Open Access

On estimates of solutions of the periodic boundary value problem for first-order functional differential equations

Boundary Value Problems20142014:119

DOI: 10.1186/1687-2770-2014-119

Received: 29 January 2014

Accepted: 6 May 2014

Published: 15 May 2014

Abstract

Inequalities for periodic solutions of first-order functional differential equations are obtained. These inequalities are best possible in a certain sense.

MSC:34K06, 34K10, 34K13.

Keywords

functional differential equations periodic solutions periodic boundary value problem estimates of solutions

1 Introduction

Periodic solutions of functional differential equations are important in different applications (see, for example, [14] and the references therein, and also works on the general theory of boundary value problems for functional differential equations [511]). Conditions for the solvability of first-order periodic problems are found in [1223]. In [15, 16] the linear case is considered, and unimprovable sufficient conditions for the solvability of the periodic problem
x ˙ ( t ) = ( T + x ) ( t ) ( T x ) ( t ) + f ( t ) , t [ a , b ] ,
(1)
x ( a ) = x ( b ) ,
(2)
are found in terms of the norms T + , T of linear positive functional operators T + , T : C L :
T 1 T < T + < 2 ( 1 + 1 T )
(3)
or
T + 1 T + < T < 2 ( 1 + 1 T + ) .
(4)

If both of these conditions are not satisfied for some norms T + , T , there exist linear positive operators T + , T with these norms such that problem (1)-(2) has no solution. As to our knowledge, similar unimprovable estimates for solutions of (1)-(2) in terms of norms T + , T are yet unknown. Here we will fill this gap. Moreover, the estimates obtained here (in Theorems 1, 2, 3) can be expanded to some non-linear functional differential equations (see Remark 1). Theorem 1 gives the best possible estimates of the norm of the Green operator for the periodic boundary value problem. In Theorem 2, we obtain unimprovable estimates of the solutions of (1)-(2) for non-negative f. In Theorem 3, unimprovable bounds of the difference between the maximum and the minimum of a solution are established.

We use the following notation: is the space of real numbers, C is the space of continuous functions x : [ a , b ] R with the norm x C = max t [ a , b ] | x ( t ) | ; L is the space of integrable functions z : [ a , b ] R with the norm z L = a b | z ( t ) | d t ; a linear bounded operator T : C L is called positive if it maps non-negative functions from C into almost everywhere non-negative functions from L.

Consider the periodic boundary value problem (1)-(2), where f L , T + , T : C L are linear positive operators with norms T + T + C L = a b ( T + 1 ) ( t ) d t , T T C L = a b ( T 1 ) ( t ) d t , 1 is the unit function. An absolutely continuous function x : [ a , b ] R is called a solution of the problem if it satisfies the periodic boundary condition (2) and equation (1) for almost all t [ a , b ] . We have to solve problem (1)-(2) if, for example, we search for periodic solutions of the equation with delay
x ˙ ( t ) = p ( t ) x ( t τ ( t ) ) + f ( t ) , t R ,
(5)
where p , f : R R are ( b a ) -periodic locally integrable functions, τ : R R is a measurable ( b a ) -periodic non-negative delay. Indeed, suppose that linear operators T + and T are defined by the equalities
( T + x ) ( t ) = p ( t ) + | p ( t ) | 2 x ( τ ˜ ( t ) ) , ( T x ) ( t ) = | p ( t ) | p ( t ) 2 x ( τ ˜ ( t ) ) , t [ a , b ] ,

where τ ˜ ( t ) = t τ ( t ) + k ( t ) ( b a ) and the integer numbers k ( t ) are such that τ ˜ ( t ) [ a , b ] for almost all t R . It is easy to show that problem (1)-(2) has a solution if and only if equation (5) has a periodic solution with the period b a .

The conditions (3), (4) for the norms of the operators T + , T : C L are well known [15]. They guarantee the existence and uniqueness of solutions of problem (1)-(2). Note that these conditions are unimprovable in the following sense: if non-negative numbers T + , T satisfy neither (3) nor (4), then problem (1)-(2) has no solution for some linear positive operators T + , T : C L with norms T + C L = T + , T C L = T and for some f L .

2 The main results

In what follows, we suppose that one of conditions (3), (4) is fulfilled. First, we formulate the results only for the simplest problem (1)-(2) with the null operator T + :
x ˙ ( t ) = ( T x ) ( t ) + f ( t ) , t [ a , b ] , x ( a ) = x ( b ) ,
(6)

where T : C L is a linear positive operator with norm T , f L . The assertions of the following Theorems 1, 2, 3 for problem (6) are as follows.

The solution x of (6) satisfies the estimates
max t [ a , b ] | x ( t ) | { 1 + T T a b | f ( t ) | d t if  0 < T 3 , 4 T ( 4 T ) a b | f ( t ) | d t if  3 < T < 4 ,
(7)
max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) { a b | f ( t ) | d t if  0 < T 1 , 1 2 T T a b | f ( t ) | d t if  1 < T < 4 .
(8)
If a function f is non-negative, the solution x of (6) satisfies the estimates
1 T T a b f ( t ) d t x ( t ) 1 + T T a b f ( t ) d t if  0 < T 2 , 1 4 T a b f ( t ) d t x ( t ) 1 + T T a b f ( t ) d t if  2 < T 3 , 1 4 T a b f ( t ) d t x ( t ) 4 T ( 4 T ) a b f ( t ) d t if  3 < T < 4 .
(9)

All estimates (7), (8) and (9), which are proved in Theorems 1, 2, 3 in the general case, are best possible (see Remarks 3, 5, 6).

Remark 1 Consider also the non-linear periodic problem
x ˙ ( t ) = ( F + x ) ( t ) ( F x ) ( t ) + f ( t ) , t [ a , b ] ,
(10)
x ( a ) = x ( b ) ,
(11)
provided there exist non-negative functions p + , p L with norms
p + L = T + , p L = T
(12)
such that the operators F + , F : C L satisfy the inequalities
p + ( t ) min t [ a , b ] x ( t ) ( F + x ) ( t ) p + ( t ) max t [ a , b ] x ( t ) for a.a.  t [ a , b ] ,
(13)
p ( t ) min t [ a , b ] x ( t ) ( F x ) ( t ) p ( t ) max t [ a , b ] x ( t ) for a.a.  t [ a , b ]
(14)

for all x C .

It follows from Lemma 3 and the proofs of Theorems 1, 2, 3 that all statements of these theorems are also valid for solutions of periodic problem (10)-(11) (if the solutions exist).

Theorem 1 If the norms T + < T of the linear positive operators T + , T : C L satisfy the conditions
3 T < 2 ( 1 + 1 T + ) , T + < 3 / 4 ,
(15)
and x is a solution of (1)-(2), then the inequality
max t [ a , b ] | x ( t ) | 1 T ( 1 T / 4 ) T + a b | f ( t ) | d t
(16)

holds.

If the norms T + < T of the operators T + , T : C L satisfy
T + 1 T + < T 3 , T + < 3 / 4 ,
(17)
and x is a solution of problem (1)-(2), then the inequality
max t [ a , b ] | x ( t ) | 1 + T T ( 1 T + ) T + a b | f ( t ) | d t
(18)

holds.

Remark 2 ([15])

If T > T + 0 and both of the conditions (15), (17) are not fulfilled, then there exist linear positive operators T , T + with norms T , T + and a function f L such that problem (1)-(2) has no solution.

Remark 3 From the proof of Theorem 1 it follows that estimates (16), (18) are best possible: if non-negative numbers T , T + satisfy (15) (or (17)), then equality holds in condition (16) (or (18)) for a unique solution x of problem (1)-(2) for some linear positive operators T , T + with norms T , T + and for some function f L , f 0 .

The estimates of solutions (1)-(2) for T < T + can be obtained in the same way.

Theorem 1 If the norms T + > T of the linear positive operators T + , T : C L satisfy the conditions
3 T + < 2 ( 1 + 1 T ) , T < 3 / 4 ,
(19)
and x is a solution of (1)-(2), then the inequality
max t [ a , b ] | x ( t ) | 1 T + ( 1 T + / 4 ) T a b | f ( t ) | d t
(20)

holds.

If the norms T + > T of the operators T + , T : C L satisfy
T 1 T < T + 3 , T < 3 / 4 ,
(21)
and x is a solution of problem (1)-(2), then the inequality
max t [ a , b ] | x ( t ) | 1 + T + T + ( 1 T ) T a b | f ( t ) | d t
(22)

holds.

Remark 2 ([15])

If T > T + 0 and both of conditions (19), (21) are not fulfilled, then there exist linear positive operators T and T + with norms T , T + and a function f L such that problem (1)-(2) has no solution.

Remark 3 From the proof of Theorem 1 it follows that estimates (20), (22) are best possible: if non-negative numbers T , T + satisfy (19) (or (21)), then equality holds in condition (20) (or (22)) for a unique solution x of problem (1)-(2) for some linear positive operators T , T + with norms T , T + and for some function f L , f 0 .

In the next statement we get the best possible lower bounds for solutions of problem (1)-(2) for non-negative f.

Theorem 2 Let x be a solution of problem (1)-(2) for some non-negative f.

If the norms T + , T of the operators T + , T : C L satisfy the conditions
max { 1 + 1 T + , T + 1 T + } < T < 2 ( 1 + 1 T + ) , T + < 3 / 4 ,
(23)
then
min t [ a , b ] x ( t ) 1 2 ( 1 + 1 T + ) T a b f ( t ) d t ;
(24)
if the norms T + , T of the operators T + , T : C L satisfy the conditions
max { 1 , T + 1 T + } < T 1 + 1 T + ,
(25)
then
min t [ a , b ] x ( t ) T 1 T T + a b f ( t ) d t ;
(26)
if the norms T + , T of the operators T + , T : C L satisfy the conditions
T + 1 T + < T 1 , T + < 1 / 2 ,
(27)
then
min t [ a , b ] x ( t ) 1 T T ( 1 + T + ) T + a b f ( t ) d t .
(28)

Remark 4 ([15])

If T > T + and all of conditions (23), (25), (27) are not fulfilled, then there exist linear positive operators T and T + with norms T , T + and a function f L such that problem (1)-(2) has no solution.

Remark 5 From the proof of Theorem 2 it follows that estimates (24), (26), (28) are best possible: if non-negative numbers T + , T satisfy (23) ((25) or (27)), then equality holds in condition (24) ((26) or (28)) for a unique solution x of problem (1)-(2) for some linear positive operators T , T + with norms T , T + and for some function f L , f 0 .

Now we estimate the difference between the maximum and the minimum of solutions.

Theorem 3 Let the solvability conditions (4) be fulfilled and x be a unique solution of (1)-(2). If
T > 1 , T + < T ( T 1 T + 1 ) 2 ,
then
max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) 1 2 T T + T a b | f ( s ) | d s ;
(29)
otherwise
max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) T T T + T T + a b | f ( s ) | d s .
(30)

Remark 6 From the proof of Theorem 3 it follows that inequalities (29) and (30) are unimprovable. It means that for every number T + , T satisfying the conditions of the theorem, equality holds in conditions (29) or (30) for the solution x of problem (1)-(2) for some positive operators T + , T : C L with norms T , T + , and for some non-negative function f L , f 0 .

Remark 7 Theorems 2, 3, as Theorem 1, can be easily reformulated for the case T + > T when the solvability condition (3) holds.

3 Proofs

We need three lemmas to prove the main theorems.

Lemma 1 Let T + , T : C L be linear positive operators, p + = T + 1 , p = T 1 , y C . Then there exist points t 1 , t 2 [ a , b ] and a function p 1 L satisfying
p ( t ) p 1 ( t ) p + ( t ) for a.a.  t [ a , b ]
(31)
such that the equality
( T + y ) ( t ) ( T y ) ( t ) = p 1 ( t ) y ( t 1 ) + ( p + ( t ) p ( t ) p 1 ( t ) ) y ( t 2 ) for a.a.  t [ a , b ]
(32)

holds.

Proof Let y ( t 1 ) = max t [ a , b ] y ( t ) , y ( t 2 ) = min t [ a , b ] y ( t ) . Since y C and the linear operators T + , T : C L are positive, we have
p + ( t ) y ( t 2 ) p ( t ) y ( t 1 ) ( T + y ) ( t ) ( T y ) ( t ) p + ( t ) y ( t 1 ) p ( t ) y ( t 2 ) for a.a.  t [ a , b ] .

Therefore, for some function p 1 L satisfying (31), equality (32) holds. □

Lemma 2 If y C , functions p + , p L are non-negative, and p 1 L satisfies (31), then there exist linear positive operators T + , T : C L with the norms
T + C L = p + L , T C L = p L
(33)

such that equality (32) holds.

Proof Let p 1 + ( t ) = ( | p 1 ( t ) | + p 1 ( t ) ) / 2 , p 1 ( t ) = ( | p 1 ( t ) | p 1 ( t ) ) / 2 , t [ a , b ] . Then the operators T + , T defined by the equalities
( T + x ) ( t ) = p 1 + ( t ) x ( t 1 ) + ( p + ( t ) p 1 + ( t ) ) x ( t 2 ) , t [ a , b ] , ( T x ) ( t ) = p 1 ( t ) x ( t 1 ) + ( p ( t ) p 1 ( t ) ) x ( t 2 ) , t [ a , b ] ,

satisfy the conditions of the lemma. □

Lemma 3 Let F + , F : C L satisfy (13)-(14), y C . Then there exist a function p 1 L satisfying (31) and points t 1 , t 2 [ a , b ] such that the equality
( F + y ) ( t ) ( F y ) ( t ) = p 1 ( t ) y ( t 1 ) + ( p + ( t ) p ( t ) p 1 ( t ) ) y ( t 2 ) for a.a.  t [ a , b ]
(34)

holds.

Proof Let y ( t 1 ) = max t [ a , b ] y ( t ) , y ( t 2 ) = min t [ a , b ] y ( t ) . Since y C and using (13), (14), we get
p + ( t ) y ( t 2 ) p ( t ) y ( t 1 ) ( F + y ) ( t ) ( F y ) ( t ) p + ( t ) y ( t 1 ) p ( t ) y ( t 2 ) for a.a.  t [ a , b ] .

Therefore, for some function p 1 L satisfying (31), equality (34) holds. □

Remark 8 It is obvious that one can choose the points t 1 and t 2 in Lemmas 1 and 3 in such a way that the solution y takes its maximum and minimum at these points.

Proofs of Theorems 1, 2, 3 If x is a solution of problem (1)-(2) ((10)-(11)), then by Lemma 1 (3) this solution satisfies the boundary value problem
x ˙ ( t ) = p 1 ( t ) x ( t 1 ) + ( p + ( t ) p ( t ) p 1 ( t ) ) x ( t 2 ) + f ( t ) , t [ a , b ] ,
(35)
x ( a ) = x ( b ) ,
(36)

where p 1 L and non-negative p + , p L satisfy (31), (33). If condition (3) or (4) holds, then problem (35)-(36) has a unique solution, which can be easily found explicitly. Since we are only interested in the maximal and minimal values of the solutions, by Remark 8, we have to obtain only representations for values x ( t 1 ) and x ( t 2 ) .

Let a t 1 < t 2 b , E [ t 1 , t 2 ] , I [ a , t 1 ] [ t 2 , b ] ,
I p 1 ( s ) d s E ( p + ( s ) p ( s ) ) d s E p 1 ( s ) d s I ( p + ( s ) p ( s ) ) d s a b ( p + ( s ) p ( s ) ) d s .
For x ( t 1 ) , x ( t 2 ) we have
x ( t 1 ) = 1 ( I f ( s ) d s E ( p + ( s ) p ( s ) p 1 ( s ) ) d s + E f ( s ) d s I ( p + ( s ) p ( s ) p 1 ( s ) ) d s + a b f ( s ) d s ) ,
(37)
x ( t 2 ) = 1 ( I f ( s ) d s E p 1 ( s ) d s E f ( s ) d s I p 1 ( s ) d s + a b f ( s ) d s )
(38)
and
x ( t 1 ) x ( t 2 ) = 1 ( I f ( s ) d s E ( p + ( s ) p ( s ) ) d s + E f ( s ) d s I ( p + ( s ) p ( s ) ) d s ) .
(39)

Suppose here that T > T + and condition (4) is fulfilled.

Define by P the set of all solutions of problem (35)-(36) for all a t 1 < t 2 b , for all functions p 1 L and non-negative p + , p L such that conditions (12), (31) hold, and for all f L with f L = 1 .

Let S be the subset of P corresponding to non-negative functions f.

From Lemmas 1 and 2, it follows that the set P coincides with the set of all solutions of problem (1)-(2) for all linear positive operators T , T + : C L with norms T + C L = T + , T C L = T and for all f L with f L = 1 . The subset S consists of all solutions of corresponding problems (1)-(2) with non-negative f.

Define the constants
M 1 max x P , t [ a , b ] | x ( t ) | , M 2 max x P ( max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) ) , N 1 max x S , t [ a , b ] x ( t ) , N 2 max x S ( max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) ) , N 3 min x S , t [ a , b ] x ( t ) .
From representations (37), (38), (39), it easily follows that all the constants are defined correctly and
M 1 = max { | N 1 | , | N 3 | } , M 2 = N 2 .
Moreover, for every solution x of (1)-(2), the following inequalities hold:
| x ( t ) | M 1 a b | f ( s ) | d s , t [ a , b ] , max t [ a , b ] x ( t ) min t [ a , b ] x ( t ) N 2 a b | f ( s ) | d s .
If f L is non-negative, then
N 3 a b f ( s ) d s x ( t ) N 1 a b f ( s ) d s , t [ a , b ] ,

where the constants N 1 , N 2 , N 3 , M 1 are best possible.

It remains to find N 1 , N 2 , N 3 .

The numerator and denominator of fractions in (37), (38), (39) are linear with respect to variables E p 1 ( s ) d s and I p 1 ( s ) d s . Therefore x ( t 1 ) , x ( t 2 ) , and x ( t 1 ) x ( t 2 ) take their minimal and maximal values at the bounds of restriction (31) with respect to variables p 1 on each of the sets E and I. Hence we have to consider only the following four different cases:
  1. (i)

    p 1 ( t ) = { p + ( t ) if  t E , p ( t ) if  t I ,

     
  2. (ii)

    p 1 ( t ) = { p ( t ) if  t E , p + ( t ) if  t I ,

     
  3. (iii)

    p 1 = p + ,

     
  4. (iv)

    p 1 = p .

     
In case (i) we have
x ( t 1 ) = 1 ( I f ( s ) d s ( E p ( s ) d s + 1 ) + E f ( s ) d s ( I p + d s + 1 ) ) , x ( t 2 ) = 1 ( I f ( s ) d s ( E p + ( s ) d s + 1 ) + E f ( s ) d s ( I p d s + 1 ) ) , x ( t 1 ) x ( t 2 ) = 1 ( I f ( s ) d s ( E ( p ( s ) p + ( s ) ) d s ) E f ( s ) d s ( I ( p ( s ) p + ( s ) ) d s ) ) , = I p ( s ) d s E p ( s ) d s I p + ( s ) d s E p + ( s ) d s + T T + .
In case (ii) we have
x ( t 1 ) = 1 ( I f ( s ) d s ( E p + ( s ) d s + 1 ) + E f ( s ) d s ( I p ( s ) d s + 1 ) ) , x ( t 2 ) = 1 ( I f ( s ) d s ( E p ( s ) d s + 1 ) + E f ( s ) d s ( I p + ( s ) d s + 1 ) ) , x ( t 1 ) x ( t 2 ) = 1 ( I f ( s ) d s ( E ( p ( s ) p + ( s ) ) d s ) E f ( s ) d s ( I ( p ( s ) p + ( s ) ) d s ) ) , = I p + ( s ) d s E p + ( s ) d s I p ( s ) d s E p ( s ) d s + T T + .
In case (iii) we have
x ( t 1 ) = 1 ( I f ( s ) d s ( E p ( s ) d s + 1 ) + E f ( s ) d s ( I p ( s ) d s + 1 ) ) , x ( t 2 ) = 1 ( I f ( s ) d s ( E p + ( s ) d s + 1 ) + E f ( s ) d s ( I p + ( s ) d s + 1 ) ) , x ( t 1 ) x ( t 2 ) = 1 ( I f ( s ) d s ( E ( p ( s ) p + ( s ) ) d s ) E f ( s ) d s ( I ( p ( s ) p + ( s ) ) d s ) ) , = I p + ( s ) d s E p ( s ) d s + I p ( s ) d s E p + ( s ) d s + T T + .
In case (iv) we have
x ( t 1 ) = 1 ( I f ( s ) d s ( E p + ( s ) d s + 1 ) + E f ( s ) d s ( I p + ( s ) d s + 1 ) ) , x ( t 2 ) = 1 ( I f ( s ) d s ( E p ( s ) d s + 1 ) + E f ( s ) d s ( I p ( s ) d s + 1 ) ) , x ( t 1 ) x ( t 2 ) = 1 ( I f ( s ) d s ( E ( p ( s ) p + ( s ) ) d s ) E f ( s ) d s ( I ( p ( s ) p + ( s ) ) d s ) ) , = I p + ( s ) d s E p ( s ) d s I p ( s ) d s E p + ( s ) d s + T T + .

Let S ( i ) , S ( ii ) , S ( iii ) , S ( iv ) be the subsets of S for p 1 corresponding to cases (i), (ii), (iii), (iv).

We can easily calculate the minimal and maximal values in every case.

In case (iv) we have
max x S ( iv ) { x ( t 1 ) , x ( t 2 ) } = T + 1 T T + T T + , min x S ( iv ) { x ( t 1 ) , x ( t 2 ) } = { 1 T T T + if  T > 1 , 1 T T T + + T T + if  T 1 , min x S ( iv ) ( x ( t 1 ) x ( t 2 ) ) = T T T + T T + , max x S ( iv ) ( x ( t 1 ) x ( t 2 ) ) = { T T T + + T T + if  T < 1 , 1 if  T 1 .
In case (iii) we have
max x S ( iii ) { x ( t 1 ) , x ( t 2 ) } = T + 1 T T + T T + , min x S ( iii ) { x ( t 1 ) , x ( t 2 ) } = { 1 T T T + if  T > 1 , 1 T T T + + T T + if  T 1 , max x S ( iii ) ( x ( t 1 ) x ( t 2 ) ) = T T T + T T + , min x S ( iii ) ( x ( t 1 ) x ( t 2 ) ) = { T T T + + T T + if  T < 1 , 1 if  T 1 .
Therefore, in cases (iii) and (iv) we have
max x S ( iii ) S ( iv ) { x ( t 1 ) , x ( t 2 ) } = T + 1 T T + T T + , min x S ( iii ) S ( iv ) { x ( t 1 ) , x ( t 2 ) } = { 1 T T T + if  T > 1 , 1 T T T + + T T + if  T 1 , max x S ( iii ) S ( iv ) | x ( t 1 ) x ( t 2 ) | = T T T + T T + .
In case (i) we have
max x S ( i ) { x ( t 1 ) , x ( t 2 ) } = T + 1 T T + ( T + ) 2 / 4 , min x S ( i ) { x ( t 1 ) , x ( t 2 ) } = 1 T T + + ( T ) 2 / 4 , max x S ( i ) | x ( t 1 ) x ( t 2 ) | = max z [ 0 , T + ] { T z T T + z ( T + z ) , z T T + z ( T + z ) } .
In case (ii) we have
max x S ( ii ) { x ( t 1 ) , x ( t 2 ) } = 1 T T + ( T ) 2 / 4 , min x S ( ii ) { x ( t 1 ) , x ( t 2 ) } = { min { K , G } if  T 1 , 1 T T T + if  1 < T 1 + 1 T + , 1 2 ( 1 + 1 T + ) T if  1 + 1 T + < T ,
where K = min z [ 0 , T + ] 1 z T T + + z ( T + z ) , G = min z [ 0 , T ] 1 z T T + + ( T + ) 2 / 4 z ( T z ) ,
max x S ( ii ) | x ( t 1 ) x ( t 2 ) | = max z [ 0 , T ] { z T T + z ( T z ) , T + z T T + z ( T z ) } .
Considering extremal values in all cases (i), (ii), (iii) and (vi), by elementary calculation, we obtain
N 1 = { 1 T T + ( T ) 2 / 4 if  T > 3 , T + 1 T T + T T + if  T 3 , N 3 = { 1 T T T + + T T + if  T 1 , 1 T T T + if  1 < T 1 + 1 T + , 1 2 ( 1 + 1 T + ) T if  1 + 1 T + < T .
If 0 T + < T ( 1 T / 4 ) , 3 T or 0 T + T ( T 1 ) 2 ( T + 1 ) 2 , 1 T 3 , then
N 2 = 1 2 T T + T .
If 0 < T + < T 1 + T , 0 < T 1 or T ( T 1 ) 2 ( T + 1 ) 2 < T + < T 1 + T , 1 < T 3 , then
N 2 = T T T + T T + .

This proves all Theorems 1, 2, 3. □

Declarations

Acknowledgements

Research was supported by the Russian Foundation for Basic Research (14-01-0033814). The author would like to thank both reviewers for their careful reading of the manuscript and valuable remarks.

Authors’ Affiliations

(1)
Scientific center ‘Functional Differential Equations’, Perm National Research Polytechnical University

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