# Existence Theory for Pseudo-Symmetric  Solution top-Laplacian Differential  Equations Involving Derivative

Authors: You-Hui Su; Weili Wu; Xingjie Yan
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We all-sidedly consider a three-point boundary value problem forp-Laplacian differential
equation with nonlinear term involving derivative. Some new sufficient conditions
are obtained for the existence of at least one, triple, or arbitrary odd positive pseudosymmetric solutions by using pseudosymmetric technique and fixed-point theory in cone. As an application, two examples are given to illustrate the main results.


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## Body

## 1. Introduction

Recent research results indicate that considerable achievement was made in the existence of positive solutions to dynamic equations; for details, please see [1–6] and the references therein. In particular, the existence of positive pseudosymmetric solutions to p-Laplacian difference and differential equations attract many researchers' attention, such as [7–11]. The reason is that the pseudosymmetry problem not only has theoretical value, such as in the study of metric manifolds [12], but also has practical value itself; for example, we can apply this characteristic into studying the chemistry structure [13]. On another hand, there are much attention paid to the positive solutions of boundary value problems (BVPs) for differential equation with the nonlinear term involved with the derivative explicitly [14–18]. Hence, it is natural to continue study pseudosymmetric solutions to p-Laplacian differential equations with the nonlinear term involved with the first-order derivative explicitly.First, let us recall some relevant results about BVPs withp-Laplacian, We would like to mention the results of Avery and Henderson [7, 8],  Ma and Ge [11] and  Sun and Ge [16]. Throughout this paper, we denote the p-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 the three-point BVPs withp-Laplacian (1.1)(φp(u′(t)))′+h(t)f(t,u(t))=0fort∈[0,1],u(0)=0,u(η)=u(1),
here, η∈(0,1) is constant, by using the five functionals fixed point theorem in a cone [19],  Avery and Henderson [8] established the existence of at least three positive pseudosymmetric solutions to BVPs (1.1). The authors also obtained the similar results in their paper [7] for the discrete case. In addition,  Ma and Ge [11] developed the existence of at least two positive pseudosymmetric solutions to BVPs (1.1) by using the monotone iterative technique.For the three-pointp-Laplacian BVPs with dependence on the first-order derivative (1.2)(φp(u′(t)))′+h(t)f(t,u(t),u′(t))=0fort∈[0,1],u(0)=0,u(η)=u(1),
Sun and Ge [16] obtained the existence of at least two positive pseudosymmetric solutions to BVPs (1.2) via the monotone iterative technique again. However, it is worth mentioning that the above-mentioned papers [7, 8, 10, 11, 16], the authors only considered results on the existence of positive pseudosymmetric solutions partly, they failed to further provide comprehensive results on the existence of positive pseudosymmetric solutions to p-Laplacian. Naturally, in this paper, we consider the existence of positive pseudosymmetric solutions for p-Laplacian differential equations in all respects.Motivated by the references [7, 8, 10, 11, 16, 18], in present paper, we consider all-sidedly p-Laplacian BVPs (1.2), using the compression and expansion fixed point theorem [20] and Avery-Peterson fixed point theorem [21]. We obtain that there exist at least one, triple or arbitrary oddpositive pseudosymmetric solutions to problem (1.2). In particular, we not only get some local properties of pseudosymmetric solutions, but also obtain that the position of pseudosymmetric solutions is determined under some conditions, which is much better than the results in papers [8, 11, 16]. Correspondingly, we generalize and improve the results in papers Avery and Henderson [8]. From the view of applications, two examples are given to illustrate the main results.Throughout this paper, we assume that(S1)
f(t,u,u′):[0,1]×[0,∞)×(-∞,+∞)→[0,∞) is continuous, does not vanish identically on interval [0,1], and f(t,u,u′) is pseudosymmetric about η on [0,1],(S2)
h(t)∈L([0,1],[0,∞)) is pseudosymmetric about η on [0,1], and does not vanish identically on any closed subinterval of [0,1]. Furthermore, 0<∫01h(t)dt<∞.


## 2. Preliminaries

In the preceding of this section, we state the definition of cone and several fixed point theorems needed later [20, 22]. In the rest of this section, we will prove that solving BVPs (1.2) is equivalent to finding the fixed points of a completely continuous operator.We first list the definition of cone and the compression and expansion fixed point theorem [20, 22].Definition 2.1.
LetE be a real Banach space. A nonempty, closed, convex set P⊂E is said to be a cone provided the following conditions are satisfied:(i)
ifx∈P and λ≥0, then λx∈P,(ii)
ifx∈P and -x∈P, then x=0.Lemma 2.2 (see [20, 22]).
LetP be a cone in a Banach space  E. Assume that  Ω1,Ω2 are open bounded 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∥,∀x∈P∩∂Ω1and∥Ax∥≥∥x∥,∀x∈P∩∂Ω2, or(ii)
∥Ax∥≥∥x∥,∀x∈P∩∂Ω1and∥Ax∥≤∥x∥,∀x∈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}.Letγ and θ be nonnegative continuous convex functionals on P, α a nonnegative continuous concave functional on P, and ψ a nonnegative continuous functional on P respectively. We define the following convex sets: (2.1)P(γ,α,b,d)={x∈P:b≤α(x),γ(x)≤d},P(γ,θ,α,b,c,d)={x∈P:b≤α(x),θ(x)≤c,γ(x)≤d},
and a closed set R(γ,ψ,a,d)={x∈P:a≤ψ(x),γ(x)≤d}.Next, we list the fixed point theorem due to Avery-Peterson [21].Lemma 2.3 (see [21]).
LetP be a cone in a real Banach space  E and  γ,θ,α,ψ defined as above; moreover,  ψ satisfies  ψ(λ′x)≤λ′ψ(x) for  0≤λ′≤1 such that for some positive numbers  h and  d,
(2.2)α(x)≤ψ(x),‖x‖≤hγ(x),
for all x∈P(γ,d)¯. Suppose that A:P(γ,d)¯→P(γ,d)¯ is completely continuous and there exist positive real numbers a,b,c with a<b such that(i)
{x∈P(γ,θ,α,b,c,d):α(x)>b}≠∅ and α(A(x))>b for x∈P(γ,θ,α,b,c,d),(ii)
α(A(x))>b for x∈P(γ,α,b,d) with θ(A(x))>c,(iii)
0∉R(γ,ψ,a,d) and ψ(A(x))<a for all x∈R(γ,ψ,a,d) with ψ(x)=a.Then, A has at least three fixed points x1,x2,x3∈P(γ,d)¯ such that
(2.3)γ(xi)≤dfori=1,2,3,b<α(x1),a<ψ(x2),α(x2)<bwithψ(x3)<a.Now, letE=C1([0,1],ℝ). Then, E is a Banach space with norm (2.4)‖u‖=max⁡{max⁡t∈[0,1]|u(t)|,max⁡t∈[0,1]|u′(t)|}.
Define a cone P⊂E by (2.5)P={u∈E∣u(0)=0,uisconcave,nonnegativeon⁡[0,1]anduissymmetricon[η,1]}.The following lemma can be founded in [11], which is necessary to prove our result.Lemma 2.4 (see [11]).
Ifu∈P, then the following statements are true:   (i)
u(t)≥(u(ω1)/ω1)min{t,1+η-t} for t∈[0,1], here ω1=(η+1)/2,(ii)
u(t)≥(η/ω1)u(ω1) for t∈[η,ω1],(iii)
maxt∈[0,1]u(t)=u(ω1).Lemma 2.5.
Ifu∈P, then the following statements are true:  (i)
u(t)≤maxt∈[0,1]|u′(t)|,(ii)
∥u(t)∥=maxt∈[0,1]|u′(t)|=max{|u′(0)|,|u′(1)|},(iii)
mint∈[0,ω1]u(t)=u(0)andmint∈[ω1,1]u(t)=u(1).Proof.
(i) Since(2.6)u(t)=u(0)+∫0tu′(t)dtfort∈[0,1],
which reduces to
(2.7)u(t)≤∫0t|u′(t)|dt≤max⁡t∈[0,1]|u′(t)|.
(ii) By usingu′′(t)≤0 for t∈[0,1], we have u′(t) is monotone decreasing function on [0,1]. Moreover,
(2.8)max⁡t∈[0,1]u(t)=u(η+12)=u(ω1),
which implies that u′(ω1)=0, so, u′(t)≥0 for t∈[0,ω1] and u′(t)≤0 for t∈[ω1,1].Now, we define the operatorA:P→E by (2.9)(Au)(t)={∫0tφq(∫sω1h(r)f(r,u(r),u′(r))dr)dsfort∈[0,ω1],w(η)+∫t1φq(∫ω1sh(r)f(r,u(r),u′(r))dr)dsfort∈[ω1,1],
here, w(η)=(Au)(η).Lemma 2.6.
A:P→P is a completely continuous operator.Proof.
In fact,(Au)(t)≥0 for t∈[0,1], (Au)(η)=(Au)(1) and (Au)(0)=0.
It is easy to see that the operatorA is pseudosymmetric about ω1 on [0,1].
In fact, fort∈[η,ω1], we have 1-t+η∈[ω1,1], and according to the integral transform, one has
(2.10)∫1-t+η1φq(∫ω1sh(r)f(r,u(r),u′(r))dr)ds=∫ηtφq(∫s1ω1h(r1)f(r1,u(r1),u′(r1))dr1)ds1,
here, s=1-s1+η,r=1-r1+η. Hence,
(2.11)(Au)(1-t+η)=w(η)+∫1-t+η1φq(∫ω1sh(r)f(r,u(r),u′(r))dr)ds=w(η)+∫ηtφq(∫s1ω1h(r1)f(r1,u(r1),u′(r1))dr1)ds1=∫0ηφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds+∫ηtφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds=∫0tφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds=(Au)(t).
Fort∈[ω1,1], we note that 1-t+η∈[η,ω1], by using the integral transform, one has
(2.12)∫η1-t+ηφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds=∫t1φq(∫ω1s1h(r1)f(r1,u(r1),u′(r1))dr1)ds1,
where s=1-s1+η,r=1-r1+η. Thus,
(2.13)(Au)(1-t+η)=∫01-t+ηφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds=w(η)+∫η1-t+ηφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds=w(η)+∫t1φq(∫ω1s1h(r1)f(r1,u(r1),u′(r1))dr1)ds1=w(η)+∫t1φq(∫ω1sh(r)f(r,u(r),u′(r))dr)ds=(Au)(t).
Hence,A is pseudosymmetric about η on [0,1].
In addition,(2.14)(Au)′(t)=φq(∫tω1h(r)f(r,u(r),u′(r))ds)≥0,t∈[0,ω1]
is continuous and nonincreasing in [0,ω1]; moreover, φq(x) is a monotone increasing continuously differentiable function
(2.15)(∫tω1h(s)f(s,u(s),u′(s))ds)′=-h(t)f(t,u(t),u′(t))≤0,t∈[0,ω1],
it is easy to obtain (Au)′′(t)≤0fort∈[0,ω1]. By using the similar way, we can deduce (Au)′′(t)≤0fort∈[ω1,1]. So, A:P→P. It is easy to obtain that A:P→P is completely continuous.Hence, the solutions of BVPs (1.2) are fixed points of the completely continuous operator A.


## 3. One Solutions

In this section, we will study the existence of one positive pseudosymmetric solution to problem (1.2) by Krasnosel'skii's fixed point theorem in a cone.Motivated by the notations in reference [23], for u∈P, let (3.1)f0=sup⁡t∈[0,1]lim⁡(u,u′)→(0,0)f(t,u,u′)φp(|u′|),f0=inf⁡t∈[0,1]lim⁡(u,u′)→(0,0)f(t,u,u′)φp(|u′|),f∞=sup⁡t∈[0,1]lim⁡(u,u′)→(∞,∞)f(t,u,u′)φp(|u′|),f∞=inf⁡t∈[0,1]lim⁡(u,u′)→(∞,∞)f(t,u,u′)φp(|u′|).In the following, we discuss the problem (1.2) under the following four possible cases.Theorem 3.1.
Iff0=0 and  f∞=∞, problem  (1.2) has at least one positive pseudosymmetric solution  u.Proof.
In view off0=0, there exists an H1>0 such that
(3.2)f(t,u,u′)≤φp(ε)φp(|u′|)=φp(ε|u′|)for(t,u,u′)∈[0,1]×(0,H1]×[-H1,H1],
here, ε>0 and satisfies
(3.3)εφq(∫0ω1h(s)ds)≤1.
Ifu∈P with ∥u∥=H1, by Lemma 2.5, we have
(3.4)u(t)≤max⁡t∈[0,1]|u′(t)|≤‖u‖=H1fort∈[0,1],
hence,
(3.5)‖Au‖=max⁡{|(Au)′(0)|,|(Au)′(1)|}=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≤εmax⁡t∈[0,1]|u′(t)|φq(∫0ω1h(s)ds)≤‖u‖.
If set ΩH1={u∈E:∥u∥<H1}, one has ∥Au∥≤∥u∥foru∈P∩∂ΩH1.
According tof∞=∞, there exists an H2′>0 such that
(3.6)f(t,u,u′)≥max⁡t∈[0,1]φp(k)φp(|u′|)=max⁡t∈[0,1]φp(k|u′|),
where (t,u,u′)∈[0,1]×[H2′,∞)×(-∞,H2′]∪[H2′,∞), k>0 and satisfies
(3.7)kφq(∫ω11h(r)dr)≥1.
Set
(3.8)H2=max⁡{2H1,ω1ηH2′},ΩH2*={u∈E:‖u‖<5H2},ΩH2={u∈ΩH2*:u(ω1)<H2}.
Foru∈P∩∂ΩH2, we have u(ω1)=H2 since u(t)≤|u′(t)|foru∈P.
Ifu∈P with u(ω1)=H2, Lemmas 2.4 and 2.5 reduce to
(3.9)min⁡t∈[ω1,1]|u′(t)|≥min⁡t∈[ω1,1]u(t)=u(1)≥ηu(ω1)ω1≥H2′.
For u∈P∩∂ΩH2, according to (3.6), (3.7) and (3.9), we get
(3.10)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≥φq(∫ω11h(r)f(r,u(r),u′(r))dr)≥kmax⁡t∈[0,1]|u′(t)|φq(∫1ω1h(r)dr)=‖u‖.
Thus, by (i) of Lemma 2.2, the problem (1.2) has at least one positive pseudosymmetric solution u in P∩(Ω¯H2∖ΩH1).Theorem 3.2.
Iff0=∞ and  f∞=0, problem  (1.2) has at least one positive pseudosymmetric solution  u.Proof.
Sincef0=∞, there exists an H3>0 such that
(3.11)f(t,u,u′)≥max⁡t∈[0,1]φp(m)φp(|u′|)=max⁡t∈[0,1]φp(m|u′|),
here, (t,u,u′)∈[0,1]×(0,H3]×[-H3,H3] and m is such that
(3.12)mφq(∫ω11h(r)dr)≥1.
If u∈P with ∥u∥=H3, Lemma 2.5 implies that
(3.13)u(t)≤max⁡t∈[0,1]|u′(t)|≤‖u‖=H3fort∈[0,1],
now, by (3.11), (3.12), and (3.13), we have
(3.14)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≥φq(∫ω11h(r)f(r,u(r),u′(r))dr)≥mmax⁡t∈[0,1]|u′(t)|φq(∫ω11h(r)dr)=‖u‖.
If let ΩH3={u∈E:∥u∥<H3}, one has ∥Au∥≥∥u∥foru∈P∩∂ΩH3.
Now, we considerf∞=0.
Suppose thatf is bounded, for some constant K>0, then
(3.15)f(t,u,u′)≤φp(K)∀(t,u,u′)∈[0,1]×[0,∞)×(-∞,∞).
Pick
(3.16)H4≥max⁡{H4′,2H3,Kφq(∫0ω1h(s)ds),Cδ},
here, C is an arbitrary positive constant and satisfy the (3.21). Let
(3.17)ΩH4={u∈E:∥u∥<H4}.
Ifu∈P∩∂ΩH4, one has ∥u∥=H4, then (3.15) and (3.16) imply that
(3.18)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≤Kφq(∫0ω1h(s)ds)≤H4=‖u‖.
Suppose thatf is unbounded.
By definition off∞=0, there exists H4′>0 such that
(3.19)f(t,u,u′)≤φp(δ)φp(|u′|)=φp(δ|u′|),
where (t,u,u′)∈[0,ω1]×[H4′,∞)×(-∞,H4′]∪[H4′,∞) and δ>0 satisfies
(3.20)δφq(∫0ω1h(s)ds)≤1.
Fromf∈C([0,1]×[0,+∞)×(-∞,∞),[0,+∞)), we have
(3.21)f(t,u,u′)≤φp(C)for(t,u,u′)∈[0,1]×[0,H4′]×[-H4′,H4′],
here, C is an arbitrary positive constant.
Then,   for(t,u,u′)∈[0,1]×[0,∞)×(-∞,∞), we have
(3.22)f(t,u,u′)≤max⁡{φp(C),φp(δ)φp(|u′|)}.
Ifu∈P∩∂ΩH4, one has ∥u∥=H4, which reduces to
(3.23)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≤max⁡{C,δ‖u′‖}φq(∫0ω1h(r)dr)≤H4=‖u‖.
Consequently, for any cases, if we take ΩH4={u∈E:∥u∥<H4}, we have ∥Au∥≤∥u∥foru∈P∩∂ΩH4. Thus, the condition (ii) of Lemma 2.2 is satisfied.
Consequently, the problem (1.2) has at least one positive pseudosymmetric solution
(3.24)u∈P∩(Ω¯H4∖ΩH3)withH3≤‖u‖≤H4.Theorem 3.3.
Suppose that the following conditions hold:(i)
there exist nonzero finite constantsc1 and  c2 such that  f0=c1 and  f∞=c2,(ii)
there exist nonzero finite constantsc3 and  c4 such that  f0=c3 and  f∞=c4.Then, problem (1.2) has at least one positive pseudosymmetric solution  u.Proof.
(i) In view off0=c1, there exists an H5>0 such that
(3.25)f(t,u,u′)≤φp(ε+c11)φp(|u′|)=φp((ε+c11)|u′|)for(t,u,u′)∈[0,1]×(0,H5]×[-H5,H5],
here, c1=φp(c11+ε),ε>0 and satisfies
(3.26)(ε+c11)φq(∫0ω1h(s)ds)≤1.
Ifu∈P with ∥u∥=H5, by Lemma 2.5, we have
(3.27)u(t)≤|u′(t)|≤‖u‖=H5fort∈[0,1],
hence,
(3.28)‖Au‖=max⁡{|(Au)′(0)|,|(Au)′(1)|}=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≤(ε+c11)max⁡t∈[0,1]|u′(t)|φq(∫0ω1h(s)ds)≤‖u‖.
If set ΩH5={u∈E:∥u∥<H5}, one has ∥Au∥≤∥u∥foru∈P∩∂ΩH5.
According tof∞=c2, there exists an H6′>0 such that
(3.29)f(t,u,u′)≥max⁡t∈[0,1]φp(c22-ε)φp(|u′|)=max⁡t∈[0,1]φp((c22-ε)|u′|),
where (t,u,u′)∈[0,1]×[H6′,∞)×(-∞,H6′]∪[H6′,∞), c2=φp(c22-ε),ε>0 and satisfies
(3.30)(c22-ε)φq(∫ω11h(r)dr)≥1.
Set
(3.31)H6=max⁡{2H5,ω1ηH6′},ΩH6*={u∈E:‖u‖<5H6},ΩH6={u∈ΩH6*:u(ω1)<H6}.
If u∈P with u(ω1)=H6, Lemmas 2.4 and 2.5 reduce to
(3.32)min⁡t∈[ω1,1]|u′(t)|≥min⁡t∈[ω1,1]u(t)=u(1)≥ηu(ω1)ω1≥H6′.
For u∈P∩∂ΩH6, according to (3.29), (3.30) and (3.32), we get
(3.33)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≥φq(∫ω11h(r)f(r,u(r),u′(r))dr)≥(c22-ε)max⁡t∈[0,1]|u′(t)|φq(∫1ω1h(r)dr)=‖u‖.
Thus, by (i) of Lemma 2.2, the problem (1.2) has at least one positive pseudosymmetric solution u in P∩(Ω¯H6∖ΩH5).
(ii) By using the similar way as to Theorem3.2, we can prove to it.


## 4. Triple Solutions

In the previous section, some results on the existence of at least one positive pseudosymmetric solutions to problem (1.2) are obtained. In this section, we will further discuss the existence criteria for at least three and arbitrary odd positive pseudosymmetric solutions of problems (1.2) by using the Avery-Peterson fixed point theorem [21].Choose ar∈(η,ω1), for the notational convenience, we denote (4.1)M=ω1φq(∫0ω1h(r)dr),N=ηφq(∫ηω1h(r)dr),W=φq(∫0ω1h(r)dr).Define the nonnegative continuous convex functionalsθ and γ, nonnegative continuous concave functional α, and nonnegative continuous functional φ, respectively, on P by (4.2)γ(u)=max⁡t∈[0,1]|u′(t)|=max⁡{u′(0),u′(1)}=‖u‖,ψ(u)=θ(u)=max⁡t∈[0,ω1]u(t)=u(ω1)≤‖u‖,α(u)=min⁡t∈[η,ω1]u(t)=u(η).Now, we state and prove the results in this section.Theorem 4.1.
Suppose that there exist constantsa*,b*, and  d* such that  0<a*<b*<(N/W)d*. In addition,  f satisfies the following conditions: (i)
f(t,u,u′)≤φp(d*/W) for (t,u,u′)∈[0,1]×[0,d*]×[-d*,d*],(ii)
f(t,u,u′)>φp(b*/N) for (t,u,u′)∈[η,ω1]×[b*,d*]×[-d*,d*],(iii)
f(t,u,u′)<φp(a*/M) for (t,u,u′)∈[0,ω1]×[0,a*]×[-d*,d*].
Then, problem (1.2) has at least three positive pseudosymmetric solutions u1,u2,andu3 such that
(4.3)‖xi‖≤d*fori=1,2,3,b*<min⁡t∈[η,ω1]u1(t),a*<max⁡t∈[0,1]u2(t),min⁡t∈[η,ω1]u2(t)<b*withmax⁡t∈[0,1]u3(t)<a*.Proof.
According to the definition of completely continuous operatorA and its properties, we need to show that all the conditions of Lemma 2.3 hold with respect to A.
It is obvious that(4.4)ψ(λ′u)=λ′u(ω1)=λ′ψ(u)for0<λ′<1,α(u)≤ψ(u)∀u∈P,‖u‖=γ(u)∀u∈P.
Firstly, we show thatA:P(γ,d*)¯→P(γ,d*)¯.
For anyu∈P(γ,d*)¯, we have
(4.5)u(t)≤max⁡t∈[0,1]|u′(t)|≤‖u‖=γ(u)≤d*fort∈[0,1],
hence, the assumption (i) implies that
(4.6)‖Au‖=max⁡{φq(∫0ω1h(r)f(r,u(r),u′(r))dr),φq(∫ω11h(r)f(r,u(r),u′(r))dr)}≤d*Wφq(∫0ω1h(r)dr)=d*.
From the above analysis, it remains to show that (i)–(iii) of Lemma 2.3 hold.
Secondly, we verify that condition (i) of Lemma2.3 holds; let u(t)≡(tb*/η)+b*,t∈[0,1], and it is easy to see that
(4.7)α(u)=u(η)=2b*>b*,θ(u)=u(ω1)=ω1b*η+b*≤ω1b*η+b*,
in addition, we have γ(u)=(b*/η)<d*, since b*<(N/W)d*. Thus
(4.8){u∈P(γ,θ,α,b*,ω1b*η+b*,d*):α(x)>b*}≠∅.
For any
(4.9)u∈P(γ,θ,α,b*,ω1b*η+b*,d*),
one has
(4.10)b*≤u(t)≤∥u∥≤d*∀t∈[η,ω1],
it follows from the assumption (ii) that
(4.11)α(Au)=(Au)(η)=∫0ηφq(∫sω1h(r)f(r,u(r),u′(r))dr)ds≥∫0ηφq(∫ηω1h(r)f(r,u(r),u′(r))dr)ds>b*Nηφq(∫ηω1h(r)dr)=b*.
Thirdly, we prove that the condition (ii) of Lemma2.3 holds. In fact,
(4.12)α(Au)=Au(η),θ(Au)=max⁡t∈[0,ω1]A(u)=Au(ω1).
For any u∈P(γ,α,b*,d*)withθ(Au)>(ω1b*/η)+b*, we have
(4.13)α(Au)=Au(η)≥ηω1Au(ω1)≥ηω1θ(Au)=b*+ω1b*η>b*.
Finally, we check condition (iii) of Lemma2.3.
Clearly, sinceψ(0)=0<a*, we have 0∉R(γ,ψ,a*,d*). If
(4.14)u∈R(γ,ψ,a*,d*)withψ(u)=max⁡t∈[0,ω1]u(t)=u(ω1)=a*,
then
(4.15)0≤u(t)≤a*∀t∈[0,ω1],max⁡t∈[0,1]|u′(t)|=‖u‖=γ(u)≤d*.
Hence, by assumption (iii), we have
(4.16)ψ(Au)=(Au)(ω1)≤∫0ω1φq(∫0ω1h(r)f(r,u(r),u′(r))dr)ds<a*Mω1φq(∫0ω1h(r)dr)=a*.
Consequently, from above, all the conditions of Lemma 2.3 are satisfied. The proof is completed.Corollary 4.2.
If the condition(i) in Theorem 4.1 is replaced by the following condition  (i′):(i′)
lim(u,u′)→(∞,∞)(f(t,u,u′)/(φp(|u′|)))≤φp(1/W), then the conclusion of Theorem 4.1 also holds.Proof.
From Theorem4.1, we only need to prove that (i′) implies that (i) holds. That is, assume that (i′) holds, then there exists a number d*≥(W/N)b* such that
(4.17)f(t,u,u′)≤φp(d*W)for(t,u,u′)∈[0,1]×[0,d*]×[-d*,d*].
Suppose on the contrary that for anyd*≥(W/N)b*, there exists (uc,u′c)∈[0,d*]×[-d*,d*] such that
(4.18)f(t,uc,uc′)>φp(d*W)fort∈[0,1].
Hence, if we choose cn*>(W/N)b*(n=1,2,…)withcn*→∞, then there exist (un,un′)∈[0,cn*]×[-cn*,cn*] such that
(4.19)f(t,un,un′)>φp(cn*W)fort∈[0,1],
and so
(4.20)lim⁡n→∞f(t,un,un′)=∞fort∈[0,1].
Since the condition (i′) holds, there exists τ>0 satisfying
(4.21)f(t,u,u′)≤φp(|u′|W)for(t,u,u′)∈[0,1]×[τ,∞)×(-∞,τ]∪[τ,∞).
Hence, we have
(4.22)un<|un′|≤τ.
Otherwise, if
(4.23)|un′|>un>τfort∈[0,1],
it follows from (4.21) that
(4.24)f(t,un,un′)≤φp(unW)≤φp(cn*W)fort∈[0,1],
which contradicts (4.19).
Let(4.25)W=max⁡(t,u,u′)∈[0,1]×[0,τ]×[-τ,τ]f(t,u,u′),
then
(4.26)f(t,un,un′)≤W(n=1,2,…),
which also contradicts (4.20).Theorem 4.3.
Suppose that there exist constantsai*,bi*,anddi* such that
(4.27)0<a1*<b1*<NWd2*<a2*<b2*<NWd3*<⋯<an*<bn*<NWdn+1*,
here, n∈ℕ and i=1,2,…,n. In addition, suppose that f satisfies the following conditions:(i)
f(t,u,u′)≤φp(di*/W) for (t,u,u′)∈[0,1]×[0,di*]×[-di*,di*],(ii)
f(t,u,u′)>φp(bi*/N) for (t,u,u′)∈[η,ω1]×[bi*,di*]×[-di*,di*],(iii)
f(t,u,u′)<φp(ai*/M) for (t,u,u′)∈[0,ω1]×[0,ai*]×[-di*,di*].Then, problem (1.2) has at least 2n-1 positive pseudosymmetric solutions.Proof.
Whenn=1, it is immediate from condition (i) that
(4.28)A:P¯a1*⟶Pa1*⊂P¯a1*.
It follows from the Schauder fixed point theorem that A has at least one fixed point
(4.29)u1∈P¯a1*,
which means that
(4.30)‖u1‖≤a1*.
Whenn=2, it is clear that Theorem 4.1 holds (with a*=a1*,b*=b1*,andd*=d2*). Then, there exists at least three positive pseudosymmetric solutions u1,u2, and u3 such that
(4.31)‖x1‖≤d2*,‖x2‖≤d2*,‖x3‖≤d2*,b*<min⁡t∈[η,ω1]u1(t),a1*<max⁡t∈[0,1]u2(t),min⁡t∈[η,ω1]u2(t)<b1*withmax⁡t∈[0,1]u3(t)<a1*.
Following this way, we finish the proof by induction. The proof is complete.


## 5. Examples

In this section, we present two simple examples to illustrate our results.Example 5.1.
Consider the following BVPs:(5.1)(φp(u′(t)))′+t(t+1+|u′(t)|p-2)=0,t∈[0,1],u(0)=0,u(0.2)=u(1).
Note that(5.2)f0=inf⁡t∈[0,1]lim⁡(u,u′)→(0,0)t+1+|u′(t)|p-2|u′(t)|p-2u′(t)=∞,f∞=sup⁡t∈[0,1]lim⁡(u,u′)→(∞,∞)t+1+|u′(t)|p-2|u′(t)|p-2u′(t)=0.
Hence,  Theorem 3.2 implies that the BVPs in (5.1) have at least one pseudosymmetric solution u.Example 5.2.
Consider the following BVPs withp=3:
(5.3)(φp(u′(t)))′+h(t)f(t,u(t),u′(t))=0,t∈[0,1],u(0)=0,u(0.2)=u(1),
where h(t)=2t and
(5.4)f(t,u,u′)={t+4+(u′5.5)2,u∈[0,0.9],t+750u-671+(u′5.5)2,u∈[0.9,1],t+79+(u′5.5)2,u∈[1,5.5],t+14.364u+(u′5.5)2,u∈[5.5,+∞).
Note thatη=0.2,ω1=0.6, then a direct calculation shows that
(5.5)M=ω1φq(∫0ω1h(r)dr)=0.6×0.6=0.36,N≈0.1131,W=0.6.
If we take a′=0.9,b′=1,d′=5.5, then a′<b′<(N/W)d′ holds; furthermore,
(5.6)f(t,u,u′)<82<84.028≈φp(d′W)for(t,u,u′)∈[0,0.6]×[0,5.5]×[-5.5,5.5],f(t,u,u′)>79>78.176≈φp(b′N)for(t,u,u′)∈[0.6,1]×[1,5.5]×[-5.5,5.5],f(t,u,u′)<6.2<6.25=φp(a′M)for(t,u,u′)∈[0,0.6]×[0,0.9]×[-5.5,5.5].
By  Theorem 4.1, we see that the BVPs in (5.3) have at least three positive pseudosymmetric solutions u1,u2 and u3 such that
(5.7)‖xi‖≤5.5fori=1,2,3,1<min⁡t∈[0.2,0.6]u1(t),0.9<max⁡t∈[0,1]u2(t),min⁡t∈[0.2,0.6]u2(t)<1withmax⁡t∈[0,1]u3(t)<0.9.



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*Source: 182831-2011-06-28.xml*