Differential equations involving causal operators with nonlinear ...
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Mathematical and Computer Modelling 48 (2008) 859–866 www.elsevier.com/locate/mcm
Differential equations involving causal operators with nonlinear periodic boundary conditionsI Fengjie Geng School of Information Engineering, China University of Geosciences (Beijing), Beijing, 100083, China Received 5 September 2006; received in revised form 26 October 2007; accepted 2 November 2007
Abstract The notion of causal operators is extended to periodic boundary value problems with nonlinear boundary conditions in this paper. By utilizing the monotone iterative technique and the method of lower and upper solutions (resp. weakly coupled lower and upper solutions), we establish the existence of the extremal solutions (resp. weakly coupled extremal quasi-solutions) for nonlinear periodic boundary value problems with causal operators. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Lower and upper solutions; Causal operators; Weakly coupled lower and upper solutions; Weakly coupled quasi-solutions; Nonlinear periodic boundary conditions
1. Introduction In this paper, we deal with the following differential equation with a causal operator: x 0 (t) = (Qx)(t),
t ∈ J := [0, T ],
g(x(0), x(T )) = 0,
(1.1)
where Q ∈ C[E, E] is a causal operator, namely, a nonanticipative operator. Some authors have focused their interest on differential equations with causal operators recently, because its theory has the powerful quality of unifying ordinary differential equations, integro differential equations, differential equations with finite or infinite delay, Volterra integral equations and neutral functional equations. Readers are referred to the papers [2–4] and references cited therein. In 2005, Drici et al. [3] extended differential equations with causal operators to the framework of an arbitrary Banach space and proved the basic results such as existence, uniqueness and global existence. Immediately after it, they discussed in [4] the periodic boundary value problems with a causal operator u 0 (t) = (Qu)(t),
u(0) = u(T ),
I Supported by National Natural Science Foundation of China (] 10671069).
E-mail address: gengfengjie [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.11.009
(1.2)
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where t ∈ J , Q ∈ C[E, E] is a causal operator. The authors developed the monotone iterative technique to obtain the existence of extremal solutions for (1.2). At the same time, periodic boundary value problems with nonlinear boundary conditions have attracted much attention, we may see [1,5,6]. Inspired by the above results, in this paper, we extend the notion of causal operators to the periodic boundary value problems with nonlinear boundary conditions. First, some comparison principles and the existence and uniqueness of the solutions for the first order linear differential equations with linear boundary conditions are presented. Next, by utilizing the monotone iterative technique and the method of lower and upper solutions, we establish the existence of extremal solutions of the problem (1.1). At last, using the notion of weakly coupled lower and upper solutions, we testify the existence of the weakly coupled minimal and maximal quasisolutions of the problem (1.1). The results obtained in this paper generalized those in [4]. 2. Preliminaries In this section, we present some definitions and lemmas which help to prove our main results. Let E = C[J, R], where J = [0, T ]. We define ku − vk(t) = max |u(t) − v(t)|. t0 ≤s≤t
Take Ω = E
T
C 1 [J, R]. A function x ∈ Ω is said to be a solution of problem (1.1) if it satisfies (1.1).
Definition 2.1. Suppose that Q ∈ C[E, E], then Q is said to be a causal map or a nonanticipative map if u(s) = v(s), t0 ≤ s ≤ t ≤ T , where u, v ∈ E, then (Qu)(s) = (Qv)(s),
t0 ≤ s ≤ t.
Lemma 2.1. Let m ∈ Ω be such that m 0 (t) ≤ −Mm(t) − (£m)(t),
m(0) ≤ λm(T ),
(2.1)
where 0 < λ ≤ 1, M ≥ 0 and £ ∈ C[E, E] is a positive linear operator, that is, £m ≥ 0 whenever m ≥ 0. Then, m(t) ≤ 0 for t ∈ J provided Z T e M T (£e−M )(t)dt ≤ λe−M T and 0 < λe−M T ≤ 1;
(2.2)
0
or T (M + (£1)(T )) ≤ λ
and
0 < λ ≤ 1.
(2.20 )
Proof. Let v(t) = e Mt m(t), then v ∈ Ω and v 0 (t) ≤ −e Mt (£e−M v)(t),
v(0) ≤ λe−M T v(T ).
(2.3)
Obviously, v(t) ≤ 0 implies m(t) ≤ 0. So, it suffices to show v(t) ≤ 0 for any t ∈ J . Suppose on the contrary, v(t) > 0 for some t ∈ J . Then there are two cases: (i) there exists t¯ ∈ J such that v(t¯) > 0 and v(t) ≥ 0 for all t ∈ J ; (ii) there exist t ∗ and t∗ such that v(t ∗ ) > 0 and v(t∗ ) < 0. First, suppose (2.2) holds. In case (i), (2.3) implies v 0 (t) ≤ 0 on J and v(t) is nonincreasing on J . If λe−M T = 1, v(0) ≤ v(T ) shows that v(t) ≡ c, then v 0 (t) = 0. On the other hand, we have 0 = v 0 (t¯) ≤ −e M t¯(£e−M v)(t¯) < 0 since v(t¯) > 0, which is a contradiction. If 0 < λe−M T < 1, then v(T ) ≤ v(0) ≤ λe−M T v(T ), which is also a contradiction. Now turning to case (ii). Let mint∈J v(t) = −γ , then γ > 0. Without loss of generality, we may suppose v(t∗ ) = −γ . First suppose t∗ < t ∗ , integrating the first inequality of (2.3) from t∗ to t ∗ one attains Z t∗ Z T Z t∗ v(t ∗ ) − v(t∗ ) = v 0 (t)dt ≤ −e Mt (£e−M v)(t) ≤ γ e M T (£e−M )(t)dt, t∗
t∗
0
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and hence 0 < v(t ) ≤ −γ + γ ∗
T
Z
e
MT
(£e
−M
)(t)dt ≤ −γ
T
Z 1−
e
0
MT
(£e
−M
)(t)dt ,
0
which yields Z T e M T (£e−M )(t)dt > 1. 0
A contradiction is then elicited due to (2.2). If t∗ > t ∗ , we have from (2.3) Z T v(T ) ≤ v(t∗ ) + γ e M T (£e−M )(t)dt
(2.4)
t∗
and v(t ) ≤ v(0) + γ ∗
Z
t∗
e M T (£e−M )(t)dt,
(2.5)
0
based on the fact v(0) ≤ λe−M T v(T ) and 0 < λe−M T ≤ 1, it then follows from (2.2), (2.4) and (2.5) that Z T Z t∗ ∗ −M T MT −M 0 < v(t ) ≤ λe −γ + γ e (£e )(t)dt + γ e M T (£e−M )(t)dt ≤ −γ λe−M T + γ ≤ −γ (λe−M T −
Z
e M T (£e−M )(t)dt + γ
t∗ T
Z
0
t∗ T
t∗
Z
e M T (£e−M )(t)dt
0
e M T (£e−M )(t)dt) ≤ 0,
0
which is a contradiction. 0 If (2.2) holds, by using (2.1) directly instead of (2.3), we may derive contradiction in a similar way. This completes the proof. Consider the following problem y 0 (t) = −M y(t) − (£y)(t) + σu (t),
t ∈ J, g(u(0), u(T )) + M1 (y(0) − u(0)) − M2 (y(T ) − u(T )) = 0,
(2.6)
where σu (t) = (Qu)(t) + Mu(t) + (£a)(t). Lemma 2.2. y ∈ Ω is a solution of problem (2.6) if and only if y is a solution of the following integral equation Z T e−M T Bu y(t) = + G(t, s)[−(£y)(s) + σu (s)]ds, t ∈ J, M1 − M2 e−M T 0 where Bu = −g(u(0), u(T )) + M1 u(0) − M2 u(T ), M, M1 , M2 are constants satisfying M ≥ 0, M1 6= M2 e−M T and M2 e−M(T +t−s) + e−M(t−s) , 0 ≤ s < t ≤ T ; −M T M − M e 1 2 (2.7) G(t, s) = M2 −M(T +t−s) e , 0 ≤ t ≤ s ≤ T. M1 − M2 e−M T The result may be obtained by a simple computation. M1 M2 e−M T Obviously, kG(t, s)k = max{| M −M |, | |}. In the remainder of the paper, we denote τ = −M T e M −M e−M T 1
kG(t, s)k = max{| M
M1 −M T 1 −M2 e
2 −M T
2e |, | M M−M 1
2
|}. e−M T
1
2
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Lemma 2.3. If M ≥ 0, M1 6= M2 e−M T and τ k£kT < 1.
(2.8)
Then problem (2.6) has a unique solution y ∈ Ω . Proof. Define an operator A : Ω → Ω by Z T e−M T Bu (Ay)(t) = G(t, s)[−(£y)(s) + σu (s)]ds, + M1 − M2 e−M T 0
t ∈ J.
For y1 , y2 ∈ Ω , one derives Z T |Ay1 − Ay2 | = G(t, s)(£(y2 − y1 ))(s)ds ≤ τ k£kT ky1 − y2 k. 0
Hence, (2.8) shows that A is a contraction. Banach fixed point theorem shows that A has a unique fixed point y ∗ such that Ay ∗ = y ∗ . It is clear that this fixed point is the solution of (2.6). The proof is then finished. 3. Main results In this section, we shall establish the existence of extremal solutions of problem (1.1). Definition 3.1. Functions α, β ∈ Ω are said to be lower and upper solutions of problem (1.1), respectively, if α 0 (t) ≤ (Qα)(t),
g(α(0), α(T )) ≤ 0
and β 0 (t) ≥ (Qβ)(t)
g(β(0), β(T )) ≥ 0.
Now we state our theorem. Theorem 3.1. Let (2.8) hold and Q ∈ C[E, E]. Suppose the following conditions hold, (H1 ) the functions α, β ∈ Ω are lower and upper solutions of problem (1.1) respectively, such that α(t) ≤ β(t) for t ∈ J ; (H2 ) whenever α(t) ≤ x2 (t) ≤ x1 (t) ≤ β(t), Q satisfies (Qx1 )(t) − (Qx2 )(t) ≥ −M(x1 (t) − x2 (t)) − (£(x1 − x2 ))(t),
t ∈ J,
where M > 0 is a constant and £ ∈ C[E, E] is a positive linear operator such that Z T M2 −M T e M T (£e−M )(t)dt ≤ e M 1 0 or T (M + (£1)(T )) ≤
M2 M1
(3.1)
(3.10 )
hold and M1 ≥ M2 > 0; (H3 ) the function g(x, y) ∈ C(R2 , R) satisfies g(x1 , x¯1 ) − g(x2 , x¯2 ) ≤ M1 (x1 − x2 ) − M2 (x¯1 − x¯2 ) whenever α(0) ≤ x2 ≤ x1 ≤ β(0), α(T ) ≤ x¯2 ≤ x¯1 ≤ β(T ), where M1 ≥ M2 > 0. Then there exist monotone sequences {αn (t)} and {βn (t)} with α0 = α, β0 = β, such that lim αn (t) = x∗ (t),
n→∞
lim βn (t) = x ∗ (t)
n→∞
uniformly on J , and x∗ (t), x ∗ (t) are the minimal and the maximal solutions of problem (1.1), respectively, such that α(t) ≤ x∗ ≤ x ∗ ≤ β(t) on J.
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Proof. For any u ∈ [α, β], where [α, β] = {u ∈ Ω | α(t) ≤ u(t) ≤ β(t), t ∈ J }. Consider the problem (2.6) with σu (t) = (Qu)(t) + Mu(t) + (£u)(t). Noticing that (2.8) holds, then by Lemma 2.3 one may see that problem (2.6) has exactly one solution y ∈ Ω . Denote y(t) = Au(t), then the operator A has the following properties: (i) α ≤ Aα, Aβ ≤ β; (ii) A is monotonically nondecreasing in [α, β], i.e. for any u 1 , u 2 ∈ [α, β], u 1 ≤ u 2 implies Au 1 ≤ Au 2 . To prove (i), set m = α − α1 , where α1 = Aα. Owing to (H1 ), we acquire m 0 (t) = α 0 (t) − α10 (t) ≤ (Qα)(t) − [−Mα1 (t) − (£α1 )(t) + (Qα)(t) + Mα(t) + (£α)(t)] = −Mm(t) − (£m)(t), t ∈ J, and 1 M2 m(0) = α(0) − − g(α(0), α(T )) + (α1 (T ) − α(T )) + α(0) M1 M1 M2 1 g(α(0), α(T )) + [α(T ) − α1 (T )] = M1 M1 M2 ≤ m(T ). M1 Based on the fact M1 ≥ M2 > 0, (3.1) (or (3.1)0 ) and Lemma 2.1, we get m(t) ≤ 0 on J , i.e. α ≤ Aα. Similar arguments may show that Aβ ≤ β. To prove (ii), take y1 = Au 1 , y2 = Au 2 , where u 1 ≤ u 2 on J and u 1 , u 2 ∈ [α, β]. Set m = y1 − y2 , employing (H2 ) and (H3 ), we achieve m 0 (t) = y10 (t) − y20 (t) = [−M y1 (t) − (£y1 )(t) + σu 1 (t)] − [−M y2 (t) − (£y2 )(t) + σu 2 (t)] = −Mm(t) − (£m)(t) + [(Qu 1 )(t) − (Qu 2 )(t) + M(u 1 (t) − u 2 (t)) + (£(u 1 − u 2 ))(t)] ≤ −Mm(t) − (£m)(t), t ∈ J, m(0) = y1 (0) − y2 (0) 1 M2 =− g(u 1 (0), u 1 (T )) + (y1 (T ) − u 1 (T )) + u 1 (0) M1 M1 1 M2 − − g(u 2 (0), u 2 (T )) + (y2 (T ) − u 2 (T )) + u 2 (0) M1 M1 M2 1 = [y1 (T ) − y2 (T )] + [g(u 2 (0), u 2 (T )) − g(u 1 (0), u 1 (T ))] M1 M1 M2 −(u 2 (0) − u 1 (0)) + [u 2 (T ) − u 1 (T )] M1 M2 ≤ m(T ). M1 By virtue of Lemma 2.1, one has m(t) ≤ 0 on J , i.e. Au 1 ≤ Au 2 . From (i) and (ii), one knows that α ≤ Aα ≤ Aβ ≤ β. Now, define the sequences {αn (t)}, {βn (t)} with α0 = α, β0 = β such that αn+1 = Aαn , βn+1 = Aβn , n = 1, 2, . . . . Due to (i) and (ii), one achieves α0 ≤ α1 ≤ α2 ≤ · · · ≤ αn ≤ · · · ≤ βn ≤ · · · ≤ β2 ≤ β1 ≤ β0
on J.
Apparently, αn , βn (n = 1, 2, . . .) satisfy 0 αn (t) = −Mαn − (£αn )(t) + (Qαn−1 )(t) + Mαn−1 (t) + (£αn−1 )(t), g(αn−1 (0), αn−1 (T )) + M1 (αn (0) − αn−1 (0)) − M2 (αn (T ) − αn−1 (T )) = 0 and
βn0 (t) = −Mβn − (£βn )(t) + (Qβn−1 )(t) + Mβn−1 (t) + (£βn−1 )(t), g(βn−1 (0), βn−1 (T )) + M1 (βn (0) − βn−1 (0)) − M2 (βn (T ) − βn−1 (T )) = 0.
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Consequently, there exist x∗ and x ∗ such that limn→∞ αn (t) = x∗ (t), limn→∞ βn (t) = x ∗ (t) uniformly on J . Apparently, x∗ , x ∗ are the solutions of problem (1.1). To show x∗ , x ∗ are extremal solutions of problem (1.1), let x(t) be any solution of (1.1) such that α(t) ≤ x(t) ≤ β(t). Assume that there exists a positive integer n such that αn (t) ≤ x(t) ≤ βn (t) on J . Based on the monotonically nondecreasing property of A, it then follows that αn+1 = Aαn ≤ Ax = x, i.e. αn+1 (t) ≤ x(t) on J . Similarly, one derives x(t) ≤ βn+1 (t) on J . Since α0 (t) ≤ x(t) ≤ β0 (t) on J , by induction we see that αn (t) ≤ x(t) ≤ βn (t) on J for every n. Therefore x∗ (t) ≤ x(t) ≤ x ∗ (t) on J as n → ∞. The proof is then finished. 4. Weakly coupled lower and upper solutions Definition 4.1. Functions α, β ∈ Ω are called weakly coupled lower and upper solutions of problem (1.1) if α 0 (t) ≤ (Qα)(t),
g(β(0), α(T )) ≤ 0
and β 0 (t) ≥ (Qβ)(t)
g(α(0), β(T )) ≥ 0.
Definition 4.2. A pair (U, V ), U, V ∈ Ω , is called a weakly coupled quasi-solution of problem (1.1) if for t ∈ J , there are U 0 (t) = (QU )(t),
g(V (0), U (T )) = 0
V 0 (t) = (QV )(t),
g(U (0), V (T )) = 0.
and
Definition 4.3. A weakly coupled quasi-solution (ρ, γ ), ρ, γ ∈ Ω , is called weakly coupled minimal and maximal quasi-solution of problem (1.1), if for any weakly coupled quasi-solution (U, V ) of (1.1), we have ρ(t) ≤ U (t), V (t) ≤ γ (t) on J . Theorem 4.1. Assume condition (H2 ) and (2.8) hold, let Q ∈ C[E, E]. In addition, suppose that (H4 ) α, β ∈ Ω are weakly coupled lower and upper solutions such that α(t) ≤ β(t); (H5 ) the function g(x, y) ∈ C(R2 , R) is nondecreasing in the first variable and g(x, y1 ) − g(x, y2 ) ≤ −M2 (y1 − y2 )
if α(T ) ≤ y2 ≤ y1 ≤ β(T ),
where M1 ≥ M2 > 0. Then there exist monotone sequences {αn (t)} and {βn (t)} with α0 = α, β0 = β, such that lim αn (t) = x∗ (t),
n→∞
lim βn (t) = x ∗ (t)
n→∞
uniformly on J , and (x∗ (t), x ∗ (t)) is the weakly coupled minimal and maximal quasi-solution of problem (1.1), respectively, such that α(t) ≤ x∗ ≤ x ∗ ≤ β(t) on J. Proof. Let 0 αn (t) = −Mαn − (£αn )(t) + (Qαn−1 )(t) + Mαn−1 (t) + (£αn−1 )(t), g(βn−1 (0), αn−1 (T )) + M1 (αn (0) − αn−1 (0)) − M2 (αn (T ) − αn−1 (T )) = 0
(4.1)
and
βn0 (t) = −Mβn − (£βn )(t) + (Qβn−1 )(t) + Mβn−1 (t) + (£βn−1 )(t), g(αn−1 (0), βn−1 (T )) + M1 (βn (0) − βn−1 (0)) − M2 (βn (T ) − βn−1 (T )) = 0
for n = 1, 2, . . . , where α0 = α, β0 = β. First, we show that α0 ≤ α1 ≤ β1 ≤ β0 .
(4.2)
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Set m = α0 − α1 , then owe to (H4 ), one arrives at m 0 (t) = α00 (t) − α10 (t) ≤ Qα0 (t) − [−Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t)] ≤ −Mm(t) − (£m)(t),
t ∈ J,
m(0) = α0 (0) − α1 (0) M2 1 g(β0 (0), α0 (T )) + (α1 (T ) − α0 (T )) + α0 (0) = α0 (0) − − M1 M1 1 M2 = g(β0 (0), α0 (T )) + (α0 (T ) − α1 (T )) M1 M1 M2 ≤ m(T ). M1 It then follows from (3.1) (or (3.1)0 ) and Lemma 2.1 that m(t) ≤ 0, that is, α0 (t) ≤ α1 (t) for t ∈ J . Similarly, we may obtain β1 (t) ≤ β0 (t) on J . Next, take m = α1 − β1 , by way of (H2 ), one attains m 0 (t) = α10 (t) − β10 (t) = −Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t) − [−Mβ1 (t) − (£β1 (t)) + (Qβ0 )(t) + Mβ0 (t) + (£β0 )(t)] ≤ −Mm(t) − (£m)(t),
t ∈ J,
noticing α0 ≤ β0 and (H5 ), we derive m(0) = α1 (0) − β1 (0) 1 M2 =− g(β0 (0), α0 (T )) + (α1 (T ) − α0 (T )) + α0 (0) M1 M1 1 M2 − − g(α0 (0), β0 (T )) + (β1 (T ) − β0 (T )) + β0 (0) M1 M1 1 M2 ≤ [g(α0 (0), β0 (T )) − g(β0 (0), α0 (T ))] + [α1 (T ) − β1 (T ) − (α0 (T ) − β0 (T ))] M1 M1 1 M2 = [g(α0 (0), β0 (T )) − g(α0 (0), α0 (T ))] + [β0 (T ) − α0 (T )] M1 M1 M2 1 [g(α0 (0), α0 (T )) − g(β0 (0), α0 (T ))] + [α1 (T ) − β1 (T )] + M1 M1 M2 ≤ m(T ). M1 (3.1) (or (3.1)0 ) and Lemma 2.1 then yield m(t) ≤ 0 on J , i.e., α1 ≤ β1 . In the following, we shall show that α1 , β1 are the weakly coupled lower and upper solutions of (1.1). Following (H2 ) and α0 ≤ α1 , we achieve α10 (t) = (Qα1 )(t) − (Qα1 )(t) − Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t) = (Qα1 )(t) + [(Qα0 )(t) − (Qα1 )(t) + M(α0 (t) − α1 (t)) + (£(α0 − α1 ))(t)] ≤ (Qα1 )(t),
(4.3)
and by means of the fact α0 ≤ α1 , β1 ≤ β0 , (H4 ) and (H5 ), one reaches g(β1 (0), α1 (T )) = g(β1 (0), α1 (T )) − g(β0 (0), α0 (T )) − M1 (α1 (0) − α0 (0)) + M2 (α1 (T ) − α0 (T )) ≤ g(β1 (0), α1 (T )) − g(β1 (0), α0 (T )) + M2 (α1 (T ) − α0 (T )) + g(β1 (0), α0 (T )) − g(β0 (0), α0 (T )) ≤ 0.
(4.4)
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Similarly, we may get β10 (t) ≥ (Qβ1 )(t),
g(α1 (0), β1 (T )) ≥ 0.
(4.5)
(4.3)–(4.5) imply that α1 , β1 are weakly coupled lower and upper solutions of (1.1). Employing the mathematical induction, one may show that there exist x∗ and x ∗ such that limn→∞ αn (t) = x∗ (t), limn→∞ βn (t) = x ∗ (t) uniformly on J . Obviously, x∗ , x ∗ are fulfilling that x∗0 (t) = (Qx∗ )(t),
g(x ∗ , x∗ ) = 0,
x ∗0 (t) = (Qx ∗ )(t),
g(x∗ , x ∗ ) = 0.
This means that the pair (x∗ , x ∗ ) is a weakly coupled quasi-solution of problem (1.1). It remains to demonstrate x∗ , x ∗ are the weakly coupled minimal and maximal quasi-solutions of problem (1.1). Let (u, v) be any weakly coupled quasi-solutions of (1.1) such that u, v ∈ [α, β]. Suppose there exist a positive integer n such that αn (t) ≤ u, v ≤ βn (t), for t ∈ J . Putting m(t) = αn+1 (t) − u(t), then for t ∈ J and utilizing αn ≤ u and (H2 ), one arrives at 0 m 0 (t) = αn+1 (t) − u 0 (t)
= −Mαn+1 (t) − (£αn+1 )(t) + (Qαn )(t) + Mαn (t) + (£αn )(t) − (Qu)(t) = (Qαn )(t) − (Qu)(t) + M(αn (t) − u(t)) + (£(αn − u))(t) − Mαn+1 (t) − (£αn+1 )(t) + Mu(t) + (£u)(t) ≤ −Mm(t) − (£m)(t),
t ∈ J,
and using the fact g(u(0), u(T )) = 0 and (H5 ), we have m(0) = αn+1 (0) − u(0) M2 1 g(βn (0), αn (T )) + (αn+1 (T ) − αn (T )) + αn (0) − u(0) =− M1 M1 1 M2 1 g(βn (0), αn (T )) + g(u(0), u(T )) + (αn+1 (T ) − αn (T )) ≤− M1 M1 M1 1 1 [g(u(0), αn (T )) − g(βn (0), αn (T ))] − [g(u(0), αn (T )) − g(u(0), u(T ))] = M1 M1 M2 [αn+1 (T ) − αn (T )] + M1 M2 M2 (u(T ) − αn (T )) + (αn+1 (T ) − αn (T )) ≤− M1 M1 M2 ≤ m(T ). M1 (3.1) (or (3.1)0 ) and Lemma 2.1 then leads to m(t) ≤ 0 for t ∈ J , i.e., αn+1 ≤ u. Similarly, we obtain v ≤ βn+1 . Following the induction, one has αn (t) ≤ u(t), v(t) ≤ βn (t) for all t ∈ J and any n, which implies x∗ ≤ u, v ≤ x ∗ . This completes the proof. References [1] L. Chen, J. Sun, Nonlinear boundary value problem of first order impulsive functional differential equations, J. Math. Anal. Appl. 318 (2006) 726–741. [2] C. Corduneanu, Some existence results for functional equations with causal operators, Nonlinear Anal. 47 (2001) 709–716. [3] Z. Drici, F.A. McRae, J.V. Devi, Differential equations with causal operators in a Banach space, Nonlinear Anal. 62 (2005) 301–313. [4] Z. Drici, F.A. McRae, J.V. Devi, Monotone iterative technique for periodic boundary value problems with causal operators, Nonlinear Anal. 64 (2006) 1271–1277. [5] D. Franco, J.J. Nieto, D. O’Regan, Existence of solutions for first order ordinary differential equations with nonlinear boundary conditions, Appl. Math. Comput. 153 (2004) 793–802. [6] T. Jankowski, Advanced differential equations with nonlinear boundary conditions, J. Math. Anal. Appl. 304 (2005) 490–503.
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Mathematical and Computer Modelling 48 (2008) 859–866 www.elsevier.com/locate/mcm
Differential equations involving causal operators with nonlinear periodic boundary conditionsI Fengjie Geng School of Information Engineering, China University of Geosciences (Beijing), Beijing, 100083, China Received 5 September 2006; received in revised form 26 October 2007; accepted 2 November 2007
Abstract The notion of causal operators is extended to periodic boundary value problems with nonlinear boundary conditions in this paper. By utilizing the monotone iterative technique and the method of lower and upper solutions (resp. weakly coupled lower and upper solutions), we establish the existence of the extremal solutions (resp. weakly coupled extremal quasi-solutions) for nonlinear periodic boundary value problems with causal operators. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Lower and upper solutions; Causal operators; Weakly coupled lower and upper solutions; Weakly coupled quasi-solutions; Nonlinear periodic boundary conditions
1. Introduction In this paper, we deal with the following differential equation with a causal operator: x 0 (t) = (Qx)(t),
t ∈ J := [0, T ],
g(x(0), x(T )) = 0,
(1.1)
where Q ∈ C[E, E] is a causal operator, namely, a nonanticipative operator. Some authors have focused their interest on differential equations with causal operators recently, because its theory has the powerful quality of unifying ordinary differential equations, integro differential equations, differential equations with finite or infinite delay, Volterra integral equations and neutral functional equations. Readers are referred to the papers [2–4] and references cited therein. In 2005, Drici et al. [3] extended differential equations with causal operators to the framework of an arbitrary Banach space and proved the basic results such as existence, uniqueness and global existence. Immediately after it, they discussed in [4] the periodic boundary value problems with a causal operator u 0 (t) = (Qu)(t),
u(0) = u(T ),
I Supported by National Natural Science Foundation of China (] 10671069).
E-mail address: gengfengjie [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.11.009
(1.2)
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where t ∈ J , Q ∈ C[E, E] is a causal operator. The authors developed the monotone iterative technique to obtain the existence of extremal solutions for (1.2). At the same time, periodic boundary value problems with nonlinear boundary conditions have attracted much attention, we may see [1,5,6]. Inspired by the above results, in this paper, we extend the notion of causal operators to the periodic boundary value problems with nonlinear boundary conditions. First, some comparison principles and the existence and uniqueness of the solutions for the first order linear differential equations with linear boundary conditions are presented. Next, by utilizing the monotone iterative technique and the method of lower and upper solutions, we establish the existence of extremal solutions of the problem (1.1). At last, using the notion of weakly coupled lower and upper solutions, we testify the existence of the weakly coupled minimal and maximal quasisolutions of the problem (1.1). The results obtained in this paper generalized those in [4]. 2. Preliminaries In this section, we present some definitions and lemmas which help to prove our main results. Let E = C[J, R], where J = [0, T ]. We define ku − vk(t) = max |u(t) − v(t)|. t0 ≤s≤t
Take Ω = E
T
C 1 [J, R]. A function x ∈ Ω is said to be a solution of problem (1.1) if it satisfies (1.1).
Definition 2.1. Suppose that Q ∈ C[E, E], then Q is said to be a causal map or a nonanticipative map if u(s) = v(s), t0 ≤ s ≤ t ≤ T , where u, v ∈ E, then (Qu)(s) = (Qv)(s),
t0 ≤ s ≤ t.
Lemma 2.1. Let m ∈ Ω be such that m 0 (t) ≤ −Mm(t) − (£m)(t),
m(0) ≤ λm(T ),
(2.1)
where 0 < λ ≤ 1, M ≥ 0 and £ ∈ C[E, E] is a positive linear operator, that is, £m ≥ 0 whenever m ≥ 0. Then, m(t) ≤ 0 for t ∈ J provided Z T e M T (£e−M )(t)dt ≤ λe−M T and 0 < λe−M T ≤ 1;
(2.2)
0
or T (M + (£1)(T )) ≤ λ
and
0 < λ ≤ 1.
(2.20 )
Proof. Let v(t) = e Mt m(t), then v ∈ Ω and v 0 (t) ≤ −e Mt (£e−M v)(t),
v(0) ≤ λe−M T v(T ).
(2.3)
Obviously, v(t) ≤ 0 implies m(t) ≤ 0. So, it suffices to show v(t) ≤ 0 for any t ∈ J . Suppose on the contrary, v(t) > 0 for some t ∈ J . Then there are two cases: (i) there exists t¯ ∈ J such that v(t¯) > 0 and v(t) ≥ 0 for all t ∈ J ; (ii) there exist t ∗ and t∗ such that v(t ∗ ) > 0 and v(t∗ ) < 0. First, suppose (2.2) holds. In case (i), (2.3) implies v 0 (t) ≤ 0 on J and v(t) is nonincreasing on J . If λe−M T = 1, v(0) ≤ v(T ) shows that v(t) ≡ c, then v 0 (t) = 0. On the other hand, we have 0 = v 0 (t¯) ≤ −e M t¯(£e−M v)(t¯) < 0 since v(t¯) > 0, which is a contradiction. If 0 < λe−M T < 1, then v(T ) ≤ v(0) ≤ λe−M T v(T ), which is also a contradiction. Now turning to case (ii). Let mint∈J v(t) = −γ , then γ > 0. Without loss of generality, we may suppose v(t∗ ) = −γ . First suppose t∗ < t ∗ , integrating the first inequality of (2.3) from t∗ to t ∗ one attains Z t∗ Z T Z t∗ v(t ∗ ) − v(t∗ ) = v 0 (t)dt ≤ −e Mt (£e−M v)(t) ≤ γ e M T (£e−M )(t)dt, t∗
t∗
0
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and hence 0 < v(t ) ≤ −γ + γ ∗
T
Z
e
MT
(£e
−M
)(t)dt ≤ −γ
T
Z 1−
e
0
MT
(£e
−M
)(t)dt ,
0
which yields Z T e M T (£e−M )(t)dt > 1. 0
A contradiction is then elicited due to (2.2). If t∗ > t ∗ , we have from (2.3) Z T v(T ) ≤ v(t∗ ) + γ e M T (£e−M )(t)dt
(2.4)
t∗
and v(t ) ≤ v(0) + γ ∗
Z
t∗
e M T (£e−M )(t)dt,
(2.5)
0
based on the fact v(0) ≤ λe−M T v(T ) and 0 < λe−M T ≤ 1, it then follows from (2.2), (2.4) and (2.5) that Z T Z t∗ ∗ −M T MT −M 0 < v(t ) ≤ λe −γ + γ e (£e )(t)dt + γ e M T (£e−M )(t)dt ≤ −γ λe−M T + γ ≤ −γ (λe−M T −
Z
e M T (£e−M )(t)dt + γ
t∗ T
Z
0
t∗ T
t∗
Z
e M T (£e−M )(t)dt
0
e M T (£e−M )(t)dt) ≤ 0,
0
which is a contradiction. 0 If (2.2) holds, by using (2.1) directly instead of (2.3), we may derive contradiction in a similar way. This completes the proof. Consider the following problem y 0 (t) = −M y(t) − (£y)(t) + σu (t),
t ∈ J, g(u(0), u(T )) + M1 (y(0) − u(0)) − M2 (y(T ) − u(T )) = 0,
(2.6)
where σu (t) = (Qu)(t) + Mu(t) + (£a)(t). Lemma 2.2. y ∈ Ω is a solution of problem (2.6) if and only if y is a solution of the following integral equation Z T e−M T Bu y(t) = + G(t, s)[−(£y)(s) + σu (s)]ds, t ∈ J, M1 − M2 e−M T 0 where Bu = −g(u(0), u(T )) + M1 u(0) − M2 u(T ), M, M1 , M2 are constants satisfying M ≥ 0, M1 6= M2 e−M T and M2 e−M(T +t−s) + e−M(t−s) , 0 ≤ s < t ≤ T ; −M T M − M e 1 2 (2.7) G(t, s) = M2 −M(T +t−s) e , 0 ≤ t ≤ s ≤ T. M1 − M2 e−M T The result may be obtained by a simple computation. M1 M2 e−M T Obviously, kG(t, s)k = max{| M −M |, | |}. In the remainder of the paper, we denote τ = −M T e M −M e−M T 1
kG(t, s)k = max{| M
M1 −M T 1 −M2 e
2 −M T
2e |, | M M−M 1
2
|}. e−M T
1
2
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Lemma 2.3. If M ≥ 0, M1 6= M2 e−M T and τ k£kT < 1.
(2.8)
Then problem (2.6) has a unique solution y ∈ Ω . Proof. Define an operator A : Ω → Ω by Z T e−M T Bu (Ay)(t) = G(t, s)[−(£y)(s) + σu (s)]ds, + M1 − M2 e−M T 0
t ∈ J.
For y1 , y2 ∈ Ω , one derives Z T |Ay1 − Ay2 | = G(t, s)(£(y2 − y1 ))(s)ds ≤ τ k£kT ky1 − y2 k. 0
Hence, (2.8) shows that A is a contraction. Banach fixed point theorem shows that A has a unique fixed point y ∗ such that Ay ∗ = y ∗ . It is clear that this fixed point is the solution of (2.6). The proof is then finished. 3. Main results In this section, we shall establish the existence of extremal solutions of problem (1.1). Definition 3.1. Functions α, β ∈ Ω are said to be lower and upper solutions of problem (1.1), respectively, if α 0 (t) ≤ (Qα)(t),
g(α(0), α(T )) ≤ 0
and β 0 (t) ≥ (Qβ)(t)
g(β(0), β(T )) ≥ 0.
Now we state our theorem. Theorem 3.1. Let (2.8) hold and Q ∈ C[E, E]. Suppose the following conditions hold, (H1 ) the functions α, β ∈ Ω are lower and upper solutions of problem (1.1) respectively, such that α(t) ≤ β(t) for t ∈ J ; (H2 ) whenever α(t) ≤ x2 (t) ≤ x1 (t) ≤ β(t), Q satisfies (Qx1 )(t) − (Qx2 )(t) ≥ −M(x1 (t) − x2 (t)) − (£(x1 − x2 ))(t),
t ∈ J,
where M > 0 is a constant and £ ∈ C[E, E] is a positive linear operator such that Z T M2 −M T e M T (£e−M )(t)dt ≤ e M 1 0 or T (M + (£1)(T )) ≤
M2 M1
(3.1)
(3.10 )
hold and M1 ≥ M2 > 0; (H3 ) the function g(x, y) ∈ C(R2 , R) satisfies g(x1 , x¯1 ) − g(x2 , x¯2 ) ≤ M1 (x1 − x2 ) − M2 (x¯1 − x¯2 ) whenever α(0) ≤ x2 ≤ x1 ≤ β(0), α(T ) ≤ x¯2 ≤ x¯1 ≤ β(T ), where M1 ≥ M2 > 0. Then there exist monotone sequences {αn (t)} and {βn (t)} with α0 = α, β0 = β, such that lim αn (t) = x∗ (t),
n→∞
lim βn (t) = x ∗ (t)
n→∞
uniformly on J , and x∗ (t), x ∗ (t) are the minimal and the maximal solutions of problem (1.1), respectively, such that α(t) ≤ x∗ ≤ x ∗ ≤ β(t) on J.
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Proof. For any u ∈ [α, β], where [α, β] = {u ∈ Ω | α(t) ≤ u(t) ≤ β(t), t ∈ J }. Consider the problem (2.6) with σu (t) = (Qu)(t) + Mu(t) + (£u)(t). Noticing that (2.8) holds, then by Lemma 2.3 one may see that problem (2.6) has exactly one solution y ∈ Ω . Denote y(t) = Au(t), then the operator A has the following properties: (i) α ≤ Aα, Aβ ≤ β; (ii) A is monotonically nondecreasing in [α, β], i.e. for any u 1 , u 2 ∈ [α, β], u 1 ≤ u 2 implies Au 1 ≤ Au 2 . To prove (i), set m = α − α1 , where α1 = Aα. Owing to (H1 ), we acquire m 0 (t) = α 0 (t) − α10 (t) ≤ (Qα)(t) − [−Mα1 (t) − (£α1 )(t) + (Qα)(t) + Mα(t) + (£α)(t)] = −Mm(t) − (£m)(t), t ∈ J, and 1 M2 m(0) = α(0) − − g(α(0), α(T )) + (α1 (T ) − α(T )) + α(0) M1 M1 M2 1 g(α(0), α(T )) + [α(T ) − α1 (T )] = M1 M1 M2 ≤ m(T ). M1 Based on the fact M1 ≥ M2 > 0, (3.1) (or (3.1)0 ) and Lemma 2.1, we get m(t) ≤ 0 on J , i.e. α ≤ Aα. Similar arguments may show that Aβ ≤ β. To prove (ii), take y1 = Au 1 , y2 = Au 2 , where u 1 ≤ u 2 on J and u 1 , u 2 ∈ [α, β]. Set m = y1 − y2 , employing (H2 ) and (H3 ), we achieve m 0 (t) = y10 (t) − y20 (t) = [−M y1 (t) − (£y1 )(t) + σu 1 (t)] − [−M y2 (t) − (£y2 )(t) + σu 2 (t)] = −Mm(t) − (£m)(t) + [(Qu 1 )(t) − (Qu 2 )(t) + M(u 1 (t) − u 2 (t)) + (£(u 1 − u 2 ))(t)] ≤ −Mm(t) − (£m)(t), t ∈ J, m(0) = y1 (0) − y2 (0) 1 M2 =− g(u 1 (0), u 1 (T )) + (y1 (T ) − u 1 (T )) + u 1 (0) M1 M1 1 M2 − − g(u 2 (0), u 2 (T )) + (y2 (T ) − u 2 (T )) + u 2 (0) M1 M1 M2 1 = [y1 (T ) − y2 (T )] + [g(u 2 (0), u 2 (T )) − g(u 1 (0), u 1 (T ))] M1 M1 M2 −(u 2 (0) − u 1 (0)) + [u 2 (T ) − u 1 (T )] M1 M2 ≤ m(T ). M1 By virtue of Lemma 2.1, one has m(t) ≤ 0 on J , i.e. Au 1 ≤ Au 2 . From (i) and (ii), one knows that α ≤ Aα ≤ Aβ ≤ β. Now, define the sequences {αn (t)}, {βn (t)} with α0 = α, β0 = β such that αn+1 = Aαn , βn+1 = Aβn , n = 1, 2, . . . . Due to (i) and (ii), one achieves α0 ≤ α1 ≤ α2 ≤ · · · ≤ αn ≤ · · · ≤ βn ≤ · · · ≤ β2 ≤ β1 ≤ β0
on J.
Apparently, αn , βn (n = 1, 2, . . .) satisfy 0 αn (t) = −Mαn − (£αn )(t) + (Qαn−1 )(t) + Mαn−1 (t) + (£αn−1 )(t), g(αn−1 (0), αn−1 (T )) + M1 (αn (0) − αn−1 (0)) − M2 (αn (T ) − αn−1 (T )) = 0 and
βn0 (t) = −Mβn − (£βn )(t) + (Qβn−1 )(t) + Mβn−1 (t) + (£βn−1 )(t), g(βn−1 (0), βn−1 (T )) + M1 (βn (0) − βn−1 (0)) − M2 (βn (T ) − βn−1 (T )) = 0.
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Consequently, there exist x∗ and x ∗ such that limn→∞ αn (t) = x∗ (t), limn→∞ βn (t) = x ∗ (t) uniformly on J . Apparently, x∗ , x ∗ are the solutions of problem (1.1). To show x∗ , x ∗ are extremal solutions of problem (1.1), let x(t) be any solution of (1.1) such that α(t) ≤ x(t) ≤ β(t). Assume that there exists a positive integer n such that αn (t) ≤ x(t) ≤ βn (t) on J . Based on the monotonically nondecreasing property of A, it then follows that αn+1 = Aαn ≤ Ax = x, i.e. αn+1 (t) ≤ x(t) on J . Similarly, one derives x(t) ≤ βn+1 (t) on J . Since α0 (t) ≤ x(t) ≤ β0 (t) on J , by induction we see that αn (t) ≤ x(t) ≤ βn (t) on J for every n. Therefore x∗ (t) ≤ x(t) ≤ x ∗ (t) on J as n → ∞. The proof is then finished. 4. Weakly coupled lower and upper solutions Definition 4.1. Functions α, β ∈ Ω are called weakly coupled lower and upper solutions of problem (1.1) if α 0 (t) ≤ (Qα)(t),
g(β(0), α(T )) ≤ 0
and β 0 (t) ≥ (Qβ)(t)
g(α(0), β(T )) ≥ 0.
Definition 4.2. A pair (U, V ), U, V ∈ Ω , is called a weakly coupled quasi-solution of problem (1.1) if for t ∈ J , there are U 0 (t) = (QU )(t),
g(V (0), U (T )) = 0
V 0 (t) = (QV )(t),
g(U (0), V (T )) = 0.
and
Definition 4.3. A weakly coupled quasi-solution (ρ, γ ), ρ, γ ∈ Ω , is called weakly coupled minimal and maximal quasi-solution of problem (1.1), if for any weakly coupled quasi-solution (U, V ) of (1.1), we have ρ(t) ≤ U (t), V (t) ≤ γ (t) on J . Theorem 4.1. Assume condition (H2 ) and (2.8) hold, let Q ∈ C[E, E]. In addition, suppose that (H4 ) α, β ∈ Ω are weakly coupled lower and upper solutions such that α(t) ≤ β(t); (H5 ) the function g(x, y) ∈ C(R2 , R) is nondecreasing in the first variable and g(x, y1 ) − g(x, y2 ) ≤ −M2 (y1 − y2 )
if α(T ) ≤ y2 ≤ y1 ≤ β(T ),
where M1 ≥ M2 > 0. Then there exist monotone sequences {αn (t)} and {βn (t)} with α0 = α, β0 = β, such that lim αn (t) = x∗ (t),
n→∞
lim βn (t) = x ∗ (t)
n→∞
uniformly on J , and (x∗ (t), x ∗ (t)) is the weakly coupled minimal and maximal quasi-solution of problem (1.1), respectively, such that α(t) ≤ x∗ ≤ x ∗ ≤ β(t) on J. Proof. Let 0 αn (t) = −Mαn − (£αn )(t) + (Qαn−1 )(t) + Mαn−1 (t) + (£αn−1 )(t), g(βn−1 (0), αn−1 (T )) + M1 (αn (0) − αn−1 (0)) − M2 (αn (T ) − αn−1 (T )) = 0
(4.1)
and
βn0 (t) = −Mβn − (£βn )(t) + (Qβn−1 )(t) + Mβn−1 (t) + (£βn−1 )(t), g(αn−1 (0), βn−1 (T )) + M1 (βn (0) − βn−1 (0)) − M2 (βn (T ) − βn−1 (T )) = 0
for n = 1, 2, . . . , where α0 = α, β0 = β. First, we show that α0 ≤ α1 ≤ β1 ≤ β0 .
(4.2)
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Set m = α0 − α1 , then owe to (H4 ), one arrives at m 0 (t) = α00 (t) − α10 (t) ≤ Qα0 (t) − [−Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t)] ≤ −Mm(t) − (£m)(t),
t ∈ J,
m(0) = α0 (0) − α1 (0) M2 1 g(β0 (0), α0 (T )) + (α1 (T ) − α0 (T )) + α0 (0) = α0 (0) − − M1 M1 1 M2 = g(β0 (0), α0 (T )) + (α0 (T ) − α1 (T )) M1 M1 M2 ≤ m(T ). M1 It then follows from (3.1) (or (3.1)0 ) and Lemma 2.1 that m(t) ≤ 0, that is, α0 (t) ≤ α1 (t) for t ∈ J . Similarly, we may obtain β1 (t) ≤ β0 (t) on J . Next, take m = α1 − β1 , by way of (H2 ), one attains m 0 (t) = α10 (t) − β10 (t) = −Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t) − [−Mβ1 (t) − (£β1 (t)) + (Qβ0 )(t) + Mβ0 (t) + (£β0 )(t)] ≤ −Mm(t) − (£m)(t),
t ∈ J,
noticing α0 ≤ β0 and (H5 ), we derive m(0) = α1 (0) − β1 (0) 1 M2 =− g(β0 (0), α0 (T )) + (α1 (T ) − α0 (T )) + α0 (0) M1 M1 1 M2 − − g(α0 (0), β0 (T )) + (β1 (T ) − β0 (T )) + β0 (0) M1 M1 1 M2 ≤ [g(α0 (0), β0 (T )) − g(β0 (0), α0 (T ))] + [α1 (T ) − β1 (T ) − (α0 (T ) − β0 (T ))] M1 M1 1 M2 = [g(α0 (0), β0 (T )) − g(α0 (0), α0 (T ))] + [β0 (T ) − α0 (T )] M1 M1 M2 1 [g(α0 (0), α0 (T )) − g(β0 (0), α0 (T ))] + [α1 (T ) − β1 (T )] + M1 M1 M2 ≤ m(T ). M1 (3.1) (or (3.1)0 ) and Lemma 2.1 then yield m(t) ≤ 0 on J , i.e., α1 ≤ β1 . In the following, we shall show that α1 , β1 are the weakly coupled lower and upper solutions of (1.1). Following (H2 ) and α0 ≤ α1 , we achieve α10 (t) = (Qα1 )(t) − (Qα1 )(t) − Mα1 (t) − (£α1 )(t) + (Qα0 )(t) + Mα0 (t) + (£α0 )(t) = (Qα1 )(t) + [(Qα0 )(t) − (Qα1 )(t) + M(α0 (t) − α1 (t)) + (£(α0 − α1 ))(t)] ≤ (Qα1 )(t),
(4.3)
and by means of the fact α0 ≤ α1 , β1 ≤ β0 , (H4 ) and (H5 ), one reaches g(β1 (0), α1 (T )) = g(β1 (0), α1 (T )) − g(β0 (0), α0 (T )) − M1 (α1 (0) − α0 (0)) + M2 (α1 (T ) − α0 (T )) ≤ g(β1 (0), α1 (T )) − g(β1 (0), α0 (T )) + M2 (α1 (T ) − α0 (T )) + g(β1 (0), α0 (T )) − g(β0 (0), α0 (T )) ≤ 0.
(4.4)
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Similarly, we may get β10 (t) ≥ (Qβ1 )(t),
g(α1 (0), β1 (T )) ≥ 0.
(4.5)
(4.3)–(4.5) imply that α1 , β1 are weakly coupled lower and upper solutions of (1.1). Employing the mathematical induction, one may show that there exist x∗ and x ∗ such that limn→∞ αn (t) = x∗ (t), limn→∞ βn (t) = x ∗ (t) uniformly on J . Obviously, x∗ , x ∗ are fulfilling that x∗0 (t) = (Qx∗ )(t),
g(x ∗ , x∗ ) = 0,
x ∗0 (t) = (Qx ∗ )(t),
g(x∗ , x ∗ ) = 0.
This means that the pair (x∗ , x ∗ ) is a weakly coupled quasi-solution of problem (1.1). It remains to demonstrate x∗ , x ∗ are the weakly coupled minimal and maximal quasi-solutions of problem (1.1). Let (u, v) be any weakly coupled quasi-solutions of (1.1) such that u, v ∈ [α, β]. Suppose there exist a positive integer n such that αn (t) ≤ u, v ≤ βn (t), for t ∈ J . Putting m(t) = αn+1 (t) − u(t), then for t ∈ J and utilizing αn ≤ u and (H2 ), one arrives at 0 m 0 (t) = αn+1 (t) − u 0 (t)
= −Mαn+1 (t) − (£αn+1 )(t) + (Qαn )(t) + Mαn (t) + (£αn )(t) − (Qu)(t) = (Qαn )(t) − (Qu)(t) + M(αn (t) − u(t)) + (£(αn − u))(t) − Mαn+1 (t) − (£αn+1 )(t) + Mu(t) + (£u)(t) ≤ −Mm(t) − (£m)(t),
t ∈ J,
and using the fact g(u(0), u(T )) = 0 and (H5 ), we have m(0) = αn+1 (0) − u(0) M2 1 g(βn (0), αn (T )) + (αn+1 (T ) − αn (T )) + αn (0) − u(0) =− M1 M1 1 M2 1 g(βn (0), αn (T )) + g(u(0), u(T )) + (αn+1 (T ) − αn (T )) ≤− M1 M1 M1 1 1 [g(u(0), αn (T )) − g(βn (0), αn (T ))] − [g(u(0), αn (T )) − g(u(0), u(T ))] = M1 M1 M2 [αn+1 (T ) − αn (T )] + M1 M2 M2 (u(T ) − αn (T )) + (αn+1 (T ) − αn (T )) ≤− M1 M1 M2 ≤ m(T ). M1 (3.1) (or (3.1)0 ) and Lemma 2.1 then leads to m(t) ≤ 0 for t ∈ J , i.e., αn+1 ≤ u. Similarly, we obtain v ≤ βn+1 . Following the induction, one has αn (t) ≤ u(t), v(t) ≤ βn (t) for all t ∈ J and any n, which implies x∗ ≤ u, v ≤ x ∗ . This completes the proof. References [1] L. Chen, J. Sun, Nonlinear boundary value problem of first order impulsive functional differential equations, J. Math. Anal. Appl. 318 (2006) 726–741. [2] C. Corduneanu, Some existence results for functional equations with causal operators, Nonlinear Anal. 47 (2001) 709–716. [3] Z. Drici, F.A. McRae, J.V. Devi, Differential equations with causal operators in a Banach space, Nonlinear Anal. 62 (2005) 301–313. [4] Z. Drici, F.A. McRae, J.V. Devi, Monotone iterative technique for periodic boundary value problems with causal operators, Nonlinear Anal. 64 (2006) 1271–1277. [5] D. Franco, J.J. Nieto, D. O’Regan, Existence of solutions for first order ordinary differential equations with nonlinear boundary conditions, Appl. Math. Comput. 153 (2004) 793–802. [6] T. Jankowski, Advanced differential equations with nonlinear boundary conditions, J. Math. Anal. Appl. 304 (2005) 490–503.
