Electronic Journal of Differential Equations, Vol. 2004(2004), No. 133, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
POSITIVE SOLUTIONS FOR SINGULAR SEMI-POSITONE
NEUMANN BOUNDARY-VALUE PROBLEMS
YONG-PING SUN, YAN SUN
Abstract. In this paper, we study the singular semi-positone Neumann boundary-value problem
−u00 + m2 u = λf (t, u) + g(t, u),
0
0 < t < 1,
0
u (0) = u (1) = 0,
where m is a positive constant. Under some suitable assumptions on the
functions f and g, for sufficiently small λ, we prove the existence of a positive
solution. Our approach is based on the Krasnasel’skii fixed point theorem in
cones.
1. Introduction
In this paper, we shall study the following singular semi-positone Neumann
boundary-value problem (NBVP)
−u00 + m2 u = λf (t, u) + g(t, u),
0
0 < t < 1,
0
u (0) = u (1) = 0,
(1.1)
where m > 0 is a constant, λ > 0 is a parameter, f : (0, 1) × [0, +∞) → [0, +∞)
and g : [0, 1] × [0, +∞) → (−∞, +∞) are continuous.
We say problem (1.1) is singular because f may be singular at t = 0 and/or t = 1.
When g(t, u) 6≡ 0, problem (1.1) is a semi-positone problem, this situation arises
naturally in chemical reaction theory [8]. In recent years, attention has been paid
to (1.1) in the case of g(t, u) ≡ 0; see, for example, [11, 12, 13] and the references
therein. Attention has been paid also to the semi-positone boundary-value problem;
see, for example, [6, 7, 9] and the references therein. As far as the authors know,
there are no existence results for the singular semi-positone NBVP. Recently, Xu [9]
studied the existence of positive solutions for the singular semi-positone boundaryvalue problem
u00 + f (t, u) + q(t) = 0,
0 < t < 1,
u(0) = u(1) = 0,
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Positive solution; semi-positone; fixed points; cone;
singular Neumann boundary-value problem.
c
2004
Texas State University - San Marcos.
Submitted October 12, 2004. Published November 16, 2004.
Supported by NSFC (19871075), NSFSP (Z2003A01) and EDZP (20040495).
1
2
Y. P. SUN & Y. SUN
EJDE-2004/133
where f : (0, 1) × [0, +∞) → [0, +∞) and q : (0, 1) → (−∞, +∞) are continuous.
Motivated by the papers mentioned above, we study the existence of positive
solutions for the singular semi-positone NBVP (1.1), and give an explicit interval
for λ. Our results can be regarded as an extension and improvement of the corresponding results of [12, 13]. The paper is organized as follows. In Section 2, we
present some lemmas that will be used to prove the main result and Krasnasel’skii
fixed point theorem in cones. In Section 3, we prove the main result of this paper.
2. Preliminaries
We consider problem in the Banach space E = C[0, 1] equipped with the norm
kuk = supt∈[0,1] |u(t)|. Let G(t, s) be the Green’s function for the Boundary-value
problem
−u00 + m2 u = 0, 0 < t < 1,
(2.1)
u0 (0) = u0 (1) = 0 .
Then
1
G(t, s) =
ρ
(
ϕ(s)ϕ(1 − t), 0 ≤ s ≤ t ≤ 1,
ϕ(t)ϕ(1 − s), 0 ≤ t ≤ s ≤ 1,
where ρ = 12 m(em −e−m ), ϕ(t) = 12 (emt +e−mt ). It is obvious that ϕ(t) is increasing
on [0, 1], and
G(t, s) ≤ G(s, s), 0 ≤ t, s ≤ 1.
Lemma 2.1. Let G(t, s) be the Green’s function for the NBVP (2.1).
(1) Assume that 0 < θ < 21 , then
G(t, s) ≥ Mθ G(s, s),
where
Mθ =
θ ≤ t ≤ 1 − θ, 0 ≤ s ≤ 1.
emθ + e−mθ
.
em + e−m
(2)
G(t, s) ≥ Cϕ(t)ϕ(1 − t)G(t0 , s),
t, t0 , s ∈ [0, 1],
where C = 1/ϕ2 (1).
Proof. (1) Let t ∈ [θ, 1 − θ]. For s ≤ t,
G(t, s)
ϕ(1 − t)
ϕ(θ)
emθ + e−mθ
=
≥
= m
= Mθ .
G(s, s)
ϕ(1 − s)
ϕ(1)
e + e−m
If t ≤ s, then
G(t, s)
ϕ(t)
ϕ(θ)
emθ + e−mθ
=
≥
= m
= Mθ .
G(s, s)
ϕ(s)
ϕ(1)
e + e−m
Thus
G(t, s) ≥ Mθ G(s, s),
θ ≤ t ≤ 1 − θ, 0 ≤ s ≤ 1.
(2) When t, t0 ≤ s,
G(t, s)
ϕ(t)ϕ(1 − s)
ϕ(t)ϕ(1 − t)
=
=
G(t0 , s)
ϕ(t0 )ϕ(1 − s)
ϕ(t0 )ϕ(1 − t)
1
≥ 2 ϕ(t)ϕ(1 − t) = Cϕ(t)ϕ(1 − t).
ϕ (1)
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POSITIVE SOLUTIONS FOR SEMI-POSITONE NEUMANN BVP
3
If t ≤ s ≤ t0 ,
G(t, s)
ϕ(t)ϕ(1 − s)
ϕ(t)ϕ(1 − t) ϕ(1 − s)
=
=
·
G(t0 , s)
ϕ(s)ϕ(1 − t0 )
ϕ(s)ϕ(1 − t) ϕ(1 − t0 )
1
≥ 2 ϕ(t)ϕ(1 − t) = Cϕ(t)ϕ(1 − t).
ϕ (1)
If t0 ≤ s ≤ t,
G(t, s)
ϕ(s)ϕ(1 − t)
ϕ(t)ϕ(1 − t) ϕ(s)
=
=
·
G(t0 , s)
ϕ(t0 )ϕ(1 − s)
ϕ(t)ϕ(1 − s) ϕ(t0 )
1
≥ 2 ϕ(t)ϕ(1 − t) = Cϕ(t)ϕ(1 − t).
ϕ (1)
For s ≤ t, t0 ,
G(t, s)
ϕ(s)ϕ(1 − t)
ϕ(t)ϕ(1 − t)
=
=
G(t0 , s)
ϕ(s)ϕ(1 − t0 )
ϕ(t)ϕ(1 − t0 )
1
≥ 2 ϕ(t)ϕ(1 − t) = Cϕ(t)ϕ(1 − t).
ϕ (1)
Therefore,
G(t, s) ≥ Cϕ(t)ϕ(1 − t)G(t0 , s),
t, t0 , s ∈ [0, 1].
This completes the proof.
Lemma 2.2. Let y ∈ C((0, 1), [0, ∞)), 0 <
R1
0
y(s)ds < ∞. Then the NBVP
−w00 + m2 w = y(t),
0
0 < t < 1,
(2.2)
0
w (0) = w (1) = 0,
has a unique solution w and there exists a constant Cy such that
Ckwkϕ(t)ϕ(1 − t) ≤ w(t) ≤ Cy ϕ(t)ϕ(1 − t),
0 ≤ t ≤ 1.
(2.3)
R1
Proof. It is obvious that w(t) = 0 G(t, s)y(s)ds is the unique solution of (2.2).
R1
First, let t0 ∈ (0, 1) such that kwk = w(t0 ) = 0 G(t0 , s)y(s)ds. By Lemma 2.1, we
have
Z 1
w(t) =
G(t, s)y(s)ds
0
Z
≥
1
Cϕ(t)ϕ(1 − t)G(t0 , s)y(s)ds
0
Z
= Cϕ(t)ϕ(1 − t)
1
G(t0 , s)y(s)ds
0
= Cϕ(t)ϕ(1 − t)kwk,
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Y. P. SUN & Y. SUN
EJDE-2004/133
which is the first inequality of (2.3). On the other hand,
Z 1
w(t) =
G(t, s)y(s)ds
0
1
=
ρ
≤
=
Z
t
ϕ(s)ϕ(1 − t)y(s)ds +
0
1
ϕ(1 − t)ϕ(t)
ρ
Z
1
ϕ(1 − t)ϕ(t)
ρ
Z
t
0
1
ρ
Z
1
ϕ(t)ϕ(1 − s)y(s)ds
t
1
y(s)ds + ϕ(t)ϕ(1 − t)
ρ
Z
1
y(s)ds
t
1
y(s)ds.
0
By setting
Z
1 1
y(s)ds,
ρ 0
then the second inequality of (2.3) is proved.
Cy =
Remark 2.3. From Lemma 2.2 we know, if y(t) ≡ M , then Cy = CM =
M
ρ .
We make the following assumptions
(H1) f (t, u) ≤ p(t)q(u), where p : (0, 1) → [0, +∞) and q : [0, +∞) → [0, +∞)
are continuous.
(H2) |g(t, u)| ≤ M , where M > 0 is a constant.
R1
(H3) 0 < 0 G(s, s)p(s)ds < +∞.
= +∞ uniformly on any compact subinterval of (0, 1).
(H4) limu→+∞ f (t,u)
u
Let
C + [0, 1] = {u ∈ C[0, 1] : u(t) ≥ 0, 0 ≤ t ≤ 1},
K = {u : u ∈ C + [0, 1],
min
θ≤t≤1−θ
u(t) ≥ Mθ kuk}.
It is obvious that C + [0, 1] and K are cones of E. Let v(t) be the solution of the
boundrary-value problem
−v 00 + m2 v = M, 0 < t < 1,
v 0 (0) = v 0 (1) = 0.
By Lemma 2.2, v(t) ≤ CM ϕ(t)ϕ(1 − t) = M
ρ ϕ(t)ϕ(1 − t). Set
(
y(t), y(t) ≥ 0,
[y(t)]∗ =
0 < t < 1,
0,
y(t) < 0,
F (t, u) = λf (t, [u − v]∗ ) + g(t, [u − v]∗ ) + M, 0 ≤ t ≤ 1.
Consider the boundary-value problem
−u00 + m2 u = F (t, u),
0
0
0 < t < 1,
u (0) = u (1) = 0.
(2.4)
It is no difficulty to prove that u = u0 − v is a positive solution of (1.1) if and only
if u0 is a positive solution of (2.4) and u0 (t) > v(t), 0 < t < 1.
Define an operator Tλ : C + [0, 1] → C + [0, 1] by
Z 1
(Tλ u)(t) = λ
G(t, s)F (s, u(s))ds, u ∈ K.
0
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POSITIVE SOLUTIONS FOR SEMI-POSITONE NEUMANN BVP
5
Lemma 2.4. Let (H1)–(H3) hold. Then Tλ : K → K is a completely continuous
operator.
Proof. For any u ∈ K, t ∈ [0, 1], we have
Z 1
Z
(Tλ u)(t) = λ
G(t, s)F (s, u(s))ds ≤ λ
0
1
G(s, s)F (s, u(s))ds.
0
thus
1
Z
kTλ uk ≤ λ
G(s, s)F (s, u(s))ds.
0
On the other hand, by Lemma 2.1,
1
Z
min (Tλ u)(t) =
min
θ≤t≤1−θ
θ≤t≤1−θ
Z
λ
G(t, s)F (s, u(s))ds
0
1
≥ Mθ λ
G(s, s)F (s, u(s))ds
0
≥ Mθ kTλ uk.
Therefore, Tλ (K) ⊂ K. For a natural number n ≥ 2, define

1
1

min{F (t, x), F ( n , x)}, 0 < t ≤ n ,
1
1
Fn (t, x) = F (t, x),
n < t < 1 − n,


1
1
min{F (t, x), F ( n , x)}, 1 − n ≤ t < 1,
and
1
Z
(Tn u)(t) = λ
G(t, s)Fn (t, u(s))ds,
∀u ∈ E.
0
It is easy to prove that Tn is completely continuous. Let D ⊂ E be a bounded
set, then there is a constant L > 0 such that kuk ≤ L for all u ∈ D, hence
[u(s) − x(s)]∗ ≤ u(s) ≤ kuk ≤ L. We have
(Tλ u)(t) − (Tn u)(t)
Z 1/n
1
≤λ
G(t, s)F (s, u(s)) − F ( , u(s))ds
n
0
Z 1
1
+λ
G(t, s)F (s, u(s)) − F (1 − , u(s))ds
1
n
1− n
Z 1
Z 1/n
≤ 2λ
G(t, s)[p(s)q(u(s)) + M ]ds +
G(t, s)[p(s)q(u(s)) + M ]ds
1
1− n
0
Z
≤ 2λ max q(x)
0≤x≤L
+ 2M
Z
1/n
Z
1/n
G(s, s)p(s)ds
1
1− n
0
0
1
G(s, s)p(s)ds +
Z
1
G(s, s)ds +
G(s, s)ds
1
1− n
→ 0(n → ∞).
Therefore, Tn converge uniformly to Tλ on any bounded subset of E. This implies
that Tλ is a completely continuous operator.
The following Krasnosel’skii fixed point theorem in a cone plays an important
role in proving the main result [10].
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Y. P. SUN & Y. SUN
EJDE-2004/133
Theorem 2.5. Let E be Banach space and K ⊂ E be a cone in E. Suppose Ω1
and Ω2 are open subset of E with 0 ∈ Ω1 and Ω1 ⊂ Ω2 . Let T : K ∩ (Ω2 \ Ω1 ) → K
be a completely continuous operator such that
(A) kT uk ≤ kuk for all u ∈ K ∩ ∂Ω1 and kT uk ≥ kuk for all u ∈ K ∩ ∂Ω2 or
(B) kT uk ≤ kuk for all u ∈ K ∩ ∂Ω2 and kT uk ≥ kuk for all u ∈ K ∩ ∂Ω1 .
Then T has a fixed point in K ∩ (Ω2 \ Ω1 ).
3. Main Result
In this section, we present and prove our main result.
Theorem 3.1. Suppose (H1)–(H4)hold, then (1.1) has at least one positive solution
u ∈ C(0, 1) ∩ C 2 [0, 1] if
Z 1
h
i−1
0 < λ ≤ max q(τ )
G(s, s)p(s)ds)
,
0≤t≤r
where r = max 1 + 2M
R1
0
G(s, s)ds,
0
M
ρC
.
Proof. By Lemma 2.4, we know that Tλ is a completely continuous operator. Let
Ω1 = {u ∈ C[0, 1] : kuk < r}.
For all u ∈ K ∩ ∂Ω1 , t ∈ [0, 1], we have
Z 1
(Tλ u)(t) = λ
G(t, s)F (t, u(s))ds
0
1
Z
G(t, s)(λf (s, [u(s) − x(s)]∗ ) + g(s, [u(s) − x(s)]∗ ) + M )ds
=
0
1
Z
G(t, s)λp(s)q([u(s) − x(s)]∗ )ds + 2M
≤
Z
0
1
G(t, s)ds
0
Z
1
≤ λ max q(τ )
0≤τ ≤r
Z
G(s, s)p(s)ds + 2M
0
Z
≤ 1 + 2M
1
G(s, s)ds
0
1
G(s, s)ds ≤ r = kuk.
0
This implies
kTλ uk ≤ kuk,
for u ∈ K ∩ ∂Ω1 .
On the other hand, choose N large enough such that
Z 1−θ
1
2
λMθ N Cϕ (θ)
G(s, s)ds ≥ 1.
2
θ
By (H4), there exists a constant B > 0 such that
f (s, u)
> N,
u
for (s, u) ∈ [θ, 1 − θ] × [B, ∞).
Set
Ω2 = {u ∈ C[0, 1] : kuk < R},
n 2M
2B o
R = max 2r,
,
.
ρC Cϕ2 (θ)
(3.1)
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POSITIVE SOLUTIONS FOR SEMI-POSITONE NEUMANN BVP
7
For any u ∈ K ∩ ∂Ω2 , s ∈ [0, 1],
M
ϕ(s)ϕ(1 − s)
ρ
M
1
1
= u(s) −
Ckukϕ(s)ϕ(1 − s) ≥ u(s) − u(s) = u(s) ≥ 0.
ρCR
2
2
u(s) − v(s) ≥ u(s) − CM ϕ(s)ϕ(1 − s) = u(s) −
Thus
min (u(s) − v(s)) ≥
θ≤s≤1−θ
≥
≥
min
θ≤s≤1−θ
min
θ≤s≤1−θ
1
u(s)
2
C
kukϕ(s)ϕ(1 − s)
2
CR 2
ϕ (θ) ≥ B.
2
Therefore, for t ∈ [θ, 1 − θ],
Z 1
(Tλ u)(t) = λ
G(t, s)F (s, u(s))
Z
=
0
1
G(t, s)(λf (s, [u(s) − v(s)]∗ ) + g(s, [u(s) − v(s)]∗ ) + M )ds
0
Z
1−θ
≥
G(t, s)λf (s, [u(s) − v(s)]∗ )ds
θ
Z
1−θ
≥ λMθ
G(s, s)N (u(s) − v(s))ds
θ
Z
1−θ
≥ λMθ
G(s, s)N
u(s)
ds
2
G(s, s)N
C
kukϕ(s)ϕ(1 − s)ds
2
θ
Z
≥ λMθ
1−θ
θ
1
≥ λMθ N Cϕ2 (θ)
2
Z
1−θ
G(s, s)dskuk ≥ kuk.
θ
Thus
kTλ uk ≥ kuk,
for u ∈ K ∩ ∂Ω2 .
(3.2)
Applying (B) of Theorem 2.5 to (3.1) and (3.2) yields that Tλ has a fixed point u0
with r ≤ ku0 k ≤ R. By Lemma 2.1 it follows that
u0 (t) ≥ Cku0 kϕ(t)ϕ(1 − t)
= Crϕ(t)ϕ(1 − t)
rρC M
=
·
ϕ(t)ϕ(1 − t)
M
ρ
M
≥
ϕ(t)ϕ(1 − t) = v(t).
ρ
Set u(t) = u0 (t) − v(t), then u(t) is a C[0, 1] ∩ C 2 (0, 1) positive solution to (1.1).
This completes the proof.
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Y. P. SUN & Y. SUN
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Yong-Ping Sun
Department of Mathematics, Qufu Normal University, Qufu, Shandong 273165, China.
Department of Fundamental Courses, Hangzhou Radio & TV University, Hangzhou,
Zhejiang 310012, China
E-mail address: syp@mail.hzrtvu.edu.cn
Yan Sun
Department of Mathematics, Qufu Normal University, Qufu, Shandong 273165, China
E-mail address: ysun@163169.net