Electronic Journal of Differential Equations, Vol. 2008(2008), No. 159, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
MULTIPLE POSITIVE SOLUTIONS OF FOURTH-ORDER
FOUR-POINT BOUNDARY-VALUE PROBLEMS WITH
CHANGING SIGN COEFFICIENT
ZHENG FANG, CHUNHONG LI, CHUANZHI BAI
Abstract. In this paper, we investigate the existence of multiple positive
solutions of the fourth-order four-point boundary-value problems
y (4) (t) = h(t)g(y(t), y 00 (t)),
0 < t < 1,
y(0) = y(1) = 0,
ay 00 (ξ1 ) − by 000 (ξ1 ) = 0,
cy 00 (ξ2 ) + dy 000 (ξ2 ) = 0,
where 0 < ξ1 < ξ2 < 1. We show the existence of three positive solutions by
applying the Avery and Peterson fixed point theorem in a cone, here h(t) may
change sign on [0, 1].
1. Introduction
Recently, several authors have studied the existence of positive solutions to
boundary-value problems for fourth-order differential equations. For details; see,
for example, [3, 4, 5, 6, 7, 8, 9, 10]. Zhong, Chen and Wang [10] investigated the
fourth-order nonlinear ordinary differential equation
y (4) (t) − f (t, y(t), y 00 (t)) = 0,
0 ≤ t ≤ 1,
(1.1)
with the four-point boundary conditions
y(0) = y(1) = 0,
00
000
cy 00 (ξ2 ) + dy 000 (ξ2 ) = 0,
ay (ξ1 ) − by (ξ1 ) = 0,
(1.2)
where f ∈ C([0, 1] × [0, ∞) × (−∞, 0], [0, ∞)), a, b, c, d are nonnegative constants,
and 0 ≤ ξ1 < ξ2 ≤ 1. Some results on the existence of at least one positive solution
to BVP (1.1)-(1.2) are obtained by using the Krasnoselskii fixed point theorem.
Their key result reads as follows.
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Four-point boundary-value problem; positive solution;
fixed point on a cone.
c
2008
Texas State University - San Marcos.
Submitted September 3, 2008. Published December 3, 2008.
1
2
Z. FANG, C. LI, C. BAI
EJDE-2008/159
Lemma 1.1 ([10, Lemma 2.2]). If α = ad+bc+ac(ξ2 −ξ1 ) 6= 0 and h(t) ∈ C[ξ1 , ξ2 ],
then the boundary-value problem
u(4) (t) = h(t),
0 < t < 1,
u(0) = u(1) = 0,
au00 (ξ1 ) − bu000 (ξ1 ) = 0,
cu00 (ξ2 ) + du000 (ξ2 ) = 0
has a unique solution
1
Z
u(t) =
Z
ξ2
G1 (t, s)
0
G2 (s, τ )h(τ ) dτ ds,
(1.3)
ξ1
where
(
s(1 − t),
G1 (t, s) =
t(1 − s),
1
G2 (t, s) =
α
0 ≤ s ≤ t ≤ 1,
0 ≤ t < s ≤ 1,
(
(a(s − ξ1 ) + b)(d + c(ξ2 − t)),
(a(t − ξ1 ) + b)(d + c(ξ2 − s)),
(1.4)
s < t ≤ 1, ξ1 ≤ s ≤ ξ2 ,
0 ≤ t ≤ s, ξ1 ≤ s ≤ ξ2 .
Unfortunately this lemma is wrong. Indeed, by [2, Lemma 2.1], expression (1.3)
should be replaced by
Z 1
Z ξ1
u(t) =
G1 (t, s)
(τ − s)h(τ ) dτ ds
0
t
(1.5)
Z
Z ξ2
1 1
+
G1 (t, s)
(a(ξ1 − s) − b)(c(ξ2 − τ ) + d)h(τ ) dτ ds,
δ 0
ξ1
where δ = ad+bc+ac(ξ2 −ξ1 ) > 0. So the conclusions in [10] should be reconsidered.
If f (t, y(t), y 00 (t)) in (1.1) are replaced by h(t)g(y(t), y 00 (t)), then (1.1) reduces to
y (4) (t) − h(t)g(y(t), y 00 (t)) = 0,
0 ≤ t ≤ 1,
(1.6)
where h ∈ C[0, 1] and g ∈ C([0, ∞) × (−∞, 0], [0, ∞)).
To the authors’ knowledge, no one has studied the existence of positive solutions
for problem (1.6), (1.2) using the assumption that h(t) changes sign. Hence, the
aim of this paper is to investigate the existence of positive solutions of the BVP
(1.6) and (1.2) by using a triple positive fixed-point theorem of Avery and Peterson
in [1].
2. Preliminaries
2
Let E = {y ∈ C [0, 1] : y(0) = y(1) = 0}. Then we have the following lemma.
Lemma 2.1 ([10]). For y ∈ E, we get
kyk∞ ≤ ky 0 k∞ ≤ ky 00 k∞ ,
where kyk∞ = supt∈[0,1] |y(t)|.
By Lemma 2.1, E is a Banach space with the norm kyk = ky 00 k∞ . We define the
operator T : E → E by
Z 1
T y(t) =
G1 (t, s)(Qy)(s)ds,
(2.1)
0
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POSITIVE SOLUTIONS OF FOURTH-ORDER FOUR-POINT BVP
3
where G1 (t, s) as in (1.4), and
Z s
(τ − s)h(τ )g y(τ ), y 00 (τ ) dτ
(Qy)(s) =
ξ1
1
+
δ
Z
(2.2)
ξ2
00
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)h(τ )g y(τ ), y (τ ) dτ.
ξ1
Here, δ = ad + bc + ac(ξ2 − ξ1 ) > 0.
From [2, Lemma 2.1], we easily know that u(t) is a solution of the four-point
boundary-value problem (1.6), (1.2) if and only if u(t) is a fixed point of the operator
T.
It is rather straightforward to show that
0 ≤ G1 (t, s) ≤ G1 (s, s),
0 ≤ t, s ≤ 1,
(2.3)
and
G1 (t, s) ≥ ωG1 (s, s),
t ∈ [ω, 1 − ω],
s ∈ [0, 1],
(2.4)
where
1
.
(2.5)
2
For the convenience of the reader, we present some definitions from the cone
theory in Banach spaces.
0 < ω < min{ξ1 , 1 − ξ2 } <
Definition. The map α is said to be a nonnegative continuous concave functional
on a cone P of a real Banach space E provided that α : P → [0, ∞) is continuous
and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y), ∀x, y ∈ P, 0 ≤ t ≤ 1.
Similarly, we say the map β is a nonnegative continuous convex functional on a
cone P of a real Banach space E provided that β : P → [0, ∞) is continuous and
β(tx + (1 − t)y) ≤ tβ(x) + (1 − t)β(y),
∀x, y ∈ P,
0 ≤ t ≤ 1.
Let γ and θ be nonnegative continuous convex functionals on P , α be a nonnegative continuous concave functional on P , and ψ be a nonnegative continuous
functional on P . Then for positive real numbers a, b, c, and d, we define the following convex sets:
P (γ, d) = {x ∈ P : γ(x) < d},
P (γ, α, b, d) = {x ∈ P : b ≤ α(x), γ(x) ≤ d},
P (γ, θ, α, b, c, d) = {x ∈ P : b ≤ α(x), θ(x) ≤ c, γ(x) ≤ d},
R(γ, ψ, a, d) = {x ∈ P : a ≤ ψ(x), γ(x) ≤ d}.
The following fixed-point theorem due to Avery and Peterson is fundamental in the
proof of our main result.
Lemma 2.2 ([1]). Let P be a cone in a real Banach space E. Let γ and θ be
nonnegative continuous convex functionals on P , α be a nonnegative continuous
concave functional on P , and ψ be a nonnegative continuous functional on P satisfying ψ(λx) ≤ λψ(x) for 0 ≤ λ ≤ 1, such that for some positive numbers M and
d,
α(x) ≤ ψ(x) and kxk ≤ M γ(x),
for all x ∈ P (γ, d). Suppose T : P (γ, d) → P (γ, d) is completely continuous and
there exist positive numbers a, b, and c with a < b such that
4
Z. FANG, C. LI, C. BAI
EJDE-2008/159
(i) {x ∈ P (γ, θ, α, b, c, d) : α(x) > b} =
6 and α(T x) > b for x ∈ P (γ, θ, α, b, c, d);
(ii) α(T x) > b for x ∈ P (γ, α, b, d) with θ(T x) > c;
(iii) 0 6∈ R(γ, ψ, a, d) and ψ(T x) < a for x ∈ R(γ, ψ, a, d) with ψ(x) = a.
Then T has at least three fixed points x1 , x2 , x3 ∈ P (γ, d), such that
γ(xi ) ≤ d for i = 1, 2, 3,
b < α(x1 ),
a < ψ(x2 ) with α(x2 ) < b,
ψ(x3 ) < a.
3. Main result
Define the cone P ⊂ E = {y ∈ C 2 [0, 1] : y(0) = y(1) = 0} by
P = {y ∈ E : y(t) ≥ 0, y is concave on [0, 1]}.
Let the nonnegative, increasing, continuous functionals γ, ψ, θ and α be
γ(y) = max |y 00 (t)|,
0≤t≤1
ψ(y) = θ(y) = max |y(t)|,
0≤t≤1
α(y) =
min
ω≤t≤1−ω
|y(t)|.
We make the following assumptions:
(H1) g : [0, ∞) × (−∞, 0] → [0, ∞) is continuous;
(H2) h ∈ C[0, 1], h(t) ≤ 0, ∀t ∈ [0, ξ1 ], h(t) ≥ 0, ∀t ∈ [ξ1 , ξ2 ], h(t) ≤ 0, ∀t ∈
[ξ2 , 1], and h(t) is not identically zero on any subinterval of [0, 1].
Lemma 3.1. Assume that (H1)–(H2) hold. If b ≥ aξ1 and d ≥ c(1 − ξ2 ), then
T : P → P is completely continuous.
Proof. For each t ∈ [0, 1], we consider three cases:
Case 1: t ∈ [0, ξ1 ]. For any y ∈ P , we have from (2.2), (H1), (H2) and b ≥ aξ1 that
Z
ξ1
(Qy)(t) =
(t − τ )h(τ )g y(τ ), y 00 (τ ) dτ
t
+
1
δ
Z
ξ2
(3.1)
(b − aξ1 + at)(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ ≥ 0.
ξ1
Case 2: t ∈ [ξ1 , ξ2 ]. For each y ∈ P , we have from (H1), (H2) and (2.2) that
Z
t
(τ − t)h(τ )g y(τ ), y 00 (τ ) dτ
(Qy)(t) =
ξ1
+
1
δ
Z
t
(b − aξ1 + at)(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
Z
1 ξ2
(3.2)
+
(b − aξ1 + at)(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
δ t
Z
1 t
=
(b + a(τ − ξ1 ))(c(ξ2 − t) + d)h(τ )g y(τ ), y 00 (τ ) dτ
δ ξ1
Z
1 ξ2
+
(b + a(t − ξ1 ))(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ ≥ 0.
δ t
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POSITIVE SOLUTIONS OF FOURTH-ORDER FOUR-POINT BVP
5
Case 3: t ∈ [ξ2 , 1]. For any y ∈ P , we have from (H1), (H2), (2.2) and d ≥ (1 − ξ2 )c
that
Z ξ2
Z t
00
(Qy)(t) =
(τ − t)h(τ )g y(τ ), y (τ ) dτ +
(τ − t)h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
ξ2
Z
1 ξ2
(b − aξ1 + at)(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
+
δ ξ1
Z ξ2
1
(b + a(τ − ξ1 ))(d − c(t − ξ2 ))h(τ )g y(τ ), y 00 (τ ) dτ
=
δ ξ1
Z t
(τ − t)h(τ )g y(τ ), y 00 (τ ) dτ ≥ 0.
+
ξ2
(3.3)
Thus, from (3.1)-(3.3), we get
(Qy)(t) ≥ 0,
t ∈ [0, 1].
(3.4)
Therefore, by (2.1), G1 (t, s) ≥ 0 and (3.4), we obtain
(T y)(t) ≥ 0,
t ∈ [0, 1].
(3.5)
Obviously, we have (T u)(0) = (T u)(1) = 0, and
(T u)00 (t) = −(Qu)(t) ≤ 0,
t ∈ [0, 1].
Hence, T : P → P . Moreover, it is easy to check by the Arzera-Ascoli theorem that
the operator T is completely continuous.
Remark 3.2. By δ = ad + bc + ac(ξ2 − ξ1 ) > 0, b ≥ aξ1 and d ≥ c(1 − ξ2 ), we have
b > 0 and d > 0.
For convenience of notation, we set
Z ξ1
Z 1
Z
bd ξ2
M=
−τ h(τ )dτ +
−(1 − τ )h(τ )dτ + ξ2 − ξ1 +
h(τ )dτ,
δ
0
ξ2
ξ1
m = min{m1 , m2 },
(3.6)
(3.7)
where
Z
Z ξ2
bd ξ2
m1 =
h(τ )dτ
G1 (ω, s)ds,
δ ξ1
ξ1
Z
Z ξ2
bd ξ2
m2 =
h(τ )dτ
G1 (1 − ω, s)ds.
δ ξ1
ξ1
We are now in a position to present and prove our main results.
(3.8)
Theorem 3.3. Let b ≥ aξ1 and d ≥ c(1 − ξ2 ). Assume (H1)–(H2) hold. Suppose
there exist constants 0 < p < q < min{ω, 81 }r such that
(H3) g(u, v) ≤ r/M , for (u, v) ∈ [0, r] × [−r, 0],
(H4) g(u, v) > q/m, for (u, v) ∈ [q, q/ω] × [−r, 0],
(H5) g(u, v) < 8p/M , for (u, v) ∈ [0, p] × [−r, 0],
where M, m are as in (3.6)-(3.7), then (1.6), (1.2) has at least three positive solutions y1 , y2 , and y3 such that
max |yi00 (t)| ≤ r,
0≤t≤1
for i = 1, 2, 3;
6
Z. FANG, C. LI, C. BAI
EJDE-2008/159
min
|y1 (t)| > q;
p < max |y2 (t)|;
min
|y2 (t)| < q;
max |y3 (t)| < p.
ω≤t≤1−ω
ω≤t≤1−ω
0≤t≤1
0≤t≤1
Proof. From Lemma 3.1, T : P → P is completely continuous. We now show that
all the conditions of Lemma 2.2 are satisfied.
If y ∈ P (γ, r), then γ(y) = max0≤t≤1 |y 00 (t)| ≤ r. By Lemma 2.1, we have
max0≤t≤1 |y(t)| ≤ r, then assumption (H3) implies g(y(t), y 00 (t)) ≤ r/M . On the
other hand, from (3.1)-(3.3), we have
Z ξ1
max (Qy)(t) ≤
−τ h(τ )g y(τ ), y 00 (τ ) dτ
0≤t≤ξ1
0
+
1
δ
Z
ξ1
Z
ξ2
b(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
(3.9)
−τ h(τ )g y(τ ), y 00 (τ ) dτ
≤
0
1
+ b(c(ξ2 − ξ1 ) + d)
δ
max (Qy)(t) ≤
ξ1 ≤t≤ξ2
Z
1
δ
ξ2
h(τ )g y(τ ), y 00 (τ ) dτ,
ξ1
t
(b + a(t − ξ1 ))(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
1
δ
Z
+
1
=
δ
≤
Z
ξ2
Z
t
ξ2
(b + a(t − ξ1 ))(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
(b + a(t − ξ1 ))(c(ξ2 − τ ) + d)h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
1
(b + a(ξ2 − ξ1 ))(c(ξ2 − ξ1 ) + d)
δ
Z
ξ2
h(τ )g y(τ ), y 00 (τ ) dτ,
ξ1
(3.10)
and
max (Qy)(t) ≤
ξ2 ≤t≤1
1
δ
Z
Z
ξ2
d(b + a(τ − ξ1 ))h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
1
+
−(1 − τ )h(τ )g y(τ ), y 00 (τ ) dτ
ξ2
≤
Z ξ2
1
d(b + a(ξ2 − ξ1 ))
h(τ )g y(τ ), y 00 (τ ) dτ
δ
ξ1
Z 1
+
−(1 − τ )h(τ )g y(τ ), y 00 (τ ) dτ.
ξ2
By (3.9)-(3.11), we get
γ(T y) = max |(T y)00 (t)| = max |(Qy)(t)|
t∈[0,1]
t∈[0,1]
n
o
= max max |(Qy)(t)|, max |(Qy)(t)|, max |(Qy)(t)|
0≤t≤ξ1
Z
≤
0
ξ1
ξ1 ≤t≤ξ2
−τ h(τ )g y(τ ), y 00 (τ ) dτ +
ξ2 ≤t≤1
Z
1
ξ2
−(1 − τ )h(τ )g y(τ ), y 00 (τ ) dτ
(3.11)
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POSITIVE SOLUTIONS OF FOURTH-ORDER FOUR-POINT BVP
7
Z ξ2
1
+ (b + a(ξ2 − ξ1 ))(c(ξ2 − ξ1 ) + d)
h(τ )g y(τ ), y 00 (τ ) dτ
(3.12)
δ
ξ1
Z 1
Z
Z
bd ξ2
r ξ1
−(1 − τ )h(τ )dτ + ξ2 − ξ1 +
≤
−τ h(τ )dτ +
h(τ )dτ
M
σ
ξ2
0
ξ1
r
=
M = r.
M
Hence, T : P (γ, r) → P (γ, r).
To check condition (i) of Lemma 2.2, we choose y(t) = q/ω, 0 ≤ t ≤ 1. It is
easy to see that y(t) = q/ω ∈ P (γ, θ, α, q, q/ω, r) and α(y) = q/ω > q, and so
{y ∈ P (γ, θ, α, q, q/ω, r) : α(y) > q} 6= ∅. Hence, if y ∈ P (γ, θ, α, q, q/ω, r), then
q ≤ y(t) ≤ q/ω, −r ≤ y 00 (t) ≤ 0 for ω ≤ t ≤ 1 − ω. From assumption (H4), we have
g(y(t), y 00 (t)) > b/m for ω ≤ t ≤ 1 − ω, and by the definitions of α and the cone P ,
we distinguish two cases as follows:
Case (1): α(T y) = (T y)(ω). By (3.4) and (3.2), we have
α(T y)
Z
= (T y)(ω) =
1
G1 (ω, s)(Qy)(s)ds
0
Z
ξ2
>
G1 (ω, s)(Qy)(s)ds
ξ1
1
≥
δ
Z
bd
=
δ
ξ2
G1 (ω, s)ds
hZ
ξ1
Z
ξ2
Z
s
bdh(τ )g y(τ ), y (τ ) dτ +
ξ1
ξ2
G1 (ω, s)ds
ξ1
00
ξ2
Z
i
bdh(τ )g y(τ ), y 00 (τ ) dτ
s
h(τ )g y(τ ), y 00 (τ ) dτ
ξ1
Z ξ2
bd q
≥
G1 (ω, s)ds
h(τ )dτ
δ m ξ1
ξ1
q
=
· m1 ≥ q.
m
Case (2): α(T y) = (T y)(1 − ω). Similarly, we obtain
Z ξ2
α(T y) = (T y)(1 − ω) >
G1 (1 − ω, s)(Qy)(s)ds
Z
ξ2
ξ1
Z
ξ2
bd q
≥
G1 (1 − ω, s)ds
δ m ξ1
q
· m2 ≥ q.
=
m
Z
ξ2
h(τ )dτ
ξ1
i.e.,
q
, r).
ω
This show that condition (i) of Lemma 2.2 is satisfied. Secondly, we have
q
α(T y) = min |(T y)(t)| ≥ ωkT yk∞ = ωθ(T y) > ω = q,
ω≤t≤1−ω
ω
α(T y) > q,
∀y ∈ P (γ, θ, α, q,
for all y ∈ P (γ, α, q, r) with θ(T y) > q/ω. Thus, condition (ii) of Lemma 2.2 is
satisfied.
8
Z. FANG, C. LI, C. BAI
EJDE-2008/159
We finally show that (iii) of Lemma 2.2 also holds. Clearly, as ψ(0) = 0 < p,
there holds that 0 6∈ R(γ, ψ, p, r). Suppose that y ∈ R(γ, ψ, p, r) with ψ(y) = p.
Then, by (H5) and (3.12), we get
Z 1
ψ(T y) = max |(T y(t)| = max
G1 (t, s)(Qy)(s)ds
0≤t≤1
0≤t≤1
0≤t≤1
Z
1
+
ξ2
Z
ξ2 ≤s≤1
1
G1 (t, s)ds
ξ2
G1 (t, s)ds
ξ1
i
ξ2
max (Qy)(s), max (Qy)(s) max
1
G1 (t, s)ds
ξ1
hZ
0
ξ2 ≤s≤1
ξ1 ≤s≤ξ2
0≤s≤ξ1
≤ max
ξ1 ≤s≤ξ2
0
max (Qy)(s),
Z
Z
G1 (t, s)ds + max (Qy)(s)
0≤s≤ξ1
Z
Z
ξ1
Z
max (Qy)(s)
+ max (Qy)(s)
0≤t≤1
G1 (t, s)(Qy)(s)ds
ξ1
G1 (t, s)(Qy)(s)ds
0≤t≤1
≤ max
ξ2
G1 (t, s)(Qy)(s)ds +
0
h
≤ max
0
ξ1
Z
= max Z
0≤t≤1
1
G1 (t, s)ds
0
−τ h(τ )g y(τ ), y 00 (τ ) dτ
0
1
−(1 − τ )h(τ )g y(τ ), y 00 (τ ) dτ
+
ξ2
Z ξ2
i
1
+ (b + a(ξ2 − ξ1 ))(c(ξ2 − ξ1 ) + d)
h(τ )g y(τ ), y 00 (τ ) dτ
δ
ξ1
Z 1
Z
Z 1
8p h ξ1
≤ max
G1 (t, s)ds ·
−τ h(τ )dτ +
−(1 − τ )h(τ )dτ
0≤t≤1 0
M 0
ξ2
Z
i
bd ξ2
+ ξ2 − ξ1 +
h(τ )dτ
σ
ξ1
1 8p
= ·
· M = p.
8 M
So, condition (iii) of Lemma 2.2 is satisfied. Therefore, an application of Lemma 2.2
imply the boundary-value problem (1.6), (1.2) has at least three positive solutions
y1 , y2 , and y3 such that
max |yi00 (t)| ≤ r,
0≤t≤1
p < max |y2 (t)|,
0≤t≤1
for i = 1, 2, 3;
min
ω≤t≤1−ω
min
ω≤t≤1−ω
|y2 (t)| < q;
|y1 (t)| > q;
max |y3 (t)| < p.
0≤t≤1
The proof is complete.
Now, we give an example to demonstrate our result. Consider the fourth-order
four-point boundary-value problem
y (4) (t) = h(t)g(y(t), y 00 (t)),
0 < t < 1,
y(0) = y(1) = 0,
1
2
2
1
y 00 ( ) − y 000 ( ) = 0, y 00 ( ) + y 000 ( ) = 0,
3
3
3
3
(3.13)
(3.14)
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POSITIVE SOLUTIONS OF FOURTH-ORDER FOUR-POINT BVP
9
where ξ1 = 13 , ξ2 = 23 , h(t) = 9π sin(3t − 1)π, and
 2
u
v 3

− ( 150
) ,
0 ≤ u ≤ 1, v ≤ 0,
2√
4
v 3
g(u, v) = 11 u − 1 − ( 150
) + 12 , 1 < u ≤ 9, v ≤ 0,

 √
v 3
11 4 8 + 12 − ( 150
) ,
u > 9, v ≤ 0.
It is easy to check that the functions h and g satisfy (H1) and (H2). Set ω = 1/3.
It follows from a direct calculation that
Z 1
h Z 1/3
M = 9π
−τ sin(3τ − 1)πdτ +
−(1 − τ ) sin(3τ − 1)πdτ
0
16
21
46
=
,
7
2/3
Z
2/3
sin(3τ − 1)πdτ
+
i
1/3
and
m = 9π ·
3
7
Z
2/3
sin(3τ − 1)πdτ · min
1/3
nZ
2/3
1/3
1
G( , s)ds,
3
Z
2/3
1/3
o 2
2
G( , s)ds = .
3
7
Choose p = 1, q = 3 and r = 130, then we have
8p
g(u, v) ≤ 1.151 < 1.21 =
, for 0 ≤ u ≤ 1, −130 ≤ v ≤ 0;
M
q
g(u, v) ≥ 14.232 > 10.5 = , for 3 ≤ u ≤ 9, −130 ≤ v ≤ 0;
m
r
g(u, v) ≤ 19.651 < 19.78 =
, for 0 ≤ u ≤ 130, −130 ≤ v ≤ 0.
M
Noticing that b > ξ1 a and d > (1 − ξ2 )c hold, then all conditions of Theorem
3.3 hold. Hence, by Theorem 3.3, BVP (3.13), (3.14) has at least three positive
solutions y1 , y2 and y3 such that
max |yi00 (t)| ≤ 130,
0≤t≤1
1 < max |y2 (t)|,
0≤t≤1
for i = 1, 2, 3;
min |y2 (t)| < 3
1
2
3 ≤t≤ 3
min |y1 (t)| > 3;
1
2
3 ≤t≤ 3
max |y3 (t)| < 1.
0≤t≤1
Acknowledgements. The authors are grateful to the anonymous referee for his
or her useful suggestions. This research was supported by grant 10771212 from the
National Natural Science Foundation of China, and by grant 06KJB110010 from
the Natural Science Foundation of Jiangsu Education Office.
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Zheng Fang
School of Science, Jiangnan University, Wuxi, Jiangsu 214122, China
E-mail address: fangzhengm@yahoo.com.cn
Chunhong Li
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223300, China
E-mail address: lichshy2006@126.com
Chuanzhi Bai
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223300, China
E-mail address: czbai8@sohu.com