Positive solutions for p -Laplacian functional dynamic equations on time scales

Positive solutions for p -Laplacian functional dynamic equations on time scales

Nonlinear Analysis 66 (2007) 1989–1998 www.elsevier.com/locate/na Positive solutions for p-Laplacian functional dynamic equations on time scales✩ Cha...

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Nonlinear Analysis 66 (2007) 1989–1998 www.elsevier.com/locate/na

Positive solutions for p-Laplacian functional dynamic equations on time scales✩ Changxiu Song a,b,∗ , Cuntao Xiao b a School of Mathematical Sciences, South China Normal University, Guangzhou 510631, PR China b School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, PR China

Received 27 November 2005; accepted 23 February 2006

Abstract In this paper, the authors study the boundary value problems for a p-Laplacian functional dynamic equation on a time scale,  [φ p (x  (t))]∇ + a(t) f (x(t), x(μ(t))) = 0, t ∈ (0, T ), x0 (t) = ψ(t), t ∈ [−r, 0], x(0) − B0 (x  (η)) = 0, x  (T ) = 0. By using the twin fixed point theorem, sufficient conditions are established for the existence of twin positive solutions. c 2006 Elsevier Ltd. All rights reserved.  MSC: 39K10; 34B15 Keywords: Time scale; Boundary value problem; p-Laplacian functional dynamic equations; Positive solutions; Fixed point theorem

1. Introduction Let T be a closed nonempty subset of R, and let T have the subspace topology inherited from the Euclidean topology on R. In some of the current literature, T is called a time scale (or measure ✩ Supported by grant 10571064 from the NNSF of China, and by grant 011471 from the NSF of Guangdong. ∗ Corresponding author at: School of Applied Mathematics, Guangdong University of Technology, Guangzhou

510006, PR China. Tel.: +86 020 37217313; fax: +86 020 85213533. E-mail address: [email protected] (C. Song). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter  doi:10.1016/j.na.2006.02.036

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C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998

chain). For notation, we shall use the convention that, for each interval of J of R, J will denote the time scales interval, that is, J := J ∩ T. In this paper, let T be a time scale such that −r, 0, T ∈ T. We are concerned with the existence of positive solutions of the p-Laplacian dynamic equation on a time scale  [φ p (x  (t))]∇ + a(t) f (x(t), x(μ(t))) = 0, t ∈ (0, T ), (1.1) x 0 (t) = ψ(t), t ∈ [−r, 0], x(0) − B0 (x  (η)) = 0, x  (T ) = 0, where φ p (u) is the p-Laplacian operator, i.e., φ p (u) = |u| p−2 u, p > 1, (φ p )−1 (u) = φq (u), where 1p + q1 = 1; η ∈ (0, ρ(T )) and (H1 ) the function f : (R+ )2 → R+ is continuous; (H2 ) the function a : T → R+ is left dense continuous (i.e., a ∈ Cld (T, R+ )) and does not vanish identically on any closed subinterval of [0, T ]; here Cld (T, R+ ) denotes the set of all left dense continuous functions from T to R+ ; (H3 ) ψ : [−r, 0] → R+ is continuous and r > 0; (H4 ) μ : [0, T ] → [−r, T ] is continuous, μ(t) ≤ t for all t; (H5 ) B0 : R → R is continuous and satisfies that there are β ≥ δ ≥ 0 such that δs ≤ B0 (s) ≤ βs

for s ∈ R+ .

p-Laplacian problems with two-, three-, m-point boundary conditions for ordinary differential equations and finite difference equations have been studied extensively; for example see [1–4] and references therein. However, there are not many concerning the p-Laplacian problems on time scales, especially for p-Laplacian functional dynamic equations on time scales. The motivation for the present work stems from many recent investigations in [5–8] and references therein. In particular, Kaufmann and Raffoul [6] considered a nonlinear functional dynamic equation on a time scale and obtained sufficient conditions for the existence of positive solutions. In this paper, we apply the twin fixed point theorem to obtain at least two positive solutions of boundary value problem (BVP for short) (1.1) when growth conditions are imposed on f . Finally, we present two corollaries, which show that under the assumption that f is superlinear or sublinear, BVP (1.1) has at least two positive solutions. Our results generalize the recent paper [8]. For convenience, we list the following well-known definitions which can be found in [9–11] and the references therein. Definition 1.1. For t < sup T and r > inf T, define the forward jump operator σ and the backward jump operator ρ, respectively: σ (t) = inf{τ ∈ T | τ > t} ∈ T, ρ(r ) = sup{τ ∈ T | τ < r } ∈ T

for all t, r ∈ T.

If σ (t) > t, t is said to be right scattered, and if ρ(r ) < r, r is said to be left scattered. If σ (t) = t, t is said to be right dense, and if ρ(r ) = r, r is said to be left dense. If T has a right scattered minimum m, define Tk = T − {m}; otherwise set Tk = T. If T has a left scattered maximum M, define Tk = T − {M}; otherwise set Tk = T. Definition 1.2. For x : T → R and t ∈ Tk , we define the delta derivative of x(t), x  (t), to be the number (when it exists), with the property that, for any ε > 0, there is a neighborhood U of t such that |[x(σ (t)) − x(s)] − x  (t)[σ (t) − s]| < ε|σ (t) − s|

for all s ∈ U.

C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998

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For x : T → R and t ∈ Tk , we define the nabla derivative of x(t), x ∇ (t), to be the number (when it exists), with the property that, for any ε > 0, there is a neighborhood V of t such that |[x(ρ(t)) − x(s)] − x ∇ (t)[ρ(t) − s]| < ε|ρ(t) − s| x  (t)

x ∇ (t)

for all s ∈ V.

x (t).

= = If T = Z , then x  (t) = x(t + 1) − x(t) is forward If T = R, then ∇ difference operator while x (t) = x(t) − x(t − 1) is the backward difference operator. t Definition 1.3. If F  (t) = f (t), then we define the delta integral by a f (s)s = F(t) − F(a).  t If Φ ∇ (t) = f (t), then we define the nabla integral by a f (s)∇s = Φ(t) − Φ(s). In the following, we provide some background materials from the theory of cones in Banach spaces, and we then state the double fixed point theorem for a cone preserving operator. Definition 1.4. Let E be a real Banach space. A nonempty, closed, convex set P ∈ E is called a cone, if it satisfies the following two conditions: (i) x ∈ P, λ ≥ 0 implies λx ∈ P; (ii) x, −x ∈ P implies x = 0. Every cone P ⊂ E induces an ordering in E given by x ≤ y if and only if y − x ∈ P. Definition 1.5. Given a cone P in a real Banach space E, a functional ω : P → R is said to be increasing on P provided ω(x) ≤ ω(y), for all x, y ∈ P with x ≤ y. Given a nonnegative continuous functional γ on a cone P of a real Banach space E, we define for each d > 0 the sets P(γ , d) = {x ∈ P : γ (x) < d}, ∂ P(γ , d) = {x ∈ P : γ (x) = d}, P(γ , d) = {x ∈ P : γ (x) ≤ d}. The following twin fixed point lemma due to [7] will play an important role in the proof of our results. Lemma 1.1. Let E be a real Banach space, P a cone of E, γ and α two nonnegative increasing continuous functionals, θ a nonnegative continuous functional, and θ (0) = 0. Suppose that there are two positive numbers c and M such that γ (x) ≤ θ (x) ≤ α(x),

x ≤ Mγ (x)

for x ∈ P(γ , c).

F : P(γ , c) → P is completely continuous. There are positive numbers 0 < a < b < c such that θ (λx) ≤ λθ (x) for all λ ∈ [0, 1] and x ∈ ∂ P(θ, b), and (i) γ (F x) > c for x ∈ ∂ P(γ , c); (ii) θ (F x) < b for x ∈ ∂ P(θ, b); (iii) α(F x) > a and P(α, a) = ∅ for x ∈ ∂ P(α, a). Then F has at least two fixed points x 1 and x 2 ∈ P(γ , c) satisfying a < α(x 1 ),

θ (x 1 ) < b,

b < θ (x 2 ),

γ (x 2 ) < c.

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2. Positive solutions We note that x(t) is a solution of (1.1) if and only if ⎧ T ⎪ ⎪ B ⎪ a(r ) f (x(r ), x(μ(r )))∇r φ 0 q ⎪ ⎪ ⎨ η T

t x(t) = ⎪ + φq a(r ) f (x(r ), x(μ(r )))∇r s, ⎪ ⎪ ⎪ 0 s ⎪ ⎩ ψ(t),

t ∈ [0, T ],

(2.1)

t ∈ [−r, 0].

Let E = Cld ([0, T ], R) be endowed with the norm x = maxt ∈[0,T ] |x(t)| and P = {x ∈ E : x is concave and nonnegative valued on [0, T ], and x  (T ) = 0}. Clearly, E is a Banach space with the norm x and P is a cone in E. For each x ∈ E, extend x(t) to [−r, T ] with x(t) = ψ(t) for t ∈ [−r, 0]. Define F : P → E as T a(r ) f (x(r ), x(μ(r )))∇r F x(t) = B0 φq η



t

+



T

φq

0

a(r ) f (x(r ), x(μ(r )))∇r s,

t ∈ [0, T ].

s

We seek a fixed point, x 1 , of F in the cone P. Define  x (t), t ∈ [0, T ], x(t) = 1 ψ(t), t ∈ [−r, 0]. Then x(t) denotes a positive solution of BVP (1.1). It follows from (2.2) that Lemma 2.1. Let F be defined by (2.2). If x ∈ P, then (i) (ii) (iii) (iv)

F(P) ⊂ P. F : P → P is completely continuous. x(t) ≥ Tt x , t ∈ [0, T ]. x(t) is increasing on [0, T ].

The proof is similar to the proofs of Lemma 3.1 and Theorem 3.1 in [8], and is omitted. Fix l ∈ T such that 0 < η < l < T , and set Y1 := {t ∈ [0, T ] : μ(t) < 0};

Y2 := {t ∈ [0, T ] : μ(t) ≥ 0}; Y3 := Y1 ∩ [l, T ].  Throughout this paper, we assume Y3 = ∅ and Y3 a(r )∇r > 0. Now, we define the nonnegative, increasing, continuous functionals γ , θ , and α on P by γ (x) = min x(t) = x(η), t ∈[η,l]

θ (x) = max x(t) = x(η), t ∈[0,η]

α(x) = min x(t) = x(l). t ∈[l,T ]

We have γ (x) = θ (x) ≤ α(x),

x ∈ P,

(2.2)

C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998

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and θ (x) = γ (x) = x(η) ≥

η

x , T

α(x) = x(l) ≥

l

x , T

for each x ∈ P.

Then

x ≤

T γ (x), η

x ≤

T α(x), l

for each x ∈ P.

(2.3)

We also see that θ (λx) = λθ (x),

∀ λ ∈ [0, 1],

x ∈ ∂ P(θ, b).

For notational convenience, we define σ1 , σ2 and ρ1 , ρ2 :



a(r )∇r ; σ2 = (δ + l)φq a(r )∇r ; σ1 = (δ + η)φq Y3 T



ρ1 = (β + η)φq

Y3



a(r )∇r ;

ρ2 = (β + l)φq

0

T

a(r )∇r .

0

Theorem 2.1. Suppose that there are positive numbers a < b < c such that σ2 ησ2 0 φ p (c/σ1 ) for c ≤ x ≤ (B) f (x, ψ(s)) < φ p (b/ρ1 ) for 0 ≤ x ≤

T η c, uniformly in s ∈ [−r, 0], T η b, uniformly in s ∈ [−r, 0];

f (x 1 , x 2 ) < φ p (b/ρ1 ) for 0 ≤ x i ≤ (C) f (x, ψ(s)) > φ p (a/σ2 ) for a ≤ x ≤

T l

T b, i = 1, 2, η

a, uniformly in s ∈ [−r, 0].

Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c. Proof. By the definition of operator F and its properties, it suffices to show that the conditions of Lemma 1.1 hold with respect to F. First, we verify that x ∈ ∂ P(γ , c) implies γ (T x) > c. Since γ (x) = x(η) = c, one gets x(t) ≥ c for t ∈ [η, T ]. Recalling that (2.3), we know c ≤ x ≤ Tη c for t ∈ [η, T ]. Then, we get T a(r ) f (x(r ), x(μ(r )))∇r γ (F x) = B0 φq

+ 0

η

η



T

φq s

a(r ) f (x(r ), x(μ(r )))∇r s

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C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998



≥ B 0 φq

T

a(r ) f (x(r ), x(μ(r )))∇r

l T

a(r ) f (x(r ), x(μ(r )))∇r l

≥ (δ + η)φq a(r ) f (x(r ), ψ(μ(r )))∇r Y 3 c > (δ + η)φq a(r )∇r = c. σ 1 Y3

+ ηφq

Secondly, we prove that x ∈ ∂ P(θ, b) implies θ (T x) < b. Since θ (x) = b implies x(η) = b, it holds that 0 ≤ x(t) ≤ b for t ∈ [0, η], and ∀ x ∈ ∂ P(θ, b) implies b ≤ x(t) ≤ x ≤

T T θ (x) = b, η η

for t ∈ [η, T ].

Then 0 ≤ x(t) ≤

T b, η

t ∈ [0, T ].

So, we have



θ (F x) = B0 φq

η

+ 0

T η



φq

a(r ) f (x(r ), x(μ(r )))∇r

T

a(r ) f (x(r ), x(μ(r )))∇r s

s

T a(r ) f (x(r ), x(μ(r )))∇r + ηφq a(r ) f (x(r ), x(μ(r )))∇r 0 0



= (β + η)φq a(r ) f (x(r ), ψ(μ(r )))∇r + a(r ) f (x(r ), x(μ(r )))∇r

< βφq

T

Y1

<

b (β + η)φq ρ1



T

Y2

a(r )∇r

= b.

0

Finally, we show that P(α, a) = ∅, α(T y) > a

for all x ∈ ∂ P(α, a).

It is obvious that P(α, a) = ∅. On the other hand, α(x) = x(l) = a and (2.3) imply a≤x≤

T a l

for t ∈ [l, T ].

Thus,

α(F x) = B0 φq

+ 0

l

T η



a(r ) f (x(r ), x(μ(r )))∇r T

φq s

a(r ) f (x(r ), x(μ(r )))∇r s

C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998



≥ B 0 φq

T

1995

a(r ) f (x(r ), x(μ(r )))∇r

l T

a(r ) f (x(r ), x(μ(r )))∇r l

≥ (δ + l)φq a(r ) f (x(r ), ψ(μ(r )))∇r Y 3 a > (δ + l)φq a(r )∇r = a. σ 2 Y3

+ lφq

By Lemma 1.1, F has at least two different fixed points x 1 and x 2 satisfying a < α(x 1 ), Let x(t) =

θ (x 1 ) < b,

 ψ(t), x i (t),

b < θ (x 2 ),

γ (x 2 ) < c.

t ∈ [−r, 0], t ∈ [0, T ], i = 1, 2, 

which are twin positive solutions of BVP (1.1). The proof is complete. In analogy to Theorem 2.1, we have the following result.

Theorem 2.2. Suppose that there are positive numbers a < b < c such that 0
lσ1 l b< c. T Tρ1

Assume f satisfies the following conditions: (A ) f (x, ψ(s)) < φ p (c/ρ1 ) for 0 ≤ x ≤

T η c,

uniformly in s ∈ [−r, 0],

f (x 1 , x 2 ) < φ p (c/ρ1 ) for 0 ≤ x i ≤ (B ) f (x, ψ(s)) > φ p (b/σ1 ) for b ≤ x ≤ (C ) f (x, ψ(s)) < φ p (a/ρ2 ) for 0 ≤ x ≤

T c, i = 1, 2, η

T η b, uniformly in s T l a, uniformly in s

f (x 1 , x 2 ) < φ p (a/ρ2 ) for 0 ≤ x i ≤

∈ [−r, 0], ∈ [−r, 0],

T a, i = 1, 2. l

Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c. Now, we give theorems which may be considered as the corollaries of Theorems 2.1 and 2.2. Let f 0 = lim

x→0+

f (x, ψ(s)) f (x, ψ(s)) ; f ∞ = lim ; f 00 = p−1 x→∞ x x p−1

and choose k1 , k2 , k3 such that

x1

lim

→0+ ;x

2

f (x 1 , x 2 ) →0+

p−1

max{x 1

p−1

, x2

}

,

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k1 σ2 > 1, k2 σ1 > 1; 0 < k3 ρ2 <

η . T

From the above we deduce that 0 < k3 ρ1 <

l T

.

Theorem 2.3. Suppose that the following conditions are satisfied: (D) f 0 > k1 p−1 , f ∞ > k2 p−1 , uniformly in s ∈ [−r, 0]; (E) there exists a p1 > 0 such that for ∀ 0 ≤ x ≤ f (x, ψ(s)) < and

f (x 1 , x 2 ) <

p1 ρ1

p1 ρ1

T η

p1 , one has

p−1 ,

uniformly in s ∈ [−r, 0],

p−1 ,

for 0 ≤ x i ≤

T p1 , i = 1, 2. η

Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c. Proof. First, choose b = p1 ; one gets f (x, ψ(s)) < φ p (b/ρ1 ) f (x 1 , x 2 ) < φ p (b/ρ1 ) p−1

Secondly, since f 0 > k1

T b, uniformly in s ∈ [−r, 0]; η T for 0 ≤ x i ≤ b, i = 1, 2. η for 0 ≤ x ≤

, there is R1 > 0 sufficiently small such that

f (x, ψ(s)) > (k1 x) p−1

for 0 ≤ x ≤ R1 .

Without loss of generality, suppose R1 ≤ a≤x≤

T l

a, we have x ≤ R1 and a <

σ2 ρ1 b.

f (x, ψ(s)) > (k1 x) p−1 ≥ (k1 a) p−1

lσ2 Tρ1 b.

Choose a > 0 so that a < (T /2l)R1 . For

Thus, a > φp σ2

for a ≤ x ≤

T a. l

Thirdly, since f ∞ > k2 p−1 , there is R2 > 0 sufficiently large such that f (x, ψ(s)) > (k2 x) p−1

for x ≥ R2 .

Without loss of generality, suppose R2 > f (x, ψ(s)) > (k2 x) p−1 ≥ (k2 c) p−1

Choose c ≥ R2 . Then c T > φp for c ≤ x ≤ c. σ1 η T η b.

ησ2 We get now 0 < a < σρ21 b < Tρ c, and then the conditions in Theorem 2.1 are all satisfied. 1 By Theorem 2.1, BVP (1.1) has at least two positive solutions. The proof is complete. 

C. Song, C. Xiao / Nonlinear Analysis 66 (2007) 1989–1998

1997

Theorem 2.4. Suppose that the following conditions are satisfied: (F) f 0 < k3 p−1 , uniformly in s ∈ [−r, 0]; f 00 < k3 p−1 ; (G) there exists a p2 > 0 such that for ∀ 0 ≤ x ≤ Tη p2 , one has p−1 p2 , uniformly in s ∈ [−r, 0]. f (x, ψ(s)) > σ1 Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c. The proof is similar to that of Theorem 2.3 and we omit it. The following corollaries are obvious. Corollary 2.1. Suppose that the following conditions are satisfied: (D ) f 0 = ∞, f∞ = ∞, uniformly in s ∈ [−r, 0]; (E) there exists a p1 > 0 such that for ∀ 0 ≤ x ≤ Tη p1 , one has f (x, ψ(s)) < and

f (x 1 , x 2 ) <

p1 ρ1

p1 ρ1

p−1 , uniformly in s ∈ [−r, 0],

p−1 ,

for 0 ≤ x i ≤

T p1 , i = 1, 2. η

Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c. Corollary 2.2. Suppose that the following conditions are satisfied: (F ) f0 = 0, uniformly in s ∈ [−r, 0], f 00 = 0; (G) there exists a p2 > 0 such that for ∀ 0 ≤ x ≤ Tη p2 , one has p−1 p2 , uniformly in s ∈ [−r, 0]. f (x, ψ(s)) > σ1 Then BVP (1.1) has at least two positive solutions of the form  ψ(t), t ∈ [−r, 0], x(t) = x i (t), t ∈ [0, T ], i = 1, 2, where a < mint ∈[l,T ] x 1 (t), maxt ∈[0,η] x 1 (t) < b and b < maxt ∈[0,η] x 2 (t), mint ∈[η,l] x 2 (t) < c.

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