Electronic Journal of Differential Equations, Vol. 2007(2007), No. 177, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF POSITIVE SOLUTIONS FOR p(x)-LAPLACIAN
PROBLEMS
GHASEM A. AFROUZI, HORIEH GHORBANI
Abstract. We consider the system of differential equations
−∆p(x) u = λ[g(x)a(u) + f (v)]
in Ω
−∆q(x) v = λ[g(x)b(v) + h(u)]
in Ω
u=v=0
on ∂Ω
where p(x) ∈ C 1 (RN ) is a radial symmetric function such that sup |∇p(x)| <
∞, 1 < inf p(x) ≤ sup p(x) < ∞, and where −∆p(x) u = − div |∇u|p(x)−2 ∇u
which is called the p(x)-Laplacian. We discuss the existence of positive solution
via sub-super-solutions without assuming sign conditions on f (0), h(0).
1. Introduction
The study of differential equations and variational problems with nonstandard
p(x)-growth conditions has been a new and interesting topic. Many results have
been obtained on this kind of problems; see for example [3, 4, 5, 6, 7, 8, 13]. In
[5, 6] Fan and Zhao give the regularity of weak solutions for differential equations
with nonstandard p(x)-growth conditions. Zhang [11] investigated the existence of
positive solutions of the system
−∆p(x) u = f (v)
in Ω
−∆p(x) v = g(u)
in Ω
(1.1)
u = v = 0 on ∂Ω
where p(x) ∈ C 1 (RN ) is a function, Ω ⊂ RN is a bounded domain. The operator
−∆p(x) u = − div |∇u|p(x)−2 ∇u) is called p(x)-Laplacian. Especially, if p(x) is a
constant p, System (1.1) is the well-known p-Laplacian system. There are many
papers on the existence of solutions for p-Laplacian elliptic systems, for example
[1, 3, 4, 5, 6, 7, 8, 9].
2000 Mathematics Subject Classification. 35J60, 35B30, 35B40.
Key words and phrases. Positive radial solutions; p(x)-Laplacian problems;
boundary value problems.
c
2007
Texas State University - San Marcos.
Submitted July 18, 2007. Published December 17, 2007.
1
2
G. A. AFROUZI AND H. GHORBANI
EJDE-2007/177
In [9] the authors consider the existence of positive weak solutions for the pLaplacian problem
−∆p u = f (v) in Ω
−∆p v = g(u)
in Ω
(1.2)
u = v = 0 on ∂Ω .
There the first eigenfunctions is used for constructing the subsolution of p-Laplacian
problems. Under the condition limu→+∞ f (M (g(u))1/(p−1) /up−1 = 0, for all M >
0, the authors show the existence of positive solutions for problem (1.2).
In this paper, at first, we consider the existence of positive solutions of the system
−∆p(x) u = F (x, u, v)
in Ω
−∆p(x) v = G(x, u, v)
in Ω
(1.3)
u = v = 0 on ∂Ω
1
N
where p(x) ∈ C (R ) is a function, F (x, u, v) = [g(x)a(u) + f (v)], G(x, u, v) =
[g(x)b(v) + h(u)], and Ω ⊂ RN is a bounded domain. Then we consider the system
−∆p(x) u = λF (x, u, v)
in Ω
−∆p(x) v = λG(x, u, v)
in Ω
(1.4)
u = v = 0 on ∂Ω
where p(x) ∈ C 1 (RN ) is a function, F (x, u, v) = [g(x)a(u) + f (v)], G(x, u, v) =
[g(x)b(v) + h(u)], λ is a positive parameter and Ω ⊂ RN is a bounded domain.
To study p(x)-Laplacian problems, we need some theory on the spaces Lp(x) (Ω),
1,p(x)
W
(Ω) and properties of p(x)-Laplacian which we will use later (see [4]). If
Ω ⊂ RN is an open domain, write
C+ (Ω) = {h : h ∈ C(Ω), h(x) > 1 for x ∈ Ω}
h+ = supx∈Ω h(x), h− = inf x∈ΩRh(x), for any h ∈ C(Ω), Lp(x) (Ω) = {u|u is a
measurable real-valued function, Ω |u|p(x) dx < ∞}.
Throughout the paper, we will assume that p ∈ C+ (Ω) and 1 < inf x∈RN p(x) ≤
supx∈RN p(x) < N . We introduce the norm on Lp(x) (Ω)by
Z
u(x) p(x)
|
dx ≤ 1},
|u|p(x) = inf{λ > 0 :
|
λ
Ω
and (Lp(x) (Ω), | · |p(x) ) becomes a Banach space, we call it generalized Lebesgue
space. The space (Lp(x) (Ω), | · |p(x) ) is a separable, reflexive and uniform convex
Banach space (see [4, Theorem 1.10, 1.14]).
The space W 1,p(x) (Ω) is defined by W 1,p(x) (Ω) = {u ∈ Lp(x) (Ω) : |∇u| ∈
p(x)
L
(Ω)}, and it is equipped with the norm
kuk = |u|p(x) + |∇u|p(x) ,
1,p(x)
∀u ∈ W 1,p(x) (Ω).
We denote by W0
(Ω) the closure of C0∞ (Ω) in W 1,p(x) (Ω). W 1,p(x) (Ω) and
1,p(x)
(Ω) are separable, reflexive and uniform convex Banach space (see [4, TheW0
orem 2.1]). We define
Z
(L(u), v) =
|∇u|p(x)−2 ∇u∇vdx, ∀u, v ∈ W 1,p(x) (Ω),
RN
EJDE-2007/177
EXISTENCE OF POSITIVE SOLUTIONS
3
then L : W 1,p(x) (Ω) → (W 1,p(x) (Ω))∗ is a continuous, bounded and is a strictly
monotone operator, and it is a homeomorphism [7, Theorem 3.11].
1,p(x)
Functions u, v in W0
(Ω), is called a weak solution of (1.4); it satisfies
Z
Z
1,p(x)
p(x)−2
|∇u|
∇u∇ξdx =
λF (x, u, v)ξdx, ∀ξ ∈ W0
(Ω),
ZΩ
ZΩ
1,p(x)
|∇v|q(x)−2 ∇v∇ξdx =
λG(x, u, v)ξdx, ∀ξ ∈ W0
(Ω).
Ω
Ω
We make the following assumptions
(H1) p(x) ∈ C 1 (RN ) is a radial symmetric and sup |∇p(x)| < ∞
(H2) Ω = B(0, R) = {x||x| < R} is a ball, where R > 0 is a sufficiently large
constant.
(H3) a, b ∈ C 1 ([0, ∞)) are nonnegative, nondecreasing functions such that
lim
u→+∞
a(u)
= 0,
uP − −1
lim
u→+∞
b(u)
= 0.
uP − −1
(H4) f, h ∈ C 1 ([0, ∞)) are nondecreasing functions, limu→+∞ f (u) = +∞,
limu→+∞ h(u) = +∞, and
1
f (M (h(u)) p− −1 )
lim
= 0,
u→+∞
up− −1
∀M > 0 .
(H5) g : [0, +∞) → (0, ∞) is a continuous function such that L1 = minx∈Ω̄ g(x),
and L2 = maxx∈Ω̄ g(x).
We shall establish the following result.
Theorem 1.1. If (H1)–(H5) hold, then (1.3) has a positive solution.
Proof. We establish this theorem by constructing a positive subsolution (φ1 , φ2 )
and supersolution (z1 , z2 ) of (1.3), such that φ1 ≤ z1 and φ2 ≤ z2 . That is (φ1 , φ2 )
and (z1 , z2 ) satisfy
Z
Z
Z
|∇φ1 |p(x)−2 ∇φ1 · ∇ξdx ≤
g(x)a(φ1 )ξdx +
f (φ2 )ξdx,
ZΩ
ZΩ
ZΩ
|∇φ2 |p(x)−2 ∇φ1 · ∇ξdx ≤
g(x)b(φ2 )ξdx +
h(φ1 )ξdx,
Ω
Z
ZΩ
ZΩ
|∇z1 |p(x)−2 ∇z1 · ∇ξdx ≥
g(x)a(z1 )ξdx +
f (z2 )ξdx,
Ω
Z
ZΩ
ZΩ
|∇z2 |p(x)−2 ∇z2 · ∇ξdx ≥
g(x)b(z2 )ξdx +
h(z1 )ξdx,
Ω
1,p(x)
for all ξ ∈ W0
Ω
Ω
(Ω) with ξ ≥ 0. Then (1.3) has a positive solution.
Step 1. We construct a subsolution of (1.3). Denote
inf p(x) − 1
R−α
, R0 =
,
4(sup |∇p(x)| + 1)
2
b = min{a(0)L1 + f (0), b(0)L1 + h(0), −1},
α=
4
G. A. AFROUZI AND H. GHORBANI
EJDE-2007/177
and let


e−k(r−R) − 1,
2R0 < r ≤ R,



p(2R0 )−1
R

2R
0

(keαk ) p(r)−1
eαk − 1 +


 (2R )N −1 r
1
π
π
0
φ(r) = ×[ rN −1 sin(ε(r − 2R0 ) + 2 )(L1 + 1)] p(r)−1 dr, 2R0 − 2ε < r ≤ 2R0 ,

p(2R0 )−1

R

eαk − 1 + 2R0 π (keαk ) p(r)−1

2R
−

0
2ε

1
 (2R0 )N −1

π
×[ rN −1 sin(ε0 (r − 2R0 ) + π2 )(L1 + 1)] p(r)−1 dr, r ≤ 2R0 − 2ε
,
where R0 is sufficiently large, ε is a small positive constant which satisfies R0 ≤
π
2R0 − 2ε
,
In the following, we will prove that (φ, φ) is a subsolution of (1.3). Since


e−k(r−R) − 1,
2R0 < r ≤ R,


p(2R0 )−1


αk p(r)−1
−(ke )
φ0 (r) =
N −1
1
π
0)

sin(ε(r − 2R0 ) + π2 )(L1 + 1)] p(r)−1 dr, 2R0 − 2ε
×[ (2R
< r ≤ 2R0 ,


r N −1


π
0,
0 ≤ r ≤ 2R0 − 2ε
,
it is easy to see that φ ≥ 0 is decreasing and φ ∈ C 1 ([0, R]), φ(x) = φ(|x|) ∈ C 1 (Ω̄).
Let r = |x|. By computation,
−∆p(x) φ = − div |∇φ(x)|p(x)−2 ∇φ(x)) = −(rN −1 |φ0 (r)|p(r)−2 φ0 (r))0 /rN −1 .
Then
 −k(r−R) p(r)−1 (ke
)
− k(p(r) − 1) + p0 (r) ln k




N
−1
0


−kp (r)(r − R) + r ,
−∆p(x) φ = ε( 2R0 )N −1 (keαk )( p(2R0 ) − 1)
r




×
cos(ε(r
− 2R0 ) + π2 )(L1 + 1),



0,
2R0 < r ≤ R,
π
2R0 − 2ε
< r ≤ 2R0 ,
π
0 ≤ r ≤ 2R0 − 2ε
,
If k is sufficiently large, when 2R0 < r ≤ R, then
ln k
N −1
+ R − r) +
] ≤ −kα.
k
kr
Since α is a constant dependent only on p(x), if k is a big enough, such that −ka < b,
and since φ(x) ≥ 0 and a, f are monotone, this implies
−∆p(x) φ ≤ −k[inf p(x) − 1 − sup |∇p(x)|(
−∆p(x) φ ≤ a(0)L1 + f (0) ≤ g(x)a(φ) + f (φ),
2R0 < |x| ≤ R .
(1.5)
If k is sufficiently large, then
a(eαk − 1) ≥ 1,
f (eαk − 1) ≥ 1,
b(eαk − 1) ≥ 1,
h(eαk − 1) ≥ 1
where k is dependent on a, f, b, h, p, and independent on R. Since
2R0 N −1 αk (
π
)
(ke ) p(2R0 ) − 1) cos(ε(r − 2R0 ) + )(L1 + 1)
r
2
π
N p+ αkp+
≤ ε(L1 + 1)2 k e
, 2R0 −
< |x| < 2R0 .
2ε
−∆p(x) φ = ε(
+
+
Let ε = 2−N k −p e−αkp . Then
−∆p(x) φ ≤ L1 + 1 ≤ g(x)a(φ) + f (φ), 2R0 −
π
< |x| < 2R0 .
2ε
(1.6)
EJDE-2007/177
EXISTENCE OF POSITIVE SOLUTIONS
5
Obviously,
−∆p(x) φ = 0 ≤ L1 + 1 ≤ g(x)a(φ) + f (φ), |x| < 2R0 −
π
.
2ε
(1.7)
Since φ(x) ∈ C 1 (Ω), combining (1.5), (1.6), (1.7), we have
−∆p(x) φ ≤ g(x)a(φ) + f (φ)
for a.e. x ∈ Ω. Similarly we have
−∆p(x) φ ≤ g(x)b(φ) + h(φ),
for a.e. x ∈ Ω. Let (φ1 , φ2 ) = (φ, φ), since φ(x) ∈ C 1 (Ω̄), it is easy to see that
(φ1 , φ2 ) is a subsolution of (1.3).
Step 2. We construct a supersolution of (1.3) Let z1 be a radial solution of
−∆p(x) z1 (x) = (L2 + 1)µ,
z1 = 0
in Ω,
on ∂Ω .
We denote z1 = z1 (r) = z1 (|x|), then z1 satisfies
−(rN −1 |z10 |p(r)−2 z10 )0 = rN −1 (L2 + 1)µ, z1 (R) = 0, z10 (0) = 0 .
Then
z10 = −|
and
Z
1
r(L2 + 1)µ p(r)−1
|
,
N
(1.8)
R
1
r(L2 + 1)µ p(r)−1
dr.
|
N
r
We denote β = β((L2 + 1)µ) = max0≤r≤R z1 (r), then
Z R
Z R
1
1
1
r(L2 + 1)µ p(r)−1
r
p(q)−1
β((L2 + 1)µ) =
|
|
dr = ((L2 + 1)µ)
| | p(r)−1 dr,
N
N
0
0
RR r
1
where q ∈ [0, 1]. Since 0 | N | p(r)−1 dr is a constant, then there exists a positive
constant C ≥ 1 such that
1
1
1
((L2 + 1)µ) p+ −1 ≤ β((L2 + 1)µ) = max z1 (r) ≤ C((L2 + 1)µ) p− −1 . (1.9)
0≤r≤R
C
We consider
z1 =
|
−∆p(x) z1 = (L2 + 1)µ in Ω
−∆p(x) z2 = (L2 + 1)h(β((L2 + 1)µ))
z1 = z2 = 0
in Ω
on ∂Ω .
Then we shall prove that (z1 , z2 ) is a supersolution for (1.3). For ξ ∈ W 1,p(x) (Ω)
with ξ ≥ 0, it is easy to see that
Z
Z
|∇z2 |p(x)−2 ∇z2 · ∇ξdx = (L2 + 1)h(β((L2 + 1)µ))ξdx
Ω
ZΩ
Z
≥
L2 h(β((L2 + 1)µ))ξdx +
h(z1 )ξdx.
Ω
Ω
Similar to (1.9), we have
1
max z2 (r) ≤ C[(L2 + 1)h(β((L2 + 1)µ))] (p− −1) .
0≤r≤R
6
G. A. AFROUZI AND H. GHORBANI
EJDE-2007/177
By (H3), for µ large enough we have
1
h(β((L2 + 1)µ)) ≥ b(C[(L2 + 1)h(β((L2 + 1)µ))] p− −1 ) ≥ b(z2 ).
Hence
Z
p(x)−2
|∇z2 |
Z
∇z2 · ∇ξdx ≥
Z
g(x)b(z2 )ξdx +
Ω
h(z1 )ξdx,
Ω
(1.10)
Ω
Also
Z
|∇z1 |p(x)−2 ∇z1 · ∇ξdx =
Ω
Z
(L2 + 1)µξdx.
Ω
By (H3), (H4), when µ is sufficiently large, according to (1.9), we have
(L2 + 1)µ ≥ [
−
1
β((L2 + 1)µ)]p −1
C
1
1
≥ L2 a(β((L2 + 1)µ)) + f [C[(L2 + 1) (p− −1) (h(β((L2 + 1)µ))) (p− −1) ]
≥ g(x)a(z1 ) + f (z2 ),
then
Z
|∇z1 |p(x)−2 ∇z1 · ∇ξdx ≥
Z
Ω
Z
g(x)a(z1 )ξdx +
f (z2 )ξdx.
Ω
(1.11)
Ω
According to (1.10) and (1.11), we can conclude that (z1 , z2 ) is a supersolution of
(1.3).
Let µ be sufficiently large, then from (1.8) and the definition of (φ1 , φ2 ), it is
easy to see that φ1 ≤ z1 and φ2 ≤ z2 . This completes the proof.
Now we consider the problem
−∆p(x) u = λF (x, u, v)
in Ω
−∆p(x) v = λG(x, u, v)
in Ω
(1.12)
u = v = 0 on ∂Ω.
If p(x) ≡ p (a constant), because of the homogenity of p-Laplacian, (1.3) and
(1.4) can be transformed into each other; but, if p(x) is a general function, since
p(x)-Laplacian is nonhomogeneous, they cannot be transformed into each other.
So we can see that p(x)-Laplacian problem is more complicated than than that of
p-Laplacian, and it is necessary to discuss the problem (1.4) separately.
Theorem 1.2. If p(x) ∈ C 1 (Ω̄), Ω = B(0, R), and (H3)–(H5) hold, then there
exists a λ∗ which is sufficiently large, such that (1.4) possesses a positive solution
for any λ ≥ λ∗ .
Proof. We construct a subsolution of (1.4). Let β ≤
|p(r1 ) − p(r2 )| ≤
R
4
satisfy
1
, ∀r1 , r2 ∈ [R − 2β, R].
2
(1.13)
In the following we denote
δ = min{
inf p(x) − 1
},
4(sup |∇p(x)| + 1)
p+
∗ =
sup
p(x),
p−
∗ =
R−2β≤|x|≤R
inf
p(x),
R−2β≤|x|≤R
b = min{a(0)L1 + f (0), b(0)L1 + h(0), −1}.
(1.14)
EJDE-2007/177
EXISTENCE OF POSITIVE SOLUTIONS
Let α ∈ (0, β], and set
 −k(r−R)
e
− 1,



R R−α αk p(R−α)−1 (R−α)N −1

αk


e −1+ r
(ke ) p(r)−1 [ rN −1


1
π
φ(r) = × sin(ε(r − (R − α)) + 2 )(L1 + 1)] p(r)−1 dr,

p(R−α)−1
R R−α

N −1


eαk − 1 + R−α− π (keαk ) p(r)−1 [ (R−α)

N −1
r

2ε

1

× sin(ε(r − (R − α)) + π2 )(L1 + 1)] p(r)−1 dr,
7
R − α < r ≤ R,
R − 2β < r ≤ R − α,
r ≤ R − 2β,
π
where ε = 2(2β−α)
which satisfies ε(R − 2β − (R − α)) + π2 = 0.
In the following, we will prove that (φ, φ) is a subsolution of (1.4). Since
 −k(r−R)
e
− 1,
R − α < r ≤ R,



p(R−α)−1

αk

p(r)−1

−(ke )
N −1
φ0 (r) = [ (R−α)
r N −1

1



× sin(ε(r − (R − α)) + π2 )(L1 + 1)] p(r)−1 dr, R − 2β < r ≤ R − α,



0,
r ≤ R − 2β.
It is easy to see that φ ≥ 0 is decreasing and φ ∈ C 1 ([0, R]), φ(x) = φ(|x|) ∈ C 1 (Ω).
Let r = |x|. By computation,
 −k(r−R) p(r)−1
(ke
)
[−k(p(r) − 1)



+p0 (r) ln k − kp0 (r)(r − R) + N −1 ], R − α < r ≤ R,


r

−∆p(x) φ(x) = ε( R−α )N −1 (keαk )(p(R−α)−1)
r



× cos(ε(r − (R − α)) + π2 )(L1 + 1), R − 2β < r ≤ R − α,



0,
r ≤ R − 2β.
If k is sufficiently large, when R − α < r ≤ R, then we have
ln k
N −1
−∆p(x) φ ≤ −k p(r) [inf p(x) − 1 − sup |∇p(x)|(
+ R − r) +
] ≤ −k p(r) δ.
k
kr
If k satisfies
−
k p∗ δ = −λb,
(1.15)
and since φ(x) ≥ 0 and a, f is monotone, it means that
−∆p(x) φ ≤ λ(a(0)L1 + f (0)) ≤ λ(g(x)a(φ) + f (φ)), R − α < |x| ≤ R.
(1.16)
From (H3), (H4) there exists a positive constant M such that a(M − 1) ≥ 1,
f (M − 1) ≥ 1, b(M − 1) ≥ 1, h(M − 1) ≥ 1. Let
αk = ln M.
(1.17)
Since
R − α N −1 αk (
π
)
(ke ) p(R − α) − 1) cos(ε(r − (R − α)) + )(L1 + 1)
r
2
N
αk p+
−1
∗
≤ ε(L1 + 1)2 (ke )
, R − 2β < |x| < R − α,
−∆p(x) φ(x) = ε(
if
+
ε2N (keαk )p∗ −1 ≤ λ,
(1.18)
then
−∆p(x) φ(x) ≤ λ(L1 + 1) ≤ λ(g(x)a(φ) + f (φ)),
R − 2β < |x| < R − α. (1.19)
8
G. A. AFROUZI AND H. GHORBANI
EJDE-2007/177
Obviously
−∆p(x) φ(x) = 0 ≤ λL1 + 1 ≤ λ(g(x)a(φ) + f (φ)),
|x| < R − 2β .
(1.20)
Combining (1.15), (1.17) and (1.18), we only need
+
−b p∗p−−1 p+
λ| ∗ M ∗ −1 ≤ λ,
δ
and according to (1.13), (1.14), we only need
ε2N |
+
+
π
−b p∗ −−1
( 2N M p∗ −1 | | p∗ )2p−
∗ ≤ λ.
β
δ
Let
+
+
−b p∗ −−1 −
π
λ∗ = ( 2N M p∗ −1 | | p∗ )2p∗ .
β
δ
If λ ≥ λ∗ is sufficiently large, then (1.18) is satisfied.
Since φ(x) = φ(|x|) ∈ C 1 (Ω), according to (1.16), (1.19) and (1.20), it is easy to
see that if λ is sufficiently large, then (φ1 , φ2 ) is a subsolution of (1.4).
Step 2. We construct a supersolution of (1.4). Similar to the proof of Theorem
1.1, we consider
−∆p(x) z1 = λ(L2 + 1)µ in Ω
−∆p(x) z2 = λ(L2 + 1)h(β(λ(L2 + 1)µ))
z1 = z2 = 0
in Ω
on ∂Ω ,
where β = β(λ(L2 + 1)µ) = max0≤r≤R z1 (r). It is easy to see that
Z
Z
|∇z2 |p(x)−2 ∇z2 · ∇ξdx =
λ(L2 + 1)h(β(λ(L2 + 1)µ))ξdx
Ω
Ω
Z
Z
≥
λL2 h(β(λ(L2 + 1)µ))ξdx +
λh(z1 )ξdx.
Ω
Ω
Similar to (1.9), we have
1
max z2 (r) ≤ C[λ(L2 + 1)h(β(λ(L2 + 1)µ))] (p− −1) .
0≤r≤R
By (H3) for µ large enough we have
1
h(β(λ(L2 + 1)µ)) ≥ b(C[λ(L2 + 1)h(β(λ(L2 + 1)µ))] p− −1 ) ≥ b(z2 ).
Hence
Z
|∇z2 |p(x)−2 ∇z2 · ∇ξdx ≥
Ω
Z
Z
λg(x)b(z2 )ξdx +
Ω
Also
Z
p(x)−2
|∇z1 |
λh(z1 )ξdx.
(1.21)
Ω
Z
∇z1 · ∇ξdx =
Ω
λ(L2 + 1)µξdx.
Ω
By (H3), (H4), when µ is sufficiently large, according to (1.9), we have
(L2 + 1)µ ≥
−
1 1
[ β(λ(L2 + 1)µ)]p −1
λ C
1
≥ L2 a(β(λ(L2 + 1)µ)) + f (C[λ(L2 + 1)h(β(λ(L2 + 1)µ))] (p− −1) ) .
EJDE-2007/177
EXISTENCE OF POSITIVE SOLUTIONS
9
Then
Z
p(x)−2
|∇z1 |
Ω
Z
∇z1 · ∇ξdx ≥
Z
λg(x)a(z1 )ξdx +
Ω
λf (z2 )ξdx.
(1.22)
Ω
According to (1.21) and (1.22), we can conclude that (z1 , z2 ) is a supersolution of
(1.4).
Similar to the proof of Theorem 1.1, if µ is sufficiently large, we have φ1 ≤ z1
and φ2 ≤ z2 . This completes the proof.
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Ghasem A. Afrouzi
Department of Mathematics, Faculty of Basic Sciences, Mazandaran University, Babolsar, Iran
E-mail address: afrouzi@umz.ac.ir
Horieh Ghorbani
Department of Mathematics, Faculty of Basic Sciences, Mazandaran University, Babolsar, Iran
E-mail address: seyed86@yahoo.com