IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation95079710.1155/2009/950797950797Research ArticleExistence and Uniqueness of Periodic Solutions for a Second-Order Nonlinear Differential Equation with Piecewise Constant ArgumentWangGen-qiang1ChengSui Sun2ObrechtEnrico1Department of Computer ScienceGuangdong Polytechnic Normal UniversityGuangzhouGuangdong 510665Chinagdin.edu.cn2Department of MathematicsTsing Hua UniversityHsinchuTaiwan 30043Chinatsinghua.edu.cn20093011200920090407200901102009021120092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

Qualitative behaviors of first-order delay differential equations with piecewise constant arguments are the subject of many investigations (see, e.g., ), 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 , 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

(r(t)x(t))=-ϕ(x[t]). Since this equation is “similar” to the harmonic oscillator equation

(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:

(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 tR 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 tR, 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:DomLXY is a linear mapping and N:XY 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:XX and Q:YY such that ImP=KerL and ImL=KerQ=Im(I-Q). It follows that L|DomLKerP:(I-P)XImL 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:Ω̅X is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQKerL.

Theorem 1 A (Mawhin's continuation theorem [<xref ref-type="bibr" rid="B18">18</xref>]).

Let L be a Fredholm mapping of index zero, and let N be L-compact on Ω̅. Suppose that

for each λ(0,1),   xΩ,  LxλNx;

for each xΩKerL,  QNx0 and deg(JQN,ΩKer,0)0.

Then the equation Lx=Nx has at least one solution in Ω̅domL.

2. Existence and Uniqueness Criteria

Our main results of this paper are as follows.

Theorem 2.1.

Suppose that there exist constants D>0 and δ0 such that

f(t,x)sgnx>0 for tR and |x|>D,

limx-max0tω(f(t,x)/x)δ (or limx+max0tω(f(t,x)/x)δ).

If ω2δ(max0tω(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.

f(t,x) is strictly monotonous in x and there exists nonnegative constant b<(4/ω2)min0tωr(t) such that

|f(t,x1)-f(t,x2)|b|x1-x2|,(t,x1),(t,x2)R2.

Theorem 2.2.

Suppose that there exist constants D>0 and δ0 such that

f(t,x)sgnx<0 for tR and |x|>D,

limx-max0tω(f(t,x)/x)-δ (or limx+max0tω(f(t,x)/x)-δ).

If ω2δ(max0tω(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.

f(t,x) is strictly monotonous in x and there exists nonnegative constant b<(4/ω2)min0tωr(t) such that (2.1) holds.

We only give the proof of Theorem 2.1, as Theorem 2.2 can be proved similarly.

First we make the simple observation that x(t) is an ω-periodic solution of the following equation:

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 x1=i=01max0iω|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 αR and h(t)Xω, and endowed with the usual linear structure as well as the norm y2=|α|+h1. Let the zero element of Xω and Yω be denoted by θ1 and θ2 respectively.

Define the mappings L:XωYω and N:XωYω, respectively, by

Lx(t)=r(t)x(t)-r(0)x(0),Nx(t)=-0t(f(s,x([s]))-p(s))ds. Let

h̅(t)=-0t(f(s,x([s]))-p(s))ds+tω0ωf(s,x([s]))ds. Since h̅Xω and h̅(0)=0, N is a well-defined operator from Xω to Yω. Let us define P:XωXω and Q:YωYω, respectively, by

Px(t)=x(0),nZ for x=x(t)Xω and

Qy(t)=αt for y(t)=αt+h(t)Yω.

Lemma 2.3.

Let the mapping L be defined by (2.3). Then KerL=R.

Proof.

It suffices to show that if x(t) is a real ω-periodic continuously differentiable function which satisfies r(t)x(t)=r(0)x(0),tR, then x(t) is a constant function. To see this, note that for such a function x=x(t),x(t)=r(0)x(0)r(t),tR. Hence by integrating both sides of the above equality from 0 to t, we see that x(t)=x(0)+r(0)x(0)0tdsr(s),tR. Since r(t) is positive, continuous, and periodic, 0dsr(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 tR. The proof is complete.

Lemma 2.4.

Let the mapping L be defined by (2.3). Then ImL={yXωy(0)=0}Yω.

Proof.

It suffices to show that for each y=y(t)Xω that satisfies y(0)=0, there is a x=x(t)Xω such that y(t)=r(t)x(t)-r(0)x(0),t0. But this is relatively easy, since we may let α=10ω(ds/r(s)),x(t)=0ty(s)r(s)ds-α0ωy(s)r(s)ds0tdsr(s),t0. Then it may easily be checked that (2.14) holds. The proof is complete.

Lemma 2.5.

The mapping L defined by (2.3) is a Fredholm mapping of index zero.

Indeed, from Lemmas 2.3 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 mapping L,P, and Q be defined by (2.3), (2.6), and (2.7), respectively. Then ImP=KerL and ImL=KerQ.

Indeed, from Lemmas 2.3 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.

Let L 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 any xΩ¯,QNx(t)=-tω0ωf(s,x([s]))ds, so that QNx2=|1ω0ωf(s,x([s]))ds|,(I-Q)Nx(t)=-0t(f(s,x([s]))-p(s))ds+tω0ωf(s,x([s]))ds,t0. These lead us to KP(I-Q)Nx(t)=-0t1r(v)dv0v(f(s,x([s]))-p(s))ds+α(0ωdvr(v)0v(f(s,x([s]))-p(s))ds)0t1r(v)dv+1ω0tvr(v)dv0ωf(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 that   g(t) is a real, bounded and continuous function on [a,b) and limxb-g(t) exists. Then there is a point ξ(a,b) such that 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 Theorem 2.1 holds. Suppose further that x(t)Xω satisfies 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 i=1ωf(ξi,x(i-1))=i=1ωi-1if(s,x([s]))ds0ωf(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 ω2. 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 -Dx(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:

r(t)x(t)-r(0)x(0)=-λ0t(f(s,x([s]))-p(s))ds, where λ(0,1).

Lemma 2.10.

Suppose that conditions (i) and (ii) of Theorem 2.1 hold. If ω2δ(max0tω(1/r(t)))<1, then there are positive constants D0 and D1 such that for any ω-periodic solution x(t) of (2.24), |x(i)(t)|Di,t[0,ω];i=0,1.

Proof.

Let x(t) be a ω-periodic solution of (2.24). By (2.24) and our assumption that 0ωp(s)ds=0, we have 0ωf(s,x([s]))ds=0. By Lemma 2.9, there is t1[0,ω] such that |x(t1)|D. Since x(t) and x(t) are with period ω, thus for any t[t1,t1+ω], we have 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],|x(t)||x(t1)|+12t1t1+ω|x(s)|ds=|x(t1)|+120ω|x(s)|ds. It is easy to see from (2.27) and (2.29) that for any t[0,ω]|x(t)||x(t1)|+120ω|x(s)|dsD+120ω|x(s)|ds. In view of the condition ω2δ(max0tω(1/r(t)))<1, we know that there is a positive number ε such that η1:=ω2(δ+ε)(max0tω1r(t))<1. From condition (ii), we see that there is a ρ>D such that for tR and x<-ρ,f(t,x)x<δ+ε. Let E1={tt[0,ω],x([t])<-ρ},E2={tt[0,ω],|x([t])|  ρ},E3=[0,ω](E1E2),M0=max0tω,|x|ρ|f(t,x)|. By (2.32) and (2.33), we have E1|f(s,x([s]))|ds(δ+ε)E1|x([s])|ds(δ+ε)ωmax0tω|x(t)|. From (2.34) and (2.36), we have E2|f(s,x([s]))|dsωM0. In view of condition (i), (2.26), (2.37), and (2.38), we get E3|f(s,x([s]))|ds=E3f(s,x([s]))ds=-E1f(s,x([s]))ds-E2f(s,x([s]))dsE1|f(s,x([s]))|ds+E2|f(s,x([s]))|ds(δ+ε)ωmax0tω|x(t)|+ωM0. It follows from (2.37), (2.38), and (2.39) that 0ω|f(s,x([s]))|ds=E1|f(s,x([s]))|ds+E2|f(s,x([s]))|ds+E3|f(s,x([s]))|ds2(δ+ε)ωmax0tω|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],|r(t)x(t)|=|r(t1)x(t1)-λt1t(f(s,x([s]))-p(s))ds|=|-λt1t(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)|ds0ω|f(s,x([s]))|ds+0ω|p(s)|ds. From (2.40) and (2.41), we see that max0tω|x(t)|(max0tω1r(t)){2(δ+ε)ωmax0tω|x(t)|+2ωM0+max0tω|p(t)|}. It follows from (2.30), (2.31), and (2.42) that max0tω|x(t)|D+120ω|x(s)|dsω2(max0tω1r(t))(δ+ε)max0tω|x(t)|+M1=η1max0tω|x(t)|+M1, where M1=D+(max0tω1r(t))(2ωM0+max0tω|p(t)|). Let D0=M1/(1-η1), then from (2.43) we have max0tω|x(t)|D0. From (2.42) and (2.45), for any t[0,ω], we have max0tω|x(t)|D1, where D1=(max0tω1r(t)){2(δ+ε)ωD0+2ωM0+max0tω|p(t)|}. The proof is complete.

Lemma 2.11.

Suppose that condition (iii) of Theorem 2.1 is satisfied. Then (1.3) has at most one ω-periodic solution.

Proof.

Suppose that x1(t) and x2(t) are two ω-periodic solutions of (1.3). Set z(t)=x1(t)-x2(t). Then we have (r(t)z(t))+f(t,x1([t]))-f(t,x2([t]))=0.Case 2 (i).

For all t[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 0ω[f(t,x1(x1([t])))-f(t,x2([t]))]dt=0. Combining condition (iii) and x1(t)>x2(t), either f(t,x1([t]))-f(t,x2([t]))>0,t[0,ω] or f(t,x1([t]))-f(t,x2([t]))<0,t[0,ω] holds. This is contrary to (2.49).

Case 2 (ii).

There exist ξ[0,ω] such that z(ξ)=0. As in the proof of (2.30) in Lemma 2.10, we have max0tω|z(t)||z(ξ)|+120ω|z(s)|ds=120ω|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],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],r(t)z(t)=r(t1)z(t1)-12t1t(f(s,(x1([s])))-f(s,x2([s])))ds+12tt1+ω(f(s,(x1([s])))-f(s,x2([s])))ds.=-12t1t(f(s,(x1([s])))-f(s,x2([s])))ds+12tt1+ω(f(s,(x1([s])))-f(s,x2([s])))ds. It follows that for any t[t1,t1+ω],|r(t)z(t)|12t1t1+ω|f(s,(x1([s])))-f(s,x2([s]))|ds120ω|f(s,(x1([s])))-f(s,x2([s]))|ds12bωmax0tω|z(t)|. We know that for any t[0,ω],|r(t)z(t)|12bωmax0tω|z(t)|. From (2.56), we have max0tω|z(t)|bω2(max0tω1r(t))max0tω|z(t)|. By (2.52), we get max0tω|z(t)|ω2max0tω|z(t)|. It is easy to see from (2.57) and (2.58) that max0tω|z(t)|bω24(max0tω1r(t))max0tω|z(t)|. By condition (iii) of Theorem 2.1, we see that (bω2/4)(max0tω(1/r(t)))<1. Thus (2.58) leads us to max0tω|z(t)|=0, which is contrary to x1x2. So (1.3) has at most one ω-periodic solution. The proof is complete.

We now turn to the proof of Theorem 2.1. Suppose ω2δ(max0tω(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

Ω={xXωx1<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 xΩ,  LxλNx. Next note that a function xΩKerL 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),

QNx(t)=-tω0ωf(s,x([s]))ds=-tω0ωf(s,x)ds, so QNxθ2. The isomorphism J:ImQKerL is defined by J(tα)=α for αR and   tR. Then

JQNx=-1ω0ωf(s,x)ds1ω0. In particular, we see that if x=D¯, then

JQNx=-1ω0ωf(s,D¯)ds<0, and if x=-D¯, then JQNx=-1ω0ωf(s,-D¯)ds>0. Consider the mapping

H(x,μ)=μx+(1-μ)JQNx,0μ1. From (2.63) and (2.65), for each μ[0,1] and x=D¯, we have

H(x,μ)=μD¯+(1-μ)-1ω0ωf(s,D¯)ds<0. Similarly, from (2.64) and (2.65), for each μ[0,1] and x=-D¯, we have

H(x,μ)=μD¯+(1-μ)-1ω0ωf(s,-D¯)ds<0. By (2.66) and (2.67), H(x,μ) is a homotopy. This shows that

deg(JQNx,ΩKerL,θ1)=deg(-x,ΩKerL,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

(x(t)exp(-2-cos2πt5))+(3-sin2πt5)arctanx([t])=cos2πt5, and we can show that it has a nontrivial 5-periodic solution. Indeed, take

r(t)=exp(2-cos2πt5),p(t)=cos2πt5,f(t,x)=1100(3-sin2πt5)arctanx. We see that min0t5r(t)=e. Let D>0 and δ=b=1/25. Then condition (i) of Theorem 2.1 is satisfied:

limx-max0tωf(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(max0tω(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).

Acknowledgment

The first author is supported by the Natural Science Foundation of Guangdong Province of China under Grant no. 9151008002000012.