# On Inequalities of Lyapunov for Two-Dimensional Nonlinear Dynamic Systems on Time Scales

Authors: Qiao-Luan Li; Wing-Sum Cheung; Xu-Yang Fu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We establish some new Lyapunov-type inequalities for two-dimensional nonlinear dynamic systems on time scales.
As for application, boundedness of the Emden-Fowler-type equation is proved.


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## Body

## 1. Introduction

In this paper, we establish some Lyapunov-type inequalities for the following two-dimensional nonlinear dynamic system:(1)xΔ1Δ2(s,t)=α1(s,t)x(σ(s),σ(t))+β1(s,t)|u(s,t)|γ-2u(s,t),uΔ1Δ2(s,t)=-β2(s,t)|x(σ(s),σ(t))|β-2x(σ(s),σ(t))-α1(s,t)u(s,t),
wheres,t∈𝕋=a time scale,Δ1denotes the delta derivative with respect tos, andΔ2denotes the delta derivative with respect tot.Lyapunov-type inequalities have proven to be very useful in the study of qualitative behavior of solutions such as oscillation, disconjugacy, and eigenvalue problems for differential equations and difference equations. Since the appearance of Lyapunov's fundamental paper [1], considerable attention has been given to various extensions and improvements of the Lyapunov-type inequality from different viewpoints [2–8]. Although Lyapunov-type inequalities are well developed for the continuous cases, their time scale versions are still in early stages and are worth due attention.Recently, He et al. in [2] considered the linear Hamiltonian system
(2)xΔ(t)=α(t)x(σ(t))+β(t)y(t),yΔ(t)=-γ(t)x(σ(t))-α(t)y(t)
and obtained several useful Lyapunov-type inequalities.Chen et al. in [3] considered the nonlinear system
(3)∂2x(s,t)∂s∂t=α1(s,t)x(s,t)+β1(s,t)|u(s,t)|γ-2u(s,t),∂2u(s,t)∂s∂t=-β2(s,t)|x(s,t)|β-2x(s,t)-α1(s,t)u(s,t)
and obtained some interesting Lyapunov-type inequalities for partial differential equations.In this paper, under the assumption of existence of a nontrivial solution(x(s,t),u(s,t))to the 2-dimensional nonlinear dynamic system (1), some new and interesting Lyapunov-type inequalities are established.


## 2. Main Results

Throughout this paper, the following mild and natural conditions are assumed:(i)
γ>1,β>1are real constants,(ii)
α1(s,t),  β1(s,t),β2(s,t):[s0,∞)𝕋×[t0,∞)𝕋→Rare rd-continuous functions such thatβ1(s,t)>0for(s,t)∈[s0,∞)𝕋×[t0,∞)𝕋, where[s0,∞)𝕋:=[s0,∞)∩𝕋, andσis the forward jump operator; that is,σ(t):=inf{s∈𝕋:s>t}.Theorem 1.
If the nonlinear dynamic system (1) has a real solution(x(s,t),u(s,t))which is not identically zero on[a,b]𝕋×[c,d]𝕋satisfyingx(a,t)=x(b,t)=x(s,c)=x(s,d)=0andxΔ1(s,σ(t))uΔ2(s,t)+xΔ2(σ(s),t)uΔ1(s,t)=0for all(s,t)∈[a,b]𝕋×[c,d]𝕋, wherea,b,c,d∈𝕋witha<b,  c<d, then
(4)2≤∫ab‍∫cd‍|α1(s,t)|ΔtΔs+Mβ/α-1(∫ab‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫ab‍∫cd‍β2+(s,t)ΔtΔs)1/α,
where1/α+1/γ=1,M:=max{|x(s,t)|:σ(a)≤s<b,σ(c)≤t<d}, andβ2+(s,t):=max{β2(s,t),0}.Proof.
From the conditionsx(a,t)=x(b,t)=x(s,c)=x(s,d)=0andx(s,t)is not identically zero on[a,b]𝕋×[c,d]𝕋, there exists(τ1,τ2)such that|x(τ1,τ2)|=M>0.
Multiplying the first equation of (1) byu(s,t)and the second one byx(σ(s),σ(t))and adding up, we get
(5)xΔ1Δ2(s,t)u(s,t)+uΔ1Δ2(s,t)x(σ(s),σ(t))=β1(s,t)|u(s,t)|γ-β2(s,t)|x(σ(s),σ(t))|β,
and, hence,
(6)(x(s,t)u(s,t))Δ1Δ2=xΔ1Δ2(s,t)u(s,t)+xΔ2(σ(s),t)uΔ1(s,t)+xΔ1(s,σ(t))uΔ2(s,t)+x(σ(s),σ(t))uΔ1Δ2(s,t)=xΔ1Δ2(s,t)u(s,t)+uΔ1Δ2(s,t)x(σ(s),σ(t))=β1(s,t)|u(s,t)|γ-β2(s,t)|x(σ(s),σ(t))|β.
Integrating the left hand side of (6) overtfromctodand then oversfromatob, we get
(7)∫ab‍∫cd‍(x(s,t)u(s,t))Δ1Δ2ΔtΔs=∫ab‍(x(s,d)u(s,d))Δ1Δs-∫ab‍(x(s,c)u(s,c))Δ1Δs=x(b,d)u(b,d)-x(a,d)u(a,d)-x(b,c)u(b,c)+x(a,c)u(a,c).
Noting thatx(a,t)=x(b,t)=0, we have
(8)∫ab‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs=∫ab‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs.
On the other hand, integrating the first equation of (1) overtfromctoτ2and then oversfromatoτ1, we get
(9)∫aτ1‍∫cτ2‍xΔ1Δ2(s,t)ΔtΔs=∫aτ1‍∫cτ2‍(|u(s,t)|γ-2α1(s,t)x(σ(s),σ(t))hhhhhhh1hh+β1(s,t)|u(s,t)|γ-2u(s,t))ΔtΔs.
By the boundary conditions onx, it is elementary to verify that
(10)∫aτ1‍∫cτ2‍(x(s,t))Δ1Δ2ΔtΔs=x(τ1,τ2),
and so
(11)x(τ1,τ2)=∫aτ1‍∫cτ2‍(|u(s,t)|γ-2α1(s,t)x(σ(s),σ(t))hhhhhhh+β1(s,t)|u(s,t)|γ-2u(s,t))ΔtΔs.
Hence,
(12)|x(τ1,τ2)|≤∫aτ1‍∫cτ2‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫aτ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-1ΔtΔs.
By similar arguments, we easily get
(13)|x(τ1,τ2)|≤∫τ1b‍∫τ2d‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫τ1b‍∫τ2d‍β1(s,t)|u(s,t)|γ-1ΔtΔs.
Summing (12) and (13) and by Hölder's inequality with indicesγandα, we obtain
(14)2|x(τ1,τ2)|≤∫ab‍∫cd‍|α1(s,t)|×|x(σ(s),σ(t))|ΔtΔs+∫ab‍∫cd‍β1(s,t)|u(s,t)|γ-1ΔtΔs≤∫ab‍∫cd‍|α1(s,t)|×|x(σ(s),σ(t))|ΔtΔs+(∫ab‍∫cd‍β1(s,t)ΔtΔs)1/γ×(∫ab‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs)1/α.
In view of (8), we have(15)2|x(τ1,τ2)|≤∫ab‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+(∫ab‍∫cd‍β1(s,t)ΔtΔs)1/γ×(∫ab‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs)1/α,
and so
(16)2≤∫ab‍∫cd‍|α1(s,t)|ΔtΔs+Mβ/α-1(∫ab‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫ab‍∫cd‍β2+(s,t)ΔtΔs)1/α.
The proof is complete.Remark 2.
It is interesting to note that when𝕋=ℝTheorem 1 reduces to Theorem2.1of [3].Theorem 3.
If the nonlinear dynamic system (1) has a real solution(x(s,t),u(s,t))which is not identically zero on[a,b]𝕋×[c,d]𝕋satisfyingx(s,c)=x(s,d)=0,x(a,t)x(σ(a),t)<0,x(b,t)x(σ(b),t)<0, andxΔ1(s,σ(t))uΔ2(s,t)+xΔ2(σ(s),t)uΔ1(s,t)=0for all(s,t)∈[a,b]𝕋×[c,d]𝕋, wherea,b,c,d∈𝕋witha<b,c<d, then
(17)2≤∫aσ(b)‍∫cd‍|α1(s,t)|ΔtΔs+Mβ/α-1(∫aσ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aσ(b)‍∫cd‍β2+(s,t)ΔtΔs)1/α,
where1/α+1/γ=1andM,β2+(s,t)are as defined in Theorem 1.Proof.
Choose(τ1,τ2)such that|x(τ1,τ2)|=M. Note thatM>0. From (6) and
(18)∫aσ(b)‍∫cd‍(x(s,t)u(s,t))Δ1Δ2ΔtΔs=x(σ(b),d)u(σ(b),d)-x(a,d)u(a,d)-x(σ(b),c)u(σ(b),c)+x(a,c)u(a,c)=0,
we have
(19)∫aσ(b)‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs=∫aσ(b)‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs.
By the boundary conditions onx, it is elementary to check that
(20)∫aτ1‍∫cτ2‍(x(s,t))Δ1Δ2ΔtΔs=x(τ1,τ2)-x(a,τ2).
So
(21)x(τ1,τ2)=x(a,τ2)+∫aτ1‍∫cτ2‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫aτ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs.
For the fixedτ2, byx(a,τ2)x(σ(a),τ2)<0, there existsξ∈(0,1)such that
(22)(1-ξ)x(a,τ2)+ξx(σ(a),τ2)=0.
Integrating the first equation of (1) overtfromctoτ2, we obtain
(23)xΔ1(s,τ2)-xΔ1(s,c)=∫cτ2‍α1(s,t)x(σ(s),σ(t))Δt+∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)Δt.
Multiplying (23) byμ(s)and noting thatx(s,c)=0, we get
(24)x(σ(s),τ2)-x(s,τ2)=μ(s)∫cτ2‍α1(s,t)x(σ(s),σ(t))Δt+μ(s)∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)Δt.
Lettings=a, we get
(25)x(σ(a),τ2)-x(a,τ2)=μ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt+μ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt.
By (22) and (25), we obtain
(26)x(a,τ2)=-ξμ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt-ξμ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt.
Substituting (26) into (21), we get
(27)x(τ1,τ2)=-ξμ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt-ξμ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt+∫aτ1‍∫cτ2‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫aτ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs=-ξμ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt+μ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt-ξμ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt+μ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt+∫σ(a)τ1‍∫cτ2‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫σ(a)τ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs=(1-ξ)μ(a)∫cτ2‍α1(a,t)x(σ(a),σ(t))Δt+(1-ξ)μ(a)∫cτ2‍β1(a,t)|u(a,t)|γ-2u(a,t)Δt+∫σ(a)τ1‍∫cτ2‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫σ(a)τ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs.
Immitating the arguments from (21) to (27) step by step, we have
(28)x(τ1,τ2)=x(b,τ2)+∫τ1b‍∫τ2d‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫τ1b‍∫τ2d‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs.
Byx(b,τ2)x(σ(b),τ2)<0, there existsη∈(0,1)such that
(29)(1-η)x(b,τ2)+ηx(σ(b),τ2)=0.
Integrating the first equation of (1) overtfromτ2todand usingx(s,d)=0, we get
(30)x(b,τ2)=ημ(b)∫τ2d‍α1(b,t)x(σ(b),σ(t))Δt+ημ(b)∫τ2d‍β1(b,t)|u(b,t)|γ-2u(b,t)Δt,
and, hence,
(31)x(τ1,τ2)=ημ(b)∫τ2d‍α1(b,t)x(σ(b),σ(t))Δt+ημ(b)∫τ2d‍β1(b,t)|u(b,t)|γ-2u(b,t)Δt+∫τ1b‍∫τ2d‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫τ1b‍∫τ2d‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs.
Let
(32)α~1(s,t)={(1-ξ)α1(a,t),s=a,α1(s,t),σ(a)≤s<b,ηα1(b,t),s=b,β~1(s,t)={(1-ξ)β1(a,t),s=a,β1(s,t),σ(a)≤s<b,ηβ1(b,t),s=b.
Then, (27) can be written as
(33)x(τ1,τ2)=∫aτ1‍∫cτ2‍α~1(s,t)x(σ(s),σ(t))ΔtΔs+∫aτ1‍∫cτ2‍β~1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs,
and (31) can be written as
(34)x(τ1,τ2)=∫τ1σ(b)‍∫τ2d‍α~1(s,t)x(σ(s),σ(t))ΔtΔs+∫τ1σ(b)‍∫τ2d‍β~1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs.
It follows from (33) and (34) that
(35)|x(τ1,τ2)|≤∫aτ1‍∫cτ2|α~1(s,t)|‍|x(σ(s),σ(t))|ΔtΔs+∫aτ1‍∫cτ2‍β~1(s,t)|u(s,t)|γ-1ΔtΔs,|x(τ1,τ2)|≤∫τ1σ(b)‍∫τ2d‍|α~1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫τ1σ(b)‍∫τ2d‍β~1(s,t)|u(s,t)|γ-1ΔtΔs.
Using (19),|α~1(s,t)|≤|α1(s,t)|,β~1(s,t)≤β1(s,t), and Hölder's inequality, we have
(36)2|x(τ1,τ2)|≤∫aσ(b)‍∫cd‍|α~1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫aσ(b)‍∫cd‍β~1(s,t)|u(s,t)|γ-1ΔtΔs≤∫aσ(b)‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫aσ(b)‍∫cd‍β1(s,t)|u(s,t)|γ-1ΔtΔs≤∫aσ(b)‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+(∫aσ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ×(∫aσ(b)‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs)1/α≤∫aσ(b)‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+(∫aσ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ×(∫aσ(b)‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs)1/α≤M∫aσ(b)‍∫cd‍|α1(s,t)|ΔtΔs+Mβ/α(∫aσ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ×(∫aσ(b)‍∫cd‍β2+(s,t)ΔtΔs)1/α.
Therefore,
(37)2≤∫aσ(b)‍∫cd‍|α1(s,t)|ΔtΔs+Mβ/α-1(∫aσ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aσ(b)‍∫cd‍β2+(s,t)ΔtΔs)1/α.
The proof is complete.Theorem 4.
Suppose thatβandγare conjugate exponents; that is,1/β+1/γ=1. If the nonlinear dynamic system (1) has a real solution(x(s,t),u(s,t))which is not identically zero on[a,b]𝕋×[c,d]𝕋such thatx(a,t)=x(s,c)=x(s,d)=0andxΔ1(s,σ(t))uΔ2(s,t)+xΔ2(σ(s),t)uΔ1(s,t)=0for all(s,t)∈[a,b]𝕋×[c,d]𝕋, wherea,b,c,d∈𝕋witha<b,c<d, then
(38)1≤∫aρ(b)‍∫cd‍|α1(s,t)|ΔtΔs+(∫aρ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aρ(b)‍∫cd‍β2+(s,t)ΔtΔs)1/β.Proof.
By (6) and the conditionsx(s,c)=x(s,d)=0for alls∈[a,b]𝕋, we have
(39)∫aρ(b)‍∫cd‍(x(s,t)u(s,t))Δ1Δ2ΔtΔs=x(ρ(b),d)u(ρ(b),d)-x(a,d)u(a,d)-x(ρ(b),c)u(ρ(b),c)+x(a,c)u(a,c)=0.
So
(40)∫aρ(b)‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs=∫aρ(b)‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs.
Fix(τ1,τ2)such that|x(τ1,τ2)|=max{|x(s,t)|:σ(a)≤s<b,σ(c)≤t<d}>0. Integrating the first equation of (1) overtfromctoτ2and then oversfromatoτ1, we get
(41)x(τ1,τ2)=∫aτ1‍∫cτ2‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫aτ1‍∫cτ2‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs,
and so by Hölder's inequality with indicesγandβwe have
(42)|x(τ1,τ2)|≤∫aρ(b)‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫aρ(b)‍∫cd‍β1(s,t)|u(s,t)|γ-1ΔtΔs≤∫aρ(b)‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+(∫aρ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aρ(b)‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs)1/β≤|x(τ1,τ2)|·∫aρ(b)‍∫cd‍|α1(s,t)|ΔtΔs+(∫aρ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aρ(b)‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs)1/β.
Hence,(43)1≤∫aρ(b)‍∫cd‍|α1(s,t)|ΔtΔs+(∫aρ(b)‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫aρ(b)‍∫cd‍β2+(s,t)ΔtΔs)1/β.
The proof is complete.Theorem 5.
Suppose thatβandγare conjugate exponents. If the nonlinear dynamic system (1) has a real solution(x(s,t),u(s,t))which is not identically zero on[a,b]𝕋×[c,d]𝕋such thatx(b,t)=x(s,c)=x(s,d)=0andxΔ1(s,σ(t))uΔ2(s,t)+xΔ2(σ(s),t)uΔ1(s,t)=0for all(s,t)∈[a,b]𝕋×[c,d]𝕋, wherea,b,c,d∈𝕋witha<b,c<d, then
(44)1≤∫σ(a)b‍∫cd‍|α1(s,t)|ΔtΔs+(∫σ(a)b‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫σ(a)b‍∫cd‍β2+(s,t)ΔtΔs)1/β.Proof.
By (6) and the assumption thatx(s,c)=x(s,d)=0for alls∈[a,b]𝕋, we get
(45)∫σ(a)b‍∫cd‍(x(s,t)u(s,t))Δ1Δ2ΔtΔs=x(b,d)u(b,d)-x(σ(a),d)u(σ(a),d)-x(b,c)u(b,c)+x(σ(a),c)u(σ(a),c)=0.
So
(46)∫σ(a)b‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs=∫σ(a)b‍∫cd‍β2(s,t)|x(σ(s),σ(t))|βΔtΔs.
Fix(τ1,τ2)such that|x(τ1,τ2)|=max{|x(s,t)|:σ(a)≤s≤ρ(b),σ(t)≤t≤ρ(b)}>0. Integrating the first equation of (1) overtfromτ2todand then oversfromτ1tob, we get
(47)x(τ1,τ2)=∫τ1b‍∫τ2d‍α1(s,t)x(σ(s),σ(t))ΔtΔs+∫τ1b‍∫τ2d‍β1(s,t)|u(s,t)|γ-2u(s,t)ΔtΔs,
and so by Hölder's inequality with indicesγandβwe have
(48)|x(τ1,τ2)|≤∫σ(a)b‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+∫σ(a)b‍∫cd‍β1(s,t)|u(s,t)|γ-1ΔtΔs≤∫σ(a)b‍∫cd‍|α1(s,t)||x(σ(s),σ(t))|ΔtΔs+(∫σ(a)b‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫σ(a)b‍∫cd‍β1(s,t)|u(s,t)|γΔtΔs)1/β≤|x(τ1,τ2)|∫σ(a)b‍∫cd‍|α1(s,t)|ΔtΔs+(∫σ(a)b‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫σ(a)b‍∫cd‍β2(s,t)|x(σ(s),σ(t)|βΔtΔs)1/β.
Hence,
(49)1≤∫σ(a)b‍∫cd‍|α1(s,t)|ΔtΔs+(∫σ(a)b‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫σ(a)b‍∫cd‍β2+(s,t)ΔtΔs)1/β.
The proof is complete.Remark 6.
Analogously, we can also consider the cases (i)x(a,t)=x(b,t)=x(s,d)=0and (ii)x(a,t)=x(b,t)=x(s,c)=0. Similar results to those in Theorem 4 and Theorem 5 can easily be arrived at. The detailed proofs are omitted here.Next, we exhibit an application of our results. Consider the following special case of (1):
(50)xΔ1Δ2(s,t)=β1(s,t)|u(s,t)|γ-2u(s,t),uΔ1Δ2(s,t)=-β2(s,t)|x(σ(s),σ(t))|β-2x(σ(s),σ(t)).Definition 7.
A nontrivial solution(x(s,t),u(s,t))of the dynamic equation (50) defined on[s0,∞)𝕋×[t0,∞)𝕋is said to be proper if
(51)sup{|x(s,t)|+|u(s,t)|:s∈[a,∞)𝕋,t∈[c,∞)𝕋}>0
for alla≥s0,  c≥t0.A proper solution(x(s,t),u(s,t))of the dynamic system (50) is called weakly oscillatory if at least one argument has a sequence of zeros diverging to∞.Theorem 8.
Assume thatu(τ,t)is bounded on[t0,∞)𝕋for a fixedτ,u(s,η)is bounded on[s0,∞)𝕋for a fixedη,
(52)∫s0∞‍∫t0∞‍β1(s,t)ΔtΔs<∞,∫s0∞‍∫t0∞‍|β2(s,t)|ΔtΔs<∞,
then every weakly oscillatory proper solution of (50) is bounded on[s0,∞)𝕋×[t0,∞)𝕋.Proof.
Let(x(s,t),u(s,t))be a nontrivial weakly oscillatory proper solution of (50) on[s0,∞)𝕋×[t0,∞)𝕋,andx(s,t)have a sequence of zeros diverging to∞. Suppose thatlimsup|x(s,t)|=∞; then, for any positive constantM0, there exists{Sn,Tn}such that|x(Sn,Tn)|>M0,n=1,2,….Sincex(s,t)is an oscillatory solution, there existsI:=(a,b)𝕋×(c,d)𝕋such thatx(a,t)=x(b,t)=x(s,c)=x(s,d)=0,|x(s,t)|>0onI, and|x(τ1,τ2)|=max{|x(s,t)|:σ(a)≤s<d,σ(c)≤t<d}:=M>M0. From (52), we can choosea,csufficiently large such that
(53)∫a∞‍∫c∞‍β1(s,t)ΔtΔs<M(α-β)/(α-1),∫a∞‍∫c∞‍|β2(s,t)|ΔtΔs<1.
By Theorem 1, we have
(54)2≤Mβ/α-1(∫ab‍∫cd‍β1(s,t)ΔtΔs)1/γ·(∫ab‍∫cd‍β2+(s,t)ΔtΔs)1/α,
and so
(55)2α≤Mβ-α(∫ab‍∫cd‍β1(s,t)ΔtΔs)α-1·(∫ab‍∫cd‍β2+(s,t)ΔtΔs)≤Mβ-α·Mα-β=1,
which contradictsα>1. Hence, there exists a positive constantKsuch that|x(s,t)|≤K.
Integrating the second equation of the dynamic equation (50) overtfromτ2totand then oversfromτ1tos, respectively, we get
(56)∫τ1s‍∫τ2t‍uΔ1Δ2(s,t)ΔtΔs=u(s,t)-u(τ1,t)-u(s,τ2)+u(τ1,τ2)=-∫τ1s‍∫τ2t‍β2(s,t)|x(σ(s),σ(t))|β-2x(σ(s),σ(t))ΔtΔs.
Sinceu(τ,t)is bounded on[t0,∞)𝕋for a fixedτandu(s,η)is bounded on[s0,∞)𝕋for a fixedη, we get
(57)|u(s,t)|≤|u(τ1,t)+u(s,τ2)-u(τ1,τ2)|+∫τ1s‍∫τ2t‍|β2(s,t)||x(σ(s),σ(t))|β-1ΔtΔs≤|u(τ1,t)+u(s,τ2)-u(τ1,τ2)|+Kβ-1∫τ1∞‍∫τ2∞‍|β2(s,t)|ΔtΔs.
Since∫s0∞‍∫t0∞‍|β2(s,t)|ΔtΔs<∞,|u(s,t)|is bounded. Solimsup{|x(s,t)|+u(s,t)|}is bounded on[s0,∞)𝕋×[t0,∞)𝕋.The proof is complete.



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*Source: 830595-2013-11-26.xml*