Electronic Journal of Differential Equations, Vol. 2010(2010), No. 52, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLICITY OF POSITIVE SOLUTIONS FOR FOUR-POINT
BOUNDARY VALUE PROBLEMS OF IMPULSIVE
DIFFERENTIAL EQUATIONS WITH p-LAPLACIAN
LI SHEN, XIPING LIU, ZHENHUA LU
Abstract. Using a fixed-point theorem in cones, we obtain sufficient conditions for the multiplicity of positive solutions for four-point boundary value
problems of third-order impulsive differential equations with p-Laplacian.
1. Introduction
Recently, there has been much attention focused on the theory of impulsive
differential equation as it is widely used in various areas such as mechanics, electromagnetism, chemistry. A lot of theories have been established to solve these
problems, see [9], [3] and the references therein. Guo [4] obtained the existence of
solutions, via cone theory, for second-order impulsive differential equation
x00 = f (t, x, T x),
t ≥ 0, t 6= tk k = 1, 2, 3, . . . ,
∆x|t=tk = Ik (x(tk )),
k = 1, 2, 3, . . . ,
0
∆x |t=tk = I k (x(tk )),
k = 1, 2, 3, . . . ,
0
x (0) = x∗0 .
x(0) = x0 ,
In [1], using Leggett-Williams fixed point theorem, authors studied the multiplicity result for second order impulsive differential equations
y 00 + φ(t)f (y(t)) = 0,
t ∈ (0, 1) \ {t1 , t2 , . . . , tm },
∆y(tk ) = Ik (y(t−
k )),
0
∆y (tk ) = Jk (y(t−
k )),
k = 1, 2, 3, . . . , m,
k = 1, 2, 3, . . . , m,
y(0) = y(1) = 0.
Kaufmann [8] studied a second-order nonlinear differential equation on an unbounded domain with solutions subject to impulsive conditions and the SturmLiouville type boundary conditions. In [5]-[7], the authors studied positive solutions
of multiple points boundary value problems for second order impulsive differential
equations.
2000 Mathematics Subject Classification. 34A37, 34B37.
Key words and phrases. p-Laplacian; impulsive; positive solutions; boundary value problem.
c
2010
Texas State University - San Marcos.
Submitted November 16, 2009. Published April 14, 2010.
Supported by grant 10ZZ93 from Innovation Program of Shanghai Municipal
Education Commission.
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L. SHEN, X. LIU, Z. LU
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All the works above concern boundary value problems with second-order impulsive equations, and there are just a few works that consider multiplicity of positive
solutions for third-order impulsive equations with p-Laplacian.
Motivated by all the works above, we concentrate on getting multiple positive solutions for four-point boundary value problems of third-order impulsive differential
equations with p-Laplacian
(φp (u00 (t)))0 = f (t, u(t), u0 (t)),
00
∆u (t)|t=tk = 0,
t ∈ (0, 1)\{t1 , t2 , . . . , tm },
k = 1, 2, . . . , m,
0
∆u (t)|t=tk = Ik (u(tk )),
k = 1, 2, . . . , m,
∆u(t)|t=tk = Jk (u(tk )),
k = 1, 2, . . . , m,
00
0
u (0) = αu (ξ) + βu0 (η),
u (0) = 0,
0
(1.1)
u(1) = δu(0),
where φp is p-Laplacian operator
φp (s) = |s|p−2 s, p > 1,
(φp )−1 = φq ,
1 1
+ = 1,
p q
tk , k = 0, 1, 2, . . . , m, m + 1, are constants which satisfy
0 = t0 < t1 < t2 < · · · < tk < · · · < tm < tm+1 = 1,
−
+
−
∆u|t=tk = u(t+
k ) − u(tk ), in which u(tk ) (u(tk ) respectively) denote the right
limit (left limit respectively) of u(t) at t = tk , and α, β > 0, α + β < 1; 0 < ξ,
η < 1; ξ, η 6= tk (k = 1, 2, . . . , m); δ > 1; f ∈ C([0, 1] × [0, +∞) × R, [0, +∞)),
Ik , Jk ∈ C([0, +∞), [0, +∞)).
2. Preliminaries
Let J = [0, 1]\{t1 , t2 , . . . , tm }, P C[0, 1] = {u : [0, 1] → R, u is continuous at
−
−
1
t 6= tk , u(t+
k ), u(tk ) exist, and u(tk ) = u(tk ), k = 1, 2, . . . , m}, P C [0, 1] = {u ∈
0
0 +
0 −
P C[0, 1] | u is continuous at t 6= tk , u (tk ), u (tk ) exist, k = 1, 2, . . . , m}, with the
norm
kukP C 1 =max {kukP C , ku0 kP C }.
kukP C = sup |u(t)|,
t∈J
t∈J
Obviously P C[0, 1] and P C 1 [0, 1] are Banach spaces.
Lemma 2.1. u ∈ P C 1 [0, 1]
T
C 3 [J] is a solution of (1.1) if and only if
s
u(t) = u(0) + u (0)t +
(t − s)φq
f (r, u(r), u0 (r))dr ds
0
X
X 0
+
(t − tk )Ik (u(tk )) +
Jk (u(tk )),
0
tk <t
Z
t
Z
tk <t
(2.1)
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where
u(0) =
α
Rs
Rη
Rs
φq ( 0 f (r, u(r), u0 (r))dr)ds + β 0 φq ( 0 f (r, u(r), u0 (r))dr)ds
(δ − 1)(1 − α − β)
P
P
α tk <ξ Ik (u(tk )) + β tk <η Ik (u(tk ))
Rξ
0
+
R1Rs
+
+
0
0
1
δ−1
(δ − 1)(1 − α − β)
Rr
φq ( 0 f (w, u(w), u0 (w))dw) dr ds
δ−1
m
X
(1 − tk )Ik (u(tk ) + Jk (u(tk )) ,
(2.2)
k=1
Rs
Rs
Rη
φq ( 0 f (r, u(r), u0 (r))dr)ds + β 0 φq ( 0 f (r, u(r), u0 (r))dr)ds
u (0) =
1−α−β
(2.3)
P
P
α tk <ξ Ik (u(tk )) + β tk <η Ik (u(tk ))
+
.
1−α−β
T
Proof. Suppose u ∈ P C 1 [0, 1] C 3 [J] is a solution of (1.1), for all k = 1, 2, . . . , m,
from Lagrange’s mean value theorem we have
α
0
Rξ
0
u(tk ) − u(tk − h) = u0 (ξk )h,
0 < h < tk − tk−1 , ξk ∈ (tk − h, tk ),
because u0 (t−
k ) exists, we get
u0− (tk ) = lim+
h→0
u(tk ) − u(tk − h)
= lim− u0 (ξk ) = u0 (t−
k ).
h
ξk →tk
Let u0 (tk ) = u0− (tk ) = u0 (t−
k ), k = 1, 2, . . . , m. We use Lagrange’s mean value
theorem again and obtain
u0 (tk ) − u0 (tk − h) = u00 (ηk )h,
we can get
u00− (tk )
0 < h < tk − tk−1 , ηk ∈ (tk − h, tk ),
00 −
exists from ∆u (t)|t=tk = u00 (t+
k ) − u (tk ) = 0, and
00
u00− (tk ) = lim+
h→0
u0 (tk ) − u0 (tk − h)
= lim− u00 (ξk ) = u00 (t−
k ).
h
ξk →tk
Let u00 (tk ) = u00 (t−
k ), k = 1, 2, . . . , m. Integrating the differential equation (1.1) we
have
Z
t
φp (u00 (t)) − φp (u00 (0)) =
f (s, u(s), u0 (s))ds,
0
By u00 (0) = 0, we have
Z t
u00 (t) = φq (
f (s, u(s), u0 (s))ds);
0
that is,
Z t
u00 (t) = φq (
f (s, u(s), u0 (s))ds),
0
and
Z
u (t1 ) = φq (
00
0
t1
f (s, u(s), u0 (s))ds).
0 ≤ t ≤ t1 .
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L. SHEN, X. LIU, Z. LU
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00 −
Since ∆u00 (t)|t=t1 = u00 (t+
1 ) − u (t1 ) = 0, for t1 < t ≤ t2 , we obtain
Z t
00
f (s, u(s), u0 (s))ds).
u (t) = φq (
0
00
Similarly, by ∆u (t)|t=tk = u
all t ∈ [0, 1],
00
− u00 (t−
k ) = 0, k = 1, 2, . . . , m, we can show for
(t+
k)
Z t
u (t) = φq (
f (s, u(s), u0 (s))ds).
00
(2.4)
0
For each t ∈ (0, 1), there exist 0 ≤ tk < tk+1 ≤ 1, such that tk < t ≤ tk+1 , by
integrating both sides of (2.4), we obtain
Z t1
Z s
0
u0 (t−
)
−
u
(0)
=
φ
(
f (r, u(r), u0 (r))dr)ds,
q
1
0
0
Z t2
Z s
0 +
u0 (t−
)
−
u
(t
)
=
φ
(
f (r, u(r), u0 (r))dr)ds,
q
2
1
t1
0
...
0 +
u0 (t−
k ) − u (tk−1 ) =
Z
φq (
tk−1
t
Z
u0 (t) − u0 (t+
k)=
tk
Z
f (r, u(r), u0 (r))dr)ds,
0
Z
φq (
tk
s
s
f (r, u(r), u0 (r))dr)ds.
0
Hence,
0
t
Z
0
u (t) = u (0) +
Z
φq (
0
s
f (r, u(r), u0 (r))dr)ds +
0
X
Ik (u(tk )).
tk <t
We have
αu0 (ξ) = αu0 (0) + α
ξ
Z
Z
φq (
0
βu0 (η) = βu0 (0) + β
Z
0
η
s
f (r, u(r), u0 (r))dr)ds + α
0
Z
φq (
X
Ik (u(tk )),
tk <ξ
s
f (r, u(r), u0 (r))dr)ds + β
0
X
Ik (u(tk )).
tk <η
It follows that
Rη
Rs
Rξ
Rs
α 0 φq ( 0 f (r, u(r), u0 (r))dr)ds + β 0 φq ( 0 f (r, u(r), u0 (r))dr)ds
0
u (0) =
1−α−β
P
P
α tk <ξ Ik (u(tk )) + β tk <η Ik (u(tk ))
+
1−α−β
from u0 (0) = αu0 (ξ) + βu0 (η).
Similarly, we get the results as follows with the method above
Z t
Z s
0
u(t) = u(0) + u (0)t +
(t − s)φq
f (r, u(r), u0 (r))dr ds
0
X
X 0
+
(t − tk )Ik (u(tk )) +
Jk (u(tk )).
tk <t
tk <t
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Note that by the boundary condition u(1) = δu(0),
Rξ
Rs
Rη
Rs
α 0 φq ( 0 f (r, u(r), u0 (r))dr)ds + β 0 φq ( 0 f (r, u(r), u0 (r))dr)ds
u(0) =
(δ − 1)(1 − α − β)
P
P
α tk <ξ Ik (u(tk )) + β tk <η Ik (u(tk ))
+
(δ − 1)(1 − α − β)
R1Rs
Rr
φq ( 0 f (w, u(w), u0 (w))dw) dr ds
+ 0 0
δ−1
m
1 X
[(1 − tk )Ik (u(tk ) + Jk (u(tk ))].
+
δ−1
k=1
T
On the other hand, let u ∈ P C 1 [0, 1] C 3 [J] be a solution of (2.1), differentiate
(2.1) when t 6= tk , we have
Z t
u00 (t) = φq (
f (s, u(s), u0 (s))ds);
0
that is,
φp (u00 (t)) =
t
Z
f (s, u(s), u0 (s))ds .
0
Differentiating again,
(φp (u00 (t)))0 = f (t, u(t), u0 (t)).
By (2.1), we can easily get
∆u00 (t)|t=tk = 0,
k = 1, 2, . . . , m,
0
00
∆u (t)|t=tk = Ik (u(tk )),
k = 1, 2, . . . , m,
∆u(t)|t=tk = Jk (u(tk )),
k = 1, 2, . . . , m,
u (0) = 0,
0
0
u (0) = αu (ξ) + βu0 (η),
u(1) = δu(0).
Next, we give the Bai-Ge fixed point theorem which is used in the proof of our
main result. Let E be a Banach space, P ⊂ E be a cone, θ, ψ : P → [0, +∞) be
nonnegative convex functions which satisfy
kxk ≤ k max{θ(x), ψ(x)},
for all x ∈ P,
(2.5)
where k is a positive constant.
Ω = {x ∈ P : θ(x) < r, ψ(x) < L} =
6 φ, where r > 0, L > 0.
(2.6)
Let r > a > 0, L > 0 be constants, θ, ψ : P → [0, +∞) be two nonnegative
continuous convex functions which satisfy (2.5) and (2.6), and γ be a nonnegative
concave function on P . We define convex sets as follows
P (θ, r; ψ, L) = {x ∈ P : θ(x) < r, ψ(x) < L},
P (θ, r; ψ, L) = {x ∈ P : θ(x) ≤ r, ψ(x) ≤ L},
P (θ, r; ψ, L; γ, a) = {x ∈ P : θ(x) < r, ψ(x) < L, γ(x) > a},
P (θ, r; ψ, L; γ, a) = {x ∈ P : θ(x) ≤ r, ψ(x) ≤ L, γ(x) ≥ a}.
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L. SHEN, X. LIU, Z. LU
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Lemma 2.2 ([2]). Let E be Banach space, P ⊂ E be a cone and r2 ≥ d > b >
r1 > 0, L2 ≥ L1 > 0 be constants. Assume θ, ψ : P → [0, +∞) are nonnegative continuous convex functions which satisfy (2.5) and (2.6). γ is a nonnegative
concave function on P such that for all x in P (θ, r2 ; ψ, L2 ) satisfies γ(x) ≤ θ(x).
T : P (θ, r2 ; ψ, L2 ) → P (θ, r2 ; ψ, L2 ) is a completely continuous operator. Suppose
(C1) {x ∈ P (θ, d; ψ, L2 ; γ, b) : γ(x) > b} =
6 φ, and γ(T x) > b, for
x ∈ P (θ, d; ψ, L2 ; γ, b);
(C2) θ(T x) < r1 , ψ(T x) < L1 , for x ∈ P (θ, r1 ; ψ, L1 );
(C3) γ(T x) > b, for x ∈ P (θ, r2 ; ψ, L2 ; γ, b) with θ(T x) > d.
Then T has at least three fixed points x1 , x2 , x3 in P (θ, r2 ; ψ, L2 ). Further,
x1 ∈ P (θ, r1 ; ψ, L1 ), x2 ∈ {P (θ, r2 ; ψ, L2 ; γ, b) : γ(x) > b},
x3 ∈ P (θ, r2 ; ψ, L2 )\ P (θ, r1 ; ψ, L1 ) ∪ P (θ, r2 ; ψ, L2 ; γ, b) .
3. Main results
Let closed cone P be defined by
P = {u ∈ P C 1 [0, 1] : u(t) ≥ 0}.
Define operator T : P → P C 1 [0, 1] by
Z t
Z s
T u(t) = u(0) + u0 (0)t +
(t − s)φq
f (r, u(r), u0 (r))dr ds
0
X 0
X
+
(t − tk )Ik (u(tk )) +
Jk (u(tk )), t ∈ [0, 1],
tk <t
tk <t
0
which u(0), u (0) are defined in (2.2), (2.3).
The nonnegative continuous convex functions θ, ψ, and nonnegative continuous
concave function γ are defined by
θ(u) = sup u(t),
ψ(u) = sup |u0 (t)|,
0≤t≤1
for all u ∈ P , where am =
l = R bm
am
R
Iu =
M1 = R 1
0
γ(u) =
3tm +tm+1
,
4
bm =
min
u(t),
t∈[am ,bm ]
0≤t≤1
tm +3tm+1
.
4
δ−1
(bm − r)φq (r − am )ds
Let
q+1
=
q(q + 1)(δ − 1)
,
(1 − tm )q+1
2
max{I1 (u), I2 (u), . . . , Im (u)},
1−α−β
φq (s)ds + m(1 − α − β) + xα + yβ
=
u ∈ [0, R],
1−α−β
,
1/q + m(1 − α − β) + xα + yβ
where x and y satisfy tx < ξ < tx+1 , ty < η < ty+1 .
Theorem 3.1. Suppose there exist constants r2 ≥ d ≥ δb > b > r1 > 0, L2 ≥ L1 >
0 such that
blδ(m + 1/q)
blδ(m + 1/q)
r2 ≥
, L2 ≥
,
(δ − 1)(1 − α − β)
1−α−β
and the following conditions hold
(H1) f (t, u, v) < φp (min{ δ−1
δ M1 r1 , M1 L1 }), (t, u, v) ∈ [0, 1] × [0, r1 ] × [−L1 , L1 ];
(H2) φp (bl) < f (t, u, v), (t, u, v) ∈ [am , bm ] × [b, d] × [−L2 , L2 ];
(H3) f (t, u, v) < φp (min{ δ−1
δ M1 r2 , M1 L2 }), (t, u, v) ∈ [0, 1) × [0, r2 ] × [−L2 , L2 ];
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(H4) tk Ik (u) > Jk (u) for u ∈ [0, r2 ], Iur1 < min{ δ−1
δ M1 r1 , M1 L1 },
Iur2 < min{ δ−1
M
r
,
M
L
}.
1 2
1 2
δ
Then boundary-value problem (1.1) has at least three positive solutions u1 , u2 , u3 ∈
P (θ, r2 ; ψ, L2 ) which satisfy
sup u1 (t) ≤ r1 ,
0≤t≤1
b<
min
t∈[am ,bm ]
sup |u01 (t)| ≤ L1 ;
0≤t≤1
u2 (t) ≤ sup u2 (t) ≤ r2 ,
0≤t≤1
sup u3 (t) ≤ δd,
0≤t≤1
sup |u02 (t)| ≤ L2 ;
0≤t≤1
sup |u03 (t)| ≤ L2 .
0≤t≤1
Proof. We need to prove δ−1
δ M1 r1 ≥ bl in order to make sure that the theorem
m+1/q
makes sense, since we have r2 ≥ blδ (δ−1)(1−α−β)
, and
(δ − 1)(1 − α − β)
δ−1
M1 r 1 =
r1
δ
δ(1/q + m(1 − α − β) + xα + yβ)
m + 1/q
δ−1 1−α−β
×
×
blδ
≥
δ
1/q + m
(δ − 1)(1 − α − β)
= bl.
Similarly, we have M1 L2 ≥ M1 r2 (δ − 1) ≥ blδ > bl, so there has no contradiction
among conditions. It is easy to see that (1.1) has a solution if and only if
Z s
Z t
T u(t) = u(0) + u0 (0)t +
(t − s)φq
f (r, u(r), u0 (r))dr ds
0
0
X
X
+
(t − tk )Ik (u(tk )) +
Jk (u(tk )), t ∈ [0, 1]
tk <t
tk <t
has a fixed point.
Next, we will check the conditions (C1), (C2) and (C3) of Lemma 2.2 are satisfied
for the operator T .
Obviously, we can get T u(t) ≥ 0, (T u)0 (t) ≥ 0, for all t ∈ [0, 1] and u ∈ P , that
also means T u is a monotone increasing function.
Firstly, we have θ(u) ≤ r2 , ψ(u) ≤ L2 for all u ∈ P (θ, r2 ; ψ, L2 ). By the condition
(H4) tk Ik (u) > Jk (u) and Iur2 < min{ δ−1
δ M1 r2 , M1 L2 }, we get
m X
k=1
m
X
δ−1
(1 − tk )Ik (u(tk )) + Jk (u(tk )) ≤
Ik (u(tk )) ≤ m
M1 r 2 .
δ
k=1
φp ( δ−1
δ M1 r2 ),
By condition (H3), f (t, u, v) <
we obtain
Z
s
δ−1
φq
f (t, u(r), u0 (r))dr ≤
M1 r2 φq (s).
δ
0
Hence,
u(0) ≤
α/q + β/q + xα + yβ
1
m
M1 r 2 + M1 r 2 + M1 r 2 .
δ(1 − α − β)
qδ
δ
Similarly,
u0 (0) ≤
(δ − 1)(α + β + q(xα + yβ))
M1 r 2 .
qδ(1 − α − β)
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Therefore, we can show that
(t − s)φq
0≤t≤1
+
X
t
Z
θ(T u) = sup (u(0) + u0 (0)t +
0
s
f (r, u(r), u0 (r))dr ds
0
(t − tk )Ik (u(tk )) +
tk <t
Z
X
Jk (u(tk )))
tk <t
α/q + β/q + xα + yβ
1
m
M1 r 2 + M 1 r 2 + M1 r 2
δ(1 − α − β)
qδ
δ
(δ − 1)(1 + mq)
(δ − 1)(α + β + q(xα + yβ))
M1 r 2 +
M1 r2
+
qδ(1 − α − β)
qδ
1/q + m(1 − α − β) + xα + yβ
=
M1 r2 .
1−α−β
≤
1−α−β
Since M1 = 1/q+m(1−α−β)+xα+yβ
, we have θ(T u) ≤ r2 .
Similarly, we have
ψ(T u) = sup |(T u)0 (t)|
0≤t≤1
= sup |u0 (0) +
Z
0≤t≤1
≤
t
Z
φq (
0
s
f (r, u(r), u0 (r))dr)ds +
0
X
Ik (u(tk ))|
tk <t
1/q + m(1 − α − β) + xα + yβ
M1 L2 = L2 .
1−α−β
Therefore, T : P (θ, r2 ; ψ, L2 ) → P (θ, r2 ; ψ, L2 ), and it is easy to see that T is a
completely continuous operator.
The proof of the condition (C2) in Lemma 2.2 is similar to the one above.
To check condition (C1) of Lemma 2.2, we choose u0 = d. It is easy to see that
u0 ∈ P (θ, d; ψ, L2 ) and γ(u) = d > b, so {x ∈ P (θ, d; ψ, L2 ; γ, b) : γ(x) > b} =
6 φ.
0
For u ∈ P (θ, d; ψ, L2 ; γ, b), we have b ≤ u(t) ≤ d, |u (t)| ≤ L2 for all t ∈ [am , bm ].
Since T u is a monotone increasing function, and (T u)(t) ≥ 0, t ∈ [0, 1], we have
γ(T u) =
0
Z
t
Z
(t − s)φq (
s
min
u(0) + u (0)t +
f (r, u(r), u0 (r))dr)ds
t∈[am ,bm ]
0
0
X
X
+
[(t − tk )Ik (u(tk ) +
Jk (u(tk ))
tk <t
tk <t
= T u(am ).
By (H2) and u(0), u0 (0) defined before, we have
φp (bl) < f (t, u, v),
(t, u, v) ∈ [am , bm ] × [b, d] × [−L2 , L2 ],
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Z s
Z am
f (r, u(r), u0 (r))dr)ds
(am − s)φq (
T u(am ) = u(0) + u0 (0)am +
0
0
X
X
+
(am − tk )Ik (u(tk )) +
Jk (u(tk ))
tk <am
tk <am
≥ u(0)
≥
>
=
=
1
δ−1
Z
1
δ−1
Z
bl
δ−1
bl
δ−1
1
s
Z
0
0
r
Z
φq (
bm
Z
s
ds
am
Z bm
f (w, u(w), u0 (w))dw) dr ds
0
Z
φq (
am
Z s
φp (bl)dw)dr
am
φq (r − am )dr
ds
am
Z bm
r
am
(bm − r)φq (r − am )dr = b.
am
Thus γ(T u) > b and the condition (C1) of Lemma 2.2 also holds.
Finally to prove (C3) of Lemma 2.2, we check γ(T u) > b to be satisfied for
all u ∈ P (θ, r2 ; ψ, L2 ; γ, b) with θ(T u) > d. Since T u is a nonnegative monotone
increasing function, we can get
θ(T u) = sup T u(t) = T u(1),
0≤t≤1
γ(T u) =
min
t∈[am ,bm ]
T u(t) = T u(am ),
T u(am ) ≥ T u(0) =
d
1
T u(1) > ≥ b;
δ
δ
that is, γ(T u) > b.
We have checked Lemma 2.2 to make sure all the conditions are satisfied with
the work we have done in the section above. Then T has at least three fixed points
u1 , u2 , u3 in P (θ, r2 ; ψ, L2 ). Further,
u1 ∈ P (θ, r1 ; ψ, L1 ),
u2 ∈ {P (θ, r2 ; ψ, L2 ; γ, b) : γ(x) > b},
u3 ∈ P (θ, r2 ; ψ, L2 )\{P (θ, r1 ; ψ, L1 ) ∪ P (θ, r2 ; ψ, L2 ; γ, b)}.
Therefore (1.1) has at least three positive solutions u1 , u2 , u3 . From the boundary
conditions we have u3 (1) = δu3 (0) and u3 is a monotone increasing function, so we
have
1
1
b > γ(u3 ) = min u3 (t) = u3 (am ) ≥ u3 (0) = u3 (1) = θ(u3 ),
am ≤t≤bm
δ
δ
so θ(u3 ) ≤ δb, that means sup0≤t≤1 u3 (t) ≤ δd, and u1 , u2 , u3 satisfy
sup u1 (t) ≤ r1 ,
0≤t≤1
b<
min
t∈[am ,bm ]
sup |u01 (t)| ≤ L1 ;
0≤t≤1
u2 (t) ≤ sup u2 (t) ≤ r2 ,
0≤t≤1
sup u3 (t) ≤ δd,
0≤t≤1
sup |u02 (t)| ≤ L2 ;
0≤t≤1
sup |u03 (t)| ≤ L2 .
0≤t≤1
10
L. SHEN, X. LIU, Z. LU
EJDE-2010/52
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Li Shen
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: eric0shen@gmail.com
Xiping Liu
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: xipingliu@163.com
Zhenhua Lu
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: ehuanglu@163.com