Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2012, Article ID 851691, 25 pages doi:10.1155/2012/851691 Research Article Study of Solutions to Some Functional Differential Equations with Piecewise Constant Arguments Juan J. Nieto1, 2 and Rosana Rodrı́guez-López1 1 Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela, 15782 Santiago de Compostela, Spain 2 Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia Correspondence should be addressed to Rosana Rodrı́guez-López, rosana.rodriguez.lopez@usc.es Received 12 October 2011; Revised 6 January 2012; Accepted 9 January 2012 Academic Editor: Josef Diblı́k Copyright q 2012 J. J. Nieto and R. Rodrı́guez-López. 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. We provide optimal conditions for the existence and uniqueness of solutions to a nonlocal boundary value problem for a class of linear homogeneous second-order functional differential equations with piecewise constant arguments. The nonlocal boundary conditions include terms of the state function and the derivative of the state function. A similar nonhomogeneous problem is also discussed. 1. Introduction In the study of second-order functional differential equations, there is a wide range of works dealing with periodic boundary value problems and piecewise continuous functional dependence. We mention, for instance, 1, where an equation independent of the first derivative is analyzed, and many other works as 2–14, where existence and stability results are provided. Most of works on this field deal with nonconstructive existence results. However, in 7, 11, explicit solutions are found for second-order functional differential equations with piecewise constant arguments, through the calculus of the Green’s function. Other works in relation with first and higher-order differential equations with delay are 15–18. In 19, a class of linear second-order differential equations with piecewise constant arguments is considered under the nonlocal conditions x0 ϕ, xT xμ ψ, where 2 Abstract and Applied Analysis ϕ, ψ ∈ R, that is, an initial position is assumed and the boundary conditions are independent of the derivative of the function. In this paper, we study the following nonlocal boundary value problem for homogeneous linear second-order functional differential equations: x t ax t bxt cx t dxt 0, xT x μ ϕ, t ∈ J 0, T , 1.1 x T x μ ψ, for a, b, c, d, ϕ, ψ ∈ R, T > 0 and μ ∈ 0, T , where the functional dependence is given by the greatest integer part t, and the nonlocal boundary conditions involve both the state function and its derivative, which is the main difference from the study in 19. A discussion for nonhomogeneous equations is also included. We consider the existence and uniqueness of solution to this problem, providing optimal conditions and calculating the exact expression of solutions. To better illustrate the significant differences between the results included in this paper and those in 19, we remark that 19 is devoted to the study of the same class of linear second-order differential equations with piecewise constant arguments but considering a nonlocal boundary value problem where the value of the unknown function x is fixed at the initial instant t 0 i.e., an initial condition is imposed and the boundary condition also involves the value of the sought solution at the right endpoint of the interval T and an intermediate point μ. None of the conditions imposed in 19 involve the value of the rate of change of the solution at any points. Thus, the nonlocal conditions in 19 can be reduced, by using the expression of the solution, to a boundary value problem affecting only the state of the solution, being independent of the rate of change of the state of the system. On the other hand, the nonlocal problem considered in this paper does not fix a certain initial position and, moreover, the boundary condition introduces the dependence, not only on the state, but also on the variation of the state of the system. Indeed, the value of the unknown function and its derivative is determined by the value of the corresponding magnitudes at an intermediate point of the interval of interest. Moreover, since the results in these two papers are also extensible to the special case where the intermediate point is identical to the initial instant μ 0, we compare the consequences of both works to explain better the implications of the study of each problem. If we consider μ 0, the results in 19 are applicable to obtain the solution to a class of linear second-order differential equations with piecewise constant arguments subject to the boundary conditions x0 ϕ, xT x0 ψ ϕ ψ and, therefore, we solve a problem with fixed values of the function at the end-points of the interval. However, again for μ 0, the results presented in this paper allow to characterize the existence of solution and provide its explicit expression imposing boundary conditions of the type xT x0 ϕ, x T x 0 ψ, which include, as a particular case, periodic boundary conditions on x and x . The main results are stated after a few preliminary results are recalled, and, finally, examples are included to show the applicability of these results. Abstract and Applied Analysis 3 2. Preliminaries Consider the equation x t ax t bxt cx t dxt 0, t ∈ J 0, T , 2.1 where a, b, c, d and T > 0. The following concepts and results come from 7. Definition 2.1. 7 Let the spaces Λ : y : J −→ R : y is continuous in J \ {1, 2, . . . , T }, and there exist y n− ∈ R, yn yn, ∀n ∈ {1, 2, . . . , T } , 2.2 E : {x : J −→ R : x, x are continuous and x ∈ Λ}. A solution to 2.1 is a function x ∈ E which satisfies 2.1, taking x n x n , for all n ∈ {0, 1, 2, . . . , T }. For the constants a, b, c, d ∈ R, we define h1 s as 1− d d s 2 1 − e−as , a a 1− d 2 s , 2 if b 0, a / 0, if b 0, a 0, d d a 0, a2 4b, 1 s e−a/2s − , if b / b 2 b 2.3 αs d βe − αeβs d − , if b / 0, a2 > 4b, 1 b β−α b ⎫ ⎧ a d −a/2s ⎨ a2 a2 ⎬ d 0, a2 < 4b. 1 sin b − s − , if b / cos b − s e ⎭ ⎩ 2 b 4 4 b 2 b − a /4 1 Consider also h2 s given by 1 c 0, 1 − e−as − cs 1 − e−as , if b 0, a / a a c s − s2 , if b 0, a 0, 2 c c a 2 1 s s − , if b e−a/2s / 0, a 4b, b 2 b 4 Abstract and Applied Analysis e−a/2s ⎧ ⎨c ⎩b βc/b − 1 eαs 1 − αc/beβs c − , β−α b cos b− 1 ac/2b a s sin 4 b − a2 /4 2 if b / 0, a2 > 4b, ⎫ a ⎬ c b− s − , 4 ⎭ b 2 2 if b / 0, a < 4b. 2.4 In the definitions of functions h1 and h2 , we denote, for b / 0, a2 > 4b, a α− 2 a 2 2 − b, a β− − 2 a 2 2 − b. 2.5 It is easy to check 7 that h1 0 1, h2 0 0, h1 0 0, h2 0 1. Theorem 2.2 see 7, Theorem 2.1. The initial value problem v t av t bvt cv t dvt 0, t ∈ 0, ∞, 2.6 v0 v0 , v 0 v0 , for v0 , v0 ∈ R, has the solution vt h1 t − n h2 t − n n v0 C1 C2 C1 C2 , v0 t ∈ n, n 1, 2.7 where n ∈ Z , C1 h1 1, C2 h2 1, C1 h1 1, C2 h2 1. 2.8 To simplify calculus, for z ∈ 0, 1, one denotes Hz h1 z h2 z h1 z h2 z C1 C2 . C H1 C1 C2 , 2.9 Abstract and Applied Analysis 5 3. Main Results First, we consider the nonlocal boundary value problem x t ax t bxt cx t dxt 0, xT x μ ϕ, x T x μ ψ, t ∈ J 0, T , 3.1 for a, b, c, d, ϕ, ψ ∈ R, T > 0 and μ ∈ 0, T . If μ 0 ϕ, the condition xT xμ ϕ is reduced to a periodic boundary condition of Dirichlet type. On the other hand, if μ 0, we obtain x T x 0 ψ, and, moreover, if ψ 0, we deduce the periodicity of the derivative of the function. ϕ Theorem 3.1. If T > 0 and μ ∈ 0, T , then problem 3.1 is solvable for each ψ in the image of the mapping F: u v μ u T −μ . −H μ− μ C −→ HT − T C v Hence, there exists a unique solution to 3.1 for every ϕ ψ 3.2 ∈ R2 if and only if the matrix HT − T CT −μ − H μ − μ Cμ 3.3 is nonsingular. In this case, the solution is given by vt h1 t − n h2 t − n C n v0 v0 , t ∈ n, n 1, n ∈ Z , 3.4 v ϕ is the inverse image of ψ by F. If the ϕ matrix 3.3 is singular, then there exists an infinite number of solutions to problem 3.1 slope in 3.4 any preimage of for every ψ in the image of F, taking as initial position andinitial ϕ ϕ 2 . by F, and there exist no solutions for the remaining pairs ∈ R ψ ψ where 0 v0 Proof. We consider the initial value problem v t av t bvt cv t dvt 0, v0 v0 , v 0 v0 , t ∈ 0, ∞, 3.5 6 Abstract and Applied Analysis for v0 , v0 ∈ R, whose solution is, by Theorem 2.2 7, Theorem 2.1, n v0 , vt h1 t − n h2 t − n C v0 t ∈ n, n 1, 3.6 where n ∈ Z . We analyze under which conditions the boundary conditions are fulfilled, taking into account that v t h1 t − n h2 t − n C n v0 v0 , t ∈ n, n 1, 3.7 where n ∈ Z . First, we consider the case T ∈ / Z, and μ ∈ / Z. To obtain the solution to 3.1, we calculate v0 and v0 from the expressions of 3.6 and 3.7, in order to satisfy vT vμ ϕ and v T v μ ψ. Hence the boundary conditions are written as μ v0 C ϕ, − h1 μ − μ h2 μ − μ h1 T − T h2 T − T C v0 T −μ μ v0 C ψ, h1 T − T h2 T − T C − h1 μ − μ h2 μ − μ v0 T −μ 3.8 that is, ϕ μ v0 T −μ HT − T C . −H μ− μ C ψ v0 3.9 The properties of functions h1 and h2 produce that the boundary conditions, in the cases where T ∈ Z, or μ ∈ Z or both, can be derived from the expression obtained in the case studied T ∈ / Z, and μ ∈ / Z. ϕ Therefore, this system has a solution only for ψ in the image of the mapping u v −→ HT − T C T −μ μ u . −H μ− μ C v 3.10 Therefore, the solution is unique if the matrix 3.3 is nonsingular, in which case, the initial condition and initial slope are given as v0 v0 HT − T CT −μ − H μ − μ Cμ Cμ −1 ϕ ψ −1 −1 ϕ T −μ . HT − T C −H μ− μ ψ 3.11 Abstract and Applied Analysis 7 ϕ On the other hand, if the matrix 3.3 is singular, for ψ in the image of F, the infinitely ϕ many solutions are calculated from 3.6 taking as initial conditions any preimage of ψ by F, which proves the result. Remark 3.2. The existence and uniqueness condition 3.3 is reduced, if μ > 0, to the nonsingularity of the matrices C and HT − T CT −μ − H μ − μ . 3.12 On the other hand, if μ 0, it is just reduced to the nonsingularity of the matrix HT − T CT − Hμ. Remark 3.3. In Theorem 3.1, if we consider μ 0, then the boundary conditions are xT x0ϕ, x T x 0ψ, and the condition of existence and uniqueness of solution is reduced to the nonsingularity of HT − T CT − I , 3.13 which coincides with condition 15 in 7, Theorem 2.2. Moreover, if ϕ ψ 0, the abovementioned nonsingularity condition provides that the unique solution to the homogeneous equation subject to periodic boundary value conditions is the trivial solution see 7, Theorem 2.2. Remark 3.4. In Theorem 3.1, the order 2 matrix 3.3 can be written in a simplified manner in the following particular cases. i If T ∈ / Z and T − T μ − μ, HT − T CT −μ − I Cμ . 3.14 HT − T CT −μ − I Cμ . 3.15 CT −μ − H μ − μ Cμ . 3.16 CT −μ − I Cμ . 3.17 ii If T ∈ / Z and μ ∈ Z, iii If T ∈ Z and μ ∈ / Z, iv If T, μ ∈ Z, 8 Abstract and Applied Analysis Remark 3.5. Summarizing, the study of the solvability of 3.1 is reduced to the discussion of system 3.9. In the case of existence of an infinite number of solutions, we must analyze the rank of the matrix in 3.3 to determine whether any pair is admissible as initial position and slope or the space of initial is one-dimensional. If the rank of 3.3 is zero, problem ϕconditions 3.1 is solvable only for ψ 00 , that is, problem x t ax t bxt cx t dxt 0, xT x μ , x T x μ , t ∈ J 0, T , 3.18 for a, b, c, d ∈ R, T > 0 and μ ∈ 0, T , has an infinite number of solutions, given by 3.4, for v0 any initial point v0 ∈ R2 . ϕ On the other hand, and denoting by G the matrix in 3.3, if rankG 1 and ψ depends linearly on each column vsolutions, v of G, then we have a one-dimensional space of ϕ 0 0 whose starting conditions V0 v0 are determined from the row of the system G v0 ψ corresponding to a nonzero minor of G. Next, with the purpose of extending Theorem 3.1 to the nonhomogeneous case, we consider the following nonlocal boundary value problem for a nonhomogeneous equation: x t ax t bxt cx t dxt σt, xT x μ ϕ, x T x μ ψ, t ∈ J 0, T , 3.19 for a, b, c, d, ϕ, ψ ∈ R, T > 0, μ ∈ 0, T , and σ ∈ Λ. Using the expression of the solution for the corresponding initial value problem provided by 19, Theorem 5.2, we prove the following existence and uniqueness result. Theorem 3.6. Consider a, b, c, d ∈ R, T > 0, μ ∈ 0, T , σ ∈ Λ and ϕ, ψ ∈ R. Then problem 3.19 has a unique solution if and only if the matrix 3.3 is nonsingular. Under this assumption, the unique solution to 3.19 is given by xt h1 t − n h2 t − n Cn V0 n−1 k1 h1 t − n h2 t − n C k0 t n k gt − sσsds, n−1−k gk 1 − s g k 1 − s t ∈ n, n 1, n ∈ Z , σsds 3.20 Abstract and Applied Analysis 9 with g defined as ⎧1 −az ⎪ ⎪ ⎪ a 1 − e , ⎪ ⎪ ⎪ ⎪ ⎪ z, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −a/2z ⎨ , ze gz ⎪ eβz − eαz ⎪ ⎪ ⎪ , ⎪ ⎪ ⎪ β−α ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ e−a/2z sin ⎩ b − a2 /4 if b 0, a / 0, if b 0, a 0, if b / 0, a2 4b, 3.21 if b / 0, a > 4b, 2 b− a2 z, 4 if b / 0, a2 < 4b, and taking, as the initial condition V0 , V0 HT − T C T −1 − H μ − μ Cμ ϕ ψ − MT,μ,σ 3.22 , where MT,μ,σ μ−1 k1 HT − T C T −1 k1 kμ − H μ − μ Cμ−1−k k k0 T −μ T T HT − T C k T −1−k gk 1 − s gk 1 − s g k 1 − s σsds g k 1 − s 3.23 σsds μ g μ−s σsds − σsds. g T − s μ g μ − s gT − s On the other hand, if the matrix 3.3 is singular, the number of solutions to 3.19 is determined by the discussion of the linear system HT − T CT − H μ − μ Cμ V0 ϕ ψ − MT,μ,σ , 3.24 and, for each solution V0 to this system (in case it exists), the expression 3.20 provides a solution to 3.19. 10 Abstract and Applied Analysis Proof. The result follows from the expression of the solution 3.20 for the corresponding initial value problem, whose derivative is given by x t h1 t − n h2 t − n Cn V0 n−1 k1 k0 t k h1 t − n g t − sσsds, h2 t − n C n−1−k gk 1 − s g k 1 − s σsds 3.25 t ∈ n, n 1, n ∈ Z , n and the fact that the restrictions represented by the boundary conditions produce, respectively, 1 0 HT − T CT − H μ − μ Cμ V0 ϕ − 1 0 MT,μ,σ , 0 1 HT − T CT − H μ − μ Cμ V0 ψ − 0 1 MT,μ,σ . 3.26 4. Examples We present some examples where different situations are analyzed in order to decide if the existence and uniqueness condition nonsingularity of 3.3 holds and, in case of nonuniqueness, the dimension of the space of solutions. The exact expression of the solution or solutions is given explicitly. We recall that, for s ∈ 0, 1, Hs is given, depending on the values of the coefficients a, b, c, d as follows. i If b 0, a / 0, ⎛ 1 d c d −as s 1 − e−as − cs 1 − e−as − e 1 − 1 ⎜ 2 a a a a ⎜ Hs ⎜ ⎝ d 1 −1 e−as a ce−as − c a a ⎞ ⎟ ⎟ ⎟, ⎠ 4.1 thus C H1 C1 C2 C1 C2 ⎛ ⎜ ⎜ ⎝ 1− ⎞ 1 d d c 2 1 − e−a 1 − e−a − c 1 − e−a ⎟ a a a a ⎟. ⎠ d 1 −a −a −1 e a ce − c a a 4.2 Abstract and Applied Analysis 11 Note that the determinant of the matrix C is equal to e−a ea − a − 1d a − a ea c a2 . detC a2 4.3 ii If b 0, a 0, ⎛ ⎞ d 2 c 2 1 − s s s − ⎜ 2 2 ⎟ Hs ⎝ ⎠, −ds 1 − cs 4.4 thus C H1 C1 C2 C1 C2 ⎛ ⎞ d c 1 − 1 − ⎜ 2 2⎟ ⎝ ⎠. −d 1 − c 4.5 The determinant of the matrix C is equal to detC d − c 1. 2 4.6 iii If b / 0, a2 4b, ⎛ c⎞ d a a d −a/2s c −a/2s 1 1 s e e 1 s s − ⎟ − ⎜ b 2 b b 2 b⎟ ⎜ $ % ⎟ Hs ⎜ 2 ⎜ ⎟, a −a2 c a d ⎝ ⎠ − s e−a/2s s− s1 e−a/2s 1 b 4 4b 2 4.7 thus ⎛ c⎞ d a a −a/2 d −a/2 c 1 e e 1 1 − ⎟ − 1 ⎜ b 2 b b 2 b⎟ ⎜ ⎟. $ % C H1 ⎜ ⎜ ⎟ 2 −a a2 a c d ⎝ ⎠ − e−a/2 − 1 e−a/2 1 b 4 4b 2 4.8 The determinant of the matrix C is equal to 2a − 4ea/2 4 d − a2 ea/2 c 4b . 4b detC e −a 4.9 12 Abstract and Applied Analysis iv If b / 0, a2 > 4b, and denoting a 2 a a 2 a − b, β− − − b, 2 2 2 2 ⎞ ⎛ d βeαs − αeβs d βc/b − 1 eαs 1 − αc/beβs c − − 1 ⎟ ⎜ b β−α b β−α b⎟ ⎜ ⎟, Hs ⎜ ⎟ ⎜ ⎠ ⎝ c − αeαs β − c eβs eαs − eβs b d β−α β−α α− 4.10 thus ⎛ ⎞ d βeα − αeβ d βc/b − 1 eα 1 − αc/beβ c − − ⎟ ⎜ 1 b β−α b β−α b⎟ ⎜ ⎜ ⎟. C H1 ⎜ α β ⎟ β α ⎝ ⎠ c − αe β − c e d b e −e 1 b β−α β−α 4.11 In this case, the determinant of the matrix C is equal to detC d 1 b βeα − αeβ α e − eβ βc − b eα b − αceβ c − αeα β − c eβ − 2 2 β−α β−α d eα − eβ d c − αeα β − c eβ c 1 − b β−α b β−α 2 d α d β d α β2 − 2b αβ 1 α β αe − βe ce − ce 1 2 e b β−α b b β−α d −a d 1 d α 1 e αc e − β c eβ . b β−α b b v If b / 0, a2 < 4b, and denoting R 4.12 b − a2 /4, & ' ⎛ ⎞ d a 1 ac/2b c d −a/2s −a/2s c cos Rs e sin Rs − e cos Rs sin Rs − 1 ⎜ b 2R b b R b⎟ ⎜ ⎟ Hs ⎜ ⎟, & ' ⎝ ⎠ a 2c 1 −a/2s −a/2s cos Rs − − b d e sin Rs e sin Rs R 2R 4.13 Abstract and Applied Analysis 13 thus ⎛ d ⎜ 1 b ⎜ C H1 ⎜ ⎝ ' & d 1 ac/2b c a e−a/2 cos R sin R − e−a/2 cos R sin R − 2R b b R ' & a 2c 1 sin R e−a/2 cos R − − b d e−a/2 sin R R 2R ⎞ c b⎟ ⎟ ⎟. ⎠ 4.14 The determinant of C is given by & ' aa 2c 2 d −a c e 1 sin R cos2 R − sin R cos R − b R 4R2 d 1 d d a 2c − e−a/2 cos R sin R 1 ce−a sin R cos R b b 2R R b b d −a ac d −a/2 c 2 1 sin 1 e e R − sin R 2b R b R2 d −a 2 ac aa 2c d −a b 1 1 e cos R 1 e sin2 R − b b 2b 4R2 R2 d d −a/2 1 da 2c − e−a/2 cos R −c 1 e sin R b R 2b b d −a d −a/2 d −a/2 1 da 2c 1 e − e −c 1 e cos R sin R. b b R 2b b detC 4.15 Example 4.1. Consider the problem x t x t dxt 0, ) ( 3 t ∈ J 0, , 2 3 1 x ϕ, 2 2 3 1 x ψ, x 2 2 x 4.16 where d, ϕ, ψ ∈ R. In this case, b c 0, a 1 / 0, T 3/2 and μ 1/2. Therefore, we d1−e−s −ds1 1−e−s get Hs so that matrix G HT − T CT −μ − Hμ − μCμ de−s −d e−s H1/2C − I is reduced to ⎛ ⎜− G⎜ ⎝ e−3/2 ⎞ √ 2 √ e d 2ed − 2d2 2 − 2ed e−3/2 ee − 1d 2 − 2ed 2e − 2 ⎟ 2 2 ⎟, ⎠ √ 2 √ −3/2 2 −3/2 e ed − d 1 − ed −e ee − 1d 1 − ed e − 1 4.17 14 Abstract and Applied Analysis whose determinant is calculated as √ e 3e − 3d2 2 − 2ed 2e − 2e2 d2 detG − 2 −3/2 e 3e − 3 2e−1 − 2 d2 e−3/2 2 − 2ed . − 2 e−2 4.18 Hence, G is nonsingular if and only 0 and d / 2/3 − 2e1/2 . In this case, problem 4.16 ϕ if d 2/ has a unique solution for every ψ ∈ R , given by see 3.4 xt d 1 − e−t − dt 1 v0 1 − e−t v0 , xt 1 − e1−t t ∈ 0, 1, e−1 d − d v0 e−1 v0 1 − e−1 d − d 1 v0 1 − e−1 v0 d 1 − e1−t − dt − 1 1 −e−t−1 4.19 ed − d2 t 2d2 − ed − e et − ed2 ed v0 e − 1dt 2 − 2ed − eet e2 − e d e v0 , ( ) 3 t ∈ 1, , 2 where v0 v0 G−1 ϕ ψ √ √ ⎞ e−2 d2 2 e−1 d2 − √ − √ 2 2 ⎟ ⎜ 2 e − 3 d 2d 2 e − 3 d 2d ⎟ ϕ ⎜ ⎟ ⎜ ⎜ √ √ √ ⎟ ψ , ⎠ ⎝ e − 2 d 2 ee − 2e 2 2 e − 1 d − 2e 2 − √ − √ e2e − 2 3 − 3e d 2e − 2 e2e − 2 3 − 3e d 2e − 2 4.20 ⎛ that is, √ √ − 2 e−1 d2 ϕ− e−2 d2 ψ v0 √ 2 e − 3 d2 2d √ √ √ −2 e − 1 d − 2e 2 ϕ − e − 2 d 2 ee − 2e 2 . v0 √ e2e − 2 3 − 3e d 2e − 2 4.21 Abstract and Applied Analysis 15 For the particular case where d 2, ϕ 1, and ψ −1, then v0 v0 −1 G 1 −1 √ √ ⎞ ⎞ ⎛ 1 2 e−1 e 1 − − − √ √ ⎟ ⎟ ⎜ ⎜ 4 ⎟ 1 4 e−4 ⎜ 4 e − 1⎟ ⎜ ⎜ ⎜ ⎟, ⎟ √ √ ⎝ 1 ⎝ e 1 ⎠ −1 −e 2 e − 1 e ⎠ − −√ √ 2 e−1 e2e − 2 − 2e 2 2e − 2 ⎛ 4.22 and hence the unique solution to x t x t 2xt 0, ) ( 3 t ∈ J 0, , 2 3 1 x 1, 2 2 3 1 x x − 1, 2 2 x 4.23 is given by √ e2t − 1 − 2et e , t ∈ 0, 1, √ −4e 8 e − 4 √ 2 e 2e 6e t − 7e2 − 21e et 4e3 12e2 xt − 4e2 24e 4et1/2 −16e2 − 16eet ) ( −6e2 − 2e t 21e2 7e et − 12e3 − 4e2 3 − , t ∈ 1, . 2 4e2 24e 4et1/2 −16e2 − 16eet xt 4.24 See Figure 1. −s and the matrix If d 0, then Hs 10 1−e e−s 1 G HT − T CT −μ − H μ − μ Cμ H C − I 2 is reduced to G 0 e−3/2 e−1 0 −e−3/2 e−1 4.25 , which has rank one. Since G is singular, there exists an infiϕ nite number of solutions to problem for every ψ in the image of the mapping given ϕ4.16 by the matrix G, that is, for every −ϕ , where ϕ ∈ R. On each of these cases, the solutions 16 Abstract and Applied Analysis 1.2 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t Figure 1: Solution to problem 4.23. are given by 3.4 taking as initial position and initial slope any preimage of initial slope can be taken as v0 e3/2 ϕ , e−1 ϕ −ϕ , that is, the 4.26 and the initial position can be chosen as any real number v0 . Hence, the solutions to the problem ) ( 3 t ∈ J 0, , 2 3 1 x x ϕ, 2 2 3 1 x x − ϕ, 2 2 x t x t 0, 4.27 where ϕ ∈ R, are given by xt v0 1 − e−t v0 , t ∈ 0, 1, 4.28 and the same expression xt 1 1−e −t−1 v0 1 − e v0 1 1 − e−1 e−1 v0 ) ( 3 t ∈ 1, , 2 0 −t v0 , 4.29 Abstract and Applied Analysis 17 2.5 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t Figure 2: Solution which generates by vertical shifts the remaining solutions to problem 4.27 with ϕ 1. where v0 ∈ R and v0 e3/2 ϕ/e − 1. Hence, for each ϕ fixed, all the vertical shifts of the function xt 1 − e−t e3/2 ϕ/e − 1, t ∈ 0, 3/2 are solutions to 4.27. In Figure 2, we show the graph of a solution which generates, by vertical shifting, all the solutions to problem 4.27 for ϕ 1. If ψ / − ϕ, there is no solution for problem 4.16 with d 0. Finally, if d 2/3 − 2e1/2 , then √ ⎞ 5 − 2 e − 2s − 2e−s −s 1 − e √ ⎟ ⎜ 3−2 e ⎟ ⎜ Hs ⎜ ⎟, ⎠ ⎝ 2e−s − 1 −s e √ 3−2 e 1 GH C − I 2 ⎞ ⎛ √ ⎛ ⎞ 4 − 2 e − 2e−1/2 −1/2 −2e−1 1 − e √ ⎟ ⎜ −1 1−e ⎟ √ ⎟⎜ ⎜ 3−2 e ⎟⎜ 3 − 2 e ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ −1 ⎟ − 1 −1/2 ⎟⎝ 2 e ⎜ ⎠ −1/2 e −1 −1 ⎠ ⎝ 2 e e −1 √ √ 3−2 e 3−2 e ⎛ ⎞ √ −2e−3/2 2e−1 6e−1/2 − 10 4 e −e−3/2 e−1 e−1/2 − 1 √ √ 2 ⎜ ⎟ 3−2 e ⎜ ⎟ 3−2 e ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ −3/2 −1/2 −3/2 −1/2 2e − 6e 4 e −e ⎝ ⎠ √ √ 2 3−2 e 3−2 e ⎛ 4.30 18 Abstract and Applied Analysis hasrank one. Hence problem 4.16, for d 2/3 − 2e1/2 , has an infinite number of solutions ϕ for ψ in the image of the mapping given by the matrix G, that is, for the values of ϕ, and ψ satisfying the system √ √ 2 −2e−3/2 2e−1 6e−1/2 − 10 4 e v0 −3e−3/2 5e−1 e−1/2 − 5 2e1/2 v0 ϕ 3 − 2 e , √ 2 2e−3/2 − 6e−1/2 4 v0 3e−3/2 − 3e−1/2 − 2e−1 2 v0 ψ 3 − 2 e , 4.31 for some values of v0 , v0 ∈ R. For the existence of these values of v0 , v0 , it is necessary and sufficient that the rank of the matrix of the system equal to one coincides with the rank of the matrix ⎛ √ √ 2 ⎞ −2e−3/2 2e−1 6e−1/2 − 10 4 e −3e−3/2 5e−1 e−1/2 − 5 2e1/2 ϕ 3 − 2 e ⎠ ⎝ √ 2 , 2e−3/2 − 6e−1/2 4 3e−3/2 − 3e−1/2 − 2e−1 2 ψ 3−2 e 4.32 that is, for instance, det −3e−3/2 5e−1 e−1/2 − 5 2e1/2 ϕ 3e−3/2 − 3e−1/2 − 2e−1 2 ψ 0, 4.33 equivalent to ψ 3e−3/2 − 3e−1/2 − 2e−1 2 ϕ. −3e−3/2 5e−1 e−1/2 − 5 2e1/2 4.34 Under this assumption, there exists an infinite number of solutions to problem x t x t ( ) 3 2 xt 0, t ∈ J 0, , 2 3 − 2e1/2 1 3 x ϕ, x 2 2 3 1 x ψ, x 2 2 4.35 Abstract and Applied Analysis 19 which are given by √ 5 − 2 e − 2t − 2e−t xt v0 1 − e−t v0 , √ 3−2 e t ∈ 0, 1, ⎞ √ 3 − 2 e − 2e−1 −1 1 − e √ ⎟ √ ⎜ ⎟ v0 ⎜ 3−2 e 5 − 2 e − 2t − 1 − 2e−t−1 −t−1 ⎟ ⎜ xt 1−e √ −1 ⎟ ⎜ 3−2 e ⎠ v0 ⎝ 2 e −1 −1 e √ 3−2 e √ √ −2e−t 4e1/2−t 4 et 4e−1 t − 6t 4e − 16 e − 8e−1 15 v0 √ 2 3−2 e √ ( ) −e−t − 2e1−t 2e1/2−t 2e−1 t − 2t − 2 e − 4e−1 7 3 v0 , t ∈ 1, , √ 2 3−2 e ⎛ 4.36 where v0 and v0 satisfy one of the equations in 4.31. On the other hand, if 4.34 fails, there is no solution to 4.35. Example 4.2. Next, consider the problem x t cx t dxt 0, ) ( 3 t ∈ J 0, , 2 3 1 x ϕ, 2 2 3 1 x x ψ, 2 2 x 4.37 where c, d, ϕ, ψ ∈ R. In this case, a b 0, T 3/2, and μ 1/2. Hence, the matrix 1 G HT − T CT −μ − H μ − μ Cμ H C − I 2 4.38 ⎞ d2 2c − 16d c − 2d 2c2 − 16c 16 ⎟ ⎜ 16 16 ⎟ ⎜ G⎜ ⎟, ⎝ d2 2c − 4d 2 c − 2d 2c − 4c ⎠ 4 4 4.39 is reduced to ⎛ 20 Abstract and Applied Analysis whose determinant is detG dd 8 − 4c/8. If d / 0 and d / 4c − 8, then G is invertible ϕ and problem 4.37 has a unique solution for every ψ ∈ R2 . Since ⎛ G−1 ⎞ 2c − 4d 4c2 − 8c c − 2d 2c2 − 16c 16 − ⎜ ⎟ 1 d 2d ⎜ ⎟ ⎜ ⎟, ⎠ d − 4c 8 ⎝ d 2c − 16 −2d 4c − 8 2 the unique solution to 4.37, for vt ϕ ψ 4.40 fixed, is given by d c 1 − t2 v0 t − t2 v0 , 2 2 t ∈ 0, 1, ⎛ ⎞ d c v 0 1 − 1 − d c vt 1 − t − 12 t − 1 − t − 12 ⎝ 2 2⎠ 2 2 v0 −d 1 − c d d c t − 1 − t − 12 −d v0 1 − t − 12 1− 2 2 2 d c c t − 1 − t − 12 1 − c v0 , 1 − t − 12 1 − 2 2 2 ( ) 3 t ∈ 1, , 2 4.41 where v0 v0 c − 2d 2c2 − 16c 16 2c − 4d 4c2 − 8c ϕ− ψ, dd 8 − 4c 2dd 8 − 4c 2d 4c − 8 d 2c − 16 v0 v . − d − 4c 8 2d − 4c 8 0 4.42 2 −1/2 s−s2 and G −11/16 , For instance, for c 2 and d 1, then Hs 1−s−s/2 1−2s 1/4 0 0 4 −1 which is invertible ϕ with inverse G −2 −11/2 . Then the unique solution to 4.37, for c 2 and d 1 and ψ fixed, is given by 1 2 7t − 11t 8 ψ, t ∈ 0, 1, vt 2 t2 − t ϕ 2 ( ) 1 1 2 3 vt − 4t2 − 12t 8 ϕ −5t 13t − 4 ψ, t ∈ 1, . 2 2 2 4.43 See Figure 3 for the solution to 4.37 with c 2, d 1, ϕ 1, and ψ −1: ⎧ 1 ⎪ ⎪ ⎪ −3 t2 7t − 8 , ⎨ 2 vt ⎪1 2 ⎪ ⎪ ⎩ t −t−4 , 2 t ∈ 0, 1, ) 3 t ∈ 1, . 2 ( 4.44 Abstract and Applied Analysis 21 −1.5 −2 −2.5 −3 −3.5 −4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t Figure 3: Solution to 4.37 for c 2, d 1, ϕ 1, and ψ −1. On the other hand, if d 0, then ⎛ Hs ⎞ cs2 2 ⎠, 0 1 − cs ⎝1 s − ⎛ ⎞ c2 − 8c 8 ⎜0 ⎟ 1 8 ⎟ GH C − I ⎜ 2 ⎝ c − 2c ⎠ 2 0 2 4.45 √ is singular with rank 1, since G12 0 if and only if c 4 ± 2 2 and G22 0 if and only if c 0 or c 2. there exists an infinite number of solutions to problem 4.37 with ϕ d 20, for Hence, ϕ every ψ in the image of the mapping given by the matrix G, that is, for every ψ ∈ R such that 4c2 − 2cϕ c2 − 8c 8ψ. Under this assumption, the solutions to 4.37 with d 0 are given by c vt v0 t − t2 v0 , 2 t ∈ 0, 1, ⎛ c ⎞ 1 1− v c 2⎠ 0 vt 1 t − 1 − t − 12 ⎝ 2 v0 0 1−c c c v0 1 − 1 − c t − 1 − t − 12 2 2 v0 , 4.46 ) 3 t ∈ 1, , 2 ( 22 Abstract and Applied Analysis where the initial position v0 can be chosen as any real number and the initial slope must satisfy v0 c2 8ϕ , − 8c 8 2ψ , v0 2 c − 2c √ if c / 4 ± 2 2, 4.47 if c / 0, c / 2. 2 /2 , so that G 0 1/8 . Therefore, For instance, if d 0 and c 1, then Hs 10 s−s 0 −1/2 1−s there exists an infinite number of solutions to problem 4.37 with d 0 and c 1, for every ϕ ψ such that ψ −4ϕ. In this case, the solutions to 4.37 with d 0 and c 1 are given by ⎧ 1 2 ⎪ ⎪ t ϕ, 8 t − ⎨ 2 vt v0 ⎪ ⎪ ⎩4ϕ, t ∈ 0, 1, ( ) 3 t ∈ 1, , 2 4.48 where v0 is any real number. Note that v0 8ϕ. All the solutions are obtained as vertical shifts of the function ⎧ 1 2 ⎪ ⎪ t ϕ, 8 t − ⎨ 2 ft ⎪ ⎪ ⎩4ϕ, t ∈ 0, 1, ( ) 3 t ∈ 1, . 2 4.49 See Figure 4 for the graph of function f for ϕ 1, which generates by vertical shifting the one-dimensional space of solutions to problem 4.37 with d 0, c 1, ϕ 1, and ψ −4. Finally, if d 4c − 8, then c ⎞ 1 − 2c − 4s2 s − s2 2 ⎠, Hs ⎝ ⎛ −4c 8s ⎛ 1 − cs ⎞ 3c2 − 18c 24 3c2 − 16c 16 ⎜ ⎟ 2 8 ⎜ ⎟ G⎜ ⎟, ⎝ ⎠ 2 − 10c 8 3c 6c2 − 24c 24 2 4.50 which has rank 1. Note that G11 / 0 for c / 2 and c / 4; G12 / 0 for c / 4/3 and c / 4; G21 / 0 for 0 for c / 2 and c / 4/3. c/ 2; and G22 / Abstract and Applied Analysis 23 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t Figure 4: Solution which generates by vertical shifting the remaining solutions to problem 4.37 with d 0, c 1, ϕ 1, and ψ −4. Therefore, problem 4.37 for d 4c − 8 has solution an infinite number of solutions if and only if 3c2 − 18c 24 ψ 6c2 − 24c 24 ϕ, 2 4.51 3c2 − 10c 8 3c2 − 16c 16 ψ ϕ, 8 2 4.52 in which case the expressions of the solutions are given by 4.41, replacing the value of d, where v0 and v0 satisfy 3c2 − 18c 24 3c2 − 16c 16 v0 v0 ϕ, 2 8 3c2 − 10c 8 6c2 − 24c 24 v0 v0 ψ. 2 4.53 4.54 Indeed, , and the problem has an infinite number of solutions for i if c 2, then G 00 −1/2 0 ψ 0 see 4.52 since 4.51 is trivially satisfied, where v0 ∈ R and v0 −2ϕ from 4.53 since 4.54 is trivially satisfied, 0 ii if c 4/3, then G 8/3 8/3 0 , and the problem has an infinite number of solutions for ϕ ψ see 4.51 since 4.52 is trivially satisfied, where v0 3/8ϕ 3/8ψ from 4.53 or 4.54 and v0 ∈ R, iii if c / 2 and c / 4/3, then the elements in the second row of G are nonzero and, hence, the problem has an infinite number of solutions for ϕ 3c2 −18c24/26c2 −24c 24ψ from 4.51 or the same expression ϕ 3c2 − 16c 16/43c2 − 10c 8ψ from 4.52. The initial position and slope are taken satisfying 4.54. 24 Abstract and Applied Analysis 0 0 If we take c 4 see the last case distinguished, then G 24 16 , and conditions 4.51 and 4.52 are reduced to ϕ 0 and 4.54 is written as 24v0 16v0 ψ. 5. Conclusions In this work, we have analyzed the existence and uniqueness of solutions to the nonlocal boundary value problem for linear second-order functional differential equations with piecewise constant arguments: x t ax t bxt cx t dxt σt, xT x μ ϕ, x T x μ ψ, t ∈ J 0, T , 5.1 where a, b, c, d, ϕ, ψ ∈ R, T > 0, μ ∈ 0, T and σ ∈ Λ, by reducing the study to the discussion of a linear system of order 2. The nonlocal boundary conditions considered involve terms of the state function and the derivative of the state function, and the problem of interest also includes, as a particular case, periodic boundary value problems for second-order linear differential equations with functional dependence given by the greatest integer part. 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