On Solvability of Periodic Boundary Value Problems for Second Order Linear Functional Differential Equations

The periodic boundary value problem for second order linear functional differential equations with pointwise restrictions (instead of integral ones) is considered. Sharp sufficient conditions for the solvability are obtained.

Here we research a rather common case when the linear functional operator T of the second-order functional differential equation is the difference of two positive operators: T = T + − T − .A new class of sufficient conditions for the existence of periodic solutions is offered.For arbitrary given non-negative functions p + , p − , we find sharp sufficient conditions for the existence of periodic solutions to all functional differential equations with linear positive operators T + , T − such that T + 1 = p + , T − 1 = p − (1 is the unit function).To the best of our knowledge, for arbitrary functions p + , p − , such conditions are new.
Our main result is that we can find all real numbers λ such that problem (1.1) is uniquely solvable for all operators T from the operator family S(p + , p − ).It allows to obtain some new sufficient conditions for the solvability.These conditions are unimprovable in a sense.It means that if our conditions are not fulfilled, then there exists an operator T ∈ S(p + , p − ) such that boundary value problem (1.1) is not uniquely solvable.
Suppose two different non-negative numbers P + , P − are given.Sharp sufficient conditions for the solvability of (1.1) for all operators T from the unify of sets S(p + , p − ) with all non-negative p + , p − satisfying the equalities 1 0 p + (s) ds = P + , 1 0 p − (s) ds = P − , are obtained in [10] (see condition (2.7)).However, if we consider the boundary value problem (1.1)only for the operator family S(p + , p − ) with given non-negative functions p + , p − , we can essentially improve the results (see Example 2.6).
Here in Section 2, Theorems 2.1, 2.2 contain conditions for the unique solvability of (1.1) for all operators T from the family S(p + , p − ) with arbitrary non-negative functions p + , p − .Further, we refine these results when p − is the zero function, and the function p + has one symmetry axis (Theorem 2.7), two symmetry axes (Theorem 2.10), or three symmetry axes (Theorem 2.12).It follows from Theorem 2.12 that the consideration of the cases with more symmetries does not improve the results.
In Section 3, all proofs are given.
The well-known integral sufficient conditions for the solvability of (1.1) from [10] gives the following result: problem (1.1) is uniquely solvable for all T ∈ S(p It is obvious that the sufficient condition for the solvability obtained in Theorem 2.2 0 < |λ| 15.4 is better.Moreover, if |λ| 15.5, then there exists an operator T ∈ S(p + , p − ) such that problem (1.1) is not uniquely solvable.
Let 0(t) = 0, t ∈ [0, 1], be the zero function. where It is easy to compute that max Further we consider symmetric functions p + and p − = 0.It makes the computation of λ * much more easier, especially in Theorem 2.12.

Theorem 2.7. Let p
(2.8) It is not difficult to show that λ * (p + , p − ) can take any value from the interval (0, +∞) for different functions p + , p − under the conditions of Theorem 2.2.Moreover, under the conditions of Theorems 2.2 and 2.7, we have λ * (p + , 0) ∈ (16, ∞).It follows from this that the periodic boundary value problem (1.1) is uniquely solvable for every This result is well known.For the first time, the best constant 16 was obtained in [15] for ordinary differential equations, and in [19] for functional differential equations (non-linear).
, where We have 1 max , where We have 1 max It is easy to show that λ * (p, 0) can take any value from the interval (16,32] under the conditions of Theorem 2.10.Theorem 2.12.Let p − = 0, p + = p, where and Then Theorems 2.2, 2.7, 2.10 and 2.12 can be reformulated in the form of sharp sufficient conditions for the solvability.In particular, Theorems 2.1 and 2.2 have the following equivalent version.

13)
If P = 0 or inequality (2.13) is not fulfilled, then there exists an operator T ∈ S(p + , p − ) such that problem (2.12) is not uniquely solvable.

Proofs
We need two lemmas for proving Theorems 2.1 and 2.2.The proof of the following lemma on the Fredholm property can be found in [28, p. 85] has only the trivial solution.
Lemma 3.2.Let T ∈ S(p + , p − ).Then for every x ∈ C[0, 1], there exist (depending on x) points t 1 , t 2 ∈ [0, 1] and functions p 1 , p 2 ∈ L[0, 1] satisfying the conditions such that the following equality holds: x(t) = x(t 1 ). Then Hence, there exists a measurable function ξ Therefore, the chosen points t 1 , t 2 and the functions p 1 , p 2 defined by the equalities satisfy the conditions of the lemma.
Remark 3.3.It follows from the proof of Lemma 3.2, that some maximum and minimum points of x can be taken as points t 1 and t 2 in Lemma 3.2.
Remark 3.5.It is obvious that if y is a solution of (3.12), then −y is also a solution.Therefore, if (3.12) has a non-trivial solution, then this problem has a solution satisfying (3.13).
Proof.By Lemma 3.4 and Remark 3.5, there exists a measurable g : [0, 1] → [τ 1 , τ 2 ] such that boundary value problem (3.15) has a non-trivial solution y satisfying (3.13).Note, that under the conditions of the lemma, a solution of (3.15) has a zero at some point t 0 .Define the intervals t 0 , otherwise.
Then, using the equality p , it is easy to prove that the boundary value problem (3.16) has the solution with a minimum at the point 1/2 and a maximum at the point 1. If satisfy (2.9), T ∈ S(p + , 0), and the homogeneous boundary value problem (3.12) have a non-trivial solution.Then there exists a measurable function h : [0, 1] → [0, 1] such that problem (3.16) has a non-zero solution with some maximum and minimum points (t 1 , t 2 ) ∈ R 2 .
Proof.By Lemmas 3.4 and Remark 3.5, there exists a measurable g : [0, 1] → [τ 1 , τ 2 ] such that the boundary value problem (3.15) has a non-trivial solution y with a minimum point τ 1 and a maximum point τ 2 > τ 1 .Under the conditions of Lemma 3.8 a solution of (3.15) has a zero at some point t 0 .
Define the intervals . By Lemma 3.7, we have to consider only three cases. If Using condition (2.9), it is easy to prove that the boundary value problem (3.16) has the solution t 0 , otherwise.
In this case, using condition (2.9), we can also show that the boundary value problem (3.16) has the solution with a minimum at the point 1/4 and a maximum at the point 3/4.So, in all cases, there exists a measurable function h with the required properties.
For every i = 1, 2, repeating the proof of Theorems 2.1 and 2.2 for the set of pairs (t 1 , t 2 ) R i instead of the set of all pairs {(t ) We introduce some notation: the points t 3 ∈ [0, t 1 ] and t 4 ∈ [t 1 , t 2 ] satisfy the equalities and where 0 θ 1 θ 2 1 and ) Thus we have

.22)
We will show that max It will prove Theorem 2.10, because
Let .25)Therefore, the points θ 3 , θ 4 are on opposite sides of the point θ 0 .Using (3.25), one can prove that