Electronic Journal of Differential Equations, Vol. 2009(2009), No. 89, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR
THIRD-ORDER BOUNDARY-VALUE PROBLEM ON THE
HALF-LINE WITH DEPENDENCE ON THE FIRST ORDER
DERIVATIVE
SIHUA LIANG, JIHUI ZHANG
Abstract. By using fixed-point theorem for operators on a cone, sufficient
conditions for the existence of multiple positive solutions for a third-order
boundary-value problem on the half-line are established. In the case of the pLaplace operator our results for p > 1 generalize previous known results. The
interesting point lies in the fact that the nonlinear term is allowed to depend
on the first order derivative u0 .
1. Introduction
Boundary value problems on the half-line arise naturally in the study of radially
symmetric solutions of nonlinear elliptic equations, see [3], and various physical phenomena [2, 5], such as an unsteady flow of gas through a semi-infinite porous media,
the theory of drain flows, plasma physics, in determining the electrical potential in
an isolated neutral atom. In recently years, the boundary-value problems on the
half-line have received a great deal of attention in literature (see [1, 7, 10, 11, 15, 16]
and references therein). However, in [6, 13, 14] the authors only studied multi-point
boundary-value problems on the finite interval. They showed that there exist multiple positive solutions by using fixed-point theorems for operators on a cone. But
so far, very few results are obtained for a third-order multi-point boundary-value
problems on the half-line. To the author’s knowledge, there are no results about
third order multi-point boundary-value problems, whose non- linear term does not
depend on the first derivative u0 . The goal of this paper is to fill the gap in this
area. In this paper, by using fixed-point theorem for operators on a cone, some
sufficient conditions for the existence of multiple positive solutions for third-order
boundary-value problem on the half-line are established, which are the complement
of previously known results. In the case of the p-Laplace operator our results for
some p > 1 generalize previous known results.
2000 Mathematics Subject Classification. 34B18, 34B40.
Key words and phrases. Third-order boundary-value problems; positive solutions; half-line;
fixed-point index theory.
c
2009
Texas State University - San Marcos.
Submitted March 26, 2009. Published July 27, 2009.
1
2
S. LIANG, J. ZHANG
EJDE-2009/89
In this paper, we study the existence of multiple positive solutions for the following third-order boundary-value problem with dependence on the first order derivative on the half-line
(ϕ(−u00 )(t))0 = a(t)f (t, u, u0 ), 0 < t < +∞,
u(0) − βu0 (0) = 0,
u0 (∞) = 0,
(1.1)
u00 (0) = 0,
where ϕ : R → R is an increasing homeomorphism and positive homomorphism with
ϕ(0) = 0 and f ∈ C([0, +∞)3 , [0, +∞)) and β ∈ (0, +∞). a(t) is a nonnegative
measurable function defined in (0, +∞) and a(t) does not identically vanish on any
subinterval of (0, +∞).
A projection ϕ : R → R is called an increasing homeomorphism and positive
homomorphism (see [8]), if the following conditions are satisfied:
(i) if x ≤ y, then ϕ(x) ≤ ϕ(y), for all x, y ∈ R;
(ii) ϕ is a continuous bijection and its inverse mapping is also continuous;
(iii) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ [0, +∞).
In above definition, we can replace the condition (iii) by the following stronger
condition:
(iv) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ R, where R = (−∞, +∞).
Remark 1.1. (1) If conditions (i), (ii) and (iv) hold, then it implies that ϕ is
homogenous generating a p-Laplace operator; i.e., ϕ(x) = |x|p−2 x, for some p > 1.
(2) It is well known that a p-Laplacian operator is odd. However, the operator
which we defined above is not necessary odd, see (5.2). We emphasize that the
results of the papers [10, 12, 13, 14] cannot be applied if ϕ is defined as above.
(3) The nonlinear term is allowed to depend on the first order derivative u0 which
is the complement of previously known results [6, 7, 13, 14].
In this article, the following hypotheses are needed:
(C1) f ∈ C([0, +∞)3 , [0, +∞)), f (t, 0, 0) 6≡ 0 on any subinterval of (0, +∞) and
when u is bounded f (t, (1 + t)u, u0 ) is bounded on [0, +∞);
(C2) a(t) satisfies the following relations:
Z +∞
Z +∞
Z s
Z s
−1
−1
ϕ
a(τ )dτ ds < +∞,
sϕ
a(τ )dτ ds < +∞.
0
0
0
0
The plan of the article is as follows. In Section 2 for the convenience of the reader
we give some background and definitions. In Section 3 we present some lemmas
in order to prove our main results. Section 4 is devoted to presenting and proving
our main results. Some examples are presented in Section 5 to demonstrate the
application of our main results.
2. Some definitions and fixed point theorems
In this section, we provide background definitions from the cone theory in Banach
spaces.
Definition 2.1. Let (E, k · k) be a real Banach space. A nonempty, closed, convex
set P ⊂ E is said to be a cone provided the following 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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EXISTENCE OF MULTIPLE POSITIVE 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 .
Definition 2.2. A map α is said to be a nonnegative, continuous, concave functional on a cone P of a real Banach space E, if α : P → [0, ∞) is continuous,
and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y)
for all x, y ∈ P and t ∈ [0, 1].
Definition 2.3. An operator is called completely continuous if it is continuous and
maps bounded sets into precompact sets.
The following fixed point theorems are fundamental and important to the proofs
of our main results.
Theorem 2.4. [4] Let E be a Banach space and P ⊂ E be
T a cone in E. Let r > 0
define Ωr = {x ∈ P : kxk < r}. Assume that T : P Ωr → P is completely
continuous operator such that T x 6= x for x ∈ ∂Ωr .
(i) If kT xk ≤ kxk for x ∈ ∂Ωr , then i(T, Ωr , P ) = 1.
(ii) If kT xk ≥ kxk for x ∈ ∂Ωr , then i(T, Ωr , P ) = 0.
Theorem 2.5. [6] Let K be a cone in a Banach space X. Let D be an open bounded
set with Dk = D ∩ K 6= ∅ and Dk 6= K. Let T : Dk → K be a compact map such
that x 6= T x for x ∈ ∂Dk . Then the following results hold:
(1) If kT xk ≤ kxk for x ∈ ∂Dk , then ik (T, Dk ) = 1.
(2) Suppose there is e ∈ K, e 6= 0 such that x 6= T x + λe for all x ∈ ∂Dk and
all λ > 0, then ik (T, Dk ) = 0.
(3) Let D1 be open in X such that D1 ⊂ Dk . If ik (T, Dk ) = 1 and ik (T, Dk1 ) =
0, then T has a fixed point in Dk \Dk1 . Then same result holds if ik (T, Dk ) =
0 and ik (T, Dk1 ) = 1.
3. Preliminaries and Lemmas
Let E be the set defined as
E = u ∈ C 1 [0, +∞) :
|u(t)|
< +∞,
0≤t<+∞ 1 + t
sup
lim u0 (t) = 0.
t→+∞
Then E is a Banach space, equipped with the norm kuk = kuk1 + kuk2 , where
0
kuk1 = sup0≤t<+∞ |u(t)|
1+t < +∞, kuk2 = sup0≤t<+∞ |u (t)|.
Also, define the cone K ⊂ E by
K = u ∈ E : u(t) ≥ 0, t ∈ [0, +∞), u(0) − βu0 (0) = 0,
u(t) is concave on [0, +∞).
To prove the main results in this paper, we will employ several lemmas.
Lemma 3.1. For any p ∈ C[0, +∞), the problem
(ϕ(−u00 (t)))0 = p(t),
0
u(0) − βu (0) = 0,
0
0 < t < +∞,
u (∞) = 0,
00
u (0) = 0
(3.1)
(3.2)
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S. LIANG, J. ZHANG
EJDE-2009/89
has a unique solution
Z +∞
Z +∞
Z s
Z s
−1
−1
p(τ )dτ ds
sϕ
p(τ )dτ ds +
(t − s)ϕ
u(t) =
0
0
0
t
Z +∞
Z s
p(τ )dτ ds.
+β
ϕ−1
(3.3)
0
0
Proof. Necessity. By taking the integral of the equation (3.1) on [0, t], we have
Z t
p(τ )dτ.
(3.4)
ϕ(−u00 (t)) − ϕ(−u00 (0)) =
0
By the boundary condition and ϕ(0) = 0, we have
Z t
00
−1
p(τ )dτ .
u (t) = −ϕ
(3.5)
0
By taking the integral of (3.5) on [t, +∞), we obtain
Z +∞
Z s
u0 (∞) − u0 (t) = −
ϕ−1
p(τ )dτ ds.
t
By the boundary condition u0 (∞) = 0, we obtain
Z +∞
Z s
u0 (t) =
ϕ−1
p(τ )dτ ds.
t
(3.7)
0
By taking the integral of (3.7) on [t, +∞), we have
Z +∞ Z +∞
Z τ
u(t) =
ϕ−1
p(η)dη dτ ds + u(0).
t
(3.6)
0
s
(3.8)
0
Substituting u(0) = βu0 (0) and integrating by parts, we obtain
Z +∞
Z +∞
Z s
Z s
−1
−1
u(t) =
(t − s)ϕ
p(τ )dτ ds +
sϕ
p(τ )dτ ds
t
0
0
0
Z +∞
Z s
+β
ϕ−1
p(τ )dτ ds.
0
0
Sufficiency: Let u be as in (3.3). Taking the derivative of (3.3), it implies that
Z +∞
Z s
u0 (t) =
ϕ−1
p(τ )dτ ds.
t
0
Furthermore, we obtain
00
u (t) = −ϕ
−1
Z
ϕ(−u00 (t)) =
Z
t
p(τ )dτ ,
0
t
p(τ )dτ,
0
taking the derivative of this expression yields (ϕ(−u00 (t)))0 = p(t). Routine calculation verifies that u satisfies the boundary value conditions, so that u given in (3.3)
is a solution of (3.1)-(3.2).
It is easy to see that the problem
(ϕ(−u00 (t)))0 = 0,
u(0) − βu0 (0) = 0,
u0 (∞) = 0,
u00 (0) = 0,
has only the trivial solution. Thus u in (3.3) is the unique solution of (3.1)-(3.2).
The proof is complete.
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EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS
5
Lemma 3.2. For any p(t) ∈ C[0, +∞) and p(t) ≥ 0, the unique solution u of
(3.1)-(3.2) satisfies
u(t) ≥ 0, for t ∈ [0, +∞).
Lemma 3.3. For any u ∈ K, it holds that βkuk2 ≤ kuk1 ≤ µkuk2 , where µ =
max{β, 1}.
Proof. Since u(t) is concave and nondecreasing, together with u0 (∞) = 0, we have
kuk2 = u0 (0) and
Z
t+β
u(t)
1 t 0
u (s)ds + βu0 (0) ≤
≤
u0 (0) ≤ µkuk2 .
1+t
1+t 0
1+t
On the other hand,
u(0)
|u(t)|
≥
= βu0 (0) = βkuk2 .
1+0
0≤t<+∞ 1 + t
kuk1 =
sup
So we can obtain the desired result.
Lemma 3.4. Let u ∈ K. Then mint∈[1/a,a] u(t) ≥ δ(t)kuk, where a > 1, δ(t) =
1
2 min{λ(t), β/a},
(
σ, t ≥ σ,
λ(t) =
t, t ≤ σ,
and σ = inf{ξ ∈ [0, +∞) : supt∈[0,+∞)
|u(t)|
1+t
=
u(ξ)
1+ξ }.
Proof. From [7, Lemma 3.2], we know that
min u(t) ≥ λ(t)kuk1 .
1
,a]
t∈[ a
On the other hand, since u ∈ K, we have
1
β
β
min
u(t) = u( ) ≥ u0 (0) = kuk2 .
1
a
a
a
t∈[ a ,a]
So (3.9) and (3.10) imply that the result of Lemma 3.4 holds.
(3.9)
(3.10)
Remark 3.5. From the definition of λ(t), we know that 0 < δ(t) < 1, for t ∈ (0, 1).
Define T : K → E by
Z
(T u)(t) =
+∞
(t − s)ϕ
−1
s
Z
t
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
0
+∞
Z
sϕ−1
+
s
Z
0
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
0
Z
+∞
+β
ϕ−1
Z
0
s
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds.
0
Obviously, we have (T u)(t) ≥ 0, for t ∈ (0, +∞), and
Z +∞
Z s
0
−1
(T u) (t) =
ϕ
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds ≥ 0.
t
0
Furthermore,
00
(T u) (t) = −ϕ
−1
Z
0
t
a(s)f (s, u(s), u0 (s))ds ≤ 0,
(3.11)
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S. LIANG, J. ZHANG
EJDE-2009/89
and (T u)(0) − β(T u)0 (0) = 0. This shows (T K) ⊂ K.
To obtain the complete continuity of T , the following lemma is needed.
Lemma 3.6 ([9]). Let W be a bounded subset of K. Then W is relatively compact in
0
E if { v(t)
1+t , v ∈ W } and {v (t), v ∈ W } are equicontinuous on any finite subinterval
of [0, +∞) and for any ε > 0, there exists T = T (ε) > 0 such that
v(t1 )
v(t2 ) < ε, |v 0 (t1 ) − v 0 (t2 )| < ε
−
1 + t1
1 + t2
uniformly with respect to v ∈ W as t1 , t2 ≥ T .
Lemma 3.7. Let (C1), (C2) hold. Then T : K → K is completely continuous.
Proof. Firstly, it is easy to check that T : K → K is well defined. Let un → u in
K, then from the definition of E, we can choose r0 such that supn∈N \{0} kun k < r0 .
Let Ar0 = sup{f (t, (1 + t)u, v), (t, u, v) ∈ [0, +∞) × [0, r0 ]2 } and we have
Z +∞
Z s
ϕ−1
a(τ )|f (τ, un , u0n ) − f (τ, u, u0 )|dτ ds
t
0
Z +∞
Z s
≤ 2ϕ−1 (Ar0 )
ϕ−1
a(τ )dτ ds.
0
0
Therefore, by the Lebesgue dominated convergence theorem,
Z +∞
Z s
0
0
−1
|(T un ) (t) − (T u) (t)| = ϕ
a(τ )(f (τ, un , u0n ) − f (τ, u, u0 ))dτ ds
t
0
Z +∞
Z s
≤
ϕ−1
a(τ )|f (τ, un , u0n ) − f (τ, u, u0 )|dτ ds
t
→ 0,
0
as n → +∞.
From Lemma 3.3, we have
kT un − T uk ≤ (1 + µ)kT un − T uk2 → 0,
as n → +∞.
Thus T is continuous.
Let Ω be any bounded subset of K. Then there exists r > 0 such that kuk ≤ r
for all u ∈ Ω and we have
Z +∞
Z s
−1
kT uk2 =
ϕ
a(τ )f (τ, u, u0 )dτ ds
0
0
Z +∞
Z s
−1
−1
≤ ϕ (Ar )
ϕ
a(τ )dτ ds < +∞.
0
0
From Lemma 3.3, we have kT uk ≤ (1 + µ)kT uk2 < +∞. So T Ω is bounded.
Moreover for any S ∈ (0, +∞) and t1 , t2 ∈ [0, S]. Without loss of generality, let
t1 ≥ t2 . Then we have
Z
Z
(T u)(t1 ) (T u)(t2 ) h +∞ −1 s
≤
−
sϕ
a(τ )f (τ, u, u0 )dτ ds
1 + t1
1 + t2
0
0
Z +∞
i 1
Z s
1 +β
ϕ−1
a(τ )f (τ, u, u0 )dτ ds −
1 + t1
1 + t2
0
0
Z +∞
Z s
t
t2 1
+
ϕ−1
a(τ )f (τ, u, u0 )dτ ds
−
1 + t1
1 + t2
t2
0
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EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS
Z
+∞
t1
Z
s
1
1 −
1 + t1
1 + t2
0
Z t1
Z s
a(τ )f (τ, u, u0 )dτ ds
sϕ−1
sϕ−1
+
7
a(τ )f (τ, u, u0 )dτ ds
1
1 + t2 t2
0
→ 0, uniformly as t1 → t2
+
and
|(T u)0 (t2 ) − (T u)0 (t1 )| =
Z
t2
ϕ−1
s
Z
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds → 0,
0
t1
uniformly as t1 → t2 . We obtain that T Ω is equicontinuous on any finite subinterval
of [0, +∞).
For any u ∈ Ω, we have
Z
Z s
(T u)(t) 1 n +∞
= lim
lim (t − s)ϕ−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
t→+∞
t→+∞ 1 + t
1+t
t
0
Z +∞
Z s
+
sϕ−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
0
0
Z +∞
Z s
o
+β
ϕ−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
0
0
Z +∞
Z s
≤ lim ϕ−1 (Ar )
ϕ−1
a(τ )dτ ds
t→+∞
t
0
=0
and
lim |(T u)0 (t)| = lim ϕ−1 (Ar )
t→+∞
t→+∞
Z
+∞
ϕ−1
Z
t
s
a(τ )dτ ds = 0.
0
So T Ω is equiconvergent at infinity. By Lemma 3.6, T Ω is relatively compact.
Therefore we know that T is compact. So T : K → K is completely continuous.
The proof is complete.
Let k > 1 be a fixed constant and choose a = k. Then define
R
Rk
s
1
−1
δ
β
ϕ
a(τ
)dτ
ds
k
1/k
1/k
1
R
,
γ = δ( )
R
k (1 + µ) +∞ ϕ−1 s a(τ )dτ ds
0
0
R
R
k
s
δ k1 β 1/k ϕ−1 1/k a(τ )dτ ds
R
,
γ1 =
R +∞
s
(1 + µ) 0 ϕ−1 0 a(τ )dτ ds
Kρ = {u ∈ K : kuk1 ≤ ρ},
u(t) < γρ} = {u ∈ K : γkuk1 ≤ min
u(t) < γρ}.
Ωρ = {u ∈ K : min
1
1
t∈[ k ,k]
t∈[ k ,k]
Lemma 3.8 ([6]). The set Ωρ has the following properties:
(a) Ωρ is open relative to K.
(b) Kγρ ⊂ Ωρ ⊂ Kρ .
(c) u ∈ ∂Ωρ if and only if mint∈[ k1 ,k] u(t) = γρ.
(d) u ∈ ∂Ωρ , then γρ ≤ u(t) ≤ ρ for t ∈ [ k1 , k].
8
S. LIANG, J. ZHANG
EJDE-2009/89
Now, we introduce the following notation. Let
f (t, (1 + t)u, v)
ρ
fγρ
= min
: t ∈ [1/k, k], u ∈ [γρ, ρ], v ∈ [0, ρ/β] ,
ϕ(ρ)
f (t, (1 + t)u, v)
f0ρ = sup
: t ∈ [0, +∞), u ∈ [0, ρ], v ∈ [0, ρ/β] ,
ϕ(ρ)
f (t, (1 + t)u, v)
: t ∈ [0, +∞) ,
f α = lim sup
u→α
ϕ(u)
f (t, (1 + t)u, v)
fα = lim min
: t ∈ [1/k, k]
(α := ∞ or 0+ ),
u→α
ϕ(u)
Z +∞
Z s
1
a(τ )dτ ds,
= (1 + µ)
ϕ−1
m
0
0
Z k
Z s
1
1
ϕ−1
a(τ )dτ ds.
=δ
β
M
k
1/k
1/k
Remark 3.9. It is easy to see that 0 < m, M < ∞ and M γ = M γ1 δ( k1 ) =
δ( k1 )m < m.
Lemma 3.10. If f satisfies the condition
f0ρ ≤ ϕ(m)
and u 6= T u f or u ∈ ∂Kρ ,
(3.12)
then ik (T, Kρ ) = 1.
Proof. By (3.11) and (3.12), for u(t) ∈ ∂Kρ , we have kuk1 = sup0≤t<+∞ |u(t)|
1+t = ρ.
Moreover, Lemma 3.3 implies that kuk2 ≤ β1 kuk1 = β1 ρ, Therefore, from definition
of f0ρ we have
f (t, u, u0 ) ≤ ϕ(ρ)ϕ(m) = ϕ(ρm).
Therefore,
kT uk = kT uk1 + kT uk2 ≤ (1 + µ)kT uk2
Z +∞
Z s
= (1 + µ)
ϕ−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds
0
0
Z +∞
Z s
≤ ρm(1 + µ)
ϕ−1
a(τ )dτ ds
0
0
≤ ρ = kuk1 ≤ kuk.
This implies that kT uk ≤ kuk for u(t) ∈ ∂Kρ . By Theorem 2.4 (1) we have
ik (T, Kρ ) = 1.
Lemma 3.11. If f satisfies the conditions
ρ
fγρ
≥ ϕ(M γ)
and u 6= T u f or u ∈ ∂Ωρ ,
(3.13)
then ik (T, Ωρ ) = 0.
Proof. Let e(t) ≡ 1 for t ∈ [0, +∞). Then e ∈ ∂K1 , and we claim that
u 6= T u + λe,
u ∈ ∂Ωρ ,
λ > 0.
If not, there exist u0 ∈ ∂Ωρ and λ0 > 0 such that u0 = T u0 + λ0 e. By (3.11) and
(3.13) we have
u0 = T u0 (t) + λ0 e
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EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS
9
1
kT u0 k1 + λ0
k
|T u0 |
1
sup
+ λ0
=δ
k t∈[0,+∞) 1 + t
1 (T u)(0)
≥δ
+ λ0
k 1+0
Z +∞
Z s
1
=δ
β
ϕ−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds + λ0
k
0
0
Z k
Z s
1
−1
a(τ )f (τ, u(τ ), u0 (τ ))dτ ds + λ0
ϕ
β
≥δ
k
1/k
1/k
Z k
Z s
1
≥δ
a(τ )dτ ds + λ0
ϕ−1
M γρβ
k
1/k
1/k
≥ γρ + λ0 .
≥δ
This implies that γρ ≥ γρ + λ0 which is a contradiction. Hence by Theorem 2.4
(2), we have ik (T, Ωρ ) = 0.
4. Main results
The main results in this articles are the following.
Theorem 4.1. Assume that one of the following conditions holds:
(C3) There exist ρ1 , ρ2 , ρ3 ∈ (0, ∞) with ρ1 < γρ2 and ρ2 < ρ3 such that
f0ρ1 ≤ ϕ(m),
ρ2
fγρ
≥ ϕ(M γ),
2
u 6= T u for u ∈ ∂Ωρ2 and f0ρ3 ≤ ϕ(m).
(C4) There exist ρ1 , ρ2 , ρ3 ∈ (0, ∞) with ρ1 < ρ2 < γρ3 such that
ρ1
fγρ
≥ ϕ(M γ),
1
f0ρ2 ≤ ϕ(m),
ρ3
u 6= T u for u ∈ ∂Kρ2 and fγρ
≥ ϕ(M γ).
3
Then (1.1) has two positive solutions in K. Moreover if in (C3) f0ρ1 ≤ ϕ(m) is
replaced by f0ρ1 < ϕ(m), then (1.1) has a third positive solution u3 ∈ Kρ1 .
Proof. The proof is similar to the one in [6, Theorem 2.10]. We omit it here.
As a special case of Theorem 4.1 we obtain the following result.
Corollary 4.2. If there exists ρ > 0 such that one of the following conditions holds:
ρ
(C5) 0 ≤ f 0 < ϕ(m), fγρ
≥ ϕ(M γ), u 6= T u for u ∈ ∂Ωρ and 0 ≤ f ∞ < ϕ(m),
ρ
(C6) ϕ(M ) < f0 ≤ ∞, f0 ≥ ϕ(m), u 6= T u for u ∈ ∂Kρ and ϕ(M ) < f∞ ≤ ∞.
Then (1.1) has two positive solutions in K.
Proof. We show that (C5) implies (C3). It is easy to verify that 0 ≤ f 0 < ϕ(m)
implies that there exist ρ1 ∈ (0, γρ) such that f0ρ1 < ϕ(m). Let a ∈ (f ∞ , ϕ(m)).
Then there exists r > ρ such that supt∈[0,+∞) f (t, (1+t)u, v) ≤ aϕ(u) for u ∈ [r, ∞)
since 0 ≤ f 0 < ϕ(m). Let
r
β = max
sup f (t, (1 + t)u, v) : 0 ≤ u ≤ r, 0 ≤ v ≤
β
t∈[0,+∞)
and
ρ3 > ϕ−1
β
.
ϕ(m) − a
10
S. LIANG, J. ZHANG
EJDE-2009/89
Then
sup
f (t, (1 + t)u, v) ≤ aϕ(u) + β ≤ aϕ(ρ3 ) + β < ϕ(m)ϕ(ρ3 )
t∈[0,+∞)
for u ∈ [0, ρ3 ]. This implies that f0ρ3 < ϕ(m) and (C3) holds. Similarly, (C6)
implies (C4).
By an argument similar to that of Theorem 4.1 we obtain the following result.
Theorem 4.3. Assume that one of the following conditions holds:
ρ2
(C7) There exist ρ1 , ρ2 ∈ (0, ∞) with ρ1 < γρ2 such that f0ρ1 ≤ ϕ(m) and fγρ
≥
2
ϕ(M γ).
ρ1
(C8) There exist ρ1 , ρ2 ∈ (0, ∞) with ρ1 < ρ2 such that fγρ
≥ ϕ(M γ) and
1
ρ2
f0 ≤ ϕ(m).
Then (1.1) has a positive solution in K.
As a special case of Theorem 4.3 we obtain the following result.
Corollary 4.4. If there exists ρ > 0 such that one of the following conditions holds:
ρ
≥ ϕ(M γ), u 6= T u for u ∈ ∂Ωρ and 0 ≤ f ∞ < ϕ(m),
(C5) 0 ≤ f 0 < ϕ(m), fγρ
ρ
(C6) ϕ(M ) < f0 ≤ ∞, f0 ≥ ϕ(m), u 6= T u for u ∈ ∂Kρ and ϕ(M ) < f∞ ≤ ∞.
Then (1.1) has two positive solutions in K.
5. Examples
As an example we mention the boundary-value problem
(ϕ(−u00 )(t))0 = a(t)f (t, u, u0 ),
0 < t < +∞,
0
u(0) − u (0) = 0,
0
u (∞) = 0,
where
(
ϕ(u) =
and
u (0) = 0,
u3
1+u2 ,
2
u 9
1+t 2 9
1+t
We choose k = 2, β = 1, δ(t) = t/2 and so
Z +∞
Z
Z s
−1
ϕ
a(τ )dτ ds = 4,
0
0
u ≤ 0,
u > 0,
u
(
10−5 | sin t| +
f (t, u, v) =
10−5 | sin t| +
(5.1)
00
+
+
1
v
100 1000 ,
1
v
100 1000 ,
2
ϕ
1
2
(5.2)
−1
Z
s
u ≤ 2,
u ≥ 2.
a(τ )dτ ds = 2.
1
2
It is easy to see by calculating that µ = 1, γ = 1/64, γ1 = 1/16 and
Z +∞
Z s
1
= (1 + µ)
ϕ−1
a(τ )dτ ds = 8,
m
0
0
Z 2
Z s
1
1
1
=δ
β
ϕ−1
a(τ )dτ ds = .
1
1
M
2
2
2
2
Thus m = 1/8, M = 2 and let ρ1 = 1/2, ρ2 = 128, ρ3 = 320. After some simple
calculation we have
1
1
1
f (t, (1 + t)u, u0 ) ≤ 10−5 +
+ × 10−5 <
= ϕ(mρ1 ) = ϕ(m)ϕ(ρ1 ),
512 2
256
EJDE-2009/89
EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS
11
for all (t, u, u0 ) ∈ [0, +∞) × [0, 1/2] × [0, 12 ]. Therefore, f0ρ1 < ϕ(m). On the other
hand,
f (t, (1 + t)u, u0 ) ≥ 29 = 512 > 16 = ϕ(M γρ2 ) = ϕ(M )ϕ(γρ2 ),
ρ2
for all (t, u, u0 ) ∈ [1/2, 2] × [2, 128] × [0, 128]. We have fγρ
> ϕ(M γ). At last
2
f (t, (1 + t)u, u0 ) ≤ 10−5 + 29 + 320 × 10−5 < 513 < 1600 = ϕ(mρ3 ) = ϕ(m)ϕ(ρ3 ),
for all (t, u, u0 ) ∈ [0, +∞) × [0, 320] × [0, 320]. Thus we have f0ρ3 < ϕ(m). Then the
condition (C3) in Theorem 4.1 is satisfied. So boundary-value problem (5.1) has at
least three positive solutions in K.
Remark 5.1. From (5.2), we can see that ϕ is not odd, therefore the boundaryvalue problem with p-Laplacian operator [10, 12, 13, 14] do not apply to (5.2). So
we generalize a p-Laplace operator for some p > 1 and the function ϕ which we
defined above is more comprehensive and general than p-Laplace operator.
Acknowledgments. The authors of this paper wish to thanks the referee for valuable suggestions regarding the original manuscript.
This project is supported by grant NSFC(10871096), Foundation of Major Project
of Science and Technology of Chinese Education Ministry, SRFDP of Higher Education, the project of Nanjing Normal University and Project of Graduate Education
Innovation of Jiangsu Province (181200000214).
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Sihua Liang
Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, 210097, Jiangsu, China
College of Mathematics, Changchun Normal University, Changchun 130032, Jilin, China
E-mail address: liangsihua@163.com
Jihui Zhang
Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, 210097, Jiangsu, China
E-mail address: jihuiz@jlonline.com