A continuation lemma and its applications to periodic solutions of Rayleigh differential equations with subquadratic potential conditions

J. Math. Anal. Appl. 385 (2012) 1107–1118

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A continuation lemma and its applications to periodic solutions of Rayleigh differential equations with subquadratic potential conditions ✩ Tiantian Ma, Zaihong Wang ∗ School of Mathematical Sciences, Capital Normal University, Beijing 100048, People’s Republic of China

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a b s t r a c t

Article history: Received 23 July 2010 Available online 23 July 2011 Submitted by J. Mawhin

In this paper, we study the existence of periodic solutions of Rayleigh equations





x + f t , x + g (x) = e (t ), where f , g, e are continuous functions and f is T -periodic with the first variable, e is T -periodic. By developing a continuation lemma for Rayleigh equations, we prove that the given equation has at least one T -periodic solution provided that f (t , y ) is sublinear with x respect to the variable y and G (x)(= 0 g (u ) du ) satisfies some subquadratic conditions. © 2011 Elsevier Inc. All rights reserved.

Keywords: Rayleigh equation Periodic solution Continuation lemma

1. Introduction We are concerned with the existence of T -periodic solutions of the Rayleigh equations





x + f t , x + g (x) = e (t ),

(1.1)

where f : R2 → R, g , e : R → R are continuous and f is T -periodic with the first variable, e is T -periodic. Eq. (1.1) can be used to model the oscillations of a clarinet reed [1]. The dynamic behaviors of Eq. (1.1) have been widely investigated due to their applications in many fields such as physics, mechanics and the engineering technique fields (see [7,8,11–17] and the references therein). When f ≡ 0, Eq. (1.1) is a conservation system

x + g (x) = e (t ).

(1.2)

In the case when g (x) satisfies the sublinear condition

g (x)

lim

|x|→+∞

x

= 0,

(1.3)

and the sign condition





sgn(x) g (x) − e¯  0,

|x|  d,

(1.4)

T

where d is a positive constant and e¯ := T1 0 e (s) ds, Lazer [6] first proved the existence of T -periodic solutions of Eq. (1.2). From then on, there have appeared many works which generalized the sublinear condition (1.3) in different styles. In ✩ Research supported by National Natural Science Foundation of China, No. 10771145 and Beijing Natural Science Foundation (Existence and multiplicity of periodic solutions in nonlinear oscillations), No. 1112006. Corresponding author. E-mail addresses: [email protected] (T. Ma), [email protected] (Z. Wang).

*

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2011.07.032

©

2011 Elsevier Inc. All rights reserved.

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T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

x

particular, some one-sided growth conditions on g (x) or G (x)(= 0 g (u ) du ) are introduced in [3,4,9] and the references therein. In [9], Mawhin and Ward proved the existence of T -periodic solutions of Eq. (1.2) by assuming, besides (1.4), that g (x) still satisfies the following condition,

lim sup

g (x) x

x→+∞



<

π

2

(1.5)

.

T

Later, Fernandes and Zanolin [3] extended the condition (1.5) to the condition as follows,

(h1 )

lim inf

2G (x) x2

x→+∞

 <

π

2 .

T

One aim of this paper is to study the existence of T -periodic solutions of Eq. (1.1) provided that G (x) satisfies the subquadratic condition (h1 ). Assume that g (x) satisfies the condition, (h2 )











sgn(x) g (x) > sup  f (t , 0) + e (t ): t ∈ R ,

|x|  d¯ ,

where d¯ is a positive constant, and f (t , y ) satisfies the sublinear condition, (h3 )

f (t , y )

lim

| y |→+∞

y

=0

uniformly for t ∈ [0, T ]. By developing a continuation lemma for Rayleigh equations, we obtain the following result. Theorem 1.1. Assume that conditions (h i ) (i = 1, 2, 3) hold. Then Eq. (1.1) has at least one T -periodic solution. When the one-sided condition (h1 ) is replaced by the double-sided condition as follows,

(h4 )

lim inf x→+∞

2G (x) x2

 

π

2

a

,

lim sup

g (x)

x→−∞

x

 

π b

2 ,

where a, b are two positive constants satisfying a + b > T , we can obtain the following result. Theorem 1.2. Assume that conditions (h i ) (i = 2, 3, 4) hold. Then Eq. (1.1) has at least one T -periodic solution. Throughout this paper, the usual norm in L p (0, T ), 1  p < +∞, will be denoted by  ·  p . That is,

T x p =

  x(t ) p dt

1p ,

x ∈ L p (0, T ),

0

and for any continuous T -periodic function x(t ), we set

  x∞ = max x(t ): 0  t  T .

2. A continuation lemma It is well known that the continuation lemma based on coincidence degree theory plays a key role in studying the existence of periodic solutions of ordinary differential equations [3–5,11]. In order to deal with the existence of periodic solutions of Eq. (1.1), we shall introduce a continuation lemma for Rayleigh equations. To this end, we consider the equivalent system of Eq. (1.1),

x = y ,





y  = − g (x) + f (t , y ) − e (t ) .

(2.1)

Now, we embed system (2.1) into a family of equations with one parameter λ ∈ [0, 1],

x = λ y ,





y  = −λ g (x) + f (t , λ y ) − e (t ) .

(2.2)

Lemma 2.1. Assume that conditions (h2 ) and (h3 ) hold. Suppose that there exists a constant ζ > 0 such that, if (x(t ), y (t )) is a T -periodic solution of system (2.2) for some λ ∈ (0, 1], then

x(t )  ζ,

t ∈ [0, T ].

Then system (2.1) has at least one T -periodic solution.

T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

1109

Proof. We apply Lemma 2 in [10] to prove this lemma. Set

¯f =

1

T e¯ =

f (s, 0) ds,

T

1

T e (s) ds.

T

0

0

Obviously, there exists t˜ ∈ [0, T ] such that

¯f − e¯ = 1

T

T



f (s, 0) − e (s) ds = f (t˜, 0) − e (t˜).

0

From (h2 ) we know that















sgn(x) g (x) > sup  f (t , 0) + e (t ): t ∈ R >  f (t˜, 0) − e (t˜) = | ¯f − e¯ |,

|x|  d¯ .

(2.3)

Consequently, the map S : (x, y ) → ( y , − g (x) − ¯f + e¯ ) does not vanish outside the rectangle [−d¯ , d¯ ] × [−σ , σ ], where σ is an arbitrary positive  constant. Furthermore, the Brouwer degree of S, d B ( S , B (0, r ), 0) is defined and equal to 1 on any ball B (0, r ), with r > d¯ 2 + σ 2 . Therefore, it is sufficient to prove that there is a constant c > 0 such that, if (x(t ), y (t )) is a T -periodic solution of (2.2) (λ ∈ (0, 1]) with x(t )  ζ , t ∈ [0, T ], then

x∞ +  y ∞  c ,

t ∈ [0, T ].

(2.4)

Accordingly, let (x(t ), y (t )) be a T -periodic solution of (2.2) satisfying the condition as follows,

x(t )  ζ,

t ∈ [0, T ].

(2.5)

Integrating the second equality of (2.2) on [0, T ] and applying the first equality of (2.2), we get

T



 g x(t ) dt = −

0

T



T

 f t , x (t ) dt + 

e (t ) dt .

0

0

Then we obtain





 g x(t ) dt =

− I−

where I −

T





 g x(t ) dt +

I+

T



 f t , x (t ) dt − 

0

T e (t ) dt , 0

= {t ∈ [0, T ]: x(t ) < −d¯ }, I + = {t ∈ [0, T ]: −d¯  x(t )  ζ }. Hence, from (2.3) we have

    g x(t )  dt =



    g x(t )  dt +

I−

0



=−

2

    g x(t )  dt

I+



 g x(t ) dt +

I−



 

    g x(t )  dt

I+

    g x(t )  dt +

I+

T

     f t , x (t )  dt +

0

T  2M 1 +

T

  e (t ) dt

0

     f t , x (t )  dt + e 1 ,

(2.6)

0

where M 1 := T · max{| g (x)|: −d¯  x  ζ }. Since condition (h3 ) holds, for any given constant δ , with 0 < ( 2Tπ + √T )δ < 1, there exists R δ > 0 such that, if | y |  R δ , 12 then we have

   f (t , y )  δ| y |,

Let



t ∈ [0, T ].





M = max  f (t , y ): | y |  R δ , t ∈ [0, T ] .

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T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

Thus, for any (t , y ) ∈ R2 , we get

   f (t , y )  δ| y | + M .

(2.7)

From (2.6) and (2.7) we know

T

    g x(t )  dt  2M 1 + δ

0

   x (t ) dt + M T + e 1 .

0

Setting M 1

T

T

= 2M 1 + M T + e 1 , we obtain

T

    g x(t )  dt  δ

0

√      x (t ) dt + M   δ T x  + M  . 1 1 2

(2.8)

0

From (2.2) we know that x(t ) satisfies the equation as follows,

 









x (t ) + λ2 f t , x (t ) + g x(t ) − e (t ) = 0. Multiplying Eq. (2.9) by x(t ) − x¯ , with x¯ = that

T

T

  2 x (t ) dt = λ2

0



1 T

T

(2.9)

x(s) ds, and integrating the equality on [0, T ], we get from (2.7) and (2.8)

0

  f t , x (t ) x(t ) − x¯ dt + λ2 

0

T 

T



  g x(t ) x(t ) − x¯ dt − λ2

0

      f t , x (t ) x(t ) − x¯  dt +

0

T

δ

    x (t )x(t ) − x¯  dt + M

0

T





e (t ) x(t ) − x¯ dt 0

     g x(t ) x(t ) − x¯  dt +

0

T

T

T

   e (t ) x(t ) − x¯  dt

0

  x(t ) − x¯  dt + x − x¯ ∞

0

T

    g x(t )  dt + e 2 x − x¯ 2

0

√       √    δ x 2 + M T + e 2 x − x¯ 2 + δ T x 2 + M 1 x − x¯ ∞ . According to Wirtinger inequality and Sobolev inequality, we have

x − x¯ 2 

2π    x  , 2 T



x − x¯ ∞ 

T   x  . 2 12

Thus, we get

T

     2     π M 2π T    x (t ) dt  2π δ x 2 + 2√ x  + √δ T x 2 . + e 2 + M 1 2 2 2 T

T

0

T

12

12

Therefore,

    x   γ x  + M 2 , 2 2 where

γ = ( 2Tπ + √T12 )δ , M 2 =

2√ πM T

+ 2Tπ e 2 + M 1



T . 12

Since 0 < γ < 1, we have

  x   M 2 := M  . 2 2 1−γ Integrating the first equation of (2.2) on [0, T ] and noticing λ ∈ (0, 1], we get

T y (t ) dt = 0

1

T

λ

x (t ) dt = 0.

0

From the mean value theorem of integrals we know that there exists t 0 ∈ [0, T ] such that y (t 0 ) = 0. Thus,

(2.10)

T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

     y (t ) =  y (t ) − y (t 0 ) 

T

1111

    y (t ) dt

0

T 

     f t , x (t )  dt +

0

T

    g x(t )  dt +

0

T δ

  e (t ) dt

0

   x (t ) dt + M T + δ

0

T

T

   x (t ) dt + M  + e 1 1

0

√    2δ T x 2 + M T + M 1 + e 1 . Hence, we get

√  y ∞  2δ T M 2 + M T + M 1 + e 1 := c 1 . Let x(t ∗ )

(2.11)

be the minimum of x(t ). Then we have x (t ∗ ) = 0, x (t ∗ )  0. Since x(t ∗ ) satisfies





x (t ∗ ) + f (t ∗ , 0) + g x(t ∗ ) = e (t ∗ ), we have





f (t ∗ , 0) + g x(t ∗ )  e (t ∗ ). Thus,























g x(t ∗ )  − f (t ∗ , 0) + e (t ∗ )   f (t ∗ , 0) + e (t ∗ )  sup  f (t , 0) + e (t ): t ∈ R . From condition (h2 ) we know

x(t ∗ ) < d¯ .

(2.12)

Let x(t ∗ ) be the maximum of x(t ). Then we have x (t ∗ ) = 0, x (t ∗ )  0. Since x(t ∗ ) satisfies

         x t ∗ + f t ∗ , 0 + g x t ∗ = e t ∗ ,

we have

       f t∗, 0 + g x t∗  e t∗ .

Thus,

  

g x t∗

               − f t ∗ , 0 + e t ∗  −  f t ∗ , 0  + e t ∗   − sup  f (t , 0) + e (t ): t ∈ R .

From condition (h2 ) we know

  x t ∗ > −d¯ .

(2.13)

According to (2.12) and (2.13), we know that there exists t  ∈ [0, T ] such that

    x t   d¯ .

Therefore, for t ∈ [0, T ],

     x(t )  x t   +

T

   x (t ) dt  d¯ +

0

T

   y (t ) dt  d¯ + c 1 T := c 2 .

0

Furthermore,

x∞  c 2 .

(2.14)

Taking c = c 1 + c 2 , we get from (2.11) and (2.14) that

x∞ +  y ∞  c . The proof is completed.

2

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T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

Remark 2.2. Assume that conditions (h2 ) and (h3 ) hold. Let (x(t ), y (t )) be any T -periodic solution of system (2.2). From the proof of Lemma 2.1 we know that, for any constant η > 0, there exists ς > 0 such that the following conclusion holds,









then x(t ) +  y (t )  ς ,

if max x(t )  η, t ∈[0, T ]

Similarly, we can prove that







t ∈ [0, T ].



then x(t ) +  y (t )  ς ,

if min x(t )  −η, t ∈[0, T ]

t ∈ [0, T ].

Remark 2.3. Assume that condition (h2 ) holds and f (t , y ) satisfies the condition

   f (t , y )  δ| y | + M ,

where δ , with 0 < ( 2Tπ + √T )δ < 1, and M are two positive constants. Then from the proof of Lemma 2.1 we know that the 12

conclusion of Lemma 2.1 still holds. 3. Periodic solutions under one-sided conditions In this section, we shall use the continuation Lemma 2.1 to prove Theorem 1.1. Proof of Theorem 1.1. Set

ρ = lim inf

2G (x) x2

x→+∞



<

π

2 .

T

Then there exists a constant ε0 > 0 such that, for 0 < ε  ε0 , ρε = ρ + ε < ( πT )2 . From [2] we know that, for any given ε ∈ (0, ε0 ], there is a sequence {xn } such that limn→+∞ xn = +∞ and









2 G (xn ) − G (x)  ρε xn2 − x2 ,

x ∈ (0, xn ).

It follows from condition (h3 ) that, for the same

   f (t , y )  ε | y | + M ε ,

(3.1)

ε ∈ (0, ε0 ], there exists M ε > 0 such that

2

(t , y ) ∈ R .

(3.2)

We claim that the condition of the continuation Lemma 2.1 holds with

ζ = xn for n sufficiently large. Let (x(t ), y (t )) be any T -periodic solution of (2.2) for some λ ∈ (0, 1] and suppose that

  x t ∗ = max x(t ) = xn > d¯ .

(3.3)

t ∈[0, T ]

Assume that x(t ∗ ) (t ∗ ∈ [0, T ]) is a local minimum of x(t ). We claim that

x(t ∗ ) < d¯ . In fact, since x (t ∗ ) = 0 and x (t ∗ )  0, we have that





f (t ∗ , 0) + g x(t ∗ )  e (t ∗ ). Therefore,























g x(t ∗ )  − f (t ∗ , 0) + e (t ∗ )   f (t ∗ , 0) + e (t ∗ )  sup  f (t , 0) + e (t ): t ∈ R .

¯ Then, there is an interval [α , β] containing t ∗ , with From (h2 ) we know that x(t ∗ ) < d. β − α < T such that x(α ) = x(β) = d¯ ;

x(t ) > d¯ ,

α = α (x, λ), β = β(x, λ), and

t ∈ (α , β)

and

  1   y t ∗ = x t ∗ = 0;

y (t ) > 0,

λ

t∈



α, t ∗ ;

y (t ) < 0,



Now, without loss of generality, we assume that t ∗ − α  T2 ; otherwise, β − t ∗  with [t ∗ , β] throughout the proof. From (2.2) we have that

 









y (t ) y  (t ) + λ f t , λ y (t ) + g x(t ) − e (t ) y (t ) = 0.



t ∈ t∗, β . T 2

and then we can replace interval [α , t ∗ ]

(3.4)

T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

Integrating both sides of (3.4) on interval [t , t ∗ ] with

   ∗ 

y 2 (t ) = 2 G x t



 − G x(t ) + 2λ

t ∗

α  t  t ∗ , we have



 f τ , λ y (τ ) y (τ ) dτ − 2λ

t

   ∗ 

2 G x t

 − G x(t ) + 2 

t ∗

t ∗ e (τ ) y (τ ) dτ t

     f τ , λ y (τ ) λ y (τ ) dτ + 2

t

   ∗ 

y (t )  2 G x t

 − G x(t ) + 2ε 

2 G x t



 − G x(t ) + 2ε

t ∗ 2

y (τ ) dτ + 2









M ε + e (τ ) λ y (τ ) dτ

t

t ∗ 2



t ∗

y (τ ) dτ + M ε t

      = 2 G x t ∗ − G x(t ) + 2ε

   e (τ )λ y (τ ) dτ .

∈ [α , t ∗ ] ,

t ∗ t

   ∗ 

t ∗ t

Since y (t ) > 0, t ∈ [α , t ∗ ], it follows from (3.2) that, for t 2

1113

x (τ ) dτ

t

t ∗

    y 2 (τ ) dτ + M ε x t ∗ − x(t ) ,

(3.5)

t

where M ε = 2( M ε + M e ), M e = max{|e (t )|, t ∈ R}. Set

t ∗ y 2 (τ ) dτ .

Φ(t ) = t

Then we get

Φ  (t ) = − y 2 (t ). Therefore, from (3.5) we have

          −Φ  (t ) − 2ε Φ(t )  2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) .

Multiplying both sides of the above inequality by e 2εt and integrating over the interval [t , t ∗ ] yields

t ∗ −



Φ(τ )e

 2ετ 

t ∗ dτ 

t

   ∗  2 G x t

      − G x(τ ) + M ε x t ∗ − x(τ ) e 2ετ dτ .

t

Since Φ(t ∗ ) = 0, we have

t ∗ Φ(t )e

2ε t



   ∗  2 G x t

      − G x(τ ) + M ε x t ∗ − x(τ ) e 2ετ dτ

t

          ∗  T 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) e 2εt . Hence, for t ∈ [α , t ∗ ], we obtain

          Φ(t )  T e 2ε T 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) .

(3.6)

It follows from (3.5) and (3.6) that, for t ∈ [α , t ∗ ],

       

            y 2 (t )  2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) + 2ε T e 2ε T 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t )             = 1 + 2ε T e 2ε T 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) .

Set

η = 2ε T e 2ε T . Obviously, η → 0 as ε → 0. From (3.7) we have that

          y 2 (t )  (1 + η) 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) .

(3.7)

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T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

Since x (t ) = λ y (t ) and y (t ) > 0, t ∈ [α , t ∗ ], we get that, for t ∈ [α , t ∗ ],

x (t ) 



          (1 + η) 2 G x t ∗ − G x(t ) + M ε x t ∗ − x(t ) .

Thus,



x (t )

 1.

(1 + η)[2(G (x(t ∗ )) − G (x(t ))) + M ε (x(t ∗ ) − x(t ))]

Integrating both sides of the above inequality over interval [α , t ∗ ] yields

x∗  d¯

dx

(1 + η)[2(G (x∗ ) − G (x)) + M ε (x∗

− x)]

 t ∗ − α,

where x∗ = x(t ∗ ) = xn . From (3.1) and (3.3) we get

xn  d¯

dx

(1 + η)[ρ

2 ε (xn

x2 ) +

−

M ε (xn

− x)]

 t ∗ − α.

(3.8)

It is easy to check that

d¯ 

lim

n→+∞ 0

dx

(1 + η)[ρ

2 ε (xn

− x2 ) + M ε (xn − x)]

= 0.

(3.9)

On the other hand,

xn  0

1

dx

(1 + η)[ρ

2 ε (xn

− x2 ) + M ε (xn − x)]



= 0

dt

(1 + η)[ρε (1 − t 2 ) +

M ε (1 − t )] xn

.

(3.10)

Recalling limn→+∞ xn = +∞, we get

1  0

1

dt

(1 + η)[ρε (1 − t 2 ) +

M

ε (1 − t )] xn



 0

dt

(1 + η)[ρε (1 − t 2 ) + ε ]

,

(3.11)

for n large enough. Meanwhile, we get by elementary computations,

1 

lim

ε →0+ 0

1

dt

(1 + η)[ρε

(1 − t 2 ) +

ε]



= 0

dt

ρ

(1 − t 2 )

π

= √ 2

ρ

>

T 2

.

(3.12)

From (3.8)–(3.12) we obtain

t∗ − α >

T 2

,

for n large enough. This is a contradiction. Then, we find ζ = xn for n sufficiently large. Consequently, from Lemma 2.1, we know that Eq. (1.1) has at least one T -periodic solution. 2 Remark 3.1. If (h1 ) is replaced by the condition as follows,

lim inf x→+∞

2G (x) x2

= 0,

and (h2 ) still holds. Moreover, f (t , y ) satisfies the inequality

   f (t , y )  δ| y | + M ,

(t , y ) ∈ R2 ,

where δ , with 0 < ( 2Tπ + √T )δ < 1, and M > 0 are two constants. By using Remark 2.3 and the similar method as in proving 12

Theorem 1.1, we can prove that Eq. (1.1) has at least one T -periodic solution. For simplicity, we omit the proof here.

T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

1115

4. Periodic solutions under double-sided conditions In this section, we also use the continuation Lemma 2.1 to prove Theorem 1.2. Proof of Theorem 1.2. Since a + b > T , there exists 0 < μ1 < min(a, b) satisfying (a − μ1 ) + (b − μ1 ) > T . For any given μ ∈ (0, μ1 ), let us set

 aμ =

2

π

 bμ =

,

a−μ

2

π

.

b−μ

From condition (h4 ) we know that there is a sequence {xn } such that limn→+∞ xn = +∞ and









2 G (xn ) − G (x)  aμ xn2 − x2 ,

x ∈ (0, xn ).

Moreover, there is a positive constant c 0 such that, for any x˜ < −c 0 , we have









2 G (˜x) − G (x)  bμ x˜ 2 − x2 ,

x ∈ (˜x, −c 0 ).

We also claim that the condition of continuation Lemma 2.1 holds with

ζ = xn for n sufficiently large. Let (x(t ), y (t )) be any T -periodic solution of (2.2) for some λ ∈ (0, 1] and suppose that

  x t ∗ = max x(t ) = xn > d¯ , t ∈[0, T ]

where d¯ is given in the proof of Theorem 1.1. Then, there is an interval [α , β] containing t ∗ , with and β − α < T such that

x(α ) = x(β) = d¯ ;

x(t ) > d¯ ,

α = α (x, λ), β = β(x, λ),

t ∈ (α , β)

and

  1   y t ∗ = x t ∗ = 0;

y (t ) > 0,

λ

t∈



α, t ∗ ;





t ∈ t∗, β .

y (t ) < 0,

In what follows, we shall estimate β − α . To this end, we will estimate t ∗ − α and β − t ∗ , respectively. Note that for any sufficiently small ε > 0, there exists M ε > 0 such that

   f (t , y )  ε | y | + M ε ,

(t , y ) ∈ R2 .

Using the same method as in proving Theorem 1.1, we can get

xn  d¯

where

dx

(1 + η

)[aμ (xn2

−

x2 ) +

M ε (xn

− x)]

 t ∗ − α,

(4.1)

η = 2ε T e 2ε T , M ε is a constant depending on ε . Since limn→∞ xn = +∞, it is easy to check that d¯ lim

n→+∞ 0

dx

 = 0. (1 + η)[aμ (xn2 − x2 ) + M ε (xn − x)]

(4.2)

On the other hand,

xn  0

1

dx

(1 + η

)[aμ (xn2

−

x2 ) +

M ε (xn

− x)]



= 0

dt

(1 + η)[aμ (1 − t 2 ) +

M ε (1 − t )] xn

.

(4.3)

Since limn→+∞ xn = +∞, we get

1  0

1

dt

(1 + η)[aμ (1 − t 2 ) +

M

ε

xn

(1 − t )]

 0

dt

 (1 + η)[aμ (1 − t 2 ) + ε ]

(4.4)

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T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

for n large enough. Meanwhile, we get, by elementary computations,

1 

lim

ε →0+ 0

1

dt

(1 + η)[aμ

(1 − t 2 ) +

ε]



=

aμ

0

dt

=

(1 − t 2 )

a−μ 2

(4.5)

.

From (4.1)–(4.5) we have that, for n sufficiently large,

t∗ − α >

a − μ1 2

.

In the same way, we can get

β − t∗ >

a − μ1 2

.

Thus, we conclude

β − α > a − μ1 .

(4.6)

Next, let us set

x∗ = x(t ∗ ) = min x(t ). t ∈[0, T ]

˜ with d˜ = max{d¯ , c 0 }. Assume that x(t˜) is a local From Remark 2.2 we know that, if n is large enough, then x∗  −d, ¯ Consequently, there is an interval [α˜ , β] ˜ containing t ∗ , with α˜ = α˜ (x, λ), maximum of x(t ). We can also prove that x(t˜) > −d. ˜ x, λ), and β˜ − α˜ < T such that β˜ = β(

˜ = −d˜ ; ˜ ) = x(β) x(α

x(t ) < −d˜ ,

˜ ˜ , β) t ∈ (α

and

y (t ∗ ) =

1  x (t ∗ ) = 0;

˜ , t ∗ ); t ∈ [α

y (t ) < 0,

λ

y (t ) > 0,

˜ t ∈ (t ∗ , β].

˜ . To this end, we first estimate t ∗ − α˜ and β˜ − t ∗ , respectively. Next, we shall estimate β˜ − α ˜ , t ∗ ] with α˜  t  t ∗ , we have that Integrating both sides of (3.4) on interval [α

 

   y (t )  2 G x(t ∗ ) − G x(t ) + 2 2

t∗

     f τ , λ y (τ ) λ y (τ ) dτ + 2

t

t∗

   e (τ )λ y (τ ) dτ .

t

˜ , t ∗ ], we get Since y (t ) < 0, t ∈ [α

 

   y (t )  2 G x(t ∗ ) − G x(t ) + 2ε 2

t∗

t∗ 2

y (τ ) dτ + 2 t

 

    2 G x(t ∗ ) − G x(t ) + 2ε

2



t∗

y (τ ) dτ − M ε

t∗





x (τ ) dτ

t





y 2 (τ ) dτ − M ε x(t ∗ ) − x(t ) ,

t

with M ε = 2( M ε + M e ), M e = max{|e (t )|, t ∈ R}. ˜ , t ∗ ], Using the same method as in proving Theorem 1.1, we have that, for t ∈ [α

 

  − M ε x(t ∗ ) − x(t ) 

    (1 + η) bμ x2 (t ∗ ) − x2 (t ) − M ε x(t ∗ ) − x(t ) . 





y 2 (t )  (1 + η) 2 G x(t ∗ ) − G x(t )

˜ , t ∗ ], Thus, we get that, for t ∈ [α

−x (t ) 

 

   (1 + η) bμ x2 (t ∗ ) − x2 (t ) − M ε x(t ∗ ) − x(t ) .

Furthermore,



−x (t ) (1 + η)[bμ (x2 (t ∗ ) − x2 (t )) − M ε (x(t ∗ ) − x(t ))]

 1.



M ε + e (τ ) λ y (τ ) dτ

t

t∗ t

     = 2 G x(t ∗ ) − G x(t ) + 2ε



T. Ma, Z. Wang / J. Math. Anal. Appl. 385 (2012) 1107–1118

1117

˜ , t ∗ ] yields Integrating both sides of the above inequality over interval [α

−d˜  x∗

dx

(1 + η)[bμ (x2∗ − x2 ) − M ε (x∗ − x)]

 t ∗ − α˜ .

From Remark 2.2 we know that x∗ → −∞ as xn → +∞. Then we can get that, for n large enough,

b − μ1

˜> t∗ − α

2

b − μ1

β˜ − t ∗ >

,

2

.

Therefore, for n large enough,

β˜ − α˜ > b − μ1 .

(4.7)

From (4.6) and (4.7) we obtain

˜ > a + b − 2μ1 , T > β − α + β˜ − α which yields a contradiction with the choice of μ1 . Then, we find ζ = xn for n sufficiently large. Hence, Eq. (1.1) has at least one T -periodic solution.

2

5. An example In this section, we shall construct an example to show the application of Theorem 1.1 obtained before. Example. Let e (t ) be a continuous 2π -periodic function. Assume g (x) is an odd function defined by



g (x) = where r ∈ (1,

lim inf

1 4

rx + x cos ln x, 0, √2

+

5

g (x)

x→+∞

x

x > 0; x = 0,

) is a constant. Obviously, g (x) satisfies condition (h2 ). By a direct calculation, we get

= r − 1 > 0,

lim sup

g (x)

x→+∞

x

= r + 1 < +∞

and

G (x) =

r 2

x2 +

x2 5

(2 cos ln x + sin ln x).

It is easy to check that

0 < lim inf

2G (x)

x→+∞

x2

2

1

=r− √ < . 5

4

Let φ : R → R be a continuous function satisfying

lim

| y |→+∞

φ( y ) y

= 0.

Define

f (t , y ) = φ( y ) sin(t + y ). Thus, conditions (h i ) (i = 1, 2, 3) hold. According to Theorem 1.1, equation x + f (t , x ) + g (x) = e (t ) has at least one T -periodic solution. Acknowledgment The authors are grateful to the referee for many valuable suggestions to make the paper more readable.

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