Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 155, pp. 1–6.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
INFINITE SEMIPOSITONE PROBLEMS WITH INDEFINITE
WEIGHT AND ASYMPTOTICALLY LINEAR GROWTH
FORCING-TERMS
GHASEM A. AFROUZI, SALEH SHAKERI
Abstract. In this work, we study the existence of positive solutions to the
singular problem
−∆p u = λm(x)f (u) − u−α
u=0
in Ω,
on ∂Ω,
where λ is positive parameter, Ω is a bounded domain with smooth boundary,
0 < α < 1, and f : [0, ∞] → R is a continuous function which is asymptotically
p-linear at ∞. The weight function is continuous satisfies m(x) > m0 > 0,
kmk∞ < ∞. We prove the existence of a positive solution for a certain range
of λ using the method of sub-supersolutions.
1. Introduction
In this article, we consider the positive solution to the boundary-value problem
1
−∆p u = λm(x)f (u) − α in Ω,
(1.1)
u
u = 0 on ∂Ω,
where λ is positive parameter, ∆p u = div(|∇u|p−2 ∇u), p > 1, Ω is a bounded
domain with smooth boundary ∂Ω, 0 < α < 1, and f : [0, ∞] → R is a continuous
function which is asymptotically p-linear at ∞. The weight function m(x) satisfies
m(x) ∈ C(Ω) and m(x) > m0 > 0 for x ∈ Ω and also kmk∞ = l < ∞. We prove
the existence of a positive solution for a certain range of λ.
We consider problem (1.1) under the following assumptions.
(H1) There exist σ1 > 0, k > 0 and s0 > 1 such that f (s) ≥ σ1 sp−1 − k for all
s ∈ [0, s0 ];
(s)
(H2) lims→+∞ sfp−1
= 0 for some σ > 0.
Let F (u) := λm(x)f (u) − u1α . The case when F (0) < 0 (and finite) is referred
to in the literature as a semipositone problem. Finding a positive solution for a
semipositone problem is well known to be challenging (see [2, 10]). Here we consider
the more challenging case when limu→0+ F (u) = −∞. which has received attention
2000 Mathematics Subject Classification. 35J55, 35J25.
Key words and phrases. Infinite semipositone problem; indefinite weight; forcing term;
asymptotically linear growth; sub-supersolution method.
c
2013
Texas State University - San Marcos.
Submitted September 17, 2012. Published July 8, 2013.
1
2
G. A. AFROUZI, S. SHAKERI
EJDE-2013/155
very recently and is referred to as an infinite semipositone problem. However, most
of these studies have concentrated on the case when the nonlinear function satisfies a
sublinear condition at ∞ (see [11, 12, 14]). The only paper to our knowledge dealing
with an infinite semipositone problem with an asymptotically linear nonlinearity
is [7], where the author is restricted to the case p = 2. Also here the existence of
a positive solution is focused near λ1 /σ , where λ1 is the first eigenvalue of−∆.
See also [1, 15], where asymptotically linear nonlinearities have been discussed in
the case of a nonsingular semipositone problem and an infinite positone problem.
Recently, in the case when m(x) = 1 problem (1.1) has been studied by Shivaji et
al [8]. The purpose of this paper is to improve the result of [8] with weight m. We
shall establish our existence results via the method of sub and super-solutions.
Definition 1.1. We say that (ψ) (resp. Z) in W 1,p (Ω) ∩ C(Ω) are called a subsolution (resp. super-solution) of (1.1), if ψ satisfies
Z
Z
1
|∇ψ(x)|p−2 ∇ψ(x).∇w(x)dx ≤ (λm(x)f (ψ) − α )w(x)dx ∀w ∈ W,
ψ
Ω
Ω
(1.2)
ψ > 0 in Ω,
ψ=0
on ∂Ω,
and
Z
Ω
|∇Z|p−2 ∇Z.∇w(x)dx ≥
Z
(λm(x)f (Z) −
Ω
Z>0
Z=0
1
)w(x)dx
Zα
in Ω,
∀w ∈ W,
(1.3)
on ∂Ω,
where W = {ξ ∈ C0∞ (Ω) : ξ ≥ 0in Ω}
The following lemma was established by Cui [3].
Lemma 1.2 ([3]). If there exist sub-supersolutions ψ and Z, respectively, such that
ψ ≤ Z on Ω, Then (1.1) has a positive solution u such that ψ ≤ u ≤ Z in Ω.
In next Section, we will state and prove the existence of a positive solution for a
certain range of λ.
2. Main result
Our main result for problem (1.1) reads as follows.
Theorem 2.1. Assume (H1)–(H2). If there exist constants s∗0 (σ, Ω), J(Ω), λ, and
λ̂ > λ such that if s0 ≥ s∗0 and m0 σ1 /(lσ) ≥ J, then (1.1) has a positive solution for
λ ∈ [λ, λ̂]. Here µ1 is the principal eigenvalue of operator −∆p with zero Dirichlet
boundary condition.
Proof. By Anti-maximum principle [13], there exists ξ = ξ(Ω) > 0 such that the
solution zµ of
−∆p z − µ|z|p−2 z = −1
z=0
in Ω,
on ∂Ω,
for µ ∈ (µ1 , µ1 + ξ), is positive in Ω and is such that
outward normal vector at ∂Ω.
∂zµ
∂ν
< 0 on ∂Ω, where ν is
EJDE-2013/155
INFINITE SEMIPOSITONE PROBLEMS
3
∂z
Since zµ > 0 in Ω and ∂νµ < 0 there exist m > 0, A > 0, and δ > 0 such that
|∇zµ | ≥ m in Ωδ and zµ ≥ A in Ω \ Ωδ , where W = {x ∈ Ω : d(x, ∂Ω) ≤ δ}.
We prove the existence of a solution by the comparison method [4]. It is easy to
see that any sub-solution of
1
uα
on ∂Ω,
−∆p u = λm0 f (u) −
u=0
in Ω,
(2.1)
is a sub-solution of (1.1). Also any super-solution of
1
in Ω,
uα
on ∂Ω,
−∆p u = λlf (u) −
u=0
(2.2)
is a super-solution of (1.1), where l is as defined above.
We first construct a supersolution for(1.1).Let Z = Mλ ep , where Mλ 1 and
ep is the unique positive solution of
−∆p ep = 1
ep = 0
in Ω,
on ∂Ω,
Let f˜(s) = maxt∈[0,s] f (t). Then f (s) ≤ f˜(s), f˜(s) is increasing, and
f˜(u)
= 0.
u→+∞ up−1
lim
Hence, we can choose Mλ 1 such that
2σ ≥
f˜(Mλ kep k∞ )
(Mλ kep k∞ )p−1
Now let λ̂ = 1/ 2lσkep kp−1
. For λ ≤ λ̂,
∞
−∆p Z = Mλp−1 ≥
1
f˜(Mλ kep k∞ )
˜
p−1 ≥ λlf (Mλ ep ) ≥ λlf (Mλ ep ) ≥ λlf (Z) − Z α .
2σkep k∞
Thus, Z is a supersolution of (2.2); therefore Z is a supersolution of (1.1) Define
p
ψ := k0 zµp−1+α ,
where k0 > 0 is such that
αp
1
k0p−1+α
p−1 o
n mp (1 − α)(p − 1)pp−1 kk α zµp−1+α p
1+ 0
,
A .
p−1 ≤ min
p
(p − 1 + α)
p−1+α
2lσkep k∞
Then
∇ψ = k0
1−α
p
zµp−1+α ∇zµ ,
p−1+α
4
G. A. AFROUZI, S. SHAKERI
EJDE-2013/155
and
−∆p ψ = − div(|∇ψ|p−2 ∇ψ)
p−1
(1−α)(p−1)
p
= −k0p−1
div(zµ p−1+α |∇zµ |p−2 ∇zµ )
p−1+α
p−1 n
(1−α)(p−1)
p
(∇zµ p−1+α ).|∇zµ |p−2 ∇zµ
= −k0p−1
p−1+α
o
(1−α)(p−1)
+ zµ p−1+α ∆p zµ
p−1 n (1 − α)(p − 1) −αp
p
= −k0p−1
zµp−1+α |∇zµ |p
p−1+α
p−1+α
o
(1−α)(p−1)
+ zµ p−1+α (1 − µzµp−1 )
p−1 p(p−1)
p−1 (1−α)(p−1)
p
p
= k0p−1
µzµp−1+α − k0p−1
zµ p−1+α
p−1+α
p−1+α
−
k0p−1 pp−1 (1 − α)(p − 1)|∇zµ |p
αp
(2.3)
.
(p − 1 + α)p zµp−1+α
p
Now we let s∗0 (σ, Ω) = k0 kzµp−1+α k∞ . If we can prove that
p(p−1)
− ∆p ψ ≤ λm0 σ1 k0p−1 zµp−1+α − λk −
1
αp
,
(2.4)
k0α zµp−1+α
then (H1) implies that −∆p ψ ≤ λm0 f (ψ) − ψ1α , and ψ will be a subsolution of
(1.1).
We will now prove (2.4) by comparing terms in (2.3) and (2.4). Let λ =
p
µ( p−1+α
)p−1
.
m0 σ1
For λ ≥ λ,
k0p−1
Also since λ ≤ λ̂ =
p−1 p(p−1)
p(p−1)
p
µzµp−1+α ≤ λm0 σ1 k0p−1 zµp−1+α
p−1+α
(2.5)
1
,
2lσkep kp−1
∞
1
λk +
1
≤
αp
k0α zµp−1+α
+
αp
k0α zµp−1+α
=
k0p−1 h
αp
zµp−1+α
k
p−1
2lσkep k∞
αp
kk0α zµp−1+α i
)
p−1+α (1 +
k0
2lσkep kp−1
∞
(2.6)
1
Now in Ωδ , we have |∇zµ | ≥ m, and by (2.2),
αp
kk0α zµp−1+α
mp (1 − α)(p − 1)pp−1
(1
+
)]
≤
.
(p − 1 + α)p
k0p−1+α
2lσkep kp−1
∞
1
Hence
λk +
1
αp
k0α zµp−1+α
≤
k0p−1 pp−1 (1 − α)(p − 1)|∇zµ |p
αp
(p − 1 + α)p zµp−1+α
From (2.5), (2.7) it can be seen that (2.4) holds in Ωδ .
in Ωδ
(2.7)
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INFINITE SEMIPOSITONE PROBLEMS
5
We will now prove (2.4) holds also in Ω \ Ωδ . Since zµ ≥ A in Ω \ Ωδ and by (2.2)
and (2.6) we obtain
λk +
1
αp
p−1+α
k0α zµ
≤
k0p−1 αp
p−1+α
zµ
p−1
p−1 (1−α)(p−1)
p
p
zµ ≤ k0p−1
zµ p−1+α
p−1+α
p−1+α
(2.8)
in Ω \ Ωδ .
From (2.5) and (2.8), (2.4) holds also in Ω \ Ωδ . Thus ψ is a positive subsolution
of(1.1) if λ ∈ [λ, λ̂]. We can now choose Mλ 1 such that ψ ≤ Z. Let J(Ω) =
p−1
p
0 σ1
2kep kp−1
. if mlσ
≥ J it is easy to see that λ ≤ λ̂, and for λ ∈ [λ, λ̂]
∞ µ p−1+α
we have a positive solution. This completes the proof
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Ghasem Alizadeh Afrouzi
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: afrouzi@umz.ac.ir
6
G. A. AFROUZI, S. SHAKERI
EJDE-2013/155
Saleh Shakeri
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: s.shakeri@umz.ac.ir