Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 27, pp. 1–16.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC

EQUATIONS WITH GENERALIZED ORLICZ GROWTH

MARIA A. SHAN, IGOR I. SKRYPNIK, MYKHAILO V. VOITOVYCH

Abstract. We prove Harnack’s inequality for bounded weak solutions to

quasilinear second order elliptic equations with generalized Orlicz growth con-

ditions. Our approach covers new cases of variable exponent and (p, q) growth
conditions.

1. Introduction and main results

This article concerns quasilinear elliptic equations of the form

div
(
g(x, |∇u|) ∇u

|∇u|

)
= 0, x ∈ Ω, (1.1)

where Ω is a bounded domain in Rn, n ≥ 2.
Throughout this article we assume that the function g(x, v) : Ω × R+ → R+,

R+ = [0,+∞) satisfies the following assumptions:

(A1) g(·, v) ∈ L1(Ω) for all v ∈ R+, g(x, ·) is continuous and non-decreasing for
almost all x ∈ Ω, limv→+0 g(x, v) = 0 and limv→+∞ g(x, v) = +∞;

(A2) there exist c1 > 0, q > 1 and b0 ≥ 0 such that

g(x,w)

g(x, v)
≤ c1

(w

v

)q−1

(1.2)

for all x ∈ Ω and for all w ≥ v > b0;
(A3) there exists p > 1 such that

g(x,w)

g(x, v)
≥
(w

v

)p−1

(1.3)

for all x ∈ Ω and for all w ≥ v > 0;

(A4) for any K > 0 and for any ball B8r(x0) ⊂ Ω there exists c2(K) > 0 such
that

g(x1, v/r) ≤ c2(K) eλ(r)g(x2, v/r)

for all x1, x2 ∈ Br(x0) and for all r ≤ v ≤ K. Here λ(r) : (0, r∗) → R+ is
a continuous, non-increasing function, satisfying the conditions described
below.

2010 Mathematics Subject Classification. 35B65, 35D30, 35J60, 49N60.
Key words and phrases. Elliptic equation; non-standard growth; non-log condition;

bounded solution; Harnack inequality.
c©2021 Texas State University.

Submitted September 7, 2020. Published April 7, 2021.

1



2 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

The following functions defined on Ω× R+ satisfy assumptions (A1)–(A4):

g(x, v) = vp(x)−1 + vq(x)−1,

g(x, v) = vp(x)−1
(
1 + ln(1 + v)

)
,

g(x, v) = vp−1 + a(x)vq−1, a(x) ≥ 0,

g(x, v) = vp−1
(
1 + b(x) ln(1 + v)

)
, b(x) ≥ 0,

where the exponents p, q, p(·), q(·), and the coefficients a(·) and b(·) satisfy the
following conditions:

(i) 1 < p < p(x) ≤ q(x) < q < +∞ for all x ∈ Ω;
(ii)

|p(x)− p(y)|+ |q(x)− q(y)| ≤ λ(|x− y|)∣∣ ln |x− y|∣∣ , x, y ∈ Ω, x 6= y, (1.4)

the function λ(r)/| ln r| is non-decreasing on (0, r∗), limr→0 λ(r)/| ln r| = 0;
(iii)

|a(x)− a(y)| ≤ A|x− y|αeλ(|x−y|), x, y ∈ Ω, x 6= y, (1.5)

A > 0, 0 < q− p ≤ α ≤ 1, the function rαeλ(r) is non-decreasing on (0, r∗),
limr→0 r

αeλ(r) = 0;
(iv)

|b(x)− b(y)| ≤ B eλ(|x−y|)∣∣ ln |x− y|∣∣ , x, y ∈ Ω, x 6= y, B > 0, (1.6)

the function eλ(r)/| ln r| is non-decreasing on (0, r∗), limr→0 e
λ(r)/| ln r| = 0.

The study of regularity of minima of functionals with non-standard growth of
(p, q)-type was initiated by Zhikov [35, 36, 37, 38, 40], Marcellini [22, 23] and
Lieberman [21]. In the last thirty years, the qualitative theory of second order
equations with so-called “log-condition” (i.e. if 0 ≤ λ(r) ≤ L < +∞) has been
actively developed; see, for instance, [1, 2, 4, 5, 8, 9, 10, 11, 12, 13, 14, 17, 18, 19,
25, 33]. These classes of equations have numerous applications in physics and have
been attracted attention for several decades; see [7, 28, 34] and references therein.

The case when conditions (1.4), (1.5), (1.6) hold differs substantially from the
log-case. To the best of our knowledge there are only a few results in this direction.
Zhikov [39] obtained a generalization of the logarithmic condition which guarantees
the denseness of smooth functions in a Sobolev space W 1,p(x)(Ω). Particularly, this
result holds if 1 < p ≤ p(x) and

|p(x)− p(y)| ≤ L | ln | ln |x− y| | |
| ln |x− y| |

, x, y ∈ Ω, x 6= y, L < p/n.

In the case when the variable exponent p(x) satisfies the condition

|p(x)− p(x0)| ≤ L ln ln ln |x− x0|−1

ln |x− x0|−1
,

0 < L < p/(n+ 1), x, x0 ∈ Ω, |x− x0| < 1/27,

(1.7)

Alkhutov and Krasheninnikova [3] proved the continuity of solutions to the p(x)-
Laplace equation at the point x0, and Surnachev [31] established the Harnack in-
equality for solutions. The continuity of solutions to the p(x)-Laplace equation up



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 3

to the boundary were proved by Alkhutov and Surnachev [6] under the additional
condition ∫

0

exp
(
− C exp

(
βλ(r)

)) dr
r

= +∞, (1.8)

where C and β are some positive constants, depending only upon the data. We
note that the function λ(r) = L ln ln ln r−1, r ∈ (0, e−e), Lβ < 1, satisfies condition
(1.8).

In [30], we attempted to systematize and unify the approach to establish the
local regularity of bounded solutions of elliptic and parabolic equations with non-
standard growth. For this, we have introduced elliptic and parabolic B1 classes,
which generalize the well-known Bp classes (p > 1) of De Giorgi, Ladyzhenskaya,
Ural’tseva [20] and cover their other numerous and scattered analogues (see ref-
erences in [30]). It was proved in [30] that functions from the B1,g,λ(Ω) class are
continuous if conditions (A2), (A4) and (1.8) are fulfilled. In addition, if con-
dition (A3) is fulfilled, then the solutions of (1.1) belong to the B1,g,λ(Ω) class.
At the same time, we do not use the specific properties of the generalized Orlicz
and Sobolev-Orlicz spaces, as was done, for example, in the papers of Harjulehto,
Hästö et al [16, 17, 18, 19]. Although it should be noted that in the case when
0 ≤ λ(r) < L < +∞, the assumptions (A2)–(A4) are almost equivalent to the
conditions (aDec)∞q , (A1-n), (aInc)p from their papers.

Returning to our paper [30], we note that there are no Harnack-type theorems
in it. Although, such type results were obtained in [1, 2, 4, 5, 27] in the log-
case and in [31] under condition (1.7). Therefore, it is natural to conjecture that
the Harnack inequality holds for bounded solutions of (1.1) under the conditions
(A1)–(A4). In this paper, we give a positive answer to this hypothesis. This
also encompasses the classic results of Moser [26], Serrin [29], Trudinger [32] and
Di Benedetto & Trudinger [15] for bounded solutions in the standard growth case,
and of course, we use some of the ideas of Moser and Trudinger in our proofs.

Before formulating the main results, let us remind the reader the definition of
a weak solution to (1.1). Moreover, throughout the article, we use the well-known
notation for sets in Rn, spaces of functions and their elements, etc. (see, e.g. [20]).
In particular, we will use the notation –

∫
E
f dx = |E|−1

∫
E
f dx for any measurable

set E ⊂ Rn with |E| 6= 0 and f ∈ L1(E), where |E| denotes the n-dimensional
Lebesgue measure of E. We set

G(x, v) = g(x, v)v for x ∈ Ω, v ≥ 0 (1.9)

and write u ∈ W 1,G(Ω) if u ∈ W 1,1(Ω) and
∫

Ω
G(x, |∇u|) dx < +∞; u ∈ W 1,G

loc (Ω)

if u ∈ W 1,G(E) for any open set E compactly embedding in Ω. We denote by

W 1,G
0 (Ω) the set of all functions u ∈W 1,G(Ω) which have a compact support in Ω.

Definition 1.1. We say that a function u : Ω → R is a bounded weak solution

(subsolution, supersolution) to (1.1) if u ∈ W 1,G
loc (Ω) ∩ L∞(Ω) and the integral

equality (inequality) ∫
Ω

g(x, |∇u|) ∇u
|∇u|

∇ϕdx = (≤ ,≥) 0 (1.10)

holds for any ϕ ∈W 1,G
0 (Ω) (for subsolutions and supersolutions, we require ϕ ≥ 0).

We refer to the parameters M = ess supΩ |u|, n, p, q, c1, c2(M) as our structural
data, and we write γ if it can be quantitatively determined a priori only in terms



4 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

of the above quantities. The generic constant γ may vary from line to line. The
main result of this paper reads as follows.

Theorem 1.2 (weak Harnack inequality). Fix a point x0 ∈ Ω and consider the ball
B8ρ(x0) ⊂ Ω. Let u be a non-negative bounded weak supersolution to (1.1) under
conditions (A1)–(A4). Then for any 0 < s < n/(n− 1) it holds:(

–

∫
B5ρ/4(x0)

gs
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx
)1/s

≤ Λ(γ, 3n, ρ) g
(
x0,

m(ρ) + 2(1 + b0)ρ

ρ

)
,

(1.11)

where m(ρ) = ess infBρ(x0) u and Λ(c, β, ρ) = exp
(
c exp

(
βλ(ρ)

))
for any c, β ∈ R

and ρ ∈ (0, r∗).

Corollary 1.3. Let u be a non-negative bounded weak solution to (1.1) under
conditions (A1)–(A4), and let ρ0 be a sufficiently small positive number such that
B8ρ0(x0) ⊂ Ω. There exist positive numbers c, β depending only on the data such
that if Λ(c, β, r) ≤ 3

2 Λ(c, β, 2r) for all 0 < r ≤ ρ/2 < ρ0/2, and additionally∫
0

Λ(−c, β, r)dr
r

= +∞ and lim
r→0

rΛ(c, β, r) = 0,

then the solution u is continuous at x0. Particularly, the function λ(r) = L ln ln ln r−1,
r ∈ (0, e−e), satisfies the above conditions if 0 < L < 1/β.

Theorem 1.4 (Moser-type sup-estimate of solutions). Fix a point x0 ∈ Ω and
consider the ball B8ρ(x0) ⊂ Ω. Let conditions (A1)–(A4) be fulfilled, and let u be a
non-negative bounded weak solution to (1.1), M(ρ) = ess supBρ(x0) u. Then

g
(
x0,

M(ρ) + 2(1 + b0)ρ

ρ

)
≤ γe2nλ(ρ)–

∫
B5ρ/4(x0)

g
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx. (1.12)

From Theorems 1.2 and 1.4 we arrive at the following theorem.

Theorem 1.5 (Harnack inequality). Let all the assumptions of Theorems 1.2, 1.4
be fulfilled. Then there exist positive constants C, c, β depending only on the data,
such that

ess supBρ(x0) u ≤ CΛ(c, β, ρ)
(

ess infBρ(x0) u+ (1 + b0)ρ
)
, (1.13)

where Λ(c, β, ρ) was defined in Theorem 1.2.

The rest of this article contains the proofs of Theorems 1.2 and 1.4.

2. Proof of Theorem 1.2 (weak Harnack inequality)

For proving Theorem 1.2, we need some inequalities and several lemmas. First,
we note simple analogues of Young’s inequality:

g(x, a)b ≤ εg(x, a)a+ g(x, b/ε)b if ε, a, b > 0, x ∈ Ω; (2.1)

g(x, a)b ≤ 1

ε
g(x, a)a+ εp−1g(x, b)b if ε ∈ (0, 1], a, b > 0, x ∈ Ω. (2.2)

In fact, if b ≤ εa, then g(x, a) b ≤ εg(x, a)a, and if b > εa, then since the function
v→ g(x, v) is increasing we have that g(x, a)b ≤ g(x, b/ε)b, which proves inequality
(2.1). Using assumption (A3) by similar arguments we arrive at inequality (2.2).



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 5

Next, we set

G(x,w) =

∫ w

0

g(x, v) dv for x ∈ Ω, w > 0. (2.3)

Then the following inequalities hold:

G(x,w) ≥ γ G(x,w) for all x ∈ Ω, w ≥ 2(1 + b0), (2.4)

G(x,w) ≥ pG(x,w) for all x ∈ Ω, w > 0. (2.5)

Indeed, if x ∈ Ω and w ≥ 2(1 + b0) then by (1.2), (1.9), and (2.3), we have

G(x,w) =

∫ w

0

g(x, v) dv ≥
∫ w

b0

g(x, v) dv ≥ g(x,w)

c1wq−1

∫ w

b0

vq−1dv ≥ 1− 2−q

c1q
G(x,w),

which implies (2.4). Now, let x ∈ Ω and w > 0 be arbitrary, then by (1.3), (1.9)
and (2.3) we obtain

G(x,w) =

∫ w

0

g(x, v) dv ≤ g(x,w)

wp−1

∫ w

0

vp−1 dv =
1

p
g(x,w)w =

1

p
G(x,w),

which yields (2.5).
The rest of the lemmas in this section are successive stages in the proof of

Theorem 1.2. The proof follows Trudinger’s strategy [32], which we adapted to
equation (1.1) under conditions (A1)–(A4).

Lemma 2.1. Let all the assumptions of Theorem 1.2 be fulfilled. Then there exists
positive constant γ depending only on the known data such that

exp
(

–

∫
B2ρ(x0)

ln
(
u+ 2(1 + b0)ρ

)
dx
)
≤ Λ(γ, 3n, ρ)

[
m(ρ) + 2(1 + b0)ρ

]
. (2.6)

Proof. We fix σ ∈ (0, 1), for any ρ ≤ r < r(1 + σ) ≤ 2ρ, we take a function
ζ ∈ C∞0 (Br(1+σ)(x0)), 0 ≤ ζ ≤ 1, ζ = 1 in Br(x0) and |∇ζ| ≤ (σr)−1. Let

w = ln
κ

u
, u = u+ 2(1 + b0)ρ, (2.7)

where the constant κ is defined by the condition (w)x0,2ρ = –
∫
B2ρ(x0)

w dx = 0, i.e.

κ = exp
(

–

∫
B2ρ(x0)

lnu dx
)
. (2.8)

We test (1.10) by ϕ = u (w−k)+
G(x0,u/ρ)

ζq, (w− k)+ = max{0, w− k}, k > 0. Since we are

dealing with bounded and non-negative solutions (supersolutions), then this and

all other test functions used in the paper belong to W 1,G
0 (Ω). This is a consequence

of conditions (A1) and (A2) and the result of Marcus and Mizel [24, Theorem 2].
So, we have ∫

Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx

+

∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

{G(x0, u/ρ)

G(x0, u/ρ)
− 1
}

(w − k)+ζ
q dx

≤ γ

σ

∫
Ak,r(1+σ)

g(x, |∇u|)
G(x0, u/ρ)

u

ρ
(w − k)+ζ

q−1 dx,



6 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

where Ak,r = Br(x0) ∩ {w > k}. By (2.5), the value in curly brackets is estimated
from below as follows:

G(x0, u/ρ)

G(x0, u/ρ)
− 1 ≥ p− 1, (2.9)

and therefore∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx+ (p− 1)

∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

(w − k)+ζ
q dx

≤ γ
∫
Ak,r(1+σ)

g(x, |∇u|)
G(x0, u/ρ)

u

σρζ
(w − k)+ζ

q dx.

(2.10)

We use inequality (2.1) with a = |∇u|, b = u
σρζ and sufficiently small ε > 0, and

then (2.4) with w = u/ρ, to estimate from above the right-hand side of (2.10):

γ

∫
Ak,r(1+σ)

g(x, |∇u|)
G(x0, u/ρ)

u

σρζ
(w − k)+ζ

q dx

≤ p− 1

2

∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

(w − k)+ζ
q dx

+
γ

σ

∫
Ak,r(1+σ)

g
(
x, γuσρζ

)
g(x0, u/ρ)

(w − k)+ζ
q−1 dx.

Combining this inequality and (2.10), we obtain∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx ≤ γ

σ

∫
Ak,r(1+σ)

g
(
x, γ uσρζ

)
g(x0, u/ρ)

(w − k)+ζ
q−1 dx. (2.11)

Since γu
σρζ ≥

u
ρ ≥ b0 and |x − x0| < r(1 + σ) ≤ 2ρ for x ∈ Ak,r(1+σ), then using

conditions (A2) and (A4), we get that for all x ∈ Ak,r(1+σ), it holds

g
(
x,
γ u

σρζ

)
≤ γ (σζ)1−q g(x, u/ρ) ≤ γ (σζ)1−q eλ(ρ)g (x0, u/ρ) .

So, from (2.11) we obtain∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx ≤ γ σ−q eλ(ρ)

∫
Ak,r(1+σ)

(w − k)+ dx. (2.12)

To estimate the term on the left-hand side of (2.12), we use (2.1) with ε = 1,
a = u/ρ, b = |∇u|, assumption (A4), the definitions of the functions G, G, w (see
equalities (1.9), (2.3) and (2.7), respectively) and (2.5):∫

Ak,r(1+σ)

|∇w|ζq dx =

∫
Ak,r(1+σ)

|∇u|
u

g(x, u/ρ)

g(x, u/ρ)
ζq dx

≤ 1

ρ
|Ak,r(1+σ)|+

1

ρ

∫
Ak,r(1+σ)

G(x, |∇u|)
G (x, u/ρ)

ζq dx

≤ 1

ρ
|Ak,r(1+σ)|+ γ

eλ(ρ)

ρ

∫
Ak,r(1+σ)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx.

(2.13)

Collecting (2.12) and (2.13), we obtain∫
Ak,r(1+σ)

|∇w|ζq dx ≤ γ e2λ(ρ)

σqρ

(
|Ak,r(1+σ)|+

∫
Ak,r(1+σ)

(w − k)+ dx
)
.



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 7

From this, using Sobolev’s embedding theorem and standard iteration arguments
(see, for instance [20, Section 2, Theorem 5.3]), and choosing k from the condition

k ≥ γ e2nλ(ρ)
(

–

∫
B2ρ(x0)

|w|
n
n−1 dx

)n−1
n

+ 1,

we obtain that

ess supBρ(x0) w ≤ γ e2nλ(ρ)
(

–

∫
B2ρ(x0)

|w|
n
n−1 dx

)n−1
n

+ 1. (2.14)

To estimate the right-hand side of (2.14) we use the Poincaré inequality by our
choice of κ (see (2.8)) we have(

–

∫
B2ρ(x0)

|w|
n
n−1 dx

)n−1
n

=
(

–

∫
B2ρ(x0)

|w − (w)x0,2ρ|
n
n−1 dx

)n−1
n

≤ γ ρ1−n
∫
B2ρ(x0)

|∇w| dx.
(2.15)

Next, similarly to (2.13), we have∫
B2ρ(x0)

|∇w| dx ≤
∫
B4ρ(x0)

|∇w|ζq dx

≤ γρn−1 + γ
eλ(ρ)

ρ

∫
B4ρ(x0)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx,

(2.16)

here we have ζ ∈ C∞0 (B4ρ(x0)), 0 ≤ ζ ≤ 1, ζ = 1 in B2ρ(x0), and |∇ζ| ≤ 2/ρ. In

addition, testing (1.10) by ϕ = uζq

G(x0,u/ρ)
, similarly to (2.12), we obtain∫

B4ρ(x0)

G(x, |∇u|)
G(x0, u/ρ)

ζq dx ≤ γρneλ(ρ). (2.17)

Now, collecting (2.14)–(2.17) and taking into account (2.7) and (2.8), we arrive at
the required inequality (2.6). The proof is complete. �

Lemma 2.2. Under the assumptions of Theorem 1.2 there exists δ0 = δ0(ρ) > 0
depending only on the data and ρ, such that(

–

∫
B3ρ/2(x0)

(
u+ 2(1 + b0)ρ

)δ0
dx
)1/δ0

≤ Λ(γ, 2n, ρ) exp
(

–

∫
B2ρ(x0)

ln
(
u+ 2(1 + b0)ρ

)
dx
)
.

(2.18)

Proof. Let us fix σ ∈ (0, 1) and for any 3ρ/2 ≤ r < r(1 + σ) ≤ 2ρ consider the
function ζ ∈ C∞0

(
Br(1+σ)(x0)

)
, 0 ≤ ζ ≤ 1, ζ = 1 in Br(x0), |∇ζ| ≤ (σr)−1. We

define

v = ln
u+ 2(1 + b0)ρ

κ
= ln

u

κ
, vµ = max{v, µ}, µ > 0.

Testing (1.10) by ϕ =
vs−1
µ uζl

G(x0,u/ρ)
, s ≥ 1, l ≥ q, and using (2.9), we have

(p− 1)

∫
Br(1+σ)(x0)

G(x, |∇u|)
G(x0, u/ρ)

vs−1
µ ζl dx

≤ (s− 1)

∫
Br(1+σ)(x0)∩{v>µ}

G(x, |∇u|)
G(x0, u/ρ)

vs−2
µ ζl dx



8 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

+ γ l

∫
Br(1+σ)(x0)

g(x, |∇u|)
G(x0, u/ρ)

u

σρζ
vs−1
µ ζl dx.

Choosing µ from the condition s
µ = p−1

2 and using inequalities (2.1), (2.4) and

conditions (A2) and (A4) similarly to the derivation of (2.12), from the previous
we obtain∫

Br(1+σ)(x0)

G(x, |∇u|)
G(x0, u/ρ)

vs−1
µ ζl dx ≤ γ lγeλ(ρ)

σq

∫
Br(1+σ)(x0)

vs−1
µ ζl−q dx. (2.19)

Estimating the term on the left-hand side of (2.19), similarly to (2.13), we obtain∫
Br(1+σ)(x0)

|∇vµ| vs−1
µ ζl dx ≤

∫
Br(1+σ)(x0)

|∇u|
u

vs−1
µ ζl dx

≤ γ lγ

σq
e2λ(ρ)

ρ

∫
Br(1+σ)(x0)

vs−1
µ ζl−q dx

≤ γ lγ

σq
e2λ(ρ)

ρ

∫
Br(1+σ)(x0)

vsµζ
l−q dx.

Using Sobolev’s embedding theorem, from this we have

–

∫
Br(x0)

v
sn
n−1
µ dx ≤

(γs e2λ(ρ)

σq
–

∫
Br(1+σ)(x0)

vsµ dx
) n
n−1

. (2.20)

For j = 0, 1, 2, . . ., we define the sequences

rj =
ρ

2
(3 + 2−j), Bj = Brj (x0),

sj =
( n

n− 1

)j+1

, µj =
2sj
p− 1

, yj = –

∫
Bj

vsjµj dx.

Then inequality (2.20) can be rewritten in the form

y
1/sj+1

j+1 ≤
(
γ2jqsj e

2λ(ρ)
)1/sj

y
1/sj
j , j = 0, 1, 2, . . . , (2.21)

and by Sobolev’s inequality and (2.17), we have

y0 ≤ γ exp
(2nλ(ρ)

n− 1

)
. (2.22)

From this by iteration, for j = 0, 1, 2, . . ., we have

y
1/sj+1

j+1 ≤ γ
∑j
i=0

1
si 2

q
∑j
i=1

i
si

( n

n− 1

)∑j
i=0

i+1
si

exp
(

2λ(ρ)

j∑
i=0

1

si

)
y
n−1
n

0

≤ γe2nλ(ρ).

(2.23)

Let m ∈ N be arbitrary, then there exists j ≥ 1 such that sj−1 < m ≤ sj . Using
Hölder’s inequality, from (2.23) we obtain

–

∫
B3ρ/2(x0)

vm+
m!

dx ≤ –

∫
B3ρ/2(x0)

vmµj
m!

dx ≤
γ y

m/sj
j

m!
≤ γm+1

m!
e2nmλ(ρ) ≤ γm+1e2nmλ(ρ).

Choosing δ0 = δ0(ρ) from the condition

δ0 =
1

2γ
e−2nλ(ρ), (2.24)



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 9

from the previous we have

–

∫
B3ρ/2(x0)

(δ0v+)m

m !
dx ≤ γ 2−m,

which implies

–

∫
B3ρ/2(x0)

eδ0v dx ≤ –

∫
B3ρ/2(x0)

eδ0v+ dx ≤
∞∑
m=0

–

∫
B3ρ/2(x0)

(δ0v+)m

m!
dx ≤ 2γ.

From this, since eδ0v = (u/κ)δ0 we have(
–

∫
B3ρ/2(x0)

uδ0 dx
)1/δ0

≤ (2γ)1/δ0κ ≤ Λ(γ, 2n, ρ)κ,

that together with (2.8) yields the desired inequality (2.18). This completes the
proof. �

The next lemma is a simple consequence of Lemmas 2.1 and 2.2.

Lemma 2.3. Let all the assumptions of Lemma 2.2 be fulfilled, and set

δ1 = δ0/(q − 1), (2.25)

where δ0 is defined by (2.24). Then(
–

∫
B3ρ/2(x0)

gδ1
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx
)1/δ1

≤ Λ(γ, 3n, ρ) g
(
x0,

m(ρ) + 2(1 + b0)ρ

ρ

)
.

(2.26)

Proof. By condition (A2) we have

–

∫
B3ρ/2(x0)

gδ1
(
x0,

u+2(1+b0)ρ
ρ

)
gδ1
(
x0,

m(ρ)+2(1+b0)ρ
ρ

) dx
≤ 1 + cδ11 –

∫
B3ρ/2(x0)∩{u>m(ρ)}

( u+ 2(1 + b0)ρ

m(ρ) + 2(1 + b0)ρ

)δ0
dx.

By Lemmas 2.1 and 2.2 the second term on the right-hand side of this inequality
is estimated from above as follows:

–

∫
B3ρ/2(x0)

( u+ 2(1 + b0)ρ

m(ρ) + 2(1 + b0)ρ

)δ0
dx ≤ Λ(γ, 3n, ρ),

which completes the proof. �

To complete the proof of Theorem 1.2 we need the following lemma.

Lemma 2.4 (Inverse Hölder inequality). Let the assumptions of Theorem 1.2 be
fulfilled, then for all δ1 ≤ s < n/(n− 1) we have(

–

∫
B5ρ/4(x0)

gs
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx
)1/s

≤ Λ(γ, 2n+ 1, ρ)
(

–

∫
B3ρ/2(x0)

gδ1
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx
)1/δ1

.

(2.27)



10 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

Proof. We set ψ(x,w) = G(x,w)
w for x ∈ Ω, w > 0, and note that by (2.4) and (2.5),

we have

g(x,w) ≤ γψ(x,w) for all x ∈ Ω, w ≥ 2(1 + b0), (2.28)

ψ(x,w) ≤ 1

p
g(x,w) for all x ∈ Ω, w > 0, (2.29)

which gives

ψ′w(x,w) ≤ γ ψ(x,w)

w
for all x ∈ Ω, w ≥ 2(1 + b0), (2.30)

ψ′w(x,w) =
g(x,w)− ψ(x,w)

w
≥ (p− 1)

ψ(x,w)

w
for all x ∈ Ω, w > 0. (2.31)

We need a Cacciopoli-type inequality for negative powers of ψ(x0, u/ρ). To
establish it, we fix σ ∈ (0, 1) and r > 0 such that 5ρ/4 ≤ r < r(1 + σ) ≤ 3ρ/2, and
take a function ζ ∈ C∞0

(
Br(1+σ)(x0)

)
, 0 ≤ ζ ≤ 1, ζ = 1 in Br(x0), |∇ζ| ≤ (σr)−1.

Testing (1.10) by ϕ = ψ−τ (x0, u/ρ)ζθ, 0 < τ < 1, θ ≥ q, and using (2.31), we
obtain

(p− 1)τ

∫
Br(1+σ)(x0)

ψ−τ (x0, u/ρ)
G(x, |∇u|)

u
ζθ dx

≤ γ θ

σρ

∫
Br(1+σ)(x0)

ψ−τ (x0, u/ρ)g(x, |∇u|)ζθ−1 dx,

which by (2.1), (A2), (A4) and (2.28) implies

∫
Br(1+σ)(x0)

ψ−τ (x0, u/ρ)
G(x, |∇u|)

u
ζθ dx

≤ γ θq

(στ)q
eλ(ρ)

ρ

∫
Br(1+σ)(x0)

ψ1−τ (x0, u/ρ)ζθ−q dx.

(2.32)

Based on inequality (2.32), we organize Moser-type iterations for the function
ψ(x0, u/ρ). To do this, we fix 0 < t < n/(n − 1) and l ≥ nq/(n − 1), then by the
Sobolev inequality and by (2.30) and (2.29), we obtain

(∫
Br(1+σ)(x0)

ψt(x0, u/ρ)ζl dx
)n−1

n

≤ γ
∫
Br(1+σ)(x0)

∣∣∇[ψ t(n−1)
n (x0, u/ρ)ζ

l(n−1)
n

]∣∣ dx
≤ γt

∫
Br(1+σ)(x0)

ψ
t(n−1)
n −1(x0, u/ρ)

g(x0, u/ρ)

u
|∇u|ζ

l(n−1)
n dx

+
γ l

σρ

∫
Br(1+σ)(x0)

ψ
t(n−1)
n (x0, u/ρ)ζ

l(n−1)
n −1 dx.

(2.33)



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 11

Using (2.1), (A4), (2.28) and (2.32) with τ = 1 − t(n − 1)/n and θ = l(n − 1)/n,
we estimate the first term on the right-hand side of (2.33) as follows:∫

Br(1+σ)(x0)

ψ
t(n−1)
n −1(x0, u/ρ)

g(x0, u/ρ)

u
|∇u|ζ

l(n−1)
n dx

≤ γeλ(ρ)

∫
Br(1+σ)(x0)

ψ
t(n−1)
n −1(x0, u/ρ)

g(x, u/ρ)

u
|∇u|ζ

l(n−1)
n dx

≤ γeλ(ρ)

∫
Br(1+σ)(x0)

ψ
t(n−1)
n −1(x0, u/ρ)

G (x, |∇u|)
u

ζ
l(n−1)
n dx

+ γ
eλ(ρ)

ρ

∫
Br(1+σ)(x0)

ψ
t(n−1)
n −1(x0, u/ρ)g(x, u/ρ)ζ

l(n−1)
n dx

≤ γ lq

σq
[
1− t(n− 1)

n

]−q e2λ(ρ)

ρ

∫
Br(1+σ)(x0)

ψ
t(n−1)
n (x0, u/ρ)ζ

l(n−1)
n −q dx.

(2.34)

Combining (2.33), (2.34), we arrive at(
–

∫
Br(x0)

ψt(x0, u/ρ)dx
)n−1

n

≤ γ lq

σq

(
1− t(n− 1)

n

)−q
e2λ(ρ)–

∫
Br(1+σ)(x0)

ψ
t(n−1)
n (x0, u/ρ)dx,

(2.35)

for 0 < t < n
n−1 and l ≥ nq

n−1 .

Now, let δ1 ≤ s < n/(n− 1), and let j be a non-negative integer such that

s
(n− 1

n

)j+1 ≤ δ1 ≤ s
(n− 1

n

)j
. (2.36)

Setting in (2.35) l = nq, r = ri = ρ
4 (6 − 2−i), r(1 + σ) = ri+1, Bi = Bri(x0) and

t = ti = s
(
n−1
n

)i
for i = 0, 1, . . . , j + 1, we have(

–

∫
Bi

ψti(x0, u/ρ)dx
)1/ti

≤
[
γ2iq

(
1− n− 1

n
s
)−q

e2λ(ρ)
]1/ti+1

(
–

∫
Bi+1

ψti+1(x0, u/ρ)dx
)1/ti+1

.

Iterating this relation and using Hölder’s inequality, we obtain(
–

∫
B5ρ/4(x0)

ψs(x0, u/ρ)dx
)1/s

=
(

–

∫
B0

ψt0(x0, u/ρ)dx
)1/t0

≤
j∏
i=0

[
γ 2iqe2λ(ρ)

(
1− n− 1

n
s
)−q]1/ti+1

(
–

∫
Bj+1

ψtj+1(x0, u/ρ)dx
)1/tj+1

≤ 2q
∑j
i=0 i/ti+1

[
γe2λ(ρ)

(
1− n− 1

n
s
)−q]∑j

i=0 1/ti+1
(
γ–

∫
B3ρ/2(x0)

ψδ1(x0, u/ρ)dx
)1/δ1

,



12 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

and by (2.36), (2.25) and (2.24) we have

j∑
i=0

1

ti+1
≤ 1

δ1

n

n− 1

∞∑
i=0

(n− 1

n

)i
=

n2

δ1(n− 1)
,

j∑
i=0

i

ti+1
≤ j

j∑
i=0

1

ti+1
≤ γ(λ(ρ) + 1)

δ1
.

From this, and recalling the definition of δ1 (see again (2.25) and (2.24)), we arrive
at the required inequality (2.27). This completes the proof. �

Combining Lemmas 2.3 and 2.4, we obtain that(
–

∫
B5ρ/4(x0)

gs
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx
)1/s

≤ Λ(γ, 3n, ρ)g
(
x0,

m(ρ) + 2(1 + b0)ρ

ρ

)
,

which proves Theorem 1.2.

3. Proof of Theorem 1.4 (sup-estimate of solutions)

Let us fix σ, σ1 ∈ (0, 1), ρ ≤ r < r(1 + σσ1) < r(1 + σ) ≤ 5ρ/4, and consider
a function ζ ∈ C∞0

(
Br(1+σσ1)(x0)

)
such that 0 ≤ ζ ≤ 1, ζ = 1 in Br(x0) and

|∇ζ| ≤ (σσ1r)
−1. Testing (1.10) by ϕ = uGs−1(x0, u/ρ)ζl, s ≥ 1, l ≥ max{q, s/2},

and using (2.5), we have

s

∫
Br(1+σσ1)(x0)

G(x, |∇u|)Gs−1(x0, u/ρ)ζl dx

≤ l
∫
Br(1+σσ1)(x0)

g(x, |∇u|) u

σσ1ρζ
Gs−1(x0, u/ρ)ζl dx.

(3.1)

Using (2.1) with ε = s
2l , a = |∇u|, b = u

σσ1ρζ
, we estimate the right-hand side of

(3.1) from above as follows:

l

∫
Br(1+σσ1)(x0)

g(x, |∇u|) u

σσ1ρζ
Gs−1(x0, u/ρ)ζl dx

≤ s

2

∫
Br(1+σσ1)(x0)

G(x, |∇u|)Gs−1(x0, u/ρ)ζl dx

+ l

∫
Br(1+σσ1)(x0)

g
(
x,

u

εσσ1ρζ

) u

σσ1ρζ
Gs−1(x0, u/ρ)ζl dx,

(3.2)

moreover, since u
εσσ1ρζ

≥ u
ρ ≥ 2(1 + b0), conditions (A2), (A4), inequality (2.4) and

ε = s/(2l) give the estimate

l

∫
Br(1+σσ1)(x0)

g
(
x,

u

εσσ1ρζ

) u

σσ1ρζ
Gs−1(x0, u/ρ)ζl dx

≤ c1l

εq−1

1

(σσ1)q

∫
Br(1+σσ1)(x0)

g
(
x,
u

ρ

)u
ρ
Gs−1(x0, u/ρ)ζl−q dx

≤ γ lqeλ(ρ)

(σσ1)q

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)ζl−q dx.

(3.3)



EJDE-2021/27 HARNACK’S INEQUALITY FOR QUASILINEAR ELLIPTIC EQUATIONS 13

Combining (3.1), (3.2), (3.3), we obtain

s

∫
Br(1+σσ1)(x0)

G(x, |∇u|)Gs−1(x0, u/ρ)ζl dx

≤ γlqeλ(ρ)

(σσ1)q

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)ζl−q dx.

In turn, using this inequality, as well as (A4), (2.1) and (2.4), we deduce that∫
Br(1+σσ1)(x0)

∣∣∇[Gs(x0, u/ρ)ζl
]∣∣ dx

≤ s

ρ

∫
Br(1+σσ1)(x0)

Gs−1(x0, u/ρ)g(x0, u/ρ)|∇u| ζl dx

+
l

σσ1ρ

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)ζl−1 dx

≤ γse
λ(ρ)

ρ

∫
Br(1+σσ1)(x0)

G(x, |∇u|)Gs−1(x0, u/ρ)ζl dx

+
γs l

σσ1ρ

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)ζl−1 dx

≤ γs lq

(σσ1)q
e2λ(ρ)

ρ

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)ζl−q dx.

Combining this and Sobolev’s inequality, we arrive to(∫
Br(x0)

G
sn
n−1 (x0, u/ρ)dx

)n−1
n ≤

∫
Br(1+σσ1)(x0)

∣∣∇[Gs(x0, u/ρ)ζl
]∣∣dx

≤ γs lq

(σσ1)q
e2λ(ρ)

ρ

∫
Br(1+σσ1)(x0)

Gs(x0, u/ρ)dx.

(3.4)

Now, for i, j = 0, 1, 2, . . ., we define the sequences

ri,j =
ρ

4
(5− 2−i) +

ρ

8
2−i−j , sj =

( n

n− 1

)j
, lj = q

( n

n− 1

)j
.

Let ζ i,j ∈ C∞0
(
Bri,j (x0)

)
, 0 ≤ ζ i,j ≤ 1, ζ i,j = 1 in Bri,j+1

(x0), |∇ζ i,j | ≤ 2i+j+4

ρ .

For i, j = 0, 1, 2, . . ., we also set ri = ri,∞, Mi = ess supBri (x0) u and

yi,j =
(

–

∫
Bri,j (x0)

Gsj (x0, u/ρ)dx
)1/sj

.

From (3.4) we obtain

yi,j+1 ≤
(
γ2(i+j)qe2λ(ρ)

)1/sj
yi,j , i, j = 0, 1, 2, . . . . (3.5)

We iterate inequality (3.5) with respect to j and use the fact that ri+1 = ri,0 to
obtain

G
(
x0,

Mi + 2(1 + b0)ρ

ρ

)
≤ γ 2iγe2nλ(ρ)–

∫
Bri+1

(x0)

G(x0, u/ρ)dx

≤ γ2iγe2nλ(ρ)Mi+1 + 2(1 + b0)ρ

ρ
–

∫
Bri+1

(x0)

g(x0, u/ρ)dx.



14 M. A. SHAN, I. I. SKRYPNIK, M. V. VOITOVYCH EJDE-2021/27

This inequality, (2.2), and (2.4) imply that, for any ε ∈ (0, 1) and i = 0, 1, 2, . . .,
the following inequalities hold:

g
(
x0,

Mi + 2(1 + b0)ρ

ρ

)
≤ εp−1g

(
x0,

Mi+1 + 2(1 + b0)ρ

ρ

)
+

1

ε
g
(
x0,

Mi + 2(1 + b0)ρ

ρ

) Mi + 2(1 + b0)ρ

Mi+1 + 2(1 + b0)ρ

≤ εp−1g
(
x0,

Mi+1 + 2(1 + b0)ρ

ρ

)
+
γ 2iγ

ε
e2nλ(ρ)–

∫
B5ρ/4(x0)

g(x0, u/ρ)dx.

Iterating the resulting inequality, for each i ≥ 1 we have

g
(
x0,

M(ρ) + 2(1 + b0)ρ

ρ

)
= g
(
x0,

M0 + 2(1 + b0)ρ

ρ

)
≤ εi(p−1)g

(
x0,

Mi + 2(1 + b0)ρ

ρ

)
+ γε−1e2nλ(ρ)

i−1∑
j=0

(εp−12γ)j–

∫
B5ρ/4(x0)

g(x0, u/ρ)dx.

Finally, choosing ε from the condition εp−12γ = 1/2 and passing i to infinity, we
arrive at

g
(
x0,

M(ρ) + 2(1 + b0)ρ

ρ

)
≤ γe2nλ(ρ)–

∫
B5ρ/4(x0)

g
(
x0,

u+ 2(1 + b0)ρ

ρ

)
dx.

This completes the proof of Theorem 1.4.

Acknowledgments. This work was supported by Ministry of Education and Sci-
ence of Ukraine (project numbers 0121U109525 and 0119U100421), and by the
Volkswagen Foundation project “From Modeling and Analysis to Approximation”.

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Maria A. Shan

Institute of Applied Mathematics and Mechanics, National Academy of Sciences of
Ukraine, Gen. Batiouk str. 19, 84116 Sloviansk, Ukraine.

Vasyl’ Stus Donetsk National University, 600-richcha str. 21, 21021 Vinnytsia, Ukraine

Email address: shan maria@ukr.net

Igor I. Skrypnik

Institute of Applied Mathematics and Mechanics, National Academy of Sciences of
Ukraine, Gen. Batiouk str. 19, 84116 Sloviansk, Ukraine.

Vasyl’ Stus Donetsk National University, 600-richcha str. 21, 21021 Vinnytsia, Ukraine

Email address: iskrypnik@iamm.donbass.com

Mykhailo V. Voitovych

Institute of Applied Mathematics and Mechanics, National Academy of Sciences of
Ukraine, Gen. Batiouk str. 19, 84116 Sloviansk, Ukraine

Email address: voitovichmv76@gmail.com


	1. Introduction and main results
	2. Proof of Theorem 1.2 (weak Harnack inequality)
	3. Proof of Theorem 1.4 (sup-estimate of solutions)
	Acknowledgments

	References