MATEMATIQKI VESNIK
originalni nauqni rad
research paper
68, 1 (2016), 12–25
March 2016
EXISTENCE OF A POSITIVE SOLUTION FOR A
THIRD-ORDER THREE POINT BOUNDARY VALUE PROBLEM
Ali Rezaiguia and Smail Kelaiaia
Abstract. By applying the Krasnoselskii fixed point theorem in cones and the fixed point
index theory, we study the existence of positive solutions of the non linear third-order three point
boundary value problem
u000 (t) + a(t)f (t, u(t)) = 0,
u0 (0) = u0 (1) = αu(η),
t ∈ (0, 1),
u(0) = βu(η),
where α, β and η are constants with α ∈ [0, η1 ), and 0 < η < 1. The results obtained here
generalize the work of Torres [Positive solution for a third-order three point boundary value
problem, Electronic J. Diff. Equ. 2013 (2013), 147, 1–11].
1. Introduction
Third order equations arise in a variety of different areas of applied mathematics and physics, as the deflection of a curved beam having a constant or varying cross
section, three layer beam, electromagnetic waves or gravity driven flows and so on
[9]. Different types of techniques have been used to study such problems: reduce
them to first and/or second order equations [5], use Green’s functions and comparison principles [2, 3, 15] (for periodic boundary value conditions), [4, 6, 7, 10, 20]
(two point ones), and [16, 19] (three point boundary conditions). A large part of
the literature on multiple solutions to boundary value problems seems to be traced
back to Krasnoseleskii’s work on nonlinear operator equations [1], especially the
part dealing with the theory of cones in Banach spaces.
In this paper, we are interested in the analysis of qualitative theory of positive
solutions of third-order differential equations. Motivated by the papers [12–14,17]
and the references therein, we concentrate on the existence of positive solutions for
the nonlinear third-order differential equation three point boundary value problem
u000 (t) + a(t)f (t, u(t)) = 0,
0
0
u (0) = u (1) = αu(η),
t ∈ (0, 1),
(1.1)
u(0) = βu(η),
(1.2)
2010 Mathematics Subject Classification: 34B10
Keywords and phrases: Third-order differential equations; three point boundary value problem; Krasnoselski fixed point in a cone; fixed point index theory
12
13
Existence of positive solution for third-order three point BVP
where α, β and η are constants with α ∈ [0, η1 ), 0 < η < 1 and β ∈ [0, 1 − αη). In
the case β = 0, Torres in [18] showed that (1.1) and (1.2) has positive solutions by
using Krasnoselskii’s fixed point theorem and the fixed point index theory.
This paper is organized as follows. In Section 2, we present some theorems
and lemmas that will be used to prove our main results. In Section 3, we discuss
the existence of single positive solutions of BVP (1.1) and (1.2). In Section 4, we
discuss the existence conditions of multiple positive solutions of BVP (1.1) and
(1.2). In Section 5, we give some examples to illustrate our results. The results
presented in this paper generalize the main results in [18].
2. Preliminaries
We shall consider the Banach space X = C[0, 1] equipped with the norm kxk =
max0≤t≤1 |x(t)|. Define a cone in X by C + [0, 1] = {x ∈ X : x(t) > 0 for t ∈ [0, 1]},
and the ordering 6 by x 6 y iff x(t) 6 y(t) for all t ∈ [0, 1].
Definition 1. A function u(t) is called a positive solution of (1.1) and (1.2)
if u ∈ C[0, 1] and u(t) > 0 for all t ∈ (0, 1).
The proof of existence of positive solutions is based on applications of the
following theorems.
Theorem 1. [8,11] Let X be a Banach space and K be a cone in X. Assume
that Ω1 and Ω2 are open subsets of X with 0 ∈ Ω1 ⊆ Ω1 ⊆ Ω2 and let
T : K ∩ (Ω2 \ Ω1 ) → K
be a completely continuous operator such that either
(i) kT uk ≤ kuk if u ∈ K ∩ ∂Ω1 and kT uk ≥ kuk if u ∈ K ∩ ∂Ω2 , or
(ii) kT uk ≥ kukif u ∈ K ∩ ∂Ω1 and kT uk ≤ kuk if u ∈ K ∩ ∂Ω2 .
Then T has a fixed point in K ∩ (Ω2 \ Ω1 ).
Theorem 2. [8,11] Let X be a Banach space and K be a cone in X. For
r > 0, define Kr = {u ∈ K : kuk ≤ r} and assume T : Kr → K is a completely
continuous operator such that T u 6= u for u ∈ ∂Kr .
(1) If kT uk ≤ kuk for all u ∈ ∂Kr then i(T, Kr , K) = 1.
(2) If kT uk ≥ kuk for all u ∈ ∂Kr then i(T, Kr , K) = 0.
Lemma 1. Assume that β ∈ [0, 1 − αη). Then for y ∈ C[0, 1] the problem
u000 (t) + y(t) = 0,
0
t ∈ (0, 1),
(2.1)
u(0) = βu(η),
(2.2)
0
u (0) = u (1) = αu(η),
has a unique solution
Z
u(t) =
1
G(t, s)y(s) ds +
0
where
½
G(t, s) =
αt + β
1 − αη − β
1
2
2 (2t − t −
1 2
2 t (1 − s),
s)s,
Z
1
G(η, s)y(s) ds,
0
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
(2.3)
14
A. Rezaiguia, S. Kelaiaia
Proof. Rewriting the differential equation as u000 (t) = −y(t) and integrating
three times, we obtain
Z
1 t
(t − s)2 y(s) ds + A1 t2 + A2 t + A3 where A1 , A2 , A3 ∈ R.
(2.4)
u(t) = −
2 0
Since u0 (0) = u0 (1),
Z
1
A2 = −
A1 =
1
2
(1 − s)y(s) ds + 2A1 + A2 ,
Z
0
1
(1 − s)y(s) ds.
0
Since u0 (0) = αu(η), we obtain
µ
¶
Z
Z
1 η
η2 1
A2 = α −
(η − s)2 y(s) ds +
(1 − s)y(s) ds + A2 η + A3 ,
2 0
2 0
Z η
Z
α
αη 2 1
(η − s)2 y(s) ds −
(1 − s)y(s) ds.
αA3 = (1 − αη)A2 +
2 0
2 0
From u(0) = βu(η), we have
µ
¶
Z
1 η
2
2
A3 = β −
(η − s) y(s) ds + A1 η + A2 η + A3 ,
2 0
Z 1
Z η
βη 2
β
A3 =
(1 − s)y(s) ds −
(η − s)2 y(s) ds,
2(1 − αη − β) 0
2(1 − αη − β) 0
Z 1
Z η
αβη 2
αβ
αA3 =
(1 − s)y(s) ds −
(η − s)2 y(s) ds.
2(1 − αη − β) 0
2(1 − αη − β) 0
Then
αη 2
A2 =
(1 − αη − β)
Z
0
1
1
α
(1 − s)y(s) ds −
2
(1 − αη − β)
Z
0
η
1
(η − s)2 y(s) ds
2
Replacing these expressions in (2.4),
Z
Z
1 t
t2 1
u(t) = −
(t − s)2 y(s) ds +
(1 − s)y(s) ds
2 0
2 0
Z 1
Z η
1
αη 2 t
1
αt
+
(1 − s)y(s) ds −
(η − s)2 y(s) ds
2 (1 − αη − β) 0
2 (1 − αη − β) 0
Z 1
Z η
β
βη 2
+
(1 − s)y(s) ds −
(η − s)2 y(s) ds
2(1 − αη − β) 0
2(1 − αη − β) 0
Z
Z
1 t
t2 1
=−
(t − s)2 y(s) ds +
(1 − s)y(s) ds
2 0
2 0
Z η
Z
1
αt + β
αt + β
η2 1
−
(η − s)2 y(s) ds +
(1 − s)y(s) ds
2 (1 − αη − β) 0
(1 − αη − β) 2 0
15
Existence of positive solution for third-order three point BVP
Z
t2 1
(t − s)2 y(s) ds +
(1 − s)y(s) ds
2 0
0
µ
¶
Z
Z
αt + β
1 η
η2 1
+
−
(η − s)2 y(s) ds +
(1 − s)y(s) ds
(1 − αη − β)
2 0
2 0
Z t
Z
2 Z t
t
t2 1
1
2
=−
(t − s) y(s) ds +
(1 − s)y(s) ds +
(1 − s)y(s) ds
2 0
2 0
2 t
µ
¶
Z η
Z
αt + β
1
η2 1
2
+
−
(η − s) y(s) ds +
(1 − s)y(s) ds
(1 − αη − β)
2 0
2 0
·Z t
¸
Z 1
1
=
(2t − t2 − s)sy(s) ds +
t2 (1 − s)y(s) ds
2 0
t
µ
¶
Z
Z
1 η
η2 1
αt + β
−
(η − s)2 y(s) ds +
(1 − s)y(s) ds
+
(1 − αη − β)
2 0
2 0
Z 1
Z 1
αt + β
=
G(t, s)y(s) ds +
G(η, s)y(s) ds
1
−
αη − β 0
0
=−
1
2
Z
t
Lemma 2. For all t and s such that 0 ≤ s ≤ 1 and 0 < τ ≤ t ≤ 1, we have
θG(1, s) ≤ G(t, s) ≤ G(1, s) =
1
(1 − s)s where θ = τ 2 .
2
(2.5)
Proof. For all t, s ∈ [0, 1], if s ≤ t,
1
1
(2t − t2 − s)s = [(1 − s) − (1 − t)2 ]s
2
2
1
≤ (1 − s)s = G(1, s),
2
G(t, s) =
and
1
1
1
(2t − t2 − s)s = st2 (1 − s) + (1 − t)[(t − s) + (1 − s)t]s
2
2
2
≥ θG(1, s);
G(t, s) =
if t ≤ s,
t2
1
(1 − s)s ≤ G(t, s) = t2 (1 − s) ≤ G(1, s).
2
2
Therefore
θG(1, s) ≤ G(t, s) ≤ G(1, s), ∀(t, s) ∈ [τ, 1] × [0, 1].
(2.6)
Lemma 3. For all y ∈ C + [0, 1], the unique solution u(t) of (2.1) and (2.2) is
nonnegative and satisfies
min u(t) ≥ θkuk.
τ ≤t≤1
16
A. Rezaiguia, S. Kelaiaia
Proof. From Lemma 1 and Lemma 2, u(t) is nonnegative,
Z 1
Z 1
αt + β
u(t) =
G(t, s)y(s) ds +
G(η, s)y(s) ds
1 − αη − β 0
0
Z
Z 1
1 1
α+β
≤
s(1 − s)y(s) ds +
G(η, s)y(s) ds;
2 0
1 − αη − β 0
then
kuk ≤
1
2
Z
1
s(1 − s)y(s) ds +
0
α+β
1 − αη − β
Z
1
G(η, s)y(s) ds.
(2.7)
0
For t ∈ [τ, 1],
Z
Z 1
αt + β
u(t) =
G(t, s)y(s) ds +
G(η, s)y(s) ds
1 − αη − β 0
0
Z 1
Z 1
1
αt2 + β
s(1 − s)y(s) ds +
G(η, s)y(s) ds
≥ τ2
1 − αη − β 0
0 2
Z 1
Z 1
1
ατ 2 + β
s(1 − s)y(s) ds +
G(η, s)y(s) ds
≥ τ2
1 − αη − β 0
0 2
Z 1
Z 1
1
α+β
≥ τ2
s(1 − s)y(s) ds + τ 2
G(η, s)y(s) ds
1 − αη − β 0
τ 2
· Z 1
¸
Z 1
1
α+β
=θ
s(1 − s)y(s) ds +
G(η, s)y(s) ds
2 0
1 − αη − β 0
≥ θkuk.
1
Therefore minτ ≤t≤1 (t) ≥ θkuk.
Now, we assume the following
(J1) f ∈ C([0, 1] × [0, ∞), [0, ∞)),
(J2) a ∈ L1 [0, 1] is nonnegative and a(t) 0 on any subinterval [0, 1].
Define the cone
½
¾
K = u ∈ C[0, 1] : u(t) ≥ 0, min u(t) ≥ θkuk ,
τ ≤t≤1
and the operator T : K → X by
Z 1
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
0
αt + β
1 − αη − β
Z
1
G(η, s)a(s)f (s, u(s)) ds.
0
(2.8)
By Lemma1 and Lemma 3, it is easy to see that the BVP (1.1) and (1.2) has a
positive solution u(t) if and only if u is a fixed point of T .
Lemma 4. The operator defined in (2.8) is completely continuous and satisfies
T (K) ⊆ K.
Proof. By Lemma 3, T (K) ⊆ K, and by Ascoli-Arzela theorem we prove that
T is a completely continuous operator.
17
Existence of positive solution for third-order three point BVP
3. Existence results
Throughout this paper, we shall use the following notations
f (t, u)
f (t, u)
, f0 = lim inf min
,
u→0 0≤t≤1
u
u
f (t, u)
f (t, u)
f ∞ = lim sup max
, f∞ = lim inf min
,
u→∞ 0≤t≤1
u
u
u→∞ 0≤t≤1
Z 1
Z 1
α+β
G(1, s)a(s) ds +
G(η, s)a(s) ds, R = A−1 ,
A=
1 − αη − β 0
0
µZ 1
¶
Z 1
α+β
G(1, s)a(s) ds +
G(η, s)a(s) ds , r = B −1 .
B = τ2
1 − αη − β τ
τ
f 0 = lim sup max
u→0
0≤t≤1
Theorem 3. Assume that f 0 = 0 and f∞ = ∞ are valid. Then the problem
(1.1) and (1.2) has at least one positive solution.
Proof. Since f 0 = 0, there exists H1 > 0 such that f (t, u) ≤ ²u, for all t ∈ [0, 1]
where 0 < u ≤ H1 and ² > 0. Then for u ∈ K ∩∂Ω1 with Ω1 = {u ∈ X : kuk < H1 },
we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
≤
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
≤
G(1, s)a(s)²u(s) ds +
G(η, s)a(s)²u(s) ds
1 − αη − β 0
0
·Z 1
¸
Z 1
α+β
≤²
G(1, s)a(s) ds +
G(η, s)a(s) ds kuk.
1 − αη − β 0
0
If ²A 6 1, then T u(t) ≤ kuk. Therefore
kT uk ≤ kuk for u ∈ K ∩ ∂Ω1 .
b 2 > 0 such that f (t, u) ≥ δu,
On the other hand, since f∞ = ∞, there exists H
b
for all t ∈ [τ, 1] where u ≥ H2 and δ > 0. Then for u ∈ K ∩ ∂Ω2 where Ω2 =
b2 }. Then u ∈ K ∩ ∂Ω implies that
{u ∈ X, kuk < H } with H = max{2H , H
2
2
b 2 , and
minτ ≤t≤1 u(t) ≥ θkuk = θH2 > H
Z 1
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
0
Z
1
≥
τ
Z
≥
τ
1
1
2
θ
α+β
1 − αη − β
Z
1
G(η, s)a(s)f (s, u(s)) ds
0
Z 1
1
α+β
s(1 − s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
2
1 − αη − β τ
Z 1
1
α+β
s(1 − s)a(s)δu(s) ds +
G(η, s)a(s)δu(s) ds
2
1 − αη − β τ
18
A. Rezaiguia, S. Kelaiaia
·Z
≥δ
τ
1
α+β
τ2
s(1 − s)a(s) ds + τ 2
2
1 − αη − β
Z
1
¸
G(η, s)a(s) ds kuk.
τ
If δB ≥ 1 then T u(1) ≥ kuk. We conclude that
kT uk ≥ kuk for u ∈ K ∩ ∂Ω2 .
Then T has a fixed point in K ∩ (Ω2 \ Ω1 ) such that H1 ≤ kuk ≤ H2 .
Theorem 4. Assume that f0 = ∞ and f ∞ = 0 are valid. Then the problem
(1.1) and (1.2) has at least one positive solution.
Proof. Since f0 = ∞, there exists H1 > 0 such that f (t, u) ≥ δu, for all t ∈ [τ, 1]
where 0 < u ≤ H1 and δ > 0. Then for u ∈ K∩∂Ω1 with Ω1 = {u ∈ X : kuk < H1 },
we have
Z 1
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
1
α+β
≥
s(1 − s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β τ
τ 2
Z 1
Z 1
1
α+β
≥
s(1 − s)a(s)δu(s) ds +
G(η, s)a(s)δu(s) ds
1 − αη − β τ
τ 2
·Z 1 2
¸
Z 1
τ
α+β
2
≥δ
s(1 − s)a(s) ds + τ
G(η, s)a(s) ds kuk.
2
1 − αη − β τ
τ
If δB ≥ 1, then T u(1) ≥ kuk. Therefore
kT uk ≥ kuk for u ∈ K ∩ ∂Ω1 .
b 2 > 0 such that f (t, u) ≤ ²u, for
On the other hand, Since f ∞ = 0, there exists H
b
all t ∈ [0, 1] where u ≥ H2 and ² > 0.
We consider two cases:
Case 1. Suppose f is bounded. Let L be such that f (t, u) ≤ L and Ω2 =
{u ∈ X : kuk < H2 } with H2 = max{2H1 , LA}. Then for u ∈ K ∩ ∂Ω2 , we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
1
α+β
≤
s(1 − s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0 2
Z 1
Z 1
1
α+β
≤
s(1 − s)a(s)L ds +
G(η, s)a(s)L ds
1 − αη − β 0
0 2
·Z 1
¸
Z 1
α+β
=L
G(1, s)a(s) ds +
G(η, s)a(s) ds
1 − αη − β 0
0
≤ H2 = kuk.
Then kT uk ≤ kuk.
Existence of positive solution for third-order three point BVP
19
Case 2. Suppose f is unbounded. Then from (J1), there exists H2 >
b 2 } such that f (t, u) ≤ f (t, H2 ) with 0 < u ≤ H2 and let Ω2 = {u ∈
max{2H1 , H
X, kuk < H2 }. Then for u ∈ K ∩ ∂Ω2 , we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
1
s(1 − s)a(s)f (s, H2 ) ds +
G(η, s)a(s)f (s, H2 ) ds
≤
1 − αη − β 0
0 2
Z 1
Z 1
1
α+β
G(η, s)a(s)²H2 ds
≤
s(1 − s)a(s)²H2 ds +
1 − αη − β 0
0 2
·Z 1
¸
Z 1
α+β
= ²H2
G(1, s)a(s) ds +
G(η, s)a(s) ds .
1 − αη − β 0
0
If ²A ≤ 1, then T u(t) ≤ H2 = kuk. Therefore kT uk ≤ kuk.
Theorem 5. Assume that 0 ≤ f 0 < R and r < f∞ ≤ ∞ hold. Then the
problem (1.1) and (1, 2) has at least one positive solution.
Proof. Since 0 ≤ f 0 < R, there exist H1 > 0 and 0 < ²1 < R such that
f (t, u) ≤ (R − ²1 )u, 0 ≤ t ≤ 1, 0 < u ≤ H1 . Let Ω1 = {u ∈ X : kuk < H1 }. So for
any u ∈ K ∩ ∂Ω1 and RA = 1 we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
1
α+β
≤
s(1 − s)a(s)(R − ²1 )u ds +
G(η, s)a(s)(R − ²1 )u ds
1 − αη − β 0
0 2
·Z 1
¸
Z 1
α+β
≤ (R − ²1 )
G(1, s)a(s) ds +
G(η, s)a(s) ds kuk
1 − αη − β 0
0
= (R − ²1 )Akuk < kuk.
Thus kT uk < kuk.
Since r ≤ f∞ ≤ ∞ there exist H 2 > 0 and 0 < ²2 such that f (t, u) ≥ (r + ²2 )u,
u ≥ H 2 and τ ≤ t ≤ 1. Let H2 > max{2H1 , Hθ2 } and Ω2 = {u ∈ X : kuk < H2 }. If
u ∈ K ∩ ∂Ω2 , then minτ ≤t≤1 u(t) ≥ θkuk = θH2 > H 2 . So by a property of G(t, s)
we have
Z 1
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
≥
G(1, s)a(s)(r + ²2 )u ds +
G(η, s)a(s)(r + ²2 )u ds
1 − αη − β τ
τ
¸
·Z 1
Z 1
α+β
2
2
τ G(1, s)a(s) ds + τ
G(η, s)a(s) ds kuk
≥ (r + ²2 )
1 − αη − β τ
τ
= (r + ²2 )Bkuk > kuk.
Thus kT uk > kuk.
20
A. Rezaiguia, S. Kelaiaia
Therefore, by Theorem 1, the operator T has at least one fixed point, which is
a positive solution of (1.1) and (1.2).
Theorem 6. Suppose that r < f0 ≤ ∞ and 0 ≤ f ∞ < R hold. Then the
problem (1.1) and (1.2) has at least one positive solution.
Proof. Since r < f0 ≤ ∞ there exist H1 > 0 and 0 < ²1 such that f (t, u) >
(r + ²1 )u, 0 < u ≤ H1 and τ ≤ t ≤ 1. Let Ω1 = {u ∈ X : kuk < H1 }. If
u ∈ K ∩ ∂Ω1 , then we have
Z 1
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
≥
1 − αη − β τ
τ
Z 1
Z 1
α+β
G(1, s)a(s)(r + ²1 )u ds +
≥
G(η, s)a(s)(r + ²1 )u ds
1 − αη − β τ
τ
·Z 1
¸
Z 1
α+β
G(η, s)a(s) ds kuk
≥ (r + ²1 )
τ 2 G(1, s)a(s) ds + τ 2
1 − αη − β τ
τ
= (r + ²1 )Bkuk > kuk.
Thus, T u(1) > kuk. Then kT uk > kuk.
Since 0 ≤ f ∞ < R, there exist H 2 > 0 and 0 < ²2 < R such that f (t, u) <
(R − ²2 )u, H 2 ≤ u and 0 ≤ t ≤ 1. Let Ω2 = {u ∈ X : kuk < H2 } where
H2 > max{2H1, Hθ2 }. If u ∈ K ∩ ∂Ω1 then minτ ≤t≤1 u(t) ≥ θkuk = θH2 > H 2 . So
by a property of G(t, s) we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
≤
G(t, s)a(s)(R − ²2 )u ds +
G(η, s)a(s)(R − ²2 )u ds
1 − αη − β 0
0
·Z 1
¸
Z 1
α+β
≤ (R − ²2 )
G(1, s)a(s) ds +
G(η, s)a(s) ds kuk
1 − αη − β 0
0
= (R − ²1 )Akuk < kuk.
Then T u(t) < kuk. Therefore, by Theorem 1, the operator T has at least one fixed
point, which is a positive solution of (1.1) and (1.2).
4. Multiplicity results
Now, we will study the problem (1.1) and (1.2) in the following cases:
(a) ∃ρ > 0 : f (t, u) < Rρ, 0 < u ≤ ρ, t ∈ [0, 1].
(b) ∃ρ > 0 : f (t, u) > rρ, 0 < u ≤ ρθ , t ∈ [τ, 1].
Theorem 7. Assume that (a) holds, f0 = ∞ and f∞ = ∞. Then the problem
(1.1) and (1.2) has at least two positive solutions.
Existence of positive solution for third-order three point BVP
21
Proof. Since f0 = ∞, there exists H1 > 0 where 0 < H1 < ρ such that
f (t, u) > ru, for all t ∈ [τ, 1] where 0 < u ≤ H1 . Then for u ∈ K ∩ ∂Ω1 with
Ω1 = {u ∈ X : kuk < H1 }, we have
Z 1
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
>
G(1, s)a(s)ru ds +
G(η, s)a(s)ru ds
1 − αη − β τ
τ
·Z 1
¸
Z 1
α+β
2
2
>r
τ G(1, s)a(s) ds + τ
G(η, s)a(s) ds kuk
1 − αη − β τ
τ
= rBkuk = kuk.
Then T u(1) > kuk. Therefore
kT uk > kuk for u ∈ K ∩ ∂Ω1 .
By Theorem 2, we have i(T, KH1 , K) = 0.
b 2 > ρ such that f (t, u) > ru, for all t ∈ [τ, 1],
Since f∞ = ∞, there exists H
b2 .
b 2 . Then u ∈ K ∩ ∂Ω2 where Ω2 = {u ∈ X : kuk < H2 } with H2 > H
where u > H
θ
b 2 and
This implies that minτ ≤t≤1 u(t) ≥ θkuk = θH2 > H
Z
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
G(η, s)a(s)ru ds
>
G(1, s)a(s)ru ds +
1
−
αη − β τ
τ
·Z 1
¸
Z 1
α+β
2
2
>r
τ G(1, s)a(s) ds + τ
G(η, s)a(s) ds kuk
1 − αη − β τ
τ
= rBkuk.
1
Then T u(1) > kuk. Therefore
kT uk > kuk for u ∈ K ∩ ∂Ω2 .
By Theorem 2, we have i(T, KH2 , K) = 0.
On the other hand, let Ω3 = {u ∈ X : kuk < ρ}. Then for u ∈ K ∩ ∂Ω3 such
that f (t, u) < Rρ for all t ∈ [0, 1] we get
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
<
G1 (t, s)a(s)Rρ ds +
G(η, s)a(s)Rρ ds
1 − αη − β 0
0
¸
·Z 1
Z 1
α+β
G(η, s)a(s) ds ρ
<R
G(1, s)a(s) ds +
1 − αη − β 0
0
= RAρ 6 kuk.
22
A. Rezaiguia, S. Kelaiaia
Then T u(t) 6 kuk. Therefore
kT uk 6 kuk for u ∈ K ∩ ∂Ω3 .
By Theorem 2, we have i(T, Kρ , K) = 1. Hence
i(T, KH2 \ K ρ , K) = i(T, KH2 , K) − i(T, Kρ , K) = −1.
i(T, Kρ \ K H1 , K) = i(T, Kρ , K) − i(T, KH1 , K) = +1.
Therefore, there exist at least two positive solutions u1 ∈ K ∩ (Ω3 \ Ω1 ) and u2 ∈
K ∩ (Ω2 \ Ω3 ) of (1.1) and (1.2) in K, such that
0 < ku1 k < ρ < ku2 k.
Theorem 8. Assume that (b) holds, f 0 = 0 and f ∞ = 0. Then the problem
(1.1) and (1, 2) has at least two positive solutions.
Proof. Since f 0 = 0 there exists H1 > 0 such that f (t, u) ≤ ²u, for all t ∈ [0, 1]
where 0 < u ≤ H1 and ² > 0. Then for u ∈ K ∩∂Ω1 with Ω1 = {u ∈ X : kuk < H1 },
we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
≤
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
1
α+β
≤
s(1 − s)a(s)²u(s) ds +
G(η, s)a(s)²u(s) ds
1 − αη − β 0
0 2
·Z 1
¸
Z 1
α+β
≤²
G(1, s)a(s) ds +
G(η, s)a(s) ds kuk.
1 − αη − β 0
0
If ²A ≤ 1, then T u(t) ≤ kuk. Therefore
kT uk ≤ kuk for u ∈ K ∩ ∂Ω1 .
By Theorem 1, we have
i(T, KH1 , K) = 1.
b 2 > 0 such that f (t, u) ≤ ²u, for all t ∈ [0, 1] where
Since f ∞ = 0, there exists H
b
u ≥ H2 and ² > 0.
We consider two cases:
Case 1. Suppose f is bounded. Let L be such that f (t, u) ≤ L and Ω2 =
{u ∈ X : kuk < H2 } with H2 = max{2H1 , LA}. Then for u ∈ K ∩ ∂Ω2 , we have
Z 1
Z 1
αt + β
G(t, s)a(s)f (s, u(s)) ds +
T u(t) =
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
1
α+β
≤
s(1 − s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0 2
Existence of positive solution for third-order three point BVP
Z
23
Z 1
α+β
1
s(1 − s)a(s)L ds +
G(η, s)a(s)L ds
1 − αη − β 0
0 2
·Z 1
¸
Z 1
α+β
=L
G(1, s)a(s) ds +
G(η, s)a(s) ds
1 − αη − β 0
0
≤ H2 = kuk.
1
≤
Then kT uk ≤ kuk.
Case 2. Suppose f is unbounded. Then from (J1), there exists H2 >
b 2 } such that f (t, u) ≤ f (t, H2 ) with 0 < u ≤ H2 . Let Ω2 = {u ∈
max{2H1 , H
X : kuk < H2 }. Then for u ∈ K ∩ ∂Ω2 , we have
Z 1
Z 1
αt + β
T u(t) =
G(t, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
1
s(1 − s)a(s)f (s, H2 ) ds +
G(η, s)a(s)f (s, H2 ) ds
≤
1 − αη − β 0
0 2
Z 1
Z 1
1
α+β
s(1 − s)a(s)²H2 ds +
G(η, s)a(s)²H2 ds
≤
1 − αη − β 0
0 2
¸
·Z 1
Z 1
α+β
= ²H2
G(1, s)a(s) ds +
G(η, s)a(s) ds
1 − αη − β 0
0
If ²A ≤ 1 then T u(t) ≤ H2 = kuk, and kT uk ≤ kuk. Therefore
kT uk ≤ kuk for u ∈ K ∩ ∂Ω2 .
By Theorem 2, we have i(T, KH2 , K) = 1.
On the other hand, let Ω3 = {u ∈ X : kuk < ρ}. Then for u ∈ K ∩ ∂Ω3 such
that f (t, u) > rρ for all t ∈ [τ, 1] we get
Z 1
Z 1
α+β
T u(1) =
G(1, s)a(s)f (s, u(s)) ds +
G(η, s)a(s)f (s, u(s)) ds
1 − αη − β 0
0
Z 1
Z 1
α+β
>
G(1, s)a(s)rρ ds +
G(η, s)a(s)rρ ds
1
−
αη − β τ
τ
·Z 1
¸
Z 1
α+β
2
2
> rρ
τ G(1, s)a(s) ds + τ
G(η, s)a(s) ds kuk
1 − αη − β τ
τ
= rρBkuk.
If ρ = 1 then T u(1) ≥ kuk. Therefore
kT uk > kuk f or u ∈ K ∩ ∂Ω3 .
By Theorem 2, we have i(T, Kρ , K) = 0. Hence
i(T, KH2 \ Kρ , K) = i(T, KH2 , K) − i(T, Kρ , K) = 1 − 0 = 1.
i(T, Kρ \ KH1 , K) = i(T, Kρ , K) − i(T, KH1 , K) = 0 − 1 = −1.
Therefore, there exist at least two positive solutions u1 ∈ K ∩ (Ω3 \ Ω1 ) and u2 ∈
K ∩ (Ω2 \ Ω3 ) of (1.1) and (1.2) in K, such that
0 < ku1 k < ρ < ku2 k.
24
A. Rezaiguia, S. Kelaiaia
5. Examples
Example 1. If f (t, u) = (u2 + t)e−u , then the condition of Theorem 3 hold
(superlinear case). If f (t, u) = αt + ch(u), α > 0, then the conditions of Theorem
4 hold (sublinear case).
Example 2. Consider the boundary value problem
u000 (t) + ch(u) = 0, t ∈ (0, 1),
1 ³1´
1 ³1´
u0 (0) = u0 (1) = u
, u(0) = u
,
2 2
2 2
(5.1)
(5.2)
where f (t, u) = f (u) = ch(u), a(t) = 1. Then f0 = ∞ and f∞ = ∞. By simple
calculation, A = 14 and R = 4.
On the other hand, we could chose ρ = 1. Then f (t, u) ≤ 12 (e+e−1 ) ≤ 4 = ρR,
for (t, u) ∈ [0, 1] × [0, ρ]. By Theorem 7, (5.1) and (5.2) have at least two positive
solutions.
Example 3. Consider the boundary value problem
u000 (t) + ueu ln(t + u) = 0, t ∈ (0, 1),
1 ³1´
1 ³1´
u0 (0) = u0 (1) = u
, u(0) = u
,
2 2
4 2
(5.3)
(5.4)
where f (t, u) = f (u) = ueu ln(t + u), a(t) = 1. Then f 0 = 0 and f∞ = ∞. By
40σ 5 +3σ 4 +14σ 2
96
7
, R = 48
and r = 40σ5 +3σ
simple calculation A = 48
4 +14σ 2 .
7 , B =
96
48
0
Therefore 0 ≤ f = 0 < 7 = R and r < f∞ ≤ ∞. By Theorem 3, (5.3) and (5.4)
have at least one positive solution.
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(received 15.01.2015; in revised form 28.08.2015; available online 10.10.2015)
A.R.: Univ. Souk Ahras Fact sci Dep MI 41000 Souk Ahras Algeria
E-mail: ali rezaig@yahoo.fr
S.K.: Department of Mathematics, Faculty of Sciences, University of Annaba, P.O. Box12 Annaba,
Algerie
E-mail: kelaiaiasmail@yahoo.fr