Electronic Journal of Differential Equations, Vol. 2010(2010), No. 22, pp. 1–20.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MONOTONE POSITIVE SOLUTIONS FOR p-LAPLACIAN
EQUATIONS WITH SIGN CHANGING COEFFICIENTS AND
MULTI-POINT BOUNDARY CONDITIONS
JIANYE XIA, YUJI LIU
Abstract. We prove the existence of three monotone positive solutions for
the second-order multi-point boundary value problem, with sign changing coefficients,
[p(t)φ(x0 (t))]0 + f (t, x(t), x0 (t)) = 0,
x0 (0) = −
l
X
ai x0 (ξi ) +
i=1
x(1) + βx0 (1) =
k
X
m
X
t ∈ (0, 1),
ai x0 (ξi ),
i=l+1
bi x(ξi ) −
i=1
m
X
bi x(ξi ) −
i=k+1
m
X
ci x0 (ξi ).
i=1
To obtain these results, we use a fixed point theorem for cones in Banach
spaces. Also we present an example that illustrates the main results.
1. Introduction
As is well known, a differential equation defined on the interval a ≤ t ≤ b having
the form
[p(t)x0 (t)]0 + (q(t) + λr(t))x(t) = 0,
a1 x(a) + a2 x0 (a) = a3 x(b) + a4 x0 (b) = 0,
is called a Sturm-Liouville boundary-value problem or Sturm-Liouville system. Here
p(t) > 0, q(t), the weighting function r(t), the constants a1 , a2 , a3 , a4 are given, and
the eigenvalue λ is an unspecified parameter.
Sturm-Liouville boundary-value problems for nonlinear second-order p-Laplacian
differential equations have been studied extensively; see for example [1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 15, 16]. The study of existence of positive solutions for such problems
is complicated since there is no Green’s function for the p-Laplacian (p 6= 2).
2000 Mathematics Subject Classification. 34B10, 34B15, 35B10.
Key words and phrases. Second order differential equation; positive solution
multi-point boundary value problem.
c
2010
Texas State University - San Marcos.
Submitted October 26, 2009. Published February 4, 2010.
Supported by grants 06JJ5008 and 7004569 from the Natural Science Foundation of
Hunan and Guangdong provinces, China.
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2
J. XIA, Y. LIU
EJDE-2010/22
Some authors have extend Sturm-Liouville boundary conditions to nonlinear
cases. For example, He, Ge and Peng [7], by means of the Leggett-Williams fixedpoint theorem, established criteria for the existence of at least three positive solutions to the one-dimensional p-Laplacian boundary-value problem
(φ(y 0 ))0 + g(t)f (t, y) = 0,
t ∈ (0, 1),
0
y(0) − B0 (y (0)) = 0,
(1.1)
0
y(1) + B1 (y (1)) = 0,
where φ(v) = |v|p−2 v with v > 1, under the assumptions xB0 (x) ≥ 0, xB1 (x) ≥ 0
and that there exist constants Mi > 0 such that
|Bi (x)| ≤ Mi |x|,
x ∈ R.
(1.2)
In [9, 10, 11, 7], the authors extended (1.1) to a more general case. They established
some existence results of at least one positive solution of Sturm-Liouville boundary
value problems of higher order differential equations. Liu [12] studied the boundaryvalue problem
[φ(x(n−1) (t))]0 = f (t, x(t), x0 (t), . . . , x(n−2) (t)),
(i)
x (0) = 0
(n−2)
x
0 < t < 1,
for i = 0, 1, . . . , n − 3,
(0) − B0 (x(n−1) (0)) = 0,
(1.3)
B1 (x(n−2) (1)) + x(n−1) (1) = 0.
which is a general case of (1.1). Liu [12] established existence of at least one positive
solution of (1.3) without assumption (1.2).
The Sturm-Liouville boundary conditions have been extended to multi-point
cases. For example, Ma [15, 16] studied the problem
[p(t)x0 (t)]0 − q(t)x(t) + f (t, x(t)) = 0,
αx(0) − βp(0)x0 (0) =
γx(1) + δp(1)x0 (1) =
m
X
i=1
m
X
t ∈ (0, 1),
ai x(ξi ),
(1.4)
bi x(ξi ),
i=1
where 0 < ξ1 < · · · < ξm < 1, α, β, γ, δ ≥ 0, ai , bi ≥ 0 with ρ = γβ + αγ + αδ >
0. By using Green’s functions (which is complicate for studying (1.1)) and GuoKrasnoselskii fixed point theorem [4, 6], the existence and multiplicity of positive
solutions for (1.4) were given. There is no paper discussing the existence of multiple
positive solutions of (1.4) by using Leggett-Williams fixed point theorem. Liu in [10,
11] also studied some Sturm-Liouville type multi-point boundary value problems.
In recent papers [8, 17], the authors studied the four-point boundary-value problem
(φ(x0 ))0 + f (t, x(t), x0 (t)) = 0, t ∈ (0, 1),
αx(0) − βx0 (ξ) = 0,
(1.5)
0
γx(1) + δx (η) = 0,
p−2
−1
where φ(x) = |x| x, p > 1, φ (x) = |x|q−2 x with 1/p + 1/q = 1, α > 0, β ≥ 0,
γ > 0, δ ≥ 0, 0 < ξ < η < 1, f is continuous and nonnegative. When ξ → 0 and
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MONOTONE POSITIVE SOLUTIONS
3
η → 1, (1.5) converges to a Sturm-Liouville boundary-value problem. So (1.5) can
be seen as a generalized Sturm-Liouville boundary-value problem.
Xu [18] proved the existence of at least one or two positive solutions of the
differential equation
[φ(x0 (t))]0 + a(t)f (x(t)) = 0,
x0 (0) =
m
X
t ∈ (0, 1),
ai x0 (ξi ),
(1.6)
i=1
x(1) =
k
X
bi x(ξi ) −
i=1
s
X
bi x(ξi ) −
m
X
ci x0 (ξi ),
i=s+1
i=k+1
where 0 < ξ1 < · · · < ξm < 1, ai , bi , ci ≥ 0, a and f are continuous functions,
φ(x) = |x|p−2 x with p > 1.
Motivated by above mentioned papers, we investigate the boundary-value problem
[p(t)φ(x0 (t))]0 + f (t, x(t), x0 (t)) = 0, t ∈ (0, 1),
x0 (0) = −
l
X
i=1
x(1) + βx0 (1) =
k
X
i=1
m
X
ai x0 (ξi ) +
bi x(ξi ) −
ai x0 (ξi ),
(1.7)
i=l+1
m
X
bi x(ξi ) −
i=k+1
m
X
ci x0 (ξi ),
i=1
where
• 0 < ξ1 < · · · < ξm < 1, β ≥ 0, 1 ≤ k, l ≤ m and ai ≥ 0, bi ≥ 0, ci ≥ 0 for all
i = 1, . . . , m;
• f is defined on [0, 1] × R × R, continuous, nonnegative with f (t, 0, 0) 6≡ 0
on each subinterval of [0, 1];
• p is defined on [0, 1], continuous and positive;
• φ is called p-Laplacian, φ(x) = |x|p−2 x with p > 1, its inverse function is
denoted by φ−1 (x) with φ−1 (x) = |x|q−2 x with 1/p + 1/q = 1.
Sufficient conditions for the existence of at least three monotone positive solutions of (1.7) are established by using a fixed point theorem for cones in Banach
spaces.
We remark that the fixed point theorem used here is different form the one in
[18]. Our results improve the the results in [18] since: Three positive solutions are
obtained, while one or two positive solutions were obtained in [18]; the nonlinearity
in (1.7) depends on t, x, x0 , while the nonlinearity in [18] depends only on t, x. The
main result for this article is presented in Section 2, and an example is given in
Section 3.
2. Main Results
In this section, we first present some background definitions and state an important fixed point theorem. Then the main results are given and proved.
Definition 2.1. Let X be a semi-ordered real Banach space. The nonempty convex
closed subset P of X is called a cone in X if x + y ∈ P and ax ∈ P for all x, y ∈ P
and a ≥ 0, and x ∈ X and −x ∈ X imply x = 0.
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J. XIA, Y. LIU
EJDE-2010/22
Definition 2.2. Let X be a semi-ordered real Banach space and P a cone in X. A
map ψ : P → [0, +∞) is a nonnegative continuous concave (or convex) functional
map provided ψ is nonnegative, continuous and satisfies
ψ(tx + (1 − t)y) ≥ (or ≤) tψ(x) + (1 − t)ψ(y)
for all x, y ∈ P, t ∈ [0, 1].
Definition 2.3. Let X be a semi-ordered real Banach space. An operator T ; X →
X is completely continuous if it is continuous and maps bounded sets into precompact sets.
Let a1 , a2 , a3 , a4 > 0 be positive constants, ψ be a nonnegative continuous functional on the cone P . Define the sets as follows:
P (β1 ; a4 ) = {x ∈ P : β1 (x) < a4 },
P (β1 , α1 ; a2 , a4 ) = {x ∈ P : α1 (x) ≥ a2 , β1 (x) ≤ a4 },
P (β1 , β2 , α1 ; a2 , a3 , a4 ) = {x ∈ P : α1 (x) ≥ a2 , β2 (x) ≤ a3 , β1 (x) ≤ a4 }
and a closed set
R(β1 , ψ; a1 , a4 ) = {x ∈ P : ψ(x) ≥ a1 , β1 (x) ≤ a4 }.
Theorem 2.4 ([3]). Let P be a cone in a real Banach space X with the norm k · k.
Suppose that
(1) T : P → P is completely continuous;
(2) β1 and β2 be nonnegative continuous convex functionals on P , α1 be a
nonnegative continuous concave functional on P , and ψ be a nonnegative
continuous functional on P satisfying ψ(λx) ≤ λψ(x) for all x ∈ P and
λ ∈ [0, 1], and α1 (x) ≤ ψ(x) and there exists a constant M > 0 such that
kxk ≤ M β1 (x) for all x ∈ P ;
(3) there exist positive numbers a1 < a2 , a3 and a4 such that
(E1) T (P (β1 ; a4 )) ⊆ P (β1 ; a4 );
(E2) α1 (T x) > a2 for all x ∈ P (β1 , α1 ; a2 , a4 ) with β2 (T x) > a3 ;
(E3) {x ∈ P (β1 , β2 , α1 ; a2 , a3 , a4 ) : α1 (x) > a2 } =
6 ∅ and α1 (T x) > b for all
x ∈ P (β1 , β2 , α1 ; a2 , a3 , a4 );
(E4) 0 6∈ R(β1 , ψ; a1 , a4 ) and ψ(T x) < a1 for all x ∈ R(β1 , ψ; a1 , a4 ) with
ψ(x) = a1 ;
Then T has at least three fixed points x1 , x2 , x3 ∈ P (β1 ; a4 ) such that
β1 (xi ) ≤ a4 ,
α1 (x1 ) > a2 ,
ψ(x2 ) > a1 ,
i = 1, 2, 3;
α1 (x2 ) < a2 ,
ψ(x3 ) < a1 .
Choose X = C 1 [0, 1]. We call x ≤ y for x, y ∈ X if x(t) ≤ y(t) for all t ∈ [0, 1],
for x ∈ X, define its norm by
kxk = max max |x(t)|, max |x0 (t)| .
t∈[0,1]
t∈[0,1]
It is easy to see that X is a semi-ordered real Banach space. Let
n
P = y ∈ X : y(t) ≥ 0 for all t ∈ [0, 1], y 0 (t) ≤ 0 is decreasing on [0, 1],
y(t) ≥ (1 − t)y(0) for all t ∈ [0, 1],
0
x(1) + βp(1)x (1) =
k
X
i=1
bi x(ξi ) −
m
X
i=k+1
bi x(ξi ) −
m
X
i=1
bi x0 (ξi ),
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MONOTONE POSITIVE SOLUTIONS
x0 (0) = −
l
X
m
X
ai x0 (ξi ) +
i=1
5
o
ai x0 (ξi ) .
i=l+1
Then P is a nonempty cone in X. For σ ∈ (0, 1/2). Define functionals from P to
R by
β1 (y) = max |y 0 (t)|,
ψ(x) = max |y(t)|,
t∈[0,1]
β2 (y) =
max |y(t)|,
t∈[0,1]
α1 (y) =
t∈[σ,1−σ]
|y(t)|,
min
y ∈ P.
t∈[σ,1−σ]
Let us list some conditions to be used in this article.
(A1) p : [0, 1] → (0, +∞) is continuous;
(A2) ai ≥ 0, bi ≥ 0, ci ≥ 0 for all i = 1, . . . , m satisfying
m
X
i=l+1
m
X
ai
p(0) p(ξi )
ai φ−1
i=l+1
−
l
X
i=1
ai
p(0)
< 1,
p(ξi )
k
X
i=1
l
1 X
1 ai φ−1
≥
,
p(ξi )
p(ξi )
i=1
s
X
bi −
bi < 1,
i=k+1
k
X
bi ≥
i=1
m
X
bi ;
i=k+1
(A2’) ai ≥ 0, bi ≥ 0 for all i = 1, . . . , m satisfying
m
X
i=l+1
m
X
i=l+1
ai φ−1
ai
p(0) p(ξi )
< 1,
k
X
bi < 1,
i=1
l
1 X
1 ≥
ai φ−1
,
p(ξi )
p(ξi )
i=1
k
X
bi ≥
i=1
m
X
bi ;
i=k+1
(A3) β ≥ 0;
(A4) f : [0, 1] × [0, +∞) × R → [0, +∞) is continuous with f (t, 0, 0) 6≡ 0 on each
sub-interval of [0, 1];
(A5) θ : [0, 1] → [0, +∞) is a continuous function and θ(t) 6≡ 0 on each subinterval of [0, 1].
It is easy to see that (A2) holds if (A2’) holds.
Lemma 2.5. Suppose that p : [0, 1] → (0, +∞) with p ∈ C 1 [0, 1], x ∈ X, x(t) ≥ 0
for all t ∈ [0, 1] and [p(t)φ(x0 (t))]0 ≤ 0 on [0, 1]. Then x is concave and
x(t) ≥ min{t, 1 − t} max x(t),
t ∈ [0, 1].
(2.1)
t∈[0,1]
Proof. Suppose x(t0 ) = maxt∈[0,1] x(t). If t0 < 1, for t ∈ (t0 , 1), since x0 (t0 ) ≤ 0,
we have p(t0 )φ(x0 (t0 )) ≤ 0. Then p(t)φ(x0 (t)) ≤ 0 for all t ∈ (t0 , 1]. It follows that
x0 (t) ≤ 0 for all t ∈ [0, 1]. Thus x(t0 ) ≥ x(t) ≥ x(1) ≥ 0. Let
R t −1 1 φ
p(s) ds
t
τ (t) = R 10
.
1
φ−1 p(s)
ds
t0
Then τ ∈ C([t0 , 1], [0, 1]) and is increasing on [t0 , 1] since
1
φ−1 p(t)
dτ
= R1
> 0,
dt
φ−1 1 ds
t0
p(s)
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J. XIA, Y. LIU
EJDE-2010/22
τ (t0 ) = 0 and τ (1) = 1. Thus
−1
dx
dx dτ
dx φ
=
=
R
dt
dτ dt
dτ 1 φ−1
1
p(t)
1
p(s)
t0
ds
which implies
dx 1
R
p(t)φ(x0 (t)) = φ
1
dτ φ
φ−1
t0
1
p(s)
.
ds
Hence
0
φ
dx dτ
R1
t0
φ−1
1
p(t)
φ−1
1
p(s)
Z 1
0
d2 x
1 =
φ
φ−1
ds p(t)φ(x0 (t)) ≤ 0
2
p(s)
ds dτ
t0
2
for all t ∈ [t0 , 1]. Note that φ0 (x) ≥ 0. It follows that ddτx2 ≤ 0. Together with
x00 (τ ) ≤ 0(τ ∈ [0, 1]). Then x0 is decreasing on [0, 1]. We get that there exist
t0 ≤ η ≤ t ≤ ξ ≤ 1 such that
x(t0 ) − x(1) x(t) − x(1)
(t − 1)[x(t0 ) − x(t)] + (t0 − t)[x(1) − x(t)]
−
=−
t0 − 1
t−1
(t − 1)(t0 − 1)
(t − 1)(t0 − t)x0 (η) + (t0 − t)(1 − t)tx0 (ξ)
=−
(t − 1)(t0 − 1)
(t − 1)(t0 − t)x0 (ξ) + (t0 − t)(1 − t)tx0 (ξ)
≤−
= 0.
(t − 1)(t0 − 1)
It follows that for t ∈ (t0 , 1),
x(t) ≥ x(1) + (t − 1)
x(t0 ) − x(1)
1−t 1−t
= x(1) 1 −
+
x(t0 ) ≥ (1 − t)x(t0 ).
t0 − 1
1 − t0
1 − t0
If t0 > 0, for t ∈ (0, t0 ), similarly to above discussion, we have
x(t) ≥ tx(t0 ), t ∈ (0, t0 ).
Then one gets that x(t) ≥ min{t, 1 − t} maxt∈[0,1] x(t) for all t ∈ [0, 1]. The proof
is complete.
Consider the boundary-value problem
[p(t)φ(x0 (t))]0 + θ(t) = 0,
x0 (0) = −
l
X
x(1) + βx0 (1) =
k
X
i=1
m
X
ai x0 (ξi ) +
i=1
bi x(ξi ) −
t ∈ (0, 1),
ai x0 (ξi ),
(2.2)
i=l+1
s
X
i=k+1
bi x(ξi ) −
m
X
ci x0 (ξi ).
i=s+1
Lemma 2.6. Suppose that (A1)–(A5) hold. If x is a solution of (2.2), then x is
concave, decreasing and positive on (0, 1).
Proof. Suppose x satisfies (2.2). It follows from the assumptions that px0 is decreasing on [0, 1]. Lemma 2.5 implies that x0 is decreasing on [0, 1]. Then x is
concave on [0, 1].
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MONOTONE POSITIVE SOLUTIONS
7
First, we prove that x0 (0) ≤ 0. One sees from (A2) and the deceasing property
of p(t)φ(x0 (t)) that
x0 (1) = −
l
X
m
X
1 −1
1 −1
φ (p(ξi )φ(x0 (ξi ))) +
φ (p(ξi )φ(x0 (ξi )))
ai
p(ξi )
p(ξi )
ai φ−1
i=1
≤−
+
l
X
i=l+1
1 −1
φ (p(ξl )φ(x0 (ξl )))
p(ξi )
ai φ−1
i=1
m
X
ai φ−1
1 −1
φ (p(ξl+1 )φ(x0 (ξl+1 )))
p(ξi )
ai φ−1
l
1 X
1 −1
φ (p(ξl+1 )φ(x0 (ξl+1 )))
−
ai φ−1
p(ξi )
p(ξ
)
i
i=1
i=l+1
=
m
X
i=l+1
+
l
X
1 −1
[φ (p(ξl+1 )φ(x0 (ξl+1 ))) − φ−1 (p(ξl )φ(x0 (ξl )))]
p(ξi )
ai φ−1
i=1
≤
m
X
ai φ−1
l
1 X
1 −1
ai φ−1
−
φ (p(ξl+1 φ(x0 (ξl+1 )))
p(ξi )
p(ξ
)
i
i=1
ai φ−1
l
1 −1
1 X
ai φ−1
−
φ (p(0)φ(x0 (0))).
p(ξi )
p(ξ
)
i
i=1
i=l+1
≤
m
X
i=l+1
It follows that
1−
m
X
i=l+1
0
ai
p(0) p(ξi )
+
l
X
ai
i=1
p(0) 0
x (0) ≤ 0.
p(ξi )
0
It follows that x (0) ≤ 0. Then x (t) ≤ 0 for all t ∈ [0, 1].
Second, we prove that x(1) ≥ 0. It follows from the boundary conditions in (2.2)
that
m
s
k
X
X
X
ci x0 (ξi )
bi x(ξi ) −
bi x(ξi ) −
x(1) + βx0 (1) =
≥
i=1
i=k+1
k
X
s
X
bi x(ξi ) −
i=1
≥
k
X
k
X
k
X
k
X
i=1
bi x(ξk+1 )
i=k+1
bi −
s
X
s
X
bi x(ξk ) +
bi [x(ξk ) − x(ξk+1 )]
i=k+1
bi −
i=1
≥
s
X
bi x(ξk ) −
i=1
≥
bi x(ξi )
i=k+1
i=1
=
i=s+1
s
X
i=k+1
bi x(ξk )
i=k+1
bi −
s
X
i=k+1
bi x(1).
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J. XIA, Y. LIU
EJDE-2010/22
Then
1−
k
X
i=1
s
X
bi +
bi x(1) + βx0 (1) ≥ 0.
i=k+1
We get x(1) ≥ 0 since x0 (1) ≤ 0 and β ≥ 0. Then x(t) > x(1) ≥ 0 for all t ∈ [0, 1).
The proof is complete.
Lemma 2.7. Suppose that (A1),(A2’), (A3)–(A5) hold. Let
1
− 1.
µ = φ Pm
−1 p(0)
i=l+1 ai φ
p(ξi )
If y is a solution of (2.2), then
Z 1
Z s
1
1
y(t) = Bθ −
φ−1
θ(u)du ds, t ∈ [0, 1],
p(0)φ(Aθ ) −
p(s)
p(s) 0
t
where
h
R 1 θ(u)du i
−1
0
,0 ,
Aθ ∈ − φ
µp(0)
Z ξi
l
1
X
1
−1
Aθ = −
ai φ
p(0)φ(Aθ ) −
θ(u)du
p(ξi )
p(ξi ) 0
i=1
Z ξi
m
X
1
1
+
ai φ−1
p(0)φ(Aθ ) −
θ(u)du
p(ξi )
p(ξi ) 0
i=l+1
and
Z 1
1
1
Bθ =
− βφ
p(0)φ(Aθ ) −
θ(u)du
Pk
Pm
p(1)
p(1) 0
1 − i=1 bi + i=k+1 bi
Z
Z
k
s
1
1
X
1
p(0)φ(Aθ ) −
θ(u)du ds
−
bi
φ−1
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X
1
−1
+
bi
φ
p(0)φ(Aθ ) −
θ(u)du ds
p(s)
p(s) 0
ξi
i=k+1
Z
m
ξi
1
i
X
1
−
ci φ−1
p(0)φ(Aθ ) −
θ(u)du ds .
p(ξi )
p(ξi ) 0
i=1
1
h
−1
Proof. Since y is solution of (2.2),
Z t
1
1
y 0 (t) = φ−1
p(0)φ(y 0 (0)) −
θ(u)du ,
p(t)
p(t) 0
Z 1
Z s
1
1
y(t) = y(1) −
φ−1
p(0)φ(y 0 (0)) −
θ(u)du ds.
p(s)
p(s) 0
t
The boundary conditions in (2.2) imply
Z ξi
l
1
X
1
y 0 (0) = −
ai φ−1
p(0)φ(y 0 (0)) −
θ(u)du
p(ξi )
p(ξi ) 0
i=1
Z ξi
m
1
X
1
−1
0
+
ai φ
p(0)φ(y (0)) −
θ(u)du
p(ξi )
p(ξi ) 0
i=l+1
EJDE-2010/22
MONOTONE POSITIVE SOLUTIONS
9
and
Z 1
1
1
θ(u)du
p(0)φ(y 0 (0)) −
p(1)
p(1) 0
Z s
Z
k
1
X
1
1
θ(u)du ds
=
p(0)φ(y 0 (0)) −
bi y(1) −
φ−1
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X 1
−
bi y(1) −
φ−1
θ(u)du ds
p(0)φ(y 0 (0)) −
p(s)
p(s) 0
ξi
i=k+1
Z ξi
m
1
X
1
θ(u)du .
−
p(0)φ(y 0 (0)) −
ci φ−1
p(ξi )
p(ξi ) 0
i=1
y(1) + βφ−1
It follows that
y(1) =
Z 1
1
1
0
p(0)φ(y
(0))
−
θ(u)du
Pk
Pm
p(1)
p(1) 0
1 − i=1 bi + i=k+1 bi
Z s
Z 1
k
1
X
1
p(0)φ(y 0 (0)) −
θ(u)du ds
−
bi
φ−1
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X
1
+
bi
φ−1
p(0)φ(y 0 (0)) −
θ(u)du ds
p(s)
p(s) 0
ξi
i=k+1
Z ξi
m
i
1
X
1
−
p(0)φ(y 0 (0)) −
θ(u)du ds .
ci φ−1
p(ξi )
p(ξi ) 0
i=1
1
h
− βφ−1
Lemma 2.6 implies that y 0 (0) ≤ 0. One finds that
l
X
Z ξi
1
1
p(0)φ(y 0 (0)) −
θ(u)du
p(ξi )
p(ξi ) 0
i=1
Z ξi
m
1
X
1
+
ai φ−1
p(0)φ(y 0 (0)) −
θ(u)du
p(ξi )
p(ξi ) 0
i=l+1
Z ξi
m
1
X
1
0
−1
≥
ai φ
p(0)φ(y (0)) −
θ(u)du .
p(ξi )
p(ξi ) 0
y 0 (0) = −
ai φ−1
i=l+1
If y 0 (0) 6= 0, one gets
Z ξi
m
1
X
1
1
−1
1−
ai φ
p(0) −
θ(u)du
≤ 0.
p(ξi )
φ(y 0 (0)) p(ξi ) 0
i=l+1
Let
G(c) = c −
m
X
i=l+1
Then
ai φ−1
Z ξi
1
1
p(0)φ(c) −
θ(u)du .
p(ξi )
p(ξi ) 0
Z ξi
m
1
X
G(c)
1
1
−1
=1−
ai φ
p(0) −
θ(u)du .
c
p(ξi )
φ(c) p(ξi ) 0
i=l+1
Note that
µ = φ Pm
1
−1 p(0)
i=l+1 ai φ
p(ξi )
− 1.
(2.3)
10
J. XIA, Y. LIU
EJDE-2010/22
It is easy to see that G(c)
is decreasing on (−∞, 0) and on (0, +∞). First, since
c
limt→0+ G(c)/c = +∞ and
m
1
X
G(c)
=1−
p(0) > 0,
lim
ai φ−1
c→+∞
c
p(ξi )
i=l+1
we get that G(c)
c > 0 for all c > 0. Second, we have
R1
θ(u)du
m
1
G − φ−1 0 µp(0)
X
R1
=1−
ai φ−1
p(0)
θ(u)du
p(ξi )
−φ−1 0 µp(0)
i=l+1
=1−
m
X
ai φ−1
i=t+1
0
Z
ξi
θ(u)du
m
X
0
ai φ−1
p(0) p(ξi )
= 0.
≤ 0. We get
R 1
0 ≥ y 0 (0) ≥ −φ−1
0
θ(u)du .
µp(0)
The proof is complete.
Define the nonlinear operator T : X → X by
Z s
Z 1
1
1
−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds,
(T x)(t) = Bx −
φ
p(s)
p(s) 0
t
for t ∈ [0, 1], where
h
−1
Aσ ∈ − φ
R1
0
φ Pm
i
f (u, x(u), x0 (u))du
,0 ,
1
− 1 p(0)
p(0)
−1
i=t+1
l
X
ai φ
p(ξi )
Z ξi
1
1
Ax = −
ai φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
p(ξi )
p(ξi ) 0
i=1
Z ξi
m
1
X
1
−1
+
ai φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ,
p(ξi )
p(ξi ) 0
−1
i=l+1
Bx =
1
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
h
1
1
× − βφ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
p(1)
p(1) 0
Z s
Z
k
1
X
1
1
f (u, x(u), x0 (u))du ds
−
bi
φ−1
p(0)φ(Ax ) −
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X
1
−1
+
bi
φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=k+1
R ξi
1
θ(u)du 1
p(0) + µ R01
p(0)
p(ξi )
θ(u)du p(ξi )
i=l+1
G(y 0 (0))
y 0 (0)
1
+ R1
p(ξ
i)
θ(u)du
p(0)µ
0
≥ 1 − φ−1 (1 + µ)
It follows from (2.3) that
limc→0− G(c)/c = −∞ and
EJDE-2010/22
−
MONOTONE POSITIVE SOLUTIONS
m
X
−1
ci φ
i=1
11
Z ξi
i
1
1
f (u, x(u), x0 (u))du ds .
p(0)φ(Ax ) −
p(ξi )
p(ξi ) 0
Lemma 2.8. Suppose that (A1),(A2’), (A3), (A4) hold. Then
(i) the following equalities hold:
[p(t)φ((T y)0 (t))]0 + f (t, y(t), y 0 (t)) = 0,
(T y)0 (0) = −
l
X
i=1
(T y)(1) + β(T y)0 (1) =
k
X
m
X
ai (T y)0 (ξi ) +
t ∈ (0, 1),
ai (T y)0 (ξi ),
i=l+1
bi (T y)(ξi ) −
i=1
m
X
bi (T y)(ξi ) −
m
X
ci (T y)0 (ξi );
i=1
i=k+1
(ii) T y ∈ P for each y ∈ P ;
(iii) x is a positive solution of (1.7) if and only if x is a solution of the operator
equation y = T y in P ;
(iv) T : P → P is completely continuous.
Proof. The proofs of (i), (ii) and (iii) are simple. To prove (iv), it suffices to prove
that T is continuous on P and T is relative compact. We divide the proof into two
steps:
Step 1. For each bounded subset D ⊂ P , and each x0 ∈ D, since f (t, u, v) is
continuous, we can prove that T is continuous at y(t).
Step 2. For each bounded subset D ⊂ P , prove that T is relative compact on
D. It is similar to that of the proof of Lemmas in [13] and are omitted.
Lemma 2.9. Suppose that (A1), (A2’), (A3), (A4) hold. Then there exists a
constant M > 0 such that
max (T x)(t) ≤ M max |(T x)0 (t)|
t∈[0,1]
for each x ∈ P.
t∈[0,1]
Proof. For x ∈ P , Lemma 2.6 implies that (T x)(t) ≥ 0 and (T x)0 (t) ≤ 0 for all
t ∈ [0, 1]. Lemma 2.8 implies
(T x)(1) + β(T x)0 (1) =
k
X
i=1
bi (T x)(ξi ) −
m
X
i=k+1
bi (T x)(ξi ) −
m
X
ci (T x)0 (ξi ).
i=1
Then there exist numbers ηi ∈ [ξi , 1] such that
Pk
Pm
(T x)(1) − i=1 bi (T x)(1) + i=k+1 bi (T x)(1)
(T x)(1) =
Pk
Pm
1 − i=1 bi + i=k+1 bi
Pm
−β(T x)0 (1) − i=1 ci (T x)0 (ξi )
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
Pk
Pm
i=k+1 bi [(T x)(ξi ) − (T x)(1)]
i=1 bi [(T x)(ξi ) − (T x)(1)] −
+
Pk
Pm
1 − i=1 bi + i=k+1 bi
Pm
0
−β(T x) (1) − i=1 ci (T x)0 (ξi )
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
12
J. XIA, Y. LIU
Pk
i=1 bi (ξi
+
EJDE-2010/22
Pm
− 1)(T x)0 (ηi ) − i=k+1 bi (ξi − 1)(T x)0 (ηi )
.
Pk
Pm
1 − i=1 bi + i=k+1 bi
It follows that
|(T x)(t)|
= |(T x)(t) − (T x)(1) + (T x)(1)|
≤ |(T x)(1)| + (1 − t)|(T x)0 (ξ)| where ξ ∈ [t, 1]
Pk
Pm
Pm
βp(1) + i=1 bi (1 − ξi ) + i=k+1 bi (1 − ξi ) + i=1 ci max |(T x)0 (t)|.
≤ 1+
Pk
Pm
t∈[0,1]
1 − i=1 bi + i=k+1 bi
Then
max |(T x)(t)|
t∈[0,1]
≤ 1+
βp(1) +
Pk
Pm
Pm
− ξi ) + i=k+1 bi (1 − ξi ) + i=1 ci max |(T x)0 (t)|.
Pk
Pm
t∈[0,1]
1 − i=1 bi + i=k+1 bi
i=1 bi (1
It follows that there exists a constant M > 0 such that for all x ∈ P ,
max (T x)(t) ≤ M β1 ((T x)) .
t∈[0,1]
Denote
1
µ = φ Pm
− 1,
p(0)
ai φ−1 p(ξi )
Pm
Pm
βp(1) + i=1 bi (1 − ξi ) + i=k+1 bi (1 − ξi ) + i=1 ci
M =1+
,
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
Z ξi
k
m
X
1 + µs X
−1 1 + µs
−1 1 + µ
bi
bi
+
φ
ds +
φ−1
ds
L1 = βφ
µp(1)
µp(s)
µp(s)
ξi
0
i=1
i=t+1
Pk
i=k+1
+
m
X
ci φ−1
i=1
L2 = βφ
−1
1 + µξi
µp(ξi )
ds + 1 −
Z 1
k
1 X
+
bi
φ−1
p(1)
ξ
i
i=1
k
X
1
φ−1
1 + µs ds,
µp(s)
Z ξi
m
X
s s ds +
bi
φ−1
ds
p(s)
p(s)
0
i=1
bi
Z
0
i=k+1
m
X
k
ξ Z 1
X
s i
−1
+
ci φ
ds + 1 −
bi
φ−1
ds.
p(ξ
)
p(s)
i
0
i=1
i=1
Theorem 2.10. Choose k ∈ (0, 1). Suppose that (A1), (A2’), (A3), (A4) hold. Let
e1 , e2 , c be positive numbers and
n c 1 − Pk bi + Pm
µp(1)φ(c) o
i=k+1 bi
i=1
Q = min φ
,
;
L1
1+µ
e2 1 − Pk bi + Pm
i=1
i=k+1 bi
W =φ
;
L2
EJDE-2010/22
MONOTONE POSITIVE SOLUTIONS
13
e1 1 − Pk bi + Pm
i=1
i=k+1 bi
E=φ
.
L1
If M c > e2 >
e1
k
> e1 > 0 and
(A6) f (t, u, v) ≤ Q for all t ∈ [0, 1], u ∈ [0, M c], v ∈ [−c, c];
(A7) f (t, u, v) ≥ W for all t ∈ [0, k], u ∈ [e2 , e2 /k], v ∈ [−c, c];
(A8) f (t, u, v) ≤ E for all t ∈ [0, 1], u ∈ [0, e1 ], v ∈ [−c, c];
then (1.7) has at least three solutions x1 , x2 , x3 such that
x1 (0) < e1 ,
x2 (k) > e2 ,
x3 (0) > e1 ,
x3 (k) < e2 .
Proof. To apply Theorem 2.4, we check that all its conditions are satisfied. By the
definitions, it is easy to see that α1 is a nonnegative continuous concave functional
on the cone P , β1 , β2 are three nonnegative continuous convex functionals on the
cone P , ψ a nonnegative continuous functional on the cone P . Lemma 2.8 implies
that x = x(t) is a positive solution of (1.7) if and only if x is a solution of the
operator equation y = T y and T is completely continuous.
For x ∈ P , one sees that ψ(λx) ≤ λψ(x) for all x ∈ P and λ ∈ [0, 1], and
α1 (x) ≤ ψ(x) for all x ∈ P . There exists a constant M > 0 such that kxk ≤ M β1 (x)
for all x ∈ P . Then (1) and (2) of Theorem 2.4 hold.
Corresponding to Theorem 2.4, we have a4 = c, a3 = ek2 , a2 = e2 , a1 = e1 . Now,
we check that (1.3) of Theorem 2.4 holds. One sees that 0 < a1 < a2 , a3 > 0,
a4 > 0. The rest is divided into four steps.
Step 1. Prove that T (P1 (β1 ; a4 )) ⊆ P1 (β1 ; a4 ); For x ∈ P1 (β1 ; a4 ), we have
β1 (x) ≤ a4 = c. Then
0 ≤ x(t) ≤ max x(t) ≤ M c for t ∈ [0, 1],
t∈[0,1]
−c ≤ x0 (t) ≤ c for all t ∈ [0, 1].
So (A6) implies that f (t, x(t), x0 (t)) < Q for t ∈ [0, 1]. Then Lemma 2.7 implies
R 1 f (u, x(u), x0 (u))du Q 0 ≥ Ax ≥ −φ−1 0
≥ −φ−1
.
µp(0)
µp(0)
Since
c ≥ max
e2
2 1 , La, φ−1
a, φ−1
a ,
σ0
p(0)
p(1)
we obtain
c φ(c)p(0)
φ(a) ≤ min φ
,
, φ(c)p(1) = Q.
L
2
Then maxt∈[0,1] (T y)(t) = (T y)(0) implies
(T x)(0)
Z
= Bx −
1
φ−1
0
=
Z s
1
1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
1
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
h
1
1
−1
× − βφ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
p(1)
p(1) 0
14
J. XIA, Y. LIU
k
X
EJDE-2010/22
Z s
1
1
f (u, x(u), x0 (u))du ds
p(0)φ(Ax ) −
p(s)
p(s) 0
ξi
i=1
Z s
Z
m
1
1
X
1
f (u, x(u), x0 (u))du ds
+
p(0)φ(Ax ) −
bi
φ−1
p(s)
p(s) 0
ξi
i=k+1
Z ξi
m
1
i
X
1
−
f (u, x(u), x0 (u))du ds
ci φ−1
p(0)φ(Ax ) −
p(ξi )
p(ξi ) 0
i=1
Z 1
Z s
1
1
φ−1
−
f (u, x(u), x0 (u))du ds
p(0)φ(Ax ) −
p(s)
p(s) 0
0
1
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
h
1
1
−1
p(0)φ(Ax ) −
× − βφ
f (u, x(u), x0 (u))du
p(1)
p(1) 0
Z s
Z
k
1
1
X
1
−1
p(0)φ(Ax ) −
−
bi
φ
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=1
Z s
Z
m
ξi
1
X
1
−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
bi
φ
−
p(s)
p(s) 0
0
i=k+1
Z
m
ξi
1
X
1
−
ci φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(ξi )
p(ξi ) 0
i=1
Z s
k
Z 1
i
1
X
1
bi
p(0)φ(Ax ) −
− 1−
φ−1
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
0
i=1
h
1
Q Q
≤
+
βφ−1
Pk
Pm
µp(1) p(1)
1 − i=1 bi + i=k+1 bi
Z 1
Z ξi
k
m
X
X
Qs Qs Q
Q
+
bi
+
ds +
bi
+
ds
φ−1
φ−1
µp(s) p(s)
µp(s) p(s)
ξi
0
i=1
−
Z
1
bi
φ−1
i=k+1
+
m
X
−1
ci φ
i=1
=
1−
k
Z 1
X
Qξi Qs i
Q
Q
+
ds + 1 −
bi
+
ds
φ−1
µp(ξi ) p(ξi )
µp(s) p(s)
0
i=1
φ−1 (Q)
Pm
i=1 bi +
i=k+1 bi
Pk
Z 1
k
h
1 + µ X
1 + µs × βφ−1
+
bi
φ−1
ds
µp(1)
µp(s)
ξi
i=1
Z ξi
m
m
1 + µs 1 + µξ X
X
i
ds +
ci φ−1
ds
+
bi
φ−1
µp(s)
µp(ξ
)
i
0
i=1
i=k+1
k
Z
X
+ 1−
bi
i=1
0
1
φ−1
1 + µs i
ds ≤ c.
µp(s)
EJDE-2010/22
MONOTONE POSITIVE SOLUTIONS
15
On the other hand, similarly to above discussion, since (T y)0 (t) is decreasing and
(T y)0 (0) ≤ 0, from Lemma 2.7 we have
Z 1
1
1
0
−1
(T x) (1) = φ
f (u, x(u), x0 (u))du ,
p(0)φ(Ax ) −
p(1)
p(1) 0
Z 1
1
1
f (u, x(u), x0 (u))du
max |(T x)0 (t)| = −φ−1
p(0)φ(Ax ) −
p(1)
p(1) 0
t∈[0,1]
Q
Q
≤ φ−1
≤ c.
+
µp(1) p(1)
It follows that kT xk = max maxt∈[0,1] |(T x)(t)|, maxt∈[0,1] |(T x)0 (t)| ≤ c. Then
T (P (β1 ; a4 )) ⊆ P (β1 ; a4 ). This completes the proof of (E1) in Theorem 2.4.
Step 2. Prove that α1 (T x) > a2 for all x ∈ P (β1 , α1 ; a2 , a4 ) with β2 (T x) > a3 ;
For y ∈ P (β1 , α1 ; a2 , a4 ) = P (β1 , α1 ; e2 , c) with β2 (T y) > a3 = ek2 , we have
e2
α1 (y) = min y(t) ≥ e2 , β1 (y) = max |y 0 (t)| ≤ c, max (T y)(t) > .
k
t∈[0,k]
t∈[0,1]
t∈[0,k]
Then
α1 (T y) = min (T y)(t) ≥ kβ2 (T y) > k
t∈[0,k]
e2
= e2 = a2 .
k
This completes the proof of (E2) in Theorem 2.4.
Step 3. Prove that {x ∈ P (β1 , β2 , α1 ; a2 , a3 , a4 ) : α1 (x) > a2 } 6= ∅ and
α1 (T x) > b for all x ∈ P (β1 , β2 , α1 ; a2 , a3 , a4 ); It is easy to check that {x ∈
P (β1 , β2 , α1 ; a2 , a3 , a4 ) : α1 (x) > a2 } =
6 ∅. For y ∈ P (β1 , β2 , α1 ; a2 , a3 , a4 ), one
has that
e2
α1 (y) = min y(t) ≥ e2 , β2 (y) = max y(t) ≤ , β1 (y) = max |y 0 (t)| ≤ c.
k
t∈[0,k]
t∈[0,k]
t∈[0,1]
Then
e2 ≤ y(t) ≤
e2
,
k
t ∈ [0, k], |y 0 (t)| ≤ c.
Thus (A7) implies
f (t, y(t), y 0 (t)) ≥ W,
t ∈ [0, k].
Since
α1 (T y) = min (T y)(t) ≥ k max (T y)(t),
t∈[0,k]
t∈[0,1]
we obtain α1 (T y) ≥ k maxt∈[0,1] (T y)(t). Then
α1 (T y) ≥ k max (T y)(t) ≥ k(T y)(0).
t∈[0,1]
From Ax ≤ 0, we obtain
h
1
α1 (T1 y) ≥ k
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
1
1
× − βφ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
p(1)
p(1) 0
Z s
Z
k
1
X
1
1
−
bi
φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X
1
−1
+
bi
φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=k+1
16
J. XIA, Y. LIU
m
X
EJDE-2010/22
Z ξi
1
1
−
f (u, x(u), x0 (u))du ds
ci φ
p(0)φ(Ax ) −
p(ξi )
p(ξi ) 0
i=1
Z 1
Z s
1
i
1
−1
φ
−
f (u, x(u), x0 (u))du ds
p(0)φ(Ax ) −
p(s)
p(s) 0
0
k
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
1
h
1
f (u, x(u), x0 (u))du
× − βφ−1
p(0)φ(Ax ) −
p(1)
p(1) 0
Z s
Z
k
1
1
X
1
f (u, x(u), x0 (u))du ds
p(0)φ(Ax ) −
−
bi
φ−1
p(s)
p(s) 0
ξi
i=1
Z
Z s
m
1
1
X
1
−1
+
bi
φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=k+1
Z ξi
m
1
X
1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
−
ci φ−1
p(ξi )
p(ξi ) 0
i=1
m
k
Z 1
1
X
X
bi
bi +
φ−1
− 1−
p(0)φ(Ax )
p(s)
0
i=1
i=k+1
Z s
i
1
−
f (u, x(u), x0 (u))du ds
p(s) 0
k
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
h
1
1
−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
× − βφ
p(1)
p(1) 0
Z
Z s
k
1
1
X
1
−1
−
bi
φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=1
Z s
Z ξi
m
1
X
1
−
bi
φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
0
i=k+1
Z
m
ξi
1
X
1
−
ci φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(ξi )
p(ξi ) 0
i=1
k
Z 1
1
X
− 1−
bi
φ−1
p(0)φ(Ax )
p(s)
0
i=1
Z s
i
1
−
f (u, x(u), x0 (u))du ds
p(s) 0
1
≥
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
Z ξi
k
m
h
X
W X
Ws Ws × βφ−1
+
bi
φ−1
ds +
bi
φ−1
ds
p(1)
p(s)
p(s)
ξi
0
i=1
−1
i=k+1
EJDE-2010/22
MONOTONE POSITIVE SOLUTIONS
+
m
X
ci φ−1
i=1
=
17
k
Z 1
X
W ξi Ws i
ds + 1 −
ds
bi
φ−1
p(ξi )
p(s)
0
i=1
φ−1 (W )
Pm
1 − i=1 bi + i=k+1 bi
Z 1
Z ξi
k
m
h
X
1 X
s s −1
−1
+
ds +
ds
× βφ
bi
φ
bi
φ−1
p(1)
p(s)
p(s)
ξi
0
i=1
Pk
i=k+1
+
m
X
ci φ−1
i=1
ξi
p(ξi )
k
Z
X
ds + 1 −
bi
i=1
1
φ−1
0
s i
ds = e2 .
p(s)
This completes the proof of (E3) in Theorem 2.4.
Step 4. Prove that 0 6∈ R(β1 , ψ; a1 , a4 ) and that ψ(T x) < a1 for all x in
R(β1 , ψ; a1 , a4 ) with ψ(x) = a1 ; It is easy to see that 0 6∈ R(β1 , ψ; a1 , a4 ). For
y ∈ R(β1 , ψ; a1 , a4 ) with ψ(x) = a1 , one has that
ψ(y) = max y(t) = a1 ,
t∈[0,1]
β1 (y) = max |y 0 (t)| ≤ a4 .
t∈[0,1]
Hence
0 ≤ y(t) ≤ a1 ,
t ∈ [0, 1];
−c ≤ y 0 (t) ≤ c,
t ∈ [0, 1].
Then (A8) implies f (t, y(t), y 0 (t)) ≤ E, t ∈ [0, 1]. So
ψ(T y) = max (T y)(t) = (T y)(0)
t∈[0,1]
1
=
Pk
Pm
1 − i=1 bi + i=k+1 bi
Z 1
h
1
1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du
× − βφ−1
p(1)
p(1) 0
Z 1
Z s
k
1
X
1
−1
bi
−
φ
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
ξi
i=1
Z s
Z
m
ξ
i
X
1
1
−
bi
φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
0
i=k+1
Z
m
ξi
1
X
1
−
ci φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(ξi )
p(ξi ) 0
i=1
Z s
k
i
Z 1
1
X
1
− 1−
bi
φ−1
p(0)φ(Ax ) −
f (u, x(u), x0 (u))du ds
p(s)
p(s) 0
0
i=1
h
E
1
E βφ−1
≤
+
Pm
Pk
µp(1) p(1)
1 − i=1 bi + i=k+1 bi
Z ξi
Z
k
m
1
E
E
X
X
Es Es −1
+
bi
φ
+
ds +
bi
φ−1
+
ds
µp(s) p(s)
µp(s) p(s)
ξi
0
i=1
i=k+1
+
m
X
i=1
ci φ−1
k
Z 1
E
X
E
Eξi Es i
+
ds + 1 −
bi
φ−1
+
ds
µp(ξi ) p(ξi )
µp(s) p(s)
0
i=1
18
J. XIA, Y. LIU
=
EJDE-2010/22
Z 1
k
1 + µ X
1 + µs h
φ−1 (E)
+
ds
βφ−1
bi
φ−1
Pm
µp(1)
µp(s)
1 − i=1 bi + i=k+1 bi
ξi
i=1
Z ξi
m
m
1 + µs 1 + µξ X
X
i
+
ds +
ds
bi
φ−1
ci φ−1
µp(s)
µp(ξ
)
i
0
i=1
Pk
i=k+1
k
Z
X
+ 1−
bi
i=1
0
1
φ−1
1 + µs i
ds ≤ a1 .
µp(s)
This completes the proof of (E4) in Theorem 2.4.
Then Theorem 2.4 implies that T has at least three fixed points x1 , x2 and x3
such that
β(x1 ) < e1 ,
α(x2 ) > e2 ,
β(x3 ) > e1 ,
α(x3 ) < e2 .
Hence (1.7) has three decreasing positive solutions x1 , x2 and x3 as needed in the
statement of the theorem; therefore, the proof is complete.
3. Examples
Now, we present an example that illustrates our main results, and that can not
be solved by results in [3, 9, 10, 15, 16].
Example 3.1. Consider the boundary-value problem
x00 (t) + f (t, x(t), x0 (t)) = 0, t ∈ (0, 1),
1
1
x0 (0) = − x0 (1/4) + x0 (1/2),
4
2
1
1
1
1
0
x(1) + 2x (1) = x(1/4) − x(1/2) − x0 (1/4) − x0 (1/2),
4
4
2
4
t|v|
and
where f (t, u, v) = f0 (u) + 25500000
2
u


 51

8


51


564000− 8

564000 + 1004−1251 (u − 1004)
f0 (u) =

564000





564000



564000eu−2000004
(3.1)
u ∈ [0, 4],
u ∈ [4, 12],
u ∈ [12, 1004],
u ∈ [1004, 4004],
u ∈ [4004, 2000004],
u ≥ 2000004.
Corresponding to (1.7), one sees that φ−1 (x) = x, ξ1 = 1/4, ξ2 = 1/2, a1 =
1/4, a2 = 1/2, b1 = 1/4, b2 = 1/4, c1 = 1/2, c2 = 1/4. It is easy to see that
(A1)–(A4) hold. Choose constants k = 21 , e1 = 2, e2 = 100 and c = 20000. One
obtains
Pm
Pm
Pk
βp(1) + i=1 bi (1 − ξi ) + i=k+1 bi (1 − ξi ) + i=1 ci
51
M =1+
=
,
Pk
Pm
4
1 − i=1 bi + i=k+1 bi
1
− 1 = 1,
µ=φ P
m
−1 p(0)
i=t+1 ai φ
p(ξi )
EJDE-2010/22
MONOTONE POSITIVE SOLUTIONS
L1 = βφ−1
1 + µ
+
k
X
Z
1
φ−1
19
1 + µs ds
µp(1)
µp(s)
ξi
i=1
Z ξi
m
m
1 + µs 1 + µξ X
X
i
+
ds +
ds
bi
φ−1
ci φ−1
µp(s)
µp(ξ
)
i
0
i=1
bi
i=k+1
k
Z
X
+ 1−
bi
1
φ−1
1 + µs ds,
µp(s)
Z 1
k
1 X
s L2 = βφ−1
+
ds
bi
φ−1
p(1)
p(s)
ξi
i=1
Z ξi
m
m
X
X
s ξi +
bi
φ−1
ds +
ds
ci φ−1
p(s)
p(ξi )
0
i=1
i=1
0
i=k+1
k
Z
X
+ 1−
bi
1
s ds,
p(s)
0
i=1
Pk
Pm
c 1 − i=1 bi + i=k+1 bi µp(1)φ(c) Q = min φ
,
;
L1
1+µ
P
P
e2 1 − ki=1 bi + m
i=k+1 bi
W =φ
;
L2
P
P
e1 1 − ki=1 bi + m
i=k+1 bi
E=φ
.
L1
It is easy to see that If M c > e2 >
e1
k
φ−1
> e1 > 0 and
f (t, u, v) ≤ Q for all t ∈ [0, 1], u ∈ [0, M c], v ∈ [−20000, 20000];
f (t, u, v) ≥ W
for all t ∈ [0, 1/2], u ∈ [100, 200], v ∈ [−20000, 20000];
f (t, u, v) ≤ E
for all t ∈ [0, 1], u ∈ [0, 2], v ∈ [−20000, 20000];
Then (A6), (A7) and (A8) hold. Theorem 2.10 implies that (3.1) has at least three
positive solutions such that
x1 (0) < 2,
x2 (k) > 100,
x3 (0) > 2,
x3 (k) < 100.
References
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3(1999), 9–14.
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Ordered Banach Spaces, Computers Mathematics with Applications, 42(2001), 313–322.
[3] Z. Bai, Z. Gui, W. Ge; Multiple positive solutions for some p-Laplacian boundary value
problems. J. Math. Anal. Appl. 300(2004), 477–490.
[4] K. Deimling; Nonlinear Functional Analysis, Springer, Berlin, Germany, 1985.
[5] W. Ge; Boundary Value Problems for Ordinary Differential Equations, Science Press, Beijing,
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value problems, Appl. Math. Letters, 15(2002)937-943.
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p-Laplacian, J. Math. Anal. Appl. 277(2003), 293–302.
[10] Y. Liu; Solutions of Sturm-Liouville type multi-point boundary value problems for higherorder differential equations, J. Appl. Math. Computing, 23(2007), 167–182.
[11] Y. Liu; Solutions of Sturm-liouville boundary value problems for higher-order differential
equations, J. Appl. Math. Computing, 24(2007),231–243.
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Problems for Higher-Order p-Laplacian Differential Equations, Rocky Mountain J. Math.,
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p-Laplacian. Electronic Journal of Diff. Eqns. 2008(2008), no. 20, pp 1–43.
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theorem, Electronic Journal of Diff. Eqns. 2008(2008), no. 96, pp 1–52.
[15] R. Ma; Multiple positive solutions for nonlinear m-point boundary value problems. Appl.
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Anal. Appl. 297(2004), 24–37.
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Jianye Xia
Department of Mathematics, Guangdong University of Finance, Guangzhou 510320,
China
E-mail address: JianyeXia@sohu.com
Yuji Liu
Department of Mathematics, Hunan Institute of Science and Technology, Yueyang
414000, China
E-mail address: liuyuji888@sohu.com