Electronic Journal of Differential Equations, Vol. 2008(2008), No. 147, pp. 1–29.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
MULTIPLE POSITIVE SOLUTIONS FOR SINGULAR m-POINT
BOUNDARY-VALUE PROBLEMS WITH NONLINEARITIES
DEPENDING ON THE DERIVATIVE
YA MA, BAOQIANG YAN
Abstract. Using the fixed point theorem in cones, this paper shows the existence of multiple positive solutions for the singular m-point boundary-value
problem
x00 (t) + a(t)f (t, x(t), x0 (t)) = 0,
x0 (0) = 0,
x(1) =
m−2
X
0 < t < 1,
ai x(ξi ),
i=1
where 0 < ξ1 < ξ2 < · · · < ξm−2 < 1, ai ∈ [0, 1), i = 1, 2, . . . , m − 2, with
P
0
0 < m−2
i=1 ai < 1 and f maybe singular at x = 0 and x = 0.
1. Introduction
The study of multi-point boundary-value problems (BVP) for linear second-order
ordinary differential equations was initiated by Il’in and Moiseev [5, 6]. Since then,
many authors have studied general nonlinear multi-point BVP; see for examples
[4, 17], and references therein. Gupta, Ntouyas and Tsamatos [4] considered the
existence of a solution in C 1 [0, 1] for the m-point boundary-value problem
x00 (t) = f (t, x(t), x0 (t)) + e(t),
x0 (0) = 0,
x(1) =
m−2
X
0 < t < 1,
ai x(ξi ),
(1.1)
i=1
where ξi ∈ (0, 1), i = 1, 2, . . . , m − 2, 0 < ξ1 < ξ2 < · · · < ξm−2 < 1, ai ∈ R,
Pm−2
i = 1, 2, . . . , m−2, have the same sign, i=1 ai 6= 1, e ∈ L1 [0, 1], f : [0, 1]×R2 → R
is a function satisfying Carathèodory conditions and a growth condition of the form
|f (t, u, v)| ≤ p1 (t)|u| + q1 (t)|v| + r1 (t), where p1 , q1 , r1 ∈ L1 [0, 1]. Recently, using
Leray-Schauder continuation theorem, Ma and O’Regan proved the existence of
positive solutions of C 1 [0, 1) solutions for the above BVP, where f : [0, 1] × R2 → R
satisfies the Carathèodory conditions (see [17]). Khan and Webb [10] obtained
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. m-point boundary-value problem; singularity; positive solutions;
fixed point theorem.
c
2008
Texas State University - San Marcos.
Submitted August 13, 2008. Published October 24, 2008.
Supported by grants 10871120 from the National Natural Science, and Y2005A07
from the Natural Science of Shandong Province, China.
1
2
Y. MA, B. YAN
EJDE-2008/147
a interesting result which presents the multiplicity of existence of at least three
solutions of a second-order three-point boundary-value problem. However, up to
now, there are a fewer results on the existence of multiple solutions to (1.1). In
view of the importance of the research on the multiplicity of positive solutions for
differential equations [1, 2, 5, 9, 10, 11, 12, 14, 15], the goal of this paper is to fill
this gap in the literature.
There are main four sections in this paper. In section 2, we give a special cone
and its properties. In section 3, using the theory of fixed point index on a cone, we
present the existence of multiple positive solutions to (1.1) with f may be singular
at x0 = 0 but not at x = 0. In section 4, under the condition f is singular at
x0 = 0 and x = 0, we present the existence of multiple positive solutions to (1.1).
In section 5, under the condition f is singular at x = 0 but not at x0 = 0, we present
the existence of multiple positive solutions to (1.1).
2. Preliminaries
Let C [0, 1] = {x : [0, 1] → R such that x(t) be continuous on [0, 1] and x0 (t)
continuous on [0, 1] } with norm kxk = max{γkxk1 , γδkxk2 }, where
1
kxk2 = max |x0 (t)|,
kxk1 = max |x(t)|,
t∈[0,1]
t∈[0,1]
Pm−2
ai (1 − ξi )
,
Pm−2
1 − i=1 ai ξi
i=1
γ=
δ=
m−2
X
ai (1 − ξi ).
i=1
Let
P = {x ∈ C 1 [0, 1] : x(t) ≥ γkxk1 , ∀t ∈ [0, 1], x(0) ≥ δkxk2 }.
Obviously, C 1 [0, 1] is a Banach space and P is a cone in C 1 [0, 1].
Lemma 2.1. Let Ω be a bounded open set in real Banach space E, θ ∈ Ω, P be a
cone in E and A : Ω̄ ∩ P → P be continuous and completely continuous. Suppose
λAx 6= x,
∀x ∈ ∂Ω ∩ P, λ ∈ (0, 1].
(2.1)
Then i(A, Ω ∩ P, P ) = 1.
Lemma 2.2. Let Ω be a bounded open set in real Banach space E, θ ∈ Ω, P be a
cone in E and A : Ω̄ ∩ P → P be continuous and completely continuous. Suppose
Ax 6≤ x,
∀x ∈ ∂Ω ∩ P.
(2.2)
Then i(A, Ω ∩ P, P ) = 0.
Let R+ = (0, +∞), R− = (−∞, 0), R = (−∞, +∞). The following conditions
will be used in this article.
a(t) ∈ C(0, 1) ∩ L1 [0, 1],
a(t) > 0, t ∈ (0, 1);
f ∈ C([0, 1] × R+ × R− , [0, +∞));
(2.3)
(2.4)
There exists g ∈ C([0, +∞) × (−∞, 0], [0, +∞)) such that
f (t, x, y) ≤ g(x, y), ∀(t, x, y) ∈ [0, 1] × R+ × R− .
(2.5)
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MULTIPLE POSITIVE SOLUTIONS
3
For x ∈ P and t ∈ [0, 1], define operator
Z
(Ax)(t) = −
t
(t − s)a(s)f (s, x(s)x0 (s))ds
0
+
−
1−
m−2
X
Z
1
Pm−2
ai
i=1
ξi
Z
ai
1
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
(2.6)
(ξi − s)a(s)f (s, x(s), x0 (s)) ds .
0
i=1
Lemma 2.3 ([17]). Assume (2.1). Then for y ∈ C[0, 1] the problem
x00 + y(t) = 0, t ∈ (0, 1)
0
x (0) =
m−2
X
0
bi x (ξi ),
x(1) =
i=1
m−2
X
ai x(ξi )
(2.7)
i=1
has a unique solution
Z
x(t) = −
t
(t − s)y(s)ds + M t + N,
(2.8)
0
where,
Rξ
bi 0 i a(s)y(s)ds
,
M=
Pm−2
i=1 bi − 1
m−2
Z 1
X Z
(1 − s)a(s)y(s)ds −
ai
Pm−2
i=1
N=
1
Pm−2
1 − i=1 ai 0
i=1
Pm−2 R ξi
m−2
X
bi 0 a(s)y(s)ds
(1 −
ai ξi ) .
− i=1Pm−2
i=1 bi − 1
i=1
ξi
(ξi − s)a(s)y(s)ds
0
Further, if y ≥ 0, for all t ∈ [0, 1], x satisfies
inf x(t) ≥ γkxk1 ,
(2.9)
t∈[0,1]
where γ =
Pm−2
i=1
Pm−2
ai (1 − ξi ) / 1 − i=1 ai ξi .
Lemma 2.4. Suppose (2.3)–(2.5)hold. Then A : P → P is a completely continuous
operator.
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Y. MA, B. YAN
EJDE-2008/147
Proof. For x ∈ P , from (2.6), one has
Z 1
(Ax)(t) ≥ −
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
+
−
1−
m−2
X
ai
i=1
ξi
Z
ai
−
0
i=1 ai
P
m−2
i=1
m−2
X
ai
(2.10)
1
Z
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
1
Z
(ξi − s)a(s)f (s, x(s), x0 (s))ds
ai
0
i=1
Pm−2
ai (1 − ξi )
Pm−2
1 − i=1 ai
i=1
≥
0
i=1
1−
(1 − s)a(s)f (s, x(s), x0 (s))ds
(ξi − s)a(s)f (s, x(s), x0 (s))ds
Pm−2
≥
1
Z
1
Pm−2
1
Z
a(s)f (s, x(s), x0 (s))ds ≥ 0,
t ∈ [0, 1],
0
1
Z
(t − s)a(s)f (s, x(s)x0 (s))ds
|(Ax)(t)| = −
0
+
−
1−
m−2
X
≤
≤
ai
i=1
ξi
Z
ai
1
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
(ξi − s)a(s)f (s, x(s), x0 (s))ds
0
i=1
≤
Z
1
Pm−2
1−
1
Pm−2
1−
1
Pm−2
1−
1
Pm−2
i=1
i=1
i=1
1
Z
ai
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
1
Z
ai
a(s)f (s, x(s), x0 (s))ds
0
1
Z
a(s)ds
ai
0
max
0≤c≤kxk,−kxk≤c0 ≤0
g(c, c0 ) < +∞,
t ∈ [0, 1],
(2.11)
Z t
|(Ax)0 (t)| = −
a(s)f (s, x(s)x0 (s))ds
0
Z t
=
a(s)f (s, x(s)x0 (s))ds
0
Z
(2.12)
1
0
≤
a(s)f (s, x(s)x (s))ds
0
Z
≤
1
a(s)ds
0
max
0≤c≤kxk,−kxk≤c0 ≤0
g(c, c0 ) < +∞,
which implies that A is well defined.
Z 1
1
0
(Ax)(0) =
Pm−2 ( (1 − s)a(s)f (s, x(s), x (s))ds
1 − i=1 ai 0
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MULTIPLE POSITIVE SOLUTIONS
−
m−2
X
Z
1−
−
1
Pm−2
i=1
m−2
X
Z
=
m−2
X
ai
(
Z
1
ai
(1 − s)a(s)f (s, x(s), x0 (s))ds
0
i=1
1
(ξi − s)a(s)f (s, x(s), x0 (s))ds)
ai
0
i=1
m−2
X
(ξi − s)a(s)f (s, x(s), x0 (s))ds)
0
i=1
≥
ξi
ai
5
Z
1
a(s)f (s, x(s), x0 (s))ds.
ai (1 − ξi )
0
i=1
On the other hand,
kAxk2 = max |(Ax)0 (t)|
t∈[0,1]
Z
= max | −
t∈[0,1]
Z
=
t
a(s)f (s, x(s)x0 (s))ds|
0
1
a(s)f (s, x(s)x0 (s))ds.
0
Then
(Ax)(0) ≥ δkAxk2 .
(2.13)
By Lemma 2.3, we have (Ax)(t) ≥ γkAxk1 . As a result Ax ∈ P , which implies
AP ⊆ P . By a standard argument, we know that A : P → P is continuous and
completely continuous.
3. Singularities at x0 = 0 but not at x = 0
In this section the nonlinearity f may be singular at x0 = 0 but not at x = 0.
We will assume that the following conditions hold.
(H1) a(t) ∈ C(0, 1) ∩ L1 [0, 1], a(t) > 0, t ∈ (0, 1)
(H2) f (t, u, z) ≤ h(u)[g(z) + r(z)], where f ∈ C([0, 1] × R+ × R− , R+ ), g(z) > 0
continuous and nondecreasing on R− , h(u) ≥ 0 continuous and nondecreasing on R+ , r(z) > 0 continuous and non-increasing on (−∞, 0];
(H3)
c
> 1,
sup Pm−2
R
a
ξ
+1
i
i
−1 (h(c) 1 a(s)ds)
c∈R+ − i=1
Pm−2
I
0
1−
a
i=1
R0
i
du
where I(z) = z g(u)+r(u)
, z ∈ R− ;
(H4) There exists a function g1 ∈ C([0, +∞) × (−∞, 0], [0, +∞)), such that
f (t, u, z) ≥ g1 (u, z), ∀(t, u, z) ∈ [0, 1] × R+ × R− , and limu→+∞ g1 (u,z)
=
u
+∞, uniformly for z ∈ R− .
(H5) There exists a function ΨH ∈ C([0, 1], R+ ) and a constant 0 ≤ δ < 1 such
that f (t, u, z) ≥ ΨH (t)uδ , for all (t, u, z) ∈ [0, 1] × [0, H] × R− .
6
Y. MA, B. YAN
EJDE-2008/147
For n ∈ {1, 2, . . . } and x ∈ P , define operator
Z t
1
(An x)(t) = −
(t − s)a(s)f (s, x(s), −|x0 (s)| − )ds
n
0
Z 1
1
1
(1 − s)a(s)f (s, x(s), −|x0 (s)| − )ds
+
Pm−2
(3.1)
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, x(s), −|x0 (s)| − )ds , t ∈ [0, 1].
n
0
i=1
Theorem 3.1. Suppose (H1)–(H5) hold. Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1) with x0,1 (t), x0,2 (t) > 0, t ∈ [0, 1].
Proof. Choose R1 > 0 such that
R1
Pm−2
R1
a
ξ
+1
i i
i=1
Pm−2
I −1 (h(R1 ) 0
− 1−
a
i
i=1
> 1.
(3.2)
a(s)ds)
From the continuity of I −1 and h, we can choose ε > 0 and ε < R1 with
R1
Pm−2
R1
a
ξ
+1
i i
i=1
Pm−2
I −1 (h(R1 ) 0
− 1−
i=1 ai
> 1,
(3.3)
a(s)ds + I(−ε))
n0 ∈ {1, 2, . . . } with n10 < min{ε, δ/2} and let N0 = {n0 , n0 + 1, . . . }.
Lemma 2.4 guarantees that for n ∈ N0 , An : P → P is a completely continuous
operator. Let
Ω1 = {x ∈ C 1 [0, 1] : kxk < R1 }.
We show that
x 6= µAn x,
∀x ∈ P ∩ ∂Ω1 , µ ∈ (0, 1], n ∈ N0 .
(3.4)
In fact, if there exists an x0 ∈ P ∩ ∂Ω1 and µ0 ∈ (0, 1] such that x0 = µ0 An x0 ,
Z t
1
x0 (t) = −µ0
(t − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
n
0
Z 1
µ0
1
+
(1 − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
Pm−2
n
1 − i=1 ai 0
Z
m−2
ξi
X
1
−
ai
(ξi − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds , t ∈ [0, 1].
n
0
i=1
Then
Z
t
1
)ds, ∀t ∈ [0, 1].
(3.5)
n
0
Obviously, x00 (t) ≤ 0, t ∈ (0, 1), and since x0 (1) > 0, x0 (t) > 0, t ∈ [0, 1]. Differentiating (3.5), we have
x00 (t)
= −µ0
a(s)f (s, x0 (s), −|x00 (s)| −
1
) = 0, 0 < t < 1,
n
m−2
X
x0 (1) =
ai x0 (ξi ).
x000 (t) + µ0 a(t)f (t, x0 (t), x00 (t) −
x00 (0) = 0,
i=1
(3.6)
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MULTIPLE POSITIVE SOLUTIONS
7
Then
1
)
n
−x000 (t) = µ0 a(t)f (t, x0 (t), x00 (t) −
≤ a(t)h(x0 (t))[g(x00 (t) −
g(x00 (t) −
−x000 (t)
1
0
n ) + r(x0 (t)
− n1 )
1
1
) + r(x00 (t) − )],
n
n
≤ a(t)h(x0 (t)),
∀t ∈ (0, 1),
∀t ∈ (0, 1).
Integrating from 0 to t, we have
I(x00 (t)
1
1
− ) − I(− ) ≤
n
n
t
Z
Z
0
1
− ) ≤ h(R1 )
n
Z
Z
t
Then
x00 (t)
≥I
−1
a(s)ds,
0
and
I(x00 (t)
t
a(s)h(x0 (s))ds ≤ h(R1 )
t
a(s)ds + I(−ε).
0
(h(R1 )
a(s)ds + I(−ε));
0
that is,
−x00 (t)
≤ −I
−1
Z
(h(R1 )
t
a(s)ds + I(−ε)),
t ∈ (0, 1).
(3.7)
0
Then
Rξ
ai 0 i −x00 (s)ds
Pm−2
1 − i=1 ai 0
1 − i=1 ai
Z 1
Z s
1
≤
−I −1 h(R1 )
a(τ )dτ + I(−ε) ds
Pm−2
1 − i=1 ai 0
0
Pm−2
Z ξi
Z s
−1
i=1 ai
−I
(h(R
)
+
a(τ )dτ + I(−ε))ds
P
1
m−2
1 − i=1 ai 0
0
Z 1
Z 1
1
−I −1 h(R1 )
≤
a(τ )dτ + I(−ε) ds
Pm−2
1 − i=1 ai 0
0
Pm−2
Z ξi
Z 1
−1
i=1 ai
−I (h(R1 )
a(τ )dτ + I(−ε))ds
+
Pm−2
1 − i=1 ai 0
0
Pm−2
Z 1
1 + i=1 ai ξi −1
−I
h(R1 )
a(s)ds + I(−ε) .
=
Pm−2
1 − i=1 ai
0
x0 (0) =
µ0
Pm−2
Z
1
−x00 (s)ds −
µ0
Pm−2
i=1
Since x0 (0) ≥ x0 (t) ≥ γkx0 k1 ≥ γx0 (0) ≥ γδkx0 k2 , x0 (0) ≥ kx0 k = R1 ,
R1
Pm−2
R1
1+ i=1
ai ξi −1
− 1−Pm−2 a I (h(R1 ) 0
i
i=1
≤ 1,
(3.8)
a(s)ds + I(−ε))
which is a contradiction to (3.3). Then (3.4) holds.
From Lemma 2.1, for n ∈ N0 ,
i(An , Ω1 ∩ P, P ) = 1.
(3.9)
Now we show that there exists a set Ω2 such that
An x 6≤ x,
∀x ∈ ∂Ω2 ∩ P.
(3.10)
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Y. MA, B. YAN
EJDE-2008/147
Choose a∗ with 0 < a∗ < 1. Let
−1
1
N∗ =
+ 1.
Pm−2
R1
i=1 ai (1−ξi )
Pm−2
γa∗ 1−
a(s)ds
0
a
i
i=1
From (H4), there exists R2 > R1 such that
g1 (x, y) ≥ N ∗ x,
1
Let Ω2 = {x ∈ C [0, 1] : kxk <
R2
a∗ }.
∀x ≥ R2 , y ∈ R− .
(3.11)
Then
Ax 6≤ x,
∀x ∈ ∂Ω2 ∩ P.
In fact, if there exists x0 ∈ ∂Ω2 ∩ P with x0 ≥ An x0 . By the definition of the cone
and Lemma 2.3, one has
R2
x0 (t) ≥ γkx0 k1 ≥ γx(0) ≥ γδkx0 k2 , x0 (t) ≥ ∗ > R2 , ∀t ∈ [0, 1],
a
from (3.11),
γx0 (t) ≥ γAn x0 (t)
Z t
1
=γ −
(t − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
n
0
Z 1
1
1
+
(1 − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
Pm−2
n
1 − i=1 ai 0
Z
m−2
ξi
X
1
ai
(ξi − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
−
n
0
i=1
Z
t
1
≥γ −
(t − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
n
0
Z 1
1
1
+
(1 − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
Pm−2
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
n
0
i=1
Pm−2 a Z t
1
i
i=1
≥γ
(t − s)a(s)f (s, x0 (s), −|x00 (s)| − )ds
P
m−2
n
1 − i=1 ai 0
Pm−2 R 1
ai 0 (ξi − s)a(s)f (s, x0 (s), −|x00 (s)| − n1 )ds − i=1
Pm−2
1 − i=1 ai
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
1
=
a(s)f (s, x0 (s), −|x00 (s)| − )ds
Pm−2
n
1 − i=1 ai
0
Pm−2
Z 1
γ i=1 ai (1 − ξi )
1
≥
a(s)g1 (x0 (s), −|x00 (s)| − )ds
Pm−2
n
1 − i=1 ai
0
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
≥
a(s)dsN ∗ x0 (s)
Pm−2
1 − i=1 ai
0
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
R2
R2
≥ a∗
a(s)dsN ∗ ∗ > ∗ .
Pm−2
a
a
1 − i=1 ai
0
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MULTIPLE POSITIVE SOLUTIONS
9
2
Then kx0 k ≥ γkx0 k1 > R
a∗ , which is a contradiction to x0 ∈ ∂Ω2 ∩ P . Then (3.10)
holds. From Lemma 2.2,
i(An , Ω2 ∩ P, P ) = 0.
(3.12)
which with (3.9) guarantee that
i(An , (Ω2 − Ω̄1 ) ∩ P, P ) = −1.
(3.13)
From this equality and (3.9), An has two fixed points with xn,1 ∈ Ω1 ∩ P, xn,2 ∈
(Ω2 − Ω̄1 ) ∩ P .
For each n ∈ N0 , there exists xn,1 ∈ Ω1 ∩ P such that xn,1 = An xn,1 ; that is,
Z t
1
xn,1 (t) = −
(t − s)a(s)f (s, xn,1 (s), −|x0n,1 (s)| − )ds
n
0
Z 1
1
1
+
(1 − s)a(s)f (s, xn,1 (s), −|x0n,1 (s)| − )ds (3.14)
Pm−2
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, xn,1 (s), −|x0n,1 (s)| − )ds .
n
0
i=1
As in the proof of (3.5), we have x0n,1 (t) ≤ 0, t ∈ (0, 1) and
Z t
1
x0n,1 (t) = −
a(s)f (s, xn,1 (s), x0n,1 (s) − )ds, n ∈ N0 , t ∈ (0, 1).
n
0
Now we consider {xn,1 (t)}n∈N0 and {x0n,1 (t)}n∈N0 . Since kxn,1 k ≤ R1 , it follows
that
{xn,1 (t)} is uniformly bounded on [0, 1],
(3.15)
{x0n,1 (t)}
(3.16)
is uniformly bounded on [0, 1].
Then
{xn,1 (t)} is equicontinuous on [0, 1].
As in the proof as (3.6),
1
x00n,1 (t) + a(t)f (t, xn,1 (t), x0n,1 (t) − ) = 0, 0 < t < 1,
n
m−2
X
x0n,1 (0) = 0, xn,1 (1) =
ai xn,1 (ξi ).
(3.17)
(3.18)
i=1
Now we show that for all t1 , t2 ∈ [0, 1],
|I(x0n,1 (t2 ) −
1
1
) − I(x0n,1 (t1 ) − )| ≤ h(R1 )|
n
n
Z
t2
a(t)dt|.
(3.19)
t1
From (3.18),
1
)
n
1
1 ≤ a(t)h(xn,1 (t)) g x0n,1 (t) −
+ r(x0n,1 (t) − ) ,
n
n
−x00n,1 (t) = a(t)f (t, xn,1 (t), x0n,1 (t) −
∀t ∈ (0, 1),
and
1
n
1
1 0
≥ −a(t)h(xn,1 (t)) g xn,1 (t) −
+ r(x0n,1 (t) − ) ,
n
n
x00n,1 (t) = −a(t)f t, xn,1 (t), x0n,1 (t) −
∀t ∈ (0, 1),
10
Y. MA, B. YAN
so
−x00n,1 (t)
g(x0n,1 (t) − n1 ) + r(x0n,1 (t) − n1 )
EJDE-2008/147
≤ a(t)h(xn,1 (t)), ∀t ∈ (0, 1),
(3.20)
and
x00n,1 (t)
g(x0n,1 (t) − n1 ) + r(x0n,1 (t) − n1 )
≥ −a(t)h(xn,1 (t)),
∀t ∈ (0, 1).
(3.21)
Then, for all t1 , t2 in [0, 1] and t1 < t2 ,
Z t2
Z t2
1
1 0
−
a(t)dt
1
1 d(xn,1 (s) − n ) ≤ h(R1 )
0
0
t1 g(xn,1 (s) − n ) + r(xn,1 (s) − n )
t1
Z t2
= h(R1 )|
a(t)dt|,
t1
Inequality (3.19) holds.
Since I −1 is uniformly continuous on [0, I(−R1 − )], for all ¯ > 0, there exists
0 > 0 such that
|I −1 (s1 ) − I −1 (s2 )| < ¯,
∀|s1 − s2 | < 0 , s1 , s2 ∈ [0, I(−R1 − )].
(3.22)
And (3.19) guarantees that for 0 > 0, there exists δ 0 > 0 such that
1
1
) − I(x0n,1 (t1 ) − )| < 0 ,
n
n
From this inequality and (3.22),
|I(x0n,1 (t2 ) −
∀ |t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1].
1
1
− (x0n,1 (t1 ) − )|
n
n
1
1
= |I −1 (I(x0n,1 (t2 ) − )) − I −1 (I(x0n,1 (t1 ) − ))|
n
n
< ¯, ∀|t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1];
(3.23)
|x0n,1 (t2 ) − x0n,1 (t1 )| = |x0n,1 (t2 ) −
(3.24)
that is,
{x0n,1 (t)} is equicontinuous on [0, 1].
(3.25)
{x0n,1 (t)}
From (3.15)–(3.17), (3.25) and the Arzela-Ascoli Theorem, {xn,1 (t)} and
are relatively compact on C[0, 1], which implies there exists a subsequence {xnj ,1 }
of {xn,1 } and function x0,1 (t) ∈ C[0, 1] such that
lim max |xnj ,1 (t) − x0,1 (t)| = 0,
j→+∞ t∈[0,1]
Since x0nj ,1 (0) = 0, xnj ,1 (1) =
(0, 1), j ∈ {1, 2, . . . },
x00,1 (0) = 0,
x0,1 (1) =
m−2
X
Pm−2
i=1
lim max |x0nj ,1 (t) − x00,1 (t)| = 0.
j→+∞ t∈[0,1]
ai xnj ,1 (ξi ), x0nj ,1 (t) < 0, xnj ,1 (t) > 0, t ∈
ai x0,1 (ξi ),
x00,1 (t) ≤ 0,
x0,1 (t) ≥ 0,
t ∈ (0, 1).
i=1
(3.26)
For (t, xnj ,1 (t), x0nj ,1 (t) − n1 ) ∈ [0, 1] × [0, R1 + ] × (−∞, 0), from (H5) there exists
a function ΨR1 ∈ C([0, 1], R+ ) such that
f (t, xnj ,1 (t), x0nj ,1 (t) −
1
)ds ≥ ΨR1 (t)(xnj ,1 (t))δ ,
nj
0 ≤ δ < 1.
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
11
Then, for n ∈ N0 ,
t
Z
xnj ,1 (t) = −
0
+
−
(t − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
1−
m−2
X
1
Pm−2
i=1
Z
0
i=1
1
Z
≥−
0
+
−
ξi
ai
Z
ai
1
(1 − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
0
(ξi − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
(1 − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
1−
m−2
X
1
Pm−2
i=1
Z
ξi
ai
0
i=1
1
)ds
nj
Z
ai
0
1
)ds
nj
1
)ds
nj
1
)ds
nj
1
(1 − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
(ξi − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
1
)ds
nj
1
)ds
nj
Pm−2
Z 1
1
i=1 ai
≥
(1 − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) − )ds
Pm−2
n
j
1 − i=1 ai 0
Pm−2
Z 1
1
i=1 ai
−
(ξi − s)a(s)f (s, xnj ,1 (s), x0nj ,1 (s) − )ds
P
m−2
n
j
1 − i=1 ai 0
Pm−2
Z 1
ai (1 − ξi )
1
= i=1 Pm−2
a(s)f (s, xnj ,1 (s), x0nj ,1 (s) − )ds
n
j
1 − i=1 ai
0
Pm−2
Z
ai (1 − ξi ) 1
a(s)ΨR1 (s)(xnj ,1 (s))δ ds
Pm−2
1 − i=1 ai
0
Pm−2
Z 1
ai (1 − ξi )
≥ i=1 Pm−2
a(s)ΨR1 (s)γ δ dskxnj ,1 kδ1 ,
1 − i=1 ai
0
≥
i=1
which implies
1
1−δ
Pm−2 a (1 − ξ ) Z 1
i
i
i=1
a(s)ΨR1 (s)γ δ ds
,
kxnj ,1 k1 ≥
Pm−2
1 − i=1 ai
0
and
xnj ,1 (t) ≥
1
Pm−2 a (1 − ξ ) Z 1
1−δ
i
i
δ
i=1
a(s)Ψ
(s)γ
ds
= a0 > 0.
Pm−2
R1
1 − i=1 ai
0
Thus
x0nj ,1 (t) = −
Z
t
a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
0
Z
1
)ds
nj
t
a(s)ΨR1 (s)(xnj ,1 (s))δ ds
≤−
0
Z
≤−
0
t
a(s)ΨR1 (s)dsaδ0 ,
t ∈ [0, 1], n ∈ N0 .
12
Y. MA, B. YAN
EJDE-2008/147
Consequently,
inf min |x0nj ,1 (s)| > 0,
j≥1 s∈[ 1 ,t]
2
inf min |x0nj ,1 (s)| > 0,
j≥1 s∈[t, 1 ]
2
1
t ∈ [ , 1),
2
1
t ∈ (0, ].
2
Since
1
x0nj ,1 (t) − x0nj ,1 ( ) = −
2
Z
t
1/2
a(s)f (s, xnj ,1 (s), x0nj ,1 (s) −
1
)ds,
nj
t ∈ (0, 1),
and
f (t, xnj ,1 (t), x0nj ,1 (t) −
1
1
1
) ≤ h(xnj ,1 (t))[g(x0nj ,1 (t) − ) + r(x0nj ,1 (t) − )]
nj
nj
nj
Z t
R1
R1
≤ h( )[g(−
a(s)ΨR1 (s)dsaδ0 ) + r(−
− )],
γ
γδ
0
letting j → +∞, the Lebesgue Dominated Convergence Theorem guarantees that
Z t
1
0
0
a(s)f (s, x0,1 (s), x00,1 (s))ds, t ∈ (0, 1).
(3.27)
x0,1 (t) − x0,1 ( ) = −
2
1/2
Differentiating, we have
x000,1 (t) + a(t)f (t, x0,1 (t), x00,1 (t)) = 0,
0 < t < 1,
and from (3.26) x0,1 (t) is a positive solution of (1.1) with x0,1 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
For the set {xn,2 }n∈N0 ⊆ (Ω2 − Ω1 ) ∩ P , the proof is as that for the set
{xn,1 }n∈N0 . We can obtain a convergent subsequence {xni ,2 }n∈N0 of {xn,2 }n∈N0
with limi→+∞ xni ,2 = x0,2 ∈ C 1 [0, 1]∩C 2 (0, 1). Moreover, x0,2 is a positive solution
to (1.1).
Example 3.1 In (1.1), let f (t, u, z) = µ[1 + (−z)−a ][1 + ub + ud ] and a(t) ≡ 1
with 0 < a < 1, b > 1, 0 < d < 1 and µ > 0. If
µ < sup
P
c(1− m−2 ai )
)
i ξi
i=1
.
1 + cb + cd
i=1
I(− 1+Pm−2
a
c∈R+
(3.28)
Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
We apply Theorem 3.1 with g(z) = (−z)−a , r(z) = 1, h(u) = µ(1 + ub +
d
u ), Ψ(t) = µ, g1 (u, z) = µub . (H1), (H2), (H4), (H5) hold. Also
sup
c∈R+
c
Pm−2
R1
i=1 ai ξi +1 −1
Pm−2
− 1−
I (h(c) 0
i=1 ai
= sup
c∈R+
a(s)ds)
c
Pm−2
i=1 ai ξi +1 −1
Pm−2
− 1−
I (µ(1
i=1 ai
and (3.28) guarantees that (H3) holds.
,
+ cb + cd ))
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
13
4. Singularities at x0 = 0 and x = 0
In this section the nonlinearity f may be singular at x0 = 0 and x = 0. We
assume that the following conditions hold.
(P1) a(t) ∈ C[0, 1], a(t) > 0, t ∈ (0, 1);
(P2) f (t, u, z) ≤ [h(u) + ω(u)][g(z) + r(z)], where f ∈ C([0, 1] × R+ × R− , R+ ),
g(z) > 0 continuous and non-increasing on (−∞, 0], ω(u) > 0 continuous
and non-increasing on [0, +∞), h(u) ≥ 0 continuous and nondecreasing on
R+ , r(z) > 0 continuous and nondecreasing on R− ;
(P3)
c
> 1,
sup Pm−2
Rc
a
ξ
+1
i i
−1 [− max
c∈R+ − i=1
Pm−2
(I
a(t)(ch(c)
+
ω(s)ds)])
t∈[0,1]
0
1−
a
i=1
i
R0
Ra
udu
where I(z) = z g(u)+r(u)
, z ∈ R− , 0 ω(s)ds < +∞;
(P4) There exists a function g1 ∈ C([0, +∞) × (−∞, 0], [0, +∞)), such that
f (t, u, z) ≥ g1 (u, z), ∀(t, u, z) ∈ [0, 1] × R+ × R− , and limu→+∞ g1 (u,z)
=
u
+∞, uniformly for z ∈ R− .
(P5) There exists a function ΨH ∈ C([0, 1], R+ ) with f (t, u, z) ≥ ΨH (t), for all
(t, u, z) ∈ [0, 1] × [0, H] × [−H, 0).
For n ∈ {1, 2, . . . }, x ∈ P , t ∈ [0, 1], define operator
Z t
1
1
(An x)(t) = −
(t − s)a(s)f (s, x(s) + , −|x0 (s)| − )ds
n
n
0
Z 1
1
1
1
+
(1 − s)a(s)f (s, x(s) + , −|x0 (s)| − )ds (4.1)
Pm−2
n
n
1 − i=1 ai 0
m−2
X Z ξi
1
1
ai
−
(ξi − s)a(s)f (s, x(s) + , −|x0 (s)| − )ds .
n
n
0
i=1
Theorem 4.1. Suppose (P1)–(P5) hold. Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1) and x0,1 (t), x0,2 (t) > 0, t ∈ [0, 1].
Proof. Choose R1 > 0 such that
R1
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [− maxt∈[0,1]
i=1 ai
a(t)(R1 h(R1 ) +
R R1
0
> 1.
(4.2)
ω(s)ds)])
From the continuity of I −1 and h, we can choose > 0 and < R1 such that
R1
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [I(−)
i=1 ai
− maxt∈[0,1] a(t)((R1 + )h(R1 + ) +
R R1 +
0
ω(s)ds)])
(4.3)
is greater than 1, n0 ∈ {1, 2, . . . } with n10 < min{, δ/2} and let N0 = {n0 , n0 +
1, . . . }. Then Lemma 2.4 guarantees that for n ∈ N0 , An : P → P is a completely
continuous operator. Let
Ω1 = {x ∈ C 1 [0, 1] : kxk < R1 }.
We show that
x 6= µAn x,
∀x ∈ P ∩ ∂Ω1 , µ ∈ (0, 1], n ∈ N0 .
(4.4)
14
Y. MA, B. YAN
EJDE-2008/147
In fact, if there exists an x0 ∈ P ∩ ∂Ω1 and µ0 ∈ (0, 1] with x0 = µ0 An x0 ,
Z t
1
1
x0 (t) = −µ0
(t − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
n
n
0
Z 1
µ0
1
1
0
+
Pm−2 ( (1 − s)a(s)f (s, x0 (s) + , −|x0 (s)| − )ds
n
n
1 − i=1 ai 0
Z
m−2
ξi
X
1
1
−
ai
(ξi − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds), t ∈ [0, 1].
n
n
0
i=1
Then
x00 (t) = −µ0
Z
t
a(s)f (s, x0 (s) +
0
1
1
, −|x00 (s)| − )ds,
n
n
∀t ∈ [0, 1].
(4.5)
Obviously, x00 (t) ≤ 0, t ∈ (0, 1), and since x0 (1) > 0, x0 (t) > 0, t ∈ [0, 1]. Differentiating (4.5), we have
1 0
1
, x (t) − ) = 0, 0 < t < 1,
n 0
n
m−2
X
x0 (1) =
ai x0 (ξi ).
x000 (t) + µ0 a(t)f (t, x0 (t) +
x00 (0) = 0,
(4.6)
i=1
Then, for t ∈ (0, 1),
1 0
1
, x (t) − )
n 0
n
1
1 1
1 ≤ a(t) h x0 (t) +
+ ω(x0 (t) + ) g x00 (t) −
+ r(x00 (t) − ) ,
n
n
n
n
−x000 (t) = µ0 a(t)f (t, x0 (t) +
and
−x000 (t)
1
1
1
1 ≤ a(t)[h(x0 (t) + n ) + ω(x0 (t) + n )],
0
0
g(x0 (t) − n ) + r(x0 (t) − n )
and
∀t ∈ (0, 1),
−x000 (t)(x00 (t) − n1 )
g(x00 (t) − n1 ) + r(x00 (t) − n1 )
1
1
1
≥ a(t)[h(x0 (t) + ) + ω(x0 (t) + )](x00 (t) − )
n
n
n
(4.7)
1
1
0
≥ a(t)[h(R1 + ) + ω(x0 (t) + (1 − t))](x0 (t) − )
n
n
1
1
1
0
= a(t)[h(R1 + )(x0 (t) − ) + ω(x0 (t) + (1 − t))(x00 (t) − )]
n
n
n
Integrating from 0 to t, we have
1
1
I(x00 (t) − ) − I(− )
n
n
Z t
1
1
1 ≥
a(s) h(R1 + )(x00 (s) − ) + ω(x0 (s) + (1 − s))(x00 (s) − ) ds
n
n
n
0
Z t
Z t
Z t
1
1
≥ max a(t)h(R1 + )
x00 (s)ds −
ds +
ω(x0 (s) + (1 − s))d(x0 (s)
n
n
t∈[0,1]
0
0
0
1
+ (1 − s))
n
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
15
R1 +
Z
≥ − max a(t)(h(R1 + )(R1 + ) +
ω(s)ds),
t∈[0,1]
0
I(x00 (t) −
1
) ≥ I(−) − max a(t)(h(R1 + )(R1 + ) +
n
t∈[0,1]
Z
R1 +
ω(s)ds).
0
Then
R1 +
Z
x00 (t) ≥ I −1 I(−) − max {a(t)} h(R1 + )(R1 + ) +
t∈[0,1]
ω(s)ds
;
0
that is,
−x00 (t)
≤ −I
−1
R1 +
Z
I(−)− max a(t)(h(R1 +)(R1 +)+
t∈[0,1]
ω(s)ds) ,
t ∈ (0, 1).
0
(4.8)
Since
x0 (0) =
≤
1−
µ0
Pm−2
i=1
Z
ai
−x00 (s)ds
−
µ0
Rξ
ai 0 i −x00 (s)ds
Pm−2
1 − i=1 ai
Pm−2
i=1
0
Z
1
Pm−2
1
1
−I −1 I(−) − max {a(t)}(h(R1 + )(R1 + )
1 − i=1 ai 0
Z R1 +
+
ω(s)ds) ds +
0
t∈[0,1]
Pm−2
1−
Z
i=1 ai
P
m−2
i=1
ξi
ai 0
Z R1 +
− max a(t)(h(R1 + )(R1 + ) +
t∈[0,1]
=
1+
−I −1 I(−)
ω(s)ds) ds
0
Pm−2
ai ξi
(−I −1 I(−) − max a(t)(h(R1 + )(R1 + )
Pi=1
m−2
t∈[0,1]
1 − i=1 ai
Z R1 +
+
ω(s)ds) .
0
Since x0 (0) ≥ x0 (t) ≥ γkx0 k1 ≥ γx0 (0) ≥ γδkx0 k2 , x0 (0) ≥ kx0 k = R1 . So
R1
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [I(−)
i=1 ai
− max{a(t)}((R1 + )h(R1 + ) +
R R1 +
0
≤ 1,
ω(s)ds)])
(4.9)
which is a contradiction to (4.3). Then (4.4) holds.
From Lemma 2.1, for n ∈ N0 ,
i(An , Ω1 ∩ P, P ) = 1.
(4.10)
Now we show that there exists a set Ω2 such that
An x 6≤ x,
∀x ∈ ∂Ω2 ∩ P.
Choose a∗ , N ∗ as in section 3. Let
Ω2 = {x ∈ C 1 [0, 1] : kxk <
R2
}.
a∗
Then
An x 6≤ x,
∀x ∈ ∂Ω2 ∩ P.
(4.11)
16
Y. MA, B. YAN
EJDE-2008/147
In fact, if there exists x0 ∈ ∂Ω2 ∩ P with x0 ≥ An x0 . By the definition of the cone
and Lemma 2.3, one has
x0 (t) ≥ γkx0 k1 ≥ γx(0) ≥ γδkx0 k2 ,
and so x0 (t) ≥
R2
a∗
> R2 , for all t ∈ [0, 1], x0 (t) +
1
n
> R2 , from (3.11),
γx0 (t) ≥ γAn x0 (t)
Z t
1
1
(t − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
=γ −
n
n
0
Z 1
1
1
1
+
(1 − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
Pm−2
n
n
1 − i=1 ai 0
Z
m−2
ξ
i
X
1
1
−
ai
(ξi − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
n
n
0
i=1
Z
t
1
1
≥γ −
(t − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
n
n
0
Z 1
1
1
1
+
(1 − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
Pm−2
n
n
1 − i=1 ai 0
m−2
X Z ξi
1
1
−
ai
(ξi − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
n
n
0
i=1
P
Z t
m−2
1
1
i=1 ai
≥γ
(t − s)a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
P
m−2
n
n
1 − i=1 ai 0
Pm−2 R 1
1
1
ai 0 (ξi − s)a(s)f s, x0 (s) + n , −|x00 (s)| − n ds − i=1
Pm−2
1 − i=1 ai
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
1
1
=
a(s)f (s, x0 (s) + , −|x00 (s)| − )ds
Pm−2
n
n
1 − i=1 ai
0
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
1
1
≥
a(s)g1 (x0 (s) + , −|x00 (s)| − )ds
Pm−2
n
n
1 − i=1 ai
0
Pm−2
Z 1
γ i=1 ai (1 − ξi )
≥
a(s)dsN ∗ x0 (s)
Pm−2
1 − i=1 ai
0
Pm−2
Z 1
R2
R2
∗γ
i=1 ai (1 − ξi )
a(s)dsN ∗ ∗ > ∗ .
≥a
Pm−2
a
a
1 − i=1 ai
0
Then kx0 k ≥ γkx0 k1 >
holds.
From Lemma 2.2,
R2
a∗ ,
which is a contradiction to x0 ∈ ∂Ω2 ∩ P . Then (4.11)
i(An , Ω2 ∩ P, P ) = 0.
(4.12)
This equality and (4.10) guarantee,
i(An , (Ω2 − Ω̄1 ) ∩ P, P ) = −1.
(4.13)
From this equality and (4.10), An has two fixed points with xn,1 ∈ Ω1 ∩ P, xn,2 ∈
(Ω2 − Ω1 ) ∩ P . For each n ∈ N0 , there exists xn,1 ∈ Ω1 ∩ P with xn,1 = An xn,1 ;
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
17
that is,
t
Z
(t − s)a(s)f (s, xn,1 (s) +
xn,1 (t) = −
0
+
−
1
Pm−2
1−
i=1
m−2
X
Z
ai
1
1
, −|x0n,1 (s)| − )ds
n
n
1
(1 − s)a(s)f (s, xn,1 (s) +
0
ξi
Z
(ξi − s)a(s)f (s, xn,1 (s) +
ai
0
i=1
1
1
, −|x0n,1 (s)| − )ds
n
n
1
1
, −|x0n,1 (s)| − )ds .
n
n
(4.14)
As in the same proof of (3.5), x0n (t) ≤ 0, t ∈ (0, 1) and
Z t
1
1
a(s)f (s, xn,1 (s) + , x0n,1 (s) − )ds,
x0n,1 (t) = −
n
n
0
n ∈ N0 , t ∈ (0, 1).
Now we consider {xn,1 (t)}n∈N0 and {x0n,1 (t)}n∈N0 , since kxn,1 k ≤ R1 , it follows
that
{xn,1 (t)} is uniformly bounded on [0, 1],
(4.15)
{x0n,1 (t)}
(4.16)
is uniformly bounded on [0, 1].
Then
{xn,1 (t)} is equicontinuous on [0, 1].
(4.17)
As in the same proof of (3.6),
1 0
1
, x (t) − ) = 0, 0 < t < 1,
n n,1
n
m−2
X
xn,1 (1) =
ai xn,1 (ξi ).
x00n,1 (t) + a(t)f (t, xn,1 (t) +
x0n,1 (0) = 0,
(4.18)
i=1
Now we show for all t1 , t2 in [0, 1],
1
1
|I(x0n,1 (t2 ) − ) − I(x0n,1 (t1 ) − )|
n
n
≤ max a(t) h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )| + |t2 − t1 |)
t∈[0,1]
Z
1
xn,1 (t2 )+ n
(1−t2 )
+|
(4.19)
ω(s)dt| .
1
xn,1 (t1 )+ n
(1−t1 )
From (4.18), it follows that for t ∈ (0, 1),
1 0
1
, x (t) − )
n n,1
n
1
1
1
1
≤ a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )][g(x0n,1 (t) − ) + r(x0n,1 (t) − )],
n
n
n
n
−x00n,1 (t) = a(t)f (t, xn,1 (t) +
1 0
1
, x (t) − )
n n,1
n
1
1
1
1
≥ −a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )][g(x0n,1 (t) − ) + r(x0n,1 (t) − )],
n
n
n
n
x00n,1 (t) = −a(t)f (t, xn,1 (t) +
18
Y. MA, B. YAN
EJDE-2008/147
and so for t ∈ (0, 1),
−x00n,1 (t)(x0n,1 (t) − n1 )
g(x0n,1 (t) − n1 ) + r(x0n,1 (t) − n1 )
1
1
1
≥ a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )](x0n,1 (t) − )
n
n
n
1
1
1
≥ a(t)[h(R1 + )(x0n,1 (t) − ) + ω(xn,1 (t) + (1 − t))(xn,1 (t) + (1 − t))0 ],
n
n
n
(4.20)
x00n,1 (t)(x0n,1 (t) − n1 )
g(x0n,1 (t) − n1 ) + r(x0n,1 (t) − n1 )
1
1
1
≤ −a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )](x0n,1 (t) − )
n
n
n
1
1
1
≤ −a(t)[h(R1 + )(x0n,1 (t) − ) + ω(xn,1 (t) + (1 − t))(xn,1 (t) + (1 − t))0 ].
n
n
n
(4.21)
Then, for all t1 , t2 ∈ [0, 1] and t1 < t2 ,
1
1
I(x0n,1 (t2 ) − ) − I(x0n,1 (t1 ) − )
n
n
Z t2
Z xn,1 (t2 )+ n1 (1−t2 )
1
0
≥ max a(t)[h(R1 + )(
xn,1 (s)dt − (t2 − t1 ) +
ω(s)dt]
1
n
t∈[0,1]
t1
(1−t1 )
xn,1 (t1 )+ n
≥ − max a(t) h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )| + |t2 − t1 |)
t∈[0,1]
Z
1
(1−t2 )
xn,1 (t2 )+ n
+|
ω(s)dt| ,
1
xn,1 (t1 )+ n
(1−t1 )
1
1
) − I(x0n,1 (t2 ) − )
n
n
≤ max a(t) h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )| + |t2 − t1 |)
I(x0n,1 (t1 ) −
t∈[0,1]
Z
1
xn,1 (t2 )+ n
(1−t2 )
+|
ω(s)dt| .
1
xn,1 (t1 )+ n
(1−t1 )
Therefore, (4.19) holds. Since I −1 is uniformly continuous on [0, I(−R1 − )], for
all ¯ > 0, there exists 0 > 0 such that
|I −1 (s1 ) − I −1 (s2 )| < ¯, ∀ |s1 − s2 | < 0 , s1 , s2 ∈ [0, I(−R1 − )].
Then (4.19) guarantees that for 0 > 0, there exists δ 0 > 0 such that
1
1
|I(x0n,1 (t2 ) − ) − I(x0n,1 (t1 ) − )| < 0 , ∀|t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1].
n
n
From this inequality and (4.22),
1
1
|x0n,1 (t2 ) − x0n,1 (t1 )| = |x0n,1 (t2 ) − − (x0n,1 (t1 ) − )|
n
n
1
1
= |I −1 (I(x0n,1 (t2 ) − )) − I −1 (I(x0n,1 (t1 ) − ))|
n
n
< ¯, ∀|t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1];
(4.22)
(4.23)
(4.24)
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
19
that is,
{x0n,1 (t)} is equicontinuous on [0, 1].
(4.25)
0
From (4.15)–(4.17), (4.25) and the Arzela-Ascoli Theorem, {xn,1 (t)} and {xn,1 (t)}
are relatively compact on C[0, 1], which implies, there exists a subsequence {xnj ,1 }
of {xn,1 } and function x0,1 (t) ∈ C[0, 1] such that
lim max |xnj ,1 (t) − x0,1 (t)| = 0,
j→+∞ t∈[0,1]
Since x0nj ,1 (0) = 0, xnj ,1 (1) =
(0, 1), j ∈ {1, 2, . . . },
x00,1 (0)
= 0, x0,1 (1) =
m−2
X
Pm−2
i=1
lim max |x0nj ,1 (t) − x00,1 (t)| = 0.
j→+∞ t∈[0,1]
ai xnj ,1 (ξi ), x0nj ,1 (t) < 0, xnj ,1 (t) > 0, t ∈
ai x0,1 (ξi ), x00,1 (t) ≤ 0, x0,1 (t) ≥ 0,
t ∈ (0, 1).
(4.26)
i=1
For (t, xnj ,1 (t) + n1j , x0nj ,1 (t) − n1j ) ∈ [0, 1] × [0, R1 + ] × (−∞, 0), from (P5) there
exists a function ΨR1 ∈ C([0, 1], R+ ) such that
1
1
f (t, xnj ,1 (t) + , x0nj ,1 (t) − )ds ≥ ΨR1 (t).
nj
nj
Then, for n ∈ N0 ,
Z t
1
1
xnj ,1 (t) = −
(t − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
nj
nj
0
Z 1
1
1
1
+
(1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
Pm−2
nj
nj
1 − i=1 ai 0
m−2
X Z ξi
1
1
−
ai
(ξi − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
nj
nj
0
i=1
Z 1
1
1
≥−
(1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
n
n
j
j
0
Z 1
1
1
1 0
+
Pm−2 ( (1 − s)a(s)f (s, xnj ,1 (s) + , xnj ,1 (s) − )ds
nj
nj
1 − i=1 ai 0
Z
m−2
ξi
X
1
1
−
ai
(ξi − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds)
n
n
j
j
0
i=1
Pm−2
Z 1
1
1
i=1 ai
≥
(1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
P
m−2
n
n
j
j
1 − i=1 ai 0
Pm−2
Z 1
1
1
i=1 ai
−
(ξi − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
P
m−2
n
n
j
j
1 − i=1 ai 0
Pm−2
Z 1
ai (1 − ξi )
1
1
a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s) − )ds
= i=1 Pm−2
n
n
j
j
1 − i=1 ai
0
Pm−2
Z 1
ai (1 − ξi )
a(s)ΨR1 (s)ds = a0 ,
≥ i=1 Pm−2
1 − i=1 ai
0
and
x0nj ,1 (t)
Z
=−
t
a(s)f (s, xnj ,1 (s) +
0
1 0
1
,x
(s) − )ds
nj nj ,1
nj
20
Y. MA, B. YAN
Z
EJDE-2008/147
t
≤−
a(s)ΨR1 + (s)ds,
t ∈ [0, 1], n ∈ N0 .
0
Thus,
inf min{ min
xnj ,1 (s), min
|x0nj ,1 (s)|} > 0,
1
1
j≥1
s∈[ 2 ,t]
s∈[ 2 ,t]
inf min{ min1 xnj ,1 (s), min1 |x0nj ,1 (s)|} > 0,
j≥1
s∈[t, 2 ]
s∈[t, 2 ]
1
t ∈ [ , 1),
2
1
t ∈ (0, ].
2
From (P2), we have
1 0
1
,x
(t) −
nj nj ,1
nj
1
1
1
1
≤ [h(xnj ,1 (t) + ) + ω(xnj ,1 (t) + )][g(x0nj ,1 (t) − ) + r(x0nj ,1 (t) − )]
nj
nj
nj
nj
Z t
R1
R1
+ ) + ω(a0 )][r(−
− )].
a(s)ΨR1 + (s)ds) + g(−
≤ [h(
γ
γδ
0
f t, xnj ,1 (t) +
and since
x0nj ,1 (t)
−
1
x0nj ,1 ( )
2
Z
t
=−
a(s)f (s, xnj ,1 (s) +
1/2
1 0
1
,x
(s) − )ds,
nj nj ,1
nj
t ∈ (0, 1),
letting j → +∞, the Lebesgue Dominated Convergence Theorem guarantees that
Z t
1
a(s)f (s, x0,1 (s), x00,1 (s))ds, t ∈ (0, 1).
(4.27)
x00,1 (t) − x00,1 ( ) = −
2
1/2
Differentiating, we have
x000,1 (t) + a(t)f (t, x0,1 (t), x00,1 (t)) = 0,
0 < t < 1,
and from (4.26), x0,1 (t) is a positive solution of (1.1) with x0,1 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
For the set {xn,2 }n∈N0 ⊆ (Ω2 − Ω1 ) ∩ P , as in proof for the set {xn,1 }n∈N0 , we
obtain a convergent subsequence {xni ,2 }n∈N0 of {xn,2 }n∈N0 with limi→+∞ xni ,2 =
x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1). Moreover, x0,2 is a positive solution to (1.1).
Example 4.1 In (1.1), let f (t, u, z) = µ[1 + (−z)−a ][1 + ub + u−d ] and a(t) ≡ 1
with 0 < a < 1, b > 1, 0 < d < 1 and µ > 0. If
µ < sup
c∈R+
P
c(1− m−2 ai )
)
i i +1
i=1
I(− Pm−2 i=1
a ξ
maxt∈[0,1] a(t)(c + c1−d +
c1+b
1+b )
.
(4.28)
Then equation (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1]∩C 2 (0, 1).
We apply Theorem 4.1 with g(z) = 1, r(z) = (−z)−a , h(u) = µu−d , ω(u) =
µ(1 + ub ), Ψ(t) = µ, g1 (u, z) = µub . Note that (P1), (P2), (P4) and (P5) hold, and
that
c
sup Pm−2
Rc
a
ξ
+1
i i
c∈R+ − i=1
P
I −1 [maxt∈[0,1] a(t)(ch(c) + 0 ω(s)ds)]
1− m−2 a
i=1
= sup
c∈R+
i
c
Pm−2
i=1 ai ξi +1 −1
Pm−2
− 1−
I [µ maxt∈[0,1]
i=1 ai
Then (4.28) guarantees that (P3) holds.
a(t)(c + c1−d +
c1+b
1+b )]
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
21
5. Singularities at x = 0 but not x0 = 0
In this section the nonlinearity f may be singular at x = 0, but not at x0 = 0.
We assume that the following conditions hold.
(S1) a(t) ∈ C[0, 1], a(t) > 0, t ∈ (0, 1);
(S2) f (t, u, z) ≤ [h(u) + ω(u)]r(z), where f ∈ C([0, 1] × R+ × R− , R+ ), ω(u) > 0
continuous and non-increasing on [0, +∞), h(u) ≥ 0 continuous and nondecreasing on R+ , r(z) > 0 continuous and nondecreasing on R− ;
(S3)
sup
c∈R+
c
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [− maxt∈[0,1]
i=1 ai
a(t)(ch(c) +
Rc
0
> 1,
ω(s)ds)])
Ra
R 0 udu
where I(z) = z r(u)
, z ∈ R− , 0 ω(s)ds < +∞, a ∈ R+ ;
(S4) There exists a function g1 ∈ C([0, +∞) × (−∞, 0], [0, +∞)), such that
f (t, u, z) ≥ g1 (u, z), ∀(t, u, z) ∈ [0, 1] × R+ × R− , and limu→+∞ g1 (u,z)
=
u
+∞, uniformly for z ∈ R− .
(S5) There exists a function ΨH ∈ C([0, 1], R+ ) and a constant 0 ≤ δ such that
f (t, u, z) ≥ ΨH (t)(−z)δ , for all (t, u, z) ∈ [0, 1] × [0, H] × [−H, 0).
For n ∈ {1, 2, . . . }, x ∈ P , define operator
Z t
1
(An x)(t) = −
(t − s)a(s)f (s, x(s) + , −|x0 (s)|)ds
n
0
Z 1
1
1
+
(1 − s)a(s)f (s, x(s) + , −|x0 (s)|)ds
Pm−2
(5.1)
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, x(s) + , −|x0 (s)|)ds , t ∈ [0, 1].
n
0
i=1
Theorem 5.1. Suppose (S1)–(S5) hold. Then (1.1) has at least two positive solutions x0,1 , x0,2 in C 1 [0, 1] ∩ C 2 (0, 1) with x0,1 (t), x0,2 (t) > 0, t ∈ [0, 1].
Proof. Choose R1 > 0 such that
R1
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [− maxt∈[0,1]
i=1 ai
a(t)(R1 h(R1 ) +
R R1
0
> 1.
(5.2)
ω(s)ds)])
From the continuity of I −1 and h, we can choose > 0 and < R1 such that
R1
Pm−2
i=1 ai ξi +1
Pm−2
− 1−
(I −1 [− maxt∈[0,1]
i=1 ai
a(t)(R1 h(R1 + ) +
R R1 +
0
> 1.
(5.3)
ω(s)ds)])
Let n0 ∈ {1, 2, . . . } with n10 < min{, δ/2} and let N0 = {n0 , n0 + 1, . . . }. Then
Lemma 2.4 guarantees that for n ∈ N0 , An : P → P is a completely continuous
operator. Let
Ω1 = {x ∈ C 1 [0, 1] : kxk < R1 }.
We show that
x 6= µAn x,
∀x ∈ P ∩ ∂Ω1 , µ ∈ (0, 1], n ∈ N0 .
(5.4)
22
Y. MA, B. YAN
EJDE-2008/147
In fact, if there exists an x0 ∈ P ∩ ∂Ω1 and µ0 ∈ (0, 1] with x0 = µ0 An x0 ,
Z t
1
x0 (t) = −µ0
(t − s)a(s)f (s, x0 (s) + , −|x00 (s)|)ds
n
0
Z 1
1
µ0
(1 − s)a(s)f (s, x0 (s) + , −|x00 (s)|)ds
+
Pm−2
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, x0 (s) + , −|x00 (s)|)ds .
n
0
i=1
Then
Z
t
1
, −|x00 (s)|)ds, ∀t ∈ [0, 1].
(5.5)
n
0
Obviously, x00 (t) ≤ 0, t ∈ (0, 1), and since x0 (1) > 0, x0 (t) > 0, t ∈ [0, 1]. Differentiating (5.5), we have
x00 (t) = −µ0
a(s)f (s, x0 (s) +
x000 (t) + µ0 a(t)f (t, x0 (t) +
x00 (0) = 0,
1
,
n
x00 (t)) = 0, 0 < t < 1,
m−2
X
x0 (1) =
(5.6)
ai x0 (ξi ).
i=1
and
−x000 (t) = µ0 a(t)f (t, x0 (t) +
≤ a(t)[h(x0 (t) +
Then
1 0
, x (t))
n 0
1
1
) + ω(x0 (t) + )]r(x00 (t)),
n
n
−x000 (t)
1
1
≤ a(t)[h(x0 (t) + ) + ω(x0 (t) + )],
0
r(x0 (t))
n
n
∀t ∈ (0, 1).
∀t ∈ (0, 1),
and
1
−x000 (t)x00 (t)
1
≥ a(t)[h(x0 (t) + ) + ω(x0 (t) + )]x00 (t)
0
r(x0 (t))
n
n
1 0
≥ a(t)[h(R1 + ) + ω(x0 (t) + )]x0 (t).
n
Integrating from 0 to t, we have
Z t
1
0
I(x0 (t)) ≥
a(s)[h(R1 + )x00 (s)) + ω(x0 (s) + )x00 (s)]ds
n
0
Z t
1
≥ max a(t)
[h(R1 + )x00 (s)ω(x0 (s) + )x00 (s)]ds
n
t∈[0,1]
0
Z t
1
1
≥ − max a(t)(h(R1 + )R1 +
ω(x0 (s) + )d(x0 (s) + ))
n
n
t∈[0,1]
0
Z x0 (t)+ n1
= − max a(t)(h(R1 + )R1 +
ω(s)ds)
t∈[0,1]
1
x0 (0)+ n
Z
≥ − max a(t)(h(R1 + )R1 +
t∈[0,1]
R1 +
ω(s)ds)
0
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
and
R1 +
Z
I(x00 (t)) ≥ − max a(t)(h(R1 + )R1 +
23
ω(s)ds).
t∈[0,1]
0
Then
x00 (t) ≥ I −1 (− max a(t)(h(R1 + )R1 +
R1 +
Z
ω(s)ds));
t∈[0,1]
0
that is,
−x00 (t) ≤ −I −1 (− max a(t)(h(R1 + )R1 +
t∈[0,1]
Z
R1 +
ω(s)ds)), t ∈ (0, 1).
(5.7)
0
Since
x0 (0)
Rξ
ai 0 i −x00 (s)ds
=
−
Pm−2
1 − i=1 ai 0
1 − i=1 ai
Z 1
Z R1 +
1
−1
≤
−I
−
max
a(t)(h(R
+
)R
+
ω(s)ds) ds
Pm−2
1
1
t∈[0,1]
1 − i=1 ai 0
0
Pm−2
Z ξi
Z R1 +
−1
i=1 ai
+
−I
I(−
max
a(t)(h(R
+
)(R
)
+
ω(s)ds)) ds
Pm−2
1
1
t∈[0,1]
1 − i=1 ai 0
0
Pm−2
Z R1 +
1 + i=1 ai ξi
(−I −1 − max a(t)(h(R1 + )R1 +
ω(s)ds) ,
=
Pm−2
t∈[0,1]
1 − i=1 ai
0
(5.8)
x0 (0) ≥ x0 (t) ≥ γkx0 k1 ≥ γx0 (0) ≥ γδkx0 k2 , x0 (0) ≥ kx0 k = R1 . So
µ0
Pm−2
Z
1
−x00 (s)ds
µ0
Pm−2
i=1
R1
Pm−2
i=1 ai ξi +1
Pm−2
(I −1 [− maxt∈[0,1]
− 1−
i=1 ai
a(t)(R1 h(R1 + ) +
R R1 +
0
≤ 1,
(5.9)
ω(s)ds)])
which is a contradiction to (5.3). Then (5.4) holds. From Lemma 2.1, for n ∈ N0 ,
i(An , Ω1 ∩ P, P ) = 1.
(5.10)
Now we show that there exists a set Ω2 such that
An x 6≤ x,
∀ x ∈ ∂Ω2 ∩ P.
(5.11)
Choose a∗ , N ∗ as in section 3. Let
Ω2 = {x ∈ C 1 [0, 1] : kxk <
R2
}.
a∗
Then
An x 6≤ x,
∀x ∈ ∂Ω2 ∩ P.
In fact, if there exists x0 ∈ ∂Ω2 ∩ P with x0 ≥ An x0 , by the definition of the cone
and Lemma 2.3, one has
x0 (t) ≥ γkx0 k1 ≥ γx(0) ≥ γδkx0 k2 ,
x0 (t) ≥
R2
a∗
> R2 for all t ∈ [0, 1]. Then x0 (t) +
1
n
> R2 . From (3.11),
γx0 (t) ≥ γAn x0 (t)
Z t
1
=γ −
(t − s)a(s)f (s, x0 (s) + , x00 (s))ds
n
0
24
Y. MA, B. YAN
+
−
1−
m−2
X
1
Pm−2
i=1
ai
1
(1 − s)a(s)f (s, x0 (s) +
0
ξi
Z
(ξi − s)a(s)f (s, x0 (s) +
ai
0
i=1
Z
t
Z
≥γ −
(t − s)a(s)f (s, x0 (s) +
0
+
−
1−
m−2
X
i=1
EJDE-2008/147
1
Pm−2
i=1
Z
ai
1
(1 − s)a(s)f (s, x0 (s) +
0
(ξi − s)a(s)f (s, x0 (s) +
ai
1 0
, x0 (s))ds
n
1 0
, x (s))ds
n 0
ξi
Z
1 0
, x (s))ds
n 0
0
1 0
, x (s))ds
n 0
1 0
, x0 (s))ds
n
Pm−2 a Z t
1
i
i=1
≥γ
(t − s)a(s)f (s, x0 (s) + , x00 (s))ds
P
m−2
n
1 − i=1 ai 0
Pm−2 R 1
ai 0 (ξi − s)a(s)f (s, x0 (s) + n1 , x00 (s) − i=1
)
Pm−2
1 − i=1 ai
Pm−2
Z
γ i=1 ai (1 − ξi ) 1
1
=
a(s)f (s, x0 (s) + , x00 (s))ds
Pm−2
n
1 − i=1 ai
0
Pm−2
Z 1
γ i=1 ai (1 − ξi )
1
≥
a(s)g1 (x0 (s) + , x00 (s))ds
Pm−2
n
1 − i=1 ai
0
Pm−2
Z 1
γ i=1 ai (1 − ξi )
≥
a(s)dsN ∗ x0 (s)
Pm−2
1 − i=1 ai
0
Pm−2
Z 1
R2
R2
∗γ
i=1 ai (1 − ξi )
≥a
a(s)dsN ∗ ∗ > ∗ ;
Pm−2
a
a
1 − i=1 ai
0
2
that is, kx0 k ≥ γkx0 k1 > R
a∗ , which is a contradiction to x0 ∈ ∂Ω2 ∩ P . Then
(5.11) holds. From Lemma 2.2,
i(An , Ω2 ∩ P, P ) = 0.
(5.12)
This equality and (5.10) guarantee
i(An , (Ω2 − Ω̄1 ∩ P, P ) = −1.
(5.13)
From this equality and (5.10), An has two fixed points with xn,1 ∈ Ω1 ∩ P, xn,2 ∈
(Ω2 − Ω̄1 ) ∩ P .
For each n ∈ N0 , there exists xn,1 ∈ Ω1 ∩ P such that xn,1 = An xn,1 ; that is, for
t ∈ [0, 1],
Z t
1
xn,1 (t) = −
(t − s)a(s)f (s, xn,1 (s) + , −|x0n,1 (s)|)ds
n
0
Z 1
1
1
+
(1 − s)a(s)f (s, xn,1 (s) + , −|x0n,1 (s)|)ds (5.14)
Pm−2
n
1 − i=1 ai 0
m−2
X Z ξi
1
−
ai
(ξi − s)a(s)f (s, xn,1 (s) + , −|x0n,1 (s)|)ds .
n
0
i=1
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
As in the proof of (3.5), x0n (t) ≤ 0, t ∈ (0, 1) and
Z t
1
x0n,1 (t) = −
a(s)f (s, xn,1 (s) + , x0n,1 (s))ds,
n
0
25
n ∈ N0 , t ∈ (0, 1).
Now we consider {xn,1 (t)}n∈N0 and {x0n,1 (t)}n∈N0 , since kxn,1 k ≤ R1 ,
{xn,1 (t)} is uniformly bounded on [0, 1],
(5.15)
{x0n,1 (t)}
(5.16)
is uniformly bounded on [0, 1].
Then
{xn,1 (t)} is equicontinuous on [0, 1].
(5.17)
As in the proof of (3.6),
1 0
, x (t)) = 0, 0 < t < 1,
n n,1
m−2
X
x0n,1 (0) = 0, xn,1 (1) =
ai xn,1 (ξi ).
x00n,1 (t) + a(t)f (t, xn,1 (t) +
(5.18)
i=1
Now we show that for all t1 , t2 ∈ [0, 1],
|I(x0n,1 (t2 )) − I(x0n,1 (t1 ))|
≤ max a(t) h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )| + |t2 − t1 |)
t∈[0,1]
Z
(5.19)
1
(1−t2 )
xn,1 (t2 )+ n
+|
ω(s)dt| .
1
xn,1 (t1 )+ n
(1−t1 )
From (5.18),
1 0
, x (t))
n n,1
1
1
≤ a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )]r(x0n,1 (t)),
n
n
−x00n,1 (t) = a(t)f (t, xn,1 (t) +
1 0
, x (t))
n n,1
1
1
≥ −a(t)[h(xn,1 (t) + ) + ω(xn,1 (t) + )]r(x0n,1 (t)),
n
n
∀t ∈ (0, 1),
x00n,1 (t) = −a(t)f (t, xn,1 (t) +
∀t ∈ (0, 1),
−x00n,1 (t)x0n,1 (t)
r(x0n,1 (t))
1
1
) + ω(xn,1 (t) + )]x0n,1 (t)
n
n
1
≥ a(t)[h(R1 + )x0n,1 (t) + ω(xn,1 (t) + (1 − t))x0n,1 (t)],
n
(5.20)
≥ a(t)[h(xn,1 (t) +
∀t ∈ (0, 1),
x00n,1 (t)x0n,1 (t)
r(x0n,1 (t))
1
1
) + ω(xn,1 (t) + )]x0n,1 (t)
n
n
1
≤ −a(t)[h(R1 + )x0n,1 (t) + ω(xn,1 (t) + (1 − t))x0n,1 (t)],
n
(5.21)
≤ −a(t)[h(xn,1 (t) +
∀t ∈ (0, 1).
26
Y. MA, B. YAN
EJDE-2008/147
Since the right-hand sides of (5.20) and (5.21) are positive, for all t1 , t2 ∈ [0, 1] and
t1 < t 2 ,
I(x0n,2 (t2 )) − I(x0n,1 (t1 ))
Z
≥ max a(t)[h(R1 + )
t∈[0,1]
t2
x0n,1 (s)dt +
t1
Z
t2
ω(xn,1 (s) +
t1
Z
1 0
)xn,1 (s)ds ]
n
1
(1−t2 )
xn,1 (t2 )+ n
≥ − max a(t)[h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )|) + |
ω(s)dt|],
t∈[0,1]
1
xn,1 (t1 )+ n
(1−t1 )
I(x0n,1 (t1 )) − I(x0n,1 (t2 ))
Z
1
xn,1 (t2 )+ n
(1−t2 )
≤ max a(t)[h(R1 + )(|xn,1 (t2 ) − xn,1 (t1 )|) + |
t∈[0,1]
ω(s)dt|] ,
1
xn,1 (t1 )+ n
(1−t1 )
(5.19) holds.
Since I −1 is uniformly continuous on [0, I(−R1 − )], for all ¯ > 0, there exists
0
> 0 such that
|I −1 (s1 ) − I −1 (s2 )| < ¯,
∀|s1 − s2 | < 0 , s1 , s2 ∈ [0, I(−R1 − )].
(5.22)
Also (5.19) guarantees that for 0 > 0, there exists δ 0 > 0 such that
|I(x0n,1 (t2 )) − I(x0n,1 (t1 ))| < 0 ,
∀ |t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1].
(5.23)
From this inequality and (5.22),
|x0n,1 (t2 ) − x0n,1 (t1 )| = |I −1 (I(x0n,1 (t2 ))) − I −1 (I(x0n,1 (t1 )))| < ¯,
∀|t1 − t2 | < δ 0 , t1 , t2 ∈ [0, 1];
(5.24)
that is,
{x0n,1 (t)} is equi-continuous on [0, 1].
(5.25)
0
From (5.15)–(5.17), (5.25) and the Arzela-Ascoli Theorem, {xn,1 (t)} and {xn,1 (t)}
are relatively compact on C 1 [0, 1]. This implies, there exists a subsequence {xnj ,1 }
of {xn,1 } and function x0,1 (t) ∈ C 1 [0, 1] such that
lim max |xnj ,1 (t) − x0,1 (t)| = 0,
j→+∞ t∈[0,1]
Since x0nj ,1 (0) = 0, xnj ,1 (1) =
(0, 1), j ∈ {1, 2, . . . },
x00,1 (0) = 0, x0,1 (1) =
m−2
X
Pm−2
i=1
lim max |x0nj ,1 (t) − x00,1 (t)| = 0.
j→+∞ t∈[0,1]
ai xnj ,1 (ξi ), x0nj ,1 (t) < 0, xnj ,1 (t) > 0, t ∈
ai x0,1 (ξi ), x00,1 (t) ≤ 0, x0,1 (t) ≥ 0, t ∈ (0, 1).
(5.26)
i=1
For (t, xnj ,1 (t) + n1j , x0nj ,1 (t)) ∈ [0, 1] × [0, R1 + ] × (−∞, 0), from (S5) there exists
a function ΨR1 ∈ C([0, 1], R+ ) such that
1
f (t, xnj ,1 (t) + , x0nj ,1 (t))ds ≥ ΨR1 (t)(−x0nj ,1 (t))δ , 0 ≤ δ < 1.
nj
Then, for n ∈ N0 ,
x0nj ,1 (t) = −
Z
t
a(s)f (s, xnj ,1 (s) +
0
Z
≤−
0
1 0
,x
(s))ds
nj nj ,1
t
a(s)ΨR1 + (s)(−x0nj ,1 (s))δ ds,
t ∈ [0, 1], n ∈ N0 ,
EJDE-2008/147
MULTIPLE POSITIVE SOLUTIONS
27
which implies
x0nj ,1 (t)
Z t
1
≤ −(1 − δ)(
a(s)ΨR1 + (s)ds) 1−δ ,
0
and
Z
xnj ,1 (t) = −
t
(t − s)a(s)f (s, xnj ,1 (s) +
0
+
−
1−
m−2
X
1
Pm−2
i=1
Z
Z 1
1
( (1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s))ds
n
j
ai 0
ξi
(ξi − s)a(s)f (s, xnj ,1 (s) +
ai
0
i=1
1
Z
≥−
(1 − s)a(s)f (s, xnj ,1 (s) +
0
+
−
1−
m−2
X
1 0
,x
(s))ds
nj nj ,1
1
Pm−2
i=1
Z
i=1
1 0
,x
(s))ds
nj nj ,1
Z 1
1
( (1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s))ds
nj
ai 0
ξi
(ξi − s)a(s)f (s, xnj ,1 (s) +
ai
1 0
,x
(s))ds)
nj nj ,1
0
1 0
,x
(s))ds)
nj nj ,1
Pm−2
Z 1
1
i=1 ai
≥
(1 − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s))ds
Pm−2
nj
1 − i=1 ai 0
Pm−2
Z 1
1
i=1 ai
−
(ξi − s)a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s))ds
P
m−2
n
j
1 − i=1 ai 0
Pm−2
Z 1
ai (1 − ξi )
1
= i=1 Pm−2
a(s)f (s, xnj ,1 (s) + , x0nj ,1 (s))ds
n
j
1 − i=1 ai
0
Pm−2
Z 1
ai (1 − ξi )
≥ i=1 Pm−2
a(s)ΨR1 + (s)(−x0nj ,1 (s))δ ds
1 − i=1 ai
0
Pm−2
Z 1
Z s
1
i=1 ai (1 − ξi )
≥
a(s)ΨR1 + (s)((1 − δ)(
a(τ )ΨR1 + (τ )dτ ) 1−δ )δ ds
Pm−2
1 − i=1 ai
0
0
=: F,
t ∈ [0, 1].
Since
1 0
,x
(t))
nj nj ,1
1
1
≤ [h(xnj ,1 (t) + ) + ω(xnj ,1 (t) + )]r(x0nj ,1 (t))
nj
nj
Z t
1
1−δ
R1
≤ [h(
+ ) + ω(F )]r((δ − 1)
a(s)ΨR1 + (s)ds
)),
γ
0
f (t, xnj ,1 (t) +
and
1
x0nj ,1 (t) − x0nj ,1 ( ) = −
2
Z
t
a(s)f (s, xnj ,1 (s) +
1/2
1 0
,x
(s))ds,
nj nj ,1
t ∈ (0, 1),
28
Y. MA, B. YAN
EJDE-2008/147
letting j → +∞, the Lebesgue Dominated Convergence Theorem guarantees that
Z t
1
x00,1 (t) − x00,1 ( ) = −
a(s)f (s, x0,1 (s), x00,1 (s))ds, t ∈ (0, 1).
(5.27)
2
1/2
Differentiating, we have
x000,1 (t) + a(t)f (t, x0,1 (t), x00,1 (t)) = 0,
0 < t < 1.
From this equality and from (5.26), x0,1 (t) is a positive solution of (1.1) with
x0,1 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
For the set {xn,2 }n∈N0 ⊆ (Ω2 −Ω1 )∩P , as in the proof for the set {xn,1 }n∈N0 , we
obtain a convergent subsequence {xni ,2 }n∈N0 of {xn,2 }n∈N0 with limi→+∞ xni ,2 =
x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1). Moreover, x0,2 is a positive solution to (1.1).
Example 5.1 In (1.1), let f (t, u, z) = µ[1 + (−z)a ][1 + ub + u−d ] and a(t) ≡ 1
with 0 ≤ a < 1, b > 1, 0 < d < 1 and µ > 0. If
µ < sup
c∈R+
P
c(1− m−2 ai )
)
i i +1
i=1
I( Pm−2 i=1
a ξ
maxt∈[0,1] a(t)(c + c1−d +
c1+b
1+b )
,
(5.28)
Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
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Department of Mathematics, Shandong Normal University, Jinan, 250014, China
E-mail address, Ya Ma: maya-0907@163.com
E-mail address, Baoqiang Yan: yanbqcn@yahoo.com.cn