J. Nonlinear Sci. Appl. 4 (2011), no. 4, 247–261
The Journal of Nonlinear Sciences and Applications
http://www.tjnsa.com
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE
PROBLEM
A. GUEZANE-LAKOUD1∗ AND S. KELAIAIA2
Abstract. In this paper we consider three point boundary value problems
of second order. We introduce new and sufficient conditions that allow us to
obtain the existence of a nontrivial solution by using Leray Schauder nonlinear
alternative. As an application, we give some examples to illustrate our results.
1. Introduction and preliminaries
This paper is devoted to the study of the existence of nontrivial solution for
the following second order three point boundary value problem (BVP):
u′′ + f (t, u) = 0, 0 < t < 1,
′
(1.1)
′
u (0) = αu (0) , u (1) = βu (η) ,
(1.2)
where η ∈ (0, 1) , α, β ∈ R, f ∈ C ([0, 1] × R, R).
The parameters α and β are arbitrary in R such that 1 + α − β ̸= 0. We do
not assume any monotonicity condition on the nonlinearity f, we assume that
f (t, 0) ̸= 0 and there exist two nonnegative functions k, h ∈ L1 [0, 1] such that
|f (t, x)| ≤ k (t) |x|p + h (t) , ∀ (t, x) ∈ [0, 1] × R,
where 0 < p ≤ 2, so our conditions are new and more general than the conditions
found in the literature. The case p > 2 still an open question. The importance of
these conditions is that they appear in the study of integrodifferential equations
and integral inequalities [2, pp. 6]. The second order equations (1.1) are used
to model various phenomena in physics, chemistry and epidemiology. In general
Date: Received: August 30, 2010; Revised: November 16, 2011.
∗
Corresponding author
c 2011 N.A.G.
⃝
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Fixed point theorem, Three point boundary value problem, Non
trivial solution.
247
248
A. GUEZANE-LAKOUD, S. KELAIAIA
nonlinearities that refer to source terms represent specific physical laws, in chemu−1
istry, for example, if f (t, u) = ug(u)e ε , then it represents Arrheninus law for
chemistry reactions, where the positive parameter ε represents the activation energy for the reaction and the continuous function g represents the concentration
of the chemical product, see [1]. The nonlocal conditions (1.2) arise in the study
of the equilibrium states of a heated bar [17], in this situation two controllers at
t = 0 and t = η alter the heat according to the temperatures detected by a sensor
at t = 1.
Many of the results involving nonlocal boundary value problems are studied
in [4-22]. Using the Leray-Schauder nonlinear alternative some results on the
existence of solutions for the equation (1.1) subject to the conditions u (0) = 0,
u (1) = αu (η) can be found in [12, 20]. Similar boundary value problem with
the conditions u (0) = αu′ (0) , u (1) = βu (η) is considered in [18]. Motived by a
work of Guidotti and Merino [11] and using fixed point index theory or Sperner’s
Lemma, Infante and Webb in [5] studied the boundary conditions u′ (0) = 0,
σu′ (1) + u(η) = 0 and Webb [20] provides explicit optimal constants for the
problem. Infante and Webb [6], Palamides, Infante and Pietramala [17] and Fan
and Ma [9] studied u′ (0) + α[u] = 0, σu′ (1) + u(η) = 0. Infante in [7] considered
the more general boundary conditions u′ (0) + α[u] = 0, σu′ (1) + u(η) = β[u],
where α[u] and β[u] are bounded linear functionals on C [0, 1]. Our aim is to
give new conditions on the nonlinearity f, then, using Leray Schauder nonlinear
alternative, we establish the existence of nontrivial solution of the BVP (1.1)(1.2). As an application, some examples to illustrate our results are given.
This paper is organized as follows. First, we list some preliminary material
to be used later. Then, in Section 3, we present and prove our main results
which consist in existence theorems and corollaries. We end our work with some
illustrating examples.
2. Preliminary Lemmas
Let E = C [0, 1] , with the norm ||y|| = max |y (t)|. Firstly we state two
t∈[0,1]
preliminary results.
Lemma 2.1. Let y ∈ E. If β ̸= α + 1, then the three point BVP
{
u′′ (t) + y (t) = 0, 0 < t < 1,
u (0) = αu′ (0) , u (1) = βu′ (η) ,
(2.1)
has a unique solution
∫
u(t) = −
0
t
t+α
(t − s) y(s)ds+
1+α−β
∫
1
0
Proof. The proof is easy, then we omit it.
t+α
(1 − s) y(s)ds−β
1+α−β
∫
η
y(s)ds.
0
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
249
We define the integral operator T : E → E, by
∫ t
∫ 1
t+α
T u(t) = −
(2.1)
(t − s) f (s, u (s)) ds +
(1 − s) f (s, u (s)) ds
1+α−β 0
0
∫ η
t+α
−β
f (s, u (s)) ds
1+α−β 0
By Lemma 2.1, the BVP (1.1)-(1.2) has a solution if and only if the operator T has a fixed point in E. By Ascoli Arzela Theorem we prove that T
is completely continuous operator. Now we cite the Leray–Schauder nonlinear
alternative.
Lemma 2.2. [3] Let F be a Banach space and Ω a bounded open subset of F ,
0 ∈ Ω. Let T : Ω → F be a completely continuous operator. Then, either there
exists x ∈ ∂Ω, λ > 1 such that T (x) = λx, or there exists a fixed point x∗ ∈ Ω of
T.
3. Main Results
In this section, we present and prove our main results.
Theorem 3.1. We assume that f (t, 0) ̸= 0, 1 + α ̸= β, 0 < p < 1, and there
exist two nonnegative functions k, h ∈ L1 [0, 1], k(t) ̸= 0 almost everywhere on
[0, 1] and such that
|f (t, x)| ≤ k (t) |x|p + h (t) , ∀ (t, x) ∈ [0, 1] × R,
∫
(
)∫ 1
|1 + α|
1 + α η
1+
(1 − s) k(s)ds + β
k(s)ds <
|1 + α − β|
1 + α − β 0
0
∫
(
)∫ 1
|1 + α|
1 + α η
(1 − s) h(s)ds + β
1+
h(s)ds >
|1 + α − β|
1 + α − β 0
0
(3.1)
1
,
2
(3.2)
1
.
2
(3.3)
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C [0, 1].
Proof. Setting
(
M = 1+
and
(
N=
|1 + α|
|1 + α − β|
|1 + α|
1+
|1 + α − β|
)∫
1
0
)∫
0
1
∫
1
+
α
(1 − s) k(s)ds + β
1 + α − β
η
k(s)ds
0
∫
1
+
α
(1 − s) h(s)ds + β
1 + α − β
η
h(s)ds.
0
N
. By
M
hypothesis (3.3) we have m ̸= 0. Set Ω = {u ∈ C [0, 1] : ||u|| < m} and assume
that there exist u ∈ ∂Ω and λ > 1 such T u = λu. Then
Since k is nonnegative and k(t) ̸= 0, a.e. t ∈ [0, 1] , then M ̸= 0, let m =
λm = λ ||u|| = ||T u|| = max |(T u) (t)|
0≤t≤1
250
A. GUEZANE-LAKOUD, S. KELAIAIA
∫
[(
)∫ 1
]
|1 + α|
1 + α η
≤ ∥u∥
1+
(1 − s) k(s)ds + β
k(s)ds
|1 + α − β|
1 + α − β 0
0
∫ η
[(
)∫ 1
]
|1 + α|
1
+
α
+ 1+
(1 − s) h(s)ds + β
h(s)ds
|1 + α − β|
1 + α − β 0
0
= M ∥u∥p + N.
p
From this we get
λ ≤ M 2−p N p−1 + M.
(3.4)
p−1
Since 0 < p < 1, M < 1/2 and N > 1/2 we have N p−1 < (1/2)
and M 2−p <
2−p
(1/2) . Consequently (3.4) becomes λ < 1; this contradicts the fact that λ > 1.
By Lemma 2.1 we conclude that operator T has a fixed point u∗ ∈ Ω and then
the BVP (1.1)-(1.2) has a nontrivial solution u∗ ∈ C [0, 1] .
Theorem 3.2. We assume that f (t, 0) ̸= 0, 1 + α ̸= β, p = 1 and there exist
nonnegative functions k, h ∈ L1 [0, 1] such that
|f (t, x)| ≤ k (t) |x| + h (t) , ∀ (t, x) ∈ [0, 1] × R,
∫ η
)∫ 1
|1 + α|
1
+
α
1+
(1 − s) k(s)ds + β
k(s)ds < 1.
|1 + α − β|
1 + α − β 0
0
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C [0, 1].
(
Proof. Since the function f is continuous and f (t, 0) ̸= 0, there exists an interval
[σ, τ ] ⊂ [0, 1] such that min |f (t, 0)| > 0 and as h (t) ≥ |f (t, 0)| , a.e. t ∈ [0, 1]
σ≤t≤τ
then N > 0. Let m =
such T u = λu. Then
N
, so m ̸= 0. Assume that there exist u ∈ ∂Ω, λ > 1
1−M
λm = λ ||u|| = ||T u|| = max |(T u) (t)| ≤ M ∥u∥ + N.
0≤t≤1
= 1. This contradicts the fact that λ > 1. By
Therefore, we have λ ≤ M + N
m
Lemma 2.1 we conclude that the operator T has a fixed point u∗ ∈ Ω and then
the BVP (1.1)-(1.2) has a nontrivial solution u∗ ∈ C [0, 1] .
Theorem 3.3. We assume that f (t, 0) ̸= 0, 1 + α ̸= β, 1 < p ≤ 2, there exist
nonnegative functions k, h ∈ L1 [0, 1] , k(t) ̸= 0, a.e. on [0, 1] and such that
|f (t, x)| ≤ k (t) |x|p + h (t) , ∀ (t, x) ∈ [0, 1] × R,
∫
(
)∫ 1
|1 + α|
1 + α η
1
1+
(1 − s) k(s)ds + β
k(s)ds < , (3.5)
|1 + α − β|
1+α−β 0
2
0
∫ η
(
)∫ 1
1 + α 1
|1 + α|
h(s)ds < . (3.6)
1+
(1 − s) h(s)ds + β
|1 + α − β|
1+α−β 0
2
0
∗
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u ∈ C [0, 1].
Proof. Since k is nonnegative and k(t) ̸= 0, a.e. t ∈ [0, 1] , then M ̸= 0. From
N
f (t, 0) ̸= 0 we deduce that m =
̸= 0. Assume that there exist u ∈ ∂Ω and
M
λ > 1 such T u = λu. Then λ ≤ M 2−p N p−1 + M. Since 1 ≤ p ≤ 2, M < 1/2 and
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
251
N < 1/2 we have M 2−p < (1/2)2−p and N p−1 < (1/2)p−1 . Consequently λ < 1;
this contradicts the fact that λ > 1. By Lemma 2.1 we conclude that operator T
has a fixed point u∗ ∈ Ω and then the BVP (1.1)-(1.2) has a nontrivial solution
u∗ ∈ C [0, 1] .
Remark 3.4. For 0 < p ≤ 2, and under the conditions of Theorems 3.1, 3.2 and 3.3,
we have proved that the BVP (1.1)-(1.2) has a nontrivial solution u∗ ∈ C [0, 1] .
The case p > 2 still an open question.
Theorem 3.5. Under the conditions of Theorem 3.1 (0 < p < 1) and if one of
the following conditions is satisfied:
(1) There exist n > 1 and r > 0 such that
(∫
) n1
1
k n (s)ds
<
1+
0
(∫
η
− 1r
h
|1+α|
|1+α−β|
)−r
(s)ds
1
(1
2(
2 1+
0
)
1+α
+ β 1+α−β
(η(1 + q))
[
>
1
+ q) q
1
q
(
1 1
+ = 1),
n q
1
|1+α|
|1+α−β|
]
1+α r+1 .
(1 − η) + β 1+α−β
η
(3.7)
(3.8)
(2) There exist two constants µ > −1, τ > −1 such that
k(s) <
|1 + α − β| (µ + 1) (µ + 2)
sµ ,
|1 + α − β| + |1 + α| + |β (1 + α)| (µ + 2) η µ+1
1
2
[(
h(s) >
2η τ +1
(τ + 1)
] τ
)
1+α s ,
|1+α|
1 + |1+α−β| (1 − η) + β 1+α−β (3.9)
(3.10)








1
(µ
+
1)
(µ
+
2)
µ
2
> 0
meas s ∈ [0, 1] , k(s) <
s
(1 + α) 1+α 



µ+1


1+
+ β 1+α−β (µ + 2) η
1 + α − β




(τ + 1)
τ
]
)
and meas s ∈ [0, 1] , h(s) > [(
s
1+α τ +1  > 0.
|1+α|

(1 − η) + β 1+α−β
2 1 + |1+α−β|
η
(3)The functions k and h satisfy
1
2 (
)
k(s)| ≤
1+α ,
(1+α) 1 + 1+α−β + β 1+α−β [(
h(s) >
2
1
)
]
1+α ,
1+α 1 + 1+α−β (1 − η) + β 1+α−β
(3.11)
(3.12)
252
A. GUEZANE-LAKOUD, S. KELAIAIA
{
}
|1 + α − β|
meas s ∈ [0, 1] , k(s) <
> 0
|(1 + α)| + |1 + α − β| + |β (1 + α)|




1
]
[(
)
and meas s ∈ [0, 1] , h(s) >
> 0.

2η 1 + |1+α| (1 − η) + β 1+α 
1
2
|1+α−β|
1+α−β
(4) The function f satisfies
1
|f (t, x)|
4 <
1+α (1+α) t∈[0,1]
|x|p
1 + 1+α−β + β 1+α−β ω1 = lim sup max
|x|→∞
(3.13)
and
ω2 = lim sup max |f (t, x)| >
|x|→∞
t∈[0,1]
[(
4η
1
)
]
1+α 1+α . (3.14)
1 + 1+α−β
(1
−
η)
+
β
1+α−β Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C [0, 1] .
Proof. Let M and N be as in the proof of Theorem 3.1. To show Theorem 3.5,
1
1
we only need to prove that M < and N > .
2
2
Suppose that condition (1) holds. By using Hölder inequality, we get
) (∫ 1
(
) 1q
) n1 (∫ 1
1+α q
n
M ≤ 1 + (1 − s) ds
k (s)ds
1 + α − β
0
0
(∫
1
+
α
+ β
1 + α − β
) n1 (∫
η
n
.
0
) n1 [(
|1 + α|
k (s)ds
1+
|1 + α − β|
0
(∫ η ) 1q ]
1+α
ds
.
+|β
|
1+α−β
0
1
ds
k (s)ds
0
Therefore we have
(∫
M ≤
) 1q
η
) (∫
1
(1 − s) ds
n
) 1q
q
0
Using (3.7) we obtain
1
(1
2
M<
[
× (1 +
(1 +
|1+α|
)
|1+α+β|
1
+ q) q
1
1+α
+ |β 1+α−β
| (η(1 + q)) q
]
1
1 1q
|1 + α|
1+α
1
)(
) + |β
|η q = .
|1 + α − β| 1 + q
1+α−β
2
On the other hand, we have
) ∫ 1
∫ η
]
(
1+α 1
+
α
h(s)ds
N = 1+
(1 − s) h(s)ds + β
1 + α − β 0
1 + α − β 0
] ∫ η
[(
)
|1 + α|
1 + α ≥
1+
h(s)ds.
(1 − η) + β
|1 + α − β|
1 + α − β 0
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
253
Using the reverse Hölder inequality and applying (3.8), we get
]
)
[(
|1 + α|
1
+
α
N ≥
1+
(1 − η) + β
|1 + α − β|
1 + α − β
[( ∫
) (∫
) ]
r+1
η
×
1
[(
1+
=
1/r+1
η
ds
0
|1 + α|
|1 + α − β|
−1/r
h
)
−r
(s)ds
0
]
(∫
r+1
1
+
α
η
(1 − η) + β
1 + α − β
η
−1/r
h
)−r
(s)ds
0
1
> .
2
Suppose that condition (2) holds. Taking into account (3.9) it yields


(µ + 1) (µ + 2)
1

M <
2 1 + 1+α + β(1+α) (µ + 2) η µ+1
1+α−β
1+α−β
∫
)∫ 1
]
[(
1 + α η µ
|1 + α|
µ
s ds
(1 − s) s ds + β
× 1+
|1 + α − β|
1 + α − β 0
0


(µ + 1) (µ + 2)
1

(
)
=
2 1 + (1+α) + β 1+α (µ + 2) η µ+1
1+α−β
1+α−β
)
]
[(
1
1
|1 + α|
1 + α η µ+1
+ β
= .
× 1+
|1 + α − β| (µ + 1) (µ + 2)
1 + α − β (µ + 1)
2
On the other hand using (3.10) we obtain
] ∫ η
[
(
) |1 + α|
1 + α N > (1 − η) 1 +
+ β
h(s)ds
|1 + α − β|
1 + α − β 0
]
) |1 + α|
1
+
α
+ β
|1 + α − β|
1 + α − β
∫ η
(τ + 1)
τ
]
)
× [(
1+α τ +1 0 s ds
|1+α|
2 1 + |1+α−β| (1 − η) + β 1+α−β η
[
(
> (1 − η) 1 +
=
1
.
2
Suppose that condition (3) holds. Using the same reasoning as in the proof of
the second statement we obtain
] ∫ 1
[(
) 1 + α |1 + α|
k(s)ds
M <
1+
+ β
|1 + α − β|
1 + α − β 0
]
[(
) |1 + α|
1
+
α
<
1+
+ β
|1 + α − β|
1 + α − β
[
]
1
|1 + α − β|
1
2
×
= .
|(1 + α)| + |1 + α − β| + |β (1 + α)|
2
254
We have
A. GUEZANE-LAKOUD, S. KELAIAIA
] ∫ η
[
(
) |1 + α|
1 + α N > (1 − η) 1 +
+ β
h(s)ds
|1 + α − β|
1 + α − β 0
]
[
(
) 1
+
α
|1 + α|
> (1 − η) 1 +
+ β
|1 + α − β|
1 + α − β
1
1
] η = .
)
× [(
2
2η 1 + |1+α| (1 − η) + β 1+α |1+α−β|
1+α−β
|f (t, x)|
we deduce
|x|→∞
t∈[0,1]
|x|p
that there exists c1 > 0 such that for |x| > c1 , we get |f (t, x)| ≤ (ω1 + ε) |x|p , ∀ε >
0. For ε = ω1 then |f (t, x)| ≤ 2ω1 |x|p . From ω2 = lim sup max |f (t, x)| , we
Suppose that condition (1) holds. From ω1 = lim sup max
|x|→∞
t∈[0,1]
deduce that there exists c2 > 0 such that for |x| > c2 we have |f (t, x)| ≤ ω2 + ε.
Choosing ε = ω2 , we have |f (t, x)| ≤ 2ω2 , ∀x ∈ R\ ]−c2 , c2 [ and consequently
|f (t, x)| ≤ 2ω1 |x|p + 2ω2 , ∀x ∈ R\ ]−c, c[ ,
where c = max(c1 , c2 ). Setting
R = sup {|f (t, x)| : (t, x) ∈ [0, 1] × (−c, c)} ,
we get, ∀ (t, x) ∈ [0, 1] × R, |f (t, x)| ≤ 2ω1 |x|p + 2ω2 + R = k(t) |x| + h(t) where
k(t) = 2ω1 and h(t) = 2ω2 + R. Using (3.13) we obtain
|1 + α − β|
|(1 + α)| + |1 + α − β| + |β (1 + α)|
1
2
k (t) = 2ω1 <
then from (3.11) we get M < 1/2. Using (3.12), we arriv to
1
]
)
[(
h(t) = 2ω2 + R ≥ 2ω2 >
1+α ,
|1+α|
(1 − η) + β 1+α−β
2η 1 + |1+α−β|
then N > 1/2. Applying the third statement we achieve the proof of Theorem
3.5.
Corollary 3.6. Under the conditions of Theorem 3.1 and if one of the following
conditions is satisfied:
(1) There exist n > 1 and r > 0 such that
(∫ 1
) n1
1 1
1
n
) , ( + = 1),
k (s)ds
< (
(3.15)
1+α |1+α|
n q
0
2 1 + |1+α−β|
+ β 1+α−β
(∫
η
− 1r
h
)−r
>
(s)ds
0
or
(∫
η
− 1r
h
0
)−r
(s)ds
>
1
1+α r+1
2 β 1+α−β η
(
2 1+
1
)
|1+α|
|1+α−β|
(1 − η) η r+1
(3.16)
.
(3.17)
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
255
(2) There exist constant µ > −1 and τ > −1 such that
k(s) <
1+
1
(µ + 1)
2 ) µ
(
(1+α) 1+α s
1+α−β + β 1+α−β and
h(s) >
or, h(s) >
(3.18)
(τ + 1)
τ
1+α s
τ
+1
2η
β 1+α−β (3.19)
(τ + 1)
) sτ ,
(
1+α
2η τ +1 (1 − η) 1 + 1+α−β (3.20)








1
(µ
+
1)
µ
2
meas s ∈ [0, 1] , k(s) <
> 0
(1 + α) 1+α s 





1+
+ β 1+α−β 1 + α − β




1
τ
[(
)
]
and meas s ∈ [0, 1] , h(s) >
> 0.
s

2η τ +1 1 + |1+α| (1 − η) + β 1+α 
|1+α−β|
1+α−β
(3) The function k and h satisfy
1
2 (
)
1+α ,
(1+α) 1 + 1+α−β + β 1+α−β (3.21)
1
1
)
,
or h(s) > 1+α 1+α 2 1 + 1+α−β (1 − η)
2 β 1+α−β
(3.22)
k(s) <
h(s) >
(
{
}
|1 + α − β|
meas s ∈ [0, 1] , k(s) <
>0
|(1 + α)| + |1 + α − β| + |β (1 + α)|




1
] > 0.
[(
)
and meas s ∈ [0, 1] , h(s) >

2η 1 + |1+α| (1 − η) + β 1+α 
1
2
|1+α−β|
1+α−β
(4)The function f satisfies
1
|f (t, x)|
4 ω1 = lim sup max
<
p
(1+α) 1+α
|x|→∞
t∈[0,1]
|x|
1 + 1+α−β + β 1+α−β (3.23)
1
)
1+α 4 1 + 1+α−β (1 − η)
(3.24)
1
1+α .
4 β 1+α−β (3.25)
and
ω2 = lim sup max |f (t, x)| >
|x|→∞
t∈[0,1]
(
or
ω2 = lim sup max |f (t, x)| >
|x|→∞
t∈[0,1]
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C [0, 1] .
256
A. GUEZANE-LAKOUD, S. KELAIAIA
Proof. Suppose that condition (1) holds. We prove that N > 12 . We have
) ∫ 1
∫
(
]
1+α 1 + α η
N = 1+
(1 − s) h(s)ds + β
h(s)ds
1 + α − β 0
1 + α − β 0
∫ η
)
∫ η
|1 + α|
1
+
α
(1 − η)
h(s)ds + β
h(s)ds
≥
1+
|1 + α − β|
1 + α − β 0
0
∫ η
1 + α ≥ β
h(s)ds.
1 + α − β 0
(
Using the reverse Hölder inequality and applying (3.16), we get
∫ η
1
+
α
N ≥ β
h(s)ds
1 + α − β 0
[( ∫ η
)r+1 (∫ η
)−r ]
1
+
α
11/r+1 ds
h−1/r (s)ds
> β
1 + α − β
0
0
(∫ η
)−r
1 + α r+1
1
−1/r
= β
η
h
(s)ds
>
.
1 + α − β
2
0
By the same reasoning we prove that if (3.17) yields then N > 21 .
Suppose that condition (2) holds. Taking into account (3.18), we obtain


1
(µ + 1)

M< 
2 1 + 1+α + β(1+α) 1+α−β
(∫
1
×
) [(
s ds
1+
µ
0
1+α−β
|1 + α|
|1 + α − β|
)

=
]
1 + α + β
1 + α − β

1
2 1 + (µ + 1)
(
) 
1+α (1+α) + β 1+α−β 1+α−β [(
1+
)
1
×
µ+1
|1 + α|
|1 + α − β|
+ β
]
1 + α 1
= .
1+α−β
2
On the other hand, using (3.19) we obtain
∫ η
1 + α (τ + 1)
1
τ
N > β
s
ds
=
.
1 + α − β 2η τ +1 β 1+α 0
2
1+α−β
The other proofs follow similarly as in Theorem 3.5.
Theorem 3.7. Under the conditions of Theorem 3.2 (p = 1) and if one of the
following conditions is satisfied:
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
257
(1) There exists a constant n > 1 such that
1
(∫ 1
) n1
(1 + q) q
1 1
n
, ( + = 1).
k (s)ds
<
1
(1+α) 1+α
n q
0
1 + 1+α−β + β 1+α−β (η(1 + q)) q
(2) There exists a constant µ > −1 such that
k(s) <
and




meas



(µ + 1) (µ + 2)
(
)
sµ
(1+α) 1+α 1 + 1+α−β + β 1+α−β (µ + 2) η µ+1
s ∈ [0, 1] , k(s) <




(µ + 1) (µ + 2)
sµ > 0.
(1 + α) 1+α 
+ β

(µ + 2) η µ+1 
1 + 1+α−β 1+α−β
(3)The function k satisfies
k(s) <
and
|1 + α − β|
|(1 + α)| + |1 + α − β| + |β (1 + α)| η
{
|1 + α − β|
meas s ∈ [0, 1] , k(s) <
|(1 + α)| + |1 + α − β| + |β (1 + α)| η
(4) The function f satisfies
f (t, x) ω = lim sup max |x|→∞
t∈[0,1]
x (
< 1/2
}
> 0.
|1 + α − β|
|(1 + α)| + |1 + α − β| + |β (1 + α)| η
)
Then the BVP (1)-(2) has at least one nontrivial solution u∗ ∈ C [0, 1] .
Proof. The proof is the same as the one in theorem 3.5, indeed M <
1
2
< 1.
Theorem 3.8. Under the conditions of Theorem 3.3 (1 < p ≤ 2) and if one of
the following conditions is satisfied:
(1) There exist two constants n, r > 1 such that
1
(∫ 1
) n1
1
(1 + q) q
1 1
n
2
+ = 1). (3.26)
k (s)ds
<
,
(
1
1+α |1+α|
n
q
q
0
1 + |1+α−β| + β 1+α−β (η(1 + q))
and
(∫
) r1
1
r
h (s)ds
0
1
2
1
(1 + l) l
1 1
<
, ( + = 1).
1
(1+α) 1+α l
1 + 1+α−β + β 1+α−β (η(1 + l)) l r
(3.27)
258
A. GUEZANE-LAKOUD, S. KELAIAIA
(2) There exist two constants µ, τ > −1 such that
1
(µ + 1) (µ + 2)
2 (
)
k(s) <
sµ ,
(1+α) 1+α µ+1
1 + 1+α−β + β 1+α−β (µ + 2) η
(3.28)
1
(τ + 1) (τ + 2)
2 (
)
sτ ,
(1+α) 1+α τ
+1
1 + 1+α−β + β 1+α−β (τ + 2) η
(3.29)
h(s) <








1
(µ
+
1)
(µ
+
2)
µ
2
s
meas s ∈ [0, 1] , k(s) <
>0
(1 + α) 1+α 



µ+1


1+
+ β 1+α−β (µ + 2) η
1 + α − β




1
(τ + 1) (τ + 2)
τ
2 (
)
and meas s ∈ [0, 1] , h(s) <
s
> 0.
(1+α) 1+α 
1 + 1+α−β
+ β 1+α−β
(τ + 2) η τ +1 
(3)The functions k and h satisfy
|1 + α − β|
,
|(1 + α)| + |1 + α − β| + |β (1 + α)|
1
|1 + α − β|
2
h(s) <
,
|(1 + α)| + |1 + α − β| + |β (1 + α)|
1
2
k(s) <
{
}
|1 + α − β|
meas s ∈ [0, 1] , k(s) <
>0
|(1 + α)| + |1 + α − β| + |β (1 + α)|
{
}
1
|1 + α − β|
2
and meas s ∈ [0, 1] , h(s) <
> 0.
|(1 + α)| + |1 + α − β| + |β (1 + α)|
1
2
Then the BVP (1)-(2) has at least one nontrivial solution u∗ ∈ C [0, 1] .
Proof. It suffices to prove that M < 1/2 and N < 1/2. The proof of M < 1/2
is the same as in Theorem 3.5. Change the role of k and h in Theorem 3.5, we
prove that N < 1/2.
Remark 3.9. It will be interesting if we can formulate in this case (1 < p ≤ 2)
similar statement as the last statement of Theorem 3.5.
Corollary 3.10. Under the conditions of Theorem 3.3 and if one of the following
conditions is satisfied:
(1) There exist two constants n, r > 1 such that
(∫ 1
) n1
1 1
1
n
) , ( + = 1),
(3.30)
k (s)ds
< (
1+α |1+α|
n q
0
+ β 1+α−β
2 1 + |1+α−β|
and
(∫
) 1r
1
hr (s)ds
0
<
(
2 1+
1
|1+α|
|1+α−β|
1 1
)
1+α , ( r + l = 1).
+ β 1+α−β (3.31)
SOLVABILITY OF A NONLINEAR BOUNDARY VALUE PROBLEM
259
(2) There exist two constants µ, τ > −1 such that
k(s) <
1
(µ + 1)
2 ) µ
(
(1+α) 1+α s ,
1 + 1+α−β + β 1+α−β (3.32)
h(s) <
1
(τ + 1)
2 ) τ
(
(1+α) 1+α s ,
1 + 1+α−β + β 1+α−β (3.33)








1
(µ + 1)
2
sµ
meas s ∈ [0, 1] , k(s) <
> 0

(1 + α) 1+α 




+ β 1+α−β 1+
1 + α − β




1
(τ + 1)
τ
2
and meas s ∈ [0, 1] , h(s) <
>0
s

1 + (1+α) + β 1+α 
1+α−β
1+α−β
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C [0, 1] .
4. Examples
In order to illustrate our results, we give some examples.
Example 4.1. Consider the three point BVP
{
√
√
4 +u2
t
0 < t < 1,
u′′ + t2 |u| cos u − t4 u4u+u
2 +3 sin u + te = 0,
(4.1)
(
)
u (0) = 12 u′ (0) , u (1) = 12 u′ 12 .
√
√
1
4
2
We have f (t, x) = t2 |x| cos x − t4 x4x+x+x2 +3 sin x + tet , so |f (t, x)| ≤ k(t)|x| 2 +
h(t), p = 12 < 1, where k(s) = (s2 + s4 ) , h(s) = ses , k, h ≥ 0. Since M =
∫
∫1
∫1
5 1
(1 − s) (s2 + s4 ) ds+ 34 02 (s2 + s4 ) ds = 0.327 6 < 1/2 and N = 25 0 (1 − s) ses ds+
2 0
∫1 s
3
2
se ds = 0.836 02 > 1/2, then from Theorem 3.1, we have that the BVP (4.1)
4 0
has at least one nontrivial solution u∗ in C[0, 1].
Example 4.2. Consider the three point BVP

(√
)
1
2 3

t
 ′′ u 3
+ arcsin t + 2 cos et + 4 sin t = 0, 0 < t < 1,
u +
(4.2)
3
2

(
)

u (0) = − 21 u′ (0) , u (1) = − 12 u′ 31 .
)
(
2 3
( 1) √
t
We have f (t, x) = x33
+ arcsin t + 2 cos et + 4 sin t. So |f (t, x)| ≤
2
)
(√
2 3
1
1
t
1
+ arcsin t |x| 3 + 2 cos et + 4 sin t = k(t) |x| 3 + h(t), 0 < p = 31 < 1. For
3
2
(∫
) 12
1
1 2
(1+q) q
1
1
1
n = 3 < 1, r = 2 , we get 0 k (s)ds = 0.256 93 < 2
1 =
|1+α|
1+α
1+ |1+α−β| +(β 1+α−β
)(η(1+q) q
260
0.494 87, and
A. GUEZANE-LAKOUD, S. KELAIAIA
(∫ 1
3
0
)− 12
= 2.3781 >
h (s)ds
−2
1
|1+α|
1+α
2(1+ |1+α−β| (1−η)+|β 1+α−β
|)ηr+1
= 1.6409.
Then condition (1) in Theorem 3.5 is satisfied and the BVP (4.2) has at least one
nontrivial solution u∗ in C[0, 1].
Example 4.3. Consider the three point BVP

7
u2
cos et (1 − sin t)
 ′′
u + √
+
= 0, 0 < t < 1,
2 3
2)
3
2
t
+
1
(1
+
u
(
)

u (0) = −2u′ (0) , u (1) = 2u′ 31 .
(4.3)
7
3
t
x2
t)
2 + h(t),
We get f (t, x) = √
.
So
|f
(t,
x)|
≤
k(t)
|x|
+ cos e (1−sin
2 3
3
2 t + 1 (1 + x2 )
t
1
t)
1 < p = 23 ≤ 2 where k(t) = √
, h(t) = cos e (1−sin
∈ L1 [0, 1]. Since
2 3
3
4 t +1
)2
(
∫1 2
∫1 3
1
k
(s)ds
=
0.05222
<
=
0.062
5
and
h (s)ds =
|1+α|
1+α
0
0
2(1+ |1+α−β| +|β 1+α−β |)
(
)3
1
0.0126 <
= 0.0156, the condition (1) of corollary 3.10 is
|1+α|
1+α
2(1+ |1+α−β| +|β 1+α−β
|)
satisfied and the BVP (4.3) has at least one nontrivial solution u∗ in C[0, 1].
Acknowledgements: The authors are thankful to the referee for his valuable
comments and suggesting several changes, which have improved this paper.
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1,2
Laboratory of Advanced Materials, Faculty of Sciences, Badji Mokhtar
University, P. O. Box 12, Annaba, Algeria.
E-mail address: a guezane@yahoo.fr
E-mail address: kelaiaiasmail@yahoo.fr