In this paper, by using fixed point theorem of cone expansion-compression type and suitable conditions, we present the existence of single positive solution for the integral boundary value problems. We derive an explicit interval of λ such that for any λ in this interval, the existence of at least one positive solution to the boundary value problems is guaranteed

Fuyi Xu

Hongling Geng

CiteSeerX

Int. Journal of Math. Analysis, Vol. 3, 2009, no. 31, 1529 - 1537
The Existence of Positive Solutions for
an Integral Boundary Value Problems
Hongling Geng and Fuyi Xu
School of Science
Shandong University of Technology, Zibo, Shandong
255049, People’s Republic of China
Abstract
In this paper, by using fixed point theorem of cone expansion-compression
type and suitable conditions, we present the existence of single positive
solution for the integral boundary value problems. We derive an explicit
interval of λ such that for any λ in this interval, the existence of at least
one positive solution to the boundary value problems is guaranteed.
Mathematics Subject Classification: 34B16
Keywords: Integral boundary value problem; Positive solutions; Fixed-
point theorem; Cone
1 Introduction
In this paper, we consider the integral boundary value problem
⎧⎪⎪⎪⎨
⎪⎪⎪⎩
(p(t)u′(t))′ − q(t)u(t) + λf(t, u) = 0, 0 ≤ t ≤ 1,
au(0) − bp(0)u′(0) = ∫Rr α(t)u(t)dt,
cu(1) + dp(1)u′(1) =
∫ R
r β(t)u(t)dt
(1.1)
where a, b, c, d ∈ [0, +∞), 0 < r < R < 1 are given constants, p, q ∈ C([0, 1], (0, +∞)),
α, β ∈ C([0, 1], [0, +∞)).
Integral boundary value problems arise in a variety of different areas of ap-
plied mathematics and physics. Boundary value problems(BVP) (1.1) becomes
a generic multi-point boundary value problem, some special cases of which have
been extensively studied in many papers in recent years. Moreover, boundary
1E-mail addresses: ghl1102@163.com(H.Geng), xfy−02@163.com(F.Xu).
1530 Hongling Geng and Fuyi Xu
value problem (1.1)includes the two-point,three-point and multi-point bound-
ary value problems as special cases. So we can see our work naturally extend
and unify some known results both for two-point boundary value problem and
for multi-point boundary value problem in the literature.
The mail tool of this paper is the following well known Krasnoselskii’s fixed
point theorem.
Theorem 1.1(See[10]). Suppose E is a Banach space, K ⊂ E is a cone, let
Ω1, Ω2 be two bounded open sets of E such that θ ∈ Ω1, Ω1 ⊂ Ω2. Let operator
T : K ∩ (Ω2 \ Ω1) −→ K be completely continuous. Suppose that one of the
following two conditions holds
(i) ‖Tx‖ ≤ ‖x‖, ∀ x ∈ K ∩ ∂Ω1, ‖Tx‖ ≥ ‖x‖, ∀ x ∈ K ∩ ∂Ω2; or
(ii) ‖Tx‖ ≥ ‖x‖, ∀ x ∈ K ∩ ∂Ω1, ‖Tx‖ ≤ ‖x‖, ∀ x ∈ K ∩ ∂Ω2.
Then T has at least one fixed point in K ∩ (Ω2 \ Ω1).
2 Preliminaries and Lemmas
In the rest of the paper, we make the following assumptions:
(H1) f ∈ C([0, +∞), [0, +∞));
(H2) h ∈ C([0, 1], [0, +∞)) and there exists x0 ∈ (0, ) such that h(x0) > 0;
(H3) a ∈ C[0, 1],b ∈ C([0, 1], (−∞, 0)).
In this section , we present some lemmas that are important to our main
results .
Lemma 2.1(See[5]). Assume that (H3) holds. Let φ1(t), φ2(t) are the positive
solutions of
⎧⎨
⎩
φ′′1(t) + a(t)φ
′
1(t) + b(t)φ1(t) = 0, 0 ≤ t ≤ 1,
φ1(0) = 0, φ1(1) = 1,
⎧⎨
⎩
φ′′2(t) + a(t)φ
′
2(t) + b(t)φ2(t) = 0, 0 ≤ t ≤ 1,
φ2(0) = 1, φ2(1) = 0.
Then
(i) φ1(t) is strictly increasing on [0,1], andφ1(t) > 0 on (0, 1] ;
(ii) φ2(t) is strictly decreasing on [0,1], andφ1(t) > 0 on [0, 1) .
Set ρ = p(0)
∣∣∣∣∣∣
φ2(0) φ1(0)
φ′2(0) φ
′
1(0)
∣∣∣∣∣∣, Δ =
∣∣∣∣∣∣
− ∫Rr α(t)φ1(t)dt ρ − ∫ Rr α(t)φ2(t)dt
ρ − ∫Rr β(t)φ1(t)dt − ∫ Rr β(t)φ2(t)dt
∣∣∣∣∣∣ .
Lemma 2.2. Assume that (H3) and (H4) hold, Δ = 0. Let y ∈ C[0, 1], then
Existence of positive solutions 1531
the problems
⎧⎪⎪⎪⎨
⎪⎪⎩
(p(t)u′(t))′ − q(t)u(t) + y(t) = 0, 0 ≤ t ≤ 1,
au(0) − bp(0)u′(0) = ∫Rr α(t)u(t)dt,
cu(1) + dp(1)u′(1) =
∫R
r β(t)u(t)dt.
(2.1)
is equivalent to the integral equation
u(t) =
∫ 1
0
G(t, s)y(s)ds + A(y)φ1(t) + B(y)φ2(t), (2.2)
where
G(t, s) =
1
ρ
⎧⎨
⎩
φ1(t)φ2(s), t ≤ s,
φ1(s)φ2(t), s ≤ t,
(2.3)
A(y) =
1
Δ
∣∣∣∣∣∣
∫R
r α(t)
∫ 1
0 G(t, s)y(s)dsdt ρ −
∫ R
r α(t)φ2(t)dt∫R
r β(t)
∫ 1
0 G(t, s)y(s)dsdt −
∫ R
r β(t)φ2(t)dt
∣∣∣∣∣∣ , (2.4)
B(y) =
1
Δ
∣∣∣∣∣∣
− ∫Rr α(t)φ1(t) ∫ Rr α(t) ∫ 10 G(t, s)y(s)dsdt
ρ − ∫Rr β(t)φ1(t)dt ∫ Rr β(t) ∫ 10 G(t, s)y(s)dsdt
∣∣∣∣∣∣ . (2.5)
Moreover, u(t) ≥ 0 on [0,1] provided y(t) ≥ 0.
Set
q(t) = min
(
φ1(t)
‖φ1‖ ,
φ2(t)
‖φ2‖
)
,
where ‖.‖ denote the supernum norm. From (2.3) we have
G(s, s) ≥ G(t, s) ≥ q(t)G(s, s), for t ∈ [0, 1].
Choose δ ∈ (0, 1
2
) such that there exists x0 ∈ (δ, 1 − δ), take
γ = min{q(t)|t ∈ [δ, 1 − δ]}.
It follows from Lemma 2.1 that 0 < γ < 1
Lemma 2.3. Let (H2), (H3) and (H4) hold. Assume Δ < 0, ρ−∫Rr α(t)φ2(t)dt >
0, ρ − ∫Rr β(t)φ1(t)dt > 0. Then y ∈ C[0, 1] with y ≥ 0, the unique solution of
the problem (2.1) satisfies u(t) ≥ 0 and
u(t) ≥ γ‖u‖, t ∈ [δ, 1 − δ].
Proof. By
0 ≤ G(t, s) ≤ G(s, s), for t ∈ [0, 1],
1532 Hongling Geng and Fuyi Xu
which implies
u(t) ≤
∫ 1
0
G(s, s)p(s)y(s)ds + Aφ1(t) + Bφ2(t).
By
G(t, s) ≥ q(t)G(s, s), for t ∈ [0, 1],
we have that for t ∈ [δ, 1 − δ],
G(t, s) ≥ γG(s, s).
Thus for t ∈ [δ, 1 − δ],
u(t) =
∫ 1
0
G(t, s)p(s)y(s)ds + Aφ1(t) + Bφ2(t)
=
∫ 1
0
G(t, s)
G(s, s)
G(s, s)p(s)y(s)ds + Aφ1(t) + Bφ2(t)
≥ γ
∫ 1
0
G(s, s)p(s)y(s)ds + Aφ1(t) + Bφ2(t)
≥ γ(
∫ 1
0
G(s, s)p(s)y(s)ds + Aφ1(t) + Bφ2(t))
≥ γ‖u‖.
3 The Main Results
Let
f0 = lim
x−→0+
min
t∈[0,1]
f(t, x)
x
, f∞ = lim
x−→∞ mint∈[0,1]
f(t, x)
x
,
f 0 = lim
x−→0+
max
t∈[0,1]
f(t, x)
x
, f∞ = lim
x−→∞ maxt∈[0,1]
f(t, x)
x
,
f0 = lim
x−→0+
inf max
t∈[0,1]
f(t, x)
x
, f∞ = lim
x−→∞ inf maxt∈[0,1]
f(t, x)
x
,
f 0 = lim
x−→0+
sup min
t∈[0,1]
f(t, x)
x
, f∞ = limx−→∞ sup mint∈[0,1]
f(t, x)
x
,
A1 =
∫ 1
0
G(s, s)ds + A0φ1(1) + B0φ2(0), A2 = max
t∈[0,1]
(∫ 1−δ
δ
G(t, s)ds
)
,
where
Existence of positive solutions 1533
A0 =
1
Δ
∣∣∣∣∣∣
∫R
r α(t)
∫ 1
0 G(s, s)dsdt ρ −
∫R
r α(t)φ2(t)dt∫ R
r β(t)
∫ 1
0 G(s, s)dsdt −
∫R
r β(t)φ2(t)dt
∣∣∣∣∣∣ ,
B0 =
1
Δ
∣∣∣∣∣∣
− ∫Rr α(t)φ1(t) ∫ Rr α(t) ∫ 10 G(s, s)dsdt
ρ − ∫Rr β(t)φ1(t)dt ∫ Rr β(t) ∫ 10 G(s, s)dsdt
∣∣∣∣∣∣ .
Let E = C[0, 1], then E is Banach space, with respect to the norm ‖u‖ =
supt∈[0,1] |u(t)|.
Now BVP(1.1) has a solution u = u(t) if and only if u is a solution of the
operator equation
(Tu)(t) = λ
(∫ 1
0
G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
,
where p and G are defined in Lemma2.1.
We define a cone in E by
P = {u ∈ E|u ≥ 0, min u(t)t∈[δ,1−δ] ≥ γ‖u‖}.
It is easy to check T (P ) ⊂ P and T is completely continuous.
In this section, we discuss the existence conditions of at least one positive
solution for BVP(1.1). We obtain the following existence results.
Theorem 3.1. Suppose conditions (H1)-(H4) hold. In addition,
(1) If A1f0 < γA2f∞, then for each λ ∈ ( 1γA2f∞ , 1A1f0 ), the BVP(1.1) has at
least one positive solution.
(2) If f0 = 0 and f∞ = ∞, then for any λ ∈ (0,∞), the BVP(1.1) has at
least one positive solution.
(3) If f∞ = ∞, 0 < f0 < ∞, then for each λ ∈ (0, 1A1f0 ), the BVP(1.1) has
at least one positive solution.
(4) If f0 = 0, 0 < f∞ < ∞, then for each λ ∈ ( 1γA2f∞ ,∞), the BVP(1.1)
has at least one positive solution.
Proof. We only prove the case (1). We construct the set ΩR1 , ΩR∗2 in order to
apply Theorem1.1. Let λ ∈ ( 1
γA2f∞ ,
1
A1f0
), and choose ε > 0 such that
1
γA2(f∞ − ε) ≤ λ ≤
1
A1(f 0 + ε)
.
By the definition of f 0, we choose R1 > 0 such that f(t, u) ≤ (f 0 + ε)u, for
(t, u) ∈ [0, 1] × [0, R1]. Thus for u ∈ P and ‖u‖ = R1, we have
(Tu)(t) = λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≤ λ
(∫ 1
0 G(s, s)f(s, u(s))ds + A(f(s, u(s)))φ1(1) + B(f(s, u(s)))φ2(0)
)
≤ λA1(f0 + ε)‖u‖ ≤ ‖u‖.
1534 Hongling Geng and Fuyi Xu
Let ΩR1 = {u ∈ E|‖u‖ < R1}, it follows that
‖Tu‖ ≤ ‖u‖, for u ∈ P ∩ ∂ΩR1 . (3.1)
Considering the definition of f∞, there exists R2 > R1 such that f(t, u) ≥
(f∞−ε)u for (t, u) ∈ [0, 1]× [R2,∞). Let R∗2 = max{2R1, R2γ } and ΩR∗2 = {u ∈
E|‖u‖ < R∗2}, then u ∈ P and ‖u‖ = R∗2 implies
min
t∈[δ,1−δ]
u(t) ≥ γ‖u‖ ≥ R2.
So we have
||(Tu)|| = maxt∈[0,1] λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≥ maxt∈[0,1] λ
(∫ 1
0 G(t, s)f(s, u(s))ds
)
≥ maxt∈[0,1] λ
(∫ 1−δ
δ G(t, s)f(s, u(s))ds
)
≥ λ(f∞ − ε)γA2‖u‖ ≥ ‖u‖.
Hence
‖Tu‖ ≥ ‖u‖, for u ∈ P ∩ ∂ΩR∗2 . (3.2)
From (3.1) and (3.2) and Theorem1.1[10], thus T has a fixed point in u ∈
P ∩ (ΩR∗2 \ ΩR1), which is a positive solution of BVP(1.1).
Theorem 3.2. Suppose condition (H1)-(H4) hold. In addition,
(1) If A1f∞ < γA2f0, then for each λ ∈ ( 1γA2f0 , 1A1f∞ ), the BVP(1.1) has at
least one positive solution.
(2) If f0 = ∞ and f∞ = 0, then for any λ ∈ (0,∞), the BVP(1.1) has at
least one positive solution.
(3) If f∞ = ∞, 0 < f0 < ∞, then for each λ ∈ (0, 1A1f0 ), the BVP(1.1) has
at least one positive solution.
(4) If f0 = 0, 0 < f∞ < ∞, then for each λ ∈ ( 1γA2f∞ ,∞), the BVP(1.1)
has at least one positive solution..
Proof. We only prove the case(1). Let λ ∈ ( 1
γA2f0
, 1
A1f∞ ), and choose ε > 0
such that
1
γA2(f0 − ε) ≤ λ ≤
1
A1(f∞ + ε)
.
By the definition of f0, there exists ρ0 ∈ (0,∞) such that f(t, u) ≥ (f0 − ε)u
for (t, u) ∈ [0, 1] × [0, ρ0]. Let Ωρ0 = {u ∈ E| ‖u‖ < ρ0}, then u ∈ P and
‖u‖ = ρ0, we have
||(Tu)|| = maxt∈[0,1] λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s))φ2(t)
)
≥ maxt∈[0,1] λ
(∫ 1−δ
δ G(t, s)f(s, u(s))ds
)
≥ λ(f0 − ε)γA2‖u‖ ≥ ‖u‖.
Existence of positive solutions 1535
So
‖Tu‖ ≥ ‖u‖, for u ∈ P ∩ ∂Ωρ0 . (3.3)
By the definition of f∞, there exists ρ1 such that f(t, u) ≤ (f∞ + ε)u for
(t, u) ∈ [0, 1] × [ρ1,∞).
We consider the following two cases:
Case(i) If f is unbounded, there exists ρ∗2 > max{2ρ0, γ−1ρ1} such that
f(t, u) ≤ f(t, ρ∗2) for (t, u) ∈ [0, 1] × [0, ρ∗2]. Then, for u ∈ P and ‖u‖ = ρ∗2, we
have
(Tu)(t) = λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(fs, (u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≤ λ
(∫ 1
0 G(s, s)f(s, u(s))ds + A(f(s, u(s)))φ1(1) + B(f(s, u(s)))φ2(0)
)
≤ λ(f∞ + ε)A1‖u‖ ≤ ‖u‖.
Case(ii) If f is bounded, say f(t, u) ≤ N1 for all (t, u) ∈ [0, 1] × [0, +∞),
taking ρ∗2 ≥ {2ρ0, λN1A1} for u ∈ P and ‖u‖ = ρ∗2, we have
(Tu)(t) = λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≤ λ
(∫ 1
0 G(s, s)f(s, u(s))ds + A(f(s, u(s)))φ1(1) + B(f(s, u(s)))φ1(0)
)
≤ λN1A1 ≤ ρ∗2 ≤ ‖u‖.
Hence, in either case, we may put Ωρ∗2 = {u ∈ E|‖u‖ < ρ∗2} such that
‖Tu‖ ≤ ‖u‖, for u ∈ P ∩ ∂Ω∗ρ2 . (3.4)
From(3.3),(3.4) and Theorem1.1[10], thus T has a fixed point in u ∈ P ∩ (Ωρ∗2 \
Ωρ0), which is a positive solution of BVP(1.1).
Theorem 3.3. Suppose condition (H1)-(H4)hold. In addition, assume there
exist two positive constants R1 = R2 such that
(1) f(t, u) ≤ R1
λA1
, ∀(t, u) ∈ [0, 1] × [0, R1];
(2) f(t, u) ≥ R2
λA2
, ∀(t, u) ∈ [0, 1] × [γR2, R2]. Then the BVP(1.1) has at
least one positive solution u such that ‖u‖ between R1 and R2.
Proof. We only consider the case R1 < R2, the case R1 > R2 follows in a
similar way. Let ΩR1 = {u ∈ E|‖u‖ < R1}, ΩR2 = {u ∈ E|‖u‖ < R2}. It
follows from (1) for any u ∈ P ∩ ΩR1 ,
(Tu)(t) = λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≤ λ
(∫ 1
0 G(s, s)f(s, u(s))ds + A(f(s, u(s)))φ1(1) + B(f(s, u(s)))φ2(0)
)
≤ λA1 R1λA1 = ‖u‖.
1536 Hongling Geng and Fuyi Xu
Therefore,
||Tu|| ≤ ||u||, for u ∈ P ∩ ΩR1 . (3.5)
On the other hand, for any u ∈ P∩ΩR2 , we have γR2 ≤ |u(t)| ≤ R2, for t ∈
[δ, 1 − δ]. It follows that for any u ∈ P ∩ ΩR2 ,
||Tu|| = maxt∈[0,1] λ
(∫ 1
0 G(t, s)f(s, u(s))ds + A(f(s, u(s)))φ1(t) + B(f(s, u(s)))φ2(t)
)
≥ maxt∈[0,1] λ
(∫ 1
0 G(t, s)f(s, u(s))ds
)
≥ maxt∈[0,1] λ
(∫ 1−δ
δ G(t, s)f(s, u(s))ds
)
≥ R2 = ‖u‖.
So
‖Tu‖ ≥ ‖u‖, for u ∈ P ∩ ∂ΩR2 . (3.6)
From(3.5),(3.6) and Theorem1.1[10], thus T has a fixed point in u ∈ P ∩(ΩR2 \
ΩR1), which is a positive solution of BVP(1.1) between R1 and R2..
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Received: February, 2009


The Existence of Positive Solutions for an Integral Boundary Value Problems

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The Existence of Positive Solutions for an Integral Boundary Value Problems

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