, distribution, and reproduction in any medium, provided the original work is properly
cited.
Abstract
We investigate regularity properties of solutions to the quasilinear
elliptic equations in Sobolev spaces with variable exponent, we prove
the Harnack’s inequality and continuity of solutions.
Mathematics Subject Classification: 35J62, 49N60, 35J60, 35J25.
Keywords: Quasilinear elliptic equation, variable exponent, Caccioppoli estimate, Moser’s iteration method, Harnack inequality
1
Introduction
In the present paper we study the regularity of boundary points for solutions
to the quasilinear elliptic equations:
− div A(x, ∇u) + B(x, u) = 0 ,
(1)
Our purpose is to establish the Harnack’s inequality
ess sup u ≤ C(ess inf u + R)
B(x,R)
B(x,R)
(2)
70
A. Baalal and A. Qabil
where C is independent of u and the ball B = B(x0 , R) for x0 ∈ Ω, R > 0 and
we prove the continuity of solutions for (1).
Harnack’s inequality and other regularity results for (1) require additional
assumptions on the function p(.); see the counterexamples in [6]. The so
called logarithmic H¨older continuity condition seems to be the right one for
our purposes. This condition was originally introduced by Zhikov [15] in the
context of the Lavrentiev phenomenon for solutions of (1), and it has turned
out to be a useful tool in regularity and other applications, see, e.g., [1, 2].
For the existence and uniqueness of solutions u ∈ W 1,p(x) (Ω) where 1 < p(x) <
d for all x ∈ Ω, of the variational Dirichlet problem associated with the quasilinear elliptic equation (1) see [4], these solutions are obtained by the p(.)obstacle problem.
A typical example for the operator A and B are A(x, ∇u) = |∇u|p(x)−2∇u and
B(x, u) = |u|p(x)−2 u respectively, for all x ∈ Rd thus Δp(x) u = div(|∇u|p(x)−2 ∇u).
Our problem has been studied in many paper see e.g. [11, 12]. Olli Toivanen
[11] proved that this problem has a solution when the operator B depends
on x, u, ∇u where δ(x) = p(x) − 1. The main aim of this section is to
generalize the condition on δ(x), We are interested in the case that the operator
B depends only on x and u, satisfying the previous hypothesis (H3), where
p(x) − 1 ≤ δ(x) < p∗ (x), knowing that the study of the case where δ satisfies
the condition 0 ≤ δ(x) < p(x) − 1 is already investigated in several articles.
The contribution of this paper is to verify the Harnack principale for a weak
solutions of quasilinear elliptic equations (1) by using the trick of modified test
functions under our assumptions (H3) below.
In the first section, we introduce some generalization and position of the problem. In second section we give some basic facts about variable exponent spaces
and a rough overview of properties of solutions of the prototype equality.
In section 3, we generalize, with detailed proofs, Harnack’s inequality (2) to
all quasilinear elliptic equations (1) with growth conditions of a non-standard
form. In last section, we present the concluding remarks.
2
Some preliminaries
We start this section with some definitions and main results of Lebesgue spaces
with variable exponent, and Sobolev spaces modeled upon them. For each
open bounded subset Ω of Rd (d ≥ 2), we define the Lebesgue space with
variable exponent Lp(.) (Ω) as the set of all measurable functions p : Ω →
Harnack inequality in W 1,p(.) (Ω)
71
]1, +∞[ called a variable exponent and we denote p− := ess inf x∈Ω p(x) and
p+ := ess supx∈Ω p(x).
We introduce also the convex modular
p(x) (u) =
|u|p(x)dx.
Ω
If the exponent is bounded, i.e., if p+ < ∞, then the expression
u
up(.) = inf{λ > 0 : p(.) ( ) ≤ 1}
λ
defines a norm in Lp(.) (Ω), called the Luxemburg norm.
One central property of Lp(.) (Ω) is that the norm and the modular topologies
coincide,i.e.,
p(.) (un ) → 0 if and only if un p(.) → 0.
1
We denote by Lp (.) (Ω) the conjugate space of Lp(.) (Ω) where p(x)
+ p 1(x) = 1.
Proposition 1 (Generalized H¨
older inequality [14])
and v ∈ Lp (.) (Ω), we have
1
1 | uvdx| ≤ − + − up(.) vp(.) .
p
p
Ω
For any u ∈ Lp(.) (Ω)
We define the variable exponent Sobolev space (see [9], [5],[8], [14]) by
W 1,p(.) (Ω) = {u ∈ Lp(.) (Ω) : ∇u ∈ Lp(.) (Ω)}.
with the norm
u1,p(.) = up(.) + ∇up(.)
∀u ∈ W 1,p(.) (Ω).
1,p(.)
The local Sobolev space Wloc (Ω) consists of functions u that belong to
1,p(.)
Wloc (U) for all open sets U compactly contained in Ω. The Sobolev space
1,p(.)
with zero boundary values, W0 (Ω), is defined as the completion of C0∞ (Ω)
in the norm of W 1,p(.) (Ω) .
Let p∗ (x) be the Sobolev conjugate exponent of p(x) defined by
dp(x)
for p(x) < d,
∗
d−p(x)
p (x) =
+∞
for p(x) ≥ d.
We assume further on that, there exist positive constant C such that the
function p satisfies logarithmic H¨
older continuity condition if :
∃C > 0 : |p(x) − p(y)| ≤ − logC|x−y| f or |x − y| < 12 ,
()
1 < p− ≤ p+ < d.
72
A. Baalal and A. Qabil
Proposition 2 (The p(.)-Poincar?e inequality)
Let Ω be a bounded open set and let p : Ω → [1, ∞[ satisfy () There exists a
constant C, depending only on p(.) and Ω, such that the inequality
up(.) ≤ C∇up(.)
1,p(.)
∀u ∈ W0
(Ω).
Lemma 2.1 (Sobolev inequality [7]) Let Ω be a bounded open set and
u in
1,p(.)
W0 (Ω). There exists a constant C such that
dp(x)
d−1
1
dp(x)
d−1
dx)
≤ C( |∇u|p(x) dx) p(x)
(3)
( |u|
Ω
Ω
Proposition 3
1,p(.)
Assuming p− > 1, the spaces W 1,p(.) (Ω) and W0 (Ω) are separable and reflexive Banach spaces.
Throughout the paper we suppose that the functions A : Rd × Rd → Rd is a
Carath´eodory function satisfying the following assumptions:
(H1) |A(x, ξ)| ≤ β[k(x) + |ξ|p(x)−1 ];
(H2) A(x, ξ)ξ ≥ ν|ξ|p(x) ;
for a.e. x ∈ Ω, all ξ ∈ Rd , where k(x) is a positive bounded function lying in
Lp (x) (Ω) and β, ν > 0 .
In this paper we suppose that the function B : Rd ×R → R is given Carath´eodory
function and the following condition is satisfied:
(H3) |B(x, ζ)| ≤ g(x) + |ζ|δ(x) ;
for a.e. x ∈ Ω, all ζ ∈ Rd , where g is a positive bounded function lying in
Lp (x) (Ω) and p(x) − 1 ≤ δ(x) < p∗ (x).
Remark 2.1 Under the assumption () Harnack’s inequality and local H¨older
continuity follow from Moser or De Giorgi-type procedure; see [10, 2, 3]. An
interesting feature of this theory is that estimates are intrinsic in the sense that
they depend on the solution itself. For example, supersolutions are assumed to
be locally bounded and Harnack-type estimates in [2] depend on this bound.
1,p(.)
Definition 2.1 We say that a u ∈ Wloc (Ω) is a p(.)-solution of (1) in Ω
1,p(.)
provided that for all ϕ ∈ W0 (Ω) if ,
A(x, ∇u) · ∇ϕdx + B(x, u)ϕdx = 0 .
Ω
Ω
Harnack inequality in W 1,p(.) (Ω)
73
1,p(.)
Definition 2.2 A function u ∈ Wloc (Ω) is termed p(.)-supersolutions of
1,p(.)
(1), if and only if, for all non-negative functions ϕ ∈ W0 (Ω) we have,
A(x, ∇u) · ∇ϕdx + B(x, u)ϕdx ≥ 0 .
Ω
Ω
A function u is a p(.)-subsolution in Ω if −u is a p(.)-supersolution in Ω,
and a solution in Ω if it is both a super- and a p(.)-subsolution in Ω.
3
Harnack inequality and continuity of solutions to quasilinear elliptic equations
The Harnack inequality is a very important estimate in the study of p(.)solutions of quasilinear elliptic equations.
3.1
Main result
We start by adapting a standard Caccioppoli type estimate for p(.)-supersolution
of (1). Then we use the Caccioppoli estimate to show that for a fixed, nonnegative p(.)-supersolution u, the inequality (2).
The Harnack inequality is indispensable as a tool in the qualitative theory
of second-order elliptic equations. In particular, it implies continuity of weak
solutions see [11, 13].
By non-linearity we mean that if p = 2 then the weak solutions do not form
a linear space. However the set of weak solutions is closed under constant
multiplication. By celebrated De Giorgi’s method and Moser’s iteration the
weak solutions are locally H¨older continuous and satisfy Harnack’s inequality.
Remark 3.1 Our notation is rather standard. Various constants are denoted
by C and the value of the constant may differ even on the same line. The
quantities on which the constants depend are given in the statements of the
theorems and lemmas.
Lemma 3.1 Let E be a measurable subset of Rd . For all nonnegative
measurable functions ψ and ϕ defined on E,
p−
E
ψϕ dx ≤
ψdx +
ψϕp(x) dx
E
E
E
74
A. Baalal and A. Qabil
The following Caccioppoli estimate is the key result of this paper; and it is a
modification of [[7], Lemma 3.2]. The new feature in the estimate is the choice
of a test function which includes the variable exponent.
Lemma 3.2 (Caccioppoli estimate)
Suppose that u is a nonnegative p(.)-supersolution in B4R . Let E be a measurable subset of B4R and η ∈ C0∞ (B4R ) such that 0 ≤ η ≤ 1. Then for every
γ0 < 0 there is a constant C depending on p and γ0 such that the inequality:
p+
p+
p−
B4R γ−1
η
u |∇u| dx ≤ C
(uγ+p(x)−1 |∇η|p(x) + η B4R uγ−1 )
(4)
E
B4R
+(uγ+p(x)−1 dx + η
p+
B
4R
−1 γ+δ(x)
u
)dx
holds for every γ < γ0 < 0 and p(x) − 1 ≤ δ(x) < p∗ (x) .
θ γ
Proof 1 Let θ = p+
B4R . We want to test with the function ϕ = η u .
1,p(.)
To this end we show that ϕ ∈ W0 (B4R ). Since η has a compact support in
B4R , it is enough to show that ϕ ∈ W 1,p(.) (Ω). We observe that ϕ ∈ Lp(x) (Ω)
since |uγ |η θ ≤ Rγ . Furthermore, we have
∇ϕ = γη θ uγ−1 ∇u + θη θ−1 uγ ∇η
Using the fact that u is a p(.)-supersolution and ϕ is a nonnegative test function
we find that
0≤
B4R
A(x, ∇u) · ∇ϕdx +
B(x, u)ϕdx
θ γ−1
θ−1 γ
γη u
A(x, ∇u) · ∇udx +
θη
u A(x, ∇u) · ∇ηdx +
=
B4R
B4R
B4R
B4R
B(x, u)η θ uγ dx
We denote the left-hand side of the next inequality by I. Since γ is a negative
number this implies by the structural conditions (H1), (H2) and (H3) that
I = |γ0 |ν
ηθ uγ−1 |∇u|p(x) dx
B4R
θηθ−1 uγ k(x) + |∇u|p(x)−1 |∇η|dx +
≤β
B4R
B4R
=I1
=I2
ηθ−1 uγ |∇u|p(x)−1 |∇η|dx +βθ
≤ βθ
B4R
g(x) + uδ(x) ηθ uγ dx
=I3
ηθ−1 uγ k(x)|∇η|dx +
B4R
(g(x) + uδ(x) )ηθ uγ dx
B4R
Using Young’s inequality, 0 < ε ≤ 1, we obtain the first estimate
p(x)
p (x)
γ+p(x)−1
θ
1 p(x)−1 θ− pθ(x) −1 γ+p(x)−1
p(x)−1
(x) γ−
p(x)
p
p(x)
I1 ≤
( )
u
|∇η|
+ε η
u
|∇u|
dx
η
B4R ε
1
η θ−p(x) uγ+p(x)−1 |∇η|p(x) dx + ε
η θ uγ−1 |∇u|p(x) dx
≤ ( )θ−1
ε
B4R
B4R
1 θ−1
γ+p(x)−1
p(x)
≤( )
u
|∇η|
dx + ε
η θ uγ−1 |∇u|p(x) dx
ε
B4R
B4R
Harnack inequality in W 1,p(.) (Ω)
75
Next we estimate the last tow integrals I2 and I3 .
To estimate the integral I2 , we denote 0 ≤ v = η + |∇η| and k is a positive
bounded function there exist a constant M > 0, and by Young’s inequality we
have
θ−1 γ
η u k(x)|∇η|dx ≤ M
vη θ−1 uγ dx
I2 =
B4R
B4R
1 θ−1
θ−p(x) γ+p(x)−1 p(x)
θ γ−1
η
u
v dx + ε
η u dx
≤M ( )
ε
B4R
B4R
1 θ−1
γ+p(x)−1 p(x)
θ γ−1
≤M ( )
u
v dx + ε
η u dx
ε
B4R
B4R
1 θ−1
γ+p(x)−1
p(x)
θ γ−1
≤M ( )
u
(1 + |∇η|) dx + ε
η u dx
ε
B4R
B4R
Using the inequality (ϕ + ψ)p(.) ≤ 2p(.)−1 (|ϕ|p(.) + |ψ|p(.) ),
for p(.) ≥ 1,
1 θ−1
γ+p(x)−1 p(x)−1
p(x)
u
2
(1 + |∇η| )dx + ε
η θ uγ−1 dx
I2 ≤ ( )
ε
B
B4R
4R
2θ−1
≤ θ−1
uγ+p(x)−1 (1 + |∇η|p(x))dx + ε
η θ uγ−1 dx
ε
B4R
B4R
2 θ−1
2 θ−1
γ+p(x)−1
p(x)
γ+p(x)−1
≤( )
u
|∇η| dx + ( )
u
+ε
η θ uγ−1 dx
ε
ε
B4R
B4R
B4R
To estimate the integral I3 , and by the assumption g is a positive bounded
function there exist a N > 0 such that
δ(x) θ γ
I3 =
(g(x) + u )η u dx ≤
(N + uδ(x) )η θ uγ dx
B4R
B4R
≤
=I3
η
θ−1 γ+δ(x)
u
dx + N
B4R
η
θ−1 γ
u dx
B4R
By Young’s inequality we have
1 θ−1
γ+p(x)−1
u
dx + ε
η θ uγ−1 dx
I3 ≤ ( )
ε
B4R
B4R
Thus
I3 ≤
η
B4R
θ−1 γ+δ(x)
u
1
dx + ( )θ−1
ε
u
B4R
γ+p(x)−1
η θ uγ−1 dx
dx + ε
B4R
76
A. Baalal and A. Qabil
Therefore
(2θ−1 + 1)
η u |∇u| dx ≤ θβ
uγ+p(x)−1 |∇η|p(x) dx
(|γ0|ν − θβε)
θ−1
ε
B4R
B4R
θ−1
θβ
+
1
2
θ γ−1
γ+p(x)−1
η u dx +
u
dx +
η θ−1 uγ+δ(x) dx
+ε(θβ + 1)
θ−1
ε
B4R
B4R
B4R
θ γ−1
p(x)
0 |ν
}
By choosing ε = min{1, |γ2θβ
η θ uγ−1 |∇u|p(x)dx ≤
B4R
γ+p(x)−1
p(x)
θ γ−1
≤ C1
u
|∇η| dx + C2
η u dx + C3
uγ+p(x)−1 dx +
B4R
B4R
B4R
+ C4
η θ−1 uγ+δ(x) dx
B4R
1
, C2 = 1 + θβ
Where C1 = (2θ−1 + 1) |γ2θβ
, C3 = (2θ−1 +
|ν
0
C4 = |γ02|ν .
We take C = Ci for i = 1, 2, 3, 4. and we have
η
p+
B
4R
u
γ−1
|∇u|
p(x)
dx ≤ C
(uγ+p(x)−1 |∇η|p(x) + η
B4R
p+
B
4R
and
uγ−1 )
B4R
+(uγ+p(x)−1 dx + η
By the lemma 3.1 we obtain
θ γ−1
p−
η u |∇u| dx ≤
B4R
θ
1
) |γ2θβ
,
θβ
|ν
0
p+
B
4R
−1 γ+δ(x)
u
)dx
θ γ−1
η u
η θ uγ−1 |∇u|p(x) dx
dx +
B4R
B4R
and using the previous inequality we obtain the claim.
p+
p+
p−
B4R γ−1
η
u |∇u| dx ≤ C
(uγ+p(x)−1 |∇η|p(x) + η B4R uγ−1 )
E
B4R
+(uγ+p(x)−1 dx + η
p+
B
4R
−1 γ+δ(x)
u
)dx
So the proof of lemma is achieved.
3.2
Weak Harnack Inequality
In this section we prove a weak Harnack inequality for p(.)-supersolutions to
(1).
Harnack inequality in W 1,p(.) (Ω)
77
Throughout this subsection we write v = u + R where u is a nonnegative
p(.)-supersolution and 0 ≤ R ≤ 1.
We start by the following technical lemma that is need later. These results are
mainly from [7]
Lemma 3.3 If the exponent p(.) is log-H¨
older continuous, then R−p(x) ≤
−p−
CR E where x ∈ E ⊂ BR and R > 0.
Lemma 3.4 Let f be a positive measurable function and assume that the
exponent p(.) is log-H¨
older continuous. Then
−
p+
p+ −p−
B4R −pB4R
f B4R B4R dx ≤ Cf Ls(B
r)
Br
for any s >
p+
B4R
−
p−
B4R
Now we have everything ready for the iteration. We write
1
f q dx) q
Φ(f, q, Br ) = (
Br
for a nonnegative measurable function f .
The point is that the Moser iteration technique used in [7] remains valid under
our consideration.
Lemma 3.5 Let u is a nonnegative p(.)-supersolution of (1) in B4R and
let R ≤ ρ < r ≤ 3R. Then the inequality
Φ(v, qτ, Br ) ≤ C
1
|τ |
(1 + |τ |)
p+
B4R
|τ |
+
r pB|τ4R
d
) | Φ(v, τ
, Bρ )
(
r−ρ
d−1
(5)
d
holds for every τ < 0 and 1 < q < d−1
. The constant C depends on d,p, and
+
q s
the L (B4R )-norm of u with s > pB4R − p−
B4R and all structure constants and
functions of (H1),(H2) and (H3) hypothesis .
−
−
−
Proof 2 Let θ+ = p+
B4R and θ = pB4R , we take γ = τ − θ + 1. In (4) of the
lemma 3.2 we have
B4R
+
−
−
η θ v τ −θ |∇u|θ dx
≤C
B4R
(6)
=I1
=I2
θ+ τ −θ −
−
−
+
−
η u
+ uτ −θ +p(x) |∇η|p(x) + uτ −θ +p(x) + η θ −1 uτ −θ +1+δ(x) dx
Now we take the test function η ∈ C0∞ (Br ) with 0 ≤ η ≤ 1, η = 1 in Bρ , and
|∇η| ≤
Cr
R(r − ρ)
78
A. Baalal and A. Qabil
Next we went to estimate the integral I2 by the integral
1q
v dx
qτ
(7)
Br
Using lemma 3.3,3.4, the first integral I1 is estimated by (7) see [7].
Finally, for the second integral I2 we have by H¨older’s inequality
−
τ −θ − +p(x)
u
dx ≤
v τ −θ +p(x) dx
Br
Br
≤C
v
q (p(x)−θ− )
1 q
dx
Br
1 q
q (θ+ −θ − )
≤ C 1 + vLq s (B )
4R
1q
v dx
qτ
Br
1q
v dx
qτ
Br
On the other hand and since p(x) − 1 ≤ δ(x) < p∗ (x)
η
θ + −1 τ −θ − +1+δ(x)
u
Br
v τ −θ
dx ≤
− +1+δ(x)
Br
≤C
v
q (δ(x)−θ− +1)
1 q
dx
Br
1q
v dx
qτ
Br
1 q
q (θ+ −θ − )
≤ C 1 + vLq s (B )
4R
dx
1q
v qτ dx
Br
Therefore, the second integral I2 is estimated by (3).
τ
θ−
In lemma 2.1 we take u = v θ− η θ− and we use the inequality:
(a + b)p(.) ≤ 2p(.)−1 (|a|p(.) + |b|p(.) ),
p(.) ≥ 1,
We obtain
v
dτ
d−1
d−1
d
Bρ
≤ CR
θ−
≤ C|τ |
≤C
v
τ
θ−
η
θ−
θ−
−
dθ
d−1
d−1
d
dx
Br
θ−
− −
∇(v θτ− η θθ− )θ dx
Br
θ + τ −θ −
η v
Br
Using inequality (4) we have
θ−
|∇u| dx + C
vτ ηθ
Br
+ −θ −
−
|∇η|θ dx
Harnack inequality in W 1,p(.) (Ω)
v
dτ
d−1
d−1
d
Bρ
≤ CR
θ−
θ−
79
d−1
−
d
τ θ− dθ
d−1
−
−
vθ ηθ
dx
≤C
Br
− −
∇(v θτ− η θθ− )θ dx
B
r θ + τ −θ −
τ −θ − +p(x)
τ θ + −θ −
+u
|∇η|
+v η
η u
≤ C|τ |
Br
τ −θ − +p(x)
θ + −1 τ −θ − +1+δ(x)
+C
+η
u
u
dx
p(x)
|∇η|
θ−
dx
Br
1q
1 θ + q
r
q (θ+ −θ − )
qτ
v dx
1 + vLq s (B )
≤ C(1 + |τ |)
4R
r−ρ
Br
θ+
Finally, since τ < 0 we have
Φ(v, qτ, Br ) ≤ C
1
|τ |
(1 + |τ |)
p+
B4R
|τ |
+
r pB|τ4R
d
) | Φ(v, τ
, Bρ )
(
r−ρ
d−1
So the proof of lemma is achieved.
The proofs of the following results can be found in [11]. and [7], respectively
Lemma 3.6 Assume that u is a nonnegative p(.)-supersolution of (1) in
−
B4R and s > p+
B4R − pB4R . Then there exist constants q0 > 0 and C depending
on d, p, and Lq s (B4R )-norm of u such that:
Φ(v, q0 , B3R ) ≤ CΦ(v, −q0 , B3R )
(8)
Theorem 3.7 (Weak Harnack inequality) Let u be a non-negative p(.)superd
solution of (1) inB4R and 1 < q < d−1
. Then
1
q0
q0
u dx) ≤ C ess inf u(x) + R
(9)
(
x∈BR
B2R
Where q0 is the exponent from Lemma 3.5 and C depends on d,p,q and the
−
Lq s (B4R )-norm of u with s > p+
B4R − pB4R and all structure constants and
functions of (H1),(H2) and (H3) hypothesis .
Theorem 3.8 Let u be a non-negative p(.)-subsolution of (1) in B4R and
d
. Then
1 < q < d−1
1
ess sup u(x) ≤ C
ut dx t + R
(10)
x∈BR
B2R
Where t > 0 and C depends on d,p,q and the Lq s (B4R )-norm of u with s >
−
p+
B4R − pB4R .
80
A. Baalal and A. Qabil
To combine (9) and (10) we obtain the crucial theorem.
Theorem 3.9 (Harnack’s inequality) Let u be a non-negative p(.)-solution
−
d
of (1) in Ω and let B4R ⊂ Ω and 1 < q < d−1
, and s > p+
B4R − pB4R with
0 ≤ R ≤ 1. Then
ess sup u(x) ≤ C ess inf u(x) + R
x∈BR
x∈BR
Where C depends on d,p,q and the Lq s (B4R )-norm of u with and all structure
constants and functions of (H1),(H2) and (H3) hypothesis .
4
Conclusion
We showed that p(.)-solutions are locally bounded, locally bounded p(.)-supersolutions
satisfy the weak Harnack inequality and locally bounded p(.)-solutions satisfy
Harnack’s inequality. In the proof of Harnack’s inequality the Caccioppoli
estimate is used for the function u + R, where R is a radius of a ball.
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Received: December 1, 2014
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