Electronic Journal of Differential Equations, Vol. 2006(2006), No. 22, pp. 1–9.
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 OF p-LAPLACIAN
FUNCTIONAL DIFFERENTIAL EQUATIONS
CHANG-XIU SONG
Abstract. In this paper, the author studies the boundary value problems
of p-Laplacian functional differential equation. Sufficient conditions for the
existence of positive solutions are established by using a fixed point theorem
in cones.
1. Introduction
For p-Laplace equations there are have many results published, but most of
them are about ordinary differential equations; see for example [2, 5, 10, 11, 13]
and references therein. As pointed out in [3, 12], the study of boundary value
problems (BVP) of functional differential equations (FDE) is of significance since
it arises and has applications in variational problems in control theory and other
areas of applied mathematics. In this paper, we shall investigate the existence of
positive solutions for p-Laplacian problem
(φp (y 0 ))0 + r(t)f (y t ) = 0, 0 < t < 1,
αy(t) − βy 0 (t) = ξ(t), −τ ≤ t ≤ 0,
(1.1)
0
γy(t) + δy (t) = η(t), 1 ≤ t ≤ 1 + a,
where y t (θ) = y(t + θ), θ ∈ [−τ, a], τ ≥ 0, a ≥ 0 are constants satisfying 0 ≤ τ + a <
1; φp (u) is the p-Laplacian operator, i.e., φp (u) = |u|p−2 u, p > 1, (φp )−1 (u) =
φq (u), p1 + 1q = 1.
For the situation that τ = a = 0, BVP (1.1) becomes the two-point BVP and
has been investigated in [6, 7, 8]. Furthermore, for τ = a = 0 and p = 2, BVP (1.1)
has been studied in [1, 9]. In this paper, we shall generalize the above-mentioned
equations. Our results include four subcases: τ = a = 0 (two point BVP); τ > 0,
a = 0 (BVP of retarded FDE); τ = 0, a > 0 (BVP of advanced FDE); τ > 0, a > 0
(BVP of mixed FDE).
2000 Mathematics Subject Classification. 34K10.
Key words and phrases. p-Laplacian boundary value problem; functional differential equation;
positive solution; fixed point theorem in cones.
c
2006
Texas State University - San Marcos.
Submitted September 19, 2005. Published February 17, 2006.
Supported by grant 10571064 from NNSF of China, and by grant 011471
from NSF of Guangdong.
1
2
C.-X. SONG
EJDE-2006/22
Let C = C([−τ, a], R) be the Banach space of continuous functions from [−τ, a]
into R with the norm kϕkC = sup−τ ≤θ≤a |ϕ(θ)| and
C + = {ϕ ∈ C : ϕ(θ) ≥ 0, θ ∈ [−τ, a]}.
Define
E = {t ∈ [0, 1] : 0 ≤ t + θ ≤ 1, −τ ≤ θ ≤ a} = [τ, 1 − a].
Noting that from the assumption that 0 ≤ τ + a < 1, we conclude meas E 6= 0.
We shall assume the following conditions:
(H1) f (ϕ) is a nonnegative continuous function defined on C + .
(H2) r(t) is a nonnegative measurable function defined on [0, 1], and satisfies
Z 1
Z
Z t
r(u)du] < +∞.
r(u)du]dt < φq [
0<
φq [
E
τ
0
(H3) α, β, γ, δ ≥ 0, ρ := γβ + αγ + αδ > 0.
(H4) ξ(t) and η(t) are continuous functions defined, respectively, on [−τ, 0] and
R0
[1, b], where b = 1 + a, η(1) = 0; ξ(t) ≥ 0 if β = 0; t e−(α/β)s ξ(s)ds ≥ 0,
Rt
ξ(0) ≥ 0 if β > 0; η(t) ≥ 0 if δ = 0; 1 e(γ/δ)s η(s)ds ≥ 0 if δ > 0.
Lemma 1.1. ([4]) Assume that X is a Banach space and K ⊂ X is a cone
in T
X; Ω1 , Ω2 are open subsets of X, and 0 ∈ Ω1 ⊂ Ω2 . Furthermore, let Φ :
K (Ω2 \ Ω1 ) → K be a completely continuous operator satisfying one of the following conditions:
T
T
(i) kΦ(x)k ≤ kxk, for all x ∈ K T ∂Ω1 ; kΦ(x)k ≥ kxk, for all x ∈ K T ∂Ω2 ;
(ii) kΦ(x)k ≤ kxk, for all x ∈ K ∂Ω2 ; kΦ(x)k ≥ kxk, for all x ∈ K ∂Ω1 .
T
Then there is a fixed point of Φ in K (Ω2 \ Ω1 ).
2. Main Results
Firstly, we give the definitions of solution and positive solution.
Definition 2.1. We say a function y(t) is a solution (1.1) if:
(1) y(t) is continuous on [0, b].
(2) y(t) = y(−τ ; t) for t ∈ [−τ, 0], where y(−τ ; t) : [−τ, 0] → [0, +∞) is defined
as
(
R0
e(α/β)t ( β1 t e−(α/β)s ξ(s)ds + y(0)), β > 0,
y(−τ ; t) = 1
β = 0,
α ξ(t),
and y(−τ ; t) satisfies αy(−τ ; 0) = αy(0) = ξ(0) if β > 0.
(3) y(t) = y(b; t) for t ∈ [1, b], where y(b; t) : [1, b] → [0, +∞) is defined as
(
Rt
e−(γ/δ)t ( 1δ 1 e(γ/δ)s η(s)ds + e(γ/δ) y(1)), δ > 0,
y(b; t) = 1
δ = 0.
γ η(t),
(4) (φp (y 0 ))0 = −r(t)f (y t ) for t ∈ (0, 1) almost everywhere.
Furthermore, a solution y(t) of (1.1) is called as a positive solution if y(t) > 0 for
t ∈ (0, 1).
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EXISTENCE OF POSITIVE SOLUTIONS
3
In what follows, we shall show the results in this paper only for the case β > 0,
δ = 0, since the other situations could be discussed similarly. Suppose that y(t) is
a solution of (1.1), then it can be written as

R
(α/β)t 1 0 −(α/β)s
(β t e
ξ(s)ds + y(0)), −τ ≤ t ≤ 0,

eR
Rs
1
u
y(t) =
(2.1)
φq [ 0 r(u)f (y )du]ds,
0 ≤ t ≤ 1,
t

1
1 ≤ t ≤ b.
γ η(t),
Suppose that x0 (t) is the solution of (1.1) with f ≡ 0, then it can be expressed as

R0

 β1 e(α/β)t t e−(α/β)s ξ(s)ds, −τ ≤ t ≤ 0,
x0 (t) =
0,
0 ≤ t ≤ 1,

 1 η(t),
1 ≤ t ≤ b.
γ
If y(t) is a solution of BVP (1.1) and x(t) = y(t) − x0 (t), noting that x(t) = y(t)
for 0 ≤ t ≤ 1, then we have from (2.1) that

(α/β)t

x(0),
−τ ≤ t ≤ 0,
eR
Rs
1
u
u
x(t) =
(2.2)
φq [ 0 r(u)f (x + x0 )du]ds, 0 ≤ t ≤ 1,
t


0,
1 ≤ t ≤ b.
}) such that
By (H2), we can choose a σ ∈ (0, min{ 41 , 1−a−τ
2
Z
Z t
T :=
φq
r(u)du dt > 0,
Eσ
σ+τ
where Eσ := {t ∈ E; σ ≤ t + θ ≤ 1 − σ for − τ ≤ θ ≤ a} = [σ + τ, 1 − σ − a] ⊂ E.
Note that when t ∈ Eσ , we have xt0 = 0 ∈ C.
Let K be a cone in the Banach space X = C[−τ, b] defined by
K = {x ∈ C[−τ, b] : x(t) ≥ g(t)kxk, t ∈ [−τ, b]},
where kxk := sup{|x(t)| : −τ ≤ t ≤ b} and

(α/β)t

, −τ ≤ t ≤ 0,
e
g(t) := 1 − t,
0 ≤ t ≤ 1,


0,
1 ≤ t ≤ b.
Define Φ : K → C[−τ, b] as

R
Rs
(α/β)t 1

φ [ r(u)f (xu + xu0 )du]ds, −τ ≤ t ≤ 0,
eR
0 q 0
R
1
s
(Φx)(t) =
φq [ 0 r(u)f (xu + xu0 )du]ds,
0 ≤ t ≤ 1,
t


0,
1 ≤ t ≤ b.
(2.3)
Under assumptions (H1)-(H4), BVP (1.1) has a solution if and only if Φx(t) = x(t).
We define kxk[0,1] = sup{|x(t)| : 0 ≤ t ≤ 1}, then we have kΦxk = kΦxk[0,1] . It
follows from (2.3) that
Z 1
Z s
Z 1
u
u
kΦxk[0,1] =
φq [
r(u)f (x + x0 )du]ds ≤ φq [
r(u)f (xu + xu0 )du]. (2.4)
0
0
0
In the following, we express Tα = {x ∈ C[0, b] : kxk < α}.
Lemma 2.2. With the above notation, Φ(K) ⊂ K.
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C.-X. SONG
EJDE-2006/22
Proof. For −τ ≤ t ≤ 0, one has 0 ≤ (Φx)(t) ≤ (Φx)(0). Thus we get kΦxk =
kΦxk[0,1] and (Φx)(t) ≥ e(α/β)t kΦxk.
For 0 ≤ t ≤ 1, x ∈ K, we obtain from (2.3) and (2.4) that (Φx)(t) ≥ 0, (Φx)(1) =
0, and
Z 1
Z s
kΦxk = (Φx)(0) =
φq [
r(u)f (xu + xu0 )du]ds.
0
0
Let U (t) = (Φx)(t)−(1−t)kΦxk, then U (0) = U (1) = 0, U 00 (t) ≤ 0, for all t ∈ [0, 1].
So U (t) ≥ 0, for all t ∈ [0, 1], i.e., (Φx)(t) ≥ (1 − t)kΦxk for t ∈ [0, 1]. For 1 ≤ t ≤ b,
x ∈ K, one has (Φx)(t) = 0 = g(t). Furthermore, for −τ ≤ t ≤ b, x ∈ K, we have
that (Φx)(t) ≥ g(t)kΦxk, that is Φ(K) ⊂ K.
Lemma 2.3. The function Φ : K → K is completely continuous.
Proof. We can obtain the continuity of Φ from the continuity of f . In fact, suppose
that xn , x ∈ K and kxn − xk → 0 as n → ∞, then we get
kxun − xu k =
sup |xn (u + θ) − x(u + θ)| → 0, u ∈ [0, 1].
−τ ≤θ≤a
Thus, for t ∈ [−τ, b] we have from (2.3) that
Z
|(Φxn )(t) − (Φx)(t)| ≤ max |f q−1 (xun + xu0 ) − f q−1 (xu + xu0 )|φq [
0≤u≤1
1
r(u)du],
0
This implies that kΦxn − Φxk → 0 as n → ∞.
Now let A ⊂ K be a bounded subset of K and M > 0 is the constant such that
kxk ≤ M for x ∈ A. Define a set S as
S = {ϕ ∈ C + ; kϕkC ≤ M }.
Then, we have
kΦxk ≤ max f q−1 (ϕ + max xu0 )
ϕ∈S
u∈[0,1]
Z
1
Z
φq [
0
s
r(u)du]ds < ∞,
0
T
which implies the boundedness of Φ(A). It is easy to see that Φx ∈ C 1 [−τ, 1] C[−τ, b].
Furthermore, we have for −τ ≤ t ≤ 0,
Z
Z s
α (α/β)t 1
0
φq [
r(u)f (xu + xu0 )du]ds
(Φx) (t) = e
β
0
0
Z 1
α
α
r(u)du] =: L1 ;
≤ kΦxk ≤ max f q−1 (ϕ + max xu0 )φq [
β
β ϕ∈S
u∈[0,1]
0
for 0 ≤ t ≤ 1,
Z t
0 ≤ |(Φx)0 (t)| = φq [
r(u)f (xu + xu0 )du]
0
Z
≤ φq [
1
r(u)f (xu + xu0 )du]
0
≤ max f
ϕ∈S
q−1
Z
(ϕ + max
u∈[0,1]
xu0 )φq [
1
r(u)du] =: L2 .
0
For 1 ≤ t ≤ b, we have that (Φx)0 (t) = 0. Thus, suppose that x ∈ A. ∀ ε > 0, let
δ = max{Lε 1 ,L2 } , for t1 , t2 ∈ [−τ, b], |t1 − t2 | < δ, one has
|(Φx)(t1 ) − (Φx)(t2 )| ≤ max{L1 , L2 }|t1 − t2 | < ε.
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EXISTENCE OF POSITIVE SOLUTIONS
That is to say that Φ : K → K is completely continuous.
5
Let
C ∗ = {ϕ ∈ C + : 0 < σkϕkC ≤ ϕ(θ), θ ∈ [−τ, a]};
Z 1
Z s
λ=[
φq [
r(u)du]ds]−1 ;
0
0
Z
Z s
1
µ = [σ
φq [
r(u)du]ds]−1 =
;
σT
Eσ
σ+τ
Z 1
r(u)du]]−1 .
ν = [2(1 + kx0 k)φq [
0
For the next theorem we set the condition
(H5)
f (ϕ)
f (ϕ)
p−1
p−1
lim
,
lim
.
p−1 > µ
p−1 < ν
ϕ∈C ∗ ,kϕkC ↓0 kϕk
kϕkC ↑∞ kϕk
C
C
Theorem 2.4. Under assumption (H5), BVP (1.1) has at least a positive solution.
Proof. From assumption (H5), there exists a ρ1 > 0 such that
f (ϕ) ≥ (µkϕkC )p−1 ,
ϕ ∈ C ∗,
kϕkC ≤ ρ1 .
u
For x ∈ K and kxk = ρ1 , we have kx kC ≤ ρ1 for u ∈ [0, 1]. Furthermore, we have
xu ∈ C ∗ for u ∈ Eσ and
kxu kC ≥ σkxk = σρ1 ,
u ∈ Eσ .
(2.5)
For u ∈ Eσ , we have xu0 = 0. Thus, we obtain
Z 1
Z s
kΦxk =
φq [
r(u)f (xu + xu0 )du]ds
0
0
Z
Z s
≥
φq [
r(u)f (xu )du]ds
Eσ
σ+τ
Z s
Z
φq [
≥µ
r(u)kxu kp−1
C du]ds
Eσ
σ+τ
Z
Z s
≥ µσρ1
φq [
r(u)du]ds
Eσ
(2.6)
σ+τ
= µσρ1 T
≥ ρ1 = kxk,
which implies that
kΦxk = kΦxk[0,1] ≥ kxk,
∀x∈K
where Ω1 = Tρ1 . On the other hand, since limkϕkC ↑∞
\
∂Ω1 ,
f (ϕ)
kϕkp−1
C
< ν p−1 , there exists
N > ρ1 , such that
f (ϕ) ≤ (νkϕkC )p−1 ,
ϕ ∈ C + , kϕkC > N.
Choose a positive constant ρ2 such that
ρ2 > 1 + max{f
q−1
Z
(ϕ) : 0 ≤ kϕkC ≤ N + kx0 k}φq [
0
1
r(u)du].
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C.-X. SONG
EJDE-2006/22
For x ∈ K, kxk = ρ2 , we have, from the facts: x0 (t) ≥ 0, x(t) ≥ 0 for t ∈ [−τ, b],
that for u ∈ [0, 1],
kxu + xu0 kC ≥ kxu kC > N, if kxu kC > N,
kxu + xu0 kC ≤ kxu kC + kxu0 kC ≤ N + kx0 k, if kxu kC ≤ N.
We have
Z
kΦxk =
1
Z
φq [
≤
r(u)f (xu + xu0 )du]ds
0
0
Z
s
1
Z
φq [
1
r(u)f (xu + xu0 )du]ds
0
0
Z 1
r(u)f (xu + xu0 )du]
=φq [
0
Z
Z
=φq [
r(u)f (xu + xu0 )du +
kxu k
0≤kxu k
C >N
u
≤ max{νkx +
xu0 kC , max{f q−1 (ϕ)
r(u)f (xu + xu0 )du]
C ≤N
Z
: 0 ≤ kϕkC ≤ N + kx0 k}}φq [
1
r(u)du]
0
Z
≤ max{νkx + x0 k, max{f q−1 (ϕ) : 0 ≤ kϕkC ≤ N + kx0 k}}φq [
1
r(u)du]
0
Z 1
1
1
≤ max{ kxk + , max{f q−1 (ϕ)φq [
r(u)du] : 0 ≤ kϕkC ≤ N + kx0 k}}
2
2
0
<kxk = ρ2 ,
which implies that
kΦxk = kΦxk[0,1] < kxk, ∀ x ∈ K
\
∂Ω2 ,
where Ω2 = Tρ2 . By the second part of Lemma 1.1, it follows that Φ has a fixed
T
point x ∈ K (Ω2 \ Ω1 ) such that
0 < ρ1 ≤ kxk = kxk[0,1] ≤ ρ2 .
T
Suppose that x(t) is the fixed point of Φ in K (Ω2 \ Ω1 ), then

(α/β)t

x(0),
−τ ≤ t ≤ 0,
eR
Rs
1
u
u
x(t) =
φ
[
r(u)f
(x
+
x
)du]ds,
0 ≤ t ≤ 1,
q
0
t
0


0,
1 ≤ t ≤ b.
Let y(t) = x(t) + x0 (t). By the facts 0 < ρ1 ≤ kxk = kxk[0,1] ≤ ρ2 , x(t) ∈ K and
x0 (t) ≥ 0, we conclude that y(t) is a positive solution of BVP (1.1).
In what follows, we shall consider the existence of multiple positive solutions for
BVP (1.1).
For the next theorem we have the following hypotheses:
(H6)
lim
ϕ∈C ∗ ,kϕkC ↓0
f (ϕ)
> µp−1 ;
kϕkp−1
C
lim
ϕ∈C ∗ ,kϕkC ↑∞
f (ϕ)
p−1
;
p−1 > µ
kϕkC
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EXISTENCE OF POSITIVE SOLUTIONS
7
(H7) There exists a p1 > 0 such that for all 0 ≤ kϕkC ≤ p1 + p0 , one has
f (ϕ) ≤ (λp1 )p−1 , where
Z
1 αβ t 0 − αβ s
1
p0 = max{ max
e
ξ(s)ds, max η(t)}.
e
−τ ≤t≤0 β
1≤t≤b
γ
t
Theorem 2.5. Under assumptions (H6)-(H7), BVP (1.1) has at least two positive
solutions y1 , y2 such that 0 < ky1 k[0,1] < p1 < ky2 k[0,1] .
Proof. (H6), there exists a r : 0 < r < p1 such that
kϕkC ≤ r, ϕ ∈ C ∗ ,
f (ϕ) ≥ (µkϕkC )p−1 ,
For x ∈ K, kxk = r, similar to (2.5) one has xu ∈ C ∗ and
r ≥ kxu kC ≥ σkxk = σr,
u ∈ Eσ .
Also we obtain an analogous inequality
kΦ(x)k ≥ r = kxk, which implies kΦxk =
T
kΦxk[0,1] ≥ kxk, for all x ∈ K ∂Tr .
On the other hand, from (H6) there exists a R > p1 such that
kϕkC ≥ σR, ϕ ∈ C ∗ .
f (ϕ) ≥ (µkϕkC )p−1 ,
For x ∈ K, kxk = R, we have
xu ∈ C ∗
and kxu kC ≥ σkxk = σR
for u ∈ Eσ .
Furthermore,
kΦ(x)k ≥ R = kxk, which implies that kΦxk ≥ kxk for ∀ x ∈
T
K ∂TR .
Now, by (H7 ), for all x ∈ ∂Tp1 , one has
Z 1
Z s
kΦxk =
φq [
r(u)f (xu + xu0 )du]ds
0
0
Z
≤λp1
0
1
Z
φq [
s
r(u)du]ds = p1 = kxk.
0
According
that Φ has two fixed points x1 , x2 such that
T
T to Lemma 1.1, it follows
x1 ∈ K T p1 \ Tr , x2 ∈ K T R \ Tp1 , that is 0 < kx1 k < p1 < kx2 k. Since
xi (i = 1, 2) ∈ K, we have xi (t) > 0, for all t ∈ (0, 1), i = 1, 2. Let y1 = x1 + x0 ,
y2 = x2 +x0 , then y1 , y2 are positive solutions of BVP (1.1) satisfying 0 < ky1 k[0,1] <
p1 < ky2 k[0,1] .
The following Corollaries are obvious.
Corollary 2.6. Under the conditions
f (ϕ)
lim
p−1 = +∞,
∗
ϕ∈C ,kϕkC ↓0 kϕk
C
f (ϕ)
p−1 = 0,
kϕkC ↑∞ kϕk
C
lim
BVP (1.1) has at least a positive solution.
Corollary 2.7. Assume the conditions:
f (ϕ)
lim
p−1 = +∞;
∗
ϕ∈C ,kϕkC ↓0 kϕk
C
lim
ϕ∈C ∗ ,kϕk
f (ϕ)
p−1 = +∞;
C ↑∞ kϕk
C
and that there exists a p1 > 0 such that for all 0 ≤ kϕkC ≤ p1 + p0 , one has
f (ϕ) ≤ (λp1 )p−1 , where
Z
1 αβ t 0 − αβ s
1
p0 = max{ max
e
e
ξ(s)ds, max η(t)}.
−τ ≤t≤0 β
1≤t≤b
γ
t
8
C.-X. SONG
EJDE-2006/22
Then BVP (1.1) has at least two positive solutions y1 , y2 such that 0 < ky1 k[0,1] <
p1 < ky2 k[0,1] .
3. Examples
Example 3.1. Consider the boundary-value problem
1
1
(|y 0 |p−2 y 0 )0 + r(t)y 2 (t − ) = 0, 0 < t < 1,
3
(3.1)
1
y(t) = ξ(t), − ≤ t ≤ 0, y(1) = 0.
3
where α = γ = 1, β = δ = 0, r > 0 is a constant, ξ(t) is continuous on [− 31 , 0],
1
ξ(t) > 0, τ = 13 , a = 0, E = [ 13 , 1], p > 32 , and f (ϕ) = ϕ 2 (− 31 ). If kϕkC → +∞ we
have
1
3−2p
kϕkC2
f (ϕ)
ϕ1/2 (−1/3)
2
→ 0.
=
≤
p−1
p−1
p−1 = kϕkC
kϕkC
kϕkC
kϕkC
f (ϕ)
kϕkp−1
C
That is to say that limkϕkC ↑∞
= 0 holds. On the other hand, suppose that
ϕ ∈ C ∗ , then ϕ(θ) ≥ σkϕkC , thus, if kϕkC → 0 we get
1
1
3−2P
σ 2 kϕkC2
1
ϕ1/2 (−1/3)
f (ϕ)
2
2
→ +∞.
=
≥
p−1
p−1
p−1 = σ kϕkC
kϕkC
kϕkC
kϕkC
That is to say that limϕ∈C ∗ ,kϕkC ↓0
f (ϕ)
kϕkp−1
C
= +∞ holds. According to Corollary
2.6, it follows that BVP (3.1) has at least a positive solution y(t).
Example 3.2. Consider the boundary-value problem
1
1
1
1
(|y 0 |p−2 y 0 )0 + r[y 9 (t − ) + y 3 (t − )] = 0, 0 < t < 1,
3
3
(3.2)
1
y(t) = ξ(t), − ≤ t ≤ 0, y(1) = 0.
3
where α = γ = 1, β = δ = 0, r is a positive constant, ξ(t) is continuous on [− 31 , 0],
7
4 1
1
ξ(t) > 0, m0 = max− 13 ≤t≤0 ξ(t) 6= 0, 10
9 < p = 6 < 3 , p + q = 1 and
1
1
f (ϕ) = ϕ1/9 (− ) + ϕ1/3 (− ).
3
3
Here, τ = 13 , a = 0, E = [ 13 , 1].
Suppose that ϕ ∈ C ∗ , then ϕ(θ) ≥ σkϕkC , thus, if kϕkC → 0 or kϕkC → +∞
we get
f (ϕ)
ϕ1/9 (−1/3) + ϕ1/3 (−1/3)
p−1 =
p−1
kϕkC
kϕkC
1
≥
1
kϕkp−1
C
1
10−9p
9
= σ 9 kϕkC
We deduce that
λ=
1
1
σ 9 kϕkC9 + σ 3 kϕkC3
hZ
0
1
Z
φq (
0
1
4−3p
+ σ 3 kϕkC 3
s
rdu)ds
i−1
=
→ +∞.
q
,
rq−1
EJDE-2006/22
EXISTENCE OF POSITIVE SOLUTIONS
9
then for all m > 0 and 0 ≤ kϕkC ≤ m + m0 , one has
2
1
1
1
0 ≤ f (ϕ) ≤ (m + m0 ) 9 + (m + m0 ) 3 = (m + m0 ) 9 (m1−p +
Define H(m) = (m + m0 )1/9 (m1−p +
(m+m0 )2/9
),
mp−1
lim H(m) = +∞,
m→0
(m + m0 ) 9
)mp−1 .
mp−1
then
lim H(m) = +∞.
m→+∞
(3.3)
Suppose that r and m0 satisfy
1
−1/6
(2m0 ) 9 (m0
−1/6
1/18
+ 22/9 m0
1/18
)<
q p−1
,
r
p−1
then H(m0 ) = (2m0 )1/9 (m0
+ 22/9 m0 ) < λp−1 = q r holds. By the continuity of H(m) and (3.3), we can find an m > 0 (for example m = m0 ) such that
f (ϕ) ≤ H(m)mp−1 < (λm)p−1 for 0 ≤ kϕkC ≤ m + m0 . By Corollary 2.7, we know
that BVP (3.2) has at least two positive solutions.
References
[1] G. Q. Chai. The existence of positive solutions singular boundary problems. Acta Math.
Scientia, 2001, 21A(4), 521-526.
[2] L. H. Erbe, H. Y. Wang. On the existence of positive solutions of ordinary differential equations. Proc. Amer. Math. Soc., 1994, 120(3), 743-748.
[3] L. H. Erbe, Q. K. Kong. Boundary value problems for singular second-order functional differential equations. J. Comput. Appl. Math., 1994, 53, 377-388.
[4] D. J. Guo. Nonlinear Functional Analysis (in Chinese). Jinan: Scientific Pressing House of
Shandong, 1985: 302-303.
[5] X. M. He, W. G. Ge. Existence of positive solutions for the one-dimensionl p-Laplacian
equations. Acta Mathematica Sinica, 2003,46(4), 805-810.
[6] B. W. Li. The existence of positive solutions of singular boundary value problems with pLaplacian. Acta Math. Scientia, 2003, 23A(3), 257-264.
[7] C. Z. Li, W. G. Ge. Positive solutions for p-Laplacian singular Sturm-Liouville boundary
value problems. Math. Appl., 2002,15(3), 13-17.
[8] B. Liu, J. S. Yu. Multiple positive solutions of singular boundary value problems with pLaplacian. Chin. Ann. Math., 2001, 22A(6), 721-728.
[9] R. Y. Ma. Positive solutions of singular second order boundary value problems. Acta Math.
Sinica, 1998, 41(6),1225-1230.
[10] R. Manasevich, F. Zanolin. Time-mapping and multiplicity of solutions for the onedimensional p-Laplacians. Nonlinear Analysis, 1993, 21(4), 269-291.
[11] J. K. Wang. The exisitence of positive solutions for the one dimensional p-Laplacian. Proc.
of Amer. Math .Soc., 1999,125(8), 2275-2283.
[12] P. X. Weng, D. Q. Jing. Existence of positive solutions for boundary value problem of secondorder FDE. Computers and Mathematics with Applications, 1999, 37, 1-9.
[13] F. H. Wong. Existence of positive solutions for m-Laplacian Bvps. Appl. Math. Letters, 1999,
12, 11-17.
Chang-Xiu Song
School of Mathematical Sciences, South China of Normal University, Guangzhou, 510631,
China;
School of Applied Mathematics, Guangdong University of Technology, Guangzhou,
510006, China
E-mail address: scx168@sohu.com