Open Access

Existence of positive solutions of advanced differential equations

Advances in Difference Equations20132013:158

https://doi.org/10.1186/1687-1847-2013-158

Received: 6 November 2012

Accepted: 20 May 2013

Published: 5 June 2013

Abstract

In this paper, we study the advanced differential equations

[ r ( t ) | x ( t ) | α 1 x ( t ) ] + i = 1 n p i ( t ) | x ( t + τ i ( t ) ) | α 1 x ( t + τ i ( t ) ) = 0

and

[ r ( t ) ( y ( t ) P ( t ) y ( t τ ) ) ] + i = 1 n p i ( t ) f ( y ( t + σ ) ) = 0 .

By using the generalized Riccati transformation and the Schauder-Tyichonoff theorem, we establish the conditions for the existence of positive solutions of the above equations.

MSC:34K11, 39A10.

Keywords

advanced differential equations positive solutions existence

1 Introduction

In the last years, oscillation and nonoscillation of differential equations attracted a considerable attention. Many results have been obtained, and we refer the reader to the papers [120].

In 2008, Luo et al. [11] investigated the existence of positive periodic solutions of the following two kinds of neutral functional differential equations:
( x ( t ) c x ( t τ ( t ) ) ) = a ( t ) x ( t ) + f ( t , x ( t τ ( t ) ) )
and
( x ( t ) c 0 Q ( r ) x ( t + r ) d r ) = a ( t ) x ( t ) + b ( t ) 0 Q ( r ) f ( t , x ( t + r ) ) d r ,

where a , b C ( R , ( 0 , ) ) , τ C ( R , R ) , f C ( R × R , R ) , and a ( t ) , b ( t ) , τ ( t ) , f ( t , x ) are ω-periodic functions, ω > 0 , Q ( r ) C ( ( , 0 ] , [ 0 , ) ) , 0 Q ( r ) d r = 1 , and ω , | c | < 1 are constants.

Péics et al. [15] obtained the existence of positive solutions of half-linear delay differential equations
[ | x ( t ) | α 1 x ( t ) ] + i = 1 n p i ( t ) | x ( t τ i ( t ) ) | α 1 x ( t τ i ( t ) ) = 0 ,

where t t 0 and α > 0 , τ i ( t ) t .

Zhang et al. [19] obtained the existence of nonoscillatory solutions of the first-order linear neutral delay differential equation
[ x ( t ) + P ( t ) x ( t τ ) ] + Q 1 ( t ) x ( t σ 1 ) Q 2 ( t ) x ( t σ 2 ) = 0 ,

where P C ( [ t 0 , ) , R ) , τ ( 0 , ) , σ 1 , σ 2 [ 0 , ) , Q 1 , Q 2 > 0 .

In this paper, we consider the advanced differential equation
[ r ( t ) | x ( t ) | α 1 x ( t ) ] + i = 1 n p i ( t ) | x ( t + τ i ( t ) ) | α 1 x ( t + τ i ( t ) ) = 0 ,
(1.1)

where t t 0 and α > 0 .

Throughout this work, we always assume that the following conditions hold:

(H1) p i C ( [ t 0 , ) , R ) , i = 1 , 2 , 3 , , n ;

(H2) τ i C ( [ t 0 , ) , R + ) , i = 1 , 2 , 3 , , n , and 0 < r ( t ) k .

For convenience, we introduce the notation
η α = | η | α 1 η , α > 0 .
(1.2)
It is convenient to rewrite (1.1) in the form
[ r ( t ) | x ( t ) | α ] + i = 1 n p i ( t ) | x ( t + τ i ( t ) ) | α = 0 .
(1.3)

Definition 1.1 A function x is said to be a solution of Eq. (1.1) if x C 1 ( [ T , ) , R ) , T t 0 , which has the property | x | α 1 x C 1 ( [ T , ) , R ) and it satisfies Eq. (1.1) for t T . We say that a solution of Eq. (1.1) is oscillatory if it has arbitrarily large zeros. Otherwise, it is nonoscillatory.

One of the most important methods of the study of nonoscillation is the method of generalized characteristic equation [6]. The method was applied to second-order half-linear equations without delay, for example, in [8, 9]. Concerning cases with advanced, let us apply the Riccati-transformation
x ( t ) = exp ( t 0 t ( ω ( s ) ) ( 1 α ) d s ) .
(1.4)
By (1.4), we have
x ( t ) = ( exp ( t 0 t ω ( s ) ( 1 α ) d s ) ) = ω ( t ) ( 1 α ) exp ( t 0 t ω ( s ) ( 1 α ) d s ) , ( x ( t + τ i ( t ) ) ) α = exp ( α t 0 t + τ i ( t ) ω ( s ) ( 1 α ) d s ) .
From (1.3), we obtain
[ r ( t ) ω ( t ) exp ( α t 0 t ω ( s ) ( 1 α ) d s ) ] + i = 1 n p i ( t ) exp ( α t 0 t + τ i ( t ) ω ( s ) ( 1 α ) d s ) = 0 .
(1.5)
Since
[ r ( t ) ω ( t ) exp ( α t 0 t ω ( s ) ( 1 α ) d s ) ] = ( r ( t ) ω ( t ) ) exp ( α t 0 t ω ( s ) ( 1 α ) d s ) + r ( t ) ω ( t ) ( exp ( α t 0 t ω ( s ) ( 1 α ) d s ) ) = ( r ( t ) ω ( t ) + r ( t ) ω ( t ) ) exp ( α t 0 t ω ( s ) ( 1 α ) d s ) + α r ( t ) | ω ( t ) | 1 + 1 α × exp ( α t 0 t ω ( s ) ( 1 α ) d s ) ,
it is convenient to rewrite (1.5) in the form
r ( t ) ω ( t ) + r ( t ) ω ( t ) + α r ( t ) | ω ( t ) | 1 + 1 α + i = 1 n p i ( t ) exp ( α t t + τ i ( t ) ω ( s ) ( 1 α ) d s ) = 0 .
(1.6)

2 Preliminaries

Lemma 2.1 Suppose that (H1) and (H2) hold. Then the following statements are equivalent:
  1. (i)

    Eq. (1.1) has an eventually positive solution;

     
  2. (ii)

    There is a function ω C 1 ( [ T , ) , R ) , T t 0 , such that ω solves the Riccati equation (1.6).

     
Proof (i) (ii). Let x be an eventually positive solution of Eq. (1.1) such that x ( t ) > 0 for t T t 0 . The function ω defined by
ω ( t ) = ( x ( t ) x ( t ) ) α , t T ,

is continuous.

We will show that it is a solution of (1.6) on [ T , ) . By (1.2) and observing that
ω ( t ) = ( x ( t ) x ( t ) ) α = | x ( t ) x ( t ) | α 1 x ( t ) x ( t ) , x ( t ) x ( t ) = | ω ( t ) | 1 α 1 ω ( t ) = ω ( t ) ( 1 α ) ,
it follows that
x ( t ) = x ( T ) exp ( T t ω ( s ) ( 1 α ) d s ) .
Dividing both sides of (1.1) by | x ( t ) | α 1 x ( t ) gives that
[ r ( t ) | x ( t ) | α 1 x ( t ) ] | x ( t ) | α 1 x ( t ) + i = 1 n p i ( t ) | x ( t + τ i ( t ) ) | α 1 x ( t + τ i ( t ) ) | x ( t ) | α 1 x ( t ) = 0 .
(2.1)
From the definition of ω, we obtain
| x ( t ) | α 1 x ( t ) = ω ( t ) | x ( t ) | α 1 x ( t ) = ω ( t ) x α ( t ) .
Further
( r ( t ) | x ( t ) | α 1 x ( t ) ) = ( r ( t ) ω ( t ) x α ( t ) ) = r ( t ) ω ( t ) x α ( t ) + r ( t ) ω ( t ) x α ( t ) + α r ( t ) ω ( t ) x α 1 ( t ) x ( t )
(2.2)
and
x ( t + τ i ( t ) ) x ( t ) = exp ( t t + τ i ( t ) ω ( s ) ( 1 α ) d s ) > 0 , | x ( t + τ i ( t ) ) x ( t ) | α = exp ( α t t + τ i ( t ) ω ( s ) ( 1 α ) d s ) .
(2.3)
By substituting (2.2), (2.3) into (2.1), we get
[ r ( t ) | x ( t ) | α 1 x ( t ) ] | x ( t ) | α 1 x ( t ) + i = 1 n p i ( t ) | x ( t + τ i ( t ) ) | α 1 x ( t + τ i ( t ) ) | x ( t ) | α 1 x ( t ) = r ( t ) ω ( t ) x α ( t ) + r ( t ) ω ( t ) x α ( t ) + α r ( t ) ω ( t ) x α 1 ( t ) x ( t ) x α ( t ) + i = 1 n p i ( t ) exp ( α t t + τ i ( t ) ω ( s ) ( 1 α ) d s ) = 0 .
We obtain (1.6), and the proof of (i) (ii) is complete.
  1. (ii)

    (i). Let ω be a continuously differentiable solution of Eq. (1.6) for t T t 0 .

     
We show that a function x defined by
x ( t ) = exp ( T t ω ( s ) ( 1 α ) d s )

is the solution of Eq. (1.1).

Since
x ( t ) x ( t ) = ω ( t ) ( 1 α ) , ( x ( t ) ) α = ( x ( t ) ) α ω ( t ) = x α ( t ) ω ( t ) .
By (1.6), we obtain
[ r ( t ) ( x ( t ) ) α ] = ( r ( t ) ω ( t ) x α ( t ) ) = r ( t ) ω ( t ) x α ( t ) + r ( t ) ω ( t ) x α ( t ) + α r ( t ) ω ( t ) x α 1 ( t ) x ( t ) = r ( t ) ω ( t ) x α ( t ) + r ( t ) ω ( t ) x α ( t ) + α r ( t ) x α ( t ) | ω ( t ) | 1 + 1 α = i = 1 n p i ( t ) exp ( α t t + τ i ( t ) ω ( s ) ( 1 α ) d s ) exp ( α T t ω ( s ) ( 1 α ) d s ) = i = 1 n p i ( t ) exp ( α T t + τ i ( t ) ω ( s ) ( 1 α ) d s ) = i = 1 n p i ( t ) x α ( t + τ i ( t ) ) ,
thus,
[ r ( t ) ( x ( t ) ) α ] + i = 1 n p i ( t ) x α ( t + τ i ( t ) ) = 0 , t T .

The proof of (ii) (i) is complete. The proof is complete. □

Lemma 2.2 Suppose that (H1) and (H2) hold. The following statements are equivalent:
  1. (a)
    There is a solution ω C 1 ( [ T , ) , R ) of the Riccati equation (1.6) for some T t 0 such that
    t [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ω ( ξ ) ( 1 α ) d ξ ) ] d s < .
    (2.4)
     
  2. (b)
    There is a function u C ( [ T , ) , R ) for some T t 0 such that
    u ( t ) = 1 r ( t ) { α t r ( s ) | u ( s ) | 1 + 1 α d s + t [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ω ( ξ ) ( 1 α ) d ξ ) ] d s } .
    (2.5)
     
Proof (a) (b). Let ω = u be a solution of Eq. (1.6) for t T t 0 and with the property (2.4). Let t 1 t T be fixed arbitrarily and integrate (1.6) over [ t , t 1 ] :
u ( t 1 ) r ( t 1 ) u ( t ) r ( t ) = α t t 1 r ( s ) | u ( s ) | 1 + 1 α d s t t 1 [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) | u ( ξ ) | 1 α 1 u ( ξ ) d ξ ) ] d s .
(2.6)
We claim that
t r ( s ) | u ( s ) | 1 + 1 α d s < .
(2.7)
Assuming the contrary, if t r ( s ) | u ( s ) | 1 + 1 α d s = , then in view of (2.6) there is T 1 t such that
u ( t 1 ) r ( t 1 ) + α T 1 t 1 r ( s ) | u ( s ) | 1 + 1 α d s = u ( t ) r ( t ) α t T 1 r ( s ) | u ( s ) | 1 + 1 α d s t t 1 [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) | u ( ξ ) | 1 α 1 u ( ξ ) d ξ ) ] d s 1
for t 1 T 1 t , or equivalently,
u ( t 1 ) r ( t 1 ) 1 + α T 1 t 1 r ( s ) | u ( s ) | 1 + 1 α d s , t 1 T 1 .
(2.8)
Then we have
u ( t 1 ) 0 .
From u ( t ) = ( x ( t ) x ( t ) ) α , it follows that x ( t 1 ) < 0 , t 1 T 1 . Dividing both sides of (2.8) by 1 + α T 1 t 1 r ( s ) | u ( s ) | 1 + 1 α d s > 0 gives that
| u ( t 1 ) | 1 + 1 α r ( t 1 ) 1 + α T 1 t 1 r ( s ) | u ( s ) | 1 + 1 α d s ( u ( t 1 ) ) 1 α = x ( t 1 ) x ( t 1 ) , t 1 T 1 .
(2.9)
Integrating the above inequality over [ T 1 , t 1 ] then yields
1 α ln ( 1 + α T 1 t 1 r ( s ) | u ( s ) | 1 + 1 α d s ) ln ( x ( T 1 ) x ( t 1 ) ) .
Combining with (2.8), we have
( r ( t 1 ) u ( t 1 ) ) 1 α x ( T 1 ) x ( t 1 ) , t 1 T 1
and
r 1 α ( t 1 ) x ( t 1 ) x ( T 1 ) .

Integrating the last inequality and using 0 < r ( t ) k , we see that lim t x ( t ) = , which contradicts the assumption that x ( t ) is eventually positive. Therefore (2.7) must hold.

Let t 1 in (2.6). Using (2.4) and (2.7), we get lim t 1 r ( t 1 ) u ( t 1 ) = 0 . So,
u ( t ) = 1 r ( t ) { α t r ( s ) | u ( s ) | 1 + 1 α d s + t [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) u ( ξ ) ( 1 α ) d ξ ) ] d s }
must hold.
  1. (b)

    (a). Assume that there is a function u ( t ) satisfying Eq. (2.5) on [ T , ) . Differentiation of (2.5) then shows that u = ω is a solution of (1.6) for t T , and it satisfies (2.4). The proof of (b) (a) is complete. □

     

3 Main results

Theorem 3.1 Assume that there exist T t 0 and functions β , γ C ( [ T , ) , R ) such that β ( t ) γ ( t ) ,
t [ i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) γ ( ξ ) ( 1 α ) d ξ ) ] d s < .
(3.1)
β ( t ) ν ( t ) γ ( t ) implies that S ν is defined and β ( t ) ( S ν ) ( t ) γ ( t )
(3.2)
for every function ν C ( [ T , ) , R ) , where
( S ν ) ( t ) = 1 r ( t ) { α t r ( s ) | ν ( s ) | 1 + 1 α d s + t [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } .
(3.3)

Then there exists a continuous solution u ( t ) of Eq. (2.5) which satisfies the inequality β ( t ) u ( t ) γ ( t ) .

Proof Let T 1 and T 2 be real numbers such that T T 1 T 2 < . Then [ T 1 , T 2 ] is an arbitrary compact subinterval of [ T , ) and set
L = max T 1 t T 2 { max { | β ( t ) | , | γ ( t ) | } } , τ = max T 1 t T 2 { max 1 i n τ i ( t ) } , L 1 = L 1 α 1 e α τ L 1 α , N = min T 1 t T 2 r ( t ) , M = max T 1 t T 2 i = 1 n | p i ( t ) | , c = k ( α + 1 ) L 1 α + M L 1 τ N .
Define
F = { ν C ( [ T , ) , R ) β ( t ) ν ( t ) γ ( t ) , t [ T , ) } .
It follows from (3.1) and (3.2), that the operator S is defined for ν F and satisfies
t r ( ζ ) | ν ( ζ ) | 1 + 1 α d ζ < .
(3.4)

By (3.2), we see that the functions in the image set SF are uniformly bounded on any finite interval of [ T , ) .

To prove that the functions in SF are equicontinuous on any finite interval of [ T , ) , we choose the finite interval [ T 1 , T 2 ] as before, and let t 1 and t 2 be two arbitrary numbers from [ T 1 , T 2 ] . Since 1 r ( t ) is continuous on [ T 1 , T 2 ] , ε > 0 , δ 1 > 0 , such that for | t 1 t 2 | < δ 1 , we have
| 1 r ( t 1 ) 1 r ( t 2 ) | < ε 2 1 r ( t 1 ) < 1 r ( t 2 ) + ε 2 .
Further,
| S ν ( t 1 ) S ν ( t 2 ) | = | 1 r ( t 1 ) { α t 1 r ( s ) | ν ( s ) | 1 + 1 α d s + t 1 [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } 1 r ( t 2 ) { α t 2 r ( s ) | ν ( s ) | 1 + 1 α d s + t 2 [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } | | 1 r ( t 2 ) { α t 1 t 2 r ( s ) | ν ( s ) | 1 + 1 α d s + t 1 t 2 [ i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } | + ε 2 | { α t 1 r ( s ) | ν ( s ) | 1 + 1 α d s + t 1 [ i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } | 1 N ( α k L 1 + 1 α + M e α τ L 1 α ) | t 1 t 2 | + ε 2 | { α t 1 r ( s ) | ν ( s ) | 1 + 1 α d s + t 1 [ i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s } | .

Due to (3.1) and (3.4), there exists δ 2 such that for | t 1 t 2 | < δ 2 , | S ν ( t 1 ) S ν ( t 2 ) | < ε , hence SF is equicontinuous.

Let the sequence { ν n ( t ) } F tend to ν ( t ) uniformly on any finite interval ( n ). In particular, the convergence is uniform on the interval [ T 1 , T 2 ] . Using the mean value theorem, we have
where | σ ( s ) | is between | ν ( s ) | and | ν n ( s ) | , and similarly

for every i = 1 , 2 , 3 , , n and T 1 s T 2 , where σ i ( s ) is between α s s + τ i ( s ) | ν ( ξ ) | 1 α 1 ν ( ξ ) d ξ and α s s + τ i ( s ) | ν n ( ξ ) | 1 α 1 ν n ( ξ ) d ξ .

Since | σ i ( s ) | α τ L 1 α for T 1 t T 2 , we obtain
Hence,
The uniform convergence ν n ( t ) ν ( t ) 0 on any finite interval of [ T , ) implies that if n is sufficiently large,
| ν ( t ) ν n ( t ) | < δ , T 1 t T 2 ,
where δ = ε T 2 , and hence we obtain
| S ν ( t ) S ν n ( t ) | 1 r ( t ) [ lim T 2 ( α + 1 ) L 1 α k δ ( T 2 t ) + M L 1 τ δ ( T 2 t ) ] lim T 2 1 N [ ( α + 1 ) L 1 α k + M L 1 τ ] δ T 2 lim T 2 1 N [ ( α + 1 ) L 1 α k + M L 1 τ ] ε = c ε

for T 1 t T 2 . Thus, S ν n ( t ) S ν ( t ) uniformly on a finite interval.

We obtained that the conditions of the Schauder-Tyichonoff theorem are satisfied, hence the mapping S has at least one fixed point ν in F, and because ν ( t ) = ( S ν ) ( t ) for t T , ν is the continuous solution of Eq. (2.5). □

Theorem 3.2 Assume that (H1), (H2) hold and there exists a positive function μ ( t ) for t T t 0 such that
1 r ( t ) t [ α r ( s ) μ 1 + 1 α ( s ) + i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) μ 1 α ( ξ ) d ξ ) ] d s μ ( t )
(3.5)

holds for t large enough. Then Eq. (1.1) has a positive solution x ( t ) with the property | x ( t ) x ( t ) | μ 1 α ( t ) .

Proof Let μ ( t ) be given such that the conditions of the theorem hold. We show that the conditions of Theorem 3.1 are satisfied with β ( t ) = μ ( t ) and γ ( t ) = μ ( t ) for t large enough.

Let ν ( t ) be a continuous function such that | ν ( t ) | μ ( t ) . It follows from (3.5) that
| S ν ( t ) | = 1 r ( t ) | α t r ( s ) | ν ( s ) | 1 + 1 α d s + t [ i = 1 n p i ( s ) exp ( α s s + τ i ( s ) ν ( ξ ) ( 1 α ) d ξ ) ] d s | 1 r ( t ) [ α t r ( s ) μ 1 + 1 α ( s ) d s + t i = 1 n | p i ( s ) | exp ( α s s + τ i ( s ) μ ( 1 α ) ( ξ ) d ξ ) d s ] μ ( t ) .

Therefore, by Theorem 3.1, Lemma 2.1 and Lemma 2.2, Eq. (1.1) has a positive solution, and the proof is complete. □

Next, we consider neutral differential equations of the form
[ r ( t ) ( y ( t ) P ( t ) y ( t τ ) ) ] + i = 1 n p i ( t ) f ( y ( t + σ ) ) = 0 , t t 0 .
(3.6)
We assume that:
  1. (i)

    τ > 0 , σ 0 ;

     
  2. (ii)

    r , P , p i C ( [ t 0 , ) , ( 0 , ) ) , i = 1 , 2 , , n ;

     
  3. (iii)

    f is nondecreasing continuous function and x f ( x ) > 0 , x 0 .

     

The following fixed point theorem will be used to prove the main results.

Lemma 3.1 (Schauder’s fixed point theorem)

Let Ω be a closed, convex and nonempty subset of a Banach space X. Let T : Ω Ω be a continuous mapping such that T Ω is a relatively compact subset of X. Then T has at least one fixed point in Ω. That is, there exists an x Ω such that T x = x .

Theorem 3.3 Suppose that
t 0 i = 1 n p i ( t ) d t =
(3.7)
and there exist ζ 0 , 0 < k 1 k 2 such that
k 2 k 1 exp [ ( k 1 k 2 ) t 0 ζ t 0 i = 1 n p i ( t ) d t ] 1 , exp ( k 2 t τ t i = 1 n p i ( t ) d t ) + exp ( k 2 t 0 ζ t τ i = 1 n p i ( s ) d s ) t 1 r ( s ) s i = 1 n p i ( ξ ) × f ( exp ( k 1 t 0 ζ ξ + σ i = 1 n p i ( z ) d z ) ) d ξ d s P ( t ) exp ( k 1 t τ t i = 1 n p i ( t ) d t ) + exp ( k 1 t 0 ζ t τ i = 1 n p i ( s ) d s ) t 1 r ( s ) s i = 1 n p i ( ξ ) × f ( exp ( k 2 t 0 ζ ξ + σ i = 1 n p i ( z ) d z ) ) d ξ d s , t t 0 .
(3.8)

Then Eq. (3.6) has a positive solution which tends to zero.

Proof First: Choose T ˜ t 0 + τ ,
u ( t ) = exp ( k 2 t 0 ζ t i = 1 n p i ( t ) d t ) , ν ( t ) = exp ( k 1 t 0 ζ t i = 1 n p i ( t ) d t ) , t t 0 .
Let C ( [ t 0 , ) , R ) be the set of all continuous functions with the norm
y ( t ) = sup t t 0 | y ( t ) | < .
Then C ( [ t 0 , ) , R ) is a Banach space. We define a closed, bounded convex subset Ω of C ( [ t 0 , ) , R ) as follows:
Ω = { y y C ( [ t 0 , ) , R ) : u ( t ) y ( t ) ν ( t ) , t t 0 } .
Define the map T : Ω C ( [ t 0 , ) , R ) :
( T y ) ( t ) = { P ( t ) y ( t τ ) t 1 r ( s ) s i = 1 n p i ( ξ ) f ( y ( ξ + σ ) ) d ξ d s , t T ˜ , ( T y ) ( T ˜ ) + ν ( t ) ν ( T ˜ ) , t 0 t T ˜ .

We can show that for any y Ω , T y Ω .

Second: We prove that T is continuous.

Third: We show that TΩ is relatively compact.

The proof is similar to Theorem 2.1 of [2], we omitted it. □

Corollary 3.1 Suppose that k > 0 , (3.7) holds and
P ( t ) = exp ( k t τ t i = 1 n p i ( t ) d t ) + exp ( k t ζ t τ i = 1 n p i ( s ) d s ) t 1 r ( s ) s i = 1 n p i ( ξ ) × f ( exp ( k t 0 ζ ξ + σ i = 1 n p i ( z ) d z ) ) d ξ d s , t t 0 .
Then Eq. (3.6) has a solution
y ( t ) = exp ( k t 0 t i = 1 n p i ( t ) d t ) , t t 0 .
Example 3.1 Consider the advanced differential equations
( x ( t ) ) + i = 1 n p i ( t ) x ( 2 t ) = 0 , t 2 ,
(3.9)
where p i C ( [ t 0 , ) , R ) and i = 1 n | p i ( t ) | = 1 8 2 t 2 . Choose μ ( t ) = 1 2 t ,
t ( 1 4 s 2 + 1 8 2 s 2 exp ( 1 2 s 2 s 1 ξ d ξ ) ) d s = t 3 8 s 2 d s 3 8 t 1 2 t .

All the conditions of Theorem 3.2 are satisfied. Equation (3.9) has a positive solution and | x ( t ) x ( t ) | 1 2 t . In fact, we can choose μ ( t ) = 1 / ( η t ) , η ( 4 2 2 , 4 + 2 2 ) , Eq. (3.9) has a positive solution with | x ( t ) x ( t ) | μ ( t ) , and the solution satisfies x ( 2 ) 2 1 / η t 1 / η x ( t ) x ( 2 ) 2 1 / η t 1 / η .

Declarations

Acknowledgements

The authors sincerely thank the anonymous referees for their valuable suggestions and comments which greatly helped improve this article. Supported by NSF of China (11071054), Natural Science Foundation of Hebei Province (A2011205012).

Authors’ Affiliations

(1)
College of Mathematics and Information Science, Hebei Normal University

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