# Existence and Uniqueness of Periodic Solutions for a Second-Order Nonlinear Differential Equation with Piecewise Constant Argument

Authors: Gen-qiang Wang; Sui Sun Cheng
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

Based on a continuation theorem of Mawhin, a unique periodic solution is found for a second-order nonlinear differential equation with piecewise constant argument.


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## Body

## 1. Introduction

Qualitative behaviors of first-order delay differential equations with piecewise constant arguments are the subject of many investigations (see, e.g., [1–19]), while those of higher-order equations are not.However, there are reasons for studying higher-order equations with piecewise constant arguments. Indeed, as mentioned in [10], a potential application of these equations is in the stabilization of hybrid control systems with feedback delay, where a hybrid system is one with a continuous plant and with a discrete (sampled) controller. As an example, suppose that a moving particle with time variable mass r(t) is subjected to a restoring controller -ϕ(x[t]) which acts at sampled time [t]. Then Newton's second law asserts that(1.1)(r(t)x′(t))′=-ϕ(x[t]).
Since this equation is “similar” to the harmonic oscillator equation(1.2)(r(t)x′(t))′+κx(t)=0,
we expect that the well-known qualitative behavior of the later equation may also be found in the former equation, provided appropriate conditions on r(t) and ϕ(x) are imposed.In this paper we study a slightly more general second-order delay differential equation with piecewise constant argument:(1.3)(r(t)x′(t))′+f(t,x([t]))=p(t),
where f(t,x) is a real continuous function defined on R2 with positive integer period ω for t;r(t) and p(t) are continuous function defined on R with period ω,r(t)>0 for t∈R and ∫0ωp(t)dt=0.By a solution of (1.3) we mean a function x(t) which is defined on R and which satisfies the following conditions: (i) x′(t) is continuous on R, (ii) r(t)x′(t) is differentiable at each point t∈R, with the possible exception of the points [t]∈R where one-sided derivatives exist, and (iii) substitution of x(t) into (1.3) leads to an identity on each interval [n,n+1)⊂R with integral endpoints.In this note, existence and uniqueness criteria for periodic solutions of (1.3) will be established. For this purpose, we will make use of a continuation theorem of Mawhin. Let X and Y be two Banach spaces and L:DomL⊂X→Y is a linear mapping and N:X→Y a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL=codimImL<+∞, and ImL is closed in Y. If L is a Fredholm mapping of index zero, there exist continuous projectors P:X→X and Q:Y→Y such that ImP=KerL and ImL=KerQ=Im(I-Q). It follows that L|DomL∩KerP:(I-P)X→ImL has an inverse which will be denoted by KP. If Ω is an open and bounded subset of X, the mapping N will be called L-compact on Ω̅ if QN(Ω̅) is bounded and KP(I-Q)N:Ω̅→X is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQ→KerL.Theorem 1 A (Mawhin's continuation theorem [18]).
LetL be a Fredholm mapping of index zero, and let N be L-compact on Ω̅. Suppose that(i)
for eachλ∈(0,1), x∈∂Ω,Lx≠λNx;(ii)
for eachx∈∂Ω∩KerL,QNx≠0 and deg(JQN,Ω∩Ker,0)≠0.Then the equation Lx=Nx has at least one solution in Ω̅∩domL.


## 2. Existence and Uniqueness Criteria

Our main results of this paper are as follows.Theorem 2.1.
Suppose that there exist constantsD>0 and δ⩾0 such that(i)
f(t,x)sgnx>0 for t∈R and |x|>D,(ii)
limx→-∞max0≤t≤ω(f(t,x)/x)≤δ (or limx→+∞max0≤t≤ω(f(t,x)/x)≤δ).If ω2δ(max0≤t≤ω(1/r(t)))<1, then (1.3) has an ω-periodic solution. Furthermore, the ω-periodic solution is unique if in addition one has the following.(iii)
f(t,x) is strictly monotonous in x and there exists nonnegative constant b<(4/ω2)min0≤t≤ωr(t) such that(2.1)|f(t,x1)-f(t,x2)|≤b|x1-x2|,(t,x1),(t,x2)∈R2.Theorem 2.2.
Suppose that there exist constantsD>0 and δ⩾0 such that(i′)
f(t,x)sgnx<0 for t∈R and |x|>D,(ii′)
limx→-∞max0≤t≤ω(f(t,x)/x)⩾-δ (or limx→+∞max0≤t≤ω(f(t,x)/x)⩾-δ).If ω2δ(max0≤t≤ω(1/r(t)))<1, then (1.3) has an ω-periodic solution. Furthermore, the ω-periodic solution is unique if in addition one has the following.(iii)
f(t,x) is strictly monotonous in x and there exists nonnegative constant b<(4/ω2)min0≤t≤ωr(t) such that (2.1) holds.We only give the proof of Theorem2.1, as Theorem 2.2 can be proved similarly.First we make the simple observation thatx(t) is an ω-periodic solution of the following equation:(2.2)r(t)x′(t)=r(0)x′(0)-∫0t(f(s,x([s]))-p(s))ds,
if, and only if, x(t) is an ω-periodic solution of (1.3). Next, let Xω be the Banach space of all real ω-periodic continuously differentiable functions of the form x=x(t) which is defined on R and endowed with the usual linear structure as well as the norm ∥x∥1=∑i=01max0≤i≤ω|x(i)(t)|. Let Yω be the Banach space of all real continuous functions of the form y=αt+h(t) such that y(0)=0, where α∈R and h(t)∈Xω, and endowed with the usual linear structure as well as the norm ∥y∥2=|α|+∥h∥1. Let the zero element of Xω and Yω be denoted by θ1 and θ2 respectively.Define the mappingsL:Xω→Yω and N:Xω→Yω, respectively, by(2.3)Lx(t)=r(t)x′(t)-r(0)x′(0),(2.4)Nx(t)=-∫0t(f(s,x([s]))-p(s))ds.
Let(2.5)h̅(t)=-∫0t(f(s,x([s]))-p(s))ds+tω∫0ωf(s,x([s]))ds.
Since h̅∈Xω and h̅(0)=0, N is a well-defined operator from Xω to Yω. Let us define P:Xω→Xω and Q:Yω→Yω, respectively, by(2.6)Px(t)=x(0),n∈Z
for x=x(t)∈Xω and(2.7)Qy(t)=αt
for y(t)=αt+h(t)∈Yω.Lemma 2.3.
Let the mappingL be defined by (2.3). Then
(2.8)KerL=R.Proof.
It suffices to show that ifx(t) is a real ω-periodic continuously differentiable function which satisfies
(2.9)r(t)x′(t)=r(0)x′(0),t∈R,
then x(t) is a constant function. To see this, note that for such a function x=x(t),(2.10)x′(t)=r(0)x′(0)r(t),t∈R.
Hence by integrating both sides of the above equality from 0 to t, we see that
(2.11)x(t)=x(0)+r(0)x′(0)∫0tdsr(s),t∈R.
Since r(t) is positive, continuous, and periodic,
(2.12)∫0∞dsr(s)=∞.
Since x(t) is bounded, we may infer from (2.11) that x′(0)=0. But then (2.9) implies x′(t)=0 for t∈R. The proof is complete.Lemma 2.4.
Let the mappingL be defined by (2.3). Then
(2.13)ImL={y∈Xω∣y(0)=0}⊂Yω.Proof.
It suffices to show that for eachy=y(t)∈Xω that satisfies y(0)=0, there is a x=x(t)∈Xω such that
(2.14)y(t)=r(t)x′(t)-r(0)x′(0),t≥0.
But this is relatively easy, since we may let
(2.15)α=1∫0ω(ds/r(s)),(2.16)x(t)=∫0ty(s)r(s)ds-α∫0ωy(s)r(s)ds∫0tdsr(s),t≥0.
Then it may easily be checked that (2.14) holds. The proof is complete.Lemma 2.5.
The mappingL defined by (2.3) is a Fredholm mapping of index zero.Indeed, from Lemmas2.3 and 2.4 and the definition of Yω,dimKerL=codimImL=1<+∞. From (2.13), we see that ImL is closed in Yω. Hence L is a Fredholm mapping of index zero.Lemma 2.6.
Let the mappingL,P, and Q be defined by (2.3), (2.6), and (2.7), respectively. Then ImP=KerL and ImL=KerQ.Indeed, from Lemmas2.3 and 2.4 and defining conditions (2.6) and (2.7), it is easy to see that ImP=KerL and ImL=KerQ.Lemma 2.7.
LetL and N be defined by (2.3) and (2.4), respectively. Suppose that  Ω is an open and bounded subset of Xω. Then N is L-compact on Ω¯.Proof.
It is easy to see that for anyx∈Ω¯,(2.17)QNx(t)=-tω∫0ωf(s,x([s]))ds,
so that
(2.18)∥QNx∥2=|1ω∫0ωf(s,x([s]))ds|,(2.19)(I-Q)Nx(t)=-∫0t(f(s,x([s]))-p(s))ds+tω∫0ωf(s,x([s]))ds,t≥0.
These lead us to
(2.20)KP(I-Q)Nx(t)=-∫0t1r(v)dv∫0v(f(s,x([s]))-p(s))ds+α(∫0ωdvr(v)∫0v(f(s,x([s]))-p(s))ds)∫0t1r(v)dv+1ω∫0tvr(v)dv∫0ωf(s,x([s]))ds-αω(∫0ωvdvr(v)∫0ωf(s,x([s]))ds)∫0t1r(v)dv,
where α is defined by (2.15). By (2.18), we see that QN(Ω̅) is bounded. Noting that (2.7) holds and N is a completely continuous mapping, by means of the Arzela-Ascoli theorem we know that KP(I-Q)N(Ω̅)¯ is relatively compact. Thus N is L-compact on Ω¯. The proof is complete.Lemma 2.8.
Suppose thatg(t) is a real, bounded and continuous function on [a,b) and limx→b-g(t) exists. Then there is a point ξ∈(a,b) such that
(2.21)∫abg(s)ds=g(ξ)(b-a).The above result is only a slight extension of the integral mean value theorem and is easily proved.Lemma 2.9.
Suppose that condition (i) in Theorem2.1 holds. Suppose further that x(t)∈Xω satisfies
(2.22)∫0ωf(s,x([s]))ds=0.
Then there is t1∈[0,ω] such that |x(t1)|≤D.Proof.
From (2.22) and Lemma 2.8,we have  ξi∈(i-1,i) for i=1,…,ω such that
(2.23)∑i=1ωf(ξi,x(i-1))=∑i=1ω∫i-1if(s,x([s]))ds∫0ωf(s,x([s]))ds=0.
In case ω=1, from the condition (i) in Theorem 2.1 and (2.23), we know that |x(0)|≤D. Suppose ω≥2. Our assertion is true if one of x(0),x(1),…,x(ω-1) has absolute value less than or equal to D. Otherwise, there should be x(η1) and x(η2) among x(0),x(1),… and x(ω-1) such that x(η1)>D and x(η2)<-D. Since x(t) is continuous, in view of the intermediate value theorem, there is x(η3) such that -D≤x(η3)≤D, (here η1>η3>η2 or η2>η3>η1). Since x(t) is periodic, there is t1∈[0,ω] such that |x(t1)|=|x(η3)|≤D. The proof is complete.Now, we consider that following equation:(2.24)r(t)x′(t)-r(0)x′(0)=-λ∫0t(f(s,x([s]))-p(s))ds,
where λ∈(0,1).Lemma 2.10.
Suppose that conditions (i) and (ii) of Theorem2.1 hold. If ω2δ(max0≤t≤ω(1/r(t)))<1, then there are positive constants D0 and D1 such that for any ω-periodic solution x(t) of (2.24),
(2.25)|x(i)(t)|≤Di,t∈[0,ω];i=0,1.Proof.
Letx(t) be a ω-periodic solution of (2.24). By (2.24) and our assumption that ∫0ωp(s)ds=0, we have
(2.26)∫0ωf(s,x([s]))ds=0.
By Lemma 2.9, there is t1∈[0,ω] such that
(2.27)|x(t1)|≤D.
Since x(t) and x′(t) are with period ω, thus for any t∈[t1,t1+ω], we have
(2.28)x(t)=x(t1)+∫t1tx′(s)ds,x(t)=x(t1+ω)+∫t1+ωtx′(s)ds=x(t1)+∫t1+ωtx′(s)ds.
From (2.28), we see that for any t∈[t1,t1+ω],(2.29)|x(t)|≤|x(t1)|+12∫t1t1+ω|x′(s)|ds=|x(t1)|+12∫0ω|x′(s)|ds.
It is easy to see from (2.27) and (2.29) that for any t∈[0,ω](2.30)|x(t)|≤|x(t1)|+12∫0ω|x′(s)|ds≤D+12∫0ω|x′(s)|ds.
In view of the condition ω2δ(max0≤t≤ω(1/r(t)))<1, we know that there is a positive number ε such that
(2.31)η1:=ω2(δ+ε)(max0≤t≤ω1r(t))<1.
From condition (ii), we see that there is a ρ>D such that for t∈R and x<-ρ,(2.32)f(t,x)x<δ+ε.
Let
(2.33)E1={t∣t∈[0,ω],x([t])<-ρ},(2.34)E2={t∣t∈[0,ω],|x([t])|≤ρ},(2.35)E3=[0,ω]∖(E1∪E2),(2.36)M0=max0≤t≤ω,|x|≤ρ|f(t,x)|.
By (2.32) and (2.33), we have
(2.37)∫E1|f(s,x([s]))|ds≤(δ+ε)∫E1|x([s])|ds≤(δ+ε)ωmax0≤t≤ω|x(t)|.
From (2.34) and (2.36), we have
(2.38)∫E2|f(s,x([s]))|ds≤ωM0.
In view of condition (i), (2.26), (2.37), and (2.38), we get
(2.39)∫E3|f(s,x([s]))|ds=∫E3f(s,x([s]))ds=-∫E1f(s,x([s]))ds-∫E2f(s,x([s]))ds≤∫E1|f(s,x([s]))|ds+∫E2|f(s,x([s]))|ds≤(δ+ε)ωmax0≤t≤ω|x(t)|+ωM0.
It follows from (2.37), (2.38), and (2.39) that
(2.40)∫0ω|f(s,x([s]))|ds=∫E1|f(s,x([s]))|ds+∫E2|f(s,x([s]))|ds+∫E3|f(s,x([s]))|ds≤2(δ+ε)ωmax0≤t≤ω|x(t)|+2ωM0.
Since x(0)=x(ω), thus there is a t1∈(0,ω) such that x′(t1)=0. In view of (2.24) and the fact that x′(t1)=0, we conclude that for any t∈[t1,t1+ω],(2.41)|r(t)x′(t)|=|r(t1)x′(t1)-λ∫t1t(f(s,x([s]))-p(s))ds|=|-λ∫t1t(f(s,x([s]))-p(s))ds|≤|∫t1t(f(s,x([s]))-p(s))ds|≤∫t1t1+ω|f(s,x([s]))|ds+∫t1t1+ω|p(s)|ds≤∫0ω|f(s,x([s]))|ds+∫0ω|p(s)|ds.
From (2.40) and (2.41), we see that
(2.42)max0≤t≤ω|x′(t)|≤(max0≤t≤ω1r(t)){2(δ+ε)ωmax0≤t≤ω|x(t)|+2ωM0+max0≤t≤ω|p(t)|}.
It follows from (2.30), (2.31), and (2.42) that
(2.43)max0≤t≤ω|x(t)|≤D+12∫0ω|x′(s)|ds≤ω2(max0≤t≤ω1r(t))(δ+ε)max0≤t≤ω|x(t)|+M1=η1max0≤t≤ω|x(t)|+M1,
where
(2.44)M1=D+(max0≤t≤ω1r(t))(2ωM0+max0≤t≤ω|p(t)|).
Let D0=M1/(1-η1), then from (2.43) we have
(2.45)max0≤t≤ω|x(t)|≤D0.
From (2.42) and (2.45), for any t∈[0,ω], we have
(2.46)max0≤t≤ω|x′(t)|≤D1,
where
(2.47)D1=(max0≤t≤ω1r(t)){2(δ+ε)ωD0+2ωM0+max0≤t≤ω|p(t)|}.
The proof is complete.Lemma 2.11.
Suppose that condition (iii) of Theorem2.1 is satisfied. Then (1.3) has at most one ω-periodic solution.Proof.
Suppose thatx1(t) and x2(t) are two ω-periodic solutions of (1.3). Set z(t)=x1(t)-x2(t). Then we have
(2.48)(r(t)z′(t))′+f(t,x1([t]))-f(t,x2([t]))=0.Case 2 (i).
For allt∈[0,ω],z(t)≠0. Without loss of generality, we assume that z(t)>0, that is, x1(t)>x2(t) for t∈[0,ω]. Integrating (2.48) from 0 to ω, we have
(2.49)∫0ω[f(t,x1(x1([t])))-f(t,x2([t]))]dt=0.
Combining condition (iii) and x1(t)>x2(t), either
(2.50)f(t,x1([t]))-f(t,x2([t]))>0,t∈[0,ω]
or
(2.51)f(t,x1([t]))-f(t,x2([t]))<0,t∈[0,ω]
holds. This is contrary to (2.49).Case 2 (ii).
There existξ∈[0,ω] such that z(ξ)=0. As in the proof of (2.30) in Lemma 2.10, we have
(2.52)max0≤t≤ω|z(t)|≤|z(ξ)|+12∫0ω|z′(s)|ds=12∫0ω|z′(s)|ds.On the other hand, since z(0)=z(ω), thus there is a t1∈(0,ω) such that z′(t1)=0. In view of (2.48), we conclude that for any t∈[t1,t1+ω],(2.53)r(t)z′(t)=r(t1)z′(t1)-∫t1t(f(s,(x1([s])))-f(s,x2([s])))ds,r(t)z′(t)=r(t1+ω)z′(t1+ω)-∫t1+ωt(f(s,(x1([s])))-f(s,x2([s])))ds=r(t1)z′(t1)-∫t1+ωt(f(s,(x1([s])))-f(s,x2([s])))ds.
By (2.53) and the fact that z′(t1)=0, we have for any t∈[t1,t1+ω],(2.54)r(t)z′(t)=r(t1)z′(t1)-12∫t1t(f(s,(x1([s])))-f(s,x2([s])))ds+12∫tt1+ω(f(s,(x1([s])))-f(s,x2([s])))ds.=-12∫t1t(f(s,(x1([s])))-f(s,x2([s])))ds+12∫tt1+ω(f(s,(x1([s])))-f(s,x2([s])))ds.
It follows that for any t∈[t1,t1+ω],(2.55)|r(t)z′(t)|≤12∫t1t1+ω|f(s,(x1([s])))-f(s,x2([s]))|ds≤12∫0ω|f(s,(x1([s])))-f(s,x2([s]))|ds≤12bωmax0≤t≤ω|z(t)|.
We know that for any t∈[0,ω],(2.56)|r(t)z′(t)|≤12bωmax0≤t≤ω|z(t)|.
From (2.56), we have
(2.57)max0≤t≤ω|z′(t)|≤bω2(max0≤t≤ω1r(t))max0≤t≤ω|z(t)|.
By (2.52), we get
(2.58)max0≤t≤ω|z(t)|≤ω2max0≤t≤ω|z′(t)|.
It is easy to see from (2.57) and (2.58) that
(2.59)max0≤t≤ω|z(t)|≤bω24(max0≤t≤ω1r(t))max0≤t≤ω|z(t)|.
By condition (iii) of Theorem 2.1, we see that (bω2/4)(max0≤t≤ω(1/r(t)))<1. Thus (2.58) leads us to max0≤t≤ω|z(t)|=0, which is contrary to x1≠x2. So (1.3) has at most one ω-periodic solution. The proof is complete.We now turn to the proof of Theorem2.1. Suppose ω2δ(max0≤t≤ω(1/r(t)))<1. Let L,N,P, and Q be defined by (2.3), (2.4), (2.6), and (2.7), respectively. By Lemma 2.10, there are positive constants D0 and D1 such that for any ω-periodic solution x(t) of (2.24) such that (2.25) holds. Set(2.60)Ω={x∈Xω∣∥x∥1<D¯},
where D¯ is a fixed number which satisfies D¯>D+D0+D1. It is easy to see that Ω is an open and bounded subset of Xω. Furthermore, in view of Lemmas 2.5 and 2.7, L is a Fredholm mapping of index zero and N is L-compact on Ω¯. Noting that D¯>D0+D1, by Lemma 2.10, for each λ∈(0,1) and x∈∂Ω,Lx≠λNx. Next note that a function x∈∂Ω∩KerL must be constant: x(t)≡D¯ or x(t)≡-D¯. Hence by (i) and (2.17), x(t)≡-D¯. Hence by conditions (i), (iii) and (2.17),(2.61)QNx(t)=-tω∫0ωf(s,x([s]))ds=-tω∫0ωf(s,x)ds,
so QNx≠θ2. The isomorphism J:ImQ→KerL is defined by J(tα)=α for α∈R and t∈R. Then(2.62)JQNx=-1ω∫0ωf(s,x)ds1ω≠0.
In particular, we see that if x=D¯, then(2.63)JQNx=-1ω∫0ωf(s,D¯)ds<0,
and if x=-D¯, then
(2.64)JQNx=-1ω∫0ωf(s,-D¯)ds>0.
Consider the mapping(2.65)H(x,μ)=μx+(1-μ)JQNx,0≤μ≤1.
From (2.63) and (2.65), for each μ∈[0,1] and x=D¯, we have(2.66)H(x,μ)=μD¯+(1-μ)-1ω∫0ωf(s,D¯)ds<0.
Similarly, from (2.64) and (2.65), for each μ∈[0,1] and x=-D¯, we have(2.67)H(x,μ)=μD¯+(1-μ)-1ω∫0ωf(s,-D¯)ds<0.
By (2.66) and (2.67), H(x,μ) is a homotopy. This shows that(2.68)deg(JQNx,Ω∩KerL,θ1)=deg(-x,Ω∩KerL,θ1)≠0.
By Theorem A, we see that equation Lx=Nx has at least one solution in Ω¯∩DomL. In other words, (1.3) has an ω-periodic solution x(t). Furthermore, if (iii) is satisfied, from Lemma 2.11, we know that (1.3) has an ω-periodic solution only. The proof is complete.


## 3. Example

Consider the equation(3.1)(x′(t)exp(-2-cos2πt5))′+(3-sin2πt5)arctanx([t])=cos2πt5,
and we can show that it has a nontrivial 5-periodic solution. Indeed, take(3.2)r(t)=exp(2-cos2πt5),p(t)=cos2πt5,f(t,x)=1100(3-sin2πt5)arctanx.
We see that min0≤t≤5r(t)=e. Let D>0 and δ=b=1/25. Then condition (i) of Theorem 2.1 is satisfied:(3.3)limx→-∞max0≤t≤ωf(t,x)x=125.
Let D>0 and δ=b=1/25. Then conditions (i), (ii) and (iii), of Theorem 2.1 are satisfied. Note further that 52δ(max0≤t≤ω(1/r(t)))=e-1<1. Therefore (3.1) has exactly one 5-periodic solution. Furthermore, it is easy to see that any solution of (3.1) must be nontrivial. We have thus shown the existence of a unique nontrivial 5-periodic solution of (3.1).



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*Source: 950797-2009-11-30.xml*