On a periodic type boundary-value problem for first order linear functional differential equations

Nonimprovable sufficient conditions are established for unique solvability of the boundary-value problem u`(t) = l(u)(t) + q(t), u(a) = λu(b) + c, as well as for nonnegativeness of its solution, where l : C([a, b]; R) → L([a, b]; R) is a linear bounded operator, q ∈ L([a, b]; R), λ ∈ R+, and c ∈ R...

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Zitieren:On a periodic type boundary-value problem for first order linear functional differential equations / R. Hakl, A. Lomtatidze, J. Šremr // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 416-433. — Бібліогр.: 27 назв. — англ.

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Digital Library of Periodicals of National Academy of Sciences of Ukraine
id nasplib_isofts_kiev_ua-123456789-175810
record_format dspace
spelling Hakl, R.
Lomtatidze, A.
Šremr, J.
2021-02-02T19:46:37Z
2021-02-02T19:46:37Z
2002
On a periodic type boundary-value problem for first order linear functional differential equations / R. Hakl, A. Lomtatidze, J. Šremr // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 416-433. — Бібліогр.: 27 назв. — англ.
1562-3076
https://nasplib.isofts.kiev.ua/handle/123456789/175810
517.929
Nonimprovable sufficient conditions are established for unique solvability of the boundary-value problem u`(t) = l(u)(t) + q(t), u(a) = λu(b) + c, as well as for nonnegativeness of its solution, where l : C([a, b]; R) → L([a, b]; R) is a linear bounded operator, q ∈ L([a, b]; R), λ ∈ R+, and c ∈ R.
Знайдено достатнi умови, що не можуть бути полiпшенi, для однозначної розв’язностi граничної задачi u`(t) = l(u)(t) + q(t), u(a) = λu(b) + c, та невiд’ємностi її розв’язку, де l : C([a, b]; R) → L([a, b]; R) — неперервний лiнiйний оператор, q ∈ L([a, b]; R), λ ∈ R+ та c ∈ R.
For the first author this work was supported by the Grant No. 201/00/D058 of the Grant Agency of the Czech Republic, for the second and third authors by the Grant No. 201/99/0295 of the Grant Agency of the Czech Republic
en
Інститут математики НАН України
Нелінійні коливання
On a periodic type boundary-value problem for first order linear functional differential equations
Про граничну задачу періодичного типу для лінійних функціональних диференціальних рівнянь першого порядку
О краевой задаче периодического типа для линейных функциональных дифференциальных уравнений первого порядка
Article
published earlier
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
title On a periodic type boundary-value problem for first order linear functional differential equations
spellingShingle On a periodic type boundary-value problem for first order linear functional differential equations
Hakl, R.
Lomtatidze, A.
Šremr, J.
title_short On a periodic type boundary-value problem for first order linear functional differential equations
title_full On a periodic type boundary-value problem for first order linear functional differential equations
title_fullStr On a periodic type boundary-value problem for first order linear functional differential equations
title_full_unstemmed On a periodic type boundary-value problem for first order linear functional differential equations
title_sort on a periodic type boundary-value problem for first order linear functional differential equations
author Hakl, R.
Lomtatidze, A.
Šremr, J.
author_facet Hakl, R.
Lomtatidze, A.
Šremr, J.
publishDate 2002
language English
container_title Нелінійні коливання
publisher Інститут математики НАН України
format Article
title_alt Про граничну задачу періодичного типу для лінійних функціональних диференціальних рівнянь першого порядку
О краевой задаче периодического типа для линейных функциональных дифференциальных уравнений первого порядка
description Nonimprovable sufficient conditions are established for unique solvability of the boundary-value problem u`(t) = l(u)(t) + q(t), u(a) = λu(b) + c, as well as for nonnegativeness of its solution, where l : C([a, b]; R) → L([a, b]; R) is a linear bounded operator, q ∈ L([a, b]; R), λ ∈ R+, and c ∈ R. Знайдено достатнi умови, що не можуть бути полiпшенi, для однозначної розв’язностi граничної задачi u`(t) = l(u)(t) + q(t), u(a) = λu(b) + c, та невiд’ємностi її розв’язку, де l : C([a, b]; R) → L([a, b]; R) — неперервний лiнiйний оператор, q ∈ L([a, b]; R), λ ∈ R+ та c ∈ R.
issn 1562-3076
url https://nasplib.isofts.kiev.ua/handle/123456789/175810
citation_txt On a periodic type boundary-value problem for first order linear functional differential equations / R. Hakl, A. Lomtatidze, J. Šremr // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 416-433. — Бібліогр.: 27 назв. — англ.
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fulltext UDC 517. 929 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR FUNCTIONAL DIFFERENTIAL EQUATIONS* ПРО ГРАНИЧНУ ЗАДАЧУ ПЕРIОДИЧНОГО ТИПУ ДЛЯ ЛIНIЙНИХ ФУНКЦIОНАЛЬНИХ ДИФЕРЕНЦIАЛЬНИХ РIВНЯНЬ ПЕРШОГО ПОРЯДКУ R. Hakl Math. Inst., Czech Acad. Sci. Žižkova 22, 616 62 Brno, Czech Republic e-mail: hakl@ipm.cz A. Lomtatidze Masaryk University Janáčkovo nám. 2a, 662 95 Brno, Czech Republic e-mail: bacho@math.muni.cz J. Šremr Masaryk University Janáčkovo nám. 2a, 662 95 Brno, Czech Republic e-mail: sremr@math.muni.c Nonimprovable sufficient conditions are established for unique solvability of the boundary-value problem u′(t) = `(u)(t) + q(t), u(a) = λu(b) + c, as well as for nonnegativeness of its solution, where ` : C([a, b];R) → L([a, b];R) is a linear bounded operator, q ∈ L([a, b];R), λ ∈ R+, and c ∈ R. Знайдено достатнi умови, що не можуть бути полiпшенi, для однозначної розв’язностi гра- ничної задачi u′(t) = `(u)(t) + q(t), u(a) = λu(b) + c, та невiд’ємностi її розв’язку, де ` : C([a, b];R) → L([a, b];R) — неперервний лiнiйний оператор, q ∈ L([a, b];R), λ ∈ R+ та c ∈ R. Introduction. The following notation is used throughout the paper. R is the set of all real numbers, R+ = [0,+∞[, R− =]−∞, 0]. C([a, b];R) is the Banach space of continuous functions u : [a, b] → R with the norm ‖u‖C = max{|u(t)| : a ≤ t ≤ b}. C([a, b];R+) = {u ∈ C([a, b];R) : u(t) ≥ 0 for t ∈ [a, b]}. C̃([a, b];R) is the set of absolutely continuous functions u : [a, b] → R. L([a, b];R) is the Banach space of Lebesgue integrable functions p : [a, b] → R with the norm ‖p‖L = b∫ a |p(s)|ds. ∗ For the first author this work was supported by the Grant No. 201/00/D058 of the Grant Agency of the Czech Republic, for the second and third authors by the Grant No. 201/99/0295 of the Grant Agency of the Czech Republic. c© R. Hakl, A. Lomtatidze, J. Šremr, 2002 416 ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 417 L([a, b];D) = {p ∈ L([a, b];R) : p : [a, b] → D}, where D ⊆ R. Mab is the set of measurable functions τ : [a, b] → [a, b]. Lab is the set of linear bounded operators ` : C([a, b];R) → L([a, b];R). Pab is the set of linear operators ` ∈ Lab transforming the set C([a, b];R+) into the set L([a, b];R+). [x]+ = 1 2 (|x|+ x), [x]− = 1 2 (|x| − x). By a solution of the equation u′(t) = `(u)(t) + q(t), (0.1) where ` ∈ Lab and q ∈ L([a, b];R), we understand a function u ∈ C̃([a, b];R) satisfying the equation (0.1) almost everywhere in [a, b]. Consider the problem on the existence and uniqueness of a solution of (0.1) satisfying the boundary condition u(a) = λu(b) + c, (0.2) where λ ∈ R+, c ∈ R. The general boundary-value problems for functional differential equations have been studi- ed very intensively. There are a lot of general results (see, e.g., [1 – 27]), but still only a few effective criteria for the solvability of special boundary-value problems for functional differenti- al equations are known even in the linear case. In the present paper, we try to fill to some extent the existing gap in a certain way. More precisely, in Section 1 we give nonimprovable effecti- ve sufficient conditions for the unique solvability of the problem (0.1), (0.2) as well as for the nonnegativeness of a solution of that problem. Sections 2 and 3 are devoted respectively to the proofs of the main results and the examples verifying their optimality. All results will be concretized for the differential equation with deviating arguments, i.e., for the case where the equation (0.1) has the form u′(t) = p(t)u(τ(t))− g(t)u(µ(t)) + q(t), (0.3) where p, g ∈ L([a, b];R+), q ∈ L([a, b];R), and τ, µ ∈ Mab. The special cases of the discussed boundary-value problem are the Cauchy problem (for λ = = 0) and the periodic boundary-value problem (for λ = 1). In these cases, the below theorems coincide with the results obtained in [4] and [10]. Along with the problem (0.1), (0.2) we consider the corresponding homogeneous problem u′(t) = `(u)(t), (0.10) u(a) = λu(b). (0.20) From the general theory of linear boundary-value problem for functional differential equati- ons, the following result is known (see, e.g., [3, 19, 27]). Theorem 0.1. The problem (0.1), (0.2) is uniquely solvable if and only if the corresponding homogeneous problem (0.10), (0.20) has only the trivial solution. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 418 R. HAKL, A. LOMTATIDZE, J. ŠREMR 1. Main results. Theorem 1.1. Let λ ∈]0, 1], the operator ` admit the representation ` = `0 − `1, where `0, `1 ∈ Pab, (1.1) and let either ‖`0(1)‖L < 1, (1.2) ‖`0(1)‖L 1− ‖`0(1)‖L − 1− λ λ < ‖`1(1)‖L < 1 + λ+ 2 √ 1− ‖`0(1)‖L, (1.3) or ‖`1(1)‖L < λ, (1.4) 1 λ− ‖`1(1)‖L − 1 < ‖`0(1)‖L < 2 + 2 √ λ− ‖`1(1)‖L. (1.5) Then the problem (0.1), (0.2) has a unique solution. Remark 1.1. For λ = 0, the first inequality in (1.3) becomes unimportant. Consequently, Theorem 1.3 in [3] can be understoond as a limit case of Theorem 1.3 as λ tends to zero. Remark 1.2. Let λ ∈ [1,+∞[ and ` = `0 − `1, where `0, `1 ∈ Pab. Define an operator ψ : L([a, b];R) → L([a, b];R) by ψ(w)(t) df = w(a+ b− t) for t ∈ [a, b]. Let ϕ be a restriction of ψ to the space C([a, b];R). Put µ = 1 λ , and ̂̀ 0(w)(t) df = ψ(`0(ϕ(w)))(t), ̂̀ 1(w)(t) df = ψ(`1(ϕ(w)))(t) for t ∈ [a, b]. It is clear that if u is a solution of the problem (0.10), (0.20), then the function v df = ϕ(u) is a solution of the problem v′(t) = ̂̀ 1(v)(t)− ̂̀0(v)(t), v(a) = µv(b), (1.6) and vice versa, if v is a solution of the problem (1.6), then the function u df = ϕ(v) is a solution of the problem (0.10), (0.20),. It is evident also that ‖̂̀0(1)‖L = ‖`0(1)‖L, ‖̂̀1(1)‖L = ‖`1(1)‖L. Therefore, Theorem 1.1 immediately yields the following theorem. Theorem 1.2. Let λ ∈ [1,+∞[, the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1), and let either ‖`1(1)‖L < 1, ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 419 ‖`1(1)‖L 1− ‖`1(1)‖L + 1− λ < ‖`0(1)‖L < 1 + 1 λ + 2 √ 1− ‖`1(1)‖L, or ‖`0(1)‖L < 1 λ , 1 1 λ − ‖`0(1)‖L − 1 < ‖`1(1)‖L < 2 + 2 √ 1 λ − ‖`0(1)‖L. Then the problem (0.1), (0.2) has a unique solution. Remark 1.3. In Section 3 we give examples (see Examples 3.1 – 3.6) showing that neither one of the strict inequalities (1.2) – (1.5) can be replaced by the nonstrict ones. According to Remark 1.2 and the above-said, the conditions of Theorem 1.2 are also nonimprovable. Theorem 1.3. Let λ ∈]0, 1], q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, and the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1). Let, moreover, ‖`0(1)‖L < 1, ‖`1(1)‖L < λ (resp. ‖`1(1)‖L ≤ λ) (1.7) and ‖`0(1)‖L 1− ‖`0(1)‖L − 1− λ λ < ‖`1(1)‖L. (1.8) Then the problem (0.1), (0.2) has a unique solution, and this solution is positive (resp. nonnega- tive). Theorem 1.4. Let λ ∈]0, 1], q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, and the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1). Let, moreover, ‖`0(1)‖L < 1 (resp. ‖`0(1)‖L ≤ 1), ‖`1(1)‖L < λ (1.9) and 1 λ− ‖`1(1)‖L − 1 < ‖`0(1)‖L. (1.10) Then the problem (0.1), (0.2) has a unique solution, and this solution is negative (resp. nonposi- tive). According to Remark 1.2, from Theorems 1.3 and 1.4 we have the following. Theorem 1.5. Let λ ∈ [1,+∞[, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, and the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1). If, moreover, ‖`1(1)‖L < 1, ‖`0(1)‖L < 1 λ ( resp. ‖`0(1)‖L ≤ 1 λ ) and ‖`1(1)‖L 1− ‖`1(1)‖L + 1− λ < ‖`0(1)‖L, ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 420 R. HAKL, A. LOMTATIDZE, J. ŠREMR then the problem (0.1), (0.2) has a unique solution, and this solution is negative (resp. nonposi- tive). Theorem 1.6. Let λ ∈ [1,+∞[, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, and the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1). If, moreover, ‖`1(1)‖L < 1 (resp. ‖`1(1)‖L ≤ 1), ‖`0(1)‖L < 1 λ and 1 1 λ − ‖`0(1)‖L − 1 < ‖`1(1)‖L, then the problem (0.1), (0.2) has a unique solution, and this solution is positive (resp. nonnega- tive). Remark 1.4. In Section 3 we give examples (see Examples 3.7 and 3.8) showing that neither one of the inequalities (1.7) – (1.10) can be weakened. According to Remark 1.2 and the above- said, the conditions of Theorems 1.5 and 1.6 are also nonimprovable. For an equation of the type (0.3), from Theorems 1.1 – 1.6 we get the following assertions. Corollary 1.1. Let λ ∈]0, 1], p, g ∈ L([a, b];R+), and let either b∫ a p(s)ds < 1, b∫ a p(s)ds 1− b∫ a p(s)ds − 1− λ λ < b∫ a g(s)ds < 1 + λ+ 2 √√√√√1− b∫ a p(s)ds, or b∫ a g(s)ds < λ, 1 λ− b∫ a g(s)ds − 1 < b∫ a p(s)ds < 2 + 2 √√√√√λ− b∫ a g(s)ds. Then the problem (0.3), (0.2) has a unique solution. Corollary 1.2. Let λ ∈ [1,+∞[, p, g ∈ L([a, b];R+), and let either b∫ a g(s)ds < 1, b∫ a g(s)ds 1− b∫ a g(s)ds + 1− λ < b∫ a p(s)ds < 1 + 1 λ + 2 √√√√√1− b∫ a g(s)ds, or b∫ a p(s)ds < 1 λ , 1 1 λ − b∫ a p(s)ds − 1 < b∫ a g(s)ds < 2 + 2 √√√√√ 1 λ − b∫ a p(s)ds. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 421 Then the problem (0.3), (0.2) has a unique solution. Corollary 1.3. Let λ ∈]0, 1], p, g, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, b∫ a p(s)ds < 1, b∫ a g(s)ds < λ resp. b∫ a g(s)ds ≤ λ  , and b∫ a p(s)ds 1− b∫ a p(s)ds − 1− λ λ < b∫ a g(s)ds. Then the problem (0.3), (0.2) has a unique solution, and this solution is positive (resp. nonnega- tive). Corollary 1.4. Let λ ∈]0, 1], p, g, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, b∫ a p(s)ds < 1 resp. b∫ a p(s)ds ≤ 1  , b∫ a g(s)ds < λ, and 1 λ− b∫ a g(s)ds − 1 < b∫ a p(s)ds. Then the problem (0.3), (0.2) has a unique solution, and this solution is negative (resp. nonposi- tive). Corollary 1.5. Let λ ∈ [1,+∞[, p, g, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, b∫ a p(s)ds < 1 λ resp. b∫ a p(s)ds ≤ 1 λ  , b∫ a g(s)ds < 1, and b∫ a g(s)ds 1− b∫ a g(s)ds + 1− λ < b∫ a p(s)ds. Then the problem (0.3), (0.2) has a unique solution, and this solution is negative (resp. nonposi- tive). Corollary 1.6. Let λ ∈ [1,+∞[, p, g, q ∈ L([a, b];R+), c ∈ R+, ‖q‖L + c 6= 0, b∫ a p(s)ds < 1 λ , b∫ a g(s)ds < 1 resp. b∫ a g(s)ds ≤ 1  , ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 422 R. HAKL, A. LOMTATIDZE, J. ŠREMR and 1 1 λ − b∫ a p(s)ds − 1 < b∫ a g(s)ds. Then the problem (0.3), (0.2) has a unique solution, and this solution is positive (resp. nonnega- tive). 2. Proofs. To prove Theorems 1.1, 1.3, and 1.4, we need the following lemmas. Lemma 2.1. Let λ ∈]0, 1], q ∈ L([a, b];R−), c ∈ R−, the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1), and let ‖`0(1)‖L < 1, ‖`0(1)‖L 1− ‖`0(1)‖L − 1− λ λ < ‖`1(1)‖L. (2.1) Then the problem (0.1), (0.2) has no nontrivial solution u satisfying the inequality u(t) ≥ 0 for t ∈ [a, b]. (2.2) Proof. Assume the contrary that the problem (0.1), (0.2) has a nontrivial solution u satisfy- ing the condition (2.2). Put M = max{u(t) : t ∈ [a, b]}, m = min{u(t) : t ∈ [a, b]} (2.3) and choose tM , tm ∈ [a, b] such that u(tM ) = M, u(tm) = m. (2.4) Obviously, M > 0, m ≥ 0 and either tM > tm, (2.5) or tM < tm. (2.6) First suppose that (2.5) holds. The integration of (0.1) from tm to tM , on account of (1.1), (2.3), (2.4), and the assumption q ∈ L([a, b];R−), results in M −m = tM∫ tm [`0(u)(s)− `1(u)(s) + q(s)]ds ≤ M tM∫ tm `0(1)(s)ds ≤ M‖`0(1)‖L. Now suppose that (2.6) is fulfilled. The integration of (0.1) from a to tM and from tm to b, in view of (1.1), (2.3), (2.4) and the assumption q ∈ L([a, b];R−), yields M − u(a) ≤ M tM∫ a `0(1)(s)ds, u(b)−m ≤ M b∫ tm `0(1)(s)ds. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 423 Summing the last two inequalities and taking into account the condition u(b)− u(a) ≥ λu(b)− u(a) = −c ≥ 0, we obtain M(1− ‖`0(1)‖L) ≤ m. (2.7) Therefore, in both cases (2.5) and (2.6), the inequality (2.7) is valid. On the other hand, the integration of (0.1) from a to b, in view of (1.1), (2.3), and the assumption q ∈ L([a, b];R−), implies u(b)− u(a) = b∫ a [`0(u)(s)− `1(u)(s) + q(s)]ds ≤ M‖`0(1)‖L −m‖`1(1)‖L. Hence, by (2.3), (0.2) and the assumptions λ ∈]0, 1], c ∈ R−, we have m‖`1(1)‖L ≤ M‖`0(1)‖L + u(a) ( 1− 1 λ ) + 1 λ c ≤ M‖`0(1)‖L +m ( 1− 1 λ ) . Thus m ( ‖`1(1)‖L + 1− λ λ ) ≤ M‖`0(1)‖L. This inequality together with (2.7) results in ‖`1(1)‖L ≤ ‖`0(1)‖L 1− ‖`0(1)‖L − 1− λ λ , which contradicts the second inequality in (2.1). Lemma 2.2. Let λ ∈]0, 1], q ∈ L([a, b];R+), c ∈ R+, the operator ` admit the representation ` = `0 − `1, where `0 and `1 satisfy the condition (1.1), and let ‖`1(1)‖L < λ, 1 λ− ‖`1(1)‖L − 1 < ‖`0(1)‖L. (2.8) Then the problem (0.1), (0.2) has no nontrivial solution u satisfying the inequality (2.2). Proof. Assume the contrary that the problem (0.1), (0.2) has a nontrivial solution u satisfy- ing the condition (2.2). Define the numbers M and m by (2.3) and choose tM , tm ∈ [a, b] such that (2.4) is fulfilled. Obviously, M > 0, m ≥ 0 and either (2.5) or (2.6) is valid. First suppose that (2.6) holds. The integration of (0.1) from tM to tm, on account of (1.1), (2.3), (2.4) and the assumptions λ ∈]0, 1] and q ∈ L([a, b];R+), results in λM −m ≤ M −m = tm∫ tM [`1(u)(s)− `0(u)(s)− q(s)]ds ≤ ≤M tm∫ tM `1(1)(s)ds ≤ M‖`1(1)‖L. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 424 R. HAKL, A. LOMTATIDZE, J. ŠREMR Now suppose that (2.5) is fulfilled. The integration of (0.1) from a to tm and from tM to b, in view of (1.1), (2.3), (2.4) and the assumptions λ ∈]0, 1], q ∈ L([a, b];R+), yields u(a)−m ≤ M tm∫ a `1(1)(s)ds, λ(M − u(b)) ≤ M − u(b) ≤ M b∫ tM `1(1)(s)ds. Summing the last two inequalities and taking into account the condition u(a)− λu(b) = c ≥ 0, we obtain M(λ− ‖`1(1)‖L) ≤ m. (2.9) Therefore, in both cases (2.5) and (2.6), the inequality (2.9) is valid. On the other hand, the integration of (0.1) from a to b, in view of (1.1), (2.3), and the assumption q ∈ L([a, b];R+), results in u(a)− u(b) = b∫ a [`1(u)(s)− `0(u)(s)− q(s)]ds ≤ M‖`1(1)‖L −m‖`0(1)‖L. Hence, by (2.3), (0.2) and the assumptions λ ∈]0, 1], c ∈ R+, we have m‖`0(1)‖L ≤ M‖`1(1)‖L + u(b) (1− λ)− c ≤ M‖`1(1)‖L +M (1− λ) . Thus m‖`0(1)‖L ≤ M (‖`1(1)‖L − λ+ 1) . This inequality together with (2.9) yields ‖`0(1)‖L ≤ 1 λ− ‖`1(1)‖L − 1, which contradicts the second inequality in (2.8). Proof of Theorem 1.1. According to Theorem 0.1, it is sufficient to show that the homogeneous problem (0.10), (0.20) has no nontrivial solution. First suppose that (1.2) and (1.3) hold. Assume the contrary that the problem (0.10), (0.20) has a nontrivial solution u. According to Lemma 2.1, u has to change its sign. Put M = max{u(t) : t ∈ [a, b]}, m = −min{u(t) : t ∈ [a, b]} (2.10) and choose tM , tm ∈ [a, b] such that u(tM ) = M, u(tm) = −m. (2.11) ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 425 Obviously, M > 0, m > 0. Without loss of generality we can assume that tm < tM . The integration of (0.10) from a to tm, from tm to tM and from tM to b, by (2.10), (2.11) and (1.1), results in u(a) +m = tm∫ a [`1(u)(s)− `0(u)(s)]ds ≤ M tm∫ a `1(1)(s)ds+m tm∫ a `0(1)(s)ds, (2.12) M +m = tM∫ tm [`0(u)(s)− `1(u)(s)]ds ≤ M tM∫ tm `0(1)(s)ds+m tM∫ tm `1(1)(s)ds, (2.13) M − u(b) = b∫ tM [`1(u)(s)− `0(u)(s)]ds ≤ M b∫ tM `1(1)(s)ds+m b∫ tM `0(1)(s)ds. (2.14) Multiplying the both sides of (2.14) by λ and taking into account (2.10) and the assumption λ ∈]0, 1], we get λM − λu(b) ≤ M b∫ tM `1(1)(s)ds+m b∫ tM `0(1)(s)ds. Summing the last inequality and (2.13), by (0.20) we obtain λM +m ≤ M ∫ J `1(1)(s)ds+m ∫ J `0(1)(s)ds, (2.15) where J = [a, tm] ∪ [tM , b]. From (2.13) and (2.15) it follows that M(1−D) ≤ m(B − 1), m(1− C) ≤ M(A− λ), (2.16) where A = ∫ J `1(1)(s)ds, B = tM∫ tm `1(1)(s)ds, C = ∫ J `0(1)(s)ds, D = tM∫ tm `0(1)(s)ds. (2.17) Due to (1.2), C < 1 and D < 1. Consequently, (2.16) implies A > λ, B > 1, and 0 < (1− C)(1−D) ≤ (A− λ)(B − 1). (2.18) ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 426 R. HAKL, A. LOMTATIDZE, J. ŠREMR Obviously, (1− C)(1−D) ≥ 1− (C +D) = 1− ‖`0(1)‖L > 0, 4(A− λ)(B − 1) ≤ [A+B − (1 + λ)]2 = [‖`1(1)‖L − (1 + λ)]2. By the last inequalities, (2.18) results in 0 < 4(1− ‖`0(1)‖L) ≤ [‖`1(1)‖L − (1 + λ)]2, which contradicts the second inequality in (1.3). Now suppose that (1.4) and (1.5) are fulfilled. Assume the contrary that the problem (0.10), (0.20) has a nontrivial solution u. According to Lemma 2.2, u has to change its sign. Define M andm by (2.10) and choose tM , tm ∈ [a, b] such that (2.11) is fulfilled. Without loss of generality we can assume that tm < tM . Analogously to the above, one can show that the inequalities (2.12) – (2.15) hold, where J = [a, tm] ∪ [tM , b]. From (2.13) and (2.15) it follows that m(1−B) ≤ M(D − 1), M(λ−A) ≤ m(C − 1), (2.19) whereA,B,C,D are defined by (2.17). According to (1.4),A < λ andB < λ ≤ 1. Consequently, (2.19) implies C > 1, D > 1 and 0 < (λ−A)(1−B) ≤ (C − 1)(D − 1). (2.20) Obviously, (λ−A)(1−B) ≥ λ− (A+B) = λ− ‖`1(1)‖L > 0, 4(C − 1)(D − 1) ≤ (C +D − 2)2 = (‖`0(1)‖L − 2)2. By the last inequalities, from (2.20) we get 0 < 4(λ− ‖`1(1)‖L) ≤ (‖`0(1)‖L − 2)2, which contradicts the second inequality in (1.5). Proof of Theorem 1.3. According to Theorem 1.1 and the conditions (1.7), (1.8), the problem (0.1), (0.2) has a unique solution u. Show that u has no zero (resp. does not change its sign). Assume the contrary that there exists t1 ∈ [a, b] (resp. t2, t3 ∈ [a, b]) such that u(t1) = 0 (resp. u(t2)u(t3) < 0). (2.21) Define numbers M and m by (2.10) and choose tM , tm ∈ [a, b] such that (2.11) is fulfilled. Obviously, M ≥ 0, m ≥ 0, M +m > 0 (2.22) ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 427( resp. M > 0, m > 0 ) , (2.23) and either (2.5) or (2.6) is valid. First suppose that (2.5) holds. The integration of (0.1) from a to tm and from tM to b, on account of (1.1), (2.10), (2.11) and the assumptions λ ∈]0, 1] and q ∈ L([a, b];R+), results in m+ u(a) ≤M tm∫ a `1(1)(s)ds+m tm∫ a `0(1)(s)ds, (2.24) λ(M − u(b)) ≤ M − u(b) ≤M b∫ tM `1(1)(s)ds+m b∫ tM `0(1)(s)ds. (2.25) Summing the last two inequalities and taking into account the condition u(a)− λu(b) = c ≥ 0, we obtain λM +m ≤ M‖`1(1)‖L +m‖`0(1)‖L, (2.26) which by (1.7) yields the contradiction λM +m < λM +m (resp. m < m). Now suppose that (2.6) is fulfilled. The integration of (0.1) from tM to tm, on account of (1.1), (2.10), (2.11) and the assumption q ∈ L([a, b];R+), results in M +m = tm∫ tM [`1(u)(s)− `0(u)(s)− q(s)]ds ≤ M‖`1(1)‖L +m‖`0(1)‖L. (2.27) Hence, by (1.7) and the assumption λ ∈]0, 1], we get the contradictionM+m < M+m. Thus u has no zero (resp. does not change its sign), and so according to Lemma 2.1, u is positive (resp. nonnegative). Proof of Theorem 1.4. According to Theorem 1.1 and the conditions (1.9), (1.10), the problem (0.1), (0.2) has a unique solution u. Show that u has no zero (resp. does not change its sign). Assume the contrary that there exists t1 ∈ [a, b] (resp. t2, t3 ∈ [a, b]) such that (2.21) is fulfilled. Define numbers M and m by (2.10) and choose tM , tm ∈ [a, b] such that (2.11) is fulfilled. Obviously, (2.22) (resp. (2.23)) is satisfied, and either (2.5) or (2.6) is valid. By the same arguments as in the proof of Theorem 1.3 one can show that the assumption (2.5) yields the contradiction λM +m < λM +m (resp. M < M), and the assumption (2.6) yields the contradiction M +m < M +m. Thus u has no zero (resp. does not change its sign), and so according to Lemma 2.2, u is negative (resp. nonpositive). 3. On Remarks 1.3 and 1.4. On Remark 1.3. Let λ ∈]0, 1[ (the case λ = 0, resp. λ = 1, is studied in [4], resp. [10], where the examples are also given verifying the optimality of the obtained results). Denote by H+, resp. H−, the set of pairs (x, y) ∈ R+ ×R+ such that x < 1, x 1− x − 1− λ λ < y < 1 + λ+ 2 √ 1− x, ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 428 R. HAKL, A. LOMTATIDZE, J. ŠREMR resp. y < λ, 1 λ− y − 1 < x < 2 + 2 √ λ− y. By Theorem 1.1, if (‖`0(1)‖L, ‖`1(1)‖L) ∈ H+ ∪H−, then the problem (0.1), (0.2) has a unique solution. (Note also that for λ ≤ 1 4 , H− = ∅.) Below we give the examples which show that for any pair (x0, y0) 6∈ H+ ∪ H−, x0 ≥ 0, y0 ≥ 0 there exist functions h ∈ L([a, b];R) and τ ∈ Mab such that b∫ a [h(s)]+ds = x0, b∫ a [h(s)]−ds = y0, (3.1) and the problem u′(t) = h(t)u(τ(t)), u(a) = λu(b) (3.2) has a nontrivial solution. Then by Theorem 0.1, there exist q ∈ L([a, b];R) and c ∈ R such that the problem (0.1), (0.2), where ` = `0 − `1, `0(w)(t) df = [h(t)]+w(τ(t)), `1(w)(t) df = [h(t)]−w(τ(t)), (3.3) either has no solution or has an infinite set of solutions. It is clear that if x0, y0 ∈ R+ and (x0, y0) 6∈ H+ ∪H−, then (x0, y0) belongs at least to one of the following sets: H1 = {(x, y) ∈ R×R : 1 ≤ x, λ ≤ y} , H2 = { (x, y) ∈ R×R : 0 ≤ x < 1, 1 + λ+ 2 √ 1− x ≤ y } , H3 = { (x, y) ∈ R×R : 0 ≤ y < λ, 2 + 2 √ λ− y ≤ x } , H4 = { (x, y) ∈ R×R : 0 ≤ y < λ, y + 1− λ ≤ x ≤ y + 1− λ λ− y } , H5 = { (x, y) ∈ R×R : 1− λ < x < 1, x λ + 1− 1 λ ≤ y ≤ x+ λ− 1 λ(1− x) } , H6 = { (x, y) ∈ R×R : 1− λ < x < 1, x− 1 + λ ≤ y ≤ x λ + 1− 1 λ } . ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 429 Example 3.1. Let (x0, y0) ∈ H1. Put a = 0, b = 4, h(t) =  −λ for t ∈ [0, 1[; x0 − 1 for t ∈ [1, 2[; λ− y0 for t ∈ [2, 3[; 1 for t ∈ [3, 4], τ(t) = { 4 for t ∈ [0, 1[∪[3, 4]; 1 for t ∈ [1, 3[. Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) =  λ(1− t) for t ∈ [0, 1[; 0 for t ∈ [1, 3[; t− 3 for t ∈ [3, 4]. Example 3.2. Let (x0, y0) ∈ H2. Put a = 0, b = 6, α = √ 1− x0, β = y0 − 1− λ− 2α, h(t) =  −λ for t ∈ [0, 1[; −β for t ∈ [1, 2[; −α for t ∈ [2, 4[; −1 for t ∈ [4, 5[; x0 for t ∈ [5, 6], τ(t) =  6 for t ∈ [0, 1[∪[2, 3[∪[5, 6]; 1 for t ∈ [1, 2[; 3 for t ∈ [3, 5[. Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) =  λ(1− t) for t ∈ [0, 1[; 0 for t ∈ [1, 2[; α(2− t) for t ∈ [2, 3[; α2(t− 3)− α for t ∈ [3, 4[; α(t− 5) + α2 for t ∈ [4, 5[; x0(t− 6) + 1 for t ∈ [5, 6]. Example 3.3. Let (x0, y0) ∈ H3. Put a = 0, b = 6, α = √ λ− y0, β = x0 − 2− 2α, h(t) =  α for t ∈ [0, 1[; −y0 for t ∈ [1, 2[; β for t ∈ [2, 3[; 1 for t ∈ [3, 4[; α for t ∈ [4, 5[; 1 for t ∈ [5, 6], τ(t) =  4 for t ∈ [0, 1[∪[3, 4[; 6 for t ∈ [1, 2[∪[4, 6]; 2 for t ∈ [2, 3[. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 430 R. HAKL, A. LOMTATIDZE, J. ŠREMR Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) =  −α2t+ λ for t ∈ [0, 1[; y0(2− t) for t ∈ [1, 2[; 0 for t ∈ [2, 3[; α(3− t) for t ∈ [3, 4[; α(t− 5) for t ∈ [4, 5[; t− 5 for t ∈ [5, 6]. Example 3.4. Let (x0, y0) ∈ H4. Put a = 0, b = 2, α = 1− λ+ y0, t0 = 1 x0 − 1 α + 2, h(t) = { −y0 for t ∈ [0, 1[; x0 for t ∈ [1, 2], τ(t) = { 2 for t ∈ [0, 1[; t0 for t ∈ [1, 2]. Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) = { −y0t+ λ for t ∈ [0, 1[; α(t− 2) + 1 for t ∈ [1, 2]. Example 3.5. Let (x0, y0) ∈ H5. Put a = 0, b = 2, α = λ+ x0 − 1 1− x0 , β = λx0 1− x0 , t0 = = ( α y0 − λ ) 1 β , h(t) = { x0 for t ∈ [0, 1[; −y0 for t ∈ [1, 2], τ(t) = { 1 for t ∈ [0, 1[; t0 for t ∈ [1, 2]. Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) = { βt+ λ for t ∈ [0, 1[; α(2− t) + 1 for t ∈ [1, 2]. Example 3.6. Let (x0, y0) ∈ H6. Put a = 0, b = 2, α = λ+ x0 − 1, t0 = α− y0 x0y0 + 2, h(t) = { −y0 for t ∈ [0, 1[; x0 for t ∈ [1, 2], τ(t) = { t0 for t ∈ [0, 1[; 2 for t ∈ [1, 2]. Then (3.1) holds, and the problem (3.2) has the nontrivial solution u(t) = { −αt+ λ for t ∈ [0, 1[; x0(t− 2) + 1 for t ∈ [1, 2]. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 ON A PERIODIC TYPE BOUNDARY-VALUE PROBLEM FOR FIRST ORDER LINEAR . . . 431 On Remark 1.4. Let λ ∈]0, 1] (the case λ = 0 is studied in [4]). Denote by G+, resp. G−, the set of pairs (x, y) ∈ R+ ×R+ such that x < 1, x 1− x − 1− λ λ < y < λ, resp. y < λ, 1 λ− y − 1 < x < 1. It is clear that G+ ⊂ H+ and G− ⊂ H−. (Note also that for λ ≤ 1 2 , G− = ∅.) By Theorem 1.3, resp. Theorem 1.4, if (‖`0(1)‖L, ‖`1(1)‖L) ∈ G+, resp. (‖`0(1)‖L, ‖`1(1)‖L) ∈ G−, then the problem (0.1), (0.2) with q ∈ L([a, b];R+), c ∈ R+, ‖q‖L+ c 6= 0 has a unique solution and this solution is positive, resp. negative. Below we give the examples which show that for any pair (x0, y0) ∈ H+ \ G+, resp. (x0, y0) ∈ H− \ G−, there exist functions h ∈ L([a, b];R), q ∈ L([a, b];R+) and τ ∈ Mab such that q 6≡ 0, (3.1) is fulfilled, and the problem u′(t) = h(t)u(τ(t)) + q(t), u(a) = λu(b), (3.4) or equivalently, the problem (0.1), (0.20) where ` = `0 − `1, and `0, `1 are defined by (3.3), has a solution which is not positive, resp. negative. From Example 3.7, resp. Example 3.8, it also follows that in Theorem 1.3, resp. Theorem 1.4, the inequality ‖`1(1)‖L ≤ λ, resp. ‖`0(1)‖L ≤ 1, in the condition (1.7), resp. (1.9), cannot be replaced by the inequality ‖`1(1)‖L ≤ λ+ ε, resp. ‖`0(1)‖L ≤ 1+ ε, no matter how small ε > 0 would be. Example 3.7. Let (x0, y0) ∈ H+ \G+. Put a = 0, b = 2, α = y0−x0−λ+1, β = 1+y0−λ, τ ≡ 2, h(t) = { −y0 for t ∈ [0, 1[; x0 for t ∈ [1, 2], q(t) = { 0 for t ∈ [0, 1[; α for t ∈ [1, 2]. Then (3.1) holds, and the problem (3.4) has the solution u(t) = { −y0t+ λ for t ∈ [0, 1[; β(t− 2) + 1 for t ∈ [1, 2] with u(1) = λ− y0 ≤ 0. Example 3.8. Let (x0, y0) ∈ H− \G−. Put a = 0, b = 2, α = x0−y0+λ−1, β = x0+λ−1, τ ≡ 2, h(t) = { −y0 for t ∈ [0, 1[; x0 for t ∈ [1, 2], q(t) = { α for t ∈ [0, 1[; 0 for t ∈ [1, 2]. ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3 432 R. HAKL, A. LOMTATIDZE, J. ŠREMR Then (3.1) holds, and the problem (3.4) has the solution u(t) = { βt− λ for t ∈ [0, 1[; x0(2− t)− 1 for t ∈ [1, 2] with u(1) = x0 − 1 ≥ 0. 1. Azbelev N. V., Maksimov V. P., Rakhmatullina L. F. Introduction to the theory of functional differential equati- ons. — Moscow: Nauka, 1991 (in Russian). 2. Azbelev N. V., Rakhmatullina L. F. Theory of linear abstract functional differential equations and applications // Mem. Different. Equat. Math. Phys. — 1996. — 8. — P. 1 – 102. 3. Bravyi E. 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On an antiperiodic type boundary-value problem for first order linear functional differential equations // Arch. Math. (to appear). 14. Hakl R., Lomtatidze A., Šremr J. On an antiperiodic type boundary-value problem for first order nonlinear functional differential equations of non-Volterra’s type // Different. and Int. Equat. (submitted). 15. Hakl R., Lomtatidze A., Šremr J. On a periodic type boundary-value problem for first order nonlinear functi- onal differential equations // Nonlinear Anal.: Theory, Meth. and Appl. (to appear). 16. Hale J. Theory of functional differential equations. — New York etc.: Springer, 1977. 17. Kiguradze I. On periodic solutions of first order nonlinear differential equations with deviating arguments // Mem. Different. Equat. Math. Phys. — 1997. — 10. — P. 134 – 137. 18. Kiguradze I. Initial and boundary-value problems for systems of ordinary differential equations I. — Tbilisi: Metsniereba, 1997 (in Russian). 19. Kiguradze I., P ◦ uža B. 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On periodic solutions of systems of differential equations with deviating arguments // Nonlinear Anal.: Theory, Meth. and Appl. — 2000. — 42, N◦ 2. — P. 229 – 242. 25. Kolmanovskii V., Myshkis A. Introduction to the theory and applications of functional differential equations. — Kluwer Acad. Publ., 1999. 26. Mawhin J. Periodic solutions of nonlinear functional differential equations // J. Different. Equat. — 1971. — 10. — P. 240 – 261. 27. Schwabik Š., Tvrdý M., Vejvoda O. Differential and integral equations: boundary-value problems and adjo- ints. — Praha: Academia, 1979. Received 15.05.2002