Volume 2008, Article ID 457028,12pages doi:10.1155/2008/457028
Research Article
Positive Solutions of Singular Initial-Boundary
Value Problems to Second-Order Functional
Differential Equations
Fengfei Jin and Baoqiang Yan
School of Mathematics Sciences, Shandong Normal University, Jinan 250014, China
Correspondence should be addressed to Baoqiang Yan,[email protected]
Received 23 August 2007; Revised 9 January 2008; Accepted 5 August 2008
Recommended by Raul Manasevich
Positive solutions to the singular initial-boundary value problemsx −ft, xt,0< t < 1, x0
0, x1 0,are obtained by applying the Schauder fixed-point theorem, wherextu xtu 0≤
t≤1on−r,0andf·,·:0,1×C\{0}→RC{x∈C−r,0, R, xt≥0,∀t∈−r,0} may be singular atϕu 0−r ≤ u≤ 0andt 0. As an application, an example is given to demonstrate our result.
Copyrightq2008 F. Jin and B. Yan. 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.
1. Introduction
Recently, in 1–4, Erbe, Kong, Jiang, Wang, and Weng considered the following singular functional differential equations:
x −ft, xτt, 0< t <1, αxt−βxt μt, a≤t≤0, γxt δxt νt, 1≤t≤b,
1.1
whereamin{0,inf{τt: 0≤t≤1}}, bmax{1,sup{τt: 0≤t≤1}},and the existence of positive solutions to1.1is obtained. Whenτt t−rin1.1, Agarwal and O’Regan in5, Lin and Xu in6discussed the existence of positive solutions to1.1also. We notice that the nonlinearitiesft, uin all the above-mentioned references depend ont, u∈0,1×R.
The more difficult case is that the termft, ϕdepends ont, ϕ∈0,1×C0,1, R
att 0 andϕ θ, there are many results on the following1.2 see7–9and references therein. Up to now, to our knowledge, there are fewer results on1.2when the termft, ϕ
is allowed to possess singularity for the termft, ϕatt0 andϕ0, which is of more actual significance.
In this paper, motivated by above results, we consider the second-order initial-boundary value problems:
x−ft, xt
, 0< t <1, x00,
x1 0,
1.2
wheref :0,1×C\0→0,∞ C {x∈C−r,0, R, xt≥0,∀t∈−r,0}, xtxt
u −r ≤u≤0. By Leray-Schauder fixed-point theorem, the existence of positive solutions to 1.2is obtained whenft, ϕis singular att0 andϕ0.
For ϕ ∈ C−r,0, R and x ∈ C−r,1, R, let ϕ maxt∈−r,0|ϕt| and x
maxt∈−r,1|xt|. Then, C−r,0, R and C−r,1, R are Banach spaces. Let C {x ∈
C−r,0, R, xt ≥ 0,∀t ∈ −r,0} and P {x ∈ C−r,1, R, xt ≥ 0,∀t ∈ −r,1}. Obviously,CandP are cones inC−r,0, RandC−r,1, R,respectively. Now, we give a new definition.
Definition 1.1. ft, ϕis said to be singular att 0 forϕ ∈C− {0},whenft, ϕsatisfies limt→0ft, ϕ ∞forϕ ∈C− {0}andft, ϕis said to be singular atϕ0 fort ∈0,1
whenft, ϕsatisfies limϕ →0ft, ϕ ∞fort∈0,1.
And one defines some functions which one has to use in this paper. Let
ht
⎧ ⎨ ⎩
0, −r ≤t≤0, t1−t, 0≤t≤1,
Gt, s
⎧ ⎨ ⎩
t1−s, 0≤t≤s≤1, s1−t, 0≤s≤t≤1,
1.3
whereGt, sis a Green’s function. It is clear that Gt, s > 0 fort, s ∈ 0,1×0,1and
hths≤Gt, s≤hson0,1×0,1.
We now introduce the definition of a solution to IBVP1.2.
Definition 1.2. A functionx is said to be a solution to IBVP1.2if it satisfies the following conditions:
1xtis continuous and nonnegative on−r,1; 2x00, x1 0;
3xtandxtexist on0,1;
Furthermore, a solutionxis said to be positive ifxt>0 on0,1. Letxbe a solution to IBVP1.2. Then, it can be represented as
xt
⎧ ⎪ ⎨ ⎪ ⎩
0, −r≤t≤0,
1
0
Gt, sfs, xs
ds, 0≤t≤1. 1.4
It is clear that
xt
1
0
Gt, sfs, xs
ds≤
1
0
hsfs, xs
ds fort∈0,1,
xt≥ht
1
0
hsfs, xs
ds≥ x ht on0,1,
1.5
for all solutions,x, to IBVP1.2, where x max0≤t≤1xt. Forξ ∈R, letξu≡ξon−r,1
throughout this paper. Obviously,ξ∈C−r,1, Randξ0ξtfor allt∈0,1.
Throughout this paper, we assume the following hypotheses hold.
H1ft, ϕis continuous on0,1×C\ {0}.
H2There existsε >0, such that
ft, ϕ≥ft,ε0
, for ϕ ≤ε,
0<
1
0
hsfs,ε0
ds <∞. 1.6
Lemma 1.3. Assume that (H1)-(H2) hold, then there exists aθ∗>0, such that
xt≥θ∗ht, on0,1, 1.7 for all solutions,x, to1.2.
Proof. Suppose that the claim is false.1.5guarantees that there exists a sequence{xmt}of
solutions to IBVP1.2such that
lim
m→∞xm0. 1.8
Without loss of generality, we may assume that
ε≥xm≥xm1 ∀m≥1. 1.9
FromH1,H2,and1.5, it follows that
xm 1
2
1
0
G
1
2, s
fs, xms
ds
≥h
1 2
1
0
hsfs, xms
ds
≥h
1 2
1
0
hsfs,εs
ds
>0,
1.10
which contradicts the assumption that limm→∞ xm 0 and hence the claim is true provided
Remark 1.4. The following inequality
1
0
hsfs,ε0ds≥θ 1.11
holds provided thatθ <min{ε, θ∗}is sufficiently small, whereθ∗is inLemma 1.3.
H3 There exist a nonnegative continuous function k· defined on 0,1 and two
nonnegative continuous functionsF1ϕ, F2ϕdefined on, respectively,C \ {0}, C, such
that
ft, ϕ≤ktF1ϕ F2ϕ
fort, ϕ∈0,1×C\ {0}, 1.12
wherekt, F1ϕ,andF2ϕsatisfy
1
0
hsksds <∞,
1
0
hsksF1
θhs
ds <∞, lim
ϕ →∞
|F2ϕ|
ϕ 0. 1.13
Furthermore,F1ϕis nonincreasing andF2ϕis nondecreasing, that is,
F1ϕ≥F1ϕ forϕu≤ϕuon−r,0,
F2ϕ≤F2ϕ forϕu≤ϕuon−r,0.
1.14
Lemma 1.5 see7. LetE be the Banach space and let X be any nonempty, convex, closed, and bounded subset ofE. IfTis a continuous mapping ofXinto itself andTXis relatively compact, then the mappingThas at least one fixed point (i.e., there exists anx∈XwithxTx) .
Using Lemma 1.5, we present the existence of at least one positive solution to1.2 whenft, ϕis singular atϕ0 andt0notice the newDefinition 1.1. To some extent, our paper complements and generalizes these in1–6,8–10.
2. Main results
Theorem 2.1. Assume that (H1)–(H3) hold. Then, the IBVP1.2has at least one positive solution.
Proof. Since limϕ →∞|F2ϕ|/ ϕ 0, we can choose anN > εsuch that
F2ϕ≤μ ϕ, for ϕ ≥N, 2.1
where the positive numberμsatisfies
0< μ
1
0
hsksdsσ <1. 2.2
Let
R
1
0
hsksF1
θhs
ds,
T
1
0
hsksF2Ns
ds,
M∗ RTN
1−σ .
For eachx∈P⊆C−r,1, R, we definex∗tby
x∗t
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0, −r≤t≤0,
θht, ifxt< θhton0,1, xt, ifθht≤xt≤M∗on0,1, M∗, ifxt> M∗on0,1,
f∗t, xt
ft, x∗t fort∈0,1.
2.4
It is obvious thatf∗t, xtsatisfies the hypothesesH1–H3andM∗> N.
We now consider the modified initial-boundary value problem:
x −f∗t, xt
, 0< t <1, x00,
x1 0.
2.5
We claim that for all solutions,x, to IBVP2.5,
xt≥θht, on−r,1. 2.6
Suppose that the claim is false. Then there existst∈0,1such that
xt< θht. 2.7
Sincext hton−r,0, there are the following three cases.
Case 1. xt< θhtfor allt∈0,1.
The solution of IBVP2.5can be represented asnoticeθ <min{ε, θ∗}Remark 1.4
xt
1
0
Gt, sf∗s, xs
ds
1
0
Gt, sfs, xs∗ds
≥ 1
0
Gt, sfs, θhs
ds
≥ht
1
0
hsfs,εs
dsnotice H2
1
0
hsfs,ε0
ds
> θht, t∈0,1,
2.8
Case 2. There exists at0∈0,1such thatxt0> θht0and x < θ.
In this case, we have
xt
1
0
Gt, sf∗s, xs
ds
≥ht
1
0
hsfs,εs
ds
ht
1
0
hsfs,ε0
ds
≥θht, t∈0,1,
2.9
which contradicts2.7.
Case 3. There exists at0∈0,1such thatxt0> θht0and x ≥θ.
From1.5, we get
xt≥ x ht≥θht, t∈0,1, 2.10
which contradicts2.7. So we have
xt≥θht on−r,1. 2.11
To prove the existence of positive solutions to IBVP2.5, we seek to transform2.5 into an integral equation via the use of Green’s function and then find a positive solution by usingLemma 1.5.
Define a nonempty convex and closed subset ofC−r,1, Rby
Dx∈C−r,1, R: 0≤xt≤M∗, t∈0,1, xt 0, t∈−r,0. 2.12
Then, we define an operatorT:D→C−r,1, Rby
Txt
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
0, if −r≤t≤0,
1
0
Gt, sf∗s, xs
ds, if 0≤t≤1.
FromH1–H3and the definition ofT, we have, for everyx∈D,
Txt∈C−r,1, Txt≥0 on0,1, 2.14
Txt
1
0
Gt, sf∗s, xs
ds
≤ 1
0
hsf∗s, xs
ds
≤ 1
0
hsfs, x∗sds
≤ 1
0
hsksF1
x∗sF2x∗s
ds
≤ 1
0
hsksF1
θhs
F2
xs∗ds
≤ 1
0
hsksF1
θhs
ds
1
0
hsksF2
x∗sds
≤R
1
0
hsksF2
x∗sds
≤R
1
0
hsksF2M∗s
ds
≤R
1
0
hsksμM∗ds
≤RσM∗
≤M∗, t∈0,1.
2.15
Together with the definition ofD, we getTD⊂D. Also,
Txt −
t
0
sf∗s, xs
ds
1
t
1−sf∗s, xsds 2.16
is continuous in0,1, and
Txt −f∗t, xt
≤0 in0,1. 2.17
From H3and2.15, we can get
1
0
htTxtdt
1
0
htf∗t, xt
dt
≤M∗<∞,
2.18
Now, we claim thatTDis equicontinuous on −r,1. We will prove the claim. For anyx∈D, we have
Txt
1
0
Gt, sf∗s, xs
ds
≤ 1
0
Gt, sksF1
θhs
F2 M∗s
ds
Ut, 0≤t≤1.
2.19
SinceUtis continuous on0,1andU0 U1 0, then for anyε0 >0, there is a
δ∈0,1/4such that
0≤Txt≤Ut< ε0
2, t∈0,2δ∪1−2δ,1. 2.20
By2.6, we have, fort∈δ,1−δ,
Txt≤−t
0
sf∗s, xs
ds
1
t
1−sf∗s, xs
ds
≤ 1−δ
0
sf∗s, xs
ds
1
δ
1−sf∗s, xs
ds
≤ 1−δ
0
sksF1
θhs
F2M∗s
ds
1
δ
1−sksF1
θhs
F2M∗s
ds
≤ 1
δ
1−δ
0
1−ssksF1
θhs
F2M∗s
ds1 δ
1
δ
s1−sksF1
θhs
F2M∗s ds ≤ 2 δ 1 0
hsksF1
θhs
F2M∗s ds 2 δK L, 2.21
whereK10hsksF1θhs F2M∗sds <∞is a constant number.
Putδ1ε0/L, then fort1, t2∈δ,1−δ,|t1−t2|< δ1,
Txt1
−Txt2≤Lt1−t2< ε0. 2.22
Setδ0min{δ, δ1}. Then fort1, t2∈0,1,|t1−t2|< δ0, and
Tx
t1
−Txt2< ε0. 2.23
SinceTxt 0 ont∈−r,0, the above inequality holds fort∈−r,1.
Let{xnt}∞n0 ⊂Dbe arbitrarily chosen and letxntconverge tox0tuniformly on
−r,1 asn→∞. Now, we claim thatx∗ntconverge tox∗0tuniformly as n→∞. From the definition ofx∗t, we get
x∗nt xnt θht
2
|xnt−θht|
2 , t∈−r,1,
x0∗t x0t θht
2
|x0t−θht|
2 , t∈−r,1.
2.24
Thus,
x∗nt−x0∗txnt θht
2
|xnt−θht|
2 −
x0t θht
2 −
|x0t−θht|
2
≤xnt−x0t
2
|xnt θht| − |x0t θht|
2
≤xnt−x0t
2
|xnt θht| − |x0t θht|
2
≤xnt−x0t
2
xnt−x0t
2
xnt−x0t, t∈−r,1,
2.25
that is, the claim is true.
Sinceft, ϕis continuous with respect toϕfort∈0,1, we have lim
n→∞Gt, sf
∗s, x
ns
Gt, sf∗s, x0s
on0,1, 2.26
for each fixedt∈0,1. From the definition off∗andH3, we know that
0≤f∗t, xnt
≤ktF1
θht
F2M∗t
, 2.27
and hence
0≤Gt, sf∗s, xns
≤hsksF1
θhs
F2M∗s
, fort, s∈0,1×0,1, 2.28
wherehsksF1θhsF2M∗sis a Lebesgue integrable function defined on0,1because
ofH3. Consequently, we apply the dominated convergence theorem to get
lim
n→∞Txn
t−Tx0
t lim
n→∞tmax∈0,1 1
0
Gt, sf∗s, xns
−f∗s, x0s ds ≤ 1 0 max
t∈0,1Gt, snlim→∞f ∗s, x
ns
−f∗s, x0sds
0,
2.29
which shows that the mappingT is continuous onD.
Then from Lemma 1.5, we get that there exists at least one positive solution, x, to IBVP2.5inD. The solution can be represented by1.4, wheref is replaced withf∗. So, 2.6holds. Furthermore, from the definition ofD, we can get
xt≤M∗. 2.30
3. Application
Example 3.1. Consider the singular IBVP3.1:
x 1
tα0
−rxtudu
βsinπt
maxxtu:−r≤u≤0γ 0, 0< t <1,
x00,
x1 0,
3.1
whereα >0, β >0,0< γ <1, αβ <1.
4. Conclusion
Equation3.1has at least one positive solution. Now, we will check thatH1–H3hold in3.1.
In IBVP3.1,ft, ϕ 1/tα0
−rϕudu β
sinπt max{ϕu:−r ≤u≤0}γ. It is clear thatf :0,1×C→0,∞is continuous and singular att 0 andϕ 0. ForH3,we
choose
kt 1
tα, F1ϕ
1
0
−rϕudu
β, F2ϕ
maxϕu:−r≤u≤0γ1, 4.1
whenα >0, β >0,0< γ <1, αβ <1; by simple computation, we can get
1
0
hsksds <∞,
1
0
hsksF1
s, θhs
ds <∞ for 0< θ <∞, lim
ϕ →∞
|F2ϕ|
ϕ 0.
4.2
It is obvious thatF1ϕis nonincreasing andF2ϕis nondecreasing.
Now, we checkH2. For anyε > 0,ϕ ∈C, ϕ ≤εnotice the definition of · , we
have
0≤
0
−r
ϕudu
β
≤
0
−r
εdu
β
rεβ, 4.3
ft, ϕ−ft,ε0
1
tα ⎡
⎣ 1
0
−rϕudu
β−
1 rεβ
⎤
⎦ ϕ γ−εγ
≥ 1
0
−rϕudu
β−
1 rεβ
ϕ γ−εγ notice3.4
≥ 1
ϕ rβ
ϕ γ−
1 rεβ ε
γ
.
4.4
We define
gx 1
rxβ x
Now, we will prove that there existsε >0 such thatg·is decreasing on0, ε. Obviously,
gx γr
βx1β−βx1−γ
rβx1−γx1β . 4.6
Putg1x γrβx1β−βx1−γ, then
g10 0,
g1x γ1βrxβ−1−γβx−γ,
lim
t→0g1
x −∞.
4.7
From the continuity ofg1x, we can findε > 0 such thatg1x < 0 on0, ε. Then,
gx<0 on0, ε. That is,gxis decreasing on0, ε. Furthermore, we have
1
0
hsfs,εs
ds
1
0
s1−sfs,εsds
1
0
s1−s
⎡ ⎣1
sα
1
0
−rεdu
βεsinπs
⎤ ⎦ds
1
0
s1−α1−s 1
rεβds
1
0
s1−sεds
1
0
s1−ssinπsds.
4.8
Thus,
0<
1
0
hsfs,εs
ds <∞, 4.9
which implies thatH2holds.
So, fromTheorem 2.1, IBVP3.1has at least one positive solution.
Acknowledgments
The research was supported by NNSF of China 10571111 and the fund of Shandong Education CommitteeJ07WH08.
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