On Periods of Non-constant Solutions to Functional Differential Equations

We show that periods of solutions to Lipschitz functional differential equations cannot be too small. The problem on such periods is closely related to the unique solvability of the periodic value problem for linear functional differential equations. Sharp bounds for periods of non-constant solutions to functional differential equations with Lipschitz nonlinearities are obtained.


Introduction
Consider a problem on periodic solutions of the differential equation with deviating argument where x(t) ∈ R m , f : R m → R m is a Lipschitz function, τ : R → R is a measurable function.
If τ(t) ≡ t, the sharp lower estimate for periods T of non-constant periodic solutions to (1.1) is obtained in [28] for n = 1 and [16] for n 1 for Lipschitz f in the Euclidian norm, and in [30] for even n and Lipschitz functions f satisfying the condition max i=1,...,m 3) The estimate (1.2) gives the minimal time required for an object described by a system of ordinary differential equations with the Lipschitz constant L to return to its initial state.

E. Bravyi
For equations (1.1) with an arbitrary piece-wise continuous deviating argument τ and Lipschitz f under condition (1.3), the best constants in the lower estimates for periods T of non-constant periodic solutions are found by A. Zevin for n = 1 [29] T 4/L, and for even n [30] T α(n)/L 1/n , where α(n) are defined with the help of solutions to some boundary value problem for an ordinary differential equation of n-th order.
Here, for all n, we find a simple representation of the best constants in the estimate for periods of non-constant periodic solutions of some more general equations than (1.1) with Lipschitz nonlinearities.Some properties of the sequence of the best constants will be obtained.It turns out that the best constants in lower estimates of periods are the Favard constants [7, § 4.2].
If equation (1.1) has a T-periodic solution x with absolutely continuous derivatives up to the order n − 1, then the restriction of x to the interval [0, T] is a solution to the periodic boundary value problem where If boundary value problem (1.4) has no non-constant solutions, then (1.1) has no T-periodic non-constant solutions either.Therefore, we can consider the equivalent periodic boundary value problem for a system of m functional differential equations of the n-th order where x belongs to the space AC n−1 ([0, T], R m ) of all functions with absolutely continuous derivatives up to order n − 1; the equality x (n) (t) = (Fx)(t) holds for almost all t ∈ [0, 1]; the continuous operator F acts from the space C([0, T], R m ) of all continuous functions into the space L ∞ ([0, T], R m ) of all measurable essentially bounded functions (with the norm z L ∞ = max i=1,...,m ess sup t∈[0,T] |z i (t)|).
We assume that there exists a positive constant L ∈ R such that the following inequality holds max i=1,...,m ess sup where τ : [0, T] → [0, T] is a measurable function (not equivalent to a constant), then condition (1.6) implies that the function f : R m → R m is Lipschitz and satisfies (1.3).
Our approach is close to the work [25], where the periodic boundary value problem is considered on the interval and a general way for obtaining the lower estimate of the periods of non-constant solutions is proposed.
Note that there are a number of papers on minimal periods of non-constant solutions for different classes of equations, in particular, [13] in Hilbert spaces, [14] in Banach spaces for equations with delay, [24,27] in Banach spaces, [5] in Banach spaces for difference equations, [17] in Banach spaces for equations with differentiable delays, [23] in spaces p and L p .
Theorem 2.2.If F satisfies inequality (1.6) and periodic problem (1.5) has a non-constant solution, then To prove Theorem 2.2, we need two lemmas. (2.5) 3) has the Fredholm property [2].Hence, this problem is uniquely solvable if and only if the homogeneous problem has only the trivial solution.Let y be a nontrivial solution of (2.6).From [15, D. 32, p. 386] (or [26]) it follows that for some constant C 1 and for all constants ξ the solution y satisfies the equality then the linear operator A in the right-hand side of (2.7) is a contraction mapping in L ∞ ([0, T], R).In this case for each C 1 , equation (2.7) has a unique solution which is a constant (we use here the equality T 0 φ n (2πt/T) dt = 0).From (2.6) it follows that this constant is zero.Therefore, problem (2.3) is uniquely solvable.
Proof of Theorem 2.2.Let (1.5) have a non-constant solution.From Lemma 2.3 it follows that the non-constant component x j (from the proof of Lemma 2.3) of the solution x to (1.5) is a solution to (2.3) with some constant C and some measurable function τ : [0, T] → [0, T].If inequality (2.5) holds, it follows from Lemma 2.4 that this solution is unique: x j (t) ≡ −C/L.Then from (1.6) it follows that each component x i of the non-constant solution x is constant.Therefore, inequality (2.5) does not hold, and inequality (2.2) is valid.Now assume that an operator F in equation (1.5) acts into the space of integrable functions L 1 ([0, T], R m ) with the norm Let there exist positive functions p i ∈ L 1 ([0, T], R), i = 1, . . ., m, such that for every x ∈ C([0, T], R m ) the inequality max i=1,...,m ess sup x i (t) (2.9) holds.If periodic problem (1.5) has a non-constant solution, then the following inequalities (2.10) are fulfilled for each i = 1, . . ., m.
To prove Theorem 2.5, we also need two lemmas.
Lemma 2.6.Let F satisfy inequality (2.9).If problem (1.5) has a non-constant solution, there exist a measurable function τ : [0, T] → [0, T] and a constant C such that one of non-constant components of the solution satisfies the scalar periodic boundary value problem (2.11) Proof.Suppose y = x j is a non-constant component of the solution x to (1.5) such that equality (2.4) holds.Then the essential diameter of the range of the function (Fx) j /p j does not exceed the length of the range of x j .So, there exist a measurable function τ : [0, T] → [0, T] and a constant C such that where p = p j .This proves the lemma. (2.12) For n = 1, n = 2, n = 3, n = 4 this lemma is proved in [8,[18][19][20], for arbitrary n in [4,9,10].
Proof of Theorem 2.5.Let (1.5) have a non-constant solution.From Lemma 2.6 it follows that a non-constant component x j (from the proof of Lemma 2.6) of the solution x to (1.5) is a solution to (2.11), where p = p j , C is some constant, τ : [0, T] → [0, T] is some measurable function.If condition (2.12) hold, from Lemma 2.7 it follows that the solution x j is unique: x j (t) ≡ −C.From (2.9) it follows that each component x i of the non-constant solution x is constant.Therefore, inequalities (2.12) do not hold, and inequalities (2.10) are valid.

The sharpness of estimates
The estimates (2.2) and (2.10) in Theorems 2.2 and 2.5 are sharp.The sharpness of (2.10) is shown in [4].The sharpness of (2.2) for even n was shown in [30] in other terms.Now for every n 1 we obtain functions τ : [0, T] → [0, T] such that the periodic boundary value problem has a non-constant solution provided that (2.2) is an identity:L = 1 K n T n .Find a solution to the auxiliary problem where h(t) = 1 for t ∈ [0, T/2] and h(t) = −1 for t ∈ (T/2, T].Since T 0 h(t) dt = 0, this problem has a solution.It is not unique and defined by the equality where C is an arbitrary constant, G(t, s) is the Green function of the problem x(0) = 0, We have a simple representation for the Green function G(t, s): where B n (t) are the Bernoulli polynomials [1, p. 804] which can be defined as unique solutions to the problems for every constant C ∈ R.
Note that these functions τ were found in [30].

Example. Equations with "maxima"
Let L be a constant, τ, θ : R → R measurable functions such that τ(t) θ(t) for all t ∈ R. From Theorem 2.2, it follows that periods T of non-constants solutions of the equation where the constants K n are defined by (2.1).Suppose p : R → R is a positive locally integrable T-periodic function: p(t + T) = p(t), p(t) > 0 for all t ∈ R. From Theorem 2.5, it follows that if there exists a T-periodic nonconstants solution to the equation

Conclusion
Now we formulate unimprovable necessary conditions for the existence of a non-constant periodic solution to (1.5) which follow from Theorems 2.2 and 2.5: if F satisfies (1.6) and there exists a non-constant solution to (1.5), then the constants L = L n satisfy the inequalities L 1 4/T, L 2 32/T 2 , L 3 132/T 3 , L 4 6144/(5T 4 ), L 5 7680/T 5 , . . .; if F satisfies (2.9) and there exists a non-constant solution to (1.5), then the constants P = P n = max i=1,...,n p i L 1 satisfy the inequalities New results on existence and uniqueness of periodic solutions for higher order functional differential equations are obtained in [11,12,21,22].Note that Theorems 2.2 and 2.5 cannot be derived from the results of these articles.
The short message on these results (without proofs) was published in the journal "Russian Mathematics" (Iz.VUZ), No. 12 (2013) in Russian.

Lemma 2 . 3 .
Let F satisfy(1.6).If problem (1.5) has a non-constant solution, there exist a measurable function τ : [0, T] → [0, T] and a constant C such that at least one of non-constant components of the solution satisfies the scalar periodic boundary problem
.1) where B n are the Bernoulli numbers, E n are the Euler numbers (see, for examples, [1, p. 804]).
Suppose y = x j is a non-constant component of the solution x to (1.5) such that Let L > 0. Periodic boundary value problem (2.3) has a unique solution for every measurable τ : [0, T] → [0, T] and for every constant C .3) E. Bravyi Proof.
are the periodic Bernoulli functions, {t} is the fractional part of t.