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Existence and uniqueness of a positive periodic solution for Rayleigh type ϕ-Laplacian equation

Abstract

By using the Manásevich-Mawhin continuation theorem and some analysis skills, we establish some sufficient condition for the existence and uniqueness of positive T-periodic solutions for a generalized Rayleigh type ϕ-Laplacian operator equation. The results of this paper are new and they complement previous known results.

MSC:34K13, 34C25.

1 Introduction

During the past few years, many researchers have discussed the periodic solutions of a Rayleigh type differential equation (see [110]). For example, in 2009, Xiao and Liu [7] studied the Rayleigh type p-Laplacian equation with a deviating argument of the form

( ϕ p ( x ( t ) ) ) +f ( t , x ( t ) ) +g ( t , x ( t τ ( t ) ) ) =e(t).

By using the coincidence degree theory, we establish new results on the existence of periodic solutions for the above equation. Afterward, Xiong and Shao [9] used the coincidence degree theory to establish new results on the existence and uniqueness of positive T-periodic solutions for the Rayleigh type p-Laplacian equation of the form

( ϕ p ( x ( t ) ) ) +f ( t , x ( t ) ) +g ( t , x ( t ) ) =e(t).

In this paper, we consider the following Rayleigh type ϕ-Laplacian operator equation:

( ϕ ( x ( t ) ) ) +f ( t , x ( t ) ) +g ( t , x ( t ) ) =e(t),
(1.1)

where the function ϕ:RR is continuous and ϕ(0)=0. f,gCar(R×R,R) is an L p -Carathéodory function and p= m m 1 , m2, which means it is measurable in the first variable and continuous in the second variable. For every 0<r<s, there exists h r , s L p [0,T] such that |g(t,x(t))| h r , s for all x[r,s] and a.e. t[0,T]; and f, g is a T-periodic function about t and f(t,0)=0. e L p ([0,T],R) and is T-periodic.

Here ϕ:RR is a continuous function and ϕ(0)=0, which satisfies

(A1) (ϕ( x 1 )ϕ( x 2 ))( x 1 x 2 )>0 for x 1 x 2 , x 1 , x 2 R;

(A2) there exists a function α:[0,+][0,+], α(s)+ as s+, such that ϕ(x)xα(|x|)|x| for xR.

It is easy to see that ϕ represents a large class of nonlinear operators, including ϕ p :RR is a p-Laplacian, i.e., ϕ p (x)= | x | p 2 x for xR.

We know that the study on ϕ-Laplacian is relatively infrequent, the main difficulty lies in the fact that the ϕ-Laplacian operator typically possesses more uncertainty than the p-Laplacian operator. For example, the key step for ϕ p to get a priori solutions, 0 T ( ϕ p ( x ( t ) ) ) x(t)dt= 0 T | x ( t ) | p dt, is no longer available for general ϕ-Laplacian. So, we need to find a new method to get over it.

By using the Manásevich-Mawhin continuation theorem and some analysis skills, we establish some sufficient condition for the existence of positive T-periodic solutions of (1.1). The results of this paper are new and they complement previous known results.

2 Main results

For convenience, define

C T 1 = { x C 1 ( R , R ) : x  is  T -periodic } ,

which is a Banach space endowed with the norm ; define x=max{ | x | 0 , | x | 0 } for all x, and

| x | 0 = max t [ 0 , T ] |x(t)|,| x | 0 = max t [ 0 , T ] | x (t)|.

For the T-periodic boundary value problem

( ϕ ( x ( t ) ) ) = f ˜ ( t , x , x ) ,
(2.1)

here f ˜ :[0,T]×R×RR is assumed to be Carathéodory.

Lemma 2.1 (Manásevich-Mawhin [11])

Let Ω be an open bounded set in C T 1 . If

  1. (i)

    for each λ(0,1), the problem

    ( ϕ ( x ) ) =λ f ˜ ( t , x , x ) ,x(0)=x(T), x (0)= x (T)

has no solution on Ω;

  1. (ii)

    the equation

    F(a):= 1 T 0 T f ˜ ( t , x , x ) dt=0

has no solution on ΩR;

  1. (iii)

    the Brouwer degree of F

    deg{F,ΩR,0}0.

Then the periodic boundary value problem (2.1) has at least one periodic solution on Ω ¯ .

Lemma 2.2 If ϕ(x) is bounded, then x is also bounded.

Proof Since ϕ(x) is bounded, then there exists a positive constant N such that |ϕ(x)|N. From (A2), we have α(|x|)|x|ϕ(x)x|ϕ(x)||x|N|x|. Hence, we can get α(|x|)N for all xR. If x is not bounded, then from the definition of α, we get α(|x|)>N for some xR, which is a contradiction. So x is also bounded. □

Lemma 2.3 Suppose that the following condition holds:

(A3) ( x 1 x 2 )(g(t, x 1 )g(t, x 2 ))<0 for all t, x 1 , x 2 R, x 1 x 2 .

Then (1.1) has at most one T-periodic solution in C T 1 .

Proof Assume that x 1 (t) and x 2 (t) are two T-periodic solutions of (1.1). Then we obtain

( ϕ ( x 1 ( t ) ) ϕ ( x 2 ( t ) ) ) +f ( t , x 1 ( t ) ) f ( t , x 2 ( t ) ) +g ( t , x 1 ( t ) ) g ( t , x 2 ( t ) ) =0.
(2.2)

Set u(t)= x 1 (t) x 2 (t). Now, we claim that

u(t)0for all tR.

In contrast, in view of x 1 , x 2 C 1 [0,T], for tR, we obtain

max t R u(t)>0.

Then there must exist t R (for convenience, we can choose t (0,T)) such that

u ( t ) = max t [ 0 , T ] u(t)= max t R u(t)>0,

which implies that

u ( t ) = x 1 ( t ) x 2 ( t ) =0

and

x 1 ( t ) x 2 ( t ) >0.

By hypothesis (A3) and (2.2), we have

( ϕ ( x 1 ( t ) ) ϕ ( x 2 ( t ) ) ) = [ f ( t , x 1 ( t ) ) f ( t , x 2 ( t ) ) ] [ g ( t , x 1 ( t ) ) g ( t , x 2 ( t ) ) ] = [ g ( t , x 1 ( t ) ) g ( t , x 2 ( t ) ) ] > 0 ,

and there exists ε>0 such that ( ϕ ( x 1 ( t ) ) ϕ ( x 2 ( t ) ) ) >0 for all t( t ε, t ]. Therefore, ϕ( x 1 (t))ϕ( x 2 (t)) is strictly increasing for t( t ε, t ], which implies that

ϕ ( x 1 ( t ) ) ϕ ( x 2 ( t ) ) <ϕ ( x 1 ( t ) ) ϕ ( x 2 ( t ) ) =0for all t ( t ε , t ) .

From (A1) we get

u (t)= x 1 (t) x 2 (t)<0for all t ( t ε , t ) .

This contradicts the definition of t . Thus,

u(t)= x 1 (t) x 2 (t)0for all tR.

By using a similar argument, we can also show that

x 2 (t) x 1 (t)0.

Therefore, we obtain

x 1 (t) x 2 (t)for all tR.

Hence, (1.1) has at most one T-periodic solution in C T 1 . The proof of Lemma 2.3 is now complete. □

For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel:

(H1) there exists a positive constant D such that g(t,x)e(t)<0 for x>D and tR, g(t,x)e(t)>0 for x0 and tR;

(H2) there exist constants σ>0 and m2 such that f(t,u)uσ | u | m for (t,u)[0,T]×R;

(H3) there exist positive constants ρ and γ such that |f(t,u)|ρ | u | m 1 +γ for (t,u)[0,T]×R;

(H4) there exist positive constants α, β, B such that

|g(t,x)|α | x | m 1 +βfor |x|B and tR.

By using Lemmas 2.1-2.3, we obtain our main results.

Theorem 2.1 Assume that conditions (H1)-(H4) and (A3) hold. Then (1.1) has a unique positive T-periodic solution if σ α T m 1 2 m 1 >0.

Proof Consider the homotopic equation of (1.1) as follows:

( ϕ ( x ( t ) ) ) +λf ( t , x ( t ) ) +λg ( t , x ( t ) ) =λe(t).
(2.3)

By Lemma 2.3, it is easy to see that (1.1) has at most one T-periodic solution in C T 1 . Thus, to prove Theorem 2.1, it suffices to show that (1.1) has at least one T-periodic solution in  C T 1 . To do this, we are going to apply Lemmas 2.1 and 2.2. Firstly, we will claim that the set of all possible T-periodic solutions of (2.3) is bounded. Let x(t) C T 1 be an arbitrary solution of (2.3) with period T. As x(0)=x(T), there exists t 0 [0,T] such that x ( t 0 )=0, while ϕ(0)=0, we see

| ϕ ( x ( t ) ) | = | t 0 t ( ϕ ( x ( s ) ) ) d s | λ 0 T | f ( t , x ( t ) ) | d t + λ 0 T | g ( t , x ( t ) ) | d t + λ 0 T | e ( t ) | d t ,
(2.4)

where t[ t 0 , t 0 +T].

We claim that there is a constant ξR such that

|x(ξ)|D.
(2.5)

Let t ¯ , t ̲ be, respectively, the global maximum point and the global minimum point of x(t) on [0,T]; then x ( t ¯ )=0, and we claim that

( ϕ ( x ( t ¯ ) ) ) 0.
(2.6)

Assume, by way of contradiction, that (2.6) does not hold. Then ( ϕ ( x ( t ¯ ) ) ) >0 and there exists ε>0 such that ( ϕ ( x ( t ) ) ) >0 for t( t ¯ ε, t ¯ +ε). Therefore ϕ( x (t)) is strictly increasing for t( t ¯ ε, t ¯ +ε). From (A1) we know that x (t) is strictly increasing for t( t ¯ ε, t ¯ +ε). This contradicts the definition of t ¯ . Thus, (2.6) is true. From f(t,0)=0, (2.3) and (2.6), we have

g ( t ¯ , x ( t ¯ ) ) e( t ¯ )0.
(2.7)

Similarly, we get

g ( t ̲ , x ( t ̲ ) ) e( t ̲ )0.
(2.8)

In view of (H1), (2.7) and (2.8) imply that

x( t ¯ )D,x( t ̲ )>0.

Case (1): If x( t ̲ )(0,D), define ξ= t ¯ , obviously, |x(ξ)|D.

Case (2): If x( t ̲ )D, from x( t ¯ )D, we know x( t ¯ )=x( t ̲ ). Define ξ= t ¯ , we have |x(ξ)|=D. This proves (2.5).

Then we have

|x(t)|=|x(ξ)+ ξ t x (s)ds|D+ ξ t | x (s)|ds,t[ξ,ξ+T]

and

|x(t)|=|x(tT)|=|x(ξ) t T ξ x (s)ds|D+ t T ξ | x (s)|ds,t[ξ,ξ+T].

Combining the above two inequalities, we obtain

| x | 0 = max t [ 0 , T ] | x ( t ) | = max t [ ξ , ξ + T ] | x ( t ) | max t [ ξ , ξ + T ] { D + 1 2 ( ξ t | x ( s ) | d s + t T ξ | x ( s ) | d s ) } D + 1 2 0 T | x ( s ) | d s .
(2.9)

Since x (t) is T-periodic, multiplying x (t) and (2.3) and then integrating it from 0 to T, we have

0 = 0 T ( ϕ ( x ( t ) ) ) x ( t ) d t = λ 0 T f ( t , x ( t ) ) x ( t ) d t λ 0 T g ( t , x ( t ) ) x ( t ) d t + λ 0 T e ( t ) x ( t ) d t .
(2.10)

In view of (2.10), we have

| 0 T f ( t , x ( t ) ) x (t)dt|=| 0 T g ( t , x ( t ) ) x (t)dt+ 0 T e(t) x (t)dt|.

From (H2), we know

| 0 T f ( t , x ( t ) ) x (t)dt|σ 0 T | x (t) | m dt.

Set

E 1 = { t [ 0 , T ] | x ( t ) | B } , E 2 = { t [ 0 , T ] | x ( t ) | B } .

From (H4), we have

σ 0 T | x ( t ) | m d t E 1 + E 2 | g ( t , x ( t ) ) | | x ( t ) | d t + 0 T | e ( t ) | | x ( t ) | d t ( E 1 | g ( t , x ( t ) ) | m m 1 d t ) m 1 m ( 0 T | x ( t ) | m d t ) 1 m + α 0 T | x ( t ) | m 1 | x ( t ) | d t + β 0 T | x ( t ) | d t + 0 T | e ( t ) | | x ( t ) | d t | g B | m m 1 ( 0 T | x ( t ) | m d t ) 1 m + α ( D + 1 2 0 T | x ( t ) | d t ) m 1 0 T | x ( t ) | d t + β T m 1 m ( 0 T | x ( t ) | m d t ) 1 m + ( 0 T | e ( t ) | m m 1 ) m 1 m ( 0 T | x ( t ) | m d t ) 1 m = | g B | m m 1 ( 0 T | x ( t ) | m d t ) 1 m + α 2 m 1 ( 2 D 0 T | x ( t ) | d t + 1 ) m 1 ( 0 T | x ( t ) | d t ) m + β T m 1 m ( 0 T | x ( t ) | m d t ) 1 m + | e | m m 1 ( 0 T | x ( t ) | m d t ) 1 m ,
(2.11)

where g B = max | x | B |g(t,x(t))|, | g B | m m 1 = ( 0 T | g B | m m 1 d t ) m 1 m .

For the constant δ>0, which is only dependent on k>0, we have

( 1 + x ) k 1+(1+k)xfor x[0,δ].

So, from (2.11), we have

σ 0 T | x ( t ) | m d t | g B | m m 1 ( 0 T | x ( t ) | m d t ) 1 m + α 2 m 1 ( 1 + 2 D m 0 T | x ( t ) | d t ) ( 0 T | x ( t ) | d t ) m + β T m 1 m ( 0 T | x ( t ) | m d t ) 1 m + | e | m m 1 ( 0 T | x ( t ) | m d t ) 1 m = | g B | m m 1 ( 0 T | x ( t ) | m d t ) 1 m + α 2 m 1 ( 0 T | x ( t ) | d t ) m + α D m 2 m 2 ( 0 T | x ( t ) | d t ) m 1 + β T m 1 m ( 0 T | x ( t ) | m d t ) 1 m + | e | m m 1 ( 0 T | x ( t ) | m d t ) 1 m | g B | m m 1 ( 0 T | x ( t ) | m d t ) 1 m + α T m 1 2 m 1 0 T | x ( t ) | m d t + α D m T ( m 1 ) 2 m 2 m 2 ( 0 T | x ( t ) | m d t ) m 1 m + β T m 1 m ( 0 T | x ( t ) | m d t ) 1 m + | e | m m 1 ( 0 T | x ( t ) | m d t ) 1 m = α T m 1 2 m 1 0 T | x ( t ) | m d t + α D m T ( m 1 ) 2 m 2 m 2 ( 0 T | x ( t ) | m d t ) m 1 m + ( | g B | m m 1 + β T m 1 m + | e | m m 1 ) ( 0 T | x ( t ) | m d t ) 1 m .

Since σ α T m 1 2 m 1 >0, so it is easy to see that there is a constant M 1 >0 (independent of λ) such that

0 T | x (t) | m dt M 1 .

By applying Hölder’s inequality and (2.9), we have

| x | 0 D+ 1 2 0 T | x (s)|dsD+ 1 2 T m 1 m ( 0 T | x ( t ) | m d t ) 1 m D+ 1 2 T m 1 m ( M 1 ) 1 m := M 1 .

In view of (2.4) and (H3), we have

| ϕ ( x ) | 0 = max t [ 0 , T ] { | ϕ ( x ( t ) ) | } = max t [ t 0 , t 0 + T ] { | t 0 t ( ϕ ( x ( s ) ) ) d s | } 0 T | f ( t , x ( t ) ) | d t + 0 T | g ( t , x ( t ) ) | d t + 0 T | e ( t ) | d t ρ 0 T | x ( t ) | m 1 d t + γ T + T 1 m ( 0 T | g ( t , x ( t ) ) | m m 1 d t ) m 1 m + T 1 m ( 0 T | e ( t ) | m m 1 d t ) m 1 m ρ T 1 m ( 0 T | x ( t ) | m d t ) m 1 m + γ T + T 1 m ( 0 T | g ( t , x ( t ) ) | m m 1 d t ) m 1 m + T 1 m ( 0 T | e ( t ) | m m 1 d t ) m 1 m ρ T 1 m ( M 1 ) m 1 m + γ T + T 1 m | g M 1 | m 1 m + T 1 m | e | m 1 m : = M 2 ,

where | g M 1 |= max | x ( t ) | M 1 |g(t,x(t))|.

Thus, from Lemma 2.2, we know that there exists some positive constant M 2 > M 2 +1 such that, for all tR,

| x (t)| M 2 .

Set M= M 1 2 + M 2 2 +1, we have

Ω= { x C T 1 ( R , R ) | x | 0 M + 1 , | x | 0 M + 1 } ,

we know that (2.4) has no solution on Ω as λ(0,1) and when x(t)ΩR, x(t)=M+1 or x(t)=M1, from (2.11) we know that M+1>D. So, from (H1) we see that

1 T 0 T { g ( t , M + 1 ) e ( t ) } d t < 0 , 1 T 0 T { g ( t , M 1 ) e ( t ) } d t > 0 .

So condition (ii) is also satisfied. Set

H(x,μ)=μx(1μ) 1 T 0 T { g ( t , x ) e ( t ) } dt,

where xΩR, μ[0,1], we have

xH(x,μ)=μ x 2 (1μ)x 1 T 0 T { g ( t , x ) e ( t ) } dt>0,

and thus H(x,μ) is a homotopic transformation and

deg { F , Ω R , 0 } = deg { 1 T 0 T { g ( t , x ) e ( t ) } d t , Ω R , 0 } = deg { x , Ω R , 0 } 0 .

So condition (iii) is satisfied. In view of Lemma 2.1, there exists at least one solution with period T.

Suppose that x(t) is the T-periodic solution of (1.1). We can easily show that (2.8) also holds. Thus,

x(t) min t [ 0 , T ] x(t)=x( t ̲ )>0for all tR,

which implies that (1.1) has a unique positive solution with period T. This completes the proof. □

We illustrate our results with some examples.

Example 2.1 Consider the following second-order p-Laplacian-like Rayleigh equation:

( ϕ p ( x ( t ) ) ) + ( 10 + 5 sin 2 t ) x (t) ( 5 x ( t ) + sin 2 t 8 ) = e cos 2 t ,
(2.12)

where ϕ p (u)= | u | p 2 u.

Comparing (2.12) to (1.1), we see that g(t,x)=5x(t) sin 2 t+8, f(t,u)=(10+5 sin 2 t)u, e(t)= e cos 2 t , T=π. Obviously, we know that ϕ p is a homeomorphism from to satisfying (A1) and (A2). Consider ( x 1 x 2 )(g(t, x 1 )g(t, x 2 ))=5 ( x 1 x 2 ) 2 <0 for x 1 x 2 , then (A3) holds. Moreover, it is easily seen that there exists a constant D=2 such that (H1) holds. Consider f(t,u)u=(10+5 sin 2 t) u 2 10 u 2 , here σ=10, m=2, and |f(t,u)|=|(10+5 sin 2 t)u|15|u|+1, here ρ=15, γ=1. So, we can get that conditions (H2) and (H3) hold. Choose B>0, we have |g(t,x)|5|x|+9, here α=5, β=9, then (H4) holds and σ α T 2 =10 5 π 2 >0. So, by Theorem 2.1, we can get that (2.12) has a unique positive periodic solution.

Example 2.2 Consider the following second-order p-Laplacian-like Rayleigh equation:

( ϕ ( x ( t ) ) ) + ( 200 + 16 cos 2 t ) ( x ( t ) ) 3 ( 20 x 3 ( t ) + 10 cos 2 ( t ) 15 ) = e sin 2 t ,
(2.13)

where ϕ(u)=u e | u | 2 .

Comparing (2.13) to (1.1), we see that g(t,x)=20 x 3 10 cos 2 t+15, f(t,v)=(200+16 cos 2 t) v 3 , e(t)= e sin 2 t , T=π. Obviously, we get

( x e | x | 2 y e | y | 2 ) (xy) ( | x | e | x | 2 | y | e | y | 2 ) ( | x | | y | ) 0

and

ϕ(x)x= | x | 2 e | x | 2 .

So, we know that (A1) and (A2) hold. Consider ( x 1 x 2 )(g(t, x 1 )g(t, x 2 ))=20 ( x 1 x 2 ) 2 ( x 1 2 + x 1 x 2 + x 2 2 )<0 for x 1 x 2 , then (A3) holds. Moreover, it is easily seen that there exists a constant D=1 such that (H1) holds. Consider f(t,v)v=(200+16 cos 2 t) v 4 200 v 4 , here σ=200, m=4, and |f(t,v)|=|(200+16 cos 2 t) v 3 |216 | v | 3 +5, here ρ=216, γ=5. So, we can get that conditions (H2) and (H3) hold. Choose B>0, we have |g(t,x)|20 | x | 3 +25, here α=20, β=25, then (H4) holds and σ α T m 1 2 m 1 =200 20 × π 3 2 3 >0. Therefore, by Theorem 2.1, we know that (2.13) has a unique positive periodic solution.

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Acknowledgements

Research is supported by the National Natural Science Foundation of China (Nos. 11326124, 11271339).

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Correspondence to Yun Xin.

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YX and ZBC worked together in the derivation of the mathematical results. All authors read and approved the final manuscript.

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Xin, Y., Cheng, Z. Existence and uniqueness of a positive periodic solution for Rayleigh type ϕ-Laplacian equation. Adv Differ Equ 2014, 225 (2014). https://doi.org/10.1186/1687-1847-2014-225

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Keywords

  • positive periodic solution
  • uniqueness
  • ϕ-Laplacian
  • Rayleigh equation