Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2009, Article ID 950797, 14 pages doi:10.1155/2009/950797 Research Article Existence and Uniqueness of Periodic Solutions for a Second-Order Nonlinear Differential Equation with Piecewise Constant Argument Gen-qiang Wang1 and Sui Sun Cheng2 1 Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China 2 Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, China Correspondence should be addressed to Sui Sun Cheng, sscheng@math.nthu.edu.tw Received 4 July 2009; Revised 1 October 2009; Accepted 2 November 2009 Recommended by Enrico Obrecht Based on a continuation theorem of Mawhin, a unique periodic solution is found for a secondorder nonlinear differential equation with piecewise constant argument. Copyright q 2009 G.-q. Wang and S. S. Cheng. This 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. 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 higherorder 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 rt is subjected to a restoring controller −φxt which acts at sampled time t. Then Newton’s second law asserts that rtx t −φxt. 1.1 Since this equation is “similar” to the harmonic oscillator equation rtx t κxt 0, 1.2 2 International Journal of Mathematics and Mathematical Sciences we expect that the well-known qualitative behavior of the later equation may also be found in the former equation, provided appropriate conditions on rt and φx are imposed. In this paper we study a slightly more general second-order delay differential equation with piecewise constant argument: rtx t ft, xt pt, 1.3 where ft, x is a real continuous function defined on R2 with positive integer period ω for t; ωrt and pt are continuous function defined on R with period ω, rt > 0 for t ∈ R and ptdt 0. 0 By a solution of 1.3 we mean a function xt which is defined on R and which satisfies the following conditions: i x t is continuous on R, ii rtx 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 xt 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 L ⊂ 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 dim Ker L codimImL < ∞, and Im L 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 Im P Ker L and Im L Ker Q ImI − Q. It follows that L| Dom L∩Ker P : I − P X → Im L 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 Ω if QNΩ is bounded and KP I − QN : Ω → X is compact. Since Im Q is isomorphic to Ker L, there exists an isomorphism J : Im Q → Ker L. Theorem A Mawhin’s continuation theorem 18. Let L be a Fredholm mapping of index zero, and let N be L-compact on Ω. Suppose that i for each λ ∈ 0, 1, x ∈ ∂Ω, Lx / λNx; ii for each x ∈ ∂Ω ∩ Ker L, QNx / 0 and degJQN, Ω ∩ Ker, 0 / 0. Then the equation Lx Nx has at least one solution in Ω ∩ dom L. 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 ft, x sgn x > 0 for t ∈ R and |x| > D, ii limx → −∞ max0≤t≤ω ft, x/x ≤ δ (or limx → ∞ max0≤t≤ω ft, x/x ≤ δ). If ω δmax0≤t≤ω 1/rt < 1, then 1.3 has an ω-periodic solution. Furthermore, the ω-periodic solution is unique if in addition one has the following. 2 iii ft, x is strictly monotonous in x and there exists nonnegative constant b 4/ω2 min0≤t≤ω rt such that ft, x1 − ft, x2 ≤ b|x1 − x2 |, t, x1 , t, x2 ∈ R2 . < 2.1 International Journal of Mathematics and Mathematical Sciences 3 Theorem 2.2. Suppose that there exist constants D > 0 and δ 0 such that i ft, x sgn x < 0 for t ∈ R and |x| > D, ii limx → −∞ max0≤t≤ω ft, x/x −δ (or limx → ∞ max0≤t≤ω ft, x/x −δ). If ω2 δmax0≤t≤ω 1/rt < 1, then 1.3 has an ω-periodic solution. Furthermore, the ω-periodic solution is unique if in addition one has the following. iii ft, x is strictly monotonous in x and there exists nonnegative constant b 4/ω2 min0≤t≤ω rt 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 xt is an ω-periodic solution of the following equation: rtx t r0x 0 − t fs, xs − ps ds, 2.2 0 if, and only if, xt 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 xt which is defined on R and endowed with the usual linear structure as well as the norm x1 1 i i0 max0≤i≤ω |x t|. Let Yω be the Banach space of all real continuous functions of the form y αt ht such that y0 0, where α ∈ R and ht ∈ 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 : Xω → Yω and N : Xω → Yω , respectively, by Lxt rtx t − r0x 0, t fs, xs − ps ds. Nxt − 2.3 2.4 0 Let ht − t 0 t fs, xs − ps ds ω ω fs, xsds. 2.5 0 Since h ∈ Xω and h0 0, N is a well-defined operator from Xω to Yω . Let us define P : Xω → Xω and Q : Yω → Yω , respectively, by P xt x0, n∈Z 2.6 for x xt ∈ Xω and Qyt αt for yt αt ht ∈ Yω . 2.7 4 International Journal of Mathematics and Mathematical Sciences Lemma 2.3. Let the mapping L be defined by 2.3. Then Ker L R. 2.8 Proof. It suffices to show that if xt is a real ω-periodic continuously differentiable function which satisfies rtx t r0x 0, t ∈ R, 2.9 then xt is a constant function. To see this, note that for such a function x xt, x t r0x 0 , rt 2.10 t ∈ R. Hence by integrating both sides of the above equality from 0 to t, we see that xt x0 r0x 0 t ds , 0 rs t ∈ R. 2.11 Since rt is positive, continuous, and periodic, ∞ 0 ds ∞. rs 2.12 Since xt 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 Im L y ∈ Xω | y0 0 ⊂ Yω . 2.13 Proof. It suffices to show that for each y yt ∈ Xω that satisfies y0 0, there is a x xt ∈ Xω such that yt rtx t − r0x 0, t ≥ 0. 2.14 But this is relatively easy, since we may let α ω 1 , ds/rs ω t t ys ys ds ds − α ds , xt rs rs rs 0 0 0 2.15 0 t ≥ 0. Then it may easily be checked that 2.14 holds. The proof is complete. 2.16 International Journal of Mathematics and Mathematical Sciences 5 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 Im L 1 < ∞. From 2.13, we see that Im L 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 Im P Ker L and Im L Ker Q. Indeed, from Lemmas 2.3 and 2.4 and defining conditions 2.6 and 2.7, it is easy to see that Im P Ker L and Im L 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 Ω. Proof. It is easy to see that for any x ∈ Ω, t QNxt − ω ω 2.17 fs, xsds, 0 so that ω 1 fs, xsds, QNx2 ω 0 t t ω fs, xsds, fs, xs − ps ds I − QNxt − ω 0 0 2.18 t ≥ 0. 2.19 These lead us to v 1 dv fs, xs − ps ds 0 rv 0 v t ω dv 1 dv fs, xs − ps ds α 0 rv 0 0 rv ω 1 t v dv fs, xsds ω 0 rv 0 ω t ω vdv 1 α dv, fs, xsds − ω 0 rv 0 0 rv KP I − QNxt − t 2.20 where α is defined by 2.15. By 2.18, we see that QNΩ 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 − QNΩ is relatively compact. Thus N is L-compact on Ω. The proof is complete. 6 International Journal of Mathematics and Mathematical Sciences Lemma 2.8. Suppose that gt is a real, bounded and continuous function on a, b and limx → b− gt exists. Then there is a point ξ ∈ a, b such that b gsds gξb − a. 2.21 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 xt ∈ Xω satisfies ω 2.22 fs, xsds 0. 0 Then there is t1 ∈ 0, ω such that |xt1 | ≤ D. Proof. From 2.22 and Lemma 2.8, we have ξi ∈ i − 1, i for i 1, . . . , ω such that ω ω ω i fξi , xi − 1 fs, xsds fs, xsds 0. i1 i−1 i1 2.23 0 In case ω 1, from the condition i in Theorem 2.1 and 2.23, we know that |x0| ≤ D. Suppose ω ≥ 2. Our assertion is true if one of x0, x1, . . . , xω − 1 has absolute value less than or equal to D. Otherwise, there should be xη1 and xη2 among x0, x1, . . . and xω − 1 such that xη1 > D and xη2 < −D. Since xt is continuous, in view of the intermediate value theorem, there is xη3 such that −D ≤ xη3 ≤ D, here η1 > η3 > η2 or η2 > η3 > η1 . Since xt is periodic, there is t1 ∈ 0, ω such that |xt1 | |xη3 | ≤ D. The proof is complete. Now, we consider that following equation: t rtx t − r0x 0 −λ fs, xs − ps ds, 2.24 0 where λ ∈ 0, 1. Lemma 2.10. Suppose that conditions (i) and (ii) of Theorem 2.1 hold. If ω2 δmax0≤t≤ω 1/rt < 1, then there are positive constants D0 and D1 such that for any ω-periodic solution xt of 2.24, i x t ≤ Di , t ∈ 0, ω; i 0, 1. 2.25 Proof. Let xt be a ω-periodic solution of 2.24. By 2.24 and our assumption that ω psds 0, we have 0 ω 0 fs, xsds 0. 2.26 International Journal of Mathematics and Mathematical Sciences 7 By Lemma 2.9, there is t1 ∈ 0, ω such that |xt1 | ≤ D. 2.27 Since xt and x t are with period ω, thus for any t ∈ t1 , t1 ω, we have xt xt1 xt xt1 ω t x sds, t1 t t1 ω x sds xt1 2.28 t t1 ω x sds. From 2.28, we see that for any t ∈ t1 , t1 ω, |xt| ≤ |xt1 | 1 2 t1 ω t1 ω x sds |xt1 | 1 2 x sds. 2.29 0 It is easy to see from 2.27 and 2.29 that for any t ∈ 0, ω 1 |xt| ≤ |xt1 | 2 ω x sds ≤ D 1 2 0 ω x sds. 2.30 0 In view of the condition ω2 δmax0≤t≤ω 1/rt < 1, we know that there is a positive number ε such that 1 < 1. η1 : ω2 δ ε max 2.31 0≤t≤ω rt From condition ii, we see that there is a ρ > D such that for t ∈ R and x < −ρ, ft, x < δ ε. x 2.32 Let E1 t | t ∈ 0, ω, xt < −ρ , E2 t | t ∈ 0, ω, |xt| ≤ ρ , 2.33 2.34 E3 0, ω \ E1 ∪ E2 , M0 max ft, x. 2.35 2.36 0≤t≤ω,|x|≤ρ By 2.32 and 2.33, we have fs, xsds ≤ δ ε |xs|ds ≤ δ εω max |xt|. E1 E1 0≤t≤ω 2.37 8 International Journal of Mathematics and Mathematical Sciences From 2.34 and 2.36, we have fs, xsds ≤ ωM0 . 2.38 E2 In view of condition i, 2.26, 2.37, and 2.38, we get fs, xsds fs, xsds E3 E3 fs, xsds − − E1 fs, xsds ≤ fs, xsds E2 E1 fs, xsds 2.39 E2 ≤ δ εωmax |xt| ωM0 . 0≤t≤ω It follows from 2.37, 2.38, and 2.39 that ω 0 fs, xsds fs, xsds E1 fs, xsds E2 fs, xsds E3 ≤ 2δ εωmax |xt| 2ωM0 . 0≤t≤ω 2.40 Since x0 xω, 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, t rtx t rt1 x t1 − λ fs, xs − ps ds t1 t −λ fs, xs − ps ds t1 t ≤ fs, xs − ps ds t1 ≤ t1 ω fs, xsds t1 ≤ ω 0 fs, xsds t1 ω ω 0 psds t1 psds. 2.41 International Journal of Mathematics and Mathematical Sciences 9 From 2.40 and 2.41, we see that max x t ≤ 1 max 0≤t≤ω rt 0≤t≤ω 2δ εω max |xt| 2ωM0 max pt . 0≤t≤ω 0≤t≤ω 2.42 It follows from 2.30, 2.31, and 2.42 that max |xt| ≤ D 0≤t≤ω 1 2 ω x sds 0 1 δ εmax |xt| M1 0≤t≤ω rt 0≤t≤ω ≤ ω2 max 2.43 η1 max |xt| M1 , 0≤t≤ω where M1 D 1 0≤t≤ω rt max 2ωM0 max pt . 0≤t≤ω 2.44 Let D0 M1 /1 − η1 , then from 2.43 we have max |xt| ≤ D0 . 2.45 0≤t≤ω From 2.42 and 2.45, for any t ∈ 0, ω, we have max x t ≤ D1 , 2.46 0≤t≤ω where D1 1 0≤t≤ω rt max 2δ εωD0 2ωM0 max pt . 0≤t≤ω 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 zt x1 t − x2 t. Then we have rtz t ft, x1 t − ft, x2 t 0. 2.48 10 International Journal of Mathematics and Mathematical Sciences Case i. For all t ∈ 0, ω, zt / 0. Without loss of generality, we assume that zt > 0, that is, x1 t > x2 t for t ∈ 0, ω. Integrating 2.48 from 0 to ω, we have ω ft, x1 x1 t − ft, x2 t dt 0. 2.49 0 Combining condition iii and x1 t > x2 t, either ft, x1 t − ft, x2 t > 0, t ∈ 0, ω 2.50 ft, x1 t − ft, x2 t < 0, t ∈ 0, ω 2.51 or holds. This is contrary to 2.49. Case ii. There exist ξ ∈ 0, ω such that zξ 0. As in the proof of 2.30 in Lemma 2.10, we have 1 max |zt| ≤ |zξ| 0≤t≤ω 2 ω z sds 1 2 0 ω z sds. 2.52 0 On the other hand, since z0 zω, 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, rtz t rt1 z t1 − t fs, x1 s − fs, x2 s ds, t1 rtz t rt1 ωz t1 ω − rt1 z t1 − t t1 ω t t1 ω fs, x1 s − fs, x2 s ds 2.53 fs, x1 s − fs, x2 s ds. By 2.53 and the fact that z t1 0, we have for any t ∈ t1 , t1 ω, 1 t rtz t rt1 z t1 − fs, x1 s − fs, x2 s ds 2 t1 1 t1 ω fs, x1 s − fs, x2 s ds. 2 t 1 t − fs, x1 s − fs, x2 s ds 2 t1 1 t1 ω fs, x1 s − fs, x2 s ds. 2 t 2.54 International Journal of Mathematics and Mathematical Sciences 11 It follows that for any t ∈ t1 , t1 ω, rtz t ≤ 1 2 ≤ ≤ 1 2 t1 ω fs, x1 s − fs, x2 sds t1 ω fs, x1 s − fs, x2 sds 2.55 0 1 bωmax |zt|. 2 0≤t≤ω We know that for any t ∈ 0, ω, rtz t ≤ 1 bωmax |zt|. 2 0≤t≤ω 2.56 bω 1 max max |zt|. max z t ≤ 0≤t≤ω 2 0≤t≤ω rt 0≤t≤ω 2.57 From 2.56, we have By 2.52, we get max |zt| ≤ 0≤t≤ω ω max z t. 2 0≤t≤ω 2.58 It is easy to see from 2.57 and 2.58 that 1 bω2 max |zt| ≤ max max |zt|. 0≤t≤ω 4 0≤t≤ω rt 0≤t≤ω 2.59 By condition iii of Theorem 2.1, we see that bω2 /4max0≤t≤ω 1/rt < 1. Thus 2.58 x2 . So 1.3 has at most one ω-periodic leads us to max0≤t≤ω |zt| 0, which is contrary to x1 / solution. The proof is complete. We now turn to the proof of Theorem 2.1. Suppose ω2 δmax0≤t≤ω 1/rt < 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 xt of 2.24 such that 2.25 holds. Set Ω x ∈ Xω | x1 < D , 2.60 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 ∩ Ker L must be 12 International Journal of Mathematics and Mathematical Sciences constant: xt ≡ D or xt ≡ −D. Hence by i and 2.17, xt ≡ −D. Hence by conditions i, iii and 2.17, QNxt − t ω ω fs, xsds − 0 t ω ω fs, xds, 2.61 0 so QNx / θ2 . The isomorphism J : Im Q → Ker L is defined by Jtα α for α ∈ R and t ∈ R. Then JQNx − 1 ω ω fs, xds 0 1 0. / ω 2.62 In particular, we see that if x D, then 1 JQNx − ω ω f s, D ds < 0, 2.63 0 and if x −D, then JQNx − 1 ω ω f s, −D ds > 0. 2.64 0 Consider the mapping H x, μ μx 1 − μ JQNx, 0 ≤ μ ≤ 1. 2.65 From 2.63 and 2.65, for each μ ∈ 0, 1 and x D, we have −1 H x, μ μD 1 − μ ω ω f s, D ds < 0. 2.66 0 Similarly, from 2.64 and 2.65, for each μ ∈ 0, 1 and x −D, we have −1 H x, μ μD 1 − μ ω ω f s, −D ds < 0. 2.67 0 By 2.66 and 2.67, Hx, μ is a homotopy. This shows that degJQNx, Ω ∩ Ker L, θ1 deg−x, Ω ∩ Ker L, θ1 / 0. 2.68 By Theorem A, we see that equation Lx Nx has at least one solution in Ω ∩ Dom L. In other words, 1.3 has an ω-periodic solution xt. Furthermore, if iii is satisfied, from Lemma 2.11, we know that 1.3 has an ω-periodic solution only. The proof is complete. International Journal of Mathematics and Mathematical Sciences 13 3. Example Consider the equation 2πt 2πt 2πt , x t exp −2 − cos 3 − sin arctan xt cos 5 5 5 3.1 and we can show that it has a nontrivial 5-periodic solution. Indeed, take 2πt 2πt rt exp 2 − cos , pt cos , 5 5 1 2πt ft, x 3 − sin arctan x. 100 5 3.2 We see that min0≤t≤5 rt e. Let D > 0 and δ b 1/25. Then condition i of Theorem 2.1 is satisfied: ft, x 1 . x → −∞ 0≤t≤ω x 25 lim max 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 52 δmax0≤t≤ω 1/rt 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. References 1 K. L. Cooke and J. 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