Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 352159, 23 pages
doi:10.1155/2012/352159
Research Article
Multiple Bounded Positive Solutions to
Integral Type BVPs for Singular Second Order
ODEs on the Whole Line
Yuji Liu
Department of Mathematics, Guangdong University of Business Studies, Guangzhou 510320, China
Correspondence should be addressed to Yuji Liu, liuyuji888@sohu.com
Received 3 June 2012; Accepted 6 August 2012
Academic Editor: Ziemowit Popowicz
Copyright q 2012 Yuji Liu. This is an open access article distributed under the Creative Commons
Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
This paper is concerned with the integral type boundary value problems of the second order
differential equations with one-dimensional p-Laplacian on the whole line. By constructing a
suitable Banach space and a operator equation, sufficient conditions to guarantee the existence of
at least three positive solutions of the BVPs are established. An example is presented to illustrate
the main results. The emphasis is put on the one-dimensional p-Laplacian term ρtΦx t
involved with the function ρ, which makes the solutions un-concave.
1. Introduction
The multipoint boundary-value problems for linear second order ordinary differential
equations ODEs for short was initiated by Il’in and Moiseev 1. Since then, more general
nonlinear multi-point boundary-value problems BVPs for short were studied by several
authors, see the text books 2–4 and the references cited therein.
Differential equations governed by nonlinear differential operators have been widely
studied. In this setting the most investigated operator is the classical one-dimensional pLaplacian, that is, Φp x |x|p−2 x with p > 1. This operator is involved in some models,
for example, in non-Newtonian fluid theory, diffusion of flows in porous media, nonlinear
elasticity, and theory of capillary surfaces. The related nonlinear differential equation has the
form
Φx f t, x, x ,
t ∈ −∞, ∞,
1.1
where Φx |x|p−2 x with p > 1 is a one dimensional p-Laplacian. For a comprehensive
bibliography on this subject, see, for example 5–9.
2
Abstract and Applied Analysis
In this paper, we consider the more generalized BVP for second order differential equation on the whole line with p-Laplacian coupled with the integral type BCs, that is the BVP
ρtΦ x t f t, xt, x t 0, t ∈ R,
∞
lim xt : x−∞ gsxsds,
t → −∞
−∞
∞
lim xt : x
∞ hsxsds,
t → ∞
1.2
−∞
where f : R3 → R is a nonnegative Caratheodory function, g, h : R → 0, ∞ satisfy
g, h ∈ L1 R, ρ ∈ CR, 0, ∞, the integrals in mentioned equations are meant in the sense of
Riemann-Stieljes, Φx |x|p−2 x with p > 1 is called a one dimensional p-Laplacian, whose
inverse function is denoted by Φ−1 .
The purpose is to establish sufficient conditions for the existence of at least three
positive solutions of BVP1.2. The result in this paper generalizes and improves some known
ones since the one-dimensional p-Laplacian term ρtΦx t involved with the function ρ,
which makes the solutions unconcave and there exists no paper concerned with the existence
of at least three positive solutions of this kind of integral BVPs on the whole lines. This paper
fills the gap.
The remainder of this paper is organized as follows: the main result Theorem 2.8 is
presented in Section 2, and the example to show the main result is given in Section 3.
2. Main Results
In this section, we first present some background definitions in Banach spaces and state an
important three fixed point theorem. Then the main results are given and proved.
Definition 2.1. Let X be a real Banach space. The nonempty convex closed subset P of X is
called a cone in X if ax ∈ P for all x ∈ P and a ≥ 0 and x ∈ X and −x ∈ X imply x 0.
Definition 2.2. A map ψ : P → 0, ∞ is a nonnegative continuous concave or convex
functional map provided ψ is nonnegative, continuous, and satisfies ψtx 1 − ty ≥
tψx 1 − tψy, or ψtx 1 − ty ≤ tψx 1 − tψy, for all x, y ∈ P and t ∈ 0, 1.
Definition 2.3. An operator T : X → X is completely continuous if it is continuous and maps
bounded sets into precompact sets.
Definition 2.4. Let a, b, c, d, h > 0 be positive constants, α, ψ be two nonnegative continuous
concave functionals on the cone P , γ, β, θ be three nonnegative continuous convex functionals
on the cone P . Define the convex sets as follows:
Pc {x ∈ P : x < c},
P γ, α; a, c x ∈ P : αx ≥ a, γx ≤ c ,
P γ, θ, α; a, b, c x ∈ P : αx ≥ a, θx ≤ b, γx ≤ c ,
Q γ, β; , d, c x ∈ P : βx ≤ d, γx ≤ c ,
Q γ, β, ψ; h, d, c x ∈ P : ψx ≥ h, βx ≤ d, γx ≤ c .
2.1
Abstract and Applied Analysis
3
Lemma 2.5 see 10. Let X be a real Banach space, P be a cone in X, α, ψ be two nonnegative
continuous concave functionals on the cone P, γ, β, θ be three nonnegative continuous convex
functionals on the cone P . Assume that there exists a constant M > 0 such that
αx ≤ βx,
x ≤ Mγx,
∀ x ∈ P.
2.2
Furthermore, suppose that h, d, a, b, c > 0 are constants with d < a. Let T : Pc → Pc be a completely
continuous operator. If
C1 {x ∈ P γ, θ, α; a, b, c | αx > a} /
∅ and αT x > a for every x ∈ P γ, θ, α; a, b, c;
C2 {x ∈ Qγ, θ, ψ; h, d, c | βx < d} /
∅ and βT x < d for every x ∈ Qγ, θ, ψ; h, d, c;
C3 αT x > a for x ∈ P γ, α; a, c with θT x > b;
C4 βT x < d for each x ∈ Qγ, β; d, c with ψT x < h,
then T has at least three fixed points x1 , x2 and x3 such that βx1 < d, αx2 > a, βx3 >
d, αx3 < a.
Let us list the assumptions
H1 g, h : R → 0, ∞ satisfy
∞
−∞
gsds < 1,
∞
−∞
hsds < 1 and
∞
∞
gt 1 −
hsds − ht 1 −
gsds ≥ 0,
−∞
H2 ρ ∈ CR, ρt > 0 for t ∈ R with
−∞
0
−∞
Φ−1 1/ρtdt ∞
0
t ∈ R.
2.3
Φ−1 1/ρtdt < ∞.
H3 ft, c, 0 /
≡ 0 on any finite subinterval of R for each c ∈ R, f : R3 → R is a
Carathédory function, that is,
i t → ft, x, 1/Φ−1 ρty is measurable for any x, y ∈ R2 ,
ii x, y → ft, x, 1/Φ−1 ρty is continuous for a.e. t ∈ R,
iii for each r > 0, there exists nonnegative function φr ∈ L1 R such that max{|u|, |v|} ≤
r implies
1
v ≤ φr t,
f t, u, −1 Φ ρt
a.e. t ∈ R.
2.4
Choose
⎫
⎧
and there exist the limits⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
lim xt,
⎪
⎪
⎪
⎪
⎪
⎪
t → −∞
⎬
⎨
1
x
∈
C
xt,
:
lim
R
X
.
t → ∞
⎪
⎪
−1
⎪
⎪
⎪
⎪
lim
Φ
ρt
x
t
⎪
⎪
⎪
⎪
t → −∞
⎪
⎪
⎪
⎪
⎩
and lim Φ−1 ρt x t ⎭
t → ∞
2.5
4
Abstract and Applied Analysis
For x ∈ X, define the norm of x by
−1
x max sup|xt|, sup Φ
t∈R
2.6
ρt x t .
t∈R
One can prove that X is a Banach space with the norm ||x|| for x ∈ X.
Let x ∈ X. Consider the following auxiliary BVP
ρtΦ y t f t, xt, x t 0,
∞
y−∞ gsysds,
t ∈ R,
2.7
−∞
y
∞ ∞
−∞
hsysds.
Lemma 2.6. Suppose that (H1)–(H3) hold. If y ∈ C1 R such that ρΦy ∈ L1 R is a solution
of BVP2.7, then
i y is bounded and nonnegative on R;
t
∞
ii yt is concave with respect to τ −∞ Φ−1 1/ρsds/ −∞ Φ−1 1/ρsds;
−k
iii for k > 0, it holds that mint∈−k,k yt ≥ μ supt∈R yt with μ −∞ Φ−1 1/ρsds/2
∞ −1
Φ 1/ρsds;
−∞
∞
iv there exists a unique constant Ax ∈ − −∞ fs, xs, x sds, 0 such that
∞ gt 1 − ∞ hsds − ht 1 − ∞ gsds
−∞
−∞
∞
1 − −∞ gsds
−∞
t
∞
1
×
Φ−1 Ax Φ−1
f u, xu, x u du ds dt
ρs
−∞
s
∞
∞
1
Φ−1 Ax Φ−1
f u, xu, x u du ds 0.
ρs
−∞
s
2.8
Proof. Since x ∈ X, we get
−1
r max sup|xt|, supΦ
t∈R
ρt x t
< ∞.
2.9
t∈R
Then there exists a nonnegative function φr ∈ L1 −∞, ∞ such that
0 ≤ f t, xt, x t f
1
−1
t, xt, −1 Φ ρt x t ≤ φr t,
Φ ρt
t ∈ R.
2.10
Abstract and Applied Analysis
5
Then
∞
−∞
f s, xs, x s ds is convergent.
2.11
i We know there exist the limits limt → −∞ ρtΦy t and limt → ∞ ρtΦy t. We
claim that there exists τ0 ∈ R such that y τ0 0. In fact, if y t > 0 for all t ∈ R, we get
y−∞ < yt < y
∞ for all t ∈ R. It follows that
y−∞ ≥
∞
−∞
y
∞ ≤
gsdsy−∞,
∞
−∞
hsdsy
∞.
2.12
Then H1 implies y−∞ ≥ 0 ≥ y
∞, which contradicts to y t > 0 for all t ∈ R. Similarly
we can prove that y t < 0 does not hold. Then there exists τ0 ∈ R such that y τ0 0.
Since ρtΦy t −ft, xt, x t ≤ 0, we know that ρtΦy t is decreasing on
R. Then ρtΦy t ≥ 0 for all t ≤ τ0 and ρtΦy t ≤ 0 for all t ≥ τ0 . Hence
y is increasing for t ∈ −∞, τ0 and decreasing for t ∈ τ0 , ∞.
2.13
⎧
t
1
⎪
−1
⎪
⎪
f u, xu, x u du : −Gt,
⎨−Φ
ρt τ0
y t τ0
⎪
1
⎪ −1
⎪
⎩Φ
f u, xu, x u du : Ht,
ρt t
2.14
One sees that
Since
∞
−∞
Φ−1 1/ρtdt < ∞ and
∞
−∞
t
Gsds,
t
t ≤ τ0 .
ft, xt, x tdt < ∞, we see that
∞
both
t ≥ τ0 ,
−∞
Hsds are convergent.
2.15
Then we get that
∞
⎧
⎪
⎪
y
∞
Gsds,
⎨
tt
yt ⎪
⎪
⎩y−∞ Hsds,
−∞
This tells us that y is bounded on R.
t ≥ τ0 ,
t ≤ τ0 .
2.16
6
Abstract and Applied Analysis
It follows from 2.7 and 2.16 that
y−∞ τ0
−∞
gsds y−∞ gt
−∞
τ0
t
−∞
gt
−∞
Hsds dt
Gsds dt,
τ0
y
∞ gsds y
∞ τ0
∞
∞
τ0
∞
t
hsds y−∞ ∞
∞
∞
ht
hsds y
∞ τ0
τ0
2.17
t
−∞
ht
−∞
Hsds dt
Gsds dt.
τ0
t
Then
∞
τ0
gsds y−∞ −
gsds y
∞
1−
−∞
t
−∞
−
τ0
τ0
gt
τ0
−∞
tHsds dt gt
Gsds dt,
1−
t
∞
2.18
hsds y
∞
τ0
t
ht
∞
τ0
hsds y−∞ τ0
−∞
−∞
∞
−∞
Hsds dt ∞
∞
ht
τ0
Gsds dt.
t
Hence
τ0
∞
∞
∞
t
−∞ gt −∞
Hsdsdt
τ gt t Gsdsdt − τ gsds 0
0
τ
∞
∞
∞
t
−∞0 ht −∞
Hsdsdt
τ ht t Gsdsdt 1− τ hsds 0
0
τ0
,
y−∞ 1− −∞ gsds − τ
∞
gsds 0
τ0
∞
− −∞ hsds 1− τ hsds 0
τ0
t
∞
∞
τ0
1− −∞ gsds −∞
gt −∞ Hsdsdt
τ gt t Gsdsdt 0
τ
t
∞
∞
τ0
− −∞0 hsds −∞
ht −∞ Hsdsdt
τ ht t Gsdsdt 0
τ0
.
y
∞ 1− −∞ gsds − τ
∞
gsds 0
τ0
∞
− −∞ hsds 1− τ hsds 2.19
0
Since
∞
τ0
∞
∞
1 −
1 −
gsds
−
gsds
gsds
−
gsds
τ0
−∞
τ0
−∞
∞
∞
∞
τ0
> 0,
−
1 −
hsds
1
−
hsds
hsds
1
−
hsds
−∞
τ0
−∞
τ0
2.20
Abstract and Applied Analysis
7
we get from 2.19 that y−∞ ≥ 0 and y
∞ ≥ 0. Hence 2.16 implies that
yt ≥ 0,
∀ t ∈ R.
2.21
ii We prove that yt is concave with respect to τ on R. It is easy to see that τ ∈
CR, 0, 1 and
dτ
1
1
Φ−1
> 0.
dt
ρt ∞ Φ−1 1/ρs ds
2.22
dy dy dτ dy −1
1
1
Φ
∞
.
−1
dt
dτ dt
dτ
ρt
Φ 1/ρs ds
2.23
dy
dy
1
ρtΦ
.
Φ
Φ ∞
dt
dτ
Φ−1 1/ρs ds
2.24
2
dy
1
dy d y dτ
Φ ∞
Φ
.
ρtΦ
dt
dτ dτ 2 dt
Φ−1 1/ρs ds
2.25
∞
ρtΦ dy/dt
d2 y
1
−1
ds
Φ
Φ
.
ρs
dτ 2
Φ dy/dτ dτ/dt
−∞
2.26
−∞
Thus
−∞
It follows that
−∞
Hence
−∞
So
Since ρtΦy t ≤ 0, Φ y > 0 y / 0 and dτ/dt > 0, we get that d2 y/dτ 2 ≤ 0. Hence
yt is concave with respect to τ on R.
iii Now, we prove that
min yt ≥ μ sup yt.
t∈−k,k
t∈R
2.27
Since dτ/dt > 0 for all t ∈ R, there exists the inverse function of τ τt. Denote the inverse
function of τ τt by t tτ.
It follows from 2.13 that supt∈R yt yτ0 . One sees
min yt min y−k, yk .
t∈−k,k
2.28
8
Abstract and Applied Analysis
If min{y−k, yk} yk ytτk, note τk < 1, then for t ∈ −k, k, one has
τk
τk
1 − τk ττ0 ττ0 .
yt ≥ ytτk y t
1 ττ0 1 − τk ττ0 1 ττ0 2.29
Noting that 1 > τk and yt is concave with respect to τ, then, for t ∈ −k, k,
τk
τk
1 − τk ττ0 y t
ytττ0 1 ττ0 1 − τk ττ0 1 ττ0 k
1
1
−1
ds ∞
≥
Φ
yτ0 −1
ρs
2 −∞ Φ 1/ρs ds
−∞
yt ≥
2.30
μ sup yt.
t∈R
Similarly, if min{y−k, yk} y−k ytτ−k, note τ−k < 1, for t ∈ −k, k, one has
yt ≥ ytτ−k
τ−k
τ−k
1 ττ0 − τ−k
ττ0 y t
1 ττ0 1 ττ0 − τ−k 1 ττ0 τ−k
τ−k
1 ττ0 − τ−k
y t
ytττ0 ≥
1 ττ0 1 ττ0 − τ−k
1 ττ0 −k
1
1
≥
ds ∞
Φ−1
yτ0 ρs
2 −∞ Φ−1 1/ρs ds
−∞
2.31
> μ supy t.
t∈R
Hence 2.27 holds.
iv Finally, we prove the uniqueness of Ax . Define
∞ gt 1 − ∞ hsds − ht 1 − ∞ gsds
−∞
−∞
Hx c ∞
1 − −∞ gsds
−∞
t
∞
1
×
Φ−1 c Φ−1
f u, xu, x u du ds dt
ρs
−∞
s
∞
∞
1
Φ−1 c Φ−1
f u, xu, x u du ds.
ρs
−∞
s
2.32
Then H1 implies that Hx ∈ CR, R is increasing on R and Hx 0 ≥ 0. Let
c−
∞
−∞
f u, xu, x u du,
2.33
Abstract and Applied Analysis
9
then Hx c ≤ 0. By mean value theorem, there exists an unique Ax ∈ c, 0 satisfying
Hx Ax 0. Then iv holds. This completes the proof of the lemma.
Choose 1 ≥ k > 0 and
−k
μ
Φ−1 1/ρs ds
.
Φ−1 1/ρs ds
−∞
∞
2 −∞
2.34
Define the cone P ⊆ X by
P
x∈X:
xt ≥ 0, t ∈ R,
min xt ≥ μ max xt
t∈−k,k
2.35
.
t∈R
It is easy to see that P is a cone in X.
Define the operator T : P → X by
∞
Φ−1 1/ρs Φ−1 Ax s fr, xr, x rdr dsdu
T xt ∞
1 − −∞ gsds
t
∞
1
Φ−1 Ax Φ−1
f r, xr, x r dr ds,
ρs
−∞
s
∞
−∞
gu
u
−∞
2.36
where Ax satisfies
∞ gu 1 − ∞ hsds − hu 1 − ∞ gsds
−∞
−∞
∞
1 − −∞ gsds
−∞
∞
u
1
Φ−1 Ax Φ−1
f r, xr, x r dr dsdu
×
ρs
−∞
s
∞
∞
1
Φ−1 Ax Φ−1
f r, xr, x r dr ds 0.
ρs
−∞
s
It follows from Lemma 2.6iv that T : P → X is well defined and Ax ∈ −
x sds, 0. It is easy to show that
ρtΦ T x t f t, xtx t 0, t ∈ R,
∞
gsT xsds,
T x−∞ −∞
T x
∞ 2.37
∞
−∞
fs, xs,
2.38
∞
−∞
hsT xsds.
It follows from Lemma 2.6i and iii that T x ∈ P for all x ∈ P . Then T : P → P is well
defined.
10
Abstract and Applied Analysis
Lemma 2.7. Suppose (H1)–(H3) hold. Then T is completely continuous.
Proof. First, we prove that T is continuous.
We claim that the function Ax : P → 0, ∞ is continuous in x.
Let {xn } ∈ P with xn → x0 ∈ P as n → ∞ in P . Let {Axn } n 1, 2, . . . be the
constants determined by
∞ gt 1 − ∞ hsds − ht 1 − ∞ gsds
−∞
−∞
∞
1 − −∞ gsds
−∞
t
∞
1
−1
−1
×
Φ
Φ
f r, xn r, xn r dr ds dt
Ax n ρs
−∞
s
∞
∞
1
Φ−1 Axn Φ−1
f r, xn r, xn r dr ds 0
ρs
−∞
s
2.39
Since xn → x0 in P as n → ∞, there exists an r > 0 such that xn ≤ r. The fact f is a
Carathédory function means that there exists a nonnegative function φr ∈ L1 −∞, ∞ such
that
0≤f
t, xn t, xn t
f
1
t, xn t, −1 Φ−1 ρt xn t
Φ ρt
≤ φr t,
t ∈ R.
2.40
Then
∞
f s, xn s, xn s ds is convergent.
2.41
∞
∞
∈ −
f s, xn s, xn s ds, 0 ⊆ −
φr sds, 0 ,
2.42
−∞
So
Ax n
−∞
−∞
which means that {Axn } is uniformly bounded.
Suppose that {Axn } does not converge to Ax0 . By the bounded property, we know that
1
2
1
2
there exist two subsequences {Axnk } and {Axnk } of {Axnk } with Axnk → c1 and Axnk → c2
and c1 /
c2 .
Abstract and Applied Analysis
11
By the construction of Axn , n 1, 2, . . ., we have
∞ gt 1 − ∞ hsds − ht 1 − ∞ gsds
−∞
−∞
∞
1
−
gsds
−∞
−∞
t
∞
1
1
−1
−1
×
Φ
Φ
f u, xnk u, xnk u du ds dt
Ax n k ρs
−∞
s
∞
∞
1
1
−1
−1
Φ
Φ
f u, xnk u, xn k u du ds 0.
Ax n k ρs
−∞
s
2.43
Let k → ∞, using Lebesgue’s dominated convergence theorem, the above equality implies
∞ gt 1 − ∞ hsds − ht 1 − ∞ gsds
−∞
−∞
∞
1 − −∞ gsds
−∞
t
∞
1
−1
−1
×
Φ
Φ
f u, x0 u, xo u du ds dt
c1 ρs
−∞
s
∞
∞
1
Φ−1 c1 Φ−1
f u, x0 u, xo u du ds 0.
ρs
−∞
s
2.44
By Lemma 2.6 iv, we get c1 Ax0 . Similarly, c2 Ax0 . Thus c1 c2 , a contradiction. So, for
any xn → x0 , one has Axn → Ax0 , which means Ax : P → R is continuous.
Since Ax is continuous, together with the continuity of x, y → ft, x, 1/Φ−1 ρ
ty, we get that T is continuous.
Second, we show that T is maps bounded subsets into bounded sets.
Let D ⊆ P be bounded. Then, there exists r > 0 such that D ⊆ {x ∈ P : x ≤ r}. Hence
−1
r max sup|xt|, supΦ
t∈R
ρt x t
< ∞.
2.45
t∈R
Then there exists a nonnegative function φr ∈ L1 −∞, ∞ such that
1
−1
t, xt, −1 Φ ρt x t ≤ φr t,
Φ ρt
0 ≤ f t, xt, x t f
t ∈ R.
2.46
Then
∞
−∞
f t, xt, x t dt ≤
∞
−∞
φr tdt : L.
2.47
12
Abstract and Applied Analysis
Thus |Ax | ≤ L for all x ∈ D. Therefore,
∞
−1 −1
A
Φ
fu,
xu,
x
ds dt
1/ρs
Φ
udu
x
−∞
−∞
s
T xt ∞
1 − −∞ gsds
t
∞
1
−1
−1
Φ
Φ
f u, xu, x u du ds
Ax ρs
−∞
s
t
∞
t
gt −∞ Φ−1 1/ρs ds dt
1
−∞
−1
ds Φ−1 2L
Φ
≤
∞
ρs
1−
gsds
−∞
∞
gt
∞
≤
−∞
t
gt
−∞
t
−∞
1−
Φ−1 1/ρs ds dt
∞
−∞
gsds
∞
−1
Φ
−∞
2.48
1
ds Φ−1 2L : M1 .
ρs
On the other hand, we have
−1
Φ
∞
−1
f u, xu, x u du ≤ Φ−1 2L : M2 .
Ax ρt T x t Φ
2.49
t
Then
−1
T x max supT xt, supΦ
t∈R
ρt T x t
< ∞.
2.50
t∈R
So, {T x : x ∈ D} is bounded.
Third, given a bounded set D ⊆ P , we prove that both {T x : x ∈ D} and
{Φ−1 ρtT x : x ∈ D} are equicontinuous on each finite subinterval on R.
Then, there exists r > 0 such that D ⊆ {x ∈ P : x ≤ r}. Hence
−1
r max sup|xt|, supΦ
t∈R
ρt x t
< ∞.
2.51
t∈R
Then there exists a nonnegative function φr ∈ L1 −∞, ∞ such that
1
−1
t, xt, −1 Φ ρt x t ≤ φr t,
Φ ρt
0 ≤ f t, xt, x t f
t ∈ R.
2.52
Then
∞
∞
Ax f t, xt, x t dt ≤
φr tdt : L.
t
−∞
2.53
Abstract and Applied Analysis
13
For any > 0, since Φ−1 is uniformly continuous on −L, L, there exists δ1 > 0 such
that
−1
Φ u − Φ−1 v < ,
u, v ∈ −L, L, |u − v| < δ1 .
2.54
For any a, b ∈ R, t1 , t2 ∈ a, b and x ∈ D with t1 ≤ t2 , we have
∞
∞
t2
f t, xt, x t dt − Ax −
f t, xt, x t dt ≤
φr sds.
Ax t1
t2
t1
2.55
Then there exists δ > 0 such that
∞
∞
f t, xt, x t dt − Ax −
f t, xt, x t dt < δ1 ,
Ax t1
t2
t1 , t2 ∈ a, b, |t1 − t2 | < δ.
2.56
Hence t1 , t2 ∈ a, b, |t1 − t2 | < δ imply that
−1 Φ ρt1 T xt1 − Φ−1 ρt2 T xt2 ∞
∞
−1
−1
Φ
Ax Ax f t, xt, x t dt − Φ
f t, xt, x t dt < .
t1
t2
2.57
It follows that {T x : x ∈ D} is equicontinuous on each finite subinterval on R.
On the other hand, we have
∞
1
Φ−1 Ax f u, xu, x u du ds
ρs
s
t1
t2
1
dsΦ−1 2L
≤
Φ−1
ρs
t1
|T xt1 − T xt2 | t2
−→ 0
Φ−1
2.58
uniformly as t1 −→ t2 .
Then {Φ−1 ρtT x : x ∈ D} is equicontinuous on each finite subinterval on R.
At last given a bounded set D ⊆ P , we show that both {T x : x ∈ D} and
{Φ−1 ρtT x : x ∈ D} are equiconvergent at ±∞, respectively.
Then, there exists r > 0 such that D ⊆ {x ∈ P : x ≤ r}. Hence
−1
r max sup|xt|, supΦ
t∈R
t∈R
ρt x t
< ∞.
2.59
14
Abstract and Applied Analysis
Then there exists a nonnegative function φr ∈ L1 −∞, ∞ such that
1
−1
t, xt, −1 Φ ρt x t ≤ φr t,
Φ ρt
0 ≤ f t, xt, x t f
t ∈ R.
2.60
Then
∞
∞
Ax f t, xt, x t dt ≤
φr tdt : L.
−∞
t
2.61
For any > 0, since Φ−1 is uniformly continuous on −L, L, there exists δ1 > 0 such
that
−1
Φ u − Φ−1 v < ,
u, v ∈ −L, L, |u − v| < δ1 .
2.62
Since
ρtΦ T x t − Ax ∞
f u, xu, x u du ≤
∞
t
φr udu −→ 0
2.63
t
uniformly as t → ∞, we get that there exists T > 0 such that
ρtΦ T x t − Ax < δ1 ,
t ≥ T.
2.64
Hence t > T implies that
−1 Φ ρt T x t − Φ−1 Ax Φ−1 ρtΦ T x t − Φ−1 Ax < .
2.65
Furthermore, we get
∞
t
∞
−1 −1
A
gt
Φ
fu,
xu,
x
ds dt
1/ρs
Φ
udu
x
−∞
−∞
s
T xt −
∞
1 − −∞ gsds
∞
−∞
∞
−1
Φ
∞
1
−1
Φ
f u, xu, x u du ds
Ax ρs
s
∞
1
Φ−1 Ax Φ
f u, xu, x u du ds
ρs
t
s
∞
1
dsΦ−1 2L
Φ−1
≤
ρs
t
−1
−→ 0 uniformly as t −→ ∞.
Hence {T x : x ∈ D} and {Φ−1 ρtT x : x ∈ D} are equiconvergent at ∞.
2.66
Abstract and Applied Analysis
15
Similarly we can how that {T x : x ∈ D} and {Φ−1 ρtT x : x ∈ D} are
equiconvergent at −∞. We omit the details.
Therefore, T : P → P is equiconvergent at ±∞. So the operator T : P → P is
completely continuous.
Define the functionals on P by
γ y sup Φ−1 ρt y t,
y ∈ P,
t∈R
β y supyt,
y ∈ P,
t∈R
θ y supyt,
y ∈ P,
t∈R
α y min yt,
y ∈ P,
ψ y min yt,
y ∈ P.
t∈−k,k
t∈−k,k
2.67
It is easy to see that α, ψ are two nonnegative continuous concave functionals on the cone P ,
γ, β, θ are three nonnegative continuous convex functionals on the cone P and αy ≤ βy for
all y ∈ P .
For e1 , e2 , c > 0 and 1 ≥ k > 0, define
∞ M max
0
1/Φ−1 ρs ds
,1 ,
∞
1 − −∞ gsds
−∞
1
−1
Φ
L1 Φ
2 |s| ds,
ρs
−k
k
√ 1
Φ−1 2 s ds,
Φ−1
L2 ρs
0
∞
1 − −∞ gsds
Φc Φc
Q min
,
Φ ∞ ,
4
π 4
π
1/Φ−1 ρs ds
−∞
1
e2
1
max Φ
W Φ
,Φ
,
μ
L1
L2
−1
E
Φe1 ,
4
π
⎧
1
⎪
⎪
|t| ≥ 1,
⎨ 2,
t
δt 1
⎪
⎪
⎩ , |t| ≤ 1,
|t|
−k −1 Φ 1/ρs ds
μ −∞
.
∞
2 −∞ Φ−1 1/ρs ds
2.68
16
Abstract and Applied Analysis
Theorem 2.8. Suppose that (H1)–(H3) hold. Given positive constants e1 , e2 , c and k ∈ 0, 1, let
Q, W, and E be as above. If
c≥
e2
> e2 > e1 > 0,
μ
Q≥W
2.69
and
A1 ft, u, 1/Φ−1 ρtv ≤ δtQ for all t ∈ R, u ∈ 0, c, v ∈ −c, c;
A2 ft, u, 1/Φ−1 ρtv ≥ δtW for all t ∈ −k, k, u ∈ e2 , e2 /μ, v ∈ −c, c;
A3 ft, u, 1/Φ−1 ρtv ≤ δtE for all t ∈ R, u ∈ 0, e1 , v ∈ −c, c;
then BVP1.2 has at least three positive solutions x1 , x2 , x3 such that
sup x1 t < e1 ,
min x2 t > e2 ,
t∈−k,k
t∈R
sup x3 t > e1 ,
t∈R
min x3 t < e2 .
t∈−k,k
2.70
Proof. We prove that all conditions in Lemma 2.5 are satisfied.
i By the definitions, it is easy to show that α, ψ are two nonnegative continuous
concave functionals on the cone P , γ, β, θ are three nonnegative continuous convex
functionals on the cone P and αy ≤ βy for all y ∈ P . One sees x ∈ P is a positive solution
of BVP 1.2 if and only if x is a solution of the operator equation x T x.
ii For y ∈ P , we have
t
yt y sds y−∞
−∞
≤
t
−∞
y sds y−∞
t
∞
1
−1 ≤
gsysds
ρs
y
s
ds
Φ
−1
ρs
−∞ Φ
−∞
∞
∞
1
−1
≤
gsds supyt.
ds sup Φ ρt y t
−1
ρs
−∞ Φ
−∞
t∈R
t∈R
2.71
It follows that
supyt ≤
t∈R
∞
−∞
1
ds sup Φ−1 ρt y t
Φ−1 ρs
t∈R
∞
−∞
gsds supyt.
2.72
t∈R
Then
supyt ≤
t∈R
∞ 1/Φ−1 ρs ds
sup Φ−1 ρt y t.
∞
1 − −∞ gsds
t∈R
−∞
2.73
Abstract and Applied Analysis
17
Hence
y max supyt, sup Φ−1 ρt y t
t∈R
t∈R
∞ ≤ max
1/Φ−1 ρs ds
−∞
, 1 sup Φ−1 ρt y t.
∞
1 − −∞ gsds
t∈R
2.74
It follows that ||y|| ≤ Mγy for all y ∈ P .
iii Corresponding to Lemma 2.5,
c c,
h μe1 ,
d e1 ,
a e2 ,
b
e2
.
μ
2.75
Now, we prove that all other conditions of Lemma 2.5 hold. One sees that 0 < d < a. The
remainder is divided into five steps.
Step 1. Prove that T : Pc → Pc 0.
For y ∈ Pc , we have ||y|| ≤ c. Then 0 ≤ yt ≤ c and −c ≤ Φ−1 ρty t ≤ c for all t ∈ R.
So A1 implies that
f t, yt, y t f
1
t, yt, −1 Φ−1 ρt y t
Φ ρt
≤ δtQ,
t ∈ R.
2.76
We have
∞
Φ−1 ρt T y t Φ−1 Ay f u, yu, y u du ≤ Φ−1
−1
≤Φ
∞
−∞
t
f u, yu, y u du
∞
−∞
2.77
δsQdr
−1
Φ QΦ−1 4 π
≤ c.
Similarly to ii, we can show that
0 ≤ T y t ≤ M sup Φ−1 ρt T y t
t∈R
≤ MΦ−1
∞
−∞
δsQdr
MΦ−1 QΦ−1 4 π
≤ c.
2.78
18
Abstract and Applied Analysis
It follows that
T y max sup T y t, sup Φ−1 ρt T y t
t∈R
≤ c.
2.79
t∈R
Then T : Pc → Pc .
Step 2. Prove that
y ∈ P γ, θ, α; a, b, c | α y > a e2
∅
y ∈ P γ, θ, α; e2 , , c | α y > e2 /
μ
2.80
and αT y > e2 for every y ∈ P γ, θ, α; e2 , e2 /μ, c.
Choose yt e2 /2μ for all t ∈ R. Then y ∈ P and
e2
> e2 ,
α y 2μ
e2
e2
≤ ,
θ y 2μ
μ
γ y 0 < c.
2.81
It follows that {y ∈ P γ, θ, α; a, b, c | αy > a} /
∅.
For y ∈ P γ, θ, α; a, b, c, one has that
α y min yt ≥ e2 ,
e2
θ y sup yt ≤ ,
μ
t∈R
t∈−k,k
γ y supy t ≤ c.
t∈R
2.82
Then
e2 ≤ yt ≤
e2
,
μ
t ∈ −k, k,
Φ−1 ρt y t ≤ c.
2.83
Thus A2 implies that
f t, yt, y t ≥ δtW,
t ∈ −k, k.
2.84
Similarly to Lemma 2.6i, we know that there exists τ0 ∈ 0, 1 such that T y τ0 0. Then
s
⎧
∞
⎪
1
⎪
−1
−1
⎪ T y ∞ Φ
Φ
f u, yu, y u du ds, t ≥ τ0 ,
⎨
ρs
2.85
T y t tt
ττ00
⎪
1
⎪
−1
−1
⎪
⎩ T y −∞ Φ
Φ
f u, yu, y u du ds, t ≤ τ0 .
ρs
−∞
s
Since
α T y min T y t ≥ μ sup T y t μ T y τ0 ,
t∈−k,k
t∈R
2.86
Abstract and Applied Analysis
19
if τ0 ≥ 0, we get
τ0
τ0
1
α T y ≥ μ T y −∞ Φ−1
Φ−1
f u, yu, y u du ds
ρs
−∞
s
"
! 0
0
1
−1
−1
Φ
>μ
Φ
f u, yu, y u du ds
ρs
−k
s
0
"
! 0
1
−1
−1
Φ
Φ
δuWdu ds
≥μ
ρs
−k
s
! 0
"
1
−1
−1
Φ
Φ
2 |s| ds Φ−1 W
≥μ
ρs
−k
2.87
≥ e2 .
If τ0 < 0, we get
s
"
!
∞
1
−1
−1
Φ
Φ
f u, yu, y u du ds
α T y ≥ μ T y ∞ ρs
τ0
τ0
s
"
! k
1
−1
−1
Φ
Φ
f u, yu, y u du ds
>μ
ρs
τ0
τ0
! k
"
s
1
−1
−1
Φ
Φ
δuWdu ds
≥μ
ρs
0
0
"
! k
√
1
−1
−1
Φ 2 t ds Φ−1 W
Φ
≥μ
ρs
0
2.88
≥ e2 .
This completes Step 2.
∅
Step 3. Prove that {y ∈ Qγ, θ, ψ; h, d, c | βy < d} {y ∈ Qγ, θ, ψ; μe1 , e1 , c|βy < e1 } /
and
β T y < e1
for every y ∈ Q γ, θ, ψ; h, d, c Q γ, θ, ψ; μe1 , e1 , c .
2.89
Choose yt μe1 . Then y ∈ P , and
ψ y μe1 ≥ h,
β y θ y μe1 < e1 d,
It follows that {y ∈ Qγ, θ, ψ; h, d, c | βy < d} /
∅.
γ y 0 ≤ c.
2.90
20
Abstract and Applied Analysis
For y ∈ Qγ, θ, ψ; h, d, c, one has that
γ y supy t ≤ c.
θ y sup yt ≤ d e1 ,
ψ y min yt ≥ h μe1 ,
t∈−k,k
t∈R
t∈R
2.91
Hence we get that
0 ≤ yt ≤ e1 ,
−c ≤ Φ−1 ρt y t ≤ c,
t ∈ R;
t ∈ R.
2.92
Then A3 implies that
f t, yt, y t ≤ δtE,
t ∈ R.
2.93
So
β T y M sup Φ−1 ρt T y t
t∈R
∞
≤ MΦ−1 Ay f u, yu, y u du −1
≤ MΦ
≤ MΦ−1
∞
−∞
t
f u, yu, y u du
∞
−∞
2.94
δuEdu
≤ MΦ−1 4 πΦ−1 E
e1 d.
This completes Step 3.
Step 4. Prove that αT y > a for y ∈ P γ, α; a, c with θT y > b;
For y ∈ P γ, α; a, c P γ, α; e2 , c with θT y > b e2 /μ, we have that αy mint∈−k,k yt ≥ e2 and γy supt∈R Φ−1 ρt|y t| ≤ c and supt∈R T yt > e2 /μ. Then
e2
α T y min T y t ≥ μβ T y > μ e2 a.
t∈−k,k
μ
2.95
This completes Step 4.
Step 5. Prove that βT y < d for each y ∈ Qγ, β; d, c with ψT y < h.
For y ∈ Qγ, β; d, c with ψT y < d, we have γy supt∈R Φ−1 ρt|y t| ≤ c and
βy supt∈R yt ≤ d e1 and ψT y mint∈−k,k T yt < h e1 μ. Then
1
1
β T y sup T y t ≤
min T y t < e1 μ e1 d.
t∈−k,k
μ
μ
t∈R
This completes the Step 5.
2.96
Abstract and Applied Analysis
21
Then Lemma 2.5 implies that T has at least three fixed points x1 , x2 , and x3 such that
βx1 < e1 ,
αx2 > e2 ,
βx3 > e1 ,
αx3 < e2 .
2.97
Hence BVP1.2 has three decreasing positive solutions x1 , x2 and x3 such that 2.70 holds.
The proof is complete.
3. Examples
Now, we present an example, whose three positive solutions cannot be obtained by theorems
in known papers, to illustrate the main results.
Example 3.1. Consider the following BVP
# 2
3 $
e3t x t
f t, xt, x t 0,
t ∈ R,
x−∞ 0,
3.1
x
∞ 0.
Corresponding to BVP1.2, one sees that φx x3 , φ−1 x x1/3 , gt ht ≡ 0, ρt e3t ,
f : R × R × R → 0, ∞ is nonnegative and continuous and is defined by
2
2 f t, x, y δt f0 x g0 et y ,
⎧
1
⎪
⎪
|t| ≥ 1,
⎨ 2,
t
δt 1
⎪
⎪
⎩ , |t| ≤ 1.
|t|
3.2
Choose k 1, e1 50, e2 250, c 40000. By direct computation, we see that Q, W,
and E are given by
∞ √
1/Φ−1 ρs ds
π
,
M max
,1 ∞
2
1 − −∞ gsds
0
0
1
2
−1
−1
1/3
L1 Φ
Φ
e−s |s|1/6 ds > 0.06,
2 |s| ds 2
ρs
−k
−1
1
k
√ 1
2
Φ−1 2 s ds 21/3 e−s s1/6 ds > 0.06,
Φ−1
L2 ρs
0
0
−∞
22
Abstract and Applied Analysis
−k
−1 −s2
Φ−1 1/ρs ds
e ds
μ
> 0.37,
−∞√
π
Φ−1 1/ρs ds
∞
1 − −∞ gsds
64 × 1012
Φc Φc
,
Φ ∞ > 8.53 × 1012 ,
Q min
−1 ρs ds
4
π 4
π
4
π
1/Φ
−∞
e2
1
1
WΦ
,Φ
< 16.673 × 675.73 < 3.43 × 1011 ,
max Φ
μ
L1
L2
−∞
∞
2 −∞
E
2500
Φe1 > 333.33.
4
π
4
π
3.3
One can show that
c≥
e2
> e2 > e1 > 0,
μ
Q ≥ W.
3.4
Suppose that
⎧
166.67,
⎪
⎪
⎪
⎪
⎪
44.36 × 1011 − 166.67
⎨
166.67 x − 50,
f0 x 250 − 50
⎪
⎪
⎪44.36 × 1011 ,
⎪
⎪
⎩
44.36 × 1011 ex−40000 ,
g0 y ≤ 10, ∀ y ∈ R.
x ∈ 0, 50,
x ∈ 50, 250,
x ∈ 250, 40000,
x ≥ 40000,
3.5
From
2
f t, x, e−t y δt f0 x g0 y ,
3.6
it is easy to show that
A1 ft, u, e−t v ≤ 8.53 × 1012 δt for all t ∈ R, u ∈ 0, 400000, v ∈ −40000, 40000;
2
A2 ft, u, e−t v ≥ 3.43 × 1011 δt for all t ∈ −1, 1, u ∈ 250, 1000, v ∈ −40000, 40000;
2
A3 ft, u, e−t v ≤ 333.33δt for all t ∈ R, u ∈ 0, 50, v ∈ −40000, 40000;
2
then Theorem 2.8 implies that BVP3.1 has at least three positive solutions x1 , x2 , x3 such that
sup x1 t < 50,
t∈R
sup x3 t > 50,
t∈R
min x2 t > 250,
t∈−1,1
min x3 t < 250.
3.7
t∈−1,1
Remark 3.2. Example 3.1 implies that there is a large number of functions that satisfy the
conditions of Theorem 2.8. In addition, the conditions of Theorem 2.8 are also easy to check.
Abstract and Applied Analysis
23
Acknowledgments
This work was supported by Natural Science Foundation of Guangdong provinces Grant no
S2011010001900 and the Foundation for High-level talents in Guangdong Higher Education
Project.
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Applied Mathematics
Journal of
Function Spaces
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Optimization
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014