R E S E A R C H
Open Access
Multiple solutions of fourth-order difference
equations with different boundary conditions
Yuhua Long
1,2*, Shaohong Wang
1,2and Jiali Chen
1,2*Correspondence:
1School of Mathematics and
Information Science, Guangzhou University, Guangzhou, P.R. China
2Center for Applied Mathematics,
Guangzhou University, Guangzhou, P.R. China
Abstract
In the present paper, a class of fourth-order nonlinear difference equations with Dirichlet boundary conditions or periodic boundary conditions are considered. Based on the invariant sets of descending flow in combination with the mountain pass lemma, we establish a series of sufficient conditions on the existence of multiple solutions for these boundary value problems. In addition, some examples are provided to demonstrate the applicability of our results.
MSC: 39A12; 39A23
Keywords: Multiple solutions; Boundary value problems; Invariant sets of descending flow; Mountain pass lemma
1 Introduction
Given an integerN > 1, [1,N] denotes the discrete interval {1, 2, . . . ,N}. Consider the
fourth-order nonlinear difference equation
4x(n– 2) =fn,x(n), n∈[1,N], (1.1)
with Dirichlet boundary conditions
x(–1) =x(0) = 0 =x(N+ 1) =x(N+ 2) (1.2)
or periodic boundary conditions
ix(–1) =ix(N– 1), i= 0, 1, 2, 3. (1.3)
Heref ∈C([1,N]×R, R),F(n,x) =0xf(n,s)dsandF(n, 0) = 0.is the forward difference operator andx(n) =x(n+ 1) –x(n),0x(n) =x(n). Fori≥1,ix(n) =(i–1x(n)).
Equation (1.1) can be regarded as a discrete analogue of the continuous versions equa-tion
x(4)(t) =ft,x(t), t∈R,
which is used to describe the stationary states of the deflection of an elastic beam [1]. As to difference equation (1.1), [2] establishes the existence of periodic solutions with
mal period by employing variational techniques and the linking theorem. Using Dancer’s global bifurcation theorem, [3] shows the existence and multiplicity of positive solutions of (1.1) in the form of
4x(n– 2) =λh(n)fx(n), n∈[2,N].
With the rapid development of the technique of computers and the theory of nonlinear difference equations, difference equations have been widely used to study discrete models in many fields such as finance insurance, computing, electrical circuit analysis, dynami-cal systems, physidynami-cal field, and biology; see [4,5] and the references therein. Importantly, much literature and many monographs deal with problems of the existence and multiplic-ity of solutions by using various methods, such as critical point theory [6–8], topological degree theory [9], fixed-point index theory [10]. Recently, some literature [11,12] studied solutions ofφc-Laplacian difference equations. For more research on solutions of differ-ence equations, we can refer to [13–21,28–31].
To the best of our knowledge, there are few studies on sign-changing solutions of fourth-order difference equations. In 2015, He, Zhou et al. [22] obtained the existence of sign-changing solutions for the following periodic boundary value problem:
⎧ ⎨ ⎩
–[p(n– 1)x(n– 1)] +q(n)x(n) =f(n,x(n)), n∈[1,N],
x(0) =x(N), x(0) =x(N),
by applying invariant sets of descending flow. Furthermore, [23–25] deal with other second-order nonlinear boundary value problems and achieve sign-changing solutions in a similar way to [22].
Motivated by the above reasons, the aims of this paper are as follows. Based on the in-variant sets of descending flow, the mountain pass lemma and variational methods, we establish a series of sufficient conditions on the existence of multiple solutions including positive solutions, negative solutions and sign-changing solutions for Dirichlet boundary value problems (1.1) with (1.2) and periodic boundary value problems (1.1) with (1.3). To demonstrate the applicability of our results, some examples are provided. Here and hereafter, a positive (negative) solutionx(n) to (1.1) with (1.2) or (1.1) with (1.3) is a se-quence{x(n)}that satisfies Eq. (1.1) and the boundary conditions (1.2) or (1.3) withx(n) > 0 (x(n) < 0) for alln∈[1,N]. While{x(n)}includes both positive and negative components, we callx(n) is a sign-changing solution.
Next, we give some known results which are critical for the proofs of our main results.
Definition 1.1 ([26]) LetH be a Banach space. The functionalI ∈C1(H, R) is said to
satisfy the Palais–Smale condition ((PS) condition for short) if any sequence {xn} ⊂H satisfying|I(xn)| ≤cfor somec> 0 andI(xn)→0 asn→ ∞possesses a convergent sub-sequence.
Definition 1.2([26]) LetHbe a Banach space, the functionalI∈C1(H, R). If any sequence
LetBrdenote the open ball inHwith radiusrand center 0, and∂Brbe its boundary.
Lemma 1.1(Mountain pass lemma [1]) Let H be a real Banach space,and I∈C1(H, R) satisfies the(PS)condition.If I(0) = 0and the following conditions hold:
(i) there exist constantsr> 0andρ> 0such thatI(x)≥ρfor allx∈∂Br; (ii) there existsx0∈H\Brsuch thatI(x0)≤0,
then I has a critical value c≥ρ,and c can be characterized as
c=inf h∈Γsmax∈[0,1]I
h(s),
here
Γ =h∈C[0, 1],H|h(0) = 0,h(1) =x0 .
Lemma 1.2([27]) Let I∈C1(H, R)be a functional defined on a Hilbert space H which satisfies the(PS)condition and I(x) =x–S(x)for all x∈H.If there are open convex subsets D1and D2of H satisfying S(∂D1)⊂D1,S(∂D2)⊂D2and D1∩D2=∅,and,moreover,there is a path h: [0, 1]→H such that
h(0)∈D1\D2, h(1)∈D2\D1,
and
inf x∈D1∩D2
I(x) > sup
τ∈[0,1] Ih(τ),
then I has at least four critical points,one in H\(D1∪D2),one in D1\D2,one in D1∩D2, and one in D2\D1.
Remark1.1 Theorem 5.1 in [26] tells us that the (PS) condition can be substituted by the weaker (C) condition in Lemma1.2.
The rest of the paper is organized as follows. In Sect.2, BVP (1.1) with (1.2) is consid-ered, and a series of sufficient conditions are established to ensure the existence of multi-ple solutions including positive solutions, negative solutions and sign-changing solutions by variational methods together with invariant sets of descending flow. In a similar way, Sect.3achieves some results on BVP (1.1) with (1.3). Finally, three examples to illustrate the applicability of our theoretical results are provided in Sect.4.
2 Multiple solutions for BVP (1.1) with (1.2)
Given a constantm≥0, define the inner product ofM1as
x,ym= N+1
n=0
2x(n– 1)2y(n– 1) +m
N
n=1
whereM1={x: [–1,N+ 2]→R|x(–1) =x(0) = 0 =x(N+ 1) =x(N+ 2)}. ThenM1is an N-dimensional Hilbert space and the induced norm is
x m=
N+1
n=0
2x(n– 1)2+m
N
n=1
x(n)2
1 2
, ∀x∈M1.
LetHbe anN-dimensional Hilbert space with the common inner product (·,·) and norm · . It follows thatM1andHare isomorphic, and the norm · mis equivalent to · .
For BVP (1.1) with (1.2), we consider the following functionalI1:H→R:
I1(x) =
1 2
N+1
n=0
2x(n– 1)2– N
n=1
Fn,x(n). (2.1)
Write
A1=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
6 –4 1 · · · 0 0 0
–4 6 –4 · · · 0 0 0
1 –4 6 · · · 0 0 0
· · · ·
0 0 0 · · · 6 –4 1
0 0 0 · · · –4 6 –4
0 0 0 · · · 1 –4 6
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
N×N ,
then (2.1) can be rewritten as
I1(x) =
1
2(A1x,x) – N
n=1
Fn,x(n),
wherex= (x(1),x(2), . . . ,x(N))τ∈H.
It is not difficult to verify thatA1is a positive define matrix. Let eigenvalues ofA1be
λ1,λ2, . . . ,λN and let us haveλj> 0 (j= 1, 2, . . . ,N). Without loss of generality, we can as-sume that
0 <λ1≤λ2≤ · · · ≤λN.
Remark2.1 Obviously, theλof the linear eigenvalue problem
⎧ ⎨ ⎩
4x(n– 2) =λx(n), n∈[1,N],
x(–1) =x(0) = 0 =x(N+ 1) =x(N+ 2), (2.2)
corresponding to BVP (1.1) with (1.2), is exactly the eigenvalue of the matrixA1.
Now we state the main results of this section.
Theorem 2.1 Assume that: (F1) maxn∈[1,N]lim supx→0|
f(n,x)
(F2) lim|x|→∞f(nx,x)=lwhere either constantl>λ2orl= +∞with|f(n,x)| ≤C(1 +|x|s–1)
for alln∈[1,N]and somes> 2,C> 0; (F3) lim|x|→∞[xf(n,x) – 2F(n,x)] =∞,∀n∈[1,N].
Then we have
(i) if(F1)–(F2)are satisfied andl∈(λ2, +∞]is not an eigenvalue of(2.2),then BVP(1.1) with(1.2)has at least three nontrivial solutions:one is sign-changing,one is positive and one is negative;
(ii) if(F1)–(F3)are satisfied,the conclusion of(i)is true even iflis an eigenvalue of(2.2).
Theorem 2.2 Iflim inf|x|→∞f(nx,x) >λ1andlim supx→0f(nx,x)<λ1,then BVP(1.1)with(1.2)
possesses at least a positive solution and a negative solution.That is to say, (i) whenlim infx→–∞f(n,x)
x >λ1andlim supx→0– f(n,x)
x <λ1,BVP(1.1)with(1.2)admits
at least a negative solution;
(ii) whenlim infx→+∞f(nx,x) >λ1andlim supx→0+ f(nx,x)<λ1,BVP(1.1)with(1.2)admits at least a positive solution.
In the following, we devote ourselves to making preparations to verify our main results. Consider the following problem to obtain Green’s function of BVP (1.1) with (1.2):
⎧ ⎨ ⎩
4x(n– 2) –m2x(n– 1) =h(n), n∈[1,N],
x(–1) =x(0) = 0 =x(N+ 1) =x(N+ 2), (2.3)
hereh: [1,N]→R. Define
B1=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
2 –1 0 · · · 0 0 0
–1 2 –1 · · · 0 0 0
0 –1 2 · · · 0 0 0
· · · ·
0 0 0 · · · 2 –1 0
0 0 0 · · · –1 2 –1
0 0 0 · · · 0 –1 2
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
N×N ,
then (2.3) and the system (A1+mB1)x=hare equivalent. Therefore, the unique solution
of (2.3) can be expressed by
x= (A1+mB1)–1h. (2.4)
Noticing that there are two solutionsr1=mandr2= 0 forX(r) =r2–mr= 0,
4x(n– 2) –m2x(n– 1) =–2L+r2
–2L+r1
x(n) =–2L+r1
–2L+r2
x(n),
wherex= (x(–1),x(0), . . . ,x(N+ 1),x(N+ 2)),Lx(n) =x(n– 1),n∈[1,N]. Now we have the following lemma.
Lemma 2.1 For i∈ {1, 2},the unique solution of BVP ⎧
⎨ ⎩
–2x(n– 1) +r
ix(n) =h(n), n∈[1,N],
has the form
x(n) = N
s=1
Gi(n,s)h(s), n∈[0,N+ 1],
with
G1(n,s) =
⎧ ⎨ ⎩
(Ps–N–PN–s+2)(Pn–P–n)
(1–P2)(PN+1–P–N–1) , 0≤n≤s≤N+ 1,
(Pn–N–PN–n+2)(Ps–P–s)
(1–P2)(PN+1–P–N–1) , 0≤s≤n≤N+ 1,
G2(n,s) =
⎧ ⎨ ⎩
n(N+1–s)
N+1 , 0≤n≤s≤N+ 1,
s(N+1–n)
N+1 , 0≤s≤n≤N+ 1,
and P=(2+m)+ √
(2+m)2–4
2 .
Proof (i) Wheni= 1, BVP (2.5) corresponds to
⎧ ⎨ ⎩
–2x(n– 1) +mx(n) =h(n), n∈[1,N],
x(0) = 0, x(N+ 1) = 0. (2.6)
Consider the homogeneous equation of (2.6),
⎧ ⎨ ⎩
–2x(n– 1) +mx(n) = 0, n∈[1,N],
x(0) = 0, x(N+ 1) = 0. (2.7)
Then the corresponding characteristic equation to (2.7) is
P2– (2 +m)P+ 1 = 0,
which means thatP1,2=(2+m)±
√
(2+m)2–4
2 are two different roots of it. NoteP1= 1
P2, so we can denote
P=(2 +m) +
(2 +m)2– 4
2 and P
–1=(2 +m) –
(2 +m)2– 4
2 .
Therefore, the general solution of (2.6) can be expressed by
x(n) =c1(n)Pn+c2(n)P–n. (2.8)
Next, to determine coefficientsc1(n) andc2(n) in (2.8), replacingx(n) in (2.6) with (2.8),
we get
⎧ ⎨ ⎩
c1(n– 1)Pn+c2(n– 1)P–n= 0,
Direct computation yields
c1(n) =c1(0) +
n
s=1
h(s)P–s+1
1 –P2 , c2(n) =c2(0) +
n
s=1
–h(s)Ps+1
1 –P2 .
Thus (2.8) can be rewritten in the form of
x(n) =c1(0)Pn+c2(0)P–n+
n
s=1
P–s+1+n–Ps+1–n
1 –P2 h(s).
Applying the boundary conditionsx(0) = 0 andx(N+ 1) = 0, we have
c2(0) = –c1(0) and c1(0) =
N
s=1
Ps–N–PN–s+2
(1 –P2)(PN+1–P–N–1)h(s),
then
x(n) = N
s=1
(Ps–N–PN–s+2)(Pn–P–n) (1 –P2)(PN+1–P–N–1) h(s) +
n
s=1
Pn–s+1–Ps–n+1
1 –P2 h(s).
Denote
G1(n,s) =
⎧ ⎨ ⎩
(Ps–N–PN–s+2)(Pn–P–n)
(1–P2)(PN+1–P–N–1) , 0≤n≤s≤N+ 1,
(Pn–N–PN–n+2)(Ps–P–s)
(1–P2)(PN+1–P–N–1) , 0≤s≤n≤N+ 1,
hence the unique solution of (2.6) is in the form of
x(n) = N
s=1
G1(n,s)h(s), n∈[0,N+ 1].
(ii) Wheni= 2, the general solution of the BVP
⎧ ⎨ ⎩
–2x(n– 1) =h(n), n∈[1,N],
x(0) = 0, x(N+ 1) = 0, (2.9)
is given by
x(n) =c1(0) +nc2(0) +
n
s=1
(s–n)h(s).
Employing the boundary conditions, we get
x(n) = N
s=1
n(N+ 1 –s)
N+ 1 h(s) + n
s=1
(s–n)h(s).
Then the unique solution of (2.9) can be written as
x(n) = N
s=1
where
G2(n,s) =
⎧ ⎨ ⎩
n(N+1–s)
N+1 , 0≤n≤s≤N+ 1,
s(N+1–n)
N+1 , 0≤s≤n≤N+ 1.
Remark2.2 Fori= 1, 2 and anyn,s∈[1,N], it is easy to verifyGi(n,s) =Gi(s,n) > 0.
Combining4x(n– 2) –m2x(n– 1) = (–2L+r
2)(–2L+r1)x(n) with Lemma2.1,
induces the following.
Lemma 2.2
x(n) = N
s=1
N
j=1
G1(n,j)G2(j,s)
h(s) =
N
s=1
N
j=1
G2(n,j)G1(j,s)
h(s), n∈[1,N],
is the unique solution of BVP(2.3).
Proof Making use of Lemma2.1, we see that both
⎧ ⎨ ⎩
–2x(n– 1) +r
1x(n) =y(n), n∈[1,N], x(0) = 0, x(N+ 1) = 0,
and
⎧ ⎨ ⎩
–2y(n– 1) +r
2y(n) =h(n), n∈[1,N], y(0) = 0, y(N+ 1) = 0,
have exactly one solution, namely,
x(n) = N
j=1
G1(n,j)y(j), n∈[0,N+ 1],
and
y(n) = N
s=1
G2(n,s)h(s), n∈[0,N+ 1],
respectively. Furthermore,
4x(n– 2) –m2x(n– 1) =–2L+r2
–2L+r1
x(n).
Thus we find that (2.3) has a unique solution,
x(n) = N
j=1 G1(n,j)
N
s=1
G2(j,s)h(s)
= N
s=1
N
j=1
G1(n,j)G2(j,s)
Similarly,
x(n) = N
s=1
N
j=1
G2(n,j)G1(j,s)
h(s), n∈[1,N].
This completes the proof of Lemma2.2.
Definefm,Km:H→Has
(fmx)(n) =f
n,x(n)–m2x(n– 1),
(Kmx)(n) = N
s=1
N
j=1
G2(n,j)G1(j,s)x(s).
In view of the definition ofKm, together with (2.4) and Lemma2.2, we have
Km= (A1+mB1)–1. (2.10)
Remark2.3 Let the completely continuous operatorSm:H→HbeSm=Kmfm. Lemma2.2 implies thatx={x(n)}N+2
n=–1satisfies BVP (1.1) with (1.2) if and only ifx={x(n)}Nn=1∈H
sat-isfiesSm(x) =x.
Lemma 2.3 The functional I1defined as(2.1)is Fréchet differentiable on H,and the critical points of I1are exactly the fixed points of Sm.
Proof For anyx,y∈H, using Lagrange mean value theorem, there existsθ(n)∈(0, 1),n∈
[1,N] such that
I1(x+y) –I1(x)
=1 2
N+1
n=0
2(x+y)(n– 1)2– N
n=1
Fn, (x+y)(n)
–1 2
N+1
n=0
2x(n– 1)2+ N
n=1
Fn,x(n)
=1 2
N+1
n=0
2y(n– 1)2+ N+1
n=0
2x(n– 1)2y(n– 1) – N
n=1
fn,x(n) +θ(n)y(n)y(n)
=1 2 y
2
m–
m
2 y
2+x,y
m–m N
n=1
x(n)y(n) – N
n=1
fn,x(n) +θ(n)y(n)y(n).
Then
I1(x+y) –I1(x) –x,ym+ N
n=1
fn,x(n)+mx(n)y(n)
=1 2 y
2
m–
m
2 y
2+
N
n=1
It follows from the continuity off that
lim ym→0
fn,x(n)–fn,x(n) +θ(n)y(n)= 0.
Therefore,
lim ym→0
I1(x+y) –I1(x) –x,ym+
N
n=1[f(n,x(n)) +mx(n)]y(n) y m
= 0,
which signifies thatI1is Fréchet differentiable onHand
I1(x),ym=x,ym– N
n=1
fn,x(n)+mx(n)y(n). (2.11)
Next, we only need to provex–Sm(x),ym=x,ym–
N
n=1[f(n,x(n)) +mx(n)]y(n) to
accomplish the proof of Lemma2.3.
For anyx,y∈H, the boundary conditions imply that
N
n=1
4x(n– 2)y(n) = N
n=1
2x(n) – 22x(n– 1) +2x(n– 2)y(n)
= N
n=1
2x(n)y(n) – 2 N
n=1
2x(n– 1)y(n) + N
n=1
2x(n– 2)y(n)
= N
n=1
2x(n– 1)2y(n– 1) +2x(N)2y(N) +2x(–1)2y(–1)
= N+1
n=0
2x(n– 1)2y(n– 1),
which implies that
Sm(x),y
m= N+1
n=0
2(Smx)(n– 1)2y(n– 1) +m N
n=1
(Smx)(n)y(n)
= N
n=1
4(Smx)(n– 2)y(n) +m N
n=1
(Smx)(n)y(n)
= N
n=1
4(Smx)(n– 2) +m(Smx)(n)
y(n)
= N
n=1
fn,x(n)+mx(n)y(n). (2.12)
Then
x–Sm(x),y
m=x,ym– N
n=1
Hence
I1(x),ym=x–Sm(x),y
m,
which implies thatI1(x) =x–Sm(x) for allx∈H.
Aiming to prove our main results, we introduce the following basic notations and nec-essary results. Let
Λ={x∈H:x≥0}, –Λ={x∈H:x≤0}
be two convex cones and
D+ε=
x∈H:distm(x,Λ) <ε , D–ε=
x∈H:distm(x, –Λ) <ε
denote two sets withdistm(x,±Λ) =infy∈±Λ x–y m andε> 0 is an arbitrary constant. It is easy to find thatD±ε are open convex subsets onHwithD+ε∩D–ε=∅. Furthermore,
H\(D+
ε∪D–ε) contains only sign-changing functions.
Lemma 2.4 Under hypotheses(F1)and(F2),there existsε0> 0such that
Sm
∂D–ε
⊂D–ε, Sm
∂D+ε
⊂D+ε,
hold forε∈(0,ε0).Let x∈D–ε (D+ε)be a nontrivial critical point of I1,then x is a negative
(positive)solution of BVP(1.1)with(1.2).
Proof The proof of the conclusions ofD–
ε is analogous to that ofD+ε, here we prove the
case ofD–
ε in detail.
Because of (F1) and (F2), we have, for allx∈R\{0}andn∈[1,N],
xf(n,x) +mx> 0. (2.13)
From the definition of · m, direct calculation leads to
x 2m= N+1
n=0
2x(n– 1)2+m
N
n=1
x(n)2= (A1x,x) +m x 2,
which means
λ1+m x ≤ x m≤
λN+m x . (2.14)
For anyx∈H, letx+=max{x, 0},x–=min{x, 0}andy=S
m(x), then
x+= inf
ψ∈–Λ x–ψ ≤
1 √
m+λ1
inf
ψ∈–Λ x–ψ m=
1 √
m+λ1
distm(x, –Λ). (2.15)
It follows from (F1) and (F2) that there exists constantτ> 0 such that
Note thaty=y++y– andy–∈–Λ, sodist
m(y, –Λ)≤ y–y– m= y+ m. In view of the property of operatorSmtogether with above inequalities, it yields
distm(y, –Λ)y+m
≤y+,y+m
=Sm
x+,y+m
= N
n=1
fn,x+(n)+mx+(n)y+(n)
≤(m+λ1–τ)x+y++Cx+s–1y+
≤
m+λ1–τ m+λ1
distm(x, –Λ) +√ C (m+λ1)s
distm(x, –Λ)s–1y+m,
that is,
distm(y, –Λ)≤
m+λ1–τ m+λ1
distm(x, –Λ) +
C
√
(m+λ1)s
distm(x, –Λ)
s–1
.
Therefore, there existsε0> 0 such that, for any 0 <ε<ε0andx∈D–ε,
distm
Sm(x), –Λ
≤2(m+λ1) –τ
2(m+λ1)
distm(x, –Λ). (2.17)
Due to 2(m+λ1)–τ
2(m+λ1) < 1, then
Sm
∂D–ε⊂D–ε.
Furthermore, ifx∈D–
εis a nontrivial critical point ofI1, soI1(x) =x–Sm(x) = 0,Sm(x) =x. Equation (2.17) implies thatdistm(x, –Λ) = 0, i.e.,x∈–Λ\ {0}. According to (2.13), we can showx(n) < 0 for alln∈[1,N], which indicates thatxis a negative solution of BVP (1.1)
with (1.2).
Lemma 2.5 Under the assumption(F2),if either
(i) l= +∞or
(ii) l< +∞is not an eigenvalue of the matrixA1, then the functional I1satisfies the(PS)condition.
Proof Let{xk} ⊂Hbe a sequence such that|I1(xk)| ≤Kfor someK> 0 andI1(xk)→0 ask→ ∞. SinceHis an N-dimensional Hilbert space, it is sufficient to show that{xk}is bounded to see that{xk}has a convergent subsequence.
(i) Whenl= +∞, we can chooseη1> 0 such that, for any (n,x)∈[1,N]×R,F(n,x)≥
λNx2–η1. Then
I1(xk)≤– 1 2λN xk
2+η
1N, (2.18)
so
1 2λN xk
that is,
xk 2≤
2(K+η1N)
λN ,
which indicates that{xk}is bounded.
(ii) When l< +∞is not an eigenvalue of matrixA1, we complete the proof by
con-tradiction. Suppose there exists a subsequence of{xk}(still denoted by{xk}) such that ρk= xk →+∞ask→ ∞. Setyk=xρk
k ∈H, then yk = 1. The completeness ofHshows
that there isy∈Hsatisfyingyk→yask→ ∞. Put
ωk=
f(1,xk(1))
xk(1)
yk(1), . . . ,
f(N,xk(N))
xk(N)
yk(N)
τ
.
Bylim|x|→∞f(nx,x)=landI1(xk) =xk–Kmfmxk, we have
I1(xk) ρk
=yk– 1 ρk
K0f0xk=yk–K0ωk→y–K0ly, k→ ∞.
Noting that I1(xk)
ρk →0 ask→ ∞, so y–K0ly= 0. In view of (2.10), we find thatlis an
eigenvalue of matrixA1, which is a contradiction. Therefore,{xk}is bounded.
In the following we verify thatI1satisfies the (C) condition under suitable assumptions.
Lemma 2.6 Under the assumption(F3),the functional I1satisfies the(C)condition.
Proof Let {xk} ⊂H be a sequence satisfying I1(xk)→ c for some c ∈ R and (1 +
xk m) I1(xk) m →0 as k→ ∞. Due to the finite-dimensionality ofH, we only need to test that{xk}is bounded.
(i) Assumelim|x|→∞[xf(n,x) – 2F(n,x)] = –∞. Because of the boundedness, there is a
constantR1> 0 such that
–R1≤I1(xk)≤R1 and
1 + xk mI1(xk)m≤R1.
Applying the Cauchy–Schwarz inequality yields
–3R1≤2I1(xk) –
1 + xk mI1(xk)m
≤2I1(xk) –
I1(xk),xk
m
= N
n=1
xk(n)f
n,xk(n)
– 2Fn,xk(n)
. (2.19)
We affirm that{xk}is bounded. Otherwise, it possesses a subsequence, still denoted by {xk}, such that, for somen0∈[1,N],|xk(n0)| →+∞ask→ ∞. Then
xk(n0)f
n0,xk(n0)
– 2Fn0,xk(n0)
Bearing in mind the continuity off and assuming (i), there exists a constantR2> 0 such
that
xf(n,x) – 2F(n,x)≤R2, ∀n∈[1,N],x∈R.
Therefore, whenk→ ∞,
N
n=1
xk(n)f
n,xk(n)
– 2Fn,xk(n)
≤xk(n0)f
n0,xk(n0)
– 2Fn0,xk(n0)
+ (N– 1)R2→–∞,
which is contrary to (2.19). Our result is proved.
(ii) Assumelim|x|→∞[xf(n,x) – 2F(n,x)] = +∞. In this case, the proof thatI1satisfies (C)
condition is similar to the above case and we omit it.
In summary,I1satisfies (C) condition under (F3) and the proof of Lemma2.6is
com-pleted.
Lemma 2.7 Let z1,z2denote eigenvectors corresponding to eigenvaluesλ1,λ2of matrix A1, respectively.If l>λ2,then
lim xm→+∞
I1(x) = –∞, ∀x∈H1= span{z1,z2}.
Proof (i) Whenl= +∞. From (2.18), it is easy to see that, for anyx∈H1,I1(x)→–∞as x m→+∞.
(ii) Supposel∈(λ2, +∞). For anyx∈H1,xcan be written asx=ε1z1+ε2z2. Without loss
of generality, we assume thatz1is orthogonal toz2, then x 2=ε12 z1 2+ε22 z2 2. Choose
such that 0 <<min{l–λi},i= 1, 2. In view oflim|x|→+∞f(nx,x) =l, it follows that we can
findη2> 0 such thatF(n,x)≥l–2x2–η2. Hence, for anyx∈H1,
I1(x) =
1
2(A1x,x) – N
n=1
Fn,x(n)
≤1
2
λ1ε12 z1 2+λ2ε22 z2 2
–l– 2 x
2+η 2N
=λ1–l+ 2 ε
2 1 z1 2+
λ2–l+
2 ε
2
2 z2 2+η2N.
ThusI1(x)→–∞as x m→+∞forλi–l+< 0,i= 1, 2.
With the above preparations, we are in a position to state the proof of Theorem2.1.
Proof of Theorem2.1 According to (2.16), we have
F(n,x) +m 2|x|
2≤1
2(m+λ1–τ)|x|
2+C s|x|
Forx∈Hands> 2 defined in (F2), there existsC1> 0 such that
|x|s:=
N
n=1
x(n)s
1
s
≤C1min
x , x m , ∀x∈H. (2.20)
Then
I1(x) =
1 2 x
2
m– N
n=1
Fn,x(n)+m 2x(n)
2
≥1
2 x
2
m–
m+λ1–τ
2 x
2–C s|x|
s s
≥ τ
2(m+λ1)
x 2m–CC s
1 s x
s m.
Recall that, for allx∈D+
ε∩D–ε, x± ≤√m1+λ1distm(x,∓Λ)≤
1
√
m+λ1ε0. Then we can find a constantc0> –∞such thatinfx∈D+
ε∩D–εI1(x) =c0. Lemma2.7indicates that there exists a constantR> 2ε0such thatI1(x) <c0– 1 holds for allx∈H1and x m=R. Last but not least, define a pathh: [0, 1]→H1as follows:
h(s) =R cos(πs)z1+sin(πs)z2
cos(πs)z1+sin(πs)z2 m .
Obviously,h(s)∈H1and h(s) m=Rwhens∈[0, 1], thenI1(h(s)) <c0– 1. Further,
h(0) =R z1 z1 m ∈
D+ε\D–ε, h(1) = –R
z1 z1 m ∈
D–ε\D+ε,
inf x∈D+ε∩D–ε
I1(x) =c0>c0– 1 > sup
s∈[0,1] I1
h(s).
Applying Lemma1.2,I1(x) possesses a critical point inH\(D+ε∪D–ε) which corresponds to
a sign-changing solution of BVP (1.1) with (1.2). Besides, there also exist a critical point in
D–
ε\D+εand a critical point inD+ε\D–εcorresponding to a negative solution and a positive
solution of BVP (1.1) with (1.2), respectively. This completes the proof of (i).
With the aid of Remark1.1and Lemma2.6, the verification for the case (ii) is similar to the case (i) and we leave it for the reader.
Here and hereafter in this section, we study positive solutions and negative solutions for BVP (1.1) with (1.2) by means of the mountain pass lemma.
Consider the following functionals:
I1±(x) =1 2x,x0–
N
n=1
Fn,x±(n), ∀x∈H.
Referring to [22], we have the following.
Lemma 2.8 Under the conditions of Theorem2.2,I1±are continuously differentiable.In addition,the critical points of I1+(I–1)are just the positive(negative)solutions of BVP(1.1)
Therefore, we have reduced the problem of looking for positive (negative) solutions of BVP (1.1) with (1.2) to that of seeking nonzero critical points of the functionalI+
1 (I–1) on H.
Lemma 2.9 Iflim inf|x|→∞f(n,x)
x >λ1for all n∈[1,N],then I
±
1 satisfy the(PS)condition.
Proof Here we test the case ofI+
1 at length, the other case can be proved analogously and
is omitted.
Let{xk} ⊂Hbe a sequence such that{I1+(xk)}is bounded andI1+(xk)→0 ask→ ∞. For simplicity, denote (f+x)(n) =f(n,x+(n)) for alln∈[1,N]. Then
x–k20≤xk,x–k
0≤
xk–K0f+xk,x–k
0=
I1+(xk),x–k
0=o(1)x –
k0,
which impliesx–k→0 ask→ ∞. Next we prove that{x+k}is bounded. If it is not, there is a subsequence of{xk}(still denoted by{xk}) satisfyingρk= x+k 0→+∞ask→ ∞. Set yk=
x+k
ρk, then yk 0= 1. Moreover, the completeness ofHmeans that there isy∈Hsuch
thatyk→yask→ ∞. Letz1> 0 denote the eigenvector corresponding toλ1, then
λ1
N
n=1
xk(n)z1(n) =
N
n=1
4xk(n– 2)z1(n)
= N+1
n=0
2xk(n– 1)2z1(n– 1)
=xk,z10
=K0f+xk,z1
0+
I1+(xk),z1
0
= N
n=1
fn,x+k(n)z1(n) +
I1+(xk),z1
0.
Dividing both sides byρk, we obtain
λ1
N
n=1
yk(n)z1(n) =
N
n=1 f(n,x+
k(n))
x+
k(n)
yk(n)z1(n) +o(1). (2.21)
It is well known that, for anyn∈[1,N], eitherx+k(n)→+∞or{x+k(n)}is bounded. Ifx+
k(n)→+∞, bylim inf|x|→∞ f(n,x)
x >λ1for alln∈[1,N], we obtain
min n∈[1,N]lim inf|x|→∞
f(n,x)
x >λ1. (2.22)
If{x+k(n)}is bounded, then f(n,x +
k(n))
ρk →0 andy(n) = 0. While the definition ofykgives
yk 0= 1,y= 0. Hence, it is not difficult to find annsuch thatxk+(n)→+∞andy(n) > 0.
Passing to the limit aboutkin (2.21) and making use of (2.22) yield
λ1
N
n=1
y(n)z1(n) = lim
k→∞
N
n=1
f(n,x+k(n))
x+k(n) yk(n)z1(n)
>λ1
N
n=1
which raises a contradiction. Therefore,I1+satisfies (PS) condition. The proof of Lemma2.9
is completed.
With the help of Lemma2.8and Lemma2.9, we can prove Theorem2.2via Lemma1.1.
Proof of Theorem2.2 According tomaxn∈[1,N]lim supx→0
f(n,x)
x <λ1, we have constantsξ1> 0 andr> 0 such that
F(n,x)≤λ1–ξ1 2 |x|
2, n∈[1,N],|x| ≤r.
For allx∈∂Br, we have
I1+(x) =1 2x,x0–
N
n=1
Fn,x+(n)
≥1
2 x
2 0–
λ1–ξ1
2 x
2
≥1
2 x
2 0–
λ1–ξ1
2λ1 x 20
=ξ1r
2
2λ1
ρ> 0.
Fromminn∈[1,N]lim inf|x|→∞f(nx,x)>λ1, we can choose a constantξ2> 0 such that
min n∈[1,N]lim inf|x|→∞
f(n,x)
x >λ1+ξ2.
Further, there exists a constantC> 0 such that
F(n,x)≥λ1+ξ2 2 |x|
2–C
is true for allx∈R. Hence, forνlarge enough
I1+(νz1) =
1
2νz1,νz10– N
n=1
Fn,νz+1(n)
≤ν2 2 z1
2 0–
λ1+ξ2
2 ν
2 z
1 2+CN
=ν
2
2 z1
2 0–
λ1+ξ2
2λ1
ν2 z1 20+CN
= –ξ2ν
2
2λ1
z1 20+CN< 0.
Based on Lemma 2.9and Lemma1.1, I1+(x) possesses a critical pointx0∈H such that I+
1(x0) = 0 andI1+(x0)≥ρ> 0. Thus
x–020≤x0,x–0
0≤
x0–K0f+x0,x–0
0=
I1+(x0),x–0
which implies thatx–
0= 0 andx0=x+0≥0. Ifx0= 0, thenI1+(x0) = 0, which is contradiction
withI+
1(x0)≥ρ> 0, sox0> 0. In view of Lemma2.8,x0> 0 is a positive solution of problem
(1.1) with (1.2).
By a little modification of the case ofI+
1, we can obtain a negative solution for the case
ofI1–.
3 Multiple solutions for BVP (1.1) with (1.3)
In this section, we discuss the multiple solutions for BVP (1.1) with (1.3).
LetM2={x: [–1,N+ 2]→R|ix(–1) =ix(N– 1),i= 0, 1, 2, 3}equipped with the inner
product
x,ym= N
n=1
2x(n– 1)2y(n– 1) +mx(n)y(n).
ThenM2is an N-dimensional Hilbert space and the induced norm is
x m=
N
n=1
2x(n– 1)2+mx(n)2
1 2 .
Consider the functionalI2:H→R
I2(x) =
1 2
N
n=1
2x(n– 1)2– N
n=1
Fn,x(n). (3.1)
For anyx= (x(1),x(2), . . . ,x(N))τ∈H, (3.1) can be rewritten as
I2(x) =
1
2(A2x,x) – N
n=1
Fn,x(n), (3.2)
whereA2is anN×Nmatrix defined as
A2=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
6 –4 1 · · · 0 1 –4
–4 6 –4 · · · 0 0 1
1 –4 6 · · · 0 0 0
· · · ·
0 0 0 · · · 6 –4 1
1 0 0 · · · –4 6 –4
–4 1 0 · · · 1 –4 6
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
N×N .
Direct computation shows thatωk= 16sin4kNπ,k= 0, 1, . . . ,N– 1 are the eigenvalues ofA2.
Then whenk= 1, 2, . . . ,N– 1, we have
ωk=ωN–k, 0 <ω1≤ω2≤ · · · ≤ω[N2].
Theorem 3.1 Suppose that: (F4) maxn∈[1,N]lim supx→0|
f(n,x)
x |<ω1; (F5) whenx= 0,x(f(n,x) +mx) > 0;
(F6) there exist constantsσ> 1,C3,C4> 0such that|f(n,x)| ≤C3|x|σ +C4;
(F7) lim|x|→∞f(nx,x)= +∞;
(F8) whenx= 0,xf(n,x) – 2F(n,x) > 0andxf(n,x) – 2F(n,x)→ ∞as|x| → ∞. Then we have:
(i) if(F4)–(F7)hold,then BVP(1.1)with(1.3)has at least three nontrivial solutions:one sign-changing,one positive and one negative;
(ii) the conclusion is still true if we replace(F7)with(F8)in the case(i).
Now we pay our attention to verifying Theorem3.1with the aid of Lemma1.2. Motivated by Sect.2, we consider following boundary value problem to obtain Green’s function corresponding to BVP (1.1) with (1.3):
⎧ ⎨ ⎩
4x(n– 2) –m2x(n– 1) =h(n), n∈[1,N],
ix(–1) =ix(N– 1), i= 0, 1, 2, 3, (3.3)
whereh: [1,N]→R.
Similar to Lemma2.1, we have the following.
Lemma 3.1 Denote
G3(n,s) =
⎧ ⎨ ⎩
Ps–n+1+N+Pn–s+1
(1–PN)(1–P2) , 0≤s≤n≤N+ 1, Pn–s+1+N+Ps–n+1
(1–PN)(1–P2) , 0≤n≤s≤N+ 1,
and
G4(n,s) =
⎧ ⎨ ⎩
(s–1)(N+1–n)
N , 0≤s≤n≤N+ 1,
(n–1)(N+1–s)
N , 0≤n≤s≤N+ 1,
where P is defined in Lemma2.1.Then
x(n) = N
s=1
Gj(n,s)h(s), n∈[0,N+ 1],j= 3, 4,
is the unique solution of
⎧ ⎨ ⎩
–2x(n– 1) +r
jx(n) =h(n), n∈[1,N],j= 3, 4, ix(–1) =ix(N– 1), i= 0, 1, 2, 3,
where r3,r4are solutions of equation r2–mr= 0.
Lemma 3.2 The linear discrete fourth-order BVP(3.3)admits a unique solution
x(n) = N
s=1
N
j=1
G3(n,j)G4(j,s)h(s)
= N
s=1
N
j=1
G4(n,j)G3(j,s)h(s)
, n∈[1,N]
satisfying x(–1) =x(N– 1),x(0) =x(N),x(1) =x(N+ 1)and x(2) =x(N+ 2).
Remark3.1 Green’s functionGj(n,s) (j= 3, 4) defined by Lemma3.1satisfies Gj(n,s) =
Gj(s,n) > 0,n,s∈[1,N]. Forx∈H, define operatorsKm∗,fm∗andS∗mas follows:
Km∗x(n) = N
s=1
N
j=1
G4(n,j)G3(j,s)x(s),
fm∗x(n) =fn,x(n)–m2x(n– 1),
S∗m=Km∗fm∗,
whereS∗m:H→H is a completely continuous operator. Then{x(n)}N+1
n=0 is a solution of
BVP (1.1) with (1.3) if and only ifx={x(n)}N
n=1∈His a fixed point ofS∗m.
With the aim to look for solutions of BVP (1.1) with (1.3) by Lemma1.2, we need the following lemma and we omit its proof because of its similarity to Lemma2.3.
Lemma 3.3 The functional I2 defined by(3.1)is Fréchet differentiable on H and I2(x) = x–Sm∗(x)for all x∈H.
According to Lemma1.2and Remark1.1, it is time for us to formulate a test.
Lemma 3.4 Let I2be defined as(3.1),
(i) I2satisfies the(PS)condition under(F7);
(ii) I2satisfies the(C)condition under(F8).
Proof (i) The proof is similar to that of Lemma2.5(i) and thus is omitted. (ii) Let{xk} ⊂Hbe a sequence, then there exists a constantC˜> 0 such that
I2(xk)≤ ˜
C– 1 2 and
1 + xk mI2(xk)m< 1.
It follows from the Cauchy–Schwarz inequality that
I2(xk),xk
m≤ xk mI2
(x
k)m≤
1 + xk mI2(xk)m< 1.
Therefore,
N
n=1
xkf(n,xk) – 2F(n,xk)
= 2I2(xk) –
I2(xk),xk
m
In view of (F8), there existsδ> 0 such that
xf(n,x) – 2F(n,x) >C˜, ∀n∈[1,N] and|x|>δ,
which contradicts (3.4). Then{xk}is bounded and the proof is finished.
Lemma 3.5 Suppose that(F4), (F5)and(F6)are true,there exists constantε˜0> 0such that, for0 <ε<ε˜0,
(i) S∗m(∂D–
ε)⊂D–ε.And ifx∈D–ε is a nontrivial critical point ofI2,thenxis a negative solution of BVP(1.1)with(1.3);
(ii) S∗m(∂D+ε)⊂D+ε.And ifx∈D+εis a nontrivial critical point ofI2,thenxis a positive solution of BVP(1.1)with(1.3).
Proof (i) Sinceω1 andω[N
2] are the minimum and maximum eigenvalues of matrixA2, respectively. Then
√
ω1+m x ≤ x m≤
ω[N
2]+m x , (3.5)
which leads to
x+= inf
ψ∈–Λ x–ψ ≤
1 √
m+ω1
inf
ψ∈–Λ x–ψ m
=√ 1
m+ω1
distm(x, –Λ), ∀x∈H. (3.6)
According to (F4) and (F6), we can find a constantτ˜> 0 such that
f(n,x) +mx≤(m+ω1–τ˜)|x|+C2|x|σ, ∀(n,x)∈[1,N]×R. (3.7)
Lety=Sm∗(x) for anyx∈H. As a matter of fact,y+=y–y–andy–∈–Λmeandistm(y, –Λ)≤
y–y–
m= y+ m. Combining (3.5), (3.6) and (3.7), it yields
distm(y, –Λ)y+m
≤y+,y+m
=S∗mx+,y+m
= N
n=1
fn,x+(n)+mx+(n)y+(n)
≤(m+ω1–τ˜)
N
n=1
x+(n)y+(n) +C2
N
n=1
x+(n)σy+(n)
≤(m+ω1–τ˜)x+y++C2x+
σ
y+
≤
m+ω1–τ˜ m+ω1
distm(x, –Λ) +
C2
(m+ω1)σ+1
distm(x, –Λ)
σ
which implies
distm(y, –Λ)≤m+ω1–τ˜
m+ω1
distm(x, –Λ) +C2distm(x, –Λ)σ,
hereC2=√ C2
(m+ω1)σ+1
. LetC2(distm(x, –Λ))σ–1=2(mτ+˜ω1), then
distm
S∗m(x), –Λ≤2(m+ω1) –τ˜ 2(m+ω1)
distm(x, –Λ). (3.8)
Obviously,2(m+ω1)–τ˜
2(m+ω1) < 1 is correct for allτ˜> 0. Consequently,
S∗m(x)∈Dε–, ∀x∈∂D–ε.
Ifx∈D–
ε is a nontrivial critical point ofI2, Lemma3.3showsI2(x) =x–S∗m(x) = 0. More-over, (3.8) indicatesx∈–Λ\{0}. Following (F5) and Remark3.1, we see thatx(n) < 0 in n∈[1,N]. Therefore,xis a negative solution of BVP (1.1) with (1.3).
Item (ii) can be discussed similarly, we omit its proof.
Now we devote our efforts to completing the proof of our main results in this section.
Proof of Theorem3.1 SinceHis a finite-dimensional Hilbert space, there exists constant ¯
C> 0 such that
|x|σ+1:=
N
n=1
x(n)σ+1
1
σ+1
≤ ¯Cmin x , x m , ∀x∈H. (3.9)
According to (3.7), we have
F(n,x) +m 2|x|
2≤1
2(m+ω1–τ˜)|x|
2+ C2
σ+ 1|x|
σ+1.
Making use of this, together with (3.5) and (3.9), it follows that
I2(x) =
1
2x,xm– N
n=1
Fn,x(n)+m 2x(n)
2
≥1
2 x
2
m–
m+ω1–τ˜
2 x
2– C2
σ+ 1|x|
σ+1
σ+1
≥ τ˜
2(m+ω1)
x 2m–C2C¯
σ+1
σ+ 1 x
σ+1
m .
Due to (3.6), x± ≤√ 1
m+ω1dist(x,∓Λ)≤
1
√
m+ω1ε˜0 holds for allx∈D
+
ε∩D–ε. Then there
existsc1> –∞such thatinfx∈D+
ε∩D–εI2(x) =c1. Letυ1,υ2be eigenvectors corresponding to eigenvaluesω1,ω2of matrixA2. DenoteH2= span{υ1,υ2}, for allx∈H2. Note thatI2(x)→
–∞as x m→+∞. Thus it is easy to look for a constantR˜> 2ε˜0such thatI2(x) <c1– 1
and x m=R˜. Define a pathg: [0, 1]→H2as
g(s) =R˜ cos(πs)υ1+sin(πs)υ2
Direct calculation gives
g(0) =R˜ υ1
υ1 m ∈
D+ε\D–ε, g(1) = –R˜ υ1
υ1 m ∈
D–ε\D+ε,
and
inf x∈D+ε∩D–ε
I2(x) > sup
s∈[0,1] I2
g(s).
Combining Lemma3.4(i) with Lemma3.5, it is easy to see that there is a critical point in
H\(D+
ε∪D–ε) corresponding to a sign-changing solution of BVP (1.1) with (1.3). In
addi-tion, we have a critical point inD+
ε\D–ε corresponding to a positive solution and have a
critical point inD–
ε\D+ε corresponding to a negative solution of BVP (1.1) with (1.3). The
proof of (i) is finished.
It follows from Lemma3.4(ii) and Remark1.1, the proof for (ii) is analogous. Thus, the
proof of Theorem3.1is complete.
4 Examples
To demonstrate the applicability of our theoretical results, three examples are provided.
Example4.1 LetN= 4, we consider BVP
⎧ ⎨ ⎩
4x(n– 2) =– 14
5x(n)
1+x2(n) + 3x(n), n∈[1, 4],
x(–1) =x(0) = 0 =x(5) =x(6). (4.1)
It is easy to see thatf(n,x) = –145x
1+x2 + 3xandλ1= 0.3944,λ2= 2.6148. Then
(1) limx→0f(nx,x)=limx→0( –145
1+x2 + 3) = 0.2 <λ1;
(2) lim|x|→∞f(nx,x)=lim|x|→∞(
–145
1+x2+ 3) = 3 >λ2;
(3) lim|x|→∞[xf(n,x) – 2F(n,x)] =lim|x|→∞145(ln(1 +x2) –
x2
1+x2) = +∞.
Hence (4.1) meets all conditions in Theorem2.1. Therefore, Theorem2.1guarantees that (4.1) admits at least a positive solution, a negative solution and a sign-changing so-lution. By direct calculation, we see that (0, 0, –0.1566, –0.2993, –0.2993, –0.1566, 0, 0) is the negative solution and (0, 0, 0.1566, 0.2993, 0.2993, 0.1566, 0, 0) is the positive solution. In addition, (0, 0, 2.9287, 1.9285, –1.9285, –2.9287, 0, 0) and (0, 0, –2.9287, –1.9285, 1.9285, 2.9287, 0, 0) are two sign-changing solutions.
Example4.2 Consider the following problem:
⎧ ⎨ ⎩
4x(n– 2) =x3(n) +1
5x(n), n∈[1, 5],
x(–1) =x(0) = 0 =x(6) =x(7). (4.2)
In fact,f(n,x) =x3+1
5xandλ1= 0.2121. Furthermore,
(1) lim|x|→∞f(n,x)
x =lim|x|→∞(x
2+1 5) >λ1;
