Volume 2009, Issue 1 950797
Research Article
Open Access

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

Gen-qiang Wang

Gen-qiang Wang

Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China gdin.edu.cn

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Sui Sun Cheng

Corresponding Author

Sui Sun Cheng

Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, China tsinghua.edu.cn

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First published: 30 November 2009
Citations: 4
Academic Editor: Enrico Obrecht

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.

1. Introduction

Qualitative behaviors of first-order delay differential equations with piecewise constant arguments are the subject of many investigations (see, e.g., [119]), 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)
Since this equation is “similar” to the harmonic oscillator equation
(1.2)
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)
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

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 : Dom LXY is a linear mapping and N : XY a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dim Ker L = 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 = Ker L and ImL = Ker Q = Im (IQ). It follows that L|Dom L∩Ker P : (IP)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 if is bounded and is compact. Since ImQ is isomorphic to Ker L, there exists an isomorphism J : ImQ → Ker L.

Theorem 1 A (Mawhin′s continuation theorem [18]). Let L be a Fredholm mapping of index zero, and let N be L-compact on . Suppose that

  • (i)

    for each λ ∈ (0,1),   xΩ,   LxλNx;

  • (ii)

    for each xΩ∩Ker L,   QNx ≠ 0 and deg (JQN, Ω∩Ker , 0) ≠ 0.

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

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

  • (i)

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

  • (ii)

    lim x→−max 0≤tω(f(t, x)/x) ≤ δ (or lim x→+max 0≤tω(f(t, x)/x) ≤ δ).

If ω2δ(max 0≤tω(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)min 0≤tωr(t) such that

(2.1)

Theorem 2.2. Suppose that there exist constants D > 0 and δ⩾0 such that

  • i f(t, x)sgn  x < 0 for tR and |x| > D,

  • ii lim x→−max 0≤tω(f(t, x)/x)⩾−δ (or lim x→+max 0≤tω(f(t, x)/x)⩾−δ).

If ω2δ(max 0≤tω(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)min 0≤tωr(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:

(2.2)
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 . 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

(2.3)
(2.4)
Let
(2.5)
Since and , N is a well-defined operator from Xω to Yω. Let us define P : XωXω and Q : YωYω, respectively, by
(2.6)
for x = x(t) ∈ Xω and
(2.7)
for y(t) = αt + h(t) ∈ Yω.

Lemma 2.3. Let the mapping L be defined by (2.3). Then

(2.8)

Proof. It suffices to show that if x(t) is a real ω-periodic continuously differentiable function which satisfies

(2.9)
then x(t) is a constant function. To see this, note that for such a function x = x(t),
(2.10)
Hence by integrating both sides of the above equality from 0 to t, we see that
(2.11)
Since r(t) is positive, continuous, and periodic,
(2.12)
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

(2.13)

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

(2.14)
But this is relatively easy, since we may let
(2.15)
(2.16)
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, dim Ker L = codim ImL = 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 = Ker L and ImL = Ker  Q.

Indeed, from Lemmas 2.3 and 2.4 and defining conditions (2.6) and (2.7), it is easy to see that ImP = Ker L and ImL = Ker  Q.

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 .

Proof. It is easy to see that for any

(2.17)
so that
(2.18)
(2.19)
These lead us to
(2.20)
where α is defined by (2.15). By (2.18), we see that is bounded. Noting that (2.7) holds and N is a completely continuous mapping, by means of the Arzela-Ascoli theorem we know that is relatively compact. Thus N is L-compact on . The proof is complete.

Lemma 2.8. Suppose that   g(t) is a real, bounded and continuous function on [a, b) and exists. Then there is a point ξ ∈ (a, b) such that

(2.21)

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

(2.22)
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)
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 −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:

(2.24)
where λ ∈ (0,1).

Lemma 2.10. Suppose that conditions (i) and (ii) of Theorem 2.1 hold. If ω2δ(max 0≤tω(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)

Proof. Let x(t) be a ω-periodic solution of (2.24). By (2.24) and our assumption that we have

(2.26)
By Lemma 2.9, there is t1 ∈ [0, ω] such that
(2.27)
Since x(t) and x(t) are with period ω, thus for any t ∈ [t1, t1 + ω], we have
(2.28)
From (2.28), we see that for any t ∈ [t1, t1 + ω],
(2.29)
It is easy to see from (2.27) and (2.29) that for any t ∈ [0, ω]
(2.30)
In view of the condition ω2δ(max 0≤tω(1/r(t))) < 1, we know that there is a positive number ε such that
(2.31)
From condition (ii), we see that there is a ρ > D such that for tR and x < −ρ,
(2.32)
Let
(2.33)
(2.34)
(2.35)
(2.36)
By (2.32) and (2.33), we have
(2.37)
From (2.34) and (2.36), we have
(2.38)
In view of condition (i), (2.26), (2.37), and (2.38), we get
(2.39)
It follows from (2.37), (2.38), and (2.39) that
(2.40)
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)
From (2.40) and (2.41), we see that
(2.42)
It follows from (2.30), (2.31), and (2.42) that
(2.43)
where
(2.44)
Let D0 = M1/(1 − η1), then from (2.43) we have
(2.45)
From (2.42) and (2.45), for any t ∈ [0, ω], we have
(2.46)
where
(2.47)
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

(2.48)

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

(2.49)
Combining condition (iii) and x1(t) > x2(t), either
(2.50)
or
(2.51)
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

(2.52)

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)
By (2.53) and the fact that z(t1) = 0, we have for any t ∈ [t1, t1 + ω],
(2.54)
It follows that for any t ∈ [t1, t1 + ω],
(2.55)
We know that for any t ∈ [0, ω],
(2.56)
From (2.56), we have
(2.57)
By (2.52), we get
(2.58)
It is easy to see from (2.57) and (2.58) that
(2.59)
By condition (iii) of Theorem 2.1, we see that (bω2/4)(max 0≤tω(1/r(t))) < 1. Thus (2.58) leads us to max 0≤tω | 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(max 0≤tω(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)
where is a fixed number which satisfies 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 . Noting that by Lemma 2.10, for each λ ∈ (0,1) and xΩ,   LxλNx. Next note that a function xΩ ∩ Ker L must be constant: or Hence by (i) and (2.17), Hence by conditions (i), (iii) and (2.17),
(2.61)
so QNxθ2. The isomorphism J : ImQ → Ker L is defined by J(tα) = α for αR and   tR. Then
(2.62)
In particular, we see that if then
(2.63)
and if then
(2.64)
Consider the mapping
(2.65)
From (2.63) and (2.65), for each μ ∈ [0,1] and we have
(2.66)
Similarly, from (2.64) and (2.65), for each μ ∈ [0,1] and we have
(2.67)
By (2.66) and (2.67), H(x, μ) is a homotopy. This shows that
(2.68)
By Theorem A, we see that equation Lx = Nx has at least one solution in 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)
and we can show that it has a nontrivial 5-periodic solution. Indeed, take
(3.2)
We see that min 0≤t≤5r(t) = e. Let D > 0 and δ = b = 1/25. Then condition (i) of Theorem 2.1 is satisfied:
(3.3)
Let D > 0 and δ = b = 1/25. Then conditions (i), (ii) and (iii), of Theorem 2.1 are satisfied. Note further that 52δ(max 0≤tω(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.

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