MASARYKOVA UNIVERZITA PŘÍRODOVĚDECKÁ FAKULTA ÚSTAV MATEMATIKY A STATISTIKY Habilitační práce Brno 2015 Sulkhan Mukhigulashvili MASARYKOVA PŘÍRODOVĚDECKÁ FAKULT ÚSTAV MATEMATIKY A STATISTIKY O některých okrajových funkcionální Sulkhan Mukhigulash MASARYKOVA UNIVERZITA ŘÍRODOVĚDECKÁ FAKULTA ÚSTAV MATEMATIKY A STATISTIKY ěkterých dvoubodových krajových úlohách p unkcionální diferenciální rovnice Habilitační práce Sulkhan Mukhigulashvili Brno 2015 UNIVERZITA ÚSTAV MATEMATIKY A STATISTIKY voubodových pro iferenciální vili Bibliografický záznam Autor: Mukhigulashvili Sulkhan , Dr. CSc Vysoké učení technické v Brně, fakulta podnikatelská, Ústav informatiky. Název práce: O některých dvoubodových okrajových ulohách pro funkcionální diferenciální rovnice Obor: Matematika- matematická analzýa Akademický rok: 2015/2016 Počet stran: Klíčová slova: Dvoubodová úloha, singulární funkcionální diferenciální rovnice, řešitelnost a jednoznačnost řešení, dvoubodová úloha v rezonančním případě, Fredholmovost úlohy. Bibliographic Entry Author Mukhigulashvili Sulkhan, Dr. CSc, Brno University Of Technology, The Faculty of Business and Management, Department Of Information Technology. Titleof Thesis: On Some Two-Point Boundary Value Problems For Functional-Differential Equations Field: Mathematics- Mathematical analysis AcademicYear: 2015/2016 Number of Pages: Keywords: Two-poin boundary value problems, singular functionaldifferential equations, solvability, unique solvability, Fredholm‘s property, BVP at resonance. Abstrakt V tomto přehledu jsou obsaženy některé výsledky z jedné monografie a čtyř časopiseckých prací, v nichž jsou vyšetřovány dvoubodové okrajové úlohy pro funkcionální diferenciální rovnice jak v regulárním, tak i sigulárním případě (např. Dirichletova, smíšená, fokální, periodická). Jsou zde uvedeny efektivní podmínky zaručující řešitelnost, jednoznačnou řešitelnost, a také Fredholmovost studovaných úloh. Uvedené výsledky byly v době jejich publikace originální, nové a rozšiřovaly znalosti v daném oboru. Abstract The present survey considers four papers and a monograph where various kinds of boundary value problems (Dirichlet, periodic, mixed, focal) for linear and nonlinear functional differential equations are studied in both regular and singular cases. The works mentioned contain results on the solvability, unique solvability and Fredholm property of the problems under consideration. The results obtained, at the time of publication, had been new and contributed essentially to the knowledge of the problems mentioned. CONTENT Introduction...................................................................................................................1 Chapter 1. Two-Point Boundary Value Problems For Regular FunctionalDifferential Equations......................................................................................................5 1.1 Section. The Dirichlet BVP for the second order nonlinear ordinary differential equation at resonance................................................................................5 1.2 Section. A periodic boundary value problem for functional differential equations of higher order............................................................................................9 Chapter 2. Two-Point Boundary Value Problems For Singular FunctionalDifferential Equations....................................................................................................13 2.1 Section. Two-point boundary value problems for second order functional-differential Equations.....................................................................................................................13 2.2 Section. Two-point boundary value problems for strongly singular higher-order linear differential equations with deviating arguments.............................................21 2.3 Section. The Dirichlet boundary value problems for strongly singular higherorder nonlinear functional-differential equations......................................................26 Bibliography....................................................................................................................32 Annexe ÚVOD Předložená práce obsahuje přehled některých mých výsledků týkajících se otázek řešitelnosti, jednoznačné řešitelnosti a korektnosti některých okrajových úloh pro funkcionálně-diferenciální rovnice. Nejprve stručně popíšeme pět vybraných publikací. V článku The Dirichlet BVP for second order nonlinear ordinary differential equation at resonance. Italian J. Of Pure and Appl. Math., 2011, No. 28, 177–204 je, narozdíl od předešlých prací uvažována nelineární diferenciální rovnice tvaru u′′ = p(t)u + f(t, u) + h(t), v níž je funkce f sublineární v druhé proměnné, v případě když homogenní Dirichletova úloha pro rovnici u′′ = p(t)u má netriviální řešení (tzv. rezonanční případ). V tomto článku se nepředpokládá, že koeficient p je konstantní funkce, což je omezení pro výsledky tohoto typu obvyklé v existující literatuře. V práci A periodic boundary value problem for functionaldifferential equations of higher order. Georgian Math. J., Vol. 16, 2009, No. 4, 651–665 (spoluautor R. Hakl), v níž jsou nalezena efektivní kritéria jednoznačné řešitelnosti periodické úlohy pro funkcionálnědiferenciální rovnice vyšších řádu s nemonotonními operátory na pravé straně rovnice, které zlepšují dřívější výsledky autorů Lasota–Opial a Kiguradze– Kusano, a které jsou nezlepšitelné pro rovnice řádu n ≤ 7. Metoda, která je zde použita k důkazu hlavních tvrzení, je vyvinuta v několika předešlých publikacích. V monografii Two-point boundary value problems for second order functional differential equations. Mem. Differential Equations Math. Phys., 20, 2000, 1–112 je studována Dirichletova a smíšená okrajová úloha pro lineární singulární funkcionálně-diferenciální rovnice druhého řádu. V první kapitole jsou uvedeny postačující podmínky zaručující jednoznačnou řešitelnost daných úloh a je ukázáno, že některé z efektivních podmínek jsou v jistém smyslu nezlepšitelné. V druhé kapitole jsou pak dokázány věty o korektnosti daných úloh. V článku Two-point boundary value problems for strongly singular higher-order linear differential equations with deviating arguments. E. J. Qualitative Theory of Diff. Equ., 2012, No. 38, 1–34 (spoluaturka N. Partsvania) jsou dokázána tvrzení typu Agawala–Kiguradzeho pro dvoubodové a fokální úlohy pro silně singulární diferenciální rovnice vyšších řádů s odkloněnými argumenty. Tato tvrzení obsahují postačující podmínky zaručující, že studované úlohy mají tzv. Fredholmovu vlastnost. Dále jsou v tomto článku nalezeny efektivní nezlepšitelná kritéria jednoznačné řešitelnosti těchto lineárních úloh. Je známo, že máme-li prostudovanou otázku jednoznačné řešitelnosti 1 dvoubodových okrajových úloh pro lineární diferenciální rovnice, je možné odvodit kritéria řešitelnosti nelineárních úloh, v nichž jsou nelineární rovnice v jistém smyslu „blízké odpovídajícím rovnicím lineárním. Výsledky tohoto typu pro nelineární funkcionálně-diferenciální rovnice jsou prezentovány v práci The Dirichlet boundary value problems for strongly singular higher-order nonlinear functional-differential equations. Czechoslovak Mathematical Journal, Vol. 63, 2013, No. 1, 235–263. Pomocí výsledků známých v lineárním případě jsou zde odvozeny efektivní postačující podmínky, zaručující jednoznačnou řešitelnost Dirichletovy úlohy pro silně singulární nelineární funkcionálně-diferenciální rovnice vyšších řádů. Pro lepší přehlednost a čitelnost textu je na začátku každého oddílu uvedeno označení, které je v něm pak jednotně používáno. 2 INTRODUCTION In the present survey I review my several studies exploring solvability, unique solvability and correctness of some boundary value problems for functional-differential equations. First I will briefly characterize five selected studies. In the article The Dirichlet BVP The second Order Nonlinear Ordinary Differential Equation At Resonance. Italian J. Of Pure and Appl. Math., 2011, No. 28, 177-204, in difference with the previous paper, the nonlinear equation u′′ (t) = p(t)u(t)+f(t, u(t))+h(t) under Dirichlet boundary value problem conditions is studied in the case when f is sublinear function in the second argument and the homogeneous linear equation u′′ (t) = p(t)u(t) under homogeneous Dirichlet boundary value conditions has a nontrivial solution, i.e. in the resonance case. It is noteworthy that unlike this article, similar problems are studied in literature only in the concrete case when p ≡ Const. In the paper A Periodic Boundary Value Problem For Functional-Differential Equations Of Higher Order (with R. Hakl). Georgian Math. J. Vol.16, (2009), No.4, 651-665, the efficient sufficient conditions guaranteeing the unique solvability of the periodic problem are established in the case of nonmonotone linear operators, which improve the results of Lasota - Opial and Kiguradze-Kusano and are optimal for n ≤ 7. The method used for the investigation of the considered problem is based on the method developed in my previous papers. In the monograph Two-point boundary value problems for second order functional differential equations. Mem. Differential Equations Math. Phys. 20 (2000), 1-112, the Dirichlet and mixed problems for second order linear singular functional-differential equations are studied. In the first chapter the sufficient conditions of unique solvability of the named problem are established and some of them are sharp in some sense. The correctness of the above mentioned problem is studied in the second chapter of the monograph. In the paper Two-point boundary value problems for strongly singular higher-order linear differential equations with deviating arguments (with N. Partsvania). E. J. Qualitative Theory of Diff. Equ., 2012, No.38, 1-34, for strongly singular higher-order differential equations with deviating arguments, under two point conjugated and right-focal boundary conditions, Agarval-Kiguradze type theorems are established, which guarantee the presence of Fredholm’s property for the above mentioned problems. Also we provide easily verifiable best possible conditions that guarantee the existence of a unique solution of the studied 3 problems. As is known, if we have studied the unique solvability of the linear functional-differential equations under some two-point boundary value problem, it simplifies study of the question of solvability of the same two-point boundary value problem for nonlinear functional-differential equations if the nonlinear equation is in a some sense ”close”to this linear equation. Results of this type for the nonlinear functional-differential equations are presented in the study The Dirichlet Boundary Value Problems For Strongly Singular Higher-Order Nonlinear Functional-Differential Equations. Czechoslovak Mathematical Journal, vol. 63 (2013), No. 1, pp. 235-263, where by using the results proved for the linear equations, the efficient sufficient conditions guaranteeing the unique solvability for Dirrichlet problem are established for the strongly singular higher-order nonlinear functional differential equations. The notation used in the survey is introduced separately for every single section at its beginning. 4 Kapitola 1 Two-point boundary value problems for regular functional-differential equations 1.1 The Dirichlet BVP For The Second Order Nonlinear Ordinary Differential Equation At Resonance In this chapter first we consider the paper [4], in which on the set I = [a, b] where the second order ordinary differential equation u′′ (t) = p(t)u(t) + f(t, u(t)) + h(t) (1.1.1) are studied under the boundary conditions u(a) = 0, u(b) = 0, (1.1.2) where h, p ∈ L([a, b]) and f ∈ K(I × R; R). By a solution of the problem (1.1.1), (1.1.2) we understand a function u ∈ C′ ([a, b]), which satisfies the equation (1.1.1) almost everywhere on I and satisfies the conditions (1.1.2). Along with (1.1.1), (1.1.2) we consider the homogeneous problem w′′ (t) = p(t)w(t) for t ∈ I, (1.1.3) 5 w(a) = 0, w(b) = 0. (1.1.4) The case when the problem (1.1.3), (1.1.4) has the nontrivial solution is still little investigated and in the majority of articles, the authors study the case with p constant in the equation (1.1.1), i.e., when the problem (1.1.1), (1.1.2) and the equation (1.1.3) are of type u′′ (t) = −λ2 u(t) + f(t, u(t)) + h(t) for t ∈ [0, π], (1.1.5) u(0) = 0, u(π) = 0, (1.1.6) and w′′ (t) = −λ2 w(t) for t ∈ [0, π] (1.1.7) respectively, with λ = 1. In this work, the solvability of the problem (1.1.1), (1.1.2) is studied in the case when the function p ∈ L([a, b]) is not necessarily constant, under the assumption that the homogeneous problem (1.1.3), (1.1.4) has the nontrivial solution with an arbitrary number of zeroes. For the equation (1.1.7), this is the case when λ is not necessarily the first eigenvalue of the problem (1.1.7), (1.1.4), with a = 0, b = π. Throughout the paper the following notation are used: K(I × R; R) is the set of functions f : I × R → R satisfying the Carathéodory conditions. Also having the function w : I → R, we put: Nw def = {t ∈ ]a, b[ : w(t) = 0}, Ω+ w def = {t ∈ I : w(t) > 0}, Ω− w def = {t ∈ I : w(t) < 0}. Also, to formulate the main results of this paper we need the following definitions: Definition 1.1.1. Let A be a finite (eventually empty) subset of I. We say that f ∈ E(A), if f ∈ K(I × R; R) and, for any measurable set G ⊆ I and an arbitrary constant r > 0, we can choose ε > 0 such that if G |f(s, x)|ds = 0 for x ≥ r (x ≤ −r) then G\Uε |f(s, x)|ds − Uε |f(s, x)|ds ≥ 0 for x ≥ r (x ≤ −r), where Uε = I ∩ ∪n k=1]tk − ε/2n, tk + ε/2n[ if A = {t1, t2, ..., tn}, and Uε = ∅ if A = ∅. 6 Remark 1.1.1. It is clear that if f(t, x) def ≡ f0(t)g0(x), where f0 ∈ L([a, b]) and g0 ∈ C(R), then f ∈ E(A) for every finite set A ⊂ I. Now we can consider the main result of our paper. The first theorem deals with a case when Nw = ∅, which for problem (1.1.7),(1.1.6) corresponds to the case λ = 1. Theorem 1.1.1. Let w be a nonzero solution of the problem (1.1.3), (1.1.4), Nw = ∅, (1.1.8) there exist a constant r > 0, nonnegative functions f− , f+ ∈ L([a, b]) and g, h0 ∈ L(I; ]0, +∞[ ) such that f(t, x)sgnx ≤ g(t)|x| + h0(t) for |x| ≥ r (1.1.9) and f(t, x) ≤ −f− (t) for x ≤ −r, f+ (t) ≤ f(t, x) for x ≥ r (1.1.10) on I. Let, moreover, there exist ε > 0 such that − b a f− (s)|w(s)|ds + ε||γr||L ≤ − b a h(s)|w(s)|ds ≤ ≤ b a f+ (s)|w(s)|ds − ε||γr||L, (1.1.111) where γr(t) = sup{|f(t, x)| : |x| ≤ r}. (1.1.12) Then the problem (1.1.1), (1.1.2) has at least one solution. Example 1.1.1. It follows from Theorem 1.1.1 that the equation u′′ (t) = −λ2 u(t) + σ|u(t)|α sgnu(t) + h(t) for 0 ≤ t ≤ π (1.1.13) where σ = 1, λ = 1, and α ∈ ]0, 1], with the conditions (1.1.6) has at least one solution for every h ∈ L([0, π]). And finally let us consider two theorems for the case when Nw is not necessarily empty set, where in the second theorem we assume, that for the function f the representation f(t, x) = f0(t)g0(x) is valid. 7 Theorem 1.1.2. Let i ∈ {0, 1}, w be a nonzero solution of the problem (1.1.3), (1.1.4), f ∈ E(Nw), there exist a constant r > 0 such that the function (−1)i f is non-decreasing in the second argument for |x| ≥ r, (−1)i f(t, x)sgnx ≥ 0 for t ∈ I, |x| ≥ r, (1.1.14) Ω+ w |f(s, r)|ds + Ω− w |f(s, −r)|ds = 0, (1.1.15) and lim |x|→+∞ 1 |x| b a |f(s, x)|ds = 0. (1.1.16) Then there exists δ > 0 such that the problem (1.1.1), (1.1.2) has at least one solution for every h satisfying the condition b a h(s)w(s)ds < δ. (1.1.17) Example 1.1.2. From Theorem 1.1.2 it follows that the problem (1.1.13), (1.1.6) with σ ∈ {−1, 1}, λ ∈ N, and α ∈ ]0, 1[ has at least one solution if h ∈ L([0, π]) is such that π 0 h(s)sinλsds = 0. Theorem 1.1.3. Let i ∈ {0, 1}, w be a nonzero solution of the problem (1.1.3),(1.1.4), f(t, x) def ≡ f0(t)g0(x) with nonnegative f0 ∈ L([a, b]), g0 ∈ C(R), there exist a constant r > 0 such that (−1)i g0 is non-decreasing for |x| ≥ r and (−1)i g0(x)sgnx ≥ 0 for |x| ≥ r. (1.1.18) Let, moreover, |g0(r)| Ω+ w f0(s)ds + |g0(−r)| Ω− w f0(s)ds = 0 (1.1.19) and lim |x|→+∞ |g0(x)| = +∞, lim |x|→+∞ g0(x) x = 0. (1.1.20) Then, for every h ∈ L(I; R), the problem (1.1.1), (1.1.2) has at least one solution. Example 1.1.3. From Theorem 1.1.3 it follows that the equation u′′ (t) = p0(t)u(t) + p1(t)|u(t)|α sgnu(t) + h(t) for t ∈ I, (1.1.21) where α ∈ ]0, 1[ and p0, p1, h ∈ L([a, b]), with the conditions (1.1.2) has at least one solution provided that p1(t) > 0 for t ∈ I. 8 1.2 A Periodic Boundary Value Problem For Functional Differential Equations Of Higher Order One of the most significant problem among two point boundary value problems is the periodic problem. In the paper [5] the problem of existence and uniqueness of solution is studied for the higher-order linear functionaldifferential equation u(n) (t) = n−1 i=0 ℓ(u(i) )(t) + q(t) (1.2.1) under the periodic boundary conditions u(j) (0) = u(j) (ω) + cj (j = 0, . . . , n − 1), (1.2.2) where n ≥ 2, ℓ : C([0, ω]) → L([0, ω]) are linear bounded operators, q ∈ L([0, ω]), and cj ∈ R (i, j = 0, . . . , n − 1). By a solution to the problem (1.2.1), (1.2.2) we understand a function u ∈ Cn−1 ([0, ω]), which satisfies the equality (1.2.1) almost everywhere in [0, ω] and the boundary condition (1.2.2). The problem on the existence of a periodic solution to ordinary and functional differential equations was studied very intensively in the past. The first important step was made for linear ordinary differential equations of the type u(n) (t) = p(t)u(t) + q(t) (1.2.3) by Lasota and Opial. They showed that the problem (1.2.3), (1.2.2) is uniquely solvable for n ≥ 4 if the function p ∈ L([0, ω]) has a constant sign, p ≡ 0, and the inequality ω o |p(s)|ds < 2 ω n−1 2 · 4 · · ·(n − 2) 1 · 3 · · ·(n − 3) (1.2.4) is fulfilled. This result is far from being optimal Below we consider conditions guaranteeing the unique solvability of the problem (1.2.1), (1.2.2), even in case when the operators ℓi are not monotone, which improve the results of Lasota – Opial and Kiguradze – Kusano and are optimal for n ≤ 7. The method used for the investigation of the considered problem is based on the method developed in our previous papers for functional differential equations. 9 Definition 1.2.1. We will say that a linear operator ℓ : C([0, ω]) → L([0, ω]) belongs to the set Pω if it is non–negative, i.e., for any non–negative x ∈ C([0, ω]) the inequality ℓ(x)(t) ≥ 0 for t ∈ [0, ω] is fulfilled. The following notations is used throughout this part of our survey: N is a set of all natural numbers. If ℓ : C([0, ω]) → L([0, ω]) is a linear bounded operator, then ℓ = sup x C ≤1 ℓ(x) L . A0 = 1, A1 = 1 15 , Aj = A1 2 m1=1 m1+1 m2=1 . . . mj−2+1 mj−1=1 1 η(m1) . . . η(mj−1) , B1 = 1 8 , Bj = A1 2 m1=1 m1+1 m2=1 . . . mj−2+1 mj−1=1 1 η(m1) . . . η(mj−1) mj−1+1 i=1 1 + 1 2i , for j ≥ 2, where η(t) = (2t + 1)(2t + 3). Let d0 = 1, d1 = 4, d2 = 32, d3 = 192, (1.2.5) and for p ∈ N put d2p+2 = 1 max (hp(t)hp(1 − t))1/2 : 0 ≤ t ≤ 1 , d2p+3 = 1 max (fp(s, t)fp(1 − s, 1 − t))1/2 : 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 , (1.2.6) where the functions fp : [0, 1] × [0, 1] → R+, hp : [0, 1] → R+ are defined as follows: fp(s, t) = p−1 j=0 αpjt2(j+1) + αppt2p+3 s, hp(t) = p j=0 βpjt2(j+1) , (1.2.7) and αpj = Aj 3 · 4j+1d2(p−j)+1 , βpj = Aj 3 · 4j+1d2(p−j) (j = 0, . . . , p − 1), αpp = Ap 3 · 4p+1 , βpp = Bp 3 · 4p+1 . (1.2.8) 10 Now we can formulate our main theorem on unique solvability of problem (1.2.1), (1.2.2). Theorem 1.2.1. Let j ∈ {0, 1}, the operator ℓ0 admit the representation ℓ0 = ℓ0,1 − ℓ0,2, where ℓ0,1, ℓ0,2 ∈ Pω, and let ℓi (i = 1, . . . , n − 1) be bounded linear operators. Let, moreover, the conditions ||ℓ0,1|| + ||ℓ0,2|| = 0 (1.2.9) ωn−1 dn−1 ||ℓ0,1+j|| + Ω < 1, (1.2.10) ||ℓ0,1+j|| 1 − Ω − ωn−1 dn−1 ||ℓ0,1+j|| ≤ ||ℓ0,2−j||, (1.2.11) ||ℓ0,2−j|| ≤ 2dn−1 ωn−1 1 − Ω + (1 − Ω) 1 − Ω − ωn−1 dn−1 ||ℓ0,1+j|| (1.2.12) hold with Ω = n−1 i=1 ωn−1−i dn−1−i ||ℓi|| (1.2.13) and di (i = 0, . . . , n − 1) be defined by (1.2.5)–(1.2.8). Then the problem (1.2.1), (1.2.2) has a unique solution. In the case when the operator ℓ0 is monotone from our theorem it follows: Corollary 1.2.1. Let σ ∈ {−1, 1} and σℓ0 ∈ Pω. Let, moreover, the condi- tions ℓ0 = 0, (1.2.14) n−1 i=1 ωn−1−i dn−1−i ℓi < 1, (1.2.15) and ℓ0 ≤ 4dn−1 ωn−1 1 − n−1 i=1 ωn−1−i dn−1−i ℓi (1.2.16) hold. Then the problem (1.2.1), (1.2.2) has a unique solution. To illustrate our theorem, we consider also one corollary for the equation u(n) (t) = ℓ0(u)(t) + q(t). (1.2.17) 11 Corollary 1.2.2. Let σ ∈ {−1, 1}, σℓ0 ∈ C([0, ω]). Let, moreover, the conditions ℓ0 = 0, (1.2.18) and ℓ0 ≤ 4dn−1 ωn−1 (1.2.19) hold. Then the problem (1.2.17), (1.2.2) has a unique solution. 12 Kapitola 2 Two-Point Boundary Value Problems For Singular Functional-Differential Equations 2.1 Two-point boundary value problems for second order functional-differential equati- ons First we consider some results from the monograph [1], where the second order linear singular functional-differential equation u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + g(u)(t) + p2(t) (2.1.1) is studied under the boundary conditions u(a) = c1, u(b) = c2 (2.1.2) or u(a) = c1, u(b) = c2, (2.1.3) and separately for the case of homogeneous conditions u(a) = 0, u(b) = 0, (2.1.4) u(a) = 0, u(b) = 0, (2.1.5) 13 where c1, c2 ∈ R pj ∈ Lloc(]a, b[) (j = 0, 1, 2) and g : C(]a, b[) → Lloc(]a, b[) is a continuous linear operator. In this short survey we consider only four theorems and its corollaries about unique solvability of problems (2.1.1), (2.1.2), and (2.1.1), (2.1.4), from twelve theorems and its corollaries proved in this monograph. We do not consider problems (2.1.1), (2.1.3), and (2.1.1), (2.1.5), and theorems on the correctness of the above mentioned problems. Throughout the work the following notations are used: [x]+ = 1 2 |x| + x , [x]− = 1 2 |x| − x . C(]a, b[) is the space of continuous and bounded functions u :]a, b[→ R with the norm ||u||C = sup{|u(t)| : a < t < b}; C(]a, b[) is the set of functions u :]a, b[→ R absolutely continuous on each subsegment of ]a, b[. C′ (]a, b[) is the set of functions u :]a, b[→ R absolutely continuous on each subsegment of ]a, b[, along with their first order derivative. L([a, b]) is the space of summable functions u : [a, b] → R with the norm ||u||L = b a |u(t)|dt. L+∞([a, b]) is the space of essentially bounded functions u : [a, b] → R with the norm ||u||+∞ = essup{|u(t)| : t ∈ [a, b]}. Lloc(]a, b[) is the set of measurable functions u : [a, b] → R, summable on each subsegment of ]a, b[. Let x, y :]a, b[→]0, +∞[ be continuous functions. Cx(]a, b[) is the space of continuous u ∈ C(]a, b[) such that ||u||C,x = sup |u(t)| x(t) : a < t < b < +∞; Ly(]a, b[) is the space of functions u ∈ Lloc(]a, b[) such that ||u||L,y = b a y(t)|u(t)|dt < +∞; L(Cx, Lx) is the set of linear operators h : Cx(]a, b[→ Ly(]a, b[) such that sup{|h(x)(·)| : ||u||C,x ≤ 1} ∈ Ly(]a, b[); 14 σ : Lloc(]a, b[) → C(]a, b[) is the operator defined by σ(p)(t) = exp t a+b 2 p(s)ds for a ≤ t ≤ b. If σ(p) ∈ L([a, b]), then we define the operators σ1 and σ2 by σ1(p)(t) = 1 σ(p)(t) t a σ(p)(s)ds b t σ(p)(s)ds σ2(p)(t) = 1 σ(p)(t) t a σ(p)(s)ds for a ≤ t ≤ b Let, f, g ∈ C(]a, b[) and c ∈ [a, b], than we write f(t) = O(g(t)) (f(t) = O∗ (g(t))) as t → c, if lim sup t→c |f(t)| |g(t)| < +∞ 0 < lim inf t→c |f(t)| |g(t)| and lim sup t→c |f(t)| |g(t)| < +∞ . Now note that the problems (2.1.1), (2.1.2), and (2.1.1), (2.1.4) are studied under the assumptions pj ∈ Lloc(]a, b[) (j = 0, 1, 2), σ(p1) ∈ L([a, b]), p0 ∈ Lσ1(p1)([a, b]), (2.1.6) by the method of Nagumo’s upper and lower functions, and we find the conditions under which Fredholm’s alternative is valid, introduce the sets of nonoscillation Vi,0, and describe sets of two-dimensional vector functions (p0, p1) :]a, b[→ R2 , and linear operators h, for which our problem is uniquely solvable. Definition 2.1.1. We will say that w ∈ C(]a, b[) is the lower (upper) function of the problem (2.1.1), (2.1.2) if: (a) w′ is of the form w′ (t) = w0(t) + w1(t), where w0 :]a, b[→ R is absolutely continuous on each segment from ]a, b[, the function w1 :]a, b[→ R is nondecreasing (nonincreasing) and its derivative is almost everiwhere equal to zero; 15 (b) almost everywhere on ]a, b[ the inequality w′′ (t) ≥ p0(t)w(t) + p1(t)w′ (t) + g(w)(t) + p2(t) (w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) + g(w)(t) + p2(t)) is satisfaed; (c) there exists the limit w′ (b−) and w(a) ≤ c1, w(b − 0) ≤ c2 (w(a) ≥ c1, w(b − 0) ≥ c2). Definition 2.1.2. We will say that two-dimensional vector function (p0, p1) : ]a, b[→ R2 belongs to the set V1,0(]a, b[) if the conditions (2.1.6) are fulfilled, the solution of the problem u′′ (t) = p0(t)u(t) + p1(t)u′ (t), (2.1.7) u(a) = 0, lim t→a u′ (t) σ(p1)(t) = 1, has no zeros in the interval ]a, b[ and u(b−) ≥ 0. Definition 2.1.3. Let h : C(]a, b[) → Lloc(]a, b[) be a continuous linear operator. We will say that a two-dimensional vector function (p0, p1) :]a, b[→ R2 belong to the set V1,0(]a, b[, h) if the conditions (2.1.6) are satisfied and the problem u′′ (t) = p0(t)u(t) + p1(t)u′ (t) − h(u)(t) u(a) = 0, u(b−) = 0 has a positive upper function w on the segment [a, b]. Definition 2.1.4. Let h : C(]a, b[) → Lloc(]a, b[) be a continuous linear operator. We will say that a two-dimensional vector function (p0, p1) :]a, b[→ R2 belong to the set V1,β(]a, b[, h) if (p0, p1) ∈ V1,0(]a, b[, h) and there exists a measurable function qβ :]a, b[→ [0, +∞[ such that b a |G(t, s)|qβ(s)ds = O∗ (xβ (t)) as t → a, b → b, where G is Green’s function of the problem (2.1.7), (2.1.4), and x(t) = t a σ(p1)(s)ds b t σ(p1)(s)ds (2.1.8) for a ≤ t ≤ b. 16 Now we can consider some basic results of our monograph. Theorem 2.1.1. Let p2 ∈ Lσ1(p1)([a, b]) (2.1.9) and the constants α, β ∈ [0, 1] connected by the inequality α + β ≤ 1 (2.1.10) be such that (p0, p1) ∈ V1,β(]a, b[, h), (2.1.11) where h ∈ L(Cxβ , L xα σ(p1) ) ∩ L(C, Lσ1(p1)) (2.1.12) is a nonnegative operator and the function x is defined by (2.1.8). Let moreover a continuous linear operator g : C(]a, b[) → Lσ1(p1)([a, b]) be such that for any function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality |g(u)(t)| ≤ h(|u|)(t) (2.1.13) is satisfied. Then the problem (2.1.1), (2.1.2) has one and only one solution. From this theorem follows a few efficient sufficient conditions of unique solvability. Let us consider one of these corollaries: Corollary 2.1.1. Let the function x is defined by (2.1.8), the constants α, β ∈ [0, 1] by connected by (2.1.10), the function pj :]a, b[→ R (j = 0, 1, 2) satisfy conditions (2.1.6), (2.1.9), [p0]− ∈ L xα σ(p1) ([a, b]), (2.1.14) and for any function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality (2.1.13) be satisfied, where a nonnegative operator h satisfies (2.1.12). 17 Let moreover, b t σ(p1)(η)dη α t a ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) s a σ(p1)(η)dη α ds+ t a σ(p1)(η)dη α b t ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) b s σ(p1)(η)dη α ds < < 4 b a σ(p1)(η)dη b a σ(p1)(η)dη 2 2(α+β) . (2.1.15) Then the problem (2.1.1), (2.1.2) has one and only one solution. The next theorem shows us that in the case of boundary conditions (2.1.4) the singularity of functions p0, p1 and operator h, can be stronger as in the case of the boundary conditions (2.1.2). Theorem 2.1.2. Let the constants α, β ∈ [0, 1] connected by the inequality (2.1.10) be such that p2 ∈ Lx1−β σ(p1) ([a, b]) (2.1.16) and the functions p0, p1 :]a, b[→ R satisfy the inclusion (2.1.11), where h ∈ L(Cxβ , L xα σ(p1) ) (2.1.17) is a nonnegative operator and the function x is defined by (2.1.8). Let moreover a continuous linear operator g : Cxβ (]a, b[) → L xα σ(p1) ([a, b]) be such that for any function u ∈ Cxβ (]a, b[) almost everywhere in the interval ]a, b[ the inequality (2.1.13) is satisfied. Then the problem (2.1.1), (2.1.4) has one and only one solution in the space Cxβ (]a, b[). Corollary 2.1.2. Let the function x is defined by (2.1.8), the constants α, β ∈ [0, 1] by connected by (2.1.10), the function pj :]a, b[→ R (j = 0, 1, 2) satisfy conditions (2.1.6), (2.1.16),(2.1.14) and for any function u ∈ Cxβ (]a, b[) almost everywhere in the interval ]a, b[ the inequality (2.1.13) be satisfied, where the nonnegative operator h satisfies the inclusion (2.1.17). Let moreover (2.1.15) be satisfied. Then the problem (2.1.1), (2.1.4) has one and only one solution in the space Cxβ (]a, b[). 18 For clearness we will give two corollaries for the equation u′′ (t) = g0(t)u(τ(t)) + p2(t), (2.1.18) the first in the case of boundary conditions (2.1.2), and the second in the case of (2.1.4). Corollary 2.1.3. Let the constants α, β ∈ [0, 1] by connected by (2.1.10), τ : [a, b] → [a, b] be a measurable function and g0, p2 ∈ Lx([a, b]), (2.1.19) where x(t) = (t − a)(b − t) for a ≤ t ≤ b. (2.1.20) Let, moreover, b a |g0(s)|[(τ(s) − a)(b − τ(s))]β [(s − a)(b − s)]α ds < < 41−α−β (b − a)2(α+β)−1 . (2.1.21) Then the problem (2.1.18), (2.1.2) has one and only one solution. Corollary 2.1.4. Let the constants α, β ∈ [0, 1] by connected by (2.1.10), τ : [a, b] → [a, b] be a measurable function and p2 ∈ Lx1−β ([a, b]), (2.1.22) where the function x be defined by (2.1.20). Let, moreover condition (2.1.21) be satisfied. Then the problem (2.1.18), (2.1.4) has one and only one solution in the space Cxβ (]a, b[). In the monograph a different method of study of our boundary value problems is also developed, the method of minimums and maximums. The next two nonimprovable theorems are proved by this method Theorem 2.1.3. Let γ ∈ [0, 1] p2 ∈ Lx([a, b]) (2.1.23) and g ∈ L(C, Lxγ ) (2.1.24) be a nonnegative operator, where the function x is defined by (2.1.20). 19 Let, moreover, there exist constants α, β ∈ [0, 1/2] such that 0 ≤ β < 1 − γ, (2.1.25) α + β ≤ 1/2 (2.1.26) and b a xα (s)g(xβ )(s)ds < 2β 16 b − a b − a 4 2(α+β) . (2.1.27) Then the problem (2.1.18), (2.1.2) has one and only one solution. Theorem 2.1.4. Let γ ∈ [0, 1], δ ∈]0, 1 − γ[, p2 ∈ Lxγ ([a, b]) (2.1.28) and g ∈ L(Cxδ , Lxγ ) (2.1.29) be a nonnegative operator, where the function x is defined by (2.1.20). Let, moreover, there exist constants α ∈ [0, 1/2], β ∈]0, 1/2] such that the conditions (2.1.26) and b a xα (s)g(xβ )(s)ds ≤ 2β 16 b − a b − a 4 2(α+β) (2.1.30) are satisfied. Then the problem (2.1.18), (2.1.4) has one and only one solution in the space Cxδ (]a, b[). The conditions (2.1.27) and (2.1.30) are unimprovable in the sense that it cannot be replaced by the conditions b a xα (s)g(xβ )(s)ds < 2β 16 b − a b − a 4 2(α+β) + ε b a xα (s)g(xβ )(s)ds ≤ 2β 16 b − a b − a 4 2(α+β) + ε . 20 2.2 Two-point boundary value problems for strongly singular higher-order linear differential equations with deviating argu- ments In this section we consider the main results from the paper [2], where the differential equation with deviating arguments u(n) (t) = m j=1 pj(t)u(j−1) (τj(t)) + q(t) for a < t < b, (2.2.1) is studied with the two-point boundary value conditions u(i−1) (a) = 0 (i = 1, · · · , m), u(j−1) (b) = 0 (j = 1, · · · , n − m), (2.2.2) u(i−1) (a) = 0 (i = 1, · · · , m), u(j−1) (b) = 0 (j = m + 1, · · · , n). (2.2.3) Here n ≥ 2, m is the integer part of n/2, −∞ < a < b < +∞, pj, q ∈ Lloc(]a, b[) (j = 1, · · · , m), and τj :]a, b[→]a, b[ are measurable functions. By u(j−1) (a) (u(j−1) (b)) we denote the right (the left) limit of the function u(j−1) at the point a (b). Problems (1.2.9), (1.2.10), and (1.2.9), (2.2.3) are said to be strong singular if some or all the coefficients of (1.2.9) are nonintegrable on [a, b], having singularities at the end-points of this segment and the conditions b a (s − a)n−1 (b − s)2m−1 [(−1)n−m p1(s)]+ds < +∞, b a (s − a)n−j (b − s)2m−j |pj(s)|ds < +∞ (j = 2, · · · , m), b a (s − a)n−m−1/2 (b − s)m−1/2 |q(s)|ds < +∞, (2.2.4) 21 in the case of conditions (2.2.2), and b a (s − a)n−1 [(−1)n−m p1(s)]+ds < +∞, b a (s − a)n−j |pj(s)|ds < +∞ (j = 2, · · · , m), b a (s − a)n−m−1/2 |q(s)|ds < +∞, (2.2.5) in the case of conditions (2.2.3), are not fulfilled. Here we consider only the case of problem (2.2.1), (2.2.2), by using the following notations R+ = [0, +∞[; [x] is an integer part of x; Lα,β(]a, b[) is the space of integrable (square integrable) with the weight (t − a)α (b − t)β functions y :]a, b[→ R, with the norm ||y||Lα,β = b a (s − a)α (b − s)β |y(s)|ds; L([a, b]) = L0,0(]a, b[), L2 ([a, b]) = L2 0,0(]a, b[); M(]a, b[) is the set of measurable functions τ :]a, b[→]a, b[; L2 α,β(]a, b[) is the Banach space of functions y ∈ Lloc(]a, b[) (Lloc(]a, b])), satisfying µ1 ≡ max t a (s − a)α t s y(ξ)dξ 2 ds 1/2 : a ≤ t ≤ a + b 2 + + max b t (b − s)β s t y(ξ)dξ 2 ds 1/2 : a + b 2 ≤ t ≤ b < +∞, The norm in this space is defined by the equality || · ||L2 α,β = µ1. Ck (]a, b[) is a set of functions u : [0, ω] → R, which are absolutely continuous together with their derivatives up to the k-th order. 22 Cn−1, m (]a, b[) is the space of functions y ∈ Cn−1 loc (]a, b[), satisfying b a |y(m) (s)|2 ds < +∞. (2.2.6) When n = 2m, we assume that pj ∈ Lloc(]a, b[) (j = 1, · · · , m), (2.2.7) and if n = 2m + 1, we assume that along with (2.2.7), the condition lim sup t→b (b − t)2m−1 t t1 p1(s)ds < +∞ (t1 = a + b 2 ) (2.2.8) is fulfilled. Along with (1.2.9), we consider the homogeneous equation v(n) (t) = m j=1 pj(t)v(j−1) (τj(t)) for a < t < b. (1.10) In the case where conditions (2.2.4) and (2.2.5) are violated, the question on the presence of the Fredholm’s property for problem (2.2.1), (2.2.2) in some subspace of the space Cn−1 loc (]a, b[) remains so far open. This question is answered in Theorem 2.2.1 formulated below which contains optimal in a certain sense conditions guaranteeing the Fredholm’s property for problem (2.2.1), (2.2.2) in the space Cn−1, m (]a, b[). A solution of problem (2.2.1), (2.2.2) is sought in the space Cn−1, m (]a, b[). In order to formulate the above-mentioned theorem we need following definitions: Let hj :]a, b[×]a, b[→ R+ and fj : R × M(]a, b[) → Cloc(]a, b[×]a, b[) (j = 1, . . . , m) be the functions and the operators, respectively, defined by the equalities h1(t, s) = t s (ξ − a)n−2m [(−1)n−m p1(ξ)]+dξ , hj(t, s) = t s (ξ − a)n−2m pj(ξ)dξ (j = 2, · · · , m), (2.2.9) 23 and, fj(c, τj)(t, s)= t s (ξ−a)n−2m |pj(ξ)| τj (ξ) ξ (ξ1−c)2(m−j) dξ1 1/2 dξ (j = 1, · · · , m). (2.2.10) Let, moreover, m!! = 1 for m ≤ 0 1 · 3 · 5 · · ·m for m ≥ 1 , if m = 2k + 1. Definition 2.2.1. We will say that problem (2.2.1), (2.2.2) has the Fredholm’s property in the space Cn−1,m (]a, b[), if the unique solvability of the corresponding homogeneous problem (1.10), (1.2.10) in that space implies the unique solvability of problem (2.2.1), (2.2.2) for every q ∈ L2 2n−2m−2, 2m−2(]a, b[). Theorem 2.2.1. Let there exist a0 ∈]a, b[, b0 ∈]a0, b[, numbers lkj > 0, γkj > 0, and functions τj ∈ M(]a, b[) (k = 0, 1, j = 1, . . . , m) such that (t − a)2m−j hj(t, s) ≤ l0j for a < t ≤ s ≤ a0, lim sup t→a (t − a)m− 1 2 −γ0j fj(a, τj)(t, s) < +∞, (2.2.11) (b − t)2m−j hj(t, s) ≤ l1j for b0 ≤ s ≤ t < b, lim sup t→b (b − t)m− 1 2 −γ1j fj(b, τj)(t, s) < +∞, (2.2.12) and m j=1 (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! lkj < 1 (k = 0, 1). (2.2.13) Let, moreover, (1.10), (2.2.2) have only the trivial solution in the space Cn−1,m (]a, b[). Then problem (2.2.1), (2.2.2) has the unique solution u for every q ∈ L2 2n−2m−2, 2m−2(]a, b[), and there exists a constant r, independent of q, such that ||u(m) ||L2 ≤ r||q||L2 2n−2m−2, 2m−2 . (2.2.14) Remark 2.2.1. There exists an example which demonstrates that strict inequality (2.2.13) is sharp because it cannot be replaced by nonstrict one. The next theorem (the theorem of unique solvability) is proved on the basis of Theorem 2.2.1 which gives us the sharp sufficient conditions under which our problem has the Fredholm’s property. 24 Theorem 2.2.2. Let there exist numbers t∗ ∈]a, b[, lkj > 0, lkj ≥ 0, and γkj > 0 (k = 0, 1; j = 1, . . . , m) such that along with B0 ≡ m j=1 (2m − j)22m−j+1 l0j (2m − 1)!!(2m − 2j + 1)!! + + 22m−j−1 (t∗ − a)γ0j l0j (2m − 2j − 1)!!(2m − 3)!! 2γ0j < 1 2 , (2.2.15) B1 ≡ m j=1 (2m − j)22m−j+1 l1j (2m − 1)!!(2m − 2j + 1)!! + + 22m−j−1 (b − t∗ )γ0j l1j (2m − 2j − 1)!!(2m − 3)!! 2γ1j < 1 2 , (2.2.16) the conditions (t−a)2m−j hj(t, s) ≤ l0j, (t−a)m−γ0j −1/2 fj(a, τj)(t, s) ≤ l0j for a 0. Cm−1 1 (]a, b[) is the Banach space of the functions y ∈ Cm−1 loc (]a, b[), such that lim sup t→a |x(i−1) (t)| (t − a)m−i+1/2 < +∞ (i = 1, · · · , m), lim sup t→b |x(i−1) (t)| (b − t)m−i+1/2 < +∞ (i = 1, · · · , n − m), (2.3.6) with the norm: ||x||Cm−1 1 = m i=1 sup |x(i−1) (t)| αi(t) : a < t < b , where αi(t) = (t − a)m−i+1/2 (b − t)m−i+1/2 . Cm−1 1 (]a, b[) is the Banach space of the functions y ∈ Cm−1 loc (]a, b[), such that conditions b a (ym (s))2 ds < +∞, and (2.3.6) hold, with the norm: ||x||Cm−1 1 = m i=1 sup |x(i−1) (t)| αi(t) : a < t < b + b a |x(m) (s)|2 ds 1/2 . Dn(]a, b[×R+ ) is the set of such functions δ :]a, b[×R+ → Ln(]a, b[) that δ(t, ·) : R+ → R+ is nondecreasing for every t ∈]a, b[, and δ(·, ρ) ∈ Ln(]a, b[) for any ρ ∈ R+ . D2n−2m−2, 2m−2(]a, b[×R+ ) is the set of such functions δ :]a, b[×R+ → L2 2n−2m−2, 2m−2(]a, b[) that δ(t, ·) : R+ → R+ is nondecreasing for every t ∈ ]a, b[, and δ(·, ρ) ∈ L2 2n−2m−2, 2m−2(]a, b[) for any ρ ∈ R+ . 27 In this paper, we prove a priori boundedness principle for the problem (2.3.1), (2.3.2) in the case where the operator has form (2.3.3). For formulate this a priori boundedness principle we have to define the set Aγ = {x ∈ Cm−1 1 (]a, b[) : ||x||Cm−1 1 ≤ γ} (2.3.7) for any γ > 0, and the operator P : Cm−1 1 (]a, b[)×Cm−1 1 (]a, b[) → Lloc(]a, b[) by the equality P(x, y)(t) = m j=1 pj(x)(t)y(j−1) (τj(t)) for a < t < b, (2.3.8) where pj : Cm−1 1 (]a, b[) → Lloc(]a, b[), τj ∈ M(]a, b[), and introduce the definition: Definition 2.3.1. Let γ0 and γ be the positive numbers. We said that the continuous operator P : Cm−1 1 (]a, b[) × Cm−1 1 (]a, b[) → Ln(]a, b[) to be γ0, γ consistent with boundary condition (2.3.2) if: i. for any x ∈ Aγ0 and almost all t ∈]a, b[ the inequality m j=1 |pj(x)(t)x(j−1) (τj(t))| ≤ δ(t, ||x||Cm−1 1 )||x||Cm−1 1 (2.3.9) holds, where δ ∈ Dn(]a, b[×R+ ). ii. for any x ∈ Aγ0 and q ∈ L2 2n−2m−2, 2m−2(]a, b[) the equation y(n) (t) = m j=1 pj(x)(t)y(j−1) (τj(t)) + q(t) (2.3.10) under boundary conditions (2.3.2), has the unique solution y in the space Cn−1, m (]a, b[) and ||y||Cm−1 1 ≤ γ||q||L2 2n−2m−2, 2m−2 . (2.3.11) In the sequel it will always be assumed that the operator Fp defined by equality Fp(x)(t) = |F(x)(t) − m j=1 pj(x)(t)x(j−1) (τj(t))(t)|, continuously acting from Cm−1 1 (]a, b[) to LL2 2n−2m−2, 2m−2 (]a, b[), and Fp(t, ρ) ≡ sup{Fp(x)(t) : ||x||Cm−1 1 ≤ ρ} ∈ L2 2n−2m−2, 2m−2(]a, b[) (2.3.12) for each ρ ∈ [0, +∞[. Then the following theorem is valid 28 Theorem 2.3.1. Let the operator P be γ0, γ consistent with boundary condition (2.3.2), and there exists a positive number ρ0 ≤ γ0, such that ||Fp( ·, min{2ρ0, γ0})||L2 2n−2m−2, 2m−2 ≤ γ0 γ . (2.3.13) Let moreover, for any λ ∈]0, 1[, an arbitrary solution x ∈ Aγ0 of the equation x(n) (t) = (1 − λ)P(x, x)(t) + λF(x)(t) (2.3.14) under the conditions (2.3.2), admits the estimate ||x||Cm−1 1 ≤ ρ0. (2.3.15) Then problem (2.3.1), (2.3.2) is solvable in the space Cn−1,m (]a, b[). On the bases of this theorem we can prove some efficient theorems. Let us consider one of them. In order to consider them we define the operators: hj : Cm−1 1 (]a, b[)×]a, b[×]a, b[→ Lloc(]a, b[×]a, b[), fj : Cm−1 1 (]a, b[) × [a, b] × M(]a, b[) → Cloc(]a, b[×]a, b[) (j = 1, . . . , m) by the equalities h1(x, t, s) = t s (ξ − a)n−2m [(−1)n−m p1(x)(ξ)]+dξ , hj(x, t, s) = t s (ξ − a)n−2m pj(x)(ξ)dξ (j = 2, · · · , m), (2.3.16) and fj(x, c, τj)(t, s) = = t s (ξ − a)n−2m |pj(x)(ξ)| τj (ξ) ξ (ξ1 − c)2(m−j) dξ1 1/2 dξ . (2.3.17) Theorem 2.3.2. Let the continuous operator P : Cm−1 1 (]a, b[)×Cm−1 1 (]a, b[) → Ln(]a, b[) admits to the condition (2.3.9) where δ ∈ Dn(]a, b[×R+ ), τj ∈ M(]a, b[) and the numbers γ0, t∗ ∈]a, b[, lkj > 0, lkj > 0, γkj > 0 (k = 1, 2; j = 1, · · · , m), be such that the inequalities (t − a)2m−j hj(x, t, s) ≤ l0j, lim sup t→a (t − a)m− 1 2 −γ0j fj(x, a, τj)(t, s) ≤ l0j (2.3.18) 29 for a < t ≤ s ≤ t∗ , ||x||Cm−1 1 ≤ γ0, (b − t)2m−j hj(x, t, s) ≤ l1j, lim sup t→b (b − t)m− 1 2 −γ1j fj(x, b, τj)(t, s) ≤ l1j (2.3.19) for t∗ ≤ s ≤ t < b, ||x||Cm−1 1 ≤ γ0, and conditions (2.2.15), (2.2.16) hold. Let moreover the operator F and function η ∈ D2n−2m−2, 2m−2(]a, b[×R+ ) be such that condition |F(x)(t) − m j=1 pj(x)(t)x(j−1) (τj(t))(t)| ≤ η(t, ||x||Cm−1 1 ), (2.3.20) and inequality ||η(·, γ0)||L2 2n−2m−2, 2m−2 < γ0 rn , (2.3.21) be fulfilled, where rn = 1 + m j=1 2m−j+1/2 (m − j)!(2m − 2j + 1)1/2(b − a)m−j+1/2 × × 2m (1 + b − a)(2n − 2m − 1) (νn − 2 max{B0, B1})(2m − 1)!! . Then problem (2.3.1), (2.3.2) is solvable in the space Cn−1,m (]a, b[). To illustrate this theorem, we consider its corollary for the equation u′′ (t) = − λ|u(t)|k [(t − a)(b − t)]2+k/2 u(τ(t)) + q(x)(t), (2.3.22) where λ, k ∈ R+ , the function τ ∈ M(]a, b[), the operator q : Cm−1 1 (]a, b[) → L2 0,0(]a, b[) is continuous and η(t, ρ) ≡ sup{|q(x)(t)| : ||x||Cm−1 1 ≤ ρ} ∈ L2 0,0(]a, b[). Than from Theorem 2.3.2 it follows Corollary 2.3.1. Let the function τ ∈ M(]a, b[), the continuous operator q : Cm−1 1 (]a, b[) → L2 0,0(]a, b[), and the numbers γ0 > 0, λ ≥ 0, k > 0, by such that |τ(t) − t| ≤ (t − a)3/2 for a < t ≤ (a + b)/2 (b − t)3/2 for (a + b)/2 ≤ t < b , (2.3.23) 30 ||η(t, γ0)||L2 0, 0 ≤ ≤ 1 + 2 b − a −1 (b − a)2 − 16λγk 0 (1 + [2(b − a)]1/4 ) 2(1 + b − a)(b − a)2 , (2.3.24) and λ < (b − a)2 32γk 0 (1 + [2(b − a)]1/4) . (2.3.25) Then the problem (2.3.22), (2.3.2) is solvable. 31 Literatura [1] Mukhigulashvili S., Two-point boundary value problems for second order functional differential equations, Mem. Differential Equations Math. Phys., 20 (2000), 1-112. [2] Mukhigulashvili S., Partsvania N., Two-point boundary value problems for strongly singular higher-order linear differential equations with deviating arguments, E. J. Qualitative Theory of Diff. Equ., 2012, No.38, 1-34. [3] Mukhigulashvili S., The Dirichlet Boundary Value Problems For Strongly Singular Higher-Order Nonlinear Functional-Differential Equations, Czechoslovak Mathematical Journal, vol. 63 (2013), No. 1, pp. 235-263. [4] Mukhigulashvili S., The Dirichlet BVP The second Order Nonlinear Ordinary Differential Equation At Resonance, Italian J. Of Pure and Appl. Math., 2011, No.28, 177-204, [5] Mukhigulashvili S., Hakl R., A Periodic Boundary Value Problem For Functional-Differential Equations Of Higher Order, Georgian Math. J. Vol.16, (2009), No.4, 651-665. The reader can see the literature used in this works in the articles attached to the survey. 32 italian journal of pure and applied mathematics – n. 28−2011 (177−204) 177 THE DIRICHLET BVP FOR THE SECOND ORDER NONLINEAR ORDINARY DIFFERENTIAL EQUATION AT RESONANCE Sulkhan Mukhigulashvili Permanent addresses: 1. Institute of Mathematics Academy of Sciences of the Czech Republic ˇZiˇzkova 22, 616 62 Brno Czech Republic 2. I. Chavchavadze State University Faculty of physics and mathematics I. Chavchavadze Str. No. 32, 0179 Tbilisi Georgia e-mail: mukhig@ipm.cz Abstract. Efficient sufficient conditions are established for the solvability of the Dirichlet problem u (t) = p(t)u(t) + f(t, u(t)) + h(t) for a ≤ t ≤ b, u(a) = 0, u(b) = 0, where h, p ∈ L([a, b]; R) and f ∈ K([a, b] × R; R), in the case where the linear problem u (t) = p(t)u(t), u(a) = 0, u(b) = 0 has nontrivial solutions. Key words and phrases: nonlinear ordinary differential equation, Dirichlet problem at resonance. 2000 Mathematics Subject Classification: 34B15, 34C15, 34C25. 1. Introduction Consider on the set I = [a, b] the second order nonlinear ordinary differential equation (1.1) u (t) = p(t)u(t) + f(t, u(t)) + h(t) for t ∈ I 178 s. mukhigulashvili with the boundary conditions (1.2) u(a) = 0, u(b) = 0, where h, p ∈ L(I; R) and f ∈ K(I × R; R). By a solution of the problem (1.1), (1.2) we understand a function u ∈ C (I, R), which satisfies the equation (1.1) almost everywhere on I and satisfies the conditions (1.2). Along with (1.1), (1.2) we consider the homogeneous problem w (t) = p(t)w(t) for t ∈ I,(1.3) w(a) = 0, w(b) = 0.(1.4) At present, the foundations of the general theory of two-point boundary value problems are already laid and problems of this type are studied by many authors and investigated in detail (see, for instance, [1], [4], [5], [8], [12],[13], [14]-[16], [17] and references therein). On the other hand, in all of these works, only the case when the homogeneous problem (1.3), (1.4) has only a trivial solution is studied. The case when the problem (1.3), (1.4) has also the nontrivial solution is still little investigated and in the majority of articles, the authors study the case with p constant in the equation (1.1), i.e., when the problem (1.1), (1.2) and the equation (1.3) are of type u (t) = −λ2 u(t) + f(t, u(t)) + h(t) for t ∈ [0, π],(1.5) u(0) = 0, u(π) = 0,(1.6) and (1.7) w (t) = −λ2 w(t) for t ∈ [0, π] respectively, with λ = 1. (see, for instance, [2], [3], [4], [6]-[11], [14]-[16], and references therein). In the present paper, we study solvability of the problem (1.1), (1.2) in the case when the function p ∈ L(I; R) is not necessarily constant, under the assumption that the homogeneous problem (1.3), (1.4) has the nontrivial solution with an arbitrary number of zeroes. For the equation (1.7), this is the case when λ is not necessarily the first eigenvalue of the problem (1.7), (1.4), with a = 0, b = π. The obtained results are new and generalize some well-known results (see,[2], [3], [4], [6], [10]). The following notation is used throughout the paper: N is the set of all natural numbers. R is the set of all real numbers, R+ = [0, +∞[. C(I; R) is the Banach space of continuous functions u : I → R with the norm u C = max{|u(t)| : t ∈ I}. C (I; R) is the set of functions u : I → R which are absolutely continuous together with their first derivatives. L(I; R) is the Banach space of Lebesgue integrable functions p : I → R with the norm p L = b a |p(s)|ds. the dirichlet bvp for the second order ... 179 K(I × R; R) is the set of functions f : I × R → R satisfying the Carath´eodory conditions, i.e., f(·, x) : I → R is a measurable function for all x ∈ R, f(t, ·) : R → R is a continuous function for almost all t ∈ I, and for every r > 0 there exists qr ∈ L(I; R+) such that |f(t, x)| ≤ qr(t) for almost all t ∈ I, |x| ≤ r. Having w : I → R, we put: Nw def = {t ∈ ]a, b[ : w(t) = 0}, Ω+ w def = {t ∈ I : w(t) > 0}, Ω− w def = {t ∈ I : w(t) < 0}, and [w(t)]+ = (|w(t)| + w(t))/2, [w(t)]− = (|w(t)| − w(t))/2 for t ∈ I. Definition 1.1. Let A be a finite (eventually empty) subset of I. We say that f ∈ E(A), if f ∈ K(I ×R; R) and, for any measurable set G ⊆ I and an arbitrary constant r > 0, we can choose ε > 0 such that if G |f(s, x)|ds = 0 for x ≥ r (x ≤ −r) then G\Uε |f(s, x)|ds − Uε |f(s, x)|ds ≥ 0 for x ≥ r (x ≤ −r), where Uε = I ∩ n k=1 ]tk − ε/2n, tk + ε/2n[ if A = {t1, t2, ..., tn}, and Uε = ∅ if A = ∅. Remark 1.1. If f ∈ K(I × R; R) then f ∈ E(∅). Remark 1.2. It is clear that if f1 ∈ L(I; R) and f(t, x) def ≡ f1(t) then f ∈ E(A) for every finite set A ⊂ I. Remark 1.3. It is clear that if f(t, x) def ≡ f0(t)g0(x), where f0 ∈ L(I; R) and g0 ∈ C(I; R), then f ∈ E(A) for every finite set A ⊂ I. The example below shows that there exists a function f ∈ K(I × R; R) such that f ∈ E({t1, ..., tk}) for some points t1, ..., tk ∈ I. Example 1.1. Let f(t, x) = |t|−1/2 g(t, x) for t ∈ [−1, 0[∪]0, 1], x ∈ R, and f(0, .) ≡ 0, where g(−t, x) = g(t, x) for t ∈] − 1, 1], x ∈ R, and g(t, x) =    x for x ≤ 1/t, t > 0 1/t for x > 1/t, t > 0 . Then f ∈ K([0, 1]×R; R) and it is clear that f ∈ E({0}) because, for every ε > 0, if x ≥ 1/ε then 1 ε f(s, x)ds − ε 0 f(s, x)ds = 4(ε−1/2 − x1/2 ) − 2 < 0. 180 s. mukhigulashvili 2. Main results Theorem 2.1. Let w be a nonzero solution of the problem (1.3), (1.4), (2.1) Nw = ∅, there exist a constant r > 0, functions f− , f+ ∈ L(I; R+) and g, h0 ∈ L(I; ]0, +∞[ ) such that (2.2) f(t, x)sgnx ≤ g(t)|x| + h0(t) for |x| ≥ r and (2.3) f(t, x) ≤ −f− (t) for x ≤ −r, f+ (t) ≤ f(t, x) for x ≥ r on I. Let, moreover, there exist ε > 0 such that − b a f− (s)|w(s)|ds + ε||γr||L ≤ − b a h(s)|w(s)|ds ≤ (2.41) ≤ b a f+ (s)|w(s)|ds − ε||γr||L, where (2.5) γr(t) = sup{|f(t, x)| : |x| ≤ r}. Then the problem (1.1), (1.2) has at least one solution. Example 2.2. It follows from Theorem 2.1 that the equation (2.6) u (t) = −λ2 u(t) + σ|u(t)|α sgnu(t) + h(t) for 0 ≤ t ≤ π where σ = 1, λ = 1, and α ∈ ]0, 1], with the conditions (1.6) has at least one solution for every h ∈ L([0, π], R). Theorem 2.2. Let w be a nonzero solution of the problem (1.3), (1.4), condition (2.1) hold, there exist a constant r > 0, functions f− , f+ ∈ L(I; R+) and q ∈ K(I × R; R+) such that q is non-decreasing in the second argument, (2.7) |f(t, x)| ≤ q(t, x) for |x| ≥ r, (2.8) f− (t) ≤ f(t, x) for x ≤ −r, f(t, x) ≤ −f+ (t) for x ≥ r on I, and (2.9) lim |x|→+∞ 1 x b a q(s, x)ds = 0. the dirichlet bvp for the second order ... 181 Let, moreover, there exist ε > 0 such that − b a f− (s)|w(s)|ds + ε||γr||L ≤ b a h(s)|w(s)|ds ≤ (2.42) ≤ b a f+ (s)|w(s)|ds − ε||γr||L, where γr is defined by (2.5). Then the problem (1.1), (1.2) has at least one solu- tion. Example 2.3. From Theorem 2.2 it follows that the problem (2.6), (1.6) with σ = −1, λ = 1, and α ∈ ]0, 1[ has at least one solution for every h ∈ L([0, π]; R). Remark 2.4. If f ≡ 0 the condition (2.4i) of Theorem 2.i (i = 1, 2) can be replaced by (2.10i) − b a f− (s)|w(s)|ds < (−1)i b a h(s)|w(s)|ds < < b a f+ (s)|w(s)|ds, because, from (2.10i) there follows the existence of a constant ε > 0 such that the condition (2.4i) is satisfied. Theorem 2.3. Let i ∈ {0, 1}, w be a nonzero solution of the problem (1.3), (1.4), f ∈ E(Nw), there exist a constant r > 0 such that the function (−1)i f is non-decreasing in the second argument for |x| ≥ r, (2.11) (−1)i f(t, x)sgnx ≥ 0 for t ∈ I, |x| ≥ r, (2.12) Ω+ w |f(s, r)|ds + Ω− w |f(s, −r)|ds = 0, and (2.13) lim |x|→+∞ 1 |x| b a |f(s, x)|ds = 0. Then there exists δ > 0 such that the problem (1.1), (1.2) has at least one solution for every h satisfying the condition (2.14) b a h(s)w(s)ds < δ. 182 s. mukhigulashvili Corollary 2.1. Let the assumptions of Theorem 2.3 be satisfied and (2.15) b a h(s)w(s)ds = 0. Then the problem (1.1), (1.2) has at least one solution. Example 2.4. From Theorem 2.3 it follows that the problem (2.6), (1.6) with σ ∈ {−1, 1}, λ ∈ N, and α ∈ ]0, 1[ has at least one solution if h ∈ L([0, π], R) is such that π 0 h(s) sin λsds = 0. Theorem 2.4. Let i ∈ {0, 1}, w be a nonzero solution of the problem (1.3),(1.4), f(t, x) def ≡ f0(t)g0(x) with f0 ∈ L(I; R+), g0 ∈ C(R; R), there exist a constant r > 0 such that (−1)i g0 is non-decreasing for |x| ≥ r and (2.16) (−1)i g0(x)sgnx ≥ 0 for |x| ≥ r. Let, moreover, (2.17) |g0(r)| Ω+ w f0(s)ds + |g0(−r)| Ω− w f0(s)ds = 0 and (2.18) lim |x|→+∞ |g0(x)| = +∞, lim |x|→+∞ g0(x) x = 0. Then, for every h ∈ L(I; R), the problem (1.1), (1.2) has at least one solution. Example 2.5. From Theorem 2.4 it follows that the equation (2.19) u (t) = p0(t)u(t) + p1(t)|u(t)|α sgnu(t) + h(t) for t ∈ I, where α ∈ ]0, 1[ and p0, p1, h ∈ L(I; R), with the conditions (1.2) has at least one solution provided that p1(t) > 0 for t ∈ I. Theorem 2.5. Let i ∈ {0, 1} and w be a nonzero solution of the problem (1.3), (1.4). Let, moreover, there exist constants r > 0, ε > 0, and functions α, f+ , f− ∈ L(I; R+) such that the conditions (2.20i) (−1)i f(t, x) ≤ −f− (t) for x ≤ −r, f+ (t) ≤ (−1)i f(t, x) for x ≥ r, (2.21) sup{|f(t, x)| : x ∈ R} ≤ α(t) the dirichlet bvp for the second order ... 183 hold on I, and let (2.22i) − b a (f+ (s)[w(s)]− + f− (s)[w(s)]+)ds + ε||α||L ≤ ≤ (−1)i+1 b a h(s)w(s)ds ≤ ≤ b a (f− (s)[w(s)]− + f+ (s)[w(s)]+)ds − ε||α||L. Then the problem (1.1), (1.2) has at least one solution. Remark 2.5. If f ≡ 0 then the condition (2.22i) (i = 1, 2) of Theorem 2.5 can be replaced by (2.23i) − b a (f+ (s)[w(s)]− + f− (s)[w(s)]+)ds < < (−1)i+1 b a h(s)w(s)ds < < b a (f− (s)[w(s)]− + f+ (s)[w(s)]+)ds. because from (2.23i) there follows the existence of a constant ε > 0 such that the condition (2.22i) is satisfied. Remark 2.6. If f(t) = min{f+ (t), f− (t)} then the condition (2.22i) of Theorem 2.5 can be replaced by b a h(s)w(s)ds ≤ b a f(s)|w(s)|ds − ε||α||L. Example 2.6. From Theorem 2.5 it follows that the equation (2.24) u (t) = −λ2 u(t) + |u(t)|α 1 + |u(t)|α sgnu(t) + h(t) for 0 ≤ t ≤ π, where λ ∈ N and α ∈ ]0, +∞[ , with the conditions (1.6) has at least one solution if h ∈ L([0, π], R) is such that |h(t)| < 1 for 0 ≤ t ≤ π. 3. Problem (1.5), (1.6). Throughout this section we will assume that a = 0, b = π, and I = [0, π]. Since the functions β sin λt (β ∈ R) are nontrivial solutions of the problem (1.7), (1.4), from Theorems 2.1–2.5 it immediately follows: Corollary 3.2. Let λ = 1 and all the assumptions of Theorem 2.1 (resp. Theorem 2.2) except (2.1) be fulfilled with w(t) = sin t. Then the problem (1.5), (1.6) has at least one solution. 184 s. mukhigulashvili Now, note that Nsin λt =    ∅ for λ = 1 {πn/λ : n = 1, ..., λ − 1} for λ ≥ 2 . Corollary 3.3. Let i ∈ {0, 1}, λ ∈ N, f ∈ E(Nsin λt), there exist a constant r > 0 such that the function (−1)i f is non-decreasing in the second argument for |x| ≥ r, and let the conditions (2.11)–(2.13) be fulfilled with w(t) = sin λt. Then there exists δ > 0 such that the problem (1.5), (1.6) has at least one solution for every h ∈ L(I; R) satisfying the condition | π 0 h(s) sin λsds| < δ. Corollary 3.4. Let i ∈ {0, 1}, λ ∈ N, and let all the assumptions of Theorem 2.4 be fulfilled with w(t) = sin λt. Then, for any h ∈ L(I; R), the problem (1.5), (1.6) has at least one solution. Corollary 3.5. Let i ∈ {0, 1}, λ ∈ N and let there exist a constant r > 0 such that (2.20i)–(2.22i) be fulfilled with w(t) = sin λt. Then the problem (1.5), (1.6) has at least one solution. Remark 3.7. If f ≡ 0 then in Corollary 3.2 (resp. Corollary 3.5), the condition (2.4i) (resp. (2.22i)) can be replaced by the condition (2.10i) (resp. (2.23i)) with w(t) = sin t (resp. w(t) = sin λt). 4. Auxiliary propositions Let un ∈ C (I; R), un C = 0 (n ∈ N), w be an arbitrary solution of the problem (1.3), (1.4), and r > 0. Then, for every n ∈ N, we define: An,1 def = {t ∈ I : |un(t)| ≤ r}, An,2 def = {t ∈ I : |un(t)| > r}, Bn,i def = {t ∈ An,2 : sgnun(t) = (−1)i−1 sgnw(t)} (i = 1, 2), Cn,1 def = {t ∈ An,2 : |w(t)| ≥ 1/n}, Cn,2 def = {t ∈ An,2 : |w(t)| < 1/n}, Dn def = {t ∈ I : |w(t)| > r||un||−1 C + 1/2n}, A± n,2 def = {t ∈ An,2 : ±un(t) > r}, B± n,i def = A± n,2 ∩ Bn,i, C± n,i def = A± n,2 ∩ Cn,i (i = 1, 2), D± n def = {t ∈ I : ±w(t) > r||un||−1 C + 1/2n}, From these definitions it is clear that, for any n ∈ N, we have An,1 ∩ An,2 = ∅, A+ n,2 ∩ A− n,2 = ∅, Bn,1 ∩ Bn,2 = ∅, Cn,1 ∩ Cn,2 = ∅, (4.1) D+ n ∩ D− n = ∅, B+ n,2 ∩ B− n,2 = ∅, C+ n,i ∩ C− n,i = ∅ (i = 1, 2), and An,1 ∪ An,2 = I, A+ n,2 ∪ A− n,2 = An,2, Bn,1 ∪ Bn,2 = An,2 \ Nw, (4.2) Cn,1 ∪ Cn,2 = An,2, B+ n,2 ∪ B− n,2 = Bn,2, C± n,1 ∪ C± n,2 = A± n,2, C+ n,i ∪ C− n,i = Cn,i (i = 1, 2), D+ n ∪ D− n = Dn. the dirichlet bvp for the second order ... 185 Lemma 4.1. Let un ∈ C (I; R) (n ∈ N), r > 0, w be an arbitrary nonzero solution of the problem (1.3), (1.4), and (4.3) ||un||C ≥ 2rn for n ∈ N, (4.4) ||vn − w||C ≤ 1/2n for n ∈ N, where vn(t) = un(t)||un||−1 C . Then, for any n0 ∈ N, we have (4.5) D+ n0 ⊂ A+ n,2, D− n0 ⊂ A− n,2 for n ≥ n0, (4.6) C+ n0,1 ⊂ D+ n C− n0,1 ⊂ D− n for n ≥ n0. Moreover (4.7) lim n→+∞ mesAn,1 = 0, lim n→+∞ mesAn,2 = mesI, (4.8) Cn,1 ⊂ Bn,1, Bn,2 ⊂ Cn,2, (4.9) B+ n,2 ⊂ C+ n,2, B− n,2 ⊂ C− n,2, (4.10) C+ n,1 ⊂ B+ n,1, C− n,1 ⊂ B− n,1, (4.11) lim n→+∞ mesCn,1 = lim n→+∞ mesBn,1 = mesI, lim n→+∞ mesCn,2 = lim n→+∞ mesBn,2 = 0, (4.12) r < |un(t)| ≤ ||un||C/2n for t ∈ Bn,2, (4.13) |un(t)| ≥ ||un||C/2n > r for t ∈ Cn,1, (4.141) C± n,2 = {t ∈ An,2 : 0 ≤ ±w(t) < 1/n}, (4.15) C± n,1 ⊂ Ω± w, lim n→+∞ mesC± n,1 = mesΩ± w. Proof. From the unique solvability of the Cauchy problem for the equation (1.3) it follows that the set Nw is finite. Consequently, we can assume that Nw = {t1, ..., tk}. Let also t0 = a, tk+1 = b and Tn def = I∩ k+1 i=0 [ti − 1/n, ti + 1/n] . We first show that, for every n0 ∈ N, there exists n1 > n0 such that (4.16) An,1 ⊆ Tn0 for n ≥ n1. 186 s. mukhigulashvili Suppose on the contrary that, for some n0 ∈ N, there exists the sequence tnj ∈ Anj,1 (j ∈ N) with nj < nj+1, such that tnj ∈ Tn0 for j ∈ N. Without loss of generality we can assume that lim j→+∞ tnj = t0. Then from the conditions (4.3), (4.4), the definition of the set An,1 and the equality w(t0) = (w(t0) − w(tnj )) + (w(tnj )−vnj (tnj ))+vnj (tnj ), we get |w(t0)| = 0, i.e., t0 ∈ {t0, t1, ..., tk+1}. But this contradicts the condition tnj ∈ Tn0 and thus (4.16) is true. Since lim n→+∞ mesTn = 0, it follows from (4.2) and (4.16) that (4.7) is valid. Let t0 ∈ D+ n0 . Then from (4.4) it follows that un(t0) ||un||C ≥ w(t0) − |vn(t0) − w(t0)| > r ||un0 ||C + 1 2n0 − 1 2n ≥ r ||un0 ||C for n ≥ n0, and thus t0 ∈ A+ n,2 for n ≥ n0, i.e., D+ n0 ⊂ A+ n,2 for n ≥ n0. The second relation of (4.5) can be proved analogously. Now suppose that t0 ∈ Cn,1 and t0 ∈ Bn,1. Then, in view of (4.1) and (4.2), it is clear that t0 ∈ Bn,2, and thus (4.17) |vn(t0) − w(t0)| = |vn(t0)| + |w(t0)| > 1/n , which contradicts (4.4). Consequently, Cn,1 ⊂ Bn,1 for n ∈ N. This, together with the relations Cn,2 = An,2 \ Cn,1, Bn,2 ⊆ An,2 \ Bn,1, implies Bn,2 ⊂ Cn,2, i.e., (4.8) holds. The conditions (4.9) and (4.10) follow immediately from (4.8). In view of the fact that lim n→+∞ mesCn,i = (2 − i)mesI, from (4.8) we get (4.11). Now, let t0 ∈ Bn,2 and suppose that |vn(t0)| > 1/2n. Then from (4.4) we obtain the contradiction 1/2n ≥ |vn(t0) − w(t0)| = |vn(t0)| + |w(t0)| > 1/2n. Thus |un(t0)| ||un||C = |vn(t0)| ≤ 1 2n and using the definitions of the sets Bn,2 and An,2 we obtain (4.12). Also, from the inequality |un(t)| ||un||C = |vn(t)| ≥ |w(t)| − |vn(t) − w(t)| by (4.3), (4.4) and the definition of the sets Cn,1 and An,2 we obtain (4.13). Let there exist t0 ∈ C+ n,2 such that t0 ∈ {t ∈ An,2 : 0 ≤ w(t) ≤ 1/n}. Then from the definition of the sets Cn,2 and the inclusion C+ n,2 ⊂ Cn,2 we get −1/n < w(t) < 0 and t0 ∈ A+ n,2. In this case the inequality (4.17) is fulfilled, which contradicts (4.4). Therefore C+ n,2 ⊂ {t ∈ An,2 : 0 ≤ w(t) ≤ 1/n}. Let now t0 ∈ {t ∈ An,2 : 0 ≤ w(t) ≤ 1/n} and t0 ∈ C+ n,2. Then from the definition of the set Cn,2 and (4.2) it is clear that t0 ∈ C− n,2, i.e., t0 ∈ A− n,2, and that the inequality (4.17) holds, which contradicts (4.4). Therefore {t ∈ An,2 : 0 ≤ w(t) ≤ 1/n} ⊂ C+ n,2. From the last two inclusions it follows that (4.141) holds for C+ n,2. From (4.2) and (4.141) for C+ n,1 it is clear that (4.141) is true for C− n,1 too. Analogously one can prove that (4.142) C± n,1 = {t ∈ An,2 : ±w(t) ≥ 1/n} for n ∈ N. From (4.142), the definition of the sets D± n and (4.3) we obtain (4.6). From the definition of the set Ω± w and (4.142) we have C± n,1 ⊂ Ω± w. Hence mesC± n,1 ≤ mesΩ± w. the dirichlet bvp for the second order ... 187 On the other hand C± n,1 = {t ∈ I : ±w(t) ≥ 1/n} \ (I \ An,2) and thus mesC± n,1 ≥ mesΩ± w − mes(I \ An,2). In view of (4.7) from last two inequalities we conclude that (4.15) holds. Lemma 4.2. Let i ∈ {1, 2}, r > 0, k ∈ N, w0 be a nonzero solution of the problem (1.3), (1.4), Nw0 = {t1, ..., tk}, the function f1 ∈ E(Nw0 ) be non-decreasing in the second argument for |x| ≥ r, and (4.18) f1(t, x)sgnx ≥ 0 for t ∈ I, |x| ≥ r. Then: a) If G ⊂ I and (4.19) G |f1(s, (−1)i r)w0(s)|ds = 0, then there exist δ0 > 0 and ε1 > 0 such that (4.20) I(G, Uε, x) def ≡ G\Uε |f1(s, x)w0(s)|ds − Uε |f1(s, x)w0(s)|ds ≥ δ0 for (−1)i x ≥ r and 0 < ε ≤ ε1, where Uε = I ∩ ∪k j=1 [tj − ε/2k, tj + ε/2k] . b) If un ∈ C (I; R) (n ∈ N), r > 0, w is an arbitrary nonzero solution of the problem (1.3), (1.4), and the condition (4.3) holds, then there exist ε2 ∈]0, ε1] and n0 ∈ N such that (4.211) I(D+ n , U+ ε , x) ≥ − δ0 2 for x ≥ r, (4.212) I(D− n , U− ε , x) ≥ − δ0 2 for x ≤ −r for n ≥ n0 and 0 < ε ≤ ε2, where U± ε = {t ∈ Uε : ±w(t) ≥ 0}. Proof. First note that, for any nonzero solution w of the problem (1.3), (1.4), there exists β = 0 such that w(t) = βw0(t) and thus Nw = Nw0 . a) For any α ∈ R+, we put G1 = ([a, a + α] ∪ [b − α, b]) ∩ G. In view of the condition (4.19), we can choose α ∈]0, (b − a)/2[ such that if G2 = G \ G1, ta = inf{G2} and tb = sup{G2}, then (4.22) a < ta, tb < b, and G1 |f1(s, (−1)i r)w0(s)|ds = 0, G2 |f1(s, (−1)i r)|ds = 0. From these inequalities, by virtue of conditions (4.18) and f1 ∈ E(Nw0 ), where f1 is non-decreasing in the second argument, there follows the existence of δ0 > 0 and ε∗ > 0 such that (4.23) G2\Uε∗ |f1(s, x)|ds − Uε∗ |f1(s, x)|ds ≥ 0 for (−1)i x ≥ r, 188 s. mukhigulashvili (4.24) G1\Uε∗ |f1(s, x)w0(s)|ds ≥ δ0 for (−1)i x ≥ r. Now we put I∗ = [t∗ a, t∗ b], where t∗ a = a + min(ta, t1) 2 and t∗ b = max(tk, tb) + b 2 . In view of (4.22), we obtain (4.25) G2 ⊂ I∗ , Nw0 ⊂ I∗ , w0(t∗ a) = 0, w0(t∗ b) = 0. Then it is clear that there exists γ1 > 0 such that, for any γ ∈]0, γ1[, the equation |w0(t)| = γ has only tγ,i, t∗ γ,i ∈ I∗ (i = 1, ..., k) solutions such that (4.26) tγ,i < ti < t∗ γ,i (i = 1, ..., k), (4.27) |w0(t)| ≤ γ for t ∈ Hγ, |w0(t)| > γ for t ∈ I∗ \ Hγ, where Hγ = k i=1 [tγ,i, t∗ γ,i], and (4.28) lim γ→+0 tγ,i = lim γ→+0 t∗ γ,i = ti (i = 1, ..., k). The relations (4.26) and (4.28) imply that there exist γ ∈ ]0, γ1] and ε1 ∈]0, ε∗ ] such that (4.29) Uε1 ⊂ Hγ ⊂ Uε∗ . Moreover, in view of the inclusion G1 ⊂ G, it is clear that G \ Uε1 = G \ G1 \ Uε1 ∪ G1 \ Uε1 , G \ G1 \ Uε1 ∩ G1 \ Uε1 = ∅, and thus I(G, Uε1 , x) = G1\Uε1 |f1(s, x)w0(s)|ds + I(G2, Uε1 , x) for (−1)i x ≥ r. By virtue of (4.23), (4.25), (4.27), and (4.29), we get I(G2, Uε1 , x) ≥ γ G2\Hγ |f1(s, x)|ds − Hγ |f1(s, x)|ds ≥ ≥ γ G2\Uε∗ |f1(s, x)|ds − Uε∗ |f1(s, x)|ds ≥ 0 for (−1)i x ≥ r. In view of the last two relations, (4.24), (4.29), and the fact that Uε ⊂ Uε1 for ε ≤ ε1, we conclude that the inequality (4.20) holds. b) First consider the case when (4.30) D+ n |f1(s, x)w0(s)|ds = 0 for x ≥ r, n ∈ N. the dirichlet bvp for the second order ... 189 From (4.3) and the definitions of the sets D± n and U± ε we get (4.31) lim n→+∞ mes(U± ε \ D± n ) = 0. Then, in view of (4.30) and the fact that for any ε > 0 and n ∈ N (4.32) U± ε = (U± ε ∩ D± n ) ∪ (U± ε \ D± n ), (U± ε ∩ D± n ) ∩ (U± ε \ D± n ) = ∅, we have U+ ε |f1(s, x)w0(s)|ds = U+ ε \D+ n |f1(s, x)w0(s)|ds for x ≥ r, n ∈ N, and ε > 0. Thus by virtue of (4.31), we get U+ ε |f1(s, x)w0(s)|ds = 0. From the last equality and (4.30) we conclude that (4.33) I(D+ n , U+ ε , x) = 0 for x ≥ r, n ∈ N, ε > 0. Therefore, in this case the condition (4.211) is true. Now consider the case when for some r1 ≥ r there exists n0 ∈ N such that (4.34) D+ n |f1(s, x)w0(s)|ds = 0 for x ≥ r1, n ≥ n0. It is clear that there exist η > 0 and ε2 ∈]0, ε1] such that U+ ε |f1(s, x)w0(s)|ds ≤ δ0 2 for r ≤ x ≤ r1 + η, ε ≤ ε2, and thus (4.35) I(D+ n , U+ ε , x) ≥ − δ0 2 for r ≤ x ≤ r1 + η, n ≥ n0, ε ≤ ε2. On the other hand, from (4.34) it is clear that D+ n0 |f1(s, r1 + η)w0(s)|ds = 0. Therefore, from the item a) of our lemma with G = D+ n , and the inclusions D+ n0 ⊂ D+ n , U+ ε ⊂ Uε for n ≥ n0, ε > 0, we get I(D+ n , U+ ε , x) ≥ δ0 for x ≥ r1 + η, n ≥ n0, 0 < ε ≤ ε2. From this inequality and (4.35) we obtain (4.211) in second case too. Analogously one can prove (4.212). Lemma 4.3. Let all the conditions of Lemma 4.1 be fulfilled and there exist r > 0 such that the condition (4.18) holds, where f1 ∈ K(I × R; R). Then (4.36) lim inf n→+∞ t s f1(ξ, un(ξ))sgnun(ξ)dξ ≥ 0 for a ≤ s < t ≤ b. Proof. Let (4.37) γ∗ r (t) def = sup{|f1(t, x)| : |x| ≤ r} for t ∈ I. Then, according to (4.1), (4.2), and (4.18), we obtain the estimate t s f1(ξ, un(ξ))sgnun(ξ)dξ ≥ ≥ − [s,t]∩An,1 γ∗ r (ξ)dξ + [s,t]∩An,2 |f1(ξ, un(ξ))|dξ for a ≤ s < t ≤ b, n ∈ N. This estimate and (4.7) imply (4.36). 190 s. mukhigulashvili Lemma 4.4. Let r > 0, the functions f1 ∈ K(I × R; R), h1 ∈ L(I; R), f+ , f− ∈ L(I; R+) be such that (4.38) f1(t, x) ≤ −f− (t) for x ≤ −r, f+ (t) ≤ f1(t, x) for x ≥ r on I, and there exist a nonzero solution w0 of the problem (1.3), (1.4) and ε > 0 such that (4.39) Nw0 = ∅ and − b a f− (s)|w0(s)|ds + ε||γ∗ r ||L ≤ − b a h1(s)|w0(s)|ds ≤ (4.40) ≤ b a f+ (s)|w0(s)|ds − ε||γ∗ r ||L, where γ∗ r is defined by (4.37). Then, for every nonzero solution w of the problem (1.3), (1.4), and functions un ∈ C (I; R) (n ∈ N) such that the conditions (4.3), (4.41) |v(i) n (t) − w(i) (t)| ≤ 1/2n for t ∈ I, n ∈ N, (i = 0, 1) where vn(t) = un(t)||un||−1 C for t ∈ I and (4.42) un(a) = 0, un(b) = 0 are fulfilled, there exists n1 ∈ N such that (4.43) Mn(w) def ≡ b a (h1(s) + f1(s, un(s)))w(s)ds ≥ 0 for n ≥ n1. Proof. First note that, for any nonzero solution w of the problem (1.3), (1.4), there exists β = 0 such that w(t) = βw0(t). Also, it is not difficult to verify that all the assumptions of Lemma 4.1 are satisfied for the function w(t) = βw0(t). From the unique solvability of the Cauchy problem for the equation (1.3) and the conditions (1.4) we conclude that w (a) = 0 and w (b) = 0. Therefore, in view of (4.41) and (4.42), there exists n2 ∈ N such that (4.44) un(t)sgnβw0(t) > 0 for n ≥ n2, a < t < b. Moreover, by (4.1) and (4.2) we get the estimate (4.45) Mn(w) |β| ≥ − An,1 γ∗ r (s)|w0(s)|ds + σ b a h1(s)w0(s)ds+ +σ An,2 f1(s, un(s))w0(s)ds, the dirichlet bvp for the second order ... 191 where γ∗ r is given by (4.37) and σ = sgnβ. Now note that f− ≡ 0, f+ ≡ 0 if f1(t, x) ≡ 0. Then by virtue of (4.7), we see that there exist ε > 0 and n1 ∈ N (n1 ≥ n2) such that b a f± (s)|w0(s)|ds − ε 2 ||γ∗ r ||L ≤ An,2 f± (s)|w0(s)|ds and ε 2 ||γ∗ r ||L ≥ An,1 γ∗ r (s)|w0(s)|ds for n ≥ n1. By these inequalities, (4.3), (4.38) and (4.44), from (4.45) we obtain Mn(w) |β| ≥ −ε||γ∗ r ||L + b a h1(s)|w0(s)|ds + b a f+ (s)|w0(s)|ds if n ≥ n1, σw0(t) ≥ 0, and Mn(w) |β| ≥ −ε||γ∗ r ||L − b a h1(s)|w0(s)|ds + b a f− (s)|w0(s)|ds if n ≥ n1, σw0(t) ≤ 0. From the last two estimates in view of (4.40) it follows that (4.43) is valid. Lemma 4.5. Let w0 be a nonzero solution of the problem (1.3), (1.4), r > 0, the function f1 ∈ E(Nw0 ) be non-decreasing in the second argument for |x| ≥ r, condition (4.18) hold, and (4.46) Ω+ w0 |f1(s, r)|ds + Ω− w0 |f1(s, −r)|ds = 0. Then there exist δ > 0 and n1 ∈ N such that if (4.47) b a h1(s)w0(s)ds < δ then, for every nonzero solution w of the problem (1.3), (1.4) and the functions un ∈ C (I; R) (n ∈ N) fulfilling the conditions (4.3), (4.41), (4.42), the inequality (4.43) holds. Proof. It is not difficult to verify that all the assumption of Lemma 4.1 are satisfied. Then, by the definition of the sets Bn,1, Bn,2, the conditions (4.1), (4.2), and (4.18), we obtain the estimate (4.48) b a f1(s, un(s))w(s)ds ≥ − An,1 γ∗ r (s)|w(s)|ds + Mn(w), where Mn(w) def ≡ − Bn,2 |f1(s, un(s))w(s)|ds + Bn,1 |f1(s, un(s))w(s)|ds. 192 s. mukhigulashvili On the other hand, from the unique solvability of the Cauchy problem for the equation (1.3) it is clear that (4.49) w (a) = 0, w (b) = 0, w (ti) = 0 for i = 1, ..., k if Nw0 = {t1, ..., tk}. Now note that, for any nonzero solution w of the problem (1.3), (1.4), there exists β = 0 such that w(t) = βw0(t). Consequently, (4.50) Ω± w = Ω± w0 if β > 0 and Ωw = Ω± w0 if β < 0. Then in view of (4.15) and (4.46), there exists n2 ≥ n0 such that (4.51) C+ n2,1 |f1(s, r)w0(s)|ds = 0 and/or C− n2,1 |f1(s, −r)w0(s)|ds = 0. From (4.51), in view of (4.6), it follows that (4.521) D+ n |f1(s, r)w0(s)|ds = 0 for n ≥ n2 and/or (4.522) D− n |f1(s, −r)w0(s)|ds = 0 for n ≥ n2. Consequently, all the assumptions of Lemma 4.2 are satisfied with G = D+ n and/or G = D− n . Therefore, there exist ε0 ∈]0, ε2[, n3 ≥ n2, and δ0 > 0 such that (4.53) I(D+ n , U+ ε0 , x) ≥ δ0 for x ≥ r, n ≥ n3, I(D− n , U− ε0 , x) ≥ −δ0/2 for x ≤ −r, n ≥ n3 if (4.521) holds, and (4.54) I(D− n , U− ε0 , x) ≥ δ0 for x ≤ −r, n ≥ n3, I(D+ n , U+ ε0 , x) ≥ −δ0/2 for x ≥ r, n ≥ n3 if (4.522) holds. On the other hand, the definition of the set Uε and (4.141), imply that there exists n4 > n3, such that (4.55) C+ n,2 ⊂ U+ ε0 , C− n,2 ⊂ U− ε0 for n ≥ n4. By these inclusions, (4.2), and (4.5) we obtain (4.56) C+ n,1 = A+ n,2 \ C+ n,2 ⊃ D+ n4 \ U+ ε0 , C− n,1 = A− n,2 \ C− n,2 ⊃ D− n4 \ U+ ε0 for n ≥ n4. First suppose that Nw0 = ∅ and there exists n ≥ n4 such that (4.57) Bn,2 = ∅. the dirichlet bvp for the second order ... 193 Then, by taking into account that f1 is non-decreasing in the second argument for |x| ≥ r, (4.3), (4.12), (4.18) and the definitions of the sets B+ n,2, B− n,2, we get (4.58) |f1(t, un(t))| = f1(t, un(t)) ≤ ≤ f1 t, ||un||C 2n = f1 t, un C 2n for t ∈ B+ n,2, |f1(t, un(t))| = −f1(t, −un(t)) ≤ ≤ −f1(t, − un C 2n ) = f1 t, − un C 2n for t ∈ B− n,2. Analogously, from (4.3), (4.13), (4.18), and the definitions of the sets C+ n,1, C− n,1, we obtain the estimates (4.59) |f1(t, un(t))| ≥ f1 t, un C 2n for t ∈ C+ n,1, |f1(t, un(t))| ≥ f1 t, − un C 2n for t ∈ C− n,1. Then from (4.1), (4.2), (4.9), (4.58) and respectively from (4.1), (4.2), (4.8), and (4.59) we have Bn,2 |f1(s, un(s))w(s)|ds ≤ (4.60) ≤ B+ n,2 f1 s, un C 2n w(s) ds + B− n,2 f1 s, − un C 2n w(s) ds ≤ ≤ C+ n,2 f1 s, un C 2n w(s) ds + C− n,2 f1 s, − un C 2n w(s) ds and respectively Bn,1 |f1(s, un(s))w(s)|ds ≥ Cn,1 |f1(s, un(s))w(s)|ds ≥ (4.61) ≥ C+ n,1 f1 s, un C 2n w(s) ds + C− n,1 f1 s, − un C 2n w(s) ds. If the condition (4.57) holds, from (4.60) and (4.61) we obtain Mn(w) |β| ≥ C+ n,1 f1 s, un C 2n w0(s) ds − C+ n,2 f1 s, un C 2n w0(s) ds + C− n,1 f1 s, − ||un||C 2n w0(s) ds − C− n,2 f1 s, − un C 2n w0(s) ds , 194 s. mukhigulashvili Whence, by (4.55) and (4.56) we get (4.62) Mn(w) |β| ≥ I D+ n4 , U+ ε0 , un C 2n + I D− n4 , U− ε0 , − un C 2n for n ≥ n4. From (4.62) by (4.53) and (4.54) we obtain (4.63) Mn(w) ≥ δ0|β| 2 for n ≥ n4. On the other hand, in view of (4.10), (4.18), the definition of the sets An,2, Bn,1, and the fact that f1 is non-decreasing in the second argument, we obtain the estimate Bn,1 |f1(s, un(s))w(s)|ds ≥ (4.64) ≥ B+ n,1 |f1(s, r)w(s)|ds + B− n,1 |f1(s, −r)w(s)|ds ≥ ≥ C+ n,1 |f1(s, r)w(s)|ds + C− n,1 |f1(s, −r)w(s)|ds. Now suppose that there exists n ≥ n4 such that (4.65) Bn,2 = ∅. Then from (4.51) and (4.64), (4.65) there follows the existence of δ∗ > 0 such that Mn(w) ≥ |β|δ∗ . From this inequality and (4.63) it follows that, in both cases when (4.57) or (4.65) are fulfilled, the inequality (4.66) Mn(w) ≥ |β|δ for n ≥ n4 holds with δ = min{δ0/2, δ∗ }. From (4.48) by (4.7) and (4.66), we see that for any ε ∈]0, δ[ there exists n1 > n4 such that b a f1(s, un(s))w(s)ds ≥ |β|(δ − ε) for n ≥ n1, and thus (4.67) Mn(w) |β| ≥ δ − ε − b a h1(s)w0(s)ds for n ≥ n1. If Nw0 = ∅ then |w(t)| > 0 for a < t < b and in view of (4.3), (4.41), (4.42) and (4.49), the condition (4.65) holds, i.e., the inequality (4.67) also holds. Consequently, since ε > 0 is arbitrary, the inequality (4.43) from (4.67) and (4.47) follows. the dirichlet bvp for the second order ... 195 Lemma 4.6. Let w0 be a nonzero solution of the problem (1.3), (1.4), r > 0, and the conditions (4.18), (4.47) hold with f1(t, x) def ≡ f0(t)g1(x), where f0 ∈ L(I; R+), b a |f0(s)|ds = 0 and a non-decreasing function g1 ∈ C(R; R) be such that (4.68) lim |x|→+∞ |g1(x)| = +∞. Then, for every nonzero solution w of the problem (1.3), (1.4) and functions un ∈ C (I; R) (n ∈ N) fulfilling the conditions (4.3), (4.41), (4.42), the inequality (4.43) holds. Proof. From the assumptions of our lemma it is clear that the relations (4.48)– (4.56), (4.58)-(4.61) and (4.64) with f1(t, x) = f0(t)g1(x) and w(t) = βw0(t) (β = 0) are fulfilled. Assuming C+ n2,1 |f1(s, r)w0(s)|ds = 0, the condition (4.521) is satisfied i.e., (4.53) holds. Now notice that from (4.15) and the equality C+ n,1 = Ω+ w \(Ω+ w \C+ n,1) it follows that there exist ε > 0 and n0 ∈ N such that (4.69) C+ n,1 |f0(s)w0(s)|ds ≥ Ω+ w |f0(s)w0(s)|ds − ε > 0 for n ≥ n0. First consider the case when there exists n ≥ n4 such that the condition (4.65) holds. Without loss of generality we can assume that n4 > n0. Then by (4.50), (4.64), (4.65) and (4.69), we obtain (4.70) Mn(w) ≥ |β||g1(r)| Θβ |f0(s)w0(s)|ds − ε > 0, where Θβ = Ω+ w0 if β > 0 and Θβ = Ω− w0 if β < 0. Consider now the case when there exists n ≥ n4 such that (4.57) holds. From (4.3) and the definition of the set D+ n it follows that D+ n ⊂ D+ n+1, and since g1 is non-decreasing, from (4.53) we obtain I(D+ n , U+ ε0 , x) ≥ |g1(r)|µ = I(D+ n4 , U+ ε0 , r) ≥ δ0 for x ≥ r, with µ = D+ n4 \U+ ε0 |f0(s)w0(s)|ds − U+ ε0 |f0(s)w0(s)|ds. By the last inequality, (4.3), (4.53), and (4.62) we get µ > 0 and (4.71) Mn(w) ≥ |β|(|g1(r)|µ − δ0/2). Applying (4.70), (4.71) in (4.48) and taking (4.7) into account, we conclude that there exist ε1 > 0 and n1 ≥ n4 such that |β| |g1(r)|µ1 − δ0 2 − ε1 ≤ b a f1(s, un(s))w(s)ds for n ≥ n1 with µ1 = min(µ, Ω+ w0 |f0(s)w0(s)|ds − ε). From (4.68) and the last inequality it is clear that, for any function h1, we can choose r > 0 such that the inequality (4.43) will be true. Analogously one can prove (4.43) in the case when C− n2,1 |f1(s, r)w0(s)|ds = 0. 196 s. mukhigulashvili Lemma 4.7. Let r > 0, there exist functions α, f− , f+ ∈ L(I, R+) such that the condition (4.38) is satisfied, (4.72) sup{|f1(t, x)| : x ∈ R} = α(t) for t ∈ I, and there exist a nonzero solution w0 of the problem (1.3), (1.4) and ε > 0 such that (4.73) − b a (f+ (s)[w0(s)]− + f− (s)[w0(s)]+)ds + ε||α||L ≤ ≤ − b a h1(s)w0(s)ds ≤ ≤ b a (f− (s)[w0(s)]− + f+ (s)[w0(s)]+)ds − ε||α||L. Then, for every nonzero solution w of the problem (1.3), (1.4) and functions un ∈ C (I; R) (n ∈ N) fulfilling the conditions (4.3), (4.41), and (4.42), there exists n1 ∈ N such that the inequality (4.43) holds. Proof. First note that, for any nonzero solution w of the problem (1.3), (1.4), there exists β = 0 such that w(t) = βw0(t). Moreover, it is not difficult to verify that all the assumptions of Lemma4.1 are satisfied for the function w(t) = βw0(t). From (4.1), (4.2), and (4.72) we get (4.74) Mn(w) ≥ − An,1∪Bn,2 α(s)|w(s)|ds + Bn,1 f1(s, un)w(s)ds+ + b a h1(s)w(s)ds. On the other hand, by the definition of the set Bn,1 we obtain (4.75) sgnun(t) = sgnw(t) for t ∈ B+ n,1 ∪ B− n,1. Hence, by (4.1), (4.2), (4.10), (4.38), and (4.75), from (4.74) we obtain the estimate Mn(w) − b a h1(s)w(s)ds ≥ − An,1∪Bn,2 α(s)|w(s)|ds+ (4.76) + B+ n,1 f+ (s)|w(s)|ds + B− n,1 f− (s)|w(s)|ds ≥ ≥ − An,1∪Bn,2 α(s)|w(s)|ds + C+ n,1 f+ (s)|w(s)|ds + C− n,1 f− (s)|w(s)|ds. Now, note that f− ≡ 0 and f+ ≡ 0 if f1(t, x) ≡ 0. Therefore by (4.7), (4.11), (4.15), and the inclusions C+ n,1 ⊂ Ω+ w, C− n,1 ⊂ Ω− w, we see that there exist ε > 0 the dirichlet bvp for the second order ... 197 and n1 ∈ N such that (4.77) 1 3 ε||α||L ≥ An,1∪Bn,2 α(s)|w0(s)|ds Ω± w f± (s)|w0(s)|ds − 1 3 ε||α||L ≤ C± n,1 f± (s)|w0(s)|ds for n ≥ n1. By virtue of (4.76) and (4.77), we obtain Mn(w) |β| ≥ −ε||α||L + Ω+ w f+ (s)|w0(s)|ds+ + Ω− w f− (s)|w0(s)|ds + σ b a h1(s)w0(s)ds for n ≥ n1, where σ = sgnβ. Now, by taking into account that Ω± w l(s)|w0(s)|ds = Ω± w0 l(s)|w0(s)|ds = b a l(s)[w0(s)]±ds if β > 0 and Ω± w l(s)|w0(s)|ds = Ωw0 l(s)|w0(s)|ds = b a l(s)[w0(s)] ds if β < 0 for an arbitrary l ∈ L(I, R), from the last inequalities we get Mn(w) |β| ≥ −ε||α||L + b a (f+ (s)[w0(s)]+ + f− (s)[w0(s)]−)ds+ + b a h1(s)w0(s)ds for n ≥ n1 if σ = 1, and Mn(w) |β| ≥ −ε||α||L + b a (f+ (s)[w0(s)]− + f− (s)[w0(s)]+)ds− − b a h1(s)w0(s)ds for n ≥ n1 if σ = −1. From the last inequalities and (4.73) we immediately obtain (4.43). Now we consider the definitions of the sets V10((a, b)) introduced and described in [12] (see [Definition 1.3, p. 2350]) Definition 4.2. We say that the function p ∈ L([a, b]) belongs to the set V10((a, b)) if for any function p∗ satisfying the inequality p∗ (t) ≥ p(t) for t ∈ I the unique solution of the initial value problem (4.78) u (t) = p∗ (t)u(t) for t ∈ I, u(a) = 0, u (a) = 1, has no zeros in the set ]a, b]. 198 s. mukhigulashvili Lemma 4.8. Let i ∈ {1, 2}, p ∈ L(I; R), pn(t) = p(t) + (−1)i /n, and wn ∈ C (I; R) (n ∈ N) be a solution of the problem (4.79n) wn(t) = pn(t)wn(t) for t ∈ I, wn(a) = 0, wn(b) = 0. Then: a) There exists n0 ∈ N such that the problem (4.79n) has only the zero solution for n ≥ n0. b) If i = 2 and Nw = ∅, where w is a solution of the problem (1.3), (1.4), then the inclusion pn ∈ V10((a, b)) for every n ∈ N holds. Proof. a) Let N∗ wn be the number of zeros of the function wn on I. Assume on the contrary that there exists a sequence {wn}+∞ n≥n0 of nonzero solutions of the problem (4.79n). If i = 1 then from the facts that pn(t) < pn+1(t) and wn ≡ 0, by Sturm’s comparison theorem, we obtain N∗ wn − N∗ wn+1 ≥ 1 (n ∈ N). Now notice that, in view of (4.79n), the inequality N∗ wn ≥ 2 holds. Hence there exist k0 ≥ 2 and n0 ≥ 2 such that N∗ wn0 = k0. Therefore, we obtain the contradiction k0 = N∗ wn0 > N∗ wn0 − N∗ wn0+k0 = (N∗ wn0 −N∗ wn0+1 )+(N∗ wn0+1 −N∗ wn0+2 )+...+(N∗ wn0+k0−1 −N∗ wn0+k0 ) ≥ k0. If i = 2, from the fact that pn−1(t) > pn(t) > p(t) and wn ≡ 0, by Sturm’s comparison theorem, we obtain N∗ wn − N∗ wn−1 ≥ 1 and N∗ w ≥ N∗ wn − 1 (n ∈ N) if w is a nonzero solution of the equation (1.3). Now notice that, in view of (4.79n), the inequality N∗ wn ≥ 2 holds for every n ∈ N. Therefore, if we denote N∗ w = k0, we obtain the contradiction k0 = N∗ w ≥ N∗ wn+k0 − 1 > N∗ wn+k0 − N∗ wn ≥ k0. The contradiction obtained proves the item a) of our lemma. b) Assume on the contrary that there exists n ∈ N such that pn ∈ V10([a, b]). If p∗ (t) ≥ pn(t) and u is a solution of the problem (4.78), then there exists t0 ∈]a, b] such that u(t0) = 0. Since p(t) < p∗ (t), by Sturm’s comparison theorem, we obtain that w, the solution of the problem (1.3), (1.4), has a zero in the interval ]a, t0[, which contradicts our assumption Nw = ∅. The contradiction obtained proves the item b) of our lemma. 5. Proof of the main results Proof of Theorem 2.1. Let pn(t) = p(t) + 1/n and, for any n ∈ N, consider the problem (5.1) un(t) = pn(t)un(t) + f(t, un(t)) + h(t) for t ∈ I, (5.2) un(a) = 0, un(b) = 0. In view of the condition (2.1) and Lemma 4.8, the inclusion pn ∈ V10((a, b)) holds for every n ∈ N. On the other hand, from the conditions (2.2) and (2.3) we find (5.3) 0 ≤ f(t, x)sgnx ≤ g(t)|x| + h0(t) for t ∈ I, |x| ≥ r. the dirichlet bvp for the second order ... 199 Then the inclusion pn ∈ V10((a, b)), as is well-known (see [12, Theorem 2.2, p.2367]), guarantees that the problem (5.1), (5.2) has at least one solution, suppose un. In view of the condition (2.2), without loss of generality we can assume that there exists ε∗ > 0 such that h0(t) ≥ ε∗ on I. Then g(t)|x| + h0(t) ≥ ε∗ for x ∈ R, t ∈ I. Consequently, it is not difficult to verify that un is also a solution of the equation (5.4) un(t) = (pn(t) + p0(t, un(t))sgnun(t))un(t) + p1(t, un(t)) with p0(t, x) = f(t, x)g(t) g(t)|x| + h0(t) , p1(t, x) = h(t) + f(t, x)h0(t) g(t)|x| + h0(t) . Now assume that (5.5) lim n→+∞ ||un||C = +∞ and vn(t) = un(t)||un||−1 C . Then (5.6) vn(t) = (pn(t) + p0(t, un(t))sgnun(t))vn(t) + 1 ||un||C p1(t, un(t)), (5.7) vn(a) = 0 vn(b) = 0, and (5.8) ||vn||C = 1 for any n ∈ N. In view of the condition (5.3), the functions p0, p1 ∈ K(I × R; R) are bounded respectively by the functions g(t) and h(t) + h0(t). Therefore, from (5.6), by virtue of (5.5), (5.7) and (5.8), we see that there exists r0 > 0 such that ||vn||C ≤ r0. Consequently in view of (5.8), by Arzela-Ascoli lemma, without loss of generality we can assume that there exists w ∈ C (I, R) such that lim n→+∞ v(i) n (t) = w(i) (t) (i = 0, 1) uniformly on I. From the last equality and (5.5) there follows the existence of an increasing sequence {αk}+∞ k=1 of a natural numbers, such that ||uαk ||C ≥ 2rk and ||v (i) αk − w(i) ||C ≤ 1/2k for k ∈ N. Without loss of generality we can suppose that un ≡ uαn and vn ≡ vαn . In this case we see that un and vn are the solutions of the problems (5.1), (5.2) and (5.6), (5.7) respectively with pn(t) = p(t) + 1/αn for t ∈ I, n ∈ N, and that the inequalities (5.9) ||un||C ≥ 2rn, ||v(i) n − w(i) ||C ≤ 1/2n for n ∈ N are fulfilled. Analogously, since the functions p0, p1 ∈ K(I ×R; R) are bounded, in view of (5.5), we can assume without loss of generality that there exists a function p ∈ L(I; R) such that (5.10j) lim n→+∞ 1 ||un||j C t a pj(s, un(s))sgnun(s)ds = (1 − j) t a p(s)ds 200 s. mukhigulashvili uniformly on I for j = 0, 1. By virtue of (5.8)–(5.10j) (j = 0, 1), from (5.6) we obtain (5.11) w (t) = (p(t) + p(t))w(t), (5.12) w(a) = 0, w(b) = 0, and (5.13) ||w||C = 1. From the conditions (2.3) and (5.9) it is clear that all the assumptions of Lemma 4.3 with f1(t, x) = f(t, x) are satisfied, and thus we obtain from (5.10j) (j = 0) the relation t s p(ξ)dξ ≥ 0 for a ≤ s < t ≤ b, i.e., (5.14) p(t) ≥ 0 for t ∈ I. Now assume that p ≡ 0 and w0 is a solution of the problem (1.3), (1.4). Then using Sturm’s comparison theorem for the equations (1.3) and (5.11), from (5.14) we see that there exists a point t0 ∈]a, b[ such that w0(t0) = 0, which contradicts (2.1). This contradiction proves that p ≡ 0. Consequently, w is a solution of the problem (1.3), (1.4). Multiplying the equations (5.1) and (1.3) respectively by w and −un, and therefore integrating their sum from a to b, in view of the conditions (5.2) and (1.4), we obtain (5.15) − 1 αn b a w(s)un(s)ds = b a (h(s) + f(s, un(s)))w(s)ds for n ≥ n0. Therefore by virtue of (5.9) we get (5.16) b a (h(s) + f(s, un(s)))w(s)ds < 0 for n ≥ n0. On the other hand, in view the conditions (2.1)-(2.41), (5.2), and (5.9) it is clear that all the assumption of Lemma 4.4 with f1(t, x) = f(t, x), h1(t) = h(t) are fulfilled. Therefore, the inequality (4.43) is true, which contradicts (5.16). This contradiction proves that (5.5) does not hold and thus there exists r1 > 0 such that ||un||C ≤ r1 for n ∈ N. Consequently, from (5.1) and (5.2) it is clear that there exists r1 > 0 such that ||un||C ≤ r1 and |un(t)| ≤ σ(t) for t ∈ I, n ∈ N, where σ(t) = (1+|p(t)|)r1+|h(t)|+γr1 (t). Hence, by Arzela-Ascoli lemma, without loss of generality we can assume that there exists a function u0 ∈ C (I; R) such that lim n→+∞ u(i) n (t) = u (i) 0 (t) (i = 0, 1) uniformly on I. Therefore, it follows from (5.1) and (5.2) that u0 is a solution of the problem (1.1), (1.2). the dirichlet bvp for the second order ... 201 Proof of Theorem 2.2. Let pn(t) = p(t) − 1/n and, for any n ∈ N, consider the problems (5.1), (5.2) and (4.79n). In view of Lemma 4.8, the problem (4.79n) has only the zero solution for every n ≥ n0. Therefore, as is well-known (see [9, Theorem 1.1, p.345]), from the conditions (2.7), (2.9) it follows that the problem (5.1), (5.2) has at least one solution, suppose un. Now assume that (5.5) holds and put vn(t) = un(t)||un||−1 C . Then the conditions (5.7) and (5.8) are fulfilled, and (5.17) vn(t) = pn(t)vn(t) + 1 ||un||C (f(t, un(t))) + h(t)). In view the conditions (2.7) and (2.9), from (5.17) there follows the existence of r0 > 0 such that ||vn||C ≤ r0. Consequently, in view (5.8) by Arzela-Ascoli lemma, without loss of generality we can assume that there exists a function w ∈ C (I, R) such that lim n→+∞ v(i) n (t) = w(i) (t) (i = 0, 1) uniformly on I. Analogously as in the proof of Theorem 2.1, we can find a sequence {αk}+∞ n=1 of natural numbers such that, if we suppose un = uαn then the conditions (5.9) will by true when the functions un and vn are the solutions of the problems (5.1), (5.2) and (5.17), (5.7) respectively with pn(t) = p(t) − 1/αn for t ∈ I, n ∈ N. From (5.17), by virtue of (5.7), (5.9) and (2.9), we obtain that w is a solution of the problem (1.3), (1.4). In a similar manner as the condition (5.15) in the proof of Theorem 2.1, we show that (5.18) 1 αn b a w(s)un(s)ds = b a (h(s) + f(s, un(s)))w(s)ds for n ≥ n0. Now note that, in view of the conditions (2.1), (2.8), (2.42), (5.2), and (5.9), all the assumptions of Lemma 4.4 with f1(t, x) = −f(t, x), h1(t) = −h(t) are satisfied. Hence, analogously as in the proof of Theorem 2.1, from (5.18) we show that the problem (1.1), (1.2) has at least one solution. Proof of Theorem 2.3. Let pn(t) = p(t) + (−1)i /n and for any n ∈ N, consider the problems (5.1), (5.2) and (4.79n). In view of the condition (2.13) and the fact that (−1)i f(t, x) is non-decreasing in the second argument for |x| ≥ r, we obtain (5.19) lim n→+∞ 1 ||zn||C b a |f(s, zn(s))|ds = 0 for an arbitrary sequence zn ∈ C(I; R) with lim n→+∞ ||zn||C = +∞. Moreover, in view of Lemma 4.8, the problem (4.79n) has only the zero solution for every n ≥ n0. Therefore, as it is well-known (see [9, Theorem 1.1, p. 345]), from the inequality (5.19) it follows that the problem (5.1), (5.2) has at least one solution, suppose un. Now assume that (5.5) is fulfilled and put vn(t) = un(t)||un||−1 C . Then (5.7), (5.8) and (5.17) are also fulfilled. Hence, by the conditions (5.8) and (5.19), from (5.17) we get the existence of r0 > 0 such that ||vn||C ≤ r0. Consequently, in view 202 s. mukhigulashvili of (5.8) by the Arzela-Ascoli lemma, without loss of generality we can assume that there exists a function w ∈ C (I, R) such that lim n→+∞ v(i) n (t) = w(i) (t) (i = 0, 1) uniformly on I. Analogously as in the proof of Theorem 2.1, we can find a sequence {αk}+∞ n=1 of natural numbers such that, assuming un = uαn , the conditions (5.9) is true and the functions un and vn are the solutions of the problems (5.1), (5.2) and (5.17), (5.7) respectively with pn(t) = p(t)+(−1)i /αn for t ∈ I, n ∈ N. From (5.17), by virtue of (5.7), (5.9) and (2.13), we obtain that w is a solution of the problem (1.3), (1.4). In a similar manner as the condition (5.15) in the proof of Theorem 2.1, we show (5.20) − 1 αn b a w(s)un(s)ds = (−1)i b a (h(s) + f(s, un(s)))w(s)ds for n ∈ N ≥ n0. Now note that, in view the conditions (2.11), (2.12), (2.14), (5.2), and (5.9), all the assumptions of Lemma 4.5 with f1(t, x) = (−1)i f(t, x), h1(t) = (−1)i h(t) are satisfied. Hence, analogously as in the proof of Theorem 2.1, from (5.20) by Lemma 4.5 we obtain that the problem (1.1), (1.2) has at least one solution. Proof of Corollary 2.1. From the condition (2.15) we immediately obtain (2.14). Therefore all the conditions of Theorem 2.3 are fulfilled. Proof of Theorem 2.4. The proof is the same as the proof of Theorem 2.3. The only difference is that we use Lemma 4.6 instead of Lemma 4.5. Proof of Theorem 2.5. From (2.21) it is clear that, for an arbitrary sequence zn ∈ C(I; R) such that lim n→+∞ ||zn||C = +∞, the equality (5.19) is holds. From (5.19) and Lemma 4.7, analogously as in the proof of Theorem 2.3, we show that the problem (1.1), (1.2) has at least one solution. Acknowledgement. The research was supported by the Academy of Sciences of the Czech Republic, Institutional Research Plan N0.AV0Z10190503 and the Grant No. 201/06/0254 of the Grant Agency of the Czech Republic. References [1] R. P. Agarwal, I. 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Math., 43, no. 2, (1988), 2340–2417. [13] I. Kiguradze, Nekotorie Singularnie Kraevie Zadachi dlja Obiknovennih Differencialnih Uravneni. Tbilisi University (1975), 1–351. [14] I. Kiguradze, On a singular two-point boundary value problem. (Russian) Differentsial’ nye Uravneniya, 5 (1969), No. 11, 2002-2016; English transl.: Differ. Equations, 5 (1969), 1493-1504. [15] I. Kiguradze, On some singular boundary value problems for nonlinear second order ordinary differential equations. (Russian) Differentsial’ nye Uravneniya 4 (1968), No. 10, 1753-1773; English transl.: Differ. Equations, 4 (1968), 901-910. 204 s. mukhigulashvili [16] I. Kiguradze, On a singular boundary value problem. J. Math. Anal. Appl., 30, no. 3, (1970), 475-489. [17] J. Kurzveil, Generalized ordinary differential equations. Czechoslovak Math. J., 8, no. 3, (1958), 360–388. Accepted: 18.06.2009 Georgian Mathematical Journal Volume 16 (2009), Number 4, 651–665 A PERIODIC BOUNDARY VALUE PROBLEM FOR FUNCTIONAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER ROBERT HAKL AND SULKHAN MUKHIGULASHVILI Abstract. On the interval [0, ω], consider the periodic boundary value prob- lem u(n) (t) = n−1 i=0 i(u(i) )(t) + q(t), u(j) (0) = u(j) (ω) + cj (j = 0, . . . , n − 1), where n ≥ 2, i : C [0, ω]; R → L [0, ω]; R (i = 0, . . . , n − 1) are linear bounded operators, q ∈ L [0, ω]; R , cj ∈ R (j = 0, . . . , n − 1). The effective sufficient conditions guaranteeing the unique solvability of the considered problem are established. 2000 Mathematics Subject Classification: 34K06, 34K10. Key words and phrases: Functional differential equation, boundary value problem, periodic solution. Statement of the Problem Consider the problem on the existence and uniqueness of a solution to the equation u(n) (t) = n−1 i=0 i(u(i) )(t) + q(t) for 0 ≤ t ≤ ω (0.1) satisfying the periodic boundary conditions u(j) (0) = u(j) (ω) + cj (j = 0, . . . , n − 1), (0.2) where n ≥ 2, i : C [0, ω]; R → L [0, ω]; R are linear bounded operators, q ∈ L [0, ω]; R , and cj ∈ R (i, j = 0, . . . , n − 1). By a solution to problem (0.1), (0.2) we understand a function u ∈ Cn−1 [0, ω]; R , which satisfies equality (0.1) almost everywhere on [0, ω] and the boundary condition (0.2). It is well-known that if the linear operators i : C [0, ω]; R → L [0, ω]; R (i = 0, . . . , n − 1) are strongly bounded, i.e., if there exist summable functions ηi : [0, ω] → [0, +∞[ such that | i(x)(t)| ≤ ηi(t) x C for 0 ≤ t ≤ ω, x ∈ C [0, ω]; R , then the following theorem on the Fredholm property is valid (see, e.g., [1,10,18]) ISSN 1072-947X / $8.00 / c Heldermann Verlag www.heldermann.de 652 R. HAKL AND S. MUKHIGULASHVILI Theorem 0.1. Problem (0.1), (0.2) is uniquely solvable iff the corresponding homogeneous problem v(n) (t) = n−1 i=0 i(v(i) )(t), (0.3) v(j) (0) = v(j) (ω) (j = 0, . . . , n − 1), (0.4) has only the trivial solution. The above-mentioned Fredholm property for functional differential equations with general bounded linear operators (i.e., not necessarily strongly bounded) had not been investigated before 2000 despite of the fact that in 1972 H. H. Schaefer [17, Theorem 4] proved that there do exist linear bounded operators : C [0, ω]; R → L [0, ω]; R which are not strongly bounded. The first important steps in this direction were made by Bravyi in [2], and later in [5], where, among others, the Fredholm property was proved for the first order boundary value problems for functional differential equations with general bounded linear operators. These results were generalized for the n-th order functional differential systems in [7]. Therefore, Theorem 0.1 is also valid if i (i = 0, . . . , n − 1) are bounded (not necessarily strongly bounded) linear operators. The problem on the existence of a periodic solution to ordinary and functional differential equations was studied very intensively in the past. The first important step was made for linear ordinary differential equations of the type u(n) (t) = p(t)u(t) + q(t) (0.5) by Lasota and Opial in [11]. They showed that problem (0.5), (0.2) is uniquely solvable for n ≥ 4 if a function p ∈ L [0, ω]; R has the constant sign, p ≡ 0, and the inequality ω 0 |p(s)|ds < 2 ω n−1 2 · 4 · · · (n − 2) 1 · 3 · · · (n − 3) (0.6) is fulfilled. This result is far from being optimal, and in [12], condition (0.6) was improved to ω 0 |p(s)|ds < 2 ω 2π ω n−2 . (0.7) The next step was made by Kiguradze and Kusano in [8], where the results of [11,12] were essentially improved. In particular, they proved following propo- sitions. Proposition 0.1. Let either n = 2m, (−1)m−1 p(t) ≥ 0 for t ∈ [0, ω], p(t) ≡ 0 or n = 2m − 1, σp(t) ≥ 0 for t ∈ [0, ω], p(t) ≡ 0, where σ ∈ {−1, 1}. Then problem (0.5), (0.2) has a unique solution. Proposition 0.2. Let n = 2m, (−1)m p(t) ≥ 0 for t ∈ [0, ω], p(t) ≡ 0 and inequality (0.7) be fulfilled. Then problem (0.5), (0.2) has a unique solution. A PERIODIC BOUNDARY VALUE PROBLEM 653 Other results on the existence of a periodic solution to differential equations of higher order can be found, e.g., in [3,9,13,15,16]. However, condition (0.7) in Proposition 0.2 is not yet optimal and, moreover, Proposition 0.1 is not true for functional differential equations, which follows from the fact that the equation with deviating argument u (t) = −| cos t|u(τ(t)) with τ(t) = π/2 for t ∈ [0, π/2[ ∪ ]3π/2, 2π] 3π/2 for t ∈ [π/2, 3π/2[ , has a nonzero 2π-periodic solution sin t. Below we will establish the new conditions guaranteeing the unique solvability of problem (0.1), (0.2), which improve the results of Lasota–Opial and Kiguradze–Kusano and are optimal for n ≤ 7. The method used for the investigation of the considered problem is based on the method developed in our previous papers (see [3,4,13–16] ) for functional differential equations. The following notation is used throughout the paper: N is a set of all natural numbers. R is a set of all real numbers, R+ = [0, +∞[ . C [0, ω]; R is a Banach space of continuous functions u : [0, ω] → R with the norm u C = max |u(t)| : t ∈ [0, ω] . L [0, ω]; R is a Banach space of Lebesgue integrable functions p : [0, ω] → R with the norm p L = ω 0 |p(s)|ds. Ck [0, ω]; R is a set of functions u : [0, ω] → R which are absolutely continuous together with their derivatives up to k-th order. If : C [0, ω]; R → L [0, ω]; R is a linear bounded operator, then = sup x C ≤1 (x) L . [x]+ = 1 2 |x| + x , [x]− = 1 2 |x| − x . [x] is an integer part of x. All equalities and inequalities between the measurable functions are understood as lying almost everywhere in an appropriate interval. Definition 0.1. We will say that a linear operator : C [0, ω]; R → L [0, ω]; R belongs to the set Pω if it is non–negative, i.e., for any non–negative x ∈ C [0, ω]; R the inequality (x)(t) ≥ 0 for 0 ≤ t ≤ ω is fulfilled. 654 R. HAKL AND S. MUKHIGULASHVILI In the sequel, the following notation is used: A0 = 1, A1 = 1 15 , Aj = A1 2 m1=1 m1+1 m2=1 . . . mj−2+1 mj−1=1 1 η(m1) . . . η(mj−1) , B1 = 1 8 , Bj = A1 2 m1=1 m1+1 m2=1 . . . mj−2+1 mj−1=1 1 η(m1) . . . η(mj−1) mj−1+1 i=1 1 + 1 2i , for j ≥ 2, where η(t) = (2t + 1)(2t + 3). Let d0 = 1, d1 = 4, d2 = 32, d3 = 192, (0.8) and for p ∈ N put d2p+2 = 1 max (hp(t)hp(1 − t))1/2 : 0 ≤ t ≤ 1 , d2p+3 = 1 max (fp(s, t)fp(1 − s, 1 − t))1/2 : 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 , (0.9) where the functions fp : [0, 1] × [0, 1] → R+, hp : [0, 1] → R+ are defined as follows: fp(s, t) = p−1 j=0 αpjt2(j+1) + αppt2p+3 s, hp(t) = p j=0 βpjt2(j+1) , (0.10) and αpj = Aj 3 · 4j+1d2(p−j)+1 , βpj = Aj 3 · 4j+1d2(p−j) (j = 0, . . . , p − 1), αpp = Ap 3 · 4p+1 , βpp = Bp 3 · 4p+1 . (0.11) Now we formulate the result from [6] in the form suitable for us. Theorem 0.2. Let k ∈ N, v ∈ Ck [0, ω]; R , v(i) (0) = v(i) (ω) (i = 0, . . . , k), and let dk (k ∈ N) be given by the equalities (0.8)–(0.11). Let, moreover, v(t) ≡ Const. Then v(i) < ωk−i dk−i v(k) (i = 0, . . . , k − 1), where v(i) = max v(i) (t) : t ∈ [0, ω] − min v(i) (t) : t ∈ [0, ω] (0.12) for i = 0, . . . , k. A PERIODIC BOUNDARY VALUE PROBLEM 655 Remark 0.1. In [6], it was shown that d4 = 211 · 3 5 , d5 = 29 · 3 · 5, d6 = 216 · 32 · 5 61 , d7 = 214 · 32 · 5 · 7 17 . 1. Main Results Theorem 1.1. Let j ∈ {0, 1}, the operator 0 admit the representation 0 = 0,1 − 0,2, where 0,1, 0,2 ∈ Pω, and let i (i = 1, . . . , n − 1) be bounded linear operators. Let, moreover, the conditions 0,1 + 0,2 = 0 (1.1) ωn−1 dn−1 0,1+j + Ω < 1, (1.2) 0,1+j 1 − Ω − ωn−1 dn−1 0,1+j ≤ 0,2−j , (1.3) 0,2−j ≤ 2dn−1 ωn−1 1 − Ω + (1 − Ω) 1 − Ω − ωn−1 dn−1 0,1+j (1.4) hold with Ω = n−1 i=1 ωn−1−i dn−1−i i (1.5) and di (i = 0, . . . , n − 1) defined by (0.8)–(0.11). Then problem (0.1), (0.2) has a unique solution. In the case, where all the operators i (i = 0, . . . , n − 1) admit the represen- tation i = i,1 − i,2 (1.6) with i,1, i,2 ∈ Pω, i.e., they are strongly bounded, the following assertion improves Theorem 1.1. Theorem 1.2. Let j ∈ {0, 1} and the operators i (i = 0, . . . , n − 1) admit representations (1.6) where i,1, i,2 ∈ Pω. Let, moreover, conditions (1.1)–(1.4) hold with Ω = n−1 i=1 ωn−1−i dn−1−i max{ i,1 , i,2 } and di (i = 0, . . . , n − 1) defined by (0.8)–(0.11). Then problem (0.1), (0.2) has a unique solution. Remark 1.1. It is clear that if i ≡ 0 (i = 1, . . . , n − 1), then Ω = 0 in Theorems 1.1 and 1.2. Corollary 1.1. Let σ ∈ {−1, 1} and σ 0 ∈ Pω. Let, moreover, the conditions 0 = 0, (1.7) 656 R. HAKL AND S. MUKHIGULASHVILI n−1 i=1 ωn−1−i dn−1−i i < 1, (1.8) and 0 ≤ 4dn−1 ωn−1 1 − n−1 i=1 ωn−1−i dn−1−i i (1.9) hold. Then problem (0.1), (0.2) has a unique solution. For the equation u(n) (t) = 0(u)(t) + q(t), 0 ≤ t ≤ ω, (1.10) with σ 0 ∈ Pω, and σ ∈ {−1, 1}, from Theorem 1.1 we immediately obtain Corollary 1.2. Let σ ∈ {−1, 1}, σ 0 ∈ Pω. Let, moreover, conditions (1.7) and 0 ≤ 4dn−1 ωn−1 (1.11) hold. Then problem (1.10), (0.2) has a unique solution. The special case of equation (0.1) is an equation with deviating argument of the form u(n) (t) = p(t)u(τ(t)) + q(t), (1.12) where p, q ∈ L [0, ω]; R and τ : [0, ω] → [0, ω] is a measurable function. The following assertion immediately follows from Theorem 1.1. Corollary 1.3. Let ω 0 |p(s)|ds = 0 and let the conditions ω 0 [σp(s)]+ds < dn−1 ωn−1 , ω 0 [σp(s)]+ds 1 − ωn−1 dn−1 ω 0 [σp(s)]+ds ≤ ω 0 [σp(s)]−ds ≤ 2dn−1 ωn−1   1+ 1− ωn−1 dn−1 ω 0 [σp(s)]+ds   , hold with σ = 1 or σ = −1. Then problem (1.12), (0.2) has a unique solution. If 0(x)(t) = p(t)x(t), then Corollary 1.2 also improves Proposition 0.2. In particular we get Corollary 1.4. Let either the assumptions of Proposition 0.1 be fulfilled or let n = 2m, (−1)m p(t) ≥ 0 for t ∈ [0, ω], p(t) ≡ 0, and let the inequality ω 0 |p(s)|ds ≤ 4dn−1 ωn−1 (1.13) hold. Then problem (0.5), (0.2) has a unique solution. A PERIODIC BOUNDARY VALUE PROBLEM 657 Remark 1.2. It is not difficult to verify that condition (1.13) improves (0.7) for n ≤ 7. Remark 1.3. Let l0 = 1 and the numbers ln (n ∈ N) be defined by the equalities l2p−1 = (−1)p+1 42p−1 p−1 i=0 (−1)i16i (2p−2i−1)!l2i , l2p = (−1)p+1 42p p−1 i=0 (−1)i16i (2p−2i)!l2i for p ∈ N. (1.14) Then the equality dn−1 = ln−1 (1.15) guarantees the optimality of condition (1.11) (and, consequently, also the optimality of (1.4)) in a sense that it cannot be replaced by the condition 0 ≤ 4dn−1 ω + ε (1.11ε) no matter how small ε ∈ ]0, 1] is. Remark 1.4. According to [6, On Remark 1.3] it follows that the equality di = li (1.16i) is true for i ≤ 7, i.e., in view of Remark 1.3, condition (1.11) (and also condition (1.4)) is optimal for n ≤ 7. In [6], it is also proved (see On Remark 1.4 therein) that if (1.16i) holds for i = 1, . . . , n − 1 and max {hp(t)hp(1 − t) : 0 ≤ t ≤ 1} = h2 p(1/2), (1.16) max {fp(s, t)fp(1 − s, 1 − t) : 0 ≤ s ≤ 1, 0 ≤ t ≤ 1} = f2 p (1/2, 1/2) (1.17) for p ≤ [n−2 2 ], where the functions fp and hp are defined by (0.10), then equality (1.16n) holds. However, in a general case (starting with p = 3, i.e., for n ≥ 8), the proof of (1.16) and (1.17) is not known to the authors. One can find more details about this problem in [6]. 2. Proofs To prove the main theorems we need two auxiliary propositions. The first is rather trivial and we omit the proof. Lemma 2.1. Let ∈ Pω. Then for an arbitrary v ∈ C [0, ω]; R the inequal- ities −m (1)(t) ≤ (v)(t) ≤ M (1)(t) for 0 ≤ t ≤ ω hold, where m = − min{v(t) : 0 ≤ t ≤ ω}, M = max{v(t) : 0 ≤ t ≤ ω}. 658 R. HAKL AND S. MUKHIGULASHVILI Lemma 2.2. Let v ∈ Cn−1 [0, ω]; R and v(t) ≡ Const, v(i) (0) = v(i) (ω) (i = 0, . . . , n − 1). (2.1) Then each of the functions v(i) (i = 1, . . . , n − 1) changes its sign and therefore v(i) C ≤ ∆ v(i) (i = 1, . . . , n − 1). Proof. It is clear that if v(k) ≡ Const and v(k) (0) = v(k) (ω) then v(k+1) ≡ 0 and ω 0 v(k+1) (s)ds = 0 for any fixed k ∈ {0, . . . , n − 1}. Thus v(k+1) changes its sign. From this fact and (2.1) it follows by mathematical induction that the functions v(i) (i = 1, . . . , n − 1) change their signs. From this fact and (0.12), the second part of the lemma immediately follows. Proof of Theorem 1.1. We will prove the theorem case when conditions (1.2)– (1.4) are fulfilled with j = 0. The case where j = 1 can be proved analogously. According to Theorem 0.1 it is sufficient to show that problem (0.3), (0.4) has only a trivial solution. Assume to the contrary that problem (0.3), (0.4) has a nontrivial solution v and put Mi = max v(i) (t) : t ∈ [0, ω] , mi = − min v(i) (t) : t ∈ [0, ω] (i = 0, . . . , n − 1). (2.2) First assume that v is still non-negative or still non-positive. Without loss of generality we can assume that v(t) ≥ 0 for t ∈ [0, ω]. Obviously, M0 > 0, m0 ≤ 0. If v ≡ Const, from (0.3) we get 0,1 = 0,2 , which contradicts (1.1)–(1.3). Thus v ≡ Const and from Lemma 2.2 it follows that Mi > 0, mi > 0 (i = 1, . . . , n − 1). (2.3) Choose t1, t2 ∈ [0, ω] such that v(n−1) (t1) = −mn−1, v(n−1) (t2) = Mn−1. (2.4) Obviously, either t1 < t2 (2.5) or t1 > t2. (2.6) Let (2.5) be fulfilled. Then, in view of (0.12), (2.2), (2.4), and Lemmas 2.1, 2.2, the integration of (0.3) from t1 to t2 yields ∆ v(n−1) = t2 t1 0,1(v)(s)ds − t2 t1 0,2(v)(s)ds + n−1 i=1 t2 t1 i(v(i) )(s)ds ≤ M0 0,1 + n−1 i=1 ∆ v(i) i . (2.7) A PERIODIC BOUNDARY VALUE PROBLEM 659 If (2.6) is fulfilled, then analogously to (2.7) the integration of (0.3) from 0 to t2 and from t1 to ω results in Mn−1 − v(n−1) (0) ≤ M0 t2 0 0,1(1)(s)ds + n−1 i=1 t2 0 | i(v(i) )(s)|ds, v(n−1) (ω) + mn−1 ≤ M0 ω t1 0,1(1)(s)ds + n−1 i=1 ω t1 | i(v(i) )(s)|ds. Summing the last two inequalities, on account of (0.4), (0.12), and (2.2) we get ∆ v(n−1) ≤ M0 0,1 + n−1 i=1 ∆ v(i) i . (2.8) Thus for both (2.5) and (2.6) inequality (2.8) is fulfilled. Furthermore, from (2.8), according to Theorem 0.2, we get ∆ v(n−1) (1 − Ω) ≤ M0 0,1 , (2.9) where Ω is defined by (1.5). Now from (2.9), again using Theorem 0.2, we obtain dn−1 ωn−1 (M0 + m0) (1 − Ω) < M0 0,1 , whence we get M0 1 − Ω − ωn−1 dn−1 0,1 < −m0(1 − Ω). (2.10) On the other hand, in view of (0.4), (2.2), and Lemmas 2.1, 2.2, the integration of (0.3) from 0 to ω yields −m0 0,2 ≤ M0 0,1 + n−1 i=1 ∆ v(i) i . (2.11) According to Theorem 0.2, from (2.11) we obtain −m0 0,2 ≤ M0 0,1 + ∆ v(n−1) Ω, (2.12) where Ω is defined by (1.5). Now, (2.12) and (2.9) result in −m0(1 − Ω) 0,2 ≤ M0 0,1 . (2.13) Multiplying the corresponding sides of the inequalities (2.10) and (2.13), in view of (1.1), (1.2) and the fact that −m0 > 0 (see (2.10)), we obtain 1 − Ω − ωn−1 dn−1 0,1 0,2 < 0,1 , which contradicts (1.3) with j = 0. Now suppose that v assumes both positive and negative values. Then according to Lemma 2.2 it follows that Mi > 0, mi > 0 (i = 0, . . . , n − 1). Choose t1, t2 ∈ [0, ω] such that (2.4) holds and without loss of generality we can assume that (2.5) is fulfilled. 660 R. HAKL AND S. MUKHIGULASHVILI In view of (2.2), and Lemmas 2.1 and 2.2, the integration of (0.3) from 0 to t1, from t1 to t2, and from t2 to ω, respectively, yields mn−1 + v(n−1) (0) ≤ M0 t1 0 0,2(1)(s)ds + m0 t1 0 0,1(1)(s)ds + n−1 i=1 t1 0 | i(v(i) )(s)|ds, (2.14) Mn−1 + mn−1 ≤ M0 t2 t1 0,1(1)(s)ds + m0 t2 t1 0,2(1)(s)ds + n−1 i=1 ∆(v(i) ) i , (2.15) Mn−1 − v(n−1) (ω) ≤ M0 ω t2 0,2(1)(s)ds + m0 ω t2 0,1(1)(s)ds + n−1 i=1 ω t2 | i(v(i) )(s)|ds. (2.16) Summing (2.14) and (2.16), on account of (0.4), (0.12), and (2.2) we get ∆ v(n−1) ≤ M0 I 0,2(1)(s)ds + m0 I 0,1(1)(s)ds + n−1 i=1 ∆(v(i) ) i , (2.17) where I = [0, t1]∪[t2, ω]. However, according to Theorem 0.2, (2.17) and (2.15) result in ∆ v(n−1) (1 − Ω) ≤ M0 I 0,2(1)(s)ds + m0 I 0,1(1)(s)ds, (2.18) ∆ v(n−1) (1 − Ω) ≤ M0 t2 t1 0,1(1)(s)ds + m0 t2 t1 0,2(1)(s)ds. (2.19) Now we note that (1.2) and Theorem 0.2 imply dn−1 ωn−1 (M0 + m0)(1 − Ω) < v(n−1) (1 − Ω). In view of (1.2) and the latter inequality, from (2.18) and (2.19) we get A PERIODIC BOUNDARY VALUE PROBLEM 661 0 < m0 1 − Ω − ωn−1 dn−1 I 0,1(1)(s)ds < M0 ωn−1 dn−1 I 0,2(1)(s)ds − (1 − Ω) , (2.20) 0 < M0 1 − Ω − ωn−1 dn−1 t2 t1 0,1(1)(s)ds < m0 ωn−1 dn−1 t2 t1 0,2(1)(s)ds − (1 − Ω) , (2.21) which immediately imply the inequalities ωn−1 dn−1 I 0,2(1)(s)ds > 1 − Ω, ωn−1 dn−1 × t2 t1 0,2(1)(s)ds > 1 − Ω, and thus ωn−1 dn−1 0,2 > 2(1 − Ω). (2.22) Multiplying the corresponding sides of the inequalities (2.20) and (2.21) we obtain 1 − Ω − ωn−1 dn−1 I 0,1(1)(s)ds 1 − Ω − ωn−1 dn−1 t2 t1 0,1(1)(s)ds < ωn−1 dn−1 I 0,2(1)(s)ds − (1 − Ω) ωn−1 dn−1 t2 t1 0,2(1)(s)ds − (1 − Ω) . (2.23) On the other hand, since (α − β)(α − γ) ≥ α(α − (β + γ)) if βγ ∈ R+, we have 1 − Ω − ωn−1 dn−1 I 0,1(1)(s)ds 1 − Ω − ωn−1 dn−1 t2 t1 0,1(1)(s)ds ≥ (1 − Ω) 1 − Ω − ωn−1 dn−1 0,1 , (2.24) and, furthermore, in view of the inequality 4αβ ≤ (α + β)2 , we have ωn−1 dn−1 I 0,2(1)(s)ds − (1 − Ω) ωn−1 dn−1 t2 t1 0,2(1)(s)ds − (1 − Ω) ≤ 1 4 ωn−1 dn−1 0,2 − 2(1 − Ω) 2 . (2.25) 662 R. HAKL AND S. MUKHIGULASHVILI Then, using (2.24) and (2.25) in (2.23), in view of (2.22) we get (1 − Ω) 1 − Ω − ωn−1 dn−1 0,1 < ωn−1 2dn−1 0,2 − (1 − Ω), which contradicts (1.4) with j = 0. Consequently, our assumptions fail, and so v ≡ 0. Proof of Theorem 1.2. First note that if conditions (2.3) with Mi and mi defined by (2.2)are fulfilled, then, in view of (0.12), for any measurable set A ⊂ [0, ω] the estimates (−1)j A i(v(i) )(s)ds = A i,1+j(v(i) )(s)ds − A i,2−j(v(i) )(s)ds Mi A i,1+j(1)(s)ds + mi A i,2−j(1)(s)ds ≤ ∆ v(i) max{ i,1 , i,2 } hold for j = 0, 1, i = 1, . . . , n − 1. Consequently, all the arguments hold true in the proof of the inequalities (2.9), (2.8), (2.12), (2.18), and (2.19) using the above estimates, and thus Theorem 1.2 can be proved in the same way as Theorem 1.1. More precisely, at the end of the proof of Theorem 1.2 we obtain the contradiction with the assumptions for Ω = n−1 i=1 ωn−1−i dn−1−i max{ i,1 , i,2 }. Proof of Corollary 1.1 immediately follows from Theorem 1.1 with j = 1+σ 2 . Proof of Corollary 1.2 immediately follows from Corollary 1.1 with i ≡ 0 (i = 1, . . . , n − 1). On Remark 1.3. Define the functions W0,k, Wi,k : [0, 1] → [0, 1] and the numbers li,k (i, k ∈ N) by W0,k(t) =    1 for 0 ≤ t ≤ 1 4 − 1 8k sin kπ(1 − 4t) for 1 4 − 1 8k < t < 1 4 + 1 8k −1 1 4 + 1 8k ≤ t ≤ 1 2 , (2.26) W0,k 1 2 + t = W0,k 1 2 − t for 0 ≤ t ≤ 1 2 , (2.27) Wm,k(t) = t 0 Wm−1,k(s)ds − δm 1/4 0 Wm−1,k(s)ds for t ∈ [0, ω], m ∈ N, where δm = 0 if m = 2µ − 1 1 if m = 2µ , µ ∈ N, l2p−1,k = 1 |W2p−1,k(1/4)| , l2p,k = 1 |W2p,k(1/2)| , p ∈ N. (2.28) A PERIODIC BOUNDARY VALUE PROBLEM 663 To show the validity of Remark 1.3 we use the properties of W0,k, Wi,k, and li,k which are proved in [6]. In particular, the following equalities are valid for i, k ∈ N (see Lemmas 2.3 and 2.4 in [6]): Wi,k(0) = Wi,k(1), (2.29) W (j) i,k (t) = Wi−j,k(t) for t ∈ [0, 1] j ≤ i, (2.30) lim k→+∞ li,k = li, (2.31) ∆(Wi,k) = 1 li,k ∆(W0,k), (2.32) Wi,k 1 2 − t = (−1)i Wi,k 1 2 + t for 0 ≤ t ≤ 1 2 , (2.33) Wi,k 1 4 − t = (−1)i−1 Wi,k 1 4 + t for 0 ≤ t ≤ 1 4 . (2.34) According to Theorem 0.2, in view of (2.29) and (2.30), we have ∆(Wi,k) < 1 di ∆(W (i) i,k ) = 1 di ∆(W0,k), whence, with respect to (2.32), we obtain li,k > di. Now, assuming that (1.15) holds, on account of (2.31), it follows that for every ε > 0 there exists k0 ∈ N such that ln−1 = dn−1 < ln−1,k ≤ dn−1 + ε 4 for k ≥ k0. (2.35) Put v0(t) = dn−1 + ε 4 Wn−1,k0 (t) for t ∈ [0, 1]. According to (2.28) and (2.35) we get v0 C > ln−1,k0 Wn−1,k0 C ≥ 1. Thus in view of (2.33) and (2.34) we have {t ∈ [0, 1] : v0(t) ≥ 1} = ∅, {t ∈ [0, 1] : v0(t) ≤ −1} = ∅, which implies the existence of t1, t2 ∈ [0, 1] such that v0(t1) = 1, v0(t2) = −1. (2.36) Now let ω = 1, 0(x)(t) = |v (n) 0 (t)|x(τ(t)) with τ(t) = t1 for W0,k0 (t) ≥ 0 t2 for W0,k0 (t) < 0 . 664 R. HAKL AND S. MUKHIGULASHVILI Then 0 ∈ Pω, 0 = 0, in view of (2.36) we have v0(τ(t)) = sgn W0,k0 (t), and from the definition of the function W0,k (see (2.26) and (2.27)) we have 0 = 1 0 |v (n) 0 (s)|ds = 4 dn−1 + ε 4 1 4 1 4 − 1 8k0 | sin πk0(1 − 4s)|ds = 4dn−1 + ε. (2.37) Thus, all the assumptions of Corollary 1.2 are satisfied except (1.11), instead of which condition (1.11ε) is fulfilled with ω = 1. On the other hand, by (2.30) we get v (n) 0 (t) = W0,k0 (t) = |W0,k0 (t)| sgn W0,k0 (t) = |v (n) 0 |v0(τ(t)) = 0(v0)(t). Therefore v0 is a nontrivial solution to the homogeneous problem v(n) (t) = 0(v)(t), v(i) (0) = v(i) (1) (i = 0, . . . , n − 1), which contradicts the conclusion of Corollary 1.2. Acknowledgement The research was supported by the Academy of Sciences of the Czech Republic, Institutional Research Plan No. A V0Z10190503, by the Grant Agency of the Czech Republic, grant No. 201/06/0254 (first author), and by the Georgian National Scientific Foundation, grant # GNSF /ST06/3 − 002 (second author). References 1. N. V. Azbelev, V. P. Maksimov, and L. F. Rakhmatullina, Introduction to the theory of functional-differential equations. Nauka, Moscow, 1991. 2. E. Bravyi, A note on the Fredholm property of boundary value problems for linear functional differential equations. Mem. Differential Equations Math. Phys. 20(2000), 133– 135. 3. R. Hakl, Periodic boundary-value problem for third order linear functional differential equations. Ukrain. Mat. Zh. 60(2008), No. 3, 413–425. 4. R. Hakl, A. 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Kusano, On periodic solutions of even-order ordinary differential equations. Ann. Mat. Pura Appl. (4) 180(2001), No. 3, 285–301. 10. I. Kiguradze and B. P˚uˇza, On boundary value problems for systems of linear functional-differential equations. Czechoslovak Math. J. 47(122)(1997), No. 2, 341–373. 11. A. Lasota and Z. Opial, Sur les solutions p´eriodiques des ´equations diff´erentielles ordinaires. Ann. Polon. Math. 16(1964), 69–94. 12. A. Lasota and F. H. Szafraniec, Sur les solutions p´eriodiques d’une ´equation diff´erentielle ordinaire d’ordre n. Ann. Polon. Math. 18(1966), 339–344. 13. A. Lomtatidze and S. Mukhigulashvili, On periodic solutions of second order functional differential equations. Mem. Differential Equations Math. Phys. 5(1995), 125–126. 14. S. Mukhigulashvili, Two-point boundary value problems for second order functional differential equations. Mem. Differential Equations Math. Phys. 20(2000), 1–112. 15. S. Mukhigulashvili, On the solvability of a periodic problem for second-order nonlinear functional-differential equations. (Russian) Differ. Uravn. 42(2006), No. 3, 356–365; English transl.: Differ. Equ. 42(2006), No. 3, 380–390. 16. S. Mukhigulashvili, On a periodic boundary value problem for third order linear functional differential equations. Nonlinear Anal. 66(2007), No. 2, 527–535. 17. H. H. Schaefer, Normed tensor products of Banach lattices. Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). Israel J. Math. 13(1972), 400–415 (1973). 18. ˇS. Schwabik, M. Tvrd´y, and O. Vejvoda, Differential and integral equations. Boundary value problems and adjoints. D. Reidel Publishing Co., Dordrecht–Boston, Mass.– London, 1979. (Received 17.07.2008) Authors’ addresses: R. Hakl Institute of Mathematics AS CR ˇZiˇzkova 22, 616 62 Brno Czech Republic E-mail: hakl@ipm.cz S. Mukhigulashvili Institute of Mathematics AS CR ˇZiˇzkova 22, 616 62 Brno Czech Republic E-mail: mukhig@ipm.cz and I. Chavchavadze State University Faculty of physics and mathematics 32, I. Chavchavadze Ave., Tbilisi 0179 Georgia Memoirs on Differential Equations and Mathematical Physics Volume 20, 2000, 1–112 S. Mukhigulashvili TWO-POINT BOUNDARY VALUE PROBLEMS FOR SECOND ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS Abstract. In the paper effective sufficient conditions are obtained for unique solvability and correctness of the mixed problem and of the Dirichlet problem for second order linear singular functional differential equations. Some of these conditions are nonimprovable and some of them generalize results which are well known for ardinary differential equations. 2000 Mathematics Subject Classification. 34K10. Key words and phrases: Functional deffirential equations, singular, unique solvability, correctness. 3 Main Notation R = ] − ∞, +∞[ , R+ = ]0, +∞[. Let α ∈ R. [α] is the integral part of the number α, [α]+ = |α| + α 2 , [α]− = |α| − α 2 . C( ]a, b[) is the space of continuous and bounded functions u : ]a, b[→ R with the norm u C = sup{|u(t)| : a < t < b}. Cloc(]a, b[) is the set of the functions u :]a, b[→ R absolutely continuous on each subsegment of ]a, b[ . C′ loc( ]a, b[) is the set of the functions u : ]a, b[ → R absolutely continuous on each subsegment of ]a, b[ along with their first order derivatives. L([a, b]) is the space of summable functions u : [a, b] → R with the norm u L = b a |u(s)|ds. L∞(]a, b]) is the space of essentially bounded functions u : ]a, b[ → R with the norm u = ess sup t∈[a,b] |u(t)|. Lloc(]a, b[) (Lloc(]a, b[)) is the set of the measurable functions u : ]a, b[ → R (u : ]a, b] → R), summable on each subsegment of ]a, b[ (]a, b]). Let x, y : ]a, b[ → ]0, +∞[ be continuous functions. Cx(]a, b[) is the space of functions u ∈ C(]a, b[) such that u C,x = sup |u(t)| x(t) : a < t < b < +∞. Ly([a, b]) is the space of the functions u ∈ L(]a, b[) such that u L,y = b a y(s)|u(s)|ds < +∞. L(Cx; Ly) is the set of the linear operators h : Cx(]a, b[)→Ly([a, b]) such that sup |h(u)(·)| : u C,x ≤ 1 ∈ Ly([a, b]). σ : Lloc(]a, b[) → Cloc(]a, b[) is the operator defined by σ(p)(t) = exp t a+b 2 p(s)ds for a ≤ t ≤ b, 4 where p ∈ Lloc(]a, b[). If σ(p) ∈ L([a, b]), then we define the operators σ1 and σ2 by σ1(p)(t) = 1 σ(p)(t) t a σ(p)(s)ds b t σ(p)(s)ds, σ2(p)(t) = 1 σ(p)(t) t a σ(p)(s)ds for a ≤ t ≤ b. Let f, g ∈ C(]a, b[) and c ∈ [a, b]. Then we write f(t) = O(g(t)) f(t) = O∗ (g(t)) as t → c, if lim t→c sup |f(t)| |g(t)| < +∞ 0 < lim t→c inf |f(t)| |g(t)| and lim t→c sup |f(t)| |g(t)| < +∞ . Let A and B be normed spaces and let U : A → B be a linear operator. Then we denote the norm of the operator U as follows: U A→B. 5 Introduction During the last two decades the boundary value problems for functional differential equations attract the attention of many mathematicians and are intensively studied. At present the foundations of the general theory of such kind of problems are already laid and many of them are investigated in detail (see [1], [2], [19]–[23], [44] and references therein). Despite this fact, there remains a wide class of boundary value problems on the solvability of which not much is known. Among them are the two-point boundary value problems for linear singular functional differential equations of second order, and we devote our work to the investigation of these problems. It should be noted that the present monograph is tightly connection with the works of I. T. Kiguradze [17], L. B. Shekhter [23] and A. G. Lomtatidze [27] in which for singular ordinary differential equations we developed the method of upper and lower Nagumo’s functions in the case of boundary value problems and found the conditions under which Fredholm’s alternative is valid in the case of linear equations. We introduced and described the set V0,i (see Definition 1.1.2). In the present work we consider the equation u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + g(u)(t) + p2(t) (0.0.1) under the boundary conditions u(a) = c1, u(b) = c2 (0.0.21) or u(a) = c1, u′ (b−) = c2, (0.0.22) and separately for the case of homogeneous conditions u(a) = 0, u(b) = 0, u(a) = 0, u′ (b−) = 0, where c1, c2 ∈ R, pj ∈ Lloc(]a, b[) (j = 0, 1, 2) and g : C(]a, b[) → Lloc(]a, b[) is a continuous linear operator. In studying these problems the use is made of the auxiliary equation u′′ (t) = p0(t)u(t) + p1(t)u′ (t) − h(u)(t), where h : C(]a, b[) → Lloc(]a, b[) is the nonnegative linear operator. The question of the unique solvability of problems (0.0.1), (0.0.2i) is studied in Chapter I. We introduced sets of two-dimensional vector functions (p0, p1) :]a, b[→ R2 , Vi,β(]a, b[; h), β ∈ [0, 1] (see Definitions 1.1.3 and 1.1.4), which were found to be useful for our investigation. In Section 1.1, in terms of the sets Vi,β(]a, b[; h) we established theorems for the unique solvability of problems (0.0.1), (0.0.23i). The question on the unique solvability of problems (0.0.1), (0.0.2i0) in the space with weight Cλ(]a, b[) is studied separately. In the same chapter we can find corollaries of basic theorems 6 and and also the effective sufficient conditions for the unique solvability of the above-mentioned problems. Among them there occur unimprovable conditions and those which generalize the well-known results for ordinary differential equations. In Chapter II we consider the question dealing with the correctness of problems (0.0.1), (0.0.2i) under the assumption that (p0, p1) ∈ Vi,β(]a, b[; h). The effective sufficient conditions guaranteeing the correctness of the abovementioned problems are presented. Everywhere in our work, special attention is given to the case, when the operator g in equation (0.0.1) is defined by the equality g(u)(t) = n k=1 gk(t)u(τk(t)), where gk ∈ Lloc(]a, b[), τk : [a, b] → [a, b] (k = 1, . . . , n) are measurable functions. 7 CHAPTER I UNIQUE SOLVABILITY OF TWO-POINT BOUNDARY VALUE PROBLEMS FOR LINEAR SINGULAR FUNCTIONAL DIFFERENTIAL EQUATIONS § 1.1. Statement of the Problem and Formulation of Basic Results In this chapter we consider the linear equation u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + g(u)(t) + p2(t) (1.1.1) under the boundary conditions u(a) = c1, u(b) = c2 (1.1.21) or u(a) = c1, u′ (b−) = c2, (1.1.22) where p0, pj ∈ Lloc(]a, b[), cj ∈ R (j = 1, 2) and g : C(]a, b[) → Lloc(]a, b[) is a continuous linear operator. The equation (1.1.1) will also be studied separately in the weighted space Cxβ (]a, b[) under the homogeneous boundary conditions u(a) = 0, u(b) = 0 (1.1.210) or u(a) = 0, u′ (b−) = 0, (1.1.220) where β ∈ ]0, 1] and x(t) = t a σ(p1)(s) ds b t σ(p1)(s) ds 2−i for a ≤ t ≤ b. When considering the problems (1.1.1), (1.1.21) and (1.1.1), (1.1.210), it will always be assumed that pj ∈ Lloc(]a, b[) (j = 0, 1, 2), σ(p1) ∈ L([a, b]), p0 ∈ Lσ1(p1)([a, b]), (1.1.31) and when considering the problems (1.1.1), (1.1.22) and (1.1.1), (1.1.220) we will assume that pj ∈ Lloc(]a, b]) (j = 0, 1, 2), σ(p1) ∈ L([a, b]), p0 ∈ Lσ2(p1)([a, b]). (1.1.32) Introduce the following definitions. 8 Definition 1.1.1. Let i ∈ {1, 2}. We will say that w ∈ C(]a, b[) is the lower (upper) function of the problem (1.1.1), (1.1.2i) if: (a) w′ is of the form w′ (t) = w0(t) + w1(t), where w0 : ]a, b[ → R is absolutely continuous on each segment from ]a, b[ , the function w1 : ]a, b[ → R is nondecreasing (nonincreasing) and its derivative is almost everywhere equal to zero; (b) almost everywhere on ]a, b[ the inequality w′′ (t) ≥ p0(t)w(t) + p1(t)w′ (t) + g(w)(t) + p2(t) w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) + g(w)(t) + p2(t) is satisfied: (c) there exists the limit w′ (b−) and w(a) ≤ c1, w(i−1) (b−) ≤ c2 w(a) ≥ c1, w(i−1) (b−) ≥ c2 . Definition 1.1.2. Let i ∈ {1, 2}. We will say that a two-dimensional vector function (p0, p1) : ]a, b[ → R2 belongs to the set Vi,0(]a, b[) if the conditions (1.1.3i) are fulfilled, the solution of the problem u′′ (t) = p0(t)u(t) + p1(t)u′ (t), (1.1.4) u(a) = 0, lim t→a u′ (t) σ(p1)(t) = 1 has no zeros in the interval ]a, b[ and u(i−1) (b−) > 0. Note that this definition is in a full agreement with that of the set Vi,0(]a, b[) given in [23] as the set of three-dimensional vector functions (p0, p11, p12) : ]a, b[ → R3 if p11(t) = p12(t) = p1(t) almost everywhere on ]a, b[ . Definition 1.1.3. Let i ∈ {1, 2} and h : C(]a, b[) → Lloc(]a, b[) be a continuous linear operator. We will say that a two-dimensional vector function (p0, p1) : ]a, b[ → R2 belongs to the set Vi,0( ]a, b[ ; h) if the conditions (1.1.3i) are satisfied and the problem u′′ (t) = p0(t)u(t) + p1(t)u′ (t) − h(u)(t) u(a) = 0, u(i−1) (b−) = 0 has a positive upper function w on the segment [a, b]. Definition 1.1.4. Let i ∈ {1, 2}, β ∈ ]0, 1] and h : C(]a, b[) → Lloc(]a, b[) be a continuous linear operator. We will say that a two-dimensional vector function (p0, p1) : ]a, b[ → R2 belongs to the set Vi,β( ]a, b[ ; h) if (p0, p1) ∈ Vi,0(]a, b[), 9 there exists a measurable function qβ : ]a, b[ → [0, +∞[ such that b a |G(t, s)|qβ(s) ds = O∗ (xβ (t)) as t → a, t → b if i = 1, and as t → b if i = 2, where G is Green’s function of the problem (1.1.4), (1.1.2i0) and x(t) = t a σ(p1)(s) ds b t σ(p1)(s) ds 2−i for a ≤ t ≤ b, and the problem u′′ (t) = p0(t)u(t) + p1(t)u′ (t) − h(u)(t) − qβ(t), u(a) = 0, u(i−1) (b−) = 0 on the interval ]a, b[ has a positive upper function w such that w(t) = O∗ (xβ (t)) as t → a, t → b if i = 1 and as t → a if i = 2. 1.1.1. Theorems on the Unique Solvability of the Problems (1.1.1), (1.1.2i) (i = 1, 2). Theorem 1.1.1i. Let i ∈ {1, 2}, p2 ∈ Lσi(p1)([a, b]) (1.1.5i) and let the constants α, β ∈ [0, 1] connected by the inequality α + β ≤ 1 (1.1.6) be such that (p0, p1) ∈ Vi,β(]a, b[ ; h), (1.1.7i) where h ∈ L Cxβ ; L xα σ(p1) ∩ L C; Lσi(p1) (1.1.8i) is a nonnegative operator and x(t) = t a σ(p1)(s) ds b t σ(p1)(s) ds 2−i for a ≤ t ≤ b. (1.1.9i) 10 Let, moreover, a continuous linear operator g:C(]a, b[)→Lσi(p1)([a, b]) be such that for any function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality |g(u)(t)| ≤ h(|u|)(t) (1.1.10) is satisfied. Then the problem (1.1.1), (1.1.2i) has one and only one solu- tion. Theorem 1.1.1i0. Let i ∈ {1, 2} and let the constants α ∈ [0, 1[, β ∈ ]0, 1] connected by the inequality (1.1.6) be such that p2 ∈ Lx1−β σ(p1) [a, b] (1.1.11) and the functions p0, p1 : ]a, b[ → R satisfy the inclusion (1.1.7i), where h ∈ L Cxβ ; L xα σ(p1) (1.1.12) is a nonnegative operator and the function x : ]a, b[→ R+ is defined by the equality (1.1.9i). Let, moreover, a continuous linear operator g:Cxβ (]a, b[)→ L xα σ(p1) ([a, b]) be such that for any function u ∈ Cxβ (]a, b[) almost everywhere in the interval ]a, b[ the inequality (1.1.10) is satisfied. Then the problem (1.1.1), (1.1.2i0) has one and only one solution in the space Cxβ (]a, b[). Remark 1.1.1i. Let i ∈ {1, 2} and all the requirements of Theorem 1.1.1i be satisfied. Then for any function v0 ∈ C(]a, b[) there exists a unique sequence vn : [a, b] → R, n ∈ N, such that for every n ∈ N, vn is a solution of the problem v′′ (t) = p0(t)v1(t) + p1(t)v′ (t) + g(vn−1)(t) + p2(t), v(a) = c1, vi−1 (b−) = c2, (1.1.13i) and uniformly on ]a, b[ lim n→∞ (vn(t) − u(t)) = 0, lim n→∞ σi(p1)(t)(v′ n(t) − u′ (t)) = 0, (1.1.14) where u is a solution of the problem (1.1.1), (1.1.2i). Remark 1.1.1i0. Let i ∈ {1, 2} and all the requirements of Theorem 1.1.1i0 be satisfied. Then for any function v0 ∈ Cxβ (]a, b[) there exists a unique sequence vn : [a, b] → R, n ∈ N, such that for every n ∈ N, vn is a solution of the problem v′′ (t) = p0(t)v(t) + p1(t)v′ (t) + g(vn−1)(t) + p2(t), v(a) = 0, vi−1 (b−) = 0, (1.1.13i0) and uniformly on ]a, b[ lim n→∞ vn(t) − u(t) xβ(t) = 0, lim n→∞ xα (t) σ(p1)(t) (v′ n(t) − u′ (t)) = 0, (1.1.15) 11 where u is a solution of the problem (1.1.1), (1.1.2i0). We can easily give examples of the operator h and the function p1 such that h ∈ L(Cxβ ; L xα σ(p1) ) and h /∈ L(C; Lσi(p1)). Example 1.1.1. Let ε > 0, p1(t) ≡ 0, h(u)(t) = [(b − t)(t − a)]−2−ε for a ≤ t ≤ b and let τ : [a, b] → {a, b} be a measurable function. Example 1.1.2. Let a = −1, b = 1, α = β = 1 5 , p1(t) ≡ 0 and h(u)(t) = (1 − t2 )−3 u(τ(t)), τ(t) = 1 − (1 − t2)10 for −1 ≤ t ≤ 1. Then it is clear that σ(p1)(t) = 1, x(t) = 1 − t2 , x1/5 (τ(t)) = (1 − t2 )2 for − 1 ≤ t ≤ 1 and α + β < 1 2 . In such a case if u1 ∈ C x 1 5 ([−1, 1]) it follows from the inequality |u1(τ(t))| ≤ δx1/5 (τ(t)) for − 1 ≤ t ≤ 1, where δ = sup u1(τ(t)) x1/5(τ(t)) : −1 < t < 1 , that 1 −1 xα (s)h(u1)(s) ds ≤ δ 1 −1 (1 − s2 )−4/5 ds < +∞, i.e., the condition (1.1.11i) is satisfied. Let now u2(t) ≡ 1. Then u2 ∈ C( ] − 1, 1[ ) and 1 −1 x(s)h(u2)(s) ds = 1 −1 (1 − s2 )−2 ds, i.e., owing to the fact that the last integral does not exist, the condition (1.1.81) is violated. Consider the case where p0(t) ≡ 0, p1(t) ≡ 0, i.e., when the equation (1.1.1) has the form u′′ (t) = g(u)(t) + p2(t). (1.1.16) Then the following theorem is valid. Theorem 1.1.21. Let γ ∈ [0, 1[, p2 ∈ Lx([a, b]) (1.1.17) and g ∈ L(C; Lxγ ) (1.1.18) 12 be a nonnegative operator, where x(t) = (t − a)(t − b) for a ≤ t ≤ b. (1.1.191) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that 0 ≤ β < 1 − γ, (1.1.20) α + β ≤ 1 2 (1.1.21) and b a xα (s)g(xβ )(s) ds < 2β 16 b − a b − a 4 2(α+β) . (1.1.22) Then the problem (1.1.16), (1.1.21) has one and only one solution. Remark 1.1.2. Theorem 1.2.21 will remain valid if we replace the conditions (1.1.20) and (1.1.22) respectively by 0 < β < 1 − γ, (1.1.23) and b a xα (s)g(xβ )(s) ds ≤ 2β 16 b − a b − a 4 2(α+β) . (1.1.241) Theorem 1.1.22. Let γ ∈ [0, 1[ and let a function p2 and a nonnegative operator g satisfy respectively the inclusions (1.1.17) and (1.1.18), where x(t) = t − a for a ≤ t ≤ b. (1.1.192) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that the conditions (1.1.20), (1.1.21) are fulfilled and b a xα (s)g(xβ )(s) ds ≤ 8 b − a b − a 4 α+β . (1.1.242) Then the problem (1.1.16), (1.1.22) has one and only one solution. Theorem 1.1.2i0. Let i ∈ {1, 2}, γ ∈ [0, 1[, δ ∈]0, 1 − γ[, p2 ∈ Lxγ ([a, b]) (1.1.25) and let g ∈ L(Cxδ ; Lxγ ) (1.1.26) 13 be a nonnegative operator, where the function x is defined by the equality (1.1.19i). Let, moreover, there exist constants α ∈ [0, 1 2 ], β ∈]0, 1 2 ], such that δ ≤ β < 1 − γ (1.1.27) and the conditions (1.1.21), (1.1.24i) are satisfied. Then the problem (1.1.16), (1.1.2i0) has in the space Cxδ (]a, b[) one and only one solution. Remark 1.1.3. The condition (1.1.22) is unimprovable in the sense that it cannot be replaced by the condition b a xα (s)g(xβ )(s) ds < 2β 16 b − a b − a 4 2(α+β) + ε (1.1.28) with no matter how small ε > 0. Indeed, let α = 0, β = 0, a = − 1 2 , b = 1 2 , λ = ε 4(16 + ε) , µ = 16λ 1 + 1 (16 + ε)2 , g0(t) =    64µ2 (16µ2 − (1 + 4t)2 )− 3 2 for t ∈ − 1 4 − λ, − 1 4 + λ 64µ2 (16µ2 − (1 − 4t)2 )− 3 2 for t ∈ 1 4 − λ, 1 4 + λ 0 for − 1 2 , − 1 4 − λ ∪ − 1 4 + λ, 1 4 − λ ∪ 1 4 + λ, 1 2 , p2(t) = 0, τ(t) = − 4 16 + ε sign t for − 1 2 ≤ t ≤ 1 2 , and g(u)(t) = g0(t)u(τ(t)). Then the problem (1.1.16), (1.1.210) can be rewritten as u′′ (t) = g0(t)u(τ(t)), (1.1.29) u − 1 2 = 0, u 1 2 = 0. (1.1.30) Note that for the operator g defined in such a way the condition (1.1.18) is satisfied for γ = 0 and 1 2 − 1 2 g(1)(s) ds = 1 2 − 1 2 g0(s) ds = 16 + ε, 14 i.e., instead of (1.1.22) the condition (1.1.28) is satisfied. In spite of this fact we can check directly that the function u(t) = c t − 1 2 s − 1 2 g0(η) sign(−η)dη ds − 4 + ε 4 t + 1 2 is for any c ∈ R a solution of the problem (1.1.29), (1.1.30), i.e., the unique solvability is violated. 1.1.2. Effective Sufficient Conditions for the Unique Solvability of the Problem (1.1.1), (1.1.2i) (i = 1, 2). Corollary 1.1.11. Let the function x be defined by (1.1.91), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.31), (1.1.51), [p0]− ∈ L xα σ(p1) ([a, b]) (1.1.31) and for every function u ∈ C(]a, b[) almost everywhere on interval ]a, b[ the inequality (1.1.10) is satisfied, where a nonnegative operator h satisfies the inclusion (1.1.81). Let, moreover, b t σ(p1)(η)dη α t a ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) s a σ(p1)(η)dη α ds + + t a σ(p1)(η)dη α b t ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) b s σ(p1)(η)dη α ds < < 4 b a σ(p1)(η)dη b a σ(p1)(η)dη 2 2(α+β) for a ≤ t ≤ b (1.1.321) Then the problem (1.1.1), (1.1.21) has one and only one solution. Corollary 1.1.12. Let the function x be defined by (1.1.92), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.32), (1.1.52), (1.1.31) and for every function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality (1.1.10) be satisfied, where a 15 nonnegative operator h satisfies (1.1.82). Let, moreover, t a ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) s a σ(p1)(η)dη α ds + + t a σ(p1)(η)dη α b t ([p0(s)]−xβ (s) + h(xβ )(s)) σ(p1)(s) ds < < b a σ(p1)(η)dη α+β−1 for a ≤ t ≤ b. (1.1.322) Then the problem (1.1.1), (1.1.22) has one and only one solution. Corollary 1.1.1i0. Let i ∈ {1, 2}, the function x be defined by (1.1.9i), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.3i), (1.1.11), (1.1.31) and for any function u ∈ Cxβ (]a, b[) almost everywhere in the interval ]a, b[ the inequality (1.1.10) be satisfied, where the nonnegative operator h satisfies the inclusion (1.1.12). Let, moreover, (1.1.32i) be satisfied. Then the problem (1.1.1), (1.1.2i0) has in the space Cxβ (]a, b[) one and only one solution. Remark 1.1.4. Corollary 1.1.1i remains valid if we replace the conditions (1.1.8i) and (1.1.32i) respectively by the conditions h ∈ L(C; Lσi(p1)), (1.1.33) and b a ([p0(s)]−xα+β (s) + xα (s)h(xβ )(s)) σ(p1)(s) ds < < 4 b a σ(p1)(η)dη      b a σ(p1)(η)dη 2      2(α+β) (1.1.341) for i = 1 or by b a ([p0(s)]−xα+β (s)+xα (s)h(xβ )(s)) σ(p1)(s) ds< b a σ(p1)(η)dη α+β−1 (1.1.342) for i = 2, where the function x is defined by (1.1.9i). Remark 1.1.40. Corollary 1.1.1i0 remains valid if we replace (1.1.32i) by (1.1.34i) and reject the condition (1.1.12) at all. 16 Consider the case where the equation (1.1.1) has the form u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + n k=1 gk(t)u(τk(t)) + p2(t). (1.1.35) Corollary 1.1.21. Let the function x be defined by (1.1.91), the constants α, β ∈ [0, 1] be defined by the inequality (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy the conditions (1.1.31), (1.1.51), (1.1.31), τk : [a, b] → [a, b] (k = 1, . . . , n) be measurable functions and gkxβ (τk) ∈ L xα σ(p1) ([a, b]), gk ∈ Lσ1(p1)([a, b]) (k = 1, . . . , n). (1.1.361) Let, moreover, b t σ(p1)(η)dη α t a ([p0(s)]−xβ (s) + n k=1 |gk(s)|xβ (τk(s))) σ(p1)(s) × × s a σ(p1)(η)dη α ds + t a σ(p1)(η)dη α × × b t ([p0(s)]−xβ (s) + n k=1 |gk(s)|xβ (τk(s))) σ(p1)(s) b s σ(p1)(η)dη α ds < < 4 b a σ(p1)(η)dη b a σ(p1)(η)dη 2 2(α+β) for a ≤ t ≤ b. (1.1.371) Then the problem (1.1.35), (1.1.21) has one and only one solution. Corollary 1.1.22. Let the function x be defined by (1.1.92), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.132), (1.1.52), (1.1.31), τk : [a, b] → [a, b] (k = 1, . . . , n) be measurable functions and gkxβ (τk) ∈ L xα σ(p1) ([0, b]), gk ∈ Lσ2(p1)([a, b]) (k = 1, . . . , n). (1.1.362) Let, moreover, t 0 [p0(s)]−xβ (s) + n k=1 |gk(s)|xβ (τk(s)) σ(p1)(s) s a σ(p1)(η)dη α ds + 17 + t a σ(p1)(η)dη α b t [p0(s)]−xβ (s) + n k=1 |gk(s)|xβ (τk(s)) σ(p1)(s) ds < < b a σ(p1)(η)dη α+β−1 for a ≤ t ≤ b. (1.1.372) Then the problem (1.1.35), (1.1.22) has one and only one solution. Corollary 1.1.2i0. Let i ∈ {1, 2}, the function x be defined by (1.1.9i), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by the inequality (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy the conditions (1.1.3i), (1.1.11), (1.1.31), τk : [a, b] → [a, b] (k = 1, . . . , n) be measurable functions and gkxβ (τk) ∈ L xα σ(p1) ([a, b]) (k = 1, . . . , n). (1.1.38) Let, moreover, the conditions (1.1.37i) be satisfied. Then the problem (1.1.35), (1.1.2i0) has in the space Cxβ (]a, b[) one and only one solution. Remark 1.1.5. Corollary 1.1.2i remains valid if we replace the conditions (1.1.36i) and (1.1.37i) respectively by the conditions gk ∈ Lσi(p1)([a, b]) (k = 1, . . . , n) (1.1.39) and b a [p0(s)]−xα+β (s) + xα n k=1 |gk(s)|xβ (τk(s)) σ(p1)(s) ds < < 4 b a σ(p1)(η)dη      b a σ(p1)(η)dη 2      2(α+β) (1.1.401) for i = 1 or by b a [p0(s)]−xα+β (s) + xα (s) n k=1 |gk(s)|xβ (τk(s)) σ(p1)(s) ds < < b a σ(p1)(η)dη α+β−1 (1.1.402) for i = 2, where the function x is defined by (1.1.9i). 18 Remark 1.1.50. Corollary 1.1.2i0 remains valid if we replace (1.1.37i) by (1.1.40i) and reject the condition (1.1.38) at all. Corollary 1.1.31. Let the function x be defined by (1.1.91), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions gk, pj : ]a, b[→ R (k = 1, . . . , n; j = 0, 1, 2) satisfy (1.1.31), (1.1.51), (1.1.361), where τk : [a, b] → [a, b] (k = 1, . . . , n) are measurable functions and p0(t) ≥ 0 for a < t < b. (1.1.41) Let, moreover, for any m ∈ {1, . . . , n} the condition n k=1 τm(t) a |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη b τk(s) σ(p1)(η)dη β × × s a σ(p1)(η)dη α ds b τm(t) σ(p1)(η)dη α + + n k=1 b τm(t) |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη b τk (s) σ(p1)(η)dη β × × b s σ(p1)(η)dη α ds τm(t) a σ(p1)(η)dη α < < 4 b a σ(p1)(η)dη      b a σ(p1)(η)dη 2      2(α+β) , a ≤ t ≤ b, (1.1.421) be valid. Then the problem (1.1.35), (1.1.21) has one and only one solution. Corollary 1.1.32. Let the function x be defined by the equality (1.1.92), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions gk, pj : ]a, b[→ R (k = 1, . . . , n; j = 0, 1, 2) satisfy the conditions (1.1.32), (1.1.52), (1.1.362), (1.1.41), where τk : [a, b] → [a, b] (k = 1, . . . , n) are measurable functions. Let, moreover, for any m ∈ {1, . . . , n} the condition n k=1 τm(t) a |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη β s a σ(p1)(η)dη α ds + + n k=1 b τm(t) |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη b ds τm(t) a σ(p1)(s) ds α < 19 < b a σ(p1)(η)dη α+β−1 , a ≤ t ≤ b, (1.1.422) be valid. Then the problem (1.1.35), (1.1.22) has one and only one solution. Corollary 1.1.3i0. Let i ∈ {1, 2}, the function x be defined by (1.1.9i), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.1.6), the functions gk, pj : ]a, b[→ R (k = 1, . . . , n; j = 0, 1, 2) satisfy (1.1.3i), (1.1.11), (1.1.38), (1.1.41), where τk : [a, b] → [a, b] (k = 1, . . . , n) are measurable functions. Let, moreover, for any m ∈ {1, . . . , n} the condition (1.1.42i) be valid. Then the problem (1.1.35), (1.1.2i0) has in the space Cxβ (]a, b[) one and only one solution. Remark 1.1.6. The condition (1.1.42i) consisting of n separate inequalities can be replaced by one inequality n k=1 t a |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη b τk(s) σ(p1)(η)dη β × × s a σ(p1)(η)dη α ds b t σ(p1)(η)dη α + + n k=1 b t |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη b τk (s) σ(p1)(η)dη β × × b s σ(p1)(η)dη α ds t a σ(p1)(η)dη α < 4 b a σ(p1)(η)dη × ×      b a σ(p1)(η)dη 2      2(α+β) for t ∈ Θτ1,...,τn (1.1.431) if i = 1 and n k=1 t a |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη β s a σ(p1)(η)dη α ds + + n k=1 b t |gk(s)| σ(p1)(s) τk(s) a σ(p1)(η)dη β ds t a σ(p1)(η)dη α < 20 < b a σ(p1)(η) α+β−1 for t ∈ Θτ1,...,τn (1.1.432) if i = 2, where Θτ1,...,τn = n ∪ k=1 τk(t)| a ≤ t ≤ b . For clearness we will give one corollary for the equation u′′ (t) = g0(t)u(τ(t)) + p2(t). (1.1.44) Corollary 1.1.4i. Let i ∈ {1, 2}, the constants α, β ∈ [0, 1] be connected by the inequality (1.1.6), τ : [a, b] → [a, b] be a measurable function and p2, g0 ∈ Lx([a, b]), (1.1.45) where x(t) = (a − t)(b − t)2−i for a ≤ t ≤ b. (1.1.46) Let, moreover, b a |g(s)| (τ(s) − a)(b − τ(s))2−i β (s − a)(b − s)2−i α ds < < 2 i 2(1−α−β) (b − a) 2 i (α+β)−1 . (1.1.47i) Then the problem (1.1.44), (1.1.2i) has one and only one solution. Corollary 1.1.4i0. Let i ∈ {1, 2}, the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.1.6), τ : [a, b] → [a, b] a be measurable function, p2 ∈ Lx1−β ([a, b]), (1.1.48) where the function x is defined by (1.1.46). Let, moreover, the condition (1.1.47i) be satisfied. Then the problem (1.1.44), (1.1.2i0) has one and only one solution in the space Cxβ (]a, b[). Remark 1.1.7. In the case of the equation u′′ (t) = g0(t)u(t) + p2(t) (1.1.49) the conditions (1.321), (1.1.341), (1.1.401), (1.1.421), (1.1.471) will take for α = β = 0 the form b a |g0(s)| ds < 4 b − a . 21 As is known, this condition is unimprovable in the sense that no matter how small ε > 0 is, the inequality b a |g0(s)| ds ≤ 4 b − a + ε does not guarantee the unique solvability of the problem (1.1.49), (1.1.21). This implies that the corollaries corresponding to the above conditions are unimprovable in the above-mentioned sense. Corollary 1.1.51. Let the function x be defined by (1.1.91), the constants α, β ∈ [0, 1] be connected by the inequality (1.1.6), the functions pj :]a, b[→ R (j = 0, 1, 2) satisfy the conditions (1.1.31), (1.1.51) and for any function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ (1.1.10) be satisfied, where the nonnegative operator h satisfies the inclusion (1.1.81). Let, moreover, in case β < 1, x(t) σ2(p1)(t) h(xβ )(t) xβ(t) − p0(t) ≤ 2β2 for a < t < b, (1.1.501) and in case β = 1, ess sup t∈]a,b[ x(t) σ2(p1)(t) h(x)(t) x(t) − p0(t) < 2 (1.1.511) be satisfied. Then the problem (1.1.1), (1.1.21) has one and only one solu- tion. Remark 1.1.8. The condition (1.1.51) is unimprovable in the sense that the validity of Corollary 1.1.51 is violated if we replace it by the condition ess sup t∈]a,b[ x(t) σ2(p1)(t) h(x)(t) x(t) − p0(t) ≤ 2β2 . (1.1.52) Indeed, let h(u) ≡ 0, p1 ≡ 0, p2 ≡ 0. Then σ(p1)(t) = 1 and x(t) = (b − t)(t − a) for a ≤ t ≤ b and the condition (1.1.52) will take the form ess sup t∈]a,b[ − (b − t)(t − a)p0(t) ≤ 2. (1.1.53) If p0(t) = − 2 (b − t)(t − a) , then the condition (1.1.53) is satisfied in the form of the equality, and at the same time, for any c ∈ R the function c(b − t)(t − a) is a solution of the equation u′′ (t) = − 2 (b − t)(t − a) u(t), (1.1.54) 22 that is, the uniqueness of solution of the problem (1.1.54), (1.1.2i0) is violated although the condition (1.1.52) along with the other requirements of Corollary 1.1.51 is satisfied. Corollary 1.1.52. Let the function x be defined by (1.1.92), the constants α, β ∈ [0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.32), (1.1.52) and for any function u ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality (1.1.10) be satisfied, where a nonnegative operator h satisfies the inclusion (1.1.82). Let, moreover, ess sup t∈]a,b[ x2−[β] (t) σ2(p1)(t) h(xβ )(t) xβ(t) − p0(t) < β(1 − β), (1.1.502) x2−β σ2(p1) [p0]− ∈ L∞([a, b]) (1.1.55) if 0 < β ≤ 1 and 0 ≤ p0(t) − h(1)(t) for a < t < b (1.1.512) if β = 0 be satisfied. Then the problem (1.1.1), (1.1.22) has one and only one solution. Remark 1.1.9. In the case β = 1, the condition (1.1.55) follows automatically from the condition (1.1.502). Corollary 1.1.510. Let the function x be defined by (1.1.91), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.31), (1.1.11) and for any function u ∈ Cxβ (]a, b[) almost everywhere on the interval ]a, b[ the inequality (1.1.10) be satisfied, where the nonnegative operator h satisfies the inclusion (1.1.12). Let, moreover, in case 0 < β < 1 the condition (1.1.501) and in case β = 1 the condition (1.1.511) be satisfied. Then the problem (1.1.1), (1.1.210) has in the space Cxβ (]a, b[) one and only one solution. Corollary 1.1.520. Let the function x be defined by (1.1.92), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.1.6), the functions pj : ]a, b[→ R (j = 0, 1, 2) satisfy (1.1.32), (1.1.11) and for any function u ∈ Cxβ (]a, b[) almost everywhere on the interval ]a, b[ the inequality (1.1.10) be satisfied, where the nonnegative operator h satisfies the inclusion (1.1.12). Let, moreover, the conditions (1.1.502) and (1.1.55) be satisfied. Then the problem (1.1.1), (1.1.220) has one and only one solution in the space Cxβ (]a, b[). Corollary 1.1.61. Let the functions τk : [a, b] → [a, b] (k = 1, . . . , n) be measurable and the functions pj, pk ∈ Lloc(]a, b[) (k = 1, . . . , n; j = 0, 1, 2) as well as the constants λl,m ∈]0, +∞[, βm ∈ [0, 1] (l, m = 1, 2), c ∈]a, b[ be such that the conditions (1.1.31), (1.1.51) are satisfied, gk ∈ Lσ1(p1)([a, b]) (1.1.561) 23 and +∞ 0 ds λ11 + λ12s + s2 > (c − a)1−β1 1 − β1 , +∞ 0 ds λ21 + λ22s + s2 > (b − c)1−β2 1 − β2 . (1.1.571) Let, moreover, (t − a)2β2 p0(t) − n k=1 |gk(t)| ≥ −λ11, (t − a)β1 p1(t) + β1 t − a − n k=1 |gk(t)|(τk(t) − t) ≥ −λ12 for a < t < c, (b − t)2β2 p0(t) − n k=1 |gk(t)| ≥ −λ12, (b − t)β2 p1(t) − β2 b − t − n k=1 |gk(t)|(τk(t) − t) ≤ λ22 for c ≤ t < b. (1.1.581) Then the problem (1.1.35), (1.1.21) has one and only one solution. Corollary 1.1.62. Let the functions τk : [a, b] → [a, b] (k = 1, . . . , n) be measurable and the functions p1, pj, gk ∈ Lloc(]a, b]) (k = 1, . . . , n; j = 0, 1, 2) as well as the constants λl,m ∈]0, +∞[, (l, m = 1, 2), βr ∈ [0, 1] (r = 1, 2, 3), c ∈ ] max(a, b−1); b], ε > 0 and the dependent on them constant α ∈ [0, 1[ be such that the conditions σ(p1) ∈ L([a, b]), pjσ2(p1) ∈ L([a, b]) (j = 0, 2), gkσ2(p1) ∈ L([a, b]) (k = 1, . . . , n) (1.1.562) and +∞ ε ds λ11 + λ12s + s2 > (c − a)1−β1 1 − β1 , +∞ 0 ds λ21 + λ22s + s2 > (b − c)1−β2 1 − β2 (1.1.572) 24 are satisfied. Let, moreover, (t − a)2β2 p0(t) − n k=1 |gk(t)| ≥ −λ11, (t − a)β1 p1(t) + β1 t − a − n k=1 |gk(t)|(τk(t) − t) ≥ −λ12 for a < t < c, (b − t)β2−β3 p0(t) − n k=1 |gk(t)| ≥ −αλ21, (b − t)β2 p1(t) + β3 b − t − n k=1 |gk(t)|(τk(t) − t) ≥ λ22 for c ≤ t < b. (1.1.582) Then for any function p1 ∈ Lloc(]a, b]) such that p1(t) ≥ p1(t) for a < t < b, (1.1.59) the problem (1.1.35), (1.1.22) has one and only one solution. Consider now corollaries of Theorems 1.1.2i and 1.1.2i0 for the equation u′′ (t) = n k=1 gk(t)u(τk(t)) + p2(t). (1.1.60) Corollary 1.1.71. Let γ ∈ [0, 1[, the function p2 : ]a, b[ → R satisfy the inclusion (1.1.17), gk ∈ Lxγ ([a, b]) (k = 1, . . . , n) (1.1.61) and gk(t) ≥ 0 (k = 1, . . . , n) for a < t < b, (1.1.62) where x(t) = (b − t)(t − a) a ≤ t ≤ b. (1.1.631) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that 0 ≤ β < 1 − γ, α + β ≤ 1 2 (1.1.64) and n k=1 b a gk(s)(b − τk(s))β (τk(s) − a)β (b − s)α (s − a)α ds < < 2β 16 b − a b − a 4 2(α+β) . (1.1.65) 25 Then the problem (1.1.60), (1.1.21) has one and only one solution. Remark 1.1.10. Corollary 1.1.71 remains valid if for β ∈]0, 1 − γ[ we replace the condition (1.1.65) by the following one: n k=1 b a gk(s)(b − τk(s))β (τk(s) − a)β (b − s)α (s − a)α ds ≤ ≤ 2β 16 b − a b − a 4 2(α+β) . (1.1.661) Corollary 1.1.72. Let γ ∈ [0, 1[, the functions p2, pk : ]a, b[ → R (k = 1, . . . , n) satisfy the conditions (1.1.17), (1.1.61), and (1.1.62), where x(t) = t − a for a ≤ t ≤ b. (1.1.632) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that the conditions (1.1.64) and n k=1 b a gk(s)(τk(s) − a)β (s − a)β ds ≤ 8 b − a b − a 4 α+β (1.1.662) are satisfied. Then the problem (1.1.60), (1.1.22) has one and only one solution. Corollary 1.1.7i0. Let i ∈ {1, 2}, γ ∈ [0, 1[, δ ∈]0, 1 − γ[, p2 ∈ Lxγ ([a, b]), gkxδ (τk) ∈ Lxγ ([a, b]) (k = 1, . . . , n), and the condition (1.1.62) be satisfied, where the function x is defined by (1.1.63i). Let, moreover, there exist constants α ∈ [0, 1 2 ], β ∈]0, 1 2 ] such that the conditions δ ≤ β < 1 − γ, α + β ≤ 1 2 and (1.1.66i) are satisfied. Then the problem (1.1.60), (1.1.2i0) has in the space Cxδ (]a, b[) one and only one solution. § 1.2. Auxiliary Propositions 1.2.1. Statement of Auxiliary Problems and Some of Their Properties. Let us consider the linear equations v′′ (t) = p0(t)v(t) + p1(t)v′ (t) − h(v)(t) + p2(t), (1.2.1) v′′ (t) = p0(t)v(t) + p1(t)v′ (t) − h(v)(t) (1.2.10) under the boundary conditions u(a) = c1, u(b) = c2, (1.2.21) 26 or u(a) = c1, u′ (b−) = c2, (1.2.22) as well as under the conditions v(a) = 0, v(b) = 0, (1.2.210) v(a) = 0, v′ (b−) = 0, (1.2.220) where c1, c2 ∈ R and h : C( ]a, b[ ) → Lloc(]a, b[) is a continuous linear operator and pj ∈Lloc(]a, b[) (j = 0, 1, 2), σ(p1)∈L([a, b]), p0 ∈Lσ1(p1)([a, b]) (1.2.31) or pj ∈Lloc(]a, b]) (j = 0, 1, 2), σ(p1)∈L([a, b]), p0 ∈Lσ2(p1)([a, b]). (1.2.32) For this purpose we will need the homogeneous equation v′′ (t) = p0(t)v(t) + p1(t)v′ (t) (1.2.4) under the initial conditions v(a) = 0, lim t→a v′ (t) σ(p1)(t) = 1, (1.2.5) v(b) = 0, lim t→b v′ (t) σ(p1)(t) = −1, (1.2.51) or v(b) = 1, v′ (b−) = 0. (1.2.52) The facts mentioned in the remarks below or their analogues have been proved in [23], pp. 110–158. Remark 1.2.1. Let measurable functions p0, p1 : ]a, b[ → R satisfy the conditions (1.2.31) and the functions v1 and v2 be respectively solutions of the problems (1.2.4), (1.2.5) and (1.2.4), (1.2.51). Then any linearly independent with vj, (j = 1, 2) solution v of the equation (1.2.4) satisfies the condition v(a) = 0 for j = 1 and v(b) = 0 for j = 2. 27 Remark 1.2.2. Let i ∈ {1, 2} and (p0, p1) ∈ Vi,0(]a, b[). (1.2.6i) Then the problem (1.2.4), (1.2.2i0) has only the trivial solution and the unique Green’s function G can be represented as: G(t, s) =    − v2(t)v1(s) v2(a)σ(p1)(s) for a ≤ s < t ≤ b, − v2(s)v1(t) v2(a)σ(p1)(s) for a ≤ t < s ≤ b, (1.2.7) where v1 and v2 are respectively the solutions of the problems (1.2.4), (1.2.5) and (1.2.4), (1.2.5i), and G(t, s) < 0 for (t, s) ∈ ]a, b[ × ]a, b[ , (1.2.8) G(a, s) = 0, G(b, s) = i − 1 for a ≤ s ≤ b. (1.2.9i) Remark 1.2.3. Let i ∈ {1, 2} and the inclusion (1.2.6i) be satisfied. Then there exist constants c∗, d∗ ∈ R+ such that the estimates d∗ ≤ v1(t) t a σ(p1)(s) ds ≤ c∗, d∗ ≤ v2(t) ( b t σ(p1)(s) ds)2−i ≤ c∗ (1.2.10i) for a < t < b, |v′ 1(t)| σ(p1)(t) ≤ 1 + c∗ t a |p0(s)|σ2(p1)(s) ds, |v′ 2(t)| σ(p1)(t) ≤ 2 − i + c∗ b t |p0(s)| σ(p1)(s) b s σ(p1)(η) dη 2−i ds for a ≤ t < b (1.2.11i) are valid, where v1 and v2 are respectively the solutions of the problems (1.2.4), (1.2.5) and (1.2.4), (1.2.5i), and ∂j−1 G(t, s) ∂tj−1 ≤ ≤ c∗ σi(p1)(s) [σi(p1)(t)]j−1 (j = 1, 2) for (t, s) ∈]a, b[ × ]a, b[ (t = s). (1.2.12i) Remark 1.2.4. Let i ∈ {1, 2}, the conditions (1.2.3i) be satisfied and the problem (1.2.4), (1.2.2i) have lower w1 and upper w2 functions such that w1(t) ≤ w2(t) for a ≤ t ≤ b. Then the problem (1.2.4), (1.2.2i) has at least one solution v such that w1(t) ≤ v(t) ≤ w2(t) for a ≤ t ≤ b. 28 Remark 1.2.5. Let i ∈ {1, 2} and the inclusion (1.2.6i) be satisfied. Then every upper function w of the problem (1.2.4), (1.2.2i0) is nonnegative in the interval ]a, b[; moreover, if w(a) + w(i−1) (b−) = 0, then w is positive on the interval ]a, b[. Remark 1.2.6. Let i ∈ {1, 2}, the functions p0, p1 : ]a, b[ → R satisfy the conditions (1.2.3i) and p0(t) ≥ 0 for a < t < b. Then the inclusion (1.2.6i) is valid. Lemma 1.2.1. Let i ∈ {1, 2} and h ∈ L(C; Lσi(p1)) (1.2.13i) where h is a nonnegative operator. Then Vi,0(]a, b[; h) ⊂ Vi,0(]a, b[). Proof. Let (p0, p1) ∈ Vi,0(]a, b[; h). Then the problem (1.2.10), (1.2.2i0) has a positive upper function w which because of the nonnegativeness of the operator h will at the same time be an upper function of the problem (1.2.4), (1.2.2i0). Consider first the case i = 1. For the equation (1.2.4) we pose the problem v(a) = 0, v(b) = w(b), (1.2.14) for which β(t) ≡ 0 and w are respectively lower and upper functions. Then by virtue of Remark 1.2.4, the problem (1.2.4), (1.2.14) has a solution v0 such that 0 ≤ v0(t) ≤ w(t) for a ≤ t ≤ b. If we assume that v0(t0) = 0 for some t0 ∈ ]a, b[ , then we will get the contradiction with the unique solvability of the Cauchy problem, i.e., v0(t) > 0 for a < t ≤ b. (1.2.15) As is seen from Remark 1.2.1 and the conditions (1.2.14) that v1 a solution of the problem (1.2.4), (1.2.51), and v0 are linearly dependent, hence by virtue of (1.2.15), v1(t) > 0 for a < t ≤ b, i.e., as is seen from Definition 1.1.2, (p0, p1) ∈ V1,0(]a, b[). Let now i = 2, and for the equation (1.2.4) we pose the initial problem v(b) = 0, v′ (b−) = −1 29 which, with regard for the conditions (1.2.32), has a unique solution v defined on the whole interval [a, b]. Then we choose ε > 0 such that the inequality εv(t) < w(t) for a < t < b (1.2.16) is satisfied; this is possible because the function w is positive. It is clear from (1.2.16) that w1(t) = w(t) − εv(t) is an upper function of the problem (1.2.4), (1.2.220) and w′ 1(b−) > 0, w1(t) > 0 for a ≤ t ≤ b. We consider now for the equation (1.2.4) the problem v(a) = 0, v′ (b−) = w′ 1(b−), (1.2.17) for which β(t) ≡ 0 and w1 are respectively lower and upper functions. Hence by virtue of Remark 1.2.4, the problem (1.2.4), (1.2.17) has a solution v0 such that 0 ≤ v0(t) ≤ w1(t) for a < t < b and v0(a) = 0, v0(b) > 0, v′ 0(b−) > 0. Reasoning in the same way as for i = 1, we see that (p0, p1) ∈ V2,0(]a, b[). Along with Lemma 1.2.1 we have proved the following Lemma 1.2.2. Let i ∈ {1, 2}, the functions p0, p1 : ]a, b[ → R satisfy the conditions (1.2.3i) and, moreover, let the problem (1.2.4), (1.2.2i0) have a positive upper function. Then the inclusion (1.2.6i) is satisfied. Lemma 1.2.3. Let i ∈ {1, 2}, the functions p0, p1 : ]a, b[ → R satisfy the inclusion (1.2.6i) and the nonnegative operator h satisfy the inclusion (1.2.13i). Let, moreover, ρ0 ∈ C(]a, b[) such that ρ0(t) > 0 for a < t < b (1.2.18) and sup 1 ρ0(t) b a |G(t, s)|h(ρ0)(s)ds : a < t < b < 1, (1.2.19) where G is Green’s function of the problem (1.2.4), (1.2.2i0). Then there exists a continuous function ρ : [a, b] → R+ such that max 1 ρ(t) b a |G(t, s)|h(ρ)(s)ds : a ≤ t ≤ b < 1. (1.2.20) 30 Proof. First of all we note that the existence of Green’s function G of the problem (1.2.4), (1.2.2i0) follows from Remark 1.2.2, and the boundedness of the integrals in the inequalities (1.2.19) and (1.2.20) for any continuous function ρ follows from the estimates (1.2.12i) and the inclusion (1.2.13i). Consider now separately the case i = 2. By virtue of the equalities (1.2.92), the inequality (1.2.19) can be satisfied only under the conditions ρ0(a) ≥ 0, ρ0(b) > 0. (1.2.21) Then (1.2.19) can be rewritten as b a |G(t, s)|h(ρ0)(s)ds < ρ0(t) for a < t ≤ b. (1.2.22) As is seen from the equalities (1.2.92), there exist positive constants r1 and δ such that b a |G(t, s)|h(1)(s)ds − 1 < 0 for a ≤ t ≤ a + δ (1.2.23) and b a |G(t, s)|h(1)(s)ds − 1 < r1 for a ≤ t ≤ b. (1.2.24) On the other hand, from (1.2.22) it follows the existence of a constant r2 > 0 such that r2 < ρ0(t) − b a |G(t, s)|h(ρ0)(s) ds for a + δ ≤ t ≤ b. (1.2.25) Then from (1.2.22)–(1.2.25) we obtain r2 r1 b a |G(t, s)|h(1)(s) ds − 1 ≤ρ0(t)− b a |G(t, s)|h(ρ0)(s) ds for a≤t≤b, which implies the validity of the inequality (1.2.20) for the function ρ(t) = ε + ρ0(t), where ε = r2 r1 . To complete the proof of the lemma we note that for i = 1, unlike the case i = 2, the inequality (1.2.19) by virtue of (1.2.9i) can be satisfied also for ρ(a) > 0, ρ(b) ≥ 0 and for ρ(a) ≥ 0, ρ(b) ≥ 0 as well. 31 In these cases the above lemma can be proved similarly to the case of the conditions (1.2.21) with the only difference that the inequality (1.2.22) will be valid for t ∈ [a, b[ or t ∈]a, b[, the inequality (1.2.23) for t ∈ [b − δ, b] or t ∈ [a+δ; b−δ], and the inequality (1.2.25) will be considered for t ∈ [a, b−δ[ or t ∈]a + δ, b − δ[. Lemma 1.2.4. Let i ∈ {1, 2}, (p0, p1) ∈ Vi,0(]a, b[; h), (1.2.26i) where the nonnegative operator h satisfies the inclusion (1.2.13i). Then there exists a continuous function ρ : [a, b] → R+ such that the inequality (1.2.20) holds, where G is Green’s function of the problem (1.2.4), (1.2.2i0). Proof. As is seen from the definition of the set Vi,0( ]a, b[ ; h), the problem (1.2.10), (1.2.2i0) has on the interval [a, b] a positive upper function w. Then we introduce a continuous operator χ : C(]a, b[) → C(]a, b[) by the equality χ(y)(t) = 1 2 |y(x)| − |w(t) − y(t)| + w(t) for a ≤ t ≤ b (1.2.27) which for any v ∈ C(]a, b[) satisfies 0 ≤ χ(v)(t) ≤ w(t) for a ≤ t ≤ b, (1.2.28) and consider the problem v′′ (t) = p0(t)v(t) + p1(t)v′ (t) − h(χ(v))(t), (1.2.29) v(a) = w(a), v(i−1) (b−) = w(i−1) (b−). (1.2.30i) Note that from Lemma 1.2.1 and Remark 1.2.2 it follows the existence of Green’s function of the problem (1.2.4), (1.2.2i). Introduce the operator H : C(]a, b[) → C(]a, b[) by the equality H(g)(t) = v0(t) + b a |G(t, s)|h(χ(y))(s) ds, where v0 is a solution of the problem (1.2.4), (1.2.30i), and consider the equation v(t) = H(v)(t) (1.2.31) which is equivalent to the problem (1.2.29), (1.2.30i). Let us show that the operator H is compact. Let c∗ be a constant mentioned in Remark 1.2.3, r = c∗ b a σi(p1)(s)h(w)(s)ds, Br = z ∈ C(]a, b[) : z − v0 C ≤ r , 32 and (xn)∞ n=1 be any sequence from Br. Then from the estimate (1.2.12i) for the sequence yn(t) = H(xn)(t), n ∈ N, we have v0 − yn C ≤ r, n ∈ N. (1.2.32) Consider separately the case i = 1. By virtue of (1.2.91), (1.2.28) and the fact that the function v0 is continuous, for any constant ε > 0 there exist a1, b1 ∈]a, b[, a1 < b1 such that max |v0(t1) − v0(t2)| : a ≤ t1 ≤ t2 ≤ a1, b1 ≤ t1 ≤ t2 ≤ b ≤ ε 4 and ε∗ ≡ max b a |G(t, s)|h(χ(xn))(s) ds : a ≤ t ≤ a1, b1 ≤ t ≤ b ≤ ε 8 . Then for any n ∈ N the estimate |yn(t1) − yn(t2)| ≤ ε 4 + 2ε∗ ≤ ε 2 , for a ≤ t1 ≤ t2 ≤ a1, b1 ≤ t1 ≤ t2 ≤ b, is valid. In the same way, by virtue of the estimates (1.2.12i), there exists a constant δ, 0 < δ < min(a1 − a, b − b1), such that for any n ∈ N |yn(t1) − yn(t2)| ≤ ≤ (1 + r) max |v′ 0(t)| + σ−1 1 (p1)(t) : a1 − δ < t < b + δ |t2 − t1| ≤ ε 2 for |t1 − t2| ≤ δ, a1 − δ ≤ tj ≤ b1 + δ (j = 1, 2). It follows from the last two estimates that if tj ∈ [a, b] (j = 1, 2) and |t1 − t2| ≤ δ, then |yn(t1) − yn(t2)| ≤ ε, n ∈ N. From this and from the inequality (1.2.32) we obtain that the sequence (yn)∞ n=1 is uniformly bounded and equicontinuous. In case i = 2, the same follows from the possibility to choose for any ε > 0, owing to (1.1.92), (1.2.28), a1 ∈]a, b[ and 0 < δ < a1 − a such that max |v0(t1) − v0(t2)| : a ≤ t1 ≤ t2 ≤ a1 ≤ ε 4 , max b a |G(t, s)|h(w)(s) ds : a ≤ t ≤ a1 ≤ ε 4 , 33 and |yn(t1) − yn(t2)| ≤ ≤ (1 + r) max |v′ 0(t)| + σ−1 2 (p1)(t) : a1 − δ ≤ t ≤ b |t1 − t2| ≤ ε 2 for |t1 − t2| ≤ δ, a1 − δ ≤ tj ≤ b (j = 1, 2). Then according to the Arzella–Ascoli lemma, the operator H which is, as it is not difficult to show, continuous, transforms the ball Br into its compact subset. In this case the equation (1.2.31), i.e., the problem (1.2.29), (1.2.30i) has at least one solution, say v. Show that 0 < v(t) ≤ w(t) for a ≤ t ≤ b. Let v1(t) = w(t) − v(t). Then from the nonnegativeness of the operator h and also from the inequality (1.2.28) we have v′′ 1 (t) ≤ p0(t)v1(t) + p1(t)v′ 1(t) − h(w − χ(v))(t) ≤ p0(t)v1(t) + p1(t)v′ 1(t) and v1(a) = 0, v (i−1) 1 (b−) = 0. Hence v1 is an upper function of the problem (1.2.4), (1.2.2i0), and due to Remark 1.2.5, v1(t) ≥ 0 for a < t < b, i.e., v(t) ≥ w(t) for a < t < b. (1.2.33) On the other hand, taking into account the inequality (1.2.28) and the fact that the operator h is nonnegative, from (1.2.29) and (1.2.30i) we conclude that v is an upper function of the problem (1.2.4), (1.2.2i0), i.e., by virtue of Remark 1.2.5, v(t) > 0 for a ≤ t ≤ b. (1.2.34) It follows from (1.2.33) and (1.2.34) that the inequality 0 < v(t) ≤ w(t) is valid and hence χ(v)(t) = v(t) for a ≤ t ≤ b, i.e., v as a solution of the equation (1.2.31) has the form v(t) = v0(t) + b a |G(t, s)|h(v)(s) ds for a ≤ t ≤ b, (1.2.35) where by Remark 1.2.5, v0(t) > 0 for a ≤ t ≤ b. (1.2.36) 34 If we introduce the notation ρ(t) = v(t) and take into consideration (1.2.36), then in view of (1.2.35) we can see that our lemma is valid. Lemma 1.2.5. Let i ∈ {1, 2}, the constants α ∈ [0, 1[ and β ∈]0, 1] be connected by the inequality α + β ≤ 1, (1.2.37) (p0, p1) ∈ Vi,β(]a, b[; h), (1.2.38i) where h ∈ L Cxβ ; L xα σ(p1) (1.2.39i) is a nonnegative operator and x(t) = t a σ(p1)(s) ds b t σ(p1)(s) ds 2−i for a ≤ t ≤ b. (1.2.40i) Then there exists a positive function ρ ∈ C(]a, b[) such that the inequality (1.2.20) is satisfied, where G is Green’s function of the problem (1.2.4), (1.2.2i) and ρ(t) = O∗ (xβ (t)) (1.2.41) as t → a, t → b if i = 1, and as t → a if i = 2. Proof. As is seen from the definition of the set Vi,β(]a, b[; h), the functions p0, p1 : ]a, b[ → R satisfy the inclusion (1.2.6i) from which by virtue of Remark 1.2.2 it follows the existence of Green’s function of the problem (1.2.4), (1.2.2i0), and there exists a measurable function qβ : ]a, b[ → [0, +∞[ such that the problem v′′ (t) = p0(t)v(t) + p1(t)v′ (t) − h(v)(t) − qβ(t), (1.2.42) v(a) = 0, v(i−1) (b−) = 0 (1.2.43i) has in the interval ]a, b[ a positive upper function w, where w(t) = O∗ (xβ (t)) and b a |G(t, s)|qβ(s) ds = O∗ (xβ (t)) (1.2.44) as t → a, t → b if i = 1, and as t → a if i = 2. Introduce the operator χ as in the previous proof and let H(y)(t) = b a |G(t, s)|(qβ(s) + h(χ(y))(s)) ds. 35 As we can see from the conditions (1.2.39i), (1.2.44), the operator χ transforms the space C(]a, b[) into Cxβ (]a, b[). Consider now the equations v′′ (t) = p0(t)v(t) + p1(t)v′ (t) − h(χ(v))(t) − qβ(t), (1.2.45) v(t) = H(v)(t) (1.2.46) and note that the problem (1.2.45), (1.2.43i) is equivalent to the equation (1.2.46). From the equality (1.2.7) by means of which Green’s function is expressed, as well as from the estimates (1.2.10i) and the conditions (1.2.44), for any y ∈ C(]a, b[) we have |H(y)(t)| ≤ r0x1−α (t) t a xα (s) σ(p1)(s) h(xβ )(s) ds + + b a |G(t, s)|qβ(s) ds < +∞ for a ≤ t ≤ b, (1.2.47) where r0 = c2 ∗ d∗ sup w(t) xβ(t) : a < t < b . It follows from (1.2.37), (1.2.44) that the operator H transforms the space C(]a, b[) into Cxβ (]a, b[). Noticing that the right-hand side of the estimate (1.2.47) is independent of the function y, we make sure that a constant r exists such that for any y ∈ C(]a, b[) H(y) C,xβ ≤ r. It is clear that this estimate is the more so valid if y belongs to the ball Br = z ∈ Cxβ (]a, b[) : z C,xβ ≤ r . Repeating now the reasoning of the previous proof, we can see that the operator H : Cxβ (]a, b[) → Cxβ (]a, b[) is compact and hence there exists a solution v of the equation (1.2.46) such that v ∈ Cxβ (]a, b[), (1.2.48) χ(v)(t) = v(t) for a ≤ t ≤ b, and v(t) > 0 for a < t < b. (1.2.49) Then the following representation is valid: v(t) = b a |G(t, s)| h(v)(s) + qβ(s) ds, (1.2.50) 36 whence with regard for (1.2.49) we obtain the inequality v(t) ≥ b a |G(t, s)|qβ(s) ds for a ≤ t ≤ b which together with the conditions (1.2.44) and (1.2.48) implies that v(t) = O∗ (xβ (t)) (1.2.51) for t → a, t → b, if i = 1, and for t → a if i = 2. If we now take into consideration that owing to the conditions (1.2.44) and (1.2.51) we have inf 1 v(t) b a |G(t, s)|qβ(s) ds : a < t < b > 0, then from (1.2.50) we obtain sup 1 v(t) b a |G(t, s)|h(v)(s) ds : a < t < b < 1. (1.2.52) Introducing the notation ρ(t) = v(t), from (1.2.49), (1.2.51) and (1.2.52) we see that our lemma is valid. Lemma 1.2.6. Let i ∈ {1, 2}, the function x be defined by (1.2.40i), the constants α ∈ [0, 1[, β ∈]0, 1] be connected by (1.2.37) and the functions p0, p1 : ]a, b[ → R satisfy (1.2.38i), where h ∈ L Cxβ ; L xα σ(p1) ∩ L C; Lσi(p1) (1.2.53i) is a nonnegative operator. Then there exists a continuous function ρ : [a, b] → R+ such that the inequality (1.2.20) is satisfied, where G is Green’s function of the problem (1.2.4), (1.2.2i0). Proof. By Lemma 1.2.5, from the fact that h ∈ L(Cxβ ; L xα σ(p1) ) it follows the existence of the function ρ0 ∈ C(]a, b[) such that ρ0(t) > 0 for a < t < b and sup 1 ρ0(t) b a |G(t, s)|h(ρ0)(s) ds : a < t < b < 1. Then, taking into account that the operator h also belongs to L(C; Lσi(p1)), we can see by Lemma 1.2.3 that our lemma is valid. 37 Lemma 1.2.7. Let i ∈ {1, 2}, the function x : ]a, b[ → R+ be defined by (1.2.40i) and the functions p0, p1 : ]a, b[ → R satisfy the inclusion (1.2.6i). Then for any β ∈]0, 1] we have b a |G(t, s)| σ2 (p1)(s) x2−β−[β](s) ds = O∗ (xβ (s)) (1.2.54) as t → a, t → b if i = 1, and as t → a if i = 2, where G is Green’s function of the problem (1.2.4), (1.2.2i0). Proof. By Remark 1.2.2 and the inclusion (1.2.6i) there exists Green’s function G of the problem (1.2.4), (1.2.2i0) which is expressed by the equality (1.2.7). Consider the case i = 1 separately and note that b t σ(p1)(s) ds ≥ b a+b 2 σ(p1)(s) ds for a ≤ t ≤ a + b 2 . (1.2.55) Then, taking into consideration (1.2.7), (1.2.10i) and (1.2.55), for any β ∈ ]0, 1[ we obtain for t ∈ [a, a+b 2 ] the estimates b a |G(t, s)| σ2 (p1)(s) x2−β(s) ds ≤ c2 ∗ v2(a) xβ (t) β b a+b 2 σ(p1)(s) ds + + t a σ(p1)(s) ds β 1 − β)( b a+b 2 σ(p1)(s) ds 1−β + b a σ(p1)(s) ds 1−β β a+b 2 a σ(p1)(s) ds 2−β xβ (t) ≤ ≤ c2 ∗ βv2(a) 1 1 − β + b a σ(p1)(s) ds 1−β a+b 2 a σ(p1)(s) ds β−2 xβ (t) and b a |G(t, s)| σ2 (p1)(s) x2−β(s) ds ≥ d2 ∗ v2(a) b t σ(p1)(s) ds β b a+b 2 σ(p1)(s) ds 1−β × × t a σ(p1)(s) ds s a σ(p1)(η) dη 1−β b s σ(p1)(η) dη 2−β ≥ 38 ≥ d2 ∗ βv2(a) b a+b 2 σ(p1)(s) ds 1−β b a σ(p1)(s) ds 2−β xβ (t). The last two estimates imply the validity of (1.2.54) as t → a. Reasoning analogously for t ∈ [a+b 2 , b], we can see that this equality is also valid as t → b. Consider the case β = 1. With regard for the equalities (1.2.7) and the estimates (1.2.101) we obtain d2 ∗ 2C∗ ≤ b a |G(t, s)|σ2 (p1)(s) ds x−1 (t) ≤ C2 ∗ 2d∗ for a < t < b. (1.2.56) It follows from (1.2.56) that our lemma is valid in the case β = 1 as well. Reasoning similarly, we can prove the lemma for i = 2. 1.2.2. Auxiliary Propositions to Theorems (1.1.2i), (1.1.2i0) (i = 1, 2). Consider in the interval ]a, b[ the equation v′′ (t) = g(v)(t), (1.2.57) where g : C(]a, b[) → Lloc(]a, b[) is a continuous linear operator. We will also need the equation v′′ (t) = 0 for a ≤ t ≤ b. (1.2.58) Note that Green’s function of the problem (1.2.58), (1.2.2i0) has the form G(t, s) =    −(s − a) b − t b − a 2−i for a ≤ s < t ≤ b, −(t − a) b − s b − a 2−i for a ≤ t < s ≤ b. (1.2.59i) Lemma 1.2.81. Let γ ∈ [0, 1[, λ ∈ [0; 1 − γ[ and g ∈ L(Cxλ ; Lxγ ) (1.2.60) be a nonnegative operator, where x(t) = (b − t)(t − a) for a ≤ t ≤ b. (1.2.611) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that λ ≤ β < 1 − γ, (1.2.62) α + β ≤ 1 2 , (1.2.63) 39 and b a xα (s)g(xβ )(s) ds < 2β 16 b − a b − a 4 2(α+β) . (1.2.641) Then the problem (1.2.57), (1.2.210) has only the zero solution in the space Cxλ (]a, b[). Proof. Suppose to the contrary that the problem (1.2.57), (1.2.2i0) has a nonzero solution v0 ∈ Cxλ (]a, b[). If v0 is a function of constant signs, then from the nonnegativeness of the operator g we obtain v′′ 0 (t) sign v0(t) ≥ 0 for a < t < b, which together with the conditions (1.2.2i0) contradicts the assumption v0(t0)≡ 0, i.e., v0 is a function of constant signs. Using Green’s function of the problem (1.2.58), (1.2.2i0), v0 can be represented as follows: v0(t) = − 1 b − a (b − t) t a (s − a)g(v0)(s) ds + (t − a) b t (b − s)g(v0)(s) ds for a ≤ t ≤ b and hence for any β the estimate v0(t) [(b − t)(t − a)]β ≤ ≤ [(b − t)(t − a)]1−(γ+β) b − a b a [(b − s)(s − a)]γ g(xλ )(s) ds v0 C,xλ for a < t < b is valid. In the above estimate, taking into account the condition (1.2.60), if β satisfies the inequality (1.2.62), we get lim t→a v0(t) [(b − t)(t − a)]β = 0, lim t→b v0(t) [(b − t)(t − a)]β = 0. These equalities imply the existence of points t1, t2 ∈]a, b[ such that v0(t1) (b − t1)β(t1 − a)β = sup v0(t) (b − t)β(t − a)β : a < t < b , v0(t2) (b − t2)β(t2 − a)β = inf v0(t) (b − t)β(t − a)β : a < t < b . 40 Without loss of generality we assume t1 < t2 and notice that by (1.2.611) which defines the function x, we have −g(xβ )(t) |v0(t2)| (b − t2)β(t2 − a)β ≤ ≤ g(v0)(t) ≤ g(xβ )(t) |v0(t1)| (b − t1)β(t1 − a)β for a < t < b. (1.2.65) Recall also one simple numerical inequality A · B ≤ (A + B)2 4 , (1.2.66) where A ≥ 0 and B ≥ 0. Suppose c ∈]t1, t2[ and v0(c) = 0. Then the following representations are valid: v0(t1) = c − t1 c − a t1 a (s − a)g(−v0)(s) ds + t1 − a c − a c t1 (c − s)g(−v0)(s) ds and |v0(t2)| = b − t2 b − c t2 c (s − c)g(v0)(s) ds + t2 − a b − c b t2 (b − s)g(v0)(s) ds. These representations with regard for the inequality (1.2.65), for any α, β satisfying the conditions of the lemma, result in v0(t1) ≤ [(c − t1)(t1 − a)]1−α (c − a)[(b − t2)(t2 − a)]β c a xα (s)g(xβ )(s) ds · |v0(t2)| < +∞ and v0(t2) ≤ [(b − t2)(t2 − c)]1−α (b − c)[(b − t1)(t1 − a)]β b c xα (s)g(xβ )(s) ds · |v0(t1)| < +∞. Multiplying the above inequalities, by means of (1.2.66) we obtain λ b a xα (s)g(xβ )(s) ds ≥ 1, (1.2.67) where λ = 1 2 [(b − t2)(t2 − c)(c − t1)(t1 − a)]1−(α+β)[(t2 − c)(c − t1)]β (b − c)(c − a)(b − t1)β(t2 − a)β . 41 Then by (1.2.66) we get the estimate λ ≤ 1 2 [(b − c)(c − a)]1−2(α+β)(t2 − t1)2β 42−2(α+β)+β[(b − t1)(t2 − a)]β , whence using once more the inequality (1.2.66) and taking into consideration the fact that (t2 − t1)2β ≤ [(b − t1)(t2 − a)]β , (1.2.68) we arrive at λ ≤ b − a 16 · 2β 4 b − a 2(α+β) . (1.2.69) Substituting the last inequality in (1.2.67), we obtain the contradiction with the condition (1.2.641), i.e., our assumption is invalid and v0(t) ≡ 0. Lemma 1.2.82. Let γ ∈ [0, 1[, λ ∈ [0, 1 −γ[ and the nonnegative operator g satisfy the inclusion (1.2.60), where x(t) = t − a for a ≤ t ≤ b. (1.2.612) Let, moreover, there exist constants α, β ∈ [0, 1 2 ] such that the conditions (1.2.62), (1.2.63) are satisfied and b a xα (s)g(xβ )(s) ds ≤ 8 b − a b − a 4 α+β . (1.2.642) Then the problem (1.2.57), (1.2.220) has only the zero solution in the space Cxλ (]a, b[). Proof. Suppose to the contrary that the problem (1.2.57), (1.2.220) has a nonzero solution v0 ∈ Cxλ (]a, b[). Similarly to the previous lemma we make sure that v0 is of constant signs and the equality lim t→a v0(t) (t − a)β = 0 is valid for any β ∈ [λ, 1 − γ[. On the other hand, in any sufficiently small neighborhood of the point b, since v′ 0(b−) = 0, the equality sign v0(t) (t − a)β ′ = − signv0(t) is satisfied. It follows from the last two equalities that the function v0(t) (t−a)β attains neither its minimum nor its maximum at the points a and b. Let max v0(t) (t − a)β : a ≤ t ≤ b = v0(t1) (t1 − a)β 42 and min v0(t) (t − a)β : a ≤ t ≤ b = v0(t2) (t2 − a)β . Then from the above-said it is clear that t1, t2 ∈]a, b[. Without loss of generality we assume t1 < t2 and let the point c ∈]t1, t2[ be such that v0(c) = 0. Then from the inequality −g(xβ )(t) |v0(t2)| (t2 − a)β ≤g(v0)(t) ≤ g(xβ )(t) |v0(t1)| (t1 − a)β for a < t < b and from the equalities v0(t1) = c − t1 c − a t1 a (s − a)g(−v0)(s) ds + t1 − a c − a c t1 (c − s)g(−v0)(s) ds, |v0(t2)| = t2 c (s − c)g(v0)(s) ds + (t2 − c) b t2 g(v0)(s) ds we obtain v0(t1) ≤ (c − t1)(t1 − a)1−α (c − a)(t2 − a)β c a xα (s)g(xβ )(s) ds · |v0(t2)| |v0(t2)| ≤ (t2 − c)1−α (t1 − a)β b c xα (s)g(xβ )(s) ds · v0(t1). Multiplying these inequalities, with regard for (1.2.66) we get λ b a xα (s)g(xα )(s) ds ≥ 1, (1.2.70) where λ = 1 2 [(t1 − a)(c − t1)]1−(α+β)(t2 − c)1−α(c − t1)α+β (c − a)(t2 − a)β . Then by (1.2.66) and t2 − a > t2 − c we have λ ≤ 1 2 (c − a)1−2(α+β)(t2 − c)1−2(α+β)[(c − t1)(t2 − c)]α+β 41−(α+β) . Applying once more (1.2.66), we can see that λ ≤ (t2 − a)1−2(α+β) (t2 − t1)α+β 2 · 41−(α+β) . (1.2.71) 43 Notice that from the conditions t1, t2 ∈]a, b[ as well as from the fact that for none of α, β ∈ [0, 1 2 ] the expressions α + β and 1 − 2(α + β) vanish simultaneously, we obtain the estimate (t2 − a)1−2(α+β) · (t2 − t1)α+β < (b − a)1−(α+β) , with regard for which in (1.2.71) we get λ < (b − a) 8 4 b − a α+β . Substituting the latter inequality in (1.2.70), we obtain the contradiction with the condition (1.2.642), i.e., our assumption is invalid and v0(t) ≡ 0. Remark 1.2.7. Lemma 1.2.81 remains valid if for β = 0 we replace the condition (1.2.641) by b a xα (s)g(xβ )(s) ds ≤ 2β 16 b − a b − a 4 2(α+β) . (1.2.72) Proof. If β = 0, then the inequality (1.2.68) will be strictly satisfied and hence the estimate (1.2.69) will take the form λ < b − a 16 · 2β 4 b − a 2(α+β) . Taking into consideration the last inequality in (1.2.67), we obtain the contradiction with the condition (1.2.72) which indicates the possibility to replace in case β = 0 the condition (1.2.641) by (1.2.72). § 1.3. Proof of Propositions on Existence and Uniqueness 1.3.1. Proof of Basic Theorems on Existence and Uniqueness of Solution of Two-Point Problems. Proof of Theorem 1.1.1i. From the inclusions (1.1.7i) and (1.1.8i) and also from the fact that the operator h is nonnegative, for β = 0 by virtue of Lemma 1.2.4 and for β > 0 by virtue of Lemma 1.2.6 it follows that there exists a function ρ ∈ C(]a, b[) such that ρ(t) > 0 for a ≤ t ≤ b (1.3.1) and sup 1 ρ(t) b a |G(t, s)|h(ρ)(s) ds : a < t < b < 1, (1.3.2) where G is Green’s function of the problem (1.2.4), (1.2.2i0). Note that for any function y ∈ Cρ(]a, b[) the inequality |y(t)| ≤ ρ(t) y C,ρ for a ≤ t ≤ b (1.3.3) 44 is valid and, owing to the estimates (1.2.10i), the representation (1.2.7) of Green’s function and the conditions (1.1.5)–(1.1.8i) and (1.1.10), we have b a G(t, s)p2(s) ds < +∞, b a G(t, s)g(y)(s) ds < +∞, b a G(t, s)h(y)(s) ds < +∞. Introduce the continuous operators U0, U : Cρ(]a, b[) → Cρ(]a, b[) by the equalities U0(y)(t) = b a G(t, s)g(y)(s) ds, U(g)(t) = u0(t) + U0(y)(t) + b a G(t, s)p2(s) ds, (1.3.4) where u0 is a solution of the problem (1.2.4), (1.2.2i). Clearly every solution of the problem (1.1.1), (1.1.2i) is a solution of the equation u(t) = U(u)(t) (1.3.5) and vice versa. From the definition of the norm of the operator it follows that U0 Cρ→Cρ = = sup b a G(t, s)g(y)(s) ds C,ρ : x ∈ Cρ(]a, b[), y C,ρ = 1 which with regard for (1.1.10), (1.3.1)–(1.3.3) implies U0 Cρ→Cρ < 1, (1.3.6) i.e., the operator U contracts the space Cρ(]a, b[) into itself for any p2 ∈ Lσi(p1)([a, b]) and any operator g satisfying (1.1.10). Then by virtue of the theorem on contracting map the equation (1.3.5) has in the space Cρ(]a, b[) and hence in C(]a, b[) a unique solution because, by (1.3.1), any function from C(]a, b[) belongs to the space Cρ(]a, b[) as well. It remains to notice that the unique solvability of the problem (1.1.1), (1.1.2i) follows from the equivalence of that problem and the equation (1.3.5). Proof of Theorem 1.1.1i0. The inclusions (1.1.7i), (1.1.8i) and the nonnegativeness of the operator h imply by virtue of Lemma 1.2.5 the existence of 45 a positive function ρ ∈ C(]a, b[) such that ρ(t) = O∗ (xβ (t)) (1.3.7) as t → a, t → b, if i = 1, and as t → a if i = 2. Moreover, the condition (1.3.2) is satisfied, where G is Green’s function of the problem (1.2.4), (1.2.2i0). It is also clear that for any y ∈ Cρ(]a, b[) the inequality (1.3.3) is satisfied, and due to the estimates (1.2.10i) and the representation (1.2.7) of Green’s functions we have b a G(t, s)h(y)(s) ds ≤ r1x1−α (t) b a xα (s) σ(p1)(s) h(xβ )(s) ds y C,xβ , b a G(t, s)p2(s) ds ≤ r1xβ (t) b a x1−β (s) σ(p1)(s) |p2(s)| ds for a ≤ t ≤ b, (1.3.8) where r1 = c2 ∗ v2(a) , and the existence of integrals follows from the conditions (1.1.6), (1.1.11), (1.1.12). From (1.3.8) and (1.1.6), (1.1.10), (1.3.7) we also have that the operators U0(y)(t) = b a G(t, s)g(y)(s) ds and U(y)(t) = U0(y)(t) + b a G(t, s)p2(s) ds transform continuously the space Cρ(]a, b[) into itself. Repeating word by word the previous proof, we can see that the problem (1.1.1) (1.1.2i0) has a unique solution u in the space Cρ(]a, b[). But as is seen from (1.3.7), u will be a unique solution in the space Cxβ (]a, b[) as well. Proof of Remark 1.1.1i. Under the conditions of Theorem 1.1.1i, as is seen from its proof, the operator U contracts the space Cρ([a, b]) into itself. Then from the theorem on contracting map it follows that for any function v0 ∈ Cρ(]a, b[) the sequence vn : [a, b] → R, where vn is the unique solution of the equation vn(t) = U(vn−1)(t) (1.3.9) tends to the unique solution u of the equation (1.3.5) with respect to the norm · C,ρ. We introduce the notation U0 Cρ→Cρ = µ and u − v1 C,ρ = ω, 46 and notice that by virtue of (1.3.6), we have µ < 1. Then, as is known, the estimate u − vn C,ρ ≤ ω µn 1 − µ , n ∈ N, (1.3.10) is valid and for any n ∈ N with regard for (1.3.3) we obtain |u(t) − vn(t)| ≤ ω µn 1 − µ ρ C for a ≤ t ≤ b. (1.3.11) Differentiating the difference of the equations (1.3.5) and (1.3.9) and taking into account the inequalities (1.1.10), (1.3.11) and the estimates (1.2.12i) of Green’s function, we obtain sup σi(p1)(t)|v′ n(t) − u′ (t)| : a < t < b ≤ ω′ µn 1 − µ , n ∈ N, (1.3.12) where ω′ = ωc∗ ρ C b a σi(p1)(s)h(1)(s) ds. The inequalities (1.3.11), (1.3.12) imply the validity of the estimates (1.1.14), and after differentiating twice the equality (1.3.9) we see that vn is a solution of the problem (1.1.13i). Proof of Remark 1.1.1i0. Let ρ be the function appearing in the proof of Theorem 1.1.1i0. Introduce the constants µ and ω and the functions vn : [a, b] → R, n ∈ N, as in the previous proof. Reasoning as above, we make sure that the estimate (1.3.10) is valid, and by virtue of the condition (1.3.7) for any n ∈ N we have |u(t) − vn(t)| xβ(t) ≤ ω µn 1 − µ sup ρ(t) xβ(t) : a < t < b . (1.3.13) On the other hand, differentiating the difference of the equations (1.3.5) and (1.3.9), with regard for the equality (1.2.7) and the estimates (1.2.10i), (1.2.11i), for any n ∈ N we obtain xα (t) σ(p1)(t) |u′ (t) − v′ n(t)| ≤ r u − vn C,ρ for a ≤ t ≤ b, (1.3.14) where r = (1 + c∗ )2 b a |p0(s)| σ(p1)(s) x(s) + σ(p1)(s) ds b a xα (s) σ(p1)(s) h(xβ )(s) ds. The inequalities (1.3.10), (1.3.13) and (1.3.14) imply the validity of the estimates (1.1.15), and having differentiated twice the equality (1.3.9) we see that v0 is a solution of the problem (1.1.13i0). 47 Proof of Theorem 1.1.2i. Let G be Green’s function of the problem (1.2.58), (1.2.2i0). Introduce the operator U0 and the function q by the equalities U0(y)(t) = b a G(t, s)g(y)(s) ds, q(t) = b a G(t, s)p2(s) ds. (1.3.15) From the representation (1.2.59i) of Green’s function and from the conditions (1.1.17), (1.1.18) it follows that the operator U0 transforms continuously the space C(]a, b[) into itself and q ∈ C(]a, b[). Consider now the equation u(t) = U0(u)(t) + u0(t) + q(t), (1.3.16) where u0(t) is a solution of the problem (1.2.58), (1.1.2i). Every its solution is a solution of the problem (1.1.16), (1.1.2i), and vice versa. Let r > 0, Br = y ∈ C(]a, b[) : y C ≤ r and choose any sequence (xn)∞ n=1 from Br. Let, moreover, yn(t)=U0(xn)(t), n ∈ N. Then yn C ≤ r1, n ∈ N, (1.3.17) where r1 = r b a b − s b − a 2−i (s − a)g(1)(s) ds. Consider the case i = 1 separately. From the definition of Green’s function G, for any ε > 0 it follows the existence of a1, b1 ∈ ]a, b[ , where a1 < b1, such that max b a |G(t, s)|g(1)(s) ds : a ≤ t ≤ a1, b1 ≤ t ≤ b ≤ ε 4 , which implies the validity of the estimate |yn(t1) − yn(t2)| ≤ ε 2 , n ∈ N, for a ≤ t1 ≤ t2 ≤ a1, b1 ≤ t1 ≤ t2 ≤ b. It is also clear that there exists a constant δ, 0 < δ < min(a1 − a, b − b1) for which the following inequality is valid: |yn(t1) − yn(t2)| ≤ ≤ r1 max 1 (b − t)(t − a) : a1 − δ ≤ t ≤ b1 + δ |t1 − t2| ≤ ε 2 for |t1 − t2| ≤ δ, a1 − δ ≤ tj ≤ b1 + δ (j = 1, 2). From the last two estimates we obtain that if tj ∈ [a, b] (j = 1, 2) and |t1 − t2| ≤ δ, 48 then |yn(t1) − yn(t2)| ≤ ε, n ∈ N. This and the inequality (1.3.17) imply that the sequence (yn)∞ n=1 is uniformly bounded and equicontinuous. In case i = 2 the same follows from the possibility of choosing for any ε > 0, a1 ∈ ]a, b[ and 0 < δ < a1 − a such that max b a |G(t, s)|g(1)(s) ds : a ≤ t ≤ a1 < ε 4 , |yn(t1) − yn(t2)| ≤ r1 max 1 + 1 t − a : a1 − δ ≤ t ≤ b |t1 − t2| ≤ ε 2 for |t1 − t2| ≤ δ, a1 − δ ≤ tj ≤ b (j = 1, 2). Then by the Arzella–Ascoli lemma we obtain that U0 is a compact operator. Consequently, taking into account Fredholm’s alternatives, the equation (1.3.16) is uniquely solvable if the homogeneous equation u(t) = U0(u)(t) (1.3.160) has only the trivial solution in the space C(]a, b[). It remains to note that by virtue of the conditions (1.1.18)–(1.1.21) and (1.1.22) if i = 1 and (1.1.242) if i = 2, all the requirement of Lemma 1.2.8i are satisfied for λ = 0, whence it follows that the problem (1.2.57), (1.2.2i0), i.e., the equation (1.3.160) has only the trivial solution in the space C(]a, b[). Proof of Remark 1.1.2 follows directly from Remark 1.2.7. Proof of Theorem 1.1.2i0. Let x be a function defined by (1.1.19i) and let G be Green’s function of the problem (1.1.58), (1.1.2i0) which is expressed by (1.2.59i). Introduce the operator U0 and the function q by the equality (1.3.15). Then for any y ∈ Cxλ (]a, b[) the estimates |U0(y)(t)| ≤ x1−γ (t) (b − a)2−i b a xγ (s)g(xλ )(s) ds y C,xλ , |q(t)| ≤ x1−γ (t) b a xγ (s)|p2(s)| ds for a ≤ t ≤ b are valid, from which by the conditions λ ∈ ]0, 1 − γ[ and (1.1.25), (1.1.26) it follows that U0 transforms continuously the space Cxλ (]a, b[) into itself and q ∈ Cxλ (]a, b[). Consider now the equation u(t) = U0(u)(t) + q(t) (1.3.18) 49 which is equivalent to the problem (1.1.16), (1.1.2i0), and the corresponding homogeneous equation (1.3.160). As is seen from Lemma 1.2.8i and Remark 1.2.7, by virtue of the conditions λ ∈ ]0, 1 − γ[ , (1.1.21), (1.1.24i) and (1.1.25)–(1.1.27) the problem (1.2.57), (1.1.2i0), i.e., the equation (1.3.160), has in the space Cxλ (]a, b[) only the trivial solution. Then according to Fredholm’s alternatives, to prove the validity of our theorem it remains to show that the operator U0 is compact. Let r > 0, Br = z ∈ Cxλ (]a, b[) : z C,xλ ≤ r (xn)∞ n=1 be a sequence from Br and yn(t) = U0(xn)(t) for n ∈ N. Then as is seen from the definition of G, for any n ∈ N the estimate |y(j) n (t)| ≤ r x1−j−γ (t) (b − a)(1−j)(2−i) b a xγ (s)g(xλ )(s) ds (j = 0, 1) (1.3.19i) for a < t < b is valid, which by virtue of the condition λ ∈ ]0, 1 − γ[ yields yn(t) C,xλ ≤ r1, (1.3.20) where r1 = r (b − a)2−i b a xγ (s)g(xλ )(s) ds max x1−(λ+γ) (t) : a ≤ t ≤ b . Consider now the case i = 1 separately. From (1.3.191) for j = 0 and for any ε > 0 follows the existence of a1, b1 ∈ ]a, b[, where a1 < b1, such that |yn(t)| ≤ ε 4 , n ∈ N, for a ≤ t ≤ a1, b1 ≤ t ≤ b, which implies the estimate |yn(t1) − yn(t2)| ≤ ε 2 , n ∈ N, for a ≤ t1 < t2 ≤ a1, b1 ≤ t1 < t2 ≤ b. Moreover, from (1.3.191) for j = 1 it follows the existence of a constant δ such that |yn(t1) − yn(t2)| ≤ r2|t1 − t2| ≤ ε 2 , n ∈ N, for a1 − δ ≤ tl ≤ b1 + δ (l = 1, 2), where r2 = r b a xγ (s)g(xλ )(s) ds max x−γ (t) : a1 − δ ≤ t ≤ b1 + δ . 50 It is clear from the last two estimates that if tl ∈ [a, b] (l = 1, 2) and |t1 − t2| ≤ δ, then for any n ∈ N |yn(t1) − yn(t2)| ≤ ε. This and the estimate (1.3.20) imply that the sequence (yn)∞ n=1 is uniformly bounded and equicontinuous. In case i = 2, by virtue of the estimates (1.3.192) the same follows from the possibility of choosing, for any ε > 0, of a1 ∈ ]a, b[ and 0 < δ < a1 − a such that |yn(t)| ≤ ε 4 , n ∈ N for a ≤ t ≤ b, and |yn(t1) − yn(t2)| ≤ r2|t1 − t2| ≤ ε 2 , n ∈ N, for a1 − δ ≤ tj ≤ b (j = 1, 2), where r2 = r b a xγ (s)g(ρ)(s) ds max x−γ (t) : a1 − δ ≤ t ≤ b . Then by the Arzella–Ascoli lemma we have that U0 is a compact opera- tor. 1.3.2. Proof of Effective Sufficient Conditions for Solvability of the Problems (1.1.1), (1.1.2i) and (1.1.1), (1.1.2i0) (i = 1, 2). Before we proceed to proving the corollaries, we note that Green’s function of the problem v′′ (t) = p1(t)v′ (t), (1.3.21) v(a) = 0, v(i−1) (b−) = 0 (1.3.22i) has the form G0(t, s) = =    − 1 σ(p1)(s) s a σ(p1)(η) dη 1 b a σ(p1)(η)dη b t σ(p1)(η)dη 2−i for a ≤ s < t ≤ b, − 1 σ(p1)(s) t a σ(p1)(η) dη 1 b a σ(p1)(η)dη b s σ(p1)(η) dη 2−i for a ≤ t < s ≤ b. (1.3.23i) 51 Proof of Corollary 1.1.11. It is clear that all the requirements of Theorem 1.1.11, except (1.1.71), follow directly from the conditions of our corollary. It remains only to show that the conditions (1.1.31), (1.1.321) imply the inclusion (1.1.71) as well. Indeed, let β > 0 and zλ(t) = b t σ(p1)(η)dη α × × t a [p0(s)]−(λ + xβ (s)) + h(xβ (s)) σ(p1)(s) s a σ(p1)(η)dη α ds + + t a σ(p1)(η)dη α b t [p0(s)]−(λ + xβ (s)) + h(xβ (s)) σ(p1)(s) b s σ(p1)(η)dη ds × × b a σ(p1)(s) ds 1−2(α+β) 22−2(α+β) . (1.3.24) Then, as is seen from the conditions (1.1.31), (1.1.321), we can choose λ > 0 such that zλ(t) < 1 for a ≤ t ≤ b (1.3.25) be satisfied. Introduce also the notation qβ(t) = σ2 (p1)(t) x2−β−[β](t) , wε(t) = ε b a |G0(t, s)|qβ(s) ds, w(t) = b a |G0(t, s)| [p0(s)]−(λ + xβ (s)) + h(xβ )(s) ds + wε(t), where ε ∈ R+ , G0 is Green’s function of the problem (1.3.21), (1.3.221) which is defined by the equality (1.3.231), and by Lemma 1.2.7, wε(t) = O∗ (xβ (t)) as t → a, t → b (1.3.26) for any ε > 0. From the conditions (1.3.25), (1.3.26) we have the possibility of choosing the constant ε > 0 such that zλ(t) + sup wε(t) xβ(t) : a < t < b < 1 for a ≤ t ≤ b. (1.3.27) By virtue of (1.3.231) we easily get the estimate 0 < w(t) ≤ zλ(t)xβ (t) + wε(t) for a < t < b 52 which with regard for (1.3.27) results in 0 < w(t) ≤ xβ (t) for a < t < b. (1.3.28) The last inequality together with (1.3.26) means that w(t) = O∗ (xβ (t)) as t → a, t → b. (1.3.29) On the other hand, it is clear that w′′ (t) = −[p0(t)]−(λ + xβ (t)) + p1(t)w′ (t) − h(xβ )(s) − qβ(t). Taking into account the inequality (1.3.28) and the fact that the operator h and the constant λ are nonnegative, the above equality results in w(t)′′ ≤ p0(t)w(t) + p1(t)w′ (t) − h(w)(t) − qβ(t). (1.3.30) If we introduce the notation w(t) = λ + w(t), then w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t), (1.3.31) where w(t) > 0 for a ≤ t ≤ b. (1.3.32) From the inequalities (1.3.31) and (1.3.32), by Lemma 1.2.2 we obtain the inclusion (p0, p1) ∈ V1,0(]a, b[). (1.3.331) Then, as is seen from Remark 1.2.2, the problem (1.2.4), (1.2.2i0) has Green’s function G which is expressed by the equality (1.2.7). Using now the inequalities (1.2.101), we arrive at d2 ∗ c∗ ≤ εw−1 ε (t) b a |G(t, s)|qβ(s) ds ≤ c2 ∗ d∗ for a ≤ t ≤ b which with regard for the equality (1.3.26) yields b a |G(t, s)|qβ(s) ds = O∗ (xβ (s)) as t → a, t → b. (1.3.34) It remains to note that the conditions (1.2.28), (1.3.29), (1.3.331), (1.3.34) and the inequality (1.3.30), owing to Definition 1.1.4, ensure the inclusion (1.2.71) for β > 0. Assume now that β = 0 and w(t) = b a |G0(t, s)| [p0(s)]− + h(1)(s) ds + εv(t), (1.3.35) 53 where v is a solution of the equation (1.3.21) under the boundary conditions v(a) = 1, v(b) = 1, and z0(t) = b t σ(p1)(η)dη α t a ([p0(s)]− + h(1)(s)) σ(p1)(s) s a σ(p1)(η)dη α ds + + t a σ(p1)(η)dη α b t ([p0(s)]− + h(1)(s)) σ(p1)(s) b s σ(p1)(η)dη α ds × × ( b a σ(p1)(s) ds)1−2α 41−α . Then, as is seen from the condition (1.1.321), z0(t) < 1 for a ≤ t ≤ b, and hence we can choose ε > 0 small enough for the inequality z0(t) + εv(t) < 1 (1.3.36) to be fulfilled for a ≤ t ≤ b. Notice that by virtue of the equalities (1.3.231), we obtain the estimate 0 < w(t) ≤ z0(t) + εv(t) for a ≤ t ≤ b which with regard for (1.3.36) implies 0 < w(t) ≤ 1 for a ≤ t ≤ b. (1.3.37) On the other hand, w′′ (t) = −[p0(t)] + p1(t)w′ (t) − h(1)(t), whence, taking into account (1.3.37) and the fact that the operator h is nonnegative, we obtain w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) − h(w)(t). Consequently, owing to Definition 1.1.3, the inclusion (p0, p1)∈V1,0(]a, b[ ; h) is valid. Proof of Corollary 1.1.12. It is clear that all the requirements of Theorem 1.1.12, except (1.1.72) follow directly from the conditions of our corollary. It remains to show that the conditions (1.1.31), (1.1.321) imply the inclusion (1.1.72) as well. 54 To this end, we introduce for β > 0 the functions zλ and w by the equalities zλ(t) = t a ([p0(s)]−(λ + xβ (s)) + h(xβ )(s)) σ(p1)(s) s a σ(p1)(η)dη α ds + + t a σ(p1)(η)dη α b t ([p0(s)]−(λ + xβ (s)) + h(xβ )(s) σ(p1)(s) ds × × b a σ(p1)(η)dη 1−(α+β) and w(t) = b a |G0(t, s)|([p0(s)]−(λ + xβ (s)) + h(xβ )(s)) ds + wε(t), where G0 is Green’s function of the problem (1.3.21), (1.3.222), and wε is defined just as in the previous proof. Then reasoning in the same manner as when proving Corollary 1.1.11, we make sure that the inclusion (1.1.72) is valid for β > 0. In the case β = 0, we consider the function zλ for λ = 0 and the function w defined by (1.3.35), where v is a solution of the equation (1.3.21) under the boundary conditions v(a) = 1, v′ (b−) = 1. Then reasoning just in the same way as in proving Corollary 1.1.11 for β = 0, we can see that the inclusion (p0, p1) ∈ V2,0(]a, b[ ; h) is valid. Proof of Corollary 1.1.1i0. Coincides completely with that of Corollary 1.1.1i for β > 0. Proof of Remark 1.1.4. Denote the left-hand side of (1.1.32i) by w. Then it is obvious that w(t) ≤ b a [p0(s)]−xα+β (s) + xα (s)h(xβ )(s) σ(p1)(s) ds for a ≤ t ≤ b, i.e., it follows from (1.1.34i) that the condition (1.1.32i) is valid. On the other hand, (1.1.34i) implies the inclusion h ∈ L Cxβ ; L xα σ(p1) which together with (1.1.33) means that (1.1.8i) is satisfied. 55 Proof of Remark 1.1.40. As is seen from the proof of Remark 1.1.4, the conditions (1.1.32i) and (1.1.12) follow simultaneously from (1.1.34i). Proof of Corollary 1.1.2i. Introduce the notation g(u)(t) = n k=1 gk(t)u(τk(t)) (1.3.38) and h(u)(t) = n k=1 |gk(t)|u(τk(t)). (1.3.39) Then for any u ∈ C(]a, b[) almost everywhere on the interval ]a, b[ the inequality (1.1.10) is satisfied, and as is seen from (1.1.36i), the inclusion (1.1.8i) is valid. It is also clear that the condition (1.1.37i) in our notation can be rewritten as (1.1.32i). Hence all the requirements of Corollary 1.1.1i are fulfilled and our corollary is valid. Proof of Corollary 1.1.2i0. Define the operators g and h by the equalities (1.3.38) and (1.3.39) and note that from the condition (1.3.38) it follows the inclusion (1.1.12). Reasoning similarly as when proving the above corollary, we can see that our corollary is valid. Proof of Remark 1.1.5. Denote the left-hand side of (1.1.37i) by w. Then it is evident that w(t) ≤ b a [p0(s)]−xα+β (s) + xα (s) n k=1 |gk(s)|xβ (τk(s)) σ(p1)(s) ds for a ≤ t ≤ b, i.e., (1.1.40i) implies the validity of the condition (1.1.37i). On the other hand, (1.1.40i) implies the inclusion gkxβ (τk) ∈ L xα σ(p1) ([a, b]) which together with (1.1.39) means that (1.1.36i) is satisfied. Proof of Remark 1.1.50. As is seen from the proof of Remark 1.1.5, the conditions (1.1.37i) and (1.1.38) follow simultaneously from (1.1.40i). Proof of Corollary 1.1.31. It is clear that all the requirements of Theorem 1.1.1i, except (1.1.7i), follow directly from the conditions of our corollary. It remains to show that the conditions (1.1.41), (1.1.421) imply the inclusion (1.1.71) as well, where h(u)(t) = n k=1 |gk(t)|u(τk(t)). 56 Indeed, let β > 0 and z(t) = n k=1 t a |gk(s)| σ(p1)(s) xβ (τk(s)) s a σ(p1)(η)dη α ds b t σ(p1)(η)dη α + + n k=1 b t |gk(s)| σ(p1)(s) xβ (τk(s)) b s σ(p1)(η)dη α ds t a σ(p1)(η)dη α × × b a σ(p1)(η)dη 1−2(α+β) 22−2(α+β) . Then as is seen from (1.1.421), for every m ∈ {1, . . ., n} z(τm(t)) < 1 for a ≤ t ≤ b. (1.3.40) Moreover, let w(t) = n k=1 b a |G0(t, s)|gk(s)xβ (τk(s)) ds + wε(t), where the function wε is defined in the same way as in proving Corollary 1.1.11, ε > 0, G0 is Green’s function of the problem (1.3.21), (1.3.221) defined by the equality (1.3.231) and by Lemma 1.2.7, wε(t) = O∗ (xβ (t)) as t → a, t → b, (1.3.41) for any ε > 0. From the conditions (1.3.40), (1.3.41) it follows that we can choose a constant ε > 0 such that for every m ∈ {1, . . . , n} z(τm(t)) + sup wε(τm(t)) xβ(τm(t)) : a < t < b < 1 for a ≤ t ≤ b. (1.3.42) Using the equality (1.3.231) we can easily obtain the estimate 0 ≤ w(t) ≤ z(t)xβ (t) + wε(t) for a ≤ t ≤ b, (1.3.43) whence by virtue of (1.3.42) for every m ∈ {1, . . ., n} the inequality 0 ≤ w(τm(t)) ≤ xβ (τm(t)) for a < t < b (1.3.44) is valid. Analogously, from (1.3.41) and (1.3.43) it follows the estimate 0 < w(t) ≤ r0xβ (t) for a < t < b, (1.3.45) where r0 = sup z(t) + wε(t) xβ(t) : a < t < b < +∞, and according to (1.3.41) we get w(t) = O∗ (xβ (t)) as t → a, t → b. (1.3.46) 57 On the other hand, it is clear that w′′ (t) = p1(t)w′ (t) − n k=1 |gk(t)|xβ (τk(t)) − qβ(t), which with regard for the conditions (1.1.41) and (1.3.44) results in w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) − n k=1 |gk(t)|w(τk(t)) − qβ(t), (1.3.47) where, as is seen from Remark 1.2.6, (p0, p1) ∈ V1,0(]a, b[). (1.3.48) Then, as we have shown in proving Corollary 1.1.11, b a |G(t, s)|qβ(s) ds = O∗ (xβ (t)) as t → a, t → b, (1.3.49) where G is Green’s function of the problem (1.2.4), (1.2.2i0). It remains to notice that the conditions (1.3.45), (1.3.46), (1.3.48), (1.3.49) and the inequality (1.3.47) by virtue of Definition 1.1.4 imply the inclusion (1.1.71) for β > 1. Suppose now that β = 0 and w(t) = n k=1 b a |G0(t, s)||gk(s)| ds + εv(t), (1.3.50) where v is a solution of the equation (1.3.21) under the boundary conditions v(a) = 1 and v(b) = 1. Then, as is seen from the condition (1.1.421), for every m ∈ {1, . . . , n} z(τm(t)) < 1 for a ≤ t ≤ b and hence for every m ∈ {1, . . ., n} we can choose ε > 0 small enough for the inequality z(τm(t)) + εv(τm(t)) ≤ 1 for a ≤ t ≤ b. (1.3.51) to be fulfilled. Note that from the positiveness of v and also from (1.3.231) we have the estimate 0 < w(t) ≤ z(t) + εv(t) for a ≤ t ≤ b which by virtue of (1.3.51) for every m ∈ {1, . . ., n} yields 0 < w(τm(t)) ≤ 1 for a ≤ t ≤ b. (1.3.52) 58 On the other hand, w′′ (t) = p1(t)w′ (t) − n k=1 |gk(t)| which with regard for (1.1.41) and (1.3.52) gives w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) − n k=1 |gk(t)|w(τk(t)). Hence, owing to Definition 1.1.3, the inclusion (p0, p1) ∈ V1,0( ]a, b[ ; h), is valid, where h(u)(t) = n k=1 |gk(t)|u(τk(t)). Proof of Corollary 1.1.32. It is clear that all the requirements of Theorem 1.1.12, except (1.1.72), follow directly from the conditions of our corollary. It remains to show that the inclusion (1.1.72) follows from the condition (1.1.41), (1.1.421) as well. To this end, we introduce for β > 0 the functions z and w by the equalities z(t) = n k=1 t a |gk(s)| σ(p1)(s) xβ (τk(s)) t a σ(p1)(η)dη α ds + + n k=1 b t |gk(s)| σ(p1)(s) xβ (τk(s)) ds t a σ(p1)(η)dη α b a σ(p1)(η)dη 1−(α+β) and w(t) = n k=1 b a |G0(t, s)| |gk(s)|xβ (τk(s)) ds + wε(t), where G0 is Green’s function of the problem (1.3.21), (1.3.222) and wε is defined in the same way as in proving Corollary 1.1.11. Reasoning just as in proving Corollary 1.1.31, we make sure that the inclusion (1.1.72) is valid for β > 0. In the case β = 0 we consider the function w defined by the equality (1.3.50), where v is a solution of the equation (1.3.21) for the boundary conditions v(a) = 1, v′ (b−) = 1. Then, reasoning analogously as in proving Corollary 1.1.31 for β = 0, we can see that the inclusion (p0, p1) ∈ V2,0(]a, b[ ; h) is valid. Proof of Corollary 1.1.3i0. Coincides completely with that of Corollary 1.1.3i for β > 0. Proof of Remark 1.1.6. If the inequality (1.1.43i) is satisfied for t ∈ θτ1,...,τn , then it will especially be satisfied on each of the sets θτm , where m ∈ {1, . . ., n}, i.e., each of the n inequalities of (1.1.42i) will be satisfied. 59 Proof of Corollary 1.1.4i (1.1.4i0). It is sufficient to substitute p0 ≡ 0, p1 ≡ 0, k = 1 in Remark 1.1.5i (1.1.5i0). Proof of Corollary 1.1.51. It is clear that all the requirements of Theorem 1.1.11, except (1.1.71), follow directly from the conditions of our corollary. It remains to show that the inclusion (1.1.71) follows from the conditions (1.1.501) for 0 ≤ β < 1 and (1.1.511) for β = 1 as well. Consider first the case 0 < β < 1. Let x be a function defined by the equality (1.1.91). Then (xβ (t))′′ = p1(t)(xβ (t))′ − 2β2 σ2 (p1)(t) x1−β(t) − −β(1 − β) σ2 (p1)(t) x2−β(t) b a σ(p1)(η)dη 2 + b t σ(p1)(η)dη 2 . (1.3.53) From the condition (1.1.501) and the fact that the operator h is nonnegative it follows that − x2−β (t) σ2(p1)(t) p0(t) ≤ 2β2 b a σ(p1)(η)dη 2(1−β) for a < t < b. Moreover, 0 ≤ λp0(t) + β(1 − β) min s a σ(p1)(η)dη 2 + + b s σ(p1)(η)dη 2 : a ≤ s ≤ b σ2 (p1)(t) x2−β(t) , (1.3.54) where λ = 1 − β 2β b a σ(p1)(η)dη −2(1−β) × × min s a σ(p1)(η)dη 2 + b s σ(p1)(η)dη 2 : a ≤ s ≤ b . Let w(t) = xβ (t) + λ, and rewrite the identity (1.3.53) as w′′ (t) = p0(t)w(t) + p1(t)w′ (t) − p0(t)xβ (t) + 2β2 σ2 (p1)(t) x1−β(t) − − λp0(t) + β(1 − β) t a σ(p1)(η)dη 2 + b t σ(p1)(η)dη 2 σ2 (p1)(t) x2−β(t) . 60 Then, taking into account the fact that the operator h is nonnegative, from the condition (1.1.501) and the inequality (1.3.54) we obtain w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t), (1.3.55) i.e., owing to Lemma 1.2.2 the inclusion (p0, p1) ∈ V1,0(]a, b[) (1.3.56) is satisfied. Then, as is seen from Remark 1.2.2, there exists Green’s function G of the problem (1.2.4), (1.2.2i0), and by Lemma 1.2.6, b a |G(t, s)|qβ(s) ds = O∗ (xβ (t)) for t → a, t → b, (1.3.57) where qβ(t) = σ2 (p1)(t) x2−β(t) . Let now ε = β(1 − β) min s a σ(p1)(η)dη 2 + + b s σ(p1)(η)dη 2 : a ≤ t ≤ b (1.3.58) and rewrite (1.3.53) in the form (xβ (t))′′ = p0(t)xβ (t) + p1(t)(xβ (t))′ − h(xβ )(t) − εqβ(t) − − p0(t)xβ (t)−h(xβ )(t)+2β2 σ2 (p1)(t) x1−β(t) − β(1−β) t a σ(p1)(η)dη 2 + + b t σ(p1)(η)dη 2 − ε σ2 (p1)(t) x2−β(t) . (1.3.59) Taking into account (1.1.501) and (1.3.58), we obtain (xβ (t))′′ ≤ p0(t)xβ (t) + p1(t)(xβ (t))′ − h(xβ )(t) − εqβ(t) (1.3.60) for a < t < b. From (1.3.56), (1.3.57), and (1.3.60), by virtue of Definition 1.1.4 we conclude that the inclusion (1.1.71) is satisfied for 0 < β < 1. Assume now that β = 0. Then the condition (1.1.501) takes the form 0 ≤ p0(t) − h(1)(t) for a < t < b, 61 from which we can see that the function w(t) ≡ 1 satisfies the inequality w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) − h(w)(t), i.e., owing to Definition 1.1.3 we can conclude that the inclusion (1.1.71) is satisfied for β = 0. Finally we consider the case β = 1 and note that x′′ (t) = p1(t)x′ (t) − 2σ2 (p1)(t). (1.3.61) It follows from (1.1.511) that there exist constants ε, µ ∈ ]0, 1[ such that ess sup t∈ ]a,b[ x(t) σ2(p1)(t) h(x)(t) x(t) − p0(t) < 2µ2 (1.3.62) and ess sup t∈ ]a,b[ x(t) σ2(p1)(t) h(x)(t) x(t) − p0(t) < 2 − ε. (1.3.63) Taking into account the fact that the operator h is nonnegative, from the condition (1.3.62) we get − x2−µ (t) σ2(p1)(t) p0(t) ≤ 2µ2 b a σ(p1)(η)dη 2(1−µ) for a < t < b. Reasoning in the same way as for 0 < β < 1, from the last inequality as well as from (1.3.62) we can see that the function w(t) = xµ (t) + λ, where λ = 1 − µ 2µ b a σ(p1)(η)dη −2(1−µ) × × min s a σ(p1)(η)dη 2 + b s σ(p1)(η)dη 2 : a ≤ s ≤ b , satisfies (1.3.55), i.e., the inclusion (1.3.56) is satisfied and there exists Green’s function G of the problem (1.2.4), (1.2.2i0). As is seen from Lemma 1.2.7, if q1(t) = σ2 (p1)(t), then b a |G(t, s)|q1(s) ds = O∗ (x(s)) as t → a, t → b. (1.3.64) We rewrite now the identity (1.3.61) as follows: x′′ (t) = p0(t)x(t) + p1(t)x′ (t) − h(x)(t) − εq1(t) + + h(x)(t) − p0(t)x(t) − (2 − ε)σ2 (p1)(t) . 62 The latter with regard for (1.3.63) yields x′′ (t) ≤ p0(t)x(t) + p1(t)x′ (t) − h(x)(t) − εq1(t) for a < t < b. (1.3.65) From (1.3.56), (1.3.64), and (1.3.65), according to Definition 1.1.4 we conclude that the inclusion (1.1.71) is satisfied for β = 1. Proof of Corollary 1.1.52. It is clear that all the requirements of Theorem 1.1.12, except (1.1.72), follow directly from the conditions of our corollary. It remains to show that the inclusion (1.1.72) follows from the conditions (1.1.502), (1.1.56) for 0 < β ≤ 1 and from (1.1.512) for β = 1. First we consider the case 0 < β < 1. Let x be the function defined by (1.1.92). Then (xβ (t))′′ = p1(t)(xβ (t))′ − β(1 − β) σ2 (p1)(t) x2−β(t) . (1.3.66) From (1.1.502) it follows the existence of a constant ε > 0 such that ess sup t∈ ]a,b[ x2 (t) σ2(p1)(t) h(xβ )(t) xβ(t) − p0(t) < β(1 − β) − ε (1.3.67) and likewise from the inclusion (1.1.55) it follows the existence of a constant λ such that −λ x2−β (t) σ2(p1)(t) p0(t) < ε for a < t < b. (1.3.68) Let w(t) = xβ (t) + λ, and rewrite the identity (1.3.66) in the form w′′ (t)=p0(t)w(t) + p1(t)w′ (t) − p0(t)xβ (t) + λp0(t) + β(1 − β) σ2 (p1)(t) x2−β(t) , whence with regard for (1.3.67), (1.3.68) and the fact that the operator h is nonnegative we can see that the inequality (1.3.55) is valid, i.e., by virtue of Lemma 1.2.2 the inclusion (p0, p1) ∈ V2,0(]a, b[) (1.3.69) is satisfied. Then, as is seen from Remark 1.2.2, there exists Green’s function G of the problem (1.2.4), (1.2.220), and by Lemma 1.2.7, b a |G(t, s)|qβ(s) ds = O∗ (xβ (s)) as t → a, (1.3.70) where qβ(t) = σ2 (p1)(t) x2−β(t) . Rewrite now (1.3.66) as (xβ (t))′′ = p0(t)xβ (t) + p1(t)(xβ (t))′ − h(xβ ) − εqβ(t) + 63 + h(xβ )(t) − p0(t)xβ (t) − (β(1 − β) − ε) σ2 (p1)(t) x2−β(t) . This equality by virtue of the condition (1.3.67) enables us to see that (1.3.60) is satisfied. From the conditions (1.3.60), (1.3.69), (1.3.70) and according to Definition 1.1.4, we can conclude that the inclusion (1.1.72) is satisfied for 0 < β < 1. Assume now that β = 1. From the condition (1.1.502) for β = 1 it follows the existence of a constant ε > 0 such that ess sup t∈ ]a,b[ x(t) σ2(p1)(t) h(x)(t) x(t) − p0(t) < −ε. (1.3.71) Then it is clear from the negativeness of the operator h that p0(t) ≥ 0 for a < t < b, i.e., by virtue of Remark 1.2.6, the inclusion (1.3.69) is satisfied and hence there exists Green’s function G of the problem (1.2.4), (1.2.220). As is seen from lemma 1.2.7, if q1(t) = σ2 (p1)(t), then b a |G(t, s)|q1(s) ds = O∗ (x(t)) as t → a. (1.3.72) Note that x′′ (t) = p0(t)x(t) + p1(t)x′ (t) − h(x)(t) − εq1(t) + + h(x)(t) − p0(t)x(t) + εσ2 (p1)(t) , whence with regard for (1.3.71) we see that (1.3.65) is satisfied. From the conditions (1.3.65), (1.3.69), (1.3.72), owing to Definition 1.1.4 we conclude that the inclusion (1.1.72) is satisfied for β = 1 as well. The proof of the given and of the previous corollary is identical for the case β = 0. Proof of Corollary 1.1.5i0. Coincides completely with that of Corollary 1.1.5i for 0 < β ≤ 1. Proof of Corollary 1.1.61. Let h(u)(t) = n k=1 |gk(t)|u(τk(t)). (1.3.73) Then we can see from (1.1.561) that the inclusion (1.1.81) is satisfied for β = 0. It is also clear that all the requirements of Theorem 1.1.11 for α = 1, β = 0, except (1.1.71), follow directly from the conditions of our corollary. It remains to show that the conditions (1.1.571), (1.1.581) imply the inclusion (1.1.71) as well. 64 Without restriction of generality we assume that c ∈]a, b[ . Then by (1.1.571) there exist γm, ηm (m = 1, 2) such that 0 ≤ γm < ηm < +∞ (m = 1, 2) and η1 γ1 ds λ11 + λ12s + s2 = (c − a)1−β1 1 − β1 , η2 γ2 ds λ21 + λ22s + s2 = (b − c)1−β2 1 − β2 . (1.3.74) Introduce the functions ϕ1 and ϕ2 by η1 ϕ1(t) ds λ11 + λ12s + s2 = (t − a)1−β1 1 − β1 for a ≤ t ≤ c and η2 ϕ2(t) ds λ21 + λ22s + s2 = (b − t)1−β2 1 − β2 for c ≤ t ≤ b. From (1.3.74) we have γ1 < ϕ1(t) < η1 for a < t < c, γ2 < ϕ2(t) < η2 for c < t < b and ϕm(c) = γm (m = 1, 2). Introduce also the function w by w(t) = exp t c (s − a)−β1 ϕ1(s) ds for a ≤ t < c, w(t) = exp c t (b − s)−β2 ϕ2(s) ds for c ≤ t ≤ b. Then w′ (t) > 0 for a < t < c, w′ (t) < 0 for c ≤ t < b, w(t) > 0 for a ≤ t ≤ b, (1.3.75) w ∈ C′ loc( ]a, c[) ∩ C′ loc( ]c; b[), w(c−) ≥ w(c+), (1.3.76) 65 and the equalities w′′ (t) = − λ11 (t − a)2β1 w(t) − λ12 (t − a)β1 + β1 t − a w′ (t) for a < t < c, w′′ (t) = − λ21 (b − t)2β2 w(t) + λ22 (b − t)β2 + β2 b − t w′ (t) for c ≤ t < b (1.3.77) are valid. From the above equalities, by virtue of (1.3.75) it follows that w′′ (t) ≤ 0 for a < t < b. (1.3.78) On the other hand, taking into account the conditions (1.1.581) in the equalities (1.3.77), we obtain w′′ (t) ≤ p0(t) − n k=1 |gk(t)| w(t) + p1(t)w′ (t) − −w′ (t) n k=1 |gk(t)| τk(t) − t for a < t < b. (1.3.79) Analogously, from (1.3.78) it follows τk(t) t w′ (s) ds ≤ w′ (t) τk(t) − t (k = 1, . . . , n) for a < t < b. Taking this inequality into consideration, from (1.3.79) we can see that w′′ (t) ≤ p0(t)w(t) + p1(t)w′ (t) − n k=1 |gk(t)|w τk(t) for a < t < b. The latter inequality together with (1.3.75), (1.3.76) and by virtue of Definition 1.1.3 shows that the inclusion (p0, p1) ∈ V1,0(]a, b[; h) is satisfied. Proof of Corollary 1.1.62. We define the operator h by the equality (1.3.73). Note also that if p1 ∈ Lloc(]a, b]), then from the conditions (1.1.56) and (1.1.59) we obtain σ(p1) ∈ L([a, b]), pjσ2(p1) ∈ L([a, b]) (j = 0, 2), gkσ2(p1) ∈ L([a, b]) (k = 1, . . . , n), i.e., the conditions (1.1.32), (1.1.52), and (1.1.82), are satisfied where β = 0, α = 1. Then just as in the previous proof it remains to show that from the conditions (1.1.572)–(1.1.59) it follows the inclusion (1.1.72) for β = 0. 66 Without restriction of generality we assume that c ∈]a, b[ . Then by virtue of (1.1.572) there exist constants γm, ηm (m = 1, 2) such that ε ≤ γ1 < η1 < +∞, 0 < γ2 < η2 < +∞ and (1.3.74) is satisfied. Introduce the functions ϕ1 and ϕ2 by η ϕ1(t) ds λ11 + λ12s + s2 = (t − a)1−β1 1 − β1 for a ≤ t < c, ϕ2(t) γ2 ds λ21 + λ22s + s2 = (b − t)1−β2 1 − β2 for c ≤ t ≤ b. From (1.3.74) we have γ1 < ϕ1(t) < η1 for a < t < c, γ2 < ϕ2(t) < η2 for c < t < b, ϕ1(c) = γ1 ϕ2(c) = η2. Introduce likewise the function w by the equalities w(t) = exp t a (s − a)−β1 ϕ1(s) ds for a ≤ t < c, w(t) = exp α t c (b − s)−β3 ϕ2(s) ds for c ≤ t ≤ b, where 0 < α < min 1; γ1 η2 (b − c)−β3 (c − a)−β1 , i.e., α ∈]0, 1[. (1.3.80) Then w′ (t) > 0 for t ∈ ]a, c[∪ ]c, b[, w(t) > 0 for a ≤ t ≤ b, (1.3.81) w ∈ C′ loc( ]a, c[) ∩ C′ loc( ]c; b[), w(c−) ≥ w(c+), w′ (b−) ≥ 0, (1.3.82) and the equalities w′′ (t) = − λ11 (t − a)2β1 w(t) − λ12 (t − a)β1 + β1 t − a w′ (t) (1.3.83) for a < t < c and w′′ (t) = − αλ21 (b − t)β2−β3 w(t) − λ22 (b − t)β2 + β3 b − t w′ (t) − −α 1 − α(b − t)β2+β3 (b − t)β3−β2 w(t)ϕ2 2(t), for c < t < b (1.3.84) 67 are valid. Note also that the condition c ∈ [max(a, b − 1); b] and (1.3.80) imply 1 − α(b − t)β2+β3 ≥ 0 for c ≤ t ≤ b. Taking this into account in the equality (1.3.84), we obtain w′′ (t) ≤ − αλ21 (b − t)β2−β3 w(t) − λ22 (b − t)β2 + β3 b − t w′ (t). (1.3.85) for a ≤ t < b. From (1.3.83) and (1.3.85), according to the condition (1.3.81), it is clear that the inequality (1.3.78) is satisfied. On the other hand, taking into account in (1.3.83) and (1.3.85) the conditions (1.1.582), we get w′′ (t) ≤ p0(t) − n k=1 |gk(t)| w(t) + p1(t)w′ (t) − − w′ (t) n k=1 |gk(t)| τk(t) − t for a < t < b, which with regard for (1.3.81) and (1.1.59) imply that (1.3.79) is satisfied. Reasoning in the same way as in the previous proof, we see that the inclusion (p0, p1) ∈ V2,0(]a, b[ ; h) is valid. Proof of Corollary 1.1.71. It is not difficult to notice that if we introduce the notation g(u)(t) = n k=1 gk(t)u τk(t) , then the inequality (1.1.22) will be satisfied, and from (1.1.61), (1.1.62) it follows that the conditions (1.1.17) and (1.1.18) are valid. That is, all the requirements of Theorem 1.1.21 are fulfilled and this implies that our corollary is valid. Proof of Remark 1.1.10. Follows directly from that of Remark 1.1.2. Corollaries 1.1.72 and 1.1.7i0 are proved analogously to Corollary 1.1.71. 68 CHAPTER II CORRECTNESS OF TWO-POINT PROBLEMS FOR LINEAR SINGULAR FUNCTIONAL DIFFERENTIAL EQUATIONS OF SECOND ORDER § 2.1. Statement of the Problem and Formulation of Main Results 2.1.1. Statement of the Problem. Let us Consider the functional differential equations u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + g(u)(t) + p2(t), (2.1.1) u′′ (t) = p0k(t)u(t) + p1k(t)u′ (t) + gk(u)(t) + p2k(t), k ∈ N, (2.1.1k) under one of the following the boundary conditions u(a) = 0, u(−b) = 0, (2.1.210) u(a) = 0, u′ (b−) = 0; (2.1.220) u(a) = c1, u(b) = c2, (2.1.21) u(a) = c1, u′ (b−) = c2; (2.1.22) u(a) = c1k, u(b) = c2k, (2.1.21k) u(a) = c1k, u′ (b−) = c2k, (2.1.22k) where cl, clk ∈ R, (l = 1, 2; k ∈ N), g, gk : C(]a, b[) → Lloc(]a, b[), k ∈ N, are continuous operators, p1, pj ∈ Lloc(]a, b[) σ(p1) ∈ L([a, b]), pj ∈ Lσ1(p1)([a, b]) (j = 0, 2) (2.1.31) if i = 1, p1, pj ∈ Lloc(]a, b]) σ(p1) ∈ L([a, b]), pj ∈ Lσ2(p1)([a, b]) (j = 0, 2) (2.1.32) if i = 2, and pjk : ]a, b[ → R (j = 0, 1, 2; k ∈ N) are measurable functions. The correctness of the problem (2.1.1), (2.1.2i) will be studied under the assumption that the inclusion (p0, p1) ∈ Vi,0(]a, b[ ; h) is satisfied. (Effective sufficient conditions for the above inclusion to be fulfilled are given in §1.1, where |g(x)(t)| ≤ h(|x|)(t) almost everywhere in the interval ]a, b[ for every x ∈ C(]a, b[).) Consider also the following linear equation u′′ (t) = p0k(t)u(t) + p1k(t)u′ (t) + p2k(t). (2.1.4k) 69 Let Gk be Green’s function of the problem (2.1.4k), (2.1.2i0) and r ∈ R+ . Then we denote the set y(t) : y(t) = α1vk(t) + b a Gk(t, s)gk(x)(s) ds, α1 ∈ [0, r], x C ≤ r by Br,k if vk is a solution of the problem (2.1.4k), (2.1.2i0), and by B′ r,k if vk is a solution of the problem (2.1.4k), (2.1.2ik). Throughout this chapter the use will also be made of the notation Ii(x)(t) = t a x(s) ds b t x(s) ds 2−i for a ≤ t ≤ b, where x ∈ L([a, b]). 2.1.2. Formulation of Main Results. Theorem 2.1.1i. Let i ∈ {1, 2}, the continuous linear operators g, gk, h : C(]a, b[) → Lloc(]a, b[) (k ∈ N), the measurable functions pj, pjk :]a, b[→ R (j = 0, 1, 2; k ∈ N) and the constants α ∈ [a, b], γ ∈]1, +∞[, β, µ ∈ R be such that 0 ≤ β < µ < γ − 1 γ − α , (2.1.5) σγ (p1) ∈ L([a, b]), b a |pj(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞ (j = 0, 2), b a h(1)(s) σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞, (2.1.6) where h is a non-negative operator and uniformly on the segment [a, b] lim k→∞ t a |p1(s) − p1k(s)| ds = 0, lim k→∞ t a pj(s) − pjk(s) σ(p1)(s) Iβ i (σα (p1))(s) ds = 0 (j = 0, 2), (2.1.7) lim k→∞ sup t a g(y)(s) − gk(y(s)) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b, y ∈ B1k = 0. (2.1.8) 70 Moreover, let (p0, p1) ∈ Vi,0( ]a, b[, h), (2.1.9) where for every x ∈ C(]a, b[) almost everywhere in the interval ]a, b[ the inequality |g(x)(t)| ≤ h(|x|)(t) (2.1.10) is satisfied. Then there exists a number k0 such that if k > k0, then the problem (2.1.1k), (2.1.2i0) has a unique solution uk and uniformly in the interval ]a, b[ lim k→∞ Iµ−1 i (σ 1−αµ 1−µ (p1))(t)(u(t) − uk(t)) = 0, (2.1.11) lim k→∞ Iµ i (σα (p1))(t) σ(p1)(t) (u′ (t) − u′ k(t)) = 0, (2.1.12) where u is the solution of the problem (2.1.1), (2.1.2i0). Theorem 2.1.2i. Let i ∈ {1, 2}, the continuous linear operators g, gk, h : C(]a, b[) → Lloc(]a, b[) (k ∈ N), the measurable functions pj, pjk : (]a, b[) → R (j = 0, 1, 2; k ∈ N) and the constans α ∈ [a, b], γ ∈]1, +∞[, cl, clk, β, µ ∈ R (l = 1, 2; k ∈ N) be such that conditions (2.1.5)–(2.1.7), (2.1.9), (2.1.10) and also lim k→∞ sup t a g(y)(s) − gk(y(s)) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b, x ∈ B′ 1k = 0 (2.1.13) and lim k→∞ clk = cl (l = 1, 2) (2.1.14) are satisfied. Then there exists a number k0 such that if k > k0, the problem (2.1.1k), (2.1.2i0) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.12) and lim k→∞ u(t) − uk(t) = 0 (2.1.15) are satisfied, where u is the solution of the problem (2.1.1), (2.1.2i0). 2.1.3. Corollaries of Theorems (2.1.1i) (2.1.2i) (i = 1, 2). Corollary 2.1.1i. Let i ∈ {1, 2}, the continuous linear operators g, gk, h : C(]a, b[) → Lloc(]a, b[) (k ∈ N), the measurable functions η, pj, pjk :]a, b[→ R (j = 0, 1, 2; k ∈ N) and the constants α ∈ [0, 1], γ ∈]1, +∞[, β, µ ∈ R+ 71 be such that the conditions (2.1.5)–(2.1.7), (2.1.9), (2.1.10) are satisfied and for every y ∈ C(]a, b[) almost everywhere on the interval ]a, b[ gk(y)(t) − g(y)(t) ≤ η(t) y C (k ∈ N) (2.1.16) and uniformly on the segment [a, b] lim k→∞ t a gk(y)(s) − g(y)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds = 0, (2.1.17) where b a η(s) σ(p1)(s) Iβ i (σα (p1))(s) ds < +∞. (2.1.18) Then there exists a number k0, such that for k > k0 the problem (2.1.1k), (2.1.2i0) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.11), (2.1.12) are satisfied, where u is the solution of the problem (2.1.1), (2.1.2i0). Corollary 2.1.2i. Let i ∈ {1, 2}, the continuous linear operators g, gk, h : C(]a, b[) → Lloc(]a, b[) (k ∈ N), the measurable functions η, pj, pjk :]a, b[→ R, (j = 0, 1, 2; k ∈ N) and constants α ∈ [0, 1], γ ∈]1, +∞[, β, µ ∈ R+ be such that the conditions (2.1.5)–(2.1.7), (2.1.9), (2.1.10), (2.1.14), and (2.1.16)–(2.1.18) are satisfied. Then there exists a number k0 such that for k > k0 the problem (2.1.1k), (2.1.2ik) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.12), (2.1.15) are satisfied, where u is the solution of the problem (2.1.1), (2.1.2i). Consider now the case where the equations (2.1.1) and (2.1.1k) are of the form u′′ (t) = p0(t)u(t) + p1(t)u′ (t) + n m=1 g0m(t)u(τ0m(t)) + p2(t) (2.1.19) and u′′ (t) = p0k(t)u(t) + p1k(t)u′ (t) + n m=1 gkm(t)u(τkm(t)) + p2k(t), (2.1.19k) where g0m, gkm : ]a, b[ → R and τ0m, τkm : [a, b] → [a, b] (m = 1, . . . , n, k ∈ N) are measurable functions. Corollary 2.1.3i. Let i ∈ {1, 2}, the measurable functions η, g0m, gkm, pj, pjk : ]a, b[ → R, τ0m, τkm : [a, b] → [a, b], (m = 1, . . . , n; j = 0, 1, 2; k ∈ 72 N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R be such that conditions (2.1.5), (2.1.7), (2.1.18) as well as σγ (p1) ∈ L([a, b]), b a |pj(s)| + n m=1 |g0m(s)| Iµ i (σα (p1))(s) σ(p1)(s) ds < +∞ (j = 0, 2), (2.1.20) n m=1 g0m(t) − gkm(t) ≤ η(t) (k ∈ N) (2.1.21) are satisfied, and uniformly on the segment [a, b] lim k→∞ n m=1 t a gkm(s) − g0m(s) σ(p1)(s) Iβ i (σα (p1))(s) ds = 0, (2.1.22) ess sup Iβ−µ i (σα (p1))(t) n m=1 τkm(t) τ0m(t) σ(p1)(s) Iµ i (σα(p1))(s) ds : a k0 the problem (2.1.19k), (2.1.2i0) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.11), (2.1.12) are satisfied, where u is the solution of the problem (2.1.19), (2.1.2i0). Corollary 2.1.4i. Let i ∈ {1, 2}, the measurable functions η, g0m, gkm, pj, pjk : ]a, b[ → R, τ0m, τkm : [a, b] → [a, b], (m = 1, . . . , n; j = 0, 1, 2; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , cl, clk, β, µ ∈ R (l = 1, 2; k ∈ N) be such that the conditions (2.1.5), (2.1.7), (2.1.9), (2.1.14), (2.1.18), (2.1.20)–(2.1.23) are satisfied, where h(x)(t) = n m=1 |g0m(t)|x(τ0m(t)). Then there exists a number k0 such that for k > k0 the problem (2.1.19k), (2.1.2ik) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.12), (2.1.15) are satisfied, where u is the solution of the problem (2.1.19), (2.1.2i). Corollary 2.1.5i. Let i ∈ {1, 2}, the measurable functions η, g0m, gkm, pj, pjk : ]a, b[ → R, τ0m, τkm : [a, b] → [a, b], (m = 1, . . . , n; j = 0, 1, 2; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R be such that the 73 conditions (2.1.5), (2.1.7), (2.1.18), (2.1.22) as well as σγ (p1) ∈ L([a, b]), b a |pj(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞ (j = 0, 2), (2.1.24) n m=1 |gkm(t)| + |g0m(t)| ≤ η(t) (k ∈ N) for a < t < b (2.1.25) and ess sup n m=1 |τ0m(t) − τkm(t)| : a ≤ t ≤ b → 0 for k → +∞ (2.1.26) are satisfied. Let also the condition (2.1.9) be satisfied, where h(x)(t) = n m=1 |g0m(t)|x(τ0m(t)). Then there exists a number k0 such that for k > k0 the problem (2.1.19k), (2.1.2i0) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.11), (2.1.12) are satisfied, where u is the solution of the problem (2.1.19), (2.1.2i0). Corollary 2.1.6i. Let i ∈ {1, 2}, the measurable functions η, g0m, gkm, pj, pjm :]a, b[ → R τ0m, τkm :[a, b] → [a, b], (m = 1, . . . , n; j = 0, 1, 2; k ∈N) and the constants α∈ [0, 1], γ ∈ ]1, +∞[ , cl, clk, β, µ ∈ R (l = 1, 2; k∈N) be such that the conditions (2.1.5), (2.1.7), (2.1.9), (2.1.14), (2.1.18), (2.1.22) and (2.1.24)–(2.1.26) are satisfied, where h(x)(t) = n m=1 |g0m(t)|x(τ0m(t)). Then there exists a number k0 such that for k > k0 the problem (2.1.19k), (2.1.2ik) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.12), (2.1.15) are satisfied, where u is the solution of the problem (2.1.19), (2.1.2i0). For more clearness, let us consider the equations u′′ (t) = g0(t)u(τ0(t)) + p2(t), (2.1.27) u′′ (t) = g0k(t)u(τk(t)) + p2k(t), (2.1.27k) where g0, g0k, p2, p2k; ]a, b[ → R, and τ0, τ0k; [a, b] → [a, b] (k ∈ N) are measurable functions. Corollary 2.1.7i. Let i ∈ {1, 2}, the measurable functions η, g0, g0k, p2, p2k : ]a, b[ → R, τ0, τk : [a, b] → [a, b], (k ∈ N) and the constants β, µ ∈ R be such that the conditions β < µ < 1, (2.1.28) |g0(t)| + |g0k(t)| ≤ η(t) for a < t < b, (2.1.29) 74 b a |p2(s)|(s − a)µ (b − s)µ(2−i) ds < +∞, b a η(s)(s − a)β (b − s)β(2−i) ds < +∞ (2.1.30) are satisfied, and uniformly on the segment [a, b] lim k→∞ t a p2(s) − p2k(s) (s − a)β (b − s)β(2−i) ds = 0, lim k→∞ t a g0(s) − g0k(s) (s − a)β (b − s)β(2−i) ds = 0 (2.1.31) and ess sup |τ0(t) − τk(t)| : a ≤ t ≤ b → 0 as k → +∞. (2.1.32) Let, moreover, the inclusion (0, 0) ∈ Vi,0(]a, b[; h) (2.1.33) be satisfied, where h(x)(t) = |g0(t)|x(τ0(t)). Then there exists a number k0, such that for k > k0, the problem (2.1.27k), (2.1.2i0) has a unique solution uk, and uniformly on the interval ]a, b[ the conditions (2.1.11), (2.1.12) are satisfied, where u is a solution of the problem (2.1.27), (2.1.2i0). Corollary 2.1.8i. Let i ∈ {1, 2}, the measurable functions η, g0m, g0k, p2, p2k : ]a, b[ → R, τ0, τk : [a, b] → [a, b], (k ∈ N) and the constants cl, clk, β, µ ∈ R (l = 1, 2; k ∈ N) be such that the conditions (2.1.14) and (2.1.28)–(2.1.33) are satisfied, where h(x)(t) = |g0(t)|x(τ0(t)). Then there exists a number k0 such that for k > k0 the problem (2.1.27k), (2.1.2ik) has a unique solution uk, and uniformly on the interval ]a, b[ the equalities (2.1.12), (2.1.15) are satisfied, where u is the solution of the problem (2.1.27), (2.1.2i). § 2.2. Auxiliary Propositions 2.2.1. Correctness of the Initial Problem for Linear Second Order Ordinary Differential Equations. Consider on the interval ]a, b[ the equations v′′ (t) = p0(t)v(t) + p1(t)u′ (t) (2.2.1) and v′′ (t) = p0k(t)v(t) + p1k(t)v′ (t), k ∈ N, (2.2.1k) 75 where p0, p1 ∈ Lloc(]a, b[), σ(p1) ∈ L([a, b]), p0 ∈ Lσ1(p1), ([a, b]) (2.2.21) p0k, p1k ∈ Lloc(]a, b[), k ∈ N, (2.2.31) or p0, p1 ∈ Lloc(]a, b]), σ(p1) ∈ L( [a, b] ), p0 ∈ Lσ2(p1)([a, b]), (2.2.22) p0k, p1k ∈ Lloc( ]a, b]), k ∈ N, (2.2.32) and the following initial conditions: v(a) = 0, lim t→a v′ (t) σ(p1)(t) = 1, (2.2.41) v(a) = 0, lim t→a v′ (t) σ(p1k)(t) = 1, (2.2.4k) v(b) = 0, lim t→b v′ (t) σ(p1)(t) = −1, (2.2.51) v(b) = 0, lim t→b v′ (t) σ(p1k)(t) = −1, (2.2.51k) v(b) = 1, v′ (b) = 0. (2.2.52) Remark 2.2.1. It has been shown in [23] that for the conditions (2.2.2i) the problems (2.2.1), (2.2.4) and (2.2.1), (2.2.5i) are uniquely solvable. Analogously, if p0k, p1k ∈ Lloc(]a, b[), σ(p1k) ∈ L([a, b]), p0k ∈ Lσ1(p1k)([a, b]), then the problems (2.2.1k), (2.2.4k) and (2.2.1k), (2.2.51k) are uniquely solvable, and if p0k, p1k ∈ Lloc(]a, b]), σ(p1k) ∈ L([a, b]), p0k ∈ Lσ2(p1k)([a, b]), then the problems (2.2.1k), (2.2.4k) and (2.2.1k), (2.2.52) are uniquely solvable as well. For brevity we introduce the notation ∆pjk(t) = pj(t) − pjk(t) (j = 0, 1, 2; k ∈ N) for a < t < b. Lemma 2.2.11. Let the measurable functions pj, pjk : ]a, b[ → R (j = 0, 1; k ∈ N) and the constants α ∈ [0, 1], γ ∈]1, +∞[, β, µ ∈ R such that 0 ≤ β < µ ≤ γ − 1 γ − α , (2.2.6) σγ (p1) ∈ L([a, b]), b a |p0(s)| σ(p1)(s) Iµ 1 (σα (p1))(s) ds < +∞ (2.2.71) 76 and uniformly on the segment [a, b] the conditions lim k→∞ t a ∆p0k(s) σ(p1)(s) Iβ 1 (σα (p1))(s) ds = 0, lim k→∞ t a |∆p1k(s)| ds = 0 (2.2.81) be satisfied. Then there exists a number k0 such that for k > k0 the problem (2.2.1k), (2.2.41k) has a unique solution v1k and the problem (2.2.1k), (2.2.51k) has a unique solution v2k, and uniformly on the interval ]a, b[ lim k→∞ v1k(t) − v1(t) t a σ(p1)(s) ds −1 = 0, (2.2.911) lim k→∞ v2k(t) − v2(t) b t σ(p1)(s) ds −1 = 0 (2.2.912) and lim k→∞ v′ 1k(t) − v′ 1(t) σ(p1)(t) b t σα (p1)(s) ds µ = 0, (2.2.1011) lim k→∞ v′ 2k(t) − v′ 2(t) σ(p1)(t) t a σα (p1)(s) ds µ = 0, (2.2.1012) where v1 and v2 are the solutions of the problems (2.2.1), (2.2.41) and (2.2.1), (2.2.51), respectively. Proof. It is clear from the definition of the constants α, β, γ, µ that β − µ < 0, 0 < 1 − αβ 1 − β < 1 − αµ 1 − µ ≤ γ. (2.2.11) Hence σα (p1), σ 1−αβ 1−β (p1), σ 1−αµ 1−µ (p1) ∈ L([a, b]). (2.2.12) Using the H¨older inequality, we obtain t2 t1 σ(p1)(s) ds ≤ t2 t1 σ 1−αµ 1−µ (p1)(s) ds 1−µ × × t2 t1 σα (p1)(s) ds µ for a ≤ t1 ≤ t2 ≤ b, (2.2.13) 77 b a σ(p1)(s) s a σα(p1)(η) dη β ds ≤ ≤ b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα (p1)(s) s a σα(p1)(η) dη β µ ds µ = = µ µ − β µ b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα (p1)(s) ds µ−β , (2.2.14) b a σ(p1)(s) b s σα(p1)(η) dη β ds ≤ ≤ µ µ − β µ b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα (p1)(s) ds µ−β , (2.2.15) where the existence of the integrals follows from (2.2.12). By means of (2.2.14), (2.2.15) we easily get b a σ(p1)(s) Iβ 1 (σα(p1))(s) ds ≤ 2 µ µ − β µ I−β 1 (σα (p1)) a + b 2 × × b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα (p1)(s) ds µ−β < +∞. (2.2.16) It is also evident that for every δ ∈ [0, 1[ b a σα (p1)(s) Iδ i (σα(p1))(s) ds < +∞. (2.2.17) By virtue of condition (2.2.81), for every ε > 1 there exists a number k0 such that for k > k0 ε−1 ≤ σ(∆p1k)(t) ≤ ε for a ≤ t ≤ b. (2.2.18) We now proceed to the proof of the lemma. Taking into account the conditions (2.2.71), (2.2.12) and the inequality (2.2.13), the inequality b a |p0(s)|σ1(p1)(s) ds ≤ b a |p0(s)| σ(p1)(s) Iµ 1 (σα (p1))(s) ds × 78 × b a σ 1−αµ 1−µ (p1)(s) ds 2(1−µ) < +∞ (2.2.19) is valid, i.e. the conditions (2.2.21) are satisfied. In this case, owing to Remark 2.2.1, the problems (2.2.1), (2.2.4) and (2.2.1), (2.2.51) are uniquely solvable. Integrating by parts and using (2.2.18), we arrive at b a p0k(s) σ(p1k)(s) Iµ 1 (σα (p1k))(s) ds ≤ ≤ b a ∆p0k(s) σ(p1k)(s) Iµ 1 (σα (p1k))(s) ds + b a |p0(s)| σ(p1k)(s) Iµ 1 (σα (p1k))(s) ds ≤ ≤ Ak b a σ(∆p1k)(s) Iµ 1 (σα (p1k))(s) Iβ 1 (σα(p1))(s) ′ ds + +ε3 b a |p0(s)| σ(p1)(s) Iµ 1 (σα (p1))(s) ds for k > k0, (2.2.20) where Ak = sup t2 t1 ∆p0k(s) σ(p1)(s) Iβ 1 (σα (p1))(s) ds : a ≤ t1 < t2 ≤ b . In view of (2.2.81) lim k→∞ Ak = 0, (2.2.21) and by virtue of (2.2.18) the estimate σ(∆p1k)(t) Iµ 1 (σα (p1k))(t) Iβ 1 (σα(p1))(t) ′ ≤ ε3 |∆p1k(t)|Iµ−β 1 (σα (p1))(t) + +(µ + β)ε3 b a σα (p1)(s) ds σα (p1)(t) I1+β−µ 1 (σα(p1))(t) for a < t < b is valid. Substituting the latter in (2.2.20) and taking into account (2.2.71), (2.2.81), (2.2.17) and (2.2.21), we can see that a constant r0 ∈ R+ exist, such that sup b a |p0k(s)| σ(p1k)(s) Iµ 1 (σα (p1k))(s) ds : k > k0 < r0. (2.2.22) 79 In the same way we get p0k ∈ Lσ1(p1k)([a, b]) for k > k0, where in view of (2.2.18) σ(p1k) ∈ L([a, b]) for k > k0, which together with the conditions (2.2.3i) and Remark 2.2.1 imply that the problems (2.2.1k), (2.2.4k) and (2.2.1k), (2.2.51k) are uniquely solvable for k > k0. Note that the function wjk(t) = vj(t) − vjk(t) (j = 1, 2; k > k0) is a solution of the equation v′′ (t) = p0k(t)v(t) + p1k(t)v′ (t) + +∆p0k(t)vj(t) + ∆p1k(t)v′ j(t) (j = 1, 2) (2.2.23) and w1k(a) = 0, lim t→a w′ 1k(t) σ(p1k)(t) = σ(∆p1k)(a) − 1, (2.2.241) w2k(b) = 0, lim t→b w′ 2k(t) σ(p1k)(t) = 1 − σ(∆p1k)(b), (2.2.242) where in view of (2.2.81), lim k→∞ 1 − σ(∆p1k) C = 0. (2.2.25) Consider first the case j = 1. From (2.2.23), (2.2.241) we have w′ 1k(t) σ(p1k)(t) = σ(∆p1k)(t) − 1 + t a ∆p0k(s) v1(s) − w1k(s) σ(p1k)(s) ds + + t a p0(s)w1k(s) + ∆p1k(s)v′ 1(s) σ(p1k)(s) ds for a < t < b, (2.2.26) where the existence of integrals follows from the estimate (1.2.101), (1.2.111) and the conditions (2.2.71), (2.2.81). From (2.2.26), integration by parts results in |w′ 1k(t)| σ(p1k)(t) ≤ 1 − σ(∆p1k)(a) + Ak t a v1(s) − w1k(s) Iβ 1 (σα(p1))(s) σ(∆p1k)(s) ′ ds + + t a |p0(s)w1k(s) + ∆p1k(s)v′ 1(s)| σ(p1k)(s) ds for a < t < b, (2.2.27) 80 where in view of (2.2.18), t a v1(s) − w1k(s) Iβ 1 (σα(p1))(s) σ(∆p1k)(s) ′ ds ≤ ≤ ε t a |w′ 1k(s)| + |v′ 1(s)| Iβ 1 (σα(p1))(s) + |w1k(s)| + |v1(s)| hk(s) ds with hk(t)= |∆p1k(t)| Iβ 1 (σα(p1))(t) +β b a σα (p1)(s) ds σα (p1)(t) I1+β 1 (σα(p1))(t) for a < t < b. Substituting the latter inequality in (2.2.27), with regard for (2.2.18) we get |w′ 1k(t)| σ(p1)(t) ≤ ε2 Ak t a |w′ 1k(s)| Iβ 1 (σα(p1))(s) ds + +ε2 1 − σ(∆p1k) C + t a fk(s)|w1k(s)| + qk(s) ds , (2.2.28) where fk(t) = |p0k(t)| σ(p1)(t) + Akhk(t), qk(t) = |v′ 1(t)| σ(p1)(t) |∆p1k(t)| + Ak σ(p1)(t) Iβ 1 (σα(p1))(t) + Akhk(t)|v1(t)| for a < t < b. From (2.2.28), using Gronwall-Bellman’s lemma, it follows that |w′ 1k(t)| ≤ rkσ(p1)(t) 1 − σ(∆p1k) C + + t a fk(s)|w1k(s)| + qk(s) ds for a < t < b, (2.2.29) where rk = ε2 1 + exp ε2 Ak b a σ(p1)(s) Iβ 1 (σα(p1))(s) ds for k > k0 and by virtue of (2.2.16), (2.2.21), sup{rk : k > k0} < +∞. (2.2.30) 81 Let us now introduce the notation zk = |w1k(t)| t a σ(p1)(s) ds −1 for a < t < b. Integrating (2.2.29) from a to t, dividing by t a σ(p1)(s) ds and using integration by parts, by virtue of the inequalities (2.2.13) and t s σ(p1)(s) ds t a σ(p1)(s) ds −1 ≤ ≤ b s σ(p1)(s) ds b a σ(p1)(s) ds −1 for a < s ≤ t < b we obtain zk(t) ≤ r t a fk(s)Iµ 1 (σα (p1))(s)zk(s) ds + rk for a < t < b, where r = sup rk : k > k0 b a σ 1−αµ 1−µ (p1)(s) ds 2(1−µ) b a σ(p1)(s) ds −1 , rk = r b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα(p1)(s) ds b a qk(s) b s σα (p1)(η) dη µ ds + + 1 + σ(∆p1k) C . Applying Gronwall–Bellman’s lemma, from the latter inequality we get zk(t) ≤ rk exp r b a fk(s)Iµ 1 (σα (p1))(s) ds for a < t < b. (2.2.31) By virtue of (2.2.18) we note that the estimate b a fk(s)Iµ 1 (σα (p1))(s) ds ≤ ε3 b a |p0k(s)| σ(p1k)(s) Iµ 1 (σα (p1k))(s) ds + 82 +Ak b a σα (p1)(s) ds 2(µ−β) b a |∆p1k(s)| ds + +β b a σα (p1)(s) ds b a σα (p1)(s) I1+β−µ 1 (σα(p1))(s) ds for k > k0 is valid, which with regard for the conditions (2.2.81), (2.2.17) with δ = 1 + β − µ and the condition (2.2.22) results in sup b a fk(s)Iµ 1 (σα (p1))(s) ds : k > k0 < +∞. (2.2.32) Just in the same way, taking into account the estimates (1.2.101), (1.2.111) and the inequality (2.2.13), we obtain b a qk(s) b s σα (p1)(η) dη µ ds ≤ ≤ b a |∆p1k(s)| + Ak σ(p1)(s) Iβ 1 (σα(p1))(s) ds b a σα (p1)(s) ds µ + +c∗ b a |p0(s)| σ(p1)(s) Iµ 1 (σα (p1))(s) ds b a σ 1−αµ 1−µ (p1)(s) ds 1−µ + +c∗ Ak b a σ 1−αµ 1−µ (p1)(s)ds 1−µ β b a σα (p1)(s)ds b a σα (p1)(s) I1+β−µ 1 (σα(p1))(s) ds+ + b a σα (p1)(s) ds 2(µ−β) b a |∆p1k(s)| ds for k > k0, By virtue of the inequalities (2.2.16), (2.2.17) with δ = 1 + β − µ and the conditions (2.2.71), (2.2.81) and (2.2.21) lim k→∞ b a qk(s) b s σα (p1)(η) dη µ ds = 0 (2.2.33) which together with (2.2.25) implies lim k→∞ rk = 0. (2.2.34) 83 Substituting (2.2.32) and (2.2.34) in (2.2.31) we get lim k→∞ zk C = 0, (2.2.35) i.e., the condition (2.2.911) is satisfied. Applying (2.2.13), we see from (2.2.29) that |w′ 1k(t)| σ(p1)(t) b t σα (p1)(s) ds µ ≤ ≤ r b a σ 1−αµ 1−µ (p1)(s) ds 1−µ zk C b a fk(s)Iµ 1 (σα (p1))(s) ds + + b a qk(s) b s σα (p1)(η) dη µ ds + +r 1 − σ(∆p1k) C b a σα (p1)(s)ds µ for a < t < b, where r = sup{rk : k > k0}. The above inequality with regard for (2.2.25), (2.2.32), (2.2.33) and (2.2.35) implies that the condition (2.2.1011) is valid. Consider now the case j = 2. Let k > k0. Then for w2k, i.e., for a solution of the problem (2.2.23), (2.2.242) the representation − w′ 2k(t) σ(p1k)(t) = σ(∆p1k)(t) − 1 + b t ∆p0k(s) v2(s) − w2k(s) σ(p1k)(s) ds + + b t p0k(s)w2k(s) + ∆p1kv′ 2(s) σ(p1k)(s) ds for a < t < b is valid. Repeating the arguments presented for j = 1, where fk, hk are defined as before, qk(t) = |∆p1k(t)| + Ak σ(p1)(t) Iβ 1 (σα(p1))(t) |v′ 2(t)| σ(p1)(t) + Akhk(t)|v2(t)|, zk(t) = |w2k(t)| b t σ(p1)(s)ds −1 84 and rk = r b a σ 1−αµ 1−µ (p1)(s) ds 1−µ b a σα(p1)(s) ds b a qk(s) s a σα (p1)(η) dη µ ds + + 1 + σ(∆p1k) C , we see that the conditions (2.2.912), (2.2.1012) are valid. Lemma 2.2.12. Let the measurable functions pj, pjk : ]a, b[ → R (j = 0, 1; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[, β, µ ∈ R be such that the conditions (2.2.6) are satisfied, σγ (p1) ∈ L([a, b]), b a |p0(s)| σ(p1)(s) Iµ 2 (σα (p1))(s) ds < +∞ (2.2.72) and uniformly on the segment [a, b] the conditions lim k→∞ t a ∆p0k(s) σ(p1)(s) Iβ 2 (σα (p1))(s) ds = 0, lim k→∞ t a |∆p1k(s)| ds = 0 (2.2.82) are satisfied. Then there exists a number k0 such that for k > k0 the problem (2.2.1k), (2.2.4k) has a unique solution v1k and the problem (2.2.1k), (2.2.52) has a unique solution v2k, and uniformly on the interval ]a, b[ lim k→∞ v1k(t) − v1(t) t a σ(p1)(s) ds −1 = 0, (2.2.921) lim k→∞ v2k(t) − v2(t) = 0 (2.2.922) and lim k→∞ v′ 1k(t) − v′ 1(t) σ(p1)(t) = 0, (2.2.1021) lim k→∞ v′ 2k(t) − v′ 2(t) σ(p1)(t) t a σα (p1)(s) ds µ = 0, (2.2.1022) where v1 and v2 are the solutions of the problems (2.2.1), (2.2.4) and (2.2.1), (2.2.52), respectively. 85 Proof. Repeating word by word the previous proof for the case j = 1 and replacing everywhere I1 by I2, we can see that the problems (2.2.1k), (2.2.4k) and (2.2.1k), (2.2.52) are uniquely solvable, the condition (2.2.921) is satisfied and for the function w1k(t) = v1(t) − v1k(t) the representation |w′ 1k(t)| σ(p1)(t) ≤ rk zk C t a fk(s) s a σα (p1)(η) dη µ ds + + t a qk(s) ds + 1 − σ(∆p1k) C for a < t ≤ b (2.2.36) is valid, where the functions fk, qk and zk are defined in the previous proof. Using the same technique as when proving the relations (2.2.25), (2.2.32), (2.2.33), we obtain sup b a fk(s)Iµ 2 (σα (p1))(s) ds : k > k0 < +∞, lim k→∞ b a qk(s) ds = 0, lim k→∞ 1 − σ(∆p1k) C = 0 and lim k→∞ zk C = 0, from which it follows with regard for (2.2.36) that the condition (2.2.1021) is valid. Note that the function w2k(t) = v2(t) − v2k(t) satisfies the conditions w2k(b) = 0, w′ 2k(b) = 0, i.e., the representation |w′ 2k(t)| σ(p1k)(t) = − b t ∆p0k(s) w2k(s) σ(p1k)(s) ds − b t ∆p0k(s) v2(s) σ(p1k)(s) ds − − b t p0(s)w1k(s) + ∆p1k(s)v′ 2(s) σ(p1k)(s) ds for a < t ≤ b is valid. Repeating the arguments taking place in the proof of Lemma 2.2.1 for j = 2, we come to the conclusion that the conditions (2.2.912) and (2.2.1022) are valid. But owing to the condition p1 ∈ Lloc(]a, b]), it follows from (2.2.912) that (2.2.922) is valid. 86 Lemma 2.2.2. Let i ∈ {1, 2}, the measurable functions pj, pjk : ]a, b[ → R and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R be such that the conditions (2.2.6), (2.2.7i), (2.2.8i) and (p0, p1) ∈ Vi,0(]a, b[) (2.2.37i) are satisfied. Then there exists a number k0 such that for k > k0 (p0k, p1k) ∈ Vi,0(]a, b[). (2.2.38i) Proof. Let i = 1 and v1, v2, v1k, v2k be solutions of the problems (2.2.1), (2.2.4), (2.2.1), (2.2.51), (2.2.1k), (2.2.4k), (2.2.1k), (2.2.51k) respectively, whose existence and uniqueness follow from Remark 2.2.1. As is seen from Definition 1.1.2 of the set V1,0(]a, b[) and Remark 1.2.1, v1(b) > 0 and v1(a) > 0. Then by virtue of Remark 1.2.5 and the inclusion (2.2.37i), v1(t) + v2(t) > 0 for a ≤ t ≤ b, hence if c = min v1(t) + v2(t) : a ≤ t ≤ b , then c > 0. (2.2.39) On the other hand, by Lemma 2.2.1i, there exists a number k0 such that for any k > k0 − c 2 < vjk(t) − vj(t) (j = 1, 2) for a ≤ t ≤ b. (2.2.40) Thus for the solution vk of the equation (2.2.1k), where vk(t) = v1k(t) + v2k(t), the estimate vk(t) = v1k(t) − v1(t) + v2k(t) − v2(t) + v1(t) + v2(t) is valid from which with regard for (2.2.39) and (2.2.40) we obtain vk(t) > 0 for a ≤ t ≤ b. This inequality by virtue of Lemma 1.2.2 means that the inclusion (2.2.38i) is true. Consider now the boundary conditions u(a) = 0, u(b) = 0 (2.2.411) and u(a) = 0, u′ (b−) = 0. (2.2.412) The following Lemma is valid. 87 Lemma 2.2.3. Let i ∈ {1, 2}, the measurable functions f, pj, pjk : ]a, b[ → R and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R satisfy the conditions (2.2.6), (2.2.7i), (2.2.8i), (2.2.37i) and b a |f(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞. (2.2.42) Then there exists a number k0 such that for k > k0 the problem (2.2.1k), (2.2.41i) has a unique Green’s function Gk, and uniformly in the interval ]a, b[ lim k→∞ Iµ−1 i (σ 1−αµ 1−µ (p1))(t) b a |G(t, s) − Gk(t, s)| |f(s)| ds = 0, (2.2.43) lim k→∞ Iµ i (σα (p1))(t) σ(p1)(t) b a ∂(G(t, s) − Gk(t, s)) ∂t |f(s)| ds = 0, (2.2.44) where G is Green’s function of the problem (2.2.1), (2.2.41i). Proof. By Lemma 2.2.2i, for k > k0 the inclusion (2.2.38i) is satisfied. Then as is seen from Remark 1.2.2, the inclusions (2.2.37i) and (2.2.38i) imply the existence of the functions G and Gk, respectively, where G is defined by the equality (1.2.7), and Gk(t, s) =    − v2k(t)v1k(s) v2k(a)σ(p1k)(s) for a ≤ s < t ≤ b, − v1k(t)v2k(s) v2k(a)σ(p1k)(s) for a ≤ t < s ≤ b, (2.2.45) where v1k is the solution of the problem (2.2.1k), (2.2.4ik) and v2k is that of the problem (2.2.1k), (2.2.51k) for i = 1 and of the problem (2.2.1k), (2.2.52) for i = 2. From the estimates (1.2.10i), (1.2.11i) and the equalities (2.2.9i1), (2.2.9i2), (2.2.10i1), (2.2.10i2) it follows the existence of constants d1 and d2, such that on the interval ]a, b[ the estimates v1k(t) t a σ(p1)(s) ds −1 ≤ d1, v2k(t) b t σ(p1)(s) ds i−2 ≤ d1 for k > k0, v1(t) t a σ(p1)(s) ds −1 ≤ d1, v2(t) b t σ(p1)(s) ds i−2 ≤ d1 (2.2.46) 88 and |v′ 1k(t)| σ(p1)(t) b t σα (p1)(s) ds µ(2−i) ≤d1, |v′ 2k(t)| σ(p1)(t) t a σα (p1)(s) ds µ ≤d1 for k > k0, (2.2.47) |v′ 1(t)| σ(p1)(t) b t σα (p1)(s) ds µ(2−i) ≤d1, |v′ 2(t)| σ(p1)(t) t a σα (p1)(s) ds µ ≤d1, as well as v2k(a) ≥ d2 for k > k0, v2(a) ≥ d2 (2.2.48) are valid. Introduce now the notation w (j) lk (t) = v (j) l (t) − v (j) lk (t) (l = 1, 2; j = 0, 1; k ∈ N) and ω1k = sup |w1k(t)| t a σ(p1)(s) ds −1 : a < t ≤ b , ω2k = sup |w2k(t)| b t σ(p1)(s) ds i−2 : a ≤ t < b , ω′ 1k = sup |w′ 1k(t)| σ(p1)(t) b t σα (p1)(s) ds (2−i)µ : a < t < b , ω′ 2k = sup |w′ 2k(t)| σ(p1)(t) t a σα (p1)(s) ds µ : a < t < b . Then as is seen from Lemma 2.2.1i, lim k→∞ ωjk = 0, lim k→∞ ω′ jk = 0 (j = 1, 2). (2.2.49) It is also clear that the equality Iµ i (σα (p1))(t) σ(p1)(t) j b a ∂j ∂tj (Gk(t, s) − G(t, s)) |f(s)| ds = = Iµ i (σα (p1))(t) σ(p1)(t) j t a v (j) 2k (t)v1k(s) v2k(a)σ(p1k)(s) − v (j) 2 (t)v1(s) v2(a)σ(p1)(s) |f(s)| ds + + b t v (j) 1k (t)v2k(s) v2k(a)σ(p1k)(s) − v (j) 1 (t)v2(s) v2(a)σ(p1)(s) |f(s)| ds (j = 0, 1) (2.2.50) for a < t < b 89 is valid. Let j = 0. With regard for the inequalities (2.2.18) and (2.2.46) we obtain the estimate t a v2k(t)v1k(s) v2k(a)σ(∆p1k)(s) − v2(t)v1(s) v2(a)σ(∆p1k)(s) |f(s)| ds ≤ ≤ ε v2k(a) |w2(t)| t a |f(s)| σ(p1)(s) |v1k(s)| ds + +|v2(t)| t a |f(s)| σ(p1)(s) |w1k(s)| ds + |w2k(a)| v2(a) t a |f(s)| σ(p1)(s) |v1(s)| ds + + 1 − σ(∆p1k) C v2(a) v2(t) t a |f(s)| σ(p1)(s) |v1(s)| ds ≤ ≤ rkI1−µ i (σ 1−αµ 1−µ (p1))(t) for a ≤ t ≤ b, where rk = ε d1 d2 ω1k + ω2k 1 + d1 d2 b a σ(p1)(s) ds + d1 ε 1 − σ(∆p1k) C × × b a |f(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds and in view of the conditions (2.2.8i), (2.2.42), and (2.2.49), lim k→∞ rk = 0. (2.2.51) Having analogously estimated the second integral in (2.2.50) for j = 0, we obtain for any k > k0 Iµ−1 i (σ 1−αµ 1−µ (p1))(t) b a |G(t, s) − Gk(t, s)| |f(s)| ds ≤ 2rk for a < t < b which in view of (2.2.51) implies the validity of the condition (2.2.43). Similarly, from the equality (2.2.50) for j = 1, with regard for (2.2.18), (2.2.46) and (2.2.47), for any k > k0 we get Iµ i (σα (p1))(t) σ(p1)(t) b a ∂(G(t, s) − Gk(t, s)) ∂t |f(s)| ds ≤ rk for a < t < b, 90 where rk = 2ε d1 d2 b a σ 1−αµ 1−µ (p1)(s) ds b a |f(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds × × ω′ 1k + ω′ 2k + ω1k + ω2k 1 + d1 d2 b a σ(p1)(s) ds + d1 ε 1 − σ(∆p1k) C . By the conditions (2.2.8i), (2.2.42), and (2.2.49), lim k→∞ rk = 0 which guarantees the validity of the condition (2.2.44). Lemma 2.2.4. Let i ∈ {1, 2}, the measurable functions f, pj, pjk : ]a, b[ → R (j = 0, 1; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R satisfy conditions (2.2.6), (2.2.7i), (2.2.8i), (2.2.37i) and b a |f(s)| σ(p1)(s) Iβ i (σα (p1))(s) ds < +∞. (2.2.52) Then there exist a constant r1 ∈ R+ and a number k0 such that for k > k0 the problem (2.2.1k), (2.2.42i) has a unique Green’s function Gk, and b a Gk(t, s)f(s) ds ≤r1 max t a f(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b × ×I1−µ i (σ 1−αµ 1−µ (p1))(t) for a ≤ t ≤ b (2.2.53) and Iµ i (σα (p1))(t) σ(p1)(t) b a ∂Gk(t, s) ∂t f(s) ds ≤ ≤ r1 max t a f(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b (2.2.54) for a < t < b. Proof. In the proof of the previous lemma it has been shown that under the conditions of that lemma the problem (2.2.1k), (2.2.42i) has a unique Green’s function Gk which is represented by the equality (2.2.45). 91 Consider separately the case i = 1. First we note that in view of (2.2.12) and (2.2.17) the inequality t2 t1 σ(p1)(s) Iβ 1 (σα(p1))(s) ds ≤ t2 t1 σ 1−αµ 1−µ (p1)(s) ds 1−µ × × b a σα (p1)(s) I β µ 1 (σα(p1))(s) ds µ < +∞ for a ≤ t1 < t2 ≤ b (2.2.55) is valid. Integrating by parts and applying (2.2.48), we get b a ∂j G(t, s) ∂tj f(s) ds ≤ ≤ 2 d2 max t a f(s) σ(p1)(s) Iβ 1 (σα (p1))(s) ds : a ≤ t ≤ b × × |v (j) 2k (t)| t a v1k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds + +|v (j) 1k (t)| b t v2k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds (j = 0, 1) for a < t < b. (2.2.56) Using now the estimates (2.2.46), (2.2.55), we obtain |v2k(t)| t a v1k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds ≤ εd1 b t σ 1−αµ 1−µ (p1)(s) ds 1−µ × × t a σ(p1)(s) Iβ 1 (σα(p1))(s) v′ 1k(s) σ(p1)(s) b s σα (p1)(η) dη µ ds + +εd2 1I1−µ 1 (σ 1−αµ 1−µ (p1))(t) b a σα (p1)(s) ds 2(µ−β) b a |∆p1k(s)| ds + + b a σα (p1)(s) ds b a σα (p1)(s) I1+β−µ 1 (σα(p1))(s) ds ≤ ≤ r1I1−µ 1 (σ 1−αµ 1−µ (p1))(t) for a ≤ t ≤ b, (2.2.57) 92 where r1 = εd2 1 b a σα (p1)(s) I β µ 1 (σα(p1))(s) ds µ + + b a σα (p1)(s) ds 2(µ−β) sup b a |∆p1k(s)| ds : k > k0 + + b a σα (p1)(s) ds b a σα (p1)(s) I1+β−µ 1 (σα(p1))(s) ds . Analogously we have |v1k(t)| b t v2k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds ≤ ≤ r1I1−µ 1 (σ 1−αµ 1−µ (p1))(t) for a ≤ t ≤ b, (2.2.58) Iµ 1 (σα (p1))(t) σ(p1)(t) |v′ 2k(t)| t a v1k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds ≤ ≤ r2 for a < t < b (2.2.59) and Iµ 1 (σα (p1))(t) σ(p1)(t) |v′ 1k(t)| b t v2k(s)σ(∆p1k)(s) Iβ 1 (σα(p1))(s) ′ ds ≤ ≤ r2 for a < t < b, (2.2.60) where r2 = εd2 1 b a σ1(p1)(s) Iβ 1 (σα(p1))(s) ds + b a σ 1−αµ 1−µ (p1)(s) ds 1−µ × × sup b a |∆p1k| ds : k > k0 b a σα (p1)(s) ds 2(µ−β) + + b a σα (p1)(s) ds b a σα (p1)(s) I1+β−µ 1 (σα(p1))(s) ds . Let us now introduce the notation r1 = 4 d2 max(r1; r2). 93 Substituting the estimates (2.2.57), (2.2.58) in (2.2.56) for j = 0, we see that the condition (2.2.53) is valid. Taking then into account (2.2.59), (2.2.60) in (2.2.56) for j = 1, we are convinced of the validity of (2.2.54). For i = 2 the lemma is proved analogously. Lemma 2.2.5. Let i ∈ {1, 2}, the measurable functions pj, pjk : ]a, b[ → R (j = 0, 1; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R satisfy the conditions (2.2.6), (2.2.7i), (2.2.8i), (2.2.37i). Then there exists a number k0 such that for k > k0 the problem (2.2.1k), (2.2.41k) has a unique Green’s function Gk for which the estimate dj Gk(t, s) dtj ≤ c′ σi(p1)(s) [σi(p1)(t)]j (j = 0, 1) for a < t, s < b, t = s, (2.2.61) is valid, where c′ is a constant. Proof. The existence of Green’s function under the given conditions has been shown in Lemma 2.2.3. Similarly, by virtue of the estimate (1.2.12i) from Remark 1.2.3, dj Gk(t, s) dtj ≤ c∗ σi(p1k)(s) [σi(p1k)(t)]j (j = 0, 1) for a < t, s < b, t = s, whence with regard for the inequalities (2.2.18) and (2.2.48) follows the validity of our lemma. Consider now the equations v′′ (t) = p0(t)v(t) + p1(t)v′ (t) + p2(t), (2.2.62) v′′ (t) = p0k(t)v(t) + p1k(t)v′ (t) + p2k(t), (2.2.62k) where p2, p2k ∈ Lloc(]a, b[) (k ∈ N) and the boundary conditions u(a) = c1, u(b) = c2 (2.2.631) or u(a) = c1, u′ (b−) = c2, (2.2.632) and u(a) = c1k, u(b) = c2k (2.2.631k) or u(a) = c1k, u′ (b−) = c2k, (2.2.632k) where cl, clk ∈ R (l = 1, 2; k ∈ N). Then the following lemma is valid. 94 Lemma 2.2.6. Let i ∈ {1, 2}, the measurable functions pj, pjk : ]a, b[ → R (j = 0, 1, 2; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R satisfy the conditions (2.2.6), (2.2.7i), (2.2.8i), (2.2.37i), b a |p2(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞ (2.2.64) and uniformly on the segment [a, b] lim k→∞ t a p2(s) − p2k(s) σ(p1)(s) Iβ i (σα (p1))(s) ds = 0. (2.2.65) Then there exists a number k0 such that for k > k0: (a) the problem (2.2.62k), (2.2.41i) has a unique solution vk, and uniformly on the interval ]a, b[ lim k→∞ Iµ−1 i (σ 1−αµ 1−µ (p1))(t) v(t) − vk(t) = 0, (2.2.66) lim k→∞ v′ (t) − v′ k(t) σ(p1)(t) Iµ i (σα (p1))(t) = 0, (2.2.67) where v is a solution of the problem (2.2.61), (2.2.41i); (b) the problem (2.2.62k), (2.2.63ik) has a unique solution vk, and if lim k→∞ clk = cl (l = 1, 2), (2.2.68) then uniformly on the interval ]a, b[ the conditions (2.2.67) and lim k→∞ v(t) − vk(t) = 0 (2.2.69) are satisfied, where v is a solution of the problem (2.2.62), (2.2.63i); (c) the sequence (vk)∞ k=1, where vk is a solution of the problem (2.2.62k), (2.2.41i), ((2.2.62k), (2.2.63ik)), is uniformly bounded and equicontinuous. Proof. First we prove the validity of proposition (a). It has been mentioned in the proof of Lemma 2.2.3 that under the above-mentioned conditions the problems (2.2.1), (2.2.41i), and (2.2.1k), (2.2.41i) for k > k0 have a unique Green’s function G and Gk, respectively. Let v(t) = b a G(t, s)p2(s) ds and vk(t) = b a Gk(t, s)p2k(s) ds. Then v(j) (t) − v (j) k (t) = b a ∂j Gk(t, s) ∂tj p2(s) − p2k(s) ds + 95 + b a ∂j ∆Gk(t, s) ∂tj p2(s) ds (j = 0, 1) for a < t < b. Taking into account the equalities (2.2.43), (2.2.44) of Lemma 2.2.3 and the equalities (2.2.53), (2.2.54) of Lemma 2.2.4, by means of the conditions (2.2.64), (2.2.65) we make sure that the equalities (2.2.66) and (2.2.67) are valid. Now we proceed to proving proposition (b). Let v0 and v0k be solutions of the problems (2.2.1), (2.2.63i) and (2.2.1k), (2.2.63ik), respectively. Then v(t) = v0(t) + b a G(t, s)p2(s) ds vk(t) = v0k(t) + b a G(t, s)p2k(s) ds and v(j) (t) − v (j) k (t) = v (j) 0 (t) − v (j) 0k (t) + b a ∂j Gk(t, s) ∂tj p2(s) − p2k(s) ds + + b a ∂j ∆Gk(t, s) ∂tj p2(s) ds (j = 0, 1) for a < t < b, where v0(t) − v0k(t) = = c1 v2(t) v2(a) − c1k v2k(t) v2k(a) + c2 v1(t) v1(b) − c2k v1k(t) v1k(b) for a ≤ t < b and vj, vjk (j = 1, 2; k ≥ k0) are the solutions mentioned in Lemma 2.2.1i. It follows from the given representation, Lemma 2.2.1i and the condition (2.2.68) that uniformly in the interval ]a, b[ lim k→∞ v0(t) − v0k(t) = 0 and lim k→∞ v′ 0(t) − v′ 0k(t) σ(p1)(t) Iµ i (σα (p1))(t) = 0. Next, reasoning analogously as in proving proposition (a), we can see that the conditions (2.2.67), (2.2.69) are valid. The validity of proposition (c) follows immediately from (2.2.66) ((2.2.69)) and also from vk(t1) − vk(t2) ≤ vk(t1) − v(t1) + vk(t2) − v(t2) + v(t1) − v(t2) ≤ ≤ 2 vk − v C + v(t1) − v(t2) , where t1, t2 ∈ [a, b]. 96 Remark 2.2.2. It is not difficult to notice that if the condition (2.1.8) is satisfied, then for any fixed r ∈ R+ the equality lim k→∞ sup t a gk(x)(s) − g(x)(s) σ(p1)(s) Iµ i (σα (p1))(s) ds : a ≤ t ≤ b, x ∈ Br,k = 0 (2.2.70) is valid. The same is true for the set B′ r,k. Lemma 2.2.7. Let i ∈ {1, 2}, the measurable functions pj, pjk : ]a, b[ → R (j = 0, 1, 2; k ∈ N) and the constants α ∈ [0, 1], γ ∈ ]1, +∞[ , β, µ ∈ R satisfy the conditions (2.2.6), (2.2.7i), (2.2.8i), (2.2.37i), (2.2.64) and (2.2.65). Moreover, let continuous linear operators g, gk : C(]a, b[) → Lloc(]a, b[), be such that the condition (2.1.8) is satisfied. Then for every fixed r ∈ R+ the sequence (zk)∞ k=1 zk(t) = αkvk(t) + b a Gk(t, s)gk(xk)(s) ds, is uniformly bounded and equicontinuous, where vk is a solution of the problem (2.2.62k), (2.2.41i), Gk is the Green’s function of that problem, and for every αk ∈ [0, r], xk ∈ Br,k (k ∈ N). Proof. Introduce the notation zk(t) = b a Gk(t, s)gk(xk)(s) ds, wk(t) = b a G(t, s)g(xk)(s) ds, where G is Green’s function of the problem (2.2.62), (2.2.41i). Similarly to the proof of Lemma 1.2.4 we see that sup wk C : k ∈ N < +∞ and for any ε > 0 there exists a constant δ > 0 such that for every k ∈ N wk(t1) − wk(t2) < ε for |t1 − t2| < δ. (2.2.71) On the other hand, from the inequality zk(t) − wk(t) ≤ b a Gk(t, s) − G(t, s) g(xk)(s) ds + + b a Gk(t, s) gk(xk)(s) − g(xk)(s) ds 97 by virtue of Lemmas 2.2.3–2.2.4 and Remark 2.2.2 with all conditions satisfied, we obtain lim k→∞ zk − wk C = 0 (2.2.72) which, owing to the inequality zk(t1) − zk(t2) ≤ zk(t1) − wk(t1) + zk(t2) − wk(t2) + +|wk(t2) − wk(t1)| ≤ 2 zk − wk C + |wk(t2) − wk(t1)| with regard for (2.2.71) and (2.2.72), implies the uniform boundedness and equicontinuity of the sequence (zk)∞ k=1. This together with proposition (c) of Lemma 2.2.5 proves our lemma. Remark 2.2.3. Lemma 2.2.7 remains valid if vk is a solution of the problem (2.2.62k), (2.2.63ik), xk ∈ B′ r,k (k ∈ N) and lim k→∞ clk = cl (l = 1, 2). Lemma 2.2.8. Let functions Vk ∈ L∞(]a, b[) and Hk ∈ L([a, b]) (k ∈ N) be such that uniformly on [a, b] lim k→∞ t a Hk(s) ds = 0, (2.2.73) ess sup |Vk(t) − V(t)| : a ≤ t ≤ b → 0 as k → +∞, (2.2.74) and let there exist a function η ∈ L([a, b]) such that everywhere on the interval ]a, b[ |Hk(t)| ≤ η(t) (k ∈ N). (2.2.75) Then uniformly on the segment [a, b] lim k→∞ t a Hk(s)Vk(s) ds = 0. This lemma is a particular case of Lemma 2.1 from [19]. § 2.3. Proof of Main Results 2.3.1. Proof of Theorems 2.1.1i, 2.1.2i (i = 1, 2). Proof of Theorem 2.1.1i. From the inclusion (2.1.9), by Lemma 1.2.1 we obtain (p0, p1) ∈ Vi,0(]a, b[), which, owing to Lemma 2.2.2 for k > k0, implies (p0k, p1k) ∈ Vi,0(]a, b[). From Remark 1.2.2 follows the unique solvability of the problems (2.2.61), (2.1.2i0) and (2.2.61k), (2.1.2i0). Denote by v, vk and G, Gk, respectively, solutions and Green’s functions of these problems. 98 Then the problems (2.1.1), (2.1.2i0) and (2.1.1k), (2.1.2i0) are equivalent, respectively, to the equations u(t) = U0(u)(t) + v(t) (2.3.1) and u(t) = Uk(u)(t) + vk(t), (2.3.1k) where the continuous linear operators Uk, U0 : C(]a, b[) → C(]a, b[) are defined by the equalities U0(x)(t) = b a G(t, s)g(x)(s) ds and Uk(x)(t) = b a Gk(t, s)gk(x)(s) ds. If ρ : [a, b] → R+ is the function mentioned in the proof of Theorem 1.1.1i, then as is seen from that proof, there exists a constant λ0 ∈ [0, 1[ such that U0 Cρ→Cρ < λ0. (2.3.2) Suppose that the equation u(t) = Uk(u)(t) (2.3.10k) has a non-zero solution u0k. Not restricting the generality, we assume that u0k C,ρ = 1 for k > k0, (2.3.3) in which case u0k C ≤ ρ C, i.e., if we introduce the notation r = ρ C, then u0k ∈ Brk for k > k0. (2.3.4) Also, from (2.3.10k), (2.3.3), by Lemma 2.2.7 it follows that the sequence (u0k)∞ k=1 is uniformly bounded and equicontinuous. Hence by the Arzella– Ascoli lemma, not restricting the generality we can assume that there exists a function u0 ∈ C(]a, b[) such that uniformly on the segment [a, b] lim k→∞ u0k(t) = u0(t). (2.3.5) It is clear from the equations (2.3.3), (2.3.5) that u0 C,ρ = 1. (2.3.6) Let us now introduce the notation ∆pjk(t) = pj(t) − pjk(t) (j = 0, 1, 2), ∆Gk(t, s) = G(t, s) − Gk(t, s), ∆gk(x)(t) = g(x)(t) − gk(x)(t) (k ∈ N). 99 For u0k, when k > k0, the representation u0k(t) = U0(u0k)(t) + b a ∆Gk(t, s)g(u0k)(s) ds + + b a Gk(t, s)∆gk(u0k)(s) ds (k ∈ N) for a ≤ t ≤ b (2.3.7) is valid. Taking into account (2.3.4), (2.3.5), Remark 2.2.2, equality the (2.2.43) of Lemma 2.2.3 and also the equality (2.2.53) of Lemma 2.2.4 with all conditions satisfied, and then passing in (2.3.7) to limit as k → +∞, we get u0(t) = U0(u0)(t) which, with regard for (2.3.2), (2.3.6), results in the estimate u0 C,ρ < 1. But this contradicts (2.3.6). Hence our assumption is invalid and the equation (2.3.10k) has only the zero solution, and because of its Fredholm property the equation (2.3.1k) is uniquely solvable. The unique solvability of the equation (2.3.1) follows from Theorem 1.1.1i. Let u and uk be respectively solutions of the equations (2.3.1) and (2.3.1k), wk(t) = u(t) − uk(t) for k > k0, λk =    uk C,ρ for uk C,ρ > 1, 1 for uk C,ρ ≤ 1, (2.3.8) uk(t) = λ−1 k uk(t) and ρk(t) = v(t) − vk(t) λk + b a ∆Gk(t, s)g(uk)(s) ds + b a Gk(t, s)∆gk(uk)(s) ds. Then for wk the representation wk(t) = U0(wk)(t) + λkρk(t) for a ≤ t ≤ b (2.3.9) is valid, and if r = ρ C, then uk ∈ Br,k. (2.3.10) 100 In such a case, taking into account proposition (a) of Lemma 1.2.6, Remark 2.2.2, the equation (2.2.43) of Lemma 2.2.3 and also the equation (2.2.53) of Lemma 2.2.4, we obtain lim k→∞ ρk C,ρ = 0. (2.3.11) On the other hand, from (2.3.9), with regard for (2.3.2), we get the estimate wk C,ρ ≤ αkλk for k > k0, (2.3.12) where αk = ρk C,ρ 1 − λ0 and by virtue of (2.3.11), lim k→∞ αk = 0. (2.3.13) Suppose now that we can extract from the sequence (λk)∞ k=1 a sequence (λkm )∞ m=1 such that λkm ≥ 1 for m ∈ N and lim m→∞ λkm = +∞, (2.3.14) and note that by our definition of the function wk the inequality λkm − u C,ρ ≤ wkm C,ρ (2.3.15) is valid. Substituting now the inequality (2.3.12) in (2.3.15) and taking into account (2.3.13), we can see that this contradicts (2.3.14), i.e., our assumption is invalid, and there exists a constant λ ∈ R+ such that λk ≤ λ for k > k0 (2.3.16) which, with regard for (2.3.12), yields lim k→∞ wk C,ρ = 0. (2.3.17) Now we notice that (2.3.9) and (2.3.16) imply |w (j) k (t)| ≤ dj dtj U0(wk)(t) + λ|ρ (j) k (t)| (j = 0, 1) for a < t < b. (2.3.18j) Applying the estimates (2.2.46)–(2.2.48) and the inequalities (2.2.13), (2.2.10), we arrive at U0(wk)(t) ≤ r′ wk C,ρI1−µ i (σ 1−αµ 1−µ (p1))(t) for a ≤ t ≤ b, (2.3.19) Iµ i (σα (p1))(t) σ(p1)(t) d dt U0(wk)(t) ≤ r′ wk C,ρ for a < t < b, (2.3.20) 101 where r′ = d2 1 d2 b a h(ρ)(s) σ(p1)(s) Iµ i (σα (p1))(s) ds. By definition of the function uk, in view of the inequality (2.1.10) and the equalities (2.2.43), (2.2.44) of Lemma 2.2.3, we make sure that uniformly on the interval ]a, b[ lim k→∞ Iµ−1 i (σ 1−αµ 1−µ (p1))(t) b a ∆Gk(t, s)g(uk)(s) ds = 0 (2.3.21) and lim k→∞ Iµ i (σα (p1))(t) σ(p1)(t) b a d∆Gk(t, s) dt g(uk)(s) ds = 0. (2.3.22) Just in the same way, taking into account the inclusion (2.3.10) and the equalities (2.2.53), (2.2.54) of Lemma 2.2.4, we can see that b a Gk(t, s)∆gk(uk)(s) ds ≤ ≤ r1 sup t a ∆gk(x)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b, x ∈ Br,k × ×I1−µ i (σα (p1))(t) for a ≤ t ≤ b, (2.3.23) Iµ i (σα (p1))(t) σ(p1)(t) b a d dt Gk(t, s)∆gk(uk)(s) ds ≤ ≤ r1 sup t a ∆gk(x)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b, x ∈ Br,k for a < t < b. (2.3.24) It is clear from the equalities (2.3.21)–(2.3.24), proposition (a) of Lemma 2.2.5 and also from the condition (2.1.8) and Remark 2.2.2 that uniformly on the interval ]a, b[ lim k→∞ Iµ−1 i (σα (p1))(t)ρk(t) = 0 (2.3.25) and lim k→∞ ρk(t) σ(p1)(t) Iµ i (σα (p1))(t) = 0. (2.3.26) 102 Multiplying (2.3.180) by Iµ−1 i (σα (p1))(t) and taking into consideration (2.3.17), (2.3.19) and (2.3.25) we see that the condition (2.1.11) is valid. Analogously, multiplying (2.3.181) by σ−1 (p1)(t)Iµ i σα (p1)(t) and taking into account (2.3.17), (2.3.20) and (2.3.26), we make sure that the condition (2.1.12) is valid. Proof of Theorem 2.1.2i. Reasoning in the same way as in the previous proof for the function wk(t) = u(t) − uk(t), where uk is a solution of the problem (2.1.1k), (2.1.2ik), using Remark 2.2.3 and proposition (b) of Lemma 2.2.6, we get the equality (2.3.17) which is the same as the condition (2.1.15). The proof of the condition (2.1.12) coincides completely with its proof in Theorem 2.1.1i. 2.3.2. Proof of Corollaries. Proof of Corollary 2.1.1i. It is sufficient to show that (2.1.8) follows from (2.1.16)–(2.1.18). Suppose to the contrary that the condition (2.1.18) is violated. Then there exist ε > 0, a sequence of positive numbers (km)∞ m=1 and a sequence of functions ym ∈ Bkm (2.3.27) such that max t a ∆gkm (ym)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b > ε. (2.3.28) From (2.3.27) it follows ym(t) = α1mvkm (t) + b a Gkm (t, s)gkm (xm)(s) ds (m ∈ N), (2.3.29) where xm ∈ C(]a, b[) (m ∈ N) and 0 ≤ α1m ≤ 1 (m ∈ N), (2.3.30) xm C ≤ 1 (m ∈ N). (2.3.31) Introduce the notation zm(t) = b a Gkm (t, s)gkm (xm)(s) ds (m ∈ N) and rewrite zm as follows: zm(t) = b a Gkm (t, s)∆gkm (xm)(s) ds + b a Gkm (t, s)g(xm)(s) ds. 103 Then according to (2.1.10), (2.1.16), and (2.1.31) the inequality |z(j) m (t)| ≤ b a ∂j ∂tj ∆Gkm (t, s) η(s) + h(1)(s) ds + + b a ∂j ∂tj G(t, s) η(s) + h(1)(s) ds (j = 0, 1) (2.3.32j) is valid. By the conditions (2.1.6) and (2.1.18), b a η(s) + h(1)(s) σ(p1)(s) Iµ i (σα (p1))(s) ds < +∞ owing to which from (2.3.320), in view of the equality (2.2.43) of Lemma 2.2.3 and by Lemma 2.2.5 we obtain the existence of a constant λ1 such that zm C < λ1 (m ∈ N). (2.3.33) Consider now the case i = 1 separately. From (2.3.32j) (j = 0, 1), by Lemmas 2.2.3 and 2.2.5 and the fact that G(a, s) = G(b, s) = 0 for a < s < b we can choose for any ε0 > 0 constants m0, a1, b1, δ, where a < a1 < b1 < b, δ < min(a1 − a, b − b1), such that |zm(t)| ≤ ε0 4 , m > m0 for a ≤ t ≤ a1, b1 ≤ t ≤ b, i.e., zm(t1)−zm(t2) ≤ ε0 2 , m>m0, for a≤t1, t2 ≤a1, b1 ≤t1, t2 ≤b, (2.3.34) and Aδ < ε0 2 , where A = sup |z′ m(t)| : a1 − δ < t < b1 + δ, m > m0 < +∞, i.e., zm(t1) − zm(t2)| ≤ A|t1 − t2| < ε0 2 , m > m0 for a1 − δ < t1, t2 < b1 + δ, |t1 − t2| < δ. (2.3.35) The uniform boundedness and equicontinuity of the sequence (zm)∞ m=1 follows from (2.3.33)–(2.3.35). Then by the Arzella–Ascoli lemma, not restricting the generality, we assume that uniformly on the segment [a, b] lim m→∞ zm(t) = z(t). (2.3.36) 104 Notice now that however close may be a1 from a and b1 from b, the inequality (2.3.35) remains valid if we choose δ sufficiently small. Therefore, passing in (2.3.35) to limit, we can see that z is absolutely continuous on any segment contained in ]a, b[ , i.e., z ∈ Cloc(]a, b[) ∩ C([a, b]). (2.3.37) On the other hand, in view of (2.3.30), not restricting the generality, we can assume that lim m→∞ α1m = α0, which together with proposition (a) of Lemma 2.2.6 implies lim m→∞ α1mvkm (t) = α0v(t) uniformly on [a, b], (2.3.38) where v is a solution of the problem (2.2.62), (2.1.2i0). Further, taking into account (2.3.36)–(2.3.38) in (2.3.29), we conclude that uniformly on the segment [a, b] lim m→∞ ym(t) = y(t), (2.3.39) where y ∈ Cloc(]a, b[) ∩ C([a, b]). (2.3.40) The same takes place in the case i = 2 owing to the fact that the relations G(a, s) = 0 and ∂ ∂t G(t, s) t=b = 1 for a < s < b follow from the inequalities zm(t1) − zm(t2) ≤ ε0 2 , m > m0 for a ≤ t1, t2 ≤ a1 and zm(t1) − zm(t2) ≤ A1|t1 − t2| ≤ ε0 2 , m > m0 for a1 − δ < t1, t2 ≤ b, |t1 − t2| < δ with A1 = sup |z′ m(t)| : a1 − δ < t < b, m > m0 < +∞, and from the condition (2.3.38). Finally, the conditions (2.1.16)–(2.1.18) and (2.3.39) imply max t a ∆gkm (ym)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b ≤ ≤ max t a ∆gkm (ym − y)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b + 105 + max t a ∆gkm (y)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b ≤ ≤ b a η(s) σ(p1)(s) Iβ i (σα (p1))(s) ds ym − y C + + max t a ∆gkm (y)(s) σ(p1)(s) Iβ i (σα (p1))(s) ds : a ≤ t ≤ b → 0 as m → +∞. But this contradicts (2.3.28) and proves the validity of our corollary. Proof of Corollary 2.1.2i. Coincides completely with that of the previous corollary with the only difference that the functions vk and v in (2.3.38) are solutions of the problems (2.2.62k), (2.2.2ik) and (2.2.62), (2.1.2i), respectively, where the validity of the equality (2.3.38) follows from proposition (b) of Lemma 2.2.6. Proof of Corollary 2.1.3i. It can be easily verified that under the notation g(x)(t) = n m=1 g0m(s)x(τ0m(t)), gk(x)(t) = n m=1 gkm(t)x(τkm(t)) (2.3.41) all the requirements of Theorem 2.1.1i, except for (2.1.8), are satisfied. First we show the existence of a constant λ1 such that sup y′ σ(p1) Iµ i (σα (p1)) C : y ∈ B1k, k > k0 ≤ λ1. (2.3.42) To this end we choose arbitrarily k1 > k0 and y1 ∈ Bk1 . Then there exist α1 < 1, x1 ∈ C(]a, b[), x1 C ≤ 1 such that y1(t) = α1vk1 (t) + b a Gk1 (t, s)gk1 (x1)(s) ds, where vk1 is a solution of the problem (2.2.62k), (2.1.2i0). Next, |y′ 1(t)| ≤ |v′ k1 (t)| + b a ∂Gk1 (t, s) ∂t η(s) ds + + b a ∂G(t, s) ∂t h(1)(s) ds for a < t < b. 106 By virtue of the equality (2.2.67) of Lemma 2.2.6, there exists a constant λ2 such that for any k ≥ k0 v′ k σ(p1) Iµ i (σα (p1)) C < λ2. (2.3.43) Taking into account (2.3.43), the representation (2.2.45) of Green’s function the estimates (2.2.46)–(2.2.48), the inequality (2.2.13) and the conditions (2.1.18), (2.2.20) and (2.2.21), we make sure that the estimate (2.3.42) is valid, where λ1 =λ2+ d2 1 d2 2 b a σ 1−αµ 1−µ (p1)(s)ds 1−µ b a η(s) + h(1)(s) σ(p1)(s) Iµ i (σα (p1))(s) ds . We now notice that if lim k→∞ sup n m=1 t a g0m(s) − gkm(s) σ(p1)(s) Iβ i (σα (p1))(s)y(τkm(s)) ds : a ≤ t ≤ b, y ∈ B1k = 0 (2.3.44) and lim k→∞ sup n m=1 t a g0m(s) σ(p1)(s) Iβ i (σα (p1))(s) τ0m(s) τkm(s) y′ (η) dη ds : a ≤ t ≤ b, y ∈ B1k = 0, (2.3.45) then the condition (2.1.8) is satisfied. Reasoning analogously to the proof of Corollary 2.1.1i, we obtain that (2.3.44) is satisfied if for any y ∈ Cloc(]a, b[) ∩ C([a, b]) lim k→∞ n m=1 t a g0m(s) − gkm(s) σ(p1)(s) Iβ i (σα (p1))(s)y(τkm(s)) ds =0. (2.3.46) On the other hand, from (2.1.23) it follows that ess sup n m=1 τ0m(t) − τkm(t) : a ≤ t ≤ b → 0 as k → +∞, and hence for every y ∈ Cloc(]a, b[) ∩ C([a, b]) ess sup n m=1 y(τkm(t)) − y(τ0m(t)) : a ≤ t ≤ b → 0 as k → +∞. (2.3.47) 107 Then (2.1.21), (2.1.22), and (2.3.47) and lemma 2.2.8 imply the validity of the equality (2.3.46). The validity of the equality (2.3.45) follows from the estimate (2.3.42), the condition (2.1.23) and the inequalities t a g0m(s) σ(p1)(s) Iβ i (σα (p1))(s) τ0m(s) τkm(s) y′ (η) dη ds ≤ ≤ b a |g0m(s)| σ(p1)(s) Iµ i (σα (p1))(s) ds × × ess sup Iβ−µ i (σα (p1))(t) τ0m(t) τkm(t) σ(p1)(s)ds Iµ i (σ(p1))(s) : a ≤ t ≤ b × × y′ σ(p1) Iµ i (σα (p1)) C (m = 1, . . . , n; k ∈ N) for a ≤ t ≤ b. Proof of Corollary 2.1.4i. Coincides with the previous proof with the only difference that in the inequality (2.3.42) we will assume that y ∈ B′ 1k, i.e., the validity of (2.3.43) with vk as a solution of the problem (2.1.4k), (2.1.2ik) will be shown by means of proposition (b) of Lemma 2.2.6. Proof of Corollary 2.1.5i. It is not difficult to notice that the conditions (2.1.18), (2.1.25) yield b a |g0m(s)| σ(p1)(s) Iβ i (σα (p1))(s)ds < +∞ (m = 1, . . . , n), (2.3.48) whence, owing to the fact that β < µ, together with (2.1.24), we obtain the validity of the conditions (2.1.20), (2.1.21). That is, as it has been shown in the proof of Lemma 2.1.3i, all the requirements of Theorem 2.1.1i, except for (2.1.8), are satisfied. On the other hand, the condition (2.1.8) under the notation (2.3.41) follows from the conditions (2.3.44), (2.3.45). Repeating now word by word the proof of Corollary 2.1.3i, by the condition (2.1.26) we can see that (2.3.42) and (2.3.44) are valid. Choosing µ1 > µ so as to satisfy µ1 < 1, 1 − αµ1 1 − µ1 ≤ δ, 108 analogously to the inequalities (2.2.15),(2.2.16) we obtain b a σ(p1)(s) Iµ i (σα(p1))(s) ds ≤ 2Iµ 1 (σα (p1)) a + b 2 2−i µ1 µ1 − µ × × b a σ 1−αµ1 1−µ1 (p1)(s) ds 1−µ1 b a σα (p1)(s) ds µ1−µ < +∞. From this and also from the condition (2.1.26), owing to the absolute continuity of the Lebesgue integral it follows that ess sup n m=1 τ0m(t) τkm(t) σ(p1)(s) Iµ i (σα(p1))(s) ds : a ≤ t ≤ b → 0 (2.3.49) for k → +∞. Then the validity of the equality (2.3.45) follows from the conditions (2.3.48), (2.3.49) and also from the estimate (2.3.42) and the inequality t a g0m(s) σ(p1)(s) Iβ i (σα (p1))(s) τ0m(s) τkm(s) y′ (η) dη ds ≤ ≤ b a |g0m(s)| σ(p1)(s) Iβ i (σα (p1))(s) ds × × ess sup τ0m(t) τkm(t) σ(p1)(s) Iµ i (σα(p1))(s) ds : a ≤ t ≤ b × × y′ σ(p1) Iµ i (σα (p1)) C (m = 1, . . . , n; k ∈ N). Proof of Corollary 2.1.6i. 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Electronic Journal of Qualitative Theory of Differential Equations 2012, No. 38, 1-34; http://www.math.u-szeged.hu/ejqtde/ Two-Point Boundary Value Problems For Strongly Singular Higher-Order Linear Differential Equations With Deviating Arguments Sulkhan Mukhigulashvili∗ and Nino Partsvania Abstract For strongly singular higher-order differential equations with deviating arguments, under two-point conjugated and right-focal boundary conditions, AgarwalKiguradze type theorems are established, which guarantee the presence of Fredholm’s property for the above mentioned problems. Also we provide easily verifiable best possible conditions that guarantee the existence of a unique solution of the studied problems. 2000 Mathematics Subject Classification: 34K06, 34K10 Key words and phrases: Higher order differential equation, linear, deviating argument, strong singularity, Fredholm’s property. 1 Statement of the main results 1.1. Statement of the problems and the basic notations. Consider the differential equations with deviating arguments u(n) (t) = m j=1 pj(t)u(j−1) (τj(t)) + q(t) for a < t < b, (1.1) with the two-point boundary conditions u(i−1) (a) = 0 (i = 1, · · · , m), u(j−1) (b) = 0 (j = 1, · · · , n − m), (1.2) u(i−1) (a) = 0 (i = 1, · · · , m), u(j−1) (b) = 0 (j = m + 1, · · · , n). (1.3) Here n ≥ 2, m is the integer part of n/2, −∞ < a < b < +∞, pj, q ∈ Lloc(]a, b[) (j = 1, · · · , m), and τj :]a, b[→]a, b[ are measurable functions. By u(j−1) (a) (u(j−1) (b)) we denote the right (the left) limit of the function u(j−1) at the point a (b). Problems (1.1), (1.2), and (1.1), (1.3) are said to be singular if some or all the coefficients of (1.1) are non-integrable on [a, b], having singularities at the end-points of this segment. ∗ Corresponding author. EJQTDE, 2012 No. 38, p. 1 The linear ordinary differential equations and differential equations with deviating arguments with boundary conditions (1.2) and (1.3), and with the conditions b a (s − a)n−1 (b − s)2m−1 [(−1)n−m p1(s)]+ds < +∞, b a (s − a)n−j (b − s)2m−j |pj(s)|ds < +∞ (j = 2, · · · , m), b a (s − a)n−m−1/2 (b − s)m−1/2 |q(s)|ds < +∞, (1.4) and b a (s − a)n−1 [(−1)n−m p1(s)]+ds < +∞, b a (s − a)n−j |pj(s)|ds < +∞ (j = 2, · · · , m), b a (s − a)n−m−1/2 |q(s)|ds < +∞, (1.5) respectively, were studied by I. Kiguradze, R. P. Agarwal and some other authors (see [1], [2], [4] - [22]). The first step in studying the linear ordinary differential equations under conditions (1.2) or (1.3), in the case when the functions pj and q have strong singularities at the points a and b, i.e. when conditions (1.4) and (1.5) are not fulfilled, was made by R. P. Agarwal and I. Kiguradze in the article [3]. In this paper the Agarwal-Kiguradze type theorems are proved which guarantee Fredholm’s property for problems (1.1), (1.2), and (1.1), (1.3) (see Definition 1.1). Moreover, we establish optimal, in some sense, sufficient conditions for the solvability of problems (1.1), (1.2), and (1.1), (1.3). Throughout the paper we use the following notation. R+ = [0, +∞[; [x]+ is the positive part of number x, that is [x]+ = x+|x| 2 ; Lloc(]a, b[) (Lloc(]a, b])) is the space of functions y :]a, b[→ R, which are integrable on [a + ε, b − ε]; ([a + ε, b]) for arbitrary small ε > 0; Lα,β(]a, b[) (L2 α,β(]a, b[)) is the space of integrable (square integrable) with the weight (t − a)α (b − t)β functions y :]a, b[→ R, with the norm ||y||Lα,β = b a (s − a)α (b − s)β |y(s)|ds ||y||L2 α,β = b a (s − a)α (b − s)β y2 (s)ds 1/2 ; L([a, b]) = L0,0(]a, b[), L2 ([a, b]) = L2 0,0(]a, b[); EJQTDE, 2012 No. 38, p. 2 M(]a, b[) is the set of measurable functions τ :]a, b[→]a, b[; L2 α,β(]a, b[) (L2 α(]a, b]) is the Banach space of functions y ∈ Lloc(]a, b[) (Lloc(]a, b])), satisfying µ1 ≡ max t a (s − a)α t s y(ξ)dξ 2 ds 1/2 : a ≤ t ≤ a + b 2 + + max b t (b − s)β s t y(ξ)dξ 2 ds 1/2 : a + b 2 ≤ t ≤ b < +∞, µ2 ≡ max t a (s − a)α t s y(ξ)dξ 2 ds 1/2 : a ≤ t ≤ b < +∞. The norm in this space is defined by the equality || · ||eL2 α,β = µ1 (|| · ||eL2 α = µ2). Cn−1, m (]a, b[) (Cn−1, m (]a, b])) is the space of functions y ∈ Cn−1 loc (]a, b[) (y ∈ Cn−1 loc (]a, b])), satisfying b a |y(m) (s)|2 ds < +∞. (1.6) When problem (1.1), (1.2) is discussed, we assume that for n = 2m, the conditions pj ∈ Lloc(]a, b[) (j = 1, · · · , m) (1.7) are fulfilled, and for n = 2m + 1, along with (1.7), the conditions lim sup t→b (b − t)2m−1 t t1 p1(s)ds < +∞ (t1 = a + b 2 ) (1.8) are fulfilled. Problem (1.1), (1.3) is discussed under the assumptions pj ∈ Lloc(]a, b]) (j = 1, · · · , m). (1.9) A solution of problem (1.1), (1.2) ((1.1), (1.3)) is sought in the space Cn−1, m (]a, b[) (Cn−1, m (]a, b])). By hj :]a, b[×]a, b[→ R+ and fj : R × M(]a, b[) → Cloc(]a, b[×]a, b[) (j = 1, . . . , m) we denote the functions and the operators, respectively, defined by the equalities h1(t, s) = t s (ξ − a)n−2m [(−1)n−m p1(ξ)]+dξ , hj(t, s) = t s (ξ − a)n−2m pj(ξ)dξ (j = 2, · · · , m), (1.10) EJQTDE, 2012 No. 38, p. 3 and, fj(c, τj)(t, s)= t s (ξ−a)n−2m |pj(ξ)| τj (ξ) ξ (ξ1−c)2(m−j) dξ1 1/2 dξ (j = 1, · · · , m). (1.11) Let, moreover, m!! = 1 for m ≤ 0 1 · 3 · 5 · · · m for m ≥ 1 , if m = 2k + 1. 1.2. Fredholm type theorems. Along with (1.1), we consider the homogeneous equation v(n) (t) = m j=1 pj(t)v(j−1) (τj(t)) for a < t < b. (1.10) In the case where conditions (1.4) and (1.5) are violated, the question on the presence of the Fredholm’s property for problem (1.1), (1.2) ((1.1), (1.3)) in some subspace of the space Cn−1,m loc (]a, b[) (Cn−1,m loc (]a, b])) remains so far open. This question is answered in Theorem 1.1 (Theorem 1.2 ) formulated below which contains optimal in a certain sense conditions guaranteeing the Fredholm’s property for problem (1.1), (1.2) ((1.1), (1.3)) in the space Cn−1, m (]a, b[) (Cn−1, m (]a, b])). Definition 1.1. We will say that problem (1.1), (1.2) ((1.1), (1.3)) has the Fredholm’s property in the space Cn−1,m (]a, b[) (Cn−1,m (]a, b])), if the unique solvability of the corresponding homogeneous problem (1.10), (1.2) ((1.10), (1.3)) in that space implies the unique solvability of problem (1.1), (1.2) ((1.1), (1.3)) for every q ∈ L2 2n−2m−2, 2m−2(]a, b[) (q ∈ L2 2n−2m−2(]a, b])). Theorem 1.1. Let there exist a0 ∈]a, b[, b0 ∈]a0, b[, numbers lkj > 0, γkj > 0, and functions τj ∈ M(]a, b[) (k = 0, 1, j = 1, . . . , m) such that (t − a)2m−j hj(t, s) ≤ l0j for a < t ≤ s ≤ a0, lim sup t→a (t − a)m− 1 2 −γ0j fj(a, τj)(t, s) < +∞, (1.12) (b − t)2m−j hj(t, s) ≤ l1j for b0 ≤ s ≤ t < b, lim sup t→b (b − t)m− 1 2 −γ1j fj(b, τj)(t, s) < +∞, (1.13) and m j=1 (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! lkj < 1 (k = 0, 1). (1.14) EJQTDE, 2012 No. 38, p. 4 Let, moreover, (1.10), (1.2) have only the trivial solution in the space Cn−1,m (]a, b[). Then problem (1.1), (1.2) has the unique solution u for every q ∈ L2 2n−2m−2, 2m−2(]a, b[), and there exists a constant r, independent of q, such that ||u(m) ||L2 ≤ r||q||eL2 2n−2m−2, 2m−2 . (1.15) Corollary 1.1. Let numbers κkj, νkj ∈ R+ be such that νk1 > 2n + 2 − 2k(2m − n), νkj > 2 (k = 0, 1; j = 2, . . . , m), (1.16) lim sup t→a |τj(t) − t| (t − a)ν0j < +∞, lim sup t→b |τj(t) − t| (b − t)ν1j < +∞, (1.17) and m j=1 22m−j+1 (2m − 1)!!(2m − 2j + 1)!! κkj < 1 (k = 0, 1). (1.18) Moreover, let κ ∈ R+ , p0j ∈ Ln−j, 2m−j(]a, b[; R+ ), and − κ [(t − a)(b − t)]2n − p01(t) ≤ (−1)n−m p1(t) ≤ ≤ κ01 (t − a)n + κ11 (t − a)n−2m(b − t)2m + p01(t), (1.19) |pj(t)| ≤ κ0j (t − a)n−j+1 + κ1j (t − a)n−2m(b − t)2m−j+1 + p0j(t) (j = 2, . . ., m). (1.20) Let, moreover, (1.10), (1.2) have only the trivial solution in the space Cn−1,m (]a, b[). Then problem (1.1), (1.2) has the unique solution u for every q ∈ L2 2n−2m−2, 2m−2(]a, b[), and there exists a constant r, independent of q, such that (1.15) holds. Theorem 1.2. Let there exist a0 ∈]a, b[, numbers l0j > 0, γ0j > 0, and functions τj ∈ M(]a, b[) such that condition (1.12) is fulfilled and m j=1 (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! l0j < 1. (1.21) Let, moreover, problem (1.10), (1.3) have only the trivial solution in the space Cn−1,m (]a, b]). Then problem (1.1), (1.3) has the unique solution u for every q ∈ L2 2n−2m−2(]a, b]), and there exists a constant r, independent of q, such that ||u(m) ||L2 ≤ r||q||eL2 2n−2m−2 . (1.22) Corollary 1.2. Let numbers κ0j, ν0j ∈ R+ be such that ν01 > 2n + 2, ν0j ≥ 2 (j = 2, . . . , m), (1.23) lim sup t→a |τj(t) − t| (t − a)ν0j < +∞, (1.24) EJQTDE, 2012 No. 38, p. 5 and m j=1 22m−j+1 (2m − 1)!!(2m − 2j + 1)!! κ0j < 1. (1.25) Let, moreover, κ ∈ R+ , p0j ∈ Ln−j, 0(]a, b]; R+ ), and − κ (t − a)2n − p01(t) ≤ (−1)n−m p1(t) ≤ κ01 (t − a)n + p01(t), (1.26) |pj(t)| ≤ κ0j (t − a)n−j+1 + p0j(t) (j = 2, . . ., m). (1.27) Let, moreover, problem (1.10), (1.3) have only the trivial solution in the space Cn−1,m (]a, b]). Then problem (1.1), (1.3) has the unique solution u for every q ∈ L2 2n−2m−2(]a, b]), and there exists a constant r, independent of q, such that (1.22) holds. Theorem 1.3. Let c1 = a, c2 = b, ess sup a 0, lkj ≥ 0, and γkj > 0 (k = 0, 1; j = 1, . . ., m) such that along with m j=1 (2m − j)22m−j+1 l0j (2m − 1)!!(2m − 2j + 1)!! + 22m−j−1 (t∗ − a)γ0j l0j (2m − 2j − 1)!!(2m − 3)!! 2γ0j < 1 2 , (1.30) m j=1 (2m − j)22m−j+1 l1j (2m − 1)!!(2m − 2j + 1)!! + 22m−j−1 (b − t∗ )γ0j l1j (2m − 2j − 1)!!(2m − 3)!! 2γ1j < 1 2 , (1.31) EJQTDE, 2012 No. 38, p. 6 the conditions (t−a)2m−j hj(t, s) ≤ l0j, (t−a)m−γ0j −1/2 fj(a, τj)(t, s) ≤ l0j for a 0, 0j ≥ 0, and γ0j > 0 (j = 1, . . . , m) such that conditions (t−a)2m−j hj(t, s) ≤ l0j, (t−a)m−γ0j −1/2 fj(a, τj)(t, s) ≤ l0j for a 0, lkj ≥ 0, and γkj > 0 (k = 0, 1; j = 1, . . ., m) such that along with (1.40) and m j=1 (2m − j)22m−j+1 l1j (2m − 1)!!(2m − 2j + 1)!! + 22m−j−1 (b − t∗ )γ0j l1j (2m − 2j − 1)!!(2m − 3)!! 2γ1j < 1, (1.41) conditions (1.32), (1.33) hold. Moreover, let τj ∈ M(]a, b[) (j = 1, . . . , n) and sign[(τj(t) − t∗ )(t − t∗ )] ≥ 0 for a < t < b. (1.42) Then for every q ∈ L2 2n−2m−2, 2m−2(]a, b[) problem (1.1), (1.2) is uniquely solvable in the space Cn−1, m (]a, b[). Also, from Theorem 1.6, with n = 2, m = 1, t∗ = (a + b)/2, γ01 = γ11 = 1/2, l01 = l11 = κ0, l01 = l11 = √ 2κ1/ √ b − a, we get Corollary 1.4. Let functions p :]a, b[→ R, τ ∈ M(]a, b[) and constants κ0 > 0, κ1 > 0 be such that along with (1.36) and (1.37) the inequalities sign[(τ(t) − a + b 2 )(t − a + b 2 )] ≥ 0 for a < t < b (1.43) and 4κ0 + κ1 < 1 (1.44) hold. Then for every q ∈ L2 0, 0(]a, b[) problem (1.34), (1.35) is uniquely solvable in the space C1, 1 (]a, b[). 2 Auxiliary propositions 2.1. Lemmas on integral inequalities. Now we formulate two lemmas which are proved in [3]. Lemma 2.1. Let ∈ Cm−1 loc (]t0, t1[) and u(j−1) (t0) = 0 (j = 1, . . ., m), t1 t0 |u(m) (s)|2 ds < +∞. (2.1) Then t t0 (u(j−1) (s))2 (s − t0)2m−2j+2 ds ≤ 2m−j+1 (2m − 2j + 1)!! 2 t t0 |u(m) (s)|2 ds for t0 ≤ t ≤ t1. (2.2) EJQTDE, 2012 No. 38, p. 8 Lemma 2.2. Let u ∈ Cm−1 loc (]t0, t1[), and u(j−1) (t1) = 0 (j = 1, . . ., m), t1 t0 |u(m) (s)|2 ds < +∞. (2.3) Then t1 t (u(j−1) (s))2 (t1 − s)2m−2j+2 ds ≤ 2m−j+1 (2m − 2j + 1)!! 2 t1 t |u(m) (s)|2 ds for t0 ≤ t ≤ t1. (2.4) Let t0, t1 ∈]a, b[, u ∈ Cm−1 loc (]t0, t1[) and τj ∈ M(]a, b[) (j = 1, . . ., m). Then we define the functions µj : [a, (a + b)/2] × [(a + b)/2, b] × [a, b] → [a, b], ρk : [t0, t1] → R+ (k = 0, 1), λj : [a, b]×]a, (a + b)/2] × [(a + b)/2, b[×]a, b[→ R+, by the equalities µj(t0, t1, t) =    τj(t) for τj(t) ∈ [t0, t1] t0 for τj(t) < t0 t1 for τj(t) > t1 , ρk(t) = tk t |u(m) (s)|2 ds , λj(c, t0, t1, t) = µj (t0,t1,t) t (s − c)2(m−j) ds 1/2 . (2.5) Let also functions αj : R3 + × [0, 1[→ R+ and βj ∈ R+ × [0, 1[→ R+ (j = 1, · · · , m) be defined by the equalities αj(x, y, z, γ) = x + 2m−j y zγ (2m − 2j − 1)!! , βj(y, γ) = 22m−j−1 (2m − 2j − 1)!!(2m − 3)!! yγ √ 2γ . (2.6) Lemma 2.3. Let a0 ∈]a, b[, t0 ∈]a, a0[, t1 ∈]a0, b[, and the function u ∈ Cm−1 loc (]t0, t1[) be such that conditions (2.1) hold. Moreover, let constants l0 j > 0, l0 j ≥ 0, γ0j > 0, and functions pj ∈ Lloc(]t0, t1[), τj ∈ M(]a, b[) be such that the inequalities (t − t0)2m−1 a0 t [p1(s)]+ds ≤ l0 1, (2.7) (t − t0)2m−j a0 t pj(s)ds ≤ l0 j (j = 2, . . ., m), (2.8) (t − t0)m− 1 2 −γ0j a0 t pj(s)λj(t0, t0, t1, s)ds ≤ l0 j (j = 1, . . . , m) (2.9) hold for t0 < t ≤ a0. Then a0 t pj(s)u(s)u(j−1) (µj(t0, t1, s))ds ≤ ≤ αj(l0j, l0j, a0 − a, γ0j)ρ 1/2 0 (τ∗ )ρ 1/2 0 (t) + l0jβj(a0 − a, γ0j)ρ 1/2 0 (τ∗ )ρ 1/2 0 (a0)+ +l0j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! ρ0(a0) for t0 < t ≤ a0, (2.10) EJQTDE, 2012 No. 38, p. 9 where τ∗ = sup{µj(t0, t1, t) : t0 ≤ t ≤ a0, j = 1, . . ., m} ≤ t1. Proof. In view of the formula of integration by parts, for t ∈ [t0, a0] we have a0 t pj(s)u(s)u(j−1) (µj(t0, t1, s))ds = a0 t pj(s)u(s)u(j−1) (s)ds + + a0 t pj(s)u(s) µj (t0,t1,s) s u(j) (ξ)dξ ds = u(t)u(j−1) (t) a0 t pj(s)ds+ + 1 k=0 a0 t a0 s pj(ξ)dξ u(k) (s)u(j−k) (s)ds + a0 t pj(s)u(s) µj (t0,t1,s) s u(j) (ξ)dξ ds (2.11) (j = 2, . . . , m), and a0 t p1(s)u(s)u(µ1(t0, t1, s))ds ≤ a0 t [p1(s)]+u2 (s)ds+ + a0 t |p1(s)u(s)| µ1(t0,t1,s) s u (ξ)dξ ds ≤ u2 (t) a0 t [p1(s)]+ds+ +2 a0 t a0 s [p1(ξ)]+dξ |u(s)u (s)|ds + a0 t |p1(s)u(s)| µ1(t0,t1,s) s u (ξ)dξ ds. (2.12) On the other hand, by conditions (2.1), the Schwartz inequality and Lemma 2.1, we deduce that |u(j−1) (t)| = 1 (m − j)! t t0 (t − s)m−j u(m) (s)ds ≤ (t − t0)m−j+1/2 ρ 1/2 0 (t) (2.13) for t0 ≤ t ≤ a0 (j = 1, . . ., m). If along with this, in the case j > 1, we take into account inequality (2.8), and lemma 2.1, for t ∈ [t0, a0], we obtain the estimates u(t)u(j−1) (t) a0 t pj(s)ds ≤ (t − t0)2m−j a0 t pj(s)ds ρ0(t) ≤ l0jρ0(t), (2.14) and 1 k=0 a0 t a0 s pj(ξ)dξ u(k) (s)u(j−k) (s)ds ≤ l0j 1 k=0 a0 t |u(k) (s)u(j−k) (s)| (s − t0)2m−j ds ≤ ≤ l0j 1 k=0 a0 t |u(k) (s)|2 ds (s − t0)2m−2k 1/2 a0 t |u(j−k) (s)|2 ds (s − t0)2m+2k−2j 1/2 ≤ ≤ l0jρ0(a0) 1 k=0 22m−j (2m − 2k − 1)!!(2m + 2k − 2j − 1)!! . (2.15) EJQTDE, 2012 No. 38, p. 10 Analogously, if j = 1, by (2.7) we obtain u2 (t) a0 t [p1(s)]+ds ≤ l01ρ0(t), 2 a0 t a0 s [p1(ξ)]+dξ |u(s)u (s)|ds ≤ l01ρ0(a0) (2m − 1)22m [(2m − 1)!!]2 (2.16) for t0 < t ≤ a0. By the Schwartz inequality, Lemma 2.1, and the fact that ρ0 is nondecreasing function, we get µj (t0,t1,s) s u(j) (ξ)dξ ≤ 2m−j (2m − 2j − 1)!! λj(t0, t0, t1, s) ρ 1/2 0 (τ∗ ) (2.17) for t0 < s ≤ a0. Also, due to (2.2), (2.9) and (2.13), we have |u(t)| a0 t |pj(s)|λj(t0, t0, t1, s)ds = (t − t0)m−1/2 ρ 1/2 0 (t) a0 t |pj(s)|λj(t0, t0, t1, s)ds ≤ ≤ l0j (t − t0)γ0j ρ 1/2 0 (t), a0 t |u (s)| a0 s |pj(ξ)|λj(t0, t0, t1, ξ)dξ ds ≤ l0j a0 t |u (s)| (s − t0)m− 1 2 −γ0j ds ≤ ≤ l0j 2m−1 (a0 − a)γ0j (2m − 3)!! 2γ0j ρ 1/2 0 (a0) for t0 < t ≤ a0. From the last three inequalities it is clear that (2m − 2j − 1)!! 2m−jρ 1/2 0 (τ∗) a0 t pj(s)u(s) µj (t0,t1,s) s u(j) (ξ)dξ ds ≤ a0 t |pj(s)u(s)|λj(t0, t0, t1, s)ds ≤ ≤ |u(t)| a0 t |pj(s)|λj(t0, t0, t1, s)ds + a0 t |u (s)| a0 s |pj(ξ)|λj(t0, t0, t1, ξ)dξ ds ≤ ≤ l0j (t − t0)γ0j ρ 1/2 0 (t) + l0j 2m−1 (a0 − a)γ0j (2m − 3)!! 2γ0j ρ 1/2 0 (a0) (2.18) for t0 < t ≤ a0. Now, note that from (2.11) and (2.12) by (2.14)-(2.16) and (2.18), it immediately follows inequality (2.10). The following lemma can be proved similarly to Lemma 2.3. EJQTDE, 2012 No. 38, p. 11 Lemma 2.4. Let b0 ∈]a, b[, t1 ∈]b0, b[, t0 ∈]a, b0[, and the function u ∈ Cm−1 loc (]t0, t1[) be such that conditions (2.3) hold. Moreover, let constants l1 j > 0, l1 j ≥ 0, γ1j > 0, and functions pj ∈ Lloc(]t0, t1[), τj ∈ M(]a, b[) be such that the inequalities (t1 − t)2m−1 t b0 [p1(s)]+ds ≤ l1 1, (2.19) (t1 − t)2m−j t b0 pj(s)ds ≤ l1 j (j = 2, . . ., m), (2.20) (t1 − t)m− 1 2 −γ1j t b0 pj(s)λj(t1, t0, t1, s)ds ≤ l1 j (j = 1, . . . , m) (2.21) hold for b0 < t ≤ t1. Then t b0 pj(s)u(s)u(j−1) (µj(t0, t1, s))ds ≤ ≤ αj(l1j, l1j, b − b0, γ1j)ρ 1/2 1 (τ∗)ρ 1/2 1 (t) + l1jβj(b − b0, γ1j)ρ 1/2 1 (τ∗)ρ 1/2 1 (b0)+ +l1j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! ρ1(b0) for b0 ≤ t < t1, (2.22) where τ∗ = inf{µj(t0, t1, t) : b0 ≤ t ≤ t1, j = 1, . . . , m} ≥ t0. 2.2. Lemma on the property of functions from the space Cn−1,m (]a, b[). Lemma 2.5. Let w(t) = n−m i=1 n−m k=i cik(t)u(n−k) (t)u(i−1) (t), where Cn−1,m (]a, b[), and each cik : [a, b] → R is an (n − k − i + 1) -times continuously differentiable function. Moreover, if u(i−1) (a) = 0 (i = 1, . . . , m), lim sup t→a |cii(t)| (t − a)n−2m < +∞ (i = 1, . . . , n − m), then lim inf t→a |w(t)| = 0, and if u(i−1) (b) = 0 (i = 1, . . . , n − m), then lim inf t→b |w(t)| = 0. EJQTDE, 2012 No. 38, p. 12 The proof of this lemma is given in [9]. 2.3. Lemmas on the sequences of solutions of auxiliary problems. Now for every natural k we consider the auxiliary boundary problems u(n) (t) = m j=1 pj(t)u(j−1) (µj(t0k, t1k, t)) + qk(t) for t0k ≤ t ≤ t1k, (2.23) u(i−1) (t0k) = 0 (i = 1, . . ., m), u(j−1) (t1k) = 0 (j = 1, . . . , n − m), (2.24) where a < t0k < t1k < b (k ∈ N), lim k→+∞ t0k = a, lim k→+∞ t1k = b, (2.25) and u(n) (t) = m j=1 pj(t)u(j−1) (µj(t0k, b, t)) + qk(t) for t0k ≤ t ≤ b, (2.26) u(i−1) (t0k) = 0 (i = 1, . . . , m), u(j−1) (b) = 0 (j = 1, . . . , n − m), (2.27) where a < t0k < b (k ∈ N), lim k→+∞ t0k = a. (2.28) Throughout this section, when problems (1.1), (1.2) and (2.23), (2.24) are discussed we assume that pj ∈ Lloc(]a, b[) (j = 1, ..., m), q, qk ∈ L2 2n−2m−2, 2m−2(]a, b[), (2.29) and for an arbitrary (m − 1)-times continuously differentiable function x :]a, b[→ R, we set Λk(x)(t) = m j=1 pj(t)x(j−1) (µj(t0k, t1k, t)), Λ(x)(t) = m j=1 pj(t)x(j−1) (τj(t)). (2.30) Problems (1.1), (1.3) and (2.26), (2.27) are considered in the case pj ∈ Lloc(]a, b]) (j = 1, ..., m), q, qk ∈ L2 2n−2m−2, 0(]a, b[), (2.31) and for an arbitrary (m − 1)-times continuously differentiable function x :]a, b] → R, we set Λk(x)(t) = m j=1 pj(t)x(j−1) (µj(t0k, b, t)), Λ(x)(t) = m j=1 pj(t)x(j−1) (τj(t)). (2.32) Remark 2.1. From the definition of the functions µj (j = 1, . . ., m), the estimate |µj(t0k, t1k, t) − τj(t)| ≤ 0 for τj(t) ∈]t0k, t1k[ max{b − t1k, t0k − a} for τj(t) ∈]t0k, t1k[ follows. Thus, if conditions (2.25) hold, then lim k→+∞ µj(t0k, t1k, t) = τj(t) (j = 1, . . ., m) uniformly in ]a, b[. (2.33) EJQTDE, 2012 No. 38, p. 13 Lemma 2.6. Let conditions (2.25) hold and the sequence of the (m−1)-times continuously differentiable functions xk :]t0k, t1k[→ R, and functions x(j−1) ∈ C([a, b]) (j = 1, . . . , m) be such that lim k→+∞ x (j−1) k (t) = x(j−1) (t) (j = 1, . . . , m) uniformly in ]a, b[ (]a, b]). (2.34) Then for any nonnegative function w ∈ C([a, b]) and t∗ ∈]a, b[, lim k→+∞ t t∗ w(s)Λk(xk)(s)ds = t t∗ w(s)Λ(x)(s)ds (2.35) uniformly in ]a, b[, where Λk and Λ are defined by equalities (2.30). Proof. We have to prove that for any δ ∈]0, min{b − t∗ , t∗ − a}[, and ε > 0, there exists a constant n0 ∈ N such that t t∗ w(s)(Λk(xk)(s) − Λ(x)(s))ds ≤ ε for t ∈ [a + δ, b − δ], k > n0. (2.36) Let, now w(t∗) = max a≤t≤b w(t), and ε1 = ε 2w(t∗) m j=1 b−δ a+δ |pj(s)|ds −1 . Then from the inclusions x (j−1) k ∈ C([a + δ, b − δ]), x(j−1) ∈ C([a, b]) (j = 1, . . . , m), conditions (2.33) and (2.34), it follows the existence of such constant n0 ∈ N that |x (j−1) k (µj(t0k, t1k, s))−x(j−1) (µj(t0k, t1k, s))| ≤ ε1, |x(j−1) (µj(t0k, t1k, s))−x(j−1) (τj(s))| ≤ ε1 for t ∈ [a + δ, b − δ], k > n0, j = 1, . . . , m. Thus from the inequality |Λk(xk)(s) − Λ(x)(s)| ≤ |Λk(xk)(s) − Λk(x)(s)| + |Λk(x)(s) − Λ(x)(s)| ≤ 2ε1 m j=1 |pj(t)|, we have (2.36). The proof of the following lemma is analogous to that of Lemma 2.6. Lemma 2.7. Let conditions (2.28) hold and the sequence of the (m−1)-times continuously differentiable functions xk :]t0k, b] → R, and functions x(j−1) ∈ C([a, b]) (j = 1, . . . , m) be such that lim l→+∞ x (j−1) k (t) = x(j−1) (t) (j = 1, . . ., m) uniformly in ]a, b]. Then for any nonnegative function w ∈ C([a, b]), and t∗ ∈]a, b], condition (2.35) holds uniformly in ]a, b], where Λk and Λ are defined by equalities (2.32). Lemma 2.8. Let condition (2.25) hold, and for every natural k, problem (2.23), (2.24) have a solution uk ∈ Cn−1 loc (]a, b[), and there exist a constant r0 > 0 such that t1k t0k |u (m) k (s)|ds ≤ r2 0 (k ∈ N) (2.37) EJQTDE, 2012 No. 38, p. 14 holds, and if n = 2m + 1, let there exist constants ρj ≥ 0, ρj ≥ 0, γ1j > 0 such that ρj = sup (b − t)2m−j t t1 (s − a)pj(s)ds : t0 ≤ t < b < +∞, ρj = sup (b − t)m−γ1j −1/2 t t1 (s − a) pj(s) λj(b, t0k, t1k, s)ds : t0 ≤ t < b < +∞, (2.38) for t1 = a+b 2 , (j = 1, ..., m). Moreover, let lim k→+∞ ||qk − q||eL2 2n−2m−2, 2m−2 = 0, (2.39) and the homogeneous problem (1.10), (1.2) have only the trivial solution in the space Cn−1,m (]a, b[). Then nonhomogeneous problem (1.1), (1.2) has a unique solution u such that ||u(m) ||L2 ≤ r0, (2.40) and lim k→+∞ u (j−1) k (t) = u(j−1) (t) (j = 1, . . ., n) uniformly in ]a, b[ (2.41) (that is, uniformly on [a + δ, b − δ] for an arbitrarily small δ > 0). Proof. Suppose t1, . . ., tn are the numbers such that a + b 2 = t1 < · · · < tn < b, (2.42) and gi(t) are the polynomials of (n − 1)-th degree, satisfying the conditions gj(tj) = 1, gj(ti) = 0 (i = j; i, j = 1, . . . , n). (2.43) Then for every natural k, for the solution uk of problem (2.23), (2.24) the representation uk(t) = n j=1 uk(tj) − 1 (n − 1)! tj t1 (tj − s)n−1 (Λk(uk)(s) + qk(s))ds gj(t)+ + 1 (n − 1)! t t1 (t − s)n−1 (Λk(uk)(s) + qk(s))ds (2.44) is valid. For an arbitrary δ ∈]0, a+b 2 [, we have t1 t (s − t)n−j (qk(s) − q(s))ds = (n − j) t1 t (s − t)n−j−1 t1 s (qk(ξ) − q(ξ))dξ ds ≤ ≤ n t1 t (s − a)2m−2j ds 1/2 t1 t (s − a)2n−2m−2 t1 s (qk(ξ) − q(ξ))dξ 2 ds 1/2 ≤ EJQTDE, 2012 No. 38, p. 15 ≤ n (t1 − a)2m−2j+1 − δ2m−2j+1 1/2 ||qk − q||eL2 2n−2m−2, 2m−2 for a + δ ≤ t ≤ t1, t t1 (t − s)n−j (qk(s) − q(s))ds ≤ n (b − t1)2n−2m−2j+1 − δ2n−2m−2j+1 1/2 × ×||qk − q||eL2 2n−2m−2, 2m−2 for t1 ≤ t ≤ b − δ (j = 1, . . ., n − 1). (2.45) Hence, by condition (2.39), we find lim k→+∞ t1 t (s − t)n−j (qk(s) − q(s))ds = 0 uniformly in ]a, b[ (j = 1, . . ., n − 1). (2.46) Analogously one can show that if t0 ∈]a, b[, then lim k→+∞ t t0 (s − t0)(qk(s) − q(s))ds = 0 uniformly on I(t0), (2.47) where I(t0) = [t0, (a+ b)/2] for t0 < (a+ b)/2 and I(t0) = [(a+ b)/2, t0] for t0 > (a+ b)/2. In view of inequalities (2.37), the identities u (j−1) k (t) = 1 (m − j)! t tik (t − s)m−j u (m) k (s)ds (2.48) for i = 0, 1; j = 1, . . . , m; k ∈ N, yield |u (j−1) k (t)| ≤ rj[(t − a)(b − t)]m−j+1/2 (2.49) for t0k ≤ t ≤ t1k (j = 1, . . . , m; k ∈ N), where rj = r0 (m − j)! (2m − 2j + 1)−1/2 2 b − a m−j+1/2 (j = 1, . . ., m). (2.50) By virtue of the Arzela-Ascoli lemma and conditions (2.37) and (2.49), the sequence {uk}+∞ k=1 contains a subsequence {ukl }+∞ l=1 such that {u (j−1) kl }+∞ l=1 (j = 1, . . . , m) are uniformly convergent in ]a, b[. Suppose lim l→+∞ ukl (t) = u(t). (2.51) Then in view of (2.49), u(j−1) ∈ C([a, b]) (j = 1, . . . , m), and lim l→+∞ u (j−1) kl (t) = u(j−1) (t) (j = 1, . . . , m) uniformly in ]a, b[. (2.52) If along with this we take into account conditions (2.25) and (2.46), from (2.44) by lemma 2.6 we find u(t) = n j=1 u(tj) − 1 (n − 1)! tj t1 (tj − s)n−1 (Λ(u)(s) + q(s))ds gj(t)+ + 1 (n − 1)! t t1 (t − s)n−1 (Λ(u)(s) + q(s))ds for a < t < b, (2.53) EJQTDE, 2012 No. 38, p. 16 |u(j−1) (t)| ≤ rj[(t − a)(b − t)]m−j+1/2 for a < t < b (j = 1, . . ., m), (2.54) u ∈ Cn−1 loc (]a, b[), and lim l→+∞ u (j−1) kl (t) = u(j−1) (t) (j = 1, . . ., n − 1) uniformly in ]a, b[. (2.55) On the other hand, for any t0 ∈]a, b[ and natural l, we have (t − t0)u (n−1) kl (t) = u (n−2) kl (t) − u (n−2) kl (t0) + t t0 (s − t0)(Λk(ukl )(s) + qkl (s))ds. (2.56) Hence, due to (2.25), (2.47), (2.55), and Lemma 2.6 we get lim l→+∞ u (n−1) kl (t) = u(n−1) (t) uniformly in ]a, b[. (2.57) Now it is clear that (2.55), (2.57), and (2.37) results in (2.40) and (2.41). Therefore, u ∈ Cn−1, m loc (]a, b[). On the other hand, from (2.53) it is obvious that u is a solution of (1.1). In the case where n = 2m, from (2.54) equalities (1.2) follow, that is, u is a solution of problem (1.1), (1.2). Let us show that u is the solution of that problem in the case n = 2m + 1 as well. In view of (2.54), it suffice to prove that u(m) (b) = 0. First we find an estimate for the sequence {uk}+∞ k=1. For this, without loss of generality we assume that t1 ≤ t1k (k ∈ N). (2.58) From (2.44), by (2.39) and (2.49), it follows the existence of a positive constant ρ0, independent of k, such that |u (m+1) k (t)| ≤ ≤ ρ0 + 1 (m − 1)! t t1 (t − s)m−1 Λk(uk)(s)ds + t t1 (t − s)m−1 qk(s)ds (2.59) for t1 ≤ t ≤ t1k, and ||qk||eL2 2n−2m−2, 2m−2 ≤ ρ0, (2.60) for k ∈ N. On the other hand, it is evident that t t1 (t − s)m−1 Λk(uk)(s)ds ≤ m j=1 t t1 (t − s)m−1 pj(s)u (j−1) k (s)ds + + m j=1 t t1 (t − s)m−1 pj(s) µj (t0k,t1k,s) s u (j) k (ξ)dξ ds (2.61) for t1 ≤ t ≤ t1k (k ∈ N). EJQTDE, 2012 No. 38, p. 17 Let, now m > 1. From Lemma 2.2 and condition (2.37) we get the estimates t t1 |u (j) k (s)|2 (b − s)2m−2j ds ≤ t1k t0 |u (j) k (s)|2 (t1k − s)2m−2j ds ≤ 22m r2 0 (2.62) for t1 ≤ t ≤ t1k (j = 1, . . . , m). Then by conditions (2.38) we find t t1 (t − s)m−1 pj(s)u (j−1) k (s)ds = = t t1 1 (b − s)2m−j ∂ ∂s (t − s)m−1 u (j−1) k (s) s − a (b − s)2m−j s t1 (ξ − a)pj(ξ)dξ ds ≤ ≤ 4mρj b − a t t1 |u (j−1) k (s)| (b − s)m−j+2 ds + t t1 |u (j) k (s)| (b − s)m−j+1 ds ≤ ≤ 4mρj b − a t t1 (u (j−1) k (s))2 (b − s)2m−2j+2 ds 1/2 + t t1 (u (j) k (s))2 (b − s)2m−2j ds 1/2 × × t t1 (b − s)−2 ds 1/2 ≤ 2m mr0ρj b − a (b − t)−1/2 (2.63) for t1 ≤ t ≤ t1k (j = 1, . . ., m). On the other hand, by the Schwartz inequality, the definition of the functions µj and (2.4) it is clear that µj (t0k,t1k,s) s u (j) k (ξ)dξ ≤ 2m−j (2m − 2j − 1)!! λj(b, t0k, t1k, s) t1k t0k |u (m) k (ξ)|2 dξ 1/2 ≤ ≤ 2m r0λj(b, t0k, t1k, s) (2.64) for t1 < s ≤ t1k (j = 1, . . . , m). Then by the integration by parts and (2.38), (2.64) we get t t1 (t − s)m−1 pj(s) µj (t0k,t1k,s) s u (j) k (ξ)dξ ds ≤ ≤ 2m r0 t t1 ∂ ∂s (t − s)m−1 s − a s t1 (ξ − a)|pj(ξ)|λj(b, t0k, t1k, ξ)dξ ds ≤ 2m r0× ×ρj t t1 ∂ ∂s (t − s)m−1 s − a (b − s)γ1j −m+1/2 ds ≤ 2m r0ρj× (2.65) EJQTDE, 2012 No. 38, p. 18 × t t1 m − 1 s − a + t − a (s − a)2 (b − s)γ1j −3/2 ds ≤ (m + 1)2m+1 r0ρj(b − a)γ1j b − a × × t t1 (b − s)−3/2 ds ≤ (m + 1)2m+2 r0(b − a)γ1j ρj b − a (b − t)−1/2 for t1 < s ≤ t1k (j = 1, . . . , m). Thus from (2.61), by (2.63) and (2.65) we have t t1 (t − s)m−1 Λk(uk)(s)ds ≤ κ0(b − t)−1/2 (2.66) for t1 ≤ t ≤ t1k, m > 1, where κ0 = r0(m+1)2m+2 b−a m j=1 (ρj + ρj(b − a)γ1j ). Let, now m = 1, then due to (2.37), (2.38), and (2.64) we obtain t t1 (t − s)m−1 Λk(uk)(s)ds = t t1 p1(s)uk(s)ds+ + t t1 p1(s) µ1(t01,t1k,s) s uk(ξ)dξ ds ≤ |uk(t)| (t − a) t t1 (s − a)p1(s)ds + + t t1 |uk(s)| (s − a)(b − s) + |uk(s)| (s − a)2(b − s) (b − s) s t1 (ξ − a)p1(ξ)dξ ds + + 2r0 t1 − a t t1 (s − a)|p1(s)|λ1(b, t01, t1k, s)ds ≤ 2ρ1 b − a |uk(t)| b − t + +r0 t t1 1 (b − s)2 ds 1/2 + 2 b − a (t − t1)1/2 t t1 u2 k(s) (b − s)2 ds 1/2 + + 4r0 ρ1 b − a (b − t)γ11−1/2 for t1 ≤ t ≤ t1k. (2.67) On the other hand, from (2.24), (2.37), and Lemma 2.2 it follow the estimates |uk(t)| = t1k t uk(s)ds ≤ (t1k − t) t1k t u 2 k (s)ds 1/2 ≤ r0(b − t)1/2 , t1k t u2 k(s) (b − s)2 ds ≤ t1k t u2 k(s) (t1k − s)2 ds ≤ 2r0, EJQTDE, 2012 No. 38, p. 19 for t1 ≤ t ≤ t1k. Then from (2.67) by these inequalities we get t t1 (t − s)m−1 Λk(uk)(s)ds ≤ 2ρ1 b − a 2r0 (b − t)1/2 + 4r0 (b − a)1/2 + + 4r0ρ1 (b − a) (b − t)γ11−1/2 ≤ κ1((b − t)−1/2 + (b − t)γ11−1/2 ) + κ2, (2.68) where κ1 = 4r0 b−a (ρ1 + ρ1), κ2 = 8r0 (b−a)3/2 ρ1. If m > 1, due to conditions (2.60) and the fact that n = 2m + 1, we have t t1 (t − s)m−1 qk(s)ds = (m − 1) t t1 (t − s)2m−n−1 (t − s)n−m−1 s t1 |qk(ξ)|dξ ds ≤ ≤ m(b − t)−1/2 ||qk||eL2 2n−2m−2,2m−2 ≤ mρ0(b − t)−1/2 for t1 ≤ t < b, (2.69) and if m = 1, t t1 s t1 qk(ξ)dξ ds ≤ (b − t)1/2 ||qk||eL2 0,0 ≤ ρ0(b − t)1/2 for t1 ≤ t < b. (2.70) Also it is clear that u (m) k (t) = t t1k u (m+1) k (s)ds, (2.71) since u (m) k (t1k) = 0. Now, from (2.59), by (2.66) and (2.69) if m > 1, and by (2.68) if m = 1, we have, respectively, |u(m+1) (t)| ≤ ρ0 + (mρ0 + κ0)(b − t)−1/2 , |u(m+1) (t)| ≤ ρ0 + κ2 + κ1[(b − t)−1/2 + (b − t)γ11−1/2 ] + t t1 |qk(s)|ds, (2.72) for t1 ≤ t ≤ t1k. From (2.71), by (2.72), and (2.70), it follows the existence of a constant ρ∗ > 0 such that |u (m) k (t)| ≤ ρ∗ [(b − t)1/2 + (b − t)γ11+1/2 ] for t1 ≤ t < t1k, m ≥ 1, from which, in view of (2.25), (2.55), and (2.57), it is evident that u(m) (b) = 0. Thus we have proved that u is the solution of problem (1.1), (1.2) also in the case n = 2m + 1. To complete the proof of the lemma, it remains to show that equality (2.41) is satisfied. First note that in the space Cn−1,m (]a, b[) problem (1.1), (1.2) does not have another solution since in that space the homogeneous problem (1.10), (1.2) has only the trivial EJQTDE, 2012 No. 38, p. 20 solution. Now assume the contrary. Then there exist δ ∈]0, b−a 2 [, ε > 0, and an increasing sequence of natural numbers {kl}+∞ l=1 such that max n j=1 |u (j−1) kl (t) − u(j−1) (t)| : a + δ ≤ t ≤ b − δ > ε (l ∈ N). (2.73) By virtue of the Arzela-Ascoli lemma and condition (2.37) the sequence {u (j−1) kl }+∞ l=1 (j = 1, . . . , m), without loss of generality, can be assumed to be uniformly converging in ]a, b[. Then, in view of what we have shown above, conditions (2.55) and (2.57) hold. But this contradicts condition (2.73). The obtained contradiction proves the validity of the lemma. Analogously we can prove the following lemma if we apply Lemma 2.7 instead of Lemma 2.6. Lemma 2.9. Let condition (2.28) hold, for every natural k problem (2.26), (2.27) have a solution uk ∈ Cn−1 loc (]a, b]), and let there exist a constant r0 > 0 such that b t0k |u (m) k (s)|ds ≤ r2 0 (k ∈ N), (2.74) lim k→+∞ ||qk − q||eL2 2n−2m−2 = 0, (2.75) and the homogeneous problem (1.10), (1.3) has only the trivial solution in the space Cn−1,m (]a, b]). Then the nonhomogeneous problem (1.1), (1.3) has a unique solution u such that inequality (2.40) holds, and lim k→+∞ u (j−1) k (t) = u(j−1) (t) (j = 1, . . ., n) uniformly in ]a, b] (2.76) (that is, uniformly on [a + δ, b] for an arbitrarily small δ > 0). To prove Lemma 2.11 we need the following proposition, which is a particular case of Lemma 4.1 in [8]. Lemma 2.10. If u ∈ Cn−1 loc (]a, b[), then for any s, t ∈]a, b[ the equality (−1)n−m t s (ξ − a)n−2m u(n) (ξ)u(ξ)dξ = wn(t) − wn(s) + νn t s |u(m) (ξ)|2 dξ (2.77) is valid, where ν2m = 1, ν2m+1 = 2m+1 2 , w2m(t) = m j=1 (−1)m+j−1 u(2m−j) (t)u(t), w2m+1(t) = m j=1 (−1)m+j [(t − a)u(2m+1−j) (t) − ju(2m−j) (t)]u(j−1) (t) − t − a 2 |u(m) (t)|2 . EJQTDE, 2012 No. 38, p. 21 Lemma 2.11. Let a0 ∈]a, b[, b0 ∈]a0, b[, the functions hj and the operators fj be given by equalities (1.10) and (1.11). Let, moreover, τj ∈ M(]a, b[), and the constants lk,j > 0, γkj > 0 (k = 0, 1; j = 1, . . ., m) be such that conditions (1.12)-(1.14) are fulfilled. Then there exist positive constants δ and r1 such that if a0 ∈]a, a + δ[, b0 ∈]b − δ, b[, t0 ∈ ]a, a0[, t1 ∈]b0, b[, and q ∈ L2 2n−2m−2, 2m−2(]a, b[), an arbitrary solution u ∈ Cn−1 loc (]a, b[) of the problem u(n) (t) = m j=1 pj(t)u(j−1) (µj(t0, t1, t)) + q(t), (2.78) u(i−1) (t0) = 0 (i = 1, . . . , m), u(j−1) (t1) = 0 (j = 1, . . . , n − m) (2.79) satisfies the inequality t1 t0 |u(m) (s)|2 ds ≤ ≤ r1 m j=1 b0 a0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds + ||q||2 eL2 2n−2m−2, 2m−2 . (2.80) Proof. From conditions (1.12) and (1.13) it follows the existence of constants kj ≥ 0 such that (t − a)m− 1 2 −γ0j fj(a, τj)(t, s) ≤ 0j for a < t ≤ s ≤ a0, (b − t)m− 1 2 −γ1j fj(b, τj)(t, s) ≤ 1j for b0 ≤ s ≤ t < b. Consequently, all the requirements of Lemma 2.3 with pj(t) = (t − a)n−2m (−1)n−m pj(t), a < t0 < a0, and Lemma 2.4 with pj(t) = (b − t)n−2m (−1)n−m pj(t), b0 < t1 < b, are fulfilled. Also from condition (1.14) and the definition of a constant νn, it follows the existence of ν ∈]0, 1[ such that (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! kj < νn − 2ν (k = 0, 1). (2.81) On the other hand, without loss of generality we can assume that a0 ∈]a, a + δ[ and b0 ∈]b − δ, b[, where δ is a constant such that m j=1 (l0jβj(δ, γ0j) + l1jβj(δ, γ1j)) < ν, (2.82) where the functions βj are defined by (2.6). Let now q ∈ L2 2n−2m−2, 2m−2(]a, b[), u be a solution of problem (2.78), (2.79), and r1 = 22m+1 (1 + b − a)2 ν−2 . (2.83) EJQTDE, 2012 No. 38, p. 22 Multiplying both sides of (2.78) by (−1)n−m (t − a)n−2m u(t) and then integrating from t0 to t1, by Lemma 2.10 we obtain (n − 2m) t0 − a 2 |u(m) (t0)|2 + νn t1 t0 |u(m) (s)|2 ds = = (−1)n−m m j=1 t1 t0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds+ +(−1)n−m t1 t0 (s − a)n−2m q(s)u(s)ds. (2.84) From Lemma 2.3 with pj(t) = (t − a)n−2m (−1)n−m pj(t), Lemma 2.4 with pj(t) = (b − t)n−2m (−1)n−m pj(t), and the equalities ρ0(t0) = ρ1(t1) = 0, by (2.81) we get (−1)n−m m j=1 a0 t0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds ≤ ≤ (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! l0jρ0(a0) + m j=1 l0jβj(a − a0, γ0j)ρ0(τ∗ ) ≤ ≤ (νn − 2ν)ρ0(a0) + m j=1 l0jβj(δ, γ0j) t1 t0 |u(m) (s)|2 ds, (2.85) (−1)n−m m j=1 t1 b0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds ≤ ≤ (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! l1jρ1(b0) + m j=1 l1jβj(b0 − b, γ1j)ρ1(τ∗) ≤ ≤ (νn − 2ν)ρ1(b0) + m j=1 l1jβj(δ, γ1j) t1 t0 |u(m) (s)|2 ds. (2.86) If along with this we take into account inequalities (2.82) and a0 ≤ b0, we find (−1)n−2m m j=1 t1 t0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds ≤ EJQTDE, 2012 No. 38, p. 23 ≤ m j=1 b0 a0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds + +(νn − 2ν) ρ0(a0) + ρ1(b0) + ν t1 t0 |u(m) (s)|2 ds ≤ (νn − ν) t1 t0 |u(m) (s)|2 ds+ + m j=1 b0 a0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, t1, s))ds . (2.87) On the other hand, if we put c = (a + b)/2, then again on the basis of Lemmas 2.1, 2.2, and Young’s inequality we get t1 t0 (s − a)n−2m q(s)u(s)ds ≤ c t0 (s − a)n−2m q(s)u(s)ds + t1 c (s − a)n−2m q(s)u(s)ds = = c t0 [(n − 2m)u(s) + (s − a)n−2m u (s)] c s q(ξ)dξ ds + + t1 c [(n − 2m)u(s) + (s − a)n−2m u (s)] s c q(ξ)dξ ds ≤ ≤ (n − 2m) c t0 u2 (s) (s − a)2m ds 1/2 + c t0 u 2 (s) (s − a)2m−2 ds 1/2 × × c t0 (s − a)2n−2m−2 c s q(ξ)dξ 2 ds 1/2 + +(1 + b − a) (n − 2m) t1 c u2 (s) (b − s)2m ds 1/2 + t1 c u 2 (s) (b − s)2m−2 ds 1/2 × × t1 c (b − s)2m−2 s c q(ξ)dξ 2 ds 1/2 ≤ 2m+1 (1 + b − a)||q||eL2 2n−2m−2, 2m−2 × × c t0 |u(m) (s)|2 ds 1/2 + t1 c |u(m) (s)|2 ds 1/2 ≤ ≤ ν 2 t1 t0 |u(m) (s)|2 ds + 22m+3 (1 + b − a)2 ν−1 ||q||2 eL2 2n−2m−2, 2m−2 . (2.88) In view of inequalities (2.87), (2.88) and notation (2.83), equality (2.84) results in estimate (2.80). EJQTDE, 2012 No. 38, p. 24 The proof of the following lemma is analogous to that of Lemma 2.11. Lemma 2.12. Let a0 ∈]a, b[, the functions hj and the operators fj be given by equalities (1.10) and (1.11). Let, moreover, τj ∈ M(]a, b]), constants l0,j > 0, γ0j > 0, (j = 1, . . . , m) be such that conditions (1.12) and (1.21) are fulfilled. Then there exists a positive constant r1 such that for any t0 ∈]a, a0[, and q ∈ L2 2n−2m−2(]a, b]), an arbitrary solution u ∈ Cn−1 loc (]a, b]) of the problem u(n) (t) = m j=1 pj(t)u(j−1) (µj(t0, b, t)) + q(t), (2.89) u(i−1) (t0) = 0 (i = 1, . . . , m), u(j−1) (b) = 0 (j = m + 1, . . . , n) (2.90) satisfies the inequality b t0 |u(m) (s)|2 ds ≤ r1 m j=1 b a0 (s − a)n−2m pj(s)u(s)u(j−1) (µj(t0, b, s))ds + ||q||2 eL2 2n−2m−2 . Lemma 2.13. Let τj ∈ M(]a, b[), a0 ∈]a, b[, b0 ∈]a0, b[, conditions (1.7), (1.12)- (1.14), hold, and let in the case when n is odd, in addition (1.8) be fulfilled, where the functions hj, βj and the operators fj are given by equalities (1.10)-(1.11), and lkj, lkj, γkj (k = 0, 1; j = 1, . . . , m) are nonnegative numbers. Moreover, let the homogeneous problem (1.10), (1.2) in the space Cn−1,m (]a, b[) have only the trivial solution. Then there exist δ ∈ ]0, b−a 2 [ and r > 0 such that for any t0 ∈]a, a+δ], t1 ∈]b+δ, b], and q ∈ L2 2n−2m−2, 2m−2(]a, b[) problem (2.78), (2.79) is uniquely solvable in the space Cn−1 (]a, b[), and its solution admits the estimate t1 t0 |u(m) (s)|2 ds 1/2 ≤ r||q||eL2 2n−2m−2, 2m−2 . (2.91) Proof. First note that all the requirements of Lemma 2.11 are fulfilled, and in view of (1.8) and (1.13), conditions (2.38) of Lemma 2.8 hold. Let, now δ ∈]0, min{b−b0, a0 −a}] be such as in Lemma 2.11 and assume that estimate (2.91) is invalid. Then for an arbitrary natural k there exist t0k ∈]a, a + δ/k[, t1k ∈]b − δ/k, b[, (2.92) and a function qk ∈ L2 2n−2m−2, 2m−2(]a, b[) such that problem (2.23), (2.24) has a solution uk ∈ Cn−1 (]a, b[), satisfying the inequality t1k t0k |u (m) k (s)|ds 1/2 > k||qk||eL2 2n−2m−2, 2m−2 . (2.93) In the case when the homogeneous equation u(n) (t) = m j=1 pj(t)u(j−1) (µj(t0k, t1k, t)) (2.330) EJQTDE, 2012 No. 38, p. 25 under the boundary conditions (2.24) has a nontrivial solution, in (2.23) we put that qk(t) ≡ 0 and assume that uk is that nontrivial solution of problem (2.330), (2.24). Let now vk(t) = t1k t0k |u (m) k (s)|ds −1/2 uk(t), q0k(t) = t1k t0k |u (m) k (s)|ds −1/2 qk(t). (2.94) Then vk is a solution of the problem v(n) (t) = m i=1 pi(t)v(i−1) (µi(t0k, t1k, t)) + q0k(t) for t0k ≤ t ≤ t1k, v(i−1) (t0k) = 0 (i = 1, . . ., m), v(i−1) (t1k) = 0 (i = 1, . . ., n − m). (2.95) Moreover, in view of (2.93), it is clear that t1k t0k |v (m) k (s)|2 ds = 1, ||q0k||eL2 2n−2m−2, 2m−2 < 1 k (k ∈ N). (2.96) On the other hand, in view of the fact that problem (1.10), (1.2) has only the trivial solution in the space Cn−1,m (]a, b[), by Lemmas 2.8, 2.11, and (2.96) we have lim t→+∞ v (j−1) k (t) = 0 uniformly in ]a, b[ (j = 1, . . .n), 1 < r0 b0 a0 (s − a)n−2m Λk(vk)(s)ds + k−2 (k ∈ N), (2.97) where r0 is a positive constant independent of k. Now, if we pass to the limit in (2.97) as k → +∞, by Lemma 2.6 we obtain the contradiction 1 < 0. Consequently, for any solution of problem (2.78), (2.79), with arbitrary q ∈ L2 2n−2m−2, 2m−2(]a, b[), estimate (2.91) holds. Thus the homogeneous equation v(n) (t) = m j=1 pj(t)v(j−1) (µj(t0, t1, t)) for t0 ≤ t ≤ t1, (2.820) under conditions (2.79), has only the trivial solution. But for arbitrarily fixed t0 ∈ ]a, a + δ[, t1 ∈]b − δ, b[, and q ∈ L([t0, t1]) problem (2.78), (2.79) is regular and has the Fredholm property in the space Cn−1 (]t0, t1[). Thus problem (2.78), (2.79) is uniquely solvable. Analogously we can prove the following lemma if we apply Lemmas 2.7 and 2.12 instead of Lemmas 2.6 and 2.11. Lemma 2.14. Let τj ∈ M(]a, b[), a0 ∈]a, b[, conditions (1.9), (1.12) and (1.21) hold, where the functions hj, βj and the operators fj are given by equalities (1.10)-(1.11), and l0j, l0j γ0j (j = 1, . . . , m) are nonnegative numbers. Let, moreover, the homogeneous problem (1.10), (1.3) in the space Cn−1 (]a, b]) have only the trivial solution. Then there exist EJQTDE, 2012 No. 38, p. 26 positive constants δ and r such that if a0 ∈]a, a + δ[, and q ∈ L2 2n−2m−2(]a, b]), problem (2.89), (2.90) is uniquely solvable in the space Cn−1 (]a, b]), and its solution admits the estimate b t0 |u(m) (s)|2 ds ≤ r||q||eL2 2n−2m−2 . Lemma 2.15. Let τj ∈ M(]a, b[), α ≥ 0, β ≥ 0, and let there exist δ ∈]0, b−a[ such that |τj(t) − t| ≤ k1(t − a)β for a < t ≤ a + δ. (2.98) Then τ(t) t (s − a)α ds ≤ k1[1 + k1δβ−1 ]α (t − a)α+β for β ≥ 1 k1[δ1−β + k1]α (t − a)αβ+β for 0 ≤ β < 1 , for a < t ≤ a + δ. Proof. First note that τ(t) t (s − a)α ds ≤ (max{τ(t), t} − a)α |τ(t) − t| for a ≤ t ≤ a + δ, and max{τ(t), t} ≤ t + |τ(t) − t| for a ≤ t ≤ a + δ. Then in view of condition (2.98) we get τ(t) t (s − a)α ds ≤ k1[(t − a) + k1(t − a)β ]α (t − a)β for a ≤ t ≤ a + δ. From this inequality it immediately follows the validity of the lemma. Analogously, one can prove Lemma 2.16. Let τj ∈ M(]a, b[), α ≥ 0, β ≥ 0 and let there exist δ ∈]0, b − a[ such that |τj(t) − t| ≤ k1(b − t)β for b − δ ≤ t < b. (2.99) Then τ(t) t (b − t)α ds ≤ k1[1 + k1δβ−1 ]α (b − t)α+β for β ≥ 1 k1[δ1−β + k1]α (b − t)αβ+β for 0 ≤ β < 1 , for b − δ ≤ t < b. 3 Proofs Proof of Theorem 1.1 (Theorem 1.2). Suppose problem (1.10), (1.2) (problem (1.10), (1.3)) has only the trivial solution, and r and δ are the numbers appearing in Lemma 2.13 (Lemma 2.14). Set t0k = a + δ/k t1k = b − δ/k (k ∈ N). (3.1) EJQTDE, 2012 No. 38, p. 27 By Lemma 2.13 (Lemma 2.14), for every natural k, problem (2.78), (2.79) in the space Cn−1 loc (]a, b[) (problem (2.89), (2.90) in the space Cn−1 loc (]a, b])) has a unique solution uk, and t1k t0k |u (m) k (s)|2 ds 1/2 ≤ r||q||eL2 2n−2m−2,2m−2 b t0k |u (m) k (s)|2 ds 1/2 ≤ r||q||eL2 2n−2m−2 , (3.2) where the constant r does not depend on q. From (3.2), by Lemma 2.8 with r0 = r||q||eL2 2n−2m−2, 2m−2 (by Lemma 2.9 with r0 = r||q||eL2 2n−2m−2 ), it follows that problem (1.1), (1.2) (problem (1.1), (1.3)) in the space Cn−1 loc (]a, b[) (Cn−1 loc (]a, b])) is uniquely solvable for an arbitrary q ∈ L2 2n−2m−2, 2m−2(]a, b[) (q ∈ L2 2n−2m−2(]a, b])). Thus that problem has Fredholm’s property, and its solution admits estimate (1.15) (estimate (1.22)). Proof of Corollary 1.1. In view of conditions (1.18), there exists a number ε > 0 such that m j=1 (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! κkj 2m − j + ε < 1 (k = 0, 1). (3.3) On the other hand, in view of conditions (1.19) and (1.20) we have (t − a)2m−j hj(t, s) ≤ κ0j 2m − j + κ1j a0 a (ξ − a)2m−j (b − ξ)2m+1−j dξ + a0 a (ξ − a)n−j p0j(ξ)dξ for a < t ≤ s ≤ a0, (b − t)2m−j hj(t, s) ≤ κ1j 2m − j + κ0j b b0 (b − ξ)2m−j (ξ − a)2m−j+1 dξ+ +(b − a)n−2m b b0 (b − ξ)2m−j p0j(ξ)dξ for b0 ≤ s ≤ t < b. (3.4) Let δ be the constant defined in Lemmas 2.15, 2.16. From (1.19) it follows the existence of a0 ∈]a, a + δ[ and b0 ∈]b − δ, b[ such that |p1(t)| ≤ κ [(t − a)(b − t)]2n + p01(t) for t ∈ [a, a0] ∪ [b0, b]. (3.5) On the other hand, from lemmas 2.15, and 2.16 by the condition (1.17) it follows the existence of a constant k0 such that τj(t) t (s − a)2(m−j) ds 1/2 ≤ k 1/2 0 (s − a)m−j+ν0j/2 for a ≤ t ≤ a0, τj (t) t (b − s)2(m−j) ds 1/2 ≤ k 1/2 0 (b − s)m−j+ν1j/2 for b0 ≤ t ≤ b. (3.6) EJQTDE, 2012 No. 38, p. 28 Consequently, if p01 ∈ Ln−j, 2m−j(]a, b[), then by (1.16) and (3.6), from (1.19) and (1.20) it follows the existence of a nonnegative constant k2 such that (t − a)m−1 fj(a, τ1)(t, s) ≤ k2(a0 − a)ε0 for a ≤ t < s ≤ a0, (b − t)m−1 fj(b, τ1)(t, s) ≤ k2(b − b0)ε0 for b0 ≤ s < t ≤ b, (3.7) where 0 < ε0 = min{νk1 −2n−2+2k(2m−n), νkj −2 : k = 0, 1; j = 2, . . . , m}. Now, from (3.4), and (3.7) it is clear that we can choose δ1 ≤ δ so that if max{b − b0, a0 − a} ≤ δ1, then (t − a)2m−j hj(t, s) ≤ κ0j 2m − j + ε for a < t ≤ s ≤ a0, (b − t)2m−j hj(t, s) ≤ κ1j 2m − j + ε for b0 ≤ s ≤ t < b, j ∈ {1, . . ., m}. From (3.7), the last inequalities and (3.3), it is clear that all the assumptions of Theorem 1.1, with kj = κkj 2m−j + ε, γkj = 1/2, and max{b − b0, a0 − a} ≤ δ1, are fulfilled, and thus the corollary is valid. Proof of Theorem 1.3. It suffice to show that if u ∈ Cn−1 loc (]a, b[) (u ∈ Cn−1 loc (]a, b])) is a solution of problem (1.10), (1.2) ((1.10), (1.3)), then b a |u(m) (s)|2 ds < +∞. (3.8) For an arbitrary t0 ∈]a, b[ we have u(m) (t) = w(t0) + 1 (n − m − 1)! t t0 (t − s)n−m−1 m j=1 pj(s)u(j−1) (s) ds, + + 1 (n − m − 1)! t t0 (t − s)n−m−1 m j=1 pj(s) τj (s) s u(j) (ξ)dξ ds, (3.9) where w(t0) = n j=m+1 (t0−a)j−m−1 (j−m−1)! u(j−1) (t0). Now note that by the equalities |u(i) (t)| = 1 (k − i − 1)! t c (t − s)k−i−1 u(k) (s)ds for a < t < b, (3.10) k = 1, . . . , m, i = 0, . . . , k − 1, with c = a, from (3.9) we get the estimate |u(m) (t)| ≤ |w(t0)| + (1 − δ1m)||u(m−1) ||C m−1 j=1 t0 t (s − a)n−j−1 |pj(s)|ds+ + t0 t (s − a)n−m−1 |pj(s)| τj (s) s (ξ − a)m−j−1 dξ ds + +||u(m−1) ||C t0 t (s − a)n−m−1 |pm(s)|ds for a < t < t0, (3.11) EJQTDE, 2012 No. 38, p. 29 where δij is Kronecker’s delta. Then conditions (1.28) yield |u(m) (t)| ≤ |w(t0)| + (1 − δ1m)||u(m−1) ||C t0 t (s − a)−1 p(s)ds+ +γ||u(m−1) ||C t0 t p(s)ds + ||u(m−1) ||C t0 t (s − a)n−m−1 |pm(s)|ds for a < t < t0, where p(t) = m j=1 (t − a)n−j |pj(t)|, γj = ess sup a 0 such that a+δ a p(s)ds < 1 2 . (3.14) From (3.9), by conditions (1.28), (3.12) and inequality (3.13), we get |u(m) (t)| ≤ |w(t0)|+ t0 t p(s)v(s) s − a ds+ m j=1 t0 t (s−a)n−m−1 |pj(s)| τj (s) s (ξ−a)m−j−1 v(ξ)dξ ds ≤ ≤ |w(t0)| + t0 t p(s)v(s) s − a ds + γ||v||C a0 a p(s)ds, for a < t < a + δ. Consequently, if w0 = |w(t0)| + γ||v||C a0 a p(s)ds, then |u(m) (t)| ≤ w0 + t0 t p(s)v(s) s − a ds for a < t < a + δ. (3.15) EJQTDE, 2012 No. 38, p. 30 From the last inequality, by the integration by parts and (3.14), we get v(t) ≤ w0(t − a) + (t − a) t0 t p(s)v(s) s − a ds + 1 2 v(t) for a < t < a + δ. The last inequality, by the Gronwall-Bellman lemma, results in v(t) t − a ≤ 2w0e2 R t0 t p(s)ds ≤ 2w0e for a < t < a + δ. Due to this inequality, from (3.15) by (3.14) we get |u(m) (t)| ≤ w0(1 + e) for a < t < a + δ. Analogously we can show that u(m) is bounded in the neighborhood of the point b. Therefore, condition (3.8) is satisfied. Proof of Theorem 1.4. From Theorem 1.1 by conditions (1.30)-(1.33) it is obvious that problem (1.1), (1.2) has Fredholm’s property. Thus to prove Theorem 1.4, it suffice to show that the homogeneous problem (1.10), (1.2) has only the trivial solution in the space Cn−1, m (]a, b[). Suppose u ∈ Cn−1, m (]a, b[) is a solution of problem (1.10), (1.2). Then from Theorem 1.1 it is clear that ρ = b a |u(m) (s)|2 ds < +∞. (3.16) Multiplying both sides of (1.10) by (−1)n−m (t − a)n−2m u(t) and integrating from t0 to t1, by Lemma 2.10 we obtain wn(t) − wn(s) + νn t s |u(m) (ξ)|2 dξ = (−1)n−m m j=1 t s (ξ − a)n−2m pj(ξ)u(j−1) (τj(ξ))u(ξ)dξ. Moreover, from Lemma 2.5 it is evident that lim inf s→a |wn(s)| = 0, lim inf t→b |wn(t)| = 0. Then by (3.16) we get νnρ = (−1)n−m m j=1 b a (ξ − a)n−2m pj(ξ)u(j−1) (τj(ξ))u(ξ)dξ. (3.17) According to (1.32), (1.33) and (3.16), all the conditions of Lemmas 2.3 and 2.4 with pj(t) = (−1)n−m (t−a)n−2m pj(t), a0 = b0 = t∗ , t0 = a, t1 = b and µj(t0, t1, t) = τj(t) hold. Consequently, due to equalities ρ0(a) = ρ1(b) = 0, we have (−1)n−m b a (ξ − a)n−2m pj(ξ)u(j−1) (τj(ξ))u(ξ)dξ ≤ ≤ l0jβj(t∗ − a, γ0j)ρ 1/2 0 (τ∗ )ρ 1/2 0 (t∗ ) + l0j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! ρ0(t∗ )+ +l1jβj(b − t∗ , γ1j)ρ 1/2 1 (τ∗)ρ 1/2 1 (t∗ ) + l1j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! ρ1(t∗ ) (3.18) EJQTDE, 2012 No. 38, p. 31 for a < t∗ < b. On the other hand, due to conditions (1.30) and (1.31), the number ν ∈]0, 1[ can be chosen such that inequalities m j=1 l0j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! + l0jβj(t∗ − a, γ0j) < νn − ν 2 , m j=1 l1j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! + l1jβj(b − t∗ , γ1j) < νn − ν 2 (3.19) are satisfied. Thus according to (3.18), (3.19), and inequalities ρ 1/2 0 (τ∗ )ρ 1/2 0 (t∗ ) ≤ ρ, ρ 1/2 1 (τ∗)ρ 1/2 1 (t∗ ) ≤ ρ, (3.17) implies the inequality νnρ ≤ (νn − ν)ρ, and consequently, ρ = 0. Hence, by |u(t)| = 1 (k − 1)! t a (t − s)m−1 u(m) (s)ds ≤ (t − a)m−1/2 ρ for a < t < b, we have u(t) ≡ 0. Proof of Theorem 1.5. The proof is analogous to that of Theorem 1.4. The only difference is that instead of Theorem 1.1, Theorem 1.2 is applied. Proof of Theorem 1.6. Let u be a nonzero solution of the problem (1.10), (1.2). Then analogously to Theorem 1.4, from conditions (1.40),(1.41), (1.32) and (1.33) it follow the validity of relations (3.16), (3.17), (3.18) and the existence of ν ∈]0, 1[ such that m j=1 l0j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! + l0jβj(t∗ − a, γ0j) < νn − ν, m j=1 l1j (2m − j)22m−j+1 (2m − 1)!!(2m − 2j + 1)!! + l1jβj(b − t∗ , γ1j) < νn − ν. (3.20) For the constants τ∗ and τ∗, appearing in inequality (3.18), which are defined in Lemmas 2.3 and 2.4 (with t0 = a, t1 = b, a0 = b0 = t∗ , and µj(t0, t1, t) = τj(t)), from the condition (1.42) we have the estimates τ∗ ≤ t∗ for a < t ≤ t∗ , t∗ ≤ τ∗ for t∗ ≤ t < b. By the last estimates, from (3.18) it immediately follows the inequality νnρ ≤ (νn − ν)ρ. Thus u ≡ 0. Acknowledgement This work is supported by the Academy of Sciences of the Czech Republic (Institutional Research Plan # AV0Z10190503) and by the Shota Rustaveli National Science Foundation (Project # GNSF/ST09 175 3-101). EJQTDE, 2012 No. 38, p. 32 References [1] R. P. Agarwal, Focal boundary value problems for differential and difference equations, Mathematics and Its applications, vol. 436, Kluwer Academic Publishers, Dordrecht, 1998. [2] R. P. Agarwal and D. O’Regan, Singular differential and integral equations with applications, Kluwer Academic Publishers, Dordrecht, 2003. [3] R. P. Agarwal, I. 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Chanturia, Asymptotic properties of solutions of nonautonomous or- dinary differential equations, Mathematics and Its Applications (Soviet Series), vol. 89, Kluwer Academic Publishers, Dordrecht, 1993, Translated from the 1985 Russian original. [9] I. Kiguradze, G. Tskhovrebadze, On two-point boundary value problems for systems of higher order ordinary differential equations with singularities, Georgian Mathematical Journal 1 (1994), no. 1, 31-45. [10] I. Kiguradze, B. P˚uˇza, On certain singular boundary value problem for linear differential equations with deviating arguments, Czechoslovak Math. J 47 (1997), no. 2, 233-244. [11] I. Kiguradze, B. P˚uˇza, On the Vallee-Poussin problem for singular differential equations with deviating arguments, Arch. Math. 33 (1997), No. 1-2, 127-138. [12] I. Kiguradze, B. P˚uˇza, On boundary value problems for systems of linear functional differential equations, Czechoslovak Math. J. 47 (1997), No. 2, 341-37 [13] I. Kiguradze, B. P˚uˇza, and I. P. 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Lomtatidze, On one boundary value problem for linear ordynary differential equations of second order with singularities, Differential’nye Uravneniya 222 (1986), No. 3, 416-426. [18] S. Mukhigulashvili, Two-point boundary value problems for second order functional -differential equations, Mem. Differential Equations Math. Phys. 20 (2000), 1-112. [19] S. Mukhigulashvili, N. Partsvania, On two-point boundary value problems for higher order functional differential equations with strong singularities, Mem. Differential Equations Math. Phys. (accepted). [20] B. P˚uˇza, On a singular two-point boundary value problem for the nonlinear mthorder differential equation with deviating arguments, Georgian Mathematical Journal 4 (1997), no. 6, 557-566. [21] B. P˚uˇza and A. Rabbimov, On a weighted boundary value problem for a system of singular functional-differential equations, Mem. Differential Equations Math. Phys. 21 (2000), 125-130. [22] ˇS. Schwabik, M. Tvrdy, and O. Vejvoda, Differential and integral equations, boundary value problems and adjoints, Academia, Praha,(1979). (Received July 12, 2011) Authors’ addresses: Sulkhan Mukhigulashvili 1. Mathematical Institute, Academy of Sciences of the Czech Republic, ˇZiˇzkova 22, 616 62 Brno, Czech Republic. 2. Ilia State University, Faculty of Physics and Mathematics, 32 I. Chavchavadze St., Tbilisi 0179, Georgia. E-mail: mukhig@ipm.cz Nino Partsvania 1. A. Razmadze Mathematical Institute of I. Javakhishvili Tbilisi State University, 2 University St., Tbilisi 0186, Georgia. 2. International Black Sea University, 2 David Agmashenebeli Alley 13km, Tbilisi 0131, Georgia. E-mail: ninopa@rmi.ge EJQTDE, 2012 No. 38, p. 34 The Dirichlet Boundary Value Problems For Strongly Singular Higher-Order Nonlinear Functional-Differential Equations S. Mukhigulashvili, Brno 15.12.2011 Abstract The a priori boundedness principle is proved for the Dirichlet boundary value problems for strongly singular higher-order nonlinear functional-differential equations. Several sufficient conditions of solvability of the Dirichlet problem under consideration are derived from the a priori boundedness principle. The proof of the a priori boundedness principle is based on the Agarwal–Kiguradze type theorems, which guarantee the existence of the Fredholm property for strongly singular higher-order linear differential equations with argument deviations under the twopoint conjugate and right-focal boundary conditions. Key words and phrases: Higher order functional-differential equations, Dirichlet boundary value problem, strong singularity, Fredholm property, a priori boundedness principle. 2000 Mathematics Subject Classification: 34K06, 34K10 1 Statement of the main results 1.1. Statement of the problem and a survey of the literature. Consider the functional differential equation (1.1) u(n) (t) = F(u)(t) with the two-point boundary conditions (1.2) u(i−1) (a) = 0 (i = 1, · · · , m), u(i−1) (b) = 0 (i = 1, · · · , n − m). Here n ≥ 2, m is the integer part of n/2, −∞ < a < b < +∞, and the operator F acting from the set of (m − 1)-th time continuously differentiable on ]a, b[ functions, to the set Lloc(]a, b[). By u(j−1) (a) (u(j−1) (b)) we denote the right (the left) limit of the function u(j−1) at the point a (b). The problem is singular in the sense that for an arbitrary x the right-hand side of equation (1.41) may have nonintegrable singularities at the points a and b. 1 Throughout the paper we use the following notations: R+ = [0, +∞[; [x]+ the positive part of number x, that is [x]+ = x+|x| 2 ; Lloc(]a, b[) (Lloc(]a, b])) is the space of functions y :]a, b[→ R, which are integrable on [a + ε, b − ε] for arbitrary small ε > 0; Lα,β(]a, b[) (L2 α,β(]a, b[)) is the space of integrable (square integrable) with the weight (t − a)α (b − t)β functions y :]a, b[→ R, with the norm ||y||Lα,β = b a (s − a)α (b − s)β |y(s)|ds ||y||L2 α,β = b a (s − a)α (b − s)β y2 (s)ds 1/2 ; L([a, b]) = L0,0(]a, b[), L2 ([a, b]) = L2 0,0(]a, b[); M(]a, b[) is the set of the measurable functions τ :]a, b[→]a, b[; L2 α,β(]a, b[) (L2 α(]a, b]) is the Banach space of y ∈ Lloc(]a, b[) (Lloc(]a, b])) functions, with the norm ||y||L2 α,β ≡ max t a (s − a)α t s y(ξ)dξ 2 ds 1/2 : a ≤ t ≤ a + b 2 + + max b t (b − s)β s t y(ξ)dξ 2 ds 1/2 : a + b 2 ≤ t ≤ b < +∞. Ln(]a, b[) is the Banach space of y ∈ Lloc(]a, b[) functions, with the norm ||y||L2 α,β = sup [(s − a)(b − t)]m−1/2 t s (ξ − a)n−2m |y(ξ)|dξ : a < s ≤ t < b < +∞. Cn−1 loc (]a, b[), (Cn−1 loc (]a, b[)) is the space of the functions y :]a, b[→ R, which are continuous (absolutely continuous) together with y′ , y′′ , · · · , y(n−1) on [a+ε, b−ε] for arbitrarily small ε > 0. Cn−1, m (]a, b[) is the space of the functions y ∈ Cn−1 loc (]a, b[), such that (1.3) b a |x(m) (s)|2 ds < +∞. Cm−1 1 (]a, b[) is the Banach space of the functions y ∈ Cm−1 loc (]a, b[), such that (1.4) lim sup t→a |x(i−1) (t)| (t − a)m−i+1/2 < +∞ (i = 1, · · · , m), lim sup t→b |x(i−1) (t)| (b − t)m−i+1/2 < +∞ (i = 1, · · · , n − m), 2 with the norm: ||x||Cm−1 1 = m i=1 sup |x(i−1) (t)| αi(t) : a < t < b , where αi(t) = (t − a)m−i+1/2 (b − t)m−i+1/2 . Cm−1 1 (]a, b[) is the Banach space of the functions y ∈ Cm−1 loc (]a, b[), such that conditions (1.7) and (1.4) hold, with the norm: ||x||Cm−1 1 = m i=1 sup |x(i−1) (t)| αi(t) : a < t < b + b a |x(m) (s)|2 ds 1/2 . Dn(]a, b[×R+ ) is the set of such functions δ :]a, b[×R+ → Ln(]a, b[) that δ(t, ·) : R+ → R+ is nondecreasing for every t ∈]a, b[, and δ(·, ρ) ∈ Ln(]a, b[) for any ρ ∈ R+ . D2n−2m−2, 2m−2(]a, b[×R+ ) is the set of such functions δ :]a, b[×R+ → L2 2n−2m−2, 2m−2(]a, b[) that δ(t, ·) : R+ → R+ is nondecreasing for every t ∈]a, b[, and δ(·, ρ) ∈ L2 2n−2m−2, 2m−2(]a, b[) for any ρ ∈ R+ . A solution of problem (1.1), (1.2) is sought in the space Cn−1, m (]a, b[). The singular ordinary differential and functional-differential equations, have been studied with sufficient completeness under different boundary conditions, see for example [1], [2], [4] – [12], [15], [21]- [25] and the references cited therein. But the equation (1.1), even under the boundary condition (1.2), is not studied in the case when the operator F has the form (1.5) F(x)(t) = m j=1 pj(t)x(j−1) (τj(t)) + f(x)(t), where the singularity of the functions pj : Lloc([a, b]) be such that the inequalities (1.6) b a (s − a)n−1 (b − s)2m−1 [(−1)n−m p1(s)]+ds < +∞, b a (s − a)n−j (b − s)2m−j |pj(s)|ds < +∞ (j = 2, · · · , m), are not fulfilled (in this case we sad that the linear part of the operator F is a strongly singular), the operator f continuously acting from Cm−1 1 (]a, b[) to LL2 2n−2m−2, 2m−2 (]a, b[), and the inclusion (1.7) sup{f(x)(t) : ||x||Cm−1 1 ≤ ρ} ∈ L2 2n−2m−2, 2m−2(]a, b[). holds. The first step in studying of the differential equations with strong singularities was made by R. P. Agarwal and I. Kiguradze in the article [3], where the linear ordinary differential equations under conditions (1.2), in the case when the functions pj have strong singularities at the points a and b, are studied. Also the ordinary differential equations with strong singularities under two-point boundary conditions are studied in the articles 3 of I. Kiguradze [13], [14], and N. Partsvania [20]. In the papers [18], [19] these results are generalized for linear differential equation with deviating arguments i.e., are proved the Agarwal-Kiguradze type theorems, which guarantee Fredholm’s property for linear differential equation with deviating arguments. In this paper, on the bases of articles [3], and [17] we prove a priori boundedness principle for the problem (1.1), (1.2) in the case where the operator has form (1.5). Now we introduce some results from the articles [18], [19], which we need for this work. Consider the equation (1.8) u(n) (t) = m j=1 pj(t)u(j−1) (τj(t)) + q(t) for a < t < b. For problem (1.8), (1.2) we assume, that when n = 2m, then the conditions (1.9) pj ∈ Lloc(]a, b[) (j = 1, · · · , m) are fulfilled and when n = 2m + 1, along with (1.9), the condition (1.10) lim sup t→b (b − t)2m−1 t t1 p1(s)ds < +∞ (t1 = a + b 2 ) holds. By hj :]a, b[×]a, b[→ R+ and fj : [a, b] × M(]a, b[) → Cloc(]a, b[×]a, b[) (j = 1, . . . , m) we denote the functions and operator, respectively defined by the equalities (1.11) h1(t, s) = t s (ξ − a)n−2m [(−1)n−m p1(ξ)]+dξ , hj(t, s) = t s (ξ − a)n−2m pj(ξ)dξ (j = 2, · · · , m), and (1.12) fj(c, τj)(t, s) = t s (ξ − a)n−2m |pj(ξ)| τj (ξ) ξ (ξ1 − c)2(m−j) dξ1 1/2 dξ . Let also k = 2k1 + 1 (k1 ∈ N), then k!! = 1 for k ≤ 0 1 · 3 · 5 · · ·k for k ≥ 1 . Now we can to introduce the main theorem of paper [18]. 4 Theorem 1.1. Let there exist the numbers t∗ ∈]a, b[, ℓkj > 0, lkj ≥ 0, and γkj > 0 (k = 0, 1; j = 1, . . . , m) such that along with (1.13) B0 ≡ m j=1 (2m − j)22m−j+1 l0j (2m − 1)!!(2m − 2j + 1)!! + 22m−j−1 (t∗ − a)γ0j l0j (2m − 2j − 1)!!(2m − 3)!! 2γ0j < 1 2 , (1.14) B1 ≡ m j=1 (2m − j)22m−j+1 l1j (2m − 1)!!(2m − 2j + 1)!! + 22m−j−1 (b − t∗ )γ0j l1j (2m − 2j − 1)!!(2m − 3)!! 2γ1j < 1 2 , the conditions (1.15) (t − a)2m−j hj(t, s) ≤ l0j, (t − a)m−γ0j −1/2 fj(a, τj)(t, s) ≤ l0j for a < t ≤ s ≤ t∗ , and (1.16) (b − t)2m−j hj(t, s) ≤ l1j, (b − t)m−γ1j −1/2 fj(b, τj)(t, s) ≤ l1j for t∗ ≤ s ≤ t < b hold. Then problem (1.8), (1.2) is uniquely solvable in the space Cn−1, m (]a, b[). Also, in [19] is proved the following theorem: Theorem 1.2. Let all the conditions of Theorem 1.1 are satisfied. Then the unique solution u of problem (1.8), (1.2) for every q ∈ L2 2n−2m−2, 2m−2(]a, b[) admit the estimate (1.17) ||u(m) ||L2 ≤ r||q||L2 2n−2m−2, 2m−2 , with r = 2m (1 + b − a)(2n − 2m − 1) (νn − 2 max{B0, B1})(2m − 1)!! , ν2m = 1, ν2m+1 = 2m + 1 2 , and thus constant r > 0 dependent only on the numbers lkj, lkj, γkj (k = 1, 2; j = 1, · · · , m), and a, b, t∗ , n. Remark 1.1. Under the conditions of Theorem 1.2, for every q ∈ L2 2n−2m−2, 2m−2(]a, b[) the unique solution u of problem (1.8), (1.2) admits the estimate (1.18) ||u(m) ||Cm−1 1 ≤ rn||q||L2 2n−2m−2, 2m−2 , with rn = 1 + m j=1 2m−j+1/2 (m − j)!(2m − 2j + 1)1/2(b − a)m−j+1/2 2m (1 + b − a)(2n − 2m − 1) (νn − 2 max{B0, B1})(2m − 1)!! . 1.2. Theorems on a solvability of problem (1.1), (1.2). Define the operator P : Cm−1 1 (]a, b[) × Cm−1 1 (]a, b[) → Lloc(]a, b[), by the equality (1.19) P(x, y)(t) = m j=1 pj(x)(t)y(j−1) (τj(t)) for a < t < b 5 where pj : Cm−1 1 (]a, b[) → Lloc(]a, b[), and τj ∈ M(]a, b[). Also for any γ > 0 define the set Aγ by the relation (1.20) Aγ = {x ∈ Cm−1 1 (]a, b[) : ||x||Cm−1 1 ≤ γ}. For formulate this a priori boundedness principle we have to introduce Definition 1.1. Let γ0 and γ be the positive numbers. We said that the continuous operator P : Cm−1 1 (]a, b[)×Cm−1 1 (]a, b[) → Ln(]a, b[) to be γ0, γ consistent with boundary condition (1.2) if: i. for any x ∈ Aγ0 and almost all t ∈]a, b[ the inequality (1.21) m j=1 |pj(x)(t)x(j−1) (τj(t))| ≤ δ(t, ||x||Cm−1 1 )||x||Cm−1 1 holds, where δ ∈ Dn(]a, b[×R+ ). ii. for any x ∈ Aγ0 and q ∈ L2 2n−2m−2, 2m−2(]a, b[) the equation (1.22) y(n) (t) = m j=1 pj(x)(t)y(j−1) (τj(t)) + q(t) under boundary conditions (1.2), has the unique solution y in the space Cn−1, m (]a, b[) and (1.23) ||y||Cm−1 1 ≤ γ||q||L2 2n−2m−2, 2m−2 . Definition 1.2. We said that the operator P to be γ consistent with boundary condition (1.2), if the operator P be γ0, γ consistent with boundary condition (1.2) for any γ0 > 0. In the sequel it will always be assumed that the operator Fp defined by equality Fp(x)(t) = |F(x)(t) − m j=1 pj(x)(t)x(j−1) (τj(t))(t)|, continuously acting from Cm−1 1 (]a, b[) to LL2 2n−2m−2, 2m−2 (]a, b[), and (1.24) Fp(t, ρ) ≡ sup{Fp(x)(t) : ||x||Cm−1 1 ≤ ρ} ∈ L2 2n−2m−2, 2m−2(]a, b[) for each ρ ∈ [0, +∞[. Then the following theorem is valid Theorem 1.3. Let the operator P be γ0, γ consistent with boundary condition (1.2), and there exist a positive number ρ0 ≤ γ0, such that (1.25) ||Fp( ·, min{2ρ0, γ0})||L2 2n−2m−2, 2m−2 ≤ γ0 γ . 6 Let moreover, for any λ ∈]0, 1[, an arbitrary solution x ∈ Aγ0 of the equation (1.26) x(n) (t) = (1 − λ)P(x, x)(t) + λF(x)(t) under the conditions (1.2), admits the estimate (1.27) ||x||Cm−1 1 ≤ ρ0. Then problem (1.1), (1.2) is solvable in the space Cn−1,m (]a, b[). From theorem 1.3 with ρ0 = γ0 immediately follows Corollary 1.1. Let the operator P be γ0, γ consistent with boundary condition (1.2), and (1.28) |F(x)(t) − m j=1 pj(x)(t)x(j−1) (τj(t))(t)| ≤ η(t, ||x||Cm−1 1 ) for x ∈ Aγ0 and almost all t ∈]a, b[, and (1.29) ||η(·, γ0)||L2 2n−2m−2, 2m−2 ≤ γ0 γ , where η ∈ D2n−2m−2, 2m−2(]a, b[×R+ ). Then problem (1.1), (1.2) is solvable in the space Cn−1,m (]a, b[). Corollary 1.2. Let the operator P be γ consistent with boundary condition (1.2), inequality (1.28) holds for x ∈ Cm−1 1 (]a, b[) and almost all t ∈]a, b[, where η(·, ρ) ∈ L2 2n−2m−2, 2m−2(]a, b[) for any ρ ∈ R+ , and (1.30) lim sup ρ→+∞ 1 ρ ||η(·, ρ)||L2 2n−2m−2, 2m−2 < 1 γ . Then the problem (1.1), (1.2) is solvable in the space Cn−1,m (]a, b[). When we discuss problem (1.41), (1.2), and n = 2m+1, we assume that the continuous operator p1 : Cm−1 1 (]a, b[) → Lloc(]a, b[), by such that (1.31) lim sup t→b (b − t)2m−1 t t1 p1(x)(s)ds < +∞ (t1 = a + b 2 ) for any x ∈ Cm−1 1 (]a, b[). Now define the operators hj : Cm−1 1 (]a, b[)×]a, b[×]a, b[→ Lloc(]a, b[×]a, b[), fj : Cm−1 1 (]a, b[) × [a, b] × M(]a, b[) → Cloc(]a, b[×]a, b[) (j = 1, . . . , m) by the equalities (1.32) h1(x, t, s) = t s (ξ − a)n−2m [(−1)n−m p1(x)(ξ)]+dξ , hj(x, t, s) = t s (ξ − a)n−2m pj(x)(ξ)dξ (j = 2, · · · , m), 7 and (1.33) fj(x, c, τj)(t, s) = t s (ξ − a)n−2m |pj(x)(ξ)| τj(ξ) ξ (ξ1 − c)2(m−j) dξ1 1/2 dξ . Theorem 1.4. Let the continuous operator P : Cm−1 1 (]a, b[) × Cm−1 1 (]a, b[) → Ln(]a, b[) admits to the condition (1.21) where δ ∈ Dn(]a, b[×R+ ), τj ∈ M(]a, b[) and the numbers γ0, t∗ ∈]a, b[, lkj > 0, lkj > 0, γkj > 0 (k = 1, 2; j = 1, · · · , m), be such that the inequalities (1.34) (t − a)2m−j hj(x, t, s) ≤ l0j, lim sup t→a (t − a)m− 1 2 −γ0j fj(x, a, τj)(t, s) ≤ l0j for a < t ≤ s ≤ t∗ , ||x||Cm−1 1 ≤ γ0, (1.35) (b − t)2m−j hj(x, t, s) ≤ l1j, lim sup t→b (b − t)m− 1 2 −γ1j fj(x, b, τj)(t, s) ≤ l1j for t∗ ≤ s ≤ t < b, ||x||Cm−1 1 ≤ γ0, and conditions (1.13), (1.14) hold. Let moreover the operator F and function η ∈ D2n−2m−2, 2m−2(]a, b[×R+ ) be such that condition (1.28) and inequality (1.36) ||η(·, γ0)||L2 2n−2m−2, 2m−2 < γ0 rn , be fulfilled, where rn = 1 + m j=1 2m−j+1/2 (m − j)!(2m − 2j + 1)1/2(b − a)m−j+1/2 2m (1 + b − a)(2n − 2m − 1) (νn − 2 max{B0, B1})(2m − 1)!! . Then problem (1.1), (1.2) is solvable in the space Cn−1,m (]a, b[). Theorem 1.5. Let the operator F and function η are such that condition (1.28), (1.30) hold and the continuous operator P : Cm−1 1 (]a, b[) × Cm−1 1 (]a, b[) → Ln(]a, b[) admits condition (1.21) where δ ∈ Dn(]a, b[×R+ ). Let moreover the measurable functions τj ∈ M(]a, b[) and the numbers t∗ ∈]a, b[, lkj > 0, lkj > 0, γkj > 0, (k = 0, 1; j = 1, · · · , m) be such that the inequalities (1.37) (t − a)2m−j hj(x, t, s) ≤ l0j, lim sup t→a (t − a)m− 1 2 −γ0j fj(x, a, τj)(t, s) ≤ l0j for a < t ≤ s ≤ t∗ , x ∈ Cm−1 1 (]a, b[), (1.38) (b − t)2m−j hj(x, t, s) ≤ l1j, lim sup t→b (b − t)m− 1 2 −γ1j fj(x, b, τj)(t, s) ≤ l1j for t∗ ≤ s ≤ t < b, x ∈ Cm−1 1 (]a, b[), and conditions (1.13), (1.14) hold. Then problem (1.1), (1.2) is solvable in the space Cn−1,m (]a, b[). 8 Remark 1.2. Let γ0 > 0, operators αj(t)pj(x)(t) (j = 1, · · · , m) continuously acting from the space Cm−1 1 (]a, b[) to the space Ln(]a, b[), exist the function δj ∈ Dn(]a, b[) such that for any x ∈ Aγ0 (1.39) |pj(x)(t)|αj(t) ≤ δj(t, ||x||Cm−1 1 ) for a < t < b, and exists constants κ > 0, ε > 0 such that (1.40) |τj(t) − t| ≤ κ(t − a) (j = 1, · · · , m) for a < t < a + ε, |τj(t) − t| ≤ κ(b − t) (j = 1, · · · , m) for b − ε < t < b, Then the operator P defined by equality (1.19), continuously acting from Aγ0 to the space Ln(]a, b[), and there exists the function δ ∈ Dn(]a, b[) such that item i of definition 1.1 holds. Now consider the equation with deviating arguments (1.41) u(n) (t) = f(t, u(τ1(t)), u′ (τ2(t)), · · · , u(m−1) (τm(t))) for a < t < b, where −∞ < a < b < +∞, f :]a, b[×Rm → R is a function, satisfying the local Caratheodory conditions and τj ∈ M(]a, b[) (j = 0, . . . , n − 1) are measurable functions. Corollary 1.3. Let the functions τj ∈ M(]a, b[) and the numbers t∗ ∈]a, b[, κ ≥ 0, ε > 0, lkj > 0, lkj > 0, γkj > 0, (k = 0, 1; j = 1, · · · , m) be such that the conditions (1.13)(1.16), (1.40) and the inclusions (1.42) αjpj ∈ Ln(]a, b[) (j = 1, · · · , m) are fulfilled. Let moreover (1.43) f(t, x(τ1(t)), x′ (τ2(t)), · · · , x(m−1) (τm(t))) − m j=1 pj(t)x(j−1) (τj(t))(t) ≤ ≤ η(t, ||x||Cm−1 1 ) for x ∈ Cm−1 1 (]a, b[) and almost all t ∈]a, b[, where η(·, ρ) ∈ L2 2n−2m−2, 2m−2(]a, b[) for any ρ ∈ R+ , and condition (1.30) holds. Then problem (1.41), (1.2) is solvable in the space Cn−1,m (]a, b[). Remark 1.3. From conditions (1.42) do not follow the conditions (1.6). Now for illustration of our results consider on ]a, b[ the second order functionaldifferential equations (1.44) u′′ (t) = − λ|u(t)|k [(t − a)(b − t)]2+k/2 u(τ(t)) + q(x)(t), (1.45) u′′ (t) = − λ| sin uk (t)| [(t − a)(b − t)]2 u(τ(t)) + q(x)(t), where λ, k ∈ R+ the function τ ∈ M(]a, b[), the operator q : Cm−1 1 (]a, b[) → L2 0,0(]a, b[) is continuous and η(t, ρ) ≡ sup{|q(x)(t)| : ||x||Cm−1 1 ≤ ρ} ∈ L2 0,0(]a, b[). Than from Theorems 1.4 and 1.5 follows 9 Corollary 1.4. Let the function τ ∈ M(]a, b[), the continuous operator q : Cm−1 1 (]a, b[) → L2 0,0(]a, b[), and the numbers γ0 > 0, λ ≥ 0, k > 0, by such that (1.46) |τ(t) − t| ≤ (t − a)3/2 for a < t ≤ (a + b)/2 (b − t)3/2 for (a + b)/2 ≤ t < b , (1.47) ||η(t, γ0)||L2 0, 0 ≤ 1 + 2 b − a −1 (b − a)2 − 16λγk 0 (1 + [2(b − a)]1/4 ) 2(1 + b − a)(b − a)2 , and (1.48) λ < (b − a)2 32γk 0 (1 + [2(b − a)]1/4) . Then the problem (1.44), (1.2) is solvable. Corollary 1.5. Let the function τ ∈ M(]a, b[), continuous operator q : Cm−1 1 (]a, b[) → L2 0,0(]a, b[), and the number λ ≥ 0 by such, that inequalities (1.30) with n = 2, (1.46) and (1.49) λ < (b − a)2 32(1 + [2(b − a)]1/4) , hold. Then the problem (1.45), (1.2) is solvable. 2 Auxiliary Propositions 2.1. Lemmas on some properties of the equation x(n) (t) = λ(t). First, we introduce two lemmas without proofs. First Lemma is proved in [3]. Lemma 2.1. Let i ∈ 1, 2, x ∈ Cm−1 loc (]t0, t1[) and (2.1) x(j−1) (ti) = 0 (j = 1, . . . , m), t1 t0 |x(m) (s)|2 ds < +∞. Then (2.2) t ti (x(j−1) (s))2 (s − ti)2m−2j+2 ds 1/2 ≤ 2m−j+1 (2m − 2j + 1)!! t ti |x(m) (s)|2 ds 1/2 for t0 ≤ t ≤ t1. This second lemma is a particular case of Lemma 4.1 in [7] 10 Lemma 2.2. If x ∈ Cn−1 loc (]a, a1]), then for any s, t ∈]a, a1] the equality (−1)n−m t s (ξ − a)n−2m x(n) (ξ)x(ξ)dξ = wn(x)(t) − wn(x)(s) + νn t s |x(m) (ξ)|2 dξ is valid, where ν2m = 1, ν2m+1 = 2m+1 2 , w2m(x)(t) = m j=1 (−1)m+j−1 x(2m−j) (t)x(t), w2m+1(x)(t) = m j=1 (−1)m+j [(t − a)x(2m+1−j) (t) − jx(2m−j) (t)]x(j−1) (t) − t − a 2 |x(m) (t)|2 . Lemma 2.3. Let the numbers a1 ∈]a, b[, t0k ∈]a, a1[, and εik, εi, βk, β ∈ R+ , k ∈ N, i = 1, · · · , n − m are such that (2.3) lim k→+∞ t0 k = a, lim k→+∞ βk = β, lim k→+∞ εi,k = εi. Let, moreover (2.4) λ ∈ L2 2n−2m−2,0(]a, a1]), is a nonnegative function, xk ∈ Cn−1, m (]a, b[) be a solution of the problem (2.5) x(n) (t) = βkλ(t), (2.6) x(i−1) (t0 k) = 0 (i = 1, · · · , m), x(i−1) (a1) = εi,k (i = 1, · · · , n − m), and x ∈ Cn−1, m (]a, b[) be a solution of the problem (2.7) x(n) (t) = βλ(t), (2.8) x(i−1) (a) = 0 (i = 1, · · · , m), x(i−1) (a1) = εi (i = 1, · · · , n − m). Then (2.9) lim k→+∞ x (j−1) k (t) = x(j−1) (t) (j = 1, . . . , n) uniformly in ]a, a1]. Proof. First, prove our lemma under the assumption that there exists the number r1 > 0 such that the estimates (2.10) a1 t0k |x (m) k (s)|2 ds ≤ r1 k ∈ N hold. Now, suppose that t1, . . . , tn are such numbers that t0k < t1 < · · · < tn < a1 (k ∈ N), and gi(t) are the polynomials of (n − 1)-th degree, satisfying the conditions gj(tj) = 1, gj(ti) = 0 (i = j; i, j = 1, . . . , n). Then if xk is a solution of the problem (2.5), (2.6), 11 and x is a solution of the problem (2.7), (2.8). For the solution x − xk of the equation dn(x(t)−xk(t)) dtn = (β − βk)λ(t), the representation (2.11) x(t) − xk(t) = n j=1 (x(tj) − xk(tj)) − β − βk (n − 1)! tj t1 (tj − s)n−1 λ(s)ds gj(t)+ + β − βk (n − 1)! t t1 (t − s)n−1 λ(s)ds k ∈ N for t0k ≤ t ≤ a1 is valid. On the other hand in view of inequality (2.10), the identities x (i−1) k (t) = 1 (m − i)! t t0 k (t − s)m−i x (m) k (s)ds (i = 1, 2, k ∈ N) by Schwartz inequality yield (2.12) |x (i−1) k (t)| ≤ r2(t − a)m−i−1/2 for t0k ≤ t ≤ a1 (i = 1, 2, k ∈ N), where r2 = r1 (m−i)! √ 2m−2i+1 . By virtue of the Arzela-Ascoli lemma and (2.3), (2.12) the sequence {xk}+∞ k=1 contains a subsequence {xkl }+∞ l=1 which is uniformly convergent in ]a, a1]. Suppose lim l→+∞ xkl (t) = x0(t). Thus from (2.11) by (2.3) it follows the existence of such r3 > 0 that |x (j−1) kl (t)| ≤ r3 + |x(j−1) (t)| (j = 1, · · · , n) for t0kl ≤ t ≤ a1, and then without loss of generality we can assume that (2.13) lim l→+∞ x (j−1) kl (t) = x (j−1) 0 (t) (j = 1, . . . , n) uniformly in ]a, a1]. Then in virtue of (2.3), (2.11), and (2.13) we have x(t) − x0(t) = n j=1 (x(tj) − x0(tj)) gj(t) for a ≤ t ≤ a1. From the last two relation by (2.10) it is clear that x(n) = x (n) 0 and x0 ∈ Cn−1, m (]a, b[). I.e., the function x0 ∈ Cn−1, m (]a, b[) is a solution of problem (2.7), (2.8). In view of (2.4) all the conditions of Theorem 1.1 are fulfilled, thus problem (2.7), (2.8) is uniquely solvable in the space Cn−1, m (]a, b[) and x = x0. Therefore from (2.13) follows (2.14) lim l→+∞ x (j−1) kl (t) = x(j−1) (t) (j = 1, . . . , n) uniformly in ]a, a1]. Now suppose that relations (2.9) are not fulfilled. Then there exist δ ∈]0, a1−a 2 [, ε > 0, and the increasing sequence of natural numbers {kl}+∞ l=1 such that (2.15) max n j=1 |x (j−1) kl (t) − x(j−1) (t)| : a + δ ≤ t ≤ a1 > ε (l ∈ N). 12 By virtue of Arcela-Ascoli lemma and condition (2.10) the sequence {x (j−1) kl }+∞ l=1 (j = 1, . . . , m), without loss of generality, can be assumed to be uniformly converging in ]a + δ, a1]. Then, in view of what we have shown above, equality (2.14) holds. But this contradicts condition (2.15). Thus (2.9) holds if the conditions (2.10) are fulfilled. Let now the conditions (2.10) are not fulfilled. Then exists the subsequence {t0kl }+∞ l=1 of the sequence {t0k}+∞ k=1, such that (2.16) a1 t0k |x (m) kl (s)|2 ds ≥ l (l ∈ N). Suppose that βl = a1 t0k |x (m) kl (s)|2 ds −1 and vl(t) = ukl (t)βl. Thus in view of (2.16) and our notations (2.17) a1 t0kl |v (m) kl (s)|2 ds = 1 (l ∈ N), lim l→+∞ βl = 0, (2.18) v (n) l (t) = βlλ(t), (2.19) v (i−1) l (t0 kl ) = 0 (i = 1, · · · , m), v (i−1) l (a1) = εi,kl βl (i = 1, · · · , n − m, l ∈ N). From the first part of our lemma by (2.17) it follows that there exists limit lim l→+∞ vl(t) ≡ v0(t), and v0 is a solution of corresponding of (2.18), (2.19) homogeneous problem. thus v0 ≡ 0. On the other hand from (2.17) it is clear that a1 t0kl |v (m) 0 (s)|2 ds = 1, which contradict with v0 ≡ 0. Thus our assumption is invalid and (2.10) holds. Analogously one can prove Lemma 2.4. Let the numbers b1 ∈]a, b[, t0k ∈]b1, b[, and εik, εi, βk, β ∈ R+ , k ∈ N, i = 1, · · · , n − m are such that lim k→+∞ t0 k = b, lim k→+∞ βk = β, lim k→+∞ εi,k = εi. Let moreover, λ ∈ L2 0, 2m−2(]b1, b]) is a nonnegative function, xk ∈ Cn−1, m (]a, b[) be a solution of the problem (2.5) under the conditions x(i−1) (b1) = εi,k (i = 1, · · · , m), x(i−1) (t0 k) = 0 (i = 1, · · · , n − m), and x ∈ Cn−1, m (]a, b[) be a solution of the equation (2.7) under the conditions (2.20) x(i−1) (b1) = εi (i = 1, · · · , m), x(i−1) (b) = 0 (i = 1, · · · , n − m). Then the equalities (2.9) hold. 13 Lemma 2.5. Let a < a1 < b1 < b, εi ∈ R+ and λ ∈ L2 2n−2m−2, 0(]a, a1]) (λ ∈ L2 0, 2m−2(]b1, b])) is nonnegative function. Then for the solution x ∈ Cn−1, m (]a, b[) of the problem (2.7), (2.8) ((2.7), (2.20)) with β = 1, the estimate (2.21) a1 a |x(m) (s)|2 ds ≤ Θ1(x, a1, λ) b b1 |x(m) (s)|2 ds ≤ Θ2(x, b1, λ) (k ∈ N) is valid, where (2.22) Θ1(x, a1, λ) = 2|wn(x)(a1)| + γ1||λ||2 L2 2n−2m−2,0(]a,a1]) , Θ2(x, b1, λ) = 2|wn(x)(b1)| + γ2||λ||2 L2 0,2m−2(]b1,b]) , and γ1 = 2m−1 (2m + 1) (2m − 1)!! 2 , γ2 = 2m−1 (2m + 1)(b − a + 1) (2m − 1)!! 2 . Proof. Suppose that xk is a solution of problem (2.5), (2.6) with βk = 1, εik = εi. Then in view of Lemma 2.3, relations (2.9) hold. On the other hand by Lemma 2.2 we get (2.23) νn a1 t0k |x (m) k (s)|2 ds ≤ −wn(xk)(a1) + a1 t0k (s − a)n−2m λ(s)|xk(s)|ds. Now, on the basis of Lemma 2.1, Schwartz’s and Young’s inequalities we get a1 t0k (s − a)n−2m λ(s)xk(s)ds = a1 t0k [(n − 2m)xk(s) + (s − a)n−2m x′ k(s)] a1 s λ(ξ)dξ ds ≤ ≤ (n − 2m) a1 t0k x2 k(s) (s − a)2m ds 1/2 + a1 t0k x′2 k (s) (s − a)2m−2 ds 1/2 ||λ||L2n−2m−2,0(]a,a1]) ≤ 2m−1 (2m + 1) (2m − 1)!! a1 t0k |x (m) k (s)|2 ds 1/2 ||λ||L2n−2m−2,0(]a,a1]) ≤ ≤ 1 2 a1 t0k |x (m) k (s)|2 ds + 1 2 2m−1 (2m + 1) (2m − 1)!! 2 ||λ||2 L2n−2m−2,0(]a,a1]) . Thus from (2.23) by the definition of the numbers νn immediately follows that estimate a1 t0k |x (m) k (s)|ds ≤ 2|wn(xk)(a1)| + 2m−1 (2m + 1) (2m − 1)!! 2 ||λ||2 L2n−2m−2,0(]a,a1]) (k ∈ N). By (2.9) from the last inequality (2.21) and (2.22) follows. Thus Lemma is proved for the problem (2.7), (2.8). Analogously, by using Lemma 2.4 one can prove the case of problem (2.7), (2.20). 14 2.2. Lemmas on Banach space Cm−1 1 (]a, b[). Definition 2.1. Let ρ ∈ R+ and the function η ∈ Lloc(]a, b[) be nonnegative. Then S(ρ, η) is a set of such y ∈ Cn−1 loc (]a, b[) that (2.24) y(i−1) a + b 2 ≤ ρ (i = 1, . . . , n), (2.25) |y(n−1) (t) − y(n−1) (s)| ≤ t s η(ξ)dξ for a < s ≤ t < b, and (2.26) y(i−1) (a) = 0 (i = 1, · · · , m), y(i−1) (b) = 0 (i = 1, · · · , n − m). Lemma 2.6. Let for the function y ∈ Cn−1,m (]a, b[), conditions (2.26) be satisfied. Then y ∈ Cm−1 1 (]a, b[) and the estimates (2.27) |y(i−1) (t)| ≤ |t − ck|m−i+1/2 (m − i)!(2m − 2i + 1)1/2 t ck |y(m) (s)|2 ds 1/2 for a < t < b, i = 1, . . . , m, hold for k = 1, 2, where c1 = a, c2 = b. Proof. First not that in view of inclusion y ∈ Cn−1,m (]a, b[), the equality (2.28) y(i−1) (t) = l j=i (t − c)j−i (j − i)! y(j−1) (c) + 1 (l − i)! t c (t − s)l−i y(l) (s)ds for a < t < b for i = 1, · · · , l, l = 1, · · · , n, holds, where 1. c ∈ [a, b] if l ≤ m; 2. c ∈ ]a, b] if l = m + 1 and n = 2m + 1; 3. c ∈ ]a, b[ if l > m, and exists r > 0 such that (2.29) b a |y(m) (s)|2 ds ≤ r. Equality (2.28), with l = m, c = a and with l = m, c = b by conditions (2.26), (2.29) and Schwartz inequality yields (2.27). From (2.27) and (2.29) it is clear that y ∈ Cm 1 (]a, b[). Lemma 2.7. Let ρ ∈ R+ , and η ∈ L2 2n−2m−2, 2m−2(]a, b[) is a nonnegative function. Then S(ρ, η) is a compact subset of the space Cm−1 1 (]a, b[). 15 Proof. Condition (2.25) yields the inequality |y(n) (t)| ≤ η(t). Thus there exists such function η1 ∈ L2 2n−2m−2, 2m−2(]a, b[) that (2.30) y(n) (t) = η1(t), for a < t < b (2.31) |η1(t)| ≤ η(t) for a < t < b From the Theorem 1.1, follows that problem (2.30), (2.26) has unique solution y ∈ Cn−1, m (]a, b[), i.e. there exists r > 0 such that the inequality (2.29) holds. For any y ∈ S(ρ, η), from equality (2.28) with l = n, by (2.24), (2.30) and (2.31)we get (2.32) |y(i−1) (t)| ≤ γi(t) for a < t < b, (i = 1, · · · , n), where γi(t) = ρi + 1 (n − i)! t c (t − s)n−i η(s)ds (i = 1, · · · , n). Let, now yk ∈ S(ρ, η) (k ∈ N). By virtue of the Arzela-Ascoli lemma and conditions (2.25), (2.32) the sequence {yk}+∞ k=1 contains a subsequence {ykℓ }+∞ ℓ=1 such that {y (i−1) kℓ }+∞ ℓ=1 (i = 1, · · · , n) are uniformly convergent on ]a, b[. Thus without loss of generality we can assume that {y (i−1) k }+∞ k=1 (i = 1, · · · , n − 1) are uniformly convergent on ]a, b[. Let lim k→+∞ yk(t) = y0(t), then y0 ∈ Cn−1 loc (]a, b[) and (2.33) lim k→+∞ y (i−1) k (t) = y (i−1) 0 (t) (i = 1, · · · , n) uniformly on ]a, b[. From (2.33) in view of the inclusions yk ∈ S(ρ, η) immediately follows that (2.34) y (i−1) 0 a + b 2 ≤ ρ (i = 1, . . . , n), (2.35) y (i−1) 0 (a) = 0 (j = 1, · · · , m), y (i−1) 0 (b) = 0 (j = 1, · · · , n − m). and (2.36) |y (n−1) 0 (t) − y (n−1) 0 (s)| ≤ t s η(ξ)dξ for a < s ≤ t < b. From (2.34)-(2.36) it is clear that y0 ∈ S(ρ, η). To finish the proof we must shove that (2.37) lim k→+∞ ||yk(t) − y0(t)||Cm−1 1 = 0, and (2.38) S(ρ, η) ⊂ Cm−1 1 (]a, b[). 16 Let, xk = y0 − yk, and a1 ∈]a, b[, b1 ∈]a1, b[. Then it is cleat that xk ∈ S(ρ′ , η′ ) where ρ′ = 2ρ, η′ = 2η. Thus for any xk exists ηk ∈ L2 2n−2m−2, 2m−2(]a, b[) such that (2.39) x (n) k (t) = ηk(t), (2.40) x (i−1) k (a) = 0 (i = 1, · · · , n), x (i−1) k (b) = 0 (i = 1, · · · , n − m). where (2.41) |ηk(t)| ≤ 2η(t) for a < t < b (k ∈ N). On the other hand, from (2.27) with y = xk, in view of (2.40) we get (2.42) |x (i−1) k (t)| ≤ t a |x (m) k (s)|2 ds 1/2 (t − a)m−i+1/2 for a < t < a1, |x (i−1) k (t)| ≤ b t |x (m) k (s)|2 ds 1/2 (b − t)m−i+1/2 for b1 < t < b, for i = 1, . . . , m. Let now wn be the operator defined in Lemma 2.2 and Θ1, Θ2 are functions defined by (2.22) with λ = ηk. Then conditions (2.33) yields (2.43) lim k→+∞ wn(xk)(a1) = 0, lim k→+∞ wn(xk)(b1) = 0 (k ∈ N), and from definition of norm || · ||L2 α,β , (2.41) and (2.43), follows that for any ε > 0 we can choose a1 ∈]a, min{a + 1, b}[, b1 ∈] max{b − 1, b}, a1[ and k0 ∈ N, such that (2.44) Θ1(xk, a1, 2η) ≤ ε 6 (b − b1)m−1/2 (k ≥ k0), Θ2(xk, b1, 2η) ≤ ε 6 (a1 − a)m−1/2 (k ≥ k0). By using lemma 2.5 for xk, in view of (2.42) and (2.44)we get (2.45) a1 a |x (m) k (s)|2 ds ≤ ε 6 b b1 |x (m) k (s)|2 ds ≤ ε 6 (k ≥ k0), (2.46) |x (i−1) k (t)| αi(t) ≤ ε 2m for t ∈]a, a1] ∪ [b1, b[, (1 ≤ i ≤ m, k ≥ k0). Also, in view of (2.33) without loss of generality we can assume that (2.47) |x (i−1) k (t)| αi(t) ≤ ε 2m for a1 ≤ t ≤ b1, (1 ≤ i ≤ m, k ≥ k0), and (2.48) b1 a1 |x (m) k (s)|2 ds ≤ ε 6 (k ≥ k0). From (2.45)-(2.48), equality (2.37) immediately follows. Let, now y ∈ S(ρ, η) and yk = δky, where lim k→+∞ δk = 0. Then by (2.33) it is clear, that y0 ≡ 0 and than from (2.37) it follows y ∈ Cm−1 1 (]a, b[), i.e. the inclusion (2.38) holds. 17 Lemma 2.8. Let τj ∈ M(]a, b[), α ≥ 0, β ≥ 0 and exists δ ∈]0, b − a[ such that (2.49) |τj(t) − t| ≤ k1(t − a)β for a < t ≤ a + δ. Then τ(t) t (s − a)α ds ≤ k1[1 + k1δβ−1 ]α (t − a)α+β for β ≥ 1 k1[δ1−β + k1]α (t − a)αβ+β for 0 ≤ β < 1 , for a < t ≤ a + δ. Proof. First note that τ(t) t (s − a)α ds ≤ (max{τ(t), t} − a)α |τ(t) − t| for a ≤ t ≤ a + δ, and max{τ(t), t} ≤ t + |τ(t) − t| for a ≤ t ≤ a + δ. Then in view of condition (2.49) we get τ(t) t (s − a)α ds ≤ k1[(t − a) + k1(t − a)β ]α (t − a)β for a ≤ t ≤ a + δ. Last inequality yields the validity of our lemma. Analogously one can prove Lemma 2.9. Let τj ∈ M(]a, b[), α ≥ 0, β ≥ 0 and exists δ ∈]0, b − a[ such that (2.50) |τj(t) − t| ≤ k1(b − t)β for b − δ ≤ t < b. Then τ(t) t (b − t)α ds ≤ k1[1 + k1δβ−1 ]α (b − t)α+β for β ≥ 1 k1[δ1−β + k1]α (b − t)αβ+β for 0 ≤ β < 1 , for b − δ ≤ t < b. 2.3. Lemmas on the solutions of auxiliary problems. Throughout of this section we assume that the operator P : Cm−1 1 (]a, b[)×Cm−1 1 (]a, b[) → Ln(]a, b[) be γ0, γ consistent with boundary condition (1.2), and operator q : Cm−1 1 (]a, b[) → L2 2n−2m−2, 2m−2(]a, b[), be continuous. Consider for any x ∈ Cm−1 1 (]a, b[) ⊂ Cm−1 1 (]a, b[) the nonhomogeneous equation (2.51) y(n) (t) = m i=1 pi(x)(t)y(i−1) (τi(t)) + q(x)(t), 18 and corresponding homogeneous equation (2.52) y(n) (t) = m i=1 pi(x)(t)y(i−1) (τi(t)), and let, En be a set of the solutions of problem (2.51), (2.26). From inequality (1.23) of item (ii) of definition 1.1, it follows that boundary problem (2.51), (2.26) has the unique solution y in the space Cn−1,m (]a, b[). But in view of Lemma 2.6 it is clear that y ∈ Cm−1 1 (]a, b[). Thus En ∩ Cm−1 1 (]a, b[) = ∅, and exists the operator U : Cm−1 1 (]a, b[) → En ∩ Cm−1 1 (]a, b[) defined by the equality U(x)(t) = y(t). Lemma 2.10. U : Cm−1 1 (]a, b[) → En ∩ Cm−1 1 (]a, b[) is a continuous operator. Proof. Let xk ∈ Cm−1 1 (]a, b[) and yk(t) = U(xk)(t) (k = 1, 2), y = y2 − y1, and the operator P is defined by (1.19). Then y(n) (t) = P(x2, y)(t) + q0(x1, x2)(t) where q0(x1, x2)(t) = P(x2, y1)(t) − P(x1, y1)(t) + q(x2)(t) − q(x1)(t). Hence, by item ii of definition 1.1 we have ||U(x2) − U(x1)||Cm−1 1 ≤ γ||q0(x1, x2)||L2 2n−2m−2, 2m−2 . Since the operators P and q are continuous, this estimate implies the continuity of the operator U. 3 Proofs Proof of remark 1.1. Let x be a solution of problem (1.8), (1.2), then from inequalities (2.27) it follows the estimate (3.1) |x(i−1) (t)| ≤ [(b − t)(t − a)]m−i+1/2 (m − i)!(2m − 2i + 1)1/2 2 b − a m−i+1/2 ||x(m) ||L2 for a ≤ t ≤ b. From this estimate, by definition of norm in the space Cm−1 (]a, b[), and estimate (1.17) immediately follows (1.18). Proof of theorem 1.3. Let δ and λ are the functions and numbers appearing in Definition 1.1. We set (3.2) η(t) = δ(t, γ0)γ0 + Fp(t, min{2ρ0, γ0}), (3.3) χ(s) =    1 for 0 ≤ s ≤ ρ0 2 − s/ρ0 for ρ0 < s < 2ρ0 0 for s ≥ 2ρ0 , 19 (3.4) q(x)(t) = χ(||x||Cm−1 1 )Fp(x)(t). From (1.24) it is clear that the nonnegative functions Fp, η, admits the inclusion (3.5) Fp(·, min{2ρ0, γ0}), η ∈ L2 2n−2m−2, 2m−2(]a, b[), and for every x ∈ Aγ0 ⊂ Cm−1 1 (]a, b[) and almost all t ∈]a, b[ the inequality (3.6) |q(x)(t)| ≤ Fp(t, min{2ρ0, γ0}) for a < t < b holds. Let U : Aγ0 → En ∩ Cm−1 1 (]a, b[) is a operator appeared in Lemma 2.10, from which it follows that U is a continuous operator. On the other hand from items i and ii of Definition 1.1, (1.25) and (3.6) it is clear, that for each x ∈ Aγ0 , the conditions ||y||Cm−1 1 ≤ γ0, |y(n−1) (t) − y(n−1) (s)| ≤ t s η(ξ)dξ for a < t < b hold. Thus in view of definition 2.1 the operator U maps the ball Aγ0 into its own subset S(ρ1, η). From lemma 2.2 follows that S(ρ1, η) is the compact subset of the ball Aγ0 ⊂ Cm−1 1 (]a, b[). i.e. the operator u maps the ball Aγ0 into its own compact subset. Therefore, owing to Schauders’s principle, there exists x ∈ S(ρ1, η) ⊂ Aγ0 , such that x(t) = U(x)(t) for a < t < b. Thus by (2.51) and notation (3.4), the function x (x ∈ Aγ0 ) is a solution of problem (1.26), (1.2), where (3.7) λ = χ(||x||Cm−1 1 ). If γ0 = ρ0 then in view of condition x ∈ Aγ0 , by (3.3) we have that λ = 1, and then in view of (2.51) and (3.4) the function x is a solution of problem (1.1), (1.2) which admits to the estimate (1.27). Let us show now, that x admits estimate (1.27) in the case when ρ0 < γ0. Assume the contrary. Then either (3.8) ρ0 < ||x||Cm−1 1 < 2ρ0, or (3.9) ||x||Cm−1 1 ≥ 2ρ0. If condition (3.8) holds, then by virtue of (3.3) and (3.7) we have that λ ∈]0, 1[, which by the conditions of our theorem guarantees the validity of estimate (1.27). But this contradict (3.8). Assume now that (3.9) is fulfilled. Then by virtue of (3.3) and (3.7) we have that λ = 0. Therefore x ∈ Aγ0 is a solution of problem (2.52), (1.2). Thus from item ii of Definition 1.1 it is obvious that x ≡ 0, because problem (2.52), (1.2) has only a trivial solution. But this contradict condition (3.9), i.e. estimate (1.27) is valid. From estimate (1.27) and (3.3) we have that λ = 1, and then in view of (2.51) and (3.4) the function x is a solution of problem (1.1), (1.2) which admits to the estimate (1.27). 20 Proof of Corollary 1.2. First note that in view of condition (1.30) exists such γ0 > 2ρ0, that condition (1.25) holds, and in view of definition 1.2 the operator P is γ0, γ consistent. On the other hand from (1.30) follows the existence of the number ρ0, such that (3.10) γ||η(·, ρ)||L2 2n−2m−2, 2m−2 < ρ for ρ > ρ0. Let x be a solution of problem (1.26), (1.2) for some λ ∈]0, 1[. Then y = x is also a solution of problem (1.22), (1.2) where q(t) = λ F(x)(t) − P(x, x)(t) . Let now ρ = ||x||Cm−1 1 and assume that (3.11) ρ > ρ0. holds. Then in view of the γ−consistency of operator p with boundary conditions (1.2), inequality (1.23) holds and thus by condition (1.28) we have ρ = ||x||Cm−1 1 ≤ γ||q(x)||L2 2n−2m−2, 2m−2 ≤ γ||η(·, ρ)||L2 2n−2m−2, 2m−2 . But the last inequality contradict (3.10). Thus assumption (3.11) is not valid and ρ ≤ ρ0. Therefore for any λ ∈]0, 1[ an arbitrary solution of the problem (1.26), (1.2) admits the estimate (1.27). Therefore all the conditions of Theorem 1.3 ar fulfilled, from which the solvability of problem (1.1), (1.2) follows. Proof of theorem 1.4. Let rn be the constant defined in Remark 1.1. First prove that operator P is γ0, rn consistent with boundary conditions (1.2). From the conditions of our theorem it is obvious that the item (i) of definition 1.1 is satisfied. Let now x be an arbitrary fixed function from the set Aγ0 and let pj(t) ≡ pj(x)(t). Thus in view of (1.34), (1.35) all the assumptions of Theorem 1.1 are satisfied, and then for any q ∈ L2 2n−2m−2, 2m−2(]a, b[) the problem (1.22), (1.2) has unique solution y. Also in view of Remark 1.1 there exists the constant rn > 0, (which depends only on the numbers lkj, lkj, γkj (k = 0, 1; j = 1, · · · , m), and a, b, t∗ , n) such that estimate (1.23) holds with γ = rn. I.e., the operator P is γ0, rn consistent with boundary conditions (1.2). Therefore all the assumptions of Corollary 1.1 are fulfilled, from which the solvability of problem (1.1), (1.2) follows. Proof of theorem 1.5. Let rn be the constant defined in Remark 1.1. First prove that operator P is rn consistent with boundary conditions (1.2). From the conditions of our theorem it is obvious that the item (i) of definition 1.1 is satisfied. Let now γ0 be an arbitrary nonnegative number, x be arbitrary fixed function from the space Aγ0 and let pj(t) ≡ pj(x)(t). Then in view of (1.37), (1.38) all the assumptions of Theorem 1.1 are satisfied and then for any q ∈ L2 2n−2m−2, 2m−2(]a, b[) the problem (1.22), (1.2) has unique solution y. Also in view of Remark 1.1 there exists the constant rn > 0, (which depends only on the numbers lkj, lkj, γkj (k = 0, 1; j = 1, · · · , m), and a, b, t∗ , n,) such that estimate (1.23) holds with γ = rn. I.e., the operator P is γ0, rn consistent with boundary conditions (1.2) for arbitrary γ0 > 0. Thus by Definition 1.1, the operator P is rn consistent with boundary conditions (1.2). Therefore all the assumptions of Corollary 1.2 are fulfilled, from which follows the solvability of problem (1.1), (1.2) follows. 21 Proof of Remarc 1.2. By the Schwartz’s inequality, definition of the norm ||y||Cm−1 1 and inequalities (1.39), (2.2) for ani x, y ∈ Aγ0 and z = y − x we have (3.12) |pj(y)(t)z(j−1) (τj(t))| = |pj(y)(t)z(j−1) (t)| + |pj(y)(t)| τj (t) t z(j) (ψ)dψ ≤ ≤ ||z||Cm−1 1 |pj(y)(t)|αj(t) 1 + 1 αj(t) τj (t) t (ψ − a)2m−2j dψ 1/2 for a < t < b. On the other hand, from the conditions (1.40) by Lemmas 2.8 and 2.9 it is cleat that α−1 j (s) τj (s) s (ξ − a)2m−2j dξ 1/2 ≤ κ(1 + κ) εm−j+1/2 for s ∈]a, a + ε] ∪ [b − ε, b[, α−1 j (s) τj (s) s (ξ − a)2m−2j dξ 1/2 ≤ ε−2m+2j−1 b a (ξ − a)2m−2j dξ 1/2 = = (b − a)m−j+1/2 √ 2m − 2j + 1ε2m−2j+1 for s ∈]a + ε, b − ε[. Then if we put (3.13) κ0 = max 1≤j≤m κ(1 + κ) εm−j+1/2 , (b − a)m−j+1/2 √ 2m − 2j + 1ε2m−2j+1 , from (3.12) by the last estimates we get the inequality (3.14) |pj(y)(t)z(j−1) (τj(t))| ≤ ||z||Cm−1 1 (1 + κ0)|pj(y)(t)|αj(t) ≤ ≤ ||z||Cm−1 1 (1 + κ0)δj(t, ||y||Cm−1 1 ) for a < t < b. Analogously we get that |(pj(y)(t) − pj(x)(t))x(j−1) (τj(t))| ≤ ||x||Cm−1 1 (1 + κ0)|pj(y)(t) − pj(x)(t)|αj(t) for a < t < b. from (3.14) and the last inequality it is obvious that the operator P defined by equality (1.19) continuously acting from Aγ0 to the space Ln(]a, b[), and the item (ii) of definition 1.1 holds, with δ(t, ρ) = (1 + κ0) m j=1 δj(t, ρ). Proof of Corollary 1.3. From conditions (1.42) and (1.40) by the Remark 1.2 we obtain that the operator P defined by equality (1.19) with pj(x)(t) = pj(t), continuously acting from Aγ0 to the space Ln(]a, b[), for any γ0 > 0, i.e., continuously acting from Cm−1 1 (]a, b[) to the space Ln(]a, b[). 22 Therefore it is clear that all the conditions of Theorem 1.5 would be satisfied with F(x)(t) = f(t, x(τ1(t)), x′ (τ2(t)), · · · , x(m−1) (τm(t))), δ(t, ρ) = (1 + κ0) m j=1 |pj(t)|, where the constant κ0 is defined by equality (3.13). Thus problem (1.41), (1.2) is solvable. Proof of Corollary 1.4. Let the operators F, p1 : Cm−1 (]a, b[) → Lloc(]a, b[), and the function η :]a, b[×R+ → R+ be defined by equalities F(x)(t) = − λ|x(t)|k [(t − a)(b − t)]2+k/2 x(τ(t)) + q(x)(t), p1(x)(t) = − λ|x(t)|k [(t − a)(b − t)]2+k/2 . Then it is easy to verify that in view of (1.46)-(1.48), conditions (1.13), (1.14), (1.28), (1.34)(1.43) are satisfied with (3.15) δ(t, ρ) = ρk λ [(t − a)(b − t)]2 , l01 = l11 = 4γk 0 λ (b − a)2 , l01 = l11 = 16γk 0 λ (b − a)2 , r2 = 1 + 2 b − a 2(1 + b − a)(b − a)2 (b − a)2 − 16λγk 0 (1 + [2(b − a)]1/4) , B0 = B1 = 16λγk 0 (b − a)2 (1 + [2(b − a)]1/4 ), t∗ = (a + b)/2, γ01 = γ11 = 1 4 . Thus all the condition of theorem 1.4 are satisfied, from which follows solvability of problem (1.44), (1.2). Proof of Corollary 1.5. Let the operators F, p1 : Cm−1 (]a, b[) → Lloc(]a, b[), and the function η :]a, b[×R+ → R+ be defined by equalities F(x)(t) = − λ| sin xk (t)| [(t − a)(b − t)]2 x(τ(t)) + q(x)(t), p1(x)(t) = − λ| sin xk (t)| [(t − a)(b − t)]2 . 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