Let T be an integer with T≥5 and let T2={2,3,…,T}. We consider the existence of positive solutions of the nonlinear boundary value problems of fourth-order difference equations
Δ4u(t−2)−ra(t)f(u(t))=0, t∈T2,
u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0, where r is a constant, a:T2→(0,∞),andf:[0,∞)→[0,∞) is continuous. Our approaches are based on the Krein-Rutman theorem and the global bifurcation theorem.

1. Introduction

An elastic beam in an equilibrium state whose both ends are simply supported can be described by the fourth-order boundary value problem of the form

y′′′′=f(x,y,y′′),x∈(0,1),y(0)=y(1)=y′′(0)=y′′(1)=0;
see Gupta [1, 2]. The existence of solutions of (1.3) and (1.4) has been extensively studied; see Gupta [1, 2], Aftabizadeh [3], Yang [4], Del Pino and Manásevich [5], Galewski [6], Yao [7], and the references therein. The existence and multiplicity of positive solutions of the boundary value problem of ordinary differential equations

y′′′′(x)-λf(x,y(x))=0,x∈(0,1),y(0)=y(1)=y′′(0)=y′′(1)=0
have also been studied by many authors; see Ma and Wang [8], Ma [9], Bai and Wang [10], Chai [11], Yao and Bai [12] for some references along this line.

Recently, the existence of solutions of boundary value problems (BVPs) of difference equations has received much attention; see Agarwal and Wong [13], Henderson [14], He and Yu [15], Zhang et al. [16], and the references therein. However, relatively little is known about the existence of positive solutions of fourth-order discrete boundary value problems. To our best knowledge, only He and Yu [15] and Zhang et al. [16] dealt with that. In [15], He and Yu studied the existence of positive solutions of the nonlinear fourth-order discrete boundary value problem

Δ4u(t-2)-ra(t)f(u(t))=0,t∈𝕋2,u(0)=u(T+2)=Δ2u(0)=Δ2u(T)=0
(where T≥3 is an integer, 𝕋2:={2,…,T}, r is a parameter, a:𝕋2→[0,∞), f∈C([0,∞),[0,∞)) satisfies some growth conditions which are not optimal!). The likely reason is that the spectrum structure of the linear eigenvalue problem

Δ4u(t-2)=λa(t)u(t),t∈𝕋2,u(0)=u(T+2)=Δ2u(0)=Δ2u(T)=0
is not clear. It has been pointed out in [15, 16] that (1.3) and (1.4) are equivalent to the integral equation of the form

u(t)=r∑s=1T+1G(t,s)∑s=2TG1(s,j)a(j)f(u(j))=:A0u(t),j∈𝕋2,
where

G(t,s)=1T+2{s(T+2-t),1≤s≤t≤T+2,t(T+2-s),0≤t≤s≤T+1,G1(t,i)=1T{(T+1-t)(i-1),2≤i≤t≤T+1,(T+1-i)(t-1),1≤t≤i≤T,
and other results on the existence of positive solutions of (1.3) and (1.4) can be found in the two papers. Notice that in the integral (1.7), two distinct Green's functions, G and G1, are used. This makes the construction of cones and the verification of strong positivity of A0 more complex and difficult. Therefore, we think that the boundary condition (1.4) is not very suitable for the study of the positive solutions of fourth order difference equations.

It is the purpose of this paper to assume the fourth-order difference equation (1.3) subject to a new boundary condition of the form

u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0.
This will make our approaches much more simple and natural, and only one Green's function is needed. However, the classical definitions of positive solutions are useless for (1.3) and (1.9) any more. We have to adopt the following new definition of positive solutions.

Definition 1.1.

Denote
𝕋1:={1,2,…,T+1},𝕋0:={0,1,…,T+1,T+2}.
A function y:𝕋0→ℝ+ is called a positive solution of (1.3) and (1.9) if y satisfies (1.3), (1.9), and y(t)≥0 on 𝕋2 and y(t)≢0 on 𝕋2.

Remark 1.2.

Notice that the fact y:𝕋0→ℝ+ is a positive solution of (1.3) and (1.9) does not mean that y(t)≥0 on 𝕋0. In fact, y satisfies

y(t)≥0 for t∈𝕋2,

y(1)=y(T+1)=0,

y(0)=-y(2),y(T+2)=-y(T).

Remark 1.3.

In [17], Eleo and Henderson defined a kind of positive solutions which actually are sign-change solutions. In Definition 1.1, a positive solution may allow to take nonpositive value at t=0 and t=T+2. We think it is useful in this case that T is large enough.

In the rest of the paper, we will use global bifurcation technique; see Dancer [18, Theorem 2] or Ma and Xu [19, Lemma 2.1], to deal with (1.3) and (1.9). To do this, we have to study the spectrum properties of (1.5) and (1.9). This will be done in Section 2. Finally, in Section 3, we will state and prove our main result.

2. Eigenvalues

Recall

𝕋2:={2,3,…,T},𝕋1:={1,2,…,T+1},𝕋0:={0,1,…,T+1,T+2}.
Let

E:={u∣u:𝕋0→ℝ}.
Then, dimE=T+3, and E is a Banach space with the norm

∥u∥E=max{|u(j)|∣j∈𝕋0}.
Let

Y:={u∣u:𝕋2→ℝ}.
Then Y is a Banach space with the norm

∥u∥Y=max{|u(j)|∣j∈𝕋2}.
Let

E0:={u∈E∣u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0}.
Then the operator χ:E0→Y

χ(-u(2),0,u(2),u(3),…,u(T),0,-u(T)):=(u(2),u(3),…,u(T)),
is a homomorphism.

In this paper, we assume that

(H1)a:𝕋2→(0,∞).

Definition 2.1.

We say that λ is an eigenvalue of linear problem
Δ4u(t-2)=λa(t)u(t),t∈𝕋2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0
if (2.8) has a nontrivial solution.

In the rest of this section, we will prove the existence of the first eigenvalue of (2.8).

Theorem 2.2.

Equation (2.8) has an algebraically simple eigenvalue λ1, with an eigenfunction φ satisfying

φ(t)>0 on 𝕋2;

φ(0)=-φ(2); φ(T+2)=-φ(T).

Moreover, there is no other eigenvalue whose eigenfunction is nonnegative on 𝕋2.

To prove Theorem 2.2, we need several preliminary results.

Lemma 2.3.

For each h=(h(2),…,h(T)), the linear problem
Δ4u(t-2)=h(t),t∈𝕋2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0
has a unique solution
u(t)=∑s=2TH(t,s)∑j=2TH(s,j)h(j),t∈𝕋1,
where
H(t,s)=1T{(t-1)(T+1-s),1≤t≤s≤T,(s-1)(T+1-t),2≤s≤t≤T+1.

Proof.

Let Δ2u(t-2)=w(t-1) for t∈𝕋2. Then (2.9) is equivalent to the system
Δ2w(t-1)=h(t),t∈𝕋2,Δ2u(t-1)=w(t),t∈𝕋2,w(1)=w(T+1)=0,u(1)=u(T+1)=0.
From Kelly and Peterson [20, Theorem 6.8 and Example 6.12], it follows that
w(t)=-∑s=2TH(t,s)h(s),t∈𝕋1u(t)=-∑s=2TH(t,s)w(s),t∈𝕋1.
Therefore, (2.10) holds.

Denote

ρ:=4sin2π2T,e(t):=sinπ(t-1)T,t∈𝕋1.
Then

Δ2e(t-1)+ρe(t)=0,t∈𝕋2,e(1)=e(T+1)=0.
Notice that

{(t-1T,e(t))∣t∈𝕋1}⊂{(x,sinπx)∣x∈[0,1]}.
From the assumption T≥5, we have

ρ≤4sin2π2·5<4sin2π6=1.

Lemma 2.4.

Let v∈E satisfy v(1)=v(T+1)=Δ2v(0)=Δ2v(T)=0 and
-γe(t)≤-Δ2v(t-1)≤γe(t),t∈𝕋2,
where γ∈(0,∞). Then
-γρe(t)≤v(t)≤γρe(t),t∈𝕋1.

Proof.

From (2.18), we get
-γ∑s=2TH(t,s)e(s)≤-∑s=2TH(t,s)Δ2v(t-1)≤γ∑s=2TH(t,s)e(s),t∈𝕋2.
This is
-γρe(t)≤v(t)≤γρe(t),t∈𝕋2.
Combining this with the boundary conditions v(1)=v(T+1)=Δ2v(0)=Δ2v(T)=0, it concludes that
-γρe(t)≤v(t)≤γρe(t),t∈𝕋1.

Let

X:={u∈E0∣-γe(t)≤-Δ2u(t-1)≤γe(t),t∈𝕋1}
for some γ∈(0,∞). Since γ<γ/ρ, from Lemma 2.4 and (2.17), we may define

∥u∥X:=inf{γρ∣-γe(t)≤-Δ2u(t-1)≤γe(t),t∈𝕋1}.
For any x,y∈X, we have from the definition of ∥·∥X that

-ρe∥x∥X≤-Δ2x(t-1)≤ρe∥x∥X,t∈𝕋1,-ρe∥y∥X≤-Δ2y(t-1)≤ρe∥y∥X,t∈𝕋1.
It follows that

-ρe(∥x∥X+∥y∥X)≤-Δ2(x(t-1)+y(t-1))≤ρe(∥x∥X+∥y∥X),t∈𝕋1.
Thus, x+y∈X, and moreover,

∥x+y∥X≤∥x∥X+∥y∥X.
Therefore, ∥·∥X is a norm of X, and (X,∥·∥X) is a normed linear space. Since dimX=T-1, (X,∥·∥X) is actually a Banach space. Let

P:={u∈X∣Δ2u(t-1)≤0fort∈𝕋2;u(t)≥0fort∈𝕋1}.
Then the cone P is normal and has nonempty interior intP.

Lemma 2.5.

For u∈X,
∥u∥∞,1≤(T+1)2∥Δ2u∥∞,2,∥Δ2u∥∞,2≤ρ∥u∥X,
where
∥u∥∞,1:=max{|u(j)|∣u∈𝕋1},∥Δ2u∥∞,2:=max{|Δ2u(j-1)|∣j∈𝕋2}.

Proof.

(i) From the relation

u(t)=∑s=2TH(t,s)(-Δ2u(s-1)),t∈𝕋1,
it follows that
∥u∥∞,1≤(T+1)2∥Δ2u∥∞,2.

(ii) By (2.14) and the fact that e(t)>0 on 𝕋2, it follows that there exists γ>0, such that

|-Δ2u(t-1)|≤γe(t),t∈𝕋2.
Let
γ0:=inf{γ∣|-Δ2u(t-1)|≤γe(t),t∈𝕋2}.
Then
|-Δ2u(t-1)|≤γ0e(t),t∈𝕋2.
This implies ∥Δ2u∥∞,2≤ρ·(γ0/ρ), and accordingly ∥Δ2u∥∞,2≤ρ∥u∥X.

Proof of Theorem <xref ref-type="statement" rid="thm2.1">2.2</xref>.

For u∈E, define a linear operator K:X→Y and J:Y→X by
Ku(t):=∑s=2TH(t,s)∑j=2TH(s,j)a(j)u(j),t∈𝕋1,J(u(1),…,u(T+1))=(-u(2),u(1),…,u(T+1),-u(T)).
Then (2.8) can be written as
u=λJ∘Ku,u∈X.
Since X is finite dimensional, we have that K:X→Y is compact. Obviously, J∘K(P)⊆P.

Next, we show that J∘K:P→P is strongly positive.

Since a(t) is positive on 𝕋2, there exists a constant k>0 such that a(t)>k on 𝕋2.

For u∈P∖{0}, we have that

∑j=2TH(s,j)a(j)u(j)≥k∑j=2TH(s,j)u(j)>0,s∈𝕋1.
It follows that there exists k1>0 such that
∑j=2TH(s,j)a(j)u(j)≥k1e(s).
Also, for u∈P∖{0}, we have from the fact J∘Ku∈P and Ku≠0 in 𝕋2 that
∑j=2TH(s,j)a(j)u(j)≤maxt∈𝕋2a(t)·∑j=2TH(s,j)u(j)≤k2e(s),s∈𝕋1,
for some constant k2>0. By (2.39) and (2.41), we get
k1e(s)≤∑j=2TH(s,j)a(j)u(j)≤k2e(s),s∈𝕋1.
Thus
k1∑s=2TH(t,s)e(s)≤∑s=2TH(t,s)∑j=2TH(s,j)a(j)u(j)≤k2∑s=2TH(t,s)e(s),t∈𝕋1.
Since
∑s=2TH(t,s)e(s)=1ρe(t),t∈𝕋1.
Using (2.43) and (2.44), it follows that
k1ρe(t)≤(Ku)(t)≤k2ρe(t),t∈𝕋1.
Therefore, J∘Ku∈intP.

Now, by the Krein-Rutman theorem [21, Theorem 7.C; 20, Theorem 19.3], K has an algebraically simple eigenvalue λ>0 with an eigenvector φ(·)∈intP. Moreover, there is no other eigenvalue with an eigenfunction in P.

3. The Main Result

In this section, we will make the following assumptions:

(H2) f:[0,∞)→[0,∞) is continuous and f(s)>0 for s>0;

(H3) f0,f∞∈(0,∞), where

f0=limu→0+f(u)u,f∞=limu→+∞f(u)u.

Remark 3.1.

It is not difficult to see that (H2) and (H3) imply that there exists a constant a0∈(0,∞) such that
f(u)≥a0u,u∈[0,∞).

Theorem 3.2.

Let (H1), (H2), and (H3) hold. Assume that either
λ1rf0<1<λ1rf∞
or
λ1rf∞<1<λ1rf0.
Then (1.3) and (1.9) have at least one positive solution.

Remark 3.3.

Recently, Ma and Xu [19] considered the nonlinear fourth-order problem
u(4)(t)=f(t,u(t),u′′(t)),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0
under some conditions involved the generalized eigenvalues of the linear problem
u′′′′=λ[A(t)u-B(t)u′′],0<t<1,u(0)=u(1)=u′′(0)=u′′(1)=0.
Our main result, Theorem 3.2, needs f0,f∞∈(0,∞); see (H3). However, in [19, Theorem 3.1], some weaker conditions of the form
a0(t)u-b0p-ξ1(t,u,p)≤f(t,u,p)≤a0(t)u-b0p+ξ2(t,u,p)
are used.

Remark 3.4.

The first eigenvalue μ1 of the linear problem
Δ2u(t-1)+μu(t)=0,t∈{2,…,T},u(1)=u(T+1)=0,
is μ1=4sin2(π/2T), and the first eigenvalue λ1 of the linear problem
Δ4u(t-1)+λu(t)=0,t∈{2,…,T},u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0,
is
λ1=16sin4π2T.
It is easy to check that the function
f(s):={2λ1s,s∈(0,1),λ1s2+32λ1,s∈[1,∞).
Then, for each r∈(1/2,2), the condition (3.3) holds.

Remark 3.5.

The condition (3.3) or (3.4) is optimal since for any ϵ>0, the linear problem
Δ4u(t-2)-(λ1-ϵ)a(t)u(t)=0,t∈𝕋2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0,
has the unique solution u(t)≡0. In fact, λ1 is the least eigenvalue of the linear problem
Δ4u(t-2)-μa(t)u(t)=0,t∈𝕋2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0.

To prove Theorem 3.2, we define L:D(L)→Y by

Lu(t):=-Δ4u(t-2),u∈D(L),t∈𝕋2
where

D(L)={u∈X∣u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0}.
It is easy to check that L-1:Y→X is compact.

ξ̃(τ)=max0≤|s|≤τ|ξ(s)|.
Then ξ̃ is nondecreasing and

limτ→∞ξ̃(τ)τ=0.

Let us consider

Lu+λra(t)f0u+λra(t)ζ(u)=0,λ>0,
as a bifurcation problem for the trivial solution u≡0. It is easy to check that (3.20) can be converted to the equivalent equation

From the proof process of Theorem 2.2, we have that for each fixed λ>0, the operator K:X→X,

Ku(t)=∑s=2T∑j=2TH(t,s)H(s,j)ra(j)f0u(j)
is compact and strongly positive. Define F:[0,∞)×X→X by

F(λ,u):=λ[∑s=2T∑j=2TH(t,s)H(s,j)ra(j)ζ(u(j))].
Then we have from (3.17) and Lemma 2.5 that

∥F(λ,u)∥X=o(∥u∥X),as∥u∥X→0,
locally uniformly in λ. Now, we have from a version of Dancer [18, Theorem 2], see Ma [9, Lemma 2.1] for details, to conclude that there exists an unbounded connected subset 𝒞 in the set

{(λ,u)∈(0,∞)×P:u=λKu+F(λ,u),u∈intP}∪{(λ1rf0,0)}
such that (λ1/(rf0),0)∈𝒞.

Proof of Theorem <xref ref-type="statement" rid="thm3.1">3.2</xref>.

It is clear that any solution of (3.20) of the form (1,u) yields a solutions u of (1.3) and (1.9). We will show that 𝒞 crosses the hyperplane {1}×X in ℝ×X. To do this, it is enough to show that 𝒞 joins (λ1/rf0,0) to (λ1/rf∞,∞). Let (μn,yn)∈𝒞 satisfy
μn+∥yn∥X→∞,n→∞.
We note that μn>0 for all n∈ℕ, since (0,0) is the only solution of (3.20) for λ=0 and 𝒞∩({0}×X)=∅.Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M227"><mml:msub><mml:mrow><mml:mi>λ</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>/</mml:mo><mml:mi>r</mml:mi><mml:msub><mml:mrow><mml:mi>f</mml:mi></mml:mrow><mml:mrow><mml:mi>∞</mml:mi></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mn>1</mml:mn><mml:mo><</mml:mo><mml:msub><mml:mrow><mml:mi>λ</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>/</mml:mo><mml:mi>r</mml:mi><mml:msub><mml:mrow><mml:mi>f</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub></mml:math></inline-formula>).

In this case, we show that

(λ1rf∞,λ1rf0)⊆{λ∈ℝ∣∃(λ,u)∈𝒞}.
We divide the proof into two steps.Step 1.

We show that if there exists a constant number M>0 such that
μn⊂(0,M],
then Ckν joins (λ1/rf0,0) to (λ1/rf∞,∞).

From (3.28), we have that ∥y∥X→∞. We divide the equation

Lyn+μnra(t)f∞yn+μnra(t)ξ(yn)=0
by ∥yn∥X and set y¯n=yn/∥y∥X. Since y¯n is bounded in X, choosing a subsequence and relabelling if necessary, we see that y¯n→y¯ for some y¯∈X with ∥y¯∥=1. Moreover, from (3.19) and the fact that ξ̃ is nondecreasing, we have that
limn→∞|ξ(yn(t))|∥y∥X=0,
sinceξ(yn(t))∥yn∥X≤ξ̃(|yn(t)|)∥yn∥X≤ξ̃(∥yn∥∞,1)∥yn∥X≤ξ̃(ρ(T+1)2∥yn∥X)∥yn∥X.
Thus,
y¯(t):=∑s=2T∑j=2TG(t,s)G(s,j)μ¯ra(j)f∞y¯(j),
where μ¯:=limn→∞μn, choosing a subsequence and relabelling if necessary. Thus,
Ly¯+μ¯ra(t)f∞y¯=0.
By Theorem 2.2, we have
μ¯=λ1rf∞.
Thus, 𝒞 joins (λ1/rf0,0) to (λ1/rf∞,∞).

Step 2.

We show that there exists a constant M such that μn∈(0,M] for all n.

By [9, Lemma 2.1], we only need to show that A has a linear minorant V and there exists a (μ,y)∈(0,∞)×P such that ∥y∥X=1 and μVy≥y.

By Remark 3.1, there exist constants a0∈(0,∞) such that

f(u)≥a0u,u∈[0,∞).
For u∈X, let
Vu(t):=∑s=2T∑j=2TG(t,s)G(s,j)a0u(s).
Then V is a linear minorant of R. Moreover,
V(e(t)ρ)=∑s=2T∑j=2TG(t,s)G(s,j)a0e(j)ρ=1ρ2a0∑s=2TG(t,s)e(s)≥cρe(t)
for some constant c>0, independent of t∈𝕋0. So,
c-1V(eρ)≥eρ.
Therefore, it follows [9, Lemma 2.1] that
|ηn|≤c-1.

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M274"><mml:msub><mml:mrow><mml:mi>λ</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>/</mml:mo><mml:mi>r</mml:mi><mml:msub><mml:mrow><mml:mi>f</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mn>1</mml:mn><mml:mo><</mml:mo><mml:msub><mml:mrow><mml:mi>λ</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>/</mml:mo><mml:mi>r</mml:mi><mml:msub><mml:mrow><mml:mi>f</mml:mi></mml:mrow><mml:mrow><mml:mi>∞</mml:mi></mml:mrow></mml:msub></mml:math></inline-formula>).

In this case, if (ηn,yn)∈𝒞 is such that
limn→∞(ηn+∥yn∥X)=∞,limn→∞ηn=∞,
then
(λ1rf0,λ1rf∞)⊂{λ∈(0,∞)∣(λ,u)∈𝒞}
and, moreover,
({1}×X)∩𝒞≠∅.
Assume that there exists M>0, such that for all n∈ℕ,
ηn∈(0,M].
Applying a similar argument to that used in Step 1 of Case 1, after taking a subsequence and relabelling if necessary, if follows that
(ηn,yn)→(λ1rf∞,∞),n→∞.
Again 𝒞 joins (λ1/rf0,0) to (λ1/rf∞,∞) and the result follows.

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This paper is supported by the NSFC (no. 10671158), the NSF of Gansu Province (no. 3ZS051-A25-016), NWNUKJCXGC- 03-17, the Spring-sun program (no. Z2004-1-62033), SRFDP (no. 20060736001), and the SRF for ROCS, SEM (2006[311]).

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