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, un*1*, un, un*−1 1.1

with boundary value conditions

Δ*u*0*A,* *uk*1*B,* 1.2

where the operator*L*is the Jacobi operator

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

*xt* *ft, xt, xt*

*,* *t*∈0*, T,*

*α*0*x*0−*α*1*x*0 *φ*∈*Cr,*

*β*0*xT* *β*1*xT* *A*∈R*n,*

1.4

where*f*:0*, T*×*Cr*×R*n* → R*n*is a continuous function,*φ*∈*Cr* *C*−*r,*0*,*R*n, A*∈R*n*, and

*xtθ* *xtθ, θ*∈−*r,*0.

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

ΔΦ*p*Δ*xt* *rtfxt* 0*,* *t*∈ {1*, . . . , T*}*,*

*x*0*φ*∈*C,* Δ*xT*1 0*,*

1.5

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

is continuous,*T _{t}*

_{}

_{τ}_{}

_{1}

*rt>*0.

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

ΔΦ*p*Δ*xt* *rtfxt, xt* 0*,* *t*∈ {1*, . . . , T*}*,*

*α*0*x*0−*α*1Δ*x*0 *h,*

*β*0*xT*1 *β*1Δ*xT*1 *A,*

1.6

where*h* ∈ *C _{τ}* {

*φ*∈

*Cτ*|

*φθ*≥ 0

*, θ*∈ {−

*τ, . . . ,*0}},

*A*∈ R, and

*α*0,

*α*1,

*β*0, and

*β*1 are

nonnegative real constants.
For*a, b*∈Nand*a < b*, let

R_{}_{{}_{x}_{|}_{x}_{∈}_{R}_{, x}_{≥}_{0}_{}}_{,}

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

*Cτ*

*φ*|*φ*:−*τ,*0 → *R,* *C _{τ}*

*φ*∈

*Cτ*|

*φθ*≥0

*, θ*∈−

*τ,*0

*.*

1.7

Then*Cτ*and*Cτ* are both Banach spaces endowed with the max-norm

_{φ}

*τ* * _{k}*max

_{∈}

_{}

_{−}

_{τ,}_{0}

_{}

*φk.*1.8

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

−Δ2_{u}_{}_{t}_{−}_{1} _{r}_{}_{t}_{}_{f}_{}_{t, u}

*t,* *t*∈1*, T,*

*u*0*αu*

*ηh,* *t*∈−*τ,*0*,*

*uT*1 *βuξ* *γ,*

1.9

where*ξ, η* ∈ 1*, T*and *ξ < η*, 0 *< τ < T*,Δ*ut* *ut*1−*ut*,Δ2_{u}_{}_{t}_{ ΔΔ}_{u}_{}_{t}_{}_{,}_{f}_{:}

R×*Cτ* → Ris a continuous function,*h*∈*Cτ* and*ht*≥*h*0≥0 for*t*∈−*τ,*0,*α*,*β*, and*γ*

are nonnegative real constants, and*rt*≥0 for*t*∈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

*u*0 *αuηC* 1.10

from ordinary diﬀerence equations. Here this condition connects the history*u*0with the single

*uη*. This is suggested by the well-posedness of BVP1.9, since the function*f*depends on
the term*ut*i.e., past values of*u*.

As usual, a sequence{*u*−*τ, . . . , uT*1}is said to be a positive solution of BVP1.9
if it satisfies BVP1.9and*uk*≥0 for*k*∈−*τ, T*with*uk>*0 for*k*∈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*

*αT*1−*η> T*1*.* 2.2

*Then*1.9*has no positive solution.*

*Proof.* FromΔ2_{u}_{}_{t}_{−}_{1} _{−}_{r}_{}_{t}_{}_{f}_{}_{t, u}

*t*≤0, we know that*ut*is convex for*t*∈0*, T*1.

Assume that*xt*is a positive solution of1.9and2.1holds.

If*xT*1*>*0, then*xξ>*0. It follows that

*xT*1−*x*0

*T*1

*βxξ*−*x*0

*T*1

*>* *xξ*
*ξ* −

*x*0

*T*1

≥ *xξ*−*x*0

*ξ* *,*

2.3

which is a contradiction to the convexity of*xt*.

If*xT*1 0, then*xξ* 0. If*x*0*>*0, then we have

*xT*1−*x*0

*T*1 −

*x*0

*T*1*,*

*xξ*−*x*0

*ξ* −
*x*0

*ξ* *.*

2.4

Hence

*xT*1−*x*0

*T*1 *>*

*xξ*−*x*0

*ξ* *,* 2.5

which is a contradiction to the convexity of*xt*. If*xt*≡0 for*t*∈1*, T*, then*xt*is a trivial
solution. So there exists a*t*0∈1*, ξ*∪*ξ, T*such that*xt*0*>*0.

We assume that*t*0∈1*, ξ*. Then

*xT*1−*xt*0

*T*1−*t*0 −

*xt*0

*T*1−*t*0

*,*

*xξ*−*xt*0

*ξ*−*t*0 −

*xt*0

*ξ*−*t*0

*.*

2.6

Hence

*xT*1−*xt*0

*T*1−*t*0

*>* *xξ*−*xt*0
*ξ*−*t*0

*,* 2.7

which is a contradiction to the convexity of*xt*.

If*t*0∈*ξ, T*, similar to the above proof, we can also get a contradiction.

Now we have

*xT*1−*x*0

*T*1

*βxξ*−*x*0 *γ*
*T*1

≥ *xξ*

*ξ* −
*x*0

*T*1

*γ*
*T*1

≥ *xξ*−*x*0

*ξ*

*γ*
*T*1

*>* *xξ*−*x*0
*ξ* *,*

2.8

which is a contradiction to the convexity of*xt*.

Assume that*xt*is a positive solution of1.9and2.2holds.

1Consider that*h*0 0.
If*xT*1*>*0, then we obtain

*xT*1−*x*0

*T*1

*x*T1−*αxη*
*T*1

*<* *xT*1
*T*1−*η* −

*αxη*
*T*1

≤ *xT*1−*x*

*η*
*T*1−*η* *,*

2.9

which is a contradiction to the convexity of*xt*.

If*xη>*0, similar to the above proof, we can also get a contradiction.

If*xT*1 *xη* 0, and so*x*0 0, then there exists a*t*0 ∈1*, η*∪*η, T*such that

*xt*0*>*0. Otherwise,*xt*≡0 is a trivial solution. Assume that*t*0 ∈1*, η*, then

*xT*1−*xt*0

*T*1−*t*0 −

*xt*0

*T*1−*t*0

*,*

*xη*−*xt*0

*η*−*t*0 −

*xt*0

*η*−*t*0

*,*

2.10

which implies that

*xT*1−*xt*0

*T*1−*t*0

*>* *x*

*η*−*xt*0

*η*−*t*0

*.* 2.11

A contradiction to the convexity of*xt*follows.
If*t*0∈*η, T*, we can also get a contradiction.

Now we obtain

*xT*1−*x*0

*T*1

*xT*1−*αxη*−*h*0

*T*1

≤ *xT*1

*T*1−*η* −
*xη*
*T*1−*η* −

*h*0

*T*1

*<* *xT*1−*x*

*η*
*T*1−*η* *,*

2.12

which is a contradiction to the convexity of*xt*.

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

−Δ2_{u}_{}_{t}_{−}_{1} _{f}_{}_{t}_{}_{,}

*u*0 *αuη,*

*uT*1 *βuξ.*

2.13

We always assume that

H10≤*α*,*β*≤1 and*αβ <*1.

Motivated by Zhao10, we have the following conclusions.

**Theorem 2.2.** *The Green’s function for second-order four-point linear BVP*2.13*is given by*

*G*1*t, s* *Gt, s*

*αT*1−*t*
*αη* 1−*αT*1×

*αηβ*1−*αt* 1−*αT*1−*βξ*

1−*βαη* 1−*αT*1−*βξ* *G*

*η, s*

*β*1−*αtαβη*

1−*βαη* 1−*αT*1−*βξGξ, s,*

2.14

*where*

*Gt, s*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩

*sT*1−*t*

*T*1 *,* 0≤*s*≤*t*−1*,*

*tT*1−*s*

*T*1 *, t*≤*s*≤*T*1*.*

2.15

*Proof.* Consider the second-order two-point BVP

−Δ2_{u}_{}_{t}_{−}_{1} _{f}_{}_{t}_{}_{,}_{t}_{∈}_{}_{1}_{, T}_{}_{,}

*u*0 0*,*

*uT*1 0*.*

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

*ut*

*T*

*s*1

*Gt, sfs,* 2.17

*u*0 0*,* *uT*1 0*,* *uη*

*T*

*s*1

*Gη, sfs.* 2.18

The three-point BVP

−Δ2_{u}_{}_{t}_{−}_{1} _{f}_{}_{t}_{}_{,}_{t}_{∈}_{}_{1}_{, T}_{}_{,}

*u*0 *αuη,* *t*∈−*τ,*0*,*

*uT*1 0

2.19

can be obtained from replacing*u*0 0 by*u*0 *αuη*in2.16. Thus we suppose that the
solution of2.19can be expressed by

*vt* *ut* *cdtuη,* 2.20

where*c*and*d*are constants that will be determined.
From2.18and2.20, we have

*v*0 *u*0 *cuη,*

*vηuηcdηuη*1*cdηuη,*

*vT*1 *uT*1 *cdT*1*uη* *cdT*1*uη.*

2.21

Putting the above equations into2.19yields

1−*αc*−*αηdα,*

*c* *T*1*d*0*.* 2.22

ByH1, we obtain*c*and*d*by solving the above equation:

*c* *αT*1
*αη* 1−*αT*1*,*

*d* −*α*

*αη* 1−*αT*1*.*

By2.19and2.20, we have

*v*0 *αvη,*

*vT*1 0*,*

*vξ* *uξ* *cdξuη.*

2.24

The four-point BVP2.13can be obtained from replacing*uT*1 0 by*uT*1 *βuξ*in

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

*wt* *vt* *abtvξ,* 2.25

where*a*and*b*are constants that will be determined.
From2.24and2.25, we get

*w*0 *v*0 *avξ* *αvηavξ,*

*wηvηabηvξ,*

*wT*1 *vT*1 *abT*1*vξ* *abT*1*vξ,*

*wξ* *vξ* *abξvξ.*

2.26

Putting the above equations into2.13yields

1−*αa*−*αηb*0*,*

1−*βaT*1−*βξbβ.* 2.27

ByH1, we can easily obtain

*a* *αβη*

1−*βαη* 1−*αT*1−*βξ,*

*b* *β*1−*α*

1−*βαη* 1−*αT*1−*βξ.*

2.28

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

*wt*

*T*

*s*1

*G*1*t, sfs,* 2.29

*Remark 2.3.* ByH1, we can see that*G*1*t, s>*0 for*t, s*∈0*, T*12. Let

*m* min

*t,s*∈1*,T*2

*G*1*t, s,* *M* max

*t,s*∈1*,T*2

*G*1*t, s.* _{2.30}

Then*M*≥*m >*0.

**Lemma 2.4.** *Assume that (H*1*) holds. Then the second-order four-point BVP* 2.13*has a unique*

*solution which is given in*2.29*.*

*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* *vt*−*wt,* *t*∈−*τ, T*1*.* 2.31

Then by2.13, we have

−Δ2_{z}_{}_{t}_{−}_{1} _{−Δ}2_{v}_{}_{t}_{−}_{1}_{ Δ}2_{w}_{}_{t}_{−}_{1}_{}_{≡}_{0}_{,}_{t}_{∈}_{}_{1}_{, T}_{}_{,}_{}_{2.32}_{}

*z*0 *v*0−*w*0 *αzη,*

*zT*1 *vT*1−*wT*1 *βzξ.* 2.33

From2.32we have, for*t*∈1*, T*,

*zt* *c*1*tc*2*,* 2.34

which implies that

*z*0 *c*2*,* *z*

*ηc*1*ηc*2*,* *zξ* *c*1*ξc*2*,* *zT*1 *c*1*T*1 *c*2*.* 2.35

Combining2.33with2.35, we obtain

*αηc*1−1−*αc*20*,*

*T*1−*βξc*1

1−*βc*2 0*.*

2.36

ConditionH1implies that2.36has a unique solution*c*1*c*20. Therefore*vt*≡*wt*for

**3. Existence of Positive Solutions**

In this section, we discuss the BVP1.9.
Assume that*h*0 0,*γ*0.
We rewrite BVP1.9as

−Δ2_{u}_{}_{t}_{−}_{1} _{r}_{}_{t}_{}_{f}_{}_{t, u}

*t,* *t*∈1*, T,*

*u*0*αu*

*ηh,* *t*∈−*τ,*0*,*

*uT*1 *βuξ*

3.1

with*h*0 0.

Suppose that*ut*is a solution of the BVP3.1. Then it can be expressed as

*ut*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩

*T*

*s*1

*G*1*t, srsfs, us, t*∈1*, T,*

*αuηht,* *t*∈−*τ,*0*,*

*βuξ,* *tT*1*.*

3.2

**Lemma 3.1**see Guo et al.11. *Assume thatEis a Banach space andK* ⊂*Eis a cone inE. Let*

*Kp*{*u*∈*K*| *up*}*. Furthermore, assume that*Φ:*K* → *Kis a completely continuous operator*
*and*Φ*u /uforu*∈*∂Kp*{*u*∈*K*| *up*}*. Thus, one has the following conclusions:*

*(1) ifu* ≤ Φ*u* *foru*∈*∂Kp, theni*Φ*, Kp, K* 0*;*
*(2) ifu* ≥ Φ*u* *foru*∈*∂Kp, theni*Φ*, Kp, K* 1*.*

Assume that*f*≡0. Then3.1may be rewritten as

−Δ2_{u}_{}_{t}_{−}_{1} _{0}_{,}_{t}_{∈}_{}_{1}_{, T}_{}_{,}

*u*0 *αu*

*ηh,*

*uT*1 *βuξ.*

3.3

Let*ut*be a solution of3.3. Then by3.2and*ξ, η*∈1*, T*, it can be expressed as

*ut*
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩

0*,* *t*∈1*, T,*

*ht, t*∈−*τ,*0*,*

0*,* *tT*1*.*

Let*ut*be a solution of BVP3.1and*yt* *ut*−*ut*. Then for*t*∈1*, T*we have

*yt*≡*ut*and

*yt*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩

*T*

*s*1

*G*1*t, srsf*

*s, ysus*

*, t*∈1*, T,*

*αyη,* *t*∈−*τ,*0*,*

*βyξ,* *tT*1*.*

3.5

Let

*u* max

*t*∈−*τ,T*1|*ut*|*,* *E*{*u*|*u*:−*τ, T*1 → *R*}*,*

*K*

*u*∈*E*| min

*t*∈1*,Tut*≥

*m*

*Mu, ut* *αu*

*η, t*∈−*τ,*0*, uT*1 *βuξ*

*.*

3.6

Then*E*is a Banach space endowed with norm · and*K*is a cone in*E*.
For*y*∈*K*, we have byH1and the definition of*K*,

*y* max

*t*∈−*τ,T*1*ytt*max∈1*,Tyt.* 3.7

For every*y* ∈ *∂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

If*T* ≥*sk*≥1, we have, by3.4,

*ususk* 0*,* *ysysk*≥ min
*t*∈1*,Tyt*≥

*m*

*My,* 3.9

hence by the definition of · *τ*, we obtain for*s*∈*τ*1*, T*

*ys _{τ}* ≥

*m*

*My.* 3.10

**Lemma 3.2.** *For everyy*∈*K, there ist*0∈*τ*1*, T, such that*

_{y}_{t}

*Proof.* For*s*∈*τ*1*, T, k* ∈−*τ,*0, and*sk*∈1*, T*, by the definitions of · *τ*and · , we

have

*ys _{τ}* max

*k*∈−*τ,*0*ysk,*

*y* max

*t*∈1*,Tyt.*

3.12

Obviously, there is a*t*0∈*τ*1*, T*, such that3.11holds.

Define an operatorΦ:*K* → *E*by

Φ*yt*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

*, t*∈1*, T,*

*α*Φ*yη,* *t*∈−*τ,*0*,*

*β*Φ*yξ,* *tT*1*.*

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 that*Φ*K*⊂*K.*

*Proof.* If*t*∈−*τ,*0and*tT*1,Φ*yt* *α*Φ*η*andΦ*yT*1 *β*Φ*ξ*, respectively. Thus,

H1yields

Φ*y* max

*t*∈−*τ,T*1Φ*y*

*t* max

*t*∈1*,T*Φ*y*

*t*Φ*y*1*,T.* _{}3.14

It follows from the definition of*K*that

min

*t*∈1*,T*

Φ*yt* min

*t*∈1*,T*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

≥*m*

*T*

*s*1

*rsfs, ysus*

≥ *m*

*M*

*T*

*s*1

max

1≤*s,t*≤*TG*1*t, s*

*rsfs, ysus*

≥ *m*

*Mt*max∈1*,T*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

*m*

*M*Φ*y,*

3.15

**Lemma 3.4.** *Suppose that (H*1*) holds. Then*Φ:*K* → *Kis completely continuous.*

We assume that

H2

_{T}

*t*1*rt>*0*,*

H3*hhτ* max

*t*∈−*τ,*0*ht>*0.

We have the following main results.

**Theorem 3.5.** *Assume that (H*1*)–(H*3*) hold. Then BVP*3.1*has at least one positive solution if the*

*following conditions are satisfied:*

H4*there exists ap*1*> hsuch that, fors*∈1*, T, ifφτ* ≤*p*1*h, thenfs, φ*≤*R*1*p*1*;*

H5*there exists ap*2*> p*1*such that, fors*∈1*, T, ifφ _{τ}*≥

*m/Mp*2

*, thenfs, φ*≥

*R*2

*p*2

*or*

H61*> α >*0*;*

H7*there exists a*0*< r*1 *< p*1*such that, fors*∈1*, T, ifφ _{τ}* ≤

*r*1

*, thenfs, φ*≥

*R*2

*r*1

*;*

H8 *there exists an* *r*2 ≥ max{*p*2*h,Mh/mα*}*, such that, for* *s* ∈ 1*, T, ifφτ* ≥

*mα/Mr*2−*h, thenfs, φ*≤*R*1*r*2*,*

*where*

*R*1≤

1

*MT _{s}*

_{}

_{1}

*rs,*

*R*2≥

1

*mT _{s}*

_{}

_{τ}_{}

_{1}

*rs.*3.16

*Proof.* Assume thatH4andH5hold. For every*y*∈*∂Kp*1, we have*ysusτ* ≤*p*1*h*, thus

_{Φ}*y*Φ*y*_{}_{1}_{,T}_{}

≤*M*

*T*

*s*1

*rsfs, ysus*

≤*MR*1*p*1

*T*

*s*1

*rs*

≤*p*1

*y,*

3.17

which implies byLemma 3.1that

*i*Φ*, Kp*1*, K*

For every*y*∈*∂Kp*2, by3.8–3.10andLemma 3.2, we have, for*s*∈*τ*1*, T*,*ysτ* ≥

*m/My* *m/Mp*2. Then by3.13andH5, we have

_{Φ}*y*Φ*y*_{}_{1}_{,T}_{}≥*m*

*T*

*sτ*1

*rsfs, ysus*

*m*

*T*

*sτ*1

*rsfs, ys*

≥*mR*2*p*2

*T*

*sτ*1

*rs*≥*p*2*y,*

3.19

which implies byLemma 3.1that

*i*Φ*, Kp*2*, K*

0*.* 3.20

So by3.18and3.20, there exists one positive fixed point*y*1of operatorΦwith*y*1 ∈*Kp*2\

*Kp*1.

Assume thatH6–H8hold, for every*y*∈*∂Kr*1and*s*∈*τ*1*, T*,*ysusτ* *ysτ* ≤

*yr*1, byH7, we have

_{Φ}*y*≥*y.* 3.21

Thus we have fromLemma 3.1that

*i*Φ*, Kr*1*, K* 0*.* 3.22

For every*y*∈*∂Kr*2, by3.8–3.10, we have*ysusτ* ≥ *ysτ*−*h*≥*mα/Mr*2−*h >*0,

Φ*y* ≤ *y.* 3.23

Thus we have fromLemma 3.1that

*i*Φ*, Kr*2*, K* 1*.* 3.24

So by 3.22 and 3.24, there exists one positive fixed point*y*2 of operator Φwith

*y*2∈*Kr*2\*Kr*1.

Consequently,*u*1*y*1*u*or*u*2*y*2*u*is a positive solution of BVP3.1.

**Theorem 3.6.** *Assume that (H*1*)–(H*3*) hold. Then BVP*3.1*has at least one positive solution if (H*4*)*

*and (H*7*) or (H*5*) and (H*8*) hold.*

**Theorem 3.7.** *Assume that (H*1*)–( H*3*) hold. Then BVP*3.1*has at least two positive solutions if*

**Theorem 3.8.** *Assume that (H*1*)–(H*3*) hold. Then BVP*3.1*has at least three positive solutions if*

*(H*4*)–(H*8*) hold.*

Assume that*h*0*>*0,*γ >*0, and

H9 1−*βh*0−1−*αγ >*0*.*

Define*Ht*:−*τ, T*1 → Ras follows:

*Ht*
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩

*ht,* *t*∈−*τ,*0*,*

0*,* *t*∈1*, T,*

*HT*1*, tT*1*,*

3.25

which satisfies

H10 1−*αHT*1−1−*βh*0*>*0.

Obviously,*Ht*exists.

Assume that*ut*is a solution of1.9. Let

*wt* *ut* *pHt* *B,* 3.26

where

*p*

1−*βh*0−1−*αγ*

1−*αHT*1−1−*βh*0*,* *B*

*h*0*γ*−*HT*1

1−*αHT*1−1−*βh*0*.* 3.27

By1.9,3.26,3.27,H7,H8, and the definition of*Ht*, we have

*w*0 *u*0 *ph*0 *B*

*αwηph*0 1−*αBh*0

*αwη,*

3.28

*wT*1 *uT*1 *phT*1 *B*

*βwξ* *pHT*1 1−*βBγ*

*βwξ,*

3.29

and, for*t*∈1*, T*,

−Δ2_{w}_{}_{t}_{−}_{1} _{−Δ}2_{u}_{}_{t}_{−}_{1}_{}_{−}_{p}_{Δ}2_{H}_{}_{t}_{−}_{1}_{}

*rtft, ut*−*p*Δ2*Ht*−1

*rtft, wt*−*pHt*−*B*

−*p*{*Ht*1−*Ht*−1}*.*

3.30

Let

*Ft, wt* *rtf*

*t, wt*−*pHt*−*B*

Then by3.27,H9,H10, and the definition of*Ht*, we have*Ft, wt* *>* 0 for*t* ∈

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

−Δ2_{w}_{}_{t}_{−}_{1} _{F}_{}_{t, w}

*t,* *t*∈1*, T,*

*w*0*αw*

*ηg,* *t*∈−*τ,*0*,*

*wT*1 *βwξ,*

3.32

with*g*−*BαhpH*0*B*∈*Cτ* and*g*0 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:

−Δ2_{u}_{}_{t}_{−}_{1} *t*

120*ft, ut,* *t*∈1*,*5*,*

*u*0*u*2 *t*
2

4*,* *t*∈−2*,*0*,*

*uT*1 1
2*u*4*.*

3.33

That is,

*T*5*,* *τ* 2*,* *α* 1

2*,* *β*1*,* *ξ*2*,* *η*4*,* *ht*

*t*2

4*,* *rt*

*t*

120*.*

3.34

Then we obtain

*h*1*,* 21

24≤*G*1*t, s*≤
163

40 *,*

5

*s*1

*rt* 1

8*,*

5

*s*3

*rt* 1

10*.* 3.35

Let

*ft, φ*
⎧
⎪
⎪
⎨
⎪
⎪
⎩

2*R*2

*p*2−*r*1

*π* arctan

*s*− *m*
*Mp*2

*R*2*p*2*,* *s*≤ *m*

*Mp*2*,*

2*R*1*r*2−*R*2*p*2

*π* arctan

*s*− *m*
*Mp*2

*R*2*p*2*, s >* *m*

*Mp*2*,*

*R*1

3

2*,* *R*2 12*,* *r*11*,* *r*2400*,* *p*14*,* *p*240*,*

3.36

where*sφτ.*

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

**4. Eigenvalue Intervals**

In this section, we consider the following BVP with parameter*λ*:

−Δ2_{u}_{}_{t}_{−}_{1} _{λr}_{}_{t}_{}_{f}_{}_{t, u}

*t,* *t*∈1*, T,*

*u*0*αu*

*ηh,* *t*∈−*τ,*0*,*

*uT*1 *βuξ*

4.1

with*h*0 0.

The BVP4.1is equivalent to the equation

*ut*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
*λ*
*T*

*s*1

*G*1*t, srsfs, us, t*∈1*, T,*

*αuηht,* *t*∈−*τ,*0*,*

*βuξ,* *tT*1*.*

4.2

Let*ut*be the solution of3.3,*yt* *ut*−*ut*. Then we have

*yt*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
*λ*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

*, t*∈1*, T,*

*αyη,* *t*∈−*τ,*0*,*

*βyξ,* *tT*1*.*

4.3

Let*E*and*K*be defined as the above. DefineΦ:*K* → *E*by

Φ*yt*
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
*λ*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

*, t*∈1*, T,*

*α*Φ*yη,* *t*∈−*τ,*0*,*

*β*Φ*yξ,* *tT*1*.*

4.4

Then solving the BVP 4.1 is equivalent to finding fixed points in *K*. Obviously Φ is
completely continuous and keeps the*K*invariant for*λ*≥0.

Define

*f*0 lim inf

*φ*τ→0*t*min∈1*,T*

*ft, φ*

*φτ*

*,* *f*_{∞}lim inf

*φ*τ→ ∞*t*min∈1*,T*

*ft, φ*

*φτ*

*,* *f*∞lim sup

*φ*τ→ ∞

max

*t*∈1*,T*

*ft, φ*

*φτ*

*,*

4.5

**Theorem 4.1.** *Assume that (H*1*), (H*2*), (H*6*),*

H11 *r* min

*t*∈1*,Trt>*0*,*

H12 min{1*/mrf*0*, M/m*2*f*0

_{T}

*sτ*1*rs*}*< λ <*1*/Mδf*∞

_{T}

*s*1*rs*

*hold, whereδ* max{1*,*1*μα*}*, then BVP*4.1*has at least one positive solution, whereμis a*
*positive constant.*

*Proof.* Assume that conditionH12holds. If *λ >*1*/mrf*0and *f*0 *<* ∞, there exists an* >* 0

suﬃciently small, such that

*λ*≥ 1
*mrf*0−

*.* 4.6

By the definition of*f*0, there is an*r*1*>*0, such that for 0*<φτ* ≤*r*1,

min

*t*∈1*,T*

*ft, φ*

*φ _{τ}*

*> f*0−

*.*4.7

It follows that, for*t*∈1*, T*and 0*<φ _{τ}* ≤

*r*1,

*ft, φ>f*0−*φ _{τ}.* 4.8

For every*y*∈*∂Kr*1and*s*∈*τ*1*, T*, by3.9, we have

*ysus _{τ}ys_{τ}* ≤

*yr*1

*.*4.9

Therefore by3.13andLemma 3.2, we have

Φ*y* max

*t*∈1*,Tλ*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

≥*λ*max

*t*∈1*,T*
*T*

*sτ*1

*G*1*t, srsf*

*s, ysus*

≥*mλrf*0−*yt*0*τ*

*mλrf*0−*y*

≥ *y.*

If *λ > M/m*2_{f}

0*Tsτ*1*rs,* then for a suﬃciently small * >* 0, we have *λ* ≥ *M/m*2*f*0 −

*T _{s}*

_{}

_{τ}_{}

_{1}

*rs.*Similar to the above, for every

*y*∈

*∂Kr*1, we obtain by3.10

Φ*y* ≥*mλ*

*T*

*sτ*1

*rsf*0−*ys _{τ}*

≥*mλ*

*T*

*sτ*1

*rsf*0−

*m*
*My*

≥ *m*2*λ*

*f*0−

*M*

*T*

*sτ*1

*rsy*

≥*y.*

4.11

If*f*0 ∞, choose*K >*0 suﬃciently large, such that

*m*2_{λK}

*M*

*T*

*sτ*1

*rs*≥1*.* 4.12

By the definition of*f*0, there is an*r*1*>*0*,*such that, for*t*∈1*, T*and 0*<φ _{τ}* ≤

*r*1

*,*

*ft, φ> Kφ _{τ}.* 4.13

For every*y*∈*∂Kr*1, by3.8–3.10and3.13, we have

Φ*y* ≥*y,* 4.14

which implies that

*i*Φ*, Kr*1*, K* 0*.* 4.15

Finally, we consider the assumption*λ <* 1*/Mδf*∞*T _{s}*

_{}

_{1}

*rs*. By the definition of

*f*∞, there is

*r > max*{*r*1*, h/μα*}, such that, for*t*∈1*, T*and*φ* ≥*r*,

*ft, φ<f*∞1*φ.* 4.16

We now show that there is *r*2 ≥ *r*, such that, for *y* ∈ *∂Kr*2,Φ*y* ≤ *y.*In fact, for

we have

_{Φ}*y*≤*y,* 4.17

which implies that

*i*Φ*, Kr*2*, K* 1*.* 4.18

**Theorem 4.2.** *Assume that (H*1*),(H*2*), and (H*11*) hold. Iff*∞ ∞*orf*0 ∞*, then there is aλ*0 *>*0

*such that for*0*< λ*≤*λ*0*, BVP*4.1*has at least one positive solution.*

*Proof.* Let*r > h*be given. Define

*L*max*ft, φ*|*t, φ*∈1*, T*×*Cr*
*τ*

*.* 4.19

Then*L >*0, where*Cr*

*τ*{*φ*∈*Cτ* | *φτ* ≤*r*}.

For every*y* ∈ *∂K _{r}*

_{−}

*, we know that*

_{h}*y*

*r*−

*h*. By the definition of operatorΦ, we obtain

_{Φ}_{y}_{}_{Φ}_{y}

1*,T*≤*λLM*
*T*

*s*1

*rs.* 4.20

It follows that we can take*λ*0 *r*−*h/ML*

_{T}

*s*1*rs>*0 such that, for all 0*< λ*≤*λ*0and all

*y*∈*∂K _{r}*

_{−}

*,*

_{h}_{Φ}*y*≤*y.* 4.21

Fix 0*< λ*≤ *λ*0. If*f*∞ ∞, for*C* 1*/λmr*, we obtain a suﬃciently large*R > r* such

that, for*φτ* ≥*R*,

min

*t*∈1*,T*

*ft, φ*

Φ*τ*

*> C.* 4.22

It follows that, for*φτ* ≥*R*and*t*∈1*, T*,

For every*y*∈*∂KR*, by the definition of·,·*τ*and the definition ofLemma 3.2, there

exists a*t*0∈*τ*1*, T*such that*yyt*0*τ* *R*and*ut*0 0, thus*yt*0*ut*0*τ* ≥*R*. Hence

Φ*y* max

*t*∈1*,Tλ*
*T*

*s*1

*G*1*t, srsf*

*s, ysus*

≥ max

*t*∈1*,TλG*1*t, t*0*rt*0*f*

*t*0*, yt*0*ut*0

≥*λmrCyt*0*τ*

≥*mCRλr*

*R*

*y.*

4.24

If*f*0 ∞*,*there is*s < r*, such that, for 0*<φτ* ≤*s*and*t*∈1*, T*,

*ft, φ> Tφ _{τ},* 4.25

where*T >*1*/λmr*.

For every*y*∈*∂Ks*, by3.8–3.10andLemma 3.2,

Φ*y* ≥*mλ*

*T*

*sτ*1

*rsfs, ys*

≥*Tmλ*

*T*

*sτ*1

*rsys _{τ}*

≥*Tmλryt*0*τ*

*Tmλry*

≥*y,*

4.26

which by combining with4.21completes the proof.

*Example 4.3.* Consider the BVP3.33inExample 3.9with

*ft, φ*
⎧
⎪
⎨
⎪
⎩

*A*arctan*s,* *s*≤ *m*
*Mp*2*,*
*A*arctan*sC*

1000 *, s >*

*m*
*Mp*2*,*

*C*1000− *m*

*Mp*2

*A*arctan*m*

*Mp*2

*,*

4.27

where*sφτ*,*A*is some positive constant,*p*240*, m* 21*/*24*,* and *M* 163*/*40.

By calculation,*f*0*A*,*f*∞*πA/*2000, and*r*1*/*120; let*δ*1. Then by Theorem4.1,

**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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