Multiple positive pseudo-symmetric solutions of p-Laplacian dynamic equations on time scales

Mathematical and Computer Modelling 49 (2009) 1664–1681

Contents lists available at ScienceDirect

Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm

Multiple positive pseudo-symmetric solutions of p-Laplacian dynamic equations on time scalesI You-Hui Su School of Mathematics and Physics, XuZhou Institute of Technology, Xuzhou, Jiangsu, 221008, People’s Republic of China

article

info

Article history: Received 28 April 2008 Received in revised form 4 October 2008 Accepted 27 October 2008 Keywords: Time scales Boundary value problem Positive pseudo-symmetric solutions p-Laplacian Fixed point theorem

a b s t r a c t Let T be a pseudo-symmetric time scale such that 0, T ∈ T. We consider a three-point boundary value problem for p-Laplacian dynamic equations on time scales T. Some new sufficient conditions are obtained for the existence of at least single, twin, triple or arbitrary odd positive pseudo-symmetric solutions of this problem by using pseudo-symmetric technique and fixed-point theorems in cone. Our results generalize and improve the results in paper by Su et al. [Y.H. Su, W.T. Li, H.R. Sun, Triple positive pseudo-symmetric solutions of three-point BVPs for p-Laplacian dynamic equations on time scales, Nonlinear Anal. TMA 68 (2008) 1442–1452]. As applications, three examples are given to illustrate the main results and their differences. These results are new for the special cases of continuous and discrete equations, as well as in the general time scale setting. © 2008 Elsevier Ltd. All rights reserved.

1. Introduction The theory of dynamic equations on time scales can, not only unify differential and difference equations, but also provide accurate information of phenomena that manifest themselves partly in continuous time and partly in discrete time. In addition, time scale calculus would allow exploration of a variety of situations in economic, biological, heat transfer, stock market and epidemic models [1–4], etc. Hence, the study of dynamic equations on time scales is worthwhile, and has theoretical and practical values [5–7]. Throughout this paper, we denote the p-Laplacian operator by ϕp (u), i.e., ϕp (u) = |u|p−2 u for p > 1 with (ϕp )−1 = ϕq and 1/p + 1/q = 1. For convenience, we make the blanket assumption that 0, T are points in T, for an interval (0, T )T we always mean (0, T ) ∩ T. Other type of intervals are defined similarly. Recently, the existence of positive pseudo-symmetric solutions for p-Laplacian difference and differential equations have attracted the attention of some researchers, such as [8–12]. The reason is that the pseudo-symmetry not only has its theoretical value, such as in studying the metric manifolds [13], but also has its practical value — for example, we can apply this characteristic into studying chemical structures [14]. We would like to mention the results of Avery et al. [9], Ma et al. [11] and Sun et al. [12]. For the three-point boundary value problem with p-Laplacian


0 ϕp (u0 (t )) + h(t )f (t , u(t )) = 0 for t ∈ [0, 1], u(0) = 0

and

u(η1 ) = u(1),

(1.1)

where η1 ∈ (0, 1) is constant. By using the five functionals fixed point theorem in a cone [15], Avery et al. [9] established the existence of at least three positive pseudo-symmetric solutions of the boundary value problem (1.1). In addition,

I Supported by Department of Education Gansu Province (0709-03).

E-mail addresses: [email protected], [email protected]. 0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.10.010

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1665

Ma et al. [11] developed the existence of at least two positive pseudo-symmetric solutions of boundary value problems (1.1) by using the monotone iterative technique. For the three-point p-Laplacian boundary value problem with dependence on the first-order derivative


0 ϕp (u0 (t )) + h(t )f (t , u(t ), u0 (t )) = 0 for t ∈ [0, 1], u(0) = 0

and u(η1 ) = u(1),

(1.2)

Sun et al. [12] obtained the existence of at least two positive pseudo-symmetric solutions of boundary value problems (1.2) via the monotone iterative technique again. On one hand, the above mentioned researchers [8–12] only considered part of the results on existence of positive pseudo-symmetric solutions. As a result, they failed to further provide comprehensible results of positive pseudo-symmetric solutions. Naturally, it is quite necessary to consider the existence of positive pseudo-symmetric solutions for p-Laplacian dynamic equations in all respects. On the other hand, for the existence problems of positive solutions of boundary value problems on time scales, some authors have obtained many results — for details, see [16–19] and the references therein. It is also noted that the above mentioned References [16–19] only considered the existence of positive solutions; they did not further provide characteristics of positive solutions, such as pseudo-symmetry. Yet, few literature resources [20] are available concerning the pseudo-symmetry of positive solutions of boundary value problems for p-Laplacian dynamic equations on time scales. Motivated by the References [8,9,20], we consider all sides of the following p-Laplacian boundary value problem on pseudo-symmetric time scales T of the form

ϕp (u∆ (t ))


∇

u(0) = 0

+ h(t )f (t , u(t )) = 0 for t ∈ [0, T ]T ,

and u(η) = u(T ),

(1.3)

where η ∈ (0, T )T and T is symmetric in [η, T ]T . By using the pseudo-symmetric technique, the Krasnosel’skii’s fixed point theorem [21], the generalized Avery–Henderson fixed point theorem [22] and its dual theorem, together with AveryPeterson fixed point theorem [23], we obtain the existence of at least single, twin, triple or arbitrary odd positive pseudosymmetric solutions of the problem (1.3). The methods used in our paper are completely different from that of Su and Li [20]. That is to say, we comprehensively consider the problem (1.3) and obtain a much broader range of results than [20], and our results generalize and improve the results in [20]. As applications, three examples are given to illustrate the main results and their differences. These results are 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 Section 2, we present several fixed point results. In Section 3, by using Krasnosel’skii’s fixed point theorem, we obtain the existence of at least one or two positive solutions of problem (1.3). In Section 4, the existence criteria for at least three positive and arbitrary odd positive solutions of problem (1.3) are established. In Section 5, we present three simple examples to illustrate our main results. Now, we present some pseudo-symmetric definitions which can be found in [20]. Definition 1.1. For any η, T ∈ T with η < T , a time scale T is said to be pseudo-symmetric if T is symmetric over the interval [η, T ]T . That is, for any given t ∈ [η, T ]T , we have T − t + η ∈ [η, T ]T . Definition 1.2. For η ∈ (0, T )T , a function u : [0, T ]T → R is said to be pseudo-symmetric if u is symmetric over the interval [η, T ]T , where T is symmetric in [η, T ]T . That is, for any given t ∈ [η, T ]T , u(t ) = u(T − (t − η)). Definition 1.3. We say u is a pseudo-symmetric solution of boundary value problem (1.3) on [0, T ]T provided u is a solution of boundary value problem (1.3) and is symmetric over the interval [η, T ]T . The following definition can be found in [17]. Definition 1.4. If u∆∇ (t ) ≤ 0 on [0, T ]T , then we say u is concave on [0, T ]Tκ ∩Tκ . Basic definitions on time scale can be found in [5,6,24]. Another excellent source on dynamical systems on measure chains is the book [25]. Throughout this paper, it is assumed that (H1) f (t , u) : [0, T ]T × [0, ∞) → [0, ∞) is continuous, and does not vanish identically, in addition, f (t , u) = f (T − (t − η), u) for t ∈ [η, T ]T ; (H2) h(t ) ∈ Cld ([0, T ]T , [0, ∞)) is a symmetric on [η, 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 At the beginning of this section, we state the definition of cone and several fixed point theorems needed later [21,26]. In the rest of this section, we prove that solving the boundary value problem (1.3) is equivalent to finding the fixed points of a completely continuous operator. We firstly list the definition of cone and Krasnosel’skii’s fixed point theorem [21,26].

1666

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Definition 2.1. Let E 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) If x ∈ P and λ ≥ 0, then λx ∈ P ; (ii) If x ∈ P and −x ∈ P, then x = 0. Lemma 2.2 ([21,26]). Let P be a cone in a Banach space E. Assume 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) kAxk ≤ kxk , ∀x ∈ P ∩ ∂ Ω1 and kAxk ≥ kxk , ∀x ∈ P ∩ ∂ Ω2 , or (ii) kAxk ≥ kxk , ∀x ∈ P ∩ ∂ Ω1 and kAxk ≤ kxk , ∀x ∈ P ∩ ∂ Ω2 . Then A has a fixed point in P ∩ (Ω2 \ Ω1 ). Given a nonnegative continuous functional γ 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 [22] in the following. Lemma 2.3 ([22]). Let P be a cone in a real Banach space E. Let α, β and γ be nondecreasing, nonnegative continuous functional on P such that for some c > 0 and H > 0, γ (x) ≤ β(x) ≤ α(x) and

kxk ≤ H γ (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) 6= ∅ 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 0 ≤ α(x1 ) < a < α(x2 ) with β(x2 ) < b < β(x3 ) and γ (x3 ) < c . Motivated by Lemma 2.3 and the two Avery–Henderson fixed point theorem [27], we obtain the following Lemma which is called the generalized dual Avery–Henderson fixed point theorem. Lemma 2.4. Let P be a cone in a real Banach space E . Let α, β and γ be nonincreasing, nonnegative continuous functional on P such that for some c > 0 and H > 0, γ (x) ≤ β(x) ≤ α(x) and

kxk ≤ H γ (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) 6= ∅ 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 0 ≤ α(x1 ) < a < α(x2 ) with β(x2 ) < b < β(x3 ) and γ (x3 ) < c . Proof. Let

Ω1 = P (γ , c ) = {x ∈ P : γ (x) < c }, Ω3 = P (α, a) = {x ∈ P : α(x) < a}.

Ω2 = P (β, b) = {x ∈ P : β(x) < b},

For arbitrary x ∈ Ω3 , we have α(x) ≤ a, hence β(x) ≤ a < b, which implies x ∈ Ω2 . For arbitrary x ∈ Ω2 , we have β(x) ≤ b, hence γ (x) ≤ b < c, which means x ∈ Ω1 . Therefore Ω3 ⊂ Ω2 and Ω2 ⊂ Ω1 . Also if x ∈ Ω1 , then kxk ≤ M γ (x) ≤ Mc. Consequently, Ω1 Ω2 and Ω3 are bounded nonempty open sets. We claim that deg{I − A, Ω1 , θ} = 1.

(2.1)

Suppose that A(x) ≥ x for certain x ∈ ∂ Ω1 , then γ (Ax) ≤ γ (x) = c, which contradicts (i) and so A(x) < x. Hence (2.1) holds.

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1667

Similarly, we get deg{I − A, Ω2 , θ} = 0,

(2.2)

deg{I − A, Ω3 , θ} = 1.

(2.3)

and Now, it is easy to see that Ω2 /Ω3 , Ω1 /Ω2 are not empty. By (2.1)–(2.3), we have deg{I − A, Ω2 − Ω3 , θ} = deg{I − A, Ω2 , θ} − deg{I − A, Ω3 , θ} = −1, deg{I − A, Ω1 − Ω2 , θ} = deg{I − A, Ω1 , θ} − deg{I − A, Ω2 , θ} = 1. Consequently, it follows the solution property of the fixed point index, is that there exist three points x1 , x2 and x3 for A such that: x 1 ∈ Ω3

which implies 0 ≤ α(x1 ) < a, which implies a < α(x2 ) with β(x2 ) < b

x 2 ∈ Ω2 − Ω3 and

which implies b < β(x3 ) with γ (x3 ) < c .

x3 ∈ Ω1 − Ω2

That is, 0 ≤ α(x1 ) < a < α(x2 ) with β(x2 ) < b < β(x3 ) and γ (x3 ) < c .



Let β 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: 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 [23]. Lemma 2.5 ([23]). Let P be a cone in a real Banach space E and β, φ, λ, ϕ be defined as above, moreover, ϕ satisfies ϕ(λ0 x) ≤ λ0 ϕ(x) for 0 ≤ λ0 ≤ 1 such that, for some positive numbers h and d,

λ(x) ≤ ϕ(x) and

kxk ≤ hφ(x)

(2.4)

for all x ∈ P (φ, d). Suppose 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} 6= ∅ 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 6∈ 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

φ(xi ) ≤ d for i = 1, 2, 3,

b < λ(x1 ),

a < ϕ(x2 ) and

λ(x2 ) < b with ϕ(x3 ) < a.

In the following, we shall show that solving boundary value problem (1.3) is equivalent to finding the fixed points of a completely continuous operator. Firstly, if t ∈ [0, ω1 ]T , then we integrate the first equality in (1.3) from s to ω1 , where ω1 = sup{t ∈ [0, T ]T : t ≤ (η + T )/2}, one obtains


ϕp u∆ (ω1 ) − ϕp (u∆ (s)) = −ϕp (u∆ (s)) = −

ω1

Z

h(r )f (r , u(r ))∇ r .

s

Hence, the second equality in (1.3) implies that u(t ) − u(0) = u(t ) =

t

Z

ϕq

ω1

Z

0

h(r )f (r , u(r ))∇ r



∆s

for t ∈ [0, ω1 ]T .

(2.5)

s

If t ∈ [ω1 , T ]T , then we integrate the first equality in (1.3) from ω1 to s, one obtains


ϕp u∆ (s) − ϕp (u∆ (ω1 )) = ϕp (u∆ (s)) = −

s

Z

ω1

h(r )f (r , u(r ))∇ r ,

which implies u( T ) − u( t ) = −

T

Z

ϕq

Z

s

ω1

t

h(r )f (r , u(r ))∇ r



∆s for t ∈ [ω1 , T ]T ,

in view of u(T ) = u(η), one gets u(t ) = u(η) +

T

Z

ϕq t

Z

s

ω1

h(r )f (r , u(r ))∇ r



∆s for t ∈ [ω1 , T ]T .

(2.6)

1668

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Now, we let E = Cld ([0, T ]T , R), then E is a Banach space with norm

kuk = sup |u(t )| . t ∈[0,T ]T

Define a cone P ⊂ E by P = u ∈ E |u(0) = 0, u is concave, nonnegative and u is symmetric on [η, T ]T .





In fact, for arbitrary u1 , u2 ∈ P and arbitrary real λ∗∗ satisfies 0 ≤ λ∗∗ ≤ 1, we have λ∗∗ u1 + (1 − λ∗∗ )u2 ∈ P, which means P is a convex set. It is clear that P is a closed set. (i) For arbitrary u ∈ P and real number λ∗ ≥ 0, it is clear that λ∗ u ∈ P; (ii) For arbitrary u ∈ P and −u ∈ P, it is clear that u = 0. Hence, P is a cone. Secondly, we define the operator A : P → E by

 Z t Z ω1   h(r )f (r , u(r ))∇ r ∆s for t ∈ [0, ω1 ]T ,  ϕq 0  Zs T Z s (Au)(t ) =   ϕq h(r )f (r , u(r ))∇ r ∆s for t ∈ [ω1 , T ]T , w(η) + ω1

t

where w(η) = (Au)(η). Obviously, (Au)(t ) ≥ 0 for t ∈ [0, T ], (Au)(η) = (Au)(T ) and (Au)(0) = 0. In addition, ∆

(Au) (t ) = ϕq

ω1

Z

h(r )f (r , u(r )) ∇ s



≥ 0,

t ∈ [0, ω1 ]T

t

is continuous and nonincreasing in [0, ω1 ]T ; moreover, ϕq (x) is a monotone increasing continuously differentiable function, ω1

Z

h(s)f (s, u(s)) ∇ s

∇

= −h(t )f (t , u(t )) ≤ 0,

t ∈ [0, ω1 ]T ,

t

then by the chain rule [6, Theorem 1.87, p31], we obtain (Au)∆∇ (t ) ≤ 0 for [0, ω1 ]T . By using a similar method, we can deduce (Au)∆∇ (t ) ≤ 0 for [ω1 , T ]T . So, A : P → P . Thirdly, we shall prove A : P → P is completely continuous. We show that A maps bounded set into bounded set. Assume c > 0 is a constant and u ∈ P c = {x ∈ P : kxk ≤ c }. Note that the continuity of f guarantees that there is a C > 0 such that f (u) ≤ ϕp (C ). Hence

 Z ω1 Z ω1 kAuk = (Au)(ω1 ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s Z ω1  Z ω1  ≤ ω1 ϕq h(r )f (r , u(r ))∇ r ≤ C ω1 ϕq h(s)∆s . 0

0

That is, AP c is uniformly bounded. For t1 , t2 ∈ [0, ω1 ]T , we have

Z t Z ω1  2 |(Au)(t1 ) − (Au)(t2 )| = ϕq h(r )f (r , u(r )) ∇ s ∆s t s 1Z t Z ω1  Z 2 ≤ C ϕq h(r )∇ r ∆s ≤ C |t1 − t2 | ϕq t1

s

ω1

h(r )∇ r



.

0

For t1 , t2 ∈ [ω1 , T ]T , in view of the similar way we can obtain the analogous result. Hence, the Arzela-Ascoli Theorem on time scales [28] tells us that AP c is relatively compact. We next claim that A : P c → P is continuous. Assume that {un }∞ n=1 ⊂ P c which converges to u0 (t ) uniformly on [0, T ]T . Hence, {(Aun ) (t )}∞ Theorem on time scales [28] n=1 is uniformly bounded and equicontinuous on [0, T ]T . The Arzela-Ascoli 
∞ implies that there exists a uniformly convergent subsequence in {(Aun ) (t )}∞ . Let Au n(m) (t ) m=1 be a subsequence n =1 which converges to v(t ) uniformly on [0, T ]T . In addition, 0 ≤ (Aun )(t ) ≤ C

t

Z

ϕq 0

ω1

Z 0

h(r )∇ r



∆ s.

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1669

Observe that

 Z t Z ω1   h ( r ) f ( r , u ( r ))∇ r ∆s for t ∈ [0, ω1 ]T , ϕ  n q 0  Zs T Z s (Aun )(t ) =   h(r )f (r , un (r ))∇ r ∆s for t ∈ [ω1 , T ]T . ϕq w(η) + ω1

t

Inserting un(m) into the above and then letting m → ∞, we obtain

 Z t Z ω1   h(r )f (r , u0 (r ))∇ r ∆s for t ∈ [0, ω1 ]T ,  ϕq 0  Zs T Z s v(t ) =   h(r )f (r , u0 (r ))∇ r ∆s for t ∈ [ω1 , T ]T , ϕq w(η) + ω1

t

here we have used the Lebesgue’s dominated convergence theorem on time scales [24]. From the definition of A, we know that v(t ) = Au0 (t ) on [0, T ]T . This shows that each subsequence of {Aun (t )}∞ n=1 uniformly converges to (Au0 )(t ). Therefore, the sequence {(Aun )(t )}∞ uniformly converges to ( Au )( t ) . This means that A is continuous at u0 ∈ P c . So, A is continuous 0 n=1 on P c since u0 is arbitrary. Thus, A is completely continuous. The proof is complete.  Fourthly, in view of (2.5) and (2.6), we have that all the fixed points of completely continuous operator A are the solutions of the boundary value problem (1.3). 3. Single or twin solutions For t ∈ [0, ω1 ]T , let f0 = lim

u→0+

f ( t , u)

and f∞ = lim

ϕp (u)

u→∞

f ( t , u)

ϕp (u)

.

Similar to [29], 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; and (vi) i0 = 2 and i∞ = 0. In the following, by using Krasnosel’skii’s fixed point theorem in a cone, we study the existence for positive pseudo-symmetric solutions of problem (1.3) under the above six possible cases. 3.1. For the case i0 = 1 and i∞ = 1 In this subsection, we discuss the existence of a single positive pseudo-symmetric solution for the problem (1.3) under i0 = 1 and i∞ = 1. Theorem 3.1. Problem (1.3) has at least one positive pseudo-symmetric solution in the case i0 = 1 and i∞ = 1. Proof. We divide the proof into two cases: Case (i) f0 = 0 and f∞ = ∞. In view of f0 = 0, there exists R ω an H1
> 0 such that f (t , u) ≤ ϕp (ε)ϕp (u) = ϕp (εu) for (t , u) ∈ [0, ω1 ]T × (0, H1 ], where ε > 0 and satisfies εω1 ϕq 0 1 h(s)∆s ≤ 1. If u ∈ P with kuk = H1 , then

 Z ω1 Z ω1 kAuk = (Au)(ω1 ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s Z ω1  Z ω1  ≤ ω1 ϕq h(r )f (r , u(r ))∇ r ≤ εω1 kuk ϕq h(s)∆s ≤ kuk . 0

0

If we let ΩH1 = {u ∈ E : kuk < H1 } , then kAuk ≤ kuk for u ∈ P ∩ ∂ ΩH1 . From f∞ = ∞, there exists an H20 > 0 such that f (t , u) ≥ ϕp (k)ϕp (u) = ϕp (ku) for (t , u) ∈ [0, ω1 ]T × [H20 , ∞), where k > 0, and satisfies the following inequality Z ω1  kηϕq h(r )∇ r ≥ 1. (3.1) η

Set



H2 = max 2H1 ,

ω1 0 H η 2



and ΩH2 = {u ∈ E : kuk < H2 } .

1670

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

If u ∈ P with kuk = H2 , then, by Lemma 3.1 in [20], one has min

t ∈[η,ω1 ]T

η kuk ≥ H20 . ω1

u(t ) = u(η) ≥

(3.2)

For u ∈ P ∩ ∂ ΩH2 , in terms of (3.1) and (3.2), we get

 Z ω1 Z ω1 kAuk = (Au)(ω1 ) = h(r )f (r , u(r ))∇ r ∆s ϕq s 0    Z ω1 Z ω1 Z ω1 Z ω1 kη k uk h(r )∇ r ∆s h(r )f (r , u(r ))∇ r ∆s ≥ ϕq ϕq ≥ ω1 η η 0 0  Z ω1 h(r )∇ r ≥ kuk . = kη kuk ϕq

(3.3)

η

Thus, by (i) of Lemma 2.2, the problem (1.3) has at least single positive pseudo-symmetric solution u in P ∩ (Ω H2 \ ΩH1 ) with H1 ≤ kuk ≤ H2 . Case (ii) f0 = ∞ and f∞ = 0. Since f0 = ∞, there exists an H3 > 0 such that f (t , u) ≥ ϕp (m)ϕp (u) = ϕp (mu) for (t , u) ∈ [0, ω1 ]T × (0, H3 ], where m is such that mηϕq

ω1

Z η

h(r )∇ r



≥ 1.

(3.4)

If u ∈ P with kuk = H3 , then by (3.2) and (3.4), one has

 Z ω1 Z ω1 kAuk = (Au)(ω1 ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s  Z ω1 Z ω1   Z ω1 Z ω1 mη k uk ≥ ϕq h(r )f (r , u(r ))∇ r ∆s ≥ ϕq h(r )∇ r ∆s ω1 0 η 0 η Z ω1  = mη kuk ϕq h(r )∇ r ≥ kuk .

(3.5)

η

If we let ΩH3 = {u ∈ E : kuk < H3 } , then kAuk ≥ kuk for u ∈ P ∩ ∂ ΩH3 . Now, we consider f∞ = 0. By definition, there exists H40 > 0 such that f (t , u) ≤ ϕp (δ)ϕp (u) = ϕp (δ u) for (t , u) ∈ [0, ω1 ]T × [H40 , ∞),

(3.6)

where δ > 0 satisfies

δω1 ϕq

ω1

Z η

h(r )∇ r



≤ 1.

(3.7)

Suppose that f is bounded, then f (t , u) ≤ ϕp (K ) for all (t , u) ∈ [0, ω1 ]T × [0, ∞) and some constant K > 0. Pick



H4 = max 2H3 , K ω1 ϕq

ω1

Z

h(s)∆s



.

0

If u ∈ P with kuk = H4 , then

kAuk = (Au)(ω1 ) =

Z 0

ω1

ϕq

ω1

Z

h(r )f (r , u(r ))∇ r



∆s ≤ K ω1 ϕq

ω1

Z

s

h(s)∆s



≤ H4 = kuk .

0

Suppose that f is unbounded. From f ∈ C ([0, T ]T × [0, +∞), [0, +∞)), we have f (t , u) ≤ C3 for (t , u) ∈ [0, ω1 ]T × [0, C4 ] , here C3 and C4 are arbitrary This implies that f (t , u) → +∞ if u → +∞. Hence, it is easy to n positive constants. o ω

know that there exists H4 ≥ max 2H3 , η1 H40 such that f (t , u) ≤ f (t , H4 ) for (t , u) ∈ [0, ω1 ]T × [0, H4 ]. If u ∈ P with kuk = H4 , then by using (3.6) and (3.7), we have

 Z ω1 Z ω1 kAuk = (Au)(ω1 ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s  Z Z ω1 Z ω1 ≤ ϕq h(r )f (r , H4 )∇ r ∆s ≤ δ H4 ω1 ϕq 0

0

0

ω1

h(r )∇ r



= k uk .

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1671

Consequently, in either case, if we take ΩH4 = {u ∈ E : kuk < H4 } , then, for u ∈ P ∩ ∂ ΩH4 , we have kAuk ≤ kuk. Thus, the condition (ii) of Lemma 2.2 is satisfied. Consequently, the problem (1.3) has at least single positive pseudo-symmetric solution u in P ∩ (Ω H4 \ ΩH3 ) with H3 ≤ kuk ≤ H4 . The proof is complete.  3.2. For the case i0 = 0 and i∞ = 0 In this subsection, we discuss the existence for the positive pseudo-symmetric solutions of problems (1.3) under i0 = 0 and i∞ = 0. First, we shall state and prove the following main result of problem (1.3). Theorem 3.2. Suppose that the following conditions hold:

> 0 such that f (t , u) ≤ ϕp (p0 Λ1 ) for (t , u) ∈ [0, ω1 ]T × [0, p0 ], where Λ1

(i) there exists constant p0

ω1 ϕq

R ω1 0

h(r )∇ r



−1

=

; > 0 such that f (t , u) ≥ ϕp (q0 Λ2 ) for (t , u) ∈ [η, ω1 ]T ×

(ii) there exists constant q0

h

η 0 0 q ,q ω1

i

, where Λ2

=

 R −1 ω ω1 ϕq η 1 h(r )∇ r , furthermore, p0 6= q0 . Then problem (1.3) has at least one positive pseudo-symmetric solution u such that kuk lies between p0 and q0 . 0 0 Proof. Without  loss of generality, we may assume that p < q . 0 Let Ωp0 = u ∈ E : kuk < p . For any u ∈ P ∩ ∂ Ωp0 , in view of condition (i), we have

kAuk = (Au)(ω1 ) =

ω1

Z

ϕq

0

≤ p0 Λ1 ω1 ϕq

h(r )f (r , u(r ))∇ r



∆s

s

ω1

Z

ω1

Z

h(r )∇ r



= p0 ,

(3.8)

0

which yields for u ∈ P ∩ ∂ Ωp0 .

kAuk ≤ kuk

(3.9)

Now, set Ωq0 = u ∈ E : kuk < q . For u ∈ P ∩ ∂ Ωq0 , Lemma 3.1 in [20] implies that Hence, by condition (ii) we get



0

kAuk = (Au)(ω1 ) ≥

ω1

Z

ϕq



ω1

Z η

0

h(r )f (r , u(r ))∇ r



∆s ≥ q Λ2 ω1 ϕq 0

ω1

Z η

η 0 q ω1

h(r )∇ r



≤ u(t ) ≤ q for t ∈ [η, ω1 ]T . 0

= q0 .

So, if we take Ωq0 = u ∈ E : kuk < q0 , then



kAuk ≥ kuk ,



u ∈ P ∩ ∂ Ωq 0 .

(3.10)

Consequently, in view of p0 < q0 , (3.9) and (3.10), it follows from Lemma 2.2 that problem (1.3) has a positive pseudosymmetric solution u in P ∩ (Ω q0 \ Ωp0 ).  3.3. For the case i0 = 1 and i∞ = 0 or i0 = 0 and i∞ = 1 In this subsection, under the conditions i0 = 1 and i∞ = 0 or i0 = 0 and i∞ = 1, we discuss the existence of positive pseudo-symmetric solutions of problem (1.3). Theorem 3.3. Suppose that f0 ∈ [0, ϕp (Λ1 )) and f∞ ∈



ϕp



ω1 Λ2 η



 , ∞ hold. Then problem (1.3) has at least one positive

pseudo-symmetric solution. Proof. It is easy to see that, under the assumptions, the conditions (i) and (ii) in Theorem 3.2 are satisfied. So the proof is easy and we omit it here.  Theorem 3.4. Suppose that f0 ∈

    
ϕp ωη1 Λ2 , ∞ and f∞ ∈ 0, ϕp (Λ1 ) hold. Then problem (1.3) has at least one positive

pseudo-symmetric solution. Proof. Firstly, let ε1 = f0 − ϕp f (t , u)

ϕp (u)

≥ f0 − ε1 = ϕp





ω1 Λ2 η

ω1 Λ2 η





> 0, there exists a sufficiently small q0 > 0 satisfies for (t , u) ∈ [0, ω1 ]T × (0, q0 ].

1672

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

h

η

i

Thus, if (t , u) ∈ [η, ω1 ]T × ω q0 , q0 , then we have 1 f (t , u) ≥ ϕp



 ω1 Λ2 ϕp (u) ≥ ϕp (Λ2 q0 ), η

which implies that the condition (ii) in Theorem 3.2 holds. Next, for ε2 = ϕp (Λ1 ) − f∞ > 0, there exists a sufficiently large p00 (> q0 ) such that f (t , u)

ϕp (u)

≤ f∞ + ε2 = ϕp (Λ1 )

for (t , u) ∈ [0, ω1 ]T × [p00 , ∞).

(3.11)

We consider two cases: Case (i): Assume that f is bounded, that is, f (t , u) ≤ ϕp ( K1 ) for (t , u) ∈ [0, ω1 ]T × [0, ∞) and some constant K1 > 0. If we take sufficiently large p0 such that p0 ≥ max K1 /Λ1 , p00 , then f (t , u) ≤ ϕp (K1 ) ≤ ϕp (Λ1 p0 ) for (t , u) ∈ [0, ω1 ]T × [0, p0 ]. Consequently, from the above inequality, the condition (i) of Theorem 3.2 is true. Case (ii): Assume that f is unbounded. From f ∈ C ([0, T ]T × [0, ∞), [0, ∞)), we have p0 > p00 such that f (t , u) ≤ f (t , p0 ) for (t , u) ∈ [0, ω1 ]T × [0, p0 ]. Since p0 > p00 , by (3.11), we get f (t , p0 ) ≤ ϕp (Λ1 p0 ), hence f (t , u) ≤ f (t , p0 ) ≤ ϕp (Λ1 p0 ) for (t , u) ∈ [0, ω1 ]T × [0, p0 ]. Thus, the condition (i) of Theorem 3.2 is fulfilled. Consequently, from Theorem 3.2, the conclusion of this theorem holds.



From Theorems 3.2–3.4, respectively, we have the following two results. Corollary 3.5. Suppose that f0 = 0 and the condition(ii) in Theorem 3.2 hold. Then problem (1.3) has at least one positive pseudo-symmetric solution. Corollary 3.6. Suppose that f∞ = 0 and the condition(ii) in Theorem 3.2 hold. Then problem (1.3) has at least one positive pseudo-symmetric solution. Theorem 3.7. Suppose that f0 ∈ 0, ϕp (Λ1 ) and f∞ = ∞ hold. Then problem (1.3) has at least one positive pseudo-symmetric solution.




Proof. First, in view of f∞ = ∞, then by inequality (3.3), we have kAuk ≥ kuk for u ∈ P ∩ ∂ ΩH2 . Next, by f0 ∈ 0, ϕp (Λ1 ) , for ε3 = ϕp (Λ1 ) − f0 > 0, there exists a sufficiently small p0 ∈ (0, H2 ) such that




f (t , u) ≤ (f0 + ε3 )ϕp (u) = ϕp (Λ1 u) ≤ ϕp (Λ1 p0 ) for (t , u) ∈ [0, ω1 ] × [0, p0 ], which implies (i) of Theorem 3.2 holds, that is, (3.8) is true, hence, we obtain kAuk ≤ kuk for u ∈ P ∩ ∂ Ωp0 . The result is obtained and the proof is complete.  Theorem 3.8. Suppose that f0 = ∞ and f∞ ∈ 0, ϕp (Λ1 ) hold. Then problem (1.3) has at least one positive pseudo-symmetric solution.




Proof. On one hand, since f0 = ∞, by the inequality (3.5), one gets kAuk ≥ kuk , 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 the condition (i) of Theorem 3.2 is satisfied, that is, inequality (3.8) holds, and one has kAuk ≤ kuk , u ∈ P ∩ ∂ Ωp0 , where p0 > H3 . Hence, problem (1.3) has at least one positive pseudo-symmetric solution, and the proof is complete.  From Theorems 3.7 and 3.8, respectively, it is easy to obtain the following two corollaries. Corollary 3.9. Assume that f∞ = ∞ and the condition (i) in Theorem 3.2 hold. Then problem (1.3) has at least one positive pseudo-symmetric solution. Corollary 3.10. Assume that f0 = ∞ and the condition (i) in Theorem 3.2 hold. Then problem (1.3) has at least one positive pseudo-symmetric solution. 3.4. For the case i0 = 0 and i∞ = 2 or i0 = 2 and i∞ = 0 In this subsection, under i0 = 0 and i∞ = 2 or i0 = 2 and i∞ = 0, we study the existence of multiple positive solutions for problems (1.3).

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1673

Combining the proof of Theorems 3.1 and 3.2, it is easy to prove the following two theorems. Theorem 3.11. Suppose that i0 = 0 and i∞ = 2 and the condition (i) of Theorem 3.2 hold, then problem (1.3) has at least two positive pseudo-symmetric solutions u1 , u2 ∈ P such that 0 < ku1 k < p0 < ku2 k . Theorem 3.12. Suppose that i0 = 2 and i∞ = 0 and the condition (ii) of Theorem 3.2 hold, then problem (1.3) has at least two positive pseudo-symmetric solutions u1 , u2 ∈ P such that 0 < ku1 k < q0 < ku2 k . From the above, we can see that the proof of four subcases all revolve around proving the conditions of Krasnosel’skii’s fixed point theorem [21,26]. That is, all that we do is to prove that the condition (i) kAxk ≤ kxk , ∀x ∈ P ∩ ∂ Ω1 and kAxk ≥ kxk , ∀x ∈ P ∩ ∂ Ω2 or (ii) kAxk ≥ kxk , ∀x ∈ P ∩ ∂ Ω1 and kAxk ≤ kxk , ∀x ∈ P ∩ ∂ Ω2 is satisfied. 4. Triple solutions In the previous section, we have obtained some results on the existence of at least one or two positive pseudo-symmetric solutions of problem (1.3). In this section, we will further discuss the existence of positive pseudo-symmetric solutions of problem (1.3) by using two different methods. One is the generalized Avery–Henderson fixed-point theorem [22] and its dual theorem. The other is the Avery-Peterson fixed point theorem [23]. Choose a r ∈ (η, ω1 )T , for notational convenience we denote Mξ = ηϕq

ω1

Z

h(r )∇ r



,

Nξ = ηϕq

η ω1

0

Lξ = r ϕq

ω1

Z

h(r )∇ r



,

Lθ = r ϕq

ω1

Z

Z

0

h(r )∇ r

h(r )∇ r





,

,

Wξ = ω1 ϕq

r

Z

ω1

h(r )∇ r



.

0

4.1. Case 1 In this subsection, in view of the generalized Avery–Henderson fixed-point theorem [22] and its dual theorem, the existence criteria for at least three positive and arbitrary odd positive pseudo-symmetric solutions of problems (1.3) are established. For u ∈ P, we define the nonnegative, continuous functionals γ2 , β2 and α2 by

γ2 (u) = max u(t ) = u(η),

β2 (u) = min u(t ) = u(η),

t ∈[0,η]T

t ∈[η,ω1 ]T

α2 (u) = max u(t ) = u(r ). t ∈[0,r ]T

It is obvious that γ2 (u) = β2 (u) ≤ α2 (u) for each u ∈ P . By Lemma 3.1 in [20], one obtains kuk ≤ ω1 γ ( u) η 2

for all u ∈ P . We now present the results in this subsection.

ω1 u(η) η

=

rN Theorem 4.1. Suppose that there are positive numbers a0 , b0 , c 0 such that a0 < ωr b0 < ω Mξ c 0 . In addition, f (t , u) satisfies the 1 1 ξ following conditions:

(i) f (t , u) < ϕp



c0 Mξ



for (t , u) ∈ [0, ω1 ]T × 0, η1 c 0 ;

(ii) f (t , u) > ϕp



b0 Nξ a0



for (t , u) ∈ [η, ω1 ]T × b0 , η1 b0 ;

(iii) f (t , u) < ϕp

  Lξ

h

ω

i

h

ω

i

ω1 0 

for (t , u) ∈ [0, ω1 ]T × 0,



r

a .

Then problem (1.3) has at least three positive pseudo-symmetric solutions u1 , u2 and u3 such that 0 < max u1 (t ) < a0 < max u2 (t ) t ∈[0,r ]T

min

t ∈[η,ω1 ]T

t ∈[0,r ]T

u2 (t ) < b0 <

min

t ∈[η,ω1 ]T

with

u3 (t ) and

max u3 (t ) < c 0 .

t ∈[0,η]T

Proof. By the definition of completely continuous operator A and its properties, it suffices to show that all the conditions of Lemma 2.3 hold with respect to A. It is easy to obtain that A : P (γ2 , c ) → P . First, we verify that if u ∈ ∂ P (γ2 , c 0 ), then γ2 (Au) < c 0 . ω ω If u ∈ ∂ P (γ2 , c 0 ), then γ2 (u) = maxt ∈[0,η]T u(t ) = u(η) = c 0 . Lemma 3.1 in [20] implies that kuk ≤ η1 u(η) = η1 c 0 , we ω

have 0 ≤ u(t ) ≤ η1 c 0 , t ∈ [0, ω1 ]T .

1674

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Thus, by the condition (i), one has

 Z η Z ω1 h(r )f (r , u(r ))∇ r ∆s ϕq γ2 (Au) = (Au)(η) = s 0  Z Z η Z ω1 c0 h(r )f (r , u(r ))∇ r ∆s < ϕq ηϕq ≤ Mξ

0

0

ω1

h(r )∇ r



= c0.

0

Second, we show that β2 (Au) > b0 for u ∈ ∂ P (β2 , b0 ). If we choose u ∈ ∂ P (β2 , b0 ), then β2 (u) = mint ∈[η,ω1 ]T u(t ) = u(η) = b0 . In view of Lemma 3.1 in [20], we have kuk ≤ ωη1 u(η) = ωη1 b0 . So b0 ≤ u(t ) ≤ ωη1 b0 for t ∈ [η, ω1 ]T . Using the condition (ii), we get

 Z η Z ω1 h(r )f (r , u(r ))∇ r ∆s ϕq β2 (Au) = (Au)(η) ≥ η 0   0  Z ω1 Z ω1 b b0 h(r )∇ r = b0 . h(r )ϕp ∇r > ηϕq > ηϕq Nξ

η

Nξ

η

Finally, we prove that P (α2 , a ) 6= ∅ and α2 (Au) < a0 for all u ∈ ∂ P (α2 , a0 ). 0 In fact, the constant function a2 ∈ P (α2 , a0 ). Moreover, for u ∈ ∂ P (α2 , a0 ), we have α2 (u) = maxt ∈[0,r ]T u(t ) = u(r ) = a0 , ω ω ω 0 which implies 0 ≤ u(t ) ≤ a for t ∈ [0, r ]T . Hence, u(t ) ≤ kuk ≤ r1 u(r ) = r1 r 0 . Therefore 0 ≤ u(t ) ≤ r1 a0 , t ∈ [0, ω1 ]T . By using assumption (iii), one has 0

 Z r Z ω1 α2 (Au) = (Au)(r ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s   0  Z r Z ω1 Z r Z ω1 a ≤ ϕq h(r )f (r , u(r ))∇ r ∆s ≤ ϕq h(r )ϕp ∇ r ∆s Lξ 0 0 0 0 Z  ω1 a0 = r ϕq h(r )∇ r = a0 . Lξ

0

Thus, all the conditions in Lemma 2.3 are satisfied. From (H1) and (H2), we have that the solutions of problem (1.3) does not vanish identically on any closed subinterval of [0, T ]T . Consequently, problem (1.3) has at least three positive pseudosymmetric solutions u1 , u2 and u3 belonging to P (γ2 , c 0 ), and satisfying 0 < max u1 (t ) < a0 < max u2 (t ) with t ∈[0,r ]T

t ∈[0,r ]T

u2 (t ) < b < 0

min

t ∈[η,ω1 ]T

The proof is complete.

min

t ∈[η,ω1 ]T

u3 ( t )

max u3 (t ) < c 0 .

and

t ∈[0,η]T



From Theorem 4.1, 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 pseudo-symmetric solutions for the problem (1.3). Theorem 4.2. Suppose that there exist positive numbers a0si , b0si , cs0i such that a0s1 <

r

ω1

b0s1 <

rNξ

ω1 Mξ

rNξ 0 r 0 rNξ 0 r 0 cs01 < a0s2 < b < c < a0s3 < · · · < a0sn < b < c , ω1 s2 ω1 Mξ s2 ω1 sn ω1 Mξ sn

here n ∈ N and i = 1, 2, . . . , n. In addition, f (t , u) satisfies the following conditions: (i) f (t , u) < ϕp (ii) f (t , u) > ϕp (iii) f (t , u) < ϕp



cs0



i

Mξ



b0s

 i

Nξ



a0s

i

Lξ



h

i

ω

for (t , u) ∈ [0, ω1 ]T × 0, η1 cs0i ;

h

i

ω

for (t , u) ∈ [η, ω1 ]T × b0si , η1 b0si ; for (t , u) ∈ [0, ω1 ]T × 0,



ω1 0 r

asi .



Then problem (1.3) has at least 2n + 1 positive pseudo-symmetric solutions. Proof. When i = 1, it is clear that Theorem 4.1 holds. Then we can obtain at least three positive pseudo-symmetric solutions u1 , u2 and u3 satisfying 0 ≤ max u1 (t ) < a0s1 < max u2 (t ) with t ∈[0,r ]T

t ∈[0,r ]T

min

t ∈[η,ω1 ]T

u2 (t ) < bs1 < 0

min

t ∈[η,ω1 ]T

u3 (t ) and

max u3 (t ) < cs01 .

t ∈[0,η]T

Following this method, we finish the proof by induction. The proof is complete.



Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1675

Using Lemma 2.4, it is easy to find the following results. Theorem 4.3. Suppose that there are positive numbers a0 , b0 , c 0 such that a0 <

ηL

b0 < ω Mθ c 0 . In addition, f (t , u) satisfies the 1 ξ

Lθ Mξ

following conditions: (i) f (t , u) > ϕp



c0 Nξ



for (t , u) ∈ [η, ω1 ]T × c 0 , η1 c 0 ;

(ii) f (t , u) < ϕp



b0 Mξ



for (t , u) ∈ 0, ω1 ]T × [0, η1 b0 ;

(iii) f (t , u) > ϕp

 0 a Lθ

h

h

ω

i

ω

i

ω1 0 

for (t , u) ∈ [r , ω1 ]T × a0 ,



a .

r

Then problem (1.3) has at least three positive pseudo-symmetric solutions u1 , u2 and u3 such that 0 < max u1 (t ) < a0 < max u2 (t ) t ∈[0,r ]T

min

t ∈[η,ω1 ]T

with

t ∈[0,r ]T

u2 (t ) < b0 <

min

t ∈[η,ω1 ]T

u3 (t ) and

max u3 (t ) < c 0 .

t ∈[0,η]T

From Theorem 4.3, we can obtain Theorem 4.4 and Corollary 4.5. Theorem 4.4. Suppose that there exist positive numbers a0λi , b0λi , cλ0 i such that a0λ1 <

Lθ Mξ

b0λ1 <

η Lθ 0 η Lθ 0 η Lθ 0 Lθ 0 Lθ 0 cλ1 < a0λ2 < bλ2 < cλ2 < a0λ3 < · · · < a0λn < bλn < c , ω1 Nξ Mξ ω1 Nξ Mξ ω1 Nξ λn

here n ∈ N and i = 1, 2, . . . , n. In addition, f (t , u) satisfies the following conditions: (i) f (t , u) > ϕp (ii) f (t , u) < ϕp (iii) f (t , u) > ϕp



cλ0

 i

Nξ



b0λ

 i

Mξ



a0λ i Lθ

h

i

ω

for (t , u) ∈ [η, ω1 ]T × cλ0 i , η1 cλ0 i ;



h

i

ω

for (t , u) ∈ [0, ω1 ]T × 0, η1 b0λi ;

h

for (t , u) ∈ [r , ω1 ]T × a0λi ,

ω1 0 a r λi

i

.

Then problem (1.3) has at least 2n + 1 positive pseudo-symmetric solutions. Corollary 4.5. Assume that f satisfies conditions (i) f0 = ∞, f∞ = ∞; (ii) there exists c0 > 0 such that f (t , u) < ϕp



η

c Mξ ω1 0



for (t , u) ∈ [0, ω1 ]T × [0, c0 ].

Then problem (1.3) has at least three positive pseudo-symmetric solutions. η

Proof. First, by the condition (ii), let b0 = ω c0 , one gets 1 f (t , u) < ϕp



b0



Mξ

 ω1 0 for (t , u) ∈ [0, ω1 ]T × 0, b , η 

which implies that (ii) of Theorem 4.3 holds. Second, choose K3 sufficiently large to satisfy K3 Lθ = K3 r ϕq

ω1

Z

h(r )∇ r



> 1.

(4.1)

r

Since f0 = ∞, there exists r1 > 0 sufficiently small such that, f (t , u) ≥ ϕp (K3 )ϕp (u) = ϕp (K3 u) for (t , u) ∈ [0, ω1 ]T × [0, r1 ]. Without loss of generality, suppose r1 ≤ a < 0

Lθ Mξ

Lθ ω1 0 b rMξ

(4.2)

. Choose a0 > 0 such that a0 <

r

r . ω1 1

For a0 ≤ u ≤

0

b . Thus, by (4.1) and (4.2), we have

f (t , u) ≥ ϕp (K3 u) ≥ ϕp (K3 a0 ) > ϕp



this implies that (iii) of Theorem 4.3 is true.

a0 Lθ



h

for (t , u) ∈ [0, ω1 ]T × a0 ,

ω1 0 i r

a

,

ω1 0 r

a , we have u ≤ r1 and

1676

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Third, choose K2 sufficiently large such that K2 Nξ = K2 ηϕq

ω1 η

R

h(r )∇ r



> 1. Since f∞ = ∞, there exists r2 > 0

sufficiently large such that, f (t , u) ≥ ϕp (K2 )ϕp (u) = ϕp (K2 u) for u ≥ r2 . ω

Without loss of generality, suppose r2 > η1 b0 . Choose c 0 = r2 . Then f (t , u) ≥ ϕp (K2 u) ≥ ϕp (K2 c 0 ) > ϕp

c0







for (t , u) ∈ [0, ω1 ]T × c 0 ,

Nξ

which means that (i) of Theorem 4.3 holds. From the above analysis, we get 0 < a0 <

 ω1 0 c , η

ηL

b0 < ω Mθ c 0 , then, all conditions in Theorem 4.3 are satisfied. Hence, 1 ξ problem (1.3) has at least three positive pseudo-symmetric solutions.  Lθ Mξ

In terms of Theorem 4.1, we also have the following corollary. Corollary 4.6. Assume that f satisfies conditions (i) f0 = 0, f∞ = 0; (ii) there exists c0 > 0 such that f (t , u) > ϕp



η c Nξ ω1 0



h

i

η

for (t , u) ∈ [η, ω1 ]T × ω c0 , c0 . 1

Then problem (1.3) has at least three positive pseudo-symmetric solutions. 4.2. Case 2 In this subsection, the existence criteria for at least three positive and arbitrary odd positive pseudo-symmetric solutions of problems (1.3) are established by using the Avery-Peterson fixed point theorem [23]. Define the nonnegative continuous convex functionals θ and γ , nonnegative continuous concave functional α , and nonnegative continuous functional ϕ respectively on P by

φ(u) = max u(t ) = u(ω1 ) = kuk,

β(u) =

t ∈[0,ω1 ]T

max

t ∈[r ,ω1 ]Tκ

|u∆ (t )| = |u∆ (r )|,

λ(u) = min u(t ) = u(η) and ϕ(u) = min u(t ) = u(η). t ∈[η,ω1 ]T

t ∈[η,ω1 ]T

Now, we list and prove the results in this subsection. η

ηN

Theorem 4.7. Suppose that there exist constants a∗ , b∗ , d∗ such that 0 < a∗ < ω b∗ < ω Wξ d∗ . In addition, suppose that 1 1 ξ   Wξ > ϕq

R ω1 η

h(s)∇ s holds, f satisfies the following conditions:

(i) f (t , u) ≤ ϕp



(ii) f (t , u) > ϕp



(iii) f (t , u) < ϕp

d∗ Wξ b∗



for (t , u) ∈ [0, ω1 ]T × [0, d∗ ];



for (t , u) ∈ [η, ω1 ]T × [b∗ , d∗ ];

 Nξ∗  a Mξ

h

ω

i

for (t , u) ∈ [0, ω1 ]T × 0, η1 a∗ .

Then problem (1.3) has at least three positive pseudo-symmetric solutions u1 , u2 , u3 such that

kxi k ≤ d∗ for i = 1, 2, 3, and

min u2 (t ) < b∗

t ∈[η,ω1 ]T

b∗ < with

min u1 (t ),

a∗ <

t ∈[η,ω1 ]T

min u2 (t )

t ∈[η,ω1 ]T

min u3 (t ) < a∗ .

t ∈[η,ω1 ]T

Proof. By the definition of completely continuous operator A and its properties, it suffices to show that all the conditions of Lemma 2.5 hold with respect to A. For all u ∈ P , λ(u) = ϕ(u) = u(η) and kuk = u(ω1 ) = φ(u). Hence, the condition (2.4) is satisfied. Firstly, we show that A : P (φ, d∗ ) → P (φ, d∗ ). For any u ∈ P (φ, d∗ ), in view of φ(u) = kuk ≤ d∗ and the assumption (i), one has

Z

ω1

Z

ω1



kAuk = (Au)(ω1 ) = ϕq h(r )f (r , u(r ))∇ r ∆s 0 s  Z Z ω1 Z ω1 d∗ ≤ ϕq h(r )f (r , u(r ))∇ r ∆s ≤ ω1 ϕq 0

0

Wξ

ω1

h(r )∇ r

0

From the above analysis, it remains to show that (i)–(iii) of Lemma 2.5 hold.



= d∗ .

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Secondly, we verify that condition (i) of Lemma 2.5 holds, let u(t ) ≡ kb∗ with k =

1677

Wξ Nξ

. 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 {u ∈ P (φ, β, λ, b∗ , kb∗ , d∗ ) : λ(x) > b∗ } 6= ∅. For any u ∈ P (φ, β, λ, b∗ , kb∗ , d∗ ), then we get b∗ ≤ u(t ) ≤ d∗ for all t ∈ [η, ω1 ]T , hence, by the assumption (ii), we have η

Z

ω1

Z



h(r )f (r , u(r ))∇ r ∆s ϕq λ(Au) = (Au)(η) = s 0  Z Z η Z ω1 b∗ h(r )f (r , u(r ))∇ r ∆s > ϕq ηϕq ≥ Nξ

η

0

ω1

h(r )∇ r

η



= b∗ .

Thirdly, we prove that the condition (ii) of Lemma 2.5 holds. For any u ∈ P (φ, λ, b∗ , d∗ ) with β(Au) > kb∗ , that is,

β(Au) = (Au)∆ (r ) = ϕq

ω1

Z

h(s)f (s, u(s))∇ s



> kb∗ .

(4.3)

r

So, in view of k =

Wξ Nξ

, Wξ > ϕq

λ(Au) = (Au)(η) =

η

Z

ω1 η

R

ϕq

ω1

Z

0

h(r )f (r , u(r ))∇ r



∆s

s

η

Z



h(s)∇ s and (4.3), one has

≥

ϕq

ω1

Z η

0

h(r )f (r , u(r ))∇ r



∆s >

η

Z

ϕq

0

ω1

Z

h(r )f (r , u(r ))∇ r



∆s > ηkb∗ > b∗ .

r

Finally, we check condition (iii) of Lemma 2.5. Clearly, since ϕ(0) = 0 < a∗ , we have 0 6∈ R(φ, ϕ, a∗ , d∗ ). If ω ω u ∈ R(φ, ϕ, a∗ , d∗ ) with ϕ(u) = mint ∈[η,ω1 ]T u(t ) = u(η) = a∗ , then, Lemma 3.1 in [20] implies that kuk ≤ η1 u(η) = η1 a∗ . ω

This yields 0 ≤ u(t ) ≤ η1 a∗ for all t ∈ [0, ω1 ]T . Hence, by assumption (iii), we have

λ(Au) = (Au)(η) ≤

η

Z

ϕq

Z

0

ω1

h(r )f (r , u(r ))∇ r



∆s <

0

a∗ Mξ

ηϕq

Z

ω1

h(r )∇ r



= a∗ .

0

Consequently, all the conditions of Lemma 2.5 are satisfied. The proof is completed.



We remark that the condition (i) in Theorem 4.7 can be replaced by the following condition (i0 ): lim

u→∞

f (t , u)

ϕp (u)



≤ ϕp

1 Wξ



,

which is a special case of (i). Corollary 4.8. If the condition (i) in Theorem 4.7 is replaced by (i0 ), then the conclusion of Theorem 4.7 also holds. 0 0 Proof. By n Theorem 4.7, we o only need to prove that  ∗(i) implies that (i) holds, that is, if (i ) holds, then there is a number ∗ω W a W 1 ξ d∗ ≥ max , N ξ b∗ such that f (t , u) ≤ ϕp Wd for u ∈ [0, d∗ ]. ηN ξ

ξ

ξ

Suppose on the contrary that for any d∗ ≥ max

ϕp



d∗ Wξ



n

a∗ ω1 Wξ

η Nξ

,

Wξ Nξ

o

b∗ , there exists uc ∈ [0, d∗ ] such that f (t , uc ) >

for t ∈ [0, ω1 ]T . Hence, if we choose

cn > max 0



a∗ ω 1 W ξ

ηNξ

Wξ ∗ , b Nξ



(n = 1, 2, . . .)

with cn0 → ∞, then there exist un ∈ [0, cn0 ] such that f (t , un ) > ϕp



cn0



Wξ

for t ∈ [0, ω1 ]T

(4.4)

and so lim f (t , un ) = ∞ for t ∈ [0, ω1 ]T .

n→∞

(4.5)

Since the condition (i0 ) holds, then there exists τ > 0 such that f (t , u) ≤ ϕp



u Wξ



for (t , u) ∈ [0, ω1 ]T × (τ , ∞).

(4.6)

1678

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Hence, we have un ≤ τ . Otherwise, if un > τ , then it follows from (4.6) that f (t , un ) ≤ ϕp



un Wξ



≤ ϕp



cn0



Wξ

for t ∈ [0, ω1 ]T ,

which contradicts (4.4). Let W = max(t ,u)∈[0,ω1 ]T ×[0,τ ] f (t , u), then f (t , un ) ≤ W (n = 1, 2, . . .), which also contradicts (4.5). The proof is complete.  Theorem 4.9. Suppose that there exist constants a∗i , b∗i , d∗i such that

ηNξ ∗ η ∗ ηNξ ∗ η ∗ ηNξ ∗ η ∗ b < d < a∗2 < b < d < a∗3 < · · · < a∗n < b < d , 0 < a∗1 < ω1 1 ω1 Wξ 1 ω1 2 ω1 Wξ 2 ω1 n ω1 Wξ n here n ∈ N and i = 1, 2, . . . , n. In addition, suppose that Wξ > ϕq (i) f (t , u) < ϕp



d∗ i



for (t , u) ∈ [0, ω1 ]T × [0, d∗i ];

(ii) f (t , u) > ϕp



Wξ b∗



for (t , u) ∈ [η, ω1 ]T × [b∗i , d∗i ];

(iii) f (t , u) < ϕp

i

 Nξ∗  ai

Mξ

h

ω

R ω1



η h(s)∇ s holds, f satisfies the following conditions:

i

for (t , u) ∈ [0, ω1 ]T × 0, η1 a∗i .

Then problem (1.3) has at least 2n + 1 positive pseudo-symmetric solutions. Proof. Similar to the proof of Theorem 4.2, we omit it here.



Note that these two methods have their advantages respectively. In the following, we show the differences of the method one and two, from two aspects. From the viewpoint of the obtained pseudo-symmetric solution position and local properties, by using method one, we only get some local properties of solutions. However, the position of solutions are not determined. For method two, we not only get some local properties of solutions but also give the position of all solutions, with regard to some subsets of the cone. In addition, by using these two different methods, the local properties of obtained solutions are obviously different. Hence, it is convenient for us to comprehensively comprehend the solutions of the models by using the two different techniques. From the viewpoint of the satisfied conditions, we take Theorems 4.1 and 4.7 as examples to illustrate. Under the same parameters conditions, the f (t , u) of (i) in Theorem 4.1 has a wider range than the f (t , u) of (i) in Theorem 4.7, the region of f (t , u) in (iii) of Theorem 4.7 is wider than that of Theorem 4.1. 5. Examples In this section, we present three simple examples to illustrate our results. In addition, the second example and third example show the differences of the two methods in Section 4. Example 5.1. Let

T = 0.1n ∪ 0.12n ∪ {0, 0.2, 0.25, 0.3, 0.9, 0.95, 1} ∪ [0.4, 0.8]









and

T = 1.

Consider the following boundary value problem

(ϕp (u4 (t )))∇ + h(t )f (t , u(t )) = 0, u(0) = 0 and u (0.2) = u(1),

t ∈ [0, 1]T ,

where h(t ) =

t + ρ(t ) for t ∈ [0, 0.6]T , 1.2 − t + ρ(1.2 − t ) for t ∈ [0.6, 1]T ,



and f (t , u(t )) =



t + ρ(t ) + |u(t )|p−2 for t ∈ [0, 0.6]T , 1.2 − t + ρ(1.2 − t ) + |u(1.2 − t )|p−2 for t ∈ [0.6, 1]T .

Note that

 t + ρ(t ) + |u(t )|p−2    lim = ∞ for t ∈ [0, 0.6] , + |u|p−2 u f0 = u→0 1.2 − t + ρ(1.2 − t ) + |u(t )|p−2    lim = ∞ for t ∈ [0.6, 1]T , + |u|p−2 u u→0

(5.1)

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1679

and

f∞

 t + ρ(t ) + |u(t )|p−2   = 0 for t ∈ [0, 0.6] ,  lim u→∞ |u|p−2 u = 1.2t + ρ(1.2 − t ) + |u(t )|p−2    lim = 0 for t ∈ [0.6, 1]T . u→∞ |u|p−2 u

It is obvious that i0 = i∞ = 1. Hence, Theorem 3.1 implies that the boundary value problem (5.1) has at least one positivesymmetric solution. Example 5.2. Let

T = 0.1n ∪ 0.12n ∪ {0, 0.2, 0.25, 0.3, 0.9, 0.95, 1} ∪ [0.4, 0.8]









and T = 1.

(5.2)

Consider the following boundary value problem with p = 3,

(ϕp (u4 (t )))∇ + h(t )f (t , u(t )) = 0, u(0) = 0 and u (0.2) = u(1),

t ∈ [0, 1]T ,

(5.3)

where h( t ) =



t + ρ(t ) for t ∈ [0, 0.6]T , 1.2 − t + ρ(1.2 − t ) for t ∈ [0.6, 1]T ,

(5.4)

and

f (t , u) =: f (u) =

 0.15,   p1 (u), 

u u u u u

80,

   s(u), 16,

∈ [0, 0.6] , ∈ [0.6, 1] , ∈ [1, 3] , ∈ [3, 8] , ∈ [8, ∞) ,

here p1 (u) and s(u) satisfy p1 (0.6) = 0.15, p1 (1) = 80, s(3) = 80, s(8) = 16,

(p∇1 (u))∇ = 0 for u ∈ (0.6, 1),

(s∇ (u))∇ = 0 for u ∈ (3, 8).

Note that η = 0.2, ω1 = 0.6. If we choose r = 0.3, then a direct calculation shows that Mξ = ηϕq

ω1

Z

h(r )∇ r



= 0.2 × 0.6 = 0.12,

Nξ ≈ 0.1131,

Lξ = 0.18

and

Wξ = 0.36.

0 rN

If we take a0 = 0.3, b0 = 1, c 0 = 3, then a0 < ωr b0 < ω Mξ c 0 holds, furthermore, 1 1 ξ f (u) = 0.15 < 2.7778 ≈ ϕp f (u) = 80 > 78.176 ≈ ϕp



a0





Lξ b0



Nξ

for u ∈ [0, 0.6], for u ∈ [1, 3].

∇ ∇ ∇ From (p∇ 1 (u)) = 0 and (s (u)) = 0, we have max p1 (u) = max s(u) = 80, Thus,

f (u) < 625 = ϕp



c0



for u ∈ [0, 9].

Mξ

By Theorem 4.1 we see that the boundary value problem (5.3) has at least three positive pseudo-symmetric solutions u1 , u2 and u3 such that 0<

max u1 (t ) < 0.3 <

t ∈[0,0.3]T

min

t ∈[0.2,0.6]T

u2 (t ) < 1 <

However, ϕq

R 0.6 0.2

max u2 (t ) with

t ∈[0,0.3]T

min

t ∈[0.2,0.6]T



h(s)∆s

u3 (t ) and

max u3 (t ) < 3.

t ∈[0,0.2]T

≈ 0.5657 > Wξ = 0.36, hence, the existence of positive solutions of boundary value

problem (5.3) is not obtained by using Theorem 4.7. Example 5.3. Let

T = 0.1n ∪ 0.12n ∪ {0, 0.6, 0.7, 0.8, 0.9, 1} ∪ [0.75, 0.85]









and

T = 1.

(5.5)

1680

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

Consider the following boundary value problem with p = 3



∇ ϕp u∆ (t ) + f (u(t )) = 0, u(0) = 0

and

t ∈ [0, 1]T ,

(5.6)

u(0.6) = u(1).

It is obvious that η = 0.6, ω1 = 0.8. Choose r  = 0.7, direct calculation shows that Nξ ≈ 0.3175, Mξ = 0.48, Lξ = R 0.8  0.56, Wξ = 0.64 and Wξ = 0.64 > 0.5291 ≈ ϕq 0.6 ∆t . Let ε be an arbitrary small positive number, a∗ , b∗ and d∗ are ηN

η

arbitrary positive numbers with a∗ < ω b∗ < ω Wξ d∗ , and 1 1 ξ

  ∗  4 a   − ε, 0 ≤ u ≤ a∗ , ϕ3   0 . 48 3   4 ∗ f (u) = k(u), a ≤ u ≤ b∗ ,  3     b∗   ϕ3 + ε, b∗ ≤ u ≤ d∗ , 0.3175  ∗   ∗ 
  b here k(u) satisfies k 43 a∗ = ϕ3 0a.48 − ε, k(b∗ ) = ϕ3 0.3175 + ε, (k∆ (u))∆ = 0 for u ∈ 43 a∗ , b∗ . It is obvious that the (i), (ii) and (iii) in Theorem 4.7 are satisfied. By Theorem 4.7, we see that the boundary value problem (5.6) has at least three positive solutions u1 , u2 and u3 such that

kxi k ≤ d∗ for i = 1, 2, 3, and

min

t ∈[0.6,0.8]T

u2 (t ) < b∗

b∗ <

min

t ∈[0.6,0.8]T

with

min

u1 (t ),

t ∈[0.6,0.8]T

a∗ <

min

t ∈[0.6,0.8]T

u2 ( t )

u3 (t ) < a∗ . rN

However, for arbitrary positive numbers a∗ , b∗ , d∗ with a∗ < ωr b∗ < ω Wξ d∗ , the condition (iii) of Theorem 4.1 is not 1 1 ξ satisfied. Therefore, Theorem 4.1 is not fit to study the boundary value problem (5.6). Note, it is obvious that it is very difficult to consider the Examples 5.2 and 5.3 by using Theorem 3.2 in [20]. Acknowledgments The author is very grateful to Professor E. Rodin and the anonymous referee for their helpful comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19]

F.M. Atici, D.C. Biles, A. Lebedinsky, An application of time scales to economics, Math. Comput. Modelling 43 (2006) 718–726. M.A. Jones, B. Song, D.M. Thomas, Controlling wound healing through debridement, Math. Comput. Modelling 40 (2004) 1057–1064. V. Spedding, Taming nature’s numbers, New Scientist (2003) 28–32. D.M. Thomas, L. Vandemuelebroeke, K. Yamaguchi, A mathematical evolution model for phytoremediation of metals, Discrete Contin. Dyn. Syst. Ser. B 5 (2005) 411–422. M. Bohner, A. Peterson, Advances in Dynamic Equation on Time Scales, Birkhäuser, Boston, 2003. M. Bohner, A. Peterson, Dynamic Equation on Time Scales, An Introduction with Applications, Birkhäuser, Boston, 2001. S. Hilger, Analysis on measure chains—A unified approach to continuous and discrete calculus, Results Math. 18 (1990) 18–56. R. Avery, J. Henderson, Existence of three positive pseudo-symmetric solutions for a one dimensional discrete p-Laplacian, J. Difference Equ. Appl. 10 (2004) 529–539. R.I. Avery, J. Henderson, Existence of three positive pseudo-symmetric solutions for a one dimensional p-Laplacian, J. Math. Anal. Appl. 277 (2003) 395–404. D. Ji, Y. Yang, W. Ge, Triple positive pseudo-symmetric solutions to a four-point boundary value problem with p-Laplacian, Appl. Math. Lett. 21 (2008) 268–274. D. Ma, W. Ge, Existence and iteration of positive pseudo-symmetric solutions for a three-point second-order p-Laplacian BVP, Appl. Math. Lett. 20 (2007) 1244–1249. B. Sun, W. Ge, Successive iteration and positive pseudo-symmetric solutions for a three-point second-order p-Laplacian boundary value problems, Appl. Math. Comput. 188 (2007) 1772–1779. J.T. Cho, J. Inoguchi, Pseudo-symmetric contact 3-manifolds II—When is the tangent sphere bundle over a surface pseudo-symmetric, Note di Math. 27 (2007) 119–129. S.W. Ng, A.D. Rae, The pseudo symmetric structure of bis(dicyclohexylammonium) bis(oxalatotriphenylstannate), Zeit. für Kristallographie 215 (2000) 199–206. R.I. Avery, A generalization of the Leggett-Williams fixed point theorem, MSR Hot-Line. 2 (1998) 9–14. W. Han, S. Kang, Multiple positive solutions of nonlinear third-order BVP for a class of p-Laplacian dynamic equations on time scales, Math. Comput. Modelling 49 (3–4) (2009) 527–535. Z. He, L. Li, Multiple positive solutions for the one-dimensional p-Laplacian dynamic equations on time scales, Math. Comput. Modelling 45 (2007) 68–79. Y.H. Su, W.T. Li, H.R. Sun, Positive solutions of singular p-Laplacian BVPs with sign changing nonlinearity on time scales, Math. Comput. Modelling 48 (2008) 845–858. H.R. Sun, W.T. Li, Existence theory for positive solutions to one-dimensional p-Laplacian boundary value problems on time scales, J. Differential Equations 240 (2007) 217–248.

Y.-H. Su / Mathematical and Computer Modelling 49 (2009) 1664–1681

1681

[20] Y.H. Su, W.T. Li, H.R. Sun, Triple positive pseudo-symmetric solutions of three-point BVPs for p-Laplacian dynamic equations on time scales, Nonlinear Anal. TMA 68 (2008) 1442–1452. [21] M. Krasnosel’skii, Positive Solutions of Operator Equations, Noordhoff, Groningen, 1964. [22] J.L. Ren, W. Ge, B.X. Ren, Existence of three solutions for quasi-nonlinear boundary value problems, Acta. Math. Appl. Sinica (Engl. Ser.) 21 (2005) 353–358. [23] R.I. Avery, A. Peterson, Three positive fixed points of nonlinear operators on ordered Banach spaces, Comput. Math. Appl. 42 (2001) 313–422. [24] B. Aulbach, L. Neidhart, Integration on measure chains, in: Proceedings of the Sixth International Conference on Difference Equations, CRC, Boca Raton, FL, 2004, pp. 239–252. [25] V. Lakshmikantham, S. Sivasundaram, B. Kaymakcalan, Dynamic Systems on Measure Chains, Kluwer Academic Publishers, Boston, 1996. [26] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, San Diego, CA, 1988. [27] R.I. Avery, J. Henderson, Two positive fixed points of nonlinear operator on ordered Banach spaces, Comm. Appl. Nonlinear Anal. 8 (2001) 27–36. [28] R.P. Agarwal, M. Bohner, P. Rehak, Half-Linear Dynamic Equations, Nonlinear Analysis and Applications: To V. Lakshmikantham on His 80th Birthday, vol. 1, Kluwer Acad. Publ, Dordrecht, 2003, pp. 1–57. [29] H. Wang, Positive periodic solutions of functional differential equations, J. Differential Equations 202 (2004) 354–366.