Draft version November 26, 2015
Novi Sad J. Math.
Vol. XX, No. Y, 20ZZ, ??-??
Existence and non-existence of positive solutions of
four-point BVPs for second order ordinary differential
equations on whole line1
Yuji Liu2 and Xiaohui Yang3
Abstract.
This paper is concerned with four-point boundary value
problems of second order singular differential equations on whole lines.
The Green’s function G(t, s) for the problem
−(ρ(t)x′ (t))′ = 0, lim ρ(t)x′ (t) − kx(ξ) = lim ρ(t)x′ (t) + lx(η) = 0
t→−∞
t→+∞
is obtained. We proved that G(t, s) ≥ 0 under some assumptions which
actually generalizes a corresponding result in [Appl. Math. Comput.
217(2)(2010) 811-819]. Sufficient conditions to guarantee the existence
and non-existence of positive solutions are established.
AMS Mathematics Subject Classification (2010): 34B10, 34B15, 35B10
Key words and phrases: Second order singular differential equation on
whole line, four-point boundary value problem, positive solution, fixed
point theorem
1.
Introduction
Nonlocal boundary value problems for ordinary differential equations was
initiated by Il’in and Moiseev [14]. Since then, more general nonlocal boundary
value problems (BVPs) were studied by several authors, see the text books
[1, 11, 13], the papers [21], and the survey papers [16, 17] and the references
cited there. However, the study on existence of positive solutions of nonlocal
boundary value problems for differential equations on whole real lines does not
seem to be sufficiently developed [2, 4, 3, 5, 6, 22] and the references therein.
In recent years, the existence of solutions of boundary value problems of
the differential equations governed by nonlinear differential operators has been
studied by many authors, see [9, 10, 18, 19, 6, 8, 7, 15] and the references
therein.
In [12], Deren and Hamal investigated the existence and multiplicity of
nonnegative solutions for the following integral boundary-value problem on the
1 Supported by the Natural Science Foundation of Guangdong province (No.
S2011010001900) and Natural science research project for colleges and universities of Guangdong Province (No: 2014KTSCX126)
2 Department of Mathematics, Guangdong University of Fiance and Economics, 510320,
P R China, e-mail: liuyuji888@sohu.com
3 Department of Computer, Guangdong Police College, 510230, P R China, e-mail:
xiaohui− yang@sohu.com
Yuji Liu, Xiaohui Yang
whole line

(p(t)x′ (t))′ + λq(t)f (t, x(t), x′ (t)) = 0, a.e., t ∈ R,





∫ +∞

a1 lim x(t) − b1 lim p(t)x′ (t) = −∞ g1 (s, x(s), x′ (s))ψ(s)ds,
(1.1)
t→−∞
t→−∞





 a2 lim x(t) + b2 lim p(t)x′ (t) = ∫ ∞ g2 (s, x(s), x′ (s))ψ(s)ds,
−∞
t→+∞
t→+∞
where λ > 0 is a parameter, f, g1 , g2 ∈ C(R, ×[0, ∞) × R, [0, ∞)), q, ψ ∈
C(R, (0, ∞)) and p ∈ C(R, (0, ∞)) ∩ C 1 (R). Here, the values of
∫ +∞
∫ +∞ ds
< +∞
g (s, x(s), x′ (s))ds, (i = 1, 2), −∞ p(s)
−∞ i
and sup ψ(s) are finite and a1 + a2 > 0, bi > 0 (i = 1, 2) satisfying D =
s∈R
∫ +∞ ds
> 0.
a2 b1 + a1 b2 + a1 a2 −∞ p(s)
Motivated by mentioned papers, we consider the following four-point boundary value problem for second order singular differential equation on the whole
line

[ρ(t)x′ (t)]′ + f (t, x(t), x′ (t)) = 0, a.e., t ∈ R,






lim ρ(t)x′ (t) − kx(ξ) = 0,
(1.2)
t→−∞





 lim ρ(t)x′ (t) + lx(η) = 0
t→+∞
where
(a) −∞ < ξ < η < +∞, k, l ≥ 0 are constants,
(b) f is a nonnegative Carathéodory function, see Definition 2.3,
(c) ρ ∈ C 0 (R, [0, ∞)) with ρ(t) > 0 for all t ̸= 0 satisfying
∫ξ
(1.3)
1−k
(1.4)
1−l
(1.5)
∆ = k + l + kl
du
−∞ ρ(u)
∫ +∞
η
du
ρ(u)
∫ +∞
≥ 0,
≥ 0,
du
−∞ ρ(u)
> 0.
The purpose is to establish sufficient conditions for the existence and nonexistence of positive solutions of BVP(1.2). Our results and methods are different from those in [2, 3, 4, 6, 7, 8, 22].
The main features of our paper are as follows. Firstly, compared with [12],
we establish the existence results of solutions of second order singular differential equation on the whole line. Secondly, we investigate the existence of
positive solutions by a different method and imposing ∫growth conditions on
+∞ 1
f . Thirdly, compared with [15], we consider the case −∞ ρ(s)
ds < +∞ in
Draft version November 26, 2015
2
Four-point boundary value problem
3
∫ +∞ 1
this paper while −∞ ρ(s)
ds = +∞ considered in [15]. Finally, the Green’s
function G(t, s) for the problem −(ρ(t)x′ (t))′ = 0, lim ρ(t)x′ (t) − kx(ξ) =
t→−∞
0, lim ρ(t)x′ (t) − lx(η) = 0 is obtained. We proved that G(t, s) ≥ 0 un-
Draft version November 26, 2015
t→+∞
der some assumptions which actually generalizes a corresponding result in
([23]Appl. Math. Comput. 217(2)(2010) 811-819), see Remark 2.1.
The remainder of this paper is organized as follows: the preliminary results
are given in Section 2, the existence result of positive solutions of BVP(1.2) is
proved in Section 3. Finally the non-existence results on positive solutions of
BVP(1.2) are presented in Section 4.
2.
Preliminary Results
In this section, we present some background definitions in Banach spaces
see [11] and state an important fixed point theorem see Theorem 2.2.11 in [13].
The preliminary results are given too.
Definition 2.1. Let X be a real Banach space. The nonempty convex
closed subset P of X is called a cone in X if ax ∈ P for all x ∈ P and a ≥ 0
and x ∈ X and −x ∈ X imply x = 0.
Definition 2.2. An operator T : X 7→ X is completely continuous if it is
continuous and maps bounded sets into relatively compact sets.
Definition 2.3. F is called a Carathédory function, that is
)
(
1
y is defined almost every where on R and is measurable
(i) t 7→ f t, x, ρ(t)
on R for any x, y ∈ R,
(
)
1
(ii) (x, y) 7→ f t, x, ρ(t)
y is uniformly continuous on R2 for all most every
t ∈ R, i.e., for each ϵ > 0 there exists δ > 0 such that
(
)
(
)
1
1
y1 − f t, x2 , ρ(t)
y2 < ϵ
f t, x1 , ρ(t)
for almost every t ∈ R, |x1 − x2 | < δ, |y1 − y2 | < δ,
(iii) for each r > 0, there exists nonnegative function ϕr ∈ L1 (R) such that
|u|, |v| ≤ r implies
(
)
f t, x, 1 y ≤ ϕr (t), a.e.t ∈ R.
ρ(t) Lemma 2.1([11, 13]. Let X be a real Banach space, P be a cone of X,
Ω1 , Ω2 be two nonempty bounded open sets of P with 0 ∈ Ω1 ⊆ Ω1 ⊆ Ω2 .
Suppose that T : Ω2 → P is a completely continuous operator, and
(E1) T x ̸= λx for all λ ∈ [0, 1) and x ∈ ∂Ω1 , T x ̸= λx for all λ ∈ (1, +∞) and
x ∈ ∂Ω2 ;
or
4
Yuji Liu, Xiaohui Yang
(E2) T x ̸= λx for all λ ∈ (1, +∞) and x ∈ ∂Ω1 , T x ̸= λx for all λ ∈ [0, 1) and
x ∈ ∂Ω2 .
Then T has at least one fixed points x ∈ Ω2 \ Ω1 .

x ∈ C 0 (R), ρx′ ∈ C 0 (R), lim x(t)


t→±∞
X = x : R 7→ R :


and lim ρ(t)x′ (t) exist and are finite





.
t→±∞
For x ∈ X, define
{
}
||x|| = max sup |x(t)|, sup ρ(t)|x′ (t)| .
t∈R
t∈R
Lemma 2.2. X is a Banach space with || · || defined.
Proof. It is easy to see that X is a normed linear space. Let {xu } be a
Cauchy sequence in X. Then ||xu − xv || → 0, u, v → +∞. It follows that
lim xu (t),
t→−∞
lim xu (t),
t→+∞
lim ρ(t)x′u (t),
t→−∞
lim ρ(t)x′u (t) exist,
t→+∞
sup |xu (t) − xv (t)| → 0, u, v → +∞,
t∈R
sup ρ(t)|x′u (t) − x′v (t)|, u, v → +∞.
t∈R
Thus there exists two functions x0 , y0 defined on R such that
lim xu (t) = x0 (t),
u→+∞
It follows that
lim ρ(t)x′u (t) = y0 (t).
u→+∞
sup |xu (t) − x0 (t)| → 0, u → +∞,
t∈R
sup |ρ(t)x′u (t) − y0 (t)| → 0, u → +∞.
t∈R
This means that functions x0 , y0 : R 7→ R are well defined.
Step 1. Prove that x0 , y0 ∈ C(R).
We have for t0 ∈ R that
|x0 (t) − x0 (t0 )| ≤ |x0 (t) − xN (t)| + |xN (t) − xN (t0 )| + |xN (t0 ) − x0 (t0 )|
≤ 2 sup |xN (t) − x0 (t)| + |xN (t) − xN (t0 )| .
t∈R
Since sup |xu (t) − x0 (t)| → 0, u → +∞ and xu (t) is continuous on R, then for
t∈R
any ϵ > 0 we can choose N and δ > 0 such that sup |xN (t) − x0 (t)| < ϵ and
t∈R
Draft version November 26, 2015
Choose
Four-point boundary value problem
5
|xN (t) − xN (t0 )| < ϵ for all |t − t0 | < δ. Thus |x0 (t) − x0 (t0 )| < 3ϵ foe all
|t − t0 | < δ. So x0 ∈ C(R). Similarly we can prove that y0 ∈ C(R).
Step 2. Prove that the limits lim x0 (t), lim x0 (t), lim y0 (t), lim y0 (t)
t→−∞
t→+∞
t→−∞
t→+∞
exist.
Suppose that lim xu (t) = A−
u is a finite real number. By sup |xu (t) −
Draft version November 26, 2015
t→−∞
t∈R
xv (t)| → 0, u, v → +∞, we know that A−
u is a Cauchy sequence.
lim A−
u exists. By sup |xu (t) − x0 (t)| → 0, u → +∞, we get that
u→+∞
Then
t∈R
lim x0 (t) = lim
t→−∞
lim xu (t) = lim
t→−∞ u→+∞
lim xu (t) = lim A−
u.
u→+∞ t→−∞
u→+∞
Hence lim x0 (t) exists. Similarly we can prove that lim x0 (t), lim y0 (t),
t→−∞
t→+∞
t→−∞
lim y0 (t) exist.
t→+∞
Step 3. Prove that y0 (t) = ρ(t)x′0 (t).
We have
∫
∫
∫
xu (t) − lim xu (t) − t y0 (s) ds = t x′u (s)ds − t
ρ(s)
−∞
−∞
−∞
t→−∞
≤
≤
So lim
′
(s) −
−∞ xu
∫t
y0 (s) ρ(s) ds
≤
∫ +∞
u→+∞
∫t
1
ds sup |ρ(t)x′u (t)
−∞ ρ(t)
t∈R
1
ds sup |ρ(t)x′u (t)
−∞ ρ(t)
t∈R
t→−∞
lim
lim xu (t) =
u→+∞ t→−∞
− y0 (t)|
− y0 (t)| → 0 as u → +∞.
(
)
∫t
xu (t) − lim xu (t) = −∞
Here c0 :=
y0 (s)
ρ(s) ds
y0 (s)
ρ(s) ds.
Then x0 (t)−c0 =
lim A−
u . It follows that
u→+∞
y0 (t)
ρ(t)
∫t
y0 (s)
ds.
−∞ ρ(s)
= x′0 (t). So
x0 ∈ X with xu → x0 as u → +∞. It follows that X is a Banach space.
Lemma 2.3. Let M be a subset of X. Then M is relatively compact if
and only if the following conditions are satisfied:
(i) both {x : x ∈ M } and {ρ(t)x′ : x ∈ M } are uniformly bounded,
(ii) both {x : x ∈ M } and {ρ(t)x′ : x ∈ M } are equicontinuous in any
subinterval [a, b] in R,
(iii) both {x : x ∈ M } and {ρ(t)x′ : x ∈ M } are equi-convergent as
t → ±∞.
Proof. ” ⇐ ”. From Lemma 2.2, we know X is a Banach space. In order
to prove that the subset M is relatively compact in X, we only need to show
M is totally bounded in X, that is for all ϵ > 0, M has a finite ϵ-net.
For any given ϵ > 0, by (i) and (iii), there exist constants Ax , Cx , T >
0, a > 0, we have
|x(t) − Ax | ≤ 3ϵ , |ρ(t)x′ (t) − Cx | < 3ϵ , t ≤ −T, x ∈ M,
|x(t1 ) − x(t2 )| ≤ 3ϵ , |ρ(t1 )x′ (t1 ) − ρ(t2 )x′ (t2 )| < 3ϵ , t1 , t2 ≥ T, x ∈ M,
|x(t1 ) − x(t2 )| ≤ 3ϵ , |ρ(t1 )x′ (t1 ) − ρ(t2 )x′ (t2 )| < 3ϵ , t1 , t2 ≤ −T, x ∈ M.
6
Yuji Liu, Xiaohui Yang
For T > 0, define X|[−T,T ] = {x : x, ρ(t)x′ ∈ C[−T, T ]} . For x ∈ X|[−T,T ] ,
define
{
}
′
||x||T = max
max |x(t)|, max ρ(t)|x (t)| .
t∈[−T,T ]
Similarly to Lemma 2.2, we can prove that X[−T,T ] is a Banach space.
Let M |[−T,T ] = {t 7→ x(t), t ∈ [−T, T ] : x ∈ M }. Then M |[−T,T ] is a
subset of X|[−T,T ] . By (i) and (ii), and Ascoli-Arzela theorem, we can know
that M |[−T,T ] is relatively compact. Thus, there exist x1 , x2 , · · · , xk ∈ M such
that, for any x ∈ M , we have that there exists some i = 1, 2, · · · , k such that
{
}
||x − xi ||T = max
sup
|x(t) − xi (t)|,
t∈[−T,T ]
sup
′
ρ(t)|x (t) −
x′i (t)|
t∈[−T,T ]
≤ 3ϵ .
Therefore, for x ∈ M , we have that
{
||x − xi ||X = max
sup
|x(t) − xi (t)|,
t∈[−T,T ]
ρ(t)|x′ (t) − x′i (t)|,
sup
t∈[−T,T ]
sup |x(t) − xi (t)|, sup ρ(t)|x′ (t) − x′i (t)|,
t≥T
t≥T
}
sup |x(t) − xi (t)|, sup ρ(t)|x′ (t) − x′i (t)|
t≤−T
t≤−T
{
≤ max
ϵ
3 , sup |x(t)
t≥T
− x(T )| + |x(T ) − xi (T )| + sup |xi (T ) − xi (t)|,
t≥T
sup |ρ(t)x′ (t) − ρ(T )x′ (T )| + |ρ(T )x′ (T ) − ρ(T )x′i (T )|
t≥T
+ sup |ρ(T )x′i (T ) − ρ(t)x′i (t)|
t≥T
sup |x(t) − x(T )| + |x(T ) − xi (T )| + sup |xi (T ) − xi (t)|,
t≤−T
t≤−T
sup |ρ(t)x′ (t) − ρ(T )x′ (T )| + |ρ(T )x′ (T ) − ρ(T )x′i (T )|
t≤−T
}
+ sup
t≤−T
|ρ(T )x′i (T )
−
ρ(t)x′i (t)|
≤ ϵ.
So, for any ϵ > 0, M has a finite ϵ-net {Ux1 , Ux2 , · · · , Uxk }, that is, M is totally
bounded in X. Hence M is relatively compact in X.
⇒. Assume that M is relatively compact, then for any ϵ > 0, there exists
a finite ϵ-net of M . Let the finite ϵ-net be {Ux1 , Ux2 , · · · , Uxk } with xi ⊂ M .
Draft version November 26, 2015
t∈[−T,T ]
Four-point boundary value problem
7
Then for any x ∈ M , there exists Uxi such that x ∈ Uxi and
{
}
|x(t)| ≤ |x(t) − xi (t)| + |xi (t)| ≤ ϵ + max sup |xi (t)| : i = 1, 2, · · · , k ,
t∈R
{
′
Draft version November 26, 2015
ρ(t)|x (t)| ≤ ϵ + max
sup ρ(t)|x′i (t)|
t∈R
}
: i = 1, 2, · · · , k .
It follows that both M and {ρ(t)x′ : x ∈ M } are uniformly bounded. Then (i)
holds. Furthermore, there exists T > 0 such that |xi (t1 ) − xi (t2 )| < ϵ for all
t1 , t2 ≥ T and all t1 , t2 ≤ −T and i = 1, 2, · · · , k. Then we have for t1 , t2 ∈ R
that
|x(t1 ) − x(t2 )| ≤ |x(t1 ) − xi (t1 )| + |xi (t1 ) − xi (t2 )| + |xi (t2 ) − x(t2 )|
≤ 3ϵ for all t1 , t2 ≥ T, t1 , t2 ≤ −T, x ∈ M.
Similarly we have that
|ρ(t1 )x′ (t1 ) − ρ(t2 )x′ (t2 )| ≤ 3ϵ for all t1 , t2 ≥ T, t1 , t2 ≤ −T, x ∈ M.
Thus (iii) is valid. Similarly we can prove that (ii) holds. Consequently,
Lemma 2.3 is proved.
Denote
G(t, s) =
1
∆

∫ ξ du
∫ ξ du ∫ η du
∫ t du

+ k u ρ(u)
+ kl u ρ(u)
1 + k ξ ρ(u)

t ρ(u) ∫

∫ η du
∫ η du ∫ t du
∫ t du

t


+l u ρ(u)
+ kl u ρ(u)
−
l

ρ(u)
ρ(u) − k u
ξ
u

∫ t du ∫ η du



−kl u ρ(u)
, s ≤ min{ξ, t},

ξ ρ(u)





∫ t du



1 + k ξ ρ(u)
, s ≥ max{η, t},






∫ ξ du
∫ ξ du ∫ η du
∫ t du



+ k u ρ(u)
+ kl u ρ(u)
1 + k ξ ρ(u)

t ρ(u)

∫ η du ∫ t du
∫ η du



 +l u ρ(u) + kl u ρ(u) ξ ρ(u) , t < s ≤ ξ < η,



































du
ρ(u)
∫ t du
∫ η du
1 + k ξ ρ(u)
+l
∫ η du ∫ t duu ρ(u)
+kl u ρ(u) ξ ρ(u) , min{t, ξ} < s ≤ η,
∫ t du
∫ η du
∫ η du ∫ t du
∫ t du
1 + k ξ ρ(u)
+ l u ρ(u)
+ kl u ρ(u)
ρ(u) − l u ρ(u)
ξ
∫ t du
∫ t du ∫ η du
−k u ρ(u)
− kl u ρ(u)
, ξ < s ≤ max{t, η},
ξ ρ(u)
∫ t du
∫ t du
∫ t du
1 + k ξ ρ(u)
− l u ρ(u)
− k u ρ(u)
∫ t du ∫ η du
−kl u ρ(u)
, ξ < η < s ≤ t.
ξ ρ(u)
Lemma 2.4. x ∈ X is a solution of BVP(1.2) if and only if
∫ +∞
(2.1)
x(t) = −∞ G(t, s)f (s, x(s), x′ (s))ds.
8
Yuji Liu, Xiaohui Yang
Proof. Since x ∈ X, f is a Caratheodory function, then
{
}
′
||x|| = max sup |x(t)|, sup ρ(t)|x (t)| = r < +∞,
t∈R
t∈R
and −∞ f (r, x(r), x′ (r))dr converges. If x is a solution of BVP(1.2), we get
from [ρ(t)x′ (t)]′ + f (t, x(t), x′ (t)) = 0 that there exist constants A, B ∈ R such
that
∫t
ρ(t)x′ (t) = A − −∞ f (s, x(s), x′ (s))ds,
(2.2)
∫t
∫ t (∫ t du )
du
x(t) = B + A −∞ ρ(u)
− −∞ s ρ(u)
f (s, x(s), x′ (s))ds.
From lim ρ(t)x′ (t) − kx(ξ) = 0 and lim ρ(t)x′ (t) + lx(η) = 0, we have
t→−∞
(
t→+∞
1−k
(
1+l
+l
∫ξ
du
−∞ ρ(u)
∫η
du
−∞ ρ(u)
(∫
∫η
−∞
)
A − kB = −k
)
η du
s ρ(u)
∫ +∞
A + lB =
)
∫ξ
−∞
(∫
−∞
ξ du
s ρ(u)
B=
(∫
∫ξ
−∞
1
∆
[(
ξ du
s ρ(u)
1−k
)
f (s, x(s), x′ (s))ds.
du
−∞ ρ(u)
du
ρ(u)
(
∫ξ
+l 1 − k −∞
du
ρ(u)
du
ρ(u)
)
f (s, x(s), x′ (s))ds
]
f (s, x(s), x′ (s))ds ,
∫ξ
(
∫η
+k 1 + l −∞
f (s, x(s), x′ (s))ds,
f (s, x(s), x′ (s))ds
It follows that
[ ∫
∫ η (∫ η
+∞
1
k −∞ f (s, x(s), x′ (s))ds + kl −∞ s
A= ∆
−kl
)
)∫
)∫
)∫
ξ
−∞
η
−∞
+∞
−∞
f (s, x(s), x′ (s))ds
(∫
)
(∫
)
ξ du
s ρ(u)
η du
s ρ(u)
f (s, x(s), x′ (s))ds
]
f (s, x(s), x′ (s))ds .
Substituting A, B into (2.2), we get that
∫ t (∫ t du )
x(t) = − −∞ s ρ(u)
f (s, x(s), x′ (s))ds
1
+∆
[(
∫ξ
1 − k −∞
du
ρ(u)
(
∫η
+k 1 + l −∞
du
ρ(u)
(
∫ξ
+l 1 − k −∞
du
ρ(u)
)∫
+∞
−∞
f (s, x(s), x′ (s))ds
)∫
(∫
)
)∫
(∫
)
ξ
−∞
η
−∞
ξ du
s ρ(u)
η du
s ρ(u)
f (s, x(s), x′ (s))ds
f (s, x(s), x′ (s))ds
]
Draft version November 26, 2015
∫ +∞
Four-point boundary value problem
[ ∫
∫ η (∫ η
+∞
1
+∆
k −∞ f (s, x(s), x′ (s))ds + kl −∞ s
−kl
Draft version November 26, 2015
=
+
+
+
=
(∫
∫ξ
1
∆
−∞
ξ du
s ρ(u)
)
f (s, x(s), x′ (s))ds
[ ∫
(
∫η
t
− −∞ k + l + kl ξ
∫ +∞ (
−∞
1+k
∫t
du
ξ ρ(u)
)
( ∫
ξ
k u
du
ρ(u)
+ kl
( ∫
η
l u
−∞
du
ρ(u)
+ kl
∫ξ
−∞
∫η
∫ +∞
−∞
du
ρ(u)
) (∫
]∫
9
du
ρ(u)
)
f (s, x(s), x′ (s))ds
t
du
−∞ ρ(u)
t du
s ρ(u)
)
f (s, x(s), x′ (s))ds
f (s, x(s), x′ (s))ds
∫ξ
∫η
)
∫η
∫t
)
du
u ρ(u)
du
u ρ(u)
du
t ρ(u)
du
ξ ρ(u)
f (s, x(s), x′ (s))ds
]
f (s, x(s), x′ (s))ds
G(t, s)f (s, x(s), x′ (s))ds.
This is just (2.1). On the other hand, if x ∈ X satisfies (2.1), it is easy to show
that x is a solution of BVP(1.2). The proof is completed.
∫ −c
Fix c > 0 such that −∞
∫ −c du ( ∫ +∞ du )−1
2 −∞ ρ(u)
. Let
−∞ ρ(u)
du
ρ(u)
< 1. For ease expression, denote µ =
{
P = x ∈ X : x(t) ≥ 0 for all t ∈ R,
Define the operator T on P by (T x)(t) =
x ∈ X.
min x(t) ≥ µ sup x(t)
t∈[−c,c]
∫ +∞
−∞
t∈R
}
.
G(t, s)f (s, x(s), x′ (s))ds for
Lemma 2.5. Suppose that (a)-(c) hold. Then
(i) T : P 7→ X is well defined,
(ii) it holds that
(2.3)

[ρ(t)(T x)′ (t)]′ + f (t, x(t), x′ (t)) = 0, a.e., t ∈ R,






lim (T x)(t) − k(T x)(ξ) = 0,
t→−∞





 lim (T x)(t) − l(T x)(η) = 0,
t→+∞
(iii) T : P 7→ P is completely continuous;
(iv) x ∈ X is a positive solution of BVP(1.2) if and only if x is a fixed
point of T in P .
10
Yuji Liu, Xiaohui Yang
Proof. (i) Since x ∈ X, f is a Carathéodory function, we know that
∫ +∞
f (s, x(s), x′ (s))dsds is convergent. By the definition of T , we have
−∞
1
+∆
(∫
∫t
−∞
[(
∫ξ
1 − k −∞
t du
s ρ(u)
du
ρ(u)
(
∫η
+k 1 + l −∞
du
ρ(u)
(
∫ξ
+l 1 − k −∞
du
ρ(u)
)∫
)
f (s, x(s), x′ (s))ds
+∞
−∞
f (s, x(s), x′ (s))ds
)∫
(∫
)
)∫
(∫
)
ξ
−∞
η
−∞
ξ du
s ρ(u)
η du
s ρ(u)
f (s, x(s), x′ (s))ds
f (s, x(s), x′ (s))ds
[ ∫
∫ η (∫ η
+∞
1
+∆
k −∞ f (s, x(s), x′ (s))ds + kl −∞ s
−kl
∫ξ
−∞
(∫
ξ du
s ρ(u)
)
]∫
t
f (s, x(s), x′ (s))ds −∞
du
ρ(u)
)
]
f (s, x(s), x′ (s))ds
du
ρ(u) .
We know
ρ(t)(T x)′ (t) = −
∫t
−∞
f (s, x(s), x′ (s))ds
[ ∫
∫ η (∫ η
+∞
1
+∆
k −∞ f (s, x(s), x′ (s))ds + kl −∞ s
−kl
∫ξ
−∞
(∫
ξ du
s ρ(u)
)
du
ρ(u)
)
f (s, x(s), x′ (s))ds
]
f (s, x(s), x′ (s))ds .
Thus T x : t 7→ (T x)(t) and ρ(T x)′ : t 7→ ρ(t)(T x)′ (t) are continuous on R.
Furthermore, lim (T x)(t) and lim ρ(t)(T x)′ (t) exist and are finite. Thus
t→±∞
T : P 7→ X is well defined.
t→±∞
(ii) By (i) and direct computation we get (2.3).
(iii) First, we prove that T : P 7→ P is well defined.
Step 1. Prove that (T x)(t) ≥ 0 for all t ∈ R. By the definition of T , it
∫ b du
suffices to prove that G(t, s) ≥ 0. For reader’s convenience, denote a ρ(u)
=
∫b
.
a
Case 1. For s ≥ max{η, t}, we see from (1.3) that
∆G(t, s) = 1 + k
∫t
ξ
≥

 0, ξ ≤ t,

1−k
∫ξ
−∞
≥ 0, ξ > t.
Draft version November 26, 2015
(T x)(t) = −
Four-point boundary value problem
11
Case 2. For s ≤ min{ξ, t}, we have from (1.4) that
∆G(t, s) = 1 + k
Draft version November 26, 2015
= 1 + kl
=1+l
∫ξ∫η
u
∫η
t
t
≥
∫t
+l
ξ
+k
∫η
t
∫ξ
u
+kl
∫ξ∫η
+kl
u
∫η∫t
u
ξ
−kl
t
+l
∫η
u
[∫ ∫
t t
u
ξ
+kl
+
∫η∫t
u
ξ
−l
∫t
u
−k
∫t
u
−kl
∫t∫η
u
ξ
∫ t ∫ η]
u
t

 0, t ≤ η,

1−l
∫ +∞
η
≥ 0, t > η.
Case 3. For t < s ≤ ξ < η, note u ≤ ξ, we have from (1.3) and (1.4) that
∆G(t, s) = 1 + k
=1+k
∫t
u
+kl
∫t
ξ
+k
∫ξ∫η
u
t
∫ξ
+l
u
+kl
∫η
u
∫ξ∫η
+kl
u
t
∫η∫t
u
ξ
+l
∫η
u
+kl
≥1+k
∫η∫t
u
∫t
ξ
+l
ξ
∫η
u
+kl
∫η∫t
u
ξ
(
∫ t) (
∫ η)
= 1+k ξ 1+l u

0, ξ ≤ t, u ≤ η,






(

∫ξ )(
∫ η)


1
−
k
1 + l u ≥ 0, ξ > t, u ≤ η,


−∞

≥
(
∫ t) (
∫ +∞ )


1
+
k
1
−
l
, ξ ≤ t, η < u,


ξ
η





(

∫ )(
∫ +∞ )

 1−k ξ
1
−
l
≥ 0, ξ > t, η < u.
−∞
η
Case 4. For min{t, ξ} < s ≤ η, we have from (1.3) and (1.4) that
∫t ∫η
∫η∫t
∆G(t, s) = 1 + k ξ +l u +kl u ξ
(
∫ t) (
∫ η)
= 1+k ξ 1+l u

0, ξ ≤ t, u ≤ η,






(

∫ξ )(
∫ η)


1
−
k
1 + l u ≥ 0, ξ > t, u ≤ η,


−∞

≥
(
∫ t) (
∫ +∞ )


1+k ξ
1−l η
, ξ ≤ t, η < u,







 (
∫ )(
∫ +∞ )

 1−k ξ
1
−
l
≥ 0, ξ > t, η < u.
−∞
η
Case 5. For ξ < s ≤ max{t, η}, note ξ ≤ u ≤ min{t, η}, we consider two
cases:
12
Yuji Liu, Xiaohui Yang
Subcase 5.1. u ≥ t. From (1.3) and (1.4), we have
=1+k
≥1+k
∫u
+l
ξ
∫t
ξ
+l
∫η
t
∫η
u
∫t
ξ
∫η
+l
+kl
∫η∫t
+kl
+kl
u
u
ξ
u
−kl
∫t
−l
ξ
u
−k
∫t
u
−kl
∫t∫η
u
ξ
∫t∫η
u
ξ
(
∫ η)
∫ t) (
= 1+k ξ 1+l u
∫η∫t
u
∫η∫t
ξ

0, ξ ≤ t, u ≤ η,






(

∫ξ )(
∫ η)


1
−
k
1 + l u ≥ 0, ξ > t, u ≤ η,


−∞

≥
(
∫ t) (
∫ +∞ )


1+k ξ
1−l η
, ξ ≤ t, η < u,







(

∫ +∞ )
∫ )(

 1−k ξ
1
−
l
≥ 0, ξ > t, η < u.
−∞
η
Subcase 5.2. u < t. Let
F (t) = 1 + k
∫u
ξ
+l
∫η
t
+kl
∫η∫t
u
ξ
−kl
∫t∫η
u
.
ξ
Using (1.3), we have
(
(
∫η
∫ξ
∫ η)
l
l
F ′ (t) = − ρ(t)
1 + k u −k ξ = − ρ(t)
1+k u
du
ρ(u)
)
≤ 0.
On the other hand, we have
F (+∞) = 1 + k
≥k
=k
=k
=k
(∫
u
ξ
(∫
u
ξ
(∫
u
ξ
∫u(
ξ
+l
+l
−l
∫u
ξ
∫ η ∫ +∞
u
ξ
∫η∫η
u
ξ
1−l
ξ
η
+kl
u
∫ η ∫ +∞
u
ξ
−kl
∫ +∞ ∫ η
u
ξ
∫ +∞ ∫ η )
u
ξ
∫ η ∫ +∞
+l
∫ +∞ )
η
∫ +∞
−l
+l
∫ +∞ ∫ η
η
−l
η
−l
∫ +∞ ∫ η )
u
ξ
∫ η ∫ +∞ )
u
η
≥ 0.
Since F is decreasing and F (+∞) ≥ 0, we get F (t) ≥ F (+∞) ≥ 0 for all
t < +∞.
Draft version November 26, 2015
∆G(t, s) = 1 + k
Four-point boundary value problem
13
Case 6. For ξ < η < s ≤ t, note u ≥ η, we have from (1.3) and (1.4) that
∫t ∫t
∫t
∫t∫η
∆G(t, s) = 1 + k ξ −l u −k u −kl u ξ
Draft version November 26, 2015
≥1+k
∫u
ξ
−l
∫t
u
−kl
∫t∫u
u
ξ
(
∫ u) (
∫ t)
1−l u
= 1+k ξ

0, ξ ≤ u, u ≥ t,






(

∫ξ )(
∫ t)


1
−
k
1
−
l
≥ 0, ξ > u, u ≥ t,


−∞
u

≥
(
∫ u) (
∫ +∞ )


1+k ξ
1−l η
≥ 0, ξ ≤ u, u < t,







 (
∫ )(
∫ +∞ )

 1−k ξ
1
−
l
≥ 0, ξ > u, u < t.
−∞
η
From cases 1-6, together with (1.5), this step is proved.
Step 2. Prove that min (T x)(t) ≥ µ sup(T x)(t).
t∈[−c,c]
t∈R
Firstly, we prove that (T x)(t) is (concave
to τ = τ (t) =
( ∫ with respect
))
∫t
+∞ du
du
1
. It is easy to see that τ ∈ C R, 0, −∞ ρ(u)
and dτ
dt = ρ(t) > 0.
−∞ ρ(u)
Thus
d(T x)
d(T x) dτ
d(T x) 1
=
=
.
dt
dτ dt
dτ ρ(t)
(2.4)
d(T x)
d[ρ(t)(T x)′ (t)]
′
= −f (t,
dτ . Since
dt(
) x(t), x (t)) ≤ 0
∫
′
+∞ du
d[ρ(t)(T x) (t)]
x)
and dτ
≤ 0 on 0, −∞ ρ(u)
. Then d(T
is
dt > 0, we get that
dτ
dτ
)
( ∫
+∞ du
decreasing with respect to τ ∈ 0, −∞ ρ(u) . Hence (T x)(t) is concave with
( ∫
)
+∞ du
respect to τ ∈ 0, −∞ ρ(u)
.
Since dτ
dt > 0 for all t ∈ R, there exists the inverse function of τ = τ (t).
x)
It follows that ρ(t) d(T
=
dt
Denote the inverse function of τ = τ (t) by t = t(τ ).
Case 1. there exists τ0 ∈ R such that sup(T x)(t) = (T x)(τ0 ). One sees
t∈R
min (T x)(t) = min {(T x)(−c), (T x)(c)} .
t∈[−c,c]
Denote τ (+∞) =
∫ +∞
du
.
−∞ ρ(u)
If
min{(T x)(−c), (T x)(c)} = (T x)(c) = (T x)(t(τ (c))),
noting τ (−c) < 1 (by definition of c), one has for t ∈ [−c, c] that
(T x)(t) ≥ (T x)(t(τ (c)))
= (T x)
( (
(c)+τ (τ0 )
τ (c)τ (+∞)
t τ (+∞)−τ
τ (+∞)+τ (τ0 )
τ (+∞)−τ (c)+τ (τ0 ) +
))
τ (c)
τ (+∞)+τ (τ0 ) τ (τ0 )
.
14
Yuji Liu, Xiaohui Yang
(
)
τ (c)
+ τ (+∞)+τ
(τ0 ) (T x) t (τ (τ0 ))
≥
≥
∫c
1
ds 2 ∫ +∞1
−∞ ρ(s)
1
ds
−∞ ρ(s)
∫ −c
1
ds 2 ∫ +∞1
−∞ ρ(s)
1
ds
−∞ ρ(s)
(T x)(τ0 )
(T x)(τ0 )
= µ sup(T x)(t).
t∈R
Similarly, if min{(T x)(−c), (T x)(c)} = (T x)(−c) = (T x)(t(τ (−c))), noting
τ (−c) < 1 (by definition of c), one has for t ∈ [−c, c] that
(T x)(t) ≥ (T x)(t(τ (−c)))
( (
(+∞)+τ (τ0 )−τ (−c) τ (+∞)τ (−c)
= (T x) t ττ (+∞)+τ
(τ0 )−τ (−c) τ (+∞)+τ (τ0 ) +
≥
τ (+∞)+τ (τ0 )−τ (−c)
(T x)
τ (+∞)+τ (τ0 )
))
τ (−c)
τ (+∞)+τ (τ0 ) τ (τ0 )
( (
))
τ (−c)τ (+∞)
t τ (+∞)+τ
(τ0 )−τ (−c)
τ (−c)
+ τ (+∞)+τ
(τ0 ) (T x) (t (τ (τ0 )))
≥
∫ −c
1
ds 2 ∫ +∞1
−∞ ρ(s)
1
ds
−∞ ρ(s)
(T x)(τ0 ) > µ sup(T x)(t).
t∈R
Case 2. sup(T x)(t) = lim (T x)(t) or lim (T x)(t)). Choose τ0 ∈ R. By
t→−∞
t∈R
t→+∞
the same methods used in Case 1, we get
min (T x)(t) ≥ µ(T x)(τ0 ).
t∈[−c,c]
Let τ0 → −∞. We get
min (T x)(t) ≥ µ lim (T x)(t) = µ sup(T x)(t) or µ lim (T x)(t)).
t∈[−c,c]
So
t→−∞
t∈R
t→+∞
min (T x)(t) ≥ µ sup(T x)(t) is proved. We see from Step 1 and Step 2
t∈[−c,c]
t∈R
that T x ∈ P . Hence T : P → P is well defined.
Step 3. we prove that T is continuous on P . Since f is a Carathéodory
function, then the result follows.
Step 4. we show that T is maps bounded subsets into bounded sets.
Draft version November 26, 2015
Noting that τ (+∞) > τ (c) and (T x)(t) is concave with respect to τ , then, for
t ∈ [−c, c],
( (
))
(c)+τ (τ0 )
τ (c)τ (+∞)
(T x)(t) ≥ τ (+∞)−τ
(T x) t τ (+∞)−τ
τ (+∞)+τ (τ0 )
(c)+τ (τ0 )
Four-point boundary value problem
15
Given a bounded set D ⊆ X. Then, there exists M > 0 such that D ⊆ {x ∈
X : ∥x∥ ≤ M }. Then there exists ϕM ∈ L1 (R) such that |f (t, x(t), x′ (t))| ≤
ϕM (t) for all t ∈ R. By the definition of G(t, s), the expressions of (T x)(t) and
ρ(t)(T x)′ (t), we get that
Draft version November 26, 2015
G(t, s) ≤
1+3k
∫ +∞
∫ +∞
du
+2l
−∞ ρ(u)
du
+3kl
−∞ ρ(u)
∫ +∞
(
du
−∞ ρ(u)
2
)
.
∆
Then
|(T x)(t)| ≤
1+3k
∫ +∞
du
+2l
−∞ ρ(u)
∫ +∞
du
+3kl
−∞ ρ(u)
∫ +∞
(
du
−∞ ρ(u)
2
) ∫ +∞
−∞
∆
ϕM (s)ds.
On the other hand, we have
[
′
ρ(t)|(T x) (t)| ≤ 1 +
Then
k+2kl
∫ +∞
du
−∞ ρ(u)
]
∫ +∞
−∞
∆
ϕM (s)ds.
{
}
∥(T x)∥ = max sup |(T x)(t)|, sup ρ(t)|(T x)′ (t)| < ∞.
t∈R
t∈R
So, {T D} is bounded.
Step 5. we prove that both {T x : x ∈ D} and {t 7→ ρ(t)(T x)′ (t) : x ∈ D}
are equi-continuous on each finite subinterval on R.
The proof is standard and is omitted. One may see [23].
Step 6. we show that both {T x : x ∈ D} and {t 7→ ρ(t)(T x)′ (t) : x ∈ D}
are equi-convergent at both +∞ and −∞ respectively.
We have that
[(
)∫
∫
+∞
1
du
f (s, x(s), x′ (s))ds
1 − k −∞ ρ(u)
(T x)(t) − ∆
−∞
(
∫η
+k 1 + l −∞
du
ρ(u)
(
∫
+l 1 − k −∞
du
ρ(u)
≤
(∫
∫t
−∞
t du
s ρ(u)
)
)∫
ξ
−∞
)∫
η
−∞
(∫
ξ du
s ρ(u)
(∫
)
η du
s ρ(u)
f (s, x(s), x′ (s))ds
)
]
f (s, x(s), x′ (s))ds |f (s, x(s), x′ (s))|ds
[ ∫
∫ η (∫ η
+∞
1
+∆
k −∞ |f (s, x(s), x′ (s))|ds + kl −∞ s
+kl
≤
1
∆
∫ξ
−∞
(∫
ξ du
s ρ(u)
[
∆ + k + kl
)
|f (s, x(s), x′ (s))|ds
∫ +∞
du
−∞ ρ(u)
+ kl
→ 0 uniformly as t → −∞.
∫ +∞
]∫
du
−∞ ρ(u)
du
ρ(u)
)
|f (s, x(s), x′ (s))|ds
t
du
−∞ ρ(u)
]∫
+∞
−∞
ΦM (s)ds
∫t
du
−∞ ρ(u)
16
Yuji Liu, Xiaohui Yang
Further more, we have that
[ ∫
+∞
1
k −∞ f (s, x(s), x′ (s))ds
ρ(t)(T x)′ (t) − ∆
≤
−∞
∫t
−∞
∫t
−∞
η du
s ρ(u)
)
f (s, x(s), x′ (s))ds − kl
(∫
∫ξ
−∞
ξ du
s ρ(u)
)
]
f (s, x(s), x′ (s))ds |f (s, x(s), x′ (s))|ds
ϕM (s)ds → 0 uniformly as t → −∞.
Hence {t 7→ ρ(t)(T x)′ (t) : x ∈ D} and {T x : x ∈ D} are equiconvergent at
−∞.
Similarly we can prove that both {T x : x ∈ D} and {t 7→ ρ(t)(T x)′ (t) : x ∈
D} are equiconvergent at +∞. The details are omitted.
From Steps 3-6, we see that T maps bounded sets into relatively compact
sets.
Therefore, the operator T : P → P is completely continuous. The proof of
(iii) is complete.
(iv) It is easy to see that x ∈ P is a positive solution of BVP(1.2) if and
only if x is a fixed point of T in P . The proof of (iv) is complete. Thus the
proof of Lemma 2.5 is ended.
Remark 2.1. From Cases 1-6 in Step 1 in the proof of Lemma 2.5, we
know G(t, s) ≥ 0 for all s, t ∈ R. This result generalizes a corresponding one
in [23].
3.
Existence of positive solutions
In this section we establish existence result on positive solutions of BVP(1.2).
∫ −c
du
−∞ ρ(u)
Fix c such that
< 1. Let µ =
∫ −c
du
−∞ ρ(u)
(∫
+∞ du
−∞ ρ(u)
)−1
. ∆ is
defined by (1.5). For ease expression, for nonnegative function ϕ ∈ L (R) and
nonnegative constants L1 , L2 and a, b, denote
{
}
a∫
a∫
∫
∫
M0 = max
, c ϕ(r)dr +∞ du ,
0
ϕ(r)dr −c du
1
−∞ ρ(u)
−c
{
}
L1
1
, ∫ +∞Lϕ(r)dr
ϕ(r)dr
−∞
0
∫0
W0 = min
ρ(u)
c
0
,
{
E0 = max
[
1+3k
∫ +∞
du
+2l
−∞ ρ(u)
∫ +∞
}
∆L2
[∆+k+2kl
∫ +∞
du
−∞ ρ(u)
]
∫ +∞
−∞
ϕ(s)ds
∆b
du
+3kl
−∞ ρ(u)
.
∫ +∞
(
du
−∞ ρ(u)
2
)
]∫
+∞
−∞
ϕ(s)ds
,
Draft version November 26, 2015
≤−
(∫
∫η
+kl
Four-point boundary value problem
17
Draft version November 26, 2015
Theorem 3.1. Suppose that (a)-(c) hold and there exists nonnegative
function ϕ ∈ L1 (R) and nonnegative constants L1 , L2 and a, b such that L1 <
L2 and a < b and
)
(
v
f t, u, ρ(t)
≥ M0 ϕ(t), t ∈ [−c, c], u ∈ [µa, a], v ∈ [−L1 , L1 ];
)
(
v
≥ W0 ϕ(t), t ∈ R, u ∈ [0, a], v ∈ [−L1 , L1 ],
f t, u, ρ(t)
(
)
v
f t, u, ρ(t)
≤ E0 ϕ(t), t ∈ R, u ∈ [0, b], v ∈ [−L2 , L2 ].
If E0 > max{M0 , W0 }, then BVP(1.2) has at least one positive solution x
satisfying
a ≤ sup x(t) ≤ b, 0 < sup ρ(t)|x′ (t)| ≤ L2
(3.1)
t∈R
t∈R
or
0 < sup x(t) ≤ b, L1 ≤ sup ρ(t)|x′ (t)| ≤ L2 .
(3.2)
t∈R
t∈R
Proof. Let X, P and the operator T be defined in section 2. By the
definition of T , Lemma 2.5, we know that T : P 7→ P is completely continuous,
x is a positive solution of BVP(1.2) if and only if x is a fixed point the T in P .
Define
ξ(x) = sup x(t), x ∈ P, η(x) = sup ρ(t)|x′ (t)|, x ∈ P,
t∈R
t∈R
{
}
{
}
Ω1 = x ∈ P : ξ(x) < a, η(x) < L1 , Ω2 = x ∈ P : ξ(x) < b, η(x) < L2 .
It is easy to see that ξ and η are continuous functionals and Ω1 and Ω2 are
bounded nonempty open subsets of P and ξ(x), η(x) ≤ ||x|| = max{ξ(x), η(x)}.
To apply (E1) in Lemma 2.1, we do the following two steps:
Step 1. We prove that
(3.3)
T x ̸= λx for all λ ∈ [0, 1) and x ∈ ∂Ω1 .
Let
C1 = {x ∈ P : ξ(x) = a, η(x) ≤ L1 }, D1 = {x ∈ P : ξ(x) ≤ a, η(x) = L1 },
C2 = {x ∈ P : ξ(x) = b, η(x) ≤ L2 }, D2 = {x ∈ P : ξ(x) ≤ b, η(x) = L2 }.
Sub-step 1.1. For x ∈ C1 , we prove that ξ(T x) ≥ a.
In fact, x ∈ C1 implies that
µa ≤ x(t) ≤ a, t ∈ [−c, c], −L1 ≤ ρ(t)x′ (t) ≤ L1 .
18
Yuji Liu, Xiaohui Yang
Then we get
(
)
ρ(t)x′ (t)
f (t, x(t), x (t)) = f t, x(t),
≥ M0 ϕ(t), t ∈ [−c, c].
ρ(t)
We consider three cases:
Case 1. ρ(t)(T x)′ (t) > 0 for all t ∈ R. In this case, we see that T x is
increasing on R and (T x)(t) ≥ 0 for all t ∈ R. Then
)
∫ −c 1 (∫ +∞
ξ(T x) ≥ (T x)(−c) = −∞ ρ(s)
f (r, x(r), x′ (r))dr ds
s
=
≥
≥
∫ +∞
−c
∫ +∞
−c
∫c
−c
≥ M0
∫ −c
f (r, x(r), x′ (r))dr
du
−∞ ρ(u)
∫c
−c
ϕ(r)dr
∫ −c ∫ r
du
f (r, x(r), x′ (r))dr
−∞ −∞ ρ(u)
∫ −c
f (r, x(r), x′ (r))dr
f (r, x(r), x′ (r))dr
+
du
−∞ ρ(u)
∫ −c
du
−∞ ρ(u)
∫ −c
≥ a.
du
−∞ ρ(u)
Case 2. ρ(t)(T x)′ (t) < 0 for all t ∈ R. In this case, we see that T x is
decreasing on R and (T x)(t) ≥ 0 for all t ∈ R. Then
∫ +∞ 1 ∫ s
′
ξ(T x) ≥ (T x)(c) ≥ c
ρ(s) −∞ f (r, x(r), x (r))drds
=
≥
∫c
−∞
∫c
−c
≥ M0
f (r, x(r), x′ (r))ds
f (r, x(r), x′ (r))ds
∫c
−c
ϕ(r)dr
∫ +∞
c
∫ +∞
c
∫ +∞
c
du
ρ(u)
du
ρ(u)
+
∫ +∞ ∫ +∞
c
r
du
′
ρ(u) f (r, x(r), x (r))dr
du
ρ(u)
≥ a.
Case 3. there exists τ0 ∈ R such that ρ(τ0 )(T x)′ (τ0 ) = 0. From (T x)(t) ≥
0, we have
 ∫t
 − τ0 f (r, x(r), x′ (r))dr, t ≥ τ0 ,
ρ(t)(T x)′ (t) =
 ∫ τ0
f (r, x(r), x′ (r))dr, t ≤ τ0 .
t
So
(T x)(t) ≥
 ∫ +∞

 t

 ∫t
1
ρ(s)
1
−∞ ρ(s)
≥
∫s
τ0
∫ τ0
s
f (r, x(r), x′ (r))drds, t ≥ τ0 ,
f (r, x(r), x′ (r))drds, t ≤ τ0
 ∫t
∫ +∞
′

 τ0 f (r, x(r), x (r))dr t
du
ρ(u) ,
t ≥ τ0 ,
du
,
−∞ ρ(u)
t ≤ τ0 .

 ∫ τ0 f (r, x(r), x′ (r))dr ∫ t
t
Draft version November 26, 2015
′
Draft version November 26, 2015
Four-point boundary value problem
19
If τ0 ≥ 0, then
∫0
∫ −c
ξ(T x) ≥ −c f (r, x(r), x′ (r))dr −∞
du
ρ(u)
≥ M0
If τ0 ≤ 0, then
∫ +∞
∫c
ξ(T x) ≥ 0 f (r, x(r), x′ (r))dr c
du
ρ(u)
≥ M0
∫0
−c
∫c
0
ϕ(r)dr
ϕ(r)dr
∫ −c
≥ a.
du
−∞ ρ(u)
∫ +∞
c
du
ρ(u)
≥ a.
Sub-step 1.2. For x ∈ D1 , we prove that η(T x) ≥ L1 .
In fact, x ∈ D1 implies that
0 ≤ x(t) ≤ a, −L1 ≤ ρ(t)x′ (t) ≤ L1 , t ∈ R.
Then we get
f (t, x(t), x′ (t)) = f
(
)
ρ(t)x′ (t)
t, x(t),
≥ W0 ϕ(t), t ∈ R.
ρ(t)
We consider three cases:
Case 1. ρ(t)(T x)′ (t) > 0 for all t ∈ R. Then
ρ(t)(T x)′ (t) = lim ρ(t)(T x)′ (t) +
∫ +∞
t
t→+∞
So
η(T x) = sup ρ(t)|(T x)′ (t)| ≥ sup
t∈R
=
∫ +∞
−∞
f (r, x(r), x′ (r))dr ≥
∫ +∞
t∈R
∫ +∞
−∞
t
f (r, x(r), x′ (r))dr.
f (r, x(r), x′ (r))dr
ϕ(r)W0 dr ≥ L1 .
Case 2. ρ(t)(T x)′ (t) < 0 for all t ∈ R. Then
ρ(t)(T x)′ (t) = lim Φ(ρ(t)(T x)′ (t)) −
t→−∞
So
∫t
−∞
f (r, x(r), x′ (r))dr.
η(T x) = sup ρ(t)|(T x)′ (t)|
t∈R
∫t
′
′
= supt∈R lim ρ(t)(T x) (t) − −∞ f (r, x(r), x (r))dr
t→−∞
≥
≥
∫ +∞
−∞
∫ +∞
−∞
f (r, x(r), x′ (r))dr
ϕ(r)W0 dr ≥ L1 .
Case 3. there exists τ0 ∈ R such that ρ(τ0 )(T x)′ (τ0 ) = 0. Then
 ∫t
 − τ0 f (r, x(r), x′ (r))dr, t ≥ τ0 ,
′
ρ(t)(T x) (t) =
 ∫ τ0
f (r, x(r), x′ (r))dr, t ≤ τ0 .
t
20
Yuji Liu, Xiaohui Yang
η(T x) = sup ρ(t)|(T x)′ (t)| ≥ sup
t∈R
≥
∫0
−∞
∫ τ0
t
t∈R
f (r, x(r), x′ (r))dr ≥
∫0
−∞
f (r, x(r), x′ (r))dr
ϕ(r)W0 dr ≥ L1 ,
If τ0 ≤ 0, then
∫t
η(T x) = sup ρ(t)|(T x)′ (t)| ≥ sup f (r, x(r), x′ (r))dr
t∈R
t∈R τ0
≥
+∞
∫
f (r, x(r), x′ (r))dr ≥
+∞
∫
0
ϕ(r)W0 dr ≥ L1 ,
0
Now we prove (3.3). It is easy to see that
∂Ω1 ⊆ C1 ∪ D1 , ∂Ω2 ⊆ C2 ∪ D2 .
If T x = λx for some λ ∈ [0, 1) and x ∈ ∂Ω1 , then either x ∈ C1 or x ∈ D1 .
If x ∈ C1 , we get from Sub-step 1.1 that ξ(T x) ≥ a. On the other hand, we
have ξ(T x) = λξ(x) < ξ(x) = a, a contradiction.
If x ∈ D1 , from Sub-step 1.2, we have η(T x) ≥ L1 . On the other hand, we
have η(T x) = λη(x) < η(x) = L1 , a contradiction too.
From above discussion, (3.3) holds.
Step 2. We prove that
T x ̸= λx for all λ ∈ (1, +∞) and x ∈ ∂Ω2 .
(3.4)
For x ∈ C2 , one has 0 ≤ x(t) ≤ b, −L2 ≤ ρ(t)x′ (t) ≤ L2 , t ∈ R. Then we
get
(
)
ρ(t)x′ (t)
f (t, x(t), x (t)) = f t, x(t),
≤ E0 ϕ(t), t ∈ R.
ρ(t)
′
Then, we have
ξ(T x) = sup
t∈R
≤
1+3k
∫ +∞
−∞
∫ +∞
G(t, s)f (s, x(s), x′ (s))ds
du
+2l
−∞ ρ(u)
∫ +∞
du
+3kl
−∞ ρ(u)
t∈R
du
−∞ ρ(u)
2
) ∫ +∞
∆
For x ∈ D2 , we have
η(T x) = sup ρ(t)|(T x)′ (t)| ≤
∫ +∞
(
1
∆
[
∆ + k + 2kl
∫ +∞
−∞
du
−∞ ρ(u)
ϕ(s)E0 ds ≤ b.
]∫
+∞
−∞
E0 ϕ(s)ds ≤ L2 .
In fact, if T x = λx for some λ ∈ (1, +∞) and x ∈ ∂Ω2 , then either x ∈ C2
or x ∈ D2 .
Draft version November 26, 2015
If τ0 ≥ 0, then
Draft version November 26, 2015
Four-point boundary value problem
21
If x ∈ C2 , we get from above discussion that ξ(T x) ≤ b. On the other hand,
we have ξ(T x) = λξ(x) > ξ(x) = b, a contradiction.
If x ∈ D2 , from above discussion, we have η(T x) ≤ L2 . On the other hand,
we have η(T x) = λη(x) > η(x) = L2 , a contradiction too.
From above discussion, (3.4) holds.
It follows from (3.3), (3.4) and (E1) in Lemma 2.1 that T has at least one
fixed point x ∈ Ω2 \ Ω1 . So BVP(1.2) has at least one positive solution x such
that x ∈ Ω2 \ Ω1 and that x satisfies (3.1) or (3.2). The proof of Theorem 3.1
is complete.
For ease expression, for nonnegative function ϕ ∈ L1 (R) and nonnegative
constants L1 , L2 and a, b, denote
M1 =
W1 =
[
∫ +∞
1+3k −∞
[∆+k+kl
du
+2l
ρ(u)
du
+3kl
−∞ ρ(u)
1 ∫
∫ +∞ ∆L
+∞
du
−∞ ρ(u)
∆a
∫ +∞
]
−∞
∫ +∞
(
E0 ϕ(s)ds
du
−∞ ρ(u)
2
)
]∫
+∞
−∞
ϕ(s)ds
,
,
{
E1 = min
}
L2∫
∫0
−c
ϕ(r)dr −∞
−c
du
ρ(u)
,
L∫2
∫c
ϕ(r)dr c+∞
0
du
ρ(u)
,
2
∫ 0 L2
, ∫ +∞Lϕ(r)dr
ϕ(r)dr
−∞
0
.
Theorem 3.2. Suppose that (a)-(c) hold and there exists a nonnegative
function ϕ ∈ L1 (R) and L1 > L2 > 0 and a > b > 0 such that
(
)
v
f t, u, ρ(t)
≤ M1 ϕ(t), t ∈ [−c, c], u ∈ [µa, a], v ∈ [−L1 , L1 ];
(
)
v
f t, u, ρ(t)
≤ W1 ϕ(t), t ∈ R, u ∈ [0, a], v ∈ [−L1 , L1 ],
)
(
v
≥ E1 ϕ(t), t ∈ R, u ∈ [0, b], v ∈ [−L2 , L2 ].
f t, u, ρ(t)
If E1 > max{M1 , W1 }, then BVP(1.2) has at least one positive solution x
satisfying
(3.5)
b ≤ sup x(t) ≤ a, 0 < sup ρ(t)|x′ (t)| ≤ L1
t∈R
t∈R
or
(3.6)
0 < sup x(t) ≤ a, L2 ≤ sup ρ(t)|x′ (t)| ≤ L1 .
t∈R
t∈R
Proof. Let X, P and the nonlinear operator T be defined in Section 2.
The proof is similar to the proof of Theorem 3.1 by using (E2) in Lemma 2.1
and is omitted.
4.
Non-existence of positive solutions
Now we establish non-existence results on positive solutions of BVP(1.2).
22
Yuji Liu, Xiaohui Yang
Theorem 4.1. Let Ω = R × [0, +∞) × R. Suppose that (a)-(c) hold and
there exists a function ϕ ∈ L1 (R) such that
(t,u,v)∈Ω
If
1
∆
(4.2)
[
∆ + k + kl
v
f (t,u, ρ(t)
)
ϕ(t)v
∫ +∞
du
−∞ ρ(u)
≤ 1.
]∫
+∞
−∞
ϕ(s)ds < 1,
then BVP(1.2) does not admit positive solutions.
Proof. Let X, P and T be defined in Section 2. From (4.1), we have
)
(
v
≤ ϕ(t)|v|, (t, u, v) ∈ R × [0, +∞) × R.
f t, u, ρ(t)
Assume that x is a positive fixed point of T . We have
f (t, x(t), x′ (t)) ≤ ϕ(t)ρ(t)x′ (t) ≤ ϕ(t) sup ρ(t)|x′ (t)|, t ∈ R.
t∈R
It follows from (4.2) that
[
∫ +∞
1
∆ + k + kl −∞
sup ρ(t)|x′ (t)| ≤ ∆
t∈R
du
ρ(u)
]∫
+∞
−∞
ϕ(s)ds sup ρ(t)|x′ (t)|
t∈R
< sup ρ(t)|x′ (t)|
t∈R
which is a contradiction. The proof is complete.
Theorem 4.2. Let Ω = R × [0, +∞) × R. Suppose that (a)-(c) hold and
there exists a function ϕ ∈ L1 (R) and a constant c > 0 such that
v
f (t,u, ρ(t)
)
ϕ(t)u
(t,u,v)∈Ω
inf
(4.3)
If
(4.4)
µ min
{∫
0
ϕ(r)dr
−c
∫ −c
∫c
du
,
−∞ ρ(u) 0
≥ 1.
ϕ(r)dr
∫ +∞
c
du
ρ(u)
}
> 1,
then BVP(1.2) does not admit positive solutions.
Proof. Let X, P and T be defined in Section 2. From (4.3), we have
)
(
v
≥ ϕ(t)u, (t, u, v) ∈ R × [0, +∞) × R.
f t, u, ρ(t)
Suppose that x is a positive solution of BVP(1.2). Then x(t) = (T x)(t) for all
t ∈ R. We have
f (t, x(t), x′ (t)) ≥ ϕ(t)x(t) ≥ µϕ(t) sup |x(t)|, t ∈ R.
t∈R
Draft version November 26, 2015
sup
(4.1)
Four-point boundary value problem
23
We consider three cases:
Case 1. ρ(t)x′ (t) > 0 for all t ∈ R. So
sup |x(t)| ≥ x(−c) =
Draft version November 26, 2015
t∈R
=
≥
≥
∫ +∞
−c
∫ +∞
−c
∫c
−c
f (r, x(r), x′ (r))dr
−c
t∈R
∫ −c
du
−∞ ρ(u)
)
f (r, x(r), x′ (r))dr ds
+
∫ −c ∫ r
du
f (r, x(r), x′ (r))dr
−∞ −∞ ρ(u)
∫ −c
du
−∞ ρ(u)
∫ −c
f (r, x(r), x′ (r))dr
∫c
+∞
s
1
−∞ ρ(s)
f (r, x(r), x′ (r))dr
≥ µ sup |x(t)|
(∫
∫ −c
du
−∞ ρ(u)
∫ −c
du
−∞ ρ(u)
ϕ(r)dr
> sup |x(t)|,
t∈R
which is a contradiction.
Case 2. ρ(t)x′ (t) < 0 for all t ∈ R. Then
sup |x(t)| ≥ (T x)(c) ≥
∫ +∞
t∈R
=
≥
∫c
−∞
∫c
−c
c
f (r, x(r), x′ (r))ds
f (r, x(r), x′ (r))ds
≥ µ sup |x(t)|
t∈R
∫c
−c
ϕ(r)dr
1
ρ(s)
∫ +∞
−∞
du
ρ(u)
c
∫ +∞
∫s
+
f (r, x(r), x′ (r))drds
∫ +∞ ∫ +∞
c
r
du
′
ρ(u) f (r, x(r), x (r))dr
du
ρ(u)
c
∫ +∞
du
ρ(u)
c
> sup |x(t)|,
t∈R
which is a contradiction.
Case 3. there exists τ0 ∈ R such that ρ(τ0 )(T x)′ (τ0 ) = 0. From (T x)(t) ≥
0, we have
 ∫t
 − τ0 f (r, x(r), x′ (r))dr, t ≥ τ0 ,
′
ρ(t)x (t) =
 ∫ τ0
f (r, x(r), x′ (r))dr, t ≤ τ0 .
t
So
x(t) ≥
 ∫ +∞

 t

 ∫t
1
ρ(s)
1
−∞ ρ(s)
≥
∫s
τ0
∫ τ0
s
f (r, x(r), x′ (r))drds, t ≥ τ0 ,
f (r, x(r), x′ (r))drds, t ≤ τ0
 ∫t
∫ +∞
′

 τ0 f (r, x(r), x (r))dr t
du
ρ(u) ,
t ≥ τ0 ,
du
,
−∞ ρ(u)
t ≤ τ0 .

 ∫ τ0 f (r, x(r), x′ (r))dr ∫ t
t
24
Yuji Liu, Xiaohui Yang
If τ0 ≥ 0, then
sup |x(t)| ≥
t∈R
∫0
−c
f (r, x(r), x′ (r))dr
∫0
≥ µ sup |x(t)|
−c
t∈R
ϕ(r)dr
∫ −c
du
−∞ ρ(u)
∫ −c
du
−∞ ρ(u)
> sup |x(t)|,
t∈R
sup |x(t)| ≥
t∈R
∫c
0
≥ µ sup |x(t)|
t∈R
f (r, x(r), x′ (r))dr
∫c
0
ϕ(r)dr
∫ +∞
c
du
ρ(u)
∫ +∞
c
du
ρ(u)
> sup |x(t)|,
t∈R
which is a contradiction.
From above discussion, we know that BVP(1.2) has no positive solution.
The proof is completed.
References
[1] R.P. Agarwal, Boundary value problems for higher order differential equations,
World Scientific, Singapore, 1986.
[2] C. Avramescu, C. Vladimirescu, Existence of Homoclinic solutions to a nonlinear
second order ODE, Dynamics of continuous, discrete and impulsive systems, Ser.
A, Math Anal. 15(2008) 481-491.
[3] C. Avramescu, C. Vladimirescu, Existence of solutions to second order ordinary
differential equations having finite limits at ±∞, Electronic Journal of differential
equations, 18(2004) 1-12.
[4] C. Avramescu and C. Vladimirescu, Limits of solutions of a perturbed linear
differential equation, E. J. Qualitative Theory of Diff. Equ., 3(2002) 1-11.
[5] A. A. Boichuk, O. A. Pokutnyi, Dichotomy and boundary value problems on the
whole line, Chaotic Modeling and Simulation (CMSIM) 2 (2013) 247-255.
[6] B. Bianconi, F. Papalini, Non-autonomous boundary value problems on the real
line, Discrete and continuous dynamical systems, 15(2006) 759-776.
[7] A. Calamai, Heteroclinic solutions of boundary value problems on the real line
involving singular Φ-Laplacian operators, Journal of Mathematical Analysis and
Applications, 378(2011) 667-679.
[8] A. Cabada, J.A. Cid, Heteroclinic solutions for non-autonomous boundary value
problems with singular Φ-Laplacian operators, Discrete Contin. Dyn. Syst. 2009,
Dynamical Systems, Differential Equations and Applications. 7th AIMS Conference, suppl., 118–122.
[9] G. Cupini, C. Marcelli, F. Papalini, Heteroclinic solutions of boundary value
problems on the real line involving general nonlinear differential operators, Differ.
Integral Equ. 24(7-8) (2011), 619-644.
Draft version November 26, 2015
which is a contradiction.
If τ0 ≤ 0, then
Four-point boundary value problem
25
[10] G. Cupini, C. Marcelli, F. Papalini, On the solvability of a boundary value
problem on the real line, Bound. Value Probl. 2011, 26 (2011).
[11] K. Deimling, Nonlinear Functional Analysis, Springer, Berlin, Germany, 1985.
Draft version November 26, 2015
[12] F. Y. Deren, N. A. Hamal, Second-order boundary-value problems with integral
boundary conditions on the real line, Electronic Journal of Differential Equations,
19 (2014) 1-13.
[13] W. Ge, Boundary Value Problems for Ordinary Differential Equations (in Chinese), Science Press, Beijing, 2007.
[14] V. A. Il’in, E. I. Moiseev, Nonlocal boundary-value problem of the second kind
for a Sturm-Liouville operator, Differential Equations 23 (1987) 979-987.
[15] Y. Liu, Multiple positive solutions to mixed boundary value problems for singular
ordinary differential equations on the whole line, Nonlinear Analysis: Modelling
and Control, 17 (4)(2012) 460-480.
[16] Y. Liu, Existence of multiple positive solutions of p-Laplacian boundary value
problems, Math. Slovak. 57(2007) 225-242.
[17] Y. Liu, Solutions to second order non-homogeneous multi-point BVPs using a
fixed-point theorem, Electronic Journal of Differential Equations, 96(2008) 1-52.
[18] C. Marcelli, Existence of solutions to some boundary-value problems governed
by general nonlinear non-autonomous differential operators, Electron. J. Differ.
Equ. 2012, 171 (2012).
[19] C. Marcelli, The role of boundary data on the solvability of some equations involving non-autonomous nonlinear differential operators, Boundary Value Problems, 2013, 2013: 252.
[20] C. Marcelli, F. Papalini, Heteroclinic connections for fully non-linear nonautonomous second-order differential equations, Journal of Differential Equations,
241(2007) 160-183.
[21] K. R. Prasad, A. K. Rao, S. N. Rao, Existence of positive solutions for the
system of higher order two-point boundary value problems, Proceedings of the
Indian Academy of Sciences Mathematical Sciences, 122(2012) 139-152.
[22] C. G. Philos, I. K. Purnaras, A boundary value problem on the whole line to
second order nonlinear differential equations, Georgian Mathematical Journal,
17(2010) 241-252.
[23] J. Zhao, F. Geng, J. Zhao, W. Ge, Positive solutions to a new kind
SturmCLiouville-like four-point boundary value problem, Appl. Math. Comput.
217(2)(2010) 811-819.