Triple positive solutions of m -point BVPs for p -Laplacian dynamic equations on time scales

Nonlinear Analysis 69 (2008) 3811–3820 www.elsevier.com/locate/na

Triple positive solutions of m-point BVPs for p-Laplacian dynamic equations on time scalesI You-Hui Su a,b,∗ , Wan-Tong Li a a School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu, 730000, People’s Republic of China b Department of Mathematics, Hexi University, Zhangye, Gansu 734000, People’s Republic of China

Received 29 August 2007; accepted 11 October 2007

Abstract This paper is concerned with the existence of positive solutions of p-Laplacian dynamic equation (ϕ p (u ∆ (t)))∇ + Pm−2 a1 (t) f (u(t)) = 0 subject to boundary conditions u(0) − B0 ( i=1 ai u ∆ (ξi )) = 0, u ∆ (T ) = 0 or u ∆ (0) = 0, u(T ) + Pm−2 B1 ( i=1 bi u ∆ (ξi )) = 0, where ϕ p (v) = |v| p−2 v with p > 1. By using the five functionals fixed-point theorem, we prove that the boundary value problem has at least three positive solutions. As an application, an example is given to illustrate the result. c 2007 Elsevier Ltd. All rights reserved.

MSC: 34B15; 39A10 Keywords: Time scales; Boundary value problem; Positive solutions; p-Laplacian; Fixed-point theorem

1. Introduction The development of the theory of time scales was initiated by Hilger in his Ph.D. thesis in 1988 as a theory can contain both differential and difference calculus in a consistent way. Since then, we have witnessed a lot of efforts in the field of time scales, especially in unifying the theory of differential equations in the continuous case and the theory of finite difference equations in the discrete case, see [3,4,9,14,16] and the references therein. Many works are concerned with the existence of positive solutions of boundary value problems with p-Laplacian in the continuous case, see [8,10,12,13]. However, very little work has been done to the existence of positive solutions of p-Laplacian dynamic equation on time scales, see [6,15,17], which motivate us to consider one-dimensional p-Laplacian boundary value problem on time scales. Throughout the remainder of the paper, 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. 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 types of intervals are defined similarly. I Supported by the NNSF of China (10571078). ∗ Corresponding author at: School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu, 730000, People’s Republic of China.

E-mail address: [email protected] (Y.-H. Su). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.10.018

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We denote the p-Laplacian operator by ϕ p (v), i.e. ϕ p (v) = |v| p−2 v, p > 1, (ϕ p )−1 = ϕq and For the two-point boundary value problem (ϕ p (u ∆ (t)))∆ + a1 (t) f (u σ (t)) = 0, ∆

u(a) − B3 (u (a)) = 0,

1 p

+

1 q

= 1.

t ∈ [a, b]T ,

∆

u (σ (b)) = 0.

Sun and Li [15] established the existence theory for positive solutions by using some fixed-point theorems [1,5,7]. For the three-point boundary value problems (ϕ p (u ∆ (t)))∇ + a1 (t) f (u(t)) = 0,

t ∈ [0, T ]T ,

(1.1)

satisfying the boundary conditions u(0) − B0 (u ∆ (η)) = 0,

u ∆ (T ) = 0,

(1.2)

u(T ) + B1 (u ∆ (η)) = 0,

(1.3)

or u ∆ (0) = 0,

where η ∈ (0, ρ(T ))T , B0 and B1 satisfy Bx ≤ Bi (x) ≤ Ax,

x ∈ R, i = 0, 1,

(1.4)

here B and A are nonnegative numbers. Under some assumptions on f , by using the fixed-point theorem due to Avery and Henderson [1], He [6] proved that the boundary value problems (1.1) and (1.2) or (1.3) has at least two positive solutions. In particular, if T = Z, then the similar result was obtained by Liu and Ge [11]. In addition, Wang [17] used the Leggett–Williams fixed-point theorem [7] and obtained the existence criteria for at least three positive solutions of boundary value problems (1.1) and (1.2) or (1.3). Motivated by [6,15,17], it is natural to consider the existence of positive solutions of problem (1.1) subject to multi-point boundary conditions. In this paper, using a new fixed-point theorem different from fixed-point theorems used in [6,11,15,17], we consider boundary value problem (1.1) satisfying the boundary conditions ! m−2 X ∆ u(0) − B0 ai u (ξi ) = 0, u ∆ (T ) = 0, (1.5) i=1

or u ∆ (0) = 0,

u(T ) + B1

m−2 X

! bi u ∆ (ξi ) = 0,

(1.6)

i=1

where ξi ∈ (0, T )T , 0 < ξ1 < ξ2 < · · · < ξm−2 < T and ai , bi ∈ [0, ∞)(i = 1, 2, . . . , m − 2). By using the five functionals fixed-point theorem in a cone [2], we prove that the boundary value problems (1.1) and (1.5) or (1.6) has at least three positive solutions. If i = 1 and a1 = b1 = 1, then our results can improve and generalize the results of Li and Shen [8] in the case T = R. As an application, an example is given to illustrate the result. Throughout this paper, it is assumed that (H1) f : R → R+ is continuous, and does not vanish identically on any closed subinterval of [0, T ]T , where R+ denotes the nonnegative real numbers; (H2) a1 : T → R+ is left dense continuous (i.e., a1 ∈ Cld (T, R+ )), and does not vanish identically on any closed subinterval of [0, T ]T , where Cld (T, R+ ) denotes the set of all left dense continuous functionals from T to R+ . For convenience, we list the following well-known definitions which can be found in [3,4]. Definition 1.1. For t < sup T and r > inf T, define the forward jump operator σ and the back 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.

Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820

3813

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 Tκ = T − {m}; otherwise set Tκ = T. If T has a left scattered maximum M, define Tκ = T − {M}; otherwise set Tκ = T. Definition 1.2. For x : T → R and t ∈ Tκ , 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 . For x : T → R and t ∈ Tκ , 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| for all s ∈ V . If T = R, then x ∆ (t) = x ∇ (t) = x 0 (t). If T = Z, then x ∆ (t) = x(t + 1) − x(t) is the forward difference operator while x ∇ (t) = x(t) − x(t − 1) is the backward difference operator. Rt Definition 1.3. If F ∆ (t) = f (t), then we define the delta integral by a f (s)∆s = F(t) − F(a). If F ∇ (t) = f (t), Rt then we define the nabla integral by a f (s)∇s = F(t) − F(a). Throughout this paper, we assume that T is a closed subset of R with 0 ∈ Tκ , T ∈ Tκ . 2. Preliminaries In this section, we provide some background materials from the theory of cones in Banach spaces that we shall use in the rest of the paper. 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. Every cone P ⊂ E induces an ordering in E given by x ≤ y if and only if y − x ∈ P. Definition 2.2. Given a cone P in a real Banach space E, a functional ψ : P → R is said to be increasing on P, provided ψ(x) ≤ ψ(y) for all x, y ∈ P with x ≤ y. Definition 2.3. A map α is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if α : P → [0, ∞) is continuous and α(t x + (1 − t)y) ≥ tα(x) + (1 − t)α(y) for all x, y ∈ P and t ∈ [0, 1]. Similarly we say that the map β is a nonnegative continuous convex functional on a cone P of a real Banach space E if β : P → [0, ∞) is continuous and β(t x + (1 − t)y) ≤ tβ(x) + (1 − t)β(y) for all x, y ∈ P and t ∈ [0, 1]. Let γ , β, θ be nonnegative continuous convex functionals on P and α, ϕ be nonnegative continuous concave functionals on P. For nonnegative real numbers h, a, b, d and c we define the following 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, ϕ, h, d, c) = {x ∈ P : h ≤ ϕ(x), β(x) ≤ d, γ (x) ≤ c}. To prove our main results, we need the following theorem, which is the five functionals fixed-point theorem [2]. Theorem 2.4 ([2]). 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 ϕ on P, and nonnegative continuous convex functionals γ , β and θ on P, with α(x) ≤ β(x)

and

kxk ≤ Mγ (x)

for all x ∈ P(γ , c). Suppose Φ : P(γ , c) → 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} 6= ∅ and α(Φ(x)) > b for x ∈ P(γ , θ, α, b, k, c);

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Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820

(ii) {x ∈ Q(γ , β, ϕ, h, a, c) : β(x) < a} 6= ∅ and β(Φ(x)) < a for x ∈ Q(γ , β, ϕ, h, a, c); (iii) α(Φ(x)) > b for x ∈ P(γ , α, b, c) with θ (Φ(x)) > k; (iv) β(Φ(x)) < a for x ∈ Q(γ , β, a, c) with ϕ(Φ(x)) < h. Then Φ has at least three fixed points x1 , x2 , x3 ∈ P(γ , c) such that β(x1 ) < a,

b < α(x2 ) and

a < β(x3 ) with α(x3 ) < b.

3. Existence results In this section, by using the five functionals fixed-point theorem, we shall obtain the existence of at least three positive solutions of problems (1.1) and (1.5) or (1.6). Let the Banach space E = Cld ([0, T ]T , R) with norm kuk = supt∈[0,T ]T |u(t)|, and define the cone, P ⊂ E, by n o P = u ∈ E|u ∆ (T ) = 0, u is concave and nonnegative on [0, T ]T . Assume that there exists l ∈ T such that ξm−2 < l < T and Lemma 3.1. If u ∈ P, then (i)

RT l

a1 (r )∇r > 0 hold, then we have the following lemma.

kuk ≤ u(t) for t ∈ [0, T ]T ; (ii) su(t) ≤ tu(s) for s, t ∈ (0, T )T and s ≤ t.

t T

Proof. From [6], we obtain that (i) is true. Now we prove (ii). If t = s, then the conclusion holds. If s < t, since u(t) is concave, nonnegative on [0, T ]T and u ∆ (T ) = 0, hence we have u(t) − u(0) u(s) − u(0) ≤ , t s thus tu(s) ≥ su(t) + (t − s)u(0) ≥ su(t). The proof is complete.



Firstly, we define the nonnegative continuous concave functionals α and ϕ and nonnegative continuous convex functionals γ , β and θ on P respectively by γ (u) = θ (u) :=

max

t∈[0,ξm−2 ]T

u(t) = u(ξm−2 ),

β(u) := max u(t) = u(l)

and

t∈[0,l]T

ϕ(u) :=

α(u) := min u(t) = u(l), t∈[l,T ]T

min

t∈[ξm−2 ,T ]T

u(t) = u(ξm−2 ).

It is clear that α(u) = β(u) for all u ∈ P. For notional convenience, we denote ! Z  m−2 T X M= A ai + ξm−2 ϕq a1 (r )∇r , 0

i=1

m=

B

m−2 X

! ai + l ϕq

T

Z



a1 (r )∇r ,

l

i=1

and λl =

A

m−2 X i=1

! ai + l ϕq

Z

T

 a1 (r )∇r .

0

Now we state and prove our main result. Theorem 3.2. Let 0 < a < lb/T < lcξm−2 /T 2 , Mb < mc and suppose that f satisfies the following conditions: (H3) f (x) < ϕ p ( Mc ) for all 0 ≤ x ≤

Tc ξm−2 ;

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Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820 T 2b ; 2 ξm−2 Ta l .

(H4) f (x) > ϕ p ( mb ) for all b ≤ x ≤ (H5) f (x) < ϕ p ( λal ) for all 0 ≤ x ≤

Then the boundary value problems (1.1) and (1.5) has at least three positive solutions u 1 , u 2 , u 3 such that max u 1 (t) < a,

b < min u 2 (t) t∈[l,T ]T

t∈[0,l]T

a < max u 3 (t)

and

with

t∈[0,l]T

min u 3 (t) < b.

t∈[l,T ]T

Proof. From (H1) and (H2), we have, the solutions of boundary value problems (1.1) and (1.5) is positive. Define a completely continuous integral operator Φ : P → E by Z T ! Z t Z T  m−2 X (Φu)(t) = B0 ai ϕq a1 (r ) f (u(r )) ∇r + ϕq a1 (r ) f (u(r ))∇r ∆s ξi

i=1

0

s

for t ∈ [0, T ]T . It is easy to obtain Φ(u) ∈ P and every fixed point of Φ is a solution of the boundary value problems (1.1) and (1.5). Let u ∈ P(γ , c), then γ (u) =

u(t) = u(ξm−2 ) ≤ c.

max

t∈[0,ξm−2 ]T

Consequently   for t ∈ 0, ξm−2 T .

0 ≤ u(t) ≤ c

In view of Lemma 3.1, one has T u(ξm−2 ) Tc ≤ , ξm−2 ξm−2

kuk = u(T ) ≤

Tc this implies 0 ≤ u(t) ≤ ξm−2 for t ∈ [0, T ]T . It follows from (H3) of Theorem 3.2 and (1.4) that

γ (Φ(u)) = (Φu)(ξm−2 ) Z m−2 X = B0 ai ϕq < A

ai ϕq

Z

m−2 X

ai ϕq

Z

c < M

A

T ξm−2

ξm−2

Z

m−2 X

0

a1 (r ) f (u(r ))∇r + ξm−2 ϕq

 a1 (r ) f (u(r ))∇r ∆s

s T

Z

T

 Z a1 (r ) f (u(r ))∇r + ξm−2 ϕq

a1 (r ) f (u(r ))∇r



a1 (r ) f (u(r ))∇r



0

Z

T



a1 (r )∇r + ξm−2 ϕq

T ξm−2

γ (u)

T

Z

a1 (r )∇r

# = c.

0

0

i=1

u(ξm−2 ) =

T

Z

0

T

ai ϕq

ϕq

+



Therefore Φ(u) ∈ P(γ , c). By Lemma 3.1, we have γ (u) = u(ξm−2 ) ≥ kuk ≤

!

0

i=1

"

T

ξi

i=1

< A

a1 (r ) f (u(r ))∇r

ξi

i=1 m−2 X

T

ξm−2 T

kuk, hence

for all u ∈ P.

Now, we show that (i)–(iv) of Theorem 2.4 are satisfied. Tb Tb Firstly, if u ≡ ξm−2 , k = ξm−2 , then α(u) = u(l) =

Tb > b, ξm−2

θ (u) = u(ξm−2 ) =

Tb = k, ξm−2

γ (u) =

Tb < c, ξm−2

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Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820

which show that {u ∈ P(γ , θ, α, b, k, c) : α(u) > b} 6= ∅. Tb , c), we have For u ∈ P(γ , θ, α, b, ξm−2

θ (u) =

max

t∈[0,ξm−2 ]T

= u(ξm−2 ) ≤

Tb ξm−2

and α(u) = min = u(l) ≥ b, t∈[l,T ]T

which imply 0 ≤ u(t) ≤

Tb ξm−2

  for all t ∈ 0, ξm−2 T ,

and b ≤ u(t) for all t ∈ [l, T ]T . By Lemma 3.1, we have kuk ≤

T 2b T u(ξm−2 ) ≤ 2 , ξm−2 ξm−2

consequently, b ≤ u(t) ≤

T 2b 2 ξm−2

for all t ∈ [l, T ]T .

By (H4) of Theorem 3.2, we obtain α(Φ(u)) = (Φu)(l) Z m−2 X = B0 ai ϕq

ξi

i=1

> B

m−2 X

ai ϕq

B

Z

T

a1 (r ) f (u(r ))∇r

!

m−2 X

ai + l ϕq

ϕq

0



a1 (r ) f (u(r ))∇r + lϕq

!

l

Z +

Z

T

T

a1 (r ) f (u(r ))∇r

T

a1 (r )∇r

l



b = b. m

So (i) of Theorem 2.4 is fulfilled. Secondly, we proved that (ii) of Theorem 2.4 is fulfilled. We take u ≡ aξm−2 < c, T

ϕ(u) = u(ξm−2 ) =

Hence {u ∈ Q (γ , β, ϕ, h, a, c) : β(u) < a} 6= ∅. If u ∈ Q(γ , β, ϕ, aξm−2 T , a, c), then β(u) := max u(t) = u(l) ≤ a, t∈[0,l]T

consequently 0 ≤ u(t) ≤ a

for t ∈ [0, l]T ,

by Lemma 3.1 kuk ≤

T u(l) Ta ≤ l l



l

Z

i=1

γ (u) = u(ξm−2 ) =

 a1 (r ) f (u(r ))∇r ∆s

s

Z

l

i=1

>

T

for t ∈ [0, T ]T ,

aξm−2 T ,h

aξm−2 = h, T

=

aξm−2 T ,

then

β(u) = u(l) =

aξm−2 < a. T

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Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820

thus Ta for t ∈ [0, T ]T . l In view of (H5) of Theorem 3.2, we have 0 ≤ u(t) ≤

β(Φ(u)) = (Φu)(l) Z m−2 X ai ϕq = B0 i=1

< A

m−2 X

ai ϕq

A

ξi

a1 (r ) f (u(r ))∇r

!

Z

m−2 X

ai + l ϕq

ϕq

Z

0



a1 (r ) f (u(r ))∇r + lϕq

!

l

+

T

 a1 (r ) f (u(r ))∇r ∆s

s T

Z

0

i=1

<

T

Z

T

a1 (r ) f (u(r ))∇r



0 T

Z

a1 (r )∇r



0

i=1

a = a. λl

Thirdly, we show that (iii) of Theorem 2.4 is satisfied. If u ∈ P(γ , α, b, c)

and θ (Φ(u)) = Φ(u(ξm−2 )) > k =

Tb , ξm−2

then α(Φ(u)) = (Φu)(l) ≥

l l l Tb lb Φ(u(l)) ≥ Φ(u(ξm−2 )) > × = > b. T T T ξm−2 ξm−2

Finally, if u ∈ Q(γ , β, a, c)

and ϕ(Φ(u)) = Φ(u(ξm−2 )) < h =

aξm−2 , T

then by Lemma 3.1 we obtain β(Φ(u)) = (Φu)(l) ≤

T T aξm−2 T (Φu(l)) ≤ Φ(u(ξm−2 )) < × =a l ξm−2 ξm−2 T

which shows that the condition (iv) of Theorem 2.4 is fulfilled. Thus, all the conditions in Theorem 2.4 are met, so the boundary value problems (1.1) and (1.5) has at least three positive solutions u 1 , u 2 , u 3 such that b < min u 2 (t)

max u 1 (t) < a,

t∈[l,T ]T

t∈[0,l]T

The proof is complete.

and a < max u 3 (t) t∈[0,l]T

with

min u 3 (t) < b.

t∈[l,T ]T



Now, we consider the boundary value problems (1.1) and (1.6). Rl Assume that there exists l1 ∈ T such that 0 < l1 < ξ1 < T and 01 a1 (r )∇r > 0 hold. Define a cone P1 ⊂ E by n o P1 = u ∈ E|u ∆ (0) = 0, u is concave and nonnegative on [0, T ]T . Similar to Lemma 3.1, we have Lemma 3.3. If u ∈ P1 , then (i) and s ≤ t.

T −t T

kuk ≤ u(t) for t ∈ [0, T ]T ; (ii) (T − t)u(s) ≤ (T − s)u(t) for s, t ∈ (0, T )T

Now define the nonnegative continuous concave functionals α1 and ϕ1 and nonnegative continuous convex functionals γ1 , β1 and θ1 on P1 respectively by γ1 (u) = θ1 (u) :=

max u(t) = u(ξ1 ),

t∈[ξ1 ,T ]T

β1 (u) := max u(t) = u(l1 ) and t∈[l1 ,T ]T

It is clear that α1 (u) = β1 (u) for all u ∈ P1 .

α1 (u) := min u(t) = u(l1 ),

ϕ1 (u) :=

t∈[0,l1 ]T

min u(t) = u(ξ1 ).

t∈[0,ξ1 ]T

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Let M1 =

A

m−2 X

!

m1 =

B

 a1 (r )∇r ,

0

i=1 m−2 X

T

Z

bi + T − ξ1 ϕq !

Z

bi + T − l1 ϕq

l1



a1 (r )∇r ,

0

i=1

and λl1 =

A

m−2 X

! bi + T − l1 ϕq

T

Z

 a1 (r )∇r .

0

i=1

We have the following result. (T −l1 )b T

Theorem 3.4. Let 0 < a < conditions:

(H6) f (x) < ϕ p ( Mc1 ) for all 0 ≤ x ≤ (H7) f (x) > ϕ p ( mb1 ) for all b ≤ x ≤ (H8) f (x) < ϕ p ( λal ) for all 0 ≤ x ≤ 1

<

(T −l1 )(T −ξ1 )c , T2

M1 b < m 1 c and suppose that f satisfies the following

Tc T −ξ1 ; T 2b ; (T −ξ1 )2 Ta T −l1 .

Then the boundary value problems (1.1) and (1.6) has at least three positive solutions u 1 , u 2 , u 3 such that max u 1 (t) < a,

t∈[l1 ,T ]T

b < min u 2 (t) t∈[0,l1 ]T

and

a < max u 3 (t) t∈[l1 ,T ]T

with

min u 3 (t) < b.

t∈[0,l1 ]T

Proof. Define a completely continuous integral operator Φ : P1 → E by Z ξi ! Z T Z s  m−2 X (Φu)(t) = B1 bi ϕq a1 (r ) f (u(r ))∇r + ϕq a1 (r ) f (u(r ))∇r ∆s. 0

i=1

t

0

The proof of Theorem 3.4 is similar to that of Theorem 3.2, we omit it here.



4. An example In this section, we present a simple example to explain our result.    N0  n o 1 1 Let T = 2 − 12 ∪ 0, 18 , 41 , 13 , 12 , 1, 32 , 2 ∪ [ 10 , 9 ] and T = 2. Consider the following boundary value problem with p = ( ) 6   ∇ X ∆ k 6−k ϕ p u (t) + t (ρ(t)) t ∇ f (u(t)) = 0,

4 3

and k ∈ N0 . t ∈ [0, 2]T ,

(4.1)

k=0

satisfying the boundary conditions      ∆ 1 ∆ 1 u(0) − 2 u +u = 0, 4 2 where  1 × 10−7 , 0 ≤ u ≤ 4,    p(u), 4 ≤ u ≤ 40, f (u) =  7 × 10−6 , 40 ≤ u ≤ 800,   s(u), u ≥ 800,

u ∆ (2) = 0,

(4.2)

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Y.-H. Su, W.-T. Li / Nonlinear Analysis 69 (2008) 3811–3820

here p(u) and s(u) satisfy p (4) = 1 × 10−7 ,

p(40) = 7 × 10−6 , s(800) = 7 × 10−6 , ( p ∇ (u))∇ = 0 for u ∈ (4, 40) , nP o 6 k 6−k t ∇ , then by Exercise 1.23 in [4], we have and s(u) : R → R+ is continuous. If a1 (t) = k=0 t (ρ(t)) (t ) = 7 ∇

( 6 X

) t (ρ(t))

t∇.

6−k

k

k=0

It is obvious that ξ2 = 12 , A = B = 2 and a1 = a2 = 1. Choose l = 1, a direct calculation shows that ) !3  Z 2 (X  6 1 k 6−k ∇ t (ρ(t)) t ∇t = 9.4372 × 106 , M = 4+ 2 0 k=0 ) !3 Z 2 (X 6 k 6−k ∇ m=5 t (ρ(t)) t ∇t = 1.0242 × 107 , 1

k=0

and λ1 = 5

Z 0

2

( 6 X

!3

) t (ρ(t))

6−k

k

∇

t ∇t

k=0

If we take a = 2, b = 40, c = 200, then 0
= 1.0486 × 107 .

lb clξ2 < 2 T T

f (u) = 1 × 10−7

lb T

2 = 20, clξ = 25. Hence, T2

mc − Mb = 1.6709 × 109 > 0,   a Ta −7 < 1.9073 × 10 = ϕ p = 4, for 0 ≤ u ≤ λ1 l and

and f (u) = 7 × 10

−6

> 3.9055 × 10

−6

= ϕp



b m

 for 40 ≤ u ≤ 640.

From ( p ∇ (u))∇ = 0, we have max p(u) = 7 × 10−6 . Thus, c  Tc f (u) ≤ 7 × 10−6 < 2.1193 × 10−5 = ϕ p = 800. for 0 ≤ u ≤ M ξ2 Therefore, all the conditions of Theorem 3.2 are satisfied. By Theorem 3.2 we see that the boundary value problems (4.1) and (4.2) has at least three positive solutions u 1 , u 2 and u 3 such that max u 1 (t) < 2,

t∈[0,1]T

40 < min u 2 (t) t∈[1,2]T

and

2 < max u 3 (t) t∈[0,1]T

with

min u 3 (t) < 40.

t∈[1,2]T

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