Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 22, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE SOLUTIONS FOR SECOND-ORDER MULTI-POINT
BOUNDARY-VALUE PROBLEMS AT RESONANCE IN BANACH
SPACES
WEIHUA JIANG, BIN WANG
Abstract. In this article, we study the existence and multiplicity of positive
solutions for a nonlinear second-order multi-point boundary-value problem at
resonance in Banach spaces. The arguments are based upon a specially constructed equivalent equation and the fixed point theory in a cone for strict set
contraction operators.
1. Introduction
The theory of ordinary differential equations in Banach spaces has become a new
important branch (see, for example, [2, 6, 7, 12] and references cited therein). In
1988, Guo and Lakshmikantham [8] discussed the existence of multiple solutions
for two-point boundary value problem of ordinary differential equations in Banach
spaces. Since then, nonlinear second-order multi-point boundary value problems
at non-resonance in Banach spaces have been studied by several authors (see, for
example, [5, 13, 14, 18] and references cited therein). Recently, the existence of
solutions for boundary value problems at resonance have been studied by many
papers, (see, for example [3, 4, 9, 10, 11, 15, 16, 19]). Using the Krasnolsel’skii-Guo
fixed point theorem, Han [9] studied a second order three-point BVP at resonance
by rewriting the original BVP as an equivalent one. Motivated by their results, in
this paper, we will discuss the existence of positive solutions for the second-order
m-point boundary value problem at resonance
y 00 (t) = f (t, y),
y 0 (0) = θ,
y(1) =
0 < t < 1,
m−2
X
ki y(ξi )
(1.1)
(1.2)
i=1
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. Banach space; positive solution; strict set contraction;
boundary value problem.
c
2011
Texas State University - San Marcos.
Submitted July 14, 2010. Published February 9, 2011.
Supported by grants 10875094 from the Natural Science Foundation of China,
A2009000664 from the Natural Science Foundation of Hebei Province, 2008153 from the
Foundation of Hebei Education Department, and XL200814 from the Foundation of
Hebei University of Science and Technology.
1
2
W. JIANG, B. WANG
EJDE-2011/22
in a real Banach space E, where θ is P
the zero element of E, 0 < ξ1 < ξ2 < · · · <
m−2
ξm−2 < 1, ki > 0, i = 1, 2, . . . , m − 2, i=1 ki = 1.
Pm−2
The boundary value problem (1.1)-(1.2) is at resonance when i=1 ki = 1; that
is, the corresponding homogeneous boundary value problem
y 00 (t) = 0,
y 0 (0) = 0,
t ∈ [0, 1],
y(1) =
m−2
X
ki y(ξi )
i=1
has nontrivial solutions.
To the best of our knowledge, no paper has considered the existence of positive
solutions for the boundary value problems at resonance in Banach spaces. We shall
fill this gap in the literature. The organization of this paper is as follows. We shall
introduce a theorem and some notations in the rest of this section. In Section 2,
we provide some necessary background. In particular, we state some properties of
Green’s function associated with the equivalent problem of (1.1)-(1.2). In Section
3, the main results will be stated and proved.
Theorem 1.1 ([1, 17]). Let K be a cone of the real Banach space X and Kr,R =
{x ∈ K|r ≤ kxk ≤ R} with R > r > 0. Assume that A : Kr,R → K is a strict set
contraction such that one of the following two conditions is satisfied
(i) Ax 6≤ x for all x ∈ K, kxk = r and Ax 6≥ x for all x ∈ K, kxk = R.
(ii) Ax ≥
6 x for all x ∈ K, kxk = r and Ax 6≤ x for all x ∈ K, kxk = R.
Then A has at least one fixed point x ∈ K satisfying r < kxk < R.
Let the real Banach space E with norm k · k be partially ordered by a normal
cone P of E; i.e., x ≤ y if and only if y − x ∈ P , and P ∗ denotes the dual cone
of P ; i.e., P ∗ = {ϕ ∈ E ∗ : ϕ(x) ≥ 0, x ∈ P }. Denote the normal constant
of P by N ; i.e., θ ≤ x ≤ y implies kxk ≤ N kyk. Take I = [0, 1]. For any
x ∈ C[I, E], evidently, (C[I, E], k·kc ) is a Banach space with kxkc = maxt∈I kx(t)k,
and Q = {x ∈ C[I, E] : x(t) ≥ θ for t ∈ I} is a cone of the Banach space C[I, E].
A function y ∈ C 2 [I, E] is called a positive solution of the boundary value problem
(1.1)-(1.2) if it satisfies (1.1)-(1.2) and y ∈ Q, y(t) 6≡ θ.
In this paper, we denote α(·) the Kuratowski measure of non-compactness of a
bounded set in E and C[I, E]. The closed balls in spaces E and C[I, E] are denoted
by Tr = {x ∈ E : kxk ≤ r}(r > 0) and Br = {y ∈ C[I, E] : kykc ≤ r}(r > 0),
respectively.
Define
F (t, y) := f (t, y) + β 2 y,
where β ∈ (0, π2 ). Obviously, y(t) is a solution of the problem (1.1)-(1.2) if and only
if it is a solution of the problem
y 00 (t) + β 2 y(t) = F (t, y(t)),
y 0 (0) = θ,
y(1) =
m−2
X
i=1
and the problem (1.3)-(1.4) is at non-resonance.
0 < t < 1,
ki y(ξi ),
(1.3)
(1.4)
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For convenience, we set
a0 =
m−2
X
ki cos βξi − cos β,
a1 = (a0 + 1) sin β +
i=1
m−2
X
ki sin βξi ,
i=1
a2 = 1 −
m−2
X
ki cos β(1 − ξi ).
i=1
In this paper, we assume the following conditions hold.
Pm−2
(H1) ki > 0, i = 1, 2, . . . , m − 2, 0 < ξ1 < ξ2 < · · · < ξm−2 < 1, i=1 ki = 1.
(H2) P is a normal cone of E and N is the normal constant; F : I × P → P ,
F (t, θ) = θ for all t ∈ I; for any r > 0, F (t, x) is uniformly continuous and
bounded on I × (P ∩ Tr ) and there exists a constant Lr with 0 ≤ Lr <
(βa0 )/(2a1 ) such that
α(F (I × D)) ≤ Lr α(D),
∀D ⊂ P ∩ Tr .
2. Preliminary lemmas
Pm−2
Lemma 2.1. Assume i=1 ki = 1, then for h(t) ∈ C[I, E], the problem
y 00 (t) + β 2 y(t) = h(t),
y 0 (0) = θ,
y(1) =
0 < t < 1,
m−2
X
ki y(ξi )
(2.1)
(2.2)
i=1
has a unique solution
Z
Z
cos βt 1
1 t
[
sin β(1 − s)h(s)ds
sin β(t − s)h(s)ds +
y(t) =
β 0
βa0 0
m−2
X Z ξi
−
ki
sin β(ξi − s)h(s)ds]
Z
i=1
1
=:
(2.3)
0
G(t, s)h(s)ds,
0
where
1
Pm−2
cos βt

 β sin β(t − s) + βa0 [sin β(1 − s) − j=i kj sin β(ξj − s)],


 if ξi−1 ≤ s ≤ min{t, ξi }, ; i = 1, 2, . . . , m − 1;
G(t, s) =
Pm−2
cos βt


[sin β(1 − s) − j=i kj sin β(ξj − s)],

βa
0


if max{ξi−1 , t} ≤ s ≤ ξi , i = 1, 2, . . . , m − 1.
The proof of the above lemma is easy, so we omit it.
Lemma 2.2. There exist c1 , c2 > 0 such that
c1 (1 − s) ≤ G(t, s) ≤ c2 (1 − s),
t, s ∈ [0, 1].
Proof. Take H(t, s) = c(1 − s) − G(t, s). We will prove that H(t, s) ≥ 0, t, s ∈ [0, 1],
when c is sufficiently large. For t, s ∈ [0, 1], we have
1
cos βt
H(t, s) ≥ c(1 − s) − sin β(t − s) −
sin β(1 − s)
β
βa0
1
sin β(1 − s)
≥ c(1 − s) − sin β(1 − s) −
β
βa0
4
W. JIANG, B. WANG
1
1
[1 + ] sin β(1 − s)
β
a0
= c(1 − s) −
≥ (c − 1 −
EJDE-2011/22
1
)(1 − s).
a0
Take c2 ≥ 1 + a10 , then H(t, s) ≥ 0, t, s ∈ [0, 1].
Now, we prove H(t, s) ≤ 0, t, s ∈ [0, 1], when c is sufficiently small. For t ∈ [0, 1],
s ∈ (ξ1 , 1], we have
H(t, s) ≤ c(1 − s) −
m−2
X
cos βt
[sin β(1 − s) −
kj sin β(ξj − s)]
βa0
j=i
≤ c(1 − s) −
m−2
X
cos β
[sin β(1 − s) −
kj sin β(ξj − s)]
βa0
j=2
≤ c(1 − s) −
k1 cos β
sin β(1 − s).
βa0
Since
(
g(x) =
sin x
x ,
1,
0 < x ≤ π/2,
x=0
is continuous on [0, π/2]. So, we obtain
min
x∈[0,π/2]
g(x) := m0 > 0;
i.e., sin x ≥ m0 x, x ∈ [0, π/2]. Therefore,
m0 k1 cos β
(1 − s)
a0
m0 k1 cos β
= (c −
)(1 − s).
a0
H(t, s) ≤ c(1 − s) −
For t ∈ [0, 1], s ∈ [0, ξ1 ], we obtain
H(t, s) ≤ c(1 − s) −
≤ c(1 − s) −
≤c−
m−2
X
cos βt
[sin β(1 − s) −
kj sin β(ξj − s)]
βa0
j=1
m−2
β(1 + ξj − 2s)
β(1 − ξj )
cos β X
2
kj cos
sin
βa0 j=1
2
2
m−2
2 cos β X
β(1 + ξi )
β(1 − ξi )
ki cos
sin
.
βa0 i=1
2
2
Take
P
β(1+ξi )
i)
m0 k1 cos β 2 cos β( m−2
sin β(1−ξ
)
i=1 ki cos
2
2
0 < c1 ≤ min
,
.
a0
βa0
Then we have c1 (1 − s) ≤ G(t, s), t, s ∈ [0, 1]. The proof is complete.
Lemma 2.3. Assume (H1) holds. If h ∈ Q, then the unique solution y of (2.1)(2.2) satisfies y(t) ≥ θ, t ∈ I and y(t) ≥ γy(s) for all t, s ∈ I, where γ = c1 /c2 .
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POSITIVE SOLUTIONS
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Proof. Obviously, y(t) ≥ θ for all t ∈ I. By Lemma 2.2, we obtain
Z
Z 1
c1 1
y(t) ≥ c1
c2 (1 − s)h(s)ds ≥ γy(r), ∀t, r ∈ I.
(1 − s)h(s)ds =
c2 0
0
The proof is complete.
Define an operator A : Q → C[I, E] as follows
Z
Z
1 t
cos βt 1
[
A(y(t)) :=
sin β(t − s)F (s, y(s))ds +
sin β(1 − s)F (s, y(s))ds
β 0
βa0 0
m−2
X Z ξi
−
ki
sin β(ξi − s)F (s, y(s))ds].
i=1
0
(2.4)
By Lemmas 2.1 and 2.3, we obtain that A : Q → C 2 [I, E] ∩ Q, and y(t) is a positive
solution of (1.1)-(1.2) if and only if y(t) ∈ C 2 [I, E] ∩ Q and y(t) 6≡ θ is a fixed point
of the operator A.
Lemma 2.4. Assume (H1), (H2) hold. Then, for any r > 0, the operator A is a
strict set contraction on Q ∩ Br .
Proof. Since F (t, x) is uniformly continuous and bounded on I × (P ∩ Tr ), we see
from (2.4) that A is continuous and bounded on Q ∩ Br . For any S ⊂ Q ∩ Br , by
(2.4), we can easily get that functions A(S) = {Ay|y ∈ S} are uniformly bounded
and equicontinuous. By [12], we have
α(A(S)) = sup α(A(S(t))),
(2.5)
t∈I
where A(S(t)) = {Ay(t) : y ∈ S, t ∈ I is fixed}. For any y ∈ C[I, E], g ∈ C[I, I],
Rt
by 0 g(s)y(s)ds ∈ co({g(t)y(t)|t ∈ I} ∪ {θ}) ⊂ co({y(t)|t ∈ I} ∪ {θ}), we obtain
α(A(S(t)))
Z
Z
1 t
cos βt h 1
sin β(1 − s)F (s, y(s))ds
= α({
sin β(t − s)F (s, y(s))ds +
β 0
βa0
0
m−2
i
X Z ξi
−
ki
sin β(ξi − s)F (s, y(s))ds : y ∈ S})
i=1
0
sin β
α(co({F (s, y(s)) : s ∈ I, y ∈ S} ∪ {θ}))
β
sin β
+
α(co({F (s, y(s)) : s ∈ I, y ∈ S} ∪ {θ}))
βa0
Pm−2
ki sin βξi
α(co({F (s, y(s)) : s ∈ I, y ∈ S} ∪ {θ}))
+ i=1
βa0
a1
=
α({F (s, y(s)) : s ∈ I, y ∈ S})
βa0
a1
≤
α(F (I × B)),
βa0
≤
where B = {y(s) : s ∈ I, y ∈ S} ⊂ P ∩ Tr .
By (H2), we obtain
α(A(S(t))) ≤
a1
Lr α(B).
βa0
(2.6)
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W. JIANG, B. WANG
EJDE-2011/22
For any given ε > 0, there exists a partition S = ∪lj=1 Sj such that
ε
(2.7)
diam(Sj ) < α(S) + , j = 1, 2, . . . , l.
3
Now, choose yj ∈ Sj , j = 1, 2, . . . , l and a partition 0 = t0 < t1 < · · · < tk = 1 such
that
ε
(2.8)
kyj (t) − yj (t)k < , ∀t, t ∈ [ti−1 , ti ], j = 1, 2, . . . , l, i = 1, 2, . . . , k.
3
Obviously, B = ∪lj=1 ∪ki=1 Bij , where Bij = {y(t) : y ∈ Sj , t ∈ [ti−1 , ti ]}. For any
y(t), y(t) ∈ Bij , by (2.7) and (2.8), we obtain
ky(t) − y(t)k ≤ ky(t) − yj (t)k + kyj (t) − yj (t)k + kyj (t) − y(t)k
ε
≤ ky − yj kc + + kyj − ykc
3
ε
≤ 2 diam(Sj ) + < 2α(S) + ε,
3
which implies diam(Bij ) ≤ 2α(S)+ε, and so, α(B) ≤ 2α(S)+ε. Since ε is arbitrary,
we obtain
α(B) ≤ 2α(S).
(2.9)
It follows from (2.5), (2.6) and (2.9) that
2a1
Lr α(S), ∀S ⊂ Q ∩ Br .
α(A(S)) ≤
βa0
By (H2), we obtain that A is a strict set contraction on Q ∩ Br .
3. Main results
Let K = {y ∈ Q : y(t) ≥ γy(s), ∀t, s ∈ I}. Clearly, K ⊂ Q is a cone of C[I, E].
By Lemmas 2.1 and 2.3, we obtain AQ ⊂ K. So, AK ⊂ K.
For convenience, for any x ∈ P and ϕ ∈ P ∗ , we set
F 0 = lim sup sup
kxk→0 t∈I
F0ϕ = lim inf inf
kxk→0 t∈I
kF (t, x)k
,
kxk
F ∞ = lim sup sup
ϕ(F (t, x))
,
ϕ(x)
ϕ
F∞
= lim inf inf
kxk→∞ t∈I
kxk→∞ t∈I
kF (t, x)k
,
kxk
ϕ(F (t, x))
ϕ(x)
and list the following assumptions:
(H3) There exists ϕ ∈ P ∗ such that ϕ(x) > 0 for any x > θ and F0ϕ >
ϕ
(H4) There exists ϕ ∈ P ∗ such that ϕ(x) > 0 for any x > θ and F∞
>
β 2 a0
γa2 .
β 2 a0
γa2 .
β 2 a0
N (1+a0 )(1−cos β) .
2
a0
F ∞ < N (1+aβ0 )(1−cos
β) .
(H5) F 0 <
(H6)
(H7) There exists r0 > 0 such that
kF (t, x)k <
sup
t∈I, x∈P
β 2 a0
r0 .
N (1 + a0 )(1 − cos β)
γr0 /N ≤kxk≤r0
(H8) There exist R0 > 0 and ϕ ∈ P ∗ with ϕ(x) > 0 for any x > θ such that
inf
t∈I,x∈P
γR0 /N ≤kxk≤R0
ϕ(F (t, x))
β 2 a0
.
>
ϕ(x)
γa2
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POSITIVE SOLUTIONS
7
Theorem 3.1. Assume (H1), (H2) hold. If one of the following conditions is
satisfied:
(i) (H4) and (H5) hold.
(ii) (H3) and (H6) hold.
Then the problem (1.1)-(1.2) has at least one positive solution.
2
Proof. (i) By (H4), we obtain that there exist constants M > βγaa20 and r1 > 0 such
that
ϕ(F (t, x)) ≥ M ϕ(x), ∀t ∈ I, x ∈ P, kxk > r1 .
(3.1)
For any R > N r1 /γ, we will show that
Ay 6≤ y,
∀y ∈ K, kykc = R.
(3.2)
In fact, if not, there exists y0 ∈ K, ky0 kc = R such that Ay0 ≤ y0 . By
y0 (t) ≥ γy0 (s) ≥ θ,
we have
ky0 (t)k ≥
∀t, s ∈ I,
γ
ky0 kc > r1 ,
N
∀t ∈ I.
(3.3)
(3.4)
By (2.4), for any t ∈ I, we have
Z
Z
1 t
cos βt h 1
sin β(1 − s)F (s, y0 (s))ds
A(y0 (t)) =
sin β(t − s)F (s, y0 (s))ds +
β 0
βa0
0
m−2
i
X Z ξi
−
ki
sin β(ξi − s)F (s, y0 (s))ds
i=1
0
Z 1
m−2
cos βt X
sin β(1 − s)F (s, y0 (s))ds.
ki
≥
βa0 i=1
ξi
This inequality, (3.1), (3.3) and (3.4), imply
Z 1
m−2
1 X
sin β(1 − s)M γϕ(y0 (0))ds
ϕ(Ay0 (0)) ≥
ki
βa0 i=1
ξi
a2
= 2 M γϕ(y0 (0)).
β a0
Considering Ay0 ≤ y0 , we obtain
ϕ(y0 (0)) ≥
γa2
M ϕ(y0 (0)).
β 2 a0
(3.5)
It is easy to see that ϕ(y0 (0)) > 0 (In fact, if ϕ(y0 (0)) = 0, by (3.3), we obtain
ϕ(y0 (0)) ≥ γϕ(y0 (s)) ≥ 0 for all s ∈ I. So, we have ϕ(y0 (s)) ≡ 0 for all s ∈ I. That
is, y0 (s) ≡ θ. This is a contradiction with ky0 kc = R). So, (3.5) contradicts with
2
M > βγaa20 . Therefore, (3.2) is true.
On the other hand, by (H5) and F (t, θ) = θ, we obtain that there exist constants
2
a0
0 < ε < N (1+aβ0 )(1−cos
β) and 0 < r2 < R such that
kF (t, x)k ≤ εkxk,
∀t ∈ I, x ∈ P, kxk < r2 .
(3.6)
For any 0 < r < r2 , we now prove that
Ay 6≥ y,
∀y ∈ K, kykc = r.
(3.7)
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W. JIANG, B. WANG
EJDE-2011/22
In fact, if not, there exists y0 ∈ K, ky0 kc = r such that Ay0 ≥ y0 . Since (2.4)
implies
Z
a0 + cos βt 1
Ay0 (t) ≤
sin β(1 − s)F (s, y0 (s))ds, ∀t ∈ I.
(3.8)
βa0
0
So, we have
Z
a0 + cos βt 1
sin β(1 − s)F (s, y0 (s))ds, ∀t ∈ I.
θ ≤ y0 (t) ≤
βa0
0
This, together with (3.6), imply
Z
N (1 + a0 )ε 1
N (1 + a0 )(1 − cos β)εky0 kc
ky0 (t)k ≤
,
sin β(1 − s)ky0 (s)kds =
βa0
β 2 a0
0
for all t ∈ I. Therefore, we obtain ε ≥ β 2 a0 /N (1 + a0 )(1 − cos β). This is a
contradiction. So, (3.7) is true.
By (3.2), (3.7), Lemma 2.4 and Theorem 1.1, we obtain that the operator A has
at least one fixed point y ∈ K satisfying r < kykc < R.
(ii) By (H3), in the same way as establishing (3.2) we can assert that there exists
r2 > 0 such that for any 0 < r < r2 ,
Ay 6≤ y,
∀y ∈ K, kykc = r.
(3.9)
On the other hand, by (H6), we obtain that there exist constants r1 > 0 and ε,
2
a0
with 0 < ε < N (1+aβ0 )(1−cos
β) , such that
kF (t, x)k ≤ εkxk,
∀t ∈ I, x ∈ P, kxk > r1 .
By (H2), we obtain
kF (t, x)k =: b < ∞.
sup
t∈I, x∈P ∩Tr1
So, we have
kF (t, x)k ≤ εkxk + b,
∀t ∈ I, x ∈ P.
(3.10)
Take
R > max r2 ,
N b(1 + a0 )(1 − cos β)
,
− N ε(1 + a0 )(1 − cos β)
β 2 a0
we will prove that
Ay 6≥ y,
∀y ∈ K, kykc = R.
(3.11)
In fact, if there exists y0 ∈ K, ky0 kc = R such that Ay0 ≥ y0 . Then, by (3.8) and
(3.10), we obtain
Z
N (a0 + cos βt) 1
sin β(1 − s)(εky0 (s)k + b)ds
ky0 (t)k ≤
βa0
0
N (1 + a0 )(1 − cos β)
≤
(εky0 kc + b), ∀t ∈ I.
β 2 a0
So, we have
N b(1 + a0 )(1 − cos β)
< R.
− N ε(1 + a0 )(1 − cos β)
A contradiction. Therefore, (3.11) holds.
By (3.9), (3.11), Lemma 2.4 and Theorem 1.1, we obtain that the operator A has
at least one fixed point y ∈ K satisfying r < kykc < R. The proof is complete. ky0 kc ≤
β 2 a0
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POSITIVE SOLUTIONS
9
Theorem 3.2. Assume (H1), (H2) hold. If one of the following conditions is
satisfied:
(i) (H3), (H4), (H7) hold.
(ii) (H5), (H6), (H8) hold.
Then (1.1)-(1.2) has at least two positive solutions.
Proof. (i) By (H3), (H4) and the proof of Theorem 3.1, we obtain that there exist
r, R with 0 < r < r0 < R such that
Ay 6≤ y,
∀y ∈ K, kykc = r,
Ay 6≤ y,
∀y ∈ K, kykc = R.
Now, we prove that
Ay 6≥ y, ∀y ∈ K, kykc = r0 .
(3.12)
In fact, if there exists y0 ∈ K, ky0 kc = r0 such that Ay0 ≥ y0 . By (3.8) and (H7),
we obtain
Z
N (1 + a0 ) 1
β 2 a0
ky0 kc <
r0 ds = r0 .
sin β(1 − s)
βa0
N (1 + a0 )(1 − cos β)
0
A contradiction. So, (3.12) is true. By Lemma 2.4 and Theorem 1.1, we obtain
that the operator A has at least two fixed points y1 , y2 ∈ K satisfying r < ky1 kc <
r0 < ky2 kc < R.
(ii) By (H5), (H6) and the proof of Theorem 3.1, we obtain that there exist r, R
with 0 < r < R0 < R such that
Ay 6≥ y,
∀y ∈ K, kykc = r,
Ay 6≥ y,
∀y ∈ K, kykc = R.
On the other hand, by (H8) and the same way as used in the proof of (3.2), we can
prove that
Ay 6≤ y, ∀y ∈ K, kykc = R0 .
(3.13)
By Lemma 2.4 and Theorem 1.1, we obtain that the operator A has at least two
fixed points y1 , y2 ∈ K satisfying r < ky1 kc < R0 < ky2 kc < R. The proof is
complete.
Similar to the proofs of Theorem 3.1 and Theorem 3.2, we can easily get the
following corollaries.
Corollary 3.3. Assume (H1), (H2) hold. If one of the following conditions is
satisfied:
(i) (H4), (H5), (H7), (H8) hold with R0 < γr0 /N .
(ii) (H3), (H6), (H7), (H8) hold with r0 < γR0 /N .
Then (1.1)-(1.2) has at least three positive solutions.
Corollary 3.4. Assume (H1), (H2) hold. If one of the following conditions is
satisfied:
(i) (H5)–(H7) hold, and there exist Ri > 0, ϕi ∈ P ∗ with ϕi (x) > 0 for x > θ,
i = 1, 2 such that
inf
t∈I, x∈P, γRi /N ≤kxk≤Ri
β 2 a0
ϕi (F (t, x))
>
,
ϕi (x)
γa2
where R1 < γr0 /N , r0 < γR2 /N .
i = 1, 2,
10
W. JIANG, B. WANG
EJDE-2011/22
(ii) (H3), (H4), (H8) hold, and there exist r1 , r2 > 0 such that
sup
kF (t, x)k <
t∈I, x∈P, γri /N ≤kxk≤ri
β 2 a0
ri ,
N (1 + a0 )(1 − cos β)
i = 1, 2,
where r1 < γR0 /N , R0 < γr2 /N .
Then (1.1)-(1.2) has at least four positive solutions.
We can prove easily the existence of multiple positive solutions for (1.1)-(1.2).
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Weihua Jiang
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
Hebei, China
E-mail address: weihuajiang@hebust.edu.cn
EJDE-2011/22
POSITIVE SOLUTIONS
11
Bin Wang
Department of Basic Courses, Hebei Professional and Technological College of Chemical and, Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei, China
E-mail address: wb@hebcpc.cn