# Electronic Journal of Differential Equations, Vol. 2009(2009), No. 150, pp.... ISSN: 1072-6691. URL: or ```Electronic Journal of Differential Equations, Vol. 2009(2009), No. 150, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
TRIPLE SOLUTIONS FOR MULTI-POINT BOUNDARY-VALUE
PROBLEM WITH p-LAPLACE OPERATOR
HAITAO LI, YANSHENG LIU
Abstract. Using a fixed point theorem due to Avery and Peterson, this article
shows the existence of solutions for multi-point boundary-value problem with
p-Laplace operator and parameters. Also, we present an example to illustrate
the results obtained.
1. Introduction
During the previous two decades, boundary-value problems for second-order differential equations with p-Laplace operator have been extensively studied and a lot
of excellent results have been established by using fixed point index theory, upper
and lower solution arguments, fixed point theorem like Leggett-Williams multiple
fixed point theorem and so on (see [2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] and references
therein). For example, Ma, Du and Ge  studied the following boundary-value
problem (BVP, for short) with p-Laplace operator
(ϕp (u0 (t)))0 + q(t)f (t, u(t)) = 0,
u0 (0) =
n
X
αi u0 (ξi ), u(1) =
i=1
n
X
t ∈ (0, 1);
βi u(ξi ),
i=1
1
1
where ϕp (s) = |s|p−2 s, p &gt; 1, ϕ−1
p = ϕq , p + q = 1, and 0 &lt; ξ1 &lt; ξ2 &lt; &middot; &middot; &middot; &lt; ξn &lt; 1.
0
The nonlinearity f is not depending on u . Using the upper and lower solutions
method, they obtained sufficient conditions for the existence of one positive solution.
Lv, O’Regan and Zhang  considered the following boundary-value problem
(BVP) with p-Laplace operator
(ϕp (y 0 (t)))0 + q(t)f (y(t)) = 0,
t ∈ [0, 1];
y(0) = y(1) = 0.
By Leggett-Williams multiple fixed point theorem, they provided sufficient conditions for the existence of multiple (at least three) positive solutions.
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Triple solutions; p-Laplace operator; fixed point theorem;
multi-point boundary-value problem.
c
2009
Texas State University - San Marcos.
Submitted April 22, 2009. Published November 25, 2009.
Supported by grants 209072 from the Key Project of Chinese Ministry of Education,
and J08LI10 from the Science and by Technology Development Funds of Shandong
Education Committee.
1
2
H. LI, Y. LIU
EJDE-2009/150
Recently Ji, Tian and Ge  studied the following boundary-value problem, in
which the nonlinearity contains u0 ,
(ϕp (u0 (t)))0 + λf (t, u(t), u0 (t)) = 0,
u0 (0) =
n
X
αi u0 (ξi ),
u(1) =
i=1
t ∈ (0, 1);
n
X
βi u(ξi ).
(1.1)
i=1
Applying Krasnosel’skii fixed point theorem, they obtained the existence of at least
one positive solution.
Wang and Ge  studied the multi-point boundary-value problem
(ϕp (u0 (t)))0 + q(t)f (t, u(t), u0 (t)) = 0,
u(0) =
n
X
αi u(ξi ),
i=1
u(1) =
n
X
t ∈ (0, 1);
βi u(ξi ).
i=1
Using the fixed point theorem due to Avery and Peterson, they provided sufficient
conditions for the existence of multiple positive solutions.
Motivated by [8, 13], we investigate (1.1). We study boundary value conditions
that are different from those in [9, 13]. We obtain three solutions by the fixed
point theorem due to Avery and Peterson, which is different from the methods in
[8, 9, 10]. To the best of our knowledge, (1.1) has not been studied via this fixed
point theorem.
is devoted to the existence of triple solutions for (1.1). Finally an example is shown
to illustrate the results obtained. Now, we give some notation which will be used
later.
Let X = C 1 [0, 1] be a Banach space with the norm
kuk = max max |u(t)|, max |u0 (t)| .
t∈[0,1]
t∈[0,1]
A function u(t) is called a positive solution of (1.1) if u ∈ X, satisfies (1.1) and
u(t) &gt; 0 for t ∈ (0, 1). Let
C ∗ [0, 1] = {u ∈ X : u(t) ≥ 0, u0 (t) ≤ 0, u0 (t) is nonincreasing for t ∈ [0, 1]},
P = {u ∈ X : u(t) ≥ 0, u0 (t) ≤ 0, u0 (t) is concave on t ∈ [0, 1]}.
It is easy to see P is a cone of X.
In this paper, we assume the following hypotheses:
Pn
Pn
(H1) αi , βi ≥ 0, 0 &lt; i=1 αi , i=1 βi &lt; 1.
(H2) f ∈ C([0, 1] &times; [0, +∞) &times; (−∞, 0], [0, +∞)).
2. Preliminaries
In this section, we provide some background definitions from the study of cone
in Banach spaces; see for example .
Let (E, k &middot; k) be a real Banach space. A nonempty, closed, convex set P ⊆ E is
said to be a cone provided the following two conditions are satisfied:
(a) if y ∈ P and λ ≥ 0, then λy ∈ P ;
(b) if y ∈ P and −y ∈ P , then y = 0.
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TRIPLE SOLUTIONS
3
If P ⊆ E is a cone, we denote the order induced by P on E by ≤, that is, x ≤ y
if and only if y − x ∈ P .
A map α is said to be a nonnegative continuous concave functional on a cone P
of a real Banach space E, provided that α : P → [0, +∞) is continuous and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y)
for all x, y ∈ P and 0 ≤ t ≤ 1.
Similarly, we say a map β is a nonnegative continuous convex functional on a
cone P of a real Banach space E, provided that β : P → [0, +∞) is continuous and
β(tx + (1 − t)y) ≤ tβ(x) + (1 − t)β(y)
for all x, y ∈ P and 0 ≤ t ≤ 1.
Let γ and θ be nonnegative continuous convex functionals on P , α be a nonnegative continuous concave functional on P , and ψ be a nonnegative continuous
functional on P . Then for positive real numbers a, b, c and d, we define the following
convex sets:
P (γ, d) = {x ∈ P |γ(x) &lt; d},
P (γ, α, b, d) = {x ∈ P |b ≤ α(x), γ(x) ≤ d},
P (γ, θ, α, b, c, d) = {x ∈ P |b ≤ α(x), θ(x) ≤ c, γ(x) ≤ d},
and a closed set
R(γ, ψ, a, d) = {x ∈ P |a ≤ ψ(x), γ(x) ≤ d}.
The following fixed point theorem is fundamental in the proofs of our main
results.
Lemma 2.1 (). Let P be a cone in a real Banach space E. Let γ and θ be
nonnegative continuous convex functionals on P , α be a nonnegative continuous
concave functional on P , and ψ be a nonnegative continuous functional on P satisfying ψ(λx) ≤ λψ(x) for 0 ≤ λ ≤ 1, such that for some positive numbers L and
d,
α(x) ≤ ψ(x) and kxk ≤ Lγ(x), ∀x ∈ P (γ, d).
Suppose T : P (γ, d) → P (γ, d) is completely continuous, and there exist positive
numbers a, b, and c with a &lt; b such that
(S1) {x ∈ P (γ, θ, α, b, c, d)|α(x) &gt; b} =
6 ∅ and α(T x) &gt; b for x ∈ P (γ, θ, α, b, c, d)
(S2) α(T x) &gt; b for x ∈ P (γ, α, b, d) with θ(T x) &gt; c
(S3) 0 ∈
/ R(γ, ψ, a, d) and ψ(T x) &lt; a for x ∈ R(γ, ψ, a, d) with ψ(x) = a.
Then T has at least three fixed points x1 , x2 , x3 ∈ P (γ, d) such that γ(xi ) ≤ d for
i = 1, 2, 3; b &lt; α(x1 ); a &lt; ψ(x2 ) with α(x2 ) &lt; b; ψ(x3 ) &lt; a.
To prove the main results in this paper, we will employ the following lemmas.
Lemma 2.2 (). Assume (H1)-(H2), and let
Pn
ϕp ( i=1 αi )
Pn
k=
.
1 − ϕp ( i=1 αi )
For x ∈ C ∗ [0, 1], if u(t) is a solution of the problem
(ϕp (u0 (t)))0 + λf (t, x(t), x0 (t)) = 0,
t ∈ (0, 1);
4
H. LI, Y. LIU
n
X
u0 (0) =
EJDE-2009/150
αi u0 (ξi ),
u(1) =
i=1
n
X
βi u(ξi ),
i=1
then
Pn
i=1
u(t) = −
βi
R1
ξi
λf (r, x(r), x0 (r))dr)ds
0
P
n
1 − i=1 βi
1
Z
s
Z
−
Rs
ϕq (Ax −
ϕq (Ax −
λf (r, x(r), x (r))dr)ds,
t
0
R1
f (s, x(s), x0 (s))ds, 0] is unique and satisfies
Z ξi
n
X
ϕq (Ax) =
αi ϕq (Ax −
λf (s, x(s), x0 (s))ds),
where Ax ∈ [−kλ
(2.1)
0
0
Define the operator T by
Pn
βi
i=1
(T u)(t) = −
R1
ξi
Z
s
ϕq (Au −
t
Rs
λf (r, u(r), u0 (r))dr)ds
0
P
n
1 − i=1 βi
ϕq (Au −
1
Z
−
(2.2)
0
i=1
λf (r, u(r), u0 (r))dr)ds.
0
Then by Lemma 2.2 it is easy to see u(t) is a solution of (1.1) if and only if
u(t) = (T u)(t).
Lemma 2.3 (). For each λ &gt; 0, the operator T : P → P is completely continuous.
Now we give an important property of Ax defined by (2.1).
Lemma 2.4. Assume (H1) holds. Then for each x ∈ C ∗ [0, 1], τ ∈ (0, ξ1 ),
Pn
Z ξ1
Z 1
ϕp ( i=1 αi )
Pn
λf (r, x(r), x0 (r))dr ≤ −Ax ≤ k
λf (r, x(r), x0 (r))dr.
1 − ϕp ( i=1 αi ) τ
0
(2.3)
Proof. By (2.1), we have
ϕq (Ax) =
n
X
λf (s, x(s), x0 (s))ds)
0
i=1
≥
ξi
Z
αi ϕq (Ax −
n
X
1
Z
λf (s, x(s), x0 (s))ds),
αi ϕq (Ax −
0
i=1
and
ϕq (Ax) =
n
X
Z
≤
i=1
λf (s, x(s), x0 (s))ds)
0
i=1
n
X
ξi
αi ϕq (Ax −
Z
αi ϕq (Ax −
ξ1
λf (s, x(s), x0 (s))ds).
τ
From the increasing property of ϕq and the two inequalities above, it is easy to get
the conclusion.
EJDE-2009/150
TRIPLE SOLUTIONS
5
Set
(m+1)ξ1
−m
+ ξ1
1
2 q−1 + 1
q−1
m :=
, l := 2
,
ξ1
2
Pn
Pn
βi (1 − ξi ) + (1 − i=1 βi )(1 − ξ1 )
Pn
.
N := i=1
1 − i=1 βi
1
Choose an τ ∈ (0, ξ1 ) such that l−1/m &lt; τ &lt; ξ1 , and define the functionals:
γ(x) = ψ(x) := kxk,
θ(x) := max |x0 (t)|,
α(x) := min x(t),
t∈[0,τ ]
∀x ∈ P. (2.4)
t∈[τ,ξ1 ]
Then it is easy to get the following lemma.
Lemma 2.5. The four functionals defined by (2.4) satisfy Lemma 2.1. In addition,
for each x ∈ P , θ(x) = −x0 (τ ), α(x) = x(ξ1 ), γ(x) = ψ(x) = x(0).
3. Main Results
First we state the following hypotheses to be used in this article.
(H3) There exists a positive constant H such that
f (t, u, v) &lt; ltm ϕp (|u| + |v|),
for t ∈ [0, 1] and (u, v) ∈ R2 satisfying 0 ≤ |u| + |v| ≤ H.
(H4) There exist positive constants b, d such that
1
1
1
1
max{
, }b &lt; d ≤ H,
,
1 − ξ1 2lq−1 N
2
f (t, u, v) &gt; ϕp (b), for (t, u, v) ∈ [τ, ξ1 ] &times; [b, d] &times; [−d, 0].
Now we are ready to state our main results.
Theorem 3.1. Assume (H1)-(H4). Let
M=
(1 −
Pn
1 − i=1 βi ξi
Pn
.
i=1 βi )ϕq (1 − ϕp (
i=1 αi ))
Pn
Then for each λ satisfying
1
ξ1 M
1
q−1
≤λ≤
1
1
1
q−1 lM
m+1 2
1
q−1
,
(3.1)
and a ∈ (0, b), Equation (1.1) has at least three solutions x1 (t), x2 (t), x3 (t) satisfying
(i) kxi k ≤ d, i = 1, 2, 3;
(ii) b &lt; min{|x1 (t)||t ∈ [0, τ ]};
(iii) kx2 k &gt; a, min{x2 (t)|t ∈ [0, τ ]} &lt; b;
(iv) kx3 k &lt; a.
Proof. We divide the proof of this theorem in four steps.
Step 1. Let us show T : P (γ, d) → P (γ, d). In fact, for any u ∈ P (γ, d), it is
not difficult to see
kT uk = max{(T u)(0), −(T u)0 (1)}.
(3.2)
From (2.3), (3.1), and (H3), we obtain
R1
Rs
Pn
0
i=1 βi ξi ϕq (Au − 0 λf (r, u(r), u (r))dr)ds
Pn
(T u)(0) = −
1 − i=1 βi
6
H. LI, Y. LIU
1
Z
−
s
Z
λf (r, u(r), u0 (r))dr)ds
ϕq (Au −
0
EJDE-2009/150
0
Rs
λf (r, u(r), u0 (r))dr − 0 λf (r, u(r), u0 (r))dr)ds
Pn
≤−
1 − i=1 βi
Z s
Z 1
Z 1
0
λf (r, u(r), u0 (r))dr)ds
λf (r, u(r), u (r))dr −
ϕq (−k
−
0
0
0
Pn
Z 1
1 − i=1 βi ξi
q−1
Pn
Pn
≤λ
ϕq
f (r, u(r), u0 (r))dr
(1 − i=1 βi )ϕq (1 − ϕp ( i=1 αi ))
0
1
&lt;(
)q−1 2λq−1 lq−1 M kuk
m+1
1
)q−1 2λq−1 lq−1 M d ≤ d
≤(
m+1
Pn
i=1 βi
R1
ϕ (−k
ξi q
R1
0
and
Z
0
−(T u) (1) = −ϕq (Au −
1
λf (r, u(r), u0 (r))dr)
0
Z
≤ ϕq (
1
λf (r, u(r), u0 (r))dr +
Z
0
s
λf (r, u(r), u0 (r))dr)
0
Z 1
1
Pn
ϕq (
f (r, u(r), u0 (r))dr)
ϕq (1 − ϕp ( i=1 αi ))
0
1
1
Pn
)q−1 2λq−1 lq−1
kuk ≤ d.
&lt;(
m+1
ϕq (1 − ϕp ( i=1 αi ))
≤ λq−1
Thus kT uk = max{(T u)(0), −(T u)0 (1)} ≤ d. Hence T : P (γ, d) → P (γ, d).
Step 2. Check condition (S1) of Lemma 2.1. Choose an integer w &gt; 0 such that
1
max{ 1−ξ
, 1 } &lt; w ≤ db . Set u(t) = wb(1 − t). Then
1 N
b &lt; θ(u) = wb, γ(u) = wb ≤ d, b &lt; α(u) = wb(1 − ξ1 ) &lt; wb.
Therefore, u(t) = wb(1 − t) ∈ P (γ, θ, α, b, wb, d), and α(u) &gt; b. This guarantees
that {u ∈ P (γ, θ, α, b, wb, d)|α(u) &gt; b} 6= ∅. For any u ∈ P (γ, θ, α, b, wb, d), it is
easy to see
b ≤ u(t) ≤ d, −d ≤ u0 (t) ≤ 0, ∀t ∈ [τ, ξ1 ].
Thus by (H4), f (t, u(t), u0 (t)) &gt; ϕp (b).
By Lemma 2.2 and Lemma 2.3, it is not difficult to see
α(T u) = (T u)(ξ1 )
R1
Rs
Pn
0
i=1 βi ξi ϕq (Au − 0 λf (r, u(r), u (r))dr)ds
Pn
=−
1 − i=1 βi
Z 1
Z s
−
ϕq (Au −
λf (r, u(r), u0 (r))dr)ds
ξ1
0
Rs
λf (r, u(r), u0 (r))dr + 0 λf (r, u(r), u0 (r))dr)ds
Pn
≥
1 − i=1 βi
Z 1
Z ξ1
Z s
+
ϕq (k
λf (r, u(r), u0 (r))dr +
λf (r, u(r), u0 (r))dr)ds
Pn
i=1
ξ1
βi
R1
ξi
ϕq (k
0
R ξ1
0
0
EJDE-2009/150
TRIPLE SOLUTIONS
≥ λq−1
&gt;
(1 −
7
Pn
Z ξ1
1 − i=1 βi ξi
Pn
ϕq (
f (r, u(r), u0 (r))dr)
β
)ϕ
(1
−
ϕ
(
α
))
q
p
τ
i=1 i
i=1 i
Pn
λq−1 ξ1q−1 M b
≥ b.
This shows that condition (S1) of Lemma (2.1) is satisfied.
Step 3. Examine (S2) of Lemma 2.1. For any u ∈ P (γ, α, b, d) with θ(T u) &gt; wb,
we know
Z τ
0
θ(T u) = −(T u) (τ ) = ϕq
λf (r, u(r), u0 (r))dr − Au &gt; wb.
(3.3)
0
Therefore by (2.3) and (3.3),
α(T u) = (T u)(ξ1 )
R1
Rs
Pn
0
i=1 βi ξi ϕq (Au − 0 λf (r, u(r), u (r))dr)ds
Pn
=−
1 − i=1 βi
Z 1
Z s
−
ϕq (Au −
λf (r, u(r), u0 (r))dr)ds
ξ1
0
R1 Rτ
Pn
0
i=1 βi ξi ϕq k 0 λf (r, u(r), u (r))dr − Au ds
Pn
≥
1 − i=1 βi
Z 1
Z τ
+
ϕq (k
λf (r, u(r), u0 (r))dr − Au)ds
ξ
Pn 1
0
Pn
Z τ
(1 − i=1 βi )(1 − ξ1 )
i=1 βi (1 − ξi ) + P
=
ϕq (
λf (r, u(r), u0 (r))dr − Au)
n
1 − i=1 βi
0
&gt; N wb &gt; b.
Thus, condition (S2) of Lemma (2.1) is satisfied.
Step 4. Finally we show (S3) of Lemma 2.1 holds. Since ψ(0) = 0 &lt; a, we
know 0 ∈
/ R(γ, ψ, a, d). For each u ∈ R(γ, ψ, a, d), ψ(u) = kuk = a, by (2.3), (3.1),
and (H3), we obtain
R1
Rs
Pn
0
i=1 βi ξi ϕq (Au − 0 λf (r, u(r), u (r))dr)ds
Pn
(T u)(0) = −
1 − i=1 βi
Z 1
Z s
−
ϕq (Au −
λf (r, u(r), u0 (r))dr)ds
0
0
Pn
Z 1
1 − i=1 βi ξi
q−1
Pn
Pn
ϕq
f (r, u(r), u0 (r))dr
≤λ
(1 − i=1 βi )ϕq (1 − ϕp ( i=1 αi ))
0
1
&lt;(
)q−1 2λq−1 lq−1 M kuk
m+1
1
)q−1 2λq−1 lq−1 M a ≤ a
=(
m+1
and
Z 1
0
−(T u) (1) = −ϕq (Au −
λf (r, u(r), u0 (r))dr)
0
Z
≤ ϕq (k
1
0
Z
λf (r, u(r), u (r))dr +
0
0
s
λf (r, u(r), u0 (r))dr)
8
H. LI, Y. LIU
≤ λq−1
EJDE-2009/150
1
Pn
ϕq
ϕq (1 − ϕp ( i=1 αi ))
Z
1
f (r, u(r), u0 (r))dr
0
1
1
Pn
kuk ≤ a.
&lt;(
)q−1 2λq−1 lq−1
m+1
ϕq (1 − ϕp ( i=1 αi ))
Therefore,
ψ(u) = kuk = max{(T u)(0), −(T u)0 (1)} &lt; a.
So condition (S3) of Lemma (2.1) is satisfied. Thus an application of Lemma
2.1 implies that the boundary value problem (1.1) has at least three solutions
x1 (t), x2 (t), x3 (t) satisfying (i)–(iv).
We remark that in Theorem 3.1, the two solutions x1 (t) and x2 (t) are positive,
while x3 (t) may be the trivial solution.
3.1. Example. Consider the differential equation
(ϕp (u0 (t)))0 + λf (t, u(t), u0 (t)) = 0,
u0 (0) =
2
X
αi u0 (ξi ),
u(1) =
i=1
2
X
t ∈ (0, 1);
βi u(ξi ),
(3.4)
i=1
where p = 3/2, q = 3, α1 = α2 = β1 = β2 = 1/4, ξ1 = 0.9, ξ2 = 0.95,
√
p
f (t, u, v) = 1.8t10( 2+1)/9 u + |v|, (t, u, v) ∈ [0, 1] &times; [0, +∞) &times; (−∞, 0].
√
Choose l = 1.835055448, m = 10( 2 + 1)/9, H = 20000, d = 10000, b = 100,
a = 50, τ = 0.88, then by simple calculations,
it is easy to show (H1)-(H4) are
√
satisfied. Therefore, by Theorem 3.1, for 9 430/430 ≤ λ ≤ 1.461370837, Equation
(3.4) has at least three solutions.
Acknowledgements. The authors want to thank the anonymous referee for the
suggestions on the paper.
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EJDE-2009/150
TRIPLE SOLUTIONS
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Haitao Li
Department of Mathematics, Shandong Normal University, Jinan, 250014, China