Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 28, 1-7; http://www.math.u-szeged.hu/ejqtde/

Three Positive Solutions for p-Laplacian Functional Dynamic Equations on Time Scales

Wen Guan Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu, 730050, People’s Republic of China wangdb@lut.cn (W. Guan)

Abstract

In this paper, existence criteria of three positive solutions to the followimg p-Laplacian functional dynamic equation on time scales

\{ \begin{matrix}[Φ_{p}(u^{△}(t))]^{▽} + a(t)f(u(t), u(μ(t))) = 0, t ∈ (0, T) , \\ u_{0}(t) = φ(t), t ∈ [−r, 0] , u(0) − B_{0}(u^{△}(η)) = 0, u^{△}(T) = 0, \end{matrix}

are established by using the well-known Five Functionals Fixed Point Theorem.

Keywords: Time scale; p-Laplacian functional dynamic equation; Boundary value problem; Positive solution; fixed point theorem

MSC: 39K10, 34B15

1 Introduction

The theory of dynamic equations on time scales has been a new important mathematical branch (see, for example, [1-5] ) since it was initiated by Hilger [6]. At the same time, boundary value problems ( BVPs ) for dynamic equation on time scales have received considerable attention [7-15]. However, to the best of our knowledge, few papers can be found in the literature for BVPs of p-Laplacian dynamic equations on time scales, especially for p-Laplacian functional dynamic equations on time scales [12, 14].

Let T be a time scale, i.e., T is a nonempty closed subset of R. Let 0, T be points in T, an interval (0, T) denote time scales interval, that is, (0, T) := (0, T) ∩ T. Other types of intervals are defined similarly. Some definitions concerning time scales can be found in [2-4].

In this paper, we are concerned with the existence of positive solutions for the p-Laplacian functional dynamic equation on time scale

\{ \begin{matrix}[Φ_{p}(u^{△}(t))]^{▽} + a(t)f(u(t), u(μ(t))) = 0, t ∈ (0, T) , \\ u_{0}(t) = φ(t), t ∈ [−r, 0] , u(0) − B_{0}(u^{△}(η)) = 0, u^{△}(T) = 0, \end{matrix}  (1.1)

where Φ_{p}(s) is p-Laplacian operator, i.e., Φ_{p}(s) = |s|^{p−2} s, p > 1, (Φ_{p})^{−1} = Φ_{q}, \frac{1}{p} + \frac{1}{q} = 1, η ∈ (0, ρ(T)) and

(C_{1}) f : (R^{+})^{2} → R^{+} is continuous ;

(C_{2}) a : T → R^{+} is left dense continuous (i.e., a ∈ C_{ld}(T, R^{+})) and does not vanish identically on any closed subinterval of [0, T] , where C_{ld}(T, R^{+}) denotes the set of all left dense continuous functions from T to R^{+}; (C_{3}) φ : [−r, 0] → R^{+} is continuous and r > 0; (C_{4}) μ : [0, T] → [−r, T] is continuous, μ(t) ≤ 0 for all t; (C_{5}) B_{0} : R → R is continuous and satisfies the condition that there are A ≥ B ≥ 0 such

that

Bv ≤ B_{0}(v) ≤ Av, for all v ≥ 0.

We note that in [16], Li and Shen studied the problem (1.1) when T = R, φ(t) = 0, t ∈ [−r, 0] and the nonlinear term is not involved u(μ(t)). They imposed conditions on f to yield at least three positive solutions to the problem (1.1), by applying the Five Functionals Fixed Point Theorem [17] (which is a generalization of the Leggett-Williams Fixed-Point Theorem [18]).

In [14], Song and Xiao considered the problem (1.1), by using a double fixed-point theorem due to Avery et al. [19] in a cone, and obtained the existence of two positive solutions.

Motivated by [14] and [16], we shall show that the problem (1.1), has at least three positive solutions by means of the Five Functionals Fixed Point Theorem.

Let γ, β, θ be nonnegative, continuous, convex functionals on P and α, ψ be nonnegative, continuous, concave functionals on P. Then, for nonnegative real numbers h, a, b, d and c, we define the convex sets

P (γ, c) = {x ∈ P : γ(x) < c} ,
P (γ, α, a, c) = {x ∈ P : a ≤ α(x), γ(x) ≤ c} ,
Q(γ, β, d, c) = {x ∈ P : β(x) ≤ d, γ(x) ≤ c} ,
P (γ, θ, α, a, b, c) = {x ∈ P : a ≤ α(x), θ(x) ≤ b, γ(x) ≤ c} and
Q(γ, β, ψ, h, d, c) = {x ∈ P : h ≤ ψ(x), β(x) ≤ d, γ(x) ≤ c} .

To prove our main results, we need the following Five Functionals Fixed Point Theorem[17].

Theorem 1.1. Let P be a cone in a real Banach space E. Suppose there exist positive numbers c and M, nonnegative, continuous, concave functionals α and ψ on P, and nonnegative, continuous, convex functionals γ, β and θ on P, with

α(x) ≤ β(x) and ∥x∥ ≤ Mγ(x)

for all x ∈ \overline{P (γ, c)}. Suppose

F : \overline{P (γ, c)} → \overline{P (γ, c)}

is completely continuous and there exist nonnegative numbers h, a, k, b, with 0 < a < b such that:

(i) {x ∈ P (γ, θ, α, b, k, c) : α(x) > b} ≠ ϕ and α(F x) > b for x ∈ P (γ, θ, α, b, k, c); (ii) {x ∈ Q(γ, β, ψ, h, a, c) : β(x) < a} ≠ ϕ and β(F x) < a for x ∈ Q(γ, β, ψ, h, a, c); (iii) α(F x) > b for x ∈ P (γ, α, b, c) with θ(F x) > k; (iv) β(F x) < a for x ∈ Q(γ, β, a, c) with ψ(F x) < h.

Then F has at least three fixed points x_{1}, x_{2}, x_{3} ∈ \overline{P (γ, c)} such that

β(x_{1}) < a, b < α(x_{2}), and a < β(x_{3}) with α(x_{3}) < b.

2 Existence of Three Positive Solutions

We note that u(t) is a solution of the BVP (1.1) if and only if

u(t) = \{ \begin{matrix}\begin{matrix}B_{0} (Φ_{q} (∫_{η}^{T} a(r)f(u(r), u(μ(r)))∇r)) \\ + ∫_{0}^{t} Φ_{q} (∫_{s}^{T} a(r)f(u(r), u(μ(r)))∇r) △s, \end{matrix} & t ∈ [0, T] , \\ φ(t), & t ∈ [−r, 0] . \end{matrix}

Let E = C_{ld}([0, T] , R) be endowed with ∥u∥ = sup_{t∈[0,T]} |u(t)| , so E is a Banach space. Define cone P ⊂ E by

P = {u ∈ E : u is concave and nonnegative valued on [0, T] , and u^{△}(T) = 0} .

For each u ∈ E, extend u(t) to [−r, T] with u(t) = φ(t) for t ∈ [−r, 0] . Define F: P → E by

\begin{aligned}(F u)(t) = & B_{0} (Φ_{q} (∫_{η}^{T} a(r)f(u(r), u(μ(r)))∇r)) \\ & + ∫_{0}^{t} Φ_{q} (∫_{s}^{T} a(r)f(u(r), u(μ(r)))∇r) △s, t ∈ [0, T] . \end{aligned}

We seek a point, u_{1}, of F in the cone P. Define

u(t) = \{ \begin{matrix}u_{1}(t), & t ∈ [0, T] , \\ φ(t), & t ∈ [−r, 0] . \end{matrix}

Then u(t) denotes a positive solution of the BVP (1.1). We have the following results

Lemma 2.1. Let u ∈ P, then

(1) u(t) ≥ \frac{t}{T} ∥u∥ for t ∈ [0, T] , and

(2) τu(ς) ≤ ςu(τ) for 0 < τ < ς < T and τ, ς ∈ T.

Proof. (1) is Lemma 3.1 of [12]. It is easy to conclude that (2) is satisfied by the concavity of u.

Set

Y_{1} = {t ∈ [0, T] : μ(t) < 0}; Y_{2} = {t ∈ [0, T] : μ(t) ≥ 0}; Y_{3} = Y_{1} ∩ [l, T] .

Throughout this paper, we assume Y_{3} ≠ ϕ and ∫_{Y_{3}} a(r)∇r > 0. Let l ∈ T be fixed such that 0 < η < l < T, and define the nonnegative, continuous, concave

functionals α, ψ and the nonnegative, continuous, convex functionals β, θ, γ on the cone P respectively as

γ(u) = θ(u) = max_{t∈[0,η]} u(t) = u(η), α(u) = min_{t∈[l,T]} u(t) = u(l),
β(u) = max_{t∈[0,l]} u(t) = u(l), ψ(u) = min_{t∈[η,T]} u(t) = u(η).

We observe that α(u) = β(u) for each u ∈ P. In addition, by Lemma 2.1, we have γ(u) = u(η) ≥ \frac{η}{T} ∥u∥ . Hence ∥u∥ ≤ \frac{T}{η} γ(u) for all u ∈ P. For convenience, we define

μ = (A + η)Φ_{q} (∫_{0}^{T} a(r)∇r) , δ = (B + l)Φ_{q} (∫_{Y_{3}} a(r)∇r) ,

λ = (A + l)Φ_{q}(∫_{0}^{T} a(r)∇r).

We now state growth conditions on f so that the BVP (1.1) has at least three positive solutions.

Theorem 2.1. Let 0 < a < \frac{l}{T} b < \frac{ηl}{T^{2}} c, μb < δc, and suppose that f satisfies the following conditions:

(H_{1}) f(u, φ(s)) < Φ_{p} (\frac{c}{μ}) , if 0 ≤ u ≤ \frac{T}{η} c, uniformly in s ∈ [−r, 0] , and f(u_{1}, u_{2}) < Φ_{p}(\frac{c}{μ}), if

0 ≤ u_{i} ≤ \frac{T}{η} c, i = 1, 2;

(H_{2}) f(u, φ(s)) > Φ_{p} (\frac{b}{δ}) , if b ≤ u ≤ (\frac{T}{η} )^{2} b, uniformly in s ∈ [−r, 0] ;

(H_{3}) f(u, φ(s)) < Φ_{p} (\frac{a}{λ}) , if 0 ≤ u ≤ \frac{T}{l} a, uniformly in s ∈ [−r, 0], and f(u_{1}, u_{2}) < Φ_{p}(\frac{a}{λ}), if 0 ≤ u_{i} ≤ \frac{T}{l} a, i = 1, 2.

Then the BVP (1.1) has at least three positive solutions of the form

u(t) = \{ \begin{matrix}u_{i}(t), & t ∈ [0, T] , i = 1, 2, 3, \\ φ(t), & t ∈ [−r, 0] , \end{matrix}

where max_{t∈[0,l]} u_{1}(t) < a, min_{t∈[l,T]} u_{2}(t) > b, and a < max_{t∈[0,l]} u_{3}(t) with min_{t∈[l,T]} u_{3}(t) < b.

Proof. By [12], it is known that F : P → P is completely continuous. Let u ∈ \overline{P (γ, c)}, then γ(u) = max_{t∈[0,η]} u(t) = u(η) ≤ c, consequently, 0 ≤ u(t) ≤ c for

t ∈ [0, η] . Since u(η) ≥ \frac{η}{T} u(T), so ∥u∥ = u(T) ≤ \frac{T}{η} u(η) ≤ \frac{T}{η} c, this implies

0 ≤ u(t) ≤ \frac{T}{η} c, for t ∈ [0, T] .

From (H_{1}), we have

γ(F u) = (F u)(η)
= B_{0} (Φ_{q} (∫_{η}^{T} a(r)f(u(r), u(μ(r)))∇r)) + ∫_{0}^{η} Φ_{q} (∫_{s}^{T} a(r)f(u(r), u(μ(r)))∇r) △s
≤ AΦ_{q} (∫_{0}^{T} a(r)f(u(r), u(μ(r)))∇r) + ηΦ_{q} (∫_{0}^{T} a(r)f(u(r), u(μ(r)))∇r)
= (A + η)Φ_{q} [∫_{Y_{1}} a(r)f(u(r), φ(μ(r)))∇r + ∫_{Y_{2}} a(r)fu(r), u(μ(r)))∇r]
< (A + η)Φ_{q} (∫_{0}^{T} a(r)∇r) \frac{c}{μ}
= c.

Therefore

F u ∈ \overline{P (γ, c)}.

We now turn to property (i) of Theorem 1.1. Choosing u ≡ \frac{T}{η} b, k = \frac{T}{η} b, it follows that

α(u) = u(l) = \frac{T}{η} b > b, θ(u) = u(η) = \frac{T}{η} b = k, γ(u) = u(η) = \frac{T}{η} b < c,

which shows that \{u ∈ P (γ, θ, α, b, k, c) : α(u) > b\} ≠ ϕ, and for u ∈ P (γ, θ, α, b, \frac{T}{η} b, c), we have

b ≤ u(t) ≤ (\frac{T}{η} )^{2} b, for t ∈ [l, T] .

From (H_{2}), we have

α(F u) = (F u)(l)
= B_{0} (Φ_{q} (∫_{η}^{T} a(r)f(u(r), u(μ(r)))∇r)) + ∫_{0}^{l} Φ_{q} (∫_{s}^{T} a(r)f(u(r), u(μ(r)))∇r) △s
≥ BΦ_{q} (∫_{l}^{T} a(r)f(u(r), u(μ(r)))∇r) + lΦ_{q} (∫_{l}^{T} a(r)f(u(r), u(μ(r)))∇r)
≥ (B + l)Φ_{q} (∫_{Y_{3}} a(r)f(u(r), φ(μ(r)))∇r)
> (B + l)Φ_{q} (∫_{Y_{3}} a(r)∇r) \frac{b}{δ}
= b.

We conclude that (i) of Theorem 1.1 is satisfied. We next address (ii) of Theorem 1.1. If we take u ≡ \frac{η}{T} a, h = \frac{η}{T} a, then

γ(u) = u(η) = \frac{η}{T} a < c, ψ(u) = u(η) = \frac{η}{T} a = h, β(u) = u(l) = \frac{η}{T} a < a.

From this we know that \{u ∈ Q(γ, β, ψ, h, a, c) : β(u) < a\} ≠ ϕ. If u ∈ Q(γ, β, ψ, \frac{η}{T} a, a, c), then

0 ≤ u(t) ≤ \frac{T}{l} a, for t ∈ [0, T] .

From (H_{3}), we have

β(F u) = (F u)(l)
= B_{0} (Φ_{q} (∫_{η}^{T} a(r)f(u(r), u(μ(r)))∇r)) + ∫_{0}^{l} Φ_{q} (∫_{s}^{T} a(r)f(u(r), u(μ(r)))∇r) △s
≤ AΦ_{q} (∫_{0}^{T} a(r)f(u(r), u(μ(r)))∇r) + lΦ_{q} (∫_{0}^{T} a(r)f(u(r), u(μ(r)))∇r)
= (A + l)Φ_{q} [∫_{Y_{1}} a(r)f(u(r), φ(μ(r)))∇r + ∫_{Y_{2}} a(r)f(u(r), u(μ(r)))∇r]
< (A + l)Φ_{q} (∫_{0}^{T} a(r)∇r) \frac{a}{λ}
= a.

Now we show that (iii) of Theorem 1.1 is satisfied. If u ∈ P (γ, α, b, c) and θ(F u) = F u(η) > \frac{T}{η} b, then

α(F u) ≥ (F u)(l) = \frac{l}{T} F u(l) ≥ \frac{l}{T} F u(η) > \frac{l}{η} b > b.

Finally, if u ∈ Q(α, β, a, c) and ψ(F u) = F u(η) < \frac{η}{T} a, then from (2) of the Lemma 2.1 we have

β(F u) = F u(l) ≤ \frac{T}{l} F u(l) ≤ \frac{T}{η} F u(η) < a.

which shows that condition (iv) of Theorem 1.1 is fulfilled.

Thus, all the conditions of Theorem 1.1 are satisfied. Hence, F has at least three fixed points u_{1}, u_{2}, u_{3} satisfying

β(u_{1}) < a, b < α(u_{2}), and a < β(u_{3}) with α(u_{3}) < b.

Let

u(t) = \{ \begin{matrix}u_{i}(t), & t ∈ [0, T] , i = 1, 2, 3, \\ φ(t), & t ∈ [−r, 0] , \end{matrix}

which are three positive solutions of the BVP (1.1).

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(Received June 9, 2008)