# Nonexistence of Homoclinic Solutions for a Class of Discrete Hamiltonian Systems

Authors: Xiaoping Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We give several sufficient conditions under which the first-order nonlinear discrete Hamiltonian systemΔx(n)=α(n)x(n+1)+β(n)|y(n)|μ-2y(n),Δy(n)=-γ(n)|x(n+1)|ν-2x(n+1)-α(n)y(n) has no solution (x(n),y(n)) satisfying condition 0<∑n=-∞+∞[|x(n)|ν+(1+β(n))|y(n)|μ]<+∞, where μ,ν>1 and 1/μ+1/ν=1 and α(n),β(n), and γ(n) are real-valued functions defined on ℤ.


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## Body

## 1. Introduction

In 1907, Lyapunov [1] established the first so-called Lyapunov inequality:
(1)(b-a)∫ab‍q(t)dt>4,
if Hill’s equation
(2)x′′(t)+q(t)x(t)=0
has a real solution x(t) such that
(3)x(a)=x(b)=0,x(t)≢0,t∈[a,b],
and the constant 4 in (1) cannot be replaced by a larger number, where q(t) is a piecewise continuous and nonnegative function defined on ℝ. Since this result has found applications in the study of various properties of solutions such as oscillation theory, disconjugacy, and eigenvalue problems of (2), a large number of Lyapunov-type inequalities were established in the literature which generalized or improved (1); see [1–20].In 1983, Cheng [3] first obtained the discrete analogy of Lyapunov inequality (1) for the second-order difference equation:
(4)Δ2x(n)+q(n)x(n+1)=0,
where, and in the sequel, Δ denotes the forward difference operator defined by Δx(n)=x(n+1)-x(n).Whena=-∞ and b=+∞, that is, system (4) has a solution x(n) satisfying lim|n|→∞x(n)=0, which is called homoclinic solution, whether one can obtain Lyapunov-type inequalities for (4)? To the best of our knowledge, there are no results.In 2003, Sh. Guseinov and Kaymakçalan [7] partly generalized the Cheng’s result to the discrete linear Hamiltonian system:(5)Δx(n)=α(n)x(n+1)+β(n)y(n),Δy(n)=-γ(n)x(n+1)-α(n)y(n),
where α(n), β(n), and γ(n) are real-valued functions defined on ℤ and a and b are not necessarily usual zeros, but rather, generalized zeros. Later, some better Lyapunov-type inequalities for system (5) were obtained in [19, 20].Very recently, He and Zhang [10] further generalized the result in [19] to the following first-order nonlinear difference system:
(6)Δx(n)=α(n)x(n+1)+β(n)|y(n)|μ-2y(n),Δy(n)=-γ(n)|x(n+1)|ν-2x(n+1)-α(n)y(n),
where μ,ν>1 and 1/μ+1/ν=1 and  α(n), β(n), and γ(n) are real-valued functions defined on ℤ.Whenμ=ν=2, system (6) reduces to (5). In addition, the special forms of system (6) contain many well-known difference equations which have been studied extensively and have much applications in the literature [21–23], such as the second-order linear difference equation:
(7)Δ[p(n)Δx(n)]+q(n)x(n+1)=0,
and the second-order half-linear difference equation:
(8)Δ[p(n)|Δx(n)|r-2Δx(n)]+q(n)|x(n+1)|r-2x(n+1)=0,
where r>1, p(n) and q(n) are real-valued functions defined on ℤ and p(n)>0. Let
(9)y(n)=p(n)|Δx(n)|r-2Δx(n),
then (8) can be written as the form of (6):
(10)Δx(n)=[p(n)]1/(1-r)|y(n)|(2-r)/(r-1)y(n),Δy(n)=-q(n)|x(n+1)|r-2x(n+1),
where μ=r/(r-1), ν=r and α(n)=0, β(n)=[p(n)]1/(1-r) and γ(n)=q(n).In this paper, we will establish several Lyapunov-type inequalities for systems (5) and (6) if they have a solution (x(n),y(n)) satisfying conditions
(11)0<∑-∞+∞‍[|x(n)|2+(1+β(n))|y(n)|2]<+∞,(12)0<∑-∞+∞‍[|x(n)|ν+(1+β(n))|y(n)|μ]<+∞,
respectively. Taking advantage of these Lyapunov-type inequalities, we are able to establish some criteria for nonexistence of homoclinic solutions of systems (5) and (6). As we know, there are no results on non-existence of homoclinic solutions for Hamiltonian systems in previous literature.


## 2.  Lyapunov-Type Inequalities for System (6)

In this section, we shall establish some Lyapunov-type inequalities for system (6). For the sake of convenience, we list some assumptions on α(n) and β(n) as follows:(A0)
α(n)<1, for all n∈ℤ,   ∏s=-∞+∞‍[1-α(s)]-1<∞;(A1)
α(n)<1, for all n∈ℤ,   ∑s=-∞+∞‍|α(s)|<+∞;(B0)
β(n)≥(≢)0, for all n∈ℤ;(B1)
∑τ=-∞0‍β(τ)∏s=τ0‍[1-α(s)]-μ+∑τ=1+∞‍β(τ)∏s=0τ-1‍[1-α(s)]μ<+∞.Denote(13)ζ(n):=[∑τ=-∞n‍β(τ)∏s=τn‍[1-α(s)]-μ]ν/μ,η(n):=[∑τ=n+1+∞‍β(τ)∏s=n+1τ-1‍[1-α(s)]μ]ν/μ.Theorem 1.
Suppose that hypotheses (A0), (B0), and (B1) are satisfied. If system (6) has a solution (x(n),y(n)) satisfying
(14)0<∑n=-∞+∞‍[|x(n)|ν+(1+β(n))|y(n)|μ]<+∞,
then one has the following inequality:
(15)∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)≥1,
where γ+(n)=max{γ(n),0}.Proof.
Hypothesis (B1) implies that functionsζ(n) and η(n) are well defined on ℤ. Without loss of generality, we can assume that
(16)∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)<+∞.
From (14) and (B0), one has
(17)lim|n|→∞|x(n)|=lim|n|→∞|y(n)|=0,(18)∑τ=-∞+∞‍β(τ)|y(τ)|μ<+∞.
It follows from (13), (18), and the Hölder inequality that
(19)∑τ=-∞n‍β(τ)|y(τ)|μ-1∏s=τn‍[1-α(s)]-1≤[∑τ=-∞n‍β(τ)∏s=τn‍[1-α(s)]-μ]1/μ[∑τ=-∞n‍β(τ)|y(τ)|μ]1/ν=[ζ(n)]1/ν[∑τ=-∞n‍β(τ)|y(τ)|μ]1/ν<+∞,∀n∈ℤ,(20)∑τ=n+1+∞‍β(τ)|y(τ)|μ-1∏s=n+1τ-1‍[1-α(s)]≤[∑τ=n+1+∞‍β(τ)∏s=n+1τ-1‍[1-α(s)]μ]1/μ[∑τ=n+1+∞‍β(τ)|y(τ)|μ]1/ν=[η(n)]1/ν[∑τ=n+1+∞‍β(τ)|y(τ)|μ]1/ν<+∞,∀n∈ℤ.
From (A0), (17), (19), (20), and the first equation of system (6), we have
(21)x(n+1)=∑τ=-∞n‍β(τ)|y(τ)|μ-2y(τ)∏s=τn‍[1-α(s)]-1,∀n∈ℤ,(22)x(n+1)=-∑τ=n+1+∞‍β(τ)|y(τ)|μ-2y(τ)∏s=n+1τ-1‍[1-α(s)],∀n∈ℤ.
Combining (19) with (21), one has
(23)|x(n+1)|ν=|∑τ=-∞n‍β(τ)|y(τ)|μ-2y(τ)∏s=τn‍[1-α(s)]-1|ν≤ζ(n)∑τ=-∞n‍β(τ)|y(τ)|μ,∀n∈ℤ.
Similarly, it follows from (20) and (22) that
(24)|x(n+1)|ν=|∑τ=n+1+∞‍β(τ)|y(τ)|μ-2y(τ)∏s=n+1τ-1‍[1-α(s)]|ν≤η(n)∑τ=n+1+∞‍β(τ)|y(τ)|μ,∀n∈ℤ.
Combining (23) with (24), one has
(25)|x(n+1)|ν≤ζ(n)η(n)ζ(n)+η(n)∑τ=-∞+∞‍β(τ)|y(τ)|μ,∀n∈ℤ.
Now, it follows from (16), (18), and (25) that
(26)∑n=-∞+∞‍γ+(n)|x(n+1)|ν≤[∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)]×∑n=-∞+∞‍β(n)|y(n)|μ<+∞.
By (6), we obtain
(27)Δ(x(n)y(n))=β(n)|y(n)|μ-γ(n)|x(n+1)|ν.
Summing the above from -∞ to +∞ and using (17) and (18), we obtain
(28)∑n=-∞+∞‍γ(n)|x(n+1)|ν=∑n=-∞+∞‍β(n)|y(n)|μ,
which, together with (26), implies that
(29)∑n=-∞+∞‍γ+(n)|x(n+1)|ν≤[∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)]∑n=-∞+∞‍β(n)|y(n)|μ=[∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)]∑n=-∞+∞‍γ(n)|x(n+1)|ν≤[∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)]∑n=-∞+∞‍γ+(n)|x(n+1)|ν.
We claim that
(30)∑n=-∞+∞‍γ+(n)|x(n+1)|ν>0.
If (30) is not true, then
(31)∑n=-∞+∞‍γ+(n)|x(n+1)|ν=0.
From (28) and (31), we have
(32)0≤∑n=-∞+∞‍β(n)|y(n)|μ=∑n=-∞+∞‍γ(n)|x(n+1)|ν≤∑n=-∞+∞‍γ+(n)|x(n+1)|ν=0.
It follows that
(33)β(n)|y(n)|μ-2y(n)≡0,∀n∈ℤ.
Combining (21) with (33), we obtain that
(34)x(n)≡0,∀n∈ℤ,
which, together with the second equation of system (6), implies that
(35)Δy(n)=-α(n)y(n),∀n∈ℤ.
Combining the above with (17), one has
(36)y(n)≡0,∀n∈ℤ.
Both (34) and (36) contradict with (14). Therefore, (30) holds. Hence, it follows from (29) and (30) that (15) holds.Corollary 2.
Suppose that hypotheses (A1), (B0), and (B1) are satisfied. If system (6) has a solution (x(n),y(n)) satisfying (14), then one has the following inequality:
(37)∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]ν/2μ≥2∏n=-∞+∞‍{Θ[α(n)]}ν/2,
where and in the sequel,
(38)Θ[α(n)]=min{1-α+(n),[1+α-(n)]-1},α+(n)=max{α(n),0},α-(n)=max{-α(n),0}.Proof.
Obviously, (A1) implies that(39)0<∏s=-∞+∞‍[1-α(s)]<+∞,
and so (A0) holds, and which, together with (B1), implies that ∑τ=-∞+∞‍β(τ)<+∞. Since
(40)ζ(n)+η(n)≥2[ζ(n)η(n)]1/2,
it follows that
(41)1≤∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)≤12∑n=-∞+∞‍[ζ(n)η(n)]1/2γ+(n)=12∑n=-∞+∞‍γ+(n){∑τ=-∞n‍β(τ)∏s=τn‍[1-α(s)]-μ×∑τ=n+1+∞‍β(τ)∏s=n+1τ-1‍[1-α(s)]μ}ν/2μ≤12∑n=-∞+∞‍γ+(n){∑τ=-∞n‍β(τ)∏s=τn‍[1-α+(s)]-μ×∑τ=n+1+∞‍β(τ)∏s=n+1τ-1‍[1+α-(s)]μ}ν/2μ≤12∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]ν/2μ×∏s=-∞n‍[1-α+(s)]-ν/2∏s=n+1+∞‍[1+α-(s)]ν/2≤12∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1b‍β(τ)]ν/2μ×∏s=-∞+∞‍{Θ[α(s)]}-ν/2,
which implies that (37) holds.Since(42)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]1/2≤12∑n=-∞+∞‍β(n),
then it follows from (37) that the following corollary is true.Corollary 3.
Suppose that hypotheses (A1), (B0), and (B1) are satisfied. If system (6) has a solution (x(n),y(n)) satisfying (14), then
(43)(∑n=-∞+∞‍β(n))1/μ(∑n=-∞+∞‍γ+(n))1/ν≥2∏n=-∞+∞‍{Θ[α(n)]}1/2.Applying Theorem1 and Corollary 2 to system (8) (i.e., (10)), we have immediately the following two corollaries.Corollary 4.
Suppose thatr>1 and p(n)>0 for n∈ℤ, and that
(44)∑τ=-∞+∞‍1[p(τ)]1/(r-1)<+∞.
If (8) has a solution x(n) satisfying
(45)0<∑n=-∞+∞‍[|x(n)|r+p(n)(1+[p(n)]1/(r-1))|Δx(n)|r]<+∞,
then (46)∑n=-∞+∞‍({∑τ=n+1+∞[p(τ)]-1(r-1)}r-1{∑τ=-∞n[p(τ)]-1/(r-1)}r-1{∑τ=n+1+∞[p(τ)]-1/(r-1)}r-1×({∑τ=-∞n[p(τ)]-1/(r-1)}r-1+{∑τ=n+1+∞[p(τ)]-1/(r-1)}r-1)-1)q+(n)≥1.Corollary 5.
Suppose thatr>1 and p(n)>0 for n∈ℤ, and that (44) holds. If (8) has a solution x(n) satisfying (45), then
(47)∑n=-∞+∞‍q+(n){∑τ=-∞n‍[p(τ)]-1/(r-1)∑τ=n+1+∞‍[p(τ)]-1/(r-1)}(r-1)/2≥2.


## 3. Lyapunov-Type Inequalities for System (5)

Whenμ=ν=2, assumption (B1) reduces the following form:(B2)
∑τ=-∞0‍β(τ)∏s=τ0‍[1-α(s)]-2+∑τ=1+∞‍β(τ)∏s=0τ-1‍[1-α(s)]2<+∞.Applying the results obtained in last section to the first-order linear Hamiltonian system (5), we have immediately the following corollaries.Corollary 66.
Suppose that hypotheses (A0), (B0), and (B2) are satisfied. If system (5) has a solution (x(n),y(n)) satisfying
(48)0<∑n=-∞+∞‍[|x(n)|2+(1+β(n))|y(n)|2]<+∞,
then
(49)∑n=-∞+∞‍({∑τ=n+1+∞[p(τ)]-1(r-1)}r-1{∑τ=-∞nβ(τ)∏s=τn[1-α(s)]-2}×{∑τ=n+1+∞β(τ)∏s=n+1τ-1[1-α(s)]2}×(∑τ=-∞ββ(τ)-1∑τ=-∞nβ(τ)∏s=τn[1-α(s)]-2+∑τ=n+1+∞β(τ)∏s=n+1τ-1[1-α(s)]2)-1)γ+(n)≥1.Corollary 7.
Suppose that hypotheses (A1), (B0), and (B2) are satisfied. If system (5) has a soldution (x(n),y(n)) satisfying (48), then
(50)∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]1/2≥2∏n=-∞+∞‍Θ[α(n)].Corollary 8.
Suppose thatp(n)>0 for n∈ℤ, and that
(51)∑τ=-∞+∞‍1p(τ)<+∞.
If (7) has a solution x(n) satisfying
(52)0<∑n=-∞+∞‍[|x(n)|2+p(n)(1+p(n))|Δx(n)|2]<+∞,
then
(53)∑n=-∞+∞‍q+(n)[∑τ=-∞n‍1p(τ)∑τ=n+1+∞‍1p(τ)]≥∑n=-∞+∞‍1p(n).


## 4. Nonexistence of Homoclinic Solutions

Applying the results obtained in Sections2 and 3, we can drive the following criteria for non-existence of homoclinic solutions of systems (5) and (6) immediately.Corollary 9.
Suppose that hypotheses (A0), (B0), and (B1) are satisfied. If(54)∑n=-∞+∞‍ζ(n)η(n)ζ(n)+η(n)γ+(n)<1,
then system (6) has no solution (x(n),y(n)) satisfying
(55)0<∑n=-∞+∞‍[|x(n)|ν+(1+β(n))|y(n)|μ]<+∞.Corollary 10.
Suppose that hypotheses (A1), (B0), and (B1) are satisfied. If(56)∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]ν/2μ<2∏n=-∞+∞‍{Θ[α(n)]}ν/2,
then system (6) has no solution (x(n),y(n)) satisfying (55).Corollary 11.
Suppose that hypotheses (A1), (B0), and (B1) are satisfied. If(57)(∑n=-∞+∞‍β(n))1/μ(∑n=-∞+∞‍γ+(n))1/ν<2∏n=-∞+∞‍{Θ[α(n)]}1/2,
then system (6) has no solution (x(n),y(n)) satisfying (55).Corollary 12.
Suppose that hypotheses (A0), (B0), and (B2) are satisfied. If(58)∑n=-∞+∞({∑τ=n+1+∞[p(τ)]-1(r-1)}r-1{∑τ=-∞nβ(τ)∏s=τn[1-α(s)]-2}‍×{∑τ=n+1+∞β(τ)∏s=n+1τ-1[1-α(s)]2}×(∑τ=-∞n-1β(τ)∑τ=-∞nβ(τ)∏s=τn[1-α(s)]-2+∑τ=n+1+∞β(τ)∏s=n+1τ-1[1-α(s)]2)-1)γ+(n)<1,
then system (5) has no solution (x(n),y(n)) satisfying
(59)0<∑n=-∞+∞‍[|x(n)|2+(1+β(n))|y(n)|2]<+∞.Corollary 13.
Suppose that hypotheses (A1), (B0), and (B2) are satisfied. If(60)∑n=-∞+∞‍γ+(n)[∑τ=-∞n‍β(τ)∑τ=n+1+∞‍β(τ)]1/2<2∏n=-∞+∞‍Θ[α(n)],
then system (5) has no solution (x(n),y(n)) satisfying (59).Corollary 14.
Suppose thatp(n)>0 for n∈ℤ, and that (51) holds. If
(61)∑n=-∞+∞‍q+(n)(∑τ=-∞n‍1p(τ)∑τ=n+1+∞‍1p(τ))<∑n=-∞+∞‍1p(n),
then (7) has no solution x(n) satisfying (52).Example 15.
Consider the second-order difference equation:(62)Δ[(1+n2)Δx(n)]+q(n)x(n+1)=0,
where q(n) is real-valued function defined on ℤ. In view of Corollary 14, if
(63)∑n=-∞+∞‍[(∑τ=-∞n‍11+τ2∑τ=n+1+∞‍11+τ2)]q+(n)<∑n=-∞+∞‍11+n2,
then (62) has no solution x(n) satisfying
(64)0<∑n=-∞+∞‍[|x(n)|2+(1+n2)2|Δx(n)|2]<+∞.



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