Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 38, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLE POSITIVE SOLUTIONS FOR SINGULAR
MULTI-POINT BOUNDARY-VALUE PROBLEMS WITH A
POSITIVE PARAMETER
CHAN-GYUN KIM, EUN KYOUNG LEE
Abstract. In this article we study the existence, nonexistence, and multiplicity of positive solutions for a singular multi-point boundary value problem
with positive parameter. We use the fixed point index theory on a cone and
a well-known theorem for the existence of a global continuum of solutions to
establish our results.
1. Introduction
Consider the singular multi-point boundary-value problem
(ϕp (u0 (t)))0 + λf (t, u(t)) = 0,
u(0) =
m−2
X
ai u(ξi ),
u(1) =
t ∈ (0, 1),
m−2
X
bi u(ξi ),
(1.1)
(1.2)
i=1
i=1
where ϕp (s) = |s|p−2 s, p > 1, λ a nonnegative real parameter, ξi ∈ (0, 1) with 0 <
Pm−2
Pm−2
ξ1 < ξ2 < · · · < ξm−2 < 1, ai , bi ∈ [0, 1) with 0 ≤ i=1 ai < 1, 0 ≤ i=1 bi < 1,
and f ∈ C((0, 1) × [0, ∞), (0, ∞)). Here, f (t, u) may be singular at t = 0 and/or 1
and satisfies the following conditions:
(F1) for all M > 0, there exists hM ∈ A such that f (t, u) ≤ hM (t), for all
u ∈ [0, M ] and all t ∈ (0, 1), where
Z 1/2
Z 1
Z 1/2
Z s
−1
A = {h :
ϕ−1
h(τ
)dτ
ds
+
ϕ
h(τ
)dτ
ds < ∞} :
p
p
0
s
1/2
1/2
(F2) there exists [α, β] ⊂ (0, 1) such that limu→∞ f (t, u)/up−1 = ∞ uniformly
in [α, β].
By a positive solution of problem (1.1)-(1.2), we mean a function u ∈ C[0, 1] ∩
C 1 (0, 1) with ϕp (u0 ) ∈ C 1 (0, 1) that satisfies (1.1)-(1.2) and u > 0 in (0, 1). Here
k · k denotes the usual maximum norm in C[0, 1].
Motivated by the work of Bitsadze [3, 4], the study of multi-point boundary value
problem for linear second-order ordinary differential equations was initially done by
2000 Mathematics Subject Classification. 34B10, 34B16.
Key words and phrases. Singular problem; multi-point boundary value problems;
positive solution; p-Laplacian; multiplicity.
c
2014
Texas State University - San Marcos.
Submitted July 3, 2013. Published February 5, 2014.
1
2
C.-G. KIM, E. K. LEE
EJDE-2014/38
Il’in and Moiseev [13, 14]. Gupta [11] studied three-point boundary value problems
for nonlinear ordinary differential equations. Since then, many researchers have
studied nonlinear second-order multi-point boundary value problems under various
conditions on the nonlinear term. We refer the reader to [2, 8, 9, 15, 19, 22, 23, 24,
25, 27, 28] and references therein.
Problem (1.1)-(1.2) is a singular boundary value problem since f is allowed to
have singularity at t = 0 and/or 1. Singular problems have been extensively studied
in the literature. For the case of two-point boundary value problems, the results
were proved in [1, 5, 6, 12, 18, 21, 26, 29, 30] and for multi-point boundary value
problems, the results were proved in [8, 9, 19, 22, 24, 28]. However, there are few
results for multi-point boundary value problems having nonlinear term which does
not satisfy L1 -Carathéodory condition. Recently, in semi-linear case, Sun et al. [24]
studied the following singular three-point boundary-value problem
y 00 + µa(t)g1 (t, y) = 0,
0
y(0) − βy (0) = 0,
t ∈ (0, 1)
y(1) = αy(η),
(1.3)
where µ > 0 is a parameter, β > 0, 0 < η < 1, 0 < αη < 1, (1 − αη) + β(1 −
R1
α) > 0, a ∈ C((0, 1), (0, ∞)) satisfies 0 < 0 (β + s)(1 − s)a(s)ds < ∞, and g1 ∈
C([0, 1] × (0, ∞), (0, ∞)) may be singular at y = 0. Without any monotone or
growth conditions imposed on the nonlinearity g1 , using fixed point index theorem,
they obtained not only the existence results of positive solutions to the problem
(1.3), but also the explicit interval about positive parameter µ. Kim [19], in pLaplacian case, presented some sufficient conditions for one or multiple positive
solutions to the problem (1.1)-(1.2), where f (t, u) = h(t)g2 (t, u), h ∈ A, g2 ∈
C([0, 1] × [0, ∞), [0, ∞)).
To the authors’ knowledge, in the case of p-Laplacian, there is no result about
the global structure of positive solutions for parameter λ ∈ (0, ∞) to multi-point
boundary-value problems with the nonlinear term admitting stronger singularity
than L1 (0, 1) at t = 0 and/or 1. The following is the main result in this paper.
Theorem 1.1. Assume that (F1) and (F2) hold. Assume in addition that f (t, u) =
h(t)g(t, u), where h ∈ A and g ∈ C((0, 1) × [0, ∞), (0, ∞)) satisfies
(A1) for all N > 0 and all > 0, there exists δ = δ(N, ) > 0 such that if
u, v ∈ [0, N ] and |u − v| < δ, then |g(t, u) − g(t, v)| < , for all t ∈ (0, 1),
(A2) inf{g(t, u) | t ∈ (0, 1), u ∈ [0, ∞)} > 0.
Then there exists λ∗ > 0 such that problem (1.1)-(1.2) has at least two positive
solutions for λ ∈ (0, λ∗ ), at least one positive solution for λ = λ∗ and no positive
solution for λ > λ∗ .
The above result is an extension of previous works for two-point boundary-value
problems by Choi [5], Wong [26], Dalmasso [6], Ha and Lee [12], Lee [21], Xu and
Ma [29], and Kim [18].
The rest of this article is organized as follows. In Section 2, the operator for
problem (1.1)-(1.2) is introduced, and well-known facts such as Picone-type identity
and Global continuation theorem are presented. In Section 3, the proofs of our
results (Theorem 3.4 and Theorem 1.1) and examples for nonlinear term to illustrate
our results are given.
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MULTIPLE POSITIVE SOLUTIONS
3
2. Preliminaries
First we introduce the operator corresponding to problem (1.1)-(1.2). Throughout this section we assume that (F1) holds. Set
K = {u ∈ C[0, 1] : ui s a nonnegative concave function on [0, 1], u satisfies (1.2)}.
Then K is an ordered cone in C[0, 1]. For (λ, u) ∈ [0, ∞) × K, we define xλ,u :
[0, 1] → R as xλ,u (t) = x1λ,u (t) − x2λ,u (t), where
Z t
m−2
hZ t
i
hZ t
i
X Z ξi
−1
λf (τ, u(τ ))dτ ds
ϕ
λf
(τ,
u(τ
))dτ
ds
+
x1λ,u (t) = A−1
ai
ϕ−1
p
p
s
0
s
0
i=1
and
x2λ,u (t)
=B
−1
m−2
X
1
Z
ϕ−1
p
bi
ξi
i=1
s
hZ
1
Z
i
ϕ−1
p
λf (τ, u(τ ))dτ ds+
t
Here
A=1−
hZ
t
m−2
X
ai ,
B =1−
m−2
X
i=1
s
i
λf (τ, u(τ ))dτ ds.
t
bi .
i=1
For λ > 0, limt→0+ xλ,u (t) < 0 and limt→1− xλ,u (t) > 0. Indeed we can rewrite
x1λ,u (t) as
x1λ,u (t)
m−2
X Z
= A−1 −
ai
ξi
ϕ−1
p
hZ
t
i=1
s
Z t
i
hZ t
i λf (τ, u(τ ))dτ ds +
ϕ−1
λf (τ, u(τ ))dτ ds .
p
t
0
s
By (F1), there exists h2 ∈ A such that
Z t
Z t
hZ t
i
hZ t
i
−1
−1
0≤
ϕp
λf (τ, u(τ ))dτ ds ≤
ϕp
h2 (τ )dτ ds,
0
s
0
and
t
Z
lim+
t→0
ϕ−1
p
0
hZ
t
s
i
λf (τ, u(τ ))dτ ds = 0.
s
Clearly limt→0+ x2λ,u (t) > 0, and thus limt→0+ xλ,u (t) < 0. In a similar manner we
can show limt→1− xλ,u (t) > 0. Since xλ,u is continuous and strictly increasing in
(0, 1), there exists a unique zero Aλ,u ∈ (0, 1) such that xλ,u (Aλ,u ) = 0. For λ = 0,
we may take A0,u = 0 since x0,u ≡ 0. Then, for (λ, u) ∈ [0, ∞) × K,
Z Aλ,u
m−2
h Z Aλ,u
i
h Z Aλ,u
i
X Z ξi
−1
−1
−1
A
ai
ϕp
λf (τ, u(τ ))dτ ds +
ϕp
λf (τ, u(τ ))dτ ds
0
i=1
= B −1
m−2
X
i=1
s
Z
1
bi
ξi
ϕ−1
p
0
hZ
s
Aλ,u
i
λf (τ, u(τ ))dτ ds +
s
Z
1
Aλ,u
ϕ−1
p
hZ
s
i
λf (τ, u(τ ))dτ ds.
Aλ,u
Define H : [0, ∞) × K → C[0, 1] as
 −1 Pm−2 R ξi −1 R Aλ,u
A
λf (τ, u(τ ))dτ ds

i=1 ai 0 ϕp
s

R Aλ,u
R


+ t ϕ−1
λf (τ, u(τ ))dτ ds,
0 ≤ t ≤ Aλ,u ,
0 p
s
H(λ, u)(t) =
R
R
P
1
s
m−2

B −1 i=1 bi ξi ϕ−1
λf (τ, u(τ ))dτ ds

p
Aλ,u


Rs
 R1
−1
+ Aλ,u ϕp
λf (τ, u(τ ))dτ ds,
Aλ,u ≤ t ≤ 1.
Aλ,u
4
C.-G. KIM, E. K. LEE
EJDE-2014/38
In view of the definition of Aλ,u , H(λ, u) is well-defined, kH(λ, u)k = H(λ, u)(Aλ,u ),
and H(λ, u) ∈ K for all (λ, u) ∈ [0, ∞) × K (see, e.g., [8, Lemma 2.2]).
Lemma 2.1. Problem (1.1)-(1.2) has a positive solution u if and only if H(λ, ·)
has a fixed point u in K for λ > 0.
Proof. We assume that u is a positive solution of problem (1.1)-(1.2). If λ = 0,
Pm−2
Pm−2
u ≡ 0 by the facts that 0 ≤ i=1 ai < 1 and 0 ≤ i=1 bi < 1. Thus λ > 0.
Since u0 is strictly decreasing in (0, 1), u ∈ K. From the fact that u satisfies
(BC), max{u(0), u(1)} < u(ξj ) for some 1 ≤ j ≤ m − 2, and there exists a unique
Au ∈ (0, 1) such that u0 (Au ) = 0. Integrating (Pλ ) from s to Au , we have
h Z Au
i
0
−1
u (s) = ϕp λ
f (τ, u(τ ))dτ .
(2.1)
s
Again integrating (2.1) from 0 to t, we have
Z t
h Z Au
i
u(t) = u(0) +
ϕ−1
λf
(τ,
u(τ
))dτ
ds,
p
0
Then u(ξi ) = u(0) +
u(0) =
R ξi
0
m−2
X
ϕ−1
p
t ∈ [0, 1).
s
R Au
s
λf (τ, u(τ ))dτ ds and
ai u(ξi )
i=1
=
m−2
X
m−2
X
ai u(0) +
i=1
−1
m−2
X
ξi
ai
ξi
Z
ai
hZ
ϕ−1
p
Au
i
λf (τ, u(τ ))dτ ds.
s
Au
hZ
0
i=1
ϕ−1
p
0
i=1
Thus
u(0) = A
Z
i
λf (τ, u(τ ))dτ ds.
s
Similarly, integrating (2.1) from t to 1,
Z 1
hZ s
i
u(t) = u(1) +
ϕ−1
λf (τ, u(τ ))dτ ds, t ∈ (0, 1]
p
t
and
u(1) = B
−1
Au
m−2
X
Z
1
bi
i=1
ϕ−1
p
ξi
hZ
s
i
λf (τ, u(τ ))dτ ds.
Au
Then, by the definition of Aλ,u , Au = Aλ,u and consequently H(λ, u) ≡ u.
Conversely, if we assume that there exists u ∈ K such that H(λ, u) = u for λ > 0,
then one can easily see that u is a positive solution of problem (1.1)-(1.2).
Lemma 2.2. Let M > 0 be given and let {(λn , un )} be a sequence in [0, ∞) × K
with |λn | + kun k ≤ M . If Aλn ,un → 0 (or 1) as n → ∞, then λn → 0 and
kH(λn , un )k → 0 as n → ∞.
Proof. We only prove the case Aλn ,un → 0 as n → ∞ since the other case can be
showed in a similar manner. By the definition of Aλ,u , we can easily know λn → 0
as n → ∞. By (F1), there exists hM ∈ A such that f (t, u) ≤ hM (t), t ∈ (0, 1),
u ∈ [0, M ]. For sufficiently large n, we have Aλn ,un < ξ1 ,
m−2
h Z Aλn ,un
i
X Z ξi
−1
−1
0 ≤ H(λn , un )(0) = A
ai
ϕp
λn f (τ, un (τ ))dτ ds
i=1
0
s
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MULTIPLE POSITIVE SOLUTIONS
≤ λn A−1
Z
ξm−2
ϕ−1
p
hZ
0
ξm−2
5
i
hM (τ )dτ ds,
s
and
Z
Aλn ,un
kH(λn , un )k = H(λn , un )(0) + λn
ϕ−1
p
hZ
0
Aλn ,un
i
hM (τ )dτ ds.
s
Thus kH(λn , un )k → 0 as n → ∞ since hM ∈ A and λn → 0 as n → ∞.
Lemma 2.3. H : [0, ∞) × K → K is completely continuous.
Proof. By Lemma 2.2, Ascoli-Arzelà theorem, and Lebesgue dominated convergence theorem, one can easily show the completely continuity of H (e.g., see [1, 19]).
Thus we omit the proof here.
Next we introduce the generalized Picone identity due to Jaros and Kusano
([16]). Let us consider the following operators:
lp [y] ≡ (ϕp (y 0 ))0 + q(t)ϕp (y),
Lp [z] ≡ (ϕp (z 0 ))0 + Q(t)ϕp (z).
Theorem 2.4 ([20, p 382]). Let q(t) and Q(t) be measurable functions on an interval I. If y and z are any functions such that y, z, ϕp (y 0 ), ϕp (z 0 ) are differentiable
a.e. on I and z(t) 6= 0 for t ∈ I, then the following holds
o
d n |y|p ϕp (z 0 )
− yϕp (y 0 )
dt
ϕp (z)
yz 0 p
z 0 |y|p
= (q − Q)|y|p − |y 0 |p + (p − 1)|
| − pϕp (y)y 0 ϕp
− ylp [y] +
Lp [z].
z
z
ϕp (z)
(2.2)
Remark 2.5. By Young’s inequality, we have
yz 0 p
z0 |y 0 |p + (p − 1)|
| − pϕp (y)y 0 ϕp
≥ 0,
z
z
and the equality holds if and only if y 0 = yz 0 /z in (a, b).
Finally we recall a well-known theorem for the existence of a global continuum
of solutions by Leray and Schauder [17].
Theorem 2.6 ([31, Corollary 14.12]). Let X be a Banach space with X 6= {0} and
let K be an ordered cone in X. Consider
x = H(µ, x),
(2.3)
where µ ∈ [0, ∞) and x ∈ K. If H : [0, ∞) × K → K is completely continuous and
H(0, x) = 0 for all x ∈ K. Then the solution component C of (2.2) in [0, ∞) × K
which contains (0, 0) is unbounded.
3. Main results
Since H(0, u) = 0 and H(λ, 0) 6= 0 if λ 6= 0, by Lemma 2.3, Theorem 2.6, we
obtain the following proposition.
Proposition 3.1. Assume that (F1) holds. Then there exists an unbounded continuum C emanating from (0, 0) in the closure of the set of positive solutions of
problem (1.1)-(1.2) in [0, ∞) × K.
6
C.-G. KIM, E. K. LEE
EJDE-2014/38
To see the shape of C, we need lemmas regarding λ-direction block and a priori
estimate. Using the generalized Picone identity (Theorem 2.4) and the properties
of the p-sine function [7, 32], we obtain the following two lemmas.
Lemma 3.2. Assume that (F1) and (F2) hold. Then there exists λ̄ > 0 such that
if problem (1.1)-(1.2) has a positive solution uλ , then λ ≤ λ̄.
Proof. Let uλ be a positive solution of problem (1.1)-(1.2). Since f (t, u) > 0 for all
(t, u) ∈ (0, 1) × [0, ∞), by (F2), there exists C1 > 0 such that
f (t, u) > C1 ϕp (u)
for u ∈ [0, ∞), t ∈ [α, β].
(3.1)
It is easy to check that w(t) = Sq (πq (t − α)/(β − α)), where Sq is the q-sine function and p1 + 1q = 1, is a solution of
π q p
(ϕp (w0 (t)))0 +
ϕp (w(t)) = 0, t ∈ (α, β),
β−α
w(α) = w(β) = 0.
p
Taking y = w, z = uλ , q(t) = (πq /(β − α)) and Q(t) = λf (t, uλ )/ϕp (uλ ) in (2.2)
and integrating (2.2) from α to β, by Remark 2.5,
Z β
πq p
f (t, uλ ) p
−λ
|w| dt ≥ 0.
β−α
ϕp (uλ )
α
It follows from (3.1) that
π q p
− λC1
β−α
and thus the proof is complete.
Z
β
|w|p dt ≥ 0,
α
Lemma 3.3. Assume that (F1) and (F2) hold, and let J = [D, E] be a compact
subset of (0, ∞). Then there exists MJ > 0 such that if u is a positive solution of
problem (1.1)-(1.2) with λ ∈ J, then kuk ≤ MJ .
Proof. Suppose on the contrary that there exists a sequence {un } of positive solutions of problem (1.1)-(1.2) with λn instead of λ, and {λn } ⊂ J = [D, E] and
kun k → ∞ as n → ∞. It follows from the concavity of un for all n that
un (t) ≥ min{α, 1 − β}kun k,
−1
t ∈ (α, β).
(3.2)
p
Take C = 2D (πq /(β − α)) > 0. By (F2), there exists K > 0 such that f (t, u) >
Cϕp (u), for t ∈ (α, β), u > K. From the assumption, we get kuN k > (min{α, 1 −
β})−1 K, for sufficiently large N . Therefore, by (3.2), we have
f (t, uN (t)) > Cϕp (uN (t)),
t ∈ (α, β).
As in the proof of Lemma 3.2, if we take y(t) = Sq (πq (t − α)/(β − α)) and z = uN ,
by Theorem 2.4 and Remark 2.5,
π q p
C ≤ D−1
.
β−α
This contradicts the choice of C, and thus the proof is complete.
Setting λ∗ = sup{µ > 0: for all λ ∈ (0, µ), there exists at least two positive
solutions of problem (1.1)-(1.2), then λ∗ > 0 is well-defined. Indeed by Proposition
3.1, C emanates from (0, 0), and problem (1.1)-(1.2) has a small solution near (0, 0)
for λ ∈ (0, s) with small s > 0. On the other hand, for any M > 0, define
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MULTIPLE POSITIVE SOLUTIONS
7
CM = {(λ, u) ∈ C : kuk ≥ M } and the projection of CM to the λ-axis as ΛM .
Then, by Lemma 3.2 and Lemma 3.3, for large M , ΛM = (0, aM ], where aM > 0
and it is decreasing in M . This implies that, for any interval (0, s) with small
s > 0, problem (1.1)-(1.2) also has a large solution for λ ∈ (0, s). Thus λ∗ > 0 is
well-defined. Moreover it follows from an easy compactness argument that problem
(1.1)-(1.2) has at least two positive solution for λ ∈ (0, λ∗ ) and at least one positive
solution for λ = λ∗ .
The following is the first result in this work.
Theorem 3.4. Assume that (F1) and (F2) hold. Then there exists λ∗ ≥ λ∗ > 0
such that problem (1.1)-(1.2) has at least two positive solutions for λ ∈ (0, λ∗ ), at
least one positive solution for λ ∈ [λ∗ , λ∗ ], and no positive solution for λ > λ∗ .
Proof. Define λ∗ = sup{λ : problem (1.1)-(1.2) has at least one positive solution}.
Then by Lemma 3.2, λ∗ ≤ λ∗ < ∞. We only consider the case λ∗ < λ∗ , since the
proof is done for the case λ∗ = λ∗ . For λ ∈ [λ∗ , λ∗ ), there exists λ̂ ∈ [λ, λ∗ ) such
that (1.1)-(1.2) with λ̂ instead of λ, has a positive solution, say û. Consider the
modified problem
(ϕp (u0 (t)))0 + λf¯(t, u(t)) = 0, t ∈ (0, 1),
u(0) =
m−2
X
ai u(ξi ),
u(1) =
m−2
X
(3.3)
bi u(ξi ),
i=1
i=1
where f¯(t, u) = f (t, γ(t, u)) and γ : (0, 1) × R → R is defined as


û(t), if u > û(t),
γ(t, u) = u,
if 0 ≤ u ≤ û(t),


0,
if u < 0.
Then all solutions u of (3.3) are concave and non-trivial. Define Tλ : C[0, 1] →
C[0, 1] as

R Â
Pm−2 R ξ

λf¯(τ, u(τ ))dτ ds
A−1 i=1 ai 0 i ϕ−1
p

s


+ R t ϕ−1 R Â λf¯(τ, u(τ ))dτ ds,
0 ≤ t ≤ Â
0 p
s
Tλ (u)(t) =
Pm−2 R 1 −1 R s ¯

−1

B
λf (τ, u(τ ))dτ ds

i=1 bi ξi ϕp
Â


+ R 1 ϕ−1 R s λf¯(τ, u(τ ))dτ ds,
 ≤ t ≤ 1.
 p
Â
where  satisfies
m−2
hZ
X Z ξi
−1
−1
A
ai
ϕp
0
i=1
= B −1
m−2
X
i=1
Â
s
Z
1
bi
ξi
ϕ−1
p
Â
Z
i
¯
λf (τ, u(τ ))dτ ds +
ϕ−1
p
hZ
0
hZ
s
Â
i
λf¯(τ, u(τ ))dτ ds +
Â
i
λf¯(τ, u(τ ))dτ ds
s
Z
1
Â
ϕ−1
p
hZ
s
i
λf¯(τ, u(τ ))dτ ds.
Â
It is easy to check that Tλ is completely continuous on C[0, 1], and u is a solution of
(3.3) if and only if u = Tλ u. It follows from the definition of γ and the continuity
of f that there exists R1 > 0 such that kTλ uk < R1 for all u ∈ C[0, 1]. Then by
Schauder fixed point theorem, there exists uλ ∈ C[0, 1] such that Tλ uλ = uλ , and
uλ is a positive solution of (Mλ ).
8
C.-G. KIM, E. K. LEE
EJDE-2014/38
We first claim that uλ (0) ≤ û(0). If the claim is not true, uλ (0) > û(0). Put
x(t) = uλ (t) − û(t). Then
0 < x(0) = uλ (0) − û(0) =
m−2
X
ai x(ξi ) ≤
i=1
m−2
X
ai x(ξj ) < x(ξj ),
i=1
where x(ξj ) = max{x(ξi )|1 ≤ i ≤ m−2}. Similarly, x(1) < x(ξj ). Thus, there exists
σ ∈ (0, 1) and a ∈ [0, σ) such that x(σ) = maxt∈[0,1] x(t) > 0, x0 (σ) = 0, x(a) = 0,
and x(t) > 0 for t ∈ (a, σ]. Since λ < λ̂, for t ∈ (a, σ], (ϕp (u0λ (t)))0 > (ϕp (û0 (t)))0
and integrating this from t to σ, u0λ (t) < û0 (t). Again integrating from a to σ, we
have x(σ) = uλ (σ) − û(σ) < uλ (a) − û(a) = x(a). This is a contradiction. Thus the
claim is proved. Similarly, we have uλ (1) ≤ û(1). Next we show that uλ (t) ≤ û(t)
for t ∈ (0, 1). If it is not true, it follows from uλ (0) ≤ û(0) and uλ (1) ≤ û(1) that
there exists an interval [t1 , t2 ] ⊂ [0, 1] such that uλ (t1 ) = û(t1 ), uλ (t2 ) = û(t2 ) and
uλ (t) > û(t) for all t ∈ (t1 , t2 ). Then
(ϕp (u0λ (t)))0 > (ϕp (û0 (t)))0 ,
t ∈ (t1 , t2 )
and we can choose an interval [b, c] ⊂ [t1 , t2 ] such that
û0 (c). Using (3.4), we can get the contradiction
u0λ (b)
(3.4)
0
> û (b) and
u0λ (c)
<
0 > [ϕp (u0λ (c)) − ϕp (u0λ (b))] − [ϕp (û0 (c)) − ϕp (û0 (b))]
Z c
=
[ϕp (u0λ (t))]0 − [ϕp (û0 (t))]0 dt > 0.
b
Therefore, by the definition of γ, uλ turns out a positive solution of problem (1.1)(1.2). Furthermore, by Lemma 3.3 and the complete continuity of H, we can show
that problem (1.1)-(1.2), with λ∗ instead of λ, has a positive solution u∗ , and thus
the proof is complete.
Now we consider f (t, u) = h(t)g(t, u) and let u∗ be a positive solution of problem
(1.1)-(1.2), with λ∗ instead of λ.
Lemma 3.5. Assume that (F1) and (F2) hold. Assume in addition that g satisfies
the conditions (A1) and (A2). Then, for all λ ∈ (0, λ∗ ), there exists δλ > 0 such
that αλ (t) = u∗ (t) + δλ satisfies
(ϕp (αλ0 (t)))0 + λh(t)g(t, αλ (t)) < 0,
t ∈ (0, 1).
(3.5)
Proof. Let λ be fixed in (0, λ∗ ). Put
1
= [λ∗ /λ − 1] inf g(t, u∗ (t)) > 0.
2
t∈(0,1)
By (A1), there exists δλ > 0 such that if u, v ∈ [0, ku∗ k + 1] and |u − v| < δλ , then
|g(t, u) − g(t, v)| < , t ∈ (0, 1). Put αλ (t) = u∗ (t) + δλ . Then
(ϕp (αλ0 (t)))0 + λf (t, αλ (t)) = (ϕp (u0∗ (t)))0 + λf (t, u∗ (t) + δλ )
= h(t)[−λ∗ g(t, u∗ (t)) + λg(t, u∗ (t) + δλ )].
From this, if αλ does not satisfy (3.5), there exists t0 ∈ (0, 1) such that
−λ∗ g(t0 , u∗ (t0 )) + λg(t0 , u∗ (t0 ) + δλ ) ≥ 0,
and then
g(t0 , u∗ (t0 ) + δλ ) ≥
λ∗
g(t0 , u∗ (t0 )).
λ
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MULTIPLE POSITIVE SOLUTIONS
9
By the choice of δλ ,
λ∗
− 1 g(t0 , u∗ (t0 )),
λ
which contradicts the choice of . This completes the proof.
≥
Proof of Theorem 1.1. Suppose on the contrary that λ∗ < λ∗ . Let λ be fixed with
λ∗ ≤ λ < λ∗ . Then by showing that (1.1)-(1.2) has at least two positive solutions
for λ ∈ [λ∗ , λ∗ ), we get a contradiction to the definition of λ∗ , which completes the
proof. By Lemma 3.5, there exists δλ > 0 such that αλ (t) = u∗ (t) + δλ satisfies
(3.5). Consider the modified problem
(ϕp (u0 (t)))0 + λh(t)g(t, γ1 (t, u(t))) = 0,
u(0) =
m−2
X
ai u(ξi ),
u(1) =
i=1
m−2
X
bi u(ξi ),
(3.6)
i=1
where γ1 : (0, 1) × R → [0, ∞) is defined as


αλ (t), if u > αλ (t),
γ1 (t, u) = u,
if 0 ≤ u ≤ αλ (t),


0,
if u < 0.
Let u be a positive solution of (3.6). Set
Ω = {u ∈ C[0, 1]| − 1 < u(t) < αλ (t),
t ∈ [0, 1]}.
Then Ω is bounded and open in C[0, 1]. We claim that if u is a positive solution of
(3.6), then u ∈ Ω ∩ K. Indeed, by the similar argument as in the proof of Theorem
3.4, 0 ≤ u(t) ≤ αλ (t), t ∈ [0, 1] and
u(0) =
m−2
X
i=1
=
m−2
X
ai u(ξi ) ≤
m−2
X
ai αλ (ξi )
i=1
ai (u∗ (ξi ) + δλ ) <
i=1
m−2
X
ai u∗ (ξi ) + δλ
i=1
= u∗ (0) + δλ = αλ (0).
Similarly, αλ (1) > u(1). If the claim is not true, then there exists [t0 , t1 ] ⊂ (0, 1)
with t0 ≤ t1 such that 0 < u(t) = αλ (t), t ∈ [t0 , t1 ] and 0 < u(t) < αλ (t),
t ∈ (t0 − δ1 , t1 + δ1 ) \ [t0 , t1 ] for some δ1 > 0. Since αλ satisfies (3.5),
max
{(ϕp (αλ0 (t)))0 + λh(t)g(t, αλ (t))} = −1 < 0.
t∈[t0 −δ1 ,t1 +δ1 ]
By condition (A1), there exists δ2 > 0 such that if |u − v| < δ2 and u, v ∈ [0, kαλ k],
then
|g(t, u) − g(t, v)| < 2 ,
where 2 = 1 [2λ maxt∈[t0 −δ1 ,t1 +δ1 ] h(t)]−1 > 0, and then there exists an interval
[a, b] ⊂ (t0 − δ1 , t1 + δ1 ) such that
(u − αλ )0 (a) > 0, (u − αλ )0 (b) < 0
and
−δ2 < γ(t, u(t)) − αλ (t) = u(t) − αλ (t) ≤ 0,
t ∈ [a, b].
10
C.-G. KIM, E. K. LEE
EJDE-2014/38
Consequently,
ϕp (u0 (a)) − ϕp (αλ0 (a)) > 0,
ϕp (u0 (b)) − ϕp (αλ0 (b)) < 0,
g(t, γ(t, u(t))) < g(t, αλ (t)) + 2 ,
t ∈ [a, b].
Then, by the choice of 2 ,
0 > ϕp (u0 (b)) − ϕp (αλ0 (b)) − ϕp (u0 (a)) + ϕp (αλ0 (a)),
= [ϕp (u0 (b)) − ϕp (u0 (a))] − [ϕp (αλ0 (b)) − ϕp (αλ0 (a))]
Z b
=
{(ϕp (u0 (t)))0 − (ϕp (αλ0 (t)))0 } dt
a
Z
b
{−λh(t)g(t, γ(t, u(t))) − (ϕp (αλ0 (t)))0 } dt
=
a
Z
b
{−λh(t)[g(t, αλ (t)) + 2 ] − (ϕp (αλ0 (t)))0 } dt
>
a
Z
b
(−λh(t)2 − [(ϕp (αλ0 (t)))0 + λh(t)g(t, αλ (t))]) dt
>
a
Z
b
≥
(−λ2 h(t) + 1 )dt ≥ 0.
a
This is a contradiction. Thus the claim is proved. Define

R Au
Pm−2 R ξ
λf (τ, γ1 (τ, u(τ )))dτ ds
A−1 i=1 ai 0 i ϕ−1

p

s
 R t −1 R Au

+ ϕ
λf (τ, γ1 (τ, u(τ )))dτ ds,
0 ≤ t ≤ Au ,
s
0 p
M u(t) =
R
R
P
1
s
m−2
−1
B −1
λf (τ, γ1 (τ, u(τ )))dτ ds

i=1 bi ξ ϕp

Au

 R 1 −1 R s i
+ Au ϕp
λf (τ, γ1 (τ, u(τ )))dτ ds,
Au ≤ t ≤ 1,
Au
where Au is defined as
A
−1
m−2
X
Z
ai
i=1
Z Au
+
ξi
=B
+
Au
i
λf (τ, γ1 (τ, u(τ )))dτ ds
s
ϕ−1
p
Au
hZ
i
λf (τ, γ1 (τ, u(τ )))dτ ds
s
m−2
X
Z
bi
1
ϕ−1
p
ξi
i=1
1
hZ s
ϕ−1
p
Au
Au
Z
hZ
0
0
−1
ϕ−1
p
hZ
s
i
λf (τ, γ1 (τ, u(τ )))dτ ds
Au
i
λf (τ, γ1 (τ, u(τ )))dτ ds.
Then M : K → K is completely continuous, and u is a positive solution of (3.6) if
and only if u = M u on K. By simple calculation, there exists R1 > 0 such that
kM uk < R1 for all u ∈ K and Ω ⊂ BR1 . Applying [10, Lemma 2.3.1] with O = BR1 ,
i(M, BR1 ∩ K, K) = 1.
By the above claim and excision property,
i(M, Ω ∩ K, K) = i(M, BR1 ∩ K, K) = 1.
Since problem (1.1)-(1.2) is equivalent to problem (3.6) on Ω ∩ K, we conclude
(1.1)-(1.2) has a positive solution in Ω ∩ K. Assume H(λ, ·) has no fixed point in
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MULTIPLE POSITIVE SOLUTIONS
11
∂Ω ∩ K, since otherwise the proof is done. Then, i(H(λ, ·), Ω ∩ K, K) is well-defined,
and
i(H(λ, ·), Ω ∩ K, K) = i(M, Ω ∩ K, K) = 1
(3.7)
since M u = H(λ, u) for u ∈ Ω ∩ K. By Lemma 3.2, (1.1)-(1.2) with λN0 instead of
λ has no solution in K for λN0 > λ. Thus, for any open subset O in X,
i(H(λN0 , ·), O ∩ K, K) = 0.
By a priori estimate (Lemma 3.3) with I = [λ, λN0 ], there exists R2 (> R1 ) such that
all possible positive solutions u of (1.1)-(1.2) with µ instead of λ for µ ∈ [λ, λN0 ],
satisfy kuk < R2 .
Define h : [0, 1] × (B R2 ∩ K) → K as
h(τ, u) = H(τ λN0 + (1 − τ )λ, u).
Then h is completely continuous on [0, 1] × K, and it satisfies that h(τ, u) 6= u for
all (τ, u) ∈ [0, 1] × (∂BR2 ∩ K). By the property of homotopy invariance,
i(H(λ, ·), BR2 ∩ K, K) = i(H(λN0 , ·), BR2 ∩ K, K) = 0.
By (3.7) and the additivity property,
i(H(λ, ·), (BR2 \ Ω) ∩ K, K) = −1.
Thus problem (1.1)-(1.2) has another positive solution in (BR2 \Ω) ∩ K. This completes the proof.
Finally, we give the examples for the nonlinear term to illustrate our results.
Example 3.6. (1) Put f1 (t, u) = [t(1 − t)]−p+1/(u+1) exp(u). Then, it is easily
verified that f1 satisfies the assumptions of Theorem 3.4.
(2) Put f2 (t, u) = (1 − t)−α1 g(t, u), where g(t, u) = c1 t−β1 + c2 (uq + 1). Then f2
satisfies the assumptions of Theorem 1.1 if α1 , β1 < p, c1 ≥ 0, c2 > 0, and q > p−1.
Acknowledgments. Chan-Gyun Kim was supported by National Research Foundation of Korea Grant funded by the Korean Government (Ministry of Education,
Science and Technology, NRF-2011-357-C00006) Eun Kyoung Lee was supported
by a 2-Year Research Grant of Pusan National University.
This work was done when the first author visited College of William and Mary.
He would like to thank Department of Mathematics, College of William and Mary
for warm hospitality, and thank Professor Junping Shi for constant encouragement
and helpful discussions.
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EJDE-2014/38
MULTIPLE POSITIVE SOLUTIONS
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Chan-Gyun Kim
Department of Mathematics, College of William and Mary, Williamsburg, Virginia
23187-8795, USA
E-mail address: cgkim75@gmail.com
Eun Kyoung Lee (corresponding author)
Department of Mathematics and Education, Pusan National University, Busan 609-735,
Korea
E-mail address: eunkyoung165@gmail.com