The following results will be used later.

Lemma 2.1 (see [12]).

Let

with

and the constant
is optimal.

Lemma 2.2 (see [12]).

Let

with the boundary value conditions

, then

Consider the periodic boundary value problem

Lemma 2.3.

Suppose that

are

-integrable

periodic function, where

satisfy the condition (H

_{2}), with

then (2.4) has only the trivial
-periodic solution
.

Proof.

If on the contrary, (2.4) has a nonzero

-periodic solution

, then using (2.4), we have

where
is undetermined.

Firstly, we prove that

has at least one zero in

. If

, we may assume

. Since

is a

-periodic solution, there exists a

with

. Then,

we could get a contradiction.

Without loss of generality, we may assume that
; then there exists a sufficiently small
such that
. Since
is a continuous function, there must exist a
with
.

Secondly, we prove that

has at least

zeros on

. Considering the initial value problem

is the solution of (2.8) and

where

with

. Since

holds under the assumptions of

, there is a

, such that

Now, let

. By the conditions (H

_{2}), (2.11), and (2.12), we have

Since

is decreasing in

, we have

. Therefore,

We also consider the initial value problem

is the solution of (2.16), where

is the same as the previous one, and

Hence, there exists a

with

, such that

From (2.12) and (2.19), it follows that

By

and (2.21), we have

Since

is decreasing on

, we have

, and

We now prove that

has a zero point in

. If on the contrary

for

, then we would have the following inequalities:

In fact, from(2.4), (2.8), and (2.15), we have

with

. Setting

, and since

Notice that

, which implies

Integrating from 0 to

, we obtain

which implies (2.24). By a similar argument, we have (2.25). Therefore,
, a contradiction, which shows that
has at least one zero in
, with
.

We let
. If
, then from a similar argument, there is a
, such that
and so on. So, we obtain that
has at least
zeros on
.

Thirdly, we prove that

has at least

zeros on

. If, on the contrary, we assume that

only has

zeros on

, we write them as

Without loss of generality, we may assume that

. Since

we obtain
, which contradicts
. Therefore,
has at least
zeros on
.

Finally, we prove Lemma 2.3. Since

has at least

zeros on

, there are two zeros

and

with

. By Lemmas 2.1 and 2.2, we have

From

, it follows that

which implies
for
. Also
. Therefore,
for
, a contradiction. The proof is complete.