1 Introduction
In the paper we are interested in the existence of positive solutions for the periodic boundary value problem (PBVP)
{
x
″
(
t
)
+
h
(
t
)
x
′
(
t
)
+
f
(
t
,
x
(
t
)
,
x
′
(
t
)
)
=
0
,
t
∈
[
0
,
T
]
,
x
(
0
)
=
x
(
T
)
,
x
′
(
0
)
=
x
′
(
T
)
,
where f
:
[
0
,
T
]
×
[
0
,
∞
)
×
R
→
R
and h
:
[
0
,
T
]
→
(
0
,
∞
)
are continuous functions. Our study is motivated by current activity of many researchers in the area of theory and applications of PVBPs; see, for example, 1234 and references therein. In particular, in a recent paper 1, Chu, Fan and Torres have studied the existence of positive periodic solutions for the singular damped differential equation
x
″
(
t
)
+
h
(
t
)
x
′
(
t
)
+
a
(
t
)
x
(
t
)
=
f
(
t
,
x
(
t
)
,
x
′
(
t
)
)
by combining the properties of the Green’s function of the PBVP
{
x
″
(
t
)
+
h
(
t
)
x
′
(
t
)
+
a
(
t
)
x
(
t
)
=
0
,
t
∈
[
0
,
T
]
,
x
(
0
)
=
x
(
T
)
,
x
′
(
0
)
=
x
′
(
T
)
,
with a nonlinear alternative of Leray-Schauder type (see, for example, 5). It should be noted that a
≢
0
was the key assumption used in 1. If a
≡
0
, then PBVP (2) has nontrivial solutions, which means that the problem we are concerned with here, that is, PBVP (1), is at resonance. There are several methods to deal with the resonant PBVPs. For example, in 6, Torres studied the existence of a positive solution for the PBVP
{
x
″
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
(
0
,
2
π
)
,
x
(
0
)
=
x
(
2
π
)
,
x
′
(
0
)
=
x
′
(
2
π
)
,
by considering the equivalent problem
{
x
″
(
t
)
+
a
(
t
)
x
(
t
)
=
f
(
t
,
x
(
t
)
)
+
a
(
t
)
x
(
t
)
,
t
∈
(
0
,
2
π
)
,
x
(
0
)
=
x
(
2
π
)
,
x
′
(
0
)
=
x
′
(
2
π
)
,
via Krasnoselskii’s theorem on cone expansion and compression. Further results in this direction can be found in 7 and 8. In 9 Rachůnková, Tvrdý and Vrkoč applied the method of upper and lower solutions and topological degree arguments to establish the existence of nonnegative and nonpositive solutions for the PBVP
{
x
″
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
(
0
,
1
)
,
x
(
0
)
=
x
(
1
)
,
x
′
(
0
)
=
x
′
(
1
)
.
The same PBVP was studied by Wang, Zhang and Wang in 10. Their existence and multiplicity results on positive solutions are based on the theory of a fixed point index for A-proper semilinear operators on cones developed by Cremins 11.
The goal of our paper is to provide sufficient conditions that ensure the existence of positive solutions of (1) with the function h positive on [
0
,
T
]
. Our general result is illustrated by two examples. The method we use in the paper is to rewrite BVP (1) as a coincidence equation L
x
=
N
x
, where L is a Fredholm operator of index zero and N is a nonlinear operator, and to apply the Leggett-Williams norm-type theorem for coincidences obtained by O’Regan and Zima 12. We would like to emphasize that the idea of results of 11 and 12, as well as these of 131415, goes back to the celebrated Mawhin’s coincidence degree theory 16. For more details on this significant tool, its modifications and wide applications, we refer the reader to 171819202122 and references therein.
In this paper, for the first time, the existence theorem from 12 is used for studying the boundary value problem with the nonlinearity f depending also on the derivative. In general, the presence of
x
′
in f makes the problem much harder to handle. We point out that, to the best of our knowledge, there are only a few papers on PBVPs that discuss such a nonlinearity; we refer the reader to 15232425 for some results of that type. We also complement several results in the literature, for example, in 126 and 27. It is evident that the existence theorems for PBVP (1) can be established by the shift method used in 6, that is, one can employ the results of 1 to the periodic problem we study here. However, the conditions imposed on f in 1 are not comparable with ours.
2 Coincidence equation
For the convenience of the reader, we begin this section by providing some background on cone theory and Fredholm operators in Banach spaces.
Definition 1 A nonempty subset C, C
≠
{
0
}
, of a real Banach space X is called a cone if C is closed, convex and
(i) λ
x
∈
C
for all x
∈
C
and λ
≥
0
,
(ii) x, −
x
∈
C
implies x
=
0
.
Every cone induces a partial ordering in X as follows: for x
,
y
∈
X
, we say that
x
⪯
y
if and only if
y
−
x
∈
C
.
The following property holds for every cone in a Banach space.
Lemma 1 28For every u
∈
C
∖
{
0
}
, there exists a positive number σ
(
u
)
such that
∥
x
+
u
∥
≥
σ
(
u
)
∥
x
∥
for all x∈C.
Consider a linear mapping L
:
dom
L
⊂
X
→
Y
and a nonlinear operator N
:
X
→
Y
, where X and Y are Banach spaces. If L is a Fredholm operator of index zero, that is, ImL is closed and dim
Ker
L
=
codim
Im
L
<
∞
, then there exist continuous projections P
:
X
→
X
and Q
:
Y
→
Y
such that Im
P
=
Ker
L
and Ker
Q
=
Im
L
(see, for example, 1416). Moreover, since dim
Im
Q
=
codim
Im
L
, there exists an isomorphism J
:
Im
Q
→
Ker
L
. Denote by
L
P
the restriction of L to Ker
P
∩
dom
L
. Then LP is an isomorphism from Ker
P
∩
dom
L
to ImL and its inverse
K
P
:
Im
L
→
Ker
P
∩
dom
L
is defined.
As a result, the coincidence equation Lx=Nx is equivalent to x
=
Ψ
x
, where
Ψ
=
P
+
J
Q
N
+
K
P
(
I
−
Q
)
N
.
Let ρ
:
X
→
C
be a retraction, that is, a continuous mapping such that ρ
(
x
)
=
x
for all x∈C. Put
Ψ
ρ
=
Ψ
∘
ρ
.
Let
Ω
1
,
Ω
2
be open bounded subsets of X with
Ω
¯
1
⊂
Ω
2
and C
∩
(
Ω
¯
2
∖
Ω
1
)
≠
∅
. Assume that
1∘ L is a Fredholm operator of index zero,
2∘ Q
N
:
X
→
Y
is continuous and bounded and
K
P
(
I
−
Q
)
N
:
X
→
X
is compact on every bounded subset of X,
3∘ L
x
≠
λ
N
x
for all x
∈
C
∩
∂
Ω
2
∩
dom
L
and λ
∈
(
0
,
1
)
,
4∘ ρ maps subsets of
Ω
¯
2
into bounded subsets of C,
5∘
d
B
(
[
I
−
(
P
+
J
Q
N
)
ρ
]
|
Ker
L
,
Ker
L
∩
Ω
2
,
0
)
≠
0
, where
d
B
stands for the Brouwer degree,
6∘ there exists
u
0
∈
C
∖
{
0
}
such that ∥
x
∥
≤
σ
(
u
0
)
∥
Ψ
x
∥
for x
∈
C
(
u
0
)
∩
∂
Ω
1
, where
C
(
u
0
)
=
{
x
∈
C
:
μ
u
0
⪯
x
for some
μ
>
0
}
and σ
(
u
0
)
is such that ∥
x
+
u
0
∥
≥
σ
(
u
0
)
∥
x
∥
for every x∈C,
7∘ (
P
+
J
Q
N
)
ρ
(
∂
Ω
2
)
⊂
C
and
Ψ
ρ
(
Ω
¯
2
∖
Ω
1
)
⊂
C
.
Theorem 1 12
Under the assumptions 1∘-7∘ the equation Lx=Nx has a solution in the set C
∩
(
Ω
¯
2
∖
Ω
1
)
.
In the next section, we use Theorem 1 to prove the existence of a positive solution for PBVP (1). For applications of Theorem 1 to nonlocal boundary value problems at resonance, we refer the reader to 22, 29 and 30.
3 Periodic boundary value problem
We now provide sufficient conditions for the existence of positive solutions for PBVP (1). For convenience and ease of exposition, we make use of the following notation:
e
(
t
)
=
exp
(
−
∫
0
t
h
(
τ
)
d
τ
)
,
φ
(
t
)
=
∫
0
t
e
(
τ
)
d
τ
,
Φ
(
t
)
=
∫
0
t
φ
(
τ
)
d
τ
,
and
ψ
(
t
)
=
1
e
(
t
)
(
1
1
−
e
(
T
)
−
φ
(
t
)
φ
(
T
)
)
,
t
∈
[
0
,
T
]
.
We observe that 0
<
ψ
(
t
)
<
1
e
(
T
)
(
1
−
e
(
T
)
)
on [0,T]. Moreover, we put
k
(
t
,
s
)
=
1
T
e
(
s
)
{
φ
(
s
)
φ
(
T
)
[
φ
(
T
)
s
−
T
φ
(
t
)
+
Φ
(
T
)
]
−
Φ
(
s
)
,
0
≤
s
≤
t
≤
T
,
φ
(
s
)
φ
(
T
)
[
φ
(
T
)
(
s
−
T
)
−
T
φ
(
t
)
+
Φ
(
T
)
]
+
T
φ
(
t
)
−
Φ
(
s
)
,
0
≤
t
≤
s
≤
T
,
and
K
(
t
,
s
)
=
k
(
t
,
s
)
+
M
−
∫
0
T
k
(
t
,
τ
)
d
τ
∫
0
T
ψ
(
τ
)
d
τ
ψ
(
s
)
,
t
,
s
∈
[
0
,
T
]
,
where M is a positive constant.
We assume that
(H1) f
:
[
0
,
T
]
×
[
0
,
∞
)
×
R
→
R
and h:[0,T]→(0,∞) are continuous functions.
We also assume that there exist R
>
0
, 0
<
α
≤
β
, 0
<
M
≤
e
(
T
)
(
1
−
e
(
T
)
)
∫
0
T
ψ
(
τ
)
d
τ
α
T
, r
∈
(
0
,
R
)
, m
∈
(
0
,
1
)
, η
∈
[
0
,
T
]
and a continuous function g
:
[
0
,
T
]
→
[
0
,
∞
)
such that
(H2) f
(
t
,
x
,
y
)
>
−
α
x
+
β
|
y
|
for (
t
,
x
,
y
)
∈
[
0
,
T
]
×
[
0
,
R
]
×
[
−
R
,
R
]
,
(H3) f
(
t
,
R
,
0
)
<
0
for t
∈
[
0
,
T
]
,
(H4) f
(
0
,
x
,
R
)
=
f
(
T
,
x
,
R
)
and f
(
0
,
x
,
−
R
)
=
f
(
T
,
x
,
−
R
)
for x
∈
[
0
,
R
]
,
(H5) f
(
t
,
x
,
−
R
)
≤
h
(
t
)
R
for t∈[0,T] and x
∈
[
0
,
R
)
,
(H6) f
(
t
,
x
,
y
)
≥
g
(
t
)
(
x
+
|
y
|
)
for (
t
,
x
,
y
)
∈
[
0
,
T
]
×
(
0
,
r
]
×
[
−
r
,
r
]
,
(H7)
1
α
T
≥
K
(
t
,
s
)
≥
0
for t
,
s
∈
[
0
,
T
]
and m
∫
0
T
K
(
η
,
s
)
g
(
s
)
d
s
≥
1
.
Theorem 2 Under the assumptions (H1)-(H7), PBVP (1) has a positive solution on [0,T].
Proof Let
∥
⋅
∥
∞
denote the supremum norm in the space C
[
0
,
T
]
, that is,
∥
x
∥
∞
=
sup
t
∈
[
0
,
T
]
|
x
(
t
)
|
. Consider the Banach spaces X
=
C
1
[
0
,
T
]
with the norm ∥
x
∥
=
max
{
∥
x
∥
∞
,
∥
x
′
∥
∞
}
, and Y
=
C
[
0
,
T
]
with the norm ∥⋅∥∞.
We write problem (1) as a coincidence equation
L
x
=
N
x
,
where
L
x
(
t
)
=
−
x
″
(
t
)
−
h
(
t
)
x
′
(
t
)
,
t
∈
[
0
,
T
]
,
and
N
x
(
t
)
=
f
(
t
,
x
(
t
)
,
x
′
(
t
)
)
,
t
∈
[
0
,
T
]
,
with dom
L
=
{
x
∈
X
:
x
″
∈
C
[
0
,
T
]
,
x
(
0
)
=
x
(
T
)
,
x
′
(
0
)
=
x
′
(
T
)
}
. Then
Ker
L
=
{
x
∈
X
:
x
(
t
)
=
c
,
t
∈
[
0
,
T
]
,
c
∈
R
}
and
Im
L
=
{
y
∈
Y
:
∫
0
T
ψ
(
s
)
y
(
s
)
d
s
=
0
}
,
where ψ is given by (5).
Clearly, ImL is closed and Y
=
Y
1
+
Im
L
with
Y
1
=
{
y
1
∈
Y
:
y
1
=
1
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
y
(
s
)
d
s
,
y
∈
Y
}
.
Since
Y
1
∩
Im
L
=
{
0
}
, we have Y
=
Y
1
⊕
Im
L
. Moreover, dim
Y
1
=
1
, which gives codim
Im
L
=
1
. Consequently, L is Fredholm of index zero, and the assumption 1∘ is satisfied.
Define the projections P:X→X by
P
x
(
t
)
=
1
T
∫
0
T
x
(
s
)
d
s
,
t
∈
[
0
,
T
]
,
and Q:Y→Y by
Q
y
(
t
)
=
1
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
y
(
s
)
d
s
,
t
∈
[
0
,
T
]
.
It is a routine matter to show that for y
∈
Im
L
, the inverse
K
P
of LP is given by
(
K
P
y
)
(
t
)
=
∫
0
T
k
(
t
,
s
)
y
(
s
)
d
s
,
t
∈
[
0
,
T
]
,
with the kernel k defined by (6). Clearly, the assumption 2∘ is satisfied. For y
∈
Im
Q
, define
J
(
y
)
=
M
y
.
Then J is an isomorphism from ImQ to KerL. Next, consider a cone
C
=
{
x
∈
X
:
x
(
t
)
≥
0
on
[
0
,
T
]
}
.
For
u
0
(
t
)
≡
1
, we have σ
(
u
0
)
=
1
and
C
(
u
0
)
=
{
x
∈
C
:
x
(
t
)
>
0
on
[
0
,
T
]
}
.
Let
Ω
1
=
{
x
∈
X
:
∥
x
∥
<
r
,
|
x
(
t
)
|
>
m
∥
x
∥
∞
and
|
x
′
(
t
)
|
>
m
∥
x
′
∥
∞
on
[
0
,
T
]
}
,
and
Ω
2
=
{
x
∈
X
:
∥
x
∥
<
R
}
.
Obviously, Ω1 and Ω2 are open bounded subsets of X, and Ω¯1⊂Ω2.
To verify 3∘, suppose that there exist
x
0
∈
C
∩
∂
Ω
2
∩
dom
L
and
λ
0
∈
(
0
,
1
)
such that L
x
0
=
λ
0
N
x
0
. Then x
(
t
)
≥
0
on [0,T], ∥
x
0
∥
=
R
,
−
x
0
″
(
t
)
−
h
(
t
)
x
0
′
(
t
)
=
λ
0
f
(
t
,
x
0
(
t
)
,
x
0
′
(
t
)
)
,
t
∈
[
0
,
T
]
,
and
x
0
(
0
)
=
x
0
(
T
)
,
x
0
′
(
0
)
=
x
0
′
(
T
)
.
There are two cases to consider.
1. If ∥
x
0
∥
=
∥
x
0
∥
∞
, then there exists
t
0
∈
[
0
,
T
]
such that x
(
t
0
)
=
R
. For
t
0
∈
(
0
,
T
)
, we get 0
≤
−
x
″
(
t
0
)
=
λ
0
f
(
t
0
,
R
,
0
)
, contrary to the assumption (H3). Similarly, if
t
0
=
0
or
t
0
=
T
, BCs (9) imply
x
′
(
0
)
=
x
′
(
T
)
=
0
. Hence, 0
≤
−
x
″
(
t
0
)
=
λ
0
f
(
t
0
,
R
,
0
)
which contradicts (H3) again.
2. If ∥
x
0
∥
=
∥
x
0
′
∥
∞
>
∥
x
0
∥
∞
, then there exists t0∈[0,T] such that |
x
′
(
t
0
)
|
=
R
. Observe that (H2) implies f
(
t
,
x
,
±
R
)
>
0
for t∈[0,T] and x∈[0,R]. Suppose that t0∈(0,T). If
x
′
(
t
0
)
=
R
, we get from (8)
−
h
(
t
0
)
R
=
λ
0
f
(
t
0
,
x
0
(
t
0
)
,
R
)
,
a contradiction. For
x
′
(
t
0
)
=
−
R
, we have
h
(
t
0
)
R
=
λ
0
f
(
t
0
,
x
0
(
t
0
)
,
−
R
)
<
f
(
t
0
,
x
0
(
t
0
)
,
−
R
)
,
contrary to (H5). By similar arguments, if t0=0 or t0=T, BCs (9) and (H4) imply either (10) or (11). Thus, 3∘ is fulfilled.
Next, for x
∈
X
, define (see 15)
ρ
x
(
t
)
=
{
x
(
t
)
if
x
(
t
)
≥
0
on
[
0
,
T
]
,
1
2
(
x
(
t
)
−
min
{
x
(
t
)
:
t
∈
[
0
,
T
]
}
)
if
x
(
t
˜
)
<
0
for some
t
˜
∈
[
0
,
T
]
.
Clearly, ρ is a retraction and maps subsets of Ω¯2 into bounded subsets of C, so 4∘ holds.
To verify 5∘, it is enough to consider, for x
∈
Ker
L
∩
Ω
2
and λ
∈
[
0
,
1
]
, the mapping
H
(
x
,
λ
)
(
t
)
=
x
(
t
)
−
λ
(
1
T
∫
0
T
(
ρ
x
)
(
s
)
d
s
+
M
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
f
(
s
,
(
ρ
x
)
(
s
)
,
(
ρ
x
)
′
(
s
)
)
d
s
)
.
Observe that if x
∈
Ker
L
∩
Ω
2
, then x
(
t
)
=
c
on [0,T] and ∥
x
∥
<
R
. Suppose H
(
x
,
λ
)
=
0
for x
∈
∂
Ω
2
. Then c
=
±
R
. For c
=
R
, we have (
ρ
x
)
(
t
)
=
x
(
t
)
and in view of (H3), we get
0
≤
R
(
1
−
λ
)
=
λ
M
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
f
(
s
,
R
,
0
)
d
s
<
0
,
which is a contradiction. If c
=
−
R
, then (
ρ
x
)
(
t
)
=
0
, hence
−
R
=
λ
M
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
f
(
s
,
0
,
0
)
d
s
,
which contradicts (H2). Thus, H
(
x
,
λ
)
≠
0
for x
∈
∂
Ω
2
and λ
∈
[
0
,
1
]
. This implies
d
B
(
H
(
x
,
0
)
,
Ker
L
∩
Ω
2
,
0
)
=
d
B
(
H
(
x
,
1
)
,
Ker
L
∩
Ω
2
,
0
)
,
and
d
B
(
[
I
−
(
P
+
J
Q
N
)
ρ
]
|
Ker
L
,
Ker
L
∩
Ω
2
,
0
)
=
d
B
(
H
(
c
,
1
)
,
Ker
L
∩
Ω
2
,
0
)
≠
0
.
We next show that 6∘ holds. Let x
∈
C
(
u
0
)
∩
∂
Ω
1
. Then for t∈[0,T], we have r
≥
x
(
t
)
≥
m
∥
x
∥
∞
>
0
, r
≥
|
x
′
(
t
)
|
≥
∥
x
′
∥
∞
, and by (H6) and (H7), we obtain
Ψ
x
(
η
)
=
1
T
∫
0
T
x
(
s
)
d
s
+
∫
0
T
K
(
η
,
s
)
f
(
s
,
x
(
s
)
,
x
′
(
s
)
)
d
s
≥
∫
0
T
K
(
η
,
s
)
g
(
s
)
[
x
(
s
)
+
|
x
′
(
s
)
|
]
d
s
≥
m
∫
0
T
K
(
η
,
s
)
g
(
s
)
[
∥
x
∥
∞
+
∥
x
′
∥
∞
]
d
s
≥
m
∥
x
∥
∫
0
T
K
(
η
,
s
)
g
(
s
)
d
s
≥
∥
x
∥
.
This implies ∥
x
∥
≤
∥
Ψ
x
∥
for x∈C(u0)∩∂Ω1, so 6∘ is satisfied.
Finally, we must check if 7∘ holds. If x
∈
∂
Ω
2
, then in view of (H2), we get
(
P
+
J
Q
N
)
(
ρ
x
)
(
t
)
=
1
T
∫
0
T
(
ρ
x
)
(
s
)
d
s
+
M
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
f
(
s
,
(
ρ
x
)
(
s
)
,
(
ρ
x
)
′
(
s
)
)
d
s
≥
1
T
∫
0
T
(
ρ
x
)
(
s
)
d
s
+
M
∫
0
T
ψ
(
s
)
d
s
∫
0
T
ψ
(
s
)
[
−
α
(
ρ
x
)
(
s
)
+
β
|
(
ρ
x
)
′
(
s
)
|
]
d
s
≥
∫
0
T
[
1
T
−
α
M
ψ
(
s
)
∫
0
T
ψ
(
τ
)
d
τ
]
(
ρ
x
)
(
s
)
d
s
≥
∫
0
T
[
1
T
−
α
M
e
(
T
)
(
1
−
e
(
T
)
)
∫
0
T
ψ
(
τ
)
d
τ
]
(
ρ
x
)
(
s
)
d
s
≥
0
.
Moreover, for x
∈
Ω
¯
2
∖
Ω
1
, we have from (H2) and (H7)
Ψ
ρ
x
(
t
)
=
1
T
∫
0
T
(
ρ
x
)
(
s
)
d
s
+
∫
0
T
K
(
t
,
s
)
f
(
s
,
(
ρ
x
)
(
s
)
,
(
ρ
x
)
′
(
s
)
)
d
s
≥
1
T
∫
0
T
(
ρ
x
)
(
s
)
d
s
+
∫
0
T
K
(
t
,
s
)
[
−
α
(
ρ
x
)
(
s
)
+
β
|
(
ρ
x
)
′
(
s
)
|
]
d
s
≥
∫
0
T
[
1
T
−
α
K
(
t
,
s
)
]
(
ρ
x
)
(
s
)
d
s
≥
0
.
Thus, 7∘ is fulfilled and the assertion follows. □
We now give two examples illustrating Theorem 2. Some calculations have been made with Mathematica. In the first example, the function h is constant, while in the second h
(
t
)
=
1
/
(
1
+
t
)
and f is independent of t.
Example 1
Consider the following PBVP:
{
x
″
(
t
)
+
x
′
(
t
)
+
(
t
(
1
−
t
)
+
1
)
(
−
2
9
x
(
t
)
+
3
4
|
x
′
(
t
)
|
+
1
)
=
0
,
t
∈
[
0
,
1
]
,
x
(
0
)
=
x
(
1
)
,
x
′
(
0
)
=
x
′
(
1
)
.
Then e
(
t
)
=
e
−
t
, φ
(
t
)
=
1
−
e
−
t
, Φ
(
t
)
=
t
+
e
−
t
−
1
, ψ
(
t
)
=
e
e
−
1
, and
k
(
t
,
s
)
=
{
−
s
+
e
s
−
t
+
1
−
e
1
−
t
e
−
1
,
0
≤
s
≤
t
≤
1
,
−
s
+
1
+
e
s
−
t
−
e
1
−
t
e
−
1
,
0
≤
t
≤
s
≤
1
.
Moreover, (7) with M
=
3
2
reads
K
(
t
,
s
)
=
{
t
−
s
+
e
s
−
t
+
1
e
−
1
,
0
≤
s
≤
t
≤
1
,
t
−
s
+
1
+
e
s
−
t
e
−
1
,
0
≤
t
≤
s
≤
1
,
and the assumptions (H2)-(H7) are met with R
=
20
, α
=
2
9
, β
=
3
4
, r
=
36
53
, m
∈
[
12
(
e
−
1
)
17
+
7
e
,
1
)
, η
=
0
and g
(
t
)
=
t
(
1
−
t
)
+
1
. By Theorem 2, problem (12) has a positive solution.
Example 2
Consider the PBVP
{
x
″
(
t
)
+
1
1
+
t
x
′
(
t
)
+
1
10
−
1
9
x
(
t
)
+
(
x
′
(
t
)
)
4
/
5
=
0
,
t
∈
[
0
,
1
2
]
,
x
(
0
)
=
x
(
1
2
)
,
x
′
(
0
)
=
x
′
(
1
2
)
.
In this case, we have e
(
t
)
=
1
1
+
t
, φ
(
t
)
=
ln
(
1
+
t
)
, Φ
(
t
)
=
−
t
+
ln
(
1
+
t
)
+
t
ln
(
1
+
t
)
and
ψ
(
t
)
=
(
1
+
t
)
(
3
−
ln
(
1
+
t
)
ln
(
3
2
)
)
.
The assumptions of Theorem 2 are fulfilled with M
=
1
, R
=
10
, α
=
1
3
, β
=
1
2
, r
=
1
100
, m
=
0.9
, η
=
1
4
and g
(
t
)
=
3
.