Electronic Journal of Qualitative Theory of Differential Equations

2008, No. 8, 1-10; http://www.math.u-szeged.hu/ejqtde/

Existence of positive solutions for nth-order

boundary value problem with sign
changing nonlinearity
Dapeng Xiea,b , Chuanzhi Baia,∗, Yang Liua,b , Chunli Wanga,b
a
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, P R China
b
Department of Mathematics, Yanbian University, Yanji, Jilin 133002, P R China

Abstract
In this paper, we investigate the existence of positive solutions for singular nth-order bound-
ary value problem

u(n) (t) + a(t) f (t, u(t)) = 0, 0 ≤ t ≤ 1,

u(i) (0) = u(n−2) (1) = 0, 0 ≤ i ≤ n − 2,
where n ≥ 2, a ∈ C((0, 1), [0, +∞)) may be singular at t = 0 and (or) t = 1 and the nonlinear
term f is continuous and is allowed to change sign. Our proofs are based on the method of lower
solution and topology degree theorem.
Keywords: Existence; Singular; Positive solution; Cone; Sign changing nonlinearity.

1. Introduction

Boundary value problems for higher order differential equations play a very impor-
tant role in both theories and applications. Existence of positive solutions for nonlinear
higher order has been studied in the literature by using the Krasnosel’skii and Guo
fixed point theorem, Leggett-Williams fixed point theorem, Lower- and upper- solu-
tions method and so on. We refer the reader to [2-9] for some recent results. However,
to the best of our knowledge, few papers can be found in the literature for nth-order
boundary value problem with sign changing nonlinearity, most papers are dealing with
the existence of positive solutions when the nonlinear term f is nonnegative. For ex-
ample, in [3], by using the Krasnosel’skii and Guo fixed point theorem, Eloe and Hen-
derson studied the existence of positive solutions for the following boundary value
problem
∗ E-mail address: czbai8@sohu.com

EJQTDE, 2008 No. 8, p. 1

 (
u(n) (t) + a(t) f (u(t)) = 0, t ∈ (0, 1),
(1.1)
u(i) (0) = u(n−2) (1) = 0, 0 ≤ i ≤ n − 2,

where
(A1 ) f : [0, ∞] → [0, ∞) is continuous;
(A2 ) a : (0, 1) → [0, ∞) is continuous and does not vanish identically on any
subinterval;
(A3 ) f is either superlinear or sublinear.
Motivated by the above works, in this paper, we study the existence of positive so-
lutions for singular nth-order boundary value problem with sign changing nonlinearity
as follows
(
u(n) (t) + a(t) f (t, u(t)) = 0, t ∈ (0, 1),
(1.2)
u(i) (0) = u(n−2) (1) = 0, 0 ≤ i ≤ n − 2.

Throughout this paper, we assume the following conditions hold.

(C1 ) f : [0, 1] × [0, ∞) → (−∞, +∞) is continuous;
R1
(C2 ) a : (0, 1) → [0, ∞) is continuous, and 0 < 0 a(t)dt < ∞, if n = 2;
R1
0 < 0 (1 − t)a(t)dt < ∞, if n ≥ 3.
The purpose of this paper is to establish the existence of positive solutions for BVP
(1.2) by constructing available operator and combining the method of lower solution
with the method of topology degree.
The rest of this paper is organized as follows: in section 2, we present some prelim-
inaries and lemmas. Section 3 is devoted to prove the existence of positive solutions
for BVP (1.2). An example is considered in section 4 to illustrate our main results.

2. Preliminary Lemmas

Lemma 2.1. Suppose that y(t) ∈ C[0, 1], then boundary value problem
( (n)
u (t) + y(t) = 0, 0 ≤ t ≤ 1,
(2.1)
u(i) (0) = u(n−2) (1) = 0, 0 ≤ i ≤ n − 2,

has a unique solution

Z 1
u(t) = G(t, s)y(s)ds,
0

where

EJQTDE, 2008 No. 8, p. 2

  n−1 n−1
1   (1 − s)t − (t − s) ,
 0 ≤ s ≤ t ≤ 1,
G(t, s) =

(n − 1)! 

 (1 − s)tn−1 ,

0 ≤ t < s ≤ 1.
Proof. The proof follows by direct calculations.
Lemma 2.2. G(t, s) has the following properties.
(1) 0 ≤ G(t, s) ≤ k(s), t, s ∈ [0, 1], where
1−s
k(s) = ;
(n − 1)!
∂i
(2) G(t, s) > 0 on (0, 1) × (0, 1), 0 ≤ i ≤ n − 2;
∂ti
∂2
(3) G00tt (t, s) := ∂t2
G(t, s) ≤ (n − 1)(n − 2)k(s), n ≥ 3.

Proof. It is easy to check that (1) and (3) hold. The proof of (2), please see [1].

Remark 2.1. By (C2 ) and Lemma 2.2, we have

Z 1 Z 1 Z 1
0< G(t, s)a(s)ds < ∞, n ≥ 2, and 0 < G00ττ (τ, s)a(s)dτds < ∞, n ≥ 3.
0 0 0

By the definition of completely continuous operator, we can check that the follow-
ing lemma holds.

Lemma 2.3. Let P is a cone of X = C[0, 1]. Suppose T : P → X is completely

continuous. Define A : T X → P by

(Ay)(t) = max{y(t), 0}, y ∈ T X.

Then, A ◦ T : P → P is also a completely continuous operator.

3. Main results

Let X = C[0, 1], P = {u ∈ X : u(t) ≥ 0, t ∈ [0, 1]} with kuk = max |u(t)|. Set
t∈[0,1]

f1 (t, u) = max{0, f (t, u)}, f2 (t, u) = max{0, − f (t, u)},

and
R1 R1
δ= 0
k(s)a(s)ds, w(t) = 0
G(t, s)a(s)ds.

For convenience, we introduce the following notations

EJQTDE, 2008 No. 8, p. 3

 f (t, u) f (t, u)
f ∞ = lim sup max , f 0 = lim sup max ,
u→∞ 0≤t≤1 u u→0+ 0≤t≤1 u
f1 (t, u) f1 (t, u)
f 1∞ = lim sup max , f 10 = lim sup max ,
u→∞ 0≤t≤1 u u→0+ 0≤t≤1 u
f2 (t, u) f2 (t, u)
f 2∞ = lim sup max , f 20 = lim sup max .
u→∞ 0≤t≤1 u u→0+ 0≤t≤1 u

Theorem 3.1. Suppose that (C1 ) and (C2 ) hold, in addition assume f 1∞ = A < +∞,
1
f 2∞ = B < +∞ (or f 10 = A < +∞, f 20 = B < +∞) with A + B < δ and there exist r, λ
with r > λ > 0 such that


 λ ≤ min f (t, λw(t)), n = 2,
 t∈[0,1]
(3.1)




 λ ≥ max f (t, λw(t)), n ≥ 3.
t∈[0,1]

Then BVP (1.2) has a positive solution u∗ (t) satisfying

0 < λw(t) ≤ u∗ (t), 0 < t < 1 and ku∗ k ≤ r.

Proof. Let
(
f (t, u(t)), u(t) ≥ λw(t),
g(t, u(t)) = (3.2)
f (t, λw(t)), u(t) ≤ λw(t).

Define the operator T : P → X by

R1
(T u)(t) = 0
G(t, s)a(s)g(s, u(s))ds, 0 ≤ t ≤ 1. (3.3)

Similar to the proof of Lemma 2.1 in [7], we can easily check that T is a completely
continuous operator.
Define the operator A : X → P by

(Au)(t) = max{u(t), 0}. (3.4)

By Lemma 2.3, we get A ◦ T : P → P is also completely continuous.

If f 1∞ = A < +∞, f 2∞ = B < +∞, then by hypothesis A + B < 1δ , we may take

A0 > A, B0 > B such that A0 + B0 < 1δ , f 1∞ < A0 and f 2∞ < B0 . Thus, we choose
L > 0 such that

f1 (t, u) < A0 u, f2 (t, u) < B0 u, if u ≥ L, t ∈ [0, 1], (3.5)

EJQTDE, 2008 No. 8, p. 4

and there exists r > L such that

f1 (t, u) < A0 r, f2 (t, u) < B0 r, if λw(t) ≤ u(t) ≤ L, t ∈ [0, 1]. (3.6)

Let Ω = {u ∈ P : kuk < r}. Then, for u ∈ ∂Ω, we have by (3.5) and (3.6) that
 

 

max f1 (t, u) = max  max f1 (t, u), max f1 (t, u) < A0 r. (3.7)
 
t∈[0,1]  t∈[0,1]
 t∈[0,1] 

u∈[λw(t),r] u∈[λw(t),L] u∈[L,r]

Similarly,
max f2 (t, u) < B0 r. (3.8)
t∈[0,1]
u∈[λw(t),r]

Thus, for each u ∈ ∂Ω, from (3.7) and (3.8), we have

R 
1
(A ◦ T )u(t) = max 0 G(t, s)a(s)g(s, u(s))ds, 0
R1
≤ 0
G(t, s)a(s)|g(s, u(s))|ds
R1
≤ max |g(t, u)| 0
k(s)a(s)ds
t∈[0,1]
u∈[0,r]

= δ max | f (t, u)|

t∈[0,1]
u∈[λw(t),r]

= δ max ( f1 (t, u) + f2 (t, u))

t∈[0,1]
u∈[λw(t),r]

< δ(A0 + B0 )r < r = kuk, (3.9)

which implies

k(A ◦ T )uk < kuk, ∀u ∈ ∂Ω.

Thus, we have

degP {I − A ◦ T, Ω, 0} = 1,

where degP means the degree on cone P. Hence, A ◦ T has a fixed point u∗ in Ω, i.e.,
(A ◦ T )(u∗ ) = u∗ , u∗ ∈ Ω.

If f 10 = A < +∞, f 20 = B < +∞, then we take A0 > A, B0 > B such that
A0 + B0 < 1δ , f 10 < A0 and f 20 < B0 . Take r > 0 such that

EJQTDE, 2008 No. 8, p. 5

 f1 (t, u) < A0 u, f2 (t, u) < B0 u, if 0 < u(t) ≤ r, t ∈ [0, 1].

Then, for u ∈ ∂Ω, similar to the proof of (3.9), we have that A ◦ T has a fixed point u∗
in Ω. Hence, in any case we always have that A ◦ T has a fixed point u∗ in Ω.
In the following, we shall show the following relation holds

(T u∗ )(t) ≥ λw(t), t ∈ [0, 1]. (3.10)

Assume the contrary, then there exists t0 ∈ [0, 1] such that

(T u∗ )(t0 ) − λw(t0 ) = min {(T u∗ )(t) − λw(t)} = M < 0. (3.11)

t∈[0,1]

Obviously, t0 , 0, so t0 ∈ (0, 1] and

(T u∗ )0 (t0 ) − λw0 (t0 ) = 0. (3.12)

There are two cases to consider.

Case 1. t0 = 1. It is obvious that T u∗ (t) −λw(t) on [0, 1] is continuous. From (3.11), we

have that there exists t1 ∈ [0, 1) such that (T u∗ )(t1 )−λw(t1 ) = 0 and (T u∗ )(t)−λw(t) < 0
for t ∈ (t1 , 1].
If n = 2, then by (3.1), (3.3) and (3.12), one has
R1
(T u∗ )0 (t) − λw0 (t) = (T u∗ )0 (1) − λw0 (1) − t
[(T u∗ )0 (s) − λw0 (s)]0 ds
R1
= t
a(s)[g(s, u∗(s)) − λ]ds
R1
= t
a(s)[ f (s, λw(s)) − λ]ds
R1
≥ [ min f (t, λw(t)) − λ] t
a(s)ds ≥ 0.
t∈[0,1]

Then, we have that (T u∗ )0 (t) − λw0 (t) ≥ 0.

If n ≥ 3, from (3.1), (3.3) and (3.12), we get
R1
(T u∗ )0 (t) − λw0 (t) = (T u∗ )0 (1) − λw0 (1) − t
[(T u∗ )0 (τ) − λw0 (τ)]0 dτ
R1R1
= t 0
G00ττ (τ, s)a(s)[λ − g(s, u∗ (s))]dsdτ
R1R1
= t 0
G00ττ (τ, s)a(s)[λ − f (s, λw(s))]dsdτ

EJQTDE, 2008 No. 8, p. 6

 R1R1
≥ [λ − max f (t, λw(t))] t 0
G00ττ (τ, s)a(s)dsdτ ≥ 0.
t∈[0,1]

Then, we have (T u∗ )0 (t) − λw0 (t) ≥ 0. Therefore, in any case we always have that
(T u∗ )0 (t) − λw0 (t) ≥ 0, which implies

(T u∗ )(t0 ) − λw(t0 ) = (T u∗ )(1) − λw(1) ≥ (T u∗ )(t1 ) − λw(t1 ) = 0.

It contradicts (3.11), so (3.10) holds.

Case 2. t0 ∈ (0, 1). Obviously, T u∗ (t) − λw(t) on [0, 1] is continuous. By (3.11) and
t0 ∈ (0, 1), we have that there exists t2 ∈ [0, t0 ) ∪ (t0 , 1] such that (T u∗ )(t2 ) − λw(t2 ) = 0
and (T u∗ )(t) − λw(t) < 0 for t ∈ (t2 , t0 ] or t ∈ [t0 , t2 ). Without loss of generality, we
assume that t ∈ (t2 , t0 ].
If n = 2, we have by (3.1), (3.3) and (3.12) that
R t0
(T u∗ )0 (t) − λw0 (t) = (T u∗ )0 (t0 ) − λw0 (t0 ) − t
[(T u∗ )0 (s) − λw0 (s)]0 ds
R t0
= t
a(s)[g(s, u∗(s)) − λ]ds
R t0
= t
a(s)[ f (s, λw(s)) − λ]ds
R t0
≥ [ min f (t, λw(t)) − λ] t
a(s)ds ≥ 0.
t∈[0,1]

Thus, (T u∗ )0 (t) − λw0 (t) ≥ 0.

If n ≥ 3, from (3.1), (3.3), (3.12) and Lemma 2.2 (2), we obtain
R t0
(T u∗ )0 (t) − λw0 (t) = (T u∗ )0 (t0 ) − λw0 (t0 ) − t
[(T u∗ )0 (τ) − λw0 (τ)]0 dτ
R t0 R 1
= t 0
G00ττ (τ, s)a(s)[λ − g(s, u∗ (s)]dsdτ
R t0 R 1
= t 0
G00ττ (τ, s)a(s)[λ − f (s, λw(s))]dsdτ
R t0 R 1
≥ [λ − max f (t, λw(t))] t 0
G00ττ (τ, s)a(s)dsdτ ≥ 0.
t∈[0,1]

Thus, (T u∗ )0 (t)−λw0 (t) ≥ 0. Hence, in any case we always have that (T u∗ )0 (t)−λw0 (t) ≥
0, which implies

(T u∗ )(t0 ) − λw(t0 ) ≥ (T u∗ )(t2 ) − λw(t2 ) = 0.

It contradicts (3.11), so (3.10) holds. Thus, (A ◦ T )u∗ = T u∗ = u∗ , u∗ ∈ Ω, i.e., BVP

EJQTDE, 2008 No. 8, p. 7

(1.2) has a positive solution u∗ (t) satisfying 0 < λw(t) ≤ u∗ (t), 0 < t < 1, and ku∗ k ≤ r.

Corollary 3.1 Suppose that (C1 ) and (C2 ) hold, in addition assume f 1∞ = 0, f 2∞ = 0 (
or f 10 = 0, f 20 = 0 ), if there exists a constant λ > 0 such that


 λ ≤ min f (t, λw(t)), n = 2,

 t∈[0,1]
 λ ≥ t∈[0,1]
max f (t, λw(t)), n ≥ 3.




Then BVP (1.2) has a positive solution.

Theorem 3.2. Suppose that (C2 ) holds, in addition assume f (t, 0) ≥ 0, a(t) f (t, 0) . 0
and f 1∞ = A < +∞ , f 2∞ = B < +∞ ( or f 10 = A < +∞, f 20 = B < +∞ ) with
A + B < 1δ . Then BVP (1.2) has a positive solution.

Proof: Similar to the proof of Theorem 3.1, we can complete the proof of Theorem
3.2, so we omit it here.

Corollary 3.4. Suppose that (C2 ) holds, in addition assume f (t, 0) ≥ 0, a(t) f (t, 0) . 0,
f 1∞ = 0 and f 2∞ = 0 ( or f 10 = 0 and f 20 = 0 ). Then BVP (1.2) has a positive
solution.

Corollary 3.3. Suppose that (C2 ) holds, in addition assume f : [0, 1]×[0, ∞) → [0, ∞),
a(t) f (t, 0) . 0 and f ∞ = A < +∞ ( or f 0 = A < +∞ ) with A < 1δ . Then BVP (1.2) has
a positive solution.

Remark 3.1. Suppose that (C2 ) holds, in addition assume f : [0, 1] × [0, ∞) → [0, ∞),
a(t) f (t, 0) . 0 and f ∞ = 0 ( or f 0 = 0 ). Then BVP (1.2) has a positive solution.

4. An example
(n − 1)!
Example 4.1. Let a(t) = , n ≥ 3, and
2(1 − t)
 1



 [1 + ln(1 + (e − 1)t)](u + u 2 + 1), (t, u) ∈ [0, 1] × [0, 1],
f (t, u) =  −2
[1 + ln(1 + (e − 1)t)][3 − (2e + 6)(u − 1)], (t, u) ∈ [0, 1] × [1, 2],


 −[1 + ln(1 + (e − 1)t)](e−u + 3 + sin πu)u,

 (t, u) ∈ [0, 1] × [2, ∞).
2

By simple calculation, we have

R1 R1 1−s (n−1)! 1
0<δ= 0
k(s)a(s)ds = 0 (n−1)!
· 2(1−s) ds = 2 < ∞.

EJQTDE, 2008 No. 8, p. 8

It is easy to see that f ∈ C([0, 1] × [0, ∞), R) and f (t, 0) > 0, a(t) f (t, 0) . 0, ∀t ∈ [0, 1].
Owing to f (t, u) > 0, (t, u) ∈ [0, 1] × [0, 1], and f (t, u) < 0, (t, u) ∈ [0, 1] × [2, ∞), we
have

f1 (t, u) = f (t, u), f2 (t, u) = 0, (t, u) ∈ [0, 1] × [0, 1],

and
f1 (t, u) = 0, f2 (t, u) = − f (t, u), (t, u) ∈ [0, 1] × [2, ∞).

By calculating, we obtain that

A = f 1∞ = 21 , B = f 2∞ = 21 ,

1
and so A + B = 1 < 2 = δ. Therefore, by Theorem 3.2, BVP (1.2) has a positive
solution.

Acknowledgment

The authors are very grateful to the referee for his/her helpful comments.

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EJQTDE, 2008 No. 8, p. 10