Electronic Journal of Differential Equations, Vol. 2006(2006), No. 68, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF POSITIVE SOLUTIONS FOR
BOUNDARY-VALUE PROBLEMS FOR SINGULAR
HIGHER-ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS
CHUANZHI BAI, QING YANG, JING GE
Abstract. We study the existence of positive solutions for the boundaryvalue problem of the singular higher-order functional differential equation
(Ly (n−2) )(t) + h(t)f (t, yt ) = 0,
y (i) (0) = 0,
for t ∈ [0, 1],
0 ≤ i ≤ n − 3,
αy (n−2) (t) − βy (n−1) (t) = η(t),
γy
(n−2)
(t) + δy
(n−1)
(t) = ξ(t),
for t ∈ [−τ, 0],
for t ∈ [1, 1 + a],
where Ly := −(py 0 )0 + qy, p ∈ C([0, 1], (0, +∞)), and q ∈ C([0, 1], [0, +∞)).
Our main tool is the fixed point theorem on a cone.
1. Introduction
As pointed out in [5], boundary-value problems associated with functional differential equations arise from problems in physics, from variational problems in control
theory, and from applied mathematics; see for example [6, 8]. Many authors have
investigated the existence of solutions for boundary-value problems of functional
differential equations; see [3, 9, 15, 18]. Recently an increasing interest in studying
the existence of positive solutions for such problems has been observed. Among
others publication, we refer to [1, 2, 10, 11, 13, 19].
In this paper, we investigate the existence of positive solutions for singular
boundary-value problems (BVP) of an n-th order (n ≥ 3) functional differential
equation (FDE) of the form
(Ly (n−2) )(t) + h(t)f (t, yt ) = 0,
(i)
y (0) = 0,
0 ≤ i ≤ n − 3,
αy (n−2) (t) − βy (n−1) (t) = η(t),
γy
(n−2)
(t) + δy
(n−1)
for t ∈ [0, 1],
(t) = ξ(t),
(1.2)
for t ∈ [−τ, 0],
(1.3)
for t ∈ [1, 1 + a],
(1.4)
2000 Mathematics Subject Classification. 34K10, 34B16.
Key words and phrases. Boundary value problem; higher-order; positive solution;
functional differential equation; fixed point.
c
2006
Texas State University - San Marcos.
Submitted April 21, 2006. Published July 6, 2006.
Supported by the Natural Science Foundation of Jiangsu Education Office and by
Jiangsu Planned Projects for Postdoctoral Research Funds.
1
(1.1)
2
C. BAI, Q. YANG, J. GE
EJDE-2006/68
where Ly := −(py 0 )0 + qy, p ∈ C([0, 1], (0, +∞)), and q ∈ C([0, 1], [0, +∞));
α, β, γ, δ ≥ 0, and αδ + αγ + βγ > 0; η ∈ C([−τ, 0], R), ξ ∈ C([1, b], R) (b = 1 + a),
and η(0) = ξ(1) = 0; h ∈ C((0, 1), R) (h(t) is allowed to have singularity at t = 0
or 1); f ∈ C([0, 1] × D, R), D = C([−τ, a], R), for every t ∈ [0, 1], yt ∈ D is defined
by yt (θ) = y(t + θ), θ ∈ [−τ, a].
The study of higher-order functional differential equation has received also some
attention; see for example [3, 10, 17]. Recently, Hong et al. [12] imposed conditions
on f (t, y t ) to yield at least one positive solution to (1.1)-(1.4) for the special case
h(t) ≡ 1, p(t) ≡ 1, and q(t) ≡ 0. They applied the Krasnosel’skii fixed-point
theorem.
The purpose of this paper is to establish the existence of positive solutions of the
singular higher-order functional differential equation (1.1) with boundary conditions
(1.2)-(1.4) under suitable conditions on f .
2. Preliminaries
To abbreviate our discussion, we assume the following hypotheses:
(H1) G(t, s) is the Green’s function of the differential equation
(Ly (n−2) )(t) = 0,
0<t<1
subject to the boundary condition (1.2)-(1.4) with τ = a = 0.
(H2) g(t, s) is the Green’s function of the differential equation
Ly(t) = 0
t ∈ (0, 1)
subject to the boundary conditions
αy(0) − βy 0 (0) = 0,
γy(1) + δy 0 (1) = 0,
where α, β, γ and δ are as in (1.3) and (1.4).
(H3) h ∈ C((0, 1), [0, +∞)) and satisfies
Z 1
0<
g(s, s)h(s)ds < +∞.
0
+
(H4) f ∈ C([0, 1] × D , [0, ∞)), where D+ = C([−τ, a], [0, +∞)).
(H5) η ∈ C([−τ, 0], [0, +∞)), ξ ∈ C([1, 1 + a], [0, +∞)), and η(0) = ξ(1) = 0.
It is easy to see that
∂ n−2
G(t, s) = g(t, s), t, s ∈ [0, 1].
∂tn−2
It is also well known that the Green’s function g(t, s) is
(
1 φ(s)ψ(t), if 0 ≤ s ≤ t ≤ 1,
g(t, s) =
c φ(t)ψ(s), if 0 ≤ t ≤ s ≤ 1,
where φ and ψ are solutions, respectively, of
Lφ = 0,
Lψ = 0,
φ(0) = β,
ψ(1) = δ,
φ0 (0) = α,
0
ψ (1) = −γ.
(2.1)
(2.2)
One can show that c = −p(t)(φ(t)ψ 0 (t) − φ0 (t)ψ(t)) > 0 and φ0 (t) > 0 on (0, 1] and
ψ 0 (t) < 0 on [0, 1). Clearly
g(t, s) ≤ g(s, s),
0 ≤ t, s ≤ 1.
(2.3)
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EXISTENCE OF POSITIVE SOLUTIONS
3
By (H3), there exists t0 ∈ (0, 1) such that h(t0 ) > 0. We may choose ε ∈ (0, 1/2)
such that t0 ∈ (ε, 1 − ε). Then for ε ≤ t ≤ 1 − ε we have φ(ε) ≤ φ(t) ≤ φ(1 − ε)
and ψ(1 − ε) ≤ ψ(t) ≤ ψ(ε). Also for (t, s) ∈ [ε, 1 − ε] × (0, 1)
ψ(1 − ε) φ(ε) ψ(1 − ε) φ(ε) g(t, s)
≥ min
,
≥ min
,
:= σ.
(2.4)
g(s, s)
ψ(s)
φ(s)
ψ(0)
φ(1)
Let E = C (n−2) ([−τ, b]; R) with a norm kuk[−τ,b] = sup−τ ≤t≤b |u(n−2) (t)| for
u ∈ E. Obviously, E is a Banach space. And let C = C (n−2) ([−τ, a], R) be a space
with norm kψk[−τ,a] = sup−τ ≤t≤a |ψ (n−2) (x)| for ψ ∈ C. Let
C + = {ψ ∈ C : ψ(x) ≥ 0, x ∈ [−τ, a]}.
It is easy to see that C + is a subspace of C.
Define a cone K ⊂ E as follows:
K = {y ∈ E : y(t) ≥ 0,
min y (n−2) (t) ≥ σkyk[−τ,b] },
(2.5)
t∈[ε,1−ε]
where σ = 1b min{ε, σ}, σ is as in (2.4).
For each ρ > 0, we define Kρ = {y ∈ K : kyk[−τ,b] < ρ}. Furthermore, we define
a set Ωρ as follows:
Ωρ = y ∈ K : min y (n−2) (t) < σρ .
ε≤t≤1−ε
Similar to the [14, Lemma 2.5], we have
Lemma 2.1. Ωρ defined above has the following properties:
(a) Ωρ is open relative to K.
(b) Kσρ ⊂ Ωρ ⊂ Kρ .
(c) y ∈ ∂Ωρ if and only if minε≤t≤1−ε y (n−2) (t) = σρ.
(d) If y ∈ ∂Ωρ , then σρ ≤ y (n−2) (t) ≤ ρ for t ∈ [ε, 1 − ε].
To obtain the positive solutions of (1.1)-(1.4), the following fixed point theorem
in cones will be fundamental.
Lemma 2.2. Let K be a cone in a Banach space E. Let D be an open bounded
subset of E with DK = D ∩ K 6= ∅ and DK 6= K. Assume that A : DK → K is a
compact map such that x 6= Ax for x ∈ ∂DK . Then the following results hold.
(1) kAxk ≤ kxk, x ∈ ∂DK , then iK (A, DK ) = 1.
(2) If there exists e ∈ K\{0} such that x 6= Ax + λe for all x ∈ ∂DK and all
λ > 0, then iK (A, DK ) = 0.
(3) Let U be an open set in E such that U ⊂ DK . If iK (A, DK ) = 1 and
iK (A, UK ) = 0, then A has a fixed point in DK \U K . The same results
holds if iK (A, DK ) = 0 and iK (A, UK ) = 1.
Suppose that y(t) is a solution of (1.1)-(1.4), then it can be written as


; t),
−τ ≤ t ≤ 0,
y(−τ
R1
y(t) =
G(t,
s)h(s)f
(s,
y
)ds,
0 ≤ t ≤ 1,
s
0


y(b; t),
1 ≤ t ≤ b,
where y(−τ ; t) and y(b; t) satisfy
( α R0 α
−βs
1
(n−2)
βt
e
e
η(s)ds
+
y
(0)
, t ∈ [−τ, 0], β 6= 0,
β t
y (n−2) (−τ ; t) =
1
t ∈ [−τ, 0], β = 0,
α η(t),
4
C. BAI, Q. YANG, J. GE
EJDE-2006/68
and
y
(n−2)
(b; t) =
( γ Rt γ
γ
e− δ t 1δ 1 e δ s ξ(s)ds + e δ y (n−2) (1) ,
t ∈ [1, b], δ 6= 0,
1
γ ξ(t),
t ∈ [1, b], δ = 0.
Throughout this paper, we assume that u0 (t) is the solution of (1.1)-(1.4) with
(n−2)
f ≡ 0, and ku0 k[−τ,b] =: M0 . Clearly, u0
(t) can be expressed as follows:
 (n−2)

(−τ ; t), −τ ≤ t ≤ 0,
u0
(n−2)
u0
(t) = 0,
0 ≤ t ≤ 1,

 (n−2)
u0
(b; t),
1 ≤ t ≤ b.
where
(
(n−2)
u0
(−τ ; t)
=
R 0 −αs
1 α
βt
e β η(s)ds,
βe
t
1
α η(t),
t ∈ [−τ, 0], β 6= 0,
t ∈ [−τ, 0], β = 0,
R
1 − γδ t t
δe
1
1
γ ξ(t),
t ∈ [1, b], δ 6= 0,
t ∈ [1, b], δ = 0.
and
(
(n−2)
u0
(b; t)
=
γ
e δ s ξ(s)ds,
Let y(t) be a solution of BVP (1.1)-(1.4) and u(t) = y(t) − u0 (t). Noting that
u(t) ≡ y(t) for 0 ≤ t ≤ 1, we have

(n−2)

(−τ ; t),
−τ ≤ t ≤ 0,
u
R1
(n−2)
u
(t) =
g(t,
s)h(s)f
(s,
u
+
(u
)
)ds,
0 ≤ t ≤ 1,
s
0 s
0

 (n−2)
u
(b; t),
1 ≤ t ≤ b,
where
u
(n−2)
( α R1
e β t 0 g(0, s)h(s)f (s, us + (u0 )s )ds,
(−τ ; t) =
0,
t ∈ [−τ, 0], β 6= 0,
t ∈ [−τ, 0], β = 0,
and
( γ
R1
e− δ (t−1) 0 g(1, s)h(s)f (s, us + (u0 )s )ds,
(n−2)
u
(b; t) =
0,
t ∈ [1, b], δ 6= 0,
t ∈ [1, b], δ = 0.
It is easy to see that y(t) is a solution of BVP (1.1)-(1.4) if and only if u(t) =
y(t) − u0 (t) is a solution of the operator equation
u(t) = Au(t)
for t ∈ [−τ, b].
(2.6)
Here, operator A : E → E is defined by


u(t),
−τ ≤ t ≤ 0,
B
R 11
Au(t) :=
G(t, s)h(s)f (s, us + (u0 )s )ds, 0 ≤ t ≤ 1,
0


B2 u(t),
1 ≤ t ≤ b,
where
B1 u(t) :=
R1
β n−2 α
e β t 0 g(0, s)h(s)f (s, us + (u0 )s )ds,
α
R1
tn−2
g(0, s)h(s)f (s, us + (u0 )s )ds,
(n−2)!
0




β=
6 0, α 6= 0,
β=
6 0, α = 0,

0,
β=0
EJDE-2006/68
EXISTENCE OF POSITIVE SOLUTIONS
5
for each t ∈ [−τ, 0], and

R
δ n−2 − γ (t−1) 1
g(1, s)h(s)f (s, us + (u0 )s )ds, δ =
6 0, γ 6= 0,

 −γ R e δ
0
1
tn−2
B2 u(t) := (n−2)!
g(1,
s)h(s)f
(s,
u
+
(u
)
)ds,
δ
6 0, γ = 0,
=
s
0 s
0


0,
δ=0
for any t ∈ [1, b]. Obviously,

(n−2)

(t),
−τ ≤ t ≤ 0,
1 u)
(B
R
1
(n−2)
(Au)
(t) :=
g(t, s)h(s)f (s, us + (u0 )s )ds, 0 ≤ t ≤ 1,
0


(B2 u)(n−2) (t),
1 ≤ t ≤ b,
where
(n−2)
(B1 u)
( α R1
e β t 0 g(0, s)h(s)f (s, us + (u0 )s )ds,
(t) :=
0,
t ∈ [−τ, 0], β 6= 0,
t ∈ [−τ, 0], β = 0,
and
(n−2)
(B2 u)
( γ
R1
e− δ (t−1) 0 g(1, s)h(s)f (s, us + (u0 )s )ds,
(t) :=
0,
t ∈ [1, b], δ 6= 0,
t ∈ [1, b], δ = 0.
Lemma 2.3. With the above notation, A(K) ⊂ K.
Proof. By the assumptions of (H1)-(H5), it is easy to know that Au ∈ E and Au ≥ 0
for any u ∈ K. Moreover, it follows from
0 ≤ (Au)(n−2) (t) ≤ (Au)(n−2) (0)
(n−2)
0 ≤ (Au)
(n−2)
(t) ≤ (Au)
(1)
for − τ ≤ t ≤ 0
for 1 ≤ t ≤ b
that kAuk[−τ,b] = kAuk[0,1] . By (2.3) we have, for any u ∈ K and t ∈ [0, 1] that
Z 1
kAuk[−τ,b] = kAuk[0,1] ≤
g(s, s)h(s)f (s, us + (u0 )s )ds.
(2.7)
0
From (2.4), we get
min (Au)(n−2) (t) =
ε≤t≤1−ε
Z
min
ε≤t≤1−ε
Z
1
g(t, s)h(s)f (s, us + (u0 )s )ds
0
1
≥σ
g(s, s)h(s)f (s, us + (u0 )s )ds
(2.8)
0
Z
≥σ
1
g(s, s)h(s)f (s, us + (u0 )s )ds.
0
In view of (2.7) and (2.8), we obtain
min (Au)(n−2) (t) ≥ σkAuk[−τ,b] ,
ε≤t≤1−ε
u ∈ K,
which implies A(K) ⊂ K.
Let
+
C[k,r]
= {ϕ ∈ C + : k ≤ kϕk[−τ,a] ≤ r},
+
C[k,∞)
= {ϕ ∈ C + : k ≤ kϕk[−τ,a] < ∞},
where 0 ≤ k < r.
6
C. BAI, Q. YANG, J. GE
EJDE-2006/68
Lemma 2.4. A : K → K is completely continuous.
Proof. We apply a truncation technique (cf. [16]). We define the function hm for
m ≥ 2, by

1
1

0<t≤ m
,
min h(t), h( m ) ,
1
1
hm (t) = h(t),
<
t
<
1
−m
,
m


m−1
m−1
min h(t), h( m ) ,
m ≤ t < 1.
It is clear that hm (t) is nonnegative and continuous on [0, 1]. We define the operator
Am by


u(t),
−τ ≤ t ≤ 0,
B
R 11m
Am u(t) :=
G(t, s)hm (s)f (s, us + (u0 )s )ds, 0 ≤ t ≤ 1,
0


B2m u(t),
1 ≤ t ≤ b,
where
 n−2
α R1
β


e β t 0 g(0, s)hm (s)f (s, us + (u0 )s )ds,
 α
R1
tn−2
B1m u(t) :=
 (n−2)! 0 g(0, s)hm (s)f (s, us + (u0 )s )ds,


0,
β 6= 0, α 6= 0,
β 6= 0, α = 0,
β=0
for each t ∈ [−τ, 0], and

R
δ n−2 − γ (t−1) 1
g(1, s)hm (s)f (s, us + (u0 )s )ds,

 −γ R e δ
0
1
tn−2
B2m u(t) := (n−2)!
g(1,
s)h
(s)f
(s, us + (u0 )s )ds,
m
0


0,
δ 6= 0, γ 6= 0,
δ 6= 0, γ = 0,
δ=0
for any t ∈ [1, b]. By Lemma 2.3, it is easy to check that Am : K → K. And, Am
is continuous, the proof is similar to that of [11, Theorem 2.1].
Next let B ⊂ K be a bounded subset of K, and M1 > 0 be a constant such
that kuk[−τ,b] ≤ M1 for u ∈ B. Noting that if xt ∈ C = C n−2 ([−τ, a], R), then
(n−2)
(n−2)
∈ C([−τ, a], R), and xt
(θ) = x(n−2) (t + θ), θ ∈ [−θ, a]. Thus
xt
kus + (u0 )s k[−τ,a] =
sup |(us + us0 )(n−2) (θ)|
−τ ≤θ≤a
≤
≤
(n−2)
sup |(u(n−2) (s + θ)| +
−τ ≤θ≤a
sup |u0
(n−2)
sup |u(n−2) (t)| + sup |u0
−τ ≤t≤b
(s + θ)|
−τ ≤θ≤a
(t)| = kuk[−τ,b] + ku0 k[−τ,b]
−τ ≤t≤b
≤ M1 + M0 := M2
(2.9)
for u ∈ B and s ∈ [0, 1]. Hence, there exists a constant M3 > 0 such that
|f (s, us + (u0 )s )| ≤ M3 ,
+
on [0, 1] × C[0,M
,
3]
since f is continuous on [0, 1] × C + . For u ∈ B we have


u)(n−2) (t),
−τ ≤ t ≤ 0,
(B
R 1 1m
(n−2)
(Am u)
(t) :=
g(t,
s)h
(s)f
(s,
u
+
(u
)
)ds,
0 ≤ t ≤ 1,
m
s
0 s
0


(n−2)
(B2m u)
(t),
1 ≤ t ≤ b,
(2.10)
EJDE-2006/68
EXISTENCE OF POSITIVE SOLUTIONS
7
where
( α R1
e β t 0 g(0, s)hm (s)f (s, us + (u0 )s )ds, t ∈ [−τ, 0], β 6= 0,
(B1m u)(n−2) (t) :=
0,
t ∈ [−τ, 0], β = 0,
and
(n−2)
(B2m u)
( γ
R1
e− δ (t−1) 0 g(1, s)hm (s)f (s, us + (u0 )s )ds, t ∈ [1, b], δ 6= 0,
(t) :=
0,
t ∈ [1, b], δ = 0.
These and (2.10) imply (Am u)(n−2) (t) is continuous and uniformly bounded for
u ∈ B. So (Am u)0 (t) is continuous and uniformly bounded for u ∈ B also. The
Ascoli-Arzela Theorem implies that Am is a completely continuous operator on K
for any m ≥ 2.
Moreover, Am converges uniformly to A as m → ∞ on any bounded subset
of K. To see this, note that if u ∈ K with kuk[−τ,b] ≤ M , then from (H3) and
0 ≤ hn (s) ≤ h(s),
(n−2)
|(Am u)
(n−2)
(t) − (Au)
Z m1
g(t, s)[h(s) − hm (s)]f (s, us + (u0 )s )ds
(t)| = 0
Z 1
g(t, s)[h(s) − hm (s)]f (s, us + (u0 )s )ds
+
m−1
m
1
m
Z
≤
g(s, s)|h(s) − hm (s)|f (s, us + (u0 )s )ds
0
Z
1
+
m−1
m
≤ 2M
hZ
g(s, s)|h(s) − hm (s)|f (s, us + (u0 )s )ds
1
m
Z
0
→0
where M := maxt∈[0,1],ϕ∈C +
1
g(s, s)h(s)ds +
g(s, s)h(s)ds
i
m−1
m
as m → ∞,
f (t, ϕ). Thus, we have
[0,M +M0 ]
kAm u − Auk[−τ,b] = kAm u − Auk[0,1] → 0,
n → ∞,
for each u ∈ K with kuk[−τ,b] ≤ M . Hence, Am converges uniformly to A as
m → ∞ and therefore A is completely continuous also. This completes the proof
of Lemma 2.4.
8
C. BAI, Q. YANG, J. GE
EJDE-2006/68
3. Main results
For convenience, we introduce the following notation. Let
Z 1−ε
Z 1
−1
−1
ω=
g(s, s)h(s)ds
; N=
min
g(t, s)h(s)ds
;
ε≤t≤1−ε
0
ε
f (t, ϕ)
ρ
+
fσρ
= inf
min
: ϕ ∈ C[σρ,ρ+M
;
]
0
ρ
t∈[ε,1−ε]
f (t, ϕ)
+
f0ρ = sup max
: ϕ ∈ C[0,ρ+M
;
]
0
ρ
t∈[0,1]
f (t, ϕ)
;
fµ =
lim
sup max
kϕk[−τ,a] →µ
t∈[0,1] kϕk[−τ,a]
fµ =
lim
kϕk[−τ,a] →µ
inf
f (t, ϕ)
,
t∈[ε,1−ε] kϕk[−τ,a]
min
(µ := ∞ or 0+ ).
Now, we impose conditions on f which we assure that iK (A, Kρ ) = 1.
Lemma 3.1. Assume that
f0ρ ≤ ω
u 6= Au
and
for u ∈ ∂Kρ .
(3.1)
Then iK (A, Kρ ) = 1.
Proof. For u ∈ ∂Kρ , we have kus + (u0 )s k[−τ,a] ≤ ρ + M0 , for all s ∈ [0, 1], i.e.,
us + (u0 )s ∈ C[0,ρ+M0 ] for any s ∈ [0, 1]. It follows from (3.1) that for t ∈ [0, 1],
Z 1
(n−2)
(Au)
(t) =
g(t, s)h(s)f (s, us + (u0 )s )ds
0
Z
1
≤
g(s, s)h(s)f (s, us + (u0 )s )ds
0
Z
1
< ρω
g(s, s)h(s)ds
0
= ρ = kuk[−τ,b] .
This implies that kAuk[−τ,b] < kuk[−τ,b] for u ∈ ∂Kρ . By Lemma 2.2 (1), we have
iK (A, Kρ ) = 1.
Let u ∈ ∂Ωρ , then for any s ∈ [ε, 1 − ε], we have by Lemma 2.1(c) that
kus + (u0 )s k[−τ,a] =
sup (u(n−2) (s + θ) + (u0 )(n−2) (s + θ))
θ∈[−τ,a]
≥
sup u(n−2) (s + θ) (since (u0 )(n−2) (t) ≥ 0 for t ∈ [−τ, b])
θ∈[−τ,a]
≥ u(n−2) (s)
≥
min u(n−2) (t) = σρ.
t∈[ε,1−ε]
By Lemma 2.1(b), we have Ωρ ⊂ K ρ , that is kuk[−τ,b] ≤ ρ. Thus, from (2.9) we get
kus + (u0 )s k[−τ,a] ≤ kuk[−τ,b] + ku0 k[−τ,b] ≤ ρ + M0 ,
∀s ∈ [0, 1].
Hence
+
us + (u0 )s ∈ C[σρ,
ρ+M0 ] ,
for u ∈ ∂Ωρ , s ∈ [ε, 1 − ε].
Next, we impose conditions on f which assure that iK (A, Ωρ ) = 0.
(3.2)
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EXISTENCE OF POSITIVE SOLUTIONS
9
Lemma 3.2. If f satisfies the condition
ρ
fσρ
≥ Nσ
u 6= Au
and
for u ∈ ∂Ωρ .
(3.3)
Then iK (A, Ωρ ) = 0.
Proof. Let

tn−1

)(n−1)! , n is odd, −τ ≤ t ≤ 0,
− (1+τ
tn
n is even , −τ ≤ t ≤ 0,
e(t) = (1+τ )2 n! ,

 tn−1
,
0
≤ t ≤ b.
b(n−1)!
It is easy to verify that e ∈ C (n−2) ([−τ, b], R), e(t) ≥ 0 for t ∈ [−τ, b], and

t
,
n is odd, −τ ≤ t ≤ 0,

− 1+τ
(n−2)
t2
e
(t) = 2(1+τ )2 , n is even, −τ ≤ t ≤ 0,

t
0 ≤ t ≤ b,
b,
which implies that e ∈ K and kek[−τ,b] = 1, that is e ∈ ∂K1 . We claim that
u 6= Au + λe,
u ∈ ∂Ωρ ,
λ > 0.
In fact, if not, there exist u ∈ ∂Ωρ and λ0 > 0 such that u = Au + λ0 e. Then by
(3.2) for t ∈ [ε, 1 − ε], we get
u(n−2) (t) = (Au)(n−2) (t) + λ0 e(n−2) (t)
Z 1
=
g(t, s)h(s)f (s, us + (u0 )s )ds + λ0 e(n−2) (t)
0
1−ε
Z
≥
g(t, s)h(s)f (s, us + (u0 )s )ds + λ0
ε
1−ε
Z
≥
g(t, s)h(s)f (s, us + (u0 )s )ds + λ0 σ
min
ε≤t≤1−ε
ε
Z
≥ ρN σ
ε
b
1−ε
min
ε≤t≤1−ε
g(t, s)h(s)ds + λ0 σ
ε
≥ σρ + λ0 σ.
This implies that σρ ≥ σρ + λ0 σ, a contradiction. Moreover, it is easy to check
that u 6= Au for u ∈ ∂Ωρ from (3.3). Hence, by Lemma 2.2 (2), it follows that
iK (A, Ωρ ) = 0.
Theorem 3.3. If one of the following conditions holds:
(H7) There exist ρ1 , ρ2 , ρ3 ∈ (0, ∞) with ρ1 < σρ2 and ρ2 < ρ3 such that
f0ρ1 ≤ ω,
ρ2
fσρ
≥ N σ,
2
u 6= Au
for u ∈ ∂Ωρ2
and
f0ρ3 ≤ ω.
(H8) There exist ρ1 , ρ2 , ρ3 ∈ (0, ∞) with ρ1 < ρ2 < σρ3 such that
ρ1
fσρ
≥ N σ,
1
f0ρ2 ≤ ω,
u 6= Au
for u ∈ ∂Kρ2
and
ρ3
fσρ
≥ N σ.
3
Then BVP (1.1)-(1.4) has two positive solutions. Moreover, if in (H7), f0ρ1 ≤ ω is
replaced by f0ρ1 < ω, then (1.1)-(1.4) has a third positive solution u3 ∈ Kρ1 .
10
C. BAI, Q. YANG, J. GE
EJDE-2006/68
Proof. Suppose that (H7) holds. We show that either A has a fixed point u1 in
∂Kρ1 or in Ωρ2 \ Kρ1 . If u 6= Au for u ∈ ∂Kρ1 ∪ ∂Kρ3 , by Lemmas 3.1-3.2, we
have iK (A, Kρ1 ) = 1, iK (A, Ωρ2 ) = 0 and iK (A, Kρ3 ) = 1. Since ρ1 < σρ2 , we have
Kρ1 ⊂ Kσρ2 ⊂ Ωρ2 by Lemma 2.1 (b). It follows from Lemma 2.2 that A has a
fixed point u1 ∈ Ωρ2 \ Kρ1 . Similarly, A has a fixed point u2 ∈ Kρ3 \ Ωρ2 . The
proof is similar when (H8) holds.
As a special case of Theorem 3.3 we obtain the following result.
Corollary 3.4. Let ξ(t) ≡ 0, η(t) ≡ 0. If there exists ρ > 0 such that one of the
following conditions holds:
ρ
≥ N σ, u 6= Au for u ∈ ∂Ωρ and 0 ≤ f ∞ < ω.
(H9) 0 ≤ f 0 < ω, fσρ
ρ
(H10) N < f0 ≤ ∞, f0 ≤ ω, u 6= Au for u ∈ ∂Kρ and N < f∞ ≤ ∞.
Then BVP (1.1)-(1.4) has two positive solutions.
Proof. From ξ(t) ≡ 0, η(t) ≡ 0, it is clear that u0 (t) ≡ 0 for t ∈ [−τ, b], thus M0 = 0.
We now show that (H9 ) implies (H7 ). It is easy to verify that 0 ≤ f 0 < ω implies
that there exists ρ1 ∈ (0, σρ) such that f0ρ1 < ω. Let k ∈ (f ∞ , ω). Then there exists
+
r > ρ such that maxt∈[0,1] f (t, ϕ) ≤ kkϕk[−τ,a] for ϕ ∈ C[r,∞)
since 0 ≤ f ∞ < ω.
Let
l
+
l = max{ max f (t, ϕ) : ϕ ∈ C[0,r]
}, and ρ3 > max
, ρ .
ω−k
t∈[0,1]
Then we have
max f (t, ϕ) ≤ kkϕk[−τ,a] + l ≤ kρ3 + l < ωρ3
t∈[0,1]
+
for ϕ ∈ C[0,ρ
.
3]
This implies that f0ρ3 < ω and (H7) holds. Similarly, (H10) implies (H8).
By a similar argument to that of Theorem 3.3, we obtain the following results
on existence of at least one positive solution of (1.1)-(1.4).
Theorem 3.5. If one of the following conditions holds:
(H11) There exist ρ1 , ρ2 ∈ (0, ∞) with ρ1 < σρ2 such that
f0ρ1 ≤ ω
and
ρ2
fσρ
≥ N σ.
2
(H12) There exist ρ1 , ρ2 ∈ (0, ∞) with ρ1 < ρ2 such that
ρ1
fσρ
≥ Nσ
1
and
f0ρ2 ≤ ω.
Then BVP (1.1)-(1.4) has a positive solution.
As a special case of Theorem 3.5, we obtain the following result.
Corollary 3.6. Let ξ(t) ≡ 0, η(t) ≡ 0. If one of the following conditions holds:
(H13) 0 ≤ f 0 < ω and N < f∞ ≤ ∞.
(H14) 0 ≤ f ∞ < ω and N < f0 ≤ ∞.
Then BVP (1.1)-(1.4) has a positive solution.
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EXISTENCE OF POSITIVE SOLUTIONS
11
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Chuanzhi Bai
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223001, China;
and Department of Mathematics, Nanjing University, Nanjing 210093, China
E-mail address: czbai8@sohu.com
Qing Yang
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223001, China
E-mail address: yangqing3511115@163.com
Jing Ge
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223001, China
E-mail address: gejing0512@163.com