Existence, nonexistence and multiplicity of positive solutions for singular boundary value problems involving φ-Laplacian
Chan-Gyun Kim
Department of Mathematics Education, Pusan National University,
Busan, 609-735, Korea
Abstract
In this paper, we establish the results on the existence, nonexistence and multiplicity of positive solutions to singular boundary value problems involving φ-Laplacian. Our approach is based on the fixed point index theory. The interesting point is that a result for the existence of three positive solutions is given.
*Keywords: three positive solutions; singular problem; φ-Laplacian *
1 Introduction
In this paper, we study the existence, nonexistence and multiplicity of positive solutions to the following problem
[TABLE]
where φ:R→R is an odd increasing homeomorphism, c,d∈C([0,1],(0,∞)), λ∈R+:=[0,∞) is a parameter, h∈C((0,1),R+)∖{0} and f∈C(R+,R+) with f(s)>0 for s>0.
Throughout this paper, the homeomorphism φ satisfies the following assumption:
there exist increasing homeomorphisms ψ1,ψ2:R+→R+ such that
φ(x)ψ1(y)≤φ(xy)≤φ(x)ψ2(y) for all x,y∈R+.
For convenience, we denote by Hξ the set
[TABLE]
where ξ:R+→R+ is an increasing homeomorphism, and we make the following notations:
f0:=s→0+limφ(s)f(s) and f∞:=s→∞limφ(s)f(s).
It is well known that
[TABLE]
and L1(0,1)∩C(0,1)⊊Hψ1⊆Hφ⊆Hψ2 (see, e.g., [1, Remark 1]).
Problem (1.1) arises naturally in studying radial solutions to the following quasilinear elliptic equation defined on an annular domain
[TABLE]
where Ω={x∈RN:R1<∣x∣<R2} with N≥2 and 0<R1<R2<∞, w∈C([R1,R2],(0,∞)) and k∈C((R1,R2),R+). Indeed, applying change of variables
v(x)=u(t) and ∣x∣=(R2−R1)t+R1,
problem (1.3) is transformed into problem (1.1) with
[TABLE]
and
[TABLE]
(see, e.g., [1, 2]).
For φ(s)=∣s∣p−2s with p>1, problem (1.1) has been extensively studied in the literature (see [3, 4, 5, 6, 7, 8, 9] for p=2 and [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20] for p>1). For example, when h∈Hφ and c≡d≡1, Agarwal, Lü and O’Regan [10] studied the existence and multiplicity of positive solutions to problem (1.1) under various assumptions on f0 and f∞. In [12], when h∈Hφ, c≡d≡1 and f(s) satisfies f(0)>0 and f∞=∞, it was shown that there exists λ∗>0 such that (1.1) has at least two positive solutions for λ∈(0,λ∗), one positive solution for λ=λ∗ and no positive solution for λ>λ∗. Recently, under the assumptions that h∈C1((0,1],(0,∞)) is strictly decreasing, h(t)≤Ct−η for some C>0 and η∈(0,1), c≡d≡1, f∈C(R+,R) is differentiable on (0,∞), f(0)<0, s→0+limsupsf′(s)<∞, f′(s)>0 for s>0, s→∞limf(s)=∞, s→∞limsp−1f(s)=0 and sqf(s) is nondecreasing on [a,∞) for some a>0 and q∈(0,p−1), Shivaji, Sim and Son [19] showed the uniqueness of positive solution to problem (1.1) for all large λ>0. For sign-changing weight h satisfying ∣h∣∈Hφ and c≡d≡1, Xu [20] studied the existence of a nontrivial solution to problem (1.1) for all small λ>0 under the assumptions that f∈C(R,R) is nondecreasing, f(0)>0 and f∞∈(0,∞).
For general φ satisfying (A), when c≡d≡1 and h∈L1(0,1)∩C(0,1), Bai and Chen [21] studied the existence of at least two positive solutions for λ belonging to an explicit interval under some assumptions on f satisfying f(0)=0. When c≡d≡1, h∈Hψ1 and either f0=f∞=∞ or f0=f∞=0, Lee and Xu [22] showed that there exist λ∗≥λ∗>0 such that (1.1) has at least two positive solutions for λ∈(0,λ∗), one positive solution for λ∈[λ∗,λ∗] and no positive solution for λ>λ∗. In [1], for nonnegative nonlinearity f=f(t,s) satisfying f(t0,0)>0 for some t0∈[0,1] and h∈Hφ, the existence of an unbounded solution component was shown and, under several assumptions on f at ∞, the existence, nonexistence and multiplicity of positive solutions were studied.
For more general φ which does not satisfy (A), when c≡d≡1 and 0≤h∈L1(0,1) with h≡0, Kaufmann and Milne [23] proved the existence of positive solution to problem (1.1) for all λ>0 under the assumptions on f which induces the sublinear nonlinearity provided φ(s)=∣s∣p−1s with p>1. For other interesting results, we refer the reader to [24, 25, 26] and the references therein.
The concavity of solutions plays a crucial role in defining a suitable cone so that the solution operator is well defined (see, e.g., [10, 12, 22] and the references therein). When c≡d≡1, it is easy to see that solutions to problem (1.1) are concave functions on [0,1]. However, it is not clear that the solutions are concave functions on [0,1], unless c≡d≡1. In order to overcome this difficulty, a lemma ([2, Lemma 2.4]) was proved, so that various results for the existence, nonexistence and multiplicity of positive solutions to problem (1.1) were proved in [2] when d is nondecreasing on [0,1] and h∈C[0,1] satisfies h≡0 on any subinterval of [0,1].
The aim of this paper is to improve on the results in [2] by assuming the weaker hypotheses on h and d than those in [2]. More precisely, the monotonicity of d is not assumed, and the weight function h may not be L1(0,1) and it can be vanished in some subinterval of (0,1). Furthermore, a result for the existence of three positive solutions is given, which does not appear in [2].
The rest of this paper is organized as follows. In Section 2, we establish preliminaries which are essential for proving our results in this paper. In Section 3, the main results are proved and an example to illustrate the results obtained in this paper is provided. Finally, the summary of this paper and future work are given in Section 4.
2 Preliminaries
In this section we give preliminaries which are essential for proving our results in this paper.
First, we introduce a solution operator related to problem (1.1). Let g∈Hφ∖{0} be given, and define a function νg:(0,1)→R by
νg(t)=νg1(t)−νg2(t) for t∈(0,1).
Here νg1 and νg2 are the functions defined by, for t∈(0,1),
[TABLE]
and
[TABLE]
It is easy to see that νg1 and νg2 are continuous and monotone functions on (0,1) satisfying
[TABLE]
Thus there exists an interval [σg1,σg2]⊊(0,1) satisfying νg(σ)=0 for all σ∈[σg1,σg2] (see [1]).
Define a function T:Hφ→C[0,1] by T(0)=0 and, for g∈Hφ∖{0},
[TABLE]
where σ=σ(g) is a zero of νg in (0,1), i.e.,
[TABLE]
We notice that, although σ=σ(g) is not necessarily unique, the right hand side of the equality in (2.1) does not depend on a particular choice of σ. In other words, T(g) is independent of the choice of σ∈[σg1,σg2] (see, e.g., [1] or [2]).
For g∈Hφ, consider the following problem
[TABLE]
For g=0, (2.3) has a unique zero solution due to the boundary conditions.
The usual maximum norm in a Banach space C[0,1] is denoted by
∥u∥∞:=t∈[0,1]max∣u(t)∣ for u∈C[0,1]
and let
[TABLE]
where c0:=t∈[0,1]minc(t)>0 and d0:=t∈[0,1]mind(t)>0. Recently, without the monotonicity of d, a result similar to [2, Lemma 2.4] was proved in [1].
Lemma 2.1**.**
([1, Lemma 2])*
Assume that (A) and g∈Hφ hold. Then T(g) is the unique solution to problem (2.3) and*
T(g)(t)≥min{t,1−t}ρ1∥T(g)∥∞* for t∈[0,1].*
From now on, we assume h∈Hφ∖{0}. Let
Ah:={x∈(0,1):h(y)=0 for all y∈(0,x)} and Bh:={x∈(0,1):h(y)=0 for all y∈(x,1)}.
For convenience, we use the following notations:
αh:=maxAh if Ah=∅ and αh:=0 if Ag=∅, where Ah={x∈(0,1):h(y)=0 for all y∈(0,x)};
βh:=minBh if Bh=∅ and βh:=1 if Bg=∅, where Bh={x∈(0,1):h(y)=0 for all y∈(x,1)};
αˉh:=max{x∈(0,1]:h(y)>0 for all y∈(αh,x)};
βˉh:=max{x∈[0,1):h(y)>0 for all y∈(x,βh)};
γh1:=4−1(3αh+αˉh) and γh2:=4−1(βˉh+3βh).
Since h≡0, it follows that
0≤αh<γh1<γh2<βh≤1.
Let K:={u∈C([0,1],R+):u(t)≥ρh∥u∥∞ for t∈[γh1,γh2]}.
Here
ρh:=ρ1min{γh1,1−γh2}∈(0,1).
Then K is a cone in C[0,1]. For r>0, let Kr:={u∈K : ∥u∥∞<r}, ∂Kr:={u∈K : ∥u∥∞=r} and Kr:=Kr∪∂Kr.
Define a function F:R+×K→C(0,1) by
F(λ,u)(t)=λh(t)f(u(t)) for (λ,u)∈R+×K and t∈(0,1).
Clearly, F(λ,u)∈Hφ for any (λ,u)∈R+×K.
Now we define an operator H:R+×K→C[0,1] by
H(λ,u)≡T(F(λ,u)) for (λ,u)∈R+×K.
That is, for (λ,u)∈R+×K,
[TABLE]
where σ=σ(λ,u) is a number satisfying
[TABLE]
Remark 2.2**.**
For any (λ,u)∈R+×K, by Lemma 2.1, H(R+×K)⊆K.
It is easy to see that (1.1) has a solution if and only if H(λ,⋅) has a fixed point in K.
Since H(0,u)=0 for all u∈K, [math] is a unique solution to problem (1.1) with λ=0.
By Lemma 2.1, any nonzero solution u to problem (1.1) is positive one, i.e., u(t)>0 for all t∈(0,1).
By the argument similar to those in the proof of [10, Lemma 3], it can be proved that H is completely continuous on R+×K (see also [27, Lemma 3.3]). We omit the proof of it.
Lemma 2.3**.**
([1, Lemma 4])*
Assume that (A) and g∈Hφ hold. Then the operator H:R+×K→K is completely continuous, i.e., compact and continuous.*
Finally, we recall a well-known theorem of the fixed point index theory.
Theorem 2.4**.**
(see, e.g., [28, 29])* Assume that, for some r>0, T1:Kr→K is completely continuous on Kr. Then*
(i)* if ∥T1(u)∥∞>∥u∥∞ for u∈∂Kr, then i(T1,Kr,K)=0;*
(ii)* if ∥T1(u)∥∞<∥u∥∞ for u∈∂Kr, then i(T1,Kr,K)=1.*
3 Main Results
Throughout this section, we assume h∈Hψ1∖{0}. For convenience, we use the following notations in this section:
γh:=2γh1+γh2,
where γh1=43αh+αˉh and γh2=4βˉh+3βh;
A1:=∥c∥∞1ψ2−1(∥d∥∞1)min{∫γh1γhψ2−1(∫sγhh(τ)dτ)ds,∫γhγh2ψ2−1(∫γhsh(τ)dτ)ds};
A2:=c01ψ1−1(d01)max{∫0γhψ1−1(∫sγhh(τ)dτ)ds,∫γh1ψ1−1(∫γhsh(τ)dτ)ds}.
Since 0≤αh<γh1<γh<γh2<βh≤1, A1>0 and A2>0. Define continuous functions R1,R2:(0,∞)→(0,∞) by
R1(m):=f∗(m)1φ(A1m)
and R2(m):=f∗(m)1φ(A2m) for m∈(0,∞).
Here, f∗(m):=min{f(y):ρhm≤y≤m} and f∗(m):=max{f(y):0≤y≤m} for m∈R+.
Remark 3.1**.**
By (1.2), ψ2−1(y)≤ψ1−1(y) for all y∈R+, and it follows that
0<A1<A2* and 0<R2(m)<R1(m) for all m∈(0,∞).*
For any l∈C(R+,R+),
la:=m→alimφ(m)l(m) for a∈{0,∞}.
It is not hard to prove that
(f∗)a=(f∗)a=0* if fa=0 and (f∗)a=(f∗)a=∞ if fa=∞.*
For convenience of readers, we give the proofs. First, we show that f0=0 implies (f∗)0=(f∗)0=0. Let ϵ>0 be given and f0=0 be assumed. Then there exists δ>0 such that for any s∈(0,δ), 0<φ(s)f(s)<ϵ. Since f∈C(R+,R+) with f(s)>0 for s∈(0,∞), by the extreme value theorem, for any s∈(0,∞), f∗(s)=f(xs) for some xs∈(0,s]. Then
0≤φ(s)f∗(s)≤φ(s)f∗(s)=φ(s)f(xs)≤φ(xs)f(xs)<ϵ* for any s∈(0,δ),*
which implies that (f∗)0=(f∗)0=0.
Next, we show that f∞=0 implies (f∗)∞=(f∗)∞=0. Indeed, let ϵ>0 be given and f∞=0 be assumed. Then there exists N>0 such that for any s≥N,
φ(s)f(s)<ϵ.
For s≥N, f∗(s)≤f∗(N)+f(xN,s), where xN,s is the point in [N,s] satisfying f(xN,s)=max{f(x):N≤x≤s}. Then
[TABLE]
Consequently, 0≤s→∞limsupφ(s)f∗(s)≤s→∞limsupφ(s)f∗(s)≤ϵ, which is true for all ϵ>0. Thus (f∗)∞=(f∗)∞=0.
Finally, we show that, for a∈{0,∞}, fa=∞ implies (f∗)a=∞. For each m∈(0,∞), by the extreme value theorem, there exists m∗∈[ρhm,m] satisfying (f∗)(m)=f(m∗), and
[TABLE]
As m→a∈{0,∞}, m∗→a, and thus (f∗)a=∞, provided fa=∞.
- (3)* By (A) and Remark 3.1 (2), for i∈{1,2} and a∈{0,∞},*
m→alimRi(m)=∞* if fa=0 and m→alimRi(m)=0 if fa=∞.*
Lemma 3.2**.**
Assume that (A) and h∈Hψ1∖{0} hold. Let m∈(0,∞) be fixed. Then, for any λ∈(R1(m),∞), ∥H(λ,v)∥∞>∥v∥ for all v∈∂Km and
[TABLE]
Proof.
Let λ∈(R1(m),∞) and v∈∂Km be fixed. Then ρhm≤v(t)≤m for t∈[γh1,γh2], and
[TABLE]
Let σ be a number satisfying H(λ,v)(σ)=∥H(λ,v)∥∞. We have two cases: either (i) σ∈[γh,1) or (ii) σ∈(0,γh). We only consider the case (i), since the case (ii) can be dealt in a similar manner.
Since λ>R1(m), it follows from (1.2) and (3.2) that
[TABLE]
By Theorem 2.4, (3.1) holds for any λ∈(R1(m),∞).
∎
Lemma 3.3**.**
Assume that (A) and h∈Hψ1∖{0} hold. Let m∈(0,∞) be fixed. Then, for any λ∈(0,R2(m)), ∥H(λ,v)∥∞<∥v∥ for all v∈∂Km and
[TABLE]
Proof.
Let λ∈(0,R2(m)) and v∈∂Km be fixed. Then f(v(t))≤R2(m)1φ(A2m) for t∈[0,1].
By the same argument as in the proof of Lemma 3.2, it follows that ∥H(λ,v)∥∞<∥v∥ for all v∈∂Km and (3.3) holds for any λ∈(0,R2(m)).
∎
By Lemma 3.2 and Lemma 3.3, we give some results for the existence and multiplicity of positive solutions to problem (1.1). Since the proofs are similar, we only give the proof of Theorem 3.4.
Theorem 3.4**.**
Assume that (A) and h∈Hψ1∖{0} hold, and that there exist
m1 and m2 such that 0<m1<m2 (resp., 0<m2<m1) and R1(m1)<R2(m2).
Then (1.1) has a positive solution u=u(λ) satisfying m1<∥u∥∞<m2 (resp., m2<∥u∥∞<m1) for any λ∈(R1(m1),R2(m2)).
Proof.
We only prove the case 0<m1<m2, since the other case is similar. Let λ∈(R1(m1),R2(m2)) be fixed.
By Lemma 3.2 and Lemma 3.3, i(H(λ,⋅),Km1,K)=1 and i(H(λ,⋅),Km2,K)=0. Since H(λ,v)=v for all v∈∂Km1, it follows from the additivity property that i(H(λ,⋅),Km2∖Km1,K)=−1. Then there exists u∈Km2∖Km1 such that H(λ,u)=u by the solution property. Thus the proof is complete.
∎
For Theorem 3.5 and Theorem 3.6, let R∗:=max{R1(m1),R1(M1)} and R∗:=min{R2(m2),R2(M2)}.
Theorem 3.5**.**
Assume that (A) and h∈Hψ1∖{0} hold, and that there exist
m1,m2 and M1 (resp., M2) such that 0<m1<m2<M1 (resp., 0<m2<m1<M2) and R∗<R2(m2) (resp., R1(m1)<R∗).
Then (1.1) has two positive solutions u1=u1(λ) and u2=u2(λ) satisfying m1<∥u1∥∞<m2<∥u2∥∞<M1 for any λ∈(R∗,R2(m2)) (resp., m2<∥u1∥∞<m1<∥u2∥∞<M2 for any λ∈(R1(m1),R∗)).
Theorem 3.6**.**
Assume that (A) and h∈Hψ1∖{0} hold, and that there exist
m1,m2,M1 and M2 such that 0<m2<m1<M2<M1 (resp., 0<m1<m2<M1<M2) and R∗<R∗.
Then (1.1) has three positive solutions u1=u1(λ),u2=u2(λ) and u3=u3(λ) satisfying m2<∥u1∥∞<m1<∥u2∥∞<M2<∥u3∥∞<M1 (resp., m1<∥u1∥∞<m2<∥u2∥∞<M1<∥u3∥∞<M2) for any λ∈(R∗,R∗).
Now we give the existence and nonexistence results for positive solutions to problem (1.1) which are analogous to [2, Theorem 1.1 and Theorem 1.2].
Theorem 3.7**.**
Assume that (A) and h∈Hψ1∖{0} hold.
If f0=0 and f∞=∞ (resp., f0=∞ and f∞=0), then (1.1) has a positive solution u(λ) for any λ∈(0,∞) satisfying ∥uλ∥∞→∞ as λ→0 and ∥uλ∥∞→0 as λ→∞ (resp., ∥uλ∥∞→0 as λ→0 and ∥uλ∥∞→∞ as λ→∞).
If f0=0 and f∞∈(0,∞) (resp., f0∈(0,∞) and f∞=0), then
there exists λ∗∈(0,∞) and m∗∈(0,∞] (resp., m∗∈R+) such that (1.1) has a positive solution u(λ) for any λ∈(λ∗,∞) satisfying ∥u(λ)∥∞∈(0,m∗) and λ→∞lim∥u(λ)∥∞=0 (resp., ∥u(λ)∥∞∈(m∗,∞) and λ→∞lim∥u(λ)∥∞=∞).
Moreover, if m∗∈(0,∞), then there exists a positive solution u(λ∗) to problem (1.1) with λ=λ∗.
If f0=∞ and f∞∈(0,∞) (resp., f0∈(0,∞) and f∞=∞), then
there exist λ∗∈(0,∞) and m∗∈(0,∞] (resp., m∗∈R+) such that problem (1.1) has a positive solution u=u(λ) for any λ∈(0,λ∗) satisfying ∥u(λ)∥∞∈(0,m∗) and λ→0lim∥u(λ)∥∞=0 (resp., ∥u(λ)∥∞∈(m∗,∞) and λ→0lim∥u(λ)∥∞=∞). Moreover, if m∗∈(0,∞), then there exists a positive solution u(λ∗) to problem (1.1) with λ=λ∗.
If f0=f∞=0, then there exist λ∗∈(0,∞) and m∗∈(0,∞) such that (1.1) has two positive solutions u1(λ) and u2(λ) for any λ∈(λ∗,∞) and it has a positive solution u(λ∗) for λ=λ∗. Moreover, u1(λ) and u2(λ) can be chosen so that 0<∥u1(λ)∥∞<m∗<∥u2(λ)∥∞, λ→∞lim∥u1(λ)∥∞=0 and λ→∞lim∥u2(λ)∥∞=∞.
If f0=f∞=∞, then
there exist λ∗∈(0,∞) and m∗∈(0,∞) such that problem (1.1) has two positive solutions u1(λ) and u2(λ) for any λ∈(0,λ∗) and it has a positive solution u(λ∗) for λ=λ∗. Moreover, u1(λ) and u2(λ) can be chosen so that 0<∥u1(λ)∥∞<m∗<∥u2(λ)∥∞, λ→0lim∥u1(λ)∥∞=0 and λ→0lim∥u2(λ)∥∞=∞.
If f0∈[0,∞) and f∞∈[0,∞), then there exists λ>0 such that (1.1) has no positive solutions for λ<λ.
If f0∈(0,∞] and f∞∈(0,∞], then there exists λ>0 such that (1.1) has no positive solutions for λ>λ.
Proof.
(1) We only give the proof of the case f0=0 and f∞=∞, since the case f0=∞ and f∞=0 can be proved in a similar manner. Since f0=0 and f∞=∞, by Remark 3.1 (3), Ri(m)→∞ as m→0 and Ri(m)→0 as m→∞ for i=1,2. For any λ∈(0,∞), there exist m1(λ) and m2(λ) such that
0<m2(λ)<m1(λ) and R1(m1(λ))<λ<R2(m2(λ)).
By Theorem 3.4, there exists a positive solution uλ to problem (1.1) satisfying
m2(λ)<∥uλ∥∞<m1(λ).
Since Ri(m)→∞ as m→0 for i=1,2, we may choose m1(λ) and m2(λ) so that 0<m2(λ)<m1(λ) and m1(λ)→0 as λ→∞. Consequently, we can choose positive solutions uλ to problem (1.1) for large λ>0 so that ∥uλ∥∞→0 as λ→∞. Similarly, since Ri(m)→0 as m→∞ for i=1,2, we can choose positive solutions uλ to problem (1.1) for small λ>0 so that ∥uλ∥∞→∞ as λ→0.
(2) We only give the proof of the case f0=0 and f∞∈(0,∞), since the case f0∈(0,∞) and f∞=0 can be proved in a similar manner. Since f0=0, by Remark 3.1 (3), Ri(m)→∞ as m→0 for i=1,2. Since
m→∞limR1(m)≥m→∞limf(m)φ(A1m)≥m→∞limf(m)φ(m)ψ1(A11)=f∞1ψ1(A11)>0,
there exist λ∗:=inf{R1(m):m∈(0,∞)}∈(0,∞) and m∗∈(0,∞] satisfying R1(m∗)=λ∗. For any λ∈(λ∗,∞), there exist m1(λ) and m2(λ) such that
0<m2(λ)<m1(λ)<m∗ and R1(m1)<λ<R2(m2).
By Theorem 3.4, there exists a positive solution uλ to problem (1.1) satisfying m2(λ)<∥u∥∞<m1(λ). Since Ri(m)→∞ as m→0 for i=1,2, we may choose m1(λ) and m2(λ) satisfying
0<m2(λ)<m1(λ) and m1(λ)→0 as λ→∞.
Consequently, we can choose positive solutions uλ to problem (1.1) for large λ>0 so that ∥uλ∥∞→0 as λ→∞.
For each n∈N, let λn:=λ∗+n1. Then we may choose m1=m1(n) and m2=m2(n) such that
R1(m1(n))<λn<R2(m2(n)) and
0<δ<m2(n)<m1(n)<m∗ for all n.
Consequently, for each n, there exists un∈K such that H(λn,un)=un and δ<∥un∥∞<m∗. Since {(λn,un)} is bounded in R+×K and H:R+×K→K is compact, there exist a subsequence {(λnk,unk)} of {(λn,un)} and u∗∈K such that H(λnk,unk)=unk→u∗ in K as k→∞. Since λn→λ∗ as n→∞ and H is continuous, H(λ∗,u∗)=u∗ and ∥u∗∥∞≥δ>0. Thus problem (1.1) has a positive solution u∗ for λ=λ∗. Thus the proof is complete.
(3) Let λ∗=sup{R2(m):m∈(0,∞)}∈(0,∞) and m∗∈[0,∞] satisfying R2(m∗)=λ∗. Then the proof is complete by the argument similar to those in the proof of Theorem 3.7 (2).
(4) Since f0=f∞=0, it follows that, for i=1,2, m→0limRi(m)=m→∞limRi(m)=∞. Then there exists m∗∈(0,∞) satisfying
R1(m∗)=min{R1(m):m∈R+}∈(0,∞).
Let λ∗=R1(m∗). For any λ∈(λ∗,∞), there exist m1(λ),m2(λ),M1(λ) and M2(λ) such that
0<m2(λ)<m1(λ)<m∗<M1(λ)<M2(λ)
and
R1(m1(λ))=R1(M1(λ))<λ<R2(m2(λ))=R2(M2(λ)).
Then the proof is complete by the argument similar to those in the proof of Theorem 3.7 (2).
(5) Since f0=f∞=∞, it follows that, for i=1,2, m→0limRi(m)=m→∞limRi(m)=0. Let λ∗=max{R2(m):m∈R+}∈(0,∞) and m∗∈(0,∞) satisfying R2(m∗)=λ∗. Then the proof is complete by the argument similar to those in the proof of Theorem 3.7 (2).
(6) Let u be a positive solution to problem (1.1) with λ>0 and let σ be a constant satisfying u(σ)=∥u∥∞. Since f0∈[0,∞) and f∞∈[0,∞), there exists C1>0 such that f(s)≤C1φ(s) for s∈R+. We only consider the case σ≤γh, since the case σ>γh can be dealt in a similar manner.
Since f(u(t))≤C1φ(u(σ)) for t∈[0,1],
[TABLE]
Here
[TABLE]
Consequently,
λ≥C1d0ψ1(h∗c0)=:λ.
(7) Let u be a positive solution to problem (1.1) with λ>0 and let σ be a constant satisfying u(σ)=∥u∥∞. Since f0∈(0,∞] and f∞∈(0,∞], there exists ϵ>0 such that f(s)>ϵφ(s) for s∈R+. We only consider the case σ≥γh, since the case σ<γh can be dealt in a similar manner.
Since u(t)≥u(γh1) for t∈[γh1,σ],
f(u(t))>ϵφ(u(γh1)) for t∈[γh1,γ]. Then
[TABLE]
Here
γ0=min{γh1,1−γh2}>0 and h∗=min{∫γh1γhh(τ)dτ,∫γhγh2h(τ)dτ}>0.
Consequently,
λ≤h∗ϵ∥d∥∞ψ2(γ0∥c∥∞)=:λ.
∎
Finally, an example to illustrate the results obtained in this paper is given.
Example 3.8**.**
Let φ be an odd function satisfying either
(i)* φ(x)=1+xx2 or (ii) φ(x)=x+x2 for x≥0.*
Then it is easy to check that (A) is satisfied for
ψ1(y)=min{y,y2}* and ψ2(y)=max{y,y2}.*
Define h:(0,1)→R+ by
h(t)=0* for t∈[0,161] and h(t)=(t−161)(1−t)−a for t∈(161,1).*
Then, since ψ1−1(s)=s for all s≥1, h∈Hψ1∖L1(0,1) for any a∈[1,2). Also, αh=βˉh=161, βh=αˉh=1, γh1=6419, γh2=6449 and γh=3217.
Let c,d be any positive continuous functions on [0,1]. Then ρ1, ρh, A1 and A2 are well defined in (0,∞). Define f:R+→R+ by
[TABLE]
Here M2 is a fixed constant satisfying
M2>max{ρh1,φ−1(φ(1)1[φ(ρhA11)]2[ψ1(A21)]−2)}.**
Clearly, f∈C(R+,R+) satisfies that f(s)>0 for s>0, f0=0 and f∞=∞.
Since f is strictly increasing on R+, f∗(m)=f(m) and f∗(m)=f(ρhm) for all m∈R+. Then, by (1.2),
R1(m)=f(ρhm)1φ(A1m)*
and
R2(m)=f(m)1φ(A2m)≥f(m)φ(m)ψ1(A21) for m∈(0,∞).*
From the choice of M2, it follows that
[TABLE]
We may choose m2 and M1 satisfying 0<m2<m1=ρh−1<M2<M1 and
R1(M1)≤R1(ρh−1)<R2(M2)≤R2(m2),**
since R2(m)→∞ as m→0 and R1(m)→0 as m→∞. By Theorem 3.6 and Theorem 3.7 (1), problem (1.1) has three positive solutions for λ∈(R1(ρh−1),R2(M2)) and it has a positive solution u(λ) for λ∈(0,∞) satisfying that λ→0lim∥u(λ)∥∞=∞ and λ→∞lim∥u(λ)∥∞=0.
4 Conclusions
In this work, we studied the existence and nonexistence of positive solutions to problem (1.1). In Theorem 3.4, Theorem 3.5, Theorem 3.6 and Theorem 3.7, various sufficient conditions on the nonlinearity f for the existence, nonexistence and multiplicity of positive solutions to problem (1.1) were given. In particular, Theorem 3.7 improves on the results in [2], since we do not assume the monotonicity of d, and the weight function h:(0,1)→R+ may not be L1(0,1) and it can be vanished in some subinterval of (0,1).
In [1], the nonlinearity f=f(t,s) should satisfy the condition
[TABLE]
so that the existence of an unbounded solution component was shown and, by examining the shape of the component according to several assumptions on f at ∞, the existence, nonexistence and multiplicity of positive solutions were studied. In this case, all nonnegative solutions are positive ones by (4.1). Compared with the results in [1], the nonlinearity f=f(s) in the present work may have the property f(0)=0. Even though the existence of unbounded solution component to problem (1.1) can be obtained by [1, Theorem 1], the existence of positive solutions cannot be shown from the solution component, since problem (1.1) has a trivial solution [math] for every λ∈R+, provided f(0)=0. Thus, the fixed point index theory was used in order to prove the main results (Theorem 3.4, Theorem 3.5, Theorem 3.6 and Theorem 3.7).
In the present work, the problem with Dirichlet boundary conditions was considered. As an extension of the results in this paper, similar results for the problem with nonlocal boundary conditions is expected. The existence, nonexistence and multiplicity of positive solutions to problem with nonlocal boundary conditions will be discussed for future work.
Acknowledgements This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education(2017R1D1A1B03035623).