AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 10.1155/2014/578672 578672 Research Article The Existence of Positive Solutions for a Fourth-Order Difference Equation with Sum Form Boundary Conditions Guo Yanping 1 Lv Xuefei 2 Ji Yude 2 Liang Yongchun 1 Latif Abdul 1 School of Electrical Engineering Hebei University of Science and Technology Shijiazhuang Hebei 050018 China hbu.cn 2 College of Sciences Hebei University of Science and Technology Shijiazhuang, Hebei 050018 China hbu.cn 2014 1672014 2014 10 03 2014 13 05 2014 17 7 2014 2014 Copyright © 2014 Yanping Guo et al. 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 fourth-order difference equation: Δ(z(k+1)Δ3u(k-1))=w(k)f(k,u(k)),  k{1,2,,n-1} subject to the boundary conditions: u(0)=u(n+2)=i=1n+1g(i)u(i), aΔ2u(0)-bz(2)Δ3u(0)=i=3n+1h(i)Δ2u(i-2), aΔ2u(n)-bz(n+1)Δ3u(n-1)=i=3n+1h(i)Δ2u(i-2), where a,b>0 and Δu(k)=u(k+1)-u(k) for k{0,1,,n-1},  f:{0,1,,n}×[0,+)[0,+) is continuous. h(i) is nonnegative i{2,3,,n+2}; g(i) is nonnegative for i{0,1,,n}. Using fixed point theorem of cone expansion and compression of norm type and Hölder’s inequality, various existence, multiplicity, and nonexistence results of positive solutions for above problem are derived, which extends and improves some known recent results.

1. Introduction

Boundary value problems (BVPs) for ordinary differential equations arise in different areas of applied mathematics and so on. The existence of solutions for second order and higher order nonlocal boundary value problems has been studied by several authors; for example, see  and the references therein. Many authors have also discussed the existence of positive solutions for higher order difference equation BVPs [12, 13], by using fixed point theorem of cone expansion and compression of norm type, sufficient conditions for the existence of positive solutions for fourth-order and third-order difference equation BVPs are established, respectively. Recently, there has been much attention on the existence of positive solutions for the fourth-order differential equations with integral boundary conditions . In , Zhang and Ge considered the differential equation BVP: (1)(u(t)x′′′(t))=w(t)f(t,x(t)),0<t<1,x(0)=x(1)=01g(s)x(s)ds,ax′′(0)-blimt0+u(t)x′′′(0)=01h(s)x′′(s)ds,ax′′(1)+blimt1-u(t)x′′′(1)=01h(s)x′′(s)ds, where  a,b>0, uC1([0,1][0,+)) is symmetric on [0,1], wLp[0,1]  for some 1p+, and it is symmetric on the interval  [0,1],  f:[0,1]×[0,+)[0,+)  is continuous, and  f(1-t,x)=f(t,x)  for all  (t,x)[0,1]×[0,+), and  g,hL1[0,1]  are nonnegative, symmetric on  [0,1]. The authors made use of fixed point theorem of cone expansion and compression of norm type and Hölder inequality to prove the existence of positive solutions for the above problem.

Motivated by the above works, we intend to study the existence and nonexistence of positive solutions of the following fourth-order difference BVP with sum form boundary conditions: (2)Δ(z(k+1)Δ3u(k-1))=w(k)f(k,u(k)),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiik{1,2,,n-1},u(0)=u(n+2)=i=1n+1g(i)u(i),aΔ2u(0)-bz(2)Δ3u(0)=i=3n+1h(i)Δ2u(i-2),aΔ2u(n)+bz(n+1)Δ3u(n-1)=i=3n+1h(i)Δ2u(i-2).

Throughout this paper, we make the following assumptions:

z  is symmetric on  {2,3,,n+1}  and  0<k=2n+1z(k)<+;

there is a  m>0  such that  w(k)m/(n-1)  for  1p+;

f:{0,1,,n}×[0,+)[0,+)  is continuous;

h(i)  is nonnegative on  {2,3,,n+2}  and  0v<a,  g(i)  is nonnegative on  {0,1,,n+2}  and  0μ<1, where  v=i=3n+1h(i),  μ=i=1n+1g(i);

a,b>0,  Q=2ab+a2j=2n+1(1/z(j))>0.

In order to establish the existence of positive solutions of the problem (2), we need the following definitions, theorem, and lemma.

Definition 1.

A function  x(t)  is said to be a solution of problem (2) if  x(t)  satisfying BVP (2).

Definition 2 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Let  E  be a real Banach space over  R. A nonempty closed set  PE  is said to be a cone provided that

au+bvP  for all  u,vP  and all  a0,b0  and

u,-uP  implies  u=0.

Every cone  PE  induces an ordering in  E  given by  xy  if and only if  y-xP.

Theorem 3 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Let  P  be a cone in a real Banach space  E. Assume that  Ω1,Ω2  are bounded open sets in  E  with  0Ω1,Ω¯1Ω2. If (3)A:P(Ω¯2Ω1)P is completely continuous such that either

Axx,  xPΩ1  and  Axx,  xPΩ2,

or

Axx,  xPΩ1  and  Axx,  xPΩ2,

then  A  has at least one fixed point in  P(Ω¯2Ω1).

Lemma 4 (Hölder).

Suppose that  u={u1,u2,,un}  is a real-valued column; let (4)up={(k=1n|uk|p)1/p,0<p<,supk{1,2,,n}|uk|,p=,p, q satisfy the condition  1/p+1/q=1  which are called conjugate exponent, and  q=  for  p=1. If  1p, then (5)uv1upvq, which can be recorded as (6)k=1n|ukvk|{(k=1n|uk|p)1/p(k=1n|vk|q)1/q,1<p<,(k=1n|uk|)(supk{1,2,,n}|vk|),p=1,(supk{1,2,,n}|uk|)(k=1n|vk|),p=.

2. Preliminaries

Let  J={0,1,,n+2};   E={u(k):{0,1,,n+2}R}  is a real Banach space with the norm  ·  defined by (7)u=maxkJ|u(k)|.

Let  K  be a cone of  E, and (8)Kr={uK:ur},Kr={uK:u=r},K¯r,R={uK:ruR}, where  0<r<R.

In our main results, we will use the following lemmas and properties.

Lemma 5.

Suppose that  (A0)  and  (A1)  hold and  v1; then, for all  yE, the BVP (9)-Δ(z(k+1)Δu(k+1))=y(k+2),k{1,2,,n+1},au(2)-bz(2)Δu(2)=i=3n+1h(i)u(i),au(n+2)+bz(n+1)Δu(n+1)=i=3n+1h(i)u(i) has unique solution  u  given by (10)u(k)=i=3n+1H(k,i)y(i), where (11)H(k,i)=G(k,i)+1a-vτ=3n+1G(i,τ)h(τ),G(k,i)=1Q{(b+aj=kn+11z(j))(b+aj=2i-11z(j)),2i<k,(b+aj=in+11z(j))(b+aj=2k-11z(j)),kin+2, where  Q=2ab+a2j=2n+1(1/z(j)),  v=i=3n+1h(i).

Proposition 6.

Assume that  0v<a; then (12)H(k,i)>0,G(k,i)>0,k,i{2,3,,n+2}.

Proposition 7.

Assume that  k,i{2,3,,n+2}; then (13)1Qb2G(k,i)G(i,i)1QD,G(n+4-k,n+4-i)=G(k,i), where (14)D=(b+aj=2n+11z(j))2,k,i{2,3,,n+2}.

Proposition 8.

Suppose that  0v<a; then (15)1Qvb2γ<H(k,i)H(i,i)1QγDa,iiiiiiiiiiiiiiiiiiiiiik,i{2,3,,n+2}, where (16)D=(b+aj=2n+11z(j))2,γ=1a-v,iiiiiiiiiiiiiiiiiiiiiiiiiik,i{2,3,,n+2}.

Proof.

From Lemma 5 and Proposition 7, we have (17)H(k,i)=G(k,i)+1a-vτ=3n+1G(i,τ)h(τ)>1a-vτ=3n+1G(i,τ)h(τ)1a-v1Qb2τ=3n+1h(τ)=1Qγb2v.

On the other hand, using  G(k,i)G(i,i)(1/Q)D, we get (18)H(k,i)=G(k,i)+1a-vτ=3n+1G(i,τ)h(τ)G(i,i)+1a-vτ=3n+1G(i,τ)h(τ)1QD+1QD1a-vτ=3n+1h(τ)1QD(1+va-v)=1QDaγ.

Lemma 9.

Suppose that  μ1, for all  yE; the BVP (19)-Δ2u(k-1)=y(k),k{1,2,,n+1},u(0)=u(n+2)=i=1n+1g(i)u(i) has unique solution  u  given by (20)u(k)=i=1n+1H1(k,i)y(i), where (21)H1(k,i)=G1(k,i)+11-μτ=1n+1G1(τ,i)g(τ),G1(k,i)={in+2(n+2-k),0i<k,kn+2(n+2-i),ki<n+2.

Proof.

From the properties of difference operator, we can get (22)-Δu(k)+Δu(k-1)=y(k); then we have (23)-Δu(1)+Δu(0)=y(1),-Δu(2)+Δu(1)=y(2),-Δu(3)+Δu(2)=y(3),-Δu(k-1)+Δu(k-2)=y(k-1). It can imply that (24)-Δu(k-1)+Δu(0)=i=1k-1y(i).

Let  Δu(0)=A; we have (25)-Δu(k-1)=A-i=1k-1y(i). That is (26)u(k)-u(k-1)=A-i=1k-1y(i); thus, (27)u(1)-u(0)=A,u(2)-u(1)=A-i=11y(i),u(3)-u(2)=A-i=12y(i),u(k)-u(k-1)=A-i=1k-1y(i).

We can get (28)u(k)=u(0)+kA-j=1k-1i=1jy(i),u(n+2)=u(0)+(n+2)A-j=1n+1i=1jy(i).

From the boundary conditions, we have (29)A=1n+2j=1n+1i=1jy(i); thus, (30)u(k)=i=1n+1g(i)u(i)+kn+2j=1n+1i=1jy(i)-j=1k-1i=1jy(i). Because (31)kn+2j=1n+1i=1jy(i)=kn+2×[y(1)+(y(1)+y(2))+(y(1)+y(2)+y(3))++(y(1)+y(2)++y(n+1))]=kn+2[(n+1)y(1)+ny(2)+(n-1)y(3)++y(n+1)]=kn+2(n+1)y(1)+kn+2ny(2)++kn+2y(n+1),j=1k-1i=1jy(i)=y(1)+(y(1)+y(2))++(y(1)+y(2)++y(k-1))=(k-1)y(1)+(k-2)y(2)++y(k-1), we have (32)kn+2j=1n+1i=1jy(i)-j=1k-1i=1jy(i)=[k(n+1)n+2-(k-1)]y(1)+[knn+2-(k-2)]y(2)++[kn+2(n-k+3)-1]y(k-1)+[kn+2(n-k+2)]y(k)++kn+2y(n+1). Thus, we get (33)u(k)=i=1n+1g(i)u(i)+i=1n+1G1(k,i)y(i), where (34)G1(k,i)={in+2(n+2-k),0i<k,kn+2(n+2-i),ki<n+2.

Multiplying the above equation with  g(k), summing them from  1  to  n+1, we have (35)k=1n+1g(k)u(k)=k=1n+1g(k)i=1n+1g(i)u(i)+k=1n+1g(k)i=1n+1G1(k,i)y(i),i=1n+1g(i)u(i)=11-k=1n+1g(k)k=1n+1g(k)i=1n+1G1(k,i)y(i). So, we can get (36)u(k)=i=1n+1G1(k,i)y(i)+11-k=1n+1g(k)k=1n+1g(k)i=1n+1G1(k,i)y(i)=i=1n+1H1(k,i)y(i).

Proposition 10.

Assume that  0μ<1; then (37)H1(k,i)>0,G1(k,i)>0,k,i{1,2,,n+1},H1(k,i)0,G1(k,i)0,k,iJ.

Proposition 11.

For  k,iJ, one has (38)1n+2e(i)e(k)G1(k,i)G1(i,i)=i(1-in+2)=e(i)n+24,G1(n+2-k,n+2-i)=G1(k,i), where  e(i)=(i/(n+2))(n+2-i).

Proposition 12.

If  0μ<1, then, for all  k,iJ, one has (39)ρe(i)H1(k,i)H1(i,i)γ*in+2(n+2-i)n+24γ*, where  γ*=1/(1-μ), ρ=τ=1n+1e(τ)g(τ)/(n+2)(1-μ).

We construct a cone on  E  by (40)K={{0,1,,n}  andminkJu(k)δ*uuE:u0,Δ2u(k)0on{0,1,,n},andminkJu(k)δ*u}, where (41)δ*=2(n+3)vb2ρ3(n+2)γ*a(b+aj=2n+1(1/z(j)))2. Obviously,  K  is a closed convex cone of  E.

Define an operator  T:EE  as (42)(Tu)(k)=τ=1n+1H1(k,τ)i=3n+1H(τ+1,i)w(i-2)f(i-2,u(i-2)). Let (43)Ψ(k,i)=τ=1n+1H1(k,τ)H(τ+1,i). Then we can obtain the following properties.

Proposition 13.

If  (A3)  and  (A4)  hold, then (44)0<Ψ(k,i)(n+2)(n+1)4(1-μ)H(i,i),iiiiiiiiiiiikJ,i{2,3,,n+2}.

Proposition 14.

If  (A3)  and  (A4)  hold, then, for all kJ,i{2,3,,n+2}, one has (45)(n+1)(n+3)6Qvb2γρ<Ψ(k,i)(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2a.

Lemma 15.

Suppose that  (A0)(A4)  hold; if  uE  is a solution of the equation (46)u(k)=(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2)), then  u  is a solution of the BVP (2).

Lemma 16.

Assume that  (A0)(A4)  hold; then  T(K)K  and  T:KK  is completely continuous.

Proof.

From above works, for all  uK, we have (47)Δ2(Tu)(k-1)iii=-i=3n+1H(k+1,i)w(i-2)f(i-2,u(i-1))0,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiik{1,2,,n+1}. Because (48)(Tu)(0)=(Tu)(n+2)=τ=1n+1H1(0,τ)×i=3n+1H(τ+1,i)w(i-2)f(i-2,u(i-2))=τ=1n+1H1(n+2,τ)×i=3n+1H(τ+1,i)w(i-2)f(i-2,u(i-2))=τ=1n+111-μj=1n+1G1(j,τ)G(j)×i=3n+1H(τ+1,i)w(i-2)f(i-2,u(i-2))0. That is to say  (Tu)(k)0  for  kJ.

On the other hand, for  kJ, we have (49)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2×ai=3n+1w(i-2)f(i-2,u(i-2)). Similarly, we can get (50)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)6Qvb2γρi=3n+1w(i-2)f(i-2,u(i-2))=δ*(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2×ai=3n+1w(i-2)f(i-2,u(i-2))δ*Tu,kJ. So,  TuK  and  T(K)K. It is easy to see that  T:KK  is completely continuous.

3. The Existence of One Positive Solution

In this part, we apply Theorem 3 and Lemma 4 to prove the existence of one positive solution for BVP (2). We need consider the following cases for:  p>1,  p=1, and  p=.

Let (51)fβ=limnβsupmaxkJf(k,u)u,fβ=limuβinfminkJf(k,u)u, where  β  denotes  0  or  , and (52)B=max{(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/q,(n+2)(n+1)4Qγγ*a×(b+aj=2n+11z(j))(i=3n+1|w|)},η=[(n+2)(n+1)γγ*(b+aj=2n+11z(j))2a]-1.

Remark 17.

If we only consider the case  p=, then we can take (53)B=(n+2)(n+1)4Qγ(b+aj=2n+11z(j))2(i=3n+1|w|).

Firstly, the following theorem deals with the case p>1.

Theorem 18.

Suppose that  (A0)(A4)  hold. If there exist two constants  r, R  with  0<r<δ*R  such that

f(k,u)B-1r  for all  (k,u)J×[δ*r,r]  and  f(k,u)(4η/mδ*)QR  for all  (k,u)J×[δ*R,R];

or

f(k,u)(4η/mδ*)QR  for all  (k,u)J×[δ*r,r]  and  f(k,u)B-1R  for all  (k,u)J×[δ*R,R], then BVP (2) has at least one positive solution.

Proof.

We only consider condition  (C1). For  uK, from the definition of  K  we obtain that (54)minkJu(k)δ*u. So, for  uKr, we have  u(k)[δ*r,r], which implies that  f(k,u(k))B-1γ. Thus, for  kJ, from  (C1)  we get (55)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4(1-μ)B-1γi=3n+1H(i,i)w(i-2)(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/qB-1γ=r=u; that is,  uKr  implies that (56)Tuu.

On the other hand, for  uKR, we have  u(k)[δ*R,R], which implies that  f(k,u(k))(4η/mδ*)QR; therefore for  kJ, from  (C1)  we have (57)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+1)(n+3)6Qvb2γρ4ηmδ*QRi=3n+1w(i-2)δ*(n+1)(n+2)4Qγ*γ×(b+aj=2n+11z(j))2a4ηmδ*QRm=R=u; that is,  uKR  implies that (58)Tuu.

From the above works, we apply (i) of Theorem 3 to yield that  T  has a fixed point  u*K¯r,R,   ru*R  and  u*(k)δ*u*>0,kJ. Thus, it follows that BVP (2) has at least one positive solution  u*.

The following theorem deals with the case  p=1.

Theorem 19.

Suppose that  (A0)(A4)  hold and (C1) or (C2) holds. Then BVP (2) has at least one positive solution.

Proof.

Let  (i=3n+1|H(i,i)|)(supi{3,4,,n+1}|w(i-2)|)  replace (i=3n+1|H(i,i)|p)1/p(i=3n+1|w(i-2)|q)1/q and repeat the argument of Theorem 18.

Finally we consider the case  p=.

Theorem 20.

Suppose that  (A0)(A4)  hold and (C1) or (C2) holds. Then BVP (2) has at least one positive solution.

Proof.

Similar to the proof of Theorem 18, for  xKr, we have (59)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2aB-1ri=3n+1w(i-2)(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2aB-1ri=3n+1|w|r=u. So, for  xKr, we have  Tuu. And from the proof of Theorem 18,  Tuu,kJ  for  uKR. Thus Theorem 20 is proved.

Theorem 21.

Assume that  (A0)(A4)  hold. If one of the following conditions is satisfied

f0>4ηQ/mδ*  and  f<1/B  (particularly,  f0=  and  f=0);

f0<1/B  and  f>4ηQ/mδ*2 (particularly,  f0=0  and  f=),

then BVP (2) has at least one positive solution.

Proof.

We only consider the case  (C3). The proof of case  (C4)  is similar to case  (C3). Considering  f0>4ηQ/mδ*2, there exists  r1>0  such that  f(k,u)(f0-ε1)u  for  kJ,  u[0,r1], where  ε1>0  satisfies  f0-ε14ηQ/mδ*2. Then, for  kJ,  uKr1, from  (C3)  we have (60)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+1)(n+3)6Qvb2γρi=3n+1w(i-2)f(i-2,u(i-2))δ*2(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2a(f0-ε1)×i=3n+1w(i-2)u(i-2)δ*2(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2a(f0-ε1)muδ*2(n+2)(n+1)4Qγ*γ(b+aj=2n+11z(j))2a4ηQmδ*2muu; that is,  xKr1  implies that  Tuu.

Next, considering  f<1/B, then there exists  R¯1>0  such that (61)f(k,u)(f+ε2)u,kJ,u(R¯1,), where  ε2>0  satisfies  B(f+ε2)<1; assume that (62)M=max0uR¯1,kJf(k,u); then (63)f(k,u)M+(f+ε2)uɛ. Choosing  R1>max{r1,R¯1,MB(1-B(f+ε2))-1}.

Thus, for  uKR1, we obtain (64)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)(M+(f+ε2)u)(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/q×(M+(f+ε2)u)B(M+(f+ε2)u)BM+(f+ε2)uB<R1-R1(f+ε2)B+(f+ε2)uBR1-(f+ε2)uB+(f+ε2)uB=R1; that is,  uKR1, we have  Tu<u.

From above works and (ii) of Theorem 3 we know that  T  has a fixed point  u*K¯r1,R1,   r1u*<R1, and  u*(k)δ*u*>0,   kJ. Thus, it implies that BVP (2) has a positive solution  u*.

Theorem 22.

Suppose that  (A0)(A4)  hold. If there exist two constants  r2,R2  with  0<r2<R2  such that

f(k,·)  is nondecreasing on  [0,R2]  for all  kJ;

f(k,δ*r2)(4ηQ/mδ*)r2  and  f(k,R2)B-1R2  for all  kJ,

then the BVP (2) has at least one positive solution.

Proof.

For  uK, from the definition of  K  we have minkJu(t)δ*u. So, for  uKr2, we have  u(k)δ*u=δ*r2,kJ, from conditions (C5) and (C6), we obtain (65)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))i=3n+1Ψ(k,i)w(i-2)f(i-2,δ*r2)(n+1)(n+3)6Qvb2γρ4ηQmδ*r2i=3n+1w(i-2)=δ*(n+2)(n+1)4Qγ*γ×(b+aj=2n+11z(j))2a4ηQmδ*r2i=3n+1w(i-2)=r2=u, that is, for  uKr2, we can imply that (66)Tuu.

On the other hand, for  uKR2, we have  u(k)R2, kJ, this together with  (C5)  and  (C6), we have (67)(Tu)(k)=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))i=3n+1Ψ(k,i)w(i-2)f(i-2,R2)(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)f(i-2,R2)(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/qB-1R2=R2=u, that is, for  uKR2, we can imply that (68)Tuu.

From above works and Theorem 3, we prove that  T  has a fixed point  u*K¯r2,R2,  r2u*R2, and  u*(k)δ*u*>0,  kJ. So the BVP (2) has at least one positive solution.

4. The Existence of Multiple Positive Solutions Theorem 23.

Assume that  (A0)(A4)  hold, and the following two conditions hold:

f0>4ηQ/mδ*2   and  f>4ηQ/mδ*2   (particularly,  f0=f=);

there exists  l>0  such that  maxkJ,uKlf(k,u)<B-1l.

Then BVP (2) has at least two positive solutions  u*(k), u**(k), which satisfy (69)0<u**<l<u*.

Proof.

We can take two constants  r, R  and suppose that  0<r<l<R. If  f0>4ηQ/mδ*2; from the proof of Theorem 21 we have (70)Tuu,uKr. If  f>4ηQ/mδ*2, similarly, we have (71)Tuu,uKR.

On the other hand, from  (C8), for  uKl, we have (72)Tu=i=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))<(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)B-1l(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/qB-1l=l; that is, (73)Tu<l=u,uKl.

Applying Theorem 3, we can prove that  T  has a fixed point  u**K¯r,l  and a fixed point  u*K¯l,R. Thus, we prove that BVP (2) has two positive solutions u*, u**. From above formula we obtain  u*l  and  u**l. So  0<u**<l<u*.

Remark 24.

From the proof of Theorem 23 we obtain that if  (C8)  holds and  f0>4ηQ/mδ*2  (or  f>4ηQ/mδ*2), then BVP (2) has at least one positive solution  u. It satisfies  0<u<l  (or  l<u).

Using a similar method we can obtain the following results.

Theorem 25.

Suppose that  (A0)(A4)  hold, and the following two conditions hold

f0<1/B  and  f<1/B;

there exists  L>0  such that  minkJ,uK2f(k,u)>(4ηQ/mδ*)L,

then BVP (2) has at least two positive solutions  u*(k),u**(k), which satisfy (74)0<u**<L<u*.

Remark 26.

If  (C10)  holds and  f0<1/B  (or  f<1/B), then BVP (2) has at least one positive solution  u  satisfying  0<u<L  (or  L<u).

Theorem 27.

Assume that  (A0)(A4)  hold. If there exist  2n  positive numbers  dt,Dt,t=1,2,,n, with  d1<δ*D1<D1<d2<δ*D2<D2<<dn<δ*Dn<Dn  such that

f(k,u)B-1dt, for  (k,u)J×(δ*dt,dt)  and  f(k,u)(4ηQ/mδ*)Dt  for  (k,u)J×[δ*Dt,Dt],k=1,2,,n; or

f(k,u)(4ηQ/mδ*)dt, for  (k,u)J×(δ*dt,dt)  and  f(k,u)B-1Dt  for  (k,u)J×[δ*Dt,Dt], t=1,2,,n,

then BVP (2) has at least  n  positive solutions  uk  satisfying  dtutDt, t=1,2,,n.

Theorem 28.

Assume that  (A0)(A4)  hold. If there exist  2n  positive numbers  dt,Dt,t=1,2,,n, with  d1<D1<d2<D2<<dn<Dn  such that

f(k,·)  is nondecreasing on  [0,Dn]  for all  kJ;

f(k,δ*dk)(4ηQ/mδ*)dt, and  f(k,Dt)B-1Dt,t=1,2,,n,

then BVP (2) has at least  n  positive solutions  ut  satisfying  dtutDt,t=1,2,,n.

5. The Nonexistence of Positive Solution Theorem 29.

Suppose that  (A0)(A4)  hold, and  Bf(k,u)<u, for all  kJ, u>0; then BVP (2) has no positive solution.

Proof.

Assume that  u(k)  is a positive solution of BVP (2). Then  uK, u(k)>0, k{0,1,,n+2}  and (75)Tu=u=maxkJu(k)=maxkJi=3n+1Ψ(k,i)w(i-2)f(i-2,u(i-2))(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)f(i-2,u(i-2))<(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)B-1u(i-2)B-1u(n+2)(n+1)4(1-μ)i=3n+1H(i,i)w(i-2)B-1u(n+2)(n+1)4(1-μ)(i=3n+1|H|p)1/p(i=3n+1|w|q)1/q=u, which is a contradiction.

Similarly, we can get the following result.

Theorem 30.

Suppose that  (A0)(A4)  hold and (mδ*2/4ηQ)f(k,u)>u, for all kJ, u>0; then BVP (2) has no positive solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work was supported by the Natural Science Foundation of China (11371120), the Natural Science Foundation of Hebei Province (A2013208147) and the Education Department of Hebei Province Science and Technology Research Project (Z2014095).

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