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Green's function for second order differential equations with piecewise constant arguments
Nonlinear Analysis 64 (2006) 1812 – 1830 www.elsevier.com/locate/na
Green’s function for second order differential equations with piecewise constant arguments夡 Pinghua Yanga, b,∗ , Yuji Liua , Weigao Gea a Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, PR China b Department of Mathematics, Mechanical Engineering College, Shijiazhuang, Hebei, 050003, PR China
Received 13 July 2005; accepted 19 July 2005
Abstract In this paper, the second order nonautonomous differential equations with piecewise constant arguments x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = (t), t ∈ [0, T ], x(0) = x(T ),
x (0) = x (T ) +
is considered. We obtain the Green’s function to express the unique solution for the system. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B15 Keywords: Green’s functions; Nonautonomous; Piecewise constant arguments
1. Introduction The study of differential equations with piecewise constant argument has been treated widely in literature. This type of equations, in which techniques of differential and difference equations are combined, models, among others, some biological phenomena, see 夡
Partially supported by the National Natural Sciences Foundation of PR China (10371006).
∗ Corresponding author. Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081,
PR China. E-mail address: [email protected] (P. Yang). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.07.019
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
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[1,3] and references therein, the stabilization of hybrid control systems with feedback discrete controller [4], or damped oscillators [10]. For the following second order differential equations with piecewise constant argument given by the greatest integer function x (t) = f (t, x(t), x (t), x([t]), x ([t])), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ), either the second order differential equations not depending on x with piecewise constant argument or the second order differential equations which are linear autonomous systems depending on x with piecewise constant argument have been studied by several authors. Second order differential equations not depending on x with piecewise constant argument have been studied widely in the literature, see [8,9,14]. For instance, in [14], it is studied the periodic problem −x (t) = f (t, x(t), x([t])), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ). Some authors have studied almost or pseudo-almost periodic problems of second order functional differential equations with piecewise constant arguments, see [5,6,11–13]. In [5], it is studied the almost periodic solutions of the following system: (x(t) + x(t − 1)) = qx([t]) + f (t). In [13], it is studied the spectrum containment of almost periodic solution to the following system d2 (x(t) + x(t − 1)) = qx([t]) + f (t). dt 2 But, to the best of our knowledge, few results are obtained on second order differential equations depending on x with piecewise constant argument. Recently, in [7], by using an analogous method in [2], it is considered the second order functional differential equations of the form x (t) + ax (t) + bx(t) + cx ([t]) + dx([t]) = (t), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ) + . The authors obtain the Green’s function to express the unique solution for the system. However, the coefficients are constants, that is, the system is autonomous. Motivated by [7], for the second order nonautonomous differential equations with piecewise constant argument x (t) + a(t)x (t) + b(t)x(t) + c(t)x ([t]) + d(t)x([t]) = (t), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ) + , we will study the special case a(t) = (t) − (t),
b(t) = −( (t) + (t)(t)),
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i.e., the second order nonautonomous differential equations with piecewise constant arguments x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) (1) + d(t)x([t]) = (t), t ∈ [0, T ], x(0) = x(T ), x (0) = x (T ) + , where T > 0, and real constant. Here, as usual, [·] denotes the greatest integer function, (t), (t), c(t), d(t) are continuously periodic functions with period 1 on R, (t) is continuously differential on R, and is continuous on [n, n + 1), for n ∈ {0, 1, 2, . . . , [T ] − 1}, and on [[T ], T ], such that there exists (t − ) ∈ R, understanding that (t) = (t + ) for all t ∈ {1, 2, . . . , [T ]}. This paper is organized as follows. In Section 2, we give the expression of the solution for some initial value problem, studying conditions which guarantee the existence of periodic solutions for those initial value problems. In Section 3, we obtain the solution of the following problems: ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(0) = x(T ), x (0− ) = x (T + ), ⎪ ⎩ x (s + ) = x (s − ) + 1 which provides the Green’s function to calculate the unique solution to system (1) under suitable conditions. 2. Initial value problem and periodic solutions We begin studying the solutions for the following particular initial value problem: x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = 0, t ∈ [n, n + 1), x(n+ ) = vn , x (n+ ) = vn , where vn , vn ∈ R. This is equivalent to x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) = −c(t)vn − d(t)vn , t ∈ [n, n + 1), x(n+ ) = vn , x (n+ ) = vn .
(2)
(3)
Lemma 1. The solution of system (3) has the following expression: x(t) = p1 (t, n)vn + p2 (t, n)vn , where
t ∈ [n, n + 1),
(4)
t t w e− n ((v)+(v)) dv p1 (t, n) = e n (v) dv 1 −
wn u (v) dv n × (n) + d(u)e du dw , p2 (t, n) = e
t
n (v) dv
n
t
e n
−
w n
((v)+(v)) dv
w
1−
c(u)e n
u n
(v) dv
du dw.
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Proof. System (3) becomes t t t ⎧ ⎨ (x (t)e n (v) dv ) − ((t)x(t)e n (v) dv ) = −(c(t)vn + d(t)vn )e n (v) dv , t ∈ [n, n + 1), ⎩ x(n+ ) = vn , x (n+ ) = vn .
(5)
Integrating the first equation of (5) from n to t, we have
t t x (t)e n (v) dv − x (n+ ) − (t)x(t)e n (v) dv + (n)x(n+ ) t s = − (c(u)vn + d(u)vn )e n (v) dv du,
n
i.e., t u t c(u)e n (v) dv du e− n (v) dv vn x (t) − (t)x(t) = 1 − n t u t − (n) + d(u)e n (v) dv du e− n (v) dv vn .
(6)
n
By using the classical theory for the initial value problem of first-order linear ordinary differential equations, we obtain
t w t u w e− n ((v)+(v)) dv (n) + d(u)e n (v) dv du dw vn x(t) = e n (v) dv 1 −
tn w n t u w e− n ((v)+(v)) dv 1 − c(u)e n (v) dv du dw vn + e n (v) dv n
n
= p1 (t, n)vn + p2 (t, n)vn .
Lemma 2. The solution of system (3) has the following expression: x(t) = h1 (t − n)vn + h2 (t − n)vn ,
t ∈ [n, n + 1),
(7)
where s h1 (s) = e 0 (v) dv 1 −
s
e−
w 0
((v)+(v)) dv
(0) +
0
s h2 (s) = e 0 (v) dv
s
e−
0
w
d(u)e
u 0
(v) dv
du dw ,
0
w 0
((v)+(v)) dv
w
1−
c(u)e
u 0
(v) dv
du dw,
0
s ∈ [0, 1). Proof. If f1 (t), f2 (t), f3 (t) are continuous periodic functions with period 1 on R, then it is not difficult to prove that (i) e
t n
f1 (v) dv
=e
t−n 0
f1 (v) dv
;
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u u t−n t (ii) n f2 (v)e n f1 (v) dv du = 0 f2 (u)e 0 f1 (v) dv du; u w w t−n w t w u (iii) n e n f3 (v) dv n f2 (u)e n f1 (v) dv du dw= 0 e 0 f3 (v) dv 0 f2 (u)e 0 f2 (v) dv du dw.
Since (t), (t), c(t), d(t) are continuous periodic functions with period 1 on R, we obtain p1 (t, n) = h1 (t − n),
p2 (t, n) = h2 (t − n).
Remark 1. The five cases of h1 (t), h2 (t) in [7] are special ones in this paper. The relations between a and b are correspondent to the relations between and under and constants. The correspondent relations are as follows: b=0 a=0
b=0 a = 0
b = 0 a 2 > 4b
b = 0 a 2 = 4b
b = 0 a 2 < 4b
==0
= 0, = −a or = 0, = −a
= 0, = 0, + = 0
= 0, = 0, +=0
= 0, = 0, + 2 < 0
The fifth case only shows that the relation between a, b and , . In this paper, we still suppose that , are real functions on R. We mention the following properties of h1 (t) and h2 (t) that will be useful later: h1 (0) = 1,
h2 (0) = 0,
h1 (0) = 0,
h2 (0) = 1.
For s ∈ J , we study the following problems: ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(0) = x(T ), x (0− ) = x (T + ), ⎪ ⎩ + x (s ) = x (s − ) + 1.
(8)
We define, for s ∈ J , Es as the class of functions y : R → R satisfying (i) y is continuous for t ∈ R; (ii) y is continuous for t ∈ R\{s}, and there exist y (s − ), y (s + ) ∈ R; (iii) y exists and is continuous for t ∈ [n, n + 1)\{s}, n ∈ Z, and there exist y (s − ), y (s + ), y (n− ) ∈ R, ∀n ∈ Z. Definition 1. For s ∈ J , y : R → R is a solution to problem (8) if y ∈ Es and satisfies conditions in (8) taking y (s) = y (s + ) and y (t) = y (t + ), for all t ∈ Z ∪ {s}. Definition 2. Let := {y : J → R : y is continuous in J \{1, 2, . . . , [T ]}, and there exist y(n− ) ∈ R, y(n+ ) = y(n), ∀n ∈ {1, 2, . . . , [T ]}}, and E := {x : J → R : x, x are continuous and x ∈ }.
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Definition 3. A function x is a solution to (1) if x ∈ E and satisfies conditions in (1), taking x (n) = x (n+ ), ∀n ∈ {0, 1, 2, . . . , [T ]}. We will study solution for problem (1) by calculating the solutions to (8) in J = [0, T ]. Theorem 2.1. The solution of the initial value problem
v (t) + ((t) − (t))v (t) − ( (t) + (t)(t))v(t) + c(t)v ([t]) + d(t)v([t]) = 0, t ∈ R+ , v(0) = v0 , v (0) = v0 ,
(9)
for v0 , v0 ∈ R, is given by
v(t) = (h1 (t − n) h2 (t − n))
C1 C1
C2 C2
n
v0 v0
,
t ∈ [n, n + 1),
(10)
where n ∈ Z+ , C1 = h1 (1),
C2 = h2 (1),
C1 = h1 (1),
C2 = h2 (1),
and h1 , h2 are expressed in Lemma 2. Proof. Problem (9) can be written in [0, +∞) as a family of initial problem in [n, n + 1), n ∈ Z+ , for the second order differential equation (2). Then, solution of (9) is given by (7). Since v is continuous, vn+1 = v(n + 1) = v(n + 1− ) = vn h1 (1) + vn h2 (1) = vn C1 + vn C2 , n ∈ Z+ .
(11)
Now, v (t) = vn h1 (t − n) + vn h2 (t − n),
t ∈ (n, n + 1),
and using the continuity of v , we have vn+1 = v (n + 1) = v (n + 1− ) = vn h1 (1) + vn h2 (1) = vn C1 + vn C2 ,
n ∈ Z+ .
(12)
Therefore, from (11) and (12), we obtain
vn+1 C 1 C2 vn = , n ∈ Z+ . vn+1 C1 C2 vn Thus
vn vn
=
C1 C1
C2 C2
vn−1 vn−1
=
C1 C1
C2 C2
n
v0 v0
,
n ∈ N.
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Then, the solution for the initial value problem (9) is
v v(t) = (h1 (t − n) h2 (t − n)) n v n
n v0 C 1 C2 = (h1 (t − n) h2 (t − n)) , C1 C2 v0 t ∈ [n, n + 1), n ∈ Z+ . We introduce the following notations. For h1 , h2 in Lemma 2, and z ∈ [0, 1], let us denote H (z) by the matrix
h1 (z) h2 (z) H (z) = , h1 (z) h2 (z) and C = H (1), so that
C 1 C2 h1 (T − [T ]) C = H (1) = , H (T − [T ]) = h1 (T − [T ]) C1 C2
Consider also V0 = vv0 , and the identity matrix I = 01 01 .
h2 (T − [T ]) . h2 (T − [T ])
0
Theorem 2.2. For T > 0, the solution v(t) of the initial value problem (9) satisfies v(T ) = v0 ,
v (T ) = v0
if and only if [I − H (T − [T ])C [T ] ]V0 =
0 . 0
(13)
Moreover, if |I − H (T − [T ])C [T ] | = 0,
(14)
then v0 = v0 = 0 and we obtain the trivial solution of (9). If |I − H (T − [T ])C [T ] | = 0,
(15)
then there exist nontrivial solutions of (9) with v(T ) = v0 and v (T ) = v0 . Proof. In solution of (10) given in Theorem 2.1, we impose that v0 = v(T ) and v0 = v (T ), then, for T ∈ / Z, v0 = v(T ) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 , and v (t) = (h1 (t − n) h2 (t − n))C n V0 ,
t ∈ (n, n + 1), n ∈ Z+ ,
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which implies that v0 = v (T ) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 . Thus, V0 = H (T − [T ])C [T ] V0 .
(16)
If T ∈ Z, we obtain v0 = v(T ) = (h1 (1) h2 (1))C T −1 V0 = (C1 C2 )C T −1 V0 , and v0 = v (T ) = (h1 (1) h2 (1))C T −1 V0 = (C1 C2 )C T −1 V0 , and the system obtained is V0 = C T V0 , that coincides with the expression in (16), by the properties of h1 , h2 , and H (T − [T ]) = H (0) = I
for T ∈ Z.
From system (16), we get [I − H (T − [T ])C
[T ]
0 ]V0 = . 0
Then, either (14) is valid, which implies that v0 = v0 = 0, and the solution is the trivial function, or (15) holds and (v0 , v0 ) is an eigenvector of the matrix H (T − [T ])C [T ] associated to the eigenvalue = 1.
Corollary 1. Problem (1) with ≡ 0 and = 0, i.e., x (t) + a(t)x (t) + b(t)x(t) + c(t)x ([t]) + d(t)x([t]) = (t), x(0) = x(T ), x (0) = x (T ),
t ∈ J,
has nontrivial solutions if and only if (15) is valid. Such solutions can be obtained from expression (10), taking the initial conditions v0 , v0 verifying (13). Corollary 2. Problem (1) with ≡ 0 and = 0 is uniquely solvable (has only trivial solution) if and only if (14) is valid. Under this hypothesis and for arbitrary ∈ and ∈ R, problem (1) has at most one solution.
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3. Solution of problem (1) Firstly, we deal with problem (8), that is, ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R+ , = x(T ), x (0− ) = x (T + ), ⎪ ⎩ x(0) + x (s ) = x (s − ) + 1 where s ∈ J . If this problem has a unique solution, for s ∈ J fixed, we denote it by K(t, s), the value of the solution for problem (8) at the point t. We are interested in the expression of the solution in the interval J = [0, T ]. Lemma 3. If (t), (t) are continuously periodic functions with period 1 on R, and t t w g(t, s) = e s (v) dv e− s ((v)+(v)) dv dw, s
then g(t, s) = e
t−n s−n
(v) dv
t−n
e−
u
s−n ((v)+(v)) dv
du := G(t − n, s − n).
s−n
The proof is obvious and is omitted. Lemma 4. If
t f (s) = e s (v) dv (h1 (s − n) h2 (s − n))C n V0 t t w (v) dv +e s e− s ((v)+(v)) dv dw(−(s) 1)H (s − n)C n V0 s t w t w u (v) dv − s ((v)+(v)) dv s −e e e s (v) dv (d(u) c(u))C n V0 du dw,
s
s
where (t), (t), c(t), d(t) are as in system (1), and h1 (t), h2 (t) are as in Lemma 2, then f (s) = (h1 (t − n) h2 (t − n))C n V0 ,
s ∈ (n, n + 1),
t ∈ [s, n + 1) ∩ [0, T ].
Proof. f (s) can become
t t t w (v) dv s f (s) = e h1 (s − n) + e− s (v) dv e w (v) dv dw(−(s)h1 (s − n) w t s t t u (v) dv w + h1 (s − n)) − e e w (v) dv d(u) du dw e s (v) dv h2 (s − n) s s t t w − s (v) dv w (v) dv + e e dw(−(s)h2 (s − n) + h2 (s − n)) s t w t u − e w (v) dv e w (v) dv c(u) du dw C n V0 s
s
:= (f1 (s) f2 (s))C n V0 .
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
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By using the property of calculus, we have
t t f1 (s) = e s (v) dv (−(s))h1 (s − n) + e s (v) dv h1 (s − n)
t
− e s (v) dv (−(s)h1 (s − n) + h1 (s − n)) t s t + e w (v) dv e w (v) dv d(s) dw s t t w + e− s (v) dv e w (v) dv (s)(−(s)h1 (s − n) + h1 s − n)) dw s t t w + e− s (v) dv e w (v) dv (− (s)h1 (s − n) s
− (s)h1 (s − n) + h1 (s − n)) dw t w t = e w (v) dv e− s (v) dv d(s) dw[h1 (s − n) + ((s) − (s))h1 (s − n) s
− ( (s) + (s)(s))h1 (s − n) + d(s)]. It is similar to obtain t w t f2 (s) = e w (v) dv e− s (v) dv d(s) dw[h2 (s − n) + ((s) − (s))h2 (s − n) s
− ( (s) + (s)(s))h2 (s − n) + c(s)]. Therefore, we have (f1 (s) f2 (s))C n V0
v = (f1 (s) f2 (s)) n vn t w t = e w (v) dv e− s (v) dv d(s) dw{[h1 (s − n)vn + h2 (s − n)vn ] s
+ ((s) − (s))[h1 (s − n)vn + h2 (s − n)vn ] − ( (s) + (s)(s))[h1 (s − n)vn + h2 (s − n)vn ] + c(s)vn + d(s)vn } = 0. But, it is easy to verify that f (t) = (f1 (t) f2 (t))C n V0 = (h1 (t − n) h2 (t − n))C n V0 . So f (s) = (h1 (t − n) h2 (t − n))C n V0 , s ∈ (n, n + 1), t ∈ [s, n + 1) ∩ [0, T ].
Theorem 3.1. For s ∈ J fixed, problem (8) has a unique solution if and only if condition (14) holds, i.e., the matrix I − H (T − [T ])C [T ] is invertible.
(17)
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In such a case, the solution to problem (8) is given in J by the following expression: • If s = 0 or s = T , K(t, s) = (h1 (t − n) h2 (t − n))C n V0 , where V0 = [I − H (T − [T ])C
[T ] −1
]
t ∈ [n, n + 1), n ∈ Z+ ,
0 . 1
• If 0 < s < T , s ∈ Z+ , (h1 (t − n) h2 (t − n))C n V0 , t ∈ [n, n + 1), n ∈ {0, 1, . . . , s − 1}, K(t, s) = 0 n−s s (h1 (t − n) h2 (t − n))C C V0 + , t ∈ [n, n + 1), ns, 1 where V0 = [I − H (T − [T ])C [T ] ]−1 H (T − [T ])C [T ]−s
0 . 1
• If 0 < s < T , s ∈ (n, n + 1), for some n ∈ Z+ , ⎧ (h1 (t − k) h2 (t − k))C k V0 , t ∈ [k, k + 1) ∩ [0, s), ⎪ ⎪ ⎪ k ∈ {0, 1, . . . , n}, ⎪ ⎪ ⎪ ⎪ ⎨ (h1 (t − n) h2 (t − n))C n V0 + g(t, s), t ∈ [s, n + 1) ∩ [0, T ], K(t, s) = (h1 (t− n − 1 − k) − k))C k
h2 (t − n − 1 ⎪ ⎪ g(n + 1, s) ⎪ ⎪ , × C n+1 V0 + ⎪ ⎪ g ⎪ 1 (n + 1, s) ⎩ t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ], k ∈ Z+ , where,
⎧ g(T , s) −1 [T ] ⎪ , ⎪ ⎪ [I − H (T − [T ])C ] g (T , s) ⎨
T < n + 1,
V0 = g(n + 1, s) −1 [T ]−n−1 [T ] ⎪ [I − H (T − [T ])C ] H (T − [T ])C , ⎪ ⎪ g1 (n + 1, s) ⎩ n + 1 T , 1
g(t, s) is as in Lemma 3, and g1 (t, s) := jg(t, s)/jt. Proof. We will consider four cases. (a) s = 0. In this case, G(t, 0) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, v = x(T ), x (0− ) = x (T + ), ⎪ ⎩ 0 = x(0) + v0 = x (0 ) = x (0− ) + 1,
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1823
so that v0 = x (0+ ) = x (T + ) + 1. From expression (10), we have K(t, 0) = (h1 (t − n) h2 (t − n))C n V0 ,
t ∈ [n, n + 1), n ∈ Z+ .
Then, for T ∈ / Z, K(T , 0) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 = v0 , and jK K(T , 0) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 = v0 − 1, jt which coincides with the system
v0 [T ] , H (T − [T ])C V0 = v0 − 1 i.e., [I − H (T − [T ])C [T ] ]V0 =
0 . 1
(18)
This system has a unique solution if and only if the matrix I −H (T −[T ])C [T ] is invertible, i.e., hypothesis (14) holds. In this case,
[T ] −1 0 V0 = [I − H (T − [T ])C ] . 1 If T ∈ Z+ , the system obtained is
0 [T ] , [I − H (T − [T ])C ] = 1 which coincides with (18) since [T ] = T and H (T − [T ]) = H (0) = I . In fact, it has a unique solution if and only if I − C T is invertible, condition that coincides with hypothesis (14). (b) s = T > 0. In this case, G(t, T ) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, v ), v0 = x (0− ) = x (T + ), ⎪ ⎩ 0 =+x(0) = x(T − x (T ) = x (T ) + 1. Using expression (10), it is not difficult to verify K(t, 0) = K(t, T ).
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
(c) 0 < s < T , s ∈ Z+ . Now K(t, s) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(T ), v0 = x (0− ) = x (T + ), v ⎪ ⎩ 0 =+x(0) = x (s ) = x (s − ) + 1. Then
K(t, s) =
(h1 (t − n) h2 (t − n))C n V0 , t ∈ [n, n + 1), n ∈ {0, 1, . . . , s − 1}, v¯0 , t ∈ [n, n + 1), n s, (h1 (t − n) h2 (t − n))C n−s v¯0
where v¯0 = K(s + , s) = K(s − , s) = (h1 (1) h2 (1))C s−1 V0 , jK + jK − v¯0 = (s , s) = (s , s) = (h1 (1) h2 (1))C s−1 V0 + 1, jt jt or, equivalently
v¯0 h1 (1) = h1 (1) v¯0
0 0 h2 (1) s−1 s C V + = C V + . 0 0 h2 (1) 1 1
The periodic boundary value conditions are equivalent to v0 = K(T , s) = (h1 (T − [T ]) h2 (T − [T ]))C and v0 =
[T ]−s
0 C V0 + 1 s
jK 0 (T , s) = (h1 (T − [T ]) h2 (T − [T ]))C [T ]−s C s V0 + , 1 jt
so that V0 = H (T − [T ])C = H (T − [T ])C
[T ]−s
0 C V0 + 1 s
V0 + H (T − [T ])C
[T ]−s
0 , 1
]V0 = H (T − [T ])C
[T ]−s
0 . 1
[T ]
i.e., [I − H (T − [T ])C
[T ]
(19)
Now, the existence of a unique solution is equivalent to the nonsingularity of the matrix I − H (T − [T ])C [T ] , i.e., condition (14). Moreover, we have
[T ] −1 [T ]−s 0 . V0 = [I − H (T − [T ])C ] H (T − [T ])C 1
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
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For T ∈ Z+ , the system is reduced to
[T ]−s 0 [T ] V 0 = C V0 + C , 1 i.e., [I − C
[T ]
]V0 = C
[T ]−s
0 , 1
which coincides with (19) since H (0) = I and [T ] = T . (d) 0 < s < T , s ∈ (n, n + 1), for some n ∈ Z+ . For all t ∈ [0, s), K(t, s) is the unique solution of the system x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = 0, t ∈ R+ , v0 = x(0), v0 = x (0), that is to say, K(t, s) = (h1 (t − k) h2 (t − k))C k V0 ,
t ∈ [k, k + 1) ∩ [0, s), k ∈ {0, 1, . . . , n}.
For t ∈ [s, n + 1) ∩ [0, T ), since [t] = n, K(t, s) is the solution of the system x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x (n) + d(t)x(n) = 0, t ∈ [s, n + 1) ∩ [0, T ], x(s + ) = vs , x (s + ) = vs , where vs = K(s − , s) = (h1 (s − n) h2 (s − n))C n V0 , and vs =
jK + jK − (s , s) = (s , s) + 1 = (h1 (s − n) h2 (s − n))C n V0 + 1, jt jt
or, equivalently,
vs 0 n = H (s − n)C V0 + . vs 1 Now, in the system (20), x(n) = (h1 (1) h2 (1))C n−1 V0 , and x (n) =
jK (n, s) = (h1 (1) h2 (1))C n−1 V0 , jt
so the system (20) can be rewritten as x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) = − c(t)x (n) − d(t)x(n) = (−d(t) − c(t))C n V0 .
(20)
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
Similar to the argument of Lemma 1, and using the results of Lemma 4, we get
t x(t) = e s (v) dv (h1 (s − n) h2 (s − n))C n V0 t t w (v) dv s +e e− s ((v)+(v)) dv dw(−(s) 1)H (s − n)C n V0 s t w t w u (v) dv −e s e− s ((v)+(v)) dv e s (v) dv (d(u) c(u))C n V0 du dw s s t t w (v) dv − ( (v)+ (v)) dv +e s e s dw
s
= f (s) + g(t, s) = (h1 (t − n) h2 (t − n))C n V0 + g(t, s).
(21)
Now, if t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ), for some k ∈ Z+ , K(t, s) is given by ⎛ ⎞ K(n + 1, s) ⎠, K(t, s) = (h1 (t − (n + 1 + k)) h2 (t − (n + 1 + k)))C k ⎝ jK (n + 1, s) jt where K(n + 1, s) = (h1 (1) h2 (1))C n V0 + g(n + 1, s) = (C1 C2 )C n V0 + g(n + 1, s), and jK (n + 1, s) = (h1 (1) h2 (1))C n V0 + g1 (n + 1, s) jt = (C1 C2 )C n V0 + g1 (n + 1, s); so
K(n + 1, s) jK (n + 1, s) jt
= (h1 (1)
h2 (1))C n V0
+
g(n + 1, s) . g1 (n + 1, s)
In consequence, for t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ), k ∈ Z+ ,
g(n + 1, s) k n+1 K(t, s) = (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C C V0 + g1 (n + 1, s) = (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C n+1+k V0
g(n + 1, s) . + (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C k g1 (n + 1, s) Next, we will verify that the solution we are seeking satisfies v0 = K(0, s) = K(T , s),
v0 =
jK jK (0, s) = (T , s). jt jt
(22)
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1827
We distinguish two cases: • If T < n + 1, then n < s < T < n + 1, [T ] = n and, using (21), we have
K(T , s) g(T , s) [T ] jK , = H (T − [T ])C V0 + V0 = (T , s) g1 (T , s) jt so [I − H (T − [T ])C
[T ]
]V0 =
g(T , s) , g1 (T , s)
which has a unique solution if and only if condition (14) holds, and
g(T , s) [T ] −1 . V0 = [I − H (T − [T ])C ] g1 (T , s) • If n + 1 T , then n + k + 1 = [T ] T < n + k + 2 = [T ] + 1, with k = [T ] − n − 1 ∈ Z+ and, using (21), we get
g(n + 1, s) , V0 = H (T − [T ])C [T ]−n−1 C n+1 V0 + g1 (n + 1, s) and, therefore, [I − H (T − [T ])C [T ] ]V0 = H (T − [T ])C [T ]−n−1
g(n + 1, s) , g1 (n + 1, s)
which has a unique solution if and only if condition (14) holds, and
g(n + 1, s) . V0 = [I − H (T − [T ])C [T ] ]−1 H (T − [T ])C [T ]−n−1 g1 (n + 1, s) The proof is completed.
Theorem 3.2. If hypothesis (14) holds, problem (1) has a unique solution, for all ∈ and ∈ R, which can be obtained by the expression
T
x(t) =
K(t, s)(s) ds + K(t, 0),
t ∈ J,
(23)
0
where, for all s ∈ J , K(·, s) is the unique solution to problem (8). Proof. Condition (14) guarantees existence of a unique solution to problem (8), by Theorem 3.1, and uniqueness for (1) by Corollary 2. We will prove that x given by (23) is a solution to system (1).
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
Let P (t) := K(t, 0), s ∈ J . Then
x(t) =
[t]−1
n+1
n=0 n [T ]−1
K(t, s)(s) ds +
n+1
+
n=[t]+1
t
[t]+1
K(t, s)(s) ds +
[t]
K(t, s)(s) ds t
K(t, s)(s) ds +
n
T [T ]
K(t, s)(s) ds + P (t)
defines a continuous function,
x (t) =
[t]−1
n+1 n
n=0
+ +
t [t]
jK (t, s)(s) ds + K(t + , t)(t) jt
[T ]−1 n+1 jK jK − (t, s)(s) ds (t, s)(s) ds − K(t , t)(t) + jt jt n=[t]+1 n T jK jK jK (t, s)(s) ds + P (t) (t, s)(s) ds + (t, 0) = jt jt jt 0
[t]+1
t
jK (t, s)(s) ds + jt
T
[T ]
is also continuous, and using that jK(t + , t)/jt − jK(t − , t)/jt = 1, we obtain
x (t) =
[t]−1 n=0
n+1 n
j2 K (t, s)(s) ds + jt 2
t [t]
j2 K (t, s)(s) ds jt 2
[T ]−1 n+1 2 j2 K j K (t, s)(s) ds + (t, s)(s) ds 2 jt jt 2 t n=[t]+1 n T 2 j K jK + jK − + (t, s)(s) ds + , t) − , t) (t) + P (t) (t (t 2 jt jt jt [T ] T 2 j K = (t, s)(s) ds + (t) + P (t), 2 jt 0
+
[t]+1
hence x ∈ . This implies that x ∈ E. Now, taking
k(t, s) =
j2 K jK (t, s) + ((t) − (t)) (t, s) − ( (t) + (t)(t))K(t, s) 2 jt jt jK ([t], s) + d(t)K([t], s), + c(t) jt
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1829
which is equal to zero since K(·, s) is the solution to (8), we obtain x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) t [t]+1 [t]−1 n+1 = k(t, s)(s) ds + k(t, s)(s) ds + k(t, s)(s) ds [t]
n=0 n [T ]−1
n+1
+
t
k(t, s)(s) ds +
n=[t]+1 n
T [T ]
k(t, s)(s) ds + (t) + [P (t) + ((t)
− (t))P (t) − ( (t) + (t)(t))P (t) + c(t)P ([t]) + d(t)P ([t])] = (t), t ∈ J . Moreover,
T
x(0) = x(T ) =
K(T , s)(s) ds − P (T )
0
0 T
=
T
K(0, s)(s) ds + P (0) −
(K(0, s) − K(T , s))(s) ds + (P (0) − P (T )) = 0,
0
and x (0) − x (T ) =
T 0
=
0
T
T jK jK (0, s)(s) ds + P (0) − (T , s)(s) ds − P (T ) jt jt 0
jK jK (0, s) − (T , s) (s) ds + (P (0) − P (T )) = . jt jt
Therefore, x is the solution to system (1).
Remark 2. If , , c, d are constants, then g(t, s) = g(t − s), where g(t − s) as in Theorem 3.1 in [7]. It shows that Theorem 3.1 in this paper is also an extension of Theorem 3.1 in [7]. References [1] S. Busenberg, K. Cooke, Vertically Transmitted Diseases, Models and Dynamics, in: Biomathematics, vol. 23, Springer, Berlin, 1993. [2] A. Cabada, J.B. Ferreiro, J.J. Nieto, Green’s function and comparison principles for first order periodic differential equations with piecewise constant arguments, J. Math. Anal. Appl. 291 (2004) 690–697. [3] J. Gouzé, T. Sari, A class of piecewise linear differential equations arising in biological models, Non-smooth dynamical systems, theory and applications, Dynam. Systems 17 (2002) 299–316 (special issue). [4] T. Küpper, R. Yuang, On quasi-periodic solutions of differential equations with piecewise constant argument, J. Math. Anal. Appl. 267 (2002) 173–193. [5] H. Li, Almost periodic solutions of second-order neutral delay-differential equations with piecewise constant arguments, J. Math. Anal. Appl. 298 (2004) 693–709. [6] J.J. Nieto, A comparison result for a linear differential equation with piecewise constant delay, Glas. Mat. Ser. III 39 (2004) 73–76. [7] J.J. Nieto, R. Rodríguez-López, Green’s function for second-order periodic boundary value problems with piecewise constant arguments, J. Math. Anal. Appl. 304 (2005) 33–57.
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[8] G. Seifert, Second order scalar functional differential equations with piecewise constant arguments, J. Differential Equations Appl. 8 (2002) 427–445. [9] G. Seifert, Second-order neutral delay-differential equations with piecewise constant time dependence, J. Math. Anal. Appl. 281 (2003) 1–9. [10] J. Wiener, V. Lakshmikantham, Differential equations with piecewise constant argument and impulsive equations, Nonlinear Stud. 7 (2000) 60–69. [11] R. Yuan, Pseudo-almost periodic solutions of second-order neutral delay differential equations with piecewise constant argument, Nonlinear Anal. 41 (2000) 871–890. [12] R. Yuan, On the second-order differential equation with piecewise constant argument and almost periodic coefficients, Nonlinear Anal. 52 (2003) 1411–1440. [13] R. Yuan, On the spectrum of almost periodic solutions of second order scalar functional differential equations with piecewise constant argument, J. Math. Anal. Appl. 303 (2005) 103–118. [14] F.Q. Zhang, Boundary value problems for second order differential equations with piecewise constant arguments, Ann. Differential Equations 9 (1993) 369–374.
Green’s function for second order differential equations with piecewise constant arguments夡 Pinghua Yanga, b,∗ , Yuji Liua , Weigao Gea a Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, PR China b Department of Mathematics, Mechanical Engineering College, Shijiazhuang, Hebei, 050003, PR China
Received 13 July 2005; accepted 19 July 2005
Abstract In this paper, the second order nonautonomous differential equations with piecewise constant arguments x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = (t), t ∈ [0, T ], x(0) = x(T ),
x (0) = x (T ) +
is considered. We obtain the Green’s function to express the unique solution for the system. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B15 Keywords: Green’s functions; Nonautonomous; Piecewise constant arguments
1. Introduction The study of differential equations with piecewise constant argument has been treated widely in literature. This type of equations, in which techniques of differential and difference equations are combined, models, among others, some biological phenomena, see 夡
Partially supported by the National Natural Sciences Foundation of PR China (10371006).
∗ Corresponding author. Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081,
PR China. E-mail address: [email protected] (P. Yang). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.07.019
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
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[1,3] and references therein, the stabilization of hybrid control systems with feedback discrete controller [4], or damped oscillators [10]. For the following second order differential equations with piecewise constant argument given by the greatest integer function x (t) = f (t, x(t), x (t), x([t]), x ([t])), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ), either the second order differential equations not depending on x with piecewise constant argument or the second order differential equations which are linear autonomous systems depending on x with piecewise constant argument have been studied by several authors. Second order differential equations not depending on x with piecewise constant argument have been studied widely in the literature, see [8,9,14]. For instance, in [14], it is studied the periodic problem −x (t) = f (t, x(t), x([t])), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ). Some authors have studied almost or pseudo-almost periodic problems of second order functional differential equations with piecewise constant arguments, see [5,6,11–13]. In [5], it is studied the almost periodic solutions of the following system: (x(t) + x(t − 1)) = qx([t]) + f (t). In [13], it is studied the spectrum containment of almost periodic solution to the following system d2 (x(t) + x(t − 1)) = qx([t]) + f (t). dt 2 But, to the best of our knowledge, few results are obtained on second order differential equations depending on x with piecewise constant argument. Recently, in [7], by using an analogous method in [2], it is considered the second order functional differential equations of the form x (t) + ax (t) + bx(t) + cx ([t]) + dx([t]) = (t), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ) + . The authors obtain the Green’s function to express the unique solution for the system. However, the coefficients are constants, that is, the system is autonomous. Motivated by [7], for the second order nonautonomous differential equations with piecewise constant argument x (t) + a(t)x (t) + b(t)x(t) + c(t)x ([t]) + d(t)x([t]) = (t), t ∈ J = [0, T ], x(0) = x(T ), x (0) = x (T ) + , we will study the special case a(t) = (t) − (t),
b(t) = −( (t) + (t)(t)),
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
i.e., the second order nonautonomous differential equations with piecewise constant arguments x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) (1) + d(t)x([t]) = (t), t ∈ [0, T ], x(0) = x(T ), x (0) = x (T ) + , where T > 0, and real constant. Here, as usual, [·] denotes the greatest integer function, (t), (t), c(t), d(t) are continuously periodic functions with period 1 on R, (t) is continuously differential on R, and is continuous on [n, n + 1), for n ∈ {0, 1, 2, . . . , [T ] − 1}, and on [[T ], T ], such that there exists (t − ) ∈ R, understanding that (t) = (t + ) for all t ∈ {1, 2, . . . , [T ]}. This paper is organized as follows. In Section 2, we give the expression of the solution for some initial value problem, studying conditions which guarantee the existence of periodic solutions for those initial value problems. In Section 3, we obtain the solution of the following problems: ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(0) = x(T ), x (0− ) = x (T + ), ⎪ ⎩ x (s + ) = x (s − ) + 1 which provides the Green’s function to calculate the unique solution to system (1) under suitable conditions. 2. Initial value problem and periodic solutions We begin studying the solutions for the following particular initial value problem: x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = 0, t ∈ [n, n + 1), x(n+ ) = vn , x (n+ ) = vn , where vn , vn ∈ R. This is equivalent to x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) = −c(t)vn − d(t)vn , t ∈ [n, n + 1), x(n+ ) = vn , x (n+ ) = vn .
(2)
(3)
Lemma 1. The solution of system (3) has the following expression: x(t) = p1 (t, n)vn + p2 (t, n)vn , where
t ∈ [n, n + 1),
(4)
t t w e− n ((v)+(v)) dv p1 (t, n) = e n (v) dv 1 −
wn u (v) dv n × (n) + d(u)e du dw , p2 (t, n) = e
t
n (v) dv
n
t
e n
−
w n
((v)+(v)) dv
w
1−
c(u)e n
u n
(v) dv
du dw.
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1815
Proof. System (3) becomes t t t ⎧ ⎨ (x (t)e n (v) dv ) − ((t)x(t)e n (v) dv ) = −(c(t)vn + d(t)vn )e n (v) dv , t ∈ [n, n + 1), ⎩ x(n+ ) = vn , x (n+ ) = vn .
(5)
Integrating the first equation of (5) from n to t, we have
t t x (t)e n (v) dv − x (n+ ) − (t)x(t)e n (v) dv + (n)x(n+ ) t s = − (c(u)vn + d(u)vn )e n (v) dv du,
n
i.e., t u t c(u)e n (v) dv du e− n (v) dv vn x (t) − (t)x(t) = 1 − n t u t − (n) + d(u)e n (v) dv du e− n (v) dv vn .
(6)
n
By using the classical theory for the initial value problem of first-order linear ordinary differential equations, we obtain
t w t u w e− n ((v)+(v)) dv (n) + d(u)e n (v) dv du dw vn x(t) = e n (v) dv 1 −
tn w n t u w e− n ((v)+(v)) dv 1 − c(u)e n (v) dv du dw vn + e n (v) dv n
n
= p1 (t, n)vn + p2 (t, n)vn .
Lemma 2. The solution of system (3) has the following expression: x(t) = h1 (t − n)vn + h2 (t − n)vn ,
t ∈ [n, n + 1),
(7)
where s h1 (s) = e 0 (v) dv 1 −
s
e−
w 0
((v)+(v)) dv
(0) +
0
s h2 (s) = e 0 (v) dv
s
e−
0
w
d(u)e
u 0
(v) dv
du dw ,
0
w 0
((v)+(v)) dv
w
1−
c(u)e
u 0
(v) dv
du dw,
0
s ∈ [0, 1). Proof. If f1 (t), f2 (t), f3 (t) are continuous periodic functions with period 1 on R, then it is not difficult to prove that (i) e
t n
f1 (v) dv
=e
t−n 0
f1 (v) dv
;
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
u u t−n t (ii) n f2 (v)e n f1 (v) dv du = 0 f2 (u)e 0 f1 (v) dv du; u w w t−n w t w u (iii) n e n f3 (v) dv n f2 (u)e n f1 (v) dv du dw= 0 e 0 f3 (v) dv 0 f2 (u)e 0 f2 (v) dv du dw.
Since (t), (t), c(t), d(t) are continuous periodic functions with period 1 on R, we obtain p1 (t, n) = h1 (t − n),
p2 (t, n) = h2 (t − n).
Remark 1. The five cases of h1 (t), h2 (t) in [7] are special ones in this paper. The relations between a and b are correspondent to the relations between and under and constants. The correspondent relations are as follows: b=0 a=0
b=0 a = 0
b = 0 a 2 > 4b
b = 0 a 2 = 4b
b = 0 a 2 < 4b
==0
= 0, = −a or = 0, = −a
= 0, = 0, + = 0
= 0, = 0, +=0
= 0, = 0, + 2 < 0
The fifth case only shows that the relation between a, b and , . In this paper, we still suppose that , are real functions on R. We mention the following properties of h1 (t) and h2 (t) that will be useful later: h1 (0) = 1,
h2 (0) = 0,
h1 (0) = 0,
h2 (0) = 1.
For s ∈ J , we study the following problems: ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(0) = x(T ), x (0− ) = x (T + ), ⎪ ⎩ + x (s ) = x (s − ) + 1.
(8)
We define, for s ∈ J , Es as the class of functions y : R → R satisfying (i) y is continuous for t ∈ R; (ii) y is continuous for t ∈ R\{s}, and there exist y (s − ), y (s + ) ∈ R; (iii) y exists and is continuous for t ∈ [n, n + 1)\{s}, n ∈ Z, and there exist y (s − ), y (s + ), y (n− ) ∈ R, ∀n ∈ Z. Definition 1. For s ∈ J , y : R → R is a solution to problem (8) if y ∈ Es and satisfies conditions in (8) taking y (s) = y (s + ) and y (t) = y (t + ), for all t ∈ Z ∪ {s}. Definition 2. Let := {y : J → R : y is continuous in J \{1, 2, . . . , [T ]}, and there exist y(n− ) ∈ R, y(n+ ) = y(n), ∀n ∈ {1, 2, . . . , [T ]}}, and E := {x : J → R : x, x are continuous and x ∈ }.
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
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Definition 3. A function x is a solution to (1) if x ∈ E and satisfies conditions in (1), taking x (n) = x (n+ ), ∀n ∈ {0, 1, 2, . . . , [T ]}. We will study solution for problem (1) by calculating the solutions to (8) in J = [0, T ]. Theorem 2.1. The solution of the initial value problem
v (t) + ((t) − (t))v (t) − ( (t) + (t)(t))v(t) + c(t)v ([t]) + d(t)v([t]) = 0, t ∈ R+ , v(0) = v0 , v (0) = v0 ,
(9)
for v0 , v0 ∈ R, is given by
v(t) = (h1 (t − n) h2 (t − n))
C1 C1
C2 C2
n
v0 v0
,
t ∈ [n, n + 1),
(10)
where n ∈ Z+ , C1 = h1 (1),
C2 = h2 (1),
C1 = h1 (1),
C2 = h2 (1),
and h1 , h2 are expressed in Lemma 2. Proof. Problem (9) can be written in [0, +∞) as a family of initial problem in [n, n + 1), n ∈ Z+ , for the second order differential equation (2). Then, solution of (9) is given by (7). Since v is continuous, vn+1 = v(n + 1) = v(n + 1− ) = vn h1 (1) + vn h2 (1) = vn C1 + vn C2 , n ∈ Z+ .
(11)
Now, v (t) = vn h1 (t − n) + vn h2 (t − n),
t ∈ (n, n + 1),
and using the continuity of v , we have vn+1 = v (n + 1) = v (n + 1− ) = vn h1 (1) + vn h2 (1) = vn C1 + vn C2 ,
n ∈ Z+ .
(12)
Therefore, from (11) and (12), we obtain
vn+1 C 1 C2 vn = , n ∈ Z+ . vn+1 C1 C2 vn Thus
vn vn
=
C1 C1
C2 C2
vn−1 vn−1
=
C1 C1
C2 C2
n
v0 v0
,
n ∈ N.
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
Then, the solution for the initial value problem (9) is
v v(t) = (h1 (t − n) h2 (t − n)) n v n
n v0 C 1 C2 = (h1 (t − n) h2 (t − n)) , C1 C2 v0 t ∈ [n, n + 1), n ∈ Z+ . We introduce the following notations. For h1 , h2 in Lemma 2, and z ∈ [0, 1], let us denote H (z) by the matrix
h1 (z) h2 (z) H (z) = , h1 (z) h2 (z) and C = H (1), so that
C 1 C2 h1 (T − [T ]) C = H (1) = , H (T − [T ]) = h1 (T − [T ]) C1 C2
Consider also V0 = vv0 , and the identity matrix I = 01 01 .
h2 (T − [T ]) . h2 (T − [T ])
0
Theorem 2.2. For T > 0, the solution v(t) of the initial value problem (9) satisfies v(T ) = v0 ,
v (T ) = v0
if and only if [I − H (T − [T ])C [T ] ]V0 =
0 . 0
(13)
Moreover, if |I − H (T − [T ])C [T ] | = 0,
(14)
then v0 = v0 = 0 and we obtain the trivial solution of (9). If |I − H (T − [T ])C [T ] | = 0,
(15)
then there exist nontrivial solutions of (9) with v(T ) = v0 and v (T ) = v0 . Proof. In solution of (10) given in Theorem 2.1, we impose that v0 = v(T ) and v0 = v (T ), then, for T ∈ / Z, v0 = v(T ) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 , and v (t) = (h1 (t − n) h2 (t − n))C n V0 ,
t ∈ (n, n + 1), n ∈ Z+ ,
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1819
which implies that v0 = v (T ) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 . Thus, V0 = H (T − [T ])C [T ] V0 .
(16)
If T ∈ Z, we obtain v0 = v(T ) = (h1 (1) h2 (1))C T −1 V0 = (C1 C2 )C T −1 V0 , and v0 = v (T ) = (h1 (1) h2 (1))C T −1 V0 = (C1 C2 )C T −1 V0 , and the system obtained is V0 = C T V0 , that coincides with the expression in (16), by the properties of h1 , h2 , and H (T − [T ]) = H (0) = I
for T ∈ Z.
From system (16), we get [I − H (T − [T ])C
[T ]
0 ]V0 = . 0
Then, either (14) is valid, which implies that v0 = v0 = 0, and the solution is the trivial function, or (15) holds and (v0 , v0 ) is an eigenvector of the matrix H (T − [T ])C [T ] associated to the eigenvalue = 1.
Corollary 1. Problem (1) with ≡ 0 and = 0, i.e., x (t) + a(t)x (t) + b(t)x(t) + c(t)x ([t]) + d(t)x([t]) = (t), x(0) = x(T ), x (0) = x (T ),
t ∈ J,
has nontrivial solutions if and only if (15) is valid. Such solutions can be obtained from expression (10), taking the initial conditions v0 , v0 verifying (13). Corollary 2. Problem (1) with ≡ 0 and = 0 is uniquely solvable (has only trivial solution) if and only if (14) is valid. Under this hypothesis and for arbitrary ∈ and ∈ R, problem (1) has at most one solution.
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
3. Solution of problem (1) Firstly, we deal with problem (8), that is, ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R+ , = x(T ), x (0− ) = x (T + ), ⎪ ⎩ x(0) + x (s ) = x (s − ) + 1 where s ∈ J . If this problem has a unique solution, for s ∈ J fixed, we denote it by K(t, s), the value of the solution for problem (8) at the point t. We are interested in the expression of the solution in the interval J = [0, T ]. Lemma 3. If (t), (t) are continuously periodic functions with period 1 on R, and t t w g(t, s) = e s (v) dv e− s ((v)+(v)) dv dw, s
then g(t, s) = e
t−n s−n
(v) dv
t−n
e−
u
s−n ((v)+(v)) dv
du := G(t − n, s − n).
s−n
The proof is obvious and is omitted. Lemma 4. If
t f (s) = e s (v) dv (h1 (s − n) h2 (s − n))C n V0 t t w (v) dv +e s e− s ((v)+(v)) dv dw(−(s) 1)H (s − n)C n V0 s t w t w u (v) dv − s ((v)+(v)) dv s −e e e s (v) dv (d(u) c(u))C n V0 du dw,
s
s
where (t), (t), c(t), d(t) are as in system (1), and h1 (t), h2 (t) are as in Lemma 2, then f (s) = (h1 (t − n) h2 (t − n))C n V0 ,
s ∈ (n, n + 1),
t ∈ [s, n + 1) ∩ [0, T ].
Proof. f (s) can become
t t t w (v) dv s f (s) = e h1 (s − n) + e− s (v) dv e w (v) dv dw(−(s)h1 (s − n) w t s t t u (v) dv w + h1 (s − n)) − e e w (v) dv d(u) du dw e s (v) dv h2 (s − n) s s t t w − s (v) dv w (v) dv + e e dw(−(s)h2 (s − n) + h2 (s − n)) s t w t u − e w (v) dv e w (v) dv c(u) du dw C n V0 s
s
:= (f1 (s) f2 (s))C n V0 .
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1821
By using the property of calculus, we have
t t f1 (s) = e s (v) dv (−(s))h1 (s − n) + e s (v) dv h1 (s − n)
t
− e s (v) dv (−(s)h1 (s − n) + h1 (s − n)) t s t + e w (v) dv e w (v) dv d(s) dw s t t w + e− s (v) dv e w (v) dv (s)(−(s)h1 (s − n) + h1 s − n)) dw s t t w + e− s (v) dv e w (v) dv (− (s)h1 (s − n) s
− (s)h1 (s − n) + h1 (s − n)) dw t w t = e w (v) dv e− s (v) dv d(s) dw[h1 (s − n) + ((s) − (s))h1 (s − n) s
− ( (s) + (s)(s))h1 (s − n) + d(s)]. It is similar to obtain t w t f2 (s) = e w (v) dv e− s (v) dv d(s) dw[h2 (s − n) + ((s) − (s))h2 (s − n) s
− ( (s) + (s)(s))h2 (s − n) + c(s)]. Therefore, we have (f1 (s) f2 (s))C n V0
v = (f1 (s) f2 (s)) n vn t w t = e w (v) dv e− s (v) dv d(s) dw{[h1 (s − n)vn + h2 (s − n)vn ] s
+ ((s) − (s))[h1 (s − n)vn + h2 (s − n)vn ] − ( (s) + (s)(s))[h1 (s − n)vn + h2 (s − n)vn ] + c(s)vn + d(s)vn } = 0. But, it is easy to verify that f (t) = (f1 (t) f2 (t))C n V0 = (h1 (t − n) h2 (t − n))C n V0 . So f (s) = (h1 (t − n) h2 (t − n))C n V0 , s ∈ (n, n + 1), t ∈ [s, n + 1) ∩ [0, T ].
Theorem 3.1. For s ∈ J fixed, problem (8) has a unique solution if and only if condition (14) holds, i.e., the matrix I − H (T − [T ])C [T ] is invertible.
(17)
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
In such a case, the solution to problem (8) is given in J by the following expression: • If s = 0 or s = T , K(t, s) = (h1 (t − n) h2 (t − n))C n V0 , where V0 = [I − H (T − [T ])C
[T ] −1
]
t ∈ [n, n + 1), n ∈ Z+ ,
0 . 1
• If 0 < s < T , s ∈ Z+ , (h1 (t − n) h2 (t − n))C n V0 , t ∈ [n, n + 1), n ∈ {0, 1, . . . , s − 1}, K(t, s) = 0 n−s s (h1 (t − n) h2 (t − n))C C V0 + , t ∈ [n, n + 1), ns, 1 where V0 = [I − H (T − [T ])C [T ] ]−1 H (T − [T ])C [T ]−s
0 . 1
• If 0 < s < T , s ∈ (n, n + 1), for some n ∈ Z+ , ⎧ (h1 (t − k) h2 (t − k))C k V0 , t ∈ [k, k + 1) ∩ [0, s), ⎪ ⎪ ⎪ k ∈ {0, 1, . . . , n}, ⎪ ⎪ ⎪ ⎪ ⎨ (h1 (t − n) h2 (t − n))C n V0 + g(t, s), t ∈ [s, n + 1) ∩ [0, T ], K(t, s) = (h1 (t− n − 1 − k) − k))C k
h2 (t − n − 1 ⎪ ⎪ g(n + 1, s) ⎪ ⎪ , × C n+1 V0 + ⎪ ⎪ g ⎪ 1 (n + 1, s) ⎩ t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ], k ∈ Z+ , where,
⎧ g(T , s) −1 [T ] ⎪ , ⎪ ⎪ [I − H (T − [T ])C ] g (T , s) ⎨
T < n + 1,
V0 = g(n + 1, s) −1 [T ]−n−1 [T ] ⎪ [I − H (T − [T ])C ] H (T − [T ])C , ⎪ ⎪ g1 (n + 1, s) ⎩ n + 1 T , 1
g(t, s) is as in Lemma 3, and g1 (t, s) := jg(t, s)/jt. Proof. We will consider four cases. (a) s = 0. In this case, G(t, 0) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, v = x(T ), x (0− ) = x (T + ), ⎪ ⎩ 0 = x(0) + v0 = x (0 ) = x (0− ) + 1,
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1823
so that v0 = x (0+ ) = x (T + ) + 1. From expression (10), we have K(t, 0) = (h1 (t − n) h2 (t − n))C n V0 ,
t ∈ [n, n + 1), n ∈ Z+ .
Then, for T ∈ / Z, K(T , 0) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 = v0 , and jK K(T , 0) = (h1 (T − [T ]) h2 (T − [T ]))C [T ] V0 = v0 − 1, jt which coincides with the system
v0 [T ] , H (T − [T ])C V0 = v0 − 1 i.e., [I − H (T − [T ])C [T ] ]V0 =
0 . 1
(18)
This system has a unique solution if and only if the matrix I −H (T −[T ])C [T ] is invertible, i.e., hypothesis (14) holds. In this case,
[T ] −1 0 V0 = [I − H (T − [T ])C ] . 1 If T ∈ Z+ , the system obtained is
0 [T ] , [I − H (T − [T ])C ] = 1 which coincides with (18) since [T ] = T and H (T − [T ]) = H (0) = I . In fact, it has a unique solution if and only if I − C T is invertible, condition that coincides with hypothesis (14). (b) s = T > 0. In this case, G(t, T ) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, v ), v0 = x (0− ) = x (T + ), ⎪ ⎩ 0 =+x(0) = x(T − x (T ) = x (T ) + 1. Using expression (10), it is not difficult to verify K(t, 0) = K(t, T ).
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
(c) 0 < s < T , s ∈ Z+ . Now K(t, s) is the unique solution of the system ⎧ x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) ⎪ ⎨ + d(t)x([t]) = 0, t ∈ R, x(T ), v0 = x (0− ) = x (T + ), v ⎪ ⎩ 0 =+x(0) = x (s ) = x (s − ) + 1. Then
K(t, s) =
(h1 (t − n) h2 (t − n))C n V0 , t ∈ [n, n + 1), n ∈ {0, 1, . . . , s − 1}, v¯0 , t ∈ [n, n + 1), n s, (h1 (t − n) h2 (t − n))C n−s v¯0
where v¯0 = K(s + , s) = K(s − , s) = (h1 (1) h2 (1))C s−1 V0 , jK + jK − v¯0 = (s , s) = (s , s) = (h1 (1) h2 (1))C s−1 V0 + 1, jt jt or, equivalently
v¯0 h1 (1) = h1 (1) v¯0
0 0 h2 (1) s−1 s C V + = C V + . 0 0 h2 (1) 1 1
The periodic boundary value conditions are equivalent to v0 = K(T , s) = (h1 (T − [T ]) h2 (T − [T ]))C and v0 =
[T ]−s
0 C V0 + 1 s
jK 0 (T , s) = (h1 (T − [T ]) h2 (T − [T ]))C [T ]−s C s V0 + , 1 jt
so that V0 = H (T − [T ])C = H (T − [T ])C
[T ]−s
0 C V0 + 1 s
V0 + H (T − [T ])C
[T ]−s
0 , 1
]V0 = H (T − [T ])C
[T ]−s
0 . 1
[T ]
i.e., [I − H (T − [T ])C
[T ]
(19)
Now, the existence of a unique solution is equivalent to the nonsingularity of the matrix I − H (T − [T ])C [T ] , i.e., condition (14). Moreover, we have
[T ] −1 [T ]−s 0 . V0 = [I − H (T − [T ])C ] H (T − [T ])C 1
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1825
For T ∈ Z+ , the system is reduced to
[T ]−s 0 [T ] V 0 = C V0 + C , 1 i.e., [I − C
[T ]
]V0 = C
[T ]−s
0 , 1
which coincides with (19) since H (0) = I and [T ] = T . (d) 0 < s < T , s ∈ (n, n + 1), for some n ∈ Z+ . For all t ∈ [0, s), K(t, s) is the unique solution of the system x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) = 0, t ∈ R+ , v0 = x(0), v0 = x (0), that is to say, K(t, s) = (h1 (t − k) h2 (t − k))C k V0 ,
t ∈ [k, k + 1) ∩ [0, s), k ∈ {0, 1, . . . , n}.
For t ∈ [s, n + 1) ∩ [0, T ), since [t] = n, K(t, s) is the solution of the system x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x (n) + d(t)x(n) = 0, t ∈ [s, n + 1) ∩ [0, T ], x(s + ) = vs , x (s + ) = vs , where vs = K(s − , s) = (h1 (s − n) h2 (s − n))C n V0 , and vs =
jK + jK − (s , s) = (s , s) + 1 = (h1 (s − n) h2 (s − n))C n V0 + 1, jt jt
or, equivalently,
vs 0 n = H (s − n)C V0 + . vs 1 Now, in the system (20), x(n) = (h1 (1) h2 (1))C n−1 V0 , and x (n) =
jK (n, s) = (h1 (1) h2 (1))C n−1 V0 , jt
so the system (20) can be rewritten as x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) = − c(t)x (n) − d(t)x(n) = (−d(t) − c(t))C n V0 .
(20)
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
Similar to the argument of Lemma 1, and using the results of Lemma 4, we get
t x(t) = e s (v) dv (h1 (s − n) h2 (s − n))C n V0 t t w (v) dv s +e e− s ((v)+(v)) dv dw(−(s) 1)H (s − n)C n V0 s t w t w u (v) dv −e s e− s ((v)+(v)) dv e s (v) dv (d(u) c(u))C n V0 du dw s s t t w (v) dv − ( (v)+ (v)) dv +e s e s dw
s
= f (s) + g(t, s) = (h1 (t − n) h2 (t − n))C n V0 + g(t, s).
(21)
Now, if t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ), for some k ∈ Z+ , K(t, s) is given by ⎛ ⎞ K(n + 1, s) ⎠, K(t, s) = (h1 (t − (n + 1 + k)) h2 (t − (n + 1 + k)))C k ⎝ jK (n + 1, s) jt where K(n + 1, s) = (h1 (1) h2 (1))C n V0 + g(n + 1, s) = (C1 C2 )C n V0 + g(n + 1, s), and jK (n + 1, s) = (h1 (1) h2 (1))C n V0 + g1 (n + 1, s) jt = (C1 C2 )C n V0 + g1 (n + 1, s); so
K(n + 1, s) jK (n + 1, s) jt
= (h1 (1)
h2 (1))C n V0
+
g(n + 1, s) . g1 (n + 1, s)
In consequence, for t ∈ [n + 1 + k, n + 2 + k) ∩ [0, T ), k ∈ Z+ ,
g(n + 1, s) k n+1 K(t, s) = (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C C V0 + g1 (n + 1, s) = (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C n+1+k V0
g(n + 1, s) . + (h1 (t − n − 1 − k) h2 (t − n − 1 − k))C k g1 (n + 1, s) Next, we will verify that the solution we are seeking satisfies v0 = K(0, s) = K(T , s),
v0 =
jK jK (0, s) = (T , s). jt jt
(22)
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1827
We distinguish two cases: • If T < n + 1, then n < s < T < n + 1, [T ] = n and, using (21), we have
K(T , s) g(T , s) [T ] jK , = H (T − [T ])C V0 + V0 = (T , s) g1 (T , s) jt so [I − H (T − [T ])C
[T ]
]V0 =
g(T , s) , g1 (T , s)
which has a unique solution if and only if condition (14) holds, and
g(T , s) [T ] −1 . V0 = [I − H (T − [T ])C ] g1 (T , s) • If n + 1 T , then n + k + 1 = [T ] T < n + k + 2 = [T ] + 1, with k = [T ] − n − 1 ∈ Z+ and, using (21), we get
g(n + 1, s) , V0 = H (T − [T ])C [T ]−n−1 C n+1 V0 + g1 (n + 1, s) and, therefore, [I − H (T − [T ])C [T ] ]V0 = H (T − [T ])C [T ]−n−1
g(n + 1, s) , g1 (n + 1, s)
which has a unique solution if and only if condition (14) holds, and
g(n + 1, s) . V0 = [I − H (T − [T ])C [T ] ]−1 H (T − [T ])C [T ]−n−1 g1 (n + 1, s) The proof is completed.
Theorem 3.2. If hypothesis (14) holds, problem (1) has a unique solution, for all ∈ and ∈ R, which can be obtained by the expression
T
x(t) =
K(t, s)(s) ds + K(t, 0),
t ∈ J,
(23)
0
where, for all s ∈ J , K(·, s) is the unique solution to problem (8). Proof. Condition (14) guarantees existence of a unique solution to problem (8), by Theorem 3.1, and uniqueness for (1) by Corollary 2. We will prove that x given by (23) is a solution to system (1).
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P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
Let P (t) := K(t, 0), s ∈ J . Then
x(t) =
[t]−1
n+1
n=0 n [T ]−1
K(t, s)(s) ds +
n+1
+
n=[t]+1
t
[t]+1
K(t, s)(s) ds +
[t]
K(t, s)(s) ds t
K(t, s)(s) ds +
n
T [T ]
K(t, s)(s) ds + P (t)
defines a continuous function,
x (t) =
[t]−1
n+1 n
n=0
+ +
t [t]
jK (t, s)(s) ds + K(t + , t)(t) jt
[T ]−1 n+1 jK jK − (t, s)(s) ds (t, s)(s) ds − K(t , t)(t) + jt jt n=[t]+1 n T jK jK jK (t, s)(s) ds + P (t) (t, s)(s) ds + (t, 0) = jt jt jt 0
[t]+1
t
jK (t, s)(s) ds + jt
T
[T ]
is also continuous, and using that jK(t + , t)/jt − jK(t − , t)/jt = 1, we obtain
x (t) =
[t]−1 n=0
n+1 n
j2 K (t, s)(s) ds + jt 2
t [t]
j2 K (t, s)(s) ds jt 2
[T ]−1 n+1 2 j2 K j K (t, s)(s) ds + (t, s)(s) ds 2 jt jt 2 t n=[t]+1 n T 2 j K jK + jK − + (t, s)(s) ds + , t) − , t) (t) + P (t) (t (t 2 jt jt jt [T ] T 2 j K = (t, s)(s) ds + (t) + P (t), 2 jt 0
+
[t]+1
hence x ∈ . This implies that x ∈ E. Now, taking
k(t, s) =
j2 K jK (t, s) + ((t) − (t)) (t, s) − ( (t) + (t)(t))K(t, s) 2 jt jt jK ([t], s) + d(t)K([t], s), + c(t) jt
P. Yang et al. / Nonlinear Analysis 64 (2006) 1812 – 1830
1829
which is equal to zero since K(·, s) is the solution to (8), we obtain x (t) + ((t) − (t))x (t) − ( (t) + (t)(t))x(t) + c(t)x ([t]) + d(t)x([t]) t [t]+1 [t]−1 n+1 = k(t, s)(s) ds + k(t, s)(s) ds + k(t, s)(s) ds [t]
n=0 n [T ]−1
n+1
+
t
k(t, s)(s) ds +
n=[t]+1 n
T [T ]
k(t, s)(s) ds + (t) + [P (t) + ((t)
− (t))P (t) − ( (t) + (t)(t))P (t) + c(t)P ([t]) + d(t)P ([t])] = (t), t ∈ J . Moreover,
T
x(0) = x(T ) =
K(T , s)(s) ds − P (T )
0
0 T
=
T
K(0, s)(s) ds + P (0) −
(K(0, s) − K(T , s))(s) ds + (P (0) − P (T )) = 0,
0
and x (0) − x (T ) =
T 0
=
0
T
T jK jK (0, s)(s) ds + P (0) − (T , s)(s) ds − P (T ) jt jt 0
jK jK (0, s) − (T , s) (s) ds + (P (0) − P (T )) = . jt jt
Therefore, x is the solution to system (1).
Remark 2. If , , c, d are constants, then g(t, s) = g(t − s), where g(t − s) as in Theorem 3.1 in [7]. It shows that Theorem 3.1 in this paper is also an extension of Theorem 3.1 in [7]. References [1] S. Busenberg, K. Cooke, Vertically Transmitted Diseases, Models and Dynamics, in: Biomathematics, vol. 23, Springer, Berlin, 1993. [2] A. Cabada, J.B. Ferreiro, J.J. Nieto, Green’s function and comparison principles for first order periodic differential equations with piecewise constant arguments, J. Math. Anal. Appl. 291 (2004) 690–697. [3] J. Gouzé, T. Sari, A class of piecewise linear differential equations arising in biological models, Non-smooth dynamical systems, theory and applications, Dynam. Systems 17 (2002) 299–316 (special issue). [4] T. Küpper, R. Yuang, On quasi-periodic solutions of differential equations with piecewise constant argument, J. Math. Anal. Appl. 267 (2002) 173–193. [5] H. Li, Almost periodic solutions of second-order neutral delay-differential equations with piecewise constant arguments, J. Math. Anal. Appl. 298 (2004) 693–709. [6] J.J. Nieto, A comparison result for a linear differential equation with piecewise constant delay, Glas. Mat. Ser. III 39 (2004) 73–76. [7] J.J. Nieto, R. Rodríguez-López, Green’s function for second-order periodic boundary value problems with piecewise constant arguments, J. Math. Anal. Appl. 304 (2005) 33–57.
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