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., [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

(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 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 [<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), x∈∂Ω,Lx≠λNx;

for each x∈∂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 constants D>0 and δ⩾0 such that

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.

f(t,x) is strictly monotonous in x and there exists nonnegative constant b<(4/ω2)min0≤t≤ωr(t) such that

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.

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 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 ∥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 mappings L:Xω→Yω and N:Xω→Yω, 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+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

Px(t)=x(0),n∈Z
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),t∈R,
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),t∈R.
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),t∈R.
Since r(t) is positive, continuous, and periodic,
∫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 mapping L be defined by (2.3). Then
ImL={y∈Xω∣y(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),t≥0.
But this is relatively easy, since we may let
α=1∫0ω(ds/r(s)),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 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 x∈Ω¯,QNx(t)=-tω∫0ωf(s,x([s]))ds,
so that
∥QNx∥2=|1ω∫0ωf(s,x([s]))ds|,(I-Q)Nx(t)=-∫0t(f(s,x([s]))-p(s))ds+tω∫0ωf(s,x([s]))ds,t≥0.
These lead us to
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 that g(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
∫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=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:

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 Theorem 2.1 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),
|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)|+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]|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
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,f(t,x)x<δ+ε.
Let
E1={t∣t∈[0,ω],x([t])<-ρ},E2={t∣t∈[0,ω],|x([t])|≤ρ},E3=[0,ω]∖(E1∪E2),M0=max0≤t≤ω,|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)ωmax0≤t≤ω|x(t)|.
From (2.34) and (2.36), we have
∫E2|f(s,x([s]))|ds≤ωM0.
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]))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
∫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],|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
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
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
M1=D+(max0≤t≤ω1r(t))(2ωM0+max0≤t≤ω|p(t)|).
Let D0=M1/(1-η1), then from (2.43) we have
max0≤t≤ω|x(t)|≤D0.
From (2.42) and (2.45), for any t∈[0,ω], we have
max0≤t≤ω|x′(t)|≤D1,
where
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 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
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],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)-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],|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],|r(t)z′(t)|≤12bωmax0≤t≤ω|z(t)|.
From (2.56), we have
max0≤t≤ω|z′(t)|≤bω2(max0≤t≤ω1r(t))max0≤t≤ω|z(t)|.
By (2.52), we get
max0≤t≤ω|z(t)|≤ω2max0≤t≤ω|z′(t)|.
It is easy to see from (2.57) and (2.58) that
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 Theorem 2.1. 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

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),

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

JQNx=-1ω∫0ωf(s,x)ds1ω≠0.
In particular, we see that if x=D¯, then

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

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

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

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

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

(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 min0≤t≤5r(t)=e. Let D>0 and δ=b=1/25. Then condition (i) of Theorem 2.1 is satisfied:

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).

Acknowledgment

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

CookeK. L.WienerJ.Retarded differential equations with piecewise constant delaysShahS. M.WienerJ.Advanced differential equations with piecewise constant argument deviationsPartheniadisE. C.Stability and oscillation of neutral delay differential equations with piecewise constant argumentAftabizadehA. R.WienerJ.XuJ.-M.Oscillatory and periodic solutions of delay differential equations with piecewise constant argumentAftabizadehA. R.WienerJ.Oscillatory and periodic solutions of an equation alternately of retarded and advanced typeWienerJ.AftabizadehA. R.Differential equations alternately of retarded and advanced typeBusenbergS.CookeL.LakshmikanthamV.Models of vertically transmitted diseases with sequential-continuous dynamicsAftabizadehA. R.WienerJ.Oscillatory and periodic solutions for systems of two first order linear differential equations with piecewise constant argumentCookeK. L.WienerJ.An equation alternately of retarded and advanced typeCookeK. L.WienerJ.A survey of differential equations with piecewise continuous argumentsGopalsamyK.KulenovićM. R. S.LadasG.On a logistic equation with piecewise constant argumentsLinL. C.WangG. Q.Oscillatory and asymptotic behaviour of first order nonlinear differential equations with retarded argument [t]HuangY. K.Oscillations and asymptotic stability of solutions of first order neutral differential equations with piecewise constant argumentPapaschinopoulosG.SchinasJ.Existence stability and oscillation of the solutions of first order neutral delay differential equations with piecewise constant argumentShenJ. H.StavroulakisI. P.Oscillatory and nonoscillatory delay equations with piecewise constant argumentWienerJ.HellerW.Oscillatory and periodic solutions to a diffusion equation of neutral typeCarvalhoL. A. V.WienerJ.A nonlinear equation with piecewise continuous argumentGainesR. E.MawhinJ. L.WangG. Q.ChengS. S.Note on the set of periodic solutions of a delay differential equation with piecewise constant argument