Two-Point Boundary Value Problems For Strongly Singular Higher-Order Linear Differential Equations With Deviating Arguments

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

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 p j (t)u (j−1) (τ j (t)) + q(t) for a < t < b, with the two-point boundary conditions Here n ≥ 2, m is the integer part of n/2, −∞ < a < b < +∞, p j , q ∈ L loc (]a, b[) (j = 1, • • • , m), and τ j :]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.
The first step in studying the linear ordinary differential equations under conditions (1.2) or (1.3), in the case when the functions p j 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].
Throughout the paper we use the following notation.

Lemma on the property of functions from the space
Lemma 2.5.Let where 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 where and where Throughout this section, when problems (1.1), (1.2) and (2.23), (2.24) are discussed we assume that and for an arbitrary (m − 1)-times continuously differentiable function and for an arbitrary (m − 1)-times continuously differentiable function From the definition of the functions µ j (j = 1, . . ., m), the estimate Proof.We have to prove that for any δ ∈]0, min{b − t * , t * − a}[, and ε > 0, there exists a constant n 0 ∈ N such that . Then from the inclusions x ), conditions (2.33) and (2.34), it follows the existence of such constant n 0 ∈ N that |x we have (2.36).
To complete the proof of the lemma, it remains to show that equality (2.41) is satisfied.First note that in the space By virtue of the Arzela-Ascoli lemma and condition (2.37) the sequence {u (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].
EJQTDE, 2012 No. 38, p. 21 Then there exist positive constants δ and r 1 such that if (2.78) satisfies the inequality (2.80) Proof.From conditions (1.12) and (1.13) it follows the existence of constants Consequently, all the requirements of Lemma 2.3 with p j (t) = (t − a) n−2m (−1) n−m p j (t), a < t 0 < a 0 , and Lemma 2.4 with p j (t) = (b − t) n−2m (−1) n−m p j (t), b 0 < t 1 < b, are fulfilled.Also from condition (1.14) and the definition of a constant ν n , it follows the existence of ν ∈]0, 1[ such that On the other hand, without loss of generality we can assume that where the functions β j are defined by (2.6).Let now q ∈ L 2 2n−2m−2, 2m−2 (]a, b[), u be a solution of problem (2.78), (2.79), and (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 t 0 to t 1 , by Lemma 2.10 we obtain (2.84) , and the equalities ρ 0 (t 0 ) = ρ 1 (t 1 ) = 0, by (2.81) we get If along with this we take into account inequalities (2.82) and a 0 ≤ b 0 , we find (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 The proof of the following lemma is analogous to that of Lemma 2.11.
Lemma 2.12.Let a 0 ∈]a, b[, the functions h j and the operators f j be given by equalities (1.10) and (1.11).Let, moreover, τ j ∈ M(]a, b]), constants l 0,j > 0, γ 0j > 0, (j = 1, . . ., m) be such that conditions (1.12) and (1.21) are fulfilled.Then there exists a positive constant r 1 such that for any t 0 ∈]a, a 0 [, and q , hold, and let in the case when n is odd, in addition (1.8) be fulfilled, where the functions h j , β j and the operators f j are given by equalities (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−b 0 , a 0 −a}] be such as in Lemma 2.11 and assume that estimate (2.91) is invalid.Then for an arbitrary natural k there exist (2.93) In the case when the homogeneous equation q k (t). (2.94) Then v k is a solution of the problem (2.95) Moreover, in view of (2.93), it is clear that where r 0 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 ∈ L 2 2n−2m−2, 2m−2 (]a, b[), estimate (2.91) holds.Thus the homogeneous equation  (2.98) for a < t ≤ a + δ.
Proof.First note that From this inequality it immediately follows the validity of the lemma.

Proofs
Proof of Theorem 1.
where the constant r does not depend on q.From (3.2), by Lemma 2.8 with Thus that problem has Fredholm's property, and its solution admits estimate (1.15) (estimate (1.22)).