Electronic Journal of Differential Equations, Vol. 2009(2009), No. 110, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE SOLUTIONS FOR NONLINEAR SECOND-ORDER
m-POINT BOUNDARY-VALUE PROBLEMS
JIQIANG JIANG, LISHAN LIU
Abstract. By constructing a special cone and applying the fixed index theory
in the cone, we prove the existence of positive solutions for a class of singular
m-point boundary-value problems.
1. Introduction
This paper considers the existence of positive solutions for the second-order mpoint boundary-value problem
(p(t)x0 (t))0 − q(t)x(t) + f (t, x(t)) = 0,
ax(0) − bp(0)x0 (0) =
m−2
X
αi x(ξi ),
t ∈ (0, 1),
cx(1) + dp(1)x0 (1) =
m−2
X
βi x(ξi ),
(1.1)
(1.2)
i=1
i=1
where a, c ∈ [0, +∞), b, d ∈ (0, +∞) with ac + ad + bc > 0, ξi ∈ (0, 1), αi , βi ∈
[0, +∞) for i ∈ {1, 2, . . . , m − 2} are given constants, p ∈ C 1 ([0, 1], (0, +∞)), q ∈
C([0, 1], (0, +∞)) and f ∈ C((0, 1) × (0, +∞), [0, +∞)), f (t, x) is allowed to be
singular at t = 0, t = 1 and x = 0.
If p ≡ 1, q ≡ 0, αi , βi = 0, (for i = 1, 2, . . . , m − 2), then (1.1)-(1.2) reduces to
the two-point boundary-value problem
x00 (t) + f (t, x(t)) = 0,
0
ax(0) − bx (0) = 0,
t ∈ (0, 1),
(1.3)
0
(1.4)
cx(1) + dx (1) = 0,
which has been intensively studied; see [5, 6].
In [7], by using the fixed index theory in a cone, positive solutions were obtained
for differential systems
−x00 (t) = f (t, y),
00
−y (t) = g(t, x),
t ∈ (0, 1),
t ∈ (0, 1),
0
α1 x(0) − β1 x (0) = γ1 x(1) + δ1 x0 (1) = 0,
2000 Mathematics Subject Classification. 34B15, 34B25.
Key words and phrases. Positive solution; singular; m-point boundary-value problem; cones.
c
2009
Texas State University - San Marcos.
Submitted December 26, 2008. Published September 10, 2009.
Supported by grants: 10771117 from the National Natural Science Foundation of China,
20060446001 from the State Ministry of Education Doctoral Foundation of China, and
Y2007A23 from the Natural Science Foundation of Shandong Province of China.
1
2
J. JIANG, L. LIU
EJDE-2009/110
α2 y(0) − β2 y 0 (0) = γ2 y(1) + δ2 y 0 (1) = 0,
where αi , βi , γi , δi ≥ 0 and ρi = αi γi + αi γi + γi βi > 0 (i = 1, 2), f (t, y) and g(t, x)
may be singular at t = 0, t = 1 and x = 0, y = 0, respectively.
In recent years, singular multi-point boundary-value problems have been extensively studied and many optimal results have been obtained, see [6, 11, 12, 13, 14]
and references therein. In addition, many papers investigated the existence of
solutions for the nonsingular multi-point boundary-value problems, for example,
[1, 4, 5, 10].
Recently, Ma [8], Ma and Thompson [9] obtained excellent results about the
existence of positive solutions for the more general m-point boundary-value problem
(1.1)-(1.2), but in the above papers there are no studies for singularity of the
nonlinearity f (t, x) at the point x = 0. Recently, by using Nonlinear Alternative
of Leray-Schauder with the properties of the associated vector field at the (u, u0 )
plane, Galanis and Palamides [2] studied the problem
−[φp (u0 )]0 = q(t)f (t, u(t)),
0<t<1
subject to
u(0) − g(u0 (0)) = 0,
u(1) − βu(η) = 0,
or to
u(0) − αu0 (η) = 0, u(1) + g(u0 (1)) = 0,
where f (t, u) is allowed to have singularity at u = 0, the obtained solutions remains
away from the origin and avoid the singularity of the nonlinear term at u = 0.
Motivated by the above mentioned papers, we consider the existence of positive
solutions for (1.1)-(1.2). Here we allow f (t, x) to have a singularity at t = 0, 1, and
at x = 0. As far as we know, there were only a few works when f has singularities
at t = 0, 1 and x = 0. This paper attempts to fill part of this gap in the literature.
This work is organized as follows. In section 2, we present some lemmas that are
used to prove our main results. Then in section 3, the existence of positive solution
for (1.1)-(1.2) will be established by using the fixed point theory in the cone, which
we state here for the convenience of the reader.
Lemma 1.1 ([3]). Let P be a cone of the real Banach space E, Ω be a bounded open
subset of E with θ ∈ Ω and T : Ω ∩ P → P is a completely continuous. Suppose
that T u 6= λu, for all u ∈ ∂Ω ∩ P , λ ≥ 1, then i(T, Ω ∩ P, P ) = 1.
Lemma 1.2 ([3]). Let P be a cone of the real Banach space E, Ω be a bounded open
subset of E with θ ∈ Ω and T : Ω ∩ P → P is a completely continuous. Suppose
that
(i) inf u∈P ∩∂Ω kT uk > 0
(ii) T u 6= λu, for all u ∈ ∂Ω ∩ P , λ ∈ (0, 1],
then i(T, Ω ∩ P, P ) = 0.
2. Preliminaries
Let E = C[0, 1] be a real Banach space, with the norm kxk = maxt∈[0,1] |x(t)|
for x ∈ C[0, 1]. Let P = {x ∈ E : x(t) ≥ 0, t ∈ [0, 1]}. Clearly P is a cone in E.
The function x is said to be a positive solution of (1.1)-(1.2) if x(t) is positive
solution on (0, 1) and satisfies the differential equation (1.1) and the boundary
conditions (1.2).
The following lemmas play an important role when proving our main results.
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3
Lemma 2.1 ([8, 9]). Assume
(H1) p ∈ C 1 ([0, 1], (0, +∞)), q ∈ C([0, 1], (0, +∞)).
Let ψ and φ be the solutions of the linear problems
(p(t)ψ 0 (t))0 (t) − q(t)ψ(t) = 0,
ψ(0) = b,
t ∈ (0, 1),
(2.1)
0
p(0)ψ (0) = a,
(2.2)
and
(p(t)φ0 (t))0 (t) − q(t)φ(t) = 0,
φ(1) = d,
t ∈ (0, 1),
(2.3)
0
p(1)φ (1) = −c,
(2.4)
respectively. Then
(i) ψ is strictly increasing on [0, 1], and ψ(t) > 0 on [0, 1];
(ii) φ is strictly decreasing on [0, 1], and φ(t) > 0 on [0, 1].
As in [9], set
Pm−2
Pm−2
− i=1 αi ψ(ξi ) ρ − i=1 αi φ(ξi )
Pm−2
Pm−2
∆ = det
,
ρ − i=1 βi ψ(ξi ) − i=1 βi φ(ξi )
ρ = p(t) det
φ(t) ψ(t)
.
φ0 (t) ψ 0 (t)
Then, by Liouville’s formula, we have
φ(0) ψ(0)
ρ = p(0) det 0
= constant.
φ (0) ψ 0 (0)
Define
1
G(t, s) =
ρ
(
φ(t)ψ(s), 0 ≤ s ≤ t ≤ 1,
φ(s)ψ(t), 0 ≤ t ≤ s ≤ 1.
(2.5)
It is easy to see that
0 ≤ G(t, s) ≤ G(s, s),
0 ≤ s, t ≤ 1.
(2.6)
Remark 2.2. By (2.5) and Lemma 2.1, for any t ∈ [0, 1], we have
( φ(t)
( d
0 ≤ s ≤ t ≤ 1,
0 ≤ s ≤ t ≤ 1,
G(t, s)
φ(s) ,
φ(0) ,
= ψ(t)
≥
b
G(s, s)
, 0 ≤ t ≤ s ≤ 1.
, 0 ≤ t ≤ s ≤ 1,
ψ(1)
ψ(s)
b
ψ(1) },
d
,
Let γ = min{ φ(0)
then G(t, s) ≥ γG(s, s), for t, s ∈ [0, 1].
d
b
Remark 2.3. Since γ = min{ φ(0)
, ψ(1)
}, according to the monotonicity of ψ(t),
we have γ ≤
b
ψ(1)
=
ψ(0)
ψ(1)
≤
ψ(t)
ψ(1) ,
monotonicity of φ(t), we have γ ≤
so ψ(t) ≥ γψ(1), for t ∈ [0, 1]. Similarly, by the
d
φ(0)
=
φ(1)
φ(0)
≤
φ(t)
φ(0) ,
so φ(t) ≥ γφ(0), for t ∈ [0, 1].
Lemma 2.4 ([8, 9]). Assume (H1) and that ∆ 6= 0. Then for any y ∈ L[0, 1], the
problem
(p(t)x0 (t))0 (t) − q(t)x(t) + y(t) = 0,
ax(0) − bp(0)x0 (0) =
m−2
X
i=1
αi x(ξi ),
t ∈ (0, 1),
cx(1) + dp(1)x0 (1) =
m−2
X
(2.7)
βi x(ξi ),
(2.8)
i=1
has a unique solution
Z
1
G(t, s)y(s)ds + A(y)ψ(t) + B(y)φ(t),
x(t) =
0
(2.9)
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J. JIANG, L. LIU
EJDE-2009/110
where
!
Pm−2 R 1
Pm−2
1
αi 0 G(ξi , s)y(s)ds ρ − i=1 αi φ(ξi )
i=1
A(y) =
det Pm−2 R 1
Pm−2
∆
− i=1 βi φ(ξi )
i=1 βi 0 G(ξi , s)y(s)ds
!
Pm−2
Pm−2 R 1
1
− i=1 αi ψ(ξi )
α
G(ξ
,
s)y(s)ds
i
i
R 01
B(y) =
det
.
Pm−2
Pi=1
m−2
∆
ρ − i=1 βi ψ(ξi )
i=1 βi 0 G(ξi , s)y(s)ds
(2.10)
(2.11)
Lemma 2.5 ([8, 9]). Assume (H1) and
Pm−2
Pm−2
(H2) ∆ < 0, ρ − i=1 αi φ(ξi ) > 0, ρ − i=1 βi ψ(ξi ) > 0.
Then for y ∈ L[0, 1] with y ≥ 0, the unique solution x of (2.7)-(2.8) satisfies
x(t) ≥ 0, for t ∈ [0, 1].
Let Q = {x ∈ P : x(t) ≥ γkxk}. It is obvious that Q is a subcone of P . With
Lemma 2.4, Problem (1.1)-(1.2) has a positive solution x = x(t) if and only if
x ∈ Q\{θ} is a solution of the nonlinear integral equation
Z 1
x(t) =
G(t, s)f (s, x(s))ds + A(f (s, x(s)))ψ(t) + B(f (s, x(s)))φ(t),
(2.12)
0
where f satisfies the condition
(H3) f ∈ C((0, 1) × (0, +∞), [0, +∞)) and there exist h ∈ C((0, 1), [0, +∞)),
g ∈ C((0, +∞), [0, +∞)) satisfying that for any t ∈ (0, 1), u ∈ (0, +∞)
implies
f (t, u) ≤ h(t)g(u), t ∈ (0, 1), u ∈ (0, +∞),
Z 1
0<
G(s, s)h(s)ds < +∞.
0
Define an operator T : Q\{θ} → P by
Z 1
(T x)(t) =
G(t, s)f (s, x(s))ds + A(f (s, x(s)))ψ(t) + B(f (s, x(s)))φ(t). (2.13)
0
It is easy to prove that the existence of solutions to (1.1)-(1.2) is equivalent to the
existence of solutions to (2.12). That is, the existence of a fixed point of operator
T.
To overcome the singularity, we consider the following approximating equation
of (2.13) with the boundary conditions (1.2).
Z 1
(Tn x)(t) =
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t),
0
(2.14)
where n is a positive integer and
1
fn (t, x) = f (t, max{ , x}).
n
(2.15)
Remark 2.6. By (H3), there exists τ ∈ (0, 21 ) such that
Z 1−τ
0<
G(s, s)h(s)ds < +∞.
τ
Lemma 2.7. Assume (H1)-(H3). Then Tn : P → P is completely continuous for
any fixed natural number n.
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POSITIVE SOLUTIONS
5
Proof. First it is easy to see that Tn maps P into P . Then we prove that Tn maps
bounded sets into bounded sets.
Suppose D ⊂ P is an arbitrary bounded set. Then there exists a constant
M1 > 0 such that kxk ≤ M1 for any x ∈ D. By (H1), for any x ∈ D and s ∈ [0, 1],
we have
Z 1
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t)
|(Tn x)(t)| =
0
Z
1
1
G(s, s)h(s)g max{ , x(s)} ds
n
0
1
+ A h(s)g max{ , x(s)} ψ(1)
n
1
+ B h(s)g max{ , x(s)} φ(0)
n
hZ 1
i
≤ M2
G(s, s)h(s)ds + A (h(s)) ψ(1) + B (h(s)) φ(0)
≤
0
Z
≤ M2 (1 + Aψ(1) + Bφ(0))
1
G(s, s)h(s)ds,
0
where M2 = supx∈[ n1 , n1 +M1 ] g(x),
Pm−2
Pm−2
1
αi ρ − i=1 αi φ(ξi )
P
A=
det Pi=1
,
m−2
m−2
∆
− i=1 βi φ(ξi )
i=1 βi
Pm−2
Pm−2 1
− i=1 αi ψ(ξi )
αi
Pm−2
Pi=1
det
B=
.
m−2
∆
ρ − i=1 βi ψ(ξi )
i=1 βi
(2.16)
(2.17)
Therefore, Tn (D) is uniformly bounded.
Now we show that Tn (D) is equicontinuous on [0, 1]. For any ε > 0, since
G(t, s), ψ(t) and φ(t) are uniformly continuous on [0, 1]×[0, 1] and [0, 1], respectively.
There exists δ > 0 such that for any t1 , t2 ∈ [0, 1], |t1 − t2 | < δ implies that
ε min0≤s≤1 G(s, s)
,
R1
3M2 0 G(s, s)h(s)ds
ε
,
|ψ(t1 ) − ψ(t2 )| <
R1
3M2 A 0 G(s, s)h(s)ds
ε
|φ(t1 ) − φ(t2 )| <
.
R1
3M2 B 0 G(s, s)h(s)ds
|G(t1 , s) − G(t2 , s)| <
Consequently, for any x ∈ D, t1 , t2 ∈ [0, 1], |t1 − t2 | < δ, we have
|Tn x(t1 ) − Tn x(t2 )|
Z 1
≤
|G(t1 , s) − G(t2 , s)|fn (s, x(s))ds
0
+ A(fn (s, x(s)))|ψ(t1 ) − ψ(t2 )| + B(fn (s, x(s)))|φ(t1 ) − φ(t2 )|
Z 1
1
|G(t1 , s) − G(t2 , s)|h(s)g max{ , x(s)} ds
≤
n
0
1
+ A h(s)g max{ , x(s)} |ψ(t1 ) − ψ(t2 )|
n
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J. JIANG, L. LIU
EJDE-2009/110
1
+ B h(s)g max{ , x(s)} |φ(t1 ) − φ(t2 )|
n
Z 1
≤ M2
|G(t1 , s) − G(t2 , s)|h(s)ds
0
+ A(h(s)M2 )|ψ(t1 ) − ψ(t2 )| + B(h(s)M2 )|φ(t1 ) − φ(t2 )|
Z 1
≤ M2
|G(t1 , s) − G(t2 , s)|h(s)ds
0
Z
1
+ M2 A|ψ(t1 ) − ψ(t2 )|
G(s, s)h(s)ds
Z
0
1
+ M2 B|φ(t1 ) − φ(t2 )|
G(s, s)h(s)ds
0
ε ε ε
+ + = ε.
3 3 3
Thus, Tn (D) is equicontinuous on [0, 1]. According to Ascoli-Arzela Theorem,
Tn (D) is a relatively compact set.
In the end, we show Tn is continuous. Suppose xm , x ∈ D, xm → x (m → +∞).
Then there exists a constant M3 > 0 such that kxk ≤ M3 , kxm k ≤ M3 (m =
1, 2, . . . ). Since fn (t, x) is uniformly continuous on [0, 1] × D for any fixed natural
number n, hence,
<
lim fn (t, xm (t)) = fn (t, x(t)),
m→+∞
uniformly on t ∈ [0, 1].
According to the Lebesgue dominated convergence theorem,
Z 1
lim
G(s, s)|fn (s, xm (s)) − fn (s, x(s))|ds = 0.
m→+∞
0
Thus for the above ε > 0, there exists a natural number M , such that m > M
implies that
Z 1
ε
.
(2.18)
G(s, s)|fn (s, xm (s)) − fn (s, x(s))|ds <
1
+
Aψ(1)
+ Bφ(0)
0
From (2.18), we obtain that for m > M ,
kTn um − Tn uk
hZ 1
= max
G(t, s)fn (s, xm (s))ds + A(fn (s, xm (s)))ψ(t) + B(fn (s, xm (s)))φ(t)
0≤t≤1
Z
−
1
0
i
G(t, s)fn (s, x(s))ds − A(fn (s, x(s)))ψ(t) − B(fn (s, x(s)))φ(t)
0
Z
≤
1
G(s, s)|fn (s, xm (s)) − fn (s, x(s))|ds
+ A |fn (s, xm (s)) − fn (s, x(s))| ψ(1)
0
+ B(|fn (s, xm (s)) − fn (s, x(s))|)φ(0)
Z
1
≤ 1 + Aψ(1) + Bφ(0)
G(s, s)|fn (s, xm (s)) − fn (s, x(s))|ds < ε.
0
Therefore, Tn : P → P is continuous. Thus Tn : P → P is a completely continuous
operator.
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POSITIVE SOLUTIONS
7
Lemma 2.8. Tn (Q) ⊂ Q.
Proof. For any x ∈ Q, (H2) and (H3) imply (Tn x)(t) ≥ 0. From (2.6), (2.14) and
the monotonicity of ψ(t) and φ(t), we have
1
Z
(Tn x)(t) ≤
G(s, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(1) + B(fn (s, x(s)))φ(0),
0
which implies
1
Z
kTn xk ≤
G(s, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(1) + B(fn (s, x(s)))φ(0). (2.19)
0
By Remarks 2.2 and 2.3, we have
1
Z
(Tn x)(t) =
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t)
0
Z
1
≥γ
G(s, s)fn (s, x(s))ds + A(fn (s, x(s)))γψ(1) + B(fn (s, x(s)))γφ(0)
0
≥γ
hZ
1
i
G(s, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(1) + B(fn (s, x(s)))φ(0) .
0
(2.20)
Then, (2.19) and (2.20) yield
(Tn x)(t) ≥ γkTn xk.
Hence Tn x ∈ Q.
3. Main results
In this section, we present our main results as follows.
Theorem 3.1. Suppose that (H1)–(H3) hold and there exist numbers R > 0 and
L > 0 such that
Z 1
R
,
(3.1)
G(s, s)h(s)ds <
f
M (1 + Aψ(1) + Bφ(0))
0
Z 1−τ
f (t, x)
Lγ 2
G(s, s)ds > 1, lim inf min
> L.
(3.2)
x→+∞ τ ≤t≤1−τ
x
τ
f = maxu∈[γR,1+R] g(u),
Then (1.1)-(1.2) has at least one positive solution, where M
γ is defined in Remark 2.2 and A, B are defined by (2.16) and (2.17), respectively.
Proof. Firstly, we shall prove that when n is sufficiently large, we have
Tn x 6= λx,
x ∈ ∂QR , λ ≥ 1,
(3.3)
where QR = {x ∈ Q : kxk < R} for R > 0. In fact, if there exists x0 ∈ ∂QR ,
and λ0 ≥ 1 such that λ0 x0 = Tn x0 , then x0 (t) ≤ Tn x0 (t) for t ∈ [0, 1] and any n.
8
J. JIANG, L. LIU
EJDE-2009/110
Choose a sufficiently large n satisfying n >
1
γR .
Then we have
x0 (t) ≤ (Tn x0 )(t)
Z 1
=
G(t, s)fn (s, x0 (s))ds + A(fn (s, x0 (s)))ψ(t) + B(fn (s, x0 (s)))φ(t)
0
1
Z
≤
G(s, s)fn (s, x0 (s))ds + A(fn (s, x0 (s)))ψ(1) + B(fn (s, x0 (s)))φ(0)
0
Z
1
≤ (1 + Aψ(1) + Bφ(0))
G(s, s)fn (s, x0 (s))ds
0
Z
1
1
G(s, s)h(s)g max{ , x0 (s)} ds
n
0
Z 1
f
G(s, s)h(s)ds < R.
≤ (1 + Aψ(1) + Bφ(0))M
≤ (1 + Aψ(1) + Bφ(0))
0
(3.4)
Therefore, by (3.4) we have kx0 k < R, which is a contradiction to x0 ∈ ∂QR . So
applying Lemma 1.1, i(Tn , QR , Q) = 1.
Next, according to (3.2), there exists R1 such that x > R1 implies
f (t, x) > Lx,
t ∈ [τ, 1 − τ ].
(3.5)
Choose R0 > {R, γ −1 R1 }. When n being sufficiently large we can claim that
Tn x 6= λx,
∀x ∈ ∂QR0 , λ ∈ (0, 1],
(3.6)
where Q0R = {x ∈ Q : kxk < R0 }. Suppose (3.6) is not true, then there exist
x1 ∈ ∂QR0 and λ0 ∈ (0, 1] such that λ0 x1 = Tn x1 . Similarly, we choose sufficiently
1
large n satisfying that n > γR
0 . Therefore, by (3.5) we have
x1 (t) ≥ (Tn x1 )(t)
Z 1
=
G(t, s)fn (s, x1 (s))ds + A(fn (s, x1 (s)))ψ(t) + B(fn (s, x1 (s)))φ(t)
0
1
Z
≥
G(t, s)fn (s, x1 (s))ds
0
1−τ
Z
≥γ
G(s, s)fn (s, x1 (s))ds
τ
Z
1−τ
≥ Lγ
G(s, s)x1 (s)ds
τ
0 2
Z
≥ LR γ
1−τ
G(s, s)ds.
τ
This is a contradiction to x1 ∈ ∂QR0 . Consequently, (3.6) holds. Furthermore, for
each x ∈ ∂QR ,
Z 1
kTn xk ≥
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t)
0
Z
≥
1
G(t, s)fn (s, x(s))ds
0
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POSITIVE SOLUTIONS
9
1−τ
Z
≥γ
G(s, s)fn (s, x(s))ds
τ
Z
1−τ
≥ Lγ
G(s, s)x(s)ds
τ
≥ LR0 γ 2
Z
1−τ
G(s, s)ds.
τ
So inf x∈∂QR0 kTn xk > 0. Thus from Lemma 1.2, i(Tn , QR0 , Q) = 0.
By the additivity of fixed point index, we know that
i(Tn , QR0 \ QR , Q) = i(Tn , QR0 , Q) − i(Tn , QR , Q) = −1.
As a result, there exist xn ∈ QR0 \ QR satisfying Tn xn = xn provided that n is
sufficiently large.
Without loss of generality, suppose Tn xn = xn , n ≥ n0 . Let D = {xn }n≥n0 be
the sequence of solutions to (2.14). It is not difficult to prove that D is uniformly
bounded. Next we show {xn }n≥n0 is equicontinuous on [0, 1]. It is obvious that
we only need to prove limt→0+ (xn (t) − xn (0)) = 0, limt→1− (xn (t) − xn (1)) = 0
uniformly with respect to n ≥ n0 and D is equicontinuous on [σ, 1 − σ] ⊂ (0, 1) for
σ ∈ (0, 1/2).
Now we prove that
lim (xn (t) − xn (0)) = 0,
t→0+
uniformly with respect to n ≥ n0 .
(3.7)
According to (2.12),
xn (t) − xn (0)
Z 1
=
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t)
0
Z
−
0
1
G(0, s)fn (s, x(s))ds − A(fn (s, x(s))ψ(0) − B(fn (s, x(s))φ(0)
Z t
Z 1
1
1
ψ(s)fn (s, x(s))ds + (ψ(t) − ψ(0))
φ(s)fn (s, x(s))ds
= φ(t)
ρ
ρ
0
t
Z t
1
− φ(0)
ψ(s)fn (s, x(s))ds + A(fn (s, x(s)))(ψ(t) − ψ(0))
ρ
0
+ B(fn (s, x(s)))(φ(t) − φ(0))
Z t
1
1
ψ(s)h(s)g max{ , x(s)} ds
≤ φ(t)
ρ
n
0
Z 1
1
1
φ(s)h(s)g max{ , x(s)} ds
+ (ψ(t) − ψ(0))
ρ
n
t
Z t
1
1
− φ(0)
ψ(s)h(s)g max{ , x(s)} ds
ρ
n
0
1
+ A h(s)g max{ , x(s)} (ψ(t) − ψ(0))
n
1
+ B h(s)g max{ , x(s)} (φ(0) − φ(t))
n
Z t
Z 1
1
1
≤ M4 φ(t)
ψ(s)h(s)ds + M4 (ψ(t) − ψ(0))
φ(s)h(s)ds
ρ
ρ
0
t
(3.8)
10
J. JIANG, L. LIU
1
+ φ(0)M4
ρ
Z
EJDE-2009/110
t
1
Z
ψ(s)h(s)ds + AM4 (ψ(t) − ψ(0))
G(s, s)h(s)ds
0
0
Z
+ BM4 (φ(0) − φ(t))
1
G(s, s)h(s)ds.
0
Since {xn (t)}n≥n0 is uniformly bounded, it follows that {g(xn )}n≥n0 is bounded.
Therefore, there exists a constant M4 such that kg(xn )k < M4 for n ≥ n0 . This
together with (H3) and (3.8) show that we need to prove only that
Z t
1
ψ(s)h(s)ds = 0,
(3.9)
lim φ(t)
t→0+ ρ
0
Z t
1
lim φ(0)
ψ(s)h(s)ds = 0,
(3.10)
t→0+ ρ
0
Z 1
1
lim (ψ(t) − ψ(0))
φ(s)h(s)ds = 0,
(3.11)
t→0+ ρ
t
lim (ψ(t) − ψ(0)) = 0,
lim (φ(0) − φ(t)) = 0.
(3.12)
t→0+
t→0+
Since ψ(t) and φ(t) are continuous on [0, 1], (3.12) holds. For all ε > 0, by the
absolutely continuity of integral function and (H3), there exists δ1 ∈ (0, 21 ) such
that t1 , t2 ∈ [0, 1], |t1 − t2 | < δ1 implies
Z
t2
G(s, s)h(s)ds < ε.
(3.13)
t1
Therefore, from (2.5) and (3.13), we have
Z t
Z t
1
φ(t)
ψ(s)h(s)ds ≤
G(s, s)h(s)ds < ε, t ∈ (0, δ1 ],
ρ
0
0
Z t
Z t
1
φ(0)
φ(0)
ψ(s)h(s)ds ≤
G(s, s)h(s)ds < ε, t ∈ (0, δ1 ];
ρ
φ(δ
1) 0
0
i.e., (3.9) and (3.10) hold.
Z 1
1
(ψ(t) − ψ(0))
φ(s)h(s)ds
ρ
t
Z δ1
Z 1
1
1
≤ ψ(t)
φ(s)h(s)ds + (ψ(t) − ψ(0))
φ(s)h(s)ds
ρ
ρ
t
δ1
Z δ1
Z
ψ(t) − ψ(0) 1
≤
G(s, s)h(s)ds +
G(s, s)h(s)ds ≤ 2ε.
ψ(δ1 )
δ1
t
That is, (3.11) holds. By (3.9)-(3.12), (3.7) holds. Since
|xn (t) − xn (1)|
Z 1
=
G(t, s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t) + B(fn (s, x(s)))φ(t)
0
Z
−
0
1
G(1, s)fn (s, x(s))ds − A(fn (s, x(s))ψ(1) − B(fn (s, x(s))φ(1)
1
≤ (φ(t) − φ(1))
ρ
Z
0
t
1
ψ(s)fn (s, x(s))ds + ψ(t)
ρ
Z
1
φ(s)fn (s, x(s))ds
t
EJDE-2009/110
POSITIVE SOLUTIONS
11
Z 1
1
− φ(1)
ψ(s)fn (s, x(s))ds + A(fn (s, x(s)))(ψ(t) − ψ(1))
ρ
t
+ B(fn (s, x(s)))(φ(t) − φ(1))
Z t
Z 1
1
1
≤ M4 (φ(t) − φ(1))
ψ(s)h(s)ds + M4 ψ(t)
φ(s)h(s)ds
ρ
ρ
0
t
Z t
Z 1
1
ψ(s)h(s)ds + AM4 (ψ(1) − ψ(t))
G(s, s)h(s)ds
+ φ(1)M4
ρ
0
0
Z 1
+ BM4 (φ(t) − φ(1))
G(s, s)h(s)ds,
0
similar to the above, we can easily prove that
lim (xn (t) − xn (1)) = 0,
uniformly with respect to n ≥ n0 .
t→1−
(3.14)
Next we prove that D is equicontinuous on [σ, 1 − σ] for any σ ∈ (0, 1/2). In
fact, for n ≥ n0 , t1 , t2 ∈ [σ, 1 − σ] with t2 > t1 , we have
|xn (t2 ) − xn (t1 )|
Z 1
=
G(t2 , s)fn (s, x(s))ds + A(fn (s, x(s)))ψ(t2 ) + B(fn (s, x(s)))φ(t2 )
0
Z
1
G(t1 , s)fn (s, x(s))ds − A(fn (s, x(s)))ψ(t1 ) − B(fn (s, x(s)))φ(t1 )
0
Z t1
Z t2
1
1
ψ(s)fn (s, x(s))ds
ψ(s)fn (s, x(s))ds + φ(t2 )
≤ (φ(t2 ) − φ(t1 ))
ρ
ρ
0
t1
Z 1
Z t2
1
1
+ (ψ(t2 ) − ψ(t1 ))
φ(s)fn (s, x(s))ds − ψ(t1 )
φ(s)fn (s, x(s))ds
ρ
ρ
t2
t1
+ A(fn (s, x(s)))(ψ(t2 ) − ψ(t1 )) + B(fn (s, x(s)))(φ(t2 ) − φ(t1 )).
(3.15)
By (2.5) and the monotonicity of ψ(t) and φ(t), we have
Z
Z
1 t1
1 1
ψ(s)h(s)ds ≤
G(s, s)h(s)ds,
(3.16)
ρ 0
d 0
Z
Z
1 1
1 1
φ(s)h(s)ds ≤
G(s, s)h(s)ds,
(3.17)
ρ t2
b 0
Z t2
Z t2
1
φ(t2 )
ψ(s)h(s)ds ≤
G(s, s)h(s)ds,
(3.18)
ρ
t1
t1
Z t2
Z t2
1
ψ(t1 )
φ(s)h(s)ds ≤
G(s, s)h(s)ds.
(3.19)
ρ
t1
t1
−
By (3.15)-(3.19), we have
|xn (t2 ) − xn (t1 )|
1
≤ M4 ( + A)(ψ(t2 ) − ψ(t1 ))
b
1
Z
G(s, s)h(s)ds
0
1
+ M4 ( + B)(φ(t1 ) − φ(t2 ))
d
Z
1
Z
t2
G(s, s)h(s)ds + 2M4
0
G(s, s)h(s)ds.
t1
12
J. JIANG, L. LIU
EJDE-2009/110
By the above inequality, (H3), (3.9)-(3.12), and continuity of ψ(t), φ(t), D is
equicontinuous on [σ, 1 − σ].
From the above proof, we can know D is equicontinuous on [0, 1]. It follows
from Ascoli-Arzela’s theorem that the sequence {xn }n≥n0 has a subsequence which
uniformly converges on [0, 1]. Without loss of generality, we assume that {xn } itself
uniformly converges to x on [0, 1]. According to the Lebesgue’s dominated theorem,
we know that x is the positive solution of (1.1)-(1.2).
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Jiqiang Jiang
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong,
China
E-mail address: qfjjq@mail.qfnu.edu.cn, qfjjq@163.com
Lishan Liu
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong,
China
E-mail address: lls@mail.qfnu.edu.cn, Tel. 86-537-4456234, Fax 86-537-4455076