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Uncountably many solutions of first-order neutral nonlinear differential equations

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Abstract

The article deals with the existence of uncountably many positive solutions which are bounded below and above by positive functions for the first-order nonlinear neutral differential equations. Some examples are included to illustrate the results presented in this article.

MSC:34K40, 34K12.

1 Introduction

In recent years, the study of existence and qualitative properties of solutions for various kinds of neutral delay differential equations has attracted much attention. For related results, we refer the reader to [113] and the references cited therein. The authors only considered the existence of solutions which are bounded by positive constants, e.g., in [8, 9, 1113]. For example, Erbe et al. [6] established a few oscillation and nonoscillation criteria for linear neutral delay differential equation

[ x ( t ) p ( t ) x ( t τ ) ] +q(t)x ( t σ ( t ) ) =0,t t 0 .

Diblík and co-autors in [14] studied the existence of positive and oscillatory solutions of differential equations with delay and nonlinear systems in view of Ważievski’s retract principle and later extended to retarded functional differential equations by Rybakowski. Zhou [12] deduced the existence of nonoscillatory solutions of the second-order nonlinear neutral differential equations and Lin et al. [9] discussed the existence of nonoscillatory solutions for a third-order nonlinear neutral delay differential equation, and by utilizing Krasnoselskii’s fixed point theorem and Schauder’s fixed point theorem, they developed some sufficient conditions for the existence of uncountably many nonoscillatory solutions bounded by positive constants. Some interesting results about the existence of nonoscillatory solutions of delay differential equations can also be found in [1, 5].

In this paper, we investigate the following nonlinear neutral differential delay differential equations:

d d t [ x ( t ) a ( t ) x ( t τ ) ] =p(t)f ( x ( t σ ) ) ,t t 0 ,
(1)

where τ>0, σ0, aC([ t 0 ,),(0,)), pC(R,(0,)), fC(R,R), f is a nondecreasing function for x>0 and f(x)>0, x>0.

By a solution of Eq. (1), we mean a function xC([ t 1 τ,),R) for some t 1 t 0 such that x(t)a(t)x(tτ) is continuously differentiable on [ t 1 ,) and such that Eq. (1) is satisfied for t t 1 .

As much as we know, in the literature there is no result for the existence of uncountably many solutions which are bounded below and above by positive functions. This problem is discussed and treated in this paper.

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

Lemma 1.1 ([6, 12] Krasnoselskii’s fixed point theorem)

Let X be a Banach space, let Ω be a bounded closed convex subset of X and let S 1 , S 2 be maps of Ω into X such that S 1 x+ S 2 yΩ for every pair x,yΩ. If S 1 is contractive and S 2 is completely continuous, then the equation

S 1 x+ S 2 x=x

has a solution in Ω.

2 The existence of positive solutions

In this section we consider the existence of uncountably many positive solutions for Eq. (1) which are bounded by two positive functions. We use the notation m=max{τ,σ}.

Theorem 2.1 Suppose that there exist bounded from below and from above by the functions u and v C 1 ([ t 0 ,),(0,)) constants c>0, K 2 > K 1 0 and t 1 t 0 +m such that

u(t)v(t),t t 0 ,
(2)
v(t)v( t 1 )u(t)+u( t 1 )0, t 0 t t 1 ,
(3)
1 u ( t τ ) ( u ( t ) K 1 + t p ( s ) f ( v ( s σ ) ) d s ) a ( t ) 1 v ( t τ ) ( v ( t ) K 2 + t p ( s ) f ( u ( s σ ) ) d s ) c < 1 , t t 1 .
(4)

Then Eq. (1) has uncountably many positive solutions which are bounded by the functions u, v.

Proof Let C([ t 0 ,),R) be the set of all continuous bounded functions with the norm x= sup t t 0 |x(t)|. Then C([ t 0 ,),R) is a Banach space. We define a closed, bounded and convex subset Ω of C([ t 0 ,),R) as follows:

Ω= { x = x ( t ) C ( [ t 0 , ) , R ) : u ( t ) x ( t ) v ( t ) , t t 0 } .

For K[ K 1 , K 2 ] we define two maps S 1 and S 2 :ΩC([ t 0 ,),R) as follows:

( S 1 x)(t)= { K + a ( t ) x ( t τ ) , t t 1 , ( S 1 x ) ( t 1 ) , t 0 t t 1 ,
(5)
( S 2 x)(t)= { t p ( s ) f ( x ( s σ ) ) d s , t t 1 , ( S 2 x ) ( t 1 ) + v ( t ) v ( t 1 ) , t 0 t t 1 .
(6)

We will show that for any x,yΩ, we have S 1 x+ S 2 yΩ. For every x,yΩ and t t 1 with regard to (4), we obtain

( S 1 x ) ( t ) + ( S 2 y ) ( t ) = K + a ( t ) x ( t τ ) t p ( s ) f ( y ( s σ ) ) d s K + a ( t ) v ( t τ ) t p ( s ) f ( u ( s σ ) ) d s K + v ( t ) K 2 v ( t ) .

For t[ t 0 , t 1 ] we have

( S 1 x ) ( t ) + ( S 2 y ) ( t ) = ( S 1 x ) ( t 1 ) + ( S 2 y ) ( t 1 ) + v ( t ) v ( t 1 ) v ( t 1 ) + v ( t ) v ( t 1 ) = v ( t ) .

Furthermore, for t t 1 we get

( S 1 x ) ( t ) + ( S 2 y ) ( t ) K + a ( t ) u ( t τ ) t p ( s ) f ( v ( s σ ) ) d s K + u ( t ) K 1 u ( t ) .
(7)

Let t[ t 0 , t 1 ]. With regard to (3), we get

v(t)v( t 1 )+u( t 1 )u(t), t 0 t t 1 .

Then, for t[ t 0 , t 1 ] and any x,yΩ, we obtain

( S 1 x ) ( t ) + ( S 2 y ) ( t ) = ( S 1 x ) ( t 1 ) + ( S 2 y ) ( t 1 ) + v ( t ) v ( t 1 ) u ( t 1 ) + v ( t ) v ( t 1 ) u ( t ) .

Thus we have proved that S 1 x+ S 2 yΩ for any x,yΩ.

We will show that S 1 is a contraction mapping on Ω. For x,yΩ and t t 1 , we have

| ( S 1 x ) ( t ) ( S 1 y ) ( t ) | = | a ( t ) | | x ( t τ ) y ( t τ ) | cxy.

This implies that

S 1 x S 1 ycxy.

Also, for t[ t 0 , t 1 ] the inequality above is valid. We conclude that S 1 is a contraction mapping on Ω.

We now show that S 2 is completely continuous. First, we show that S 2 is continuous. Let x k = x k (t)Ω be such that x k (t)x(t) as k. Because Ω is closed, x=x(t)Ω. For t t 1 we have

| ( S 2 x k ) ( t ) ( S 2 x ) ( t ) | | t p ( s ) [ f ( x k ( s σ ) ) f ( x ( s σ ) ) ] d s | t 1 p ( s ) | f ( x k ( s σ ) ) f ( x ( s σ ) ) | d s .

According to (7), we get

t 1 p(s)f ( v ( s σ ) ) ds<.
(8)

Since |f( x k (sσ))f(x(sσ))|0 as k, by applying the Lebesgue dominated convergence theorem, we obtain that

lim k ( S 2 x k ) ( t ) ( S 2 x ) ( t ) =0.

This means that S 2 is continuous.

We now show that S 2 Ω is relatively compact. It is sufficient to show by the Arzela-Ascoli theorem that the family of functions { S 2 x:xΩ} is uniformly bounded and equicontinuous on [ t 0 ,). The uniform boundedness follows from the definition of Ω. For the equicontinuity, we only need to show, according to the Levitan result [7], that for any given ε>0, the interval [ t 0 ,) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have a change of amplitude less than ε. Then, with regard to condition (8), for xΩ and any ε>0, we take t t 1 large enough so that

t p(s)f ( x ( s σ ) ) ds< ε 2 .

Then, for xΩ, T 2 > T 1 t , we have

| ( S 2 x ) ( T 2 ) ( S 2 x ) ( T 1 ) | T 2 p ( s ) f ( x ( s σ ) ) d s + T 1 p ( s ) f ( x ( s σ ) ) d s < ε 2 + ε 2 = ε .

For xΩ and t 1 T 1 < T 2 t , we get

| ( S 2 x ) ( T 2 ) ( S 2 x ) ( T 1 ) | T 1 T 2 p ( s ) f ( x ( s σ ) ) d s max t 1 s t { p ( s ) f ( x ( s σ ) ) } ( T 2 T 1 ) .

Thus there exists δ 1 = ε M , where M= max t 1 s t {p(s)f(x(sσ))}, such that

| ( S 2 x ) ( T 2 ) ( S 2 x ) ( T 1 ) | <εif 0< T 2 T 1 < δ 1 .

Finally, for any xΩ, t 0 T 1 < T 2 t 1 , there exists a δ 2 >0 such that

| ( S 2 x ) ( T 2 ) ( S 2 x ) ( T 1 ) | = | v ( T 1 ) v ( T 2 ) | = | T 1 T 2 v ( s ) d s | max t 0 s t 1 { | v ( s ) | } ( T 2 T 1 ) < ε if  0 < T 2 T 1 < δ 2 .

Then { S 2 x:xΩ} is uniformly bounded and equicontinuous on [ t 0 ,), and hence S 2 Ω is a relatively compact subset of C([ t 0 ,),R). By Lemma 1.1 there is an x 0 Ω such that S 1 x 0 + S 2 x 0 = x 0 . We conclude that x 0 (t) is a positive solution of (1).

Next we show that Eq. (1) has uncountably many bounded positive solutions in Ω. Let the constant K ¯ [ K 1 , K 2 ] be such that K ¯ K. We infer similarly that there exist mappings S ¯ 1 , S ¯ 2 satisfying (5), (6), where K, S 1 , S 2 are replaced by K ¯ , S ¯ 1 , S ¯ 2 , respectively. We assume that x,yΩ, S 1 x+ S 2 x=x, S ¯ 1 y+ S ¯ 2 y=y, which are the bounded positive solutions of Eq. (1), that is,

x ( t ) = K + a ( t ) x ( t τ ) t p ( s ) f ( x ( s σ ) ) d s , t t 1 , y ( t ) = K ¯ + a ( t ) y ( t τ ) t p ( s ) f ( y ( s σ ) ) d s , t t 1 .

From condition (8) it follows that there exists a t 2 > t 1 satisfying

t 2 p(s) [ f ( x ( s σ ) ) + f ( y ( s σ ) ) ] ds<|K K ¯ |.
(9)

In order to prove that the set of bounded positive solutions of Eq. (1) is uncountable, it is sufficient to verify that xy. For t t 2 we get

| x ( t ) y ( t ) | = | K + a ( t ) x ( t τ ) t p ( s ) f ( x ( s σ ) ) d s K ¯ a ( t ) y ( t τ ) + t p ( s ) f ( y ( s σ ) ) d s | | K K ¯ + a ( t ) [ x ( t τ ) y ( t τ ) ] t p ( s ) [ f ( x ( s σ ) ) f ( y ( s σ ) ) ] d s | | K K ¯ | a ( t ) x y | t p ( s ) [ f ( x ( s σ ) ) f ( y ( s σ ) ) ] d s | | K K ¯ | c x y t p ( s ) [ f ( x ( s σ ) ) + f ( y ( s σ ) ) ] d s .

Then we have

(1+c)xy|K K ¯ | t p(s) [ f ( x ( s σ ) ) + f ( y ( s σ ) ) ] ds,t t 2 .

According to (9) we get that xy. Since the interval [ K 1 , K 2 ] contains uncountably many constants, then Eq. (1) has uncountably many positive solutions which are bounded by the functions u(t), v(t). This completes the proof. □

Corollary 2.1 Suppose that there exist bounded from below and from above by the functions u and v C 1 ([ t 0 ,),(0,)) constants c>0, K 2 > K 1 0 and t 1 t 0 +m such that (2), (4) hold and

v (t) u (t)0, t 0 t t 1 .
(10)

Then Eq. (1) has uncountably many positive solutions which are bounded by the functions u, v.

Proof We only need to prove that condition (10) implies (3). Let t[ t 0 , t 1 ] and set

H(t)=v(t)v( t 1 )u(t)+u( t 1 ).

Then, with regard to (10), it follows that

H (t)= v (t) u (t)0, t 0 t t 1 .

Since H( t 1 )=0 and H (t)0 for t[ t 0 , t 1 ], this implies that

H(t)=v(t)v( t 1 )u(t)+u( t 1 )0, t 0 t t 1 .

Thus all the conditions of Theorem 2.1 are satisfied. □

Example 2.1 Consider the nonlinear neutral differential equation

[ x ( t ) a ( t ) x ( t 2 ) ] =p(t) x 3 (t1),t t 0 ,
(11)

where p(t)= e t . We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)=0.5, v(t)=2 satisfy (2) and also condition (10) for t[ t 0 , t 1 ]=[0,4]. For the constants K 1 =0.5, K 2 =1, condition (4) has the form

16 e t a(t) 1 2 + 1 16 e t ,t t 1 =4.
(12)

If the function a(t) satisfies (12), then Eq. (11) has uncountably many positive solutions which are bounded by the functions u, v.

Example 2.2 Consider the nonlinear neutral differential equation

[ x ( t ) a ( t ) x ( t τ ) ] =p(t) x 2 (tσ),t t 0 ,
(13)

where τ,σ(0,), p(t)= e 3 t . We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)= e 2 t , v(t)= e τ + e t , t1, satisfy (2) and since

v (t) u (t)= e t ( 2 e t 1 ) <0for t[1,2],

condition (10) is also satisfied. For the constants K 1 =0, K 2 = e τ 1, condition (4) has the form

e 2 τ + 1 3 e t + 1 2 e 2 t + σ τ + 1 5 e 3 t + 2 ( σ τ ) a(t) e τ + e 7 t + 4 σ τ 7 ( 1 + e t ) ,t2.

For τ=σ=1 we get

e 2 + 1 3 e t + 1 2 e 2 t + 1 5 e 3 t a(t) e 1 + e 7 t + 3 7 ( 1 + e t ) ,t t 1 =2.
(14)

If the function a(t) satisfies (14), then Eq. (13) has uncountably many solutions which are bounded by the functions u, v.

Example 2.3 Consider the nonlinear neutral differential equation

[ x ( t ) a ( t ) x ( t τ ) ] =p(t) x 3 (tσ),t t 0 ,
(15)

where τ,σ(0,), p(t)= e t . We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)= e t , v(t)= e τ +2 e t , t1 satisfy (2) and also (10)

v (t) u (t)= e t <0for t[1,3.2].

For the constants K 1 =1, K 2 = e τ 1, where τ>ln2, t3.2, condition (4) has the form

e τ ( 1 e t + e 3 τ + 3 e 2 τ + σ t + 4 e τ + 2 σ 2 t + 2 e 3 σ 3 t ) a(t) e τ + e 3 σ τ 4 t 4 ( 1 + 2 e t ) .

For τ=σ=1 and t3.2, we have

e 1 ( 1 e t + e 3 + 3 e 3 t + 4 e 3 2 t + 2 e 3 ( 1 t ) ) <0.

Then for a(t), which satisfies the inequalities

0<a(t) e 1 + e 2 ( 1 2 t ) 4 ( 1 + 2 e t ) ,t t 1 3.2,
(16)

Eq. (11) has uncountably many solutions which are bounded by the functions u, v.

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Acknowledgements

The research was supported by the grant 1/0090/09 of the Scientific Grant Agency of the Ministry of Education of the Slovak Republic.

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Correspondence to Božena Dorociaková.

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Keywords

  • neutral differential equation
  • nonlinear
  • existence
  • uncountably many positive solutions
  • Banach space