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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 621621, 19 pages
doi:10.1155/2008/621621
Research Article
The Method of Subsuper Solutions for Weighted
pr-Laplacian Equation Boundary Value Problems
Qihu Zhang,
1, 2
Xiaopin Liu,
2
and Zhimei Qiu
2
1
Department of Mathematics and Information Science, Zhengzhou University of Light Industry,
Zhengzhou, Henan 450002, China
2
School of Mathematics Science, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China
Correspondence should be addressed to Zhimei Qiu,
Received 23 May 2008; Accepted 21 August 2008
Recommended by Marta Garcia-Huidobro
This paper investigates the existence of solutions for weighted pr-Laplacian ordinary boundary
value problems. Our method is based on Leray-Schauder degree. As an application, we give the
existence of weak solutions for px-Laplacian partial differential equations.
Copyright q 2008 Qihu Zhang et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
In this paper, we consider the existence of solutions for the following weighted pr-Laplacian
ordinary equation with right-hand terms depending on the first-order derivative:
−
wr
u
pr−2
u
f
r, u,
wr
1/pr−1
u
0, ∀r ∈
T
1
,T
2
, P
with one of the following boundary value conditions:
u
T
1
c, u
T
2
d, 1.1
g
u
T
1
,
w
T
1
1/pT
1
−1
u
T
1
0,u
T
2
d, 1.2
g
u
T
1
,
w
T
1
1/pT
1
−1
u
T
1
0,h
u
T
2
,
w
T
2
1/pT
2
−1
u
T
2
0, 1.3
u
T
1
u
T
2
,w
T
1
u
T
1
pT
1
−2
u
T
1
w
T
2
u
T
2
pT
2
−2
u
T
2
, 1.4
where p ∈ CT
1
,T
2
, R and pr > 1; w ∈ CT
1
,T
2
, R satisfies 0 <wr, ∀r ∈ T
1
,T
2
,
and wr
−1/pr−1
∈ L
1
T
1
,T
2
; −wr|u
|
pr−2
u
is called the weighted pr-Laplacian; the
2 Journal of Inequalities and Applications
notation wT
1
1/pT
1
−1
u
T
1
means lim
r → T
1
wr
1/pr−1
u
r exists and
w
T
1
1/pT
1
−1
u
T
1
: lim
r → T
1
wr
1/pr−1
u
r, 1.5
similarly
w
T
2
1/pT
2
−1
u
T
2
: lim
r → T
−
2
wr
1/pr−1
u
r; 1.6
where gx, y and hx, y are continuous and increasing in y for any fixed x, respectively.
The study of differential equations and variational problems with nonstandard pr-
growth conditions is a new and interesting topic. Many results have been obtained on these
kinds of problem, for example, 1–18.Ifwr ≡ pr ≡ p a constant, P is the well-known
p-Laplacian problem. Because of the nonhomogeneity of px-Laplacian, px-Laplacian
problems are more complicated than those of p-Laplacian, many methods and results for
p-Laplacian problems are invalid for px-Laplacian problems. For example,
1 if Ω ⊂ R
n
is an open bounded domain, then the Rayleigh quotient
λ
px
inf
u∈W
1,px
0
Ω\{0}
Ω
1/px
|∇u|
px
dx
Ω
1/px
|u|
px
dx
1.7
is zero in general, and only under some special conditions λ
px
> 0 see 4, but the fact that
λ
p
> 0 is very important in the study of p-Laplacian problems. In 19, the author considers
the existence and nonexistence of positive weak solution to the following quasilinear elliptic
system:
−Δ
p
u λfu, vλu
α
v
γ
in Ω,
−Δ
q
v λgu, vλu
δ
v
β
in Ω,
u v 0on∂Ω,
S
the fi rst eigenfunction is used to constructing the subsolution of problem S successfully.
On the px-Laplacian problems, maybe px-Laplacian does not have the first eigenvalue
and the first eigenfunction. Because of the nonhomogeneity of px-Laplacian, the first
eigenfunction cannot be used to construct the subsolution of px-Laplacian problems, even
if the first eigenfunction of px-Laplacian exists.On the existence of solutions for px-
Laplacian equations Dirichlet problems via subsuper solution methods, we refer to 13, 14;
2 if wr ≡ pr ≡ p a constant and −Δ
p
u>0, then u is concave, this property is
used extensively in the study of one-dimensional p-Laplacian problems, but it is invalid for
−Δ
pr
. It is another difference on −Δ
p
and −Δ
pr
: −|u
|
pr−2
u
;
3 on the existence of solutions of the typical pr-Laplacian problem:
−
u
pr−2
u
|u|
qr−2
u C, r ∈ 0, 1, 1.8
because of the nonhomogeneity of pt-Laplacian, when we use critical point theory to deal
with the existence of solutions, we usually need the corresponding functional is coercive or
satisfy Palais-Smale conditions. If 1 ≤ max
r∈0,1
qr < min
r∈0,1
pr, then the corresponding
functional is coercive, if max
r∈0,1
pr < min
r∈0,1
qr, then the corresponding functional
Qihu Zhang et al. 3
satisfies Palais-Smale conditions see 3. But if min
r∈0,1
pr ≤ qr ≤ max
r∈0,1
pr,
one can see that the corresponding functional is neither coercive nor satisfying Palais-Smale
conditions, the results on this case are rare.
There are many papers on the existence of solutions for p-Laplacian boundary value
problems via subsuper solution method see 20–24. But results on the sub-super-solution
method for px-Laplacian equations and systems are rare. In this paper, when pr is
a general function, we establish several sub-super-solution theorems for the existence of
solutions for weighted pr-Laplacian equation with Dirichlet, Robin, and Periodic boundary
value conditions. Moreover, the case of min
r∈0,1
pr ≤ qr ≤ max
r∈0,1
pr is discussed.
Our results partially generalize the results of 13, 14, 20, 25.
Let T
1
<T
2
and I T
1
,T
2
, the function f : I × R × R → R is assumed to be
Caratheodory, by this we mean the following:
i for almost every t ∈ I, the function ft, ·, · is continuous;
ii for each x, y ∈ R × R, the function f·,x,y is measurable on I;
iii for each ρ>0, there is a α
ρ
∈ L
1
I,R such that, for almost every t ∈ I and every
x, y ∈ R × R with |x|≤ρ, |y|≤ρ, one has
ft, x, y
≤ α
ρ
t. 1.9
We set C CI,R, C
1
{u ∈ C | u
is continuous in T
1
,T
2
, lim
r → T
1
wr|u
|
pr−2
u
r
and lim
r → T
−
2
wr|u
|
pr−2
u
r exist}. Denote u
0
sup
r∈T
1
,T
2
|ur| and u
1
u
0
wr
1/pr−1
u
0
. The spaces C and C
1
will be equipped with the norm ·
0
and ·
1
,
respectively.
We say a function u : I → R is a solution of P,ifu ∈ C
1
and wr|u
|
pr−2
u
r is
absolutely continuous and satisfies P almost every on I.
Functions α, β ∈ C
1
are called subsolution and supersolution of P,if|α
|
pr−2
α
r and
|β
|
pr−2
β
r are absolutely continuous and satisfy
−
wr
α
pr−2
α
f
r, α,
wr
1/pr−1
α
≤ 0, a.e. on I,
−
wr
β
pr−2
β
f
r, β,
wr
1/pr−1
β
≥ 0, a.e. on I.
1.10
Throughout this paper, we assume that α ≤ β are subsolution and supersolution,
respectively. Denote
Ω
0
t, x | t ∈ I, x ∈
αt,βt
,
Ω
1
t, x, y | t ∈ I, x ∈
αt,βt
,y∈ R
.
1.11
We also assume that
H
1
|ft, x, y|≤A
1
t, xK
1
t, x, yA
2
t, xK
2
t, x, y, for all t, x, y ∈ Ω
1
, where
A
i
t, xi 1, 2 are positive value and continuous on Ω
0
, K
i
t, x, yi 1, 2 are positive
value and continuous on Ω
1
.
H
2
There exist positive numbers M
1
and M
2
such that K
1
t, x, y ≤|y|φ|y|,
K
2
t, x, y ≤ M
1
φ|y|, for |y|≥M
2
, where φ ∈ C1, ∞, 1, ∞ is increasing and satisfies
∞
1
1/φy
1/p
−
−1
dy ∞, where p
−
min
r∈I
pr.
4 Journal of Inequalities and Applications
Our main results are as the following theorem.
Theorem 1.1. If f is Caratheodory and satisfies (H
1
) and (H
2
), α and β satisfy αT
1
≤ c ≤ βT
1
,
αT
2
≤ d ≤ βT
2
,thenP with 1.1 possesses a solution.
Theorem 1.2. If f is Caratheodory and satisfies (H
1
) and (H
2
), α and β satisfy αT
2
≤ d ≤ βT
2
,
and
g
α
T
1
,
w
T
1
1/pT
1
−1
α
T
1
≥ 0 ≥ g
β
T
1
,
w
T
1
1/pT
1
−1
β
T
1
, 1.12
then P with 1.2 possesses a solution.
Theorem 1.3. If f is Caratheodory and satisfies (H
1
) and (H
2
), α and β satisfy
g
α
T
1
,
w
T
1
1/pT
1
−1
α
T
1
≥ 0 ≥ g
β
T
1
,
w
T
1
1/pT
1
−1
β
T
1
,
h
α
T
2
,
w
T
2
1/pT
2
−1
α
T
2
≤ 0 ≤ h
β
T
2
,
w
T
2
1/pT
2
−1
β
T
2
,
1.13
then P with 1.3 possesses a solution.
Theorem 1.4. If f is Caratheodory and satisfies (H
1
) and (H
2
), α and β satisfy
α
T
1
α
T
2
<β
T
1
β
T
2
,
w
T
1
α
T
1
pT
1
−2
α
T
1
≥ w
T
2
α
T
2
pT
2
−2
α
T
2
,
w
T
1
β
T
1
pT
1
−2
β
T
1
≤ w
T
2
β
T
2
pT
2
−2
β
T
2
,
1.14
then P with 1.4 possesses a solution.
As an application, we consider the existence of weak solutions for the following px-
Laplacian partial differential equation:
−div
|∇u|
px−2
∇u
f
x, u, |x|
n−1/px−1
|∇u|
0, ∀x ∈ Ω, 1.15
where Ω is a bounded symmetric domain in R
n
, p ∈ CΩ; R is radially symmetric. We will
write pxp|x|pr,andpr satisfies 1 <pr ∈ C, f ∈ C
Ω × R × R, R is radially
symmetric with respect to x, namely, fx, u, vf|x|,u,vfr, u, v,andf satisfies the
Caratheodory condition.
2. Preliminary
Denote ϕr, x|x|
pr−2
x, ∀r, x ∈ I × R. Obviously, ϕ has the following properties.
Lemma 2.1. ϕ is a continuous function and satisfies
i for any r ∈ T
1
,T
2
, ϕr, · is strictly increasing;
ii ϕr, · is a homeomorphism from R to R for any fixed r ∈ I.
Qihu Zhang et al. 5
For any fixed r ∈ I, denote ϕ
−1
r, · as
ϕ
−1
r, x|x|
2−pr/pr−1
x, for x ∈ R \{0},ϕ
−1
r, 00. 2.1
It is clear that ϕ
−1
r, · is continuous and send bounded sets into bounded sets.
Let us now consider the simple problem
wrϕ
r, u
r
fr, 2.2
with boundary value condition 1.1, where f ∈ L
1
.Ifu is a solution of 2.2 with 1.1,by
integrating 2.2 from T
1
to r,wefindthat
wrϕ
r, u
r
w
T
1
ϕ
T
1
,u
T
1
r
T
1
ftdt. 2.3
Denote
Ffr
r
T
1
ftdt, a w
T
1
ϕ
T
1
,u
T
1
, 2.4
then
uru
T
1
r
T
1
ϕ
−1
r,
wr
−1
a Ffr
dr. 2.5
The boundary conditions imply that
T
2
T
1
ϕ
−1
r,
wr
−1
a Ffr
dr d − c. 2.6
For fixed h ∈ C, we denote
Λ
h
a
T
2
T
1
ϕ
−1
r,
wr
−1
a hr
dr c − d. 2.7
We have the following lemma.
Lemma 2.2. The function Λ
h
has the following properties. i For any fixed h ∈ C, the equation
Λ
h
a0 2.8
has a unique solution ah ∈ R.
ii The function a : C → R, defined in (i), is continuous and sends bounded sets to bounded
sets.
Proof. i Obviously, for any fixed h ∈ C, Λ
h
· is continuous and strictly increasing, then, if
2.8 has a solution, it is unique.
Since wr
−1/pr−1
∈ L
1
T
1
,T
2
and h ∈ C,itiseasytoseethat
lim
a → ∞
Λ
h
a∞, lim
a →−∞
Λ
h
a−∞. 2.9
6 Journal of Inequalities and Applications
It means the existence of solutions of Λ
h
a0.
In this way, we define a function ah : CT
1
,T
2
→ R, which satisfies
T
2
T
1
ϕ
−1
r,
wr
−1
ahhr
dr 0. 2.10
ii We claim that
ah
≤
|c − d|
T
2
T
1
ϕ
−1
r,
wr
−1
dr
1
p
1
h
0
, ∀h ∈ C. 2.11
If it is false. Without loss of generality, we may assume that there are some h ∈ C such
that
ah >
|c − d|
T
2
T
1
ϕ
−1
r,
wr
−1
dr
1
p
1
h
0
, 2.12
then
ahh>
|c − d|
T
2
T
1
ϕ
−1
r,
wr
−1
dr
1
p
1
,
T
2
T
1
ϕ
−1
r,
wr
−1
ahhr
dr d − c
>
|c − d|
T
2
T
1
ϕ
−1
r,
wr
−1
dr
1
T
2
T
1
ϕ
−1
r,
wr
−1
dr d − c
|c − d|
T
2
T
1
ϕ
−1
r,
wr
−1
dr d − c
> 0.
2.13
It is a contradiction. Thus, 2.11 is valid. It mens that a sends bounded sets to bounded
sets.
Finally, to show the continuity of a,let{u
n
} be a convergent sequence in C and u
n
→
u,asn → ∞. Obviously, {au
n
} is a bounded sequence, then it contains a convergent
subsequence {au
n
j
}.Letau
n
j
→ a
0
as j → ∞. Since
T
2
T
1
ϕ
−1
r,
wr
−1
a
u
n
j
u
n
j
r
dr 0, 2.14
letting j → ∞, we have
T
2
T
1
ϕ
−1
r,
wr
−1
a
0
ur
dr 0, 2.15
from i,wegeta
0
au, it means a is continuous.
This completes the proof.
Qihu Zhang et al. 7
Now, we define a : L
1
→ R is defined by
aha
Fh
. 2.16
It is clear that a is a continuous function which send bounded sets of L
1
into bounded
sets of R, and hence it is a complete continuous mapping.
We continue now with our argument previous to Lemma 2.2. By solving for u
in 2.3
and integrating, we find
uru
T
1
F
ϕ
−1
r,
wr
−1
afFfr
r. 2.17
Let us define
KhtF
ϕ
−1
r,
wr
−1
ahFh
t, ∀t ∈
T
1
,T
2
. 2.18
We denote by N
f
u : C
1
× T
1
,T
2
→ L
1
, the Nemytsky operator associated to f
defined by
N
f
urf
r, ur,
wr
1/pr−1
u
r
, a.e. on I. 2.19
It is easy to see the following lemma.
Lemma 2.3. u is a solution of P with boundary value condition 1.1 if and only if u is a solution
of the following abstract equation:
u c K
N
f
u
. 2.20
Lemma 2.4. The operator K is continuous and sends equi-integrable sets in L
1
into relatively compact
sets in C
1
.
Proof. It is easy to check that Kht ∈ C
1
. Since wr
−1/pr−1
∈ L
1
,and
wt
1/pt−1
Kh
tϕ
−1
t,
ahFh
, ∀t ∈
T
1
,T
2
, 2.21
it is easy to check that K is a continuous operator from L
1
to C
1
.
Let now U be an equi-integrable set in L
1
, then there exists ρ ∈ L
1
, such that
ut
≤ ρt a.e. in I, for any u ∈ U. 2.22
We want to show that
KU ⊂ C
1
is a compact set.
Let {u
n
} be a sequence in KU, then there exist a sequence {h
n
}∈U such that u
n
Kh
n
. For t
1
,t
2
∈ I, we have that
F
h
n
t
1
− F
h
n
t
2
≤
t
2
t
1
ρtdt
. 2.23
Hence, the sequence {Fh
n
} is uniformly bounded and equicontinuous, then there
exists a subsequence of {Fh
n
} which is convergent in C, and we name the same. Since
the operator a is bounded and continuous, we can choose a subsequence of {ah
n
Fh
n
}
which we still denote {ah
n
Fh
n
} that is convergent in C, then
wtϕ
t,
Kh
n
t
a
h
n
F
h
n
2.24
is convergent in C. Since
K
h
n
tF
wr
−1/pr−1
ϕ
−1
r,
a
h
n
F
h
n
t, ∀t ∈
T
1
,T
2
, 2.25
according to the continuous of ϕ
−1
and the integrability of wr
−1/pr−1
in L
1
, then Kh
n
is convergent in C. Then, we can conclude that {u
n
} convergent in C
1
.
8 Journal of Inequalities and Applications
Lemma 2.5. Let α, β ∈ C
1
be subsolution and supersolution of P, respectively, which satisfies
αt ≤ βt for any t ∈ T
1
,T
2
, then there exists a positive constant L such that, for any solution x of
P with 1.1 whichsatisfies αt ≤ xt ≤ βt,one has wt
1/pt−1
x
0
≤ L.
Proof. We denote
μ
0
T
2
T
1
A
1
t, xt
A
2
t, xt
dt, a
0
max
wr
1/pr−1
| r ∈
T
1
,T
2
,
σ max
βs − αt | t, s ∈ T
1
,T
2
,
γ max
wt
1/pt−1
A
1
t, x | t, x ∈ Ω
0
,
2.26
then there exists a t
0
∈ T
1
,T
2
such that
w
t
0
1/pt
0
−1
x
t
0
≤ a
0
x
t
0
≤ a
0
σ
T
2
− T
1
. 2.27
From H
2
, there exist positive numbers σ
1
and N
1
such that
N
1
≥ σ
1
≥ max
r∈I
M
2
a
0
σ
T
2
− T
1
1
pr
,
N
1
σ
1
1
φ
y
1/pr−1
dy > γσ M
1
μ
0
, for r ∈
T
1
,T
2
uniformly.
2.28
Assume that our conclusion is not true, combining 2.27, then there exists t
1
,t
2
⊂
T
1
,T
2
such that wr
1/pr−1
x
keeps the same sign on t
1
,t
2
, and
w
t
1
x
pt
1
−2
x
t
1
σ
1
,w
t
2
x
pt
2
−2
x
t
2
N
1
, 2.29
or inversely
w
t
1
x
pt
1
−2
x
t
1
−σ
1
,w
t
2
x
pt
2
−2
x
t
2
−N
1
. 2.30
For simplicity, we assume that the former appears. Hence,
γσ M
1
μ
0
<
N
1
σ
1
1
φ
y
1/pr−1
dy
t
2
t
1
wr
x
pr−1
φ
wr
x
pr−1
1/pr−1
dr
t
2
t
1
f
r, x,
wr
1/pr−1
x
φ
wr
1/pr−1
x
dr
≤
t
2
t
1
wr
1/pr−1
A
1
r, xr
x
dr M
1
μ
0
≤ γσ M
1
μ
0
,
2.31
which is impossible. T he proof is completed.
Qihu Zhang et al. 9
Let us consider the auxiliary SBVP of the form
wr
u
pr−2
u
f
r, Rr, u,R
1
wr
1/pr−1
u
R
2
r, u
def
fr, u,r∈
T
1
,T
2
,
2.32
where
Rt, u
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
βt,ut >βt,
u, αt ≤ ut ≤ βt,
αt,ut <αt,
R
1
y
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
L
1
,y>L
1
,
y, |y|≤L
1
,
−L
1
,y<−L
1
,
2.33
where
L
1
1 max
L, sup
r∈T
1
,T
2
wr
1/pr−1
β
r
, sup
r∈T
1
,T
2
wr
1/pr−1
α
r
, 2.34
where L is defined in Lemma 2.5,and
R
2
t, u
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
et, u
u − βt
1 u
2
ut >βt,
0,αt ≤ ut ≤ βt,
et, u
u − αt
1 u
2
ut <αt,
2.35
where et, u1 A
1
t, Rt, u A
2
t, Rt, u.
Lemma 2.6. Let the conditions of Lemma 2.5 hold, and let ut be any solution of SBVP with 1.1
satisfies αT
1
≤ c ≤ βT
1
and αT
2
≤ d ≤ βT
2
,thenαt ≤ ut ≤ βt, for any t ∈ T
1
,T
2
.
Proof. We will only prove that ut ≤ βt for any t ∈ T
1
,T
2
. The argument of the case of
αt ≤ ut for any t ∈ T
1
,T
2
is similar.
Assume that ut >βt for some t ∈ T
1
,T
2
, then there exist a t
0
∈ T
1
,T
2
and a
positive number δ such that ut
0
βt
0
δ, ut ≤ βtδ, for any t ∈ T
1
,T
2
. Hence,
w
t
0
1/pt
0
−1
u
t
0
w
t
0
1/pt
0
−1
β
t
0
. 2.36
There exists a positive number η such that ut >βt, for any t ∈ J :t
0
− η, t
0
η ⊂
T
1
,T
2
. From the definition of β, u, and
f we conclude that
wr
β
pr−2
β
≤ f
r, β,
wr
1/pr−1
β
fr, β <
fr, u on
t
0
− η
1
,t
0
η
1
, 2.37
10 Journal of Inequalities and Applications
where η
1
∈ 0,η is small enough. For any r ∈ t
0
,t
0
η
1
, we have
r
t
0
wr
β
pr−2
β
dr <
r
t
0
fr, udr
r
t
0
wr
u
pr−2
u
dr. 2.38
From 2.36 and 2.38, we have
β
pr−2
β
<
u
pr−2
u
on
t
0
,t
0
η
1
, 2.39
it means that
β δ
<u
on
t
0
,t
0
η
1
. 2.40
It is a contradiction to the definition of t
0
,sout ≤ βt, for any t ∈ T
1
,T
2
.
3. Proofs of main results
In this section, we will deal with the proofs of main results.
Proof of Theorem 1.1. From Lemmas 2.5 and 2.6, we only need to prove the existence of
solutions f or SBVP with 1.1. Obviously, u is a solution of SBVP with 1.1 if and only if
u is a solution of
u Φ
f
u : c K
N
f
u
. 3.1
We set
C
1
c,d
u ∈ C
1
| u
T
1
c, u
T
2
d
. 3.2
Obviously, N
f
u sends C
1
into equi-integrable sets in L
1
. Similar to the proof of
Lemma 2.4, we can conclude that K sends equi-integrable sets in L
1
into relatively compact
sets in C
1
, then Φ
f
u is compact continuous.
Obviously, for any u ∈ C
1
, we have Φ
f
u ∈ C
1
c,d
,andΦ
f
C
1
is bounded. By virtue
of Schauder fixed point theorem, Φ
f
u has at least one fixed point u in C
1
c,d
. Then, u is a
solution of SBVP with 1.1. This completes the proof.
Proof of Theorem 1.2. Let d with αT
2
≤ d ≤ βT
2
be fixed. According to Theorem 1.1, P with
the following boundary value condition:
u
1
T
1
α
T
1
,u
1
T
2
d, 3.3
possesses a solution u
1
such that
αt ≤ u
1
t ≤ βt, ∀t ∈
T
1
,T
2
. 3.4
Since lim
r → T
1
wr
u
1
pr−2
u
1
r exists, we have
u
1
r − u
1
T
1
r
T
1
wt
−1/pt−1
wt
1/pt−1
u
1
t
dt
w
T
1
1/pT
1
−1
u
1
T
1
r
T
1
wt
−1/pt−1
1 o1
dt.
3.5
Qihu Zhang et al. 11
Similarly,
αr − α
T
1
w
T
1
1/pT
1
−1
α
T
1
r
T
1
wt
−1/pt−1
1 o1
dt. 3.6
Obviously
0 ≤ lim
r → T
1
u
1
r − αr
r
T
1
wt
−1/pt−1
dt
w
T
1
1/pT
1
−1
u
1
T
1
−
w
T
1
1/pT
1
−1
α
T
1
, 3.7
then, we can conclude that
w
T
1
1/pT
1
−1
u
1
T
1
≥
w
T
1
1/pT
1
−1
α
T
1
. 3.8
Since u
1
T
1
αT
1
,andgx, y is increasing in y, we have
g
u
1
T
1
,
w
T
1
1/pT
1
−1
u
1
T
1
≥ g
α
T
1
,
w
T
1
1/pT
1
−1
α
T
1
≥ 0. 3.9
We may assume that gu
1
T
1
, wT
1
1/pT
1
−1
u
1
T
1
> 0, or we get a solution for P
with 1.2.
Since u
1
is a solution of P, it is also a subsolution of P. Similarly, P with boundary
value condition
v
1
T
1
β
T
1
,v
1
T
2
d, 3.10
possesses a solution v
1
such that
u
1
t ≤ v
1
t ≤ βt, ∀t ∈
T
1
,T
2
, 3.11
which satisfies
w
T
1
1/pT
1
−1
v
1
T
1
≤
w
T
1
1/pT
1
−1
β
T
1
, 3.12
then
g
v
1
T
1
,
w
T
1
1/pT
1
−1
v
1
T
1
≤ g
β
T
1
,
w
T
1
1/pT
1
−1
β
T
1
≤ 0. 3.13
Obviously, u
1
t and v
1
t are subsolution and supersolution of P with 1.2,
respectively. According to Theorem 1.1, P with boundary value condition
x
T
1
u
1
T
1
v
1
T
1
2
,x
T
2
d, 3.14
possesses a solution x such that
u
1
t ≤ xt ≤ v
1
t, ∀t ∈
T
1
,T
2
. 3.15
We may assume that gxT
1
, wT
1
1/pT
1
−1
x
T
1
/
0, or we get a solution for P
with 1.2.
12 Journal of Inequalities and Applications
If gxT
1
, wT
1
1/pT
1
−1
x
T
1
> 0, then denote u
2
txt and v
2
tv
1
t;if
gxT
1
, wT
1
1/pT
1
−1
x
T
1
< 0, then denote v
2
txt and u
2
tu
1
t.Itiseasyto
see that u
2
t and v
2
t both are solutions of P and satisfy
g
u
2
T
1
,
w
T
1
1/pT
1
−1
u
2
T
1
> 0 >g
v
2
T
1
,
w
T
1
1/pT
1
−1
v
2
T
1
,
u
2
t ≤ v
2
t, ∀t ∈
T
1
,T
2
,
u
2
t,v
2
t
⊆
u
1
t,v
1
t
, ∀t ∈
T
1
,T
2
,
u
2
T
2
d v
2
T
2
,
v
2
T
1
− u
2
T
1
v
1
T
1
− u
1
T
1
2
.
3.16
Repeated the step, we get two sequences {u
n
} and {v
n
}, all are solutions of P,and
satisfy
g
u
n
T
1
,
w
T
1
1/pT
1
−1
u
n
T
1
> 0 >g
v
n
T
1
,
w
T
1
1/pT
1
−1
v
n
T
1
, 3.17
u
n
t ≤ v
n
t, ∀t ∈
T
1
,T
2
,
u
n1
t,v
n1
t
⊆
u
n
t,v
n
t
, ∀t ∈
T
1
,T
2
, 3.18
u
n
T
2
d v
n
T
2
, 3.19
v
n1
T
1
− u
n1
T
1
v
n
T
1
− u
n
T
1
2
. 3.20
According to Lemma 2.5, {u
n
t} and {v
n
t} both are bounded in C
1
, then
{wT
1
1/pT
1
−1
u
n
T
1
} is a bounded set and has a convergent subsequence. Note that
{u
n
t} are solutions of P and satisfy
wrϕ
r, u
n
r
a
n
F
N
f
u
n
r, 3.21
where
F
N
f
u
n
r
r
T
1
N
f
u
n
dt, a
n
w
T
1
ϕ
T
1
,u
n
T
1
. 3.22
Similar to t he proof of Lemma 2.4, {u
n
t} possesses a convergent subsequence {u
n
i
t}
in C
1
, and then {a
n
} is bounded. From 2, we can see that {u
n
t} and {v
n
t} have uniform
C
1,α
regularity. We may assume that u
n
i
t → ut in C
1
and v
n
j
t → vt in C
1
.
It is easy to see that ut ≤ vt both are solutions of P. From the definition of {u
n
t}
and {v
n
t}, we can see that
u
T
2
d v
T
2
. 3.23
Combining 3.18 and 3.20, we have
ut ≤ vt, ∀t ∈
T
1
,T
2
,
uT
1
lim
j →∞
u
n
i
T
1
lim
j →∞
v
n
i
T
1
v
T
1
.
3.24
Similar to 3.7, we have
w
T
1
1/pT
1
−1
u
T
1
≤
w
T
1
1/pT
1
−1
v
T
1
. 3.25
Qihu Zhang et al. 13
From 3.17 and the continuity of g, we can see that
g
u
T
1
,
w
T
1
1/pT
1
−1
u
T
1
≥ 0 ≥ g
vT
1
,
w
T
1
1/pT
1
−1
v
T
1
. 3.26
From 3.25, 3.26 , and the increasing property of gx, y with respect to y, we have
g
u
T
1
,
w
T
1
1/pT
1
−1
u
T
1
0 g
v
T
1
,
w
T
1
1/pT
1
−1
v
T
1
. 3.27
Thus, u and v both are solutions of P with 1.2. This completes the proof.
Proof of Theorem 1.3. According to Theorem 1.2, P possesses a solution u
1
such that
g
u
1
T
1
,
w
T
1
1/pT
1
−1
u
1
T
1
0,
u
1
T
2
α
T
2
,
αt ≤ u
1
t ≤ βt, ∀t ∈
T
1
,T
2
.
3.28
Similar to the proof of 3.7, we have
w
T
2
1/pT
2
−1
u
1
T
2
≤
w
T
2
1/pT
2
−1
α
T
2
. 3.29
Obviously, hu
1
T
2
, wT
2
1/pT
2
−1
u
1
T
2
≤ 0. We may assume that
hu
1
T
2
, wT
2
1/pT
2
−1
u
1
T
2
< 0, 3.30
or we get a solution for P with 1.3, then u
1
is a subsolution of P with 1.3.
According to Theorem 1.2, P possesses a solution v
1
such that
g
v
1
T
1
,
w
T
1
1/pT
1
−1
v
1
T
1
0,
v
1
T
2
β
T
2
,
u
1
t ≤ v
1
t ≤ βt, ∀t ∈
T
1
,T
2
.
3.31
Similarly, hv
1
T
2
, wT
2
1/pT
2
−1
v
1
T
2
≥ 0. We may assume that
h
v
1
T
2
,
w
T
2
1/pT
2
−1
v
1
T
2
> 0, 3.32
or we get a solution for P with 1.3, then v
1
is a supersolution of P with 1.3.
According to Theorem 1.2, P possesses a solution x such that
g
x
T
1
,
w
T
1
1/pT
1
−1
x
T
1
0,x
T
2
u
1
T
2
v
1
T
2
2
,
u
1
t ≤ xt ≤ v
1
t, ∀t ∈
T
1
,T
2
.
3.33
We may assume that hxT
2
, wT
2
1/pT
2
−1
x
T
2
/
0, or we get a solution for P
with 1.3.IfhxT
2
, wT
2
1/pT
2
−1
x
T
2
> 0, then denote v
2
txt and u
2
tu
1
t,
14 Journal of Inequalities and Applications
if hxT
2
, wT
2
1/pT
2
−1
x
T
2
< 0, then denote v
2
tv
1
t and u
2
txt.Itiseasyto
see that u
2
t and v
2
t both are solutions of P and satisfy
h
u
2
T
2
,
w
T
2
1/pT
2
−1
u
2
T
2
< 0 <h
v
2
T
2
,
w
T
2
1/pT
2
−1
v
2
T
2
,
u
2
t ≤ v
2
t, ∀t ∈
T
1
,T
2
,
u
2
t,v
2
t
⊆
u
1
t,v
1
t
, ∀t ∈
T
1
,T
2
,
v
2
T
2
− u
2
T
2
v
1
T
2
− u
1
T
2
2
.
3.34
Repeating the step, similar to the proof of Theorem 1.2, we get two sequences {u
n
} and
{v
n
}, all are solutions of P, and satisfy
g
u
n
T
1
,
w
T
1
1/pT
1
−1
u
n
T
1
0 g
v
n
T
1
,
w
T
1
1/pT
1
−1
v
n
T
1
,
h
u
n
T
2
,
w
T
2
1/pT
2
−1
u
n
T
2
< 0 <h
v
n
T
2
,
w
T
2
1/pT
2
−1
v
n
T
2
,
u
n
t ≤ v
n
t, ∀t ∈
T
1
,T
2
,
u
n1
t,v
n1
t
⊆
u
n
t,v
n
t
, ∀t ∈
T
1
,T
2
.
v
n1
T
2
− u
n1
T
2
v
n
T
2
− u
n
T
2
2
.
3.35
Similar to the proof of Theorem 1.2, {u
n
t} and {v
n
t} possess convergent
subsequence {u
n
i
t} and {v
n
j
t} in C
1
, respectively. We may assume that u
n
i
t → ut
in C
1
, and similar v
n
j
t → vt in C
1
. It is easy to see that ut ≤ vt both are solutions of
P with 1.3. This completes the proof.
Proof of Theorem 1.4. According to Theorem 1.1, P possesses solution u
1
which satisfies
u
1
T
1
α
T
1
,u
1
T
2
α
T
2
,αt ≤ u
1
t ≤ βt,t∈
T
1
,T
2
. 3.36
We may assume that wT
1
ϕT
1
,u
1
T
1
/
wT
2
ϕT
2
,u
1
T
2
, or we get a solution for
P with 1.4, then wT
1
ϕT
1
,u
1
T
1
>wT
2
ϕT
2
,u
1
T
2
,andu
1
is a subsolution of P.
According to Theorem 1.1, P possesses solutions v
1
which satisfies
v
1
T
1
β
T
1
,v
1
T
2
β
T
2
,u
1
t ≤ v
1
t ≤ βt,t∈
T
1
,T
2
. 3.37
We may assume that wT
1
ϕT
1
,v
1
T
1
/
wT
2
ϕT
2
,v
1
T
2
, or we get a solution for
P with 1.4, then wT
1
ϕT
1
,v
1
T
1
<wT
2
ϕT
2
,v
1
T
2
,andv
1
is a supersolution of P.
According to Theorem 1.1, P possesses solutions x and satisfies
x
T
1
u
1
T
1
v
1
T
1
2
x
T
2
,u
1
t ≤ xt ≤ v
1
t,t∈
T
1
,T
2
. 3.38
Similar to the proof of Theorem 1.2,weobtainu and v that are solutions of P, which
satisfy
ut ≤ vt,t∈
T
1
,T
2
, 3.39
u
T
1
u
T
2
v
T
1
v
T
2
, 3.40
w
T
1
ϕ
T
1
,u
T
1
≥ w
T
2
ϕ
T
2
,u
T
2
, 3.41
w
T
1
ϕ
T
1
,v
T
1
≤ w
T
2
ϕ
T
2
,v
T
2
. 3.42
Qihu Zhang et al. 15
From 3.39 and 3.40, we have
w
T
1
ϕ
T
1
,u
T
1
≤ w
T
1
ϕ
T
1
,v
T
1
,
w
T
2
ϕ
T
2
,u
T
2
≥ w
T
2
ϕ
T
2
,v
T
2
.
3.43
From 3.41, 3.42,and3.43, we can conclude that P with 1.4 possesses a solution.
This completes the proof.
On the case of min
r∈−R,R
pr ≤ qr ≤ max
r∈−R,R
pr, we consider
−|u
|
pr−2
u
C|u|
qr−2
u er r ∈ −R, R,
u−RuR0,
I
where qr,er ∈ C−R, R, R
,min
r∈−R,R
pr ≤ qr ≤ max
r∈−R,R
pr, C is a positive
constant. Denote
p
max
r∈−R,R
pr,p
−
min
r∈−R,R
pr. 3.44
We have the following corollary.
Corollary 3.1. If p ∈ CR, 1, ∞ is even, R satisfies
R ≤
1 C max
r∈−R,R
er
−p
−1/p
p
−
−1
, 3.45
then I possesses at least a nontrivial solution.
Proof. It is easy to see that α ≡ 0 is a subsolution of I. Denote
βr1 −
r
0
|μs|
1/ps−1−1
μs ds, 3.46
where μ is a positive constant satisfying βR0. Since p is even, then β−R0. It is easy to
see that 0 ≤ βr ≤ 1, ∀r ∈ −R, R,and
−
β
pr−2
β
μ
R
0
|s|
1/ps−1
ds
1−pξ
≥
R
0
|s|
1/p
−1
ds
1−pξ
≥
R
0
|s|
1/p
−1
ds
1−p
−
≥ 1 C max
r∈−R,R
er ≥ C|β|
qr−2
β er,
3.47
where ξ ∈ −R, R. Then, β is a supersolution of I.FromTheorem 1.1, one can see that I
possesses at least a nontrivial solution.
16 Journal of Inequalities and Applications
4. Applications in PDE
Let Ω ⊂ R
n
be an open bounded domain. In this section, we always denote
p
max
x∈Ω
px,p
−
min
x∈Ω
px. 4.1
Let us now consider 1.15 with one of the following boundary value conditions:
u|
∂Ω
0, 4.2
∇u 0, ∀x ∈ ∂Ω. 4.3
If u is a radial solution of 1.15, then it can be transformed into
−
r
n−1
u
pr−2
u
r
n−1
f
r, u, |r|
n−1/pr−1
u
0,r∈
T
1
,T
2
, where T
1
≥ 0, 4.4
and the boundary value condition will be transformed into 1.1, 1.2,or1.3, respectively.
Theorem 4.1. If 4.4 has subsolution and supersolution α and β, respectively, satisfying αt ≤ βt
for any t ∈ T
1
,T
2
, and f is continuous and satisfies (H
1
)-(H
2
), in each of the following cases:
i 0 <T
1
<T
2
, Ω{x ∈ R
n
| T
1
< |x| <T
2
}, αT
1
≤ 0 ≤ βT
1
, and αT
2
≤ 0 ≤ βT
2
;
ii 0 T
1
<T
2
, Ω{x ∈ R
n
| T
1
< |x| <T
2
} B0; T
2
\{0}, and p
−
>n; αT
1
≤ 0 ≤
βT
1
, αT
2
≤ 0 ≤ βT
2
;
iii 0 T
1
<T
2
, Ω{x ∈ R
n
||x| <T
2
} B0; T
2
, and p
−
>n; wT
1
1/pT
1
−1
α
T
1
≥
0 ≥ wT
1
1/pT
1
−1
β
T
1
, αT
2
≤ 0 ≤ βT
2
;
then 1.15 with 4.2 has at least one weak radially symmetric solution u.
Proof. Notice that r
n−1
−1/pr−1
∈ L
1
0,T
2
and satisfies 0 <r
n−1
, ∀r ∈ 0,T
2
. We can
conclude the existence of solutions for 4.4 with 1.1, 1.2,or1.3, from Theorems 1.1,
1.2,and1.3. If lim
r → 0
r
n−1
|u
|
pr−2
u
r0, notice that
u
pr−2
u
r
≤ r
1−n
r
0
t
n−1
f
t, u, |t|
n−1/pt−1
u
dt
≤
r
0
f
t, u, |t|
n−1/pt−1
u
dt −→ 0 as r −→ 0,
4.5
then we have u
00. This completes the proof.
Similarly, we have the following theorem.
Theorem 4.2. If 4.4 has subsolution and supersolution α and β, respectively, satisfying αt ≤ βt
for any t ∈ T
1
,T
2
, and
w
T
1
1/pT
1
−1
α
T
1
≥ 0 ≥
w
T
1
1/pT
1
−1
β
T
1
,
w
T
2
1/pT
2
−1
α
T
2
≤ 0 ≤
w
T
2
1/pT
2
−1
β
T
2
,
4.6
Qihu Zhang et al. 17
and f is continuous and satisfies (H
1
)-(H
2
), in each of the following cases:
i 0 <T
1
<T
2
; Ω{x ∈ R
n
| T
1
< |x| <T
2
};
ii 0 T
1
<T
2
; Ω{x ∈ R
n
| T
1
< |x| <T
2
} B0; T
2
\{0} or ΩB0; T
2
; p ∈ C
1
Ω; R
and p
−
>n;
then 1.15 with 4.3 has at least one weak radially symmetric solution u.
On the case of p
−
≤ qx ≤ p
, we consider
−div
|∇u|
px−2
∇u
C|u − 1|
qx−2
u ex,x∈ Ω,
ux0,x∈ ∂Ω,
II
where Ω{x ∈ R
n
| 0 < |x| <R}, qx,ex ∈ CΩ, R
,2≤ n<p
−
≤ qx ≤ p
, C is a
positive constant.
We have the following corollary.
Corollary 4.3. If p ∈ CR
n
, 1, ∞ is radial, and R satisfies
R ≤ min
1,
1 −
1
2
p
−
− 1
p
−1
n − 3/2
1 C max
x∈Ω
ex
1/p
−
−3/2
, 4.7
then II possesses at least a nontrivial solution.
Proof. It is easy to see that α ≡ 0 is a subsolution of II. Denote
βr1 −
r
0
μs
−1/2
1/ps−1
ds, 4.8
where μ is a positive constant satisfying βR0. It is easy to see that 0 ≤ βr ≤ 1, ∀r ∈ 0,R,
and
−
r
n−1
β
pr−2
β
μ
n −
3
2
r
n−5/2
R
0
|s|
−1/2ps−1
ds
1−pξ
n −
3
2
r
n−5/2
≥
R
0
|s|
−1/2p
−
−1
ds
1−pξ
n −
3
2
r
n−1
≥
R
−1/2p
−
−11
1−p
−
1 −
1
2
p
−
− 1
p
−1
n −
3
2
r
n−1
≥ r
n−1
1 C max
x∈Ω
ex
≥ r
n−1
C|β − 1|
qx−2
β ex
,
4.9
where ξ ∈
Ω. Then, β is a supersolution of II.FromTheorem 4.1, one can see that II
possesses at least a nontrivial solution.
18 Journal of Inequalities and Applications
Acknowledgments
This work is partly supported by the National Science Foundation of China 10701066 and
10671084, China Postdoctoral Science Foundation 20070421107, and the Natural Science
Foundation of Henan Education Committee 2007110037.
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