# Minimal period problem for second-order Hamiltonian systems with asymptotically linear nonlinearities

• Juhong Kuang and Weiyi Chen
From the journal Open Mathematics

## Abstract

By applying the combination of discrete variational method and approximation, developed in a previous study [J. Kuang, W. Chen, and Z. Guo, Periodic solutions with prescribed minimal period for second-order even Hamiltonian systems, Commun. Pure Appl. Anal. 21 (2022), no. 1, 47–59], we overcome some difficulties in the absence of Ambrosetti-Rabinowitz condition and obtain new sufficient conditions for the existence of periodic solutions with prescribed minimal period for second-order Hamiltonian systems with asymptotically linear nonlinearities.

MSC 2010: 34A12; 39A23; 39A70

## 1 Introduction and main result

In the pioneering work [1] of 1978, Rabinowitz conjectured that the first- and second-order Hamiltonian systems have nonconstant solutions with any prescribed minimal period. In 1985, Ekeland and Hofer [2] made important progress and confirmed the conjecture for a first-order Hamiltonian system with strictly convex assumptions. Since then, the minimal periodic problem of the Hamiltonian system has been extensively studied in the literature. The reader may refer to [3,4, 5,6,7, 8,9,10, 11,12,13, 14,15,16, 17,18,19, 20,21]. In particular, Kuang et al. [8] recently introduced a discrete variational method and approximation to study the minimal period problem for second-order even Hamiltonian systems under the assumption of the Ambrosetti-Rabinowitz condition. Note that discrete variational methods are very effective tools for difference equations, to mention a few, see [22,23, 24,25,26, 27,28,29, 30,31,32, 33,34,35, 36,37].

Fan and Zhang [3] dealt with the minimal periodic problem for a first-order Hamiltonian system under the assumption of asymptotically linear nonlinearities. However, little work has been done that has referred to the minimal periodic problem for second-order Hamiltonian systems with such assumptions. Therefore, in this article, we mainly consider the minimal periodic problem for the classical second-order Hamiltonian system

(1) x + f ( x ) = 0 ,

where x R and f ( x ) is asymptotically linear at infinity.

Now, we state our main result.

## Theorem 1.1

Assume that the following conditions hold:

1. f C ( R , R ) and f ( x ) = f ( x ) for all x R .

2. lim x f ( x ) x = d > 0 and lim x 0 f ( x ) x = 0 .

3. There exist two positive constants η and β such that

(2) f ( x ) x 2 F ( x ) β f ( x )

for x η , where F ( x ) = 0 x f ( s ) d s for x R .

Then, for each T > 2 π / d , problem (1) has at least one nonconstant periodic solution x with minimal period T .

## Remark 1.1

The famous Ambrosetti-Rabinowitz condition in the case when x is scalar, there exist two positive constants η and μ > 2 such that

x f ( x ) μ F ( x ) > 0 for all x η ,

which implies

x f ( x ) 2 F ( x ) = 1 2 μ x f ( x ) + 2 μ [ x f ( x ) μ F ( x ) ] 1 2 μ η f ( x )

for x η , that is, ( f 3 ) in Theorem 1.1 is weaker than that in the Ambrosetti-Rabinowitz condition.

## Corollary 1.1

Assume that conditions ( f 1 ) ( f 3 ) hold and d = + . Then, for each T > 0 , problem (1) has at least one nonconstant periodic solution x with minimal period T .

## Example 1.1

Let f ( x ) be odd with

(3) f ( x ) = c 1 x c 2 x m , for x 1 , ( c 1 c 2 ) x r , for 0 x 1 ,

where c 1 > c 2 > 0 , r > 1 and 0 < m < 1 . It is easy to check that ( f 1 ) ( f 3 ) are satisfied; however, the Ambrosetti-Rabinowitz condition is not fulfilled. Then, for each T > 2 π / c 1 , problem (1) has at least one nonconstant periodic solution x with minimal period T by Theorem 1.1.

The rest of this article is organized as follows. In Section 2, we give some preliminary results. In Section 3, we prove Theorem 1.1 by using discrete variational method and approximation.

## 2 Preliminary results

Let Z + = { k > 0 : k Z } and R = { x 0 : x R } . For each T > 0 and x C ( [ 0 , T / 2 ] , R ) , we define x by

(4) x ( t ) = x ( t k T ) , for t [ k T , k T + T / 2 ] , x ( k T + T t ) , for t [ k T + T / 2 , k T + T ]

for all k Z , provided that x C ( [ 0 , T / 2 ] , R ) is a nontrivial solution of the following system:

(5) x + f ( x ) = 0 , x ( 0 ) = x ( T / 2 ) = 0 .

Then, it follows from ( f 1 ) that x is a nonconstant solution with minimal period T of problem (1).

For each fixed positive integer n , we denote by h n = T 4 n and u ( k ) = x ( k h n ) for all k [ 0 , 2 n ] Z , where [ 0 , 2 n ] Z denotes the discrete interval [ 0 , 1 , , 2 n ] . The following system can be regarded as a discrete analog of (5)

(6) Δ 2 u ( k 1 ) = h n 2 f ( u ( k ) ) for k [ 1 , 2 n 1 ] Z , u ( 0 ) = u ( 2 n ) = 0 ,

where Δ is the forward difference operator defined by Δ u ( k ) = u ( k + 1 ) u ( k ) and Δ 2 u ( k ) = Δ ( Δ u ( k ) ) . Define

(7) K n , j = u n ( j + 1 ) u n ( j ) h n for 0 j 2 n 1 .

Then, (6) can be written as

(8) K n , 0 K n , 1 = h n f ( u n ( 1 ) ) , K n , 1 K n , 2 = h n f ( u n ( 2 ) ) , , K n , 2 n 2 K n , 2 n 1 = h n f ( u n ( 2 n 1 ) ) ,

where u n ( 0 ) = u n ( 2 n ) = 0 .

Now, we present the variational framework. Let E n = { u = { u ( k ) } k = 0 2 n : u ( 0 ) = u ( 2 n ) = 0 } and E n = { u E n : u ( k ) 0 for 0 k 2 n } . So, E n is isomorphic to R 2 n 1 . Then, E n can be equipped with the inner product , n and norm n as

u , v n = k = 1 2 n 1 u ( k ) v ( k ) , u , v E n

and

u n = k = 1 2 n 1 u ( k ) 2 1 / 2 ,

respectively. For the convenience of notations, we will identify u = ( 0 , u ( 1 ) , , u ( 2 n 1 ) , 0 ) with ( u ( 1 ) , , u ( 2 n 1 ) ) R ( 2 n 1 ) .

Consider the functional I n defined by

(9) I n ( u ) = k = 0 2 n 1 u ( k + 1 ) u ( k ) 2 2 h n 2 k = 1 2 n 1 F ( u ( k ) ) .

Then, we find that I n is Fréchet differentiable in E n , and its Fréchet derivative is given by

(10) I n ( u ) , v = k = 1 2 n 1 [ Δ 2 u ( k 1 ) h n 2 f ( u ( k ) ) ] v ( k )

for u and v E n . In view of (9) and (10), it is easy to obtain that a critical point of the functional I n in E n is a solution of problem (6).

In order to construct an equivalent form for the functional I n , define the matrix B by

B = 2 1 0 0 0 1 2 1 0 0 0 1 2 0 0 0 0 0 2 1 0 0 0 1 2 .

Then, it is easy to see that

(11) I n ( u ) = 1 2 u B u T h n 2 k = 1 2 n 1 F ( u ( k ) )

and

(12) Δ 2 u ( 0 ) Δ 2 u ( 1 ) Δ 2 u ( 2 n 2 ) = B u ( 1 ) u ( 2 ) u ( 2 n 1 ) ,

for u E n . By direct computations, we obtain that the eigenvalues of B are given by

λ k = 4 sin 2 k π 4 n , k = 1 , 2 , , 2 n 1 ,

and the eigenvector ζ k corresponding to λ k is given by

ζ k = sin k π 2 n , sin 2 k π 2 n , , sin ( 2 n 1 ) k π 2 n T .

Now, we recall the definition of the Palais-Smale (PS) condition and the classical mountain pass lemma needed in the proof of our Theorem 1.1. As usual, let H be a real Banach space, we denote by B r the open ball in H with radius r and center 0, and B r its boundary.

## Definition 2.1

Let H be a real Banach space. A functional I C 1 ( H , R ) is said to satisfy the PS condition if every sequence { x j } in H , such that { I ( x j ) } is bounded and I ( x j ) 0 as j + , has a convergent subsequence.

## Lemma 2.1

(Mountain pass lemma) Let H be a real Banach space and I C 1 ( H , R ) satisfies the PS condition. Assume that I ( 0 ) = 0 and the following two conditions hold.

1. There exist constants a > 0 and ρ > 0 such that I B ρ a ;

2. There exists e H \ B ρ such that I ( e ) 0 .

Then I possesses a critical value α a . Moreover, α can be characterized as

α = inf φ Γ max s [ 0 , 1 ] I ( φ ( s ) ) ,

where

Γ = { φ C ( [ 0 , 1 ] , H ) : φ ( 0 ) = 0 , φ ( 1 ) = e } .

## 3 Proof of Theorem 1.1

In order to prove Theorem 1.1, we need the following lemma.

## Lemma 3.1

Define E n = { u E n : u ( k ) 0 for all 0 k 2 n } . If conditions ( f 1 ) ( f 3 ) are satisfied, then (6) has at least a nontrivial solution u n in E n . Moreover, there exist positive numbers M and n such that

u n ( k ) T M

for n n and 1 k 2 n 1 , where M is independent of n .

## Proof

We divide the proof of Lemma 3.1 into three steps.

Step 1. We prove that problem (6) has a nonzero solution. For this, we first prove that I n satisfies the PS condition. Let { I n ( u j ) } be a bounded sequence and I n ( u j ) 0 as j + , where { u j } is a sequence in E n . Then, there exists a positive constant M n such that I n ( u j ) M n for all j Z + . By ( f 2 ) , we have

lim u F ( u ) u 2 = 1 2 lim u f ( u ) u = 1 2 d ( d = + is also valid ) .

Since T > 2 π / d , we can choose L > 0 such that d > L > 4 π 2 / T 2 . Then,

lim u F ( u ) u 2 = 1 2 d > 1 2 L and lim u f ( u ) u = d > L ( d = + is also valid ) .

Hence, there exists η 0 > 0 such that

(13) F ( u ) > L 2 u 2 and f ( u ) > L u for all u > η 0 .

Let

η ¯ = max { η , η 0 }

and

M 0 = max F ( u ) L 2 u 2 : 0 u η ¯ .

Then, we have

(14) F ( u ) L 2 u 2 M 0 for all u R .

By the continuity of f , there exist two positive constants σ and P such that

(15) f ( u ) σ and f ( u ) u 2 F ( u ) P

for u η ¯ , which, combined with ( f 1 ) , ( f 3 ) , and (13), produces

(16) k = 1 2 n 1 [ f ( u j ( k ) ) u j ( k ) 2 F ( u j ( k ) ) ] = u j ( k ) η ¯ + u j ( k ) > η ¯ [ f ( u j ( k ) ) u j ( k ) 2 F ( u j ( k ) ) ] ( 2 n 1 ) P + u j ( k ) > η ¯ [ f ( u j ( k ) ) u j ( k ) 2 F ( u j ( k ) ) ] ( 2 n 1 ) P + β u j ( k ) > η ¯ f ( u j ( k ) ) ( 2 n 1 ) P + L β u j ( k ) > η ¯ u j ( k ) 2 n P + L β u j ( k ) > η ¯ u j ( k ) ,

for each j Z + .

Denote u j = max { u j ( k ) : 1 k 2 n 1 } , then u j u j n / 2 n 1 . Now, we claim that u j n is bounded. Arguing by the contradiction, assume up to a subsequence which we still denote by { u j } , that u j n + as j + . By I n ( u j ) 0 as j + , we can choose j large enough such that

I n ( u j ) < ε < L h n 2 β 2 n 1 ,

which, together with (16), produces

(17) 2 M n + ε u j n 2 I n ( u j ) I n ( u j ) , u j = h n 2 k = 1 2 n 1 [ f ( u j ( k ) ) u j ( k ) 2 F ( u j ( k ) ) ] 2 n h n 2 P + L β h n 2 u j ( k ) > η ¯ u j ( k ) = 2 n h n 2 P + L β h n 2 u j ( k ) > η ¯ + u j ( k ) η ¯ u j ( k ) L β h n 2 u j ( k ) η ¯ u j ( k ) 2 n h n 2 P + L β h n 2 u j 2 L β h n 2 n η ¯ 2 n h n 2 P + L h n 2 β 2 n 1 u j n 2 L β h n 2 n η ¯ ,

which contradicts the fact that u j n + as j + . Hence, I n satisfies the PS condition for each n Z + .

Next, we show that I n satisfies ( J 1 ) in Lemma 2.1. It follows from ( f 2 ) that there exist two positive constants b 0 < π 2 / T 2 and ρ such that

(18) f ( x ) b 0 x and F ( x ) b 0 x 2 for all x ρ .

By

lim n sin π 4 n h n = π T ,

there exists a positive integer n such that

(19) π 2 2 T 2 sin π 4 n h n 2 π 2 T 2 for n n .

Hence, for all u B ρ and n n ,

(20) I n ( u ) 2 sin 2 π 4 n u n 2 b 0 h n 2 u n 2 π 2 T 2 b 0 h n 2 ρ 2 > 0 .

Thus, I n satisfies ( J 1 ) in Lemma 2.1.

Finally, we prove that I n satisfies ( J 2 ) of Lemma 2.1 for n n . Let e n E n with

(21) e n = d sin π 2 n , sin 2 π 2 n , , sin ( 2 n 1 ) π 2 n T = d ζ 1 .

We can choose d large enough such that d > M 0 / L 4 π 2 T 2 , where the fixed number d is independent of n . By (14), we have

(22) I n ( e n ) 2 d 2 sin 2 π 4 n k = 1 2 n 1 sin 2 k π 2 n h n 2 k = 1 2 n 1 L 2 d sin k π 2 n 2 M 0 .

In view of

(23) k = 1 2 n 1 sin 2 k π 2 n = 1 2 k = 1 2 n 1 1 cos k π n = 2 n 1 2 2 cos π n sin π 2 n + 2 cos 2 π n sin π 2 n + + 2 cos 2 n π π n sin π 2 n 4 sin π 2 n = n ,

which, combined with (22), gives us

(24) I n ( e n ) 2 n d 2 h n 2 π 2 T 2 L n d 2 h n 2 2 + h n 2 2 n M 0 2 n h n 2 d 2 L 4 π 2 T 2 + M 0 .

Then, it follows from d > M 0 / L 4 π 2 T 2 that I n ( e n ) < 0 . Hence, we have verified all assumptions of Lemma 2.1. For n n , we know that I n possesses a critical value α n , where

α n = inf φ Γ max s [ 0 , 1 ] I n ( φ ( s ) ) , Γ = { φ C ( [ 0 , 1 ] , E n ) : φ ( 0 ) = 0 , φ ( 1 ) = e n } .

Thus, (6) has at least a nontrivial solution in E n .

Step 2. We prove that problem (6) has a nonzero solution u n E n for any n n . For each φ Γ with

φ ( t ) = ( φ 1 ( t ) , φ 2 ( t ) , , φ 2 n 1 ( t ) )

satisfying φ ( 0 ) = 0 and φ ( 1 ) = e n , where e n is given in (21), we denote by φ ,

φ ( t ) = ( φ 1 ( t ) , φ 2 ( t ) , , φ 2 n 1 ( t ) )

for t [ 0 , 1 ] . It is easy to check that φ Γ , where

Γ = { φ C ( [ 0 , 1 ] , E n ) : φ ( 0 ) = 0 , φ ( 1 ) = e n } Γ .

By ( f 1 ) , it is easy to obtain that

max s [ 0 , 1 ] I n ( φ ( s ) ) max s [ 0 , 1 ] I n ( φ ( s ) )

and

α n = inf φ Γ max s [ 0 , 1 ] I n ( φ ( s ) ) = inf φ Γ max s [ 0 , 1 ] I n ( φ ( s ) ) .

Thus, there exists a nonzero critical point u n E n and I n ( u n ) = α n > 0 for n n .

Step 3. For n n , we claim that there exists a positive constant M , independent of n , such that 0 u n ( k ) M for all 0 k 2 n . Choose

φ ¯ ( t ) = d t sin π 2 n , d t sin 2 π 2 n , , d t sin ( 2 n 1 ) π 2 n T Γ ,

using the similar argument as (24), we have

(25) α n max s [ 0 , 1 ] I n ( φ ¯ ( s ) ) 2 n h n 2 M 0 = 1 2 h n T M 0 .

On the other hand, it follows from (9) and (10) that

(26) α n = I n ( u n ) 1 2 I n ( u n ) , u n = h n 2 k = 1 2 n 1 1 2 f ( u n ( k ) ) u n ( k ) F ( u n ( k ) ) .

Combining ( f 3 ) , (15), (25), and (26) together, we obtain that

(27) h n k = 1 j f ( u n ( k ) ) h n u n ( k ) η ¯ f ( u n ( k ) ) + h n u n ( k ) > η ¯ f ( u n ( k ) ) 1 2 σ T + 2 β h n u n ( k ) > η ¯ 1 2 f ( u n ( k ) ) u n ( k ) F ( u n ( k ) ) 1 2 σ T + 2 β h n u n ( k ) > η ¯ 1 2 f ( u n ( k ) ) u n ( k ) F ( u n ( k ) ) + 2 β h n u n ( k ) η ¯ 1 2 f ( u n ( k ) ) u n ( k ) F ( u n ( k ) ) + 1 β h n u n ( k ) η ¯ P 1 2 σ T + 2 α n β h n + 1 2 β T P 1 2 σ T + M 0 T β + 1 2 β T P ,

for each 1 j 2 n 1 . Let

(28) M = 1 2 σ T + M 0 T β + 1 2 β T P ,

then, it is easy to check that M is independent of n . By Step 2, u n E n is a nonzero solution to (6), then

(29) 0 K n , 0 M and 0 K n , 4 n 1 M ,

which, combined with (8) and (27), gives us

(30) K n , j = K n , 0 h n k = 1 j f ( u n ( k ) ) K n , 0 + h n k = 1 j f ( u n ( k ) ) 2 M

for all 1 j 2 n 1 . Hence,

u n ( k ) h n = ( u n ( k ) u n ( k 1 ) ) + + ( u n ( 1 ) u n ( 0 ) ) h n 4 n M

and

u n ( k ) 4 n M h n = T M

for all n n and 0 < k 2 n , where T and M are independent of n , n is given in (19). This completes the proof of the lemma.□

By Lemma 3.1, there exists a nonzero solution u n E n to (6) for n n . We define x n by

(31) x n ( t ) = t K n , 0 , for 0 t h n , t K n , 1 K n , 1 h n + u n ( 1 ) , for h n t 2 h n , t K n , 2 2 K n , 2 h n + u n ( 2 ) , for 2 h n t 3 h n , , t K n , 2 n 1 ( 2 n 1 ) h n K n , 2 n 1 + u n ( 2 n 1 ) , for ( 2 n 1 ) h n t T / 2 .

Then, x n C ( [ 0 , T / 2 ] , R ) with x n ( 0 ) = x n ( T / 2 ) = 0 for n n .

Now, we are in a position to prove Theorem 1.1.

## Proof

First, we prove that there exists a subsequence of { x n } n = n + , which converges uniformly on [ 0 , T / 2 ] . For any s and t with 0 s t T / 2 , there exist two integers j 1 and j 2 such that

j 1 h n s t j 2 h n ,

satisfying 0 s j 1 h n < h n and 0 j 2 h n t < h n . Then, by (29) and (30), we have

(32) x n ( t ) x n ( s ) = 0 2 M ( t s ) , for j 2 = j 1 , K n , j 1 ( t s ) 2 M ( t s ) , for j 2 = j 1 + 1 , x n ( t ) x n ( j 1 h n + h n ) + x n ( j 1 h n + h n ) x n ( s ) 2 M ( t s ) , for j 2 = j 1 + 2 , x n ( t ) x n ( j 2 h n h n ) + + x n ( j 1 h n + h n ) x n ( s ) 2 M ( t s ) , for j 2 j 1 + 3 .

Moreover,

(33) x n ( t ) = x n ( t ) x n ( 0 ) 2 M t M T

for 0 t T / 2 and n n . By Arzela-Ascoli theorem, we can choose a subsequence, still denoted by { x n } n = n + , such that { x n ( t ) } n = n + converges uniformly to x ( t ) on [ 0 , T / 2 ] , where x C ( [ 0 , T / 2 ] , R ) with x ( 0 ) = x ( T / 2 ) = 0 .

Next, we claim that, for each n n , there exists a positive integer k [ 1 , 2 n 1 ] Z such that

(34) u n ( k ) > ρ .

Suppose by contrary that u n ( k ) ρ for all k [ 1 , 2 n 1 ] Z . By (18), we have

k = 1 2 n 1 f ( u n ( k ) ) u n ( k ) b 0 u n n 2 ,

which, together with (19), produces

(35) I n ( u n ) , u n 4 sin 2 π 4 n u n n 2 h n 2 k = 1 2 n 1 f ( u n ( k ) ) u n ( k ) 2 π 2 T 2 b 0 h n 2 u n n 2 > 0

for n n , which contradicts the fact that I n ( u n ) , u n = 0 . Thus, our claim is proved.

Now, we claim that x is a nonzero function. Arguing by the contradiction, assume that x = 0 . Since { x n } n = n + converges uniformly to x on [ 0 , T / 2 ] for n n , there exists a positive integer n ¯ > n such that

(36) x n ( t ) ρ for all n n ¯ , 0 t T / 2 ,

which contradicts (34). Hence, x is a nonzero function.

Finally, we show that x C ( [ 0 , T / 2 ] , R ) is a nonzero solution of problem (5) and x is a nonconstant periodic solution with prescribed minimal period T of problem (1), where x is given in (4). It follows from (29) that there exists a subsequence of { x n } n = n + , still denoted by { x n } n = n + , such that

K n , 0 K as n + .

It is easy to check that K 0 . By the definition of { x n } n = n + , we have that x n ( t ) is left differentiable on ( 0 , T / 2 ] for n n . For every t ( 0 , T / 2 ] , there exists j [ 0 , 2 n 1 ] Z such that j h n < t ( j + 1 ) h n . Then, the left derivative x n ( t ) is given by

(37) x n ( t ) = K n , j = K n , 0 h n k = 1 j f ( u n ( k ) ) = K n , 0 0 t f ( x n ( s ) ) d s + τ n ( t ) ,

where

(38) τ n ( t ) = 0 t f ( x n ( s ) ) d s h n k = 1 j f ( u n ( k ) ) = 0 j h n f ( x n ( s ) ) d s h n k = 1 j f ( x n ( k h n ) ) + j h n t f ( x n ( s ) ) d s .

By (33) and the continuity of f , there exists a positive constant M ¯ such that

M ¯ f ( x n ( t ) ) M ¯ ,

then,

(39) M ¯ h n j h n t f ( x n ( s ) ) d s M ¯ h n

for all t [ 0 , T / 2 ] and n n . Since { x n ( t ) } n = n + converges uniformly to x ( t ) on [ 0 , T / 2 ] , we have

(40) 0 j h n f ( x n ( s ) ) d s h n k = 1 j f ( x n ( k h n ) ) 0 j h n f ( x ( s ) ) d s + 0 j h n f ( x n ( s ) ) d s + h n k = 1 j f ( x ( k h n ) ) h n k = 1 j f ( x n ( k h n ) ) + 0 j h n f ( x ( s ) ) d s h n k = 1 j f ( x ( k h n ) ) 0 as n + ,

which, together with (39), produces

(41) τ n ( t ) 0 as n + .

Similarly, we can prove that x n ( t ) is right differentiable on [ 0 , T / 2 ) , and the right derivative is given by

(42) x n ( t ) = K n , 0 0 t f ( x n ( s ) ) d s + γ n ( t ) ,

where γ n ( t ) 0 as n . Given t > 0 and δ < 0 , for any s , satisfying 0 t + δ < s t T / 2 , there exist two integers j 3 and j 4 such that

j 3 h n < s t j 4 h n ,

satisfying 0 < s j 3 h n h n and 0 j 4 h n t < h n . By (8) and (30), it is easy to see that, j 4 h n j 3 h n t s + 2 h n δ + 2 h n and

x n ( t ) x n ( s ) = x n ( j 4 h n ) x n ( ( j 3 + 1 ) h n ) = K n , j 4 1 K n , j 3 2 ( j 4 j 3 1 ) h n M 2 M ( δ + h n ) ,

then,

(43) x n ( t ) 2 M ( δ + h n ) x n ( s ) x n ( t ) + 2 M ( δ + h n ) .

Similarly, for any t > 0 and δ < 0 , satisfying 0 t + δ < t T / 2 , there exist two integers j 5 and j 6 such that

t + δ < j 5 h n j 6 h n t ,

satisfying 0 < j 5 h n ( t + δ ) h n and 0 t j 6 h n < h n . By direct computations, we have

x n ( t + δ ) x n ( t ) δ = x n ( t + δ ) x n ( j 5 h n ) δ + + x n ( j 6 h n ) x n ( t ) δ = x n ( j 5 h n ) ( t + δ j 6 h n ) δ + + x n ( t ) ( j 6 h n t ) δ ,

which, combined with (43), gives us

(44) x n ( t ) 2 M ( δ + h n ) x n ( t + δ ) x n ( t ) δ x n ( t ) + 2 M ( δ + h n ) .

Taking the limit n + implies

(45) K 0 t f ( x ( s ) ) d s + 2 M δ x ( t + δ ) x ( t ) δ K 0 t f ( x ( s ) ) d s 2 M δ .

Then,

(46) lim δ 0 x ( t + δ ) x ( t ) δ = K 0 t f ( x ( s ) ) d s

for 0 < t T / 2 . Using a similar argument as above, we have

(47) lim δ 0 + x ( t + δ ) x ( t ) δ = K 0 t f ( x ( s ) ) d s

for 0 t < T / 2 . Thus, x C 1 ( [ 0 , T / 2 ] , R ) , x ( 0 ) = x ( T / 2 ) = 0 , and

(48) x ( t ) K = 0 t f ( x ( s ) ) d s

for 0 t T / 2 . Hence, x C ( [ 0 , T / 2 ] , R ) is a nontrivial solution of (5). By ( f 1 ) , it is easy to obtain that x is a nonconstant periodic solution with prescribed minimal period T of problem (1), where x is given in (4). This completes the proof of Theorem 1.1.□

1. Funding information: This work was supported by the National Natural Science Foundation of China (No. 11901438) and the Natural Science Foundation of Guangdong Province, China (No. 2021A1515010062).

2. Conflict of interest: The authors state no conflict of interest.

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