# Multiple Positive Symmetric Solutions top-Laplacian Dynamic Equations on Time Scales

Authors: You-Hui Su; Can-Yun Huang
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

This paper makes a study on the existence of positive solution top-Laplacian dynamic
equations on time scales 𝕋. Some new sufficient conditions are obtained for the existence of at least single or twin positive solutions by using Krasnosel'skii's fixed point theorem and new sufficient conditions are also obtained for the existence of at least triple or arbitrary odd number positive solutions by using generalized Avery-Henderson fixed point theorem and Avery-Peterson fixed point theorem. As applications,
two examples are given to illustrate the main results and their differences. These results are even new for the special cases of continuous and discrete equations, as well as in the general time-scale setting.


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## Body

## 1. Introduction

Initiated by Hilger in his Ph.D. thesis [1] in 1988, the theory of time scales has been improved ever since, especially in the unification of the theory of differential equations in the continuous case and the theory of difference equations in the discrete case. For the time being, it remains a field of vitality and attracts attention of many distinguished scholars worldwide. In particular, the theory is also widely applied to biology, heat transfer, stock market, wound healing, epidemic models [2–5], and so forth.Recent research results indicate that considerable achievement has been made in the existence problems of positive solutions to dynamic equations on time scales. For details, please see [6–13] and the references therein. Symmetry and pseudosymmetry have been widely used in science and engineering [14]. The reason is that symmetry and pseudosymmetry are not only of its theoretical value in studying the metric manifolds [15] and symmetric graph [16, 17], and so forth, but also of its practical value, for example, we can apply this characteristic to study graph structure [18, 19] and chemistry structure [20]. Yet, few literature resource [21, 22] is available concerning the characteristics of positive solutions to p-Laplacian dynamic equations on time scales.Throughout this paper, we denote thep-Laplacian operator by φp(u), that is, φp(u)=|u|p-2u for p>1 with (φp)-1=φq and 1/p+1/q=1.For convenience, we think of the blanket as an assumption thata,b are points in 𝕋, for an interval (a,b)𝕋 we always mean (a,b)∩𝕋. Other type of intervals is defined similarly.We would like to mention the results of Sun and Li [11, 12]. In [12], Sun and Li considered the two-point BVP(1.1)(φp(uΔ(t)))Δ+h(t)f(uσ(t))=0,t∈[a,b]𝕋,u(a)-B0(uΔ(a))=0,uΔ(σ(b))=0,
and established the existence theory for positive solutions of the above problem. They [11] also considered the m-point boundary value problem with p-Laplacian(1.2)(φp(uΔ(t)))∇+h(t)f(t,u(t))=0,t∈(0,T)𝕋,uΔ(0)=0,u(T)=∑i=1m-2ciu(ξi),
and gave the existence of single or multiple positive solutions to the above problem. The main tools used in these two papers are some fixed-point theorems [23–25].It is also noted that the researchers mentioned above [11, 12] only considered the existence of positive solutions. As a results, they failed to further provide characteristics of solutions, such as, symmetry. Naturally, it is quite necessary to consider the characteristics of solutions to p-Laplacian dynamic equations on time scales.Let𝕋 be a symmetric time scale such that 0,T∈𝕋. we consider the following p-Laplacian boundary value problem on time scales 𝕋 of the form:(1.3)(φp(uΔ(t)))∇+h(t)f(u(t))=0,t∈[0,T]𝕋,u(0)=u(T)=0,uΔ(0)=-uΔ(T).
By using symmetric technique, the Krasnosel'skii's fixed  point theorem [24], the generalized Avery-Henderson fixed point theorem [26], and Avery-Peterson fixed point theorem [27], we obtain the existence of at least single, twin, triple, or arbitrary odd positive symmetric solutions of  problem (1.3). As applications, two examples are given to illustrate the main results and their differences. These results are even new for the special cases of continuous and discrete equations  as well as in the general time-scale setting.The rest of the paper is organized as follows. In Section2, we present several fixed point results. In Section 3, by using Krasnosel'skii's fixed point theorem, we obtain the existence of at least single or twin positive symmetric solutions to problem (1.3). In Section 4, the existence criteria for at least triple positive or arbitrary odd positive symmetric solutions to problem (1.3) are established. In Section 5, we present two simple examples to illustrate our results.For convenience, we now give some symmetric definitions.Definition 1.1.
The interval[0,T]𝕋  is said to be symmetric if any given t∈[0,T]𝕋, we have T-t∈[0,T]𝕋.We note that such a symmetric time scale𝕋 exists. For example, let(1.4)𝕋={0,0.05,0.1,0.15}∪[0.22,0.44]∪{0.5,0.85,0.9,0.95,1}∪[0.56,0.78].
It is obvious that 𝕋 is a symmetric time scale.Definition 1.2.
A functionu:[0,T]𝕋→ℝ is said to be symmetric if u is symmetric over the interval [0,T]𝕋. That is, u(t)=u(T-t), for any given t∈[0,T]𝕋.Definition 1.3.
We sayu is a symmetric solution to problem (1.3) on [0,T]𝕋 provided that  u is a solution to boundary value problem (1.3) and is symmetric over the interval [0,T]𝕋.Basic definitions on time scale can be found in [6, 7, 28]. Another excellent sources on dynamical systems on measure chains are the book in [29].Throughout this paper, it is assumed that(H1)
f:[0,∞)→[0,∞) is continuous, and does not vanish identically;(H2)
h∈Cld([0,T]𝕋,[0,∞)) is symmetric over the interval [0,T]𝕋  and does not vanish identically on any closed subinterval of [0,T]𝕋, where Cld([0,T]𝕋,[0,∞)) denotes the set of all left dense continuous functions from [0,T]𝕋 to [0,∞).


## 2. Preliminaries

LetE=Cld([0,T]𝕋,ℝ) and equip norm(2.1)∥u∥=supt∈[0,T]𝕋|u(t)|,
then E is a Banach space. Define a cone P⊂E by(2.2)P={u∈E∣u(0)=u(T)=0,uissymmetric,nonnegative,andconcaveontheinterval[0,T]𝕋}.Assume thatr,η∈(0,T/2)𝕋 with η<r. By using the symmetric and concave characters of u∈P and u(0)=u(T)=0, it is easy to obtain the following results.Lemma 2.1.
Assume thatr,η∈(0,T/2)𝕋 with η<r. If u∈P, then  (i)
u(η)≥(η/r)u(r);(ii)
(T/2)u(r)≥ru(T/2).From  the previous   lemma we know that∥u∥=u(T/2) for u∈P.The operatorA:P→E is defined by(2.3)Au(t)={∫0tφq(∫sT/2h(r)f(u(r))∇r)Δs,t∈[0,T2]𝕋,∫tTφq(∫T/2sh(r)f(u(r))∇r)Δs,t∈[T2,T]𝕋.
It is obvious that A is completely continuous operator and all the fixed points of A are the solutions to the boundary value problem (1.3).In addition, it is easy to see that the operatorA is symmetric. In fact, for t∈[0,T/2]𝕋, we have T-t∈[T/2,T]𝕋, by using the integral transform, we have(2.4)Au(T-t)=∫T-tTφq(∫T/2sh(r)f(u(r))∇r)Δs=s→T-s∫0tφq(∫T/2T-sh(r)f(u(r))∇r)Δs=r→T-r∫0tφq(∫sT/2h(T-r)f(u(T-r))∇r)Δs=∫0tφq(∫sT/2h(r)f(u(r))∇r)Δs=Au(t).
Hence, A is symmetric.Now, we provide some background material from the theory of cones in Banach spaces [24, 26, 27, 30], and then state several fixed point theorems needed later.Firstly, we list the Krasnosel'skii's fixed point theorem [24].Lemma 2.2 (see [24]).
LetP be a cone in a Banach space E. Assume that  Ω1, Ω2 are open subsets of E with 0∈Ω1,  Ω¯1⊂Ω2. If A:P∩(Ω¯2∖Ω1)→P is a completely continuous operator such that either(i)
∥Ax∥≤∥x∥,  forallx∈P∩∂Ω1 and ∥Ax∥≥∥x∥, forallx∈P∩∂Ω2  or(ii)
∥Ax∥≥∥x∥, forallx∈P∩∂Ω1 and ∥Ax∥≤∥x∥, forallx∈P∩∂Ω2,then A has a fixed point in P∩(Ω2¯∖Ω1).Given a nonnegative continuous functionalγ on a cone P of a real Banach space E, we define, for each d>0, the set P(γ,d)={x∈P:γ(x)<d}.Secondly, we state the generalized Avery-Henderson fixed point theorem [26].Lemma 2.3 (see [26]).
LetP be a cone in a real Banach space E. Let α, β, and γ be increasing, nonnegative continuous functional on P such that for some c>0 and H>0,γ(x)≤β(x)≤α(x) and ∥x∥≤Hγ(x) for all x∈P(γ,c)¯. Suppose that there exist positive numbers a and b with a<b<c and A:P(γ,c)¯→P is a completely continuous operator such that (i)
γ(Ax)<c for all x∈∂P(γ,c);(ii)
β(Ax)>b for all x∈∂P(β,b);(iii)
P(α,a)≠∅ and α(Ax)<a for x∈∂P(α,a),then A has at least three fixed points x1,  x2, and x3 belonging to P(γ,c)¯ such that
(2.5)0≤α(x1)<a<α(x2)withβ(x2)<b<β(x3),γ(x3)<c.The following lemma can be found in [21].Lemma 2.4 (see [21]).
LetP be a cone in a real Banach space E. Let α,  β, and γ be increasing, nonnegative continuous functional on P such that for some c>0 and H>0,γ(x)≤β(x)≤α(x) and ∥x∥≤Hγ(x) for all x∈P(γ,c)¯. Suppose that there exist positive numbers a and b with a<b<c and A:P(γ,c)¯→P is a completely continuous operator such that:(i)
γ(Ax)>c for all x∈∂P(γ,c);(ii)
β(Ax)<b for all x∈∂P(β,b);(iii)
P(α,a)≠∅ and α(Ax)>a for x∈∂P(α,a),then A has at least three fixed points x1,x2, and x3 belonging to P(γ,c)¯ such that
(2.6)0≤α(x1)<a<α(x2)withβ(x2)<b<β(x3),γ(x3)<c.Letβ and ϕ be nonnegative continuous convex functionals on P, λ is a nonnegative continuous concave functional on P, and φ is a nonnegative continuous functional, respectively on P. We define the following convex sets:(2.7)P(ϕ,λ,b,d)={x∈P:b≤λ(x),ϕ(x)≤d},P(ϕ,β,λ,b,c,d)={x∈P:b≤λ(x),β(x)≤c,ϕ(x)≤d},
and a closed set R(ϕ,φ,a,d)={x∈P:a≤φ(x),ϕ(x)≤d}.Finally, we list the fixed point theorem due to Avery-Peterson [27].Lemma 2.5 (see [27]).
LetP be a cone in a real Banach space E and β,ϕ,λ,φ   defined as above, moreover, φ satisfies φ(λ′x)≤λ′φ(x) for 0≤λ′≤1 such that, for some positive numbers h and d,
(2.8)λ(x)≤φ(x),∥x∥≤hϕ(x)
for all x∈P(ϕ,d)¯. Suppose that  A:P(ϕ,d)¯→P(ϕ,d)¯ is completely continuous and there exist positive real numbers a, b, c, with a<b such that (i)
{x∈P(ϕ,β,λ,b,c,d):λ(x)>b}≠∅  and  λ(A(x))>b for x∈P(ϕ,β,λ,b,c,d);(ii)
λ(A(x))>b for x∈P(ϕ,λ,b,d) with β(A(x))>c;(iii)
0∉R(ϕ,φ,a,d) and λ(A(x))<a for all x∈R(ϕ,φ,a,d) with φ(x)=a,then A has at least three fixed points x1,x2,x3∈P(ϕ,d)¯ such that
(2.9)ϕ(xi)≤dfori=1,2,3,b<λ(x1),a<φ(x2),λ(x2)<bwithφ(x3)<a.


## 3. Single or Twin Solutions

Let(3.1)f0=limu→0+f(u)φp(u),f∞=limu→∞f(u)φp(u).
We define i0=number of zeros in the set {f0,f∞} and i∞=number of infinities in the set {f0,f∞}. Clearly,  i0, i∞=0,1, or 2 and there exist six possible cases: (i)  i0=1 and i∞=1; (ii) i0=0 and i∞=0; (iii) i0=0 and i∞=1; (iv) i0=0 and i∞=2; (v)  i0=1 and i∞=0;   (vi) i0=2 and i∞=0. In the following, by using Krasnosel'skii's fixed point theorem in a cone, we study the existence of positive symmetric solutions to problem (1.3) under the above six possible cases.


### 3.1. For the Casei0=1 and i∞=1

In this subsection, we discuss the existence of single positive symmetric solution of the problem (1.3) under i0=1 and i∞=1.Theorem 3.1.
Problem (1.3) has at least one positive symmetric solution in the case i0=1 and i∞=1.Proof.
We divide the proof into two cases.
Case 1 (f0=0 and f∞=∞).
In view off0=0, there exists an H1>0 such that f(u)≤φp(ε)φp(u)=φp(εu) for u∈(0,H1], where ε arbitrary small and satisfies 0<(εT/2)φq(∫0T/2h(r)∇r)≤1.
Ifu∈P with ∥u∥=H1, then
(3.2)∥Au∥=supt∈[0,T]𝕋|Au|=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤ε∥u∥∫0T/2φq(∫0T/2h(r)∇r)Δs≤∥u∥εT2φq(∫0T/2h(r)∇r)≤∥u∥.
We let ΩH1={u∈E:∥u∥<H1}, then ∥Au∥≤∥u∥ for u∈P∩∂ΩH1.
Fromf∞=∞, there exists an H2′>0 such that f(u)≥φp(k)φp(u)=φp(ku) for u∈[H2′,∞), where k>0, and satisfies the following inequality:
(3.3)2kη2Tφq(∫ηT/2h(r)∇r)≥1.
Set
(3.4)H2=max{2H1,H2′T2η},ΩH2={u∈E:∥u∥<H2}.
If u∈P with ∥u∥=H2, then, by Lemma 2.1, one has
(3.5)mint∈[η,T/2]𝕋u(t)≥2ηTu(T2)≥H2′.
For u∈P∩∂ΩH2, in terms of (3.3) and (3.5), we get
(3.6)∥Au∥=supt∈[0,T]𝕋|Au|≥Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)φp(ku(r))∇r)Δs≥2kη2T∥u∥φq(∫ηT/2h(r)∇r)≥∥u∥.
Thus, by (i) of Lemma 2.2,   problem (1.3) has at least single positive symmetric solution u in  P∩(Ω¯H2∖ΩH1) with H1≤∥u∥≤H2.Case 2 (f0=∞ and f∞=0).
Sincef0=∞, there exists an H3>0 such that f(u)≥φp(m)φp(u)=φp(mu) for u∈(0,H3], where m is such that
(3.7)2mη2Tφq(∫ηT/2h(r)∇r)≥1.
If u∈P with ∥u∥=H3, then, by (3.7), one has
(3.8)∥Au∥=supt∈[0,T]𝕋|Au|≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥ηφq(∫ηT/2h(r)φp(mu(r))∇r)≥ηm2ηT∥u∥φq(∫ηT/2h(r)∇r)≥∥u∥.
If we let ΩH3={u∈E:∥u∥<H3}, then ∥Au∥≥∥u∥ for u∈P∩∂ΩH3.
Now, we considerf∞=0. By definition, there exists H4′>0 such that
(3.9)f(u)≤φp(δ)φp(u)=φp(δu)foru∈[H4′,∞),
where δ>0 satisfies
(3.10)δT2φq(∫0T/2h(r)∇r)≤1.
Suppose thatf is bounded, then f(u)≤φp(K) for all u∈[0,∞) and some constant K>0. Pick
(3.11)H4=max{2H3,KT2φq(∫0T/2h(s)Δs)}.
If u∈P with ∥u∥=H4, then
(3.12)∥Au∥=Au(T2)≤∫0T/2φq(∫0T/2h(r)f(u(r))∇r)Δs≤KT2φq(∫0T/2h(s)Δs)≤H4=∥u∥.
Suppose thatf is unbounded. From f∈C([0,+∞),[0,+∞)), we have f(u)≤C3 for arbitrary u∈[0,C4], here C3 and C4 are arbitrary positive constants. This implies that f(u)→+∞ if u→+∞. Hence, it is easy to know that there exists H4≥max{2H3,(T/2η)H4′} such that f(u)≤f(H4) for u∈[0,H4]. If u∈P with ∥u∥=H4, then by using (3.9) and (3.10), we have
(3.13)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0T/2φq(∫0T/2h(r)f(H4)∇r)Δs≤δH4T2φq(∫0T/2h(r)∇r)≤∥u∥.
Consequently, in either case, if we take ΩH4={u∈E:∥u∥<H4}, then, for u∈P∩∂ΩH4, we have ∥Au∥≤∥u∥. Thus,   condition (ii) of Lemma 2.2 is satisfied. Consequently,   problem (1.3) has at least single positive symmetric solution u in P∩(Ω¯H4∖ΩH3) with H3≤∥u∥≤H4. The proof is complete.


### 3.2. For the Casei0=0 and i∞=0

In this subsection, we discuss the existence of positive symmetric solutions to problems (1.3) under i0=0 and i∞=0 .First, we will state and prove the following main result of problem (1.3).Theorem 3.2.
Suppose that the following conditions hold:(i)
there exists constantp′>0 such that f(u)≤φp(p′Λ1) for u∈[0,p′], where Λ1={(T/2)φq(∫0T/2h(r)∇r)}-1;(ii)
there exists constantq′>0 such that f(u)≥φp(q′Λ2) for u∈[(2η/T)q′,q′], where Λ2={ηφq(∫ηT/2h(r)∇r)}-1, furthermore, p′≠q′,then problem (1.3) has at least one positive symmetric solution u such that ∥u∥ lies between p′ and q′.Proof.
Without loss of generality, we may assume thatp′<q′.
LetΩp′={u∈E:∥u∥<p′}. For any u∈P∩∂Ωp′, in view of condition (i), we have
(3.14)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤p′Λ1T2φq(∫0T/2h(r)∇r)=p′,
which yields
(3.15)∥Au∥≤∥u∥foru∈P∩∂Ωp′.
Now, setΩq′={u∈E:∥u∥<q′}. For u∈P∩∂Ωq′, Lemma 2.1 implies that
(3.16)2ηTq′≤u(t)≤q′fort∈[η,T2]𝕋.
Hence, by condition (ii) we get
(3.17)∥Au∥=Au(T2)≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥q′Λ2ηφq(∫ηT/2h(r)∇r)=q′.
So, if we take Ωq′={u∈E:∥u∥<q′}, then
(3.18)∥Au∥≥∥u∥,u∈P∩∂Ωq′.
Consequently, in view of p′<q′, (3.15) and (3.18), it follows from Lemma 2.2 that problem (1.3) has a positive symmetric solution u in P∩(Ω¯q′∖Ωp′). The proof is complete.


### 3.3. For the Casei0=1 and i∞=0 or i0=0 and i∞=1

In this subsection, under the conditionsi0=1 and i∞=0 or i0=0 and i∞=1, we discuss the existence of positive symmetric solutions to problem (1.3).Theorem 3.3.
Suppose thatf0∈[0,φp(Λ1)) and f∞∈(φp((T/2η)Λ2),∞) hold. Then problem (1.3) has at least one positive symmetric solution.Proof.
It is easy to see that under the assumptions,   conditions (i) and (ii) in   Theorem3.2 are satisfied. So the proof is easy and we omit it here.Theorem 3.4.
Suppose thatf0∈(φp((T/2η)Λ2),∞) and f∞∈[0,φp(Λ1)) hold, then problem (1.3) has at least one positive symmetric solution.Proof.
Firstly, letε1=f0-φp((T/2η)Λ2)>0, there exists a sufficiently small q′>0 that  satisfies
(3.19)f(u)φp(u)≥f0-ε1=φp(T2ηΛ2)foru∈(0,q′].
Thus, u∈[(2η/T)q′,q′], we have
(3.20)f(u)≥φp(T2ηΛ2)φp(u)≥φp(Λ2q′),
which implies that   condition (ii) in Theorem 3.2 holds.
Nextly, forε2=φp(Λ1)-f∞>0, there exists a sufficiently large p′′(>q′) such that
(3.21)f(u)φp(u)≤f∞+ε2=φp(Λ1)foru∈[p′′,∞).
We consider two cases.
Case 1.
Assume thatf is bounded, that is,
(3.22)f(u)≤φp(K1)foru∈[0,∞),
here K1>0 some constant. If we take sufficiently large p′ such that p′≥max{K1/Λ1,p′′}, then
(3.23)f(u)≤φp(K1)≤φp(Λ1p′)foru∈[0,p′].
Consequently, from the above inequality,   condition (i) of Theorem 3.2 is true.Case 2.
Assume thatf is unbounded.
Fromf∈C([0,∞),[0,∞)), there exists p′>p′′ such that
(3.24)f(u)≤f(p′)foru∈[0,p′].
Since p′>p′′, by (3.21), we get f(p′)≤φp(Λ1p′), hence
(3.25)f(u)≤f(p′)≤φp(Λ1p′)foru∈[0,p′].
Thus,   condition (i) of Theorem 3.2 is fulfilled.
Consequently, Theorem3.2 implies that the conclusion of this theorem holds.From the proof of Theorems3.1 and 3.2, respectively, we have the following two results.Corollary 3.5.
Suppose thatf0=0 and  condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Corollary 3.6.
Suppose thatf∞=0 and   condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Theorem 3.7.
Suppose thatf0∈(0,φp(Λ1)) and f∞=∞ hold, then problem (1.3) has at least one positive symmetric solution.Proof.
First, in view off∞=∞, then by inequality (3.7), we have ∥Au∥≥∥u∥ for u∈P∩∂ΩH2. Next, by f0∈(0,φp(Λ1)), for ε3=φp(Λ1)-f0>0, there exists a sufficiently small p′∈(0,H2) such that
(3.26)f(u)≤(f0+ε3)φp(u)=φp(Λ1u)≤φp(Λ1p′)foru∈[0,p′],
which implies  that  (i) of Theorem 3.2 holds, that is, (3.14) is true. Hence, we obtain ∥Au∥≤∥u∥ for u∈P∩∂Ωp′. The result is obtained and the proof is complete.Theorem 3.8.
Suppose thatf0=∞ and f∞∈(0,φp(Λ1)) hold,  then problem (1.3) has at least one positive symmetric solution.Proof.
On one hand, sincef0=∞, by   inequality (3.9), one gets ∥Au∥≥∥u∥, u∈P∩∂ΩH3. On the other hand, since f∞∈(0,φp(Λ1)), from the technique similar to the second part proof in Theorem 3.4, one obtains that   condition (i) of Theorem 3.2 is satisfied, that is, inequality (3.14) holds, one has ∥Au∥≤∥u∥, u∈P∩∂Ωp′, where p′>H3. Hence, problem (1.3) has at least one positive symmetric solution. The proof is complete.From Theorems3.7 and 3.8, respectively, it is easy to obtain the following two corollaries.Corollary 3.9.
Assume thatf∞=∞ and   condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Corollary 3.10.
Assume thatf0=∞ and  condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.


### 3.4. For the Casei0=0 and i∞=2 or i0=2 and i∞=0

In this subsection, underi0=0 and i∞=2 or i0=2 and i∞=0, we study the existence of multiple positive solutions to problems (1.3).Combining the proofs of Theorems3.1 and 3.2, it is easy to prove the following two theorems.Theorem 3.11.
Suppose thati0=0 and i∞=2 and   condition (i) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1,u2∈P such that 0<∥u1∥<p′<∥u2∥.Theorem 3.12.
Suppose thati0=2 and i∞=0 and  condition (ii) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1,u2∈P such that 0<∥u1∥<q′<∥u2∥.


## 3.1. For the Casei0=1 and i∞=1

In this subsection, we discuss the existence of single positive symmetric solution of the problem (1.3) under i0=1 and i∞=1.Theorem 3.1.
Problem (1.3) has at least one positive symmetric solution in the case i0=1 and i∞=1.Proof.
We divide the proof into two cases.
Case 1 (f0=0 and f∞=∞).
In view off0=0, there exists an H1>0 such that f(u)≤φp(ε)φp(u)=φp(εu) for u∈(0,H1], where ε arbitrary small and satisfies 0<(εT/2)φq(∫0T/2h(r)∇r)≤1.
Ifu∈P with ∥u∥=H1, then
(3.2)∥Au∥=supt∈[0,T]𝕋|Au|=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤ε∥u∥∫0T/2φq(∫0T/2h(r)∇r)Δs≤∥u∥εT2φq(∫0T/2h(r)∇r)≤∥u∥.
We let ΩH1={u∈E:∥u∥<H1}, then ∥Au∥≤∥u∥ for u∈P∩∂ΩH1.
Fromf∞=∞, there exists an H2′>0 such that f(u)≥φp(k)φp(u)=φp(ku) for u∈[H2′,∞), where k>0, and satisfies the following inequality:
(3.3)2kη2Tφq(∫ηT/2h(r)∇r)≥1.
Set
(3.4)H2=max{2H1,H2′T2η},ΩH2={u∈E:∥u∥<H2}.
If u∈P with ∥u∥=H2, then, by Lemma 2.1, one has
(3.5)mint∈[η,T/2]𝕋u(t)≥2ηTu(T2)≥H2′.
For u∈P∩∂ΩH2, in terms of (3.3) and (3.5), we get
(3.6)∥Au∥=supt∈[0,T]𝕋|Au|≥Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)φp(ku(r))∇r)Δs≥2kη2T∥u∥φq(∫ηT/2h(r)∇r)≥∥u∥.
Thus, by (i) of Lemma 2.2,   problem (1.3) has at least single positive symmetric solution u in  P∩(Ω¯H2∖ΩH1) with H1≤∥u∥≤H2.Case 2 (f0=∞ and f∞=0).
Sincef0=∞, there exists an H3>0 such that f(u)≥φp(m)φp(u)=φp(mu) for u∈(0,H3], where m is such that
(3.7)2mη2Tφq(∫ηT/2h(r)∇r)≥1.
If u∈P with ∥u∥=H3, then, by (3.7), one has
(3.8)∥Au∥=supt∈[0,T]𝕋|Au|≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥ηφq(∫ηT/2h(r)φp(mu(r))∇r)≥ηm2ηT∥u∥φq(∫ηT/2h(r)∇r)≥∥u∥.
If we let ΩH3={u∈E:∥u∥<H3}, then ∥Au∥≥∥u∥ for u∈P∩∂ΩH3.
Now, we considerf∞=0. By definition, there exists H4′>0 such that
(3.9)f(u)≤φp(δ)φp(u)=φp(δu)foru∈[H4′,∞),
where δ>0 satisfies
(3.10)δT2φq(∫0T/2h(r)∇r)≤1.
Suppose thatf is bounded, then f(u)≤φp(K) for all u∈[0,∞) and some constant K>0. Pick
(3.11)H4=max{2H3,KT2φq(∫0T/2h(s)Δs)}.
If u∈P with ∥u∥=H4, then
(3.12)∥Au∥=Au(T2)≤∫0T/2φq(∫0T/2h(r)f(u(r))∇r)Δs≤KT2φq(∫0T/2h(s)Δs)≤H4=∥u∥.
Suppose thatf is unbounded. From f∈C([0,+∞),[0,+∞)), we have f(u)≤C3 for arbitrary u∈[0,C4], here C3 and C4 are arbitrary positive constants. This implies that f(u)→+∞ if u→+∞. Hence, it is easy to know that there exists H4≥max{2H3,(T/2η)H4′} such that f(u)≤f(H4) for u∈[0,H4]. If u∈P with ∥u∥=H4, then by using (3.9) and (3.10), we have
(3.13)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0T/2φq(∫0T/2h(r)f(H4)∇r)Δs≤δH4T2φq(∫0T/2h(r)∇r)≤∥u∥.
Consequently, in either case, if we take ΩH4={u∈E:∥u∥<H4}, then, for u∈P∩∂ΩH4, we have ∥Au∥≤∥u∥. Thus,   condition (ii) of Lemma 2.2 is satisfied. Consequently,   problem (1.3) has at least single positive symmetric solution u in P∩(Ω¯H4∖ΩH3) with H3≤∥u∥≤H4. The proof is complete.


## 3.2. For the Casei0=0 and i∞=0

In this subsection, we discuss the existence of positive symmetric solutions to problems (1.3) under i0=0 and i∞=0 .First, we will state and prove the following main result of problem (1.3).Theorem 3.2.
Suppose that the following conditions hold:(i)
there exists constantp′>0 such that f(u)≤φp(p′Λ1) for u∈[0,p′], where Λ1={(T/2)φq(∫0T/2h(r)∇r)}-1;(ii)
there exists constantq′>0 such that f(u)≥φp(q′Λ2) for u∈[(2η/T)q′,q′], where Λ2={ηφq(∫ηT/2h(r)∇r)}-1, furthermore, p′≠q′,then problem (1.3) has at least one positive symmetric solution u such that ∥u∥ lies between p′ and q′.Proof.
Without loss of generality, we may assume thatp′<q′.
LetΩp′={u∈E:∥u∥<p′}. For any u∈P∩∂Ωp′, in view of condition (i), we have
(3.14)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤p′Λ1T2φq(∫0T/2h(r)∇r)=p′,
which yields
(3.15)∥Au∥≤∥u∥foru∈P∩∂Ωp′.
Now, setΩq′={u∈E:∥u∥<q′}. For u∈P∩∂Ωq′, Lemma 2.1 implies that
(3.16)2ηTq′≤u(t)≤q′fort∈[η,T2]𝕋.
Hence, by condition (ii) we get
(3.17)∥Au∥=Au(T2)≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs≥q′Λ2ηφq(∫ηT/2h(r)∇r)=q′.
So, if we take Ωq′={u∈E:∥u∥<q′}, then
(3.18)∥Au∥≥∥u∥,u∈P∩∂Ωq′.
Consequently, in view of p′<q′, (3.15) and (3.18), it follows from Lemma 2.2 that problem (1.3) has a positive symmetric solution u in P∩(Ω¯q′∖Ωp′). The proof is complete.


## 3.3. For the Casei0=1 and i∞=0 or i0=0 and i∞=1

In this subsection, under the conditionsi0=1 and i∞=0 or i0=0 and i∞=1, we discuss the existence of positive symmetric solutions to problem (1.3).Theorem 3.3.
Suppose thatf0∈[0,φp(Λ1)) and f∞∈(φp((T/2η)Λ2),∞) hold. Then problem (1.3) has at least one positive symmetric solution.Proof.
It is easy to see that under the assumptions,   conditions (i) and (ii) in   Theorem3.2 are satisfied. So the proof is easy and we omit it here.Theorem 3.4.
Suppose thatf0∈(φp((T/2η)Λ2),∞) and f∞∈[0,φp(Λ1)) hold, then problem (1.3) has at least one positive symmetric solution.Proof.
Firstly, letε1=f0-φp((T/2η)Λ2)>0, there exists a sufficiently small q′>0 that  satisfies
(3.19)f(u)φp(u)≥f0-ε1=φp(T2ηΛ2)foru∈(0,q′].
Thus, u∈[(2η/T)q′,q′], we have
(3.20)f(u)≥φp(T2ηΛ2)φp(u)≥φp(Λ2q′),
which implies that   condition (ii) in Theorem 3.2 holds.
Nextly, forε2=φp(Λ1)-f∞>0, there exists a sufficiently large p′′(>q′) such that
(3.21)f(u)φp(u)≤f∞+ε2=φp(Λ1)foru∈[p′′,∞).
We consider two cases.
Case 1.
Assume thatf is bounded, that is,
(3.22)f(u)≤φp(K1)foru∈[0,∞),
here K1>0 some constant. If we take sufficiently large p′ such that p′≥max{K1/Λ1,p′′}, then
(3.23)f(u)≤φp(K1)≤φp(Λ1p′)foru∈[0,p′].
Consequently, from the above inequality,   condition (i) of Theorem 3.2 is true.Case 2.
Assume thatf is unbounded.
Fromf∈C([0,∞),[0,∞)), there exists p′>p′′ such that
(3.24)f(u)≤f(p′)foru∈[0,p′].
Since p′>p′′, by (3.21), we get f(p′)≤φp(Λ1p′), hence
(3.25)f(u)≤f(p′)≤φp(Λ1p′)foru∈[0,p′].
Thus,   condition (i) of Theorem 3.2 is fulfilled.
Consequently, Theorem3.2 implies that the conclusion of this theorem holds.From the proof of Theorems3.1 and 3.2, respectively, we have the following two results.Corollary 3.5.
Suppose thatf0=0 and  condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Corollary 3.6.
Suppose thatf∞=0 and   condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Theorem 3.7.
Suppose thatf0∈(0,φp(Λ1)) and f∞=∞ hold, then problem (1.3) has at least one positive symmetric solution.Proof.
First, in view off∞=∞, then by inequality (3.7), we have ∥Au∥≥∥u∥ for u∈P∩∂ΩH2. Next, by f0∈(0,φp(Λ1)), for ε3=φp(Λ1)-f0>0, there exists a sufficiently small p′∈(0,H2) such that
(3.26)f(u)≤(f0+ε3)φp(u)=φp(Λ1u)≤φp(Λ1p′)foru∈[0,p′],
which implies  that  (i) of Theorem 3.2 holds, that is, (3.14) is true. Hence, we obtain ∥Au∥≤∥u∥ for u∈P∩∂Ωp′. The result is obtained and the proof is complete.Theorem 3.8.
Suppose thatf0=∞ and f∞∈(0,φp(Λ1)) hold,  then problem (1.3) has at least one positive symmetric solution.Proof.
On one hand, sincef0=∞, by   inequality (3.9), one gets ∥Au∥≥∥u∥, u∈P∩∂ΩH3. On the other hand, since f∞∈(0,φp(Λ1)), from the technique similar to the second part proof in Theorem 3.4, one obtains that   condition (i) of Theorem 3.2 is satisfied, that is, inequality (3.14) holds, one has ∥Au∥≤∥u∥, u∈P∩∂Ωp′, where p′>H3. Hence, problem (1.3) has at least one positive symmetric solution. The proof is complete.From Theorems3.7 and 3.8, respectively, it is easy to obtain the following two corollaries.Corollary 3.9.
Assume thatf∞=∞ and   condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.Corollary 3.10.
Assume thatf0=∞ and  condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.


## 3.4. For the Casei0=0 and i∞=2 or i0=2 and i∞=0

In this subsection, underi0=0 and i∞=2 or i0=2 and i∞=0, we study the existence of multiple positive solutions to problems (1.3).Combining the proofs of Theorems3.1 and 3.2, it is easy to prove the following two theorems.Theorem 3.11.
Suppose thati0=0 and i∞=2 and   condition (i) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1,u2∈P such that 0<∥u1∥<p′<∥u2∥.Theorem 3.12.
Suppose thati0=2 and i∞=0 and  condition (ii) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1,u2∈P such that 0<∥u1∥<q′<∥u2∥.


## 4. Triple Solutions

In the previous section, we have obtained some results on the existence of at least single or twin positive symmetric solutions to problem (1.3). In this section, we will further discuss the existence of positive symmetric solutions to problem (1.3) by using two different methods. And the conclusions we will arrive at are different with their own distinctive advantages.Based on the obtained symmetric solution position and local properties, we can only get some local properties of solutions by using method one; however, the position of solutions is not determined. In contrast, by means of method two, we cannot only get some local properties of solutions but also give the position of all solutions, with regard to some subsets of the cone, which has to meet some conditions which are stronger than those of method one. Obviously, the local properties of obtained solutions are different by using the two different methods. Hence, it is convenient for us to comprehensively comprehend the solutions of the models by using the two different techniques.In Section5, two examples are given to illustrate the differences of the results obtained by the two different methods.For the notational convenience, we denote(4.1)Mξ=ηφq(∫0T/2h(r)∇r),Nξ=ηφq(∫ηT/2h(r)∇r),Lξ=rφq(∫0T/2h(r)∇r),Lθ=rφq(∫rT/2h(r)∇r),Wξ=T2φq(∫0T/2h(r)∇r).


### 4.1. Result 1

In this subsection, in view of the generalized Avery-Henderson fixed-point theorem [26], the existence criteria for at least triple and arbitrary odd positive symmetric solutions to problems (1.3)  are established.Foru∈P, we define the nonnegative, increasing, continuous functionals γ,β, and α by(4.2)γ(u)=maxt∈[0,η]𝕋u(t)=u(η),β(u)=mint∈[η,T/2]𝕋u(t)=u(η),α(u)=maxt∈[0,r]𝕋u(t)=u(r).
It is obvious that γ(u)≤β(u)≤α(u)  for each u∈P. By Lemma 2.1, one obtains ∥u∥≤C*γ(u) for all u∈P,  here C*=T/2η.We now present the results in this subsection.Theorem 4.1.
If there are positive numbersa′, b′, c′ such that a′<(2r/T)b′<(2r/T)(c′Nξ/Mξ). In addition, f(u) satisfies the following conditions:(i)
f(u)<φp(c′/Mξ) for u∈[0,(T/2η)c′];(ii)
f(u)>φp(b′/Nξ) for u∈[b′,(T/2η)b′];(iii)
f(u)<φp(a′/Lξ) for u∈[0,(T/2r)a′].Then problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 such that
(4.3)0<maxt∈[0,r]𝕋u1(t)<a′<maxt∈[0,r]𝕋u2(t),mint∈[η,T/2]𝕋u2(t)<b′<mint∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<c′.Proof.
By the definition of completely continuous operatorA and its properties, it has to be demonstrated that all the conditions of Lemma 2.3 hold with respect to A. It is easy to obtain that A:P(γ,c)¯→P.
Firstly, we verify that ifu∈∂P(γ,c′), then γ(Au)<c′.
Ifu∈∂P(γ,c′), then
(4.4)γ(u)=maxt∈[0,η]𝕋u(t)=u(η)=c′.
Lemma 2.1 implies that
(4.5)∥u∥≤T2ηu(η)=T2ηc′,
we have
(4.6)0≤u(t)≤T2ηc′,t∈[0,T2]𝕋.
Thus, by   condition (i), one has
(4.7)γ(Au)=maxt∈[0,η]𝕋Au(t)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0ηφq(∫0T/2h(r)f(u(r))∇r)Δs<ηφq(∫0T/2h(r)φp(c′Mξ)∇r)=c′Mξηφq(∫0T/2h(r)∇r)=c′.
Secondly, we show thatβ(Au)>b′  for u∈∂P(β,b′).
If we chooseu∈∂P(β,b′), then β(u)=mint∈[η,T/2]𝕋u(t)=b′. In view of Lemma 2.1, we have
(4.8)∥u∥≤T2ηu(η)=T2ηb′.
So
(4.9)b′≤u(t)≤T2ηb′,t∈[η,T2]𝕋.
Using   condition (ii), we get
(4.10)β(Au)=Au(η)≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>ηφq(∫ηT/2h(r)φp(b′Nξ)∇r)>b′Nξηφq(∫ηT/2h(r)∇r)=b′.
Finally, we prove thatP(α,a′)≠∅ and α(Au)<a′ for all u∈∂P(α,a′).
In fact, the constant functiona′/2∈P(α,a′). Moreover, for u∈∂P(α,a′), we have α(u)=maxt∈[0,r]𝕋u(t)=a′, which implies 0≤u(t)≤a′ for t∈[0,r]𝕋. Hence, ∥u∥≤(T/2r)u(r). Therefore
(4.11)0≤u(t)≤T2ra′,t∈[0,T2]𝕋.
By using  assumption (iii), one has
(4.12)α(Au)=(Au)(r)=∫0rφq(∫sT/2h(r)f(u(r)))∇rΔs≤∫0rφq(∫0T/2h(r)f(u(r))∇r)Δs≤∫0rφq(∫0T/2h(r)φp(a′Lξ)∇r)Δs=a′Lξrφq(∫0T/2h(r)∇r)=a′.
Thus, all the conditions in Lemma 2.3 are satisfied. From (H1) and (H2), we have that the solutions to problem (1.3) do  not vanish identically on any closed subinterval of [0,T]𝕋. Consequently, problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 belonging to P(γ,c′)¯, and satisfying (4.3). The proof is complete.From Theorem4.1, we see that, when assumptions as (i), (ii), and (iii) are imposed appropriately on f, we can establish the existence of an arbitrary odd number of positive symmetric solutions to problem (1.3).Theorem 4.2.
Letl=1,2,…,n. Suppose that there exist positive numbers asl′, bsl′, csl′ such that
(4.13)as1′<2rTbs1′<2rTNξMξcs1′<as2′<2Trbs2′<2rTNξMξcs2′<as3′<⋯<asl′.
In addition, f(u) satisfies the following conditions:(i)
f(u)<φp(csl′/Mξ) for u∈[0,(T/2η)csl′];(ii)
f(u)>φp(bsl′/Nξ) for u∈[bsl′,(T/2η)bsl′];(iii)
f(u)<φp(asl′/Lξ) for u∈[0,(T/2r)asl′].Then problem (1.3) has at least  2l+1 positive symmetric solutions.Proof.
Whenl=1, it is clear that Theorem 4.1 holds. Then we can obtain at least three positive symmetric solutions u1,u2, and u3 satisfying
(4.14)0<maxt∈[0,r]𝕋u1(t)<as1<maxt∈[0,r]𝕋u2(t),minti∈[η,T/2]𝕋u2(t)<bs1<minti∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<cs1.
Following this way, we finish the proof by induction. The proof is complete.Using Lemma2.4, it is easy to have the following results.Theorem 4.3.
Suppose that there are positive numbersa′, b′, c′ such that a′<(Lθ/Mξ)b′<(2η/T)(Lθ/Mξ)c′. In addition, f(u) satisfies the following conditions:(i)
f(u)>φp(c′/Nξ) for u∈[c′,(T/2η)c′];(ii)
f(u)<φp(b′/Mξ) for u∈[0,(T/2η)b′];(iii)
f(u)>φp(a′/Lθ) for u∈[a′,(T/2r)a′].Then problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 such that
(4.15)0<maxt∈[0,r]𝕋u1(t)<a′<maxt∈[0,r]𝕋u2(t),mint∈[η,T/2]𝕋u2(t)<b′<mint∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<c′.From Theorem4.3, we can obtain Theorem 4.4 and Corollary 4.5.Theorem 4.4.
Letl=1,2,…,n. Suppose that there existence positive numbers aλl′, bλl′, cλl′ such that
(4.16)aλ1′<LθMξbλ1′<2ηTLθcλ1′Nξ<aλ2′<LθMξbλ2′<2ηTLθcλ2′Nξ<aλ3′<⋯<aλl′.
In addition, f(u) satisfies the following conditions:(i)
f(u)>φp(cλl′/Nξ) for u∈[cλl′,(T/2η)cλl′];(ii)
f(u)<φp(bλl′/Mξ) for u∈[0,(T/2η)bλl′];(iii)
f(u)>φp(aλl′/Lθ) for u∈[aλl′,(T/2r)aλl′].Then problem (1.3) has at least 2l+1 positive symmetric solutions.Corollary 4.5.
Assume thatf satisfies the following conditions:(i)
f0=∞,f∞=∞,(ii)
there existsc0>0 such that f(u)<φp((2η/T)(c0/Mξ))  for u∈[0,c0],then problem (1.3) has at least three positive symmetric solutions.Proof.
First, by   condition (ii), letb′=(2η/T)c0, one gets
(4.17)f(u)<φp(b′Mξ)foru∈[0,T2ηb′],
which implies that (ii) of   Theorem 4.3   holds.
Second, chooseK3 sufficiently large to satisfy
(4.18)K3Lθ=K3rφq(∫rT/2h(r)∇r)>1.
Since f0=∞, there exists r1′>0 sufficiently small such that
(4.19)f(u)≥φp(K3)φp(u)=φp(K3u)foru∈[0,r1′].
Without loss of generality, suppose r1′≤(LθT/2rMξ)b′. Choose a′>0 such that a′<(2r/T)r1′. For a′≤u≤(T/2r)a′, we have u≤r1′ and a′<(Lθ/Mξ)b′. Thus, by (4.18) and (4.19), we have
(4.20)f(u)≥φp(K3u)≥φp(K3a′)>φp(a′Lθ)foru∈[a′,T2ra′],
this implies that (iii) of Theorem 4.3 is true.
Third, chooseK2 sufficiently large such that
(4.21)K2Nξ=K2ηφq(∫ηT/2h(r)∇r)>1.
Since f∞=∞, there exists r2′>0 sufficiently large such that
(4.22)f(u)≥φp(K2)φp(u)=φp(K2u)foru≥r2′.
Without loss of generality, suppose r2′>(T/2η)b′. Choose c′=r2′. Then
(4.23)f(u)≥φp(K2u)≥φp(K2c′)>φp(c′Nξ)foru∈[c′,T2ηc′],
which means that (i) of Theorem 4.3 holds.
From above analysis, we get(4.24)0<a′<LθMξb′<2ηTLθc′Mξ,
then, all conditions in Theorem 4.3 are satisfied. Hence, problem (1.3) has at least three positive symmetric solutions.In terms of Theorem4.1, we also have the following corollary.Corollary 4.6.
Assume thatf satisfies conditions(i)
f0=0,f∞=0;(ii)
there existsc0>0 such that f(u)>φp((2η/T)(c0/Nξ)) for u∈[(2η/T)c0,c0],then problem (1.3) has at least three positive symmetric solutions.


### 4.2. Result 2

In this subsection, the existence criteria for at leasttriple positive or arbitrary odd positive symmetric solutions to problems (1.3) are established by using the Avery-Peterson fixed point theorem [27].Define the nonnegative continuous convex functionalsϕ and β, nonnegative continuous concave functional λ, and nonnegative continuous functional φ, respectively, on P by(4.25)ϕ(u)=maxt∈[0,T/2]𝕋u(t)=u(T2),β(u)=maxt∈[r,T/2]𝕋κ|uΔ(t)|=|uΔ(r)|,λ(u)=φ(u)=mint∈[η,T/2]𝕋u(t)=u(η).Now, we list and prove the results in this subsection.Theorem 4.7.
Suppose that there exist constantsa*, b*, d* such that 0<a*<(2η/T)b*<(2η/T)(Nξd*/Wξ). In addition, suppose that Wξ>φq(∫ηT/2h(s)∇s) holds, f satisfies the following conditions:(i)
f(u)≤φp(d*/Wξ) for u∈[0,d*];(ii)
f(u)>φp(b*/Nξ) for u∈[b*,d*];(iii)
f(u)<φp(a*/Mξ) for u∈[0,(T/2η)a*],then problem (1.3) has at least three positive symmetric solutions u1,u2,u3 such that
(4.26)∥ui∥≤d*,i=1,2,3,b*<u1(η),u2(η)<b*,u3(η)<a*.Proof.
By the definition of completely continuous operatorA and its properties, it suffices to show that all the conditions of Lemma 2.5 hold with respect to A.
For allu∈P,λ(u)=φ(u)  and  ∥u∥=u(T/2)=ϕ(u). Hence,   condition (2.8) is satisfied.
Firstly, we show thatA:P(ϕ,d*)¯→P(ϕ,d*)¯.
For anyu∈P(ϕ,d*)¯, in view of ϕ(u)=∥u∥≤d* and   assumption (i), one has
(4.27)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0T/2φq(∫0T/2h(r)f(u(r))∇r)Δs≤d*WξT2φq(∫0T/2h(r)∇r)=d*.
From the above analysis, it remains to show that (i)–(iii) of Lemma 2.5 hold.
Secondly, we verify that condition (i) of Lemma2.5 holds, let u(t)≡kb* with k=Wξ/Nξ>1. From the definitions of Nξ,Wξ, and β(u), respectively, it is easy to see that u(t)=kb*>b* and β(u)=0<kb*. In addition, in view of b*<(Nξ/Wξ)d*, we have ϕ(u)=kb*<d*. Thus
(4.28){u∈P(ϕ,β,λ,b*,kb*,d*):λ(x)>b*}≠∅.
For any u∈P(ϕ,β,λ,b*,kb*,d*), then we get b*≤u(t)≤d* for all t∈[η,T/2]𝕋. Hence, by   assumption (ii), we have
(4.29)λ(Au)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>b*Nξηφq(∫ηT/2h(r)∇r)=b*.
Thirdly, we prove that   condition (ii) of Lemma2.5 holds. For any u∈P(ϕ,λ,b*,d*) with β(Au)>kb*, that is,
(4.30)β(Au)=|(Au)Δ(r)|=φq(∫rT/2h(s)f(u(s))∇s)>kb*.
So, in view of k=Wξ/Nξ,  Wξ>φq(∫ηT/2h(s)∇s) and (4.30), one has
(4.31)λ(Au)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>∫0ηφq(∫rT/2h(r)f(u(r))∇r)Δs>ηkb*>b*.
Finally, we check condition (iii) of Lemma2.5. Clearly, since φ(0)=0<a*, we have 0∉R(ϕ,φ,a*,d*). If u∈R(ϕ,φ,a*,d*) with φ(u)=mint∈[η,T/2]𝕋u(t)=a*, then  Lemma 2.1 implies that
(4.32)∥u∥≤T2ηu(η)=T2ηa*.
This yields 0≤u(t)≤(T/2η)a* for all t∈[0,T/2]𝕋. Hence, by assumption (iii), we have
(4.33)λ(Au)=Au(η)≤∫0ηφq(∫0T/2h(r)f(u(r))∇r)Δs<a*Mξηφq(∫0T/2h(r)∇r)=a*.
Consequently, all   conditions of Lemma 2.5 are satisfied. The proof is completed.We remark that   condition (i) in Theorem4.7 can be replaced by the following condition (i′):(4.34)limu→∞f(u)φp(u)≤φp(1Wξ),
which is a special case of (i).Corollary 4.8.
If   condition (i) in Theorem4.7 is replaced by (i′), then the conclusion of Theorem 4.7 also holds.Proof.
By Theorem4.7, we only need to prove that (i') implies that (i) holds, that is, if (i') holds, then there is a number d*≥max{(T/2η)a*,(Wξ/Nξ)b*} such that f(u)≤φp(d*/Wξ) for u∈[0,d*].
Suppose on the contrary that for anyd*≥max{(T/2η)a*,(Wξ/Nξ)b*}, there exists uc∈[0,d*] such that f(uc)>φp(d*/Wξ). Hence, if we choose
(4.35)cn′>max{T2ηa*,WξNξb*}(n=1,2,…)
with cn′→∞, then there exist un∈[0,cn′] such that
(4.36)f(un)>φp(cn′Wξ),
and so
(4.37)limn→∞f(un)=∞.
Since  condition (i′) holds, then there exists τ>0 such that
(4.38)f(u)≤φp(uWξ)foru∈(τ,∞).
Hence, we have un≤τ. Otherwise, if un>τ, then it follows from (4.38) that
(4.39)f(un)≤φp(unWξ)≤φp(cn′Wξ),
which contradicts (4.36).
LetW=maxu∈[0,τ]f(u), then f(un)≤W(n=1,2,…), which also contradicts (4.37). The proof is complete.Theorem 4.9.
Letl=1,2,…,n. Suppose that there exist constants al*, bl*, dl* such that
(4.40)0<a1*<2ηTb1*<2ηTd1*NξWξ<a2*<2ηiTb2*<2ηTd2*NξWξ<a3*<⋯<al*.
In addition, suppose that Wξ>φq(∫ηT/2h(s)∇s) holds, then f satisfies the following conditions:(i)
f(u)<φp(dl*/Wξ) for u∈[0,dl*];(ii)
f(u)>φp(bl*/Nξ) for u∈[bl*,dl*];(iii)
f(u)<φp(al*/Mξ) for u∈[0,(T/2η)al*],then problem (1.3) has at least 2l+1 positive symmetric solutions.Proof.
Similar to the proof of Theorem4.2, we omit it here.


## 4.1. Result 1

In this subsection, in view of the generalized Avery-Henderson fixed-point theorem [26], the existence criteria for at least triple and arbitrary odd positive symmetric solutions to problems (1.3)  are established.Foru∈P, we define the nonnegative, increasing, continuous functionals γ,β, and α by(4.2)γ(u)=maxt∈[0,η]𝕋u(t)=u(η),β(u)=mint∈[η,T/2]𝕋u(t)=u(η),α(u)=maxt∈[0,r]𝕋u(t)=u(r).
It is obvious that γ(u)≤β(u)≤α(u)  for each u∈P. By Lemma 2.1, one obtains ∥u∥≤C*γ(u) for all u∈P,  here C*=T/2η.We now present the results in this subsection.Theorem 4.1.
If there are positive numbersa′, b′, c′ such that a′<(2r/T)b′<(2r/T)(c′Nξ/Mξ). In addition, f(u) satisfies the following conditions:(i)
f(u)<φp(c′/Mξ) for u∈[0,(T/2η)c′];(ii)
f(u)>φp(b′/Nξ) for u∈[b′,(T/2η)b′];(iii)
f(u)<φp(a′/Lξ) for u∈[0,(T/2r)a′].Then problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 such that
(4.3)0<maxt∈[0,r]𝕋u1(t)<a′<maxt∈[0,r]𝕋u2(t),mint∈[η,T/2]𝕋u2(t)<b′<mint∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<c′.Proof.
By the definition of completely continuous operatorA and its properties, it has to be demonstrated that all the conditions of Lemma 2.3 hold with respect to A. It is easy to obtain that A:P(γ,c)¯→P.
Firstly, we verify that ifu∈∂P(γ,c′), then γ(Au)<c′.
Ifu∈∂P(γ,c′), then
(4.4)γ(u)=maxt∈[0,η]𝕋u(t)=u(η)=c′.
Lemma 2.1 implies that
(4.5)∥u∥≤T2ηu(η)=T2ηc′,
we have
(4.6)0≤u(t)≤T2ηc′,t∈[0,T2]𝕋.
Thus, by   condition (i), one has
(4.7)γ(Au)=maxt∈[0,η]𝕋Au(t)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0ηφq(∫0T/2h(r)f(u(r))∇r)Δs<ηφq(∫0T/2h(r)φp(c′Mξ)∇r)=c′Mξηφq(∫0T/2h(r)∇r)=c′.
Secondly, we show thatβ(Au)>b′  for u∈∂P(β,b′).
If we chooseu∈∂P(β,b′), then β(u)=mint∈[η,T/2]𝕋u(t)=b′. In view of Lemma 2.1, we have
(4.8)∥u∥≤T2ηu(η)=T2ηb′.
So
(4.9)b′≤u(t)≤T2ηb′,t∈[η,T2]𝕋.
Using   condition (ii), we get
(4.10)β(Au)=Au(η)≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>ηφq(∫ηT/2h(r)φp(b′Nξ)∇r)>b′Nξηφq(∫ηT/2h(r)∇r)=b′.
Finally, we prove thatP(α,a′)≠∅ and α(Au)<a′ for all u∈∂P(α,a′).
In fact, the constant functiona′/2∈P(α,a′). Moreover, for u∈∂P(α,a′), we have α(u)=maxt∈[0,r]𝕋u(t)=a′, which implies 0≤u(t)≤a′ for t∈[0,r]𝕋. Hence, ∥u∥≤(T/2r)u(r). Therefore
(4.11)0≤u(t)≤T2ra′,t∈[0,T2]𝕋.
By using  assumption (iii), one has
(4.12)α(Au)=(Au)(r)=∫0rφq(∫sT/2h(r)f(u(r)))∇rΔs≤∫0rφq(∫0T/2h(r)f(u(r))∇r)Δs≤∫0rφq(∫0T/2h(r)φp(a′Lξ)∇r)Δs=a′Lξrφq(∫0T/2h(r)∇r)=a′.
Thus, all the conditions in Lemma 2.3 are satisfied. From (H1) and (H2), we have that the solutions to problem (1.3) do  not vanish identically on any closed subinterval of [0,T]𝕋. Consequently, problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 belonging to P(γ,c′)¯, and satisfying (4.3). The proof is complete.From Theorem4.1, we see that, when assumptions as (i), (ii), and (iii) are imposed appropriately on f, we can establish the existence of an arbitrary odd number of positive symmetric solutions to problem (1.3).Theorem 4.2.
Letl=1,2,…,n. Suppose that there exist positive numbers asl′, bsl′, csl′ such that
(4.13)as1′<2rTbs1′<2rTNξMξcs1′<as2′<2Trbs2′<2rTNξMξcs2′<as3′<⋯<asl′.
In addition, f(u) satisfies the following conditions:(i)
f(u)<φp(csl′/Mξ) for u∈[0,(T/2η)csl′];(ii)
f(u)>φp(bsl′/Nξ) for u∈[bsl′,(T/2η)bsl′];(iii)
f(u)<φp(asl′/Lξ) for u∈[0,(T/2r)asl′].Then problem (1.3) has at least  2l+1 positive symmetric solutions.Proof.
Whenl=1, it is clear that Theorem 4.1 holds. Then we can obtain at least three positive symmetric solutions u1,u2, and u3 satisfying
(4.14)0<maxt∈[0,r]𝕋u1(t)<as1<maxt∈[0,r]𝕋u2(t),minti∈[η,T/2]𝕋u2(t)<bs1<minti∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<cs1.
Following this way, we finish the proof by induction. The proof is complete.Using Lemma2.4, it is easy to have the following results.Theorem 4.3.
Suppose that there are positive numbersa′, b′, c′ such that a′<(Lθ/Mξ)b′<(2η/T)(Lθ/Mξ)c′. In addition, f(u) satisfies the following conditions:(i)
f(u)>φp(c′/Nξ) for u∈[c′,(T/2η)c′];(ii)
f(u)<φp(b′/Mξ) for u∈[0,(T/2η)b′];(iii)
f(u)>φp(a′/Lθ) for u∈[a′,(T/2r)a′].Then problem (1.3) has at least three positive symmetric solutions u1, u2, and u3 such that
(4.15)0<maxt∈[0,r]𝕋u1(t)<a′<maxt∈[0,r]𝕋u2(t),mint∈[η,T/2]𝕋u2(t)<b′<mint∈[η,T/2]𝕋u3(t),maxt∈[0,η]𝕋u3(t)<c′.From Theorem4.3, we can obtain Theorem 4.4 and Corollary 4.5.Theorem 4.4.
Letl=1,2,…,n. Suppose that there existence positive numbers aλl′, bλl′, cλl′ such that
(4.16)aλ1′<LθMξbλ1′<2ηTLθcλ1′Nξ<aλ2′<LθMξbλ2′<2ηTLθcλ2′Nξ<aλ3′<⋯<aλl′.
In addition, f(u) satisfies the following conditions:(i)
f(u)>φp(cλl′/Nξ) for u∈[cλl′,(T/2η)cλl′];(ii)
f(u)<φp(bλl′/Mξ) for u∈[0,(T/2η)bλl′];(iii)
f(u)>φp(aλl′/Lθ) for u∈[aλl′,(T/2r)aλl′].Then problem (1.3) has at least 2l+1 positive symmetric solutions.Corollary 4.5.
Assume thatf satisfies the following conditions:(i)
f0=∞,f∞=∞,(ii)
there existsc0>0 such that f(u)<φp((2η/T)(c0/Mξ))  for u∈[0,c0],then problem (1.3) has at least three positive symmetric solutions.Proof.
First, by   condition (ii), letb′=(2η/T)c0, one gets
(4.17)f(u)<φp(b′Mξ)foru∈[0,T2ηb′],
which implies that (ii) of   Theorem 4.3   holds.
Second, chooseK3 sufficiently large to satisfy
(4.18)K3Lθ=K3rφq(∫rT/2h(r)∇r)>1.
Since f0=∞, there exists r1′>0 sufficiently small such that
(4.19)f(u)≥φp(K3)φp(u)=φp(K3u)foru∈[0,r1′].
Without loss of generality, suppose r1′≤(LθT/2rMξ)b′. Choose a′>0 such that a′<(2r/T)r1′. For a′≤u≤(T/2r)a′, we have u≤r1′ and a′<(Lθ/Mξ)b′. Thus, by (4.18) and (4.19), we have
(4.20)f(u)≥φp(K3u)≥φp(K3a′)>φp(a′Lθ)foru∈[a′,T2ra′],
this implies that (iii) of Theorem 4.3 is true.
Third, chooseK2 sufficiently large such that
(4.21)K2Nξ=K2ηφq(∫ηT/2h(r)∇r)>1.
Since f∞=∞, there exists r2′>0 sufficiently large such that
(4.22)f(u)≥φp(K2)φp(u)=φp(K2u)foru≥r2′.
Without loss of generality, suppose r2′>(T/2η)b′. Choose c′=r2′. Then
(4.23)f(u)≥φp(K2u)≥φp(K2c′)>φp(c′Nξ)foru∈[c′,T2ηc′],
which means that (i) of Theorem 4.3 holds.
From above analysis, we get(4.24)0<a′<LθMξb′<2ηTLθc′Mξ,
then, all conditions in Theorem 4.3 are satisfied. Hence, problem (1.3) has at least three positive symmetric solutions.In terms of Theorem4.1, we also have the following corollary.Corollary 4.6.
Assume thatf satisfies conditions(i)
f0=0,f∞=0;(ii)
there existsc0>0 such that f(u)>φp((2η/T)(c0/Nξ)) for u∈[(2η/T)c0,c0],then problem (1.3) has at least three positive symmetric solutions.


## 4.2. Result 2

In this subsection, the existence criteria for at leasttriple positive or arbitrary odd positive symmetric solutions to problems (1.3) are established by using the Avery-Peterson fixed point theorem [27].Define the nonnegative continuous convex functionalsϕ and β, nonnegative continuous concave functional λ, and nonnegative continuous functional φ, respectively, on P by(4.25)ϕ(u)=maxt∈[0,T/2]𝕋u(t)=u(T2),β(u)=maxt∈[r,T/2]𝕋κ|uΔ(t)|=|uΔ(r)|,λ(u)=φ(u)=mint∈[η,T/2]𝕋u(t)=u(η).Now, we list and prove the results in this subsection.Theorem 4.7.
Suppose that there exist constantsa*, b*, d* such that 0<a*<(2η/T)b*<(2η/T)(Nξd*/Wξ). In addition, suppose that Wξ>φq(∫ηT/2h(s)∇s) holds, f satisfies the following conditions:(i)
f(u)≤φp(d*/Wξ) for u∈[0,d*];(ii)
f(u)>φp(b*/Nξ) for u∈[b*,d*];(iii)
f(u)<φp(a*/Mξ) for u∈[0,(T/2η)a*],then problem (1.3) has at least three positive symmetric solutions u1,u2,u3 such that
(4.26)∥ui∥≤d*,i=1,2,3,b*<u1(η),u2(η)<b*,u3(η)<a*.Proof.
By the definition of completely continuous operatorA and its properties, it suffices to show that all the conditions of Lemma 2.5 hold with respect to A.
For allu∈P,λ(u)=φ(u)  and  ∥u∥=u(T/2)=ϕ(u). Hence,   condition (2.8) is satisfied.
Firstly, we show thatA:P(ϕ,d*)¯→P(ϕ,d*)¯.
For anyu∈P(ϕ,d*)¯, in view of ϕ(u)=∥u∥≤d* and   assumption (i), one has
(4.27)∥Au∥=Au(T2)=∫0T/2φq(∫sT/2h(r)f(u(r))∇r)Δs≤∫0T/2φq(∫0T/2h(r)f(u(r))∇r)Δs≤d*WξT2φq(∫0T/2h(r)∇r)=d*.
From the above analysis, it remains to show that (i)–(iii) of Lemma 2.5 hold.
Secondly, we verify that condition (i) of Lemma2.5 holds, let u(t)≡kb* with k=Wξ/Nξ>1. From the definitions of Nξ,Wξ, and β(u), respectively, it is easy to see that u(t)=kb*>b* and β(u)=0<kb*. In addition, in view of b*<(Nξ/Wξ)d*, we have ϕ(u)=kb*<d*. Thus
(4.28){u∈P(ϕ,β,λ,b*,kb*,d*):λ(x)>b*}≠∅.
For any u∈P(ϕ,β,λ,b*,kb*,d*), then we get b*≤u(t)≤d* for all t∈[η,T/2]𝕋. Hence, by   assumption (ii), we have
(4.29)λ(Au)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>b*Nξηφq(∫ηT/2h(r)∇r)=b*.
Thirdly, we prove that   condition (ii) of Lemma2.5 holds. For any u∈P(ϕ,λ,b*,d*) with β(Au)>kb*, that is,
(4.30)β(Au)=|(Au)Δ(r)|=φq(∫rT/2h(s)f(u(s))∇s)>kb*.
So, in view of k=Wξ/Nξ,  Wξ>φq(∫ηT/2h(s)∇s) and (4.30), one has
(4.31)λ(Au)=Au(η)=∫0ηφq(∫sT/2h(r)f(u(r))∇r)Δs≥∫0ηφq(∫ηT/2h(r)f(u(r))∇r)Δs>∫0ηφq(∫rT/2h(r)f(u(r))∇r)Δs>ηkb*>b*.
Finally, we check condition (iii) of Lemma2.5. Clearly, since φ(0)=0<a*, we have 0∉R(ϕ,φ,a*,d*). If u∈R(ϕ,φ,a*,d*) with φ(u)=mint∈[η,T/2]𝕋u(t)=a*, then  Lemma 2.1 implies that
(4.32)∥u∥≤T2ηu(η)=T2ηa*.
This yields 0≤u(t)≤(T/2η)a* for all t∈[0,T/2]𝕋. Hence, by assumption (iii), we have
(4.33)λ(Au)=Au(η)≤∫0ηφq(∫0T/2h(r)f(u(r))∇r)Δs<a*Mξηφq(∫0T/2h(r)∇r)=a*.
Consequently, all   conditions of Lemma 2.5 are satisfied. The proof is completed.We remark that   condition (i) in Theorem4.7 can be replaced by the following condition (i′):(4.34)limu→∞f(u)φp(u)≤φp(1Wξ),
which is a special case of (i).Corollary 4.8.
If   condition (i) in Theorem4.7 is replaced by (i′), then the conclusion of Theorem 4.7 also holds.Proof.
By Theorem4.7, we only need to prove that (i') implies that (i) holds, that is, if (i') holds, then there is a number d*≥max{(T/2η)a*,(Wξ/Nξ)b*} such that f(u)≤φp(d*/Wξ) for u∈[0,d*].
Suppose on the contrary that for anyd*≥max{(T/2η)a*,(Wξ/Nξ)b*}, there exists uc∈[0,d*] such that f(uc)>φp(d*/Wξ). Hence, if we choose
(4.35)cn′>max{T2ηa*,WξNξb*}(n=1,2,…)
with cn′→∞, then there exist un∈[0,cn′] such that
(4.36)f(un)>φp(cn′Wξ),
and so
(4.37)limn→∞f(un)=∞.
Since  condition (i′) holds, then there exists τ>0 such that
(4.38)f(u)≤φp(uWξ)foru∈(τ,∞).
Hence, we have un≤τ. Otherwise, if un>τ, then it follows from (4.38) that
(4.39)f(un)≤φp(unWξ)≤φp(cn′Wξ),
which contradicts (4.36).
LetW=maxu∈[0,τ]f(u), then f(un)≤W(n=1,2,…), which also contradicts (4.37). The proof is complete.Theorem 4.9.
Letl=1,2,…,n. Suppose that there exist constants al*, bl*, dl* such that
(4.40)0<a1*<2ηTb1*<2ηTd1*NξWξ<a2*<2ηiTb2*<2ηTd2*NξWξ<a3*<⋯<al*.
In addition, suppose that Wξ>φq(∫ηT/2h(s)∇s) holds, then f satisfies the following conditions:(i)
f(u)<φp(dl*/Wξ) for u∈[0,dl*];(ii)
f(u)>φp(bl*/Nξ) for u∈[bl*,dl*];(iii)
f(u)<φp(al*/Mξ) for u∈[0,(T/2η)al*],then problem (1.3) has at least 2l+1 positive symmetric solutions.Proof.
Similar to the proof of Theorem4.2, we omit it here.


## 5. Examples

In this section, we give two simple examples to illustrate that the conclusions we will arrive at are different with their own distinctive advantages.Example 5.1.
Let(5.1)𝕋={0,0.05,0.1,0.15,0.45,0.48,0.5,0.55,0.52,0.85,0.9,0.95,1}∪[0.22,0.44]∪[0.56,0.78].
Consider the following boundary value problem withp=3:
(5.2)(φp(uΔ(t)))∇+h(t)f(u(t))=0,t∈(0,1)𝕋,u(0)=u(1)=0,uΔ(0)=-uΔ(1),
where
(5.3)h(t)={t+ρ(t),t∈[0,0.5]𝕋,1-t+ρ(1-t),t∈[0.5,1]𝕋,f(u)={0.16,u∈[0,0.6],1049.6u-629.6,u∈[0.6,1],420,u∈[1,5],2.8364u+405.82,u∈[5,60],576,u∈[60,∞).
Note that T=1. If we  choose η=0.1, r=0.25, then direct calculation shows that
(5.4)Mξ=ηφq(∫01/2h(r)∇r)=0.05,
and Nξ≈0.0489,Lξ=0.125,Wξ=0.2500. If we take a′=0.3, b′=1, c′=12, then
(5.5)a′=0.3<2rTb′≈0.9714<2rTc′NξMξ≈9.173.
Furthermore,
(5.6)f(u)=0.16<5.76=φp(a′Lξ),u∈[0,0.6],f(u)=420>418.2≈φp(b′Nξ),u∈[1,5],f(u)≤576<57600=φp(c′Mξ),u∈[0,60].
By  Theorem 4.1 we see that the boundary value problem (5.2) has at least three positive symmetric solutions u1,u2, and u3 such that
(5.7)0<u1(0.25)<0.3<u2(0.25),u2(0.1)<1<u3(0.1),u3(0.1)<12.
Yet, φq(∫0.11/2h(s)∇s)≈0.489>Wξ≈0.25, hence, the existence of positive solutions of boundary value problem (5.2) is not obtained by using  Theorem 4.7.Example 5.2.
Let(5.8)𝕋={0,0.05,0.1,0.15,0.45,0.46,0.48,0.5,0.52,0.54,0.55,0.85,0.9,0.95,1}∪[0.22,0.44]∪[0.56,0.78].
Consider the following boundary value problem:(5.9)(φp(uΔ(t)))∇+h(t)f(u(t))=0,t∈(0,1)𝕋,u(0)=u(1)=0,uΔ(0)=-uΔ(1),
where
(5.10)h(t)={t+ρ(t),t∈[0,0.5]𝕋,1-t+ρ(1-t),t∈[0.5,1]𝕋.
Note that T=1. If we choose η=0.45, r=0.48, then a direct calculation shows that
(5.11)Mξ=ηφq(∫01/2h(r)∇r)=0.225,Nξ≈0.0981,Lξ≈0.24,Wξ≈0.25,Wξ≈0.25>0.218≈φq(∫0.450.5h(r)∇r).
Let ε be an arbitrary small positive number, a*,b*,  and d* are arbitrary positive numbers with
(5.12)0<a*<2ηb*<2ηNξd*Wξ.
That is
(5.13)0<a*<0.9b*<0.3924d*,f(u)={φp(a*0.225)-ε,0≤u≤a*,l(u),a*≤u≤b*,φp(b*0.0981)+ε,b*≤u≤d*,
where  l(u) satisfy l(a*)=φp(a*/0.225)-ε, l(b*)=φp(b*/0.0981)+ε, (lΔ(u))Δ=0, u∈[a*,b*].
It is obvious that   (i),  (ii), and (iii)  in  Theorem4.7 are satisfied. By  Theorem 4.7, we see that the boundary value problem (5.9) has at least three positive solutions u1,u2, and u3 such that
(5.14)∥ui∥≤d*,i=1,2,3,b*<u1(0.45),a*<u2(0.45),u3(0.45)<a*.
However, for arbitrary positive numbersa*, b*, d*,   condition (iii) of  Theorem 4.1 is not satisfied. Therefore,  Theorem 4.1 is not fit to study the boundary value problem (5.9).



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*Source: 141929-2009-12-15.xml*