Solutions and Green's Functions for Boundary Value Problems of Second-Order Four-Point Functional Difference Equations

Full text

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Volume 2010, Article ID 973731,22pages doi:10.1155/2010/973731

Research Article

Solutions and Green’s Functions for

Boundary Value Problems of Second-Order

Four-Point Functional Difference Equations

Yang Shujie and Shi Bao

Institute of Systems Science and Mathematics, Naval Aeronautical and Astronautical University, Yantai, Shandong 264001, China

Correspondence should be addressed to Yang Shujie,yangshujie@163.com

Received 23 April 2010; Accepted 11 July 2010

Academic Editor: Irena Rach ˚unkov´a

Copyrightq2010 Y. Shujie and S. Bao. 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.

We consider the Green’s functions and the existence of positive solutions for a second-order functional difference equation with four-point boundary conditions.

1. Introduction

In recent years, boundary value problems BVPs of differential and difference equations have been studied widely and there are many excellent resultssee Gai et al.1, Guo and Tian2, Henderson and Peterson3, and Yang et al.4. By using the critical point theory, Deng and Shi5studied the existence and multiplicity of the boundary value problems to a class of second-order functional difference equations

Lunfn, un1, un, un−1 1.1

with boundary value conditions

Δu0A, uk1B, 1.2

where the operatorLis the Jacobi operator

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Ntouyas et al.6and Wong7investigated the existence of solutions of a BVP for functional differential equations

xt ft, xt, xt

, t∈0, T,

α0x0−α1x0 φCr,

β0xT β1xT A∈Rn,

1.4

wheref:0, T×Cr×Rn → Rnis a continuous function,φCr Cr,0,Rn, A∈Rn, and

xtθ xtθ, θ∈−r,0.

Weng and Guo8considered the following two-point BVP for a nonlinear functional difference equation withp-Laplacian operator

ΔΦpΔxt rtfxt 0, t∈ {1, . . . , T},

x0φC, ΔxT1 0,

1.5

whereΦpu |u|p−2u,p >1,φ0 0,T,τ ∈N,C{φ|φk≥0, k∈−τ,0},f:C → R

is continuous,Ttτ1rt>0.

Yang et al. 9 considered two-point BVP of the following functional difference equation withp-Laplacian operator:

ΔΦpΔxt rtfxt, xt 0, t∈ {1, . . . , T},

α0x0−αx0 h,

β0xT1 βxT1 A,

1.6

wherehCτ {φ | φθ ≥ 0, θ ∈ {−τ, . . . ,0}},A ∈ R, andα0,α1,β0, andβ1 are

nonnegative real constants. Fora, b∈Nanda < b, let

R{x|xR, x0},

a, b {a, a1, . . . , b}, a, b {a, a1, . . . , b−1}, a,∞ {a, a1, . . . ,},

φ|φ:−τ,0 → R, Cτ φ|φθ≥0, θ∈−τ,0

.

1.7

Thenand are both Banach spaces endowed with the max-norm

φ

τ kmaxτ,0φk. 1.8

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In this paper, we consider the following second-order four-point BVP of a nonlinear functional difference equation:

−Δ2ut1 rtft, u

t, t∈1, T,

u0αu

ηh, t∈−τ,0,

uT1 βuξ γ,

1.9

whereξ, η ∈ 1, Tand ξ < η, 0 < τ < Tut ut1−ut,Δ2ut ΔΔut,f :

→ Ris a continuous function,h andhth0≥0 fort∈−τ,0,α,β, andγ

are nonnegative real constants, andrt≥0 fort∈1, T.

At this point, it is necessary to make some remarks on the first boundary condition in

1.9. This condition is a generalization of the classical condition

u0 αuηC 1.10

from ordinary difference equations. Here this condition connects the historyu0with the single

. This is suggested by the well-posedness of BVP1.9, since the functionfdepends on the termuti.e., past values ofu.

As usual, a sequence{uτ, . . . , uT1}is said to be a positive solution of BVP1.9 if it satisfies BVP1.9anduk≥0 fork∈−τ, Twithuk>0 fork∈1, T.

2. The Green’s Function of

1.9

First we consider the nonexistence of positive solutions of1.9. We have the following result.

Lemma 2.1. Assume that

βξ > T 1, 2.1

or

αT1−η> T1. 2.2

Then1.9has no positive solution.

Proof. FromΔ2ut1 rtft, u

t≤0, we know thatutis convex fort∈0, T1.

Assume thatxtis a positive solution of1.9and2.1holds.

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IfxT1>0, thenxξ>0. It follows that

xT1−x0

T1

βxξx0

T1

> ξ

x0

T1

x0

ξ ,

2.3

which is a contradiction to the convexity ofxt.

IfxT1 0, then 0. Ifx0>0, then we have

xT1−x0

T1 −

x0

T1,

x0

ξx0

ξ .

2.4

Hence

xT1−x0

T1 >

x0

ξ , 2.5

which is a contradiction to the convexity ofxt. Ifxt≡0 fort∈1, T, thenxtis a trivial solution. So there exists at0∈1, ξξ, Tsuch thatxt0>0.

We assume thatt0∈1, ξ. Then

xT1−xt0

T1−t0 −

xt0

T1−t0

,

xt0

ξt0 −

xt0

ξt0

.

2.6

Hence

xT1−xt0

T1−t0

> xt0 ξt0

, 2.7

which is a contradiction to the convexity ofxt.

Ift0∈ξ, T, similar to the above proof, we can also get a contradiction.

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Now we have

xT1−x0

T1

βxξx0 γ T1

ξx0

T1

γ T1

x0

ξ

γ T1

> x0 ξ ,

2.8

which is a contradiction to the convexity ofxt.

Assume thatxtis a positive solution of1.9and2.2holds.

1Consider thath0 0. IfxT1>0, then we obtain

xT1−x0

T1

xT1−αxη T1

< xT1 T1−η

αxη T1

xT1−x

η T1−η ,

2.9

which is a contradiction to the convexity ofxt.

Ifxη>0, similar to the above proof, we can also get a contradiction.

IfxT1 0, and sox0 0, then there exists at0 ∈1, ηη, Tsuch that

xt0>0. Otherwise,xt≡0 is a trivial solution. Assume thatt0 ∈1, η, then

xT1−xt0

T1−t0 −

xt0

T1−t0

,

xt0

ηt0 −

xt0

ηt0

,

2.10

which implies that

xT1−xt0

T1−t0

> x

ηxt0

ηt0

. 2.11

A contradiction to the convexity ofxtfollows. Ift0∈η, T, we can also get a contradiction.

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Now we obtain

xT1−x0

T1

xT1−αxηh0

T1

xT1

T1−η T1−η

h0

T1

< xT1−x

η T1−η ,

2.12

which is a contradiction to the convexity ofxt.

Next, we consider the existence of the Green’s function of equation

−Δ2ut1 ft,

u0 αuη,

uT1 βuξ.

2.13

We always assume that

H10≤α,β≤1 andαβ <1.

Motivated by Zhao10, we have the following conclusions.

Theorem 2.2. The Green’s function for second-order four-point linear BVP2.13is given by

G1t, s Gt, s

αT1−t αη 1−αT

αηβ1−αt 1−αT1−βξ

1−βαη 1−αT1−βξ G

η, s

β1−αtαβη

1−βαη 1−αT1−βξGξ, s,

2.14

where

Gt, s ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

sT1−t

T1 , 0≤st−1,

tT1−s

T1 , tsT1.

2.15

Proof. Consider the second-order two-point BVP

−Δ2ut1 ft, t1, T,

u0 0,

uT1 0.

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It is easy to find that the solution of BVP2.16is given by

ut

T

s1

Gt, sfs, 2.17

u0 0, uT1 0,

T

s1

Gη, sfs. 2.18

The three-point BVP

−Δ2ut1 ft, t1, T,

u0 αuη, t∈−τ,0,

uT1 0

2.19

can be obtained from replacingu0 0 byu0 αuηin2.16. Thus we suppose that the solution of2.19can be expressed by

vt ut cdtuη, 2.20

wherecanddare constants that will be determined. From2.18and2.20, we have

v0 u0 cuη,

vηuηcdηuη1cdηuη,

vT1 uT1 cdT1 cdT1uη.

2.21

Putting the above equations into2.19yields

1−αcαηdα,

c T1d0. 2.22

ByH1, we obtaincanddby solving the above equation:

c αT1 αη 1−αT1,

dα

αη 1−αT1.

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By2.19and2.20, we have

v0 αvη,

vT1 0,

cdξuη.

2.24

The four-point BVP2.13can be obtained from replacinguT1 0 byuT1 βuξin

2.19. Thus we suppose that the solution of2.13can be expressed by

wt vt abtvξ, 2.25

whereaandbare constants that will be determined. From2.24and2.25, we get

w0 v0 avξ αvηavξ,

wηvηabηvξ,

wT1 vT1 abT1 abT1vξ,

abξvξ.

2.26

Putting the above equations into2.13yields

1−αaαηb0,

1−βaT1−βξbβ. 2.27

ByH1, we can easily obtain

a αβη

1−βαη 1−αT1−βξ,

b β1−α

1−βαη 1−αT1−βξ.

2.28

Then by2.17,2.20,2.23,2.25, and2.28, the solution of BVP2.13can be expressed by

wt

T

s1

G1t, sfs, 2.29

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Remark 2.3. ByH1, we can see thatG1t, s>0 fort, s∈0, T12. Let

m min

t,s∈1,T2

G1t, s, M max

t,s∈1,T2

G1t, s. 2.30

ThenMm >0.

Lemma 2.4. Assume that (H1) holds. Then the second-order four-point BVP 2.13has a unique

solution which is given in2.29.

Proof. We need only to show the uniqueness.

Obviously, wt in 2.29 is a solution of BVP 2.13. Assume that vt is another solution of BVP2.13. Let

zt vtwt, t∈−τ, T1. 2.31

Then by2.13, we have

−Δ2zt1 −Δ2vt1 Δ2wt10, t1, T, 2.32

z0 v0−w0 αzη,

zT1 vT1−wT1 βzξ. 2.33

From2.32we have, fort∈1, T,

zt c1tc2, 2.34

which implies that

z0 c2, z

ηc1ηc2, c1ξc2, zT1 c1T1 c2. 2.35

Combining2.33with2.35, we obtain

αηc1−1−αc20,

T1−βξc1

1−βc2 0.

2.36

ConditionH1implies that2.36has a unique solutionc1c20. Thereforevtwtfor

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3. Existence of Positive Solutions

In this section, we discuss the BVP1.9. Assume thath0 0,γ0. We rewrite BVP1.9as

−Δ2ut1 rtft, u

t, t∈1, T,

u0αu

ηh, t∈−τ,0,

uT1 βuξ

3.1

withh0 0.

Suppose thatutis a solution of the BVP3.1. Then it can be expressed as

ut ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

T

s1

G1t, srsfs, us, t∈1, T,

αuηht, t∈−τ,0,

βuξ, tT1.

3.2

Lemma 3.1see Guo et al.11. Assume thatEis a Banach space andKEis a cone inE. Let

Kp{uK| up}. Furthermore, assume thatΦ:KKis a completely continuous operator andΦu /uforu∂Kp{uK| up}. Thus, one has the following conclusions:

(1) ifu ≤ Φu foru∂Kp, theniΦ, Kp, K 0; (2) ifu ≥ Φu foru∂Kp, theniΦ, Kp, K 1.

Assume thatf≡0. Then3.1may be rewritten as

−Δ2ut1 0, t1, T,

u0 αu

ηh,

uT1 βuξ.

3.3

Letutbe a solution of3.3. Then by3.2andξ, η∈1, T, it can be expressed as

ut ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

0, t∈1, T,

ht, t∈−τ,0,

0, tT1.

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Letutbe a solution of BVP3.1andyt utut. Then fort∈1, Twe have

ytutand

yt ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

T

s1

G1t, srsf

s, ysus

, t∈1, T,

αyη, t∈−τ,0,

βyξ, tT1.

3.5

Let

u max

t∈−τ,T1|ut|, E{u|u:−τ, T1 → R},

K

uE| min

t∈1,Tut

m

Mu, ut αu

η, t∈−τ,0, uT1 βuξ

.

3.6

ThenEis a Banach space endowed with norm · andKis a cone inE. ForyK, we have byH1and the definition ofK,

y max

t∈−τ,T1yttmax∈1,Tyt. 3.7

For everyy∂Kp,s ∈ 1, T,and k ∈ −τ,0, by the definition of K and 3.5, if

sk≤0, we have

ysysk αy

η. 3.8

IfTsk≥1, we have, by3.4,

ususk 0, ysysk≥ min t∈1,Tyt

m

My, 3.9

hence by the definition of · τ, we obtain forsτ1, T

ysτm

My. 3.10

Lemma 3.2. For everyyK, there ist0∈τ1, T, such that

yt

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Proof. Forsτ1, T, k ∈−τ,0, andsk∈1, T, by the definitions of · τand · , we

have

ysτ max

k∈−τ,0ysk,

y max

t∈1,Tyt.

3.12

Obviously, there is at0∈τ1, T, such that3.11holds.

Define an operatorΦ:KEby

Φyt ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ T

s1

G1t, srsf

s, ysus

, t∈1, T,

αΦyη, t∈−τ,0,

βΦyξ, tT1.

3.13

Then we may transform our existence problem of positive solutions of BVP3.1into a fixed point problem of operator3.13.

Lemma 3.3. Consider thatΦKK.

Proof. Ift∈−τ,0andtT1,Φyt αΦηandΦyT1 βΦξ, respectively. Thus,

H1yields

Φy max

t∈−τ,Ty

t max

t∈1,TΦy

tΦy1,T. 3.14

It follows from the definition ofKthat

min

t∈1,T

Φyt min

t∈1,T T

s1

G1t, srsf

s, ysus

m

T

s1

rsfs, ysus

m

M

T

s1

max

1≤s,tTG1t, s

rsfs, ysus

m

Mtmax∈1,T T

s1

G1t, srsf

s, ysus

m

MΦy,

3.15

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Lemma 3.4. Suppose that (H1) holds. ThenΦ:KKis completely continuous.

We assume that

H2

T

t1rt>0,

H3hhτ max

t∈−τ,0ht>0.

We have the following main results.

Theorem 3.5. Assume that (H1)–(H3) hold. Then BVP3.1has at least one positive solution if the

following conditions are satisfied:

H4there exists ap1> hsuch that, fors∈1, T, ifφτp1h, thenfs, φR1p1;

H5there exists ap2> p1such that, fors∈1, T, ifφτm/Mp2, thenfs, φR2p2

or

H61> α >0;

H7there exists a0< r1 < p1such that, fors∈1, T, ifφτr1, thenfs, φR2r1;

H8 there exists an r2 ≥ max{p2h,Mh/mα}, such that, for s ∈ 1, T, ifφτ

mα/Mr2−h, thenfs, φR1r2,

where

R1≤

1

MTs1rs, R2≥

1

mTsτ1rs. 3.16

Proof. Assume thatH4andH5hold. For everyy∂Kp1, we haveysusτp1h, thus

ΦyΦy1,T

M

T

s1

rsfs, ysus

MR1p1

T

s1

rs

p1

y,

3.17

which implies byLemma 3.1that

iΦ, Kp1, K

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For everyy∂Kp2, by3.8–3.10andLemma 3.2, we have, forsτ1, T,ysτ

m/My m/Mp2. Then by3.13andH5, we have

ΦyΦy1,Tm

T

1

rsfs, ysus

m

T

1

rsfs, ys

mR2p2

T

1

rsp2y,

3.19

which implies byLemma 3.1that

iΦ, Kp2, K

0. 3.20

So by3.18and3.20, there exists one positive fixed pointy1of operatorΦwithy1 ∈Kp2\

Kp1.

Assume thatH6–H8hold, for everyy∂Kr1andsτ1, T,ysusτ ysτ

yr1, byH7, we have

Φyy. 3.21

Thus we have fromLemma 3.1that

iΦ, Kr1, K 0. 3.22

For everyy∂Kr2, by3.8–3.10, we haveysusτysτhmα/Mr2−h >0,

Φyy. 3.23

Thus we have fromLemma 3.1that

iΦ, Kr2, K 1. 3.24

So by 3.22 and 3.24, there exists one positive fixed pointy2 of operator Φwith

y2∈Kr2\Kr1.

Consequently,u1y1uoru2y2uis a positive solution of BVP3.1.

Theorem 3.6. Assume that (H1)–(H3) hold. Then BVP3.1has at least one positive solution if (H4)

and (H7) or (H5) and (H8) hold.

Theorem 3.7. Assume that (H1)–( H3) hold. Then BVP3.1has at least two positive solutions if

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Theorem 3.8. Assume that (H1)–(H3) hold. Then BVP3.1has at least three positive solutions if

(H4)–(H8) hold.

Assume thath0>0,γ >0, and

H9 1−βh0−1−αγ >0.

DefineHt:−τ, T1 → Ras follows:

Ht ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

ht, t∈−τ,0,

0, t∈1, T,

HT1, tT1,

3.25

which satisfies

H10 1−αHT1−1−βh0>0.

Obviously,Htexists.

Assume thatutis a solution of1.9. Let

wt ut pHt B, 3.26

where

p

1−βh0−1−αγ

1−αHT1−1−βh0, B

h0γHT1

1−αHT1−1−βh0. 3.27

By1.9,3.26,3.27,H7,H8, and the definition ofHt, we have

w0 u0 ph0 B

αwηph0 1−αBh0

αwη,

3.28

wT1 uT1 phT1 B

βwξ pHT1 1−βBγ

βwξ,

3.29

and, fort∈1, T,

−Δ2wt1 −Δ2ut1pΔ2Ht1

rtft, utpΔ2Ht−1

rtft, wtpHtB

p{Ht1−Ht−1}.

3.30

Let

Ft, wt rtf

t, wtpHtB

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Then by3.27,H9,H10, and the definition ofHt, we haveFt, wt > 0 fort

1, T.Thus, the BVP1.9can be changed into the following BVP:

−Δ2wt1 Ft, w

t, t∈1, T,

w0αw

ηg, t∈−τ,0,

wT1 βwξ,

3.32

withgBαhpH0B andg0 0.

Similar to the above proof, we can show that1.9has at least one positive solution. Consequently,1.9has at least one positive solution.

Example 3.9. Consider the following BVP:

−Δ2ut1 t

120ft, ut, t∈1,5,

u0u2 t 2

4, t∈−2,0,

uT1 1 2u4.

3.33

That is,

T5, τ 2, α 1

2, β1, ξ2, η4, ht

t2

4, rt

t

120.

3.34

Then we obtain

h1, 21

24≤G1t, s≤ 163

40 ,

5

s1

rt 1

8,

5

s3

rt 1

10. 3.35

Let

ft, φ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

2R2

p2−r1

π arctan

sm Mp2

R2p2, sm

Mp2,

2R1r2−R2p2

π arctan

sm Mp2

R2p2, s > m

Mp2,

R1

3

2, R2 12, r11, r2400, p14, p240,

3.36

wheresφτ.

By calculation, we can see thatH4–H8hold, then byTheorem 3.8, the BVP3.33

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4. Eigenvalue Intervals

In this section, we consider the following BVP with parameterλ:

−Δ2ut1 λrtft, u

t, t∈1, T,

u0αu

ηh, t∈−τ,0,

uT1 βuξ

4.1

withh0 0.

The BVP4.1is equivalent to the equation

ut ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ λ T

s1

G1t, srsfs, us, t∈1, T,

αuηht, t∈−τ,0,

βuξ, tT1.

4.2

Letutbe the solution of3.3,yt utut. Then we have

yt ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ λ T

s1

G1t, srsf

s, ysus

, t∈1, T,

αyη, t∈−τ,0,

βyξ, tT1.

4.3

LetEandKbe defined as the above. DefineΦ:KEby

Φyt ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ λ T

s1

G1t, srsf

s, ysus

, t∈1, T,

αΦyη, t∈−τ,0,

βΦyξ, tT1.

4.4

Then solving the BVP 4.1 is equivalent to finding fixed points in K. Obviously Φ is completely continuous and keeps theKinvariant forλ≥0.

Define

f0 lim inf

φτ→0tmin∈1,T

ft, φ

φτ

, flim inf

φτ→ ∞tmin∈1,T

ft, φ

φτ

, f∞lim sup

φτ→ ∞

max

t∈1,T

ft, φ

φτ

,

4.5

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Theorem 4.1. Assume that (H1), (H2), (H6),

H11 r min

t∈1,Trt>0,

H12 min{1/mrf0, M/m2f0

T

1rs}< λ <1/Mδf

T

s1rs

hold, whereδ max{1,1μα}, then BVP4.1has at least one positive solution, whereμis a positive constant.

Proof. Assume that conditionH12holds. If λ >1/mrf0and f0 < ∞, there exists an > 0

sufficiently small, such that

λ≥ 1 mrf0−

. 4.6

By the definition off0, there is anr1>0, such that for 0<φτr1,

min

t∈1,T

ft, φ

φτ > f0−. 4.7

It follows that, fort∈1, Tand 0τr1,

ft, φ>f0−φτ. 4.8

For everyy∂Kr1andsτ1, T, by3.9, we have

ysusτysτyr1. 4.9

Therefore by3.13andLemma 3.2, we have

Φy max

t∈1,Tλ T

s1

G1t, srsf

s, ysus

λmax

t∈1,T T

1

G1t, srsf

s, ysus

mλrf0−yt0τ

mλrf0−y

y.

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If λ > M/m2f

0Tsτ1rs, then for a sufficiently small > 0, we have λM/m2f0 −

Tsτ1rs.Similar to the above, for everyy∂Kr1, we obtain by3.10

Φy

T

1

rsf0−ysτ

T

1

rsf0−

m My

m2λ

f0−

M

T

1

rsy

y.

4.11

Iff0 ∞, chooseK >0 sufficiently large, such that

m2λK

M

T

1

rs≥1. 4.12

By the definition off0, there is anr1>0,such that, fort∈1, Tand 0τr1,

ft, φ> Kφτ. 4.13

For everyy∂Kr1, by3.8–3.10and3.13, we have

Φyy, 4.14

which implies that

iΦ, Kr1, K 0. 4.15

Finally, we consider the assumptionλ < 1/MδfTs1rs. By the definition off∞, there is

r > max{r1, h/μα}, such that, fort∈1, Tandφr,

ft, φ<f1φ. 4.16

We now show that there is r2 ≥ r, such that, for y∂Kr2,Φyy.In fact, for

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we have

Φyy, 4.17

which implies that

iΦ, Kr2, K 1. 4.18

Theorem 4.2. Assume that (H1),(H2), and (H11) hold. Iff∞ ∞orf0 ∞, then there is aλ0 >0

such that for0< λλ0, BVP4.1has at least one positive solution.

Proof. Letr > hbe given. Define

Lmaxft, φ|t, φ∈1, T×Cr τ

. 4.19

ThenL >0, whereCr

τ{φ | φτr}.

For everyy∂Krh, we know thaty rh. By the definition of operatorΦ, we obtain

ΦyΦy

1,TλLM T

s1

rs. 4.20

It follows that we can takeλ0 rh/ML

T

s1rs>0 such that, for all 0< λλ0and all

y∂Krh,

Φyy. 4.21

Fix 0< λλ0. Iff∞ ∞, forC 1/λmr, we obtain a sufficiently largeR > r such

that, forφτR,

min

t∈1,T

ft, φ

Φτ

> C. 4.22

It follows that, forφτRandt∈1, T,

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For everyy∂KR, by the definition of·,·τand the definition ofLemma 3.2, there

exists at0∈τ1, Tsuch thatyyt0τ Randut0 0, thusyt0ut0τR. Hence

Φy max

t∈1,Tλ T

s1

G1t, srsf

s, ysus

≥ max

t∈1,TλG1t, t0rt0f

t0, yt0ut0

λmrCyt0τ

mCRλr

R

y.

4.24

Iff0 ∞,there iss < r, such that, for 0<φτsandt∈1, T,

ft, φ> Tφτ, 4.25

whereT >1/λmr.

For everyy∂Ks, by3.8–3.10andLemma 3.2,

Φy

T

1

rsfs, ys

Tmλ

T

1

rsysτ

Tmλryt0τ

Tmλry

y,

4.26

which by combining with4.21completes the proof.

Example 4.3. Consider the BVP3.33inExample 3.9with

ft, φ ⎧ ⎪ ⎨ ⎪ ⎩

Aarctans, sm Mp2, AarctansC

1000 , s >

m Mp2,

C1000− m

Mp2

Aarctanm

Mp2

,

4.27

wheresφτ,Ais some positive constant,p240, m 21/24, and M 163/40.

By calculation,f0A,fπA/2000, andr1/120; letδ1. Then by Theorem4.1,

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Acknowledgments

The authors would like to thank the editor and the reviewers for their valuable comments and suggestions which helped to significantly improve the paper. This work is supported by Distinguished Expert Science Foundation of Naval Aeronautical and Astronautical University.

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