Multiple positive solutions for n th-order m -point boundary value problems with all derivatives

Nonlinear Analysis 68 (2008) 1064–1072 www.elsevier.com/locate/na

Multiple positive solutions for nth-order m-point boundary value problems with all derivativesI Weihua Jiang ∗ College of Mathematics and Science of Information, Hebei Normal University, Shijiazhuang, 050016, Hebei, PR China College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, PR China Received 1 September 2006; accepted 27 November 2006

Abstract We investigate the existence of positive solutions for the boundary value problem u (n) (t) + f (t, u(t), u 0 (t), . . . , u (n−2) (t), u (n−1) (t)) = 0, u(0) = u 0 (0) = · · · = u (n−2) (0) = 0,

u (n−1) (1) =

0 < t < 1,

m−2 X

ki u (n−1) (ξi ),

i=1

Pm−2 where f : [0, 1] × (R + )n → R + is continuous, ki > 0 (i = 1, 2, . . . , m − 2), 0 < i=1 ki < 1. We give at first the associated Green’s function and some of its properties. Then, imposing growth conditions on f , we obtain the existence of at least three positive solutions for the boundary value problem by using the five-functional-fixed-point theorem. Finally, we give an example to demonstrate our result. c 2006 Elsevier Ltd. All rights reserved.

MR subject classification: 34B10; 34B15 Keywords: Green’s function; The five-functional-fixed-point theorem; m-point boundary value problem

1. Introduction Recently, there has been much attention focused on the study of multi-point boundary value problems; we refer the reader to [1–17] and their references. In [1], by applying the fixed point theorem in cones due to Krasnoselkii and Guo, Eloe et al. established the existence of positive solutions for the nth-order three-point boundary value problem u (n) (t) + a(t) f (u) = 0, u(0) = 0,

u (0) = 0, 0

0 < t < 1, u (0) = 0, . . . , u 00

(1.1) (n−2)

(0) = 0,

αu(η) = u(1),

(1.2)

I This project was supported by the Natural Science Foundation of Hebei Province (A2006000298), and the Doctoral Program Foundation of Hebei Province (B2004204) and the Foundation of Hebei University of Science and Technology (XI2004060). ∗ Corresponding address: College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, PR China. E-mail address: [email protected].

c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.11.040

W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

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where 0 < η < 1, 0 < αηn−1 < 1. In [2], by using the five-functional-fixed-point theorem, Zhang et al. proved the existence of at least three positive solutions for the 2nth-order two-point boundary value problem y (2n) (t) = f (y(t), y 0 (t), . . . , y (2n−2) (t), y (2n−1) (t)), y

(2i)

(0) = y

(2i)

(1) = 0,

0 < t < 1,

(1.3)

0 ≤ i ≤ n − 1,

(1.4)

where (−1)n f : R 2n → [0, ∞) is continuous, (−1)n f (0) > 0, and f is even about odd-order derivatives of y. Motivated by their results, in this paper we study the existence of multiple positive solutions for the nth-order m-point boundary value problem u (n) (t) + f (t, u(t), u 0 (t), u 00 (t), . . . , u (n−1) (t)) = 0, u(0) = u 0 (0) = u 00 (0) = · · · = u (n−2) (0) = 0,

0 < t < 1,

u (n−1) (1) =

m−2 X

(1.5) ki u (n−1) (ξi ),

(1.6)

i=1

Pm−2 where f : [0, 1] × (R + )n → R + is continuous, ki > 0, 0 < i=1 ki < 1. In order to obtain our result, we give at first the associated Green’s function, which is constructed for a first-order initial value problem, and its properties. Then, by applying the five-functional-fixed-point theorem, we discuss the existence of at least three positive solutions for (1.5) and (1.6). Finally, we give an example to demonstrate our result. This result extends the result from Ref. [1]. The approach that constructs the Green’s function is novel. We will always suppose that the following conditions are satisfied throughout this paper. (H1 ) f : [0, 1] × (R + )n → R + is continuous and f (t, 0, . . . , 0) 6≡ 0, t ∈ [0, 1]; m−2 X (H2 ) ki > 0, (i = 1, 2, . . . , m − 2), 0 < ki < 1, 0 = ξ0 < ξ1 < ξ2 < · · · < ξm−2 < ξm−1 = 1. i=1

2. Background and definitions Definition 2.1. Suppose K is a cone in a Banach space. The map α is a nonnegative continuous concave function on K provided α : K → [0, ∞) is continuous and α(t x + (1 − t)y) ≥ tα(x) + (1 − t)α(y) for all x, y ∈ K and 0 ≤ t ≤ 1. Similarly we say β is a nonnegative continuous convex function on K provided β : K → [0, ∞) is continuous and β(t x + (1 − t)y) ≤ tβ(x) + (1 − t)β(y) for all x, y ∈ K and 0 ≤ t ≤ 1. Let γ , β, θ be nonnegative continuous convex functions on K and let α, ψ be nonnegative continuous concave functions on K . Then for nonnegative numbers h, a, b, d, and c, we define the following convex sets: P(γ , c) = {x ∈ K |γ (x) < c}, P(γ , α, a, c) = {x ∈ K |a ≤ α(x), γ (x) ≤ c}, Q(γ , β, d, c) = {x ∈ K |β(x) ≤ d, γ (x) ≤ c}, P(γ , θ, α, a, b, c) = {x ∈ K |a ≤ α(x), θ (x) ≤ b, γ (x) ≤ c}, Q(γ , β, ψ, h, d, c) = {x ∈ K |h ≤ ψ(x), β(x) ≤ d, γ (x) ≤ c}. Theorem 2.1 ([18]). Let K be a cone in a real Banach space E. Suppose α and ψ are nonnegative continuous concave functions on K and γ , β, and θ are nonnegative continuous convex functions on K such that for some positive numbers c and m, α(x) ≤ β(x),

and

kxk ≤ mγ (x)

for all x ∈ P(γ , c).

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W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

Suppose further that A : P(γ , c) → P(γ , c) is completely continuous and there exist h, d, a, b ≥ 0 with 0 < d < a such that each of the following is satisfied: (C1 ) (C2 ) (C3 ) (C4 )

{x ∈ P(γ , θ, α, a, b, c)|α(x) > a} 6= ∅ and α(Ax) > a for x ∈ P(γ , θ, α, a, b, c), {x ∈ Q(γ , β, ψ, h, d, c)|β(x) < d} 6= ∅ and β(Ax) < d for x ∈ Q(γ , β, ψ, h, d, c), α(Ax) > a provided x ∈ P(γ , α, a, c) with θ (Ax) > b, β(Ax) < d provided x ∈ Q(γ , β, d, c) with ψ(Ax) < h.

Then A has at least three fixed points x1 , x2 , x3 ∈ P(γ , c) such that β(x1 ) < d,

a < α(x2 ),

d < β(x3 ) with α(x3 ) < a.

and

3. Preliminary results Lemma 3.1. If y(t) ∈ C[0, 1], then the problem v 0 (t) + y(t) = 0, v(1) =

m−2 X

0 < t < 1,

(3.1)

ki v(ξi )

(3.2)

i=1

has a unique solution: Z t v(t) = − y(s) ds + 0

1−

1 m−2 P

1

Z

y(s) ds − 0

ki

1−

i=1

1 m−2 P

m−2 X

ki

ξi

Z

y(s) ds.

ki 0

i=1

i=1

Proof. From (3.1) we get Z t v(t) = − y(s) ds + c. 0

By (3.2), we get c= 1−

1 m−2 P

1

Z

y(s) ds − ki

0

1−

i=1

1 m−2 P

m−2 X

ki

ξi

Z

y(s) ds.

ki

i=1

0

i=1

So, we get v(t) = −

t

Z

y(s) ds + 0

1−

1 m−2 P

1

Z

y(s) ds − ki

0

i=1

1−

1 m−2 P

m−2 X

ki

i=1

ξi

Z

y(s) ds. 

ki 0

i=1

Lemma 3.2. For y(t) ∈ C[0, 1], and y(t) ≥ 0, if v(t) is the solution of (3.1) and (3.2), then we have v(t) ≥ 0, v 0 (t) ≤ 0. This is obvious by the expression for v(t) and (3.1). Lemma 3.3. The Green’s function for the following problem: −v 0 (t) = 0, v(1) =

m−2 X i=1

0 < t < 1,

ki v(ξi )

(3.3) (3.4)

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W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

is given as  j−1 P   ki    i=1   ,   m−2 P    1 − k i 

s ≤ t,

ξ j−1 < s ≤ ξ j ,

j = 1, 2, . . . , m − 1,

m−2 P    1− ki    i= j   ,   m−2  P  1 − ki

s > t,

ξ j−1 < s ≤ ξ j ,

j = 1, 2, . . . , m − 1,

i=1

G(t, s) =

i=1

where

Pl 0

= 0, for l 0 < l.

i=l ki

Proof. For 0 ≤ t ≤ ξ1 , the unique solution of (3.1) and (3.2) can be expressed as

v(t) =

ξ1

Z

y(s) ds + t

1−

m−2 X Z ξj

ki

i= j

ξ j−1

j=2

m−2 P

1−

m−2 P

1

Z y(s) ds +

1

ξm−2

ki

1−

i=1

m−2 P

y(s) ds. ki

i=1

For ξl−1 < t ≤ ξl , l = 2, 3, . . . , m − 2, the unique solution of (3.1) and (3.2) can be expressed as

v(t) =

j−1 P

l−1 Z ξ j X j=2 ξ j−1

l−1 P

ki

i=1 m−2 P

1−

t

Z y(s) ds + ki

i=1 m−2 P

ξl−1

1−

i=1

m−2 X Z

+

ξj

j=l+1 ξ j−1

1−

ξl

Z

1−

m−2 P

1−

i=l m−2 P

y(s) ds + t

ki

i=1 m−2 P

ki

i= j

1−

ki

m−2 P

Z y(s) ds +

ki

y(s) ds ki

i=1

1

1

ξm−2

ki

m−2 P

1−

i=1

y(s) ds. ki

i=1

For ξm−2 < t ≤ 1, the unique solution of (3.1) and (3.2) can be expressed as

v(t) =

m−2 XZ j=2

ξj ξ j−1

j−1 P

m−2 P

ki

i=1 m−2 P

1−

i=1

Z y(s) ds + ki

t ξm−2

ki i=1 m−2 P

1−

1

Z

1

y(s) ds + ki

i=1

t

1−

m−2 P

y(s) ds. ki

i=1

Therefore, the unique solution of (3.1) and (3.2) can be expressed as Z 1 v(t) = G(t, s)y(s) ds. 0

Lemma 3.3 is now proved.



= v(t), t ∈ [0, 1]; by conditions u(0) = u 0 (0) = u 00 (0) = · · · = u (n−2) (0) = 0, we get Z t (t − s)k (n−k−2) v(s) ds, k = 0, 1, . . . , n − 2. u (t) = k! 0

Let

u (n−1) (t)

Define an operator Ak : C[0, 1] → C[0, 1] by Z t (t − s)k Ak v(t) = v(s) ds, k = 0, 1, . . . , n − 2. k! 0

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W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

Then, for problem (1.5) and (1.6), we have v 0 (t) + f (t, An−2 v(t), An−3 v(t), . . . , A0 v(t), v(t)) = 0, v(1) =

m−2 X

(3.5)

ki v(ξi ).

(3.6)

i=1

Define an operator T : C[0, 1] → C[0, 1] by T v(t) =

1

Z

G(t, s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds,

t ∈ [0, 1].

(3.7)

0

Then, it is obvious that problem (1.5) and (1.6) has a solution u(t) = An−2 v(t) if and only if the operator T has a fixed point v(t) = u (n−1) (t), and when v(t) ≥ 0, we have u(t) ≥ 0. Take X = C[0, 1] with the maximum norm, and define the cone K in X by K = {v ∈ X |v(t) ≥ 0, v(t) is monotone nonincreasing in [0, 1]}. Define the nonnegative continuous concave functions α, ψ, and the nonnegative continuous convex functions β, θ, γ on K by α(v) = min v(t) = v(t2 ),

θ (v) = max v(t) = v(t1 ),

ψ(v) = min v(t) = v(1),

β(v) = max v(t) = v(ξ1 ),

t∈[t1 ,t2 ]

t∈[t1 ,t2 ]

t∈[ξ1 ,1]

t∈[ξ1 ,1]

γ (v) = kvk = v(0), where ξe−1 ≤ t1 ≤ ξe , ξe0 −1 ≤ t2 ≤ ξe0 , t1 < t2 , 2 ≤ e ≤ e0 ≤ m − 1. It is clear that for every v ∈ K kvk = γ (v),

α(v) ≤ β(v).

4. Main results We will make use of the following properties of G(t, s): min

s∈(ξ1 ,1)

G(b, s) G(b, a + 0) = , G(a, s) G(a, a + 0)

max

s∈(ξ1 ,1)

G(a, s) G(a, a + 0) = , G(b, s) G(b, a + 0)

where ξ1 ≤ a < b ≤ 1.

Let G(t2 , s) τ1 = min = s∈(ξ1 ,1) G(t1 , s)

e−1 P

ki

i=1 m−2 P

G(ξ1 , s) = τ2 = max s∈(ξ1 ,1) G(1, s)

,

1−

ki

1−

i=2

k1

i=e

1−

1

Z

G(0, s) ds =

M1 = 0

m−2 P

ki ξi

i=1 m−2 P

1−

, ki

i=1

Z

t2

M2 = t1

G(t2 , s) ds = 1−

1 m−2 P i=1

" ki

0 −1 eX

i=1

ki t2 −

e−1 X i=1

ki t1 −

0 −1 eX

i=e

m−2 P

# ki ξi ,

ki ,

W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

Z M3 =

1

1−

m−2 P

ki ξi

i=1 m−2 P

G(ξ1 , s) ds =

0

1069

1−

− ξ1 . ki

i=1

The following theorem is the main result in this paper. Theorem 4.1. Suppose (H1 )–(H2 ) hold. In addition, suppose there exist constants d, a, c such that 0 < h = d < a = τ1 b < b ≤ c, and a <

M2 M1 c.

d τ2

<

We also suppose that f satisfies the following growth conditions: h i Q c (H3 ) f (t, u n−2 , u n−3 , . . . , u 0 , u) ≤ Mc1 , (t, u n−2 , u n−3 , . . . , u 0 , u) ∈ [0, 1] × 0k=n−2 0, (k+1)! × [0, c],  k+1  Q t1 t2k+1 a, (k+1)! c × [a, b], (H4 ) f (t, u n−2 , u n−3 , . . . , u 0 , u) > Ma2 , (t, u n−2 , u n−3 , . . . , u 0 , u) ∈ [t1 , t2 ] × 0k=n−2 (k+1)!  k+1  Q0 ξ1 h d c (H5 ) f (t, u n−2 , u n−3 , . . . , u 0 , u) < M3 , (t, u n−2 , u n−3 , . . . , u 0 , u) ∈ [ξ1 , 1] × k=n−2 (k+1)! , (k+1)! × [h, d]. Then the problem (1.5) and (1.6) has at least three positive solutions u 1 , u 2 , u 3 satisfying (n−1)

ku i

k ≤ c,

i = 1, 2, 3.

(n−1) max u (t) t∈[ξ ,1] 1

< d,

(n−1) min u (t) t∈[t1 ,t2 ] 2

> a,

(n−1) max u (t) t∈[ξ1 ,1] 3

> d,

(n−1) min u (t) t∈[t1 ,t2 ] 3

< a.

1

and

Proof. From Lemma 3.2, we can get that T : K → K is completely continuous. To show T : P(γ , c) → P(γ , c), let v ∈ P(γ , c); then γ (v) = kvk ≤ c. For t ∈ [0, 1] we get Z t (t − s)k t k+1 c v(s) ds ≤ c≤ , k = 0, 1, 2, . . . , n − 2. Ak v(t) = k! (k + 1)! (k + 1)! 0 From (H3 ), we get γ (T v) = kT vk =

Z

1

G(0, s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds

0

≤

c M1

1

Z

G(0, s) ds = c. 0

Thus T v ∈ P(γ , c). Next we show that conditions (C1 )–(C4 ) in Theorem 2.1 are satisfied for T . It is easy to see that {v ∈ P(γ , θ, α, a, b, c)|α(v) > a} 6= ∅, {v ∈ Q(γ , β, ψ, h, d, c)|β(v) < d} 6= ∅ are valid. To prove the second part of (C1 ) holds, let v ∈ P(γ , θ, α, a, b, c); then a ≤ v(t) ≤ b, for t ∈ [t1 , t2 ], and kvk ≤ c. Z t t k+1 (t − s)k t k+1 Ak v(t) = v(s) ds ≤ c≤ 2 c, t ∈ [t1 , t2 ], k = 0, 1, 2, . . . , n − 2. k! (k + 1)! (k + 1)! 0 Z t t k+1 (t − s)k t k+1 Ak v(t) = v(s) ds ≥ a≥ 1 a, t ∈ [t1 , t2 ], k = 0, 1, 2, . . . , n − 2. k! (k + 1)! (k + 1)! 0

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W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

From (H4 ), we get Z 1 G(t2 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds α(T v) = 0 Z t2 G(t2 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds ≥ t1

a > M2

t2

Z

G(t2 , s) ds = a.

t1

To show that the second part of (C2 ) holds, let v ∈ Q(γ , β, ψ, h, d, c); then kvk ≤ c, h ≤ v(t) ≤ d, for t ∈ [ξ1 , 1]. Z t (t − s)k t k+1 c Ak v(t) = v(s) ds ≤ c≤ , t ∈ [ξ1 , 1], k = 0, 1, 2, . . . , n − 2. k! (k + 1)! (k + 1)! 0 Z t ξ k+1 (t − s)k t k+1 v(s) ds ≥ h≥ 1 h, t ∈ [ξ1 , 1], k = 0, 1, 2, . . . , n − 2. Ak v(t) = k! (k + 1)! (k + 1)! 0 From (H5 ), we get Z 1 β(T v) = G(ξ1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds 0 1

Z =

ξ1

d < M3

G(ξ1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds Z

1

ξ1

d G(ξ1 , s) ds = M3

1

Z

G(ξ1 , s) ds = d.

0

To prove (C3 ), let v(t) ∈ P(γ , α, a, c) with θ (T v) > b; we get Z 1 α(T v) = G(t2 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds 0 1

Z =

ξ1 Z 1

G(t2 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds

G(t2 , s) G(t1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds G(t 1 , s) ξ1 Z 1 ≥ τ1 · G(t1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds

=

ξ1

> τ1 b = a. Finally we show (C4 ). Let v(t) ∈ Q(γ , β, d, c) with ψ(T v) < h; we get Z 1 β(T v) = G(ξ1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds 0

Z

1

= ξ1 Z 1

G(ξ1 , s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds

G(ξ1 , s) G(1, s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds ξ1 G(1, s) Z 1 ≤ τ2 · G(1, s) f (s, An−2 v(s), An−3 v(s), . . . , A0 v(s), v(s)) ds =

ξ1

< τ2 · h = d. By Theorem 2.1, T has at least three fixed points vi , i = 1, 2, 3, satisfying kvi k ≤ c,

vi (t) ≥ 0,

i = 1, 2, 3.

W. Jiang / Nonlinear Analysis 68 (2008) 1064–1072

max v1 (t) < d,

t∈[ξ1 ,1]

1071

min v2 (t) > a,

t∈[t1 ,t2 ]

and max v3 (t) > d,

t∈[ξ1 ,1]

min v3 (t) < a.

t∈[t1 ,t2 ]

So, u i (t) = An−2 vi (t), i = 1, 2, 3, are three positive solutions of problem (1.5) and (1.6). The proof is completed.



Remark. If f (t, 0, . . . , 0) ≡ 0, t ∈ [0, 1], and the other conditions of Theorem 4.1 do not change, then (1.5) and (1.6) has at least two positive solutions. 5. Example Example 1. Consider the following boundary value problem: 00

u 000 (t) = f (t, u(t), u 0 (t), u (t)), u(0) = u 0 (0) = 0,

(5.1)

      1 00 1 1 00 1 1 00 3 00 u (1) = u + u + u 4 4 4 2 4 4

(5.2)

where  2 t     18     t2      18   2 t f (t, u 1 , u 0 , u) =  18     t2      18    2  t 18

+ + + + +

tu 1 u 0 3750 tu 1 u 0 3750 tu 1 u 0 3750 tu 1 u 0 3750 tu 1 u 0 3750

4u 2 , (t, u 1 , u 0 , u) ∈ [0, 1] × [0, ∞) × [0, ∞) × [0, 1), 9 4 + , (t, u 1 , u 0 , u) ∈ [0, 1] × [0, ∞) × [0, ∞) × [1, 2], 9u 35 + 8 − (u − 4)2 , (t, u 1 , u 0 , u) ∈ [0, 1] × [0, ∞) × [0, ∞) × (2, 4), 18 +

+ 4 log2 u, 1

+ u 3 + 10,

(t, u 1 , u 0 , u) ∈ [0, 1] × [0, ∞) × [0, ∞) × [4, 8], (t, u 1 , u 0 , u) ∈ [0, 1] × [0, ∞) × [0, ∞) × (8, ∞).

Corresponding to BVP (1.5) and (1.6) we have n = 3, m = 5, k1 = k2 = k3 = 41 , ξ1 = 14 , ξ2 = 12 , ξ3 = 34 . Let t1 = 12 , t2 = 43 , c = 50, a = 4, d = 2; then τ1 = 12 , τ2 = 2, M1 = 25 , M2 = 12 , M3 = 94 . So we have b = 8, h = 1. It is easy to check that f satisfies the conditions (H1 ), (H3 ), (H4 ), (H5 ). So, problem (5.1) and (5.2) has at least three positive solutions u 1 , u 2 , u 3 satisfying ku i00 k ≤ 50, 00 max h i u 1 (t) 1 t∈ 4 ,1

i = 1, 2, 3, < 2,

00 min i u 2 (t) > 4, h

t∈

1 3 2,4

and 00

max h i u 3 (t) > 2,

t∈

1 4 ,1

00 min h i u 3 (t) < 4.

t∈

1 3 2,4

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